Motion in one dimension
October 30, 2017 | Author: Anonymous | Category: N/A
Short Description
313. 14 Conservation of momentum. 343. 15 Conservation of angular momentum. 375. 16 Thermodynamics. 411. Vibrations and&...
Description
Fullerton, California www.lightandmatter.com
copyright 1998-2010 Benjamin Crowell
rev. June 7, 2013
This book is licensed under the Creative Commons Attribution-ShareAlike license, version 3.0, http://creativecommons.org/licenses/by-sa/3.0/, except for those photographs and drawings of which I am not the author, as listed in the photo credits. If you agree to the license, it grants you certain privileges that you would not otherwise have, such as the right to copy the book, or download the digital version free of charge from www.lightandmatter.com.
Brief Contents 0 1
Introduction and review Scaling and estimation
15 41
Motion in one dimension 2 3 4 5
Velocity and relative motion Acceleration and free fall Force and motion Analysis of forces
67 91 123 147
Motion in three dimensions 6 7 8 9 10
Newton’s laws in three dimensions Vectors Vectors and motion Circular motion Gravity
179 191 207 229 245
Conservation laws 11 12 13 14 15 16
Conservation of energy Simplifying the energy zoo Work: the transfer of mechanical energy Conservation of momentum Conservation of angular momentum Thermodynamics
277 299
Vibrations and waves 17 18 19 20
435 449 473 501
Relativity and electromagnetism 21 22 23 24 25 26 27
Electricity and circuits The nonmechanical universe Relativity and magnetism Electromagnetism Capacitance and inductance The atom and E=mc2 General relativity
553 609 643 667 701 719 781
Optics 28 29 30 31 32
313 343 375 411
Vibrations Resonance Free waves Bounded waves
The ray model of light Images by reflection Images, quantitatively Refraction Wave optics
801 819 837 855 877
The modern revolution in physics 33 34 35 36
Rules of randomness Light as a particle Matter as a wave The atom
905 929 947 973
Contents 0 Introduction and review 0.1 The scientific method . . . . . . 0.2 What is physics? . . . . . . . .
15 18
Isolated systems and reductionism, 20.
0.3 How to learn physics. . . . . . . 0.4 Self-evaluation . . . . . . . . . 0.5 Basics of the metric system . . . .
1.4 Order-of-magnitude estimates . . Summary . . . . . . . . . . . . Problems . . . . . . . . . . . . Exercise 1: Scaling applied to leaves.
. . . .
55 58 59 64
2.1 Types of motion . . . . . . . . .
67
21 23 24
The metric system, 24.—The second, 25.— The meter, 25.—The kilogram, 26.— Combinations of metric units, 26.— Checking units, 26.
0.6 0.7 0.8 0.9
The Newton, the metric unit of force Less common metric prefixes . . . Scientific notation . . . . . . . . Conversions . . . . . . . . . .
28 28 29 30
Should that exponent be positive, or negative?, 31.
0.10 Significant figures . . . . . Summary . . . . . . . . . . . Problems . . . . . . . . . . . Exercise 0: Models and idealization
. . . .
. . . .
32 35 37 39
Motion in one dimension 2 Velocity and relative motion Rigid-body motion distinguished from motion that changes an object’s shape, 67.—Center-of-mass motion as opposed to rotation, 67.—Center-of-mass motion in one dimension, 71.
1 Scaling and estimation 1.1 Introduction . . . . . . . . . .
41
Area and volume, 41.
1.2 Scaling of area and volume . . . .
43
Galileo on the behavior of nature on large and small scales, 44.—Scaling of area and volume for irregularly shaped objects, 47.
1.3 Scaling applied to biology . . . . Organisms of different sizes with the same shape, 51.—Changes in shape to accommodate changes in size, 53.
6
51
2.2 Describing distance and time . . .
71
A point in time as opposed to duration, 72.
2.3 Graphs of motion; velocity . . . . Motion with constant velocity, Motion with changing velocity, Conventions about graphing, 76.
74
74.— 75.—
2.4 The principle of inertia . . . . . . Physical effects relate only to a change in velocity, 78.—Motion is relative, 79.
78
2.5 Addition of velocities . . . . . . .
81
Addition of velocities to describe relative motion, 81.—Negative velocities in relative motion, 81.
2.6 Graphs of velocity versus time. . . 2.7 Applications of calculus. . . . .
84 84
Summary . . . . . . . . . . . . .
86
Problems . . . . . . . . . . . . .
88
3 Acceleration and free fall 3.1 The motion of falling objects
. . .
91
How the speed of a falling object increases with time, 93.—A contradiction in Aristotle’s reasoning, 94.—What is gravity?, 94.
3.2 Acceleration . . . . . . . . . .
4 Force and motion 95
Definition of acceleration for linear v − t graphs, 95.—The acceleration of gravity is different in different locations., 96.
3.3 Positive and negative acceleration .
98
3.4 Varying acceleration . . . . . . . 102 3.5 The area under the velocity-time graph . . . . . . . . . . . . . . . 105 3.6 Algebraic results for constant acceleration . . . . . . . . . . . . 107 3.7 A test of the principle of inertia . . 110 3.8 Applications of calculus. . . . . 111 Summary . . . . . . . . . . . . . 112 Problems . . . . . . . . . . . . . 113
4.1 Force . . . . . . . . . . . . . 124 We need only explain changes in motion, not motion itself., 124.—Motion changes due to an interaction between two objects., 125.—Forces can all be measured on the same numerical scale., 125.—More than one force on an object, 126.—Objects can exert forces on each other at a distance., 126.—Weight, 127.—Positive and negative signs of force, 127.
4.2 Newton’s first law . . . . . . . . 127 More general combinations of forces, 129.
4.3 Newton’s second law
. . . . . . 131
A generalization, 132.—The relationship between mass and weight, 133.
4.4 What force is not . . . . . . . . 136 1. Force is not a property of one object., 136.—2. Force is not a measure of an object’s motion., 136.—3. Force is not energy., 136.—4. Force is not stored or used up., 137.—5. Forces need not be exerted by living things or machines., 137.— 6. A force is the direct cause of a change in motion., 137.
4.5 Inertial and noninertial frames of reference . . . . . . . . . . . . . 138 Summary . . . . . . . . . . . . . 141 Problems . . . . . . . . . . . . . 142 Exercise 4: Force and motion . . . . . 145
7
5 Analysis of forces 5.1 Newton’s third law . . . . . . . . 147 A mnemonic for using Newton’s third law correctly, 150.
5.2 Classification and behavior of forces 152 Normal forces, 155.—Gravitational forces, 155.—Static and kinetic friction, 156.— Fluid friction, 160.
5.3 Analysis of forces . . . . . . . . 161 5.4 Transmission of forces by low-mass objects . . . . . . . . . . . . . . 164 5.5 Objects under strain . . . . . . . 166 5.6 Simple Machines: the pulley . . . 167 Summary . . . . . . . . . . . . . 169 Problems . . . . . . . . . . . . . 171
components, 198.—Addition of vectors given their magnitudes and directions, 198.—Graphical addition of vectors, 198.
7.4 Unit vector notation . 7.5 Rotational invariance . Summary . . . . . . . . Problems . . . . . . . .
. . . .
. . . .
. . . .
. . . .
. . . .
200 201 203 204
8 Vectors and motion 8.1 The velocity vector . . . . . . . 208 8.2 The acceleration vector . . . . . 210 8.3 The force vector and simple machines213 8.4 Calculus with vectors . . . . . 215 Summary . . . . . . . . . . . . . 219 Problems . . . . . . . . . . . . . 220 Exercise 8: Vectors and motion . . . . 226
9 Circular motion 9.1 Conceptual framework . . . . . . 229 Circular motion does not produce an outward force, 229.—Circular motion does not persist without a force, 230.—Uniform and nonuniform circular motion, 231.—Only an inward force is required for uniform circular motion., 232.—In uniform circular motion, the acceleration vector is inward., 233.
Motion in three dimensions 6 Newton’s dimensions
laws
in
three
9.2 Uniform circular motion. . 9.3 Nonuniform circular motion Summary . . . . . . . . . Problems . . . . . . . . .
. . . .
. . . .
. . . .
. . . .
236 239 240 241
6.1 Forces have no perpendicular effects 179 Relationship to relative motion, 181.
6.2 Coordinates and components . . . 182 Projectiles move along parabolas., 185.
6.3 Newton’s laws in three dimensions . 185 Summary . . . . . . . . . . . . . 187 Problems . . . . . . . . . . . . . 188
7 Vectors
10 Gravity
7.1 Vector notation . . . . . . . . . 191
10.1 Kepler’s laws . . . . . . . . . 246 10.2 Newton’s law of gravity . . . . . 248
Drawing vectors as arrows, 194.
7.2 Calculations with magnitude and direction. . . . . . . . . . . . . . 195 7.3 Techniques for adding vectors . . . 198 Addition
8
of
vectors
given
their
The sun’s force on the planets obeys an inverse square law., 248.—The forces between heavenly bodies are the same type of force as terrestrial gravity., 249.—Newton’s
law of gravity, 250.
10.3 Apparent weightlessness . . . . 254 10.4 Vector addition of gravitational forces . . . . . . . . . . . . . . . 254 10.5 Weighing the earth . . . . . . . 257 10.6 Dark energy . . . . . . . . . 259 10.7 A gravitational test of Newton’s first law . . . . . . . . . . . . . . . . 261 Summary . . . . . . . . . . . . . 263 Problems . . . . . . . . . . . . . 265 Exercise 10: The shell theorem . . . . 273
13 Work: the transfer of mechanical energy 13.1 Work: the transfer of mechanical energy . . . . . . . . . . . . . . 313 The concept of work, 313.—Calculating work as force multiplied by distance, 314.—Machines can increase force, but not work., 317.—No work is done without motion., 317.—Positive and negative work, 318.
13.2 Work in three dimensions . . . . 320 A force perpendicular to the motion does no work., 320.—Forces at other angles, 321.
13.3 Varying force . . . . . . . . . 324 13.4 Applications of calculus . . . . 327 13.5 Work and potential energy . . . . 328 13.6 When does work equal force times distance? . . . . . . . . . . . . . 330 13.7 The dot product . . . . . . . 332 Summary . . . . . . . . . . . . . 334 Problems . . . . . . . . . . . . . 336
Conservation laws 11 Conservation of energy 11.1 The search for a perpetual Machine . . . . . . . . . . . 11.2 Energy . . . . . . . . . 11.3 A numerical scale of energy
motion . . . 277 . . . 278 . . . 282
14 Conservation of momentum 14.1 Momentum . . . . . . . . . . 344 A conserved quantity of motion, 344.— Momentum, 345.—Generalization of the momentum concept, 347.—Momentum compared to kinetic energy, 348.
How new forms of energy are discovered, 285.
14.2 Collisions in one dimension . . . 351
11.4 Kinetic energy . . . . . . . . . 287
14.3 Relationship of momentum to the center of mass . . . . . . . . . . . 355
Energy and relative motion, 288.
11.5 Power . . . . . . . . . . . . 289 Summary . . . . . . . . . . . . . 292 Problems . . . . . . . . . . . . . 294
The discovery of the neutron, 353.
Momentum in different frames of reference, 357.—The center of mass frame of reference, 357.
14.4 Momentum transfer. . . . . . . 358
12 Simplifying the energy zoo 12.1 Heat is kinetic energy . . . . . . 300 12.2 Potential energy: energy of distance or closeness . . . . . . . . . . . . 302 An equation for gravitational potential energy, 303.
12.3 All energy is potential or kinetic . . 306 Summary . . . . . . . . . . . . . 308 Problems . . . . . . . . . . . . . 309
The rate of change of momentum, 358.— The area under the force-time graph, 360.
14.5 Momentum in three dimensions . 361 The center of mass, 362.—Counting equations and unknowns, 363.—Calculations with the momentum vector, 364.
14.6 Applications of calculus . . . . 366 Summary . . . . . . . . . . . . . 368 Problems . . . . . . . . . . . . . 370
9
15 Conservation momentum
of
angular
Problems . . . . . . . . . . . . . 429
15.1 Conservation of angular momentum377 Restriction to rotation in a plane, 381.
15.2 Angular momentum in planetary motion . . . . . . . . . . . . . . 381 15.3 Two theorems about angular momentum . . . . . . . . . . . . 383 15.4 Torque: the rate of transfer of angular momentum . . . . . . . . . . . 388 Torque distinguished from force, 388.— Relationship between force and torque, 389.—The torque due to gravity, 391.
15.5 Statics . . . . . . . . . . . . 395 Equilibrium, 395.—Stable and unstable equilibria, 398.
15.6 Simple Machines: the lever . . . 399 15.7 Proof of Kepler’s elliptical orbit law401 Summary . . . . . . . . . . . . . 403 Problems . . . . . . . . . . . . . 405 Exercise 15: Torque . . . . . . . . . 410
Vibrations and waves 17 Vibrations 17.1 Period, frequency, and amplitude . 436 17.2 Simple harmonic motion. . . . . 439 Why are sine-wave vibrations so common?, 439.—Period is approximately independent of amplitude, if the amplitude is small., 440.
17.3 Proofs . . . . Summary . . . . . . Problems . . . . . . Exercise 17: Vibrations
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
442 444 445 447
Energy in vibrations . . . . Energy lost from vibrations. . Putting energy into vibrations Proofs . . . . . . . . .
. . . .
. . . .
450 452 454 462
18 Resonance 18.1 18.2 18.3 18.4
16 Thermodynamics 16.1 Pressure and temperature . . . . 412 Pressure, 412.—Temperature, 416.
16.2 Microscopic description of an ideal gas . . . . . . . . . . . . . . . . 419 Evidence for the kinetic theory, 419.— Pressure, volume, and temperature, 419.
16.3 Entropy. . . . . . . . . . . . 423 Efficiency and grades of energy, 423.— Heat engines, 423.—Entropy, 425.
10
Statement 2: maximum amplitude at resonance, 463.—Statement 3: amplitude at resonance proportional to Q, 463.— Statement 4: FWHM related to Q, 464.
Summary . . . . . . . . . . . . . 465 Problems . . . . . . . . . . . . . 467 Exercise 18: Resonance . . . . . . . 471
19 Free waves 19.1 Wave motion . . . . . . . . . 475 1. Superposition, 475.—2. The medium is not transported with the wave., 477.—3. A wave’s velocity depends on the medium., 478.—Wave patterns, 479.
19.2 Waves on a string . . . . . . . 480
Photo credits for volume 1 . . . . . . 547
Intuitive ideas, 480.—Approximate treatment, 481.—Rigorous derivation using calculus (optional), 482.—Significance of the result, 484.
Relativity and electromagnetism
19.3 Sound and light waves . . . . . 484 Sound waves, 484.—Light waves, 485
21 Electricity and circuits 21.1 The quest for the atomic force . . 554 21.2 Electrical forces . . . . . . . . 555 Charge, 555.—Conservation of charge, 557.—Electrical forces involving neutral objects, 558.
21.3 Current . . . . . . . . . . . . 558 Unity of all types of electricity, 558.— Electric current, 559.
21.4 Circuits . . . . . . . . . . . . 561 21.5 Voltage . . . . . . . . . . . . 563 .
The volt unit, 563.—The voltage concept in general, 563.
19.4 Periodic waves. . . . . . . . . 487
21.6 Resistance . . . . . . . . . . 568
Period and frequency of a periodic wave, 487.—Graphs of waves as a function of position, 487.—Wavelength, 488.— Wave velocity related to frequency and wavelength, 488.—Sinusoidal waves, 490.
Resistance, 568.—Superconductors, 570.— Constant voltage throughout a conductor, 570.—Short circuits, 572.—Resistors, 573.
19.5 The Doppler effect . . . . . . . 491 The Big Bang, 493.—What the Big Bang is not, 495.
Summary . . . . . . . . . . . . . 497 Problems . . . . . . . . . . . . . 499
20 Bounded waves 20.1 Reflection, transmission, and absorption . . . . . . . . . . . . . 502 Reflection and transmission, 502.— Inverted and uninverted reflections, 505.— Absorption, 505.
20.2 Quantitative treatment of reflection509 Why reflection occurs, 509.—Intensity of reflection, 510.—Inverted and uninverted reflections in general, 511.
20.3 Interference effects . . . . . . . 512 20.4 Waves bounded on both sides . . 515 Musical applications, 517.—Standing waves, 517.—Standing-wave patterns of air columns, 519.
Summary . . . . . . . . . . . . . 521 Problems . . . . . . . . . . . . . 522 Three essential mathematical skills . . 524
21.7 Applications of calculus . . . . 575 21.8 Series and parallel circuits . . . . 576 Schematics, 576.—Parallel resistances and the junction rule, 577.—Series resistances, 582.
Summary . . . . . . . . . . . . . Problems . . . . . . . . . . . . . Exercise 21A: Electrical measurements. Exercise 21B: Voltage and current . . . Exercise 21C: Reasoning about circuits
589 593 601 602 607
22 The nonmechanical universe 22.1 The stage and the actors . . . . 610 Newton’s instantaneous action at a distance, 610.—No absolute time, 610.— Causality, 611.—Time delays in forces exerted at a distance, 611.—More evidence that fields of force are real: they carry energy., 612.
22.2 The gravitational field . . . . . . 614 Sources and sinks, 615.—Superposition of fields, 615.—Gravitational waves, 616.
22.3 The electric field . . . . . . . . 617 Definition, 617.—Dipoles, 620.— Alternative definition of the electric field, 621.—Voltage related to electric field, 621.
22.4 Calculating energy in fields . . . 623
11
36.2 Angular momentum in three dimensions . . . . . . . . . . . . 975 Three-dimensional angular momentum in classical physics, 975.—Three-dimensional angular momentum in quantum physics, 976.
36.3 The hydrogen atom. . . . . . . 977 36.4 Energies of states in hydrogen . 980 History,
980.—Approximate treatment,
980.
36.5 Electron spin . . . . . . . . . 982 36.6 Atoms with more than one electron 983 Deriving the periodic table, 985.
Summary . . . . . . . . . . . . . 987 Problems . . . . . . . . . . . . . 989 Exercise 36: Quantum versus classical randomness . . . . . . . . . . . . 992 Photo credits for volume 2 . . . . . . 1002
15
16
The Mars Climate Orbiter is prepared for its mission. The laws of physics are the same everywhere, even on Mars, so the probe could be designed based on the laws of physics as discovered on earth. There is unfortunately another reason why this spacecraft is relevant to the topics of this chapter: it was destroyed attempting to enter Mars’ atmosphere because engineers at Lockheed Martin forgot to convert data on engine thrusts from pounds into the metric unit of force (newtons) before giving the information to NASA. Conversions are important!
Chapter 0
Introduction and review If you drop your shoe and a coin side by side, they hit the ground at the same time. Why doesn’t the shoe get there first, since gravity is pulling harder on it? How does the lens of your eye work, and why do your eye’s muscles need to squash its lens into different shapes in order to focus on objects nearby or far away? These are the kinds of questions that physics tries to answer about the behavior of light and matter, the two things that the universe is made of.
0.1 The scientific method Until very recently in history, no progress was made in answering questions like these. Worse than that, the wrong answers written by thinkers like the ancient Greek physicist Aristotle were accepted without question for thousands of years. Why is it that scientific knowledge has progressed more since the Renaissance than it had in all the preceding millennia since the beginning of recorded history? Undoubtedly the industrial revolution is part of the answer. Building its centerpiece, the steam engine, required improved tech-
15
niques for precise construction and measurement. (Early on, it was considered a major advance when English machine shops learned to build pistons and cylinders that fit together with a gap narrower than the thickness of a penny.) But even before the industrial revolution, the pace of discovery had picked up, mainly because of the introduction of the modern scientific method. Although it evolved over time, most scientists today would agree on something like the following list of the basic principles of the scientific method:
a / Science is a cycle of theory and experiment.
(1) Science is a cycle of theory and experiment. Scientific theories 1 are created to explain the results of experiments that were created under certain conditions. A successful theory will also make new predictions about new experiments under new conditions. Eventually, though, it always seems to happen that a new experiment comes along, showing that under certain conditions the theory is not a good approximation or is not valid at all. The ball is then back in the theorists’ court. If an experiment disagrees with the current theory, the theory has to be changed, not the experiment. (2) Theories should both predict and explain. The requirement of predictive power means that a theory is only meaningful if it predicts something that can be checked against experimental measurements that the theorist did not already have at hand. That is, a theory should be testable. Explanatory value means that many phenomena should be accounted for with few basic principles. If you answer every “why” question with “because that’s the way it is,” then your theory has no explanatory value. Collecting lots of data without being able to find any basic underlying principles is not science. (3) Experiments should be reproducible. An experiment should be treated with suspicion if it only works for one person, or only in one part of the world. Anyone with the necessary skills and equipment should be able to get the same results from the same experiment. This implies that science transcends national and ethnic boundaries; you can be sure that nobody is doing actual science who claims that their work is “Aryan, not Jewish,” “Marxist, not bourgeois,” or “Christian, not atheistic.” An experiment cannot be reproduced if it is secret, so science is necessarily a public enterprise.
b / A satirical drawing of an alchemist’s laboratory. H. Cock, after a drawing by Peter Brueghel the Elder (16th century).
16
Chapter 0
As an example of the cycle of theory and experiment, a vital step toward modern chemistry was the experimental observation that the chemical elements could not be transformed into each other, e.g., lead could not be turned into gold. This led to the theory that chemical reactions consisted of rearrangements of the elements in 1
The term “theory” in science does not just mean “what someone thinks,” or even “what a lot of scientists think.” It means an interrelated set of statements that have predictive value, and that have survived a broad set of empirical tests. Thus, both Newton’s law of gravity and Darwinian evolution are scientific theories. A “hypothesis,” in contrast to a theory, is any statement of interest that can be empirically tested. That the moon is made of cheese is a hypothesis, which was empirically tested, for example, by the Apollo astronauts.
Introduction and review
different combinations, without any change in the identities of the elements themselves. The theory worked for hundreds of years, and was confirmed experimentally over a wide range of pressures and temperatures and with many combinations of elements. Only in the twentieth century did we learn that one element could be transformed into one another under the conditions of extremely high pressure and temperature existing in a nuclear bomb or inside a star. That observation didn’t completely invalidate the original theory of the immutability of the elements, but it showed that it was only an approximation, valid at ordinary temperatures and pressures. self-check A A psychic conducts seances in which the spirits of the dead speak to the participants. He says he has special psychic powers not possessed by other people, which allow him to “channel” the communications with the spirits. What part of the scientific method is being violated here? Answer, p. 540
The scientific method as described here is an idealization, and should not be understood as a set procedure for doing science. Scientists have as many weaknesses and character flaws as any other group, and it is very common for scientists to try to discredit other people’s experiments when the results run contrary to their own favored point of view. Successful science also has more to do with luck, intuition, and creativity than most people realize, and the restrictions of the scientific method do not stifle individuality and self-expression any more than the fugue and sonata forms stifled Bach and Haydn. There is a recent tendency among social scientists to go even further and to deny that the scientific method even exists, claiming that science is no more than an arbitrary social system that determines what ideas to accept based on an in-group’s criteria. I think that’s going too far. If science is an arbitrary social ritual, it would seem difficult to explain its effectiveness in building such useful items as airplanes, CD players, and sewers. If alchemy and astrology were no less scientific in their methods than chemistry and astronomy, what was it that kept them from producing anything useful? Discussion questions Consider whether or not the scientific method is being applied in the following examples. If the scientific method is not being applied, are the people whose actions are being described performing a useful human activity, albeit an unscientific one? A Acupuncture is a traditional medical technique of Asian origin in which small needles are inserted in the patient’s body to relieve pain. Many doctors trained in the west consider acupuncture unworthy of experimental study because if it had therapeutic effects, such effects could not be explained by their theories of the nervous system. Who is being more scientific, the western or eastern practitioners?
Section 0.1
The scientific method
17
B Goethe, a German poet, is less well known for his theory of color. He published a book on the subject, in which he argued that scientific apparatus for measuring and quantifying color, such as prisms, lenses and colored filters, could not give us full insight into the ultimate meaning of color, for instance the cold feeling evoked by blue and green or the heroic sentiments inspired by red. Was his work scientific? C A child asks why things fall down, and an adult answers “because of gravity.” The ancient Greek philosopher Aristotle explained that rocks fell because it was their nature to seek out their natural place, in contact with the earth. Are these explanations scientific? D Buddhism is partly a psychological explanation of human suffering, and psychology is of course a science. The Buddha could be said to have engaged in a cycle of theory and experiment, since he worked by trial and error, and even late in his life he asked his followers to challenge his ideas. Buddhism could also be considered reproducible, since the Buddha told his followers they could find enlightenment for themselves if they followed a certain course of study and discipline. Is Buddhism a scientific pursuit?
0.2 What is physics? Given for one instant an intelligence which could comprehend all the forces by which nature is animated and the respective positions of the things which compose it...nothing would be uncertain, and the future as the past would be laid out before its eyes. Pierre Simon de Laplace Physics is the use of the scientific method to find out the basic principles governing light and matter, and to discover the implications of those laws. Part of what distinguishes the modern outlook from the ancient mind-set is the assumption that there are rules by which the universe functions, and that those laws can be at least partially understood by humans. From the Age of Reason through the nineteenth century, many scientists began to be convinced that the laws of nature not only could be known but, as claimed by Laplace, those laws could in principle be used to predict everything about the universe’s future if complete information was available about the present state of all light and matter. In subsequent sections, I’ll describe two general types of limitations on prediction using the laws of physics, which were only recognized in the twentieth century. Matter can be defined as anything that is affected by gravity, i.e., that has weight or would have weight if it was near the Earth or another star or planet massive enough to produce measurable gravity. Light can be defined as anything that can travel from one place to another through empty space and can influence matter, but has no weight. For example, sunlight can influence your body by heating it or by damaging your DNA and giving you skin cancer. The physicist’s definition of light includes a variety of phenomena
18
Chapter 0
Introduction and review
that are not visible to the eye, including radio waves, microwaves, x-rays, and gamma rays. These are the “colors” of light that do not happen to fall within the narrow violet-to-red range of the rainbow that we can see. self-check B At the turn of the 20th century, a strange new phenomenon was discovered in vacuum tubes: mysterious rays of unknown origin and nature. These rays are the same as the ones that shoot from the back of your TV’s picture tube and hit the front to make the picture. Physicists in 1895 didn’t have the faintest idea what the rays were, so they simply named them “cathode rays,” after the name for the electrical contact from which they sprang. A fierce debate raged, complete with nationalistic overtones, over whether the rays were a form of light or of matter. What would they have had to do in order to settle the issue? Answer, p. 540
Many physical phenomena are not themselves light or matter, but are properties of light or matter or interactions between light and matter. For instance, motion is a property of all light and some matter, but it is not itself light or matter. The pressure that keeps a bicycle tire blown up is an interaction between the air and the tire. Pressure is not a form of matter in and of itself. It is as much a property of the tire as of the air. Analogously, sisterhood and employment are relationships among people but are not people themselves. Some things that appear weightless actually do have weight, and so qualify as matter. Air has weight, and is thus a form of matter even though a cubic inch of air weighs less than a grain of sand. A helium balloon has weight, but is kept from falling by the force of the surrounding more dense air, which pushes up on it. Astronauts in orbit around the Earth have weight, and are falling along a curved arc, but they are moving so fast that the curved arc of their fall is broad enough to carry them all the way around the Earth in a circle. They perceive themselves as being weightless because their space capsule is falling along with them, and the floor therefore does not push up on their feet. Optional Topic: Modern Changes in the Definition of Light and Matter Einstein predicted as a consequence of his theory of relativity that light would after all be affected by gravity, although the effect would be extremely weak under normal conditions. His prediction was borne out by observations of the bending of light rays from stars as they passed close to the sun on their way to the Earth. Einstein’s theory also implied the existence of black holes, stars so massive and compact that their intense gravity would not even allow light to escape. (These days there is strong evidence that black holes exist.) Einstein’s interpretation was that light doesn’t really have mass, but that energy is affected by gravity just like mass is. The energy in a light
Section 0.2
c / This telescope picture shows two images of the same distant object, an exotic, very luminous object called a quasar. This is interpreted as evidence that a massive, dark object, possibly a black hole, happens to be between us and it. Light rays that would otherwise have missed the earth on either side have been bent by the dark object’s gravity so that they reach us. The actual direction to the quasar is presumably in the center of the image, but the light along that central line doesn’t get to us because it is absorbed by the dark object. The quasar is known by its catalog number, MG1131+0456, or more informally as Einstein’s Ring.
What is physics?
19
beam is equivalent to a certain amount of mass, given by the famous equation E = mc 2 , where c is the speed of light. Because the speed of light is such a big number, a large amount of energy is equivalent to only a very small amount of mass, so the gravitational force on a light ray can be ignored for most practical purposes. There is however a more satisfactory and fundamental distinction between light and matter, which should be understandable to you if you have had a chemistry course. In chemistry, one learns that electrons obey the Pauli exclusion principle, which forbids more than one electron from occupying the same orbital if they have the same spin. The Pauli exclusion principle is obeyed by the subatomic particles of which matter is composed, but disobeyed by the particles, called photons, of which a beam of light is made. Einstein’s theory of relativity is discussed more fully in book 6 of this series.
The boundary between physics and the other sciences is not always clear. For instance, chemists study atoms and molecules, which are what matter is built from, and there are some scientists who would be equally willing to call themselves physical chemists or chemical physicists. It might seem that the distinction between physics and biology would be clearer, since physics seems to deal with inanimate objects. In fact, almost all physicists would agree that the basic laws of physics that apply to molecules in a test tube work equally well for the combination of molecules that constitutes a bacterium. (Some might believe that something more happens in the minds of humans, or even those of cats and dogs.) What differentiates physics from biology is that many of the scientific theories that describe living things, while ultimately resulting from the fundamental laws of physics, cannot be rigorously derived from physical principles. Isolated systems and reductionism To avoid having to study everything at once, scientists isolate the things they are trying to study. For instance, a physicist who wants to study the motion of a rotating gyroscope would probably prefer that it be isolated from vibrations and air currents. Even in biology, where field work is indispensable for understanding how living things relate to their entire environment, it is interesting to note the vital historical role played by Darwin’s study of the Gal´apagos Islands, which were conveniently isolated from the rest of the world. Any part of the universe that is considered apart from the rest can be called a “system.” Physics has had some of its greatest successes by carrying this process of isolation to extremes, subdividing the universe into smaller and smaller parts. Matter can be divided into atoms, and the behavior of individual atoms can be studied. Atoms can be split apart
20
Chapter 0
d / Reductionism.
Introduction and review
into their constituent neutrons, protons and electrons. Protons and neutrons appear to be made out of even smaller particles called quarks, and there have even been some claims of experimental evidence that quarks have smaller parts inside them. This method of splitting things into smaller and smaller parts and studying how those parts influence each other is called reductionism. The hope is that the seemingly complex rules governing the larger units can be better understood in terms of simpler rules governing the smaller units. To appreciate what reductionism has done for science, it is only necessary to examine a 19th-century chemistry textbook. At that time, the existence of atoms was still doubted by some, electrons were not even suspected to exist, and almost nothing was understood of what basic rules governed the way atoms interacted with each other in chemical reactions. Students had to memorize long lists of chemicals and their reactions, and there was no way to understand any of it systematically. Today, the student only needs to remember a small set of rules about how atoms interact, for instance that atoms of one element cannot be converted into another via chemical reactions, or that atoms from the right side of the periodic table tend to form strong bonds with atoms from the left side. Discussion questions A I’ve suggested replacing the ordinary dictionary definition of light with a more technical, more precise one that involves weightlessness. It’s still possible, though, that the stuff a lightbulb makes, ordinarily called “light,” does have some small amount of weight. Suggest an experiment to attempt to measure whether it does. B Heat is weightless (i.e., an object becomes no heavier when heated), and can travel across an empty room from the fireplace to your skin, where it influences you by heating you. Should heat therefore be considered a form of light by our definition? Why or why not? C
Similarly, should sound be considered a form of light?
0.3 How to learn physics For as knowledges are now delivered, there is a kind of contract of error between the deliverer and the receiver; for he that delivereth knowledge desireth to deliver it in such a form as may be best believed, and not as may be best examined; and he that receiveth knowledge desireth rather present satisfaction than expectant inquiry. Francis Bacon Many students approach a science course with the idea that they can succeed by memorizing the formulas, so that when a problem
Section 0.3
How to learn physics
21
is assigned on the homework or an exam, they will be able to plug numbers in to the formula and get a numerical result on their calculator. Wrong! That’s not what learning science is about! There is a big difference between memorizing formulas and understanding concepts. To start with, different formulas may apply in different situations. One equation might represent a definition, which is always true. Another might be a very specific equation for the speed of an object sliding down an inclined plane, which would not be true if the object was a rock drifting down to the bottom of the ocean. If you don’t work to understand physics on a conceptual level, you won’t know which formulas can be used when. Most students taking college science courses for the first time also have very little experience with interpreting the meaning of an equation. Consider the equation w = A/h relating the width of a rectangle to its height and area. A student who has not developed skill at interpretation might view this as yet another equation to memorize and plug in to when needed. A slightly more savvy student might realize that it is simply the familiar formula A = wh in a different form. When asked whether a rectangle would have a greater or smaller width than another with the same area but a smaller height, the unsophisticated student might be at a loss, not having any numbers to plug in on a calculator. The more experienced student would know how to reason about an equation involving division — if h is smaller, and A stays the same, then w must be bigger. Often, students fail to recognize a sequence of equations as a derivation leading to a final result, so they think all the intermediate steps are equally important formulas that they should memorize. When learning any subject at all, it is important to become as actively involved as possible, rather than trying to read through all the information quickly without thinking about it. It is a good idea to read and think about the questions posed at the end of each section of these notes as you encounter them, so that you know you have understood what you were reading. Many students’ difficulties in physics boil down mainly to difficulties with math. Suppose you feel confident that you have enough mathematical preparation to succeed in this course, but you are having trouble with a few specific things. In some areas, the brief review given in this chapter may be sufficient, but in other areas it probably will not. Once you identify the areas of math in which you are having problems, get help in those areas. Don’t limp along through the whole course with a vague feeling of dread about something like scientific notation. The problem will not go away if you ignore it. The same applies to essential mathematical skills that you are learning in this course for the first time, such as vector addition. Sometimes students tell me they keep trying to understand a
22
Chapter 0
Introduction and review
certain topic in the book, and it just doesn’t make sense. The worst thing you can possibly do in that situation is to keep on staring at the same page. Every textbook explains certain things badly — even mine! — so the best thing to do in this situation is to look at a different book. Instead of college textbooks aimed at the same mathematical level as the course you’re taking, you may in some cases find that high school books or books at a lower math level give clearer explanations. Finally, when reviewing for an exam, don’t simply read back over the text and your lecture notes. Instead, try to use an active method of reviewing, for instance by discussing some of the discussion questions with another student, or doing homework problems you hadn’t done the first time.
0.4 Self-evaluation The introductory part of a book like this is hard to write, because every student arrives at this starting point with a different preparation. One student may have grown up outside the U.S. and so may be completely comfortable with the metric system, but may have had an algebra course in which the instructor passed too quickly over scientific notation. Another student may have already taken calculus, but may have never learned the metric system. The following self-evaluation is a checklist to help you figure out what you need to study to be prepared for the rest of the course. If you disagree with this statement. . . I am familiar with the basic metric units of meters, kilograms, and seconds, and the most common metric prefixes: milli- (m), kilo- (k), and centi- (c). I know about the newton, a unit of force I am familiar with these less common metric prefixes: mega- (M), micro- (μ), and nano- (n). I am comfortable with scientific notation. I can confidently do metric conversions. I understand the purpose and use of significant figures.
you should study this section: section 0.5 Basic of the Metric System
section 0.6 The newton, the Metric Unit of Force section 0.7 Less Common Metric Prefixes section 0.8 Scientific Notation section 0.9 Conversions section 0.10 Significant Figures
It wouldn’t hurt you to skim the sections you think you already know about, and to do the self-checks in those sections.
Section 0.4
Self-evaluation
23
0.5 Basics of the metric system The metric system Units were not standardized until fairly recently in history, so when the physicist Isaac Newton gave the result of an experiment with a pendulum, he had to specify not just that the string was 37 7 / inches long but that it was “37 7 / London inches long.” The 8 8 inch as defined in Yorkshire would have been different. Even after the British Empire standardized its units, it was still very inconvenient to do calculations involving money, volume, distance, time, or weight, because of all the odd conversion factors, like 16 ounces in a pound, and 5280 feet in a mile. Through the nineteenth century, schoolchildren squandered most of their mathematical education in preparing to do calculations such as making change when a customer in a shop offered a one-crown note for a book costing two pounds, thirteen shillings and tuppence. The dollar has always been decimal, and British money went decimal decades ago, but the United States is still saddled with the antiquated system of feet, inches, pounds, ounces and so on. Every country in the world besides the U.S. uses a system of units known in English as the “metric system.2 ” This system is entirely decimal, thanks to the same eminently logical people who brought about the French Revolution. In deference to France, the system’s official name is the Syst`eme International, or SI, meaning International System. (The phrase “SI system” is therefore redundant.) The wonderful thing about the SI is that people who live in countries more modern than ours do not need to memorize how many ounces there are in a pound, how many cups in a pint, how many feet in a mile, etc. The whole system works with a single, consistent set of Greek and Latin prefixes that modify the basic units. Each prefix stands for a power of ten, and has an abbreviation that can be combined with the symbol for the unit. For instance, the meter is a unit of distance. The prefix kilo- stands for 103 , so a kilometer, 1 km, is a thousand meters. The basic units of the metric system are the meter for distance, the second for time, and the gram for mass. The following are the most common metric prefixes. You should memorize them. prefix meaning example 60 kg = a person’s mass kilok 103 28 cm = height of a piece of paper centi- c 10−2 1 ms = time for one vibration of a guitar milli- m 10−3 string playing the note D 2
Liberia and Myanmar have not legally adopted metric units, but use them in everyday life.
24
Chapter 0
Introduction and review
The prefix centi-, meaning 10−2 , is only used in the centimeter; a hundredth of a gram would not be written as 1 cg but as 10 mg. The centi- prefix can be easily remembered because a cent is 10−2 dollars. The official SI abbreviation for seconds is “s” (not “sec”) and grams are “g” (not “gm”). The second When I stated briefly above that the second was a unit of time, it may not have occurred to you that this was not much of a definition. We can make a dictionary-style definition of a term like “time,” or give a general description like Isaac Newton’s: “Absolute, true, and mathematical time, of itself, and from its own nature, flows equably without relation to anything external. . . ” Newton’s characterization sounds impressive, but physicists today would consider it useless as a definition of time. Today, the physical sciences are based on operational definitions, which means definitions that spell out the actual steps (operations) required to measure something numerically. In an era when our toasters, pens, and coffee pots tell us the time, it is far from obvious to most people what is the fundamental operational definition of time. Until recently, the hour, minute, and second were defined operationally in terms of the time required for the earth to rotate about its axis. Unfortunately, the Earth’s rotation is slowing down slightly, and by 1967 this was becoming an issue in scientific experiments requiring precise time measurements. The second was therefore redefined as the time required for a certain number of vibrations of the light waves emitted by a cesium atoms in a lamp constructed like a familiar neon sign but with the neon replaced by cesium. The new definition not only promises to stay constant indefinitely, but for scientists is a more convenient way of calibrating a clock than having to carry out astronomical measurements. self-check C What is a possible operational definition of how strong a person is? Answer, p. 540
The meter The French originally defined the meter as 10−7 times the distance from the equator to the north pole, as measured through Paris (of course). Even if the definition was operational, the operation of traveling to the north pole and laying a surveying chain behind you was not one that most working scientists wanted to carry out. Fairly soon, a standard was created in the form of a metal bar with two scratches on it. This was replaced by an atomic standard in 1960, and finally in 1983 by the current definition, which is that the meter is the distance traveled by light in a vacuum over a period of (1/299792458) seconds.
Section 0.5
e / The original the meter.
Basics of the metric system
definition
of
25
The kilogram The third base unit of the SI is the kilogram, a unit of mass. Mass is intended to be a measure of the amount of a substance, but that is not an operational definition. Bathroom scales work by measuring our planet’s gravitational attraction for the object being weighed, but using that type of scale to define mass operationally would be undesirable because gravity varies in strength from place to place on the earth.
f / A duplicate of the Paris kilogram, maintained at the Danish National Metrology Institute.
There’s a surprising amount of disagreement among physics textbooks about how mass should be defined, but here’s how it’s actually handled by the few working physicists who specialize in ultra-highprecision measurements. They maintain a physical object in Paris, which is the standard kilogram, a cylinder made of platinum-iridium alloy. Duplicates are checked against this mother of all kilograms by putting the original and the copy on the two opposite pans of a balance. Although this method of comparison depends on gravity, the problems associated with differences in gravity in different geographical locations are bypassed, because the two objects are being compared in the same place. The duplicates can then be removed from the Parisian kilogram shrine and transported elsewhere in the world. It would be desirable to replace this at some point with a universally accessible atomic standard rather than one based on a specific artifact, but as of 2010 the technology for automated counting of large numbers of atoms has not gotten good enough to make that work with the desired precision.
Combinations of metric units Just about anything you want to measure can be measured with some combination of meters, kilograms, and seconds. Speed can be measured in m/s, volume in m3 , and density in kg/m3 . Part of what makes the SI great is this basic simplicity. No more funny units like a cord of wood, a bolt of cloth, or a jigger of whiskey. No more liquid and dry measure. Just a simple, consistent set of units. The SI measures put together from meters, kilograms, and seconds make up the mks system. For example, the mks unit of speed is m/s, not km/hr.
Checking units A useful technique for finding mistakes in one’s algebra is to analyze the units associated with the variables.
26
Chapter 0
Introduction and review
Checking units example 1 1 Jae starts from the formula V = 3 Ah for the volume of a cone, where A is the area of its base, and h is its height. He wants to find an equation that will tell him how tall a conical tent has to be in order to have a certain volume, given its radius. His algebra goes like this:
[1] [2] [3] [4]
1 Ah 3 A = πr 2 1 V = πr 2 h 3 πr 2 h= 3V
V =
Is his algebra correct? If not, find the mistake. Line 4 is supposed to be an equation for the height, so the units of the expression on the right-hand side had better equal meters. The pi and the 3 are unitless, so we can ignore them. In terms of units, line 4 becomes
m=
m2 1 = 3 m m
.
This is false, so there must be a mistake in the algebra. The units of lines 1, 2, and 3 check out, so the mistake must be in the step from line 3 to line 4. In fact the result should have been
h=
3V πr 2
.
Now the units check: m = m3 /m2 . Discussion question A Isaac Newton wrote, “. . . the natural days are truly unequal, though they are commonly considered as equal, and used for a measure of time. . . It may be that there is no such thing as an equable motion, whereby time may be accurately measured. All motions may be accelerated or retarded. . . ” Newton was right. Even the modern definition of the second in terms of light emitted by cesium atoms is subject to variation. For instance, magnetic fields could cause the cesium atoms to emit light with a slightly different rate of vibration. What makes us think, though, that a pendulum clock is more accurate than a sundial, or that a cesium atom is a more accurate timekeeper than a pendulum clock? That is, how can one test experimentally how the accuracies of different time standards compare?
Section 0.5
Basics of the metric system
27
0.6 The Newton, the metric unit of force A force is a push or a pull, or more generally anything that can change an object’s speed or direction of motion. A force is required to start a car moving, to slow down a baseball player sliding in to home base, or to make an airplane turn. (Forces may fail to change an object’s motion if they are canceled by other forces, e.g., the force of gravity pulling you down right now is being canceled by the force of the chair pushing up on you.) The metric unit of force is the Newton, defined as the force which, if applied for one second, will cause a 1-kilogram object starting from rest to reach a speed of 1 m/s. Later chapters will discuss the force concept in more detail. In fact, this entire book is about the relationship between force and motion. In section 0.5, I gave a gravitational definition of mass, but by defining a numerical scale of force, we can also turn around and define a scale of mass without reference to gravity. For instance, if a force of two Newtons is required to accelerate a certain object from rest to 1 m/s in 1 s, then that object must have a mass of 2 kg. From this point of view, mass characterizes an object’s resistance to a change in its motion, which we call inertia or inertial mass. Although there is no fundamental reason why an object’s resistance to a change in its motion must be related to how strongly gravity affects it, careful and precise experiments have shown that the inertial definition and the gravitational definition of mass are highly consistent for a variety of objects. It therefore doesn’t really matter for any practical purpose which definition one adopts. Discussion question A Spending a long time in weightlessness is unhealthy. One of the most important negative effects experienced by astronauts is a loss of muscle and bone mass. Since an ordinary scale won’t work for an astronaut in orbit, what is a possible way of monitoring this change in mass? (Measuring the astronaut’s waist or biceps with a measuring tape is not good enough, because it doesn’t tell anything about bone mass, or about the replacement of muscle with fat.)
0.7 Less common metric prefixes g / This is a mnemonic to help you remember the most important metric prefixes. The word “little” is to remind you that the list starts with the prefixes used for small quantities and builds upward. The exponent changes by 3, except that of course that we do not need a special prefix for 100 , which equals one.
28
Chapter 0
The following are three metric prefixes which, while less common than the ones discussed previously, are well worth memorizing. prefix mega- M micro- μ nano- n
meaning 106 10−6 10−9
6.4 Mm 10 μm 0.154 nm
example = radius of the earth = size of a white blood cell = distance between carbon nuclei in an ethane molecule
Note that the abbreviation for micro is the Greek letter mu, μ — a common mistake is to confuse it with m (milli) or M (mega).
Introduction and review
There are other prefixes even less common, used for extremely large and small quantities. For instance, 1 femtometer = 10−15 m is a convenient unit of distance in nuclear physics, and 1 gigabyte = 109 bytes is used for computers’ hard disks. The international committee that makes decisions about the SI has recently even added some new prefixes that sound like jokes, e.g., 1 yoctogram = 10−24 g is about half the mass of a proton. In the immediate future, however, you’re unlikely to see prefixes like “yocto-” and “zepto-” used except perhaps in trivia contests at science-fiction conventions or other geekfests. self-check D Suppose you could slow down time so that according to your perception, a beam of light would move across a room at the speed of a slow walk. If you perceived a nanosecond as if it was a second, how would you perceive a microsecond? Answer, p. 540
0.8 Scientific notation Most of the interesting phenomena in our universe are not on the human scale. It would take about 1,000,000,000,000,000,000,000 bacteria to equal the mass of a human body. When the physicist Thomas Young discovered that light was a wave, it was back in the bad old days before scientific notation, and he was obliged to write that the time required for one vibration of the wave was 1/500 of a millionth of a millionth of a second. Scientific notation is a less awkward way to write very large and very small numbers such as these. Here’s a quick review. Scientific notation means writing a number in terms of a product of something from 1 to 10 and something else that is a power of ten. For instance, 32 = 3.2 × 101 320 = 3.2 × 102 3200 = 3.2 × 103
...
Each number is ten times bigger than the previous one. Since 101 is ten times smaller than 102 , it makes sense to use the notation 100 to stand for one, the number that is in turn ten times smaller than 101 . Continuing on, we can write 10−1 to stand for 0.1, the number ten times smaller than 100 . Negative exponents are used for small numbers:
3.2 = 3.2 × 100 0.32 = 3.2 × 10−1 0.032 = 3.2 × 10−2
...
Section 0.8
Scientific notation
29
A common source of confusion is the notation used on the displays of many calculators. Examples:
3.2 × 106 3.2E+6 3.26
(written notation) (notation on some calculators) (notation on some other calculators)
The last example is particularly unfortunate, because 3.26 really stands for the number 3.2 × 3.2 × 3.2 × 3.2 × 3.2 × 3.2 = 1074, a totally different number from 3.2 × 106 = 3200000. The calculator notation should never be used in writing. It’s just a way for the manufacturer to save money by making a simpler display. self-check E A student learns that 104 bacteria, standing in line to register for classes at Paramecium Community College, would form a queue of this size: The student concludes that 102 bacteria would form a line of this length:
Why is the student incorrect?
Answer, p. 540
0.9 Conversions Conversions are one of the three essential mathematical skills, summarized on pp.524-525, that you need for success in this course. I suggest you avoid memorizing lots of conversion factors between SI units and U.S. units, but two that do come in handy are: 1 inch = 2.54 cm An object with a weight on Earth of 2.2 pounds-force has a mass of 1 kg. The first one is the present definition of the inch, so it’s exact. The second one is not exact, but is good enough for most purposes. (U.S. units of force and mass are confusing, so it’s a good thing they’re not used in science. In U.S. units, the unit of force is the poundforce, and the best unit to use for mass is the slug, which is about 14.6 kg.) More important than memorizing conversion factors is understanding the right method for doing conversions. Even within the SI, you may need to convert, say, from grams to kilograms. Different people have different ways of thinking about conversions, but the method I’ll describe here is systematic and easy to understand. The idea is that if 1 kg and 1000 g represent the same mass, then
30
Chapter 0
Introduction and review
we can consider a fraction like 103 g 1 kg to be a way of expressing the number one. This may bother you. For instance, if you type 1000/1 into your calculator, you will get 1000, not one. Again, different people have different ways of thinking about it, but the justification is that it helps us to do conversions, and it works! Now if we want to convert 0.7 kg to units of grams, we can multiply kg by the number one: 0.7 kg ×
103 g 1 kg
If you’re willing to treat symbols such as “kg” as if they were variables as used in algebra (which they’re really not), you can then cancel the kg on top with the kg on the bottom, resulting in 0.7 kg ×
103 g = 700 g 1 kg
.
To convert grams to kilograms, you would simply flip the fraction upside down. One advantage of this method is that it can easily be applied to a series of conversions. For instance, to convert one year to units of seconds,
× 1 year
24 60 days hours min 365 60 s × × × = 1 year 1 day 1 hour 1 min = 3.15 × 107 s
.
Should that exponent be positive, or negative? A common mistake is to write the conversion fraction incorrectly. For instance the fraction 103 kg 1g
(incorrect)
does not equal one, because 103 kg is the mass of a car, and 1 g is the mass of a raisin. One correct way of setting up the conversion factor would be 10−3 kg (correct) . 1g You can usually detect such a mistake if you take the time to check your answer and see if it is reasonable. If common sense doesn’t rule out either a positive or a negative exponent, here’s another way to make sure you get it right. There are big prefixes and small prefixes:
Section 0.9
Conversions
31
big prefixes: small prefixes:
k m
M μ
n
(It’s not hard to keep straight which are which, since “mega” and “micro” are evocative, and it’s easy to remember that a kilometer is bigger than a meter and a millimeter is smaller.) In the example above, we want the top of the fraction to be the same as the bottom. Since k is a big prefix, we need to compensate by putting a small number like 10−3 in front of it, not a big number like 103 . Solved problem: a simple conversion
page 37, problem 6
Solved problem: the geometric mean
page 38, problem 8
Discussion question A Each of the following conversions contains an error. In each case, explain what the error is. (a) 1000 kg × (b) 50 m ×
1 kg 1000 g
1 cm 100 m
=1g
= 0.5 cm
(c) “Nano” is 10−9 , so there are 10−9 nm in a meter. (d) “Micro” is 10−6 , so 1 kg is 106 μg.
0.10 Significant figures An engineer is designing a car engine, and has been told that the diameter of the pistons (which are being designed by someone else) is 5 cm. He knows that 0.02 cm of clearance is required for a piston of this size, so he designs the cylinder to have an inside diameter of 5.04 cm. Luckily, his supervisor catches his mistake before the car goes into production. She explains his error to him, and mentally puts him in the “do not promote” category. What was his mistake? The person who told him the pistons were 5 cm in diameter was wise to the ways of significant figures, as was his boss, who explained to him that he needed to go back and get a more accurate number for the diameter of the pistons. That person said “5 cm” rather than “5.00 cm” specifically to avoid creating the impression that the number was extremely accurate. In reality, the pistons’ diameter was 5.13 cm. They would never have fit in the 5.04-cm cylinders. The number of digits of accuracy in a number is referred to as the number of significant figures, or “sig figs” for short. As in the example above, sig figs provide a way of showing the accuracy of a number. In most cases, the result of a calculation involving several pieces of data can be no more accurate than the least accurate piece of data. In other words, “garbage in, garbage out.” Since the 5 cm diameter of the pistons was not very accurate, the result of the engineer’s calculation, 5.04 cm, was really not as accurate as he
32
Chapter 0
Introduction and review
thought. In general, your result should not have more than the number of sig figs in the least accurate piece of data you started with. The calculation above should have been done as follows:
5 cm +0.04 cm =5 cm
(1 sig fig) (1 sig fig) (rounded off to 1 sig fig)
The fact that the final result only has one significant figure then alerts you to the fact that the result is not very accurate, and would not be appropriate for use in designing the engine. Note that the leading zeroes in the number 0.04 do not count as significant figures, because they are only placeholders. On the other hand, a number such as 50 cm is ambiguous — the zero could be intended as a significant figure, or it might just be there as a placeholder. The ambiguity involving trailing zeroes can be avoided by using scientific notation, in which 5 × 101 cm would imply one sig fig of accuracy, while 5.0 × 101 cm would imply two sig figs. self-check F The following quote is taken from an editorial by Norimitsu Onishi in the New York Times, August 18, 2002. Consider Nigeria. Everyone agrees it is Africa’s most populous nation. But what is its population? The United Nations says 114 million; the State Department, 120 million. The World Bank says 126.9 million, while the Central Intelligence Agency puts it at 126,635,626. What should bother you about this?
Answer, p. 540
Dealing correctly with significant figures can save you time! Often, students copy down numbers from their calculators with eight significant figures of precision, then type them back in for a later calculation. That’s a waste of time, unless your original data had that kind of incredible precision. The rules about significant figures are only rules of thumb, and are not a substitute for careful thinking. For instance, $20.00 + $0.05 is $20.05. It need not and should not be rounded off to $20. In general, the sig fig rules work best for multiplication and division, and we also apply them when doing a complicated calculation that involves many types of operations. For simple addition and subtraction, it makes more sense to maintain a fixed number of digits after the decimal point. When in doubt, don’t use the sig fig rules at all. Instead, intentionally change one piece of your initial data by the maximum amount by which you think it could have been off, and recalculate the final result. The digits on the end that are completely reshuffled are the ones that are meaningless, and should be omitted.
Section 0.10
Significant figures
33
self-check G How many significant figures are there in each of the following measurements? (1) 9.937 m (2) 4.0 s (3) 0.0000000000000037 kg
34
Chapter 0
Introduction and review
Answer, p. 540
Summary Selected vocabulary matter . . . . . . Anything that is affected by gravity. light . . . . . . . . Anything that can travel from one place to another through empty space and can influence matter, but is not affected by gravity. operational defi- A definition that states what operations nition . . . . . . . should be carried out to measure the thing being defined. Syst`eme Interna- A fancy name for the metric system. tional . . . . . . . mks system . . . The use of metric units based on the meter, kilogram, and second. Example: meters per second is the mks unit of speed, not cm/s or km/hr. mass . . . . . . . A numerical measure of how difficult it is to change an object’s motion. significant figures Digits that contribute to the accuracy of a measurement. Notation m . . . . kg . . . . s . . . . . M- . . . . k- . . . . m- . . . . μ- . . . . n- . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
meter, the metric distance unit kilogram, the metric unit of mass second, the metric unit of time the metric prefix mega-, 106 the metric prefix kilo-, 103 the metric prefix milli-, 10−3 the metric prefix micro-, 10−6 the metric prefix nano-, 10−9
Summary Physics is the use of the scientific method to study the behavior of light and matter. The scientific method requires a cycle of theory and experiment, theories with both predictive and explanatory value, and reproducible experiments. The metric system is a simple, consistent framework for measurement built out of the meter, the kilogram, and the second plus a set of prefixes denoting powers of ten. The most systematic method for doing conversions is shown in the following example: 370 ms ×
10−3 s = 0.37 s 1 ms
Mass is a measure of the amount of a substance. Mass can be defined gravitationally, by comparing an object to a standard mass on a double-pan balance, or in terms of inertia, by comparing the effect of a force on an object to the effect of the same force on a standard mass. The two definitions are found experimentally to be proportional to each other to a high degree of precision, so we
Summary
35
usually refer simply to “mass,” without bothering to specify which type. A force is that which can change the motion of an object. The metric unit of force is the Newton, defined as the force required to accelerate a standard 1-kg mass from rest to a speed of 1 m/s in 1 s. Scientific notation means, for example, writing 3.2 × 105 rather than 320000. Writing numbers with the correct number of significant figures correctly communicates how accurate they are. As a rule of thumb, the final result of a calculation is no more accurate than, and should have no more significant figures than, the least accurate piece of data.
36
Chapter 0
Introduction and review
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
74658 on a cal1 Correct use of a calculator: (a) Calculate 53222+97554 culator. [Self-check: The most common mistake results in 97555.40.] √
(b) Which would be more like the price of a TV, and which would be more like the price of a house, $3.5 × 105 or $3.55 ? 2 Compute the following things. If they don’t make sense because of units, say so. (a) 3 cm + 5 cm (b) 1.11 m + 22 cm (c) 120 miles + 2.0 hours (d) 120 miles / 2.0 hours 3 Your backyard has brick walls on both ends. You measure a distance of 23.4 m from the inside of one wall to the inside of the other. Each wall is 29.4 cm thick. How far is it from the outside of one wall to the outside of the other? Pay attention to significant figures. 4 The speed of light is 3.0 × 108 m/s. Convert this to furlongs per fortnight. A furlong is 220 yards, and a fortnight is 14 days. An √ inch is 2.54 cm. 5 Express each of the following quantities in micrograms: (a) 10 mg, (b) 104 g, (c) 10 kg, (d) 100 × 103 g, (e) 1000 ng.
√
6 Convert 134 mg to units of kg, writing your answer in scientific notation. Solution, p. 526 7 In the last century, the average age of the onset of puberty for girls has decreased by several years. Urban folklore has it that this is because of hormones fed to beef cattle, but it is more likely to be because modern girls have more body fat on the average and possibly because of estrogen-mimicking chemicals in the environment from the breakdown of pesticides. A hamburger from a hormoneimplanted steer has about 0.2 ng of estrogen (about double the amount of natural beef). A serving of peas contains about 300 ng of estrogen. An adult woman produces about 0.5 mg of estrogen per day (note the different unit!). (a) How many hamburgers would a girl have to eat in one day to consume as much estrogen as an adult woman’s daily production? (b) How many servings of peas? √
Problems
37
8 The usual definition of the mean (average) of two numbers a and b is (a+b)/2. This is called the arithmetic mean. The geometric mean, however, is defined as (ab)1/2 (i.e., the square root of ab). For the sake of definiteness, let’s say both numbers have units of mass. (a) Compute the arithmetic mean of two numbers that have units of grams. Then convert the numbers to units of kilograms and recompute their mean. Is the answer consistent? (b) Do the same for the geometric mean. (c) If a and b both have units of grams, what should we call the units of ab? Does your answer make sense when you take the square root? (d) Suppose someone proposes to you a third kind of mean, called the superduper mean, defined as Solution, p. 526 (ab)1/3 . Is this reasonable? 9 In an article on the SARS epidemic, the May 7, 2003 New York Times discusses conflicting estimates of the disease’s incubation period (the average time that elapses from infection to the first symptoms). “The study estimated it to be 6.4 days. But other statistical calculations ... showed that the incubation period could be as long as 14.22 days.” What’s wrong here? 10 The photo shows the corner of a bag of pretzels. What’s wrong here? √ 11 The distance to the horizon is given by the expression 2rh, where r is the radius of the Earth, and h is the observer’s height above the Earth’s surface. (This can be proved using the Pythagorean theorem.) Show that the units of this expression make sense. (See example 1 on p. 27 for an example of how to do this.) Don’t try to prove the result, just check its units.
Problem 10.
12 (a) Based on the definitions of the sine, cosine, and tangent, what units must they have? (b) A cute formula from trigonometry lets you find any angle of a triangle if you know the lengths of its sides. Using the notation shown in the figure, and letting s = (a + b + c)/2 be half the perimeter, we have (s − b)(s − c) . tan A/2 = s(s − a) Show that the units of this equation make sense. In other words, check that the units of the right-hand side are the same as your answer to part a of the question. Solution, p. 526
Problem 12.
38
Chapter 0
Introduction and review
Exercise 0: Models and idealization Equipment: coffee filters ramps (one per group) balls of various sizes sticky tape vacuum pump and “guinea and feather” apparatus (one) The motion of falling objects has been recognized since ancient times as an important piece of physics, but the motion is inconveniently fast, so in our everyday experience it can be hard to tell exactly what objects are doing when they fall. In this exercise you will use several techniques to get around this problem and study the motion. Your goal is to construct a scientific model of falling. A model means an explanation that makes testable predictions. Often models contain simplifications or idealizations that make them easier to work with, even though they are not strictly realistic. 1. One method of making falling easier to observe is to use objects like feathers that we know from everyday experience will not fall as fast. You will use coffee filters, in stacks of various sizes, to test the following two hypotheses and see which one is true, or whether neither is true: Hypothesis 1A: When an object is dropped, it rapidly speeds up to a certain natural falling speed, and then continues to fall at that speed. The falling speed is proportional to the object’s weight. (A proportionality is not just a statement that if one thing gets bigger, the other does too. It says that if one becomes three times bigger, the other also gets three times bigger, etc.) Hypothesis 1B: Different objects fall the same way, regardless of weight. Test these hypotheses and discuss your results with your instructor. 2. A second way to slow down the action is to let a ball roll down a ramp. The steeper the ramp, the closer to free fall. Based on your experience in part 1, write a hypothesis about what will happen when you race a heavier ball against a lighter ball down the same ramp, starting them both from rest. Hypothesis: Show your hypothesis to your instructor, and then test it. You have probably found that falling was more complicated than you thought! Is there more than one factor that affects the motion of a falling object? Can you imagine certain idealized situations that are simpler? Try to agree verbally with your group on an informal model of falling that can make predictions about the experiments described in parts 3 and 4. 3. You have three balls: a standard “comparison ball” of medium weight, a light ball, and a heavy ball. Suppose you stand on a chair and (a) drop the light ball side by side with the comparison ball, then (b) drop the heavy ball side by side with the comparison ball, then (c) join the light and heavy balls together with sticky tape and drop them side by side with the comparison ball. Use your model to make a prediction: Test your prediction.
Exercise 0: Models and idealization
39
4. Your instructor will pump nearly all the air out of a chamber containing a feather and a heavier object, then let them fall side by side in the chamber. Use your model to make a prediction:
40
Chapter 0
Introduction and review
Life would be very different if you were the size of an insect.
Chapter 1
Scaling and estimation 1.1 Introduction Why can’t an insect be the size of a dog? Some skinny stretchedout cells in your spinal cord are a meter tall — why does nature display no single cells that are not just a meter tall, but a meter wide, and a meter thick as well? Believe it or not, these are questions that can be answered fairly easily without knowing much more about physics than you already do. The only mathematical technique you really need is the humble conversion, applied to area and volume. Area and volume
a / Amoebas this size seldom encountered.
are
Area can be defined by saying that we can copy the shape of interest onto graph paper with 1 cm × 1 cm squares and count the number of squares inside. Fractions of squares can be estimated by eye. We then say the area equals the number of squares, in units of square cm. Although this might seem less “pure” than computing areas using formulae like A = πr2 for a circle or A = wh/2 for a triangle, those formulae are not useful as definitions of area because they cannot be applied to irregularly shaped areas. Units of square cm are more commonly written as cm2 in science. Of course, the unit of measurement symbolized by “cm” is not an
41
algebra symbol standing for a number that can be literally multiplied by itself. But it is advantageous to write the units of area that way and treat the units as if they were algebra symbols. For instance, if you have a rectangle with an area of 6m2 and a width of 2 m, then calculating its length as (6 m2 )/(2 m) = 3 m gives a result that makes sense both numerically and in terms of units. This algebra-style treatment of the units also ensures that our methods of converting units work out correctly. For instance, if we accept the fraction 100 cm 1m as a valid way of writing the number one, then one times one equals one, so we should also say that one can be represented by 100 cm 100 cm × 1m 1m
,
which is the same as
10000 cm2 . 1 m2 That means the conversion factor from square meters to square centimeters is a factor of 104 , i.e., a square meter has 104 square centimeters in it. All of the above can be easily applied to volume as well, using one-cubic-centimeter blocks instead of squares on graph paper. To many people, it seems hard to believe that a square meter equals 10000 square centimeters, or that a cubic meter equals a million cubic centimeters — they think it would make more sense if there were 100 cm2 in 1 m2 , and 100 cm3 in 1 m3 , but that would be incorrect. The examples shown in figure b aim to make the correct answer more believable, using the traditional U.S. units of feet and yards. (One foot is 12 inches, and one yard is three feet.)
b / Visualizing conversions of area and volume using traditional U.S. units.
self-check A Based on figure b, convince yourself that there are 9 ft2 in a square yard, and 27 ft3 in a cubic yard, then demonstrate the same thing symbolically (i.e., with the method using fractions that equal one). Answer, p. 540
42
Chapter 1
Scaling and estimation
Solved problem: converting mm2 to cm2
page 59, problem 10
Solved problem: scaling a liter
page 60, problem 19
Discussion question A How many square centimeters are there in a square inch? (1 inch = 2.54 cm) First find an approximate answer by making a drawing, then derive the conversion factor more accurately using the symbolic method.
c / Galileo Galilei (1564-1642) was a Renaissance Italian who brought the scientific method to bear on physics, creating the modern version of the science. Coming from a noble but very poor family, Galileo had to drop out of medical school at the University of Pisa when he ran out of money. Eventually becoming a lecturer in mathematics at the same school, he began a career as a notorious troublemaker by writing a burlesque ridiculing the university’s regulations — he was forced to resign, but found a new teaching position at Padua. He invented the pendulum clock, investigated the motion of falling bodies, and discovered the moons of Jupiter. The thrust of his life’s work was to discredit Aristotle’s physics by confronting it with contradictory experiments, a program that paved the way for Newton’s discovery of the relationship between force and motion. In chapter 3 we’ll come to the story of Galileo’s ultimate fate at the hands of the Church.
1.2 Scaling of area and volume Great fleas have lesser fleas Upon their backs to bite ’em. And lesser fleas have lesser still, And so ad infinitum. Jonathan Swift Now how do these conversions of area and volume relate to the questions I posed about sizes of living things? Well, imagine that you are shrunk like Alice in Wonderland to the size of an insect. One way of thinking about the change of scale is that what used to look like a centimeter now looks like perhaps a meter to you, because you’re so much smaller. If area and volume scaled according to most people’s intuitive, incorrect expectations, with 1 m2 being the same as 100 cm2 , then there would be no particular reason why nature should behave any differently on your new, reduced scale. But nature does behave differently now that you’re small. For instance, you will find that you can walk on water, and jump to many times your own height. The physicist Galileo Galilei had the basic insight that the scaling of area and volume determines how natural phenomena behave differently on different scales. He first reasoned about mechanical structures, but later extended his insights to living things, taking the then-radical point of view that at the fundamental level, a living organism should follow the same laws
Section 1.2
Scaling of area and volume
43
of nature as a machine. We will follow his lead by first discussing machines and then living things. Galileo on the behavior of nature on large and small scales One of the world’s most famous pieces of scientific writing is Galileo’s Dialogues Concerning the Two New Sciences. Galileo was an entertaining writer who wanted to explain things clearly to laypeople, and he livened up his work by casting it in the form of a dialogue among three people. Salviati is really Galileo’s alter ego. Simplicio is the stupid character, and one of the reasons Galileo got in trouble with the Church was that there were rumors that Simplicio represented the Pope. Sagredo is the earnest and intelligent student, with whom the reader is supposed to identify. (The following excerpts are from the 1914 translation by Crew and de Salvio.)
d / The small boat holds up just fine.
e / A larger boat built with the same proportions as the small one will collapse under its own weight.
f / A boat this large needs to have timbers that are thicker compared to its size.
S AGREDO : Yes, that is what I mean; and I refer especially to his last assertion which I have always regarded as false. . . ; namely, that in speaking of these and other similar machines one cannot argue from the small to the large, because many devices which succeed on a small scale do not work on a large scale. Now, since mechanics has its foundations in geometry, where mere size [ is unimportant], I do not see that the properties of circles, triangles, cylinders, cones and other solid figures will change with their size. If, therefore, a large machine be constructed in such a way that its parts bear to one another the same ratio as in a smaller one, and if the smaller is sufficiently strong for the purpose for which it is designed, I do not see why the larger should not be able to withstand any severe and destructive tests to which it may be subjected. Salviati contradicts Sagredo: S ALVIATI : . . . Please observe, gentlemen, how facts which at first seem improbable will, even on scant explanation, drop the cloak which has hidden them and stand forth in naked and simple beauty. Who does not know that a horse falling from a height of three or four cubits will break his bones, while a dog falling from the same height or a cat from a height of eight or ten cubits will suffer no injury? Equally harmless would be the fall of a grasshopper from a tower or the fall of an ant from the distance of the moon. The point Galileo is making here is that small things are sturdier in proportion to their size. There are a lot of objections that could be raised, however. After all, what does it really mean for something to be “strong”, to be “strong in proportion to its size,” or to be strong “out of proportion to its size?” Galileo hasn’t given operational definitions of things like “strength,” i.e., definitions that spell out how to measure them numerically.
44
Chapter 1
Scaling and estimation
Also, a cat is shaped differently from a horse — an enlarged photograph of a cat would not be mistaken for a horse, even if the photo-doctoring experts at the National Inquirer made it look like a person was riding on its back. A grasshopper is not even a mammal, and it has an exoskeleton instead of an internal skeleton. The whole argument would be a lot more convincing if we could do some isolation of variables, a scientific term that means to change only one thing at a time, isolating it from the other variables that might have an effect. If size is the variable whose effect we’re interested in seeing, then we don’t really want to compare things that are different in size but also different in other ways. S ALVIATI : . . . we asked the reason why [shipbuilders] employed stocks, scaffolding, and bracing of larger dimensions for launching a big vessel than they do for a small one; and [an old man] answered that they did this in order to avoid the danger of the ship parting under its own heavy weight, a danger to which small boats are not subject? After this entertaining but not scientifically rigorous beginning, Galileo starts to do something worthwhile by modern standards. He simplifies everything by considering the strength of a wooden plank. The variables involved can then be narrowed down to the type of wood, the width, the thickness, and the length. He also gives an operational definition of what it means for the plank to have a certain strength “in proportion to its size,” by introducing the concept of a plank that is the longest one that would not snap under its own weight if supported at one end. If you increased its length by the slightest amount, without increasing its width or thickness, it would break. He says that if one plank is the same shape as another but a different size, appearing like a reduced or enlarged photograph of the other, then the planks would be strong “in proportion to their sizes” if both were just barely able to support their own weight.
g / Galileo discusses planks made of wood, but the concept may be easier to imagine with clay. All three clay rods in the figure were originally the same shape. The medium-sized one was twice the height, twice the length, and twice the width of the small one, and similarly the large one was twice as big as the medium one in all its linear dimensions. The big one has four times the linear dimensions of the small one, 16 times the cross-sectional area when cut perpendicular to the page, and 64 times the volume. That means that the big one has 64 times the weight to support, but only 16 times the strength compared to the smallest one. h / 1. This plank is as long as it can be without collapsing under its own weight. If it was a hundredth of an inch longer, it would collapse. 2. This plank is made out of the same kind of wood. It is twice as thick, twice as long, and twice as wide. It will collapse under its own weight.
Section 1.2
Scaling of area and volume
45
Also, Galileo is doing something that would be frowned on in modern science: he is mixing experiments whose results he has actually observed (building boats of different sizes), with experiments that he could not possibly have done (dropping an ant from the height of the moon). He now relates how he has done actual experiments with such planks, and found that, according to this operational definition, they are not strong in proportion to their sizes. The larger one breaks. He makes sure to tell the reader how important the result is, via Sagredo’s astonished response: S AGREDO : My brain already reels. My mind, like a cloud momentarily illuminated by a lightning flash, is for an instant filled with an unusual light, which now beckons to me and which now suddenly mingles and obscures strange, crude ideas. From what you have said it appears to me impossible to build two similar structures of the same material, but of different sizes and have them proportionately strong. In other words, this specific experiment, using things like wooden planks that have no intrinsic scientific interest, has very wide implications because it points out a general principle, that nature acts differently on different scales. To finish the discussion, Galileo gives an explanation. He says that the strength of a plank (defined as, say, the weight of the heaviest boulder you could put on the end without breaking it) is proportional to its cross-sectional area, that is, the surface area of the fresh wood that would be exposed if you sawed through it in the middle. Its weight, however, is proportional to its volume.1 How do the volume and cross-sectional area of the longer plank compare with those of the shorter plank? We have already seen, while discussing conversions of the units of area and volume, that these quantities don’t act the way most people naively expect. You might think that the volume and area of the longer plank would both be doubled compared to the shorter plank, so they would increase in proportion to each other, and the longer plank would be equally able to support its weight. You would be wrong, but Galileo knows that this is a common misconception, so he has Salviati address the point specifically: S ALVIATI : . . . Take, for example, a cube two inches on a side so that each face has an area of four square inches and the total area, i.e., the sum of the six faces, amounts to twenty-four square inches; now imagine this cube to be sawed through three times [with cuts in three perpendicular planes] so as to divide it into eight smaller cubes, each one inch on the side, each face one inch square, and the total 1
Galileo makes a slightly more complicated argument, taking into account the effect of leverage (torque). The result I’m referring to comes out the same regardless of this effect.
46
Chapter 1
Scaling and estimation
surface of each cube six square inches instead of twentyfour in the case of the larger cube. It is evident therefore, that the surface of the little cube is only one-fourth that of the larger, namely, the ratio of six to twenty-four; but the volume of the solid cube itself is only one-eighth; the volume, and hence also the weight, diminishes therefore much more rapidly than the surface. . . You see, therefore, Simplicio, that I was not mistaken when . . . I said that the surface of a small solid is comparatively greater than that of a large one. The same reasoning applies to the planks. Even though they are not cubes, the large one could be sawed into eight small ones, each with half the length, half the thickness, and half the width. The small plank, therefore, has more surface area in proportion to its weight, and is therefore able to support its own weight while the large one breaks. Scaling of area and volume for irregularly shaped objects You probably are not going to believe Galileo’s claim that this has deep implications for all of nature unless you can be convinced that the same is true for any shape. Every drawing you’ve seen so far has been of squares, rectangles, and rectangular solids. Clearly the reasoning about sawing things up into smaller pieces would not prove anything about, say, an egg, which cannot be cut up into eight smaller egg-shaped objects with half the length. Is it always true that something half the size has one quarter the surface area and one eighth the volume, even if it has an irregular shape? Take the example of a child’s violin. Violins are made for small children in smaller size to accomodate their small bodies. Figure i shows a full-size violin, along with two violins made with half and 3/4 of the normal length.2 Let’s study the surface area of the front panels of the three violins. Consider the square in the interior of the panel of the full-size violin. In the 3/4-size violin, its height and width are both smaller by a factor of 3/4, so the area of the corresponding, smaller square becomes 3/4×3/4 = 9/16 of the original area, not 3/4 of the original area. Similarly, the corresponding square on the smallest violin has half the height and half the width of the original one, so its area is 1/4 the original area, not half. The same reasoning works for parts of the panel near the edge, such as the part that only partially fills in the other square. The entire square scales down the same as a square in the interior, and in each violin the same fraction (about 70%) of the square is full, so the contribution of this part to the total area scales down just the same. 2 The customary terms “half-size” and “3/4-size” actually don’t describe the sizes in any accurate way. They’re really just standard, arbitrary marketing labels.
Section 1.2
i / The area of a shape is proportional to the square of its linear dimensions, even if the shape is irregular.
Scaling of area and volume
47
Since any small square region or any small region covering part of a square scales down like a square object, the entire surface area of an irregularly shaped object changes in the same manner as the surface area of a square: scaling it down by 3/4 reduces the area by a factor of 9/16, and so on. In general, we can see that any time there are two objects with the same shape, but different linear dimensions (i.e., one looks like a reduced photo of the other), the ratio of their areas equals the ratio of the squares of their linear dimensions: 2 L1 A1 = . A2 L2 Note that it doesn’t matter where we choose to measure the linear size, L, of an object. In the case of the violins, for instance, it could have been measured vertically, horizontally, diagonally, or even from the bottom of the left f-hole to the middle of the right f-hole. We just have to measure it in a consistent way on each violin. Since all the parts are assumed to shrink or expand in the same manner, the ratio L1 /L2 is independent of the choice of measurement. It is also important to realize that it is completely unnecessary to have a formula for the area of a violin. It is only possible to derive simple formulas for the areas of certain shapes like circles, rectangles, triangles and so on, but that is no impediment to the type of reasoning we are using. Sometimes it is inconvenient to write all the equations in terms of ratios, especially when more than two objects are being compared. A more compact way of rewriting the previous equation is j / The muffin comes out of the oven too hot to eat. Breaking it up into four pieces increases its surface area while keeping the total volume the same. It cools faster because of the greater surface-to-volume ratio. In general, smaller things have greater surface-to-volume ratios, but in this example there is no easy way to compute the effect exactly, because the small pieces aren’t the same shape as the original muffin.
A ∝ L2
.
The symbol “∝” means “is proportional to.” Scientists and engineers often speak about such relationships verbally using the phrases “scales like” or “goes like,” for instance “area goes like length squared.” All of the above reasoning works just as well in the case of volume. Volume goes like length cubed: V ∝ L3
.
self-check B When a car or truck travels over a road, there is wear and tear on the road surface, which incurs a cost. Studies show that the cost C per kilometer of travel is related to the weight per axle w by C ∝ w 4 . Translate this into a statement about ratios. Answer, p. 540
If different objects are made of the same material with the same density, ρ = m/V , then their masses, m = ρV , are proportional to L3 . (The symbol for density is ρ, the lower-case Greek letter “rho.”)
48
Chapter 1
Scaling and estimation
An important point is that all of the above reasoning about scaling only applies to objects that are the same shape. For instance, a piece of paper is larger than a pencil, but has a much greater surface-to-volume ratio. Scaling of the area of a triangle example 1 In figure k, the larger triangle has sides twice as long. How many times greater is its area? Correct solution #1: Area scales in proportion to the square of the linear dimensions, so the larger triangle has four times more area (22 = 4). Correct solution #2: You could cut the larger triangle into four of the smaller size, as shown in fig. (b), so its area is four times greater. (This solution is correct, but it would not work for a shape like a circle, which can’t be cut up into smaller circles.)
k / Example 1. The big triangle has four times more area than the little one.
Correct solution #3: The area of a triangle is given by A = bh/2, where b is the base and h is the height. The areas of the triangles are A1 = b1 h1 /2 A2 = b2 h2 /2 = (2b1 )(2h1 )/2 = 2b1 h1
l / A tricky way of solving example 1, explained in solution #2.
A2 /A1 = (2b1 h1 )/(b1 h1 /2) =4 (Although this solution is correct, it is a lot more work than solution #1, and it can only be used in this case because a triangle is a simple geometric shape, and we happen to know a formula for its area.) Correct solution #4: The area of a triangle is A = bh/2. The comparison of the areas will come out the same as long as the ratios of the linear sizes of the triangles is as specified, so let’s just say b1 = 1.00 m and b2 = 2.00 m. The heights are then also h1 = 1.00 m and h2 = 2.00 m, giving areas A1 = 0.50 m2 and A2 = 2.00 m2 , so A2 /A1 = 4.00. (The solution is correct, but it wouldn’t work with a shape for whose area we don’t have a formula. Also, the numerical calculation might make the answer of 4.00 appear inexact, whereas solution #1 makes it clear that it is exactly 4.) Incorrect solution: The area of a triangle is A = bh/2, and if you plug in b = 2.00 m and h = 2.00 m, you get A = 2.00 m2 , so the bigger triangle has 2.00 times more area. (This solution is incorrect because no comparison has been made with the smaller triangle.) Section 1.2
Scaling of area and volume
49
Scaling of the volume of a sphere example 2 In figure m, the larger sphere has a radius that is five times greater. How many times greater is its volume? Correct solution #1: Volume scales like the third power of the linear size, so the larger sphere has a volume that is 125 times greater (53 = 125). Correct solution #2: The volume of a sphere is V = (4/3)πr 3 , so m / Example 2. The big sphere has 125 times more volume than the little one.
4 3 πr 3 1 4 V2 = πr23 3 4 = π(5r1 )3 3 500 3 = πr1 3 4 3 500 3 V2 /V1 = πr1 / πr 3 3 1 V1 =
= 125
Incorrect solution: The volume of a sphere is V = (4/3)πr 3 , so 4 3 πr 3 1 4 V2 = πr23 3 4 = π · 5r13 3 20 3 = πr1 3 4 3 20 3 V2 /V1 = πr / πr 3 1 3 1 V1 =
=5
(The solution is incorrect because (5r1 )3 is not the same as 5r13 .)
n / Example 3. The 48-point “S” has 1.78 times more area than the 36-point “S.”
Scaling of a more complex shape example 3 The first letter “S” in figure n is in a 36-point font, the second in 48-point. How many times more ink is required to make the larger “S”? (Points are a unit of length used in typography.) Correct solution: The amount of ink depends on the area to be covered with ink, and area is proportional to the square of the linear dimensions, so the amount of ink required for the second “S” is greater by a factor of (48/36)2 = 1.78. Incorrect solution: The length of the curve of the second “S” is longer by a factor of 48/36 = 1.33, so 1.33 times more ink is required. (The solution is wrong because it assumes incorrectly that the width of the curve is the same in both cases. Actually both the
50
Chapter 1
Scaling and estimation
width and the length of the curve are greater by a factor of 48/36, so the area is greater by a factor of (48/36)2 = 1.78.) Reasoning about ratios and proportionalities is one of the three essential mathematical skills, summarized on pp.524-525, that you need for success in this course. Solved problem: a telescope gathers light
page 59, problem 11
Solved problem: distance from an earthquake
page 59, problem 12
Discussion questions A A toy fire engine is 1/30 the size of the real one, but is constructed from the same metal with the same proportions. How many times smaller is its weight? How many times less red paint would be needed to paint it? B Galileo spends a lot of time in his dialog discussing what really happens when things break. He discusses everything in terms of Aristotle’s now-discredited explanation that things are hard to break, because if something breaks, there has to be a gap between the two halves with nothing in between, at least initially. Nature, according to Aristotle, “abhors a vacuum,” i.e., nature doesn’t “like” empty space to exist. Of course, air will rush into the gap immediately, but at the very moment of breaking, Aristotle imagined a vacuum in the gap. Is Aristotle’s explanation of why it is hard to break things an experimentally testable statement? If so, how could it be tested experimentally?
1.3 Scaling applied to biology Organisms of different sizes with the same shape The left-hand panel in figure o shows the approximate validity of the proportionality m ∝ L3 for cockroaches (redrawn from McMahon and Bonner). The scatter of the points around the curve indicates that some cockroaches are proportioned slightly differently from others, but in general the data seem well described by m ∝ L3 . That means that the largest cockroaches the experimenter could raise (is there a 4-H prize?) had roughly the same shape as the smallest ones.
Another relationship that should exist for animals of different sizes shaped in the same way is that between surface area and body mass. If all the animals have the same average density, then body mass should be proportional to the cube of the animal’s linear size, m ∝ L3 , while surface area should vary proportionately to L2 . Therefore, the animals’ surface areas should be proportional to m2/3 . As shown in the right-hand panel of figure o, this relationship appears to hold quite well for the dwarf siren, a type of salamander. Notice how the curve bends over, meaning that the surface area does not increase as quickly as body mass, e.g., a salamander with eight
Section 1.3
Scaling applied to biology
51
o / Geometrical scaling of animals.
times more body mass will have only four times more surface area.
This behavior of the ratio of surface area to mass (or, equivalently, the ratio of surface area to volume) has important consequences for mammals, which must maintain a constant body temperature. It would make sense for the rate of heat loss through the animal’s skin to be proportional to its surface area, so we should expect small animals, having large ratios of surface area to volume, to need to produce a great deal of heat in comparison to their size to avoid dying from low body temperature. This expectation is borne out by the data of the left-hand panel of figure p, showing the rate of oxygen consumption of guinea pigs as a function of their body mass. Neither an animal’s heat production nor its surface area is convenient to measure, but in order to produce heat, the animal must metabolize oxygen, so oxygen consumption is a good indicator of the rate of heat production. Since surface area is proportional to m2/3 , the proportionality of the rate of oxygen consumption to m2/3 is consistent with the idea that the animal needs to produce heat at a rate in proportion to its surface area. Although the smaller animals
52
Chapter 1
Scaling and estimation
p / Scaling of animals’ bodies related to metabolic rate and skeletal strength.
metabolize less oxygen and produce less heat in absolute terms, the amount of food and oxygen they must consume is greater in proportion to their own mass. The Etruscan pigmy shrew, weighing in at 2 grams as an adult, is at about the lower size limit for mammals. It must eat continually, consuming many times its body weight each day to survive. Changes in shape to accommodate changes in size Large mammals, such as elephants, have a small ratio of surface area to volume, and have problems getting rid of their heat fast enough. An elephant cannot simply eat small enough amounts to keep from producing excessive heat, because cells need to have a certain minimum metabolic rate to run their internal machinery. Hence the elephant’s large ears, which add to its surface area and help it to cool itself. Previously, we have seen several examples of data within a given species that were consistent with a fixed shape, scaled up and down in the cases of individual specimens. The elephant’s ears are an example of a change in shape necessitated by a change in scale.
Section 1.3
Scaling applied to biology
53
Large animals also must be able to support their own weight. Returning to the example of the strengths of planks of different sizes, we can see that if the strength of the plank depends on area while its weight depends on volume, then the ratio of strength to weight goes as follows: strength/weight ∝ A/V ∝ 1/L
.
Thus, the ability of objects to support their own weights decreases inversely in proportion to their linear dimensions. If an object is to be just barely able to support its own weight, then a larger version will have to be proportioned differently, with a different shape.
q / Galileo’s original showing how larger bones must be greater eter compared to their
drawing, animals’ in diamlengths.
Since the data on the cockroaches seemed to be consistent with roughly similar shapes within the species, it appears that the ability to support its own weight was not the tightest design constraint that Nature was working under when she designed them. For large animals, structural strength is important. Galileo was the first to quantify this reasoning and to explain why, for instance, a large animal must have bones that are thicker in proportion to their length. Consider a roughly cylindrical bone such as a leg bone or a vertebra. The length of the bone, L, is dictated by the overall linear size of the animal, since the animal’s skeleton must reach the animal’s whole length. We expect the animal’s mass to scale as L3 , so the strength of the bone must also scale as L3 . Strength is proportional to crosssectional area, as with the wooden planks, so if the diameter of the bone is d, then d2 ∝ L3 or d ∝ L3/2
.
If the shape stayed the same regardless of size, then all linear dimensions, including d and L, would be proportional to one another. If our reasoning holds, then the fact that d is proportional to L3/2 , not L, implies a change in proportions of the bone. As shown in the right-hand panel of figure p, the vertebrae of African Bovidae follow the rule d ∝ L3/2 fairly well. The vertebrae of the giant eland are as chunky as a coffee mug, while those of a Gunther’s dik-dik are as slender as the cap of a pen. Discussion questions A Single-celled animals must passively absorb nutrients and oxygen from their surroundings, unlike humans who have lungs to pump air in and out and a heart to distribute the oxygenated blood throughout their bodies. Even the cells composing the bodies of multicellular animals must absorb oxygen from a nearby capillary through their surfaces. Based on these facts, explain why cells are always microscopic in size. B The reasoning of the previous question would seem to be contradicted by the fact that human nerve cells in the spinal cord can be as much as a meter long, although their widths are still very small. Why is this possible?
54
Chapter 1
Scaling and estimation
1.4 Order-of-magnitude estimates It is the mark of an instructed mind to rest satisfied with the degree of precision that the nature of the subject permits and not to seek an exactness where only an approximation of the truth is possible. Aristotle It is a common misconception that science must be exact. For instance, in the Star Trek TV series, it would often happen that Captain Kirk would ask Mr. Spock, “Spock, we’re in a pretty bad situation. What do you think are our chances of getting out of here?” The scientific Mr. Spock would answer with something like, “Captain, I estimate the odds as 237.345 to one.” In reality, he could not have estimated the odds with six significant figures of accuracy, but nevertheless one of the hallmarks of a person with a good education in science is the ability to make estimates that are likely to be at least somewhere in the right ballpark. In many such situations, it is often only necessary to get an answer that is off by no more than a factor of ten in either direction. Since things that differ by a factor of ten are said to differ by one order of magnitude, such an estimate is called an order-of-magnitude estimate. The tilde, ∼, is used to indicate that things are only of the same order of magnitude, but not exactly equal, as in odds of survival ∼ 100 to one
.
The tilde can also be used in front of an individual number to emphasize that the number is only of the right order of magnitude. Although making order-of-magnitude estimates seems simple and natural to experienced scientists, it’s a mode of reasoning that is completely unfamiliar to most college students. Some of the typical mental steps can be illustrated in the following example. Cost of transporting tomatoes (incorrect solution) example 4 Roughly what percentage of the price of a tomato comes from the cost of transporting it in a truck? The following incorrect solution illustrates one of the main ways you can go wrong in order-of-magnitude estimates. Incorrect solution: Let’s say the trucker needs to make a $400 profit on the trip. Taking into account her benefits, the cost of gas, and maintenance and payments on the truck, let’s say the total cost is more like $2000. I’d guess about 5000 tomatoes would fit in the back of the truck, so the extra cost per tomato is 40 cents. That means the cost of transporting one tomato is comparable to the cost of the tomato itself. Transportation really adds a lot to the cost of produce, I guess. The problem is that the human brain is not very good at estimating area or volume, so it turns out the estimate of 5000 tomatoes
Section 1.4
Order-of-magnitude estimates
55
fitting in the truck is way off. That’s why people have a hard time at those contests where you are supposed to estimate the number of jellybeans in a big jar. Another example is that most people think their families use about 10 gallons of water per day, but in reality the average is about 300 gallons per day. When estimating area or volume, you are much better off estimating linear dimensions, and computing volume from the linear dimensions. Here’s a better solution to the problem about the tomato truck:
r / Can you guess how many jelly beans are in the jar? If you try to guess directly, you will almost certainly underestimate. The right way to do it is to estimate the linear dimensions, then get the volume indirectly. See problem 26, p. 62.
Cost of transporting tomatoes (correct solution) example 5 As in the previous solution, say the cost of the trip is $2000. The dimensions of the bin are probably 4 m × 2 m × 1 m, for a volume of 8 m3 . Since the whole thing is just an order-of-magnitude estimate, let’s round that off to the nearest power of ten, 10 m3 . The shape of a tomato is complicated, and I don’t know any formula for the volume of a tomato shape, but since this is just an estimate, let’s pretend that a tomato is a cube, 0.05 m × 0.05 m × 0.05 m, for a volume of 1.25 × 10−4 m3 . Since this is just a rough estimate, let’s round that to 10−4 m3 . We can find the total number of tomatoes by dividing the volume of the bin by the volume of one tomato: 10 m3 /10−4 m3 = 105 tomatoes. The transportation cost per tomato is $2000/105 tomatoes=$0.02/tomato. That means that transportation really doesn’t contribute very much to the cost of a tomato. Approximating the shape of a tomato as a cube is an example of another general strategy for making order-of-magnitude estimates. A similar situation would occur if you were trying to estimate how many m2 of leather could be produced from a herd of ten thousand cattle. There is no point in trying to take into account the shape of the cows’ bodies. A reasonable plan of attack might be to consider a spherical cow. Probably a cow has roughly the same surface area as a sphere with a radius of about 1 m, which would be 4π(1 m)2 . Using the well-known facts that pi equals three, and four times three equals about ten, we can guess that a cow has a surface area of about 10 m2 , so the herd as a whole might yield 105 m2 of leather.
s / Consider a spherical cow.
Estimating mass indirectly example 6 Usually the best way to estimate mass is to estimate linear dimensions, then use those to infer volume, and then get the mass based on the volume. For example, Amphicoelias, shown in the figure, may have been the largest land animal ever to live. Fossils tell us the linear dimensions of an animal, but we can only indirectly guess its mass. Given the length scale in the figure, let’s estimate the mass of an Amphicoelias. Its torso looks like it can be approximated by a rectangular box with dimensions 10 m×5 m×3 m, giving about 2×102 m3 . Living things are mostly made of water, so we assume the animal to have the density of water, 1 g/cm3 , which converts to 103 kg/m3 .
56
Chapter 1
Scaling and estimation
This gives a mass of about 2 × 105 kg, or 200 metric tons.
The following list summarizes the strategies for getting a good order-of-magnitude estimate. 1. Don’t even attempt more than one significant figure of precision. 2. Don’t guess area, volume, or mass directly. Guess linear dimensions and get area, volume, or mass from them. 3. When dealing with areas or volumes of objects with complex shapes, idealize them as if they were some simpler shape, a cube or a sphere, for example. 4. Check your final answer to see if it is reasonable. If you estimate that a herd of ten thousand cattle would yield 0.01 m2 of leather, then you have probably made a mistake with conversion factors somewhere.
Section 1.4
Order-of-magnitude estimates
57
Summary Notation ∝ . . . . . . . . . ∼ . . . . . . . . .
is proportional to on the order of, is on the order of
Summary Nature behaves differently on large and small scales. Galileo showed that this results fundamentally from the way area and volume scale. Area scales as the second power of length, A ∝ L2 , while volume scales as length to the third power, V ∝ L3 . An order of magnitude estimate is one in which we do not attempt or expect an exact answer. The main reason why the uninitiated have trouble with order-of-magnitude estimates is that the human brain does not intuitively make accurate estimates of area and volume. Estimates of area and volume should be approached by first estimating linear dimensions, which one’s brain has a feel for.
58
Chapter 1
Scaling and estimation
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 How many cubic inches are there in a cubic foot? The answer √ is not 12. 2 Assume a dog’s brain is twice as great in diameter as a cat’s, but each animal’s brain cells are the same size and their brains are the same shape. In addition to being a far better companion and much nicer to come home to, how many times more brain cells does a dog have than a cat? The answer is not 2. 3 The population density of Los Angeles is about 4000 people/km2 . That of San Francisco is about 6000 people/km2 . How many times farther away is the average person’s nearest neighbor in LA than in √ San Francisco? The answer is not 1.5. 4 A hunting dog’s nose has about 10 square inches of active surface. How is this possible, since the dog’s nose is only about 1 in × 1 in × 1 in = 1 in3 ? After all, 10 is greater than 1, so how can it fit? 5
Estimate the number of blades of grass on a football field.
6 In a computer memory chip, each bit of information (a 0 or a 1) is stored in a single tiny circuit etched onto the surface of a silicon chip. The circuits cover the surface of the chip like lots in a housing development. A typical chip stores 64 Mb (megabytes) of data, where a byte is 8 bits. Estimate (a) the area of each circuit, and (b) its linear size. 7 Suppose someone built a gigantic apartment building, measuring 10 km × 10 km at the base. Estimate how tall the building would have to be to have space in it for the entire world’s population to live. 8 A hamburger chain advertises that it has sold 10 billion Bongo Burgers. Estimate the total mass of feed required to raise the cows used to make the burgers. 9 10
Estimate the volume of a human body, in cm3 . How many cm2 is 1 mm2 ?
Solution, p. 526
11 Compare the light-gathering powers of a 3-cm-diameter telescope and a 30-cm telescope. Solution, p. 526 12 One step on the Richter scale corresponds to a factor of 100 in terms of the energy absorbed by something on the surface of the Earth, e.g., a house. For instance, a 9.3-magnitude quake would release 100 times more energy than an 8.3. The energy spreads out
Problems
59
from the epicenter as a wave, and for the sake of this problem we’ll assume we’re dealing with seismic waves that spread out in three dimensions, so that we can visualize them as hemispheres spreading out under the surface of the earth. If a certain 7.6-magnitude earthquake and a certain 5.6-magnitude earthquake produce the same amount of vibration where I live, compare the distances from my house to the two epicenters. Solution, p. 527 13 In Europe, a piece of paper of the standard size, called A4, is a little narrower and taller than its American counterpart. The ratio of the height to the width is the square root of 2, and this has some useful properties. For instance, if you cut an A4 sheet from left to right, you get two smaller sheets that have the same proportions. You can even buy sheets of this smaller size, and they’re called A5. There is a whole series of sizes related in this way, all with the same proportions. (a) Compare an A5 sheet to an A4 in terms of area and linear size. (b) The series of paper sizes starts from an A0 sheet, which has an area of one square meter. Suppose we had a series of boxes defined in a similar way: the B0 box has a volume of one cubic meter, two B1 boxes fit exactly inside an B0 box, and so√ on. What would be the dimensions of a B0 box? 14 Estimate the mass of one of the hairs in Albert Einstein’s moustache, in units of kg.
Albert Einstein, and his moustache, problem 14.
15 According to folklore, every time you take a breath, you are inhaling some of the atoms exhaled in Caesar’s last words. Is this true? If so, how many? 16 The Earth’s surface is about 70% water. Mars’s diameter is about half the Earth’s, but it has no surface water. Compare√the land areas of the two planets. 17 The traditional Martini glass is shaped like a cone with the point at the bottom. Suppose you make a Martini by pouring vermouth into the glass to a depth of 3 cm, and then adding gin to bring the depth to 6 cm. What are the proportions of gin and vermouth? Solution, p. 527 18 The central portion of a CD is taken up by the hole and some surrounding clear plastic, and this area is unavailable for storing data. The radius of the central circle is about 35% of the outer radius of the data-storing area. What percentage of the CD’s √ area is therefore lost?
Problem 19.
60
19 The one-liter cube in the photo has been marked off into smaller cubes, with linear dimensions one tenth those of the big one. What is the volume of each of the small cubes? Solution, p. 527
Chapter 1
Scaling and estimation
20
[This problem is now problem 0-12 on p. 38.]
21 Estimate the number of man-hours required for building the Great Wall of China. Solution, p. 527 22 (a) Using the microscope photo in the figure, estimate the mass of a one cell of the E. coli bacterium, which is one of the most common ones in the human intestine. Note the scale at the lower right corner, which is 1 μm. Each of the tubular objects in the column is one cell. (b) The feces in the human intestine are mostly bacteria (some dead, some alive), of which E. coli is a large and typical component. Estimate the number of bacteria in your intestines, and compare with the number of human cells in your body, which is believed to be roughly on the order of 1013 . (c) Interpreting your result from part b, what does this tell you about the size of a typical human cell compared to the size of a typical bacterial cell?
Problem 22.
23 A taxon (plural taxa) is a group of living things. For example, Homo sapiens and Homo neanderthalensis are both taxa — specifically, they are two different species within the genus Homo. Surveys by botanists show that the number of plant taxa native to a given contiguous land area A is usually approximately proportional to A1/3 . (a) There are 70 different species of lupine native to Southern California, which has an area of about 200, 000 km2 . The San Gabriel Mountains cover about 1, 600 km2 . Suppose that you wanted to learn to identify all the species of lupine in the San Gabriels. Approximately how many species would you have to fa√ miliarize yourself with? Answer, p. 545 (b) What is the interpretation of the fact that the exponent, 1/3, is less than one?
Problems
61
24 X-ray images aren’t only used with human subjects but also, for example, on insects and flowers. In 2003, a team of researchers at Argonne National Laboratory used x-ray imagery to find for the first time that insects, although they do not have lungs, do not necessarily breathe completely passively, as had been believed previously; many insects rapidly compress and expand their trachea, head, and thorax in order to force air in and out of their bodies. One difference between x-raying a human and an insect is that if a medical x-ray machine was used on an insect, virtually 100% of the x-rays would pass through its body, and there would be no contrast in the image produced. Less penetrating x-rays of lower energies have to be used. For comparison, a typical human body mass is about 70 kg, whereas a typical ant is about 10 mg. Estimate the ratio of the thicknesses of tissue that must be penetrated by x-rays √ in one case compared to the other. 25 Radio was first commercialized around 1920, and ever since then, radio signals from our planet have been spreading out across our galaxy. It is possible that alien civilizations could detect these signals and learn that there is life on earth. In the 90 years that the signals have been spreading at the speed of light, they have created a sphere with a radius of 90 light-years. To show an idea of the size of this sphere, I’ve indicated it in the figure as a tiny white circle on an image of a spiral galaxy seen edge on. (We don’t have similar photos of our own Milky Way galaxy, because we can’t see it from the outside.) So far we haven’t received answering signals from aliens within this sphere, but as time goes on, the sphere will expand as suggested by the dashed outline, reaching more and more stars that might harbor extraterrestrial life. Approximately what year will it be when the sphere has expanded to fill a volume√100 times greater than the volume it fills today in 2010?
Problem 25.
26
27 At the grocery store you will see oranges packed neatly in stacks. Suppose we want to pack spheres as densely as possible, so that the greatest possible fraction of the space is filled by the spheres themselves, not by empty space. Let’s call this fraction f . Mathematicians have proved that the best possible result is f ≈ 0.7405, which requires a systematic pattern of stacking. If you buy ball bearings or golf balls, however, the seller is probably not going to go to the trouble of stacking them neatly. Instead they will probably pour the balls into a box and vibrate the box vigorously for a while to make them settle. This results in a random packing. The closest random packing has f ≈ 0.64. Suppose that golf balls, with a standard diameter of 4.27 cm, are sold in bulk with the closest random packing. What is the diameter of the largest ball that could be sold in boxes of the same size, packed systematically, √ so that there would be the same number of balls per box?
Problem 27.
62
Estimate the number of jellybeans in figure r on p. 56. Solution, p. 527
Chapter 1
Scaling and estimation
28 Plutonium-239 is one of a small number of important longlived forms of high-level radioactive nuclear waste. The world’s waste stockpiles have about 103 metric tons of plutonium. Drinking water is considered safe by U.S. government standards if it contains less than 2 × 10−13 g/cm3 of plutonium. The amount of radioactivity to which you were exposed by drinking such water on a daily basis would be very small compared to the natural background radiation that you are exposed to every year. Suppose that the world’s inventory of plutonium-239 were ground up into an extremely fine dust and then dispersed over the world’s oceans, thereby becoming mixed uniformly into the world’s water supplies over time. Estimate the resulting concentration of plutonium, and compare with the government standard.
Problems
63
Exercise 1: Scaling applied to leaves Equipment: leaves of three sizes, having roughly similar proportions of length, width, and thickness balance Each group will have one leaf, and should measure its surface area and volume, and determine its surface-to-volume ratio. For consistency, every group should use units of cm2 and cm3 , and should only find the area of one side of the leaf. The area can be found by tracing the area of the leaf on graph paper and counting squares. The volume can be found by weighing the leaf and assuming that its density is 1 g/cm3 (the density of water). What implications do your results have for the plants’ abilities to survive in different environments?
64
Chapter 1
Scaling and estimation
Motion in one dimension
65
Chapter 2
Velocity and relative motion 2.1 Types of motion If you had to think consciously in order to move your body, you would be severely disabled. Even walking, which we consider to be no great feat, requires an intricate series of motions that your cerebrum would be utterly incapable of coordinating. The task of putting one foot in front of the other is controlled by the more primitive parts of your brain, the ones that have not changed much since the mammals and reptiles went their separate evolutionary ways. The thinking part of your brain limits itself to general directives such as “walk faster,” or “don’t step on her toes,” rather than micromanaging every contraction and relaxation of the hundred or so muscles of your hips, legs, and feet. Physics is all about the conscious understanding of motion, but we’re obviously not immediately prepared to understand the most complicated types of motion. Instead, we’ll use the divide-andconquer technique. We’ll first classify the various types of motion, and then begin our campaign with an attack on the simplest cases. To make it clear what we are and are not ready to consider, we need to examine and define carefully what types of motion can exist.
a / Rotation.
b / Simultaneous rotation motion through space.
and
Rigid-body motion distinguished from motion that changes an object’s shape Nobody, with the possible exception of Fred Astaire, can simply glide forward without bending their joints. Walking is thus an example in which there is both a general motion of the whole object and a change in the shape of the object. Another example is the motion of a jiggling water balloon as it flies through the air. We are not presently attempting a mathematical description of the way in which the shape of an object changes. Motion without a change in shape is called rigid-body motion. (The word “body” is often used in physics as a synonym for “object.”) Center-of-mass motion as opposed to rotation A ballerina leaps into the air and spins around once before landing. We feel intuitively that her rigid-body motion while her feet are off the ground consists of two kinds of motion going on simul-
c / One person might say that the tipping chair was only rotating in a circle about its point of contact with the floor, but another could describe it as having both rotation and motion through space.
67
taneously: a rotation and a motion of her body as a whole through space, along an arc. It is not immediately obvious, however, what is the most useful way to define the distinction between rotation and motion through space. Imagine that you attempt to balance a chair and it falls over. One person might say that the only motion was a rotation about the chair’s point of contact with the floor, but another might say that there was both rotation and motion down and to the side.
e / No matter what point you hang the pear from, the string lines up with the pear’s center of mass. The center of mass can therefore be defined as the intersection of all the lines made by hanging the pear in this way. Note that the X in the figure should not be interpreted as implying that the center of mass is on the surface — it is actually inside the pear.
d / The leaping dancer’s motion is complicated, but the motion of her center of mass is simple.
f / The circus performers hang with the ropes passing through their centers of mass.
It turns out that there is one particularly natural and useful way to make a clear definition, but it requires a brief digression. Every object has a balance point, referred to in physics as the center of mass. For a two-dimensional object such as a cardboard cutout, the center of mass is the point at which you could hang the object from a string and make it balance. In the case of the ballerina (who is likely to be three-dimensional unless her diet is particularly severe), it might be a point either inside or outside her body, depending on how she holds her arms. Even if it is not practical to attach a string to the balance point itself, the center of mass can be defined as shown in figure e. Why is the center of mass concept relevant to the question of classifying rotational motion as opposed to motion through space? As illustrated in figures d and g, it turns out that the motion of an object’s center of mass is nearly always far simpler than the motion of any other part of the object. The ballerina’s body is a large object with a complex shape. We might expect that her motion would be much more complicated than the motion of a small, simply-shaped
68
Chapter 2
Velocity and relative motion
object, say a marble, thrown up at the same angle as the angle at which she leapt. But it turns out that the motion of the ballerina’s center of mass is exactly the same as the motion of the marble. That is, the motion of the center of mass is the same as the motion the ballerina would have if all her mass was concentrated at a point. By restricting our attention to the motion of the center of mass, we can therefore simplify things greatly.
g / The same leaping dancer, viewed from above. Her center of mass traces a straight line, but a point away from her center of mass, such as her elbow, traces the much more complicated path shown by the dots.
We can now replace the ambiguous idea of “motion as a whole through space” with the more useful and better defined concept of “center-of-mass motion.” The motion of any rigid body can be cleanly split into rotation and center-of-mass motion. By this definition, the tipping chair does have both rotational and center-of-mass motion. Concentrating on the center of mass motion allows us to make a simplified model of the motion, as if a complicated object like a human body was just a marble or a point-like particle. Science really never deals with reality; it deals with models of reality.
h / An improperly balanced wheel has a center of mass that is not at its geometric center. When you get a new tire, the mechanic clamps little weights to the rim to balance the wheel.
Note that the word “center” in “center of mass” is not meant to imply that the center of mass must lie at the geometrical center of an object. A car wheel that has not been balanced properly has a center of mass that does not coincide with its geometrical center. An object such as the human body does not even have an obvious geometrical center. It can be helpful to think of the center of mass as the average location of all the mass in the object. With this interpretation, we can see for example that raising your arms above your head raises your center of mass, since the higher position of the arms’ mass raises the average. We won’t be concerned right now with calculating centers of mass mathematically; the relevant equations are in ch. 14.
Ballerinas and professional basketball players can create an illusion of flying horizontally through the air because our brains intuitively expect them to have rigid-body motion, but the body does not stay rigid while executing a grand jete or a slam dunk. The legs are low at the beginning and end of the jump, but come up higher at
Section 2.1
i / This toy was intentionally designed so that the mushroomshaped piece of metal on top would throw off the center of mass. When you wind it up, the mushroom spins, but the center of mass doesn’t want to move, so the rest of the toy tends to counter the mushroom’s motion, causing the whole thing to jump around.
Types of motion
69
j / A fixed point on the dancer’s body follows a trajectory that is flatter than what we expect, creating an illusion of flight.
the middle. Regardless of what the limbs do, the center of mass will follow the same arc, but the low position of the legs at the beginning and end means that the torso is higher compared to the center of mass, while in the middle of the jump it is lower compared to the center of mass. Our eye follows the motion of the torso and tries to interpret it as the center-of-mass motion of a rigid body. But since the torso follows a path that is flatter than we expect, this attempted interpretation fails, and we experience an illusion that the person is flying horizontally. k / Example 1.
The center of mass as an average example 1 Explain how we know that the center of mass of each object is at the location shown in figure k. The center of mass is a sort of average, so the height of the centers of mass in 1 and 2 has to be midway between the two squares, because that height is the average of the heights of the two squares. Example 3 is a combination of examples 1 and 2, so we can find its center of mass by averaging the horizontal positions of their centers of mass. In example 4, each square has been skewed a little, but just as much mass has been moved up as down, so the average vertical position of the mass hasn’t changed. Example 5 is clearly not all that different from example 4, the main difference being a slight clockwise rotation, so just as
70
Chapter 2
Velocity and relative motion
in example 4, the center of mass must be hanging in empty space, where there isn’t actually any mass. Horizontally, the center of mass must be between the heels and toes, or else it wouldn’t be possible to stand without tipping over. Another interesting example from the sports world is the high jump, in which the jumper’s curved body passes over the bar, but the center of mass passes under the bar! Here the jumper lowers his legs and upper body at the peak of the jump in order to bring his waist higher compared to the center of mass. Later in this course, we’ll find that there are more fundamental reasons (based on Newton’s laws of motion) why the center of mass behaves in such a simple way compared to the other parts of an object. We’re also postponing any discussion of numerical methods for finding an object’s center of mass. Until later in the course, we will only deal with the motion of objects’ centers of mass. Center-of-mass motion in one dimension In addition to restricting our study of motion to center-of-mass motion, we will begin by considering only cases in which the center of mass moves along a straight line. This will include cases such as objects falling straight down, or a car that speeds up and slows down but does not turn.
l / The high-jumper’s body passes over the bar, but his center of mass passes under it.
Note that even though we are not explicitly studying the more complex aspects of motion, we can still analyze the center-of-mass motion while ignoring other types of motion that might be occurring simultaneously . For instance, if a cat is falling out of a tree and is initially upside-down, it goes through a series of contortions that bring its feet under it. This is definitely not an example of rigidbody motion, but we can still analyze the motion of the cat’s center of mass just as we would for a dropping rock. self-check A Consider a person running, a person pedaling on a bicycle, a person coasting on a bicycle, and a person coasting on ice skates. In which cases is the center-of-mass motion one-dimensional? Which cases are examples of rigid-body motion? Answer, p. 540 self-check B The figure shows a gymnast holding onto the inside of a big wheel. From inside the wheel, how could he make it roll one way or the other? Answer, p. 541
m / Self-check B.
2.2 Describing distance and time Center-of-mass motion in one dimension is particularly easy to deal with because all the information about it can be encapsulated in two variables: x, the position of the center of mass relative to the origin, and t, which measures a point in time. For instance, if someone
Section 2.2
Describing distance and time
71
supplied you with a sufficiently detailed table of x and t values, you would know pretty much all there was to know about the motion of the object’s center of mass. A point in time as opposed to duration In ordinary speech, we use the word “time” in two different senses, which are to be distinguished in physics. It can be used, as in “a short time” or “our time here on earth,” to mean a length or duration of time, or it can be used to indicate a clock reading, as in “I didn’t know what time it was,” or “now’s the time.” In symbols, t is ordinarily used to mean a point in time, while Δt signifies an interval or duration in time. The capital Greek letter delta, Δ, means “the change in...,” i.e. a duration in time is the change or difference between one clock reading and another. The notation Δt does not signify the product of two numbers, Δ and t, but rather one single number, Δt. If a matinee begins at a point in time t = 1 o’clock and ends at t = 3 o’clock, the duration of the movie was the change in t, Δt = 3 hours − 1 hour = 2 hours
.
To avoid the use of negative numbers for Δt, we write the clock reading “after” to the left of the minus sign, and the clock reading “before” to the right of the minus sign. A more specific definition of the delta notation is therefore that delta stands for “after minus before.” Even though our definition of the delta notation guarantees that Δt is positive, there is no reason why t can’t be negative. If t could not be negative, what would have happened one second before t = 0? That doesn’t mean that time “goes backward” in the sense that adults can shrink into infants and retreat into the womb. It just means that we have to pick a reference point and call it t = 0, and then times before that are represented by negative values of t. An example is that a year like 2007 A.D. can be thought of as a positive t value, while one like 370 B.C. is negative. Similarly, when you hear a countdown for a rocket launch, the phrase “t minus ten seconds” is a way of saying t = −10 s, where t = 0 is the time of blastoff, and t > 0 refers to times after launch. Although a point in time can be thought of as a clock reading, it is usually a good idea to avoid doing computations with expressions such as “2:35” that are combinations of hours and minutes. Times can instead be expressed entirely in terms of a single unit, such as hours. Fractions of an hour can be represented by decimals rather than minutes, and similarly if a problem is being worked in terms of minutes, decimals can be used instead of seconds. self-check C Of the following phrases, which refer to points in time, which refer to time intervals, and which refer to time in the abstract rather than as a measurable number?
72
Chapter 2
Velocity and relative motion
(1) “The time has come.” (2) “Time waits for no man.” (3) “The whole time, he had spit on his chin.”
Answer, p. 541
Position as opposed to change in position As with time, a distinction should be made between a point in space, symbolized as a coordinate x, and a change in position, symbolized as Δx. As with t, x can be negative. If a train is moving down the tracks, not only do you have the freedom to choose any point along the tracks and call it x = 0, but it’s also up to you to decide which side of the x = 0 point is positive x and which side is negative x. Since we’ve defined the delta notation to mean “after minus before,” it is possible that Δx will be negative, unlike Δt which is guaranteed to be positive. Suppose we are describing the motion of a train on tracks linking Tucson and Chicago. As shown in the figure, it is entirely up to you to decide which way is positive.
n / Two equally valid ways of describing the motion of a train from Tucson to Chicago. In example 1, the train has a positive Δx as it goes from Enid to Joplin. In 2, the same train going forward in the same direction has a negative Δx .
Note that in addition to x and Δx, there is a third quantity we could define, which would be like an odometer reading, or actual distance traveled. If you drive 10 miles, make a U-turn, and drive back 10 miles, then your Δx is zero, but your car’s odometer reading has increased by 20 miles. However important the odometer reading is to car owners and used car dealers, it is not very important in physics, and there is not even a standard name or notation for it. The change in position, Δx, is more useful because it is so much easier to calculate: to compute Δx, we only need to know the beginning and ending positions of the object, not all the information about how it got from one position to the other. self-check D A ball falls vertically, hits the floor, bounces to a height of one meter, falls, and hits the floor again. Is the Δx between the two impacts equal to zero, one, or two meters? Answer, p. 541
Section 2.2
Describing distance and time
73
Frames of reference The example above shows that there are two arbitrary choices you have to make in order to define a position variable, x. You have to decide where to put x = 0, and also which direction will be positive. This is referred to as choosing a coordinate system or choosing a frame of reference. (The two terms are nearly synonymous, but the first focuses more on the actual x variable, while the second is more of a general way of referring to one’s point of view.) As long as you are consistent, any frame is equally valid. You just don’t want to change coordinate systems in the middle of a calculation.
o / Motion locity.
with
constant
ve-
Have you ever been sitting in a train in a station when suddenly you notice that the station is moving backward? Most people would describe the situation by saying that you just failed to notice that the train was moving — it only seemed like the station was moving. But this shows that there is yet a third arbitrary choice that goes into choosing a coordinate system: valid frames of reference can differ from each other by moving relative to one another. It might seem strange that anyone would bother with a coordinate system that was moving relative to the earth, but for instance the frame of reference moving along with a train might be far more convenient for describing things happening inside the train.
2.3 Graphs of motion; velocity Motion with constant velocity
p / Motion that decreases x is represented with negative values of Δx and v .
In example o, an object is moving at constant speed in one direction. We can tell this because every two seconds, its position changes by five meters. In algebra notation, we’d say that the graph of x vs. t shows the same change in position, Δx = 5.0 m, over each interval of Δt = 2.0 s. The object’s velocity or speed is obtained by calculating v = Δx/Δt = (5.0 m)/(2.0 s) = 2.5 m/s. In graphical terms, the velocity can be interpreted as the slope of the line. Since the graph is a straight line, it wouldn’t have mattered if we’d taken a longer time interval and calculated v = Δx/Δt = (10.0 m)/(4.0 s). The answer would still have been the same, 2.5 m/s. Note that when we divide a number that has units of meters by another number that has units of seconds, we get units of meters per second, which can be written m/s. This is another case where we treat units as if they were algebra symbols, even though they’re not.
q / Motion with changing velocity. How can we find the velocity at the time indicated by the dot?
74
Chapter 2
In example p, the object is moving in the opposite direction: as time progresses, its x coordinate decreases. Recalling the definition of the Δ notation as “after minus before,” we find that Δt is still positive, but Δx must be negative. The slope of the line is therefore
Velocity and relative motion
negative, and we say that the object has a negative velocity, v = Δx/Δt = (−5.0 m)/(2.0 s) = −2.5 m/s. We’ve already seen that the plus and minus signs of Δx values have the interpretation of telling us which direction the object moved. Since Δt is always positive, dividing by Δt doesn’t change the plus or minus sign, and the plus and minus signs of velocities are to be interpreted in the same way. In graphical terms, a positive slope characterizes a line that goes up as we go to the right, and a negative slope tells us that the line went down as we went to the right. Solved problem: light-years
page 88, problem 4
Motion with changing velocity Now what about a graph like figure q? This might be a graph of a car’s motion as the driver cruises down the freeway, then slows down to look at a car crash by the side of the road, and then speeds up again, disappointed that there is nothing dramatic going on such as flames or babies trapped in their car seats. (Note that we are still talking about one-dimensional motion. Just because the graph is curvy doesn’t mean that the car’s path is curvy. The graph is not like a map, and the horizontal direction of the graph represents the passing of time, not distance.) Example q is similar to example o in that the object moves a total of 25.0 m in a period of 10.0 s, but it is no longer true that it makes the same amount of progress every second. There is no way to characterize the entire graph by a certain velocity or slope, because the velocity is different at every moment. It would be incorrect to say that because the car covered 25.0 m in 10.0 s, its velocity was 2.5 m/s. It moved faster than that at the beginning and end, but slower in the middle. There may have been certain instants at which the car was indeed going 2.5 m/s, but the speedometer swept past that value without “sticking,” just as it swung through various other values of speed. (I definitely want my next car to have a speedometer calibrated in m/s and showing both negative and positive values.) We assume that our speedometer tells us what is happening to the speed of our car at every instant, but how can we define speed mathematically in a case like this? We can’t just define it as the slope of the curvy graph, because a curve doesn’t have a single well-defined slope as does a line. A mathematical definition that corresponded to the speedometer reading would have to be one that attached a different velocity value to a single point on the curve, i.e., a single instant in time, rather than to the entire graph. If we wish to define the speed at one instant such as the one marked with a dot, the best way to proceed is illustrated in r, where we have drawn the line through that point called the tangent line, the line that “hugs the curve.” We can then adopt the following definition of velocity:
Section 2.3
r / The velocity at any given moment is defined as the slope of the tangent line through the relevant point on the graph.
s / Example: finding the velocity at the point indicated with the dot.
t / Reversing motion.
the
Graphs of motion; velocity
direction
of
75
definition of velocity The velocity of an object at any given moment is the slope of the tangent line through the relevant point on its x − t graph. One interpretation of this definition is that the velocity tells us how many meters the object would have traveled in one second, if it had continued moving at the same speed for at least one second. To some people the graphical nature of this definition seems “inaccurate” or “not mathematical.” The equation by itself, however, is only valid if the velocity is constant, and so cannot serve as a general definition. The slope of the tangent line example 2 What is the velocity at the point shown with a dot on the graph? First we draw the tangent line through that point. To find the slope of the tangent line, we need to pick two points on it. Theoretically, the slope should come out the same regardless of which two points we pick, but in practical terms we’ll be able to measure more accurately if we pick two points fairly far apart, such as the two white diamonds. To save work, we pick points that are directly above labeled points on the t axis, so that Δt = 4.0 s is easy to read off. One diamond lines up with x ≈ 17.5 m, the other with x ≈ 26.5 m, so Δx = 9.0 m. The velocity is Δx/Δt = 2.2 m/s. Conventions about graphing The placement of t on the horizontal axis and x on the upright axis may seem like an arbitrary convention, or may even have disturbed you, since your algebra teacher always told you that x goes on the horizontal axis and y goes on the upright axis. There is a reason for doing it this way, however. In example s, we have an object that reverses its direction of motion twice. It can only be in one place at any given time, but there can be more than one time when it is at a given place. For instance, this object passed through x = 17 m on three separate occasions, but there is no way it could have been in more than one place at t = 5.0 s. Resurrecting some terminology you learned in your trigonometry course, we say that x is a function of t, but t is not a function of x. In situations such as this, there is a useful convention that the graph should be oriented so that any vertical line passes through the curve at only one point. Putting the x axis across the page and t upright would have violated this convention. To people who are used to interpreting graphs, a graph that violates this convention is as annoying as fingernails scratching on a chalkboard. We say that this is a graph of “x versus t.” If the axes were the other way around, it would be a graph of “t versus x.” I remember the “versus” terminology by visualizing the labels on the x and t axes and remembering that when you read, you go from left to right and from top to bottom.
76
Chapter 2
Velocity and relative motion
Discussion questions A Park is running slowly in gym class, but then he notices Jenna watching him, so he speeds up to try to impress her. Which of the graphs could represent his motion?
B The figure shows a sequence of positions for two racing tractors. Compare the tractors’ velocities as the race progresses. When do they have the same velocity? [Based on a question by Lillian McDermott.]
C If an object had an x − t graph that was a straight line with Δx =0 and Δt = 0, what would you say about its velocity? Sketch an example of such a graph. What about Δt =0 and Δx = 0? D If an object has a wavy motion graph like the one in figure t on p. 75, what are the times at which the object reverses its direction? Describe the object’s velocity at these points. E
Discuss anything unusual about the following three graphs.
F I have been using the term “velocity” and avoiding the more common English word “speed,” because introductory physics texts typically define them to mean different things. They use the word “speed,” and the symbol “s” to mean the absolute value of the velocity, s = |v |. Although I’ve chosen not to emphasize this distinction in technical vocabulary, there are clearly two different concepts here. Can you think of an example of a graph of x -versus-t in which the object has constant speed, but not constant velocity? G For the graph shown in the figure, describe how the object’s velocity changes.
Section 2.3
Discussion question G.
Graphs of motion; velocity
77
H Two physicists duck out of a boring scientific conference. On the street, they witness an accident in which a pedestrian is injured by a hitand-run driver. A criminal trial results, and they must testify. In her testimony, Dr. Transverz Waive says, “The car was moving along pretty fast, I’d say the velocity was +40 mi/hr. They saw the old lady too late, and even though they slammed on the brakes they still hit her before they stopped. Then they made a U turn and headed off at a velocity of about -20 mi/hr, I’d say.” Dr. Longitud N.L. Vibrasheun says, “He was really going too fast, maybe his velocity was -35 or -40 mi/hr. After he hit Mrs. Hapless, he turned around and left at a velocity of, oh, I’d guess maybe +20 or +25 mi/hr.” Is their testimony contradictory? Explain.
2.4 The principle of inertia Physical effects relate only to a change in velocity Consider two statements of a kind that was at one time made with the utmost seriousness: People like Galileo and Copernicus who say the earth is rotating must be crazy. We know the earth can’t be moving. Why, if the earth was really turning once every day, then our whole city would have to be moving hundreds of leagues in an hour. That’s impossible! Buildings would shake on their foundations. Gale-force winds would knock us over. Trees would fall down. The Mediterranean would come sweeping across the east coasts of Spain and Italy. And furthermore, what force would be making the world turn? All this talk of passenger trains moving at forty miles an hour is sheer hogwash! At that speed, the air in a passenger compartment would all be forced against the back wall. People in the front of the car would suffocate, and people at the back would die because in such concentrated air, they wouldn’t be able to expel a breath. Some of the effects predicted in the first quote are clearly just based on a lack of experience with rapid motion that is smooth and free of vibration. But there is a deeper principle involved. In each case, the speaker is assuming that the mere fact of motion must have dramatic physical effects. More subtly, they also believe that a force is needed to keep an object in motion: the first person thinks a force would be needed to maintain the earth’s rotation, and the second apparently thinks of the rear wall as pushing on the air to keep it moving. Common modern knowledge and experience tell us that these people’s predictions must have somehow been based on incorrect reasoning, but it is not immediately obvious where the fundamental flaw lies. It’s one of those things a four-year-old could infuriate you by demanding a clear explanation of. One way of getting at the fundamental principle involved is to consider how the modern
78
Chapter 2
Velocity and relative motion
concept of the universe differs from the popular conception at the time of the Italian Renaissance. To us, the word “earth” implies a planet, one of the nine planets of our solar system, a small ball of rock and dirt that is of no significance to anyone in the universe except for members of our species, who happen to live on it. To Galileo’s contemporaries, however, the earth was the biggest, most solid, most important thing in all of creation, not to be compared with the wandering lights in the sky known as planets. To us, the earth is just another object, and when we talk loosely about “how fast” an object such as a car “is going,” we really mean the carobject’s velocity relative to the earth-object.
v / Why does Aristotle look so sad? Has he realized that his entire system of physics is wrong?
u / This Air Force doctor volunteered to ride a rocket sled as a medical experiment. The obvious effects on his head and face are not because of the sled’s speed but because of its rapid changes in speed: increasing in 2 and 3, and decreasing in 5 and 6. In 4 his speed is greatest, but because his speed is not increasing or decreasing very much at this moment, there is little effect on him.
w / The earth spins. People in Shanghai say they’re at rest and people in Los Angeles are moving. Angelenos say the same about the Shanghainese.
Motion is relative According to our modern world-view, it really isn’t that reasonable to expect that a special force should be required to make the air in the train have a certain velocity relative to our planet. After all, the “moving” air in the “moving” train might just happen to have zero velocity relative to some other planet we don’t even know about. Aristotle claimed that things “naturally” wanted to be at rest, lying on the surface of the earth. But experiment after exper-
Section 2.4
x / The jets are at rest. The Empire State Building is moving.
The principle of inertia
79
iment has shown that there is really nothing so special about being at rest relative to the earth. For instance, if a mattress falls out of the back of a truck on the freeway, the reason it rapidly comes to rest with respect to the planet is simply because of friction forces exerted by the asphalt, which happens to be attached to the planet. Galileo’s insights are summarized as follows: The principle of inertia No force is required to maintain motion with constant velocity in a straight line, and absolute motion does not cause any observable physical effects.
Discussion question A.
There are many examples of situations that seem to disprove the principle of inertia, but these all result from forgetting that friction is a force. For instance, it seems that a force is needed to keep a sailboat in motion. If the wind stops, the sailboat stops too. But the wind’s force is not the only force on the boat; there is also a frictional force from the water. If the sailboat is cruising and the wind suddenly disappears, the backward frictional force still exists, and since it is no longer being counteracted by the wind’s forward force, the boat stops. To disprove the principle of inertia, we would have to find an example where a moving object slowed down even though no forces whatsoever were acting on it. Over the years since Galileo’s lifetime, physicists have done more and more precise experiments to search for such a counterexample, but the results have always been negative. Three such tests are described on pp. 110, 233, and 261. self-check E What is incorrect about the following supposed counterexamples to the principle of inertia?
Discussion question B.
(1) When astronauts blast off in a rocket, their huge velocity does cause a physical effect on their bodies — they get pressed back into their seats, the flesh on their faces gets distorted, and they have a hard time lifting their arms. (2) When you’re driving in a convertible with the top down, the wind in your face is an observable physical effect of your absolute motion. Answer, p. 541 Solved problem: a bug on a wheel
page 88, problem 7
Discussion questions A A passenger on a cruise ship finds, while the ship is docked, that he can leap off of the upper deck and just barely make it into the pool on the lower deck. If the ship leaves dock and is cruising rapidly, will this adrenaline junkie still be able to make it? B You are a passenger in the open basket hanging under a helium balloon. The balloon is being carried along by the wind at a constant
80
Chapter 2
Velocity and relative motion
velocity. If you are holding a flag in your hand, will the flag wave? If so, which way? [Based on a question from PSSC Physics.] C Aristotle stated that all objects naturally wanted to come to rest, with the unspoken implication that “rest” would be interpreted relative to the surface of the earth. Suppose we go back in time and transport Aristotle to the moon. Aristotle knew, as we do, that the moon circles the earth; he said it didn’t fall down because, like everything else in the heavens, it was made out of some special substance whose “natural” behavior was to go in circles around the earth. We land, put him in a space suit, and kick him out the door. What would he expect his fate to be in this situation? If intelligent creatures inhabited the moon, and one of them independently came up with the equivalent of Aristotelian physics, what would they think about objects coming to rest?
Discussion question D.
D The glass is sitting on a level table in a train’s dining car, but the surface of the water is tilted. What can you infer about the motion of the train?
2.5 Addition of velocities Addition of velocities to describe relative motion Since absolute motion cannot be unambiguously measured, the only way to describe motion unambiguously is to describe the motion of one object relative to another. Symbolically, we can write vP Q for the velocity of object P relative to object Q. Velocities measured with respect to different reference points can be compared by addition. In the figure below, the ball’s velocity relative to the couch equals the ball’s velocity relative to the truck plus the truck’s velocity relative to the couch: vBC = vBT + vT C = 5 cm/s + 10 cm/s = 15 cm/s The same equation can be used for any combination of three objects, just by substituting the relevant subscripts for B, T, and C. Just remember to write the equation so that the velocities being added have the same subscript twice in a row. In this example, if you read off the subscripts going from left to right, you get BC . . . = . . . BTTC. The fact that the two “inside” subscripts on the right are the same means that the equation has been set up correctly. Notice how subscripts on the left look just like the subscripts on the right, but with the two T’s eliminated.
Negative velocities in relative motion My discussion of how to interpret positive and negative signs of velocity may have left you wondering why we should bother. Why not just make velocity positive by definition? The original reason
Section 2.5
Addition of velocities
81
y / These two highly competent physicists disagree on absolute velocities, but they would agree on relative velocities. Purple Dino considers the couch to be at rest, while Green Dino thinks of the truck as being at rest. They agree, however, that the truck’s velocity relative to the couch is vT C = 10 cm/s, the ball’s velocity relative to the truck is vBT = 5 cm/s, and the ball’s velocity relative to the couch is vBC = vBT + vT C = 15 cm/s.
why negative numbers were invented was that bookkeepers decided it would be convenient to use the negative number concept for payments to distinguish them from receipts. It was just plain easier than writing receipts in black and payments in red ink. After adding up your month’s positive receipts and negative payments, you either got a positive number, indicating profit, or a negative number, showing a loss. You could then show that total with a high-tech “+” or “−” sign, instead of looking around for the appropriate bottle of ink. Nowadays we use positive and negative numbers for all kinds of things, but in every case the point is that it makes sense to add and subtract those things according to the rules you learned in grade school, such as “minus a minus makes a plus, why this is true we need not discuss.” Adding velocities has the significance of comparing relative motion, and with this interpretation negative and positive velocities can be used within a consistent framework. For example, the truck’s velocity relative to the couch equals the truck’s velocity relative to the ball plus the ball’s velocity relative to the couch: vT C = vT B + vBC = −5 cm/s + 15 cm/s = 10 cm/s
82
Chapter 2
Velocity and relative motion
If we didn’t have the technology of negative numbers, we would have had to remember a complicated set of rules for adding velocities: (1) if the two objects are both moving forward, you add, (2) if one is moving forward and one is moving backward, you subtract, but (3) if they’re both moving backward, you add. What a pain that would have been. Solved problem: two dimensions
page 89, problem 10
Airspeed example 3 On June 1, 2009, Air France flight 447 disappeared without warning over the Atlantic Ocean. All 232 people aboard were killed. Investigators believe the disaster was triggered because the pilots lost the ability to accurately determine their speed relative to the air. This is done using sensors called Pitot tubes, mounted outside the plane on the wing. Automated radio signals showed that these sensors gave conflicting readings before the crash, possibly because they iced up. For fuel efficiency, modern passenger jets fly at a very high altitude, but in the thin air they can only fly within a very narrow range of speeds. If the speed is too low, the plane stalls, and if it’s too high, it breaks up. If the pilots can’t tell what their airspeed is, they can’t keep it in the safe range. Many people’s reaction to this story is to wonder why planes don’t just use GPS to measure their speed. One reason is that GPS tells you your speed relative to the ground, not relative to the air. Letting P be the plane, A the air, and G the ground, we have vPG = vPA + vAG
,
where vPG (the “true ground speed”) is what GPS would measure, vPA (“airspeed”) is what’s critical for stable flight, and vAG is the velocity of the wind relative to the ground 9000 meters below. Knowing vPG isn’t enough to determine vPA unless vAG is also known.
z / Example 3. 1. The aircraft before the disaster. 4. Wreckage being recovered.
2. A Pitot tube.
3. The flight path of flight 447.
Discussion questions A
Interpret the general rule vAB = −vBA in words.
Section 2.5
Addition of velocities
83
B Wa-Chuen slips away from her father at the mall and walks up the down escalator, so that she stays in one place. Write this in terms of symbols.
2.6 Graphs of velocity versus time Since changes in velocity play such a prominent role in physics, we need a better way to look at changes in velocity than by laboriously drawing tangent lines on x-versus-t graphs. A good method is to draw a graph of velocity versus time. The examples on the left show the x − t and v − t graphs that might be produced by a car starting from a traffic light, speeding up, cruising for a while at constant speed, and finally slowing down for a stop sign. If you have an air freshener hanging from your rear-view mirror, then you will see an effect on the air freshener during the beginning and ending periods when the velocity is changing, but it will not be tilted during the period of constant velocity represented by the flat plateau in the middle of the v − t graph. Students often mix up the things being represented on these two types of graphs. For instance, many students looking at the top graph say that the car is speeding up the whole time, since “the graph is becoming greater.” What is getting greater throughout the graph is x, not v.
aa / Graphs of x and v versus t for a car accelerating away from a traffic light, and then stopping for another red light.
Similarly, many students would look at the bottom graph and think it showed the car backing up, because “it’s going backwards at the end.” But what is decreasing at the end is v, not x. Having both the x − t and v − t graphs in front of you like this is often convenient, because one graph may be easier to interpret than the other for a particular purpose. Stacking them like this means that corresponding points on the two graphs’ time axes are lined up with each other vertically. However, one thing that is a little counterintuitive about the arrangement is that in a situation like this one involving a car, one is tempted to visualize the landscape stretching along the horizontal axis of one of the graphs. The horizontal axes, however, represent time, not position. The correct way to visualize the landscape is by mentally rotating the horizon 90 degrees counterclockwise and imagining it stretching along the upright axis of the x-t graph, which is the only axis that represents different positions in space.
2.7
Applications of calculus
The integral symbol, , in the heading for this section indicates that it is meant to be read by students in calculus-based physics. Students in an algebra-based physics course should skip these sections. The calculus-related sections in this book are meant to be usable by students who are taking calculus concurrently, so at this early
84
Chapter 2
Velocity and relative motion
point in the physics course I do not assume you know any calculus yet. This section is therefore not much more than a quick preview of calculus, to help you relate what you’re learning in the two courses. Newton was the first person to figure out the tangent-line definition of velocity for cases where the x − t graph is nonlinear. Before Newton, nobody had conceptualized the description of motion in terms of x − t and v − t graphs. In addition to the graphical techniques discussed in this chapter, Newton also invented a set of symbolic techniques called calculus. If you have an equation for x in terms of t, calculus allows you, for instance, to find an equation for v in terms of t. In calculus terms, we say that the function v(t) is the derivative of the function x(t). In other words, the derivative of a function is a new function that tells how rapidly the original function was changing. We now use neither Newton’s name for his technique (he called it “the method of fluxions”) nor his notation. The more commonly used notation is due to Newton’s German contemporary Leibnitz, whom the English accused of plagiarizing the calculus from Newton. In the Leibnitz notation, we write v=
dx dt
to indicate that the function v(t) equals the slope of the tangent line of the graph of x(t) at every time t. The Leibnitz notation is meant to evoke the delta notation, but with a very small time interval. Because the dx and dt are thought of as very small Δx’s and Δt’s, i.e., very small differences, the part of calculus that has to do with derivatives is called differential calculus. Differential calculus consists of three things: • The concept and definition of the derivative, which is covered in this book, but which will be discussed more formally in your math course. • The Leibnitz notation described above, which you’ll need to get more comfortable with in your math course. • A set of rules that allows you to find an equation for the derivative of a given function. For instance, if you happened to have a situation where the position of an object was given by the equation x = 2t7 , you would be able to use those rules to find dx/dt = 14t6 . This bag of tricks is covered in your math course.
Section 2.7
Applications of calculus
85
Summary Selected vocabulary center of mass . . the balance point of an object velocity . . . . . . the rate of change of position; the slope of the tangent line on an x − t graph. Notation x. . . . . . . . . . t . . . . . . . . . . Δ . . . . . . . . . Δx . . . . . . . .
Δt . . . . . . . . . v . . . . . . . . . . vAB . . . . . . . .
a point in space a point in time, a clock reading “change in;” the value of a variable afterwards minus its value before a distance, or more precisely a change in x, which may be less than the distance traveled; its plus or minus sign indicates direction a duration of time velocity the velocity of object A relative to object B
Other terminology and notation displacement . . a name for the symbol Δx speed . . . . . . . the absolute value of the velocity, i.e., the velocity stripped of any information about its direction Summary An object’s center of mass is the point at which it can be balanced. For the time being, we are studying the mathematical description only of the motion of an object’s center of mass in cases restricted to one dimension. The motion of an object’s center of mass is usually far simpler than the motion of any of its other parts. It is important to distinguish location, x, from distance, Δx, and clock reading, t, from time interval Δt. When an object’s x − t graph is linear, we define its velocity as the slope of the line, Δx/Δt. When the graph is curved, we generalize the definition so that the velocity is the slope of the tangent line at a given point on the graph. Galileo’s principle of inertia states that no force is required to maintain motion with constant velocity in a straight line, and absolute motion does not cause any observable physical effects. Things typically tend to reduce their velocity relative to the surface of our planet only because they are physically rubbing against the planet (or something attached to the planet), not because there is anything special about being at rest with respect to the earth’s surface. When it seems, for instance, that a force is required to keep a book sliding across a table, in fact the force is only serving to cancel the contrary force of friction. Absolute motion is not a well-defined concept, and if two observers are not at rest relative to one another they will disagree about the absolute velocities of objects. They will, however, agree
86
Chapter 2
Velocity and relative motion
about relative velocities. If object A is in motion relative to object B, and B is in motion relative to C, then A’s velocity relative to C is given by vAC = vAB + vBC . Positive and negative signs are used to indicate the direction of an object’s motion.
Summary
87
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 The graph shows the motion of a car stuck in stop-and-go freeway traffic. (a) If you only knew how far the car had gone during this entire time period, what would you think its velocity √ was? (b) What is the car’s maximum velocity? 2 (a) Let θ be the latitude of a point on the Earth’s surface. Derive an algebra equation for the distance, L, traveled by that point during one rotation of the Earth about its axis, i.e., over one day, expressed in terms of θ and R, the radius of the earth. Check: Your equation should give L = 0 for the North Pole. (b) At what speed is Fullerton, at latitude θ = 34◦ , moving with the rotation of the Earth about its axis? Give your answer in units of mi/h. [See the table in the back of the book for the relevant √ data.]
Problem 1.
3 A person is parachute jumping. During the time between when she leaps out of the plane and when she opens her chute, her altitude is given by the equation y = (10000 m) − (50 m/s) t + (5.0 s)e−t/5.0 s . Find her velocity at t = 7.0 s. (This can be done on a calculator, without knowing calculus.) Because of air resistance, her velocity does not increase at a steady rate as it would for an object falling √ in vacuum. 4 A light-year is a unit of distance used in astronomy, and defined as the distance light travels in one year. The speed of light is 3.0×108 m/s. Find how many meters there are in one light-year, expressing your answer in scientific notation. Solution, p. 527 5 You’re standing in a freight train, and have no way to see out. If you have to lean to stay on your feet, what, if anything, does that tell you about the train’s velocity? Explain. Solution, p. 527 6 A honeybee’s position as a function of time is given by x = 10t − t3 , where t is in seconds and x in meters. What is its√velocity at t = 3.0 s? 7 The figure shows the motion of a point on the rim of a rolling wheel. (The shape is called a cycloid.) Suppose bug A is riding on the rim of the wheel on a bicycle that is rolling, while bug B is on the spinning wheel of a bike that is sitting upside down on the floor. Bug A is moving along a cycloid, while bug B is moving in a circle. Both wheels are doing the same number of revolutions per minute. Which bug has a harder time holding on, or do they find it equally
Problem 7.
88
Chapter 2
Velocity and relative motion
difficult?
Solution, p. 528
8 Peanut plants fold up their leaves at night. Estimate the top speed of the tip of one of the leaves shown in the figure, expressing √ your result in scientific notation in SI units. 9 (a) Translate the following information into symbols, using the notation with two subscripts introduced in section 2.5. Eowyn is riding on her horse at a velocity of 11 m/s. She twists around in her saddle and fires an arrow backward. Her bow fires arrows at 25 m/s. (b) Find the velocity of the arrow relative to the ground. 10 Our full discussion of two- and three-dimensional motion is postponed until the second half of the book, but here is a chance to use a little mathematical creativity in anticipation of that generalization. Suppose a ship is sailing east at a certain speed v, and a passenger is walking across the deck at the same speed v, so that his track across the deck is perpendicular to the ship’s center-line. What is his speed relative to the water, and in what direction is he moving relative to the water? Solution, p. 528 11 Freddi Fish(TM) has a position as a function of time given by x = a/(b + t2 ). (a) Infer the units of the constants a and b. (b) Find her maximum speed. (c) Check that your answer has the right √ units. 12 Driving along in your car, you take your foot off the gas, and your speedometer shows a reduction in speed. Describe a frame of reference in which your car was speeding up during that same period of time. (The frame of reference should be defined by an observer who, although perhaps in motion relative to the earth, is not changing her own speed or direction of motion.)
Problem 8.
Problems
89
Problem 13.
13 The figure shows the motion of a bluefin tuna, as measured by a radio tag (Block et al., Nature, v. 434, p. 1121, 2005), over the course of several years. Until this study, it had been believed that the populations of the fish in the eastern and western Atlantic were separate, but this particular fish was observed to cross the entire Atlantic Ocean, from Virginia to Ireland. Points A, B, and C show a period of one month, during which the fish made the most rapid progress. Estimate its speed during that month, in units of √ kilometers per hour. 14 Sometimes doors are built with mechanisms that automatically close them after they have been opened. The designer can set both the strength of the spring and the amount of friction. If there is too much friction in relation to the strength of the spring, the door takes too long to close, but if there is too little, the door will oscillate. For an optimal design, we get can motion of the form x = cte−bt , where x is the position of some point on the door, and c and b are positive constants. (Similar systems are used for other mechanical devices, such as stereo speakers and the recoil mechanisms of guns.) In this example, the door moves in the positive direction up until a certain time, then stops and settles back in the negative direction, eventually approaching x = 0. This would be the type of motion we would get if someone flung a door open and the door closer then brought it back closed again. (a) Infer the units of the constants b and c. (b) Find the door’s maximum speed (i.e., the greatest absolute value √ of its velocity) as it comes back to the closed position. (c) Show that your answer has units that make sense. 15 At a picnic, someone hands you a can of beer. The ground is uneven, and you don’t want to spill your drink. You reason that it will be more stable if you drink some of it first in order to lower its center of mass. How much should you drink in order to make the center of mass as low as possible? [Based on a problem by Walter van B. Roberts and Martin Gardner.]
90
Chapter 2
Velocity and relative motion
Galileo’s contradiction of Aristotle had serious consequences. He was interrogated by the Church authorities and convicted of teaching that the earth went around the sun as a matter of fact and not, as he had promised previously, as a mere mathematical hypothesis. He was placed under permanent house arrest, and forbidden to write about or teach his theories. Immediately after being forced to recant his claim that the earth revolved around the sun, the old man is said to have muttered defiantly “and yet it does move.” The story is dramatic, but there are some omissions in the commonly taught heroic version. There was a rumor that the Simplicio character represented the Pope. Also, some of the ideas Galileo advocated had controversial religious overtones. He believed in the existence of atoms, and atomism was thought by some people to contradict the Church’s doctrine of transubstantiation, which said that in the Catholic mass, the blessing of the bread and wine literally transformed them into the flesh and blood of Christ. His support for a cosmology in which the earth circled the sun was also disreputable because one of its supporters, Giordano Bruno, had also proposed a bizarre synthesis of Christianity with the ancient Egyptian religion.
Chapter 3
Acceleration and free fall 3.1 The motion of falling objects The motion of falling objects is the simplest and most common example of motion with changing velocity. The early pioneers of
91
physics had a correct intuition that the way things drop was a message directly from Nature herself about how the universe worked. Other examples seem less likely to have deep significance. A walking person who speeds up is making a conscious choice. If one stretch of a river flows more rapidly than another, it may be only because the channel is narrower there, which is just an accident of the local geography. But there is something impressively consistent, universal, and inexorable about the way things fall. Stand up now and simultaneously drop a coin and a bit of paper side by side. The paper takes much longer to hit the ground. That’s why Aristotle wrote that heavy objects fell more rapidly. Europeans believed him for two thousand years. Now repeat the experiment, but make it into a race between the coin and your shoe. My own shoe is about 50 times heavier than the nickel I had handy, but it looks to me like they hit the ground at exactly the same moment. So much for Aristotle! Galileo, who had a flair for the theatrical, did the experiment by dropping a bullet and a heavy cannonball from a tall tower. Aristotle’s observations had been incomplete, his interpretation a vast oversimplification. It is inconceivable that Galileo was the first person to observe a discrepancy with Aristotle’s predictions. Galileo was the one who changed the course of history because he was able to assemble the observations into a coherent pattern, and also because he carried out systematic quantitative (numerical) measurements rather than just describing things qualitatively.
a / Galileo dropped a cannonball and a musketball simultaneously from a tower, and observed that they hit the ground at nearly the same time. This contradicted Aristotle’s long-accepted idea that heavier objects fell faster. (The experiment is described in a biography by Galileo’s pupil Viviani, and may belong more to legend than to history. It was definitely performed by various others of the same generation, such as Simon Stevin.)
92
Chapter 3
Why is it that some objects, like the coin and the shoe, have similar motion, but others, like a feather or a bit of paper, are different? Galileo speculated that in addition to the force that always pulls objects down, there was an upward force exerted by the air. Anyone can speculate, but Galileo went beyond speculation and came up with two clever experiments to probe the issue. First, he experimented with objects falling in water, which probed the same issues but made the motion slow enough that he could take time measurements with a primitive pendulum clock. With this technique, he established the following facts: • All heavy, streamlined objects (for example a steel rod dropped point-down) reach the bottom of the tank in about the same amount of time, only slightly longer than the time they would take to fall the same distance in air. • Objects that are lighter or less streamlined take a longer time to reach the bottom. This supported his hypothesis about two contrary forces. He imagined an idealized situation in which the falling object did not
Acceleration and free fall
have to push its way through any substance at all. Falling in air would be more like this ideal case than falling in water, but even a thin, sparse medium like air would be sufficient to cause obvious effects on feathers and other light objects that were not streamlined. Today, we have vacuum pumps that allow us to suck nearly all the air out of a chamber, and if we drop a feather and a rock side by side in a vacuum, the feather does not lag behind the rock at all. How the speed of a falling object increases with time Galileo’s second stroke of genius was to find a way to make quantitative measurements of how the speed of a falling object increased as it went along. Again it was problematic to make sufficiently accurate time measurements with primitive clocks, and again he found a tricky way to slow things down while preserving the essential physical phenomena: he let a ball roll down a slope instead of dropping it vertically. The steeper the incline, the more rapidly the ball would gain speed. Without a modern video camera, Galileo had invented a way to make a slow-motion version of falling.
b / Velocity increases more gradually on the gentle slope, but the motion is otherwise the same as the motion of a falling object.
Although Galileo’s clocks were only good enough to do accurate experiments at the smaller angles, he was confident after making a systematic study at a variety of small angles that his basic conclusions were generally valid. Stated in modern language, what he found was that the velocity-versus-time graph was a line. In the language of algebra, we know that a line has an equation of the form y = ax + b, but our variables are v and t, so it would be v = at + b. (The constant b can be interpreted simply as the initial velocity of the object, i.e., its velocity at the time when we started our clock, which we conventionally write as vo .)
c / The v − t graph of a falling object is a line.
self-check A An object is rolling down an incline. After it has been rolling for a short time, it is found to travel 13 cm during a certain one-second interval. During the second after that, it goes 16 cm. How many cm will it travel in the second after that? Answer, p. 541
Section 3.1
The motion of falling objects
93
A contradiction in Aristotle’s reasoning
d / Galileo’s experiments show that all falling objects have the same motion if air resistance is negligible.
Galileo’s inclined-plane experiment disproved the long-accepted claim by Aristotle that a falling object had a definite “natural falling speed” proportional to its weight. Galileo had found that the speed just kept on increasing, and weight was irrelevant as long as air friction was negligible. Not only did Galileo prove experimentally that Aristotle had been wrong, but he also pointed out a logical contradiction in Aristotle’s own reasoning. Simplicio, the stupid character, mouths the accepted Aristotelian wisdom: S IMPLICIO : There can be no doubt but that a particular body . . . has a fixed velocity which is determined by nature. . . S ALVIATI : If then we take two bodies whose natural speeds are different, it is clear that, [according to Aristotle], on uniting the two, the more rapid one will be partly held back by the slower, and the slower will be somewhat hastened by the swifter. Do you not agree with me in this opinion? S IMPLICIO :
e / 1. Aristotle said that heavier objects fell faster than lighter ones. 2. If two rocks are tied together, that makes an extraheavy rock, which should fall faster. 3. But Aristotle’s theory would also predict that the light rock would hold back the heavy rock, resulting in a slower fall.
94
Chapter 3
You are unquestionably right.
S ALVIATI : But if this is true, and if a large stone moves with a speed of, say, eight [unspecified units] while a smaller moves with a speed of four, then when they are united, the system will move with a speed less than eight; but the two stones when tied together make a stone larger than that which before moved with a speed of eight. Hence the heavier body moves with less speed than the lighter; an effect which is contrary to your supposition. Thus you see how, from your assumption that the heavier body moves more rapidly than the lighter one, I infer that the heavier body moves more slowly. What is gravity? The physicist Richard Feynman liked to tell a story about how when he was a little kid, he asked his father, “Why do things fall?” As an adult, he praised his father for answering, “Nobody knows why things fall. It’s a deep mystery, and the smartest people in the world don’t know the basic reason for it.” Contrast that with the average person’s off-the-cuff answer, “Oh, it’s because of gravity.” Feynman liked his father’s answer, because his father realized that simply giving a name to something didn’t mean that you understood it. The radical thing about Galileo’s and Newton’s approach to science was that they concentrated first on describing mathematically what really did happen, rather than spending a lot of time on untestable speculation such as Aristotle’s statement that “Things fall because they are trying to reach their natural place in contact with the earth.” That doesn’t mean that science can never answer the “why” questions. Over the next month or two as you delve deeper into physics, you will learn that there are more fundamental reasons why all falling objects have v − t graphs with the same slope, regardless
Acceleration and free fall
of their mass. Nevertheless, the methods of science always impose limits on how deep our explanation can go.
3.2 Acceleration Definition of acceleration for linear v − t graphs Galileo’s experiment with dropping heavy and light objects from a tower showed that all falling objects have the same motion, and his inclined-plane experiments showed that the motion was described by v = at+vo . The initial velocity vo depends on whether you drop the object from rest or throw it down, but even if you throw it down, you cannot change the slope, a, of the v − t graph. Since these experiments show that all falling objects have linear v − t graphs with the same slope, the slope of such a graph is apparently an important and useful quantity. We use the word acceleration, and the symbol a, for the slope of such a graph. In symbols, a = Δv/Δt. The acceleration can be interpreted as the amount of speed gained in every second, and it has units of velocity divided by time, i.e., “meters per second per second,” or m/s/s. Continuing to treat units as if they were algebra symbols, we simplify “m/s/s” to read “m/s2 .” Acceleration can be a useful quantity for describing other types of motion besides falling, and the word and the symbol “a” can be used in a more general context. We reserve the more specialized symbol “g” for the acceleration of falling objects, which on the surface of our planet equals 9.8 m/s2 . Often when doing approximate calculations or merely illustrative numerical examples it is good enough to use g = 10 m/s2 , which is off by only 2%.
f / Example 1.
Finding final speed, given time example 1 A despondent physics student jumps off a bridge, and falls for three seconds before hitting the water. How fast is he going when he hits the water? Approximating g as 10 m/s2 , he will gain 10 m/s of speed each second. After one second, his velocity is 10 m/s, after two seconds it is 20 m/s, and on impact, after falling for three seconds, he is moving at 30 m/s. Extracting acceleration from a graph example 2 The x − t and v − t graphs show the motion of a car starting from a stop sign. What is the car’s acceleration? Acceleration is defined as the slope of the v-t graph. The graph rises by 3 m/s during a time interval of 3 s, so the acceleration is (3 m/s)/(3 s) = 1 m/s2 . Incorrect solution #1: The final velocity is 3 m/s, and acceleration is velocity divided by time, so the acceleration is (3 m/s)/(10 s) = 0.3 m/s2 .
g / Example 2.
Section 3.2
Acceleration
95
The solution is incorrect because you can’t find the slope of a graph from one point. This person was just using the point at the right end of the v-t graph to try to find the slope of the curve. Incorrect solution #2: Velocity is distance divided by time so v = (4.5 m)/(3 s) = 1.5 m/s. Acceleration is velocity divided by time, so a = (1.5 m/s)/(3 s) = 0.5 m/s2 . The solution is incorrect because velocity is the slope of the tangent line. In a case like this where the velocity is changing, you can’t just pick two points on the x-t graph and use them to find the velocity. Converting g to different units What is g in units of cm/s2 ?
example 3
The answer is going to be how many cm/s of speed a falling object gains in one second. If it gains 9.8 m/s in one second, then it gains 980 cm/s in one second, so g = 980 cm/s2 . Alternatively, we can use the method of fractions that equal one: m 9.8 100 cm 980 cm × = 2 1 m s s2 What is g in units of miles/hour2 ? 9.8 m 1 mile 3600 s 2 × = 7.9 × 104 mile/hour2 × 1600 m 1 hour s2 This large number can be interpreted as the speed, in miles per hour, that you would gain by falling for one hour. Note that we had to square the conversion factor of 3600 s/hour in order to cancel out the units of seconds squared in the denominator. What is g in units of miles/hour/s? 1 mile 3600 s 9.8 m × × = 22 mile/hour/s 2 1600 m 1 hour s This is a figure that Americans will have an intuitive feel for. If your car has a forward acceleration equal to the acceleration of a falling object, then you will gain 22 miles per hour of speed every second. However, using mixed time units of hours and seconds like this is usually inconvenient for problem-solving. It would be like using units of foot-inches for area instead of ft2 or in2 . The acceleration of gravity is different in different locations. Everyone knows that gravity is weaker on the moon, but actually it is not even the same everywhere on Earth, as shown by the sampling of numerical data in the following table.
96
Chapter 3
Acceleration and free fall
location north pole Reykjavik, Iceland Guayaquil, Ecuador Mt. Cotopaxi, Ecuador Mt. Everest
latitude 90◦ N 64◦ N 2◦ S 1◦ S 28◦ N
elevation (m) 0 0 0 5896 8848
g (m/s2 ) 9.8322 9.8225 9.7806 9.7624 9.7643
The main variables that relate to the value of g on Earth are latitude and elevation. Although you have not yet learned how g would be calculated based on any deeper theory of gravity, it is not too hard to guess why g depends on elevation. Gravity is an attraction between things that have mass, and the attraction gets weaker with increasing distance. As you ascend from the seaport of Guayaquil to the nearby top of Mt. Cotopaxi, you are distancing yourself from the mass of the planet. The dependence on latitude occurs because we are measuring the acceleration of gravity relative to the earth’s surface, but the earth’s rotation causes the earth’s surface to fall out from under you. (We will discuss both gravity and rotation in more detail later in the course.)
Much more spectacular differences in the strength of gravity can be observed away from the Earth’s surface:
Section 3.2
h / This false-color map shows variations in the strength of the earth’s gravity. Purple areas have the strongest gravity, yellow the weakest. The overall trend toward weaker gravity at the equator and stronger gravity at the poles has been artificially removed to allow the weaker local variations to show up. The map covers only the oceans because of the technique used to make it: satellites look for bulges and depressions in the surface of the ocean. A very slight bulge will occur over an undersea mountain, for instance, because the mountain’s gravitational attraction pulls water toward it. The US government originally began collecting data like these for military use, to correct for the deviations in the paths of missiles. The data have recently been released for scientific and commercial use (e.g., searching for sites for off-shore oil wells).
Acceleration
97
location asteroid Vesta (surface) Earth’s moon (surface) Mars (surface) Earth (surface) Jupiter (cloud-tops) Sun (visible surface) typical neutron star (surface) black hole (center)
g (m/s2 ) 0.3 1.6 3.7 9.8 26 270 1012 infinite according to some theories, on the order of 1052 according to others
A typical neutron star is not so different in size from a large asteroid, but is orders of magnitude more massive, so the mass of a body definitely correlates with the g it creates. On the other hand, a neutron star has about the same mass as our Sun, so why is its g billions of times greater? If you had the misfortune of being on the surface of a neutron star, you’d be within a few thousand miles of all its mass, whereas on the surface of the Sun, you’d still be millions of miles from most of its mass. Discussion questions A
What is wrong with the following definitions of g ?
(1) “g is gravity.” (2) “g is the speed of a falling object.” (3) “g is how hard gravity pulls on things.” B When advertisers specify how much acceleration a car is capable of, they do not give an acceleration as defined in physics. Instead, they usually specify how many seconds are required for the car to go from rest to 60 miles/hour. Suppose we use the notation “a” for the acceleration as defined in physics, and “acar ad ” for the quantity used in advertisements for cars. In the US’s non-metric system of units, what would be the units of a and acar ad ? How would the use and interpretation of large and small, positive and negative values be different for a as opposed to acar ad ? C Two people stand on the edge of a cliff. As they lean over the edge, one person throws a rock down, while the other throws one straight up with an exactly opposite initial velocity. Compare the speeds of the rocks on impact at the bottom of the cliff.
3.3 Positive and negative acceleration Gravity always pulls down, but that does not mean it always speeds things up. If you throw a ball straight up, gravity will first slow it down to v = 0 and then begin increasing its speed. When I took physics in high school, I got the impression that positive signs of acceleration indicated speeding up, while negative accelerations represented slowing down, i.e., deceleration. Such a definition would be inconvenient, however, because we would then have to say that the same downward tug of gravity could produce either a positive
98
Chapter 3
Acceleration and free fall
or a negative acceleration. As we will see in the following example, such a definition also would not be the same as the slope of the v − t graph Let’s study the example of the rising and falling ball. In the example of the person falling from a bridge, I assumed positive velocity values without calling attention to it, which meant I was assuming a coordinate system whose x axis pointed down. In this example, where the ball is reversing direction, it is not possible to avoid negative velocities by a tricky choice of axis, so let’s make the more natural choice of an axis pointing up. The ball’s velocity will initially be a positive number, because it is heading up, in the same direction as the x axis, but on the way back down, it will be a negative number. As shown in the figure, the v − t graph does not do anything special at the top of the ball’s flight, where v equals 0. Its slope is always negative. In the left half of the graph, there is a negative slope because the positive velocity is getting closer to zero. On the right side, the negative slope is due to a negative velocity that is getting farther from zero, so we say that the ball is speeding up, but its velocity is decreasing! To summarize, what makes the most sense is to stick with the original definition of acceleration as the slope of the v − t graph, Δv/Δt. By this definition, it just isn’t necessarily true that things speeding up have positive acceleration while things slowing down have negative acceleration. The word “deceleration” is not used much by physicists, and the word “acceleration” is used unblushingly to refer to slowing down as well as speeding up: “There was a red light, and we accelerated to a stop.” Numerical calculation of a negative acceleration example 4 In figure i, what happens if you calculate the acceleration between t = 1.0 and 1.5 s?
i / The ball’s acceleration stays the same — on the way up, at the top, and on the way back down. It’s always negative.
Reading from the graph, it looks like the velocity is about −1 m/s at t = 1.0 s, and around −6 m/s at t = 1.5 s. The acceleration, figured between these two points, is a=
Δv (−6 m/s) − (−1 m/s) = = −10 m/s2 Δt (1.5 s) − (1.0 s)
.
Even though the ball is speeding up, it has a negative acceleration. Another way of convincing you that this way of handling the plus and minus signs makes sense is to think of a device that measures acceleration. After all, physics is supposed to use operational definitions, ones that relate to the results you get with actual measuring devices. Consider an air freshener hanging from the rear-view mirror of your car. When you speed up, the air freshener swings backward. Suppose we define this as a positive reading. When you slow down, the air freshener swings forward, so we’ll call this a negative reading
Section 3.3
Positive and negative acceleration
99
on our accelerometer. But what if you put the car in reverse and start speeding up backwards? Even though you’re speeding up, the accelerometer responds in the same way as it did when you were going forward and slowing down. There are four possible cases: motion of car
accelerometer slope of swings v-t graph
forward, speeding up forward, slowing down backward, speeding up backward, slowing down
backward forward forward backward
+ − − +
direction of force acting on car forward backward backward forward
Note the consistency of the three right-hand columns — nature is trying to tell us that this is the right system of classification, not the left-hand column. Because the positive and negative signs of acceleration depend on the choice of a coordinate system, the acceleration of an object under the influence of gravity can be either positive or negative. Rather than having to write things like “g = 9.8 m/s2 or −9.8 m/s2 ” every time we want to discuss g’s numerical value, we simply define g as the absolute value of the acceleration of objects moving under the influence of gravity. We consistently let g = 9.8 m/s2 , but we may have either a = g or a = −g, depending on our choice of a coordinate system. Acceleration with a change in direction of motion example 5 A person kicks a ball, which rolls up a sloping street, comes to a halt, and rolls back down again. The ball has constant acceleration. The ball is initially moving at a velocity of 4.0 m/s, and after 10.0 s it has returned to where it started. At the end, it has sped back up to the same speed it had initially, but in the opposite direction. What was its acceleration? By giving a positive number for the initial velocity, the statement of the question implies a coordinate axis that points up the slope of the hill. The “same” speed in the opposite direction should therefore be represented by a negative number, -4.0 m/s. The acceleration is a = Δv /Δt = (vf − vo )/10.0 s = [(−4.0 m/s) − (4.0 m/s)]/10.0s = −0.80 m/s2
.
The acceleration was no different during the upward part of the roll than on the downward part of the roll. Incorrect solution: Acceleration is Δv /Δt, and at the end it’s not moving any faster or slower than when it started, so Δv=0 and
100
Chapter 3
Acceleration and free fall
a = 0. The velocity does change, from a positive number to a negative number.
Discussion question B.
Discussion questions A A child repeatedly jumps up and down on a trampoline. Discuss the sign and magnitude of his acceleration, including both the time when he is in the air and the time when his feet are in contact with the trampoline. B The figure shows a refugee from a Picasso painting blowing on a rolling water bottle. In some cases the person’s blowing is speeding the bottle up, but in others it is slowing it down. The arrow inside the bottle shows which direction it is going, and a coordinate system is shown at the bottom of each figure. In each case, figure out the plus or minus signs of the velocity and acceleration. It may be helpful to draw a v − t graph in each case. C Sally is on an amusement park ride which begins with her chair being hoisted straight up a tower at a constant speed of 60 miles/hour. Despite stern warnings from her father that he’ll take her home the next time she misbehaves, she decides that as a scientific experiment she really needs to release her corndog over the side as she’s on the way up. She does not throw it. She simply sticks it out of the car, lets it go, and watches it against the background of the sky, with no trees or buildings as reference points. What does the corndog’s motion look like as observed by Sally? Does its speed ever appear to her to be zero? What acceleration does she observe it to have: is it ever positive? negative? zero? What would her enraged father answer if asked for a similar description of its motion as it appears to him, standing on the ground?
Discussion question C.
D Can an object maintain a constant acceleration, but meanwhile reverse the direction of its velocity? E Can an object have a velocity that is positive and increasing at the same time that its acceleration is decreasing?
Section 3.3
Positive and negative acceleration
101
3.4 Varying acceleration So far we have only been discussing examples of motion for which the v − t graph is linear. If we wish to generalize our definition to v-t graphs that are more complex curves, the best way to proceed is similar to how we defined velocity for curved x − t graphs: definition of acceleration The acceleration of an object at any instant is the slope of the tangent line passing through its v-versus-t graph at the relevant point. A skydiver example 6 The graphs in figure k show the results of a fairly realistic computer simulation of the motion of a skydiver, including the effects of air friction. The x axis has been chosen pointing down, so x is increasing as she falls. Find (a) the skydiver’s acceleration at t = 3.0 s, and also (b) at t = 7.0 s. The solution is shown in figure l. I’ve added tangent lines at the two points in question. (a) To find the slope of the tangent line, I pick two points on the line (not necessarily on the actual curve): (3.0 s, 28m/s) and (5.0 s, 42 m/s). The slope of the tangent line is (42 m/s−28 m/s)/(5.0 s− 3.0 s) = 7.0 m/s2 . (b) Two points on this tangent line are (7.0 s, 47 m/s) and (9.0 s, 52 m/s). The slope of the tangent line is (52 m/s−47 m/s)/(9.0 s−7.0 s) = 2.5 m/s2 . Physically, what’s happening is that at t = 3.0 s, the skydiver is not yet going very fast, so air friction is not yet very strong. She therefore has an acceleration almost as great as g. At t = 7.0 s, she is moving almost twice as fast (about 100 miles per hour), and air friction is extremely strong, resulting in a significant departure from the idealized case of no air friction.
k / Example 6.
In example 6, the x−t graph was not even used in the solution of the problem, since the definition of acceleration refers to the slope of the v − t graph. It is possible, however, to interpret an x − t graph to find out something about the acceleration. An object with zero acceleration, i.e., constant velocity, has an x − t graph that is a straight line. A straight line has no curvature. A change in velocity requires a change in the slope of the x − t graph, which means that it is a curve rather than a line. Thus acceleration relates to the curvature of the x − t graph. Figure m shows some examples.
102
Chapter 3
Acceleration and free fall
l / The solution to example 6.
In example 6, the x − t graph was more strongly curved at the beginning, and became nearly straight at the end. If the x − t graph is nearly straight, then its slope, the velocity, is nearly constant, and the acceleration is therefore small. We can thus interpret the acceleration as representing the curvature of the x − t graph, as shown in figure m. If the “cup” of the curve points up, the acceleration is positive, and if it points down, the acceleration is negative.
m / Acceleration relates to the curvature of the x − t graph.
Section 3.4
Varying acceleration
103
Since the relationship between a and v is analogous to the relationship between v and x, we can also make graphs of acceleration as a function of time, as shown in figure n.
n / Examples of graphs of x , v , and a versus t . 1. An object in free fall, with no friction. 2. A continuation of example 6, the skydiver.
Solved problem: Drawing a v − t graph.
page 115, problem 14
Solved problem: Drawing v − t and a − t graphs. page 116, problem 20
Figure o summarizes the relationships among the three types of graphs. o / How position, velocity, and acceleration are related.
104
Chapter 3
Discussion questions A Describe in words how the changes in the a − t graph in figure n/2 relate to the behavior of the v − t graph.
Acceleration and free fall
B Explain how each set of graphs contains inconsistencies, and fix them.
C In each case, pick a coordinate system and draw x − t , v − t , and a − t graphs. Picking a coordinate system means picking where you want x = 0 to be, and also picking a direction for the positive x axis. (1) An ocean liner is cruising in a straight line at constant speed. (2) You drop a ball. Draw two different sets of graphs (a total of 6), with one set’s positive x axis pointing in the opposite direction compared to the other’s. (3) You’re driving down the street looking for a house you’ve never been to before. You realize you’ve passed the address, so you slow down, put the car in reverse, back up, and stop in front of the house.
3.5 The area under the velocity-time graph A natural question to ask about falling objects is how fast they fall, but Galileo showed that the question has no answer. The physical law that he discovered connects a cause (the attraction of the planet Earth’s mass) to an effect, but the effect is predicted in terms of an acceleration rather than a velocity. In fact, no physical law predicts a definite velocity as a result of a specific phenomenon, because velocity cannot be measured in absolute terms, and only changes in velocity relate directly to physical phenomena. The unfortunate thing about this situation is that the definitions of velocity and acceleration are stated in terms of the tangent-line technique, which lets you go from x to v to a, but not the other way around. Without a technique to go backwards from a to v to x, we cannot say anything quantitative, for instance, about the x − t graph of a falling object. Such a technique does exist, and I used it to make the x − t graphs in all the examples above.
Section 3.5
The area under the velocity-time graph
105
First let’s concentrate on how to get x information out of a v − t graph. In example p/1, an object moves at a speed of 20 m/s for a period of 4.0 s. The distance covered is Δx = vΔt = (20 m/s) × (4.0 s) = 80 m. Notice that the quantities being multiplied are the width and the height of the shaded rectangle — or, strictly speaking, the time represented by its width and the velocity represented by its height. The distance of Δx = 80 m thus corresponds to the area of the shaded part of the graph. The next step in sophistication is an example like p/2, where the object moves at a constant speed of 10 m/s for two seconds, then for two seconds at a different constant speed of 20 m/s. The shaded region can be split into a small rectangle on the left, with an area representing Δx = 20 m, and a taller one on the right, corresponding to another 40 m of motion. The total distance is thus 60 m, which corresponds to the total area under the graph. An example like p/3 is now just a trivial generalization; there is simply a large number of skinny rectangular areas to add up. But notice that graph p/3 is quite a good approximation to the smooth curve p/4. Even though we have no formula for the area of a funny shape like p/4, we can approximate its area by dividing it up into smaller areas like rectangles, whose area is easier to calculate. If someone hands you a graph like p/4 and asks you to find the area under it, the simplest approach is just to count up the little rectangles on the underlying graph paper, making rough estimates of fractional rectangles as you go along.
That’s what I’ve done in figure q. Each rectangle on the graph paper is 1.0 s wide and 2 m/s tall, so it represents 2 m. Adding up all the numbers gives Δx = 41 m. If you needed better accuracy, you could use graph paper with smaller rectangles. It’s important to realize that this technique gives you Δx, not x. The v − t graph has no information about where the object was when it started. The following are important points to keep in mind when applying this technique: • If the range of v values on your graph does not extend down to zero, then you will get the wrong answer unless you compensate by adding in the area that is not shown.
p / The area under the v − t graph gives Δx .
106
Chapter 3
• As in the example, one rectangle on the graph paper does not necessarily correspond to one meter of distance. • Negative velocity values represent motion in the opposite direction, so as suggested by figure r, area under the t axis should be subtracted, i.e., counted as “negative area.”
Acceleration and free fall
q / An example using estimation of fractions of a rectangle.
• Since the result is a Δx value, it only tells you xaf ter − xbef ore , which may be less than the actual distance traveled. For instance, the object could come back to its original position at the end, which would correspond to Δx=0, even though it had actually moved a nonzero distance. Finally, note that one can find Δv from an a − t graph using an entirely analogous method. Each rectangle on the a − t graph represents a certain amount of velocity change.
r / Area underneath the is considered negative.
axis
Discussion question A Roughly what would a pendulum’s v − t graph look like? What would happen when you applied the area-under-the-curve technique to find the pendulum’s Δx for a time period covering many swings?
3.6 Algebraic results for constant acceleration Although the area-under-the-curve technique can be applied to any graph, no matter how complicated, it may be laborious to carry out, and if fractions of rectangles must be estimated the result will only be approximate. In the special case of motion with constant acceleration, it is possible to find a convenient shortcut which produces exact results. When the acceleration is constant, the v − t graph
Section 3.6
Algebraic results for constant acceleration
107
is a straight line, as shown in the figure. The area under the curve can be divided into a triangle plus a rectangle, both of whose areas can be calculated exactly: A = bh for a rectangle and A = bh/2 for a triangle. The height of the rectangle is the initial velocity, vo , and the height of the triangle is the change in velocity from beginning to end, Δv. The object’s Δx is therefore given by the equation Δx = vo Δt + ΔvΔt/2. This can be simplified a little by using the definition of acceleration, a = Δv/Δt, to eliminate Δv, giving
1 Δx = vo Δt + aΔt2 2
.
[motion with constant acceleration]
Since this is a second-order polynomial in Δt, the graph of Δx versus Δt is a parabola, and the same is true of a graph of x versus t — the two graphs differ only by shifting along the two axes. Although I have derived the equation using a figure that shows a positive vo , positive a, and so on, it still turns out to be true regardless of what plus and minus signs are involved. Another useful equation can be derived if one wants to relate the change in velocity to the distance traveled. This is useful, for instance, for finding the distance needed by a car to come to a stop. For simplicity, we start by deriving the equation for the special case of vo = 0, in which the final velocity vf is a synonym for Δv. Since velocity and distance are the variables of interest, not time, we take the equation Δx = 12 aΔt2 and use Δt = Δv/a to eliminate Δt. This gives Δx = (Δv)2 /2a, which can be rewritten as s / The shaded area tells us how far an object moves while accelerating at a constant rate.
vf2 = 2aΔx
.
[motion with constant acceleration, vo = 0]
For the more general case where vo = 0, we skip the tedious algebra leading to the more general equation,
vf2 = vo2 + 2aΔx
108
Chapter 3
Acceleration and free fall
.
[motion with constant acceleration]
To help get this all organized in your head, first let’s categorize the variables as follows: Variables that change during motion with constant acceleration: x ,v, t Variable that doesn’t change: a If you know one of the changing variables and want to find another, there is always an equation that relates those two:
The symmetry among the three variables is imperfect only because the equation relating x and t includes the initial velocity. There are two main difficulties encountered by students in applying these equations: • The equations apply only to motion with constant acceleration. You can’t apply them if the acceleration is changing. • Students are often unsure of which equation to use, or may cause themselves unnecessary work by taking the longer path around the triangle in the chart above. Organize your thoughts by listing the variables you are given, the ones you want to find, and the ones you aren’t given and don’t care about. Saving an old lady example 7 You are trying to pull an old lady out of the way of an oncoming truck. You are able to give her an acceleration of 20 m/s2 . Starting from rest, how much time is required in order to move her 2 m? First we organize our thoughts: Variables given: Δx, a, vo Variables desired: Δt Irrelevant variables: vf Consulting the triangular chart above, the equation we need is clearly Δx = vo Δt + 12 aΔt 2 , since it has the four variables of interest and omits the irrelevant one. Eliminating the vo term and solving for Δt gives Δt = 2Δx/a = 0.4 s.
Section 3.6
Algebraic results for constant acceleration
109
Solved problem: A stupid celebration
page 115, problem 15
Solved problem: Dropping a rock on Mars
page 115, problem 16
Solved problem: The Dodge Viper
page 116, problem 18
Solved problem: Half-way sped up
page 116, problem 22
Discussion questions A In chapter 1, I gave examples of correct and incorrect reasoning about proportionality, using questions about the scaling of area and volume. Try to translate the incorrect modes of reasoning shown there into mistakes about the following question: If the acceleration of gravity on Mars is 1/3 that on Earth, how many times longer does it take for a rock to drop the same distance on Mars? B Check that the units make sense in the three equations derived in this section.
3.7 A test of the principle of inertia Historically, the first quantitative and well documented experimental test of the principle of inertia (p. 80) was performed by Galileo around 1590 and published decades later when he managed to find a publisher in the Netherlands that was beyond the reach of the Roman Inquisition.1 It was ingenious but somewhat indirect, and required a layer of interpretation and extrapolation on top of the actual observations. As described on p. 93, he established that objects rolling on inclined planes moved according to mathematical laws that we would today describe as in section 3.6. He knew that his rolling balls were subject to friction, as well as random errors due to the limited precision of the water clock that he used, but he took the approximate agreement of his equations with experiment to indicate that they gave the results that would be exact in the absence of friction. He also showed, purely empirically, that when a ball went up or down a ramp inclined at an angle θ, its acceleration was proportional to sin θ. Again, this required extrapolation to idealized conditions of zero friction. He then reasoned that if a ball was rolled on a horizontal ramp, with θ = 0, its acceleration would be zero. This is exactly what is required by the principle of inertia: in the absence of friction, motion continues indefinitely.
1
Galileo, Discourses and Mathematical Demonstrations Relating to Two New Sciences, 1638. The experiments are described in the Third Day, and their support for the principle of inertia is discussed in the Scholium following Theorems I-XIV. Another experiment involving a ship is described in Galileo’s 1624 reply to a letter from Fr. Ingoli, but although Galileo vigorously asserts that he really did carry it out, no detailed description or quantitative results are given.
110
Chapter 3
Acceleration and free fall
3.8
Applications of calculus
In section 2.7, I discussed how the slope-of-the-tangent-line idea related to the calculus concept of a derivative, and the branch of calculus known as differential calculus. The other main branch of calculus, integral calculus, has to do with the area-under-the-curve concept discussed in section 3.5. Again there is a concept, a notation, and a bag of tricks for doing things symbolically rather than graphically. In calculus, the area under the v − t graph between t = t1 and t = t2 is notated like this: t2 vdt . area under curve = Δx = t1
The expression on the right is called an integral, and the s-shaped symbol, the integral sign, is read as “integral of . . . ” Integral calculus and differential calculus are closely related. For instance, if you take the derivative of the function x(t), you get the function v(t), and if you integrate the function v(t), you get x(t) back again. In other words, integration and differentiation are inverse operations. This is known as the fundamental theorem of calculus. On an unrelated topic, there is a special notation for taking the derivative of a function twice. The acceleration, for instance, is the second (i.e., double) derivative of the position, because differentiating x once gives v, and then differentiating v gives a. This is written as d2 x a= 2 . dt The seemingly inconsistent placement of the twos on the top and bottom confuses all beginning calculus students. The motivation for this funny notation is that acceleration has units of m/s2 , and the notation correctly suggests that: the top looks like it has units of meters, the bottom seconds2 . The notation is not meant, however, to suggest that t is really squared.
Section 3.8
Applications of calculus
111
Summary Selected vocabulary gravity . . . . . . A general term for the phenomenon of attraction between things having mass. The attraction between our planet and a human-sized object causes the object to fall. acceleration . . . The rate of change of velocity; the slope of the tangent line on a v − t graph. Notation vo . . . . vf . . . . a. . . . . g . . . . .
. . . .
. . . .
. . . .
. . . .
. . . .
initial velocity final velocity acceleration the acceleration of objects in free fall; the strength of the local gravitational field
Summary Galileo showed that when air resistance is negligible all falling bodies have the same motion regardless of mass. Moreover, their v − t graphs are straight lines. We therefore define a quantity called acceleration as the slope, Δv/Δt, of an object’s v −t graph. In cases other than free fall, the v −t graph may be curved, in which case the definition is generalized as the slope of a tangent line on the v − t graph. The acceleration of objects in free fall varies slightly across the surface of the earth, and greatly on other planets. Positive and negative signs of acceleration are defined according to whether the v −t graph slopes up or down. This definition has the advantage that a force with a given sign, representing its direction, always produces an acceleration with the same sign. The area under the v − t graph gives Δx, and analogously the area under the a − t graph gives Δv. For motion with constant acceleration, the following three equations hold: 1 Δx = vo Δt + aΔt2 2 vf2 = vo2 + 2aΔx Δv a= Δt They are not valid if the acceleration is changing.
112
Chapter 3
Acceleration and free fall
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 The graph represents the velocity of a bee along a straight line. At t = 0, the bee is at the hive. (a) When is the bee farthest from the hive? (b) How far is the bee at its farthest point from the hive? (c) At t = 13 s, how far is the bee from the hive? [Hint:√Try problem 19 first.]
2 A rock is dropped into a pond. Draw plots of its position versus time, velocity versus time, and acceleration versus time. Include its whole motion, starting from the moment it is dropped, and continuing while it falls through the air, passes through the water, and ends up at rest on the bottom of the pond. Do your work on a photocopy or a printout of page 121. 3 In an 18th-century naval battle, a cannon ball is shot horizontally, passes through the side of an enemy ship’s hull, flies across the galley, and lodges in a bulkhead. Draw plots of its horizontal position, velocity, and acceleration as functions of time, starting while it is inside the cannon and has not yet been fired, and ending when it comes to rest. There is not any significant amount of friction from the air. Although the ball may rise and fall, you are only concerned with its horizontal motion, as seen from above. Do your work on a photocopy or a printout of page 121.
Problem 3.
4 Draw graphs of position, velocity, and acceleration as functions of time for a person bunjee jumping. (In bunjee jumping, a person has a stretchy elastic cord tied to his/her ankles, and jumps off of a high platform. At the bottom of the fall, the cord brings the person up short. Presumably the person bounces up a little.) Do your work on a photocopy or a printout of page 121.
Problems
113
5 A ball rolls down the ramp shown in the figure, consisting of a curved knee, a straight slope, and a curved bottom. For each part of the ramp, tell whether the ball’s velocity is increasing, decreasing, or constant, and also whether the ball’s acceleration is increasing, decreasing, or constant. Explain your answers. Assume there is no air friction or rolling resistance. Hint: Try problem 20 first. [Based on a problem by Hewitt.] 6 A toy car is released on one side of a piece of track that is bent into an upright U shape. The car goes back and forth. When the car reaches the limit of its motion on one side, its velocity is zero. Is its acceleration also zero? Explain using a v − t graph. [Based on a problem by Serway and Faughn.]
Problem 5.
7 What is the acceleration of a car that moves at a steady velocity of 100 km/h for 100 seconds? Explain your answer. [Based on a problem by Hewitt.] 8 A physics homework question asks, “If you start from rest and accelerate at 1.54 m/s2 for 3.29 s, how far do you travel by the end of that time?” A student answers as follows: 1.54 × 3.29 = 5.07 m His Aunt Wanda is good with numbers, but has never taken physics. She doesn’t know the formula for the distance traveled under constant acceleration over a given amount of time, but she tells her nephew his answer cannot be right. How does she know? 9 You are looking into a deep well. It is dark, and you cannot see the bottom. You want to find out how deep it is, so you drop a rock in, and you hear a splash 3.0 seconds later. How deep is√the well? 10 You take a trip in your spaceship to another star. Setting off, you increase your speed at a constant acceleration. Once you get half-way there, you start decelerating, at the same rate, so that by the time you get there, you have slowed down to zero speed. You see the tourist attractions, and then head home by the same method. (a) Find a formula for the time, T , required for the round trip, in terms of d, the distance from our sun to the star, and a, the magnitude of the acceleration. Note that the acceleration is not constant over the whole trip, but the trip can be broken up into constantacceleration parts. (b) The nearest star to the Earth (other than our own sun) is Proxima Centauri, at a distance of d = 4 × 1016 m. Suppose you use an acceleration of a = 10 m/s2 , just enough to compensate for the lack of true gravity and make you feel comfortable. How long does the round trip take, in years? (c) Using the same numbers for d and a, find your maximum speed. Compare this to the speed of light, which is 3.0 × 108 m/s. (Later in this course, you will learn that there are some new things going
114
Chapter 3
Acceleration and free fall
on in physics when one gets close to the speed of light, and that it is impossible to exceed the speed of light. For now, though, just use √ the simpler ideas you’ve learned so far.) 11 You climb half-way up a tree, and drop a rock. Then you climb to the top, and drop another rock. How many times greater is the velocity of the second rock on impact? Explain. (The answer is not two times greater.) 12 Alice drops a rock off a cliff. Bubba shoots a gun straight down from the edge of the same cliff. Compare the accelerations of the rock and the bullet while they are in the air on the way down. [Based on a problem by Serway and Faughn.] 13 A person is parachute jumping. During the time between when she leaps out of the plane and when she opens her chute, her altitude is given by an equation of the form
y = b − c t + ke−t/k , where e is the base of natural logarithms, and b, c, and k are constants. Because of air resistance, her velocity does not increase at a steady rate as it would for an object falling in vacuum. (a) What units would b, c, and k have to have for the equation to make sense? (b) Find the person’s velocity, v, as a function of time. [You will √ need to use the chain rule, and the fact that d(ex )/dx = ex .] (c) Use your answer from part (b) to get an interpretation of the constant c. [Hint: e−x approaches zero for large values of x.] √ (d) Find the person’s acceleration, a, as a function of time. (e) Use your answer from part (d) to show that if she waits long enough to open her chute, her acceleration will become very small. 14 The top part of the figure shows the position-versus-time graph for an object moving in one dimension. On the bottom part of the figure, sketch the corresponding v-versus-t graph. Solution, p. 528 15 On New Year’s Eve, a stupid person fires a pistol straight up. The bullet leaves the gun at a speed of 100 m/s. How long does it take before the bullet hits the ground? Solution, p. 528
Problem 14.
16 If the acceleration of gravity on Mars is 1/3 that on Earth, how many times longer does it take for a rock to drop the same distance on Mars? Ignore air resistance. Solution, p. 528 17 A honeybee’s position as a function of time is given by x = 10t − t3 , where t is in seconds and x in meters. What is its acceleration at t = 3.0 s? Solution, p. 528
Problems
115
18 In July 1999, Popular Mechanics carried out tests to find which car sold by a major auto maker could cover a quarter mile (402 meters) in the shortest time, starting from rest. Because the distance is so short, this type of test is designed mainly to favor the car with the greatest acceleration, not the greatest maximum speed (which is irrelevant to the average person). The winner was the Dodge Viper, with a time of 12.08 s. The car’s top (and presumably final) speed was 118.51 miles per hour (52.98 m/s). (a) If a car, starting from rest and moving with constant acceleration, covers a quarter mile in this time interval, what is its acceleration? (b) What would be the final speed of a car that covered a quarter mile with the constant acceleration you found in part a? (c) Based on the discrepancy between your answer in part b and the actual final speed of the Viper, what do you conclude about how its acceleration changed over time? Solution, p. 528 19 The graph represents the motion of a ball that rolls up a hill and then back down. When does the ball return to the location it had at t = 0? Solution, p. 529 20 (a) The ball is released at the top of the ramp shown in the figure. Friction is negligible. Use physical reasoning to draw v − t and a − t graphs. Assume that the ball doesn’t bounce at the point where the ramp changes slope. (b) Do the same for the case where the ball is rolled up the slope from the right side, but doesn’t quite have enough speed to make it over the top. Solution, p. 529
Problem 19.
21 You throw a rubber ball up, and it falls and bounces several times. Draw graphs of position, velocity, and acceleration as functions of time. Solution, p. 529 22 Starting from rest, a ball rolls down a ramp, traveling a distance L and picking up a final speed v. How much of the distance did the ball have to cover before achieving a speed of v/2? [Based on a problem by Arnold Arons.] Solution, p. 530
Problem 20.
23 The graph shows the acceleration of a chipmunk in a TV cartoon. It consists of two circular arcs and two line segments. At t = 0.00 s, the chipmunk’s velocity is −3.10 m/s. What is its √ velocity at t = 10.00 s? 24 Find the error in the following calculation. A student wants to find the distance traveled by a car that accelerates from rest for 5.0 s with an acceleration of 2.0 m/s2 . First he solves a = Δv/Δt for Δv = 10 m/s. Then he multiplies to find (10 m/s)(5.0 s) = 50 m. Do not just recalculate the result by a different method; if that was all you did, you’d have no way of knowing which calculation was correct, yours or his.
Problem 23.
116
Chapter 3
Acceleration and free fall
25 Acceleration could be defined either as Δv/Δt or as the slope of the tangent line on the v − t graph. Is either one superior as a definition, or are they equivalent? If you say one is better, give an example of a situation where it makes a difference which one you use. 26 If an object starts accelerating from rest, we have v 2 = 2aΔx for its speed after it has traveled a distance Δx. Explain in words why it makes sense that the equation has velocity squared, but distance only to the first power. Don’t recapitulate the derivation in the book, or give a justification based on units. The point is to explain what this feature of the equation tells us about the way speed increases as more distance is covered. 27 The figure shows a practical, simple experiment for determining g to high precision. Two steel balls are suspended from electromagnets, and are released simultaneously when the electric current is shut off. They fall through unequal heights Δx1 and Δx2 . A computer records the sounds through a microphone as first one ball and then the other strikes the floor. From this recording, we can accurately determine the quantity T defined as T = Δt2 − Δt1 , i.e., the time lag between the first and second impacts. Note that since the balls do not make any sound when they are released, we have no way of measuring the individual times Δt2 and Δt1 . (a) Find an equation for g in terms of the measured quantities T√, Δx1 and Δx2 . (b) Check the units of your equation. (c) Check that your equation gives the correct result in the case where Δx1 is very close to zero. However, is this case realistic? (d) What happens when Δx1 = Δx2 ? Discuss this both mathematically and physically.
Problem 27.
28 The speed required for a low-earth orbit is 7.9 × 103 m/s (see ch. 10). When a rocket is launched into orbit, it goes up a little at first to get above almost all of the atmosphere, but then tips over horizontally to build up to orbital speed. Suppose the horizontal acceleration is limited to 3g to keep from damaging the cargo (or hurting the crew, for a crewed flight). (a) What is the minimum distance the rocket must travel downrange before it reaches orbital speed? How much does it matter whether you take into account the initial eastward velocity due to the rotation of the earth? (b) Rather than a rocket ship, it might be advantageous to use a railgun design, in which the craft would be accelerated to orbital speeds along a railroad track. This has the advantage that it isn’t necessary to lift a large mass of fuel, since the energy source is external. Based on your answer to part a, comment on the feasibility of this design for crewed launches from the earth’s surface.
Problems
117
Problem 29. This spectacular series of photos from a 2011 paper by Burrows and Sutton (“Biomechanics of jumping in the flea,” J. Exp. Biology 214:836) shows the flea jumping at about a 45-degree angle, but for the sake of this estimate just consider the case of a flea jumping vertically.
29 Some fleas can jump as high as 30 cm. The flea only has a short time to build up speed — the time during which its center of mass is accelerating upward but its feet are still in contact with the ground. Make an order-of-magnitude estimate of the acceleration the flea needs to have while straightening its legs, and state your answer in units of g, i.e., how many “g’s it pulls.” (For comparison, fighter pilots black out or die if they exceed about 5 or 10 g’s.) 30 Consider the following passage from Alice in Wonderland, in which Alice has been falling for a long time down a rabbit hole: Down, down, down. Would the fall never come to an end? “I wonder how many miles I’ve fallen by this time?” she said aloud. “I must be getting somewhere near the center of the earth. Let me see: that would be four thousand miles down, I think” (for, you see, Alice had learned several things of this sort in her lessons in the schoolroom, and though this was not a very good opportunity for showing off her knowledge, as there was no one to listen to her, still it was good practice to say it over)... Alice doesn’t know much physics, but let’s try to calculate the amount of time it would take to fall four thousand miles, starting from rest with an acceleration of 10 m/s2 . This is really only a lower limit; if there really was a hole that deep, the fall would actually take a longer time than the one you calculate, both because there is air friction and because gravity gets weaker as you get deeper (at the center of the earth, g is zero, because the earth is pulling√you equally in every direction at once). 31 The photo shows Apollo 16 astronaut John Young jumping on the moon and saluting at the top of his jump. The video footage of the jump shows him staying aloft for 1.45 seconds. Gravity on the moon is 1/6 as strong as on the earth. Compute the height of √ the jump.
Problem 31.
118
Chapter 3
Acceleration and free fall
32 Most people don’t know that Spinosaurus aegyptiacus, not Tyrannosaurus rex, was the biggest theropod dinosaur. We can’t put a dinosaur on a track and time it in the 100 meter dash, so we can only infer from physical models how fast it could have run. When an animal walks at a normal pace, typically its legs swing more or less like pendulums of the same length . As a further simplification of this model, let’s imagine that the leg simply moves at a fixed acceleration as it falls to the ground. That is, we model the time for a quarter of a stride cycle as being the same as the time required for free fall from a height . S. aegyptiacus had legs about four times longer than those of a human. (a) Compare the time √ required for a human’s stride cycle to that for S. aegyptiacus. √ (b) Compare their running speeds.
Problem 32.
33 Engineering professor Qingming Li used sensors and video cameras to study punches delivered in the lab by British former welterweight boxing champion Ricky “the Hitman” Hatton. For comparison, Li also let a TV sports reporter put on the gloves and throw punches. The time it took for Hatton’s best punch to arrive, i.e., the time his opponent would have had to react, was about 0.47 of that for the reporter. Let’s assume that the fist starts from rest and moves with constant acceleration all the way up until impact, at some fixed distance (arm’s length). Compare Hatton’s acceleration √ to the reporter’s. 34 Aircraft carriers originated in World War I, and the first landing on a carrier was performed by E.H. Dunning in a Sopwith Pup biplane, landing on HMS Furious. (Dunning was killed the second time he attempted the feat.) In such a landing, the pilot slows down to just above the plane’s stall speed, which is the minimum speed at which the plane can fly without stalling. The plane then lands and is caught by cables and decelerated as it travels the length of the flight deck. Comparing a modern US F-14 fighter jet landing on an Enterprise-class carrier to Dunning’s original exploit, the stall speed is greater by a factor of 4.8, and to accomodate this, the length of the flight deck is greater by a factor of 1.9. Which deceleration is √ greater, and by what factor? 35 In college-level women’s softball in the U.S., typically a pitcher is expected to be at least 1.75 m tall, but Virginia Tech pitcher Jasmin Harrell is 1.62 m. Although a pitcher actually throws by stepping forward and swinging her arm in a circle, let’s make a simplified physical model to estimate how much of a disadvantage Harrell has had to overcome due to her height. We’ll pretend that the pitcher gives the ball a constant acceleration in a straight line, and that the length of this line is proportional to the pitcher’s height. Compare the acceleration Harrell would have to supply with the acceleration that would suffice for a pitcher of the nominal minimum √ height, if both were to throw a pitch at the same speed.
Problems
119
36 When the police engage in a high-speed chase on city streets, it can be extremely dangerous both to the police and to other motorists and pedestrians. Suppose that the police car must travel at a speed that is limited by the need to be able to stop before hitting a baby carriage, and that the distance at which the driver first sees the baby carriage is fixed. Tests show that in a panic stop from high speed, a police car based on a Chevy Impala has a deceleration 9% greater than that of a Dodge Intrepid. Compare the maximum√safe speeds for the two cars.
120
Chapter 3
Acceleration and free fall
Problems
121
122
Chapter 3
Acceleration and free fall
Isaac Newton
Chapter 4
Force and motion If I have seen farther than others, it is because I have stood on the shoulders of giants. Newton, referring to Galileo Even as great and skeptical a genius as Galileo was unable to make much progress on the causes of motion. It was not until a generation later that Isaac Newton (1642-1727) was able to attack the problem successfully. In many ways, Newton’s personality was the opposite of Galileo’s. Where Galileo agressively publicized his ideas,
123
Newton had to be coaxed by his friends into publishing a book on his physical discoveries. Where Galileo’s writing had been popular and dramatic, Newton originated the stilted, impersonal style that most people think is standard for scientific writing. (Scientific journals today encourage a less ponderous style, and papers are often written in the first person.) Galileo’s talent for arousing animosity among the rich and powerful was matched by Newton’s skill at making himself a popular visitor at court. Galileo narrowly escaped being burned at the stake, while Newton had the good fortune of being on the winning side of the revolution that replaced King James II with William and Mary of Orange, leading to a lucrative post running the English royal mint. Newton discovered the relationship between force and motion, and revolutionized our view of the universe by showing that the same physical laws applied to all matter, whether living or nonliving, on or off of our planet’s surface. His book on force and motion, the Mathematical Principles of Natural Philosophy, was uncontradicted by experiment for 200 years, but his other main work, Optics, was on the wrong track, asserting that light was composed of particles rather than waves. Newton was also an avid alchemist, a fact that modern scientists would like to forget.
4.1 Force We need only explain changes in motion, not motion itself.
a / Aristotle said motion had to be caused by a force. To explain why an arrow kept flying after the bowstring was no longer pushing on it, he said the air rushed around behind the arrow and pushed it forward. We know this is wrong, because an arrow shot in a vacuum chamber does not instantly drop to the floor as it leaves the bow. Galileo and Newton realized that a force would only be needed to change the arrow’s motion, not to make its motion continue.
So far you’ve studied the measurement of motion in some detail, but not the reasons why a certain object would move in a certain way. This chapter deals with the “why” questions. Aristotle’s ideas about the causes of motion were completely wrong, just like all his other ideas about physical science, but it will be instructive to start with them, because they amount to a road map of modern students’ incorrect preconceptions. Aristotle thought he needed to explain both why motion occurs and why motion might change. Newton inherited from Galileo the important counter-Aristotelian idea that motion needs no explanation, that it is only changes in motion that require a physical cause. Aristotle’s needlessly complex system gave three reasons for motion: Natural motion, such as falling, came from the tendency of objects to go to their “natural” place, on the ground, and come to rest. Voluntary motion was the type of motion exhibited by animals, which moved because they chose to. Forced motion occurred when an object was acted on by some other object that made it move.
124
Chapter 4
Force and motion
Motion changes due to an interaction between two objects. In the Aristotelian theory, natural motion and voluntary motion are one-sided phenomena: the object causes its own motion. Forced motion is supposed to be a two-sided phenomenon, because one object imposes its “commands” on another. Where Aristotle conceived of some of the phenomena of motion as one-sided and others as two-sided, Newton realized that a change in motion was always a two-sided relationship of a force acting between two physical objects. The one-sided “natural motion” description of falling makes a crucial omission. The acceleration of a falling object is not caused by its own “natural” tendencies but by an attractive force between it and the planet Earth. Moon rocks brought back to our planet do not “want” to fly back up to the moon because the moon is their “natural” place. They fall to the floor when you drop them, just like our homegrown rocks. As we’ll discuss in more detail later in this course, gravitational forces are simply an attraction that occurs between any two physical objects. Minute gravitational forces can even be measured between human-scale objects in the laboratory. The idea of natural motion also explains incorrectly why things come to rest. A basketball rolling across a beach slows to a stop because it is interacting with the sand via a frictional force, not because of its own desire to be at rest. If it was on a frictionless surface, it would never slow down. Many of Aristotle’s mistakes stemmed from his failure to recognize friction as a force. The concept of voluntary motion is equally flawed. You may have been a little uneasy about it from the start, because it assumes a clear distinction between living and nonliving things. Today, however, we are used to having the human body likened to a complex machine. In the modern world-view, the border between the living and the inanimate is a fuzzy no-man’s land inhabited by viruses, prions, and silicon chips. Furthermore, Aristotle’s statement that you can take a step forward “because you choose to” inappropriately mixes two levels of explanation. At the physical level of explanation, the reason your body steps forward is because of a frictional force acting between your foot and the floor. If the floor was covered with a puddle of oil, no amount of “choosing to” would enable you to take a graceful stride forward.
b / “Our eyes receive blue light reflected from this painting because Monet wanted to represent water with the color blue.” This is a valid statement at one level of explanation, but physics works at the physical level of explanation, in which blue light gets to your eyes because it is reflected by blue pigments in the paint.
Forces can all be measured on the same numerical scale. In the Aristotelian-scholastic tradition, the description of motion as natural, voluntary, or forced was only the broadest level of classification, like splitting animals into birds, reptiles, mammals, and amphibians. There might be thousands of types of motion, each of which would follow its own rules. Newton’s realization that all changes in motion were caused by two-sided interactions made
Section 4.1
Force
125
it seem that the phenomena might have more in common than had been apparent. In the Newtonian description, there is only one cause for a change in motion, which we call force. Forces may be of different types, but they all produce changes in motion according to the same rules. Any acceleration that can be produced by a magnetic force can equally well be produced by an appropriately controlled stream of water. We can speak of two forces as being equal if they produce the same change in motion when applied in the same situation, which means that they pushed or pulled equally hard in the same direction. The idea of a numerical scale of force and the newton unit were introduced in chapter 0. To recapitulate briefly, a force is when a pair of objects push or pull on each other, and one newton is the force required to accelerate a 1-kg object from rest to a speed of 1 m/s in 1 second. More than one force on an object As if we hadn’t kicked poor Aristotle around sufficiently, his theory has another important flaw, which is important to discuss because it corresponds to an extremely common student misconception. Aristotle conceived of forced motion as a relationship in which one object was the boss and the other “followed orders.” It therefore would only make sense for an object to experience one force at a time, because an object couldn’t follow orders from two sources at once. In the Newtonian theory, forces are numbers, not orders, and if more than one force acts on an object at once, the result is found by adding up all the forces. It is unfortunate that the use of the English word “force” has become standard, because to many people it suggests that you are “forcing” an object to do something. The force of the earth’s gravity cannot “force” a boat to sink, because there are other forces acting on the boat. Adding them up gives a total of zero, so the boat accelerates neither up nor down. Objects can exert forces on each other at a distance. Aristotle declared that forces could only act between objects that were touching, probably because he wished to avoid the type of occult speculation that attributed physical phenomena to the influence of a distant and invisible pantheon of gods. He was wrong, however, as you can observe when a magnet leaps onto your refrigerator or when the planet earth exerts gravitational forces on objects that are in the air. Some types of forces, such as friction, only operate between objects in contact, and are called contact forces. Magnetism, on the other hand, is an example of a noncontact force. Although the magnetic force gets stronger when the magnet is closer to your refrigerator, touching is not required.
126
Chapter 4
Force and motion
Weight In physics, an object’s weight, FW , is defined as the earth’s gravitational force on it. The SI unit of weight is therefore the Newton. People commonly refer to the kilogram as a unit of weight, but the kilogram is a unit of mass, not weight. Note that an object’s weight is not a fixed property of that object. Objects weigh more in some places than in others, depending on the local strength of gravity. It is their mass that always stays the same. A baseball pitcher who can throw a 90-mile-per-hour fastball on earth would not be able to throw any faster on the moon, because the ball’s inertia would still be the same. Positive and negative signs of force We’ll start by considering only cases of one-dimensional centerof-mass motion in which all the forces are parallel to the direction of motion, i.e., either directly forward or backward. In one dimension, plus and minus signs can be used to indicate directions of forces, as shown in figure c. We can then refer generically to addition of forces, rather than having to speak sometimes of addition and sometimes of subtraction. We add the forces shown in the figure and get 11 N. In general, we should choose a one-dimensional coordinate system with its x axis parallel the direction of motion. Forces that point along the positive x axis are positive, and forces in the opposite direction are negative. Forces that are not directly along the x axis cannot be immediately incorporated into this scheme, but that’s OK, because we’re avoiding those cases for now.
c / Forces are applied to a saxophone. In this example, positive signs have been used consistently for forces to the right, and negative signs for forces to the left. (The forces are being applied to different places on the saxophone, but the numerical value of a force carries no information about that.)
Discussion questions A In chapter 0, I defined 1 N as the force that would accelerate a 1-kg mass from rest to 1 m/s in 1 s. Anticipating the following section, you might guess that 2 N could be defined as the force that would accelerate the same mass to twice the speed, or twice the mass to the same speed. Is there an easier way to define 2 N based on the definition of 1 N?
4.2 Newton’s first law We are now prepared to make a more powerful restatement of the principle of inertia.1 Newton’s first law If the total force acting on an object is zero, its center of mass continues in the same state of motion.
In other words, an object initially at rest is predicted to remain at rest if the total force acting on it is zero, and an object in motion 1
Page 80 lists places in this book where we describe experimental tests of the principle of inertia and Newton’s first law.
Section 4.2
Newton’s first law
127
remains in motion with the same velocity in the same direction. The converse of Newton’s first law is also true: if we observe an object moving with constant velocity along a straight line, then the total force on it must be zero. In a future physics course or in another textbook, you may encounter the term “net force,” which is simply a synonym for total force. What happens if the total force on an object is not zero? It accelerates. Numerical prediction of the resulting acceleration is the topic of Newton’s second law, which we’ll discuss in the following section. This is the first of Newton’s three laws of motion. It is not important to memorize which of Newton’s three laws are numbers one, two, and three. If a future physics teacher asks you something like, “Which of Newton’s laws are you thinking of?,” a perfectly acceptable answer is “The one about constant velocity when there’s zero total force.” The concepts are more important than any specific formulation of them. Newton wrote in Latin, and I am not aware of any modern textbook that uses a verbatim translation of his statement of the laws of motion. Clear writing was not in vogue in Newton’s day, and he formulated his three laws in terms of a concept now called momentum, only later relating it to the concept of force. Nearly all modern texts, including this one, start with force and do momentum later. An elevator example 1 An elevator has a weight of 5000 N. Compare the forces that the cable must exert to raise it at constant velocity, lower it at constant velocity, and just keep it hanging. In all three cases the cable must pull up with a force of exactly 5000 N. Most people think you’d need at least a little more than 5000 N to make it go up, and a little less than 5000 N to let it down, but that’s incorrect. Extra force from the cable is only necessary for speeding the car up when it starts going up or slowing it down when it finishes going down. Decreased force is needed to speed the car up when it gets going down and to slow it down when it finishes going up. But when the elevator is cruising at constant velocity, Newton’s first law says that you just need to cancel the force of the earth’s gravity. To many students, the statement in the example that the cable’s upward force “cancels” the earth’s downward gravitational force implies that there has been a contest, and the cable’s force has won, vanquishing the earth’s gravitational force and making it disappear. That is incorrect. Both forces continue to exist, but because they add up numerically to zero, the elevator has no center-of-mass acceleration. We know that both forces continue to exist because they both have side-effects other than their effects on the car’s center-of-
128
Chapter 4
Force and motion
mass motion. The force acting between the cable and the car continues to produce tension in the cable and keep the cable taut. The earth’s gravitational force continues to keep the passengers (whom we are considering as part of the elevator-object) stuck to the floor and to produce internal stresses in the walls of the car, which must hold up the floor. Terminal velocity for falling objects example 2 An object like a feather that is not dense or streamlined does not fall with constant acceleration, because air resistance is nonnegligible. In fact, its acceleration tapers off to nearly zero within a fraction of a second, and the feather finishes dropping at constant speed (known as its terminal velocity). Why does this happen? Newton’s first law tells us that the total force on the feather must have been reduced to nearly zero after a short time. There are two forces acting on the feather: a downward gravitational force from the planet earth, and an upward frictional force from the air. As the feather speeds up, the air friction becomes stronger and stronger, and eventually it cancels out the earth’s gravitational force, so the feather just continues with constant velocity without speeding up any more. The situation for a skydiver is exactly analogous. It’s just that the skydiver experiences perhaps a million times more gravitational force than the feather, and it is not until she is falling very fast that the force of air friction becomes as strong as the gravitational force. It takes her several seconds to reach terminal velocity, which is on the order of a hundred miles per hour. More general combinations of forces It is too constraining to restrict our attention to cases where all the forces lie along the line of the center of mass’s motion. For one thing, we can’t analyze any case of horizontal motion, since any object on earth will be subject to a vertical gravitational force! For instance, when you are driving your car down a straight road, there are both horizontal forces and vertical forces. However, the vertical forces have no effect on the center of mass motion, because the road’s upward force simply counteracts the earth’s downward gravitational force and keeps the car from sinking into the ground. Later in the book we’ll deal with the most general case of many forces acting on an object at any angles, using the mathematical technique of vector addition, but the following slight generalization of Newton’s first law allows us to analyze a great many cases of interest: Suppose that an object has two sets of forces acting on it, one set along the line of the object’s initial motion and another set perpendicular to the first set. If both sets of forces cancel, then the object’s center of mass continues in the same state of motion.
Section 4.2
Newton’s first law
129
A passenger riding the subway example 3 Describe the forces acting on a person standing in a subway train that is cruising at constant velocity. No force is necessary to keep the person moving relative to the ground. He will not be swept to the back of the train if the floor is slippery. There are two vertical forces on him, the earth’s downward gravitational force and the floor’s upward force, which cancel. There are no horizontal forces on him at all, so of course the total horizontal force is zero. Forces on a sailboat example 4 If a sailboat is cruising at constant velocity with the wind coming from directly behind it, what must be true about the forces acting on it? The forces acting on the boat must be canceling each other out. The boat is not sinking or leaping into the air, so evidently the vertical forces are canceling out. The vertical forces are the downward gravitational force exerted by the planet earth and an upward force from the water. The air is making a forward force on the sail, and if the boat is not accelerating horizontally then the water’s backward frictional force must be canceling it out.
d / Example 4.
Contrary to Aristotle, more force is not needed in order to maintain a higher speed. Zero total force is always needed to maintain constant velocity. Consider the following made-up numbers:
forward force of the wind on the sail . . . backward force of the water on the hull . . . total force on the boat . . .
boat moving at a low, constant velocity 10,000 N
boat moving at a high, constant velocity 20,000 N
−10, 000 N
−20, 000 N
0N
0N
The faster boat still has zero total force on it. The forward force on it is greater, and the backward force smaller (more negative), but that’s irrelevant because Newton’s first law has to do with the total force, not the individual forces. This example is quite analogous to the one about terminal velocity of falling objects, since there is a frictional force that increases with speed. After casting off from the dock and raising the sail, the boat will accelerate briefly, and then reach its terminal velocity, at which the water’s frictional force has become as great as the wind’s force on the sail.
130
Chapter 4
Force and motion
A car crash example 5 If you drive your car into a brick wall, what is the mysterious force that slams your face into the steering wheel? Your surgeon has taken physics, so she is not going to believe your claim that a mysterious force is to blame. She knows that your face was just following Newton’s first law. Immediately after your car hit the wall, the only forces acting on your head were the same canceling-out forces that had existed previously: the earth’s downward gravitational force and the upward force from your neck. There were no forward or backward forces on your head, but the car did experience a backward force from the wall, so the car slowed down and your face caught up. Discussion questions A Newton said that objects continue moving if no forces are acting on them, but his predecessor Aristotle said that a force was necessary to keep an object moving. Why does Aristotle’s theory seem more plausible, even though we now believe it to be wrong? What insight was Aristotle missing about the reason why things seem to slow down naturally? Give an example. B In the figure what would have to be true about the saxophone’s initial motion if the forces shown were to result in continued one-dimensional motion of its center of mass? C This figure requires an ever further generalization of the preceding discussion. After studying the forces, what does your physical intuition tell you will happen? Can you state in words how to generalize the conditions for one-dimensional motion to include situations like this one?
Discussion question B.
4.3 Newton’s second law What about cases where the total force on an object is not zero, so that Newton’s first law doesn’t apply? The object will have an acceleration. The way we’ve defined positive and negative signs of force and acceleration guarantees that positive forces produce positive accelerations, and likewise for negative values. How much acceleration will it have? It will clearly depend on both the object’s mass and on the amount of force.
Discussion question C.
Experiments with any particular object show that its acceleration is directly proportional to the total force applied to it. This may seem wrong, since we know of many cases where small amounts of force fail to move an object at all, and larger forces get it going. This apparent failure of proportionality actually results from forgetting that there is a frictional force in addition to the force we apply to move the object. The object’s acceleration is exactly proportional to the total force on it, not to any individual force on it. In the absence of friction, even a very tiny force can slowly change the velocity of a very massive object.
Section 4.3
Newton’s second law
131
Experiments (e.g., the one described in example 10 on p. 135) also show that the acceleration is inversely proportional to the object’s mass, and combining these two proportionalities gives the following way of predicting the acceleration of any object: Newton’s second law
a = Ftotal /m
,
where m is an object’s mass Ftotal is the sum of the forces acting on it, and a is the acceleration of the object’s center of mass.
We are presently restricted to the case where the forces of interest are parallel to the direction of motion. An accelerating bus example 6 A VW bus with a mass of 2000 kg accelerates from 0 to 25 m/s (freeway speed) in 34 s. Assuming the acceleration is constant, what is the total force on the bus? We solve Newton’s second law for Ftotal = ma, and substitute Δv /Δt for a, giving Ftotal = mΔv /Δt = (2000 kg)(25 m/s − 0 m/s)/(34 s) = 1.5 kN
.
A generalization As with the first law, the second law can be easily generalized to include a much larger class of interesting situations: Suppose an object is being acted on by two sets of forces, one set lying parallel to the object’s initial direction of motion and another set acting along a perpendicular line. If the forces perpendicular to the initial direction of motion cancel out, then the object accelerates along its original line of motion according to a = F /m, where F is the sum of the forces parallel to the line. A coin sliding across a table example 7 Suppose a coin is sliding to the right across a table, e, and let’s choose a positive x axis that points to the right. The coin’s velocity
132
Chapter 4
Force and motion
is positive, and we expect based on experience that it will slow down, i.e., its acceleration should be negative. Although the coin’s motion is purely horizontal, it feels both vertical and horizontal forces. The Earth exerts a downward gravitational force F2 on it, and the table makes an upward force F3 that prevents the coin from sinking into the wood. In fact, without these vertical forces the horizontal frictional force wouldn’t exist: surfaces don’t exert friction against one another unless they are being pressed together. Although F2 and F3 contribute to the physics, they do so only indirectly. The only thing that directly relates to the acceleration along the horizontal direction is the horizontal force: a = F1 /m.
e / A coin slides across a table. Even for motion in one dimension, some of the forces may not lie along the line of the motion.
The relationship between mass and weight Mass is different from weight, but they’re related. An apple’s mass tells us how hard it is to change its motion. Its weight measures the strength of the gravitational attraction between the apple and the planet earth. The apple’s weight is less on the moon, but its mass is the same. Astronauts assembling the International Space Station in zero gravity cannot just pitch massive modules back and forth with their bare hands; the modules are weightless, but not massless. We have already seen the experimental evidence that when weight (the force of the earth’s gravity) is the only force acting on an object, its acceleration equals the constant g, and g depends on where you are on the surface of the earth, but not on the mass of the object. Applying Newton’s second law then allows us to calculate the magnitude of the gravitational force on any object in terms of its mass: |FW | = mg .
f / A simple double-pan balance works by comparing the weight forces exerted by the earth on the contents of the two pans. Since the two pans are at almost the same location on the earth’s surface, the value of g is essentially the same for each one, and equality of weight therefore also implies equality of mass.
(The equation only gives the magnitude, i.e. the absolute value, of FW , because we’re defining g as a positive number, so it equals the absolute value of a falling object’s acceleration.) Solved problem: Decelerating a car
page 143, problem 7
Weight and mass example 8 Figure g shows masses of one and two kilograms hung from a spring scale, which measures force in units of newtons. Explain the readings. Let’s start with the single kilogram. It’s not accelerating, so evidently the total force on it is zero: the spring scale’s upward force on it is canceling out the earth’s downward gravitational force. The spring scale tells us how much force it is being obliged to supply, but since the two forces are equal in strength, the spring scale’s reading can also be interpreted as measuring the strength of the gravitational force, i.e., the weight of the one-
Section 4.3
g / Example 8.
Newton’s second law
133
kilogram mass. The weight of a one-kilogram mass should be FW = mg = (1.0 kg)(9.8 m/s2 ) = 9.8 N
,
and that’s indeed the reading on the spring scale. Similarly for the two-kilogram mass, we have FW = mg = (2.0 kg)(9.8 m/s2 ) = 19.6 N
.
Calculating terminal velocity example 9 Experiments show that the force of air friction on a falling object such as a skydiver or a feather can be approximated fairly well with the equation |Fair | = cρAv 2 , where c is a constant, ρ is the density of the air, A is the cross-sectional area of the object as seen from below, and v is the object’s velocity. Predict the object’s terminal velocity, i.e., the final velocity it reaches after a long time. As the object accelerates, its greater v causes the upward force of the air to increase until finally the gravitational force and the force of air friction cancel out, after which the object continues at constant velocity. We choose a coordinate system in which positive is up, so that the gravitational force is negative and the force of air friction is positive. We want to find the velocity at which Fair + FW = 0
,
cρAv 2 − mg = 0
.
i.e.,
Solving for v gives vter minal =
mg cρA
self-check A It is important to get into the habit of interpreting equations. This may be difficult at first, but eventually you will get used to this kind of reasoning. (1) Interpret the equation vter minal = mg /c ρA in the case of ρ=0. (2) How would the terminal velocity of a 4-cm steel ball compare to that of a 1-cm ball? (3) In addition to teasing out the mathematical meaning of an equation, we also have to be able to place it in its physical context. How generally important is this equation? Answer, p. 541
134
Chapter 4
Force and motion
A test of the second law example 10 Because the force mg of gravity on an object of mass m is proportional to m, the acceleration predicted by Newton’s second law is a = F /m = mg/m = g, in which the mass cancels out. It is therefore an ironclad prediction of Newton’s laws of motion that free fall is universal: in the absence of other forces such as air resistance, heavier objects do not fall with a greater acceleration than lighter ones. The experiment by Galileo at the Leaning Tower of Pisa (p. 92) is therefore consistent with Newton’s second law. Since Galileo’s time, experimental methods have had several centuries in which to improve, and the second law has been subjected to similar tests with exponentially improving precision. For such an experiment in 1993,2 physicists at the University of Pisa (!) built a metal disk out of copper and tungsten semicircles joined together at their flat edges. They evacuated the air from a vertical shaft and dropped the disk down it 142 times, using lasers to measure any tiny rotation that would result if the accelerations of the copper and tungsten were very slightly different. The results were statistically consistent with zero rotation, and put an upper limit of 1 × 10−9 on the fractional difference in acceleration |gcopper − gtungsten |/g. Discussion questions A Show that the Newton can be reexpressed in terms of the three basic mks units as the combination kg·m/s2 . B
What is wrong with the following statements?
(1) “g is the force of gravity.” (2) “Mass is a measure of how much space something takes up.” C
Criticize the following incorrect statement:
“If an object is at rest and the total force on it is zero, it stays at rest. There can also be cases where an object is moving and keeps on moving without having any total force on it, but that can only happen when there’s no friction, like in outer space.” D Table h gives laser timing data for Ben Johnson’s 100 m dash at the 1987 World Championship in Rome. (His world record was later revoked because he tested positive for steroids.) How does the total force on him change over the duration of the race?
x (m) 10 20 30 40 50 60 70 80 90 100
t (s) 1.84 2.86 3.80 4.67 5.53 6.38 7.23 8.10 8.96 9.83
h / Discussion question D.
2
Carusotto et al., “Limits on the violation of g-universality with a Galileotype experiment,” Phys Lett A183 (1993) 355. Freely available online at researchgate.net.
Section 4.3
Newton’s second law
135
4.4 What force is not Violin teachers have to endure their beginning students’ screeching. A frown appears on the woodwind teacher’s face as she watches her student take a breath with an expansion of his ribcage but none in his belly. What makes physics teachers cringe is their students’ verbal statements about forces. Below I have listed six dicta about what force is not. 1. Force is not a property of one object. A great many of students’ incorrect descriptions of forces could be cured by keeping in mind that a force is an interaction of two objects, not a property of one object. Incorrect statement: “That magnet has a lot of force.” If the magnet is one millimeter away from a steel ball bearing, they may exert a very strong attraction on each other, but if they were a meter apart, the force would be virtually undetectable. The magnet’s strength can be rated using certain electrical units (ampere − meters2 ), but not in units of force.
2. Force is not a measure of an object’s motion. If force is not a property of a single object, then it cannot be used as a measure of the object’s motion. Incorrect statement: “The freight train rumbled down the tracks with awesome force.” Force is not a measure of motion. If the freight train collides with a stalled cement truck, then some awesome forces will occur, but if it hits a fly the force will be small.
3. Force is not energy. There are two main approaches to understanding the motion of objects, one based on force and one on a different concept, called energy. The SI unit of energy is the Joule, but you are probably more familiar with the calorie, used for measuring food’s energy, and the kilowatt-hour, the unit the electric company uses for billing you. Physics students’ previous familiarity with calories and kilowatthours is matched by their universal unfamiliarity with measuring forces in units of Newtons, but the precise operational definitions of the energy concepts are more complex than those of the force concepts, and textbooks, including this one, almost universally place the force description of physics before the energy description. During the long period after the introduction of force and before the careful definition of energy, students are therefore vulnerable to situations in which, without realizing it, they are imputing the properties of energy to phenomena of force. Incorrect statement: “How can my chair be making an upward force on my rear end? It has no power!” Power is a concept related to energy, e.g., a 100-watt lightbulb uses
136
Chapter 4
Force and motion
up 100 joules per second of energy. When you sit in a chair, no energy is used up, so forces can exist between you and the chair without any need for a source of power.
4. Force is not stored or used up. Because energy can be stored and used up, people think force also can be stored or used up. Incorrect statement: “If you don’t fill up your tank with gas, you’ll run out of force.” Energy is what you’ll run out of, not force.
5. Forces need not be exerted by living things or machines. Transforming energy from one form into another usually requires some kind of living or mechanical mechanism. The concept is not applicable to forces, which are an interaction between objects, not a thing to be transferred or transformed. Incorrect statement: “How can a wooden bench be making an upward force on my rear end? It doesn’t have any springs or anything inside it.” No springs or other internal mechanisms are required. If the bench didn’t make any force on you, you would obey Newton’s second law and fall through it. Evidently it does make a force on you!
6. A force is the direct cause of a change in motion. I can click a remote control to make my garage door change from being at rest to being in motion. My finger’s force on the button, however, was not the force that acted on the door. When we speak of a force on an object in physics, we are talking about a force that acts directly. Similarly, when you pull a reluctant dog along by its leash, the leash and the dog are making forces on each other, not your hand and the dog. The dog is not even touching your hand. self-check B Which of the following things can be correctly described in terms of force? (1) A nuclear submarine is charging ahead at full steam. (2) A nuclear submarine’s propellers spin in the water. (3) A nuclear submarine needs to refuel its reactor periodically. Answer, p. 541
Discussion questions A Criticize the following incorrect statement: “If you shove a book across a table, friction takes away more and more of its force, until finally it stops.” B You hit a tennis ball against a wall. Explain any and all incorrect ideas in the following description of the physics involved: “The ball gets some force from you when you hit it, and when it hits the wall, it loses part of that force, so it doesn’t bounce back as fast. The muscles in your arm are the only things that a force can come from.”
Section 4.4
What force is not
137
4.5 Inertial and noninertial frames of reference One day, you’re driving down the street in your pickup truck, on your way to deliver a bowling ball. The ball is in the back of the truck, enjoying its little jaunt and taking in the fresh air and sunshine. Then you have to slow down because a stop sign is coming up. As you brake, you glance in your rearview mirror, and see your trusty companion accelerating toward you. Did some mysterious force push it forward? No, it only seems that way because you and the car are slowing down. The ball is faithfully obeying Newton’s first law, and as it continues at constant velocity it gets ahead relative to the slowing truck. No forces are acting on it (other than the same canceling-out vertical forces that were always acting on it).3 The ball only appeared to violate Newton’s first law because there was something wrong with your frame of reference, which was based on the truck.
i / 1. In a frame of reference that moves with the truck, the bowling ball appears to violate Newton’s first law by accelerating despite having no horizontal forces on it. 2. In an inertial frame of reference, which the surface of the earth approximately is, the bowling ball obeys Newton’s first law. It moves equal distances in equal time intervals, i.e., maintains constant velocity. In this frame of reference, it is the truck that appears to have a change in velocity, which makes sense, since the road is making a horizontal force on it.
How, then, are we to tell in which frames of reference Newton’s laws are valid? It’s no good to say that we should avoid moving frames of reference, because there is no such thing as absolute rest or absolute motion. All frames can be considered as being either at rest or in motion. According to an observer in India, the strip mall that constituted the frame of reference in panel (b) of the figure was moving along with the earth’s rotation at hundreds of miles per hour. The reason why Newton’s laws fail in the truck’s frame of refer3
138
Chapter 4
Let’s assume for simplicity that there is no friction.
Force and motion
ence is not because the truck is moving but because it is accelerating. (Recall that physicists use the word to refer either to speeding up or slowing down.) Newton’s laws were working just fine in the moving truck’s frame of reference as long as the truck was moving at constant velocity. It was only when its speed changed that there was a problem. How, then, are we to tell which frames are accelerating and which are not? What if you claim that your truck is not accelerating, and the sidewalk, the asphalt, and the Burger King are accelerating? The way to settle such a dispute is to examine the motion of some object, such as the bowling ball, which we know has zero total force on it. Any frame of reference in which the ball appears to obey Newton’s first law is then a valid frame of reference, and to an observer in that frame, Mr. Newton assures us that all the other objects in the universe will obey his laws of motion, not just the ball. Valid frames of reference, in which Newton’s laws are obeyed, are called inertial frames of reference. Frames of reference that are not inertial are called noninertial frames. In those frames, objects violate the principle of inertia and Newton’s first law. While the truck was moving at constant velocity, both it and the sidewalk were valid inertial frames. The truck became an invalid frame of reference when it began changing its velocity. You usually assume the ground under your feet is a perfectly inertial frame of reference, and we made that assumption above. It isn’t perfectly inertial, however. Its motion through space is quite complicated, being composed of a part due to the earth’s daily rotation around its own axis, the monthly wobble of the planet caused by the moon’s gravity, and the rotation of the earth around the sun. Since the accelerations involved are numerically small, the earth is approximately a valid inertial frame. Noninertial frames are avoided whenever possible, and we will seldom, if ever, have occasion to use them in this course. Sometimes, however, a noninertial frame can be convenient. Naval gunners, for instance, get all their data from radars, human eyeballs, and other detection systems that are moving along with the earth’s surface. Since their guns have ranges of many miles, the small discrepancies between their shells’ actual accelerations and the accelerations predicted by Newton’s second law can have effects that accumulate and become significant. In order to kill the people they want to kill, they have to add small corrections onto the equation a = Ftotal /m. Doing their calculations in an inertial frame would allow them to use the usual form of Newton’s second law, but they would have to convert all their data into a different frame of reference, which would require cumbersome calculations. Discussion question A
If an object has a linear x − t graph in a certain inertial frame,
Section 4.5
Inertial and noninertial frames of reference
139
what is the effect on the graph if we change to a coordinate system with a different origin? What is the effect if we keep the same origin but reverse the positive direction of the x axis? How about an inertial frame moving alongside the object? What if we describe the object’s motion in a noninertial frame?
140
Chapter 4
Force and motion
Summary Selected vocabulary weight . . . . . . . the force of gravity on an object, equal to mg inertial frame . . a frame of reference that is not accelerating, one in which Newton’s first law is true noninertial frame an accelerating frame of reference, in which Newton’s first law is violated Notation FW . . . . . . . .
weight
Other terminology and notation net force . . . . . another way of saying “total force” Summary Newton’s first law of motion states that if all the forces acting on an object cancel each other out, then the object continues in the same state of motion. This is essentially a more refined version of Galileo’s principle of inertia, which did not refer to a numerical scale of force. Newton’s second law of motion allows the prediction of an object’s acceleration given its mass and the total force on it, acm = Ftotal /m. This is only the one-dimensional version of the law; the full-three dimensional treatment will come in chapter 8, Vectors. Without the vector techniques, we can still say that the situation remains unchanged by including an additional set of vectors that cancel among themselves, even if they are not in the direction of motion. Newton’s laws of motion are only true in frames of reference that are not accelerating, known as inertial frames.
Summary
141
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 An object is observed to be moving at constant speed in a certain direction. Can you conclude that no forces are acting on it? Explain. [Based on a problem by Serway and Faughn.] 2 At low speeds, every car’s acceleration is limited by traction, not by the engine’s power. Suppose that at low speeds, a certain car is normally capable of an acceleration of 3 m/s2 . If it is towing a trailer with half as much mass as the car itself, what acceleration can it achieve? [Based on a problem from PSSC Physics.] 3 (a) Let T be the maximum tension that an elevator’s cable can withstand without breaking, i.e., the maximum force it can exert. If the motor is programmed to give the car an acceleration a (a > 0 is upward), what is the maximum mass that the car can √ have, including passengers, if the cable is not to break? (b) Interpret the equation you derived in the special cases of a = 0 and of a downward acceleration of magnitude g. (“Interpret” means to analyze the behavior of the equation, and connect that to reality, as in the self-check on page 134.)
4 A helicopter of mass m is taking off vertically. The only forces acting on it are the earth’s gravitational force and the force, Fair , of the air pushing up on the propeller blades. (a) If the helicopter lifts off at t = 0, what is its vertical speed at time t? (b) Check that the units of your answer to part a make sense. (c) Discuss how your answer to part a depends on all three variables, and show that it makes sense. That is, for each variable, discuss what would happen to the result if you changed it while keeping the other two variables constant. Would a bigger value give a smaller result, or a bigger result? Once you’ve figured out this mathematical relationship, show that it makes sense physically. (d) Plug numbers into your equation from part a, using m = 2300 √ kg, Fair = 27000 N, and t = 4.0 s. 5 In the 1964 Olympics in Tokyo, the best men’s high jump was 2.18 m. Four years later in Mexico City, the gold medal in the same event was for a jump of 2.24 m. Because of Mexico City’s altitude (2400 m), the acceleration of gravity there is lower than that in Tokyo by about 0.01 m/s2 . Suppose a high-jumper has a mass of 72 kg. (a) Compare his mass and weight in the two locations. (b) Assume that he is able to jump with the same initial vertical
142
Chapter 4
Force and motion
velocity in both locations, and that all other conditions are the same except for gravity. How much higher should he be able to jump in √ Mexico City? (Actually, the reason for the big change between ’64 and ’68 was the introduction of the “Fosbury flop.”) 6 A blimp is initially at rest, hovering, when at t = 0 the pilot turns on the motor of the propeller. The motor cannot instantly get the propeller going, but the propeller speeds up steadily. The steadily increasing force between the air and the propeller is given by the equation F = kt, where k is a constant. If the mass of the blimp is m, find its position as a function of time. (Assume that during the period of time you’re dealing with, the blimp is not yet moving fast enough to cause a significant backward force due to air √ resistance.)
Problem 6.
7 A car is accelerating forward along a straight road. If the force of the road on the car’s wheels, pushing it forward, is a constant 3.0 kN, and the car’s mass is 1000 kg, then how long will the car take to go from 20 m/s to 50 m/s? Solution, p. 530 8 Some garden shears are like a pair of scissors: one sharp blade slices past another. In the “anvil” type, however, a sharp blade presses against a flat one rather than going past it. A gardening book says that for people who are not very physically strong, the anvil type can make it easier to cut tough branches, because it concentrates the force on one side. Evaluate this claim based on Newton’s laws. [Hint: Consider the forces acting on the branch, and the motion of the branch.] 9 A uranium atom deep in the earth spits out an alpha particle. An alpha particle is a fragment of an atom. This alpha particle has initial speed v, and travels a distance d before stopping in the earth. (a) Find the force, F , from the dirt that stopped the particle, in terms of v, d, and its mass, m. Don’t plug in any numbers yet. √ Assume that the force was constant. (b) Show that your answer has the right units. (c) Discuss how your answer to part a depends on all three variables, and show that it makes sense. That is, for each variable, discuss what would happen to the result if you changed it while keeping the other two variables constant. Would a bigger value give a smaller result, or a bigger result? Once you’ve figured out this mathematical relationship, show that it makes sense physically. (d) Evaluate your result for m = 6.7 × 10−27 kg, v = 2.0 × 104 km/s, √ and d = 0.71 mm.
Problems
143
Problem 10, part c.
10 You are given a large sealed box, and are not allowed to open it. Which of the following experiments measure its mass, and which measure its weight? [Hint: Which experiments would give different results on the moon?] (a) Put it on a frozen lake, throw a rock at it, and see how fast it scoots away after being hit. (b) Drop it from a third-floor balcony, and measure how loud the sound is when it hits the ground. (c) As shown in the figure, connect it with a spring to the wall, and watch it vibrate. Solution, p. 530 11 While escaping from the palace of the evil Martian emperor, Sally Spacehound jumps from a tower of height h down to the ground. Ordinarily the fall would be fatal, but she fires her blaster rifle straight down, producing an upward force of magnitude FB . This force is insufficient to levitate her, but it does cancel out some of the force of gravity. During the time t that she is falling, Sally is unfortunately exposed to fire from the emperor’s minions, and can’t dodge their shots. Let m be her mass, and g the strength of gravity on Mars. (a) Find the time t in terms of the other variables. (b) Check the units of your answer to part a. (c) For sufficiently large values of FB , your answer to part a becomes √ nonsense — explain what’s going on. 12 When I cook rice, some of the dry grains always stick to the measuring cup. To get them out, I turn the measuring cup upsidedown and hit the “roof” with my hand so that the grains come off of the “ceiling.” (a) Explain why static friction is irrelevant here. (b) Explain why gravity is negligible. (c) Explain why hitting the cup works, and why its success depends on hitting the cup hard enough.
144
Chapter 4
Force and motion
Exercise 4: Force and motion Equipment: 1-meter pieces of butcher paper wood blocks with hooks string masses to put on top of the blocks to increase friction spring scales (preferably calibrated in Newtons) Suppose a person pushes a crate, sliding it across the floor at a certain speed, and then repeats the same thing but at a higher speed. This is essentially the situation you will act out in this exercise. What do you think is different about her force on the crate in the two situations? Discuss this with your group and write down your hypothesis:
1. First you will measure the amount of friction between the wood block and the butcher paper when the wood and paper surfaces are slipping over each other. The idea is to attach a spring scale to the block and then slide the butcher paper under the block while using the scale to keep the block from moving with it. Depending on the amount of force your spring scale was designed to measure, you may need to put an extra mass on top of the block in order to increase the amount of friction. It is a good idea to use long piece of string to attach the block to the spring scale, since otherwise one tends to pull at an angle instead of directly horizontally. First measure the amount of friction force when sliding the butcher paper as slowly as possible: Now measure the amount of friction force at a significantly higher speed, say 1 meter per second. (If you try to go too fast, the motion is jerky, and it is impossible to get an accurate reading.) Discuss your results. Why are we justified in assuming that the string’s force on the block (i.e., the scale reading) is the same amount as the paper’s frictional force on the block? 2. Now try the same thing but with the block moving and the paper standing still. Try two different speeds. Do your results agree with your original hypothesis? If not, discuss what’s going on. How does the block “know” how fast to go?
Exercise 4: Force and motion
145
146
Chapter 4
Force and motion
What forces act on the girl?
Chapter 5
Analysis of forces 5.1 Newton’s third law Newton created the modern concept of force starting from his insight that all the effects that govern motion are interactions between two objects: unlike the Aristotelian theory, Newtonian physics has no phenomena in which an object changes its own motion.
147
Is one object always the “order-giver” and the other the “orderfollower”? As an example, consider a batter hitting a baseball. The bat definitely exerts a large force on the ball, because the ball accelerates drastically. But if you have ever hit a baseball, you also know that the ball makes a force on the bat — often with painful results if your technique is as bad as mine!
a / Two magnets exert forces on each other.
b / Two people’s hands forces on each other.
exert
How does the ball’s force on the bat compare with the bat’s force on the ball? The bat’s acceleration is not as spectacular as the ball’s, but maybe we shouldn’t expect it to be, since the bat’s mass is much greater. In fact, careful measurements of both objects’ masses and accelerations would show that mball aball is very nearly equal to −mbat abat , which suggests that the ball’s force on the bat is of the same magnitude as the bat’s force on the ball, but in the opposite direction. Figures a and b show two somewhat more practical laboratory experiments for investigating this issue accurately and without too much interference from extraneous forces. In experiment a, a large magnet and a small magnet are weighed separately, and then one magnet is hung from the pan of the top balance so that it is directly above the other magnet. There is an attraction between the two magnets, causing the reading on the top scale to increase and the reading on the bottom scale to decrease. The large magnet is more “powerful” in the sense that it can pick up a heavier paperclip from the same distance, so many people have a strong expectation that one scale’s reading will change by a far different amount than the other. Instead, we find that the two changes are equal in magnitude but opposite in direction: the force of the bottom magnet pulling down on the top one has the same strength as the force of the top one pulling up on the bottom one.
c / Rockets work by pushing exhaust gases out the back. Newton’s third law says that if the rocket exerts a backward force on the gases, the gases must make an equal forward force on the rocket. Rocket engines can function above the atmosphere, unlike propellers and jets, which work by pushing against the surrounding air.
148
Chapter 5
In experiment b, two people pull on two spring scales. Regardless of who tries to pull harder, the two forces as measured on the spring scales are equal. Interposing the two spring scales is necessary in order to measure the forces, but the outcome is not some artificial result of the scales’ interactions with each other. If one person slaps another hard on the hand, the slapper’s hand hurts just as much as the slappee’s, and it doesn’t matter if the recipient of the slap tries to be inactive. (Punching someone in the mouth causes just as much force on the fist as on the lips. It’s just that the lips are more delicate. The forces are equal, but not the levels of pain and injury.) Newton, after observing a series of results such as these, decided that there must be a fundamental law of nature at work:
Analysis of forces
Newton’s third law Forces occur in equal and opposite pairs: whenever object A exerts a force on object B, object B must also be exerting a force on object A. The two forces are equal in magnitude and opposite in direction.
Two modern, high-precision tests of the third law are described on p. 771. In one-dimensional situations, we can use plus and minus signs to indicate the directions of forces, and Newton’s third law can be written succinctly as FA on B = −FB on A .
d / A swimmer doing the breast stroke pushes backward against the water. By Newton’s third law, the water pushes forward on him.
self-check A Figure d analyzes swimming using Newton’s third law. Do a similar analysis for a sprinter leaving the starting line. Answer, p. 541
There is no cause and effect relationship between the two forces in Newton’s third law. There is no “original” force, and neither one is a response to the other. The pair of forces is a relationship, like marriage, not a back-and-forth process like a tennis match. Newton came up with the third law as a generalization about all the types of forces with which he was familiar, such as frictional and gravitational forces. When later physicists discovered a new type of force, such as the force that holds atomic nuclei together, they had to check whether it obeyed Newton’s third law. So far, no violation of the third law has ever been discovered, whereas the first and second laws were shown to have limitations by Einstein and the pioneers of atomic physics. The English vocabulary for describing forces is unfortunately rooted in Aristotelianism, and often implies incorrectly that forces are one-way relationships. It is unfortunate that a half-truth such as “the table exerts an upward force on the book” is so easily expressed, while a more complete and correct description ends up sounding awkward or strange: “the table and the book interact via a force,” or “the table and book participate in a force.” To students, it often sounds as though Newton’s third law implies nothing could ever change its motion, since the two equal and opposite forces would always cancel. The two forces, however, are always on two different objects, so it doesn’t make sense to add them in the first place — we only add forces that are acting on the same object. If two objects are interacting via a force and no other forces are involved, then both objects will accelerate — in opposite directions!
Section 5.1
e / Newton’s third law does not mean that forces always cancel out so that nothing can ever move. If these two figure skaters, initially at rest, push against each other, they will both move.
Newton’s third law
149
f / It doesn’t make sense for the man to talk about using the woman’s money to cancel out his bar tab, because there is no good reason to combine his debts and her assets. Similarly, it doesn’t make sense to refer to the equal and opposite forces of Newton’s third law as canceling. It only makes sense to add up forces that are acting on the same object, whereas two forces related to each other by Newton’s third law are always acting on two different objects.
A mnemonic for using Newton’s third law correctly Mnemonics are tricks for memorizing things. For instance, the musical notes that lie between the lines on the treble clef spell the word FACE, which is easy to remember. Many people use the mnemonic “SOHCAHTOA” to remember the definitions of the sine, cosine, and tangent in trigonometry. I have my own modest offering, POFOSTITO, which I hope will make it into the mnemonics hall of fame. It’s a way to avoid some of the most common problems with applying Newton’s third law correctly:
A book lying on a table example 1 A book is lying on a table. What force is the Newton’s-third-law partner of the earth’s gravitational force on the book? Answer: Newton’s third law works like “B on A, A on B,” so the partner must be the book’s gravitational force pulling upward on the planet earth. Yes, there is such a force! No, it does not cause the earth to do anything noticeable. Incorrect answer: The table’s upward force on the book is the Newton’s-third-law partner of the earth’s gravitational force on the book. This answer violates two out of three of the commandments of POFOSTITO. The forces are not of the same type, because the table’s upward force on the book is not gravitational. Also, three
150
Chapter 5
Analysis of forces
objects are involved instead of two: the book, the table, and the planet earth. Pushing a box up a hill example 2 A person is pushing a box up a hill. What force is related by Newton’s third law to the person’s force on the box? The box’s force on the person. Incorrect answer: The person’s force on the box is opposed by friction, and also by gravity. This answer fails all three parts of the POFOSTITO test, the most obvious of which is that three forces are referred to instead of a pair. If we could violate Newton’s third law. . . example 3 If we could violate Newton’s third law, we could do strange and wonderful things. Newton’s third laws says that the unequal magnets in figure a on p. 148 should exert equal forces on each other, and this is what we actually find when we do the experiment shown in that figure. But suppose instead that it worked as most people intuitively expect. What if the third law was violated, so that the big magnet made more force on the small one than the small one made on the big one? To make the analysis simple, we add some extra nonmagnetic material to the small magnet in figure g/1, so that it has the same mass and size as the big one. We also attach springs. When we release the magnets, g/2, the weak one is accelerated strongly, while the strong one barely moves. If we put them inside a box, g/3, the recoiling strong magnet bangs hard against the side of the box, and the box mysteriously accelerates itself. The process can be repeated indefinitely for free, so we have a magic box that propels itself without needing fuel. We can make it into a perpetual-motion car, g/4. If Newton’s third law was violated, we’d never have to pay for gas!
Optional Topic: Newton’s Third Law and Action at a Distance Newton’s third law is completely symmetric in the sense that neither force constitutes a delayed response to the other. Newton’s third law does not even mention time, and the forces are supposed to agree at any given instant. This creates an interesting situation when it comes to noncontact forces. Suppose two people are holding magnets, and when one person waves or wiggles her magnet, the other person feels an effect on his. In this way they can send signals to each other from opposite sides of a wall, and if Newton’s third law is correct, it would seem that the signals are transmitted instantly, with no time lag. The signals are indeed transmitted quite quickly, but experiments with electrically controlled magnets show that the signals do not leap the gap instantly: they travel at the same speed as light, which is an extremely high speed but not an infinite one. Is this a contradiction to Newton’s third law? Not really. According to current theories, there are no true noncontact forces. Action at a distance does not exist. Although it appears that the wiggling of one magnet affects the other with no need for anything to be in contact with anything, what really happens is that wiggling a magnet creates a ripple in the magnetic field pattern that exists even in empty space. The magnet shoves the ripplies out with a kick and receives a kick in return, in strict obedience to Newton’s third law. The ripples spread out in all directions, and the ones that hit the other magnet then interact with it, again obeying Newton’s third law.
g / Example 3. This doesn’t actually happen!
Section 5.1
Newton’s third law
151
Solved problem: More about example 2
page 174, problem 20
Solved problem: Why did it accelerate?
page 174, problem 18
Discussion questions A When you fire a gun, the exploding gases push outward in all directions, causing the bullet to accelerate down the barrel. What thirdlaw pairs are involved? [Hint: Remember that the gases themselves are an object.] B Tam Anh grabs Sarah by the hand and tries to pull her. She tries to remain standing without moving. A student analyzes the situation as follows. “If Tam Anh’s force on Sarah is greater than her force on him, he can get her to move. Otherwise, she’ll be able to stay where she is.” What’s wrong with this analysis? C You hit a tennis ball against a wall. Explain any and all incorrect ideas in the following description of the physics involved: “According to Newton’s third law, there has to be a force opposite to your force on the ball. The opposite force is the ball’s mass, which resists acceleration, and also air resistance.”
5.2 Classification and behavior of forces One of the most basic and important tasks of physics is to classify the forces of nature. I have already referred informally to “types” of forces such as friction, magnetism, gravitational forces, and so on. Classification systems are creations of the human mind, so there is always some degree of arbitrariness in them. For one thing, the level of detail that is appropriate for a classification system depends on what you’re trying to find out. Some linguists, the “lumpers,” like to emphasize the similarities among languages, and a few extremists have even tried to find signs of similarities between words in languages as different as English and Chinese, lumping the world’s languages into only a few large groups. Other linguists, the “splitters,” might be more interested in studying the differences in pronunciation between English speakers in New York and Connecticut. The splitters call the lumpers sloppy, but the lumpers say that science isn’t worthwhile unless it can find broad, simple patterns within the seemingly complex universe. Scientific classification systems are also usually compromises between practicality and naturalness. An example is the question of how to classify flowering plants. Most people think that biological classification is about discovering new species, naming them, and classifying them in the class-order-family-genus-species system according to guidelines set long ago. In reality, the whole system is in a constant state of flux and controversy. One very practical way of classifying flowering plants is according to whether their petals are separate or joined into a tube or cone — the criterion is so clear that it can be applied to a plant seen from across the street. But here practicality conflicts with naturalness. For instance, the begonia has
152
Chapter 5
Analysis of forces
separate petals and the pumpkin has joined petals, but they are so similar in so many other ways that they are usually placed within the same order. Some taxonomists have come up with classification criteria that they claim correspond more naturally to the apparent relationships among plants, without having to make special exceptions, but these may be far less practical, requiring for instance the examination of pollen grains under an electron microscope. In physics, there are two main systems of classification for forces. At this point in the course, you are going to learn one that is very practical and easy to use, and that splits the forces up into a relatively large number of types: seven very common ones that we’ll discuss explicitly in this chapter, plus perhaps ten less important ones such as surface tension, which we will not bother with right now. Physicists, however, are obsessed with finding simple patterns, so recognizing as many as fifteen or twenty types of forces strikes them as distasteful and overly complex. Since about the year 1900, physics has been on an aggressive program to discover ways in which these many seemingly different types of forces arise from a smaller number of fundamental ones. For instance, when you press your hands together, the force that keeps them from passing through each other may seem to have nothing to do with electricity, but at the atomic level, it actually does arise from electrical repulsion between atoms. By about 1950, all the forces of nature had been explained as arising from four fundamental types of forces at the atomic and nuclear level, and the lumping-together process didn’t stop there. By the 1960’s the length of the list had been reduced to three, and some theorists even believe that they may be able to reduce it to two or one. Although the unification of the forces of nature is one of the most beautiful and important achievements of physics, it makes much more sense to start this course with the more practical and easy system of classification. The unified system of four forces will be one of the highlights of the end of your introductory physics sequence.
h/A scientific system.
classification
The practical classification scheme which concerns us now can be laid out in the form of the tree shown in figure i. The most specific types of forces are shown at the tips of the branches, and it is these types of forces that are referred to in the POFOSTITO mnemonic. For example, electrical and magnetic forces belong to the same general group, but Newton’s third law would never relate an electrical force to a magnetic force. The broadest distinction is that between contact and noncontact forces, which has been discussed in ch. 4. Among the contact forces, we distinguish between those that involve solids only and those that have to do with fluids, a term used in physics to include both gases and liquids.
Section 5.2
Classification and behavior of forces
153
i / A practical classification scheme for forces.
It should not be necessary to memorize this diagram by rote. It is better to reinforce your memory of this system by calling to mind your commonsense knowledge of certain ordinary phenomena. For instance, we know that the gravitational attraction between us and the planet earth will act even if our feet momentarily leave the ground, and that although magnets have mass and are affected by gravity, most objects that have mass are nonmagnetic. Hitting a wall example 4 A bullet, flying horizontally, hits a steel wall. What type of force is there between the bullet and the wall? Starting at the bottom of the tree, we determine that the force is a contact force, because it only occurs once the bullet touches the wall. Both objects are solid. The wall forms a vertical plane. If the nose of the bullet was some shape like a sphere, you might imagine that it would only touch the wall at one point. Realistically, however, we know that a lead bullet will flatten out a lot on impact, so there is a surface of contact between the two, and its
154
Chapter 5
Analysis of forces
orientation is vertical. The effect of the force on the bullet is to stop the horizontal motion of the bullet, and this horizontal acceleration must be produced by a horizontal force. The force is therefore perpendicular to the surface of contact, and it’s also repulsive (tending to keep the bullet from entering the wall), so it must be a normal force. Diagram i is meant to be as simple as possible while including most of the forces we deal with in everyday life. If you were an insect, you would be much more interested in the force of surface tension, which allowed you to walk on water. I have not included the nuclear forces, which are responsible for holding the nuclei of atoms, because they are not evident in everyday life. You should not be afraid to invent your own names for types of forces that do not fit into the diagram. For instance, the force that holds a piece of tape to the wall has been left off of the tree, and if you were analyzing a situation involving scotch tape, you would be absolutely right to refer to it by some commonsense name such as “sticky force.” On the other hand, if you are having trouble classifying a certain force, you should also consider whether it is a force at all. For instance, if someone asks you to classify the force that the earth has because of its rotation, you would have great difficulty creating a place for it on the diagram. That’s because it’s a type of motion, not a type of force! Normal forces A normal force, FN , is a force that keeps one solid object from passing through another. “Normal” is simply a fancy word for “perpendicular,” meaning that the force is perpendicular to the surface of contact. Intuitively, it seems the normal force magically adjusts itself to provide whatever force is needed to keep the objects from occupying the same space. If your muscles press your hands together gently, there is a gentle normal force. Press harder, and the normal force gets stronger. How does the normal force know how strong to be? The answer is that the harder you jam your hands together, the more compressed your flesh becomes. Your flesh is acting like a spring: more force is required to compress it more. The same is true when you push on a wall. The wall flexes imperceptibly in proportion to your force on it. If you exerted enough force, would it be possible for two objects to pass through each other? No, typically the result is simply to strain the objects so much that one of them breaks. Gravitational forces As we’ll discuss in more detail later in the course, a gravitational force exists between any two things that have mass. In everyday life,
Section 5.2
Classification and behavior of forces
155
the gravitational force between two cars or two people is negligible, so the only noticeable gravitational forces are the ones between the earth and various human-scale objects. We refer to these planetearth-induced gravitational forces as weight forces, and as we have already seen, their magnitude is given by |FW | = mg. Solved problem: Weight and mass
page 175, problem 26
Static and kinetic friction
j / A model that correctly explains many properties of friction. The microscopic bumps and holes in two surfaces dig into each other.
k / Static friction: the tray doesn’t slip on the waiter’s fingers.
If you have pushed a refrigerator across a kitchen floor, you have felt a certain series of sensations. At first, you gradually increased your force on the refrigerator, but it didn’t move. Finally, you supplied enough force to unstick the fridge, and there was a sudden jerk as the fridge started moving. Once the fridge was unstuck, you could reduce your force significantly and still keep it moving. While you were gradually increasing your force, the floor’s frictional force on the fridge increased in response. The two forces on the fridge canceled, and the fridge didn’t accelerate. How did the floor know how to respond with just the right amount of force? Figure j shows one possible model of friction that explains this behavior. (A scientific model is a description that we expect to be incomplete, approximate, or unrealistic in some ways, but that nevertheless succeeds in explaining a variety of phenomena.) Figure j/1 shows a microscopic view of the tiny bumps and holes in the surfaces of the floor and the refrigerator. The weight of the fridge presses the two surfaces together, and some of the bumps in one surface will settle as deeply as possible into some of the holes in the other surface. In j/2, your leftward force on the fridge has caused it to ride up a little higher on the bump in the floor labeled with a small arrow. Still more force is needed to get the fridge over the bump and allow it to start moving. Of course, this is occurring simultaneously at millions of places on the two surfaces. Once you had gotten the fridge moving at constant speed, you found that you needed to exert less force on it. Since zero total force is needed to make an object move with constant velocity, the floor’s rightward frictional force on the fridge has apparently decreased somewhat, making it easier for you to cancel it out. Our model also gives a plausible explanation for this fact: as the surfaces slide past each other, they don’t have time to settle down and mesh with one another, so there is less friction.
l / Kinetic friction: the car skids.
156
Chapter 5
Even though this model is intuitively appealing and fairly successful, it should not be taken too seriously, and in some situations it is misleading. For instance, fancy racing bikes these days are made with smooth tires that have no tread — contrary to what we’d expect from our model, this does not cause any decrease in friction. Machinists know that two very smooth and clean metal
Analysis of forces
surfaces may stick to each other firmly and be very difficult to slide apart. This cannot be explained in our model, but makes more sense in terms of a model in which friction is described as arising from chemical bonds between the atoms of the two surfaces at their points of contact: very flat surfaces allow more atoms to come in contact. Since friction changes its behavior dramatically once the surfaces come unstuck, we define two separate types of frictional forces. Static friction is friction that occurs between surfaces that are not slipping over each other. Slipping surfaces experience kinetic friction. The forces of static and kinetic friction, notated Fs and Fk , are always parallel to the surface of contact between the two objects. self-check B 1. When a baseball player slides in to a base, is the friction static, or kinetic? 2. A mattress stays on the roof of a slowly accelerating car. Is the friction static, or kinetic? 3. Does static friction create heat? Kinetic friction?
Answer, p. 541
The maximum possible force of static friction depends on what kinds of surfaces they are, and also on how hard they are being pressed together. The approximate mathematical relationships can be expressed as follows: Fs,max = μs FN
,
where μs is a unitless number, called the coefficient of static friction, which depends on what kinds of surfaces they are. The maximum force that static friction can supply, μs FN , represents the boundary between static and kinetic friction. It depends on the normal force, which is numerically equal to whatever force is pressing the two surfaces together. In terms of our model, if the two surfaces are being pressed together more firmly, a greater sideways force will be required in order to make the irregularities in the surfaces ride up and over each other. Note that just because we use an adjective such as “applied” to refer to a force, that doesn’t mean that there is some special type of force called the “applied force.” The applied force could be any type of force, or it could be the sum of more than one force trying to make an object move. self-check C The arrows in figure m show the forces of the tree trunk on the partridge. Describe the forces the bird makes on the tree. Answer, p. 541
m / Many landfowl, even those that are competent fliers, prefer to escape from a predator by running upward rather than by flying. This partridge is running up a vertical tree trunk. Humans can’t walk up walls because there is no normal force and therefore no frictional force; when FN = 0, the maximum force of static friction Fs,max = μs FN is also zero. The partridge, however, has wings that it can flap in order to create a force between it and the air. Typically when a bird flaps its wings, the resulting force from the air is in the direction that would tend to lift the bird up. In this situation, however, the partridge changes its style of flapping so that the direction is reversed. The normal force between the feet and the tree allows a nonzero static frictional force. The mechanism is similar to that of a spoiler fin on a racing car. Some evolutionary biologists believe that when vertebrate flight first evolved, in dinosaurs, there was first a stage in which the wings were used only as an aid in running up steep inclines, and only later a transition to flight. (Redrawn from a figure by K.P. Dial.)
The force of kinetic friction on each of the two objects is in the direction that resists the slippage of the surfaces. Its magnitude is
Section 5.2
Classification and behavior of forces
157
usually well approximated as F k = μ k FN where μk is the coefficient of kinetic friction. Kinetic friction is usually more or less independent of velocity.
n / We choose a coordinate system in which the applied force, i.e., the force trying to move the objects, is positive. The friction force is then negative, since it is in the opposite direction. As you increase the applied force, the force of static friction increases to match it and cancel it out, until the maximum force of static friction is surpassed. The surfaces then begin slipping past each other, and the friction force becomes smaller in absolute value. self-check D Can a frictionless surface exert a normal force? Can a frictional force exist without a normal force? Answer, p. 542
If you try to accelerate or decelerate your car too quickly, the forces between your wheels and the road become too great, and they begin slipping. This is not good, because kinetic friction is weaker than static friction, resulting in less control. Also, if this occurs while you are turning, the car’s handling changes abruptly because the kinetic friction force is in a different direction than the static friction force had been: contrary to the car’s direction of motion, rather than contrary to the forces applied to the tire. Most people respond with disbelief when told of the experimental evidence that both static and kinetic friction are approximately independent of the amount of surface area in contact. Even after doing a hands-on exercise with spring scales to show that it is true, many students are unwilling to believe their own observations, and insist that bigger tires “give more traction.” In fact, the main reason why you would not want to put small tires on a big heavy car is that the tires would burst! Although many people expect that friction would be proportional to surface area, such a proportionality would make predictions contrary to many everyday observations. A dog’s feet, for example, have very little surface area in contact with the ground compared to a human’s feet, and yet we know that a dog can often win a tug-of-war with a person. The reason a smaller surface area does not lead to less friction
158
Chapter 5
Analysis of forces
is that the force between the two surfaces is more concentrated, causing their bumps and holes to dig into each other more deeply. self-check E Find the direction of each of the forces in figure o.
Answer, p. 542
o / 1. The cliff’s normal force on the climber’s feet. 2. The track’s static frictional force on the wheel of the accelerating dragster. 3. The ball’s normal force on the bat.
Locomotives example 5 Looking at a picture of a locomotive, p, we notice two obvious things that are different from an automobile. Where a car typically has two drive wheels, a locomotive normally has many — ten in this example. (Some also have smaller, unpowered wheels in front of and behind the drive wheels, but this example doesn’t.) Also, cars these days are generally built to be as light as possible for their size, whereas locomotives are very massive, and no effort seems to be made to keep their weight low. (The steam locomotive in the photo is from about 1900, but this is true even for modern diesel and electric trains.) p / Example 5.
The reason locomotives are built to be so heavy is for traction. The upward normal force of the rails on the wheels, FN , cancels the downward force of gravity, FW , so ignoring plus and minus signs, these two forces are equal in absolute value, FN = FW . Given this amount of normal force, the maximum force of static friction is Fs = μs FN = μs FW . This static frictional force, of the rails pushing forward on the wheels, is the only force that can accelerate the train, pull it uphill, or cancel out the force of air resistance while cruising at constant speed. The coefficient of static friction for steel on steel is about 1/4, so no locomotive can pull with a force greater than about 1/4 of its own weight. If the engine is capable of supplying more than that amount of force, the result will be simply to break static friction and spin the wheels.
Section 5.2
Classification and behavior of forces
159
The reason this is all so different from the situation with a car is that a car isn’t pulling something else. If you put extra weight in a car, you improve the traction, but you also increase the inertia of the car, and make it just as hard to accelerate. In a train, the inertia is almost all in the cars being pulled, not in the locomotive. The other fact we have to explain is the large number of driving wheels. First, we have to realize that increasing the number of driving wheels neither increases nor decreases the total amount of static friction, because static friction is independent of the amount of surface area in contact. (The reason four-wheeldrive is good in a car is that if one or more of the wheels is slipping on ice or in mud, the other wheels may still have traction. This isn’t typically an issue for a train, since all the wheels experience the same conditions.) The advantage of having more driving wheels on a train is that it allows us to increase the weight of the locomotive without crushing the rails, or damaging bridges. Fluid friction Try to drive a nail into a waterfall and you will be confronted with the main difference between solid friction and fluid friction. Fluid friction is purely kinetic; there is no static fluid friction. The nail in the waterfall may tend to get dragged along by the water flowing past it, but it does not stick in the water. The same is true for gases such as air: recall that we are using the word “fluid” to include both gases and liquids.
q / Fluid friction depends on the fluid’s pattern of flow, so it is more complicated than friction between solids, and there are no simple, universally applicable formulas to calculate it. From top to bottom: supersonic wind tunnel, vortex created by a crop duster, series of vortices created by a single object, turbulence.
160
Chapter 5
Unlike kinetic friction between solids, fluid friction increases rapidly with velocity. It also depends on the shape of the object, which is why a fighter jet is more streamlined than a Model T. For objects of the same shape but different sizes, fluid friction typically scales up with the cross-sectional area of the object, which is one of the main reasons that an SUV gets worse mileage on the freeway than a compact car.
Analysis of forces
Discussion questions A A student states that when he tries to push his refrigerator, the reason it won’t move is because Newton’s third law says there’s an equal and opposite frictional force pushing back. After all, the static friction force is equal and opposite to the applied force. How would you convince him he is wrong? B Kinetic friction is usually more or less independent of velocity. However, inexperienced drivers tend to produce a jerk at the last moment of deceleration when they stop at a stop light. What does this tell you about the kinetic friction between the brake shoes and the brake drums? C Some of the following are correct descriptions of types of forces that could be added on as new branches of the classification tree. Others are not really types of forces, and still others are not force phenomena at all. In each case, decide what’s going on, and if appropriate, figure out how you would incorporate them into the tree. sticky force opposite force flowing force surface tension horizontal force motor force canceled force
makes tape stick to things the force that Newton’s third law says relates to every force you make the force that water carries with it as it flows out of a hose lets insects walk on water a force that is horizontal the force that a motor makes on the thing it is turning a force that is being canceled out by some other force
r / What do the golf ball and the shark have in common? Both use the same trick to reduce fluid friction. The dimples on the golf ball modify the pattern of flow of the air around it, counterintuitively reducing friction. Recent studies have shown that sharks can accomplish the same thing by raising, or “bristling,” the scales on their skin at high speeds.
5.3 Analysis of forces Newton’s first and second laws deal with the total of all the forces exerted on a specific object, so it is very important to be able to figure out what forces there are. Once you have focused your attention on one object and listed the forces on it, it is also helpful to describe all the corresponding forces that must exist according to Newton’s third law. We refer to this as “analyzing the forces” in which the object participates.
Section 5.3
s / The wheelbases of the Hummer H3 and the Toyota Prius are surprisingly similar, differing by only 10%. The main difference in shape is that the Hummer is much taller and wider. It presents a much greater cross-sectional area to the wind, and this is the main reason that it uses about 2.5 times more gas on the freeway.
Analysis of forces
161
A barge example 6 A barge is being pulled to the right along a canal by teams of horses on the shores. Analyze all the forces in which the barge participates. force acting on barge force related to it by Newton’s third law ropes’ normal forces on barge, → barge’s normal force on ropes, ← water’s fluid friction force on barge, ← barge’s fluid friction force on water, → planet earth’s gravitational force on barge, ↓ barge’s gravitational force on earth, ↑ water’s “floating” force on barge, ↑ barge’s “floating” force on water, ↓ Here I’ve used the word “floating” force as an example of a sensible invented term for a type of force not classified on the tree on p. 154. A more formal technical term would be “hydrostatic force.” Note how the pairs of forces are all structured as “A’s force on B, B’s force on A”: ropes on barge and barge on ropes; water on barge and barge on water. Because all the forces in the left column are forces acting on the barge, all the forces in the right column are forces being exerted by the barge, which is why each entry in the column begins with “barge.”
Often you may be unsure whether you have forgotten one of the forces. Here are three strategies for checking your list:
1. See what physical result would come from the forces you’ve found so far. Suppose, for instance, that you’d forgotten the “floating” force on the barge in the example above. Looking at the forces you’d found, you would have found that there was a downward gravitational force on the barge which was not canceled by any upward force. The barge isn’t supposed to sink, so you know you need to find a fourth, upward force.
2. Another technique for finding missing forces is simply to go through the list of all the common types of forces and see if any of them apply.
3. Make a drawing of the object, and draw a dashed boundary line around it that separates it from its environment. Look for points on the boundary where other objects come in contact with your object. This strategy guarantees that you’ll find every contact force that acts on the object, although it won’t help you to find non-contact forces.
Fifi example 7 Fifi is an industrial espionage dog who loves doing her job and looks great doing it. She leaps through a window and lands at initial horizontal speed vo on a conveyor belt which is itself moving at the greater speed vb . Unfortunately the coefficient of kinetic friction μk between her foot-pads and the belt is fairly low, so she skids for a time Δt, during which the effect on her coiffure is un ´ desastre. Find Δt.
t / Example 7.
162
Chapter 5
Analysis of forces
We analyze the forces: force acting on Fifi planet earth’s gravitational ↓ force FW = mg on Fifi, belt’s kinetic frictional force Fk on Fifi, → belt’s normal force FN on Fifi, ↑
force related to it by Newton’s third law Fifi’s gravitational force on earth, ↑ Fifi’s kinetic frictional force on belt, ← Fifi’s normal force on belt, ↓
Checking the analysis of the forces as described on p. 162: (1) The physical result makes sense. The left-hand column consists of forces ↓→↑. We’re describing the time when she’s moving horizontally on the belt, so it makes sense that we have two vertical forces that could cancel. The rightward force is what will accelerate her until her speed matches that of the belt. (2) We’ve included every relevant type of force from the tree on p. 154. (3) We’ve included forces from the belt, which is the only object in contact with Fifi. The purpose of the analysis is to let us set up equations containing enough information to solve the problem. Let positive x be to the right. Newton’s second law gives (→)
a = Fk /m
Although it’s the horizontal motion we care about, the only way to find Fk is via the relation Fk = μk FN , and the only way to find FN is from the ↑↓ forces. If the vertical forces are to cancel, they must be of equal strength: (↑↓)
FN = mg
Using the constant-acceleration equation a = Δv /Δt, we have Δv a v − vo = b μk mg/m vb − v o = μk g
Δt =
.
The units check out: s=
m/s m/s2
,
where μk is omitted as a factor because it’s unitless. We should also check that the dependence on the variables also makes sense. If Fifi puts on her rubber ninja booties, increasing μk , then dividing by a larger number gives a smaller result
Section 5.3
Analysis of forces
163
for Δt; this makes sense physically, because the greater friction will cause her to come up to the belt’s speed more quickly. The dependence on g is similar; more gravity would press her harder against the belt, improving her traction. Increasing vb increases Δt, which makes sense because it will take her longer to get up to a bigger speed. Since vo is subtracted, the dependence of Δt on it is the other way around, and that makes sense too, because if she can land with a greater speed, she has less speeding up left to do. Discussion questions A In the example of the barge going down the canal, I referred to a “floating” or “hydrostatic” force that keeps the boat from sinking. If you were adding a new branch on the force-classification tree to represent this force, where would it go? B The earth’s gravitational force on you, i.e., your weight, is always equal to mg , where m is your mass. So why can you get a shovel to go deeper into the ground by jumping onto it? Just because you’re jumping, that doesn’t mean your mass or weight is any greater, does it?
5.4 Transmission of forces by low-mass objects You’re walking your dog. The dog wants to go faster than you do, and the leash is taut. Does Newton’s third law guarantee that your force on your end of the leash is equal and opposite to the dog’s force on its end? If they’re not exactly equal, is there any reason why they should be approximately equal? If there was no leash between you, and you were in direct contact with the dog, then Newton’s third law would apply, but Newton’s third law cannot relate your force on the leash to the dog’s force on the leash, because that would involve three separate objects. Newton’s third law only says that your force on the leash is equal and opposite to the leash’s force on you, FyL = −FLy , and that the dog’s force on the leash is equal and opposite to its force on the dog FdL = −FLd . Still, we have a strong intuitive expectation that whatever force we make on our end of the leash is transmitted to the dog, and viceversa. We can analyze the situation by concentrating on the forces that act on the leash, FdL and FyL . According to Newton’s second law, these relate to the leash’s mass and acceleration: FdL + FyL = mL aL .
164
Chapter 5
Analysis of forces
The leash is far less massive then any of the other objects involved, and if mL is very small, then apparently the total force on the leash is also very small, FdL + FyL ≈ 0, and therefore FdL ≈ −FyL
.
Thus even though Newton’s third law does not apply directly to these two forces, we can approximate the low-mass leash as if it was not intervening between you and the dog. It’s at least approximately as if you and the dog were acting directly on each other, in which case Newton’s third law would have applied. In general, low-mass objects can be treated approximately as if they simply transmitted forces from one object to another. This can be true for strings, ropes, and cords, and also for rigid objects such as rods and sticks.
u / If we imagine dividing a taut rope up into small segments, then any segment has forces pulling outward on it at each end. If the rope is of negligible mass, then all the forces equal +T or −T , where T , the tension, is a single number.
If you look at a piece of string under a magnifying glass as you pull on the ends more and more strongly, you will see the fibers straightening and becoming taut. Different parts of the string are apparently exerting forces on each other. For instance, if we think of the two halves of the string as two objects, then each half is exerting a force on the other half. If we imagine the string as consisting of many small parts, then each segment is transmitting a force to the next segment, and if the string has very little mass, then all the forces are equal in magnitude. We refer to the magnitude of the forces as the tension in the string, T . Although the tension is measured in units of Newtons, it is not itself a force. There are many forces within the string, some in one direction and some in the other direction, and their magnitudes are only approximately equal. The concept of tension only makes sense as a general, approximate statement of how big all the forces are. If a rope goes over a pulley or around some other object, then the tension throughout the rope is approximately equal so long as the pulley has negligible mass and there is not too much friction. A rod or stick can be treated in much the same way as a string, but it is possible to have either compression or tension.
v / The Golden Gate Bridge’s roadway is held up by the tension in the vertical cables.
Since tension is not a type of force, the force exerted by a rope on some other object must be of some definite type such as static
Section 5.4
Transmission of forces by low-mass objects
165
friction, kinetic friction, or a normal force. If you hold your dog’s leash with your hand through the loop, then the force exerted by the leash on your hand is a normal force: it is the force that keeps the leash from occupying the same space as your hand. If you grasp a plain end of a rope, then the force between the rope and your hand is a frictional force. A more complex example of transmission of forces is the way a car accelerates. Many people would describe the car’s engine as making the force that accelerates the car, but the engine is part of the car, so that’s impossible: objects can’t make forces on themselves. What really happens is that the engine’s force is transmitted through the transmission to the axles, then through the tires to the road. By Newton’s third law, there will thus be a forward force from the road on the tires, which accelerates the car. Discussion question A When you step on the gas pedal, is your foot’s force being transmitted in the sense of the word used in this section?
5.5 Objects under strain A string lengthens slightly when you stretch it. Similarly, we have already discussed how an apparently rigid object such as a wall is actually flexing when it participates in a normal force. In other cases, the effect is more obvious. A spring or a rubber band visibly elongates when stretched. Common to all these examples is a change in shape of some kind: lengthening, bending, compressing, etc. The change in shape can be measured by picking some part of the object and measuring its position, x. For concreteness, let’s imagine a spring with one end attached to a wall. When no force is exerted, the unfixed end of the spring is at some position xo . If a force acts at the unfixed end, its position will change to some new value of x. The more force, the greater the departure of x from xo .
Back in Newton’s time, experiments like this were considered cutting-edge research, and his contemporary Hooke is remembered today for doing them and for coming up with a simple mathematical generalization called Hooke’s law: F ≈ k(x − xo )
.
[force required to stretch a spring; valid for small forces only]
Here k is a constant, called the spring constant, that depends on how stiff the object is. If too much force is applied, the spring exhibits more complicated behavior, so the equation is only a good approximation if the force is sufficiently small. Usually when the
166
Chapter 5
Analysis of forces
w / Defining the quantities F , x , and xo in Hooke’s law.
force is so large that Hooke’s law is a bad approximation, the force ends up permanently bending or breaking the spring. Although Hooke’s law may seem like a piece of trivia about springs, it is actually far more important than that, because all solid objects exert Hooke’s-law behavior over some range of sufficiently small forces. For example, if you push down on the hood of a car, it dips by an amount that is directly proportional to the force. (But the car’s behavior would not be as mathematically simple if you dropped a boulder on the hood!) Solved problem: Combining springs
page 173, problem 14
Solved problem: Young’s modulus
page 173, problem 16
Discussion question A A car is connected to its axles through big, stiff springs called shock absorbers, or “shocks.” Although we’ve discussed Hooke’s law above only in the case of stretching a spring, a car’s shocks are continually going through both stretching and compression. In this situation, how would you interpret the positive and negative signs in Hooke’s law?
5.6 Simple Machines: the pulley Even the most complex machines, such as cars or pianos, are built out of certain basic units called simple machines. The following are some of the main functions of simple machines: transmitting a force: The chain on a bicycle transmits a force from the crank set to the rear wheel. changing the direction of a force: If you push down on a seesaw, the other end goes up. changing the speed and precision of motion: When you make the “come here” motion, your biceps only moves a couple of centimeters where it attaches to your forearm, but your arm
Section 5.6
Simple Machines: the pulley
167
moves much farther and more rapidly. changing the amount of force: A lever or pulley can be used to increase or decrease the amount of force. You are now prepared to understand one-dimensional simple machines, of which the pulley is the main example.
x / Example 8.
A pulley example 8 Farmer Bill says this pulley arrangement doubles the force of his tractor. Is he just a dumb hayseed, or does he know what he’s doing? To use Newton’s first law, we need to pick an object and consider the sum of the forces on it. Since our goal is to relate the tension in the part of the cable attached to the stump to the tension in the part attached to the tractor, we should pick an object to which both those cables are attached, i.e., the pulley itself. The tension in a string or cable remains approximately constant as it passes around an idealized pulley. 1 There are therefore two leftward forces acting on the pulley, each equal to the force exerted by the tractor. Since the acceleration of the pulley is essentially zero, the forces on it must be canceling out, so the rightward force of the pulley-stump cable on the pulley must be double the force exerted by the tractor. Yes, Farmer Bill knows what he’s talking about.
1 This was asserted in section 5.4 without proof. Essentially it holds because of symmetry. E.g., if the U-shaped piece of rope in figure x had unequal tension in its two legs, then this would have to be caused by some asymmetry between clockwise and counterclockwise rotation. But such an asymmetry can only be caused by friction or inertia, which we assume don’t exist.
168
Chapter 5
Analysis of forces
Summary Selected vocabulary repulsive . . . . . describes a force that tends to push the two participating objects apart attractive . . . . describes a force that tends to pull the two participating objects together oblique . . . . . . describes a force that acts at some other angle, one that is not a direct repulsion or attraction normal force . . . the force that keeps two objects from occupying the same space static friction . . a friction force between surfaces that are not slipping past each other kinetic friction . a friction force between surfaces that are slipping past each other fluid . . . . . . . . a gas or a liquid fluid friction . . . a friction force in which at least one of the object is is a fluid spring constant . the constant of proportionality between force and elongation of a spring or other object under strain Notation FN . . . . Fs . . . . Fk . . . . μs . . . .
. . . .
. . . .
. . . .
. . . .
. . . .
μk . . . . . . . . .
k. . . . . . . . . .
a normal force a static frictional force a kinetic frictional force the coefficient of static friction; the constant of proportionality between the maximum static frictional force and the normal force; depends on what types of surfaces are involved the coefficient of kinetic friction; the constant of proportionality between the kinetic frictional force and the normal force; depends on what types of surfaces are involved the spring constant; the constant of proportionality between the force exerted on an object and the amount by which the object is lengthened or compressed
Summary Newton’s third law states that forces occur in equal and opposite pairs. If object A exerts a force on object B, then object B must simultaneously be exerting an equal and opposite force on object A. Each instance of Newton’s third law involves exactly two objects, and exactly two forces, which are of the same type. There are two systems for classifying forces. We are presently using the more practical but less fundamental one. In this system, forces are classified by whether they are repulsive, attractive, or oblique; whether they are contact or noncontact forces; and whether
Summary
169
the two objects involved are solids or fluids. Static friction adjusts itself to match the force that is trying to make the surfaces slide past each other, until the maximum value is reached, Fs,max = μs FN . Once this force is exceeded, the surfaces slip past one another, and kinetic friction applies, F k = μ k FN
.
Both types of frictional force are nearly independent of surface area, and kinetic friction is usually approximately independent of the speed at which the surfaces are slipping. The direction of the force is in the direction that would tend to stop or prevent slipping. A good first step in applying Newton’s laws of motion to any physical situation is to pick an object of interest, and then to list all the forces acting on that object. We classify each force by its type, and find its Newton’s-third-law partner, which is exerted by the object on some other object. When two objects are connected by a third low-mass object, their forces are transmitted to each other nearly unchanged. Objects under strain always obey Hooke’s law to a good approximation, as long as the force is small. Hooke’s law states that the stretching or compression of the object is proportional to the force exerted on it, F ≈ k(x − xo ) .
170
Chapter 5
Analysis of forces
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 A little old lady and a pro football player collide head-on. Compare their forces on each other, and compare their accelerations. Explain.
2 The earth is attracted to an object with a force equal and opposite to the force of the earth on the object. If this is true, why is it that when you drop an object, the earth does not have an acceleration equal and opposite to that of the object?
Problem 1.
3 When you stand still, there are two forces acting on you, the force of gravity (your weight) and the normal force of the floor pushing up on your feet. Are these forces equal and opposite? Does Newton’s third law relate them to each other? Explain.
In problems 4-8, analyze the forces using a table in the format shown in section 5.3. Analyze the forces in which the italicized object participates.
4 Some people put a spare car key in a little magnetic box that they stick under the chassis of their car. Let’s say that the box is stuck directly underneath a horizontal surface, and the car is parked. (See instructions above.)
Problem 6.
5 Analyze two examples of objects at rest relative to the earth that are being kept from falling by forces other than the normal force. Do not use objects in outer space, and do not duplicate problem 4 or 8. (See instructions above.) 6 A person is rowing a boat, with her feet braced. She is doing the part of the stroke that propels the boat, with the ends of the oars in the water (not the part where the oars are out of the water). (See instructions above.)
7 A farmer is in a stall with a cow when the cow decides to press him against the wall, pinning him with his feet off the ground. Analyze the forces in which the farmer participates. (See instructions above.)
Problem 7.
Problems
171
8 A propeller plane is cruising east at constant speed and altitude. (See instructions above.) 9 Today’s tallest buildings are really not that much taller than the tallest buildings of the 1940’s. One big problem with making an even taller skyscraper is that every elevator needs its own shaft running the whole height of the building. So many elevators are needed to serve the building’s thousands of occupants that the elevator shafts start taking up too much of the space within the building. An alternative is to have elevators that can move both horizontally and vertically: with such a design, many elevator cars can share a few shafts, and they don’t get in each other’s way too much because they can detour around each other. In this design, it becomes impossible to hang the cars from cables, so they would instead have to ride on rails which they grab onto with wheels. Friction would keep them from slipping. The figure shows such a frictional elevator in its vertical travel mode. (The wheels on the bottom are for when it needs to switch to horizontal motion.) (a) If the coefficient of static friction between rubber and steel is μs , and the maximum mass of the car plus its passengers is M , how much force must there be pressing each wheel against the rail in order to keep the car from slipping? (Assume the car is not √ accelerating.) (b) Show that your result has physically reasonable behavior with respect to μs . In other words, if there was less friction, would the wheels need to be pressed more firmly or less firmly? Does your equation behave that way?
Problem 8.
Problem 9.
10 Unequal masses M and m are suspended from a pulley as shown in the figure. (a) Analyze the forces in which mass m participates, using a table in the format shown in section 5.3. [The forces in which the other mass participates will of course be similar, but not numerically the same.] (b) Find the magnitude of the accelerations of the two masses. [Hints: (1) Pick a coordinate system, and use positive and negative signs consistently to indicate the directions of the forces and accelerations. (2) The two accelerations of the two masses have to be equal in magnitude but of opposite signs, since one side eats up rope at the same rate at which the other side pays it out. (3) You need to apply Newton’s second law twice, once to each mass, and then solve the two equations for the unknowns: the acceleration, a, √ and the tension in the rope, T .] (c) Many people expect that in the special case of M = m, the two masses will naturally settle down to an equilibrium position side by side. Based on your answer from part b, is this correct? √ (d) Find the tension in the rope, T . (e) Interpret your equation from part d in the special case where one of the masses is zero. Here “interpret” means to figure out what happens mathematically, figure out what should happen physically, and
Problem 10.
172
Chapter 5
Analysis of forces
connect the two. 11 A tugboat of mass m pulls a ship of mass M , accelerating it. The speeds are low enough that you can ignore fluid friction acting on their hulls, although there will of course need to be fluid friction acting on the tug’s propellers. (a) Analyze the forces in which the tugboat participates, using a table in the format shown in section 5.3. Don’t worry about vertical forces. (b) Do the same for the ship. (c) If the force acting on the tug’s propeller is F , what is the tension, T , in the cable connecting the two ships? [Hint: Write down two equations, one for Newton’s second law applied to each object. Solve √ these for the two unknowns T and a.] (d) Interpret your answer in the special cases of M = 0 and M = ∞. 12 Someone tells you she knows of a certain type of Central American earthworm whose skin, when rubbed on polished diamond, has μk > μs . Why is this not just empirically unlikely but logically suspect? 13 In the system shown in the figure, the pulleys on the left and right are fixed, but the pulley in the center can move to the left or right. The two masses are identical. Show that the mass on the left will have an upward acceleration equal to g/5. Assume all the ropes and pulleys are massless and frictionless.
14 The figure shows two different ways of combining a pair of identical springs, each with spring constant k. We refer to the top setup as parallel, and the bottom one as a series arrangement. (a) For the parallel arrangement, analyze the forces acting on the connector piece on the left, and then use this analysis to determine the equivalent spring constant of the whole setup. Explain whether the combined spring constant should be interpreted as being stiffer or less stiff. (b) For the series arrangement, analyze the forces acting on each spring and figure out the same things. Solution, p. 530 15 Generalize the results of problem 14 to the case where the two spring constants are unequal.
Problem 13.
Problem 14.
16 (a) Using the solution of problem 14, which is given in the back of the book, predict how the spring constant of a fiber will depend on its length and cross-sectional area. (b) The constant of proportionality is called the Young’s modulus, E, and typical values of the Young’s modulus are about 1010 to 1011 . What units would the Young’s modulus have in the SI (meterkilogram-second) system? Solution, p. 531
Problems
173
17 This problem depends on the results of problems 14 and 16, whose solutions are in the back of the book. When atoms form chemical bonds, it makes sense to talk about the spring constant of the bond as a measure of how “stiff” it is. Of course, there aren’t really little springs — this is just a mechanical model. The purpose of this problem is to estimate the spring constant, k, for a single bond in a typical piece of solid matter. Suppose we have a fiber, like a hair or a piece of fishing line, and imagine for simplicity that it is made of atoms of a single element stacked in a cubical manner, as shown in the figure, with a center-to-center spacing b. A typical value for b would be about 10−10 m. (a) Find an equation for k in terms of b, and in terms of the Young’s modulus, E, defined in problem 16 and its solution. (b) Estimate k using the numerical data given in problem 16. (c) Suppose you could grab one of the atoms in a diatomic molecule like H2 or O2 , and let the other atom hang vertically below it. Does the bond stretch by any appreciable fraction due to gravity?
Problem 17.
18 In each case, identify the force that causes the acceleration, and give its Newton’s-third-law partner. Describe the effect of the partner force. (a) A swimmer speeds up. (b) A golfer hits the ball off of the tee. (c) An archer fires an arrow. (d) A locomotive slows down. Solution, p. 531 19 Ginny has a plan. She is going to ride her sled while her dog Foo pulls her, and she holds on to his leash. However, Ginny hasn’t taken physics, so there may be a problem: she may slide right off the sled when Foo starts pulling. (a) Analyze all the forces in which Ginny participates, making a table as in section 5.3. (b) Analyze all the forces in which the sled participates. (c) The sled has mass m, and Ginny has mass M . The coefficient of static friction between the sled and the snow is μ1 , and μ2 is the corresponding quantity for static friction between the sled and her snow pants. Ginny must have a certain minimum mass so that she will not slip off the sled. Find this in terms of the other three √ variables. (d) Interpreting your equation from part c, under what conditions will there be no physically realistic solution for M ? Discuss what this means physically.
Problem 19.
20 Example 2 on page 151 involves a person pushing a box up a hill. The incorrect answer describes three forces. For each of these three forces, give the force that it is related to by Newton’s third law, and state the type of force. Solution, p. 531 21 Example 8 on page 168 describes a force-doubling setup involving a pulley. Make up a more complicated arrangement, using two pulleys, that would multiply the force by four. The basic idea is to take the output of one force doubler and feed it into the input
174
Chapter 5
Analysis of forces
of a second one. 22 Pick up a heavy object such as a backpack or a chair, and stand on a bathroom scale. Shake the object up and down. What do you observe? Interpret your observations in terms of Newton’s third law. 23 A cop investigating the scene of an accident measures the length L of a car’s skid marks in order to find out its speed v at the beginning of the skid. Express v in terms of L and any other √ relevant variables. 24 The following reasoning leads to an apparent paradox; explain what’s wrong with the logic. A baseball player hits a ball. The ball and the bat spend a fraction of a second in contact. During that time they’re moving together, so their accelerations must be equal. Newton’s third law says that their forces on each other are also equal. But a = F/m, so how can this be, since their masses are unequal? (Note that the paradox isn’t resolved by considering the force of the batter’s hands on the bat. Not only is this force very small compared to the ball-bat force, but the batter could have just thrown the bat at the ball.) 25
This problem has been deleted.
26 (a) Compare the mass of a one-liter water bottle on earth, on the moon, and in interstellar space. Solution, p. 532 (b) Do the same for its weight. 27 An ice skater builds up some speed, and then coasts across the ice passively in a straight line. (a) Analyze the forces, using a table the format shown in section 5.3. (b) If his initial speed is v, and the coefficient of kinetic friction is μk , find the maximum theoretical distance he can glide before coming √ to a stop. Ignore air resistance. (c) Show that your answer to part b has the right units. (d) Show that your answer to part b depends on the variables in a way that makes sense physically. (e) Evaluate your answer numerically for μk = 0.0046, and a worldrecord speed of 14.58 m/s. (The coefficient of friction was measured by De Koning et al., using special skates worn by real speed skaters.) √ (f) Comment on whether your answer in part e seems realistic. If it doesn’t, suggest possible reasons why.
Problems
175
28 Mountain climbers with masses m and M are roped together while crossing a horizontal glacier when a vertical crevasse opens up under the climber with mass M . The climber with mass m drops down on the snow and tries to stop by digging into the snow with the pick of an ice ax. Alas, this story does not have a happy ending, because this doesn’t provide enough friction to stop. Both m and M continue accelerating, with M dropping down into the crevasse and m being dragged across the snow, slowed only by the kinetic friction with coefficient μk acting between the ax and the snow. There is no significant friction between the rope and the lip of the crevasse. √ (a) Find the acceleration a. (b) Check the units of your result. (c) Check the dependence of your equation on the variables. That means that for each variable, you should determine what its effect on a should be physically, and then what your answer from part a says its effect would be mathematically.
Problem 28.
176
Chapter 5
Analysis of forces
Motion in three dimensions
177
Chapter 6
Newton’s laws in three dimensions 6.1 Forces have no perpendicular effects Suppose you could shoot a rifle and arrange for a second bullet to be dropped from the same height at the exact moment when the first left the barrel. Which would hit the ground first? Nearly everyone expects that the dropped bullet will reach the dirt first,
179
and Aristotle would have agreed. Aristotle would have described it like this. The shot bullet receives some forced motion from the gun. It travels forward for a split second, slowing down rapidly because there is no longer any force to make it continue in motion. Once it is done with its forced motion, it changes to natural motion, i.e. falling straight down. While the shot bullet is slowing down, the dropped bullet gets on with the business of falling, so according to Aristotle it will hit the ground first.
a / A bullet is shot from a gun, and another bullet is simultaneously dropped from the same height. 1. Aristotelian physics says that the horizontal motion of the shot bullet delays the onset of falling, so the dropped bullet hits the ground first. 2. Newtonian physics says the two bullets have the same vertical motion, regardless of their different horizontal motions.
Luckily, nature isn’t as complicated as Aristotle thought! To convince yourself that Aristotle’s ideas were wrong and needlessly complex, stand up now and try this experiment. Take your keys out of your pocket, and begin walking briskly forward. Without speeding up or slowing down, release your keys and let them fall while you continue walking at the same pace. You have found that your keys hit the ground right next to your feet. Their horizontal motion never slowed down at all, and the whole time they were dropping, they were right next to you. The horizontal motion and the vertical motion happen at the same time, and they are independent of each other. Your experiment proves that the horizontal motion is unaffected by the vertical motion, but it’s also true that the vertical motion is not changed in any way by the horizontal motion. The keys take exactly the same amount of time to get to the ground as they would have if you simply dropped them, and the same is true of the bullets: both bullets hit the ground
180
Chapter 6
Newton’s laws in three dimensions
simultaneously. These have been our first examples of motion in more than one dimension, and they illustrate the most important new idea that is required to understand the three-dimensional generalization of Newtonian physics: Forces have no perpendicular effects. When a force acts on an object, it has no effect on the part of the object’s motion that is perpendicular to the force.
In the examples above, the vertical force of gravity had no effect on the horizontal motions of the objects. These were examples of projectile motion, which interested people like Galileo because of its military applications. The principle is more general than that, however. For instance, if a rolling ball is initially heading straight for a wall, but a steady wind begins blowing from the side, the ball does not take any longer to get to the wall. In the case of projectile motion, the force involved is gravity, so we can say more specifically that the vertical acceleration is 9.8 m/s2 , regardless of the horizontal motion. self-check A In the example of the ball being blown sideways, why doesn’t the ball take longer to get there, since it has to travel a greater distance? Answer, p. 542
Relationship to relative motion These concepts are directly related to the idea that motion is relative. Galileo’s opponents argued that the earth could not possibly be rotating as he claimed, because then if you jumped straight up in the air you wouldn’t be able to come down in the same place. Their argument was based on their incorrect Aristotelian assumption that once the force of gravity began to act on you and bring you back down, your horizontal motion would stop. In the correct Newtonian theory, the earth’s downward gravitational force is acting before, during, and after your jump, but has no effect on your motion in the perpendicular (horizontal) direction. If Aristotle had been correct, then we would have a handy way to determine absolute motion and absolute rest: jump straight up in the air, and if you land back where you started, the surface from which you jumped must have been in a state of rest. In reality, this test gives the same result as long as the surface under you is an inertial frame. If you try this in a jet plane, you land back on the same spot on the deck from which you started, regardless of whether the plane is flying at 500 miles per hour or parked on the runway. The method would in fact only be good for detecting whether the
Section 6.1
Forces have no perpendicular effects
181
plane was accelerating. Discussion questions A The following is an incorrect explanation of a fact about target shooting: “Shooting a high-powered rifle with a high muzzle velocity is different from shooting a less powerful gun. With a less powerful gun, you have to aim quite a bit above your target, but with a more powerful one you don’t have to aim so high because the bullet doesn’t drop as fast.” Explain why it’s incorrect. What is the correct explanation?
B You have thrown a rock, and it is flying through the air in an arc. If the earth’s gravitational force on it is always straight down, why doesn’t it just go straight down once it leaves your hand? C Consider the example of the bullet that is dropped at the same moment another bullet is fired from a gun. What would the motion of the two bullets look like to a jet pilot flying alongside in the same direction as the shot bullet and at the same horizontal speed?
6.2 Coordinates and components ’Cause we’re all Bold as love, Just ask the axis. Jimi Hendrix How do we convert these ideas into mathematics? Figure b shows a good way of connecting the intuitive ideas to the numbers. In one dimension, we impose a number line with an x coordinate on a certain stretch of space. In two dimensions, we imagine a grid of squares which we label with x and y values, as shown in figure b.
c / The shadow on the wall shows the ball’s y motion, the shadow on the floor its x motion.
182
Chapter 6
But of course motion doesn’t really occur in a series of discrete hops like in chess or checkers. Figure c shows a way of conceptualizing the smooth variation of the x and y coordinates. The ball’s shadow on the wall moves along a line, and we describe its position with a single coordinate, y, its height above the floor. The wall shadow has a constant acceleration of -9.8 m/s2 . A shadow on the floor, made by a second light source, also moves along a line, and we describe its motion with an x coordinate, measured from the wall.
Newton’s laws in three dimensions
b / This object experiences a force that pulls it down toward the bottom of the page. In each equal time interval, it moves three units to the right. At the same time, its vertical motion is making a simple pattern of +1, 0, −1, −2, −3, −4, . . . units. Its motion can be described by an x coordinate that has zero acceleration and a y coordinate with constant acceleration. The arrows labeled x and y serve to explain that we are defining increasing x to the right and increasing y as upward.
The velocity of the floor shadow is referred to as the x component of the velocity, written vx . Similarly we can notate the acceleration of the floor shadow as ax . Since vx is constant, ax is zero. Similarly, the velocity of the wall shadow is called vy , its acceleration ay . This example has ay = −9.8 m/s2 . Because the earth’s gravitational force on the ball is acting along the y axis, we say that the force has a negative y component, Fy , but Fx = Fz = 0. The general idea is that we imagine two observers, each of whom perceives the entire universe as if it was flattened down to a single line. The y-observer, for instance, perceives y, vy , and ay , and will infer that there is a force, Fy , acting downward on the ball. That is, a y component means the aspect of a physical phenomenon, such as velocity, acceleration, or force, that is observable to someone who can only see motion along the y axis. All of this can easily be generalized to three dimensions. In the example above, there could be a z-observer who only sees motion toward or away from the back wall of the room.
Section 6.2
Coordinates and components
183
A car going over a cliff example 1 The police find a car at a distance w = 20 m from the base of a cliff of height h = 100 m. How fast was the car going when it went over the edge? Solve the problem symbolically first, then plug in the numbers. Let’s choose y pointing up and x pointing away from the cliff. The car’s vertical motion was independent of its horizontal motion, so we know it had a constant vertical acceleration of a = −g = −9.8 m/s2 . The time it spent in the air is therefore related to the vertical distance it fell by the constant-acceleration equation
d / Example 1.
Δy =
1 ay Δt 2 2
−h =
1 (−g)Δt 2 2
,
or .
Solving for Δt gives Δt =
2h g
.
Since the vertical force had no effect on the car’s horizontal motion, it had ax = 0, i.e., constant horizontal velocity. We can apply the constant-velocity equation vx =
Δx Δt
,
vx =
w Δt
.
i.e.,
We now substitute for Δt to find vx = w/ which simplifies to
vx = w
2h g
,
g 2h
.
Plugging in numbers, we find that the car’s speed when it went over the edge was 4 m/s, or about 10 mi/hr.
184
Chapter 6
Newton’s laws in three dimensions
Projectiles move along parabolas. What type of mathematical curve does a projectile follow through space? To find out, we must relate x to y, eliminating t. The reasoning is very similar to that used in the example above. Arbitrarily choosing x = y = t = 0 to be at the top of the arc, we conveniently have x = Δx, y = Δy, and t = Δt, so 1 y = a y t2 2 x = vx t
(ay < 0)
We solve the second equation for t = x/vx and eliminate t in the first equation: 2 x 1 . y = ay 2 vx Since everything in this equation is a constant except for x and y, we conclude that y is proportional to the square of x. As you may or may not recall from a math class, y ∝ x2 describes a parabola. Solved problem: A cannon
page 188, problem 5
Discussion question A At the beginning of this section I represented the motion of a projectile on graph paper, breaking its motion into equal time intervals. Suppose instead that there is no force on the object at all. It obeys Newton’s first law and continues without changing its state of motion. What would the corresponding graph-paper diagram look like? If the time interval represented by each arrow was 1 second, how would you relate the graph-paper diagram to the velocity components vx and vy ?
e / A parabola can be defined as the shape made by cutting a cone parallel to its side. A parabola is also the graph of an equation of the form y ∝ x 2 .
B Make up several different coordinate systems oriented in different ways, and describe the ax and ay of a falling object in each one.
6.3 Newton’s laws in three dimensions It is now fairly straightforward to extend Newton’s laws to three dimensions: Newton’s first law If all three components of the total force on an object are zero, then it will continue in the same state of motion.
f / Each water droplet follows a parabola. The faster drops’ parabolas are bigger.
Newton’s second law The components of an object’s acceleration are predicted by the equations ax = Fx,total /m
,
ay = Fy,total /m
,
az = Fz,total /m
.
and
Newton’s third law
Section 6.3
Newton’s laws in three dimensions
185
If two objects A and B interact via forces, then the components of their forces on each other are equal and opposite: FA on B,x = −FB on A,x
,
FA on B,y = −FB on A,y
,
FA on B,z = −FB on A,z
.
and
Forces in perpendicular directions on the same objectexample 2 An object is initially at rest. Two constant forces begin acting on it, and continue acting on it for a while. As suggested by the two arrows, the forces are perpendicular, and the rightward force is stronger. What happens? Aristotle believed, and many students still do, that only one force can “give orders” to an object at one time. They therefore think that the object will begin speeding up and moving in the direction of the stronger force. In fact the object will move along a diagonal. In the example shown in the figure, the object will respond to the large rightward force with a large acceleration component to the right, and the small upward force will give it a small acceleration component upward. The stronger force does not overwhelm the weaker force, or have any effect on the upward motion at all. The force components simply add together:
g / Example 2.
* F2, Fx,total = F1,x + x 0 >+ F F1, Fy ,total = y 2,y
0
Discussion question A The figure shows two trajectories, made by splicing together lines and circular arcs, which are unphysical for an object that is only being acted on by gravity. Prove that they are impossible based on Newton’s laws.
186
Chapter 6
Newton’s laws in three dimensions
Summary Selected vocabulary component . . . . the part of a velocity, acceleration, or force that would be perceptible to an observer who could only see the universe projected along a certain one-dimensional axis parabola . . . . . the mathematical curve whose graph has y proportional to x2 Notation x, y, z . . . . . . vx , vy , vz . . . . .
ax , ay , az . . . . .
an object’s positions along the x, y, and z axes the x, y, and z components of an object’s velocity; the rates of change of the object’s x, y, and z coordinates the x, y, and z components of an object’s acceleration; the rates of change of vx , vy , and vz
Summary A force does not produce any effect on the motion of an object in a perpendicular direction. The most important application of this principle is that the horizontal motion of a projectile has zero acceleration, while the vertical motion has an acceleration equal to g. That is, an object’s horizontal and vertical motions are independent. The arc of a projectile is a parabola. Motion in three dimensions is measured using three coordinates, x, y, and z. Each of these coordinates has its own corresponding velocity and acceleration. We say that the velocity and acceleration both have x, y, and z components Newton’s second law is readily extended to three dimensions by rewriting it as three equations predicting the three components of the acceleration, ax = Fx,total /m
,
ay = Fy,total /m
,
az = Fz,total /m
,
and likewise for the first and third laws.
Summary
187
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 (a) A ball is thrown straight up with velocity v. Find an √ equation for the height to which it rises. (b) Generalize your equation for a ball thrown at an angle θ above horizontal, in which case its initial velocity components are vx = √ v cos θ and vy = v sin θ. 2 At the 2010 Salinas Lettuce Festival Parade, the Lettuce Queen drops her bouquet while riding on a float moving toward the right. Sketch the shape of its trajectory in her frame of reference, and compare with the shape seen by one of her admirers standing on the sidewalk. 3 Two daredevils, Wendy and Bill, go over Niagara Falls. Wendy sits in an inner tube, and lets the 30 km/hr velocity of the river throw her out horizontally over the falls. Bill paddles a kayak, adding an extra 10 km/hr to his velocity. They go over the edge of the falls at the same moment, side by side. Ignore air friction. Explain your reasoning. (a) Who hits the bottom first? (b) What is the horizontal component of Wendy’s velocity on impact? (c) What is the horizontal component of Bill’s velocity on impact? (d) Who is going faster on impact? 4 A baseball pitcher throws a pitch clocked at vx =73.3 mi/h. He throws horizontally. By what amount, d, does the ball drop by the time it reaches home plate, L=60.0 ft away? √ (a) First find a symbolic answer in terms of L, vx , and g. (b) Plug in and find a numerical answer. Express your answer in units of ft. (Note: 1 ft=12 in, 1 mi=5280 ft, and 1 in=2.54 √cm)
Problem 4.
5 A cannon standing on a flat field fires a cannonball with a muzzle velocity v, at an angle θ above horizontal. The cannonball
188
Chapter 6
Newton’s laws in three dimensions
thus initially has velocity components vx = v cos θ and vy = v sin θ. (a) Show that the cannon’s range (horizontal distance to where the cannonball falls) is given by the equation R = (2v 2 /g) sin θ cos θ . (b) Interpret your equation in the cases of θ = 0 and θ = 90◦ . Solution, p. 532 6 Assuming the result of problem 5 for the range of a projectile, R = (2v 2 /g) sin θ cos θ, show that the maximum range is for θ =45◦ . 7 Two cars go over the same speed bump in a parking lot, Maria’s Maserati at 25 miles per hour and Park’s Porsche at 37. How many times greater is the vertical acceleration of the Porsche? Hint: Remember that acceleration depends both on how much√the velocity changes and on how much time it takes to change.
Problems
189
190
Chapter 6
Newton’s laws in three dimensions
a / Vectors are used in aerial navigation.
Chapter 7
Vectors 7.1 Vector notation The idea of components freed us from the confines of one-dimensional physics, but the component notation can be unwieldy, since every one-dimensional equation has to be written as a set of three separate equations in the three-dimensional case. Newton was stuck with the component notation until the day he died, but eventually someone sufficiently lazy and clever figured out a way of abbreviating three equations as one. (a)
→ − → − F A on B = − F B on A
stands for
(b)
→ − → − → − F total = F 1 + F 2 + . . .
stands for
(c)
→ − a =
→ Δ− v Δt
stands for
FA on B,x = −FB on A,x FA on B,y = −FB on A,y FA on B,z = −FB on A,z Ftotal,x = F1,x + F2,x + . . . Ftotal,y = F1,y + F2,y + . . . Ftotal,z = F1,z + F2,z + . . . ax = Δvx /Δt ay = Δvy /Δt az = Δvz /Δt
Example (a) shows both ways of writing Newton’s third law. Which would you rather write?
191
The idea is that each of the algebra symbols with an arrow written on top, called a vector, is actually an abbreviation for three different numbers, the x, y, and z components. The three components are referred to as the components of the vector, e.g., Fx is the → − x component of the vector F . The notation with an arrow on top is good for handwritten equations, but is unattractive in a printed book, so books use boldface, F, to represent vectors. After this point, I’ll use boldface for vectors throughout this book. Quantities can be classified as vectors or scalars. In a phrase like to the northeast,” it makes sense to fill in the blank with “a “force” or “velocity,” which are vectors, but not with “mass” or “time,” which are scalars. Any nonzero vector has both a direction and an amount. The amount is called its magnitude. The notation for the magnitude of a vector A is |A|, like the absolute value sign used with scalars. Often, as in example (b), we wish to use the vector notation to represent adding up all the x components to get a total x component, etc. The plus sign is used between two vectors to indicate this type of component-by-component addition. Of course, vectors are really triplets of numbers, not numbers, so this is not the same as the use of the plus sign with individual numbers. But since we don’t want to have to invent new words and symbols for this operation on vectors, we use the same old plus sign, and the same old addition-related words like “add,” “sum,” and “total.” Combining vectors this way is called vector addition. Similarly, the minus sign in example (a) was used to indicate negating each of the vector’s three components individually. The equals sign is used to mean that all three components of the vector on the left side of an equation are the same as the corresponding components on the right. Example (c) shows how we abuse the division symbol in a similar manner. When we write the vector Δv divided by the scalar Δt, we mean the new vector formed by dividing each one of the velocity components by Δt. It’s not hard to imagine a variety of operations that would combine vectors with vectors or vectors with scalars, but only four of them are required in order to express Newton’s laws:
192
Chapter 7
Vectors
operation vector + vector vector − vector vector · scalar vector/scalar
definition Add component by component to make a new set of three numbers. Subtract component by component to make a new set of three numbers. Multiply each component of the vector by the scalar. Divide each component of the vector by the scalar.
As an example of an operation that is not useful for physics, there just aren’t any useful physics applications for dividing a vector by another vector component by component. In optional section 7.5, we discuss in more detail the fundamental reasons why some vector operations are useful and others useless. We can do algebra with vectors, or with a mixture of vectors and scalars in the same equation. Basically all the normal rules of algebra apply, but if you’re not sure if a certain step is valid, you should simply translate it into three component-based equations and see if it works. Order of addition example 1 If we are adding two force vectors, F + G, is it valid to assume as in ordinary algebra that F + G is the same as G + F? To tell if this algebra rule also applies to vectors, we simply translate the vector notation into ordinary algebra notation. In terms of ordinary numbers, the components of the vector F + G would be Fx + Gx , Fy + Gy , and Fz + Gz , which are certainly the same three numbers as Gx + Fx , Gy + Fy , and Gz + Fz . Yes, F + G is the same as G + F. It is useful to define a symbol r for the vector whose components are x, y, and z, and a symbol Δr made out of Δx, Δy, and Δz. Although this may all seem a little formidable, keep in mind that it amounts to nothing more than a way of abbreviating equations! Also, to keep things from getting too confusing the remainder of this chapter focuses mainly on the Δr vector, which is relatively easy to visualize. self-check A Translate the equations vx = Δx /Δt , vy = Δy /Δt , and vz = Δz /Δt for motion with constant velocity into a single equation in vector notation. Answer, p. 542
Section 7.1
Vector notation
193
Drawing vectors as arrows A vector in two dimensions can be easily visualized by drawing an arrow whose length represents its magnitude and whose direction represents its direction. The x component of a vector can then be visualized as the length of the shadow it would cast in a beam of light projected onto the x axis, and similarly for the y component. Shadows with arrowheads pointing back against the direction of the positive axis correspond to negative components. In this type of diagram, the negative of a vector is the vector with the same magnitude but in the opposite direction. Multiplying a vector by a scalar is represented by lengthening the arrow by that factor, and similarly for division. self-check B Given vector Q represented by an arrow in figure c, draw arrows representing the vectors 1.5Q and −Q. Answer, p. 542
b / The x an y components of a vector can be thought of as the shadows it casts onto the x and y axes.
This leads to a way of defining vectors and scalars that reflects how physicists think in general about these things:
definition of vectors and scalars A general type of measurement (force, velocity, . . . ) is a vector if it can be drawn as an arrow so that rotating the paper produces the same result as rotating the actual quantity. A type of quantity that never changes at all under rotation is a scalar.
c / Self-check B.
For example, a force reverses itself under a 180-degree rotation, but a mass doesn’t. We could have defined a vector as something that had both a magnitude and a direction, but that would have left out zero vectors, which don’t have a direction. A zero vector is a legitimate vector, because it behaves the same way under rotations as a zero-length arrow, which is simply a dot. A remark for those who enjoy brain-teasers: not everything is a vector or a scalar. An American football is distorted compared to a sphere, and we can measure the orientation and amount of that distortion quantitatively. The distortion is not a vector, since a 180-degree rotation brings it back to its original state. Something similar happens with playing cards, figure d. For some subatomic particles, such as electrons, 360 degrees isn’t even enough; a 720degree rotation is needed to put them back the way they were! d / A playing card returns to its original state when rotated by 180 degrees.
194
Chapter 7
Discussion questions A You drive to your friend’s house. How does the magnitude of your Δr vector compare with the distance you’ve added to the car’s odometer?
Vectors
7.2 Calculations with magnitude and direction If you ask someone where Las Vegas is compared to Los Angeles, they are unlikely to say that the Δx is 290 km and the Δy is 230 km, in a coordinate system where the positive x axis is east and the y axis points north. They will probably say instead that it’s 370 km to the northeast. If they were being precise, they might give the direction as 38◦ counterclockwise from east. In two dimensions, we can always specify a vector’s direction like this, using a single angle. A magnitude plus an angle suffice to specify everything about the vector. The following two examples show how we use trigonometry and the Pythagorean theorem to go back and forth between the x−y and magnitude-angle descriptions of vectors. Finding magnitude and angle from components example 2 Given that the Δr vector from LA to Las Vegas has Δx = 290 km and Δy = 230 km, how would we find the magnitude and direction of Δr? We find the magnitude of Δr from the Pythagorean theorem:
|Δr| = Δx 2 + Δy 2 = 370 km We know all three sides of the triangle, so the angle θ can be found using any of the inverse trig functions. For example, we know the opposite and adjacent sides, so θ = tan−1 = 38◦
e / Examples 2 and 3.
Δy Δx .
Finding components from magnitude and angle example 3 Given that the straight-line distance from Los Angeles to Las Vegas is 370 km, and that the angle θ in the figure is 38◦ , how can the x and y components of the Δr vector be found? The sine and cosine of θ relate the given information to the information we wish to find: Δx cos θ = |Δr| Δy sin θ = |Δr| Solving for the unknowns gives Δx = |Δr| cos θ = 290 km
and
Δy = |Δr| sin θ = 230 km
.
Section 7.2
Calculations with magnitude and direction
195
The following example shows the correct handling of the plus and minus signs, which is usually the main cause of mistakes. Negative components example 4 San Diego is 120 km east and 150 km south of Los Angeles. An airplane pilot is setting course from San Diego to Los Angeles. At what angle should she set her course, measured counterclockwise from east, as shown in the figure? If we make the traditional choice of coordinate axes, with x pointing to the right and y pointing up on the map, then her Δx is negative, because her final x value is less than her initial x value. Her Δy is positive, so we have Δx = −120 km Δy = 150 km
f / Example 4.
.
If we work by analogy with example 2, we get Δy Δx = tan−1 (−1.25)
θ = tan−1 = −51◦
.
According to the usual way of defining angles in trigonometry, a negative result means an angle that lies clockwise from the x axis, which would have her heading for the Baja California. What went wrong? The answer is that when you ask your calculator to take the arctangent of a number, there are always two valid possibilities differing by 180◦ . That is, there are two possible angles whose tangents equal -1.25: tan 129◦ = −1.25 tan −51◦ = −1.25 You calculator doesn’t know which is the correct one, so it just picks one. In this case, the one it picked was the wrong one, and it was up to you to add 180◦ to it to find the right answer.
196
Chapter 7
Vectors
A shortcut example 5 A split second after nine o’clock, the hour hand on a clock dial has moved clockwise past the nine-o’clock position by some imperceptibly small angle φ. Let positive x be to the right and positive y up. If the hand, with length , is represented by a Δr vector going from the dial’s center to the tip of the hand, find this vector’s Δx. The following shortcut is the easiest way to work out examples like these, in which a vector’s direction is known relative to one of the axes. We can tell that Δr will have a large, negative x component and a small, positive y . Since Δx < 0, there are really only two logical possibilities: either Δx = − cos φ, or Δx = − sin φ. Because φ is small, cos φ is large and sin φ is small. We conclude that Δx = − cos φ. A typical application of this technique to force vectors is given in example 6 on p. 214.
Discussion question A In example 4, we dealt with components that were negative. Does it make sense to classify vectors as positive and negative?
Section 7.2
Calculations with magnitude and direction
197
7.3 Techniques for adding vectors Vector addition is one of the three essential mathematical skills, summarized on pp.524-525, that you need for success in this course.
Addition of vectors given their components The easiest type of vector addition is when you are in possession of the components, and want to find the components of their sum.
Adding components example 6 Given the Δx and Δy values from the previous examples, find the Δx and Δy from San Diego to Las Vegas. Δxtotal = Δx1 + Δx2 = −120 km + 290 km = 170 km Δytotal = Δy1 + Δy2 = 150 km + 230 km = 380 Note how the signs of the x components take care of the westward and eastward motions, which partially cancel.
g / Example 6.
Addition of vectors given their magnitudes and directions
h / Vectors can be added graphically by placing them tip to tail, and then drawing a vector from the tail of the first vector to the tip of the second vector.
In this case, you must first translate the magnitudes and directions into components, and the add the components. In our San Diego-Los Angeles-Las Vegas example, we can simply string together the preceding examples; this is done on p. 525.
Graphical addition of vectors Often the easiest way to add vectors is by making a scale drawing on a piece of paper. This is known as graphical addition, as opposed to the analytic techniques discussed previously. (It has nothing to do with x − y graphs or graph paper. “Graphical” here simply means drawing. It comes from the Greek verb “grapho,” to write, like related English words including “graphic.”)
198
Chapter 7
Vectors
LA to Vegas, graphically example 7 Given the magnitudes and angles of the Δr vectors from San Diego to Los Angeles and from Los Angeles to Las Vegas, find the magnitude and angle of the Δr vector from San Diego to Las Vegas. Using a protractor and a ruler, we make a careful scale drawing, as shown in figure i. The protractor can be conveniently aligned with the blue rules on the notebook paper. A scale of 1 mm → 2 km was chosen for this solution because it was as big as possible (for accuracy) without being so big that the drawing wouldn’t fit on the page. With a ruler, we measure the distance from San Diego to Las Vegas to be 206 mm, which corresponds to 412 km. With a protractor, we measure the angle θ to be 65◦ . i / Example 7.
Even when we don’t intend to do an actual graphical calculation with a ruler and protractor, it can be convenient to diagram the addition of vectors in this way. With Δr vectors, it intuitively makes sense to lay the vectors tip-to-tail and draw the sum vector from the tail of the first vector to the tip of the second vector. We can do the same when adding other vectors such as force vectors. self-check C How would you subtract vectors graphically?
Answer, p. 542
Section 7.3
Techniques for adding vectors
199
Discussion questions A If you’re doing graphical addition of vectors, does it matter which vector you start with and which vector you start from the other vector’s tip? B If you add a vector with magnitude 1 to a vector of magnitude 2, what magnitudes are possible for the vector sum? C Which of these examples of vector addition are correct, and which are incorrect?
7.4 Unit vector notation When we want to specify a vector by its components, it can be cumbersome to have to write the algebra symbol for each component:
Δx = 290 km, Δy = 230 km
A more compact notation is to write
Δr = (290 km)ˆ x + (230 km)ˆ y
,
ˆ, y ˆ , and z ˆ, called the unit vectors, are defined where the vectors x as the vectors that have magnitude equal to 1 and directions lying along the x, y, and z axes. In speech, they are referred to as “x-hat” and so on. A slightly different, and harder to remember, version of this notation is unfortunately more prevalent. In this version, the unit ˆ vectors are called ˆi, ˆj, and k:
Δr = (290 km)ˆi + (230 km)ˆj
200
Chapter 7
Vectors
.
7.5 Rotational invariance Let’s take a closer look at why certain vector operations are useful and others are not. Consider the operation of multiplying two vectors component by component to produce a third vector: Rx = Px Qx R y = Py Q y R z = Pz Q z As a simple example, we choose vectors P and Q to have length 1, and make them perpendicular to each other, as shown in figure j/1. If we compute the result of our new vector operation using the coordinate system in j/2, we find: Rx = 0 Ry = 0 Rz = 0 The x component is zero because Px = 0, the y component is zero because Qy = 0, and the z component is of course zero because both vectors are in the x − y plane. However, if we carry out the same operations in coordinate system j/3, rotated 45 degrees with respect to the previous one, we find Rx = 1/2 Ry = −1/2 Rz = 0 The operation’s result depends on what coordinate system we use, and since the two versions of R have different lengths (one being zero and the other nonzero), they don’t just represent the same answer expressed in two different coordinate systems. Such an operation will never be useful in physics, because experiments show physics works the same regardless of which way we orient the laboratory building! The useful vector operations, such as addition and scalar multiplication, are rotationally invariant, i.e., come out the same regardless of the orientation of the coordinate system. Calibrating an electronic compass example 8 Some smart phones and GPS units contain electronic compasses that can sense the direction of the earth’s magnetic field vector, notated B. Because all vectors work according to the same rules, you don’t need to know anything special about magnetism in order to understand this example. Unlike a traditional compass that uses a magnetized needle on a bearing, an electronic compass has no moving parts. It contains two sensors oriented perpendicular to one another, and each sensor is only sensitive to the component of the earth’s field that lies along its own axis. Because a
Section 7.5
j / Component-by-component multiplication of the vectors in 1 would produce different vectors in coordinate systems 2 and 3.
Rotational invariance
201
choice of coordinates is arbitrary, we can take one of these sensors as defining the x axis and the other the y . Given the two components Bx and By , the device’s computer chip can compute the angle of magnetic north relative to its sensors, tan−1 (By /Bx ). All compasses are vulnerable to errors because of nearby magnetic materials, and in particular it may happen that some part of the compass’s own housing becomes magnetized. In an electronic compass, rotational invariance provides a convenient way of calibrating away such effects by having the user rotate the device in a horizontal circle. Suppose that when the compass is oriented in a certain way, it measures Bx = 1.00 and By = 0.00 (in certain units). We then expect that when it is rotated 90 degrees clockwise, the sensors will detect Bx = 0.00 and By = 1.00. But imagine instead that we get Bx = 0.20 and By = 0.80. This would violate rotational invariance, since rotating the coordinate system is supposed to give a different description of the same vector. The magnitude appears to have changed from 1.00 to √ 2 0.20 + 0.802 = 0.82, and a vector can’t change its magnitude just because you rotate it. The compass’s computer chip figures out that some effect, possibly a slight magnetization of its housing, must be adding an erroneous 0.2 units to all the Bx readings, because subtracting this amount from all the Bx values gives vectors that have the same magnitude, satisfying rotational invariance.
202
Chapter 7
Vectors
Summary Selected vocabulary vector . . . . . . . a quantity that has both an amount (magnitude) and a direction in space magnitude . . . . the “amount” associated with a vector scalar . . . . . . . a quantity that has no direction in space, only an amount Notation A . . . . → − A . . . . |A| . . . r . . . . . Δr . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
ˆ, y ˆ, z ˆ . . . . . . x ˆi, ˆj, k ˆ . . . . . . .
a vector with components Ax , Ay , and Az handwritten notation for a vector the magnitude of vector A the vector whose components are x, y, and z the vector whose components are Δx, Δy, and Δz (optional topic) unit vectors; the vectors with magnitude 1 lying along the x, y, and z axes a harder to remember notation for the unit vectors
Other terminology and notation displacement vec- a name for the symbol Δr tor . . . . . . . . . speed . . . . . . . the magnitude of the velocity vector, i.e., the velocity stripped of any information about its direction Summary A vector is a quantity that has both a magnitude (amount) and a direction in space, as opposed to a scalar, which has no direction. The vector notation amounts simply to an abbreviation for writing the vector’s three components. In two dimensions, a vector can be represented either by its two components or by its magnitude and direction. The two ways of describing a vector can be related by trigonometry. The two main operations on vectors are addition of a vector to a vector, and multiplication of a vector by a scalar. Vector addition means adding the components of two vectors to form the components of a new vector. In graphical terms, this corresponds to drawing the vectors as two arrows laid tip-to-tail and drawing the sum vector from the tail of the first vector to the tip of the second one. Vector subtraction is performed by negating the vector to be subtracted and then adding. Multiplying a vector by a scalar means multiplying each of its components by the scalar to create a new vector. Division by a scalar is defined similarly.
Summary
203
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 The figure shows vectors A and B. As in figure h on p. 198, graphically calculate the following: A + B, A − B, B − A, −2B, A − 2B No numbers are involved.
Problem 1.
2 Phnom Penh is 470 km east and 250 km south of Bangkok. Hanoi is 60 km east and 1030 km north of Phnom Penh. (a) Choose a coordinate system, and translate these data into Δx and Δy values with the proper plus and minus signs. (b) Find the components of the Δr vector pointing from Bangkok √ to Hanoi.
3 If you walk 35 km at an angle 25◦ counterclockwise from east, and then 22 km at 230◦ counterclockwise from east, find the distance √ and direction from your starting point to your destination.
4 A machinist is drilling holes in a piece of aluminum according to the plan shown in the figure. She starts with the top hole, then moves to the one on the left, and then to the one on the right. Since this is a high-precision job, she finishes by moving in the direction and at the angle that should take her back to the top hole, and checks that she ends up in the same place. What are the distance √ and direction from the right-hand hole to the top one?
Problem 4.
204
Chapter 7
Vectors
5 Suppose someone proposes a new operation in which a vector A and a scalar B are added together to make a new vector C like this: Cx = Ax + B Cy = Ay + B Cz = Az + B Prove that this operation won’t be useful in physics, because it’s not rotationally invariant.
Problems
205
206
Chapter 7
Vectors
Chapter 8
Vectors and motion In 1872, capitalist and former California governor Leland Stanford asked photographer Eadweard Muybridge if he would work for him on a project to settle a $25,000 bet (a princely sum at that time). Stanford’s friends were convinced that a galloping horse always had at least one foot on the ground, but Stanford claimed that there was a moment during each cycle of the motion when all four feet were in the air. The human eye was simply not fast enough to settle the question. In 1878, Muybridge finally succeeded in producing what amounted to a motion picture of the horse, showing conclusively that all four feet did leave the ground at one point. (Muybridge was a colorful figure in San Francisco history, and his acquittal for the murder of his wife’s lover was considered the trial of the century in California.) The losers of the bet had probably been influenced by Aristotelian reasoning, for instance the expectation that a leaping horse would lose horizontal velocity while in the air with no force to push it forward, so that it would be more efficient for the horse to run without leaping. But even for students who have converted whole-
207
heartedly to Newtonianism, the relationship between force and acceleration leads to some conceptual difficulties, the main one being a problem with the true but seemingly absurd statement that an object can have an acceleration vector whose direction is not the same as the direction of motion. The horse, for instance, has nearly constant horizontal velocity, so its ax is zero. But as anyone can tell you who has ridden a galloping horse, the horse accelerates up and down. The horse’s acceleration vector therefore changes back and forth between the up and down directions, but is never in the same direction as the horse’s motion. In this chapter, we will examine more carefully the properties of the velocity, acceleration, and force vectors. No new principles are introduced, but an attempt is made to tie things together and show examples of the power of the vector formulation of Newton’s laws.
8.1 The velocity vector For motion with constant velocity, the velocity vector is v = Δr/Δt
.
[only for constant velocity]
The Δr vector points in the direction of the motion, and dividing it by the scalar Δt only changes its length, not its direction, so the velocity vector points in the same direction as the motion. When the velocity is not constant, i.e., when the x − t, y − t, and z − t graphs are not all linear, we use the slope-of-the-tangent-line approach to define the components vx , vy , and vz , from which we assemble the velocity vector. Even when the velocity vector is not constant, it still points along the direction of motion. Vector addition is the correct way to generalize the one-dimensional concept of adding velocities in relative motion, as shown in the following example: Velocity vectors in relative motion example 1 You wish to cross a river and arrive at a dock that is directly across from you, but the river’s current will tend to carry you downstream. To compensate, you must steer the boat at an angle. Find the angle θ, given the magnitude, |vW L |, of the water’s velocity relative to the land, and the maximum speed, |vBW |, of which the boat is capable relative to the water. The boat’s velocity relative to the land equals the vector sum of its velocity with respect to the water and the water’s velocity with respect to the land,
a / Example 1.
vBL = vBW + vW L
.
If the boat is to travel straight across the river, i.e., along the y axis, then we need to have vBL,x = 0. This x component equals the sum of the x components of the other two vectors, vBL,x = vBW ,x + vW L,x
208
Chapter 8
Vectors and motion
,
or 0 = −|vBW | sin θ + |vW L |
.
Solving for θ, we find sin θ = |vW L |/|vBW |
,
|vW L | |vBW |
.
so θ = sin−1
Solved problem: Annie Oakley
page 221, problem 8
Discussion questions A Is it possible for an airplane to maintain a constant velocity vector but not a constant |v|? How about the opposite – a constant |v| but not a constant velocity vector? Explain. B New York and Rome are at about the same latitude, so the earth’s rotation carries them both around nearly the same circle. Do the two cities have the same velocity vector (relative to the center of the earth)? If not, is there any way for two cities to have the same velocity vector?
Section 8.1
The velocity vector
209
8.2 The acceleration vector When all three acceleration components are constant, i.e., when the vx − t, vy − t, and vz − t graphs are all linear, we can define the acceleration vector as a = Δv/Δt
,
[only for constant acceleration]
which can be written in terms of initial and final velocities as a = (vf − vi )/Δt
b / A change in the magnitude of the velocity vector implies an acceleration.
.
[only for constant acceleration]
If the acceleration is not constant, we define it as the vector made out of the ax , ay , and az components found by applying the slopeof-the-tangent-line technique to the vx − t, vy − t, and vz − t graphs. Now there are two ways in which we could have a nonzero acceleration. Either the magnitude or the direction of the velocity vector could change. This can be visualized with arrow diagrams as shown in figures b and c. Both the magnitude and direction can change simultaneously, as when a car accelerates while turning. Only when the magnitude of the velocity changes while its direction stays constant do we have a Δv vector and an acceleration vector along the same line as the motion. self-check A (1) In figure b, is the object speeding up, or slowing down? (2) What would the diagram look like if vi was the same as vf ? (3) Describe how the Δv vector is different depending on whether an object is speeding up or slowing down. Answer, p. 542
c / A change in the direction of the velocity vector also produces a nonzero Δv vector, and thus a nonzero acceleration vector, Δv/Δt .
If this all seems a little strange and abstract to you, you’re not alone. It doesn’t mean much to most physics students the first time someone tells them that acceleration is a vector, and that the acceleration vector does not have to be in the same direction as the velocity vector. One way to understand those statements better is to imagine an object such as an air freshener or a pair of fuzzy dice hanging from the rear-view mirror of a car. Such a hanging object, called a bob, constitutes an accelerometer. If you watch the bob as you accelerate from a stop light, you’ll see it swing backward. The horizontal direction in which the bob tilts is opposite to the direction of the acceleration. If you apply the brakes and the car’s acceleration vector points backward, the bob tilts forward. After accelerating and slowing down a few times, you think you’ve put your accelerometer through its paces, but then you make a right turn. Surprise! Acceleration is a vector, and needn’t point in the same direction as the velocity vector. As you make a right turn, the bob swings outward, to your left. That means the car’s acceleration vector is to your right, perpendicular to your velocity vector. A useful definition of an acceleration vector should relate in a systematic way to the actual physical effects produced by the
210
Chapter 8
Vectors and motion
acceleration, so a physically reasonable definition of the acceleration vector must allow for cases where it is not in the same direction as the motion. self-check B In projectile motion, what direction does the acceleration vector have? Answer, p. 542
d / Example 2.
Rappelling example 2 In figure d, the rappeller’s velocity has long periods of gradual change interspersed with short periods of rapid change. These correspond to periods of small acceleration and force, and periods of large acceleration and force.
The galloping horse example 3 Figure e on page 212 shows outlines traced from the first, third, fifth, seventh, and ninth frames in Muybridge’s series of photographs of the galloping horse. The estimated location of the horse’s center of mass is shown with a circle, which bobs above and below the horizontal dashed line. If we don’t care about calculating velocities and accelerations in any particular system of units, then we can pretend that the time between frames is one unit. The horse’s velocity vector as it moves from one point to the next can then be found simply by drawing an arrow to connect one position of the center of mass to the next. This produces a series of velocity vectors which alter-
Section 8.2
The acceleration vector
211
e / Example 3.
nate between pointing above and below horizontal. The Δv vector is the vector which we would have to add onto one velocity vector in order to get the next velocity vector in the series. The Δv vector alternates between pointing down (around the time when the horse is in the air, B) and up (around the time when the horse has two feet on the ground, D). Discussion questions A When a car accelerates, why does a bob hanging from the rearview mirror swing toward the back of the car? Is it because a force throws it backward? If so, what force? Similarly, describe what happens in the other cases described above. B Superman is guiding a crippled spaceship into port. The ship’s engines are not working. If Superman suddenly changes the direction of his force on the ship, does the ship’s velocity vector change suddenly? Its acceleration vector? Its direction of motion?
212
Chapter 8
Vectors and motion
8.3 The force vector and simple machines Force is relatively easy to intuit as a vector. The force vector points in the direction in which it is trying to accelerate the object it is acting on. Since force vectors are so much easier to visualize than acceleration vectors, it is often helpful to first find the direction of the (total) force vector acting on an object, and then use that to find the direction of the acceleration vector. Newton’s second law tells us that the two must be in the same direction. A component of a force vector example 4 Figure f, redrawn from a classic 1920 textbook, shows a boy pulling another child on a sled. His force has both a horizontal component and a vertical one, but only the horizontal one accelerates the sled. (The vertical component just partially cancels the force of gravity, causing a decrease in the normal force between the runners and the snow.) There are two triangles in the figure. One triangle’s hypotenuse is the rope, and the other’s is the magnitude of the force. These triangles are similar, so their internal angles are all the same, but they are not the same triangle. One is a distance triangle, with sides measured in meters, the other a force triangle, with sides in newtons. In both cases, the horizontal leg is 93% as long as the hypotenuse. It does not make sense, however, to compare the sizes of the triangles — the force triangle is not smaller in any meaningful sense. Pushing a block up a ramp example 5 Figure g shows a block being pushed up a frictionless ramp at constant speed by an applied force FA . How much force is required, in terms of the block’s mass, m, and the angle of the ramp, θ? Figure h shows the other two forces acting on the block: a normal force, FN , created by the ramp, and the weight force, FW , created by the earth’s gravity. Because the block is being pushed up at constant speed, it has zero acceleration, and the total force on it must be zero. From figure i, we find |FA | = |FW | sin θ = mg sin θ
.
Since the sine is always less than one, the applied force is always less than mg, i.e., pushing the block up the ramp is easier than lifting it straight up. This is presumably the principle on which the pyramids were constructed: the ancient Egyptians would have had a hard time applying the forces of enough slaves to equal the full weight of the huge blocks of stone. Essentially the same analysis applies to several other simple machines, such as the wedge and the screw.
Section 8.3
f / Example 4.
g / The applied force FA pushes the block up the frictionless ramp.
h / Three forces act on the block. Their vector sum is zero.
i / If the block is to move at constant velocity, Newton’s first law says that the three force vectors acting on it must add up to zero. To perform vector addition, we put the vectors tip to tail, and in this case we are adding three vectors, so each one’s tail goes against the tip of the previous one. Since they are supposed to add up to zero, the third vector’s tip must come back to touch the tail of the first vector. They form a triangle, and since the applied force is perpendicular to the normal force, it is a right triangle.
The force vector and simple machines
213
j / Example 6 and problem 18 on p. 224.
A layback example 6 The figure shows a rock climber using a technique called a layback. He can make the normal forces FN1 and FN2 large, which has the side-effect of increasing the frictional forces FF 1 and FF 2 , so that he doesn’t slip down due to the gravitational (weight) force FW . The purpose of the problem is not to analyze all of this in detail, but simply to practice finding the components of the forces based on their magnitudes. To keep the notation simple, let’s write FN1 for |FN1 |, etc. The crack overhangs by a small, positive angle θ ≈ 9◦ . In this example, we determine the x component of FN1 . The other nine components are left as an exercise to the reader (problem 18, p. 224). The easiest method is the one demonstrated in example 5 on p. 197. Casting vector FN1 ’s shadow on the ground, we can tell that it would point to the left, so its x component is negative. The only two possibilities for its x component are therefore −FN1 cos θ or −FN1 sin θ. We expect this force to have a large x component and a much smaller y . Since θ is small, cos θ ≈ 1, while sin θ is small. Therefore the x component must be −FN1 cos θ. Solved problem: A cargo plane
214
Chapter 8
page 221, problem 9
Solved problem: The angle of repose
page 222, problem 11
Solved problem: A wagon
page 221, problem 10
Vectors and motion
Discussion questions A The figure shows a block being pressed diagonally upward against a wall, causing it to slide up the wall. Analyze the forces involved, including their directions. B The figure shows a roller coaster car rolling down and then up under the influence of gravity. Sketch the car’s velocity vectors and acceleration vectors. Pick an interesting point in the motion and sketch a set of force vectors acting on the car whose vector sum could have resulted in the right acceleration vector.
8.4
Discussion question A.
Calculus with vectors
Using the unit vector notation introduced in section 7.4, the definitions of the velocity and acceleration components given in chapter 6 can be translated into calculus notation as v=
dy dz dx ˆ+ ˆ+ ˆ x y z dt dt dt
k / Discussion
question
A.
and a=
dvy dvx dvz ˆ+ ˆ+ ˆ x y z dt dt dt
.
To make the notation less cumbersome, we generalize the concept of the derivative to include derivatives of vectors, so that we can abbreviate the above equations as v=
dr dt
a=
dv dt
and .
In words, to take the derivative of a vector, you take the derivatives of its components and make a new vector out of those. This definition means that the derivative of a vector function has the familiar properties d(cf ) df =c dt dt
[c is a constant]
and df dg d(f + g) = + dt dt dt
.
The integral of a vector is likewise defined as integrating component by component.
Section 8.4
Calculus with vectors
215
The second derivative of a vector example 7 Two objects have positions as functions of time given by the equations r1 = 3t 2 xˆ + t yˆ and r2 = 3t 4 xˆ + t yˆ
.
Find both objects’ accelerations using calculus. Could either answer have been found without calculus? Taking the first derivative of each component, we find v1 = 6t xˆ + yˆ v2 = 12t 3 xˆ + yˆ
,
and taking the derivatives again gives acceleration, a1 = 6ˆx a2 = 36t 2 xˆ
.
The first object’s acceleration could have been found without calculus, simply by comparing the x and y coordinates with the constant-acceleration equation Δx = vo Δt + 12 aΔt 2 . The second equation, however, isn’t just a second-order polynomial in t, so the acceleration isn’t constant, and we really did need calculus to find the corresponding acceleration. The integral of a vector example 8 Starting from rest, a flying saucer of mass m is observed to vary its propulsion with mathematical precision according to the equation . F = bt 42 xˆ + ct 137 yˆ (The aliens inform us that the numbers 42 and 137 have a special religious significance for them.) Find the saucer’s velocity as a function of time. From the given force, we can easily find the acceleration F m b c = t 42 xˆ + t 137 yˆ m m
a=
.
The velocity vector v is the integral with respect to time of the acceleration, v = a dt b 42 c 137 , t xˆ + t yˆ dt = m m
216
Chapter 8
Vectors and motion
and integrating component by component gives c 137 b 42 dt yˆ = t dt xˆ + t m m c b 43 = , t xˆ + t 138 yˆ 43m 138m
where we have omitted the constants of integration, since the saucer was starting from rest. A fire-extinguisher stunt on ice example 9 Prof. Puerile smuggles a fire extinguisher into a skating rink. Climbing out onto the ice without any skates on, he sits down and pushes off from the wall with his feet, acquiring an initial velocity vo yˆ . At t = 0, he then discharges the fire extinguisher at a 45degree angle so that it applies a force to him that is backward and to the left, i.e., along the negative y axis and the positive x axis. The fire extinguisher’s force is strong at first, but then dies down according to the equation |F| = b − ct, where b and c are constants. Find the professor’s velocity as a function of time. Measured counterclockwise from the x axis, the angle of the force vector becomes 315◦ . Breaking the force down into x and y components, we have Fx = |F| cos 315◦ = (b − ct) Fy = |F| sin 315◦ = (−b + ct)
.
In unit vector notation, this is F = (b − ct)ˆx + (−b + ct)ˆy
.
Newton’s second law gives a = F/m b − ct −b + ct = √ xˆ + √ yˆ 2m 2m
.
To find the velocity vector as a function of time, we need to integrate the acceleration vector with respect to time, v = a dt b − ct −b + ct √ = xˆ + √ yˆ dt 2m 2m 1 (b − ct) xˆ + (−b + ct) yˆ dt =√ 2m
Section 8.4
Calculus with vectors
217
A vector function can be integrated component by component, so this can be broken down into two integrals, xˆ yˆ √ √ v= (b − ct) dt + (−b + ct) dt 2m 2m bt − 12 ct 2 −bt + 12 ct 2 √ √ = + constant #1 xˆ + + constant #2 yˆ 2m 2m Here the physical significance of the two constants of integration is that they give the initial velocity. Constant #1 is therefore zero, and constant #2 must equal vo . The final result is bt − 12 ct 2 −bt + 12 ct 2 √ √ + vo yˆ v= . xˆ + 2m 2m
218
Chapter 8
Vectors and motion
Summary The velocity vector points in the direction of the object’s motion. Relative motion can be described by vector addition of velocities. The acceleration vector need not point in the same direction as the object’s motion. We use the word “acceleration” to describe any change in an object’s velocity vector, which can be either a change in its magnitude or a change in its direction. An important application of the vector addition of forces is the use of Newton’s first law to analyze mechanical systems.
Summary
219
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
Problem 1.
1 A dinosaur fossil is slowly moving down the slope of a glacier under the influence of wind, rain and gravity. At the same time, the glacier is moving relative to the continent underneath. The dashed lines represent the directions but not the magnitudes of the velocities. Pick a scale, and use graphical addition of vectors to find the magnitude and the direction of the fossil’s velocity relative to √ the continent. You will need a ruler and protractor. 2 Is it possible for a helicopter to have an acceleration due east and a velocity due west? If so, what would be going on? If not, why not? 3 A bird is initially flying horizontally east at 21.1 m/s, but one second later it has changed direction so that it is flying horizontally and 7◦ north of east, at the same speed. What are the magnitude and direction of its acceleration vector during that one second √ time interval? (Assume its acceleration was roughly constant.)
Problem 4.
4 A person of mass M stands in the middle of a tightrope, which is fixed at the ends to two buildings separated by a horizontal distance L. The rope sags in the middle, stretching and lengthening the rope slightly. (a) If the tightrope walker wants the rope to sag vertically by no
220
Chapter 8
Vectors and motion
more than a height h, find the minimum tension, T , that the rope must be able to withstand without breaking, in terms of h, g, M√, and L. (b) Based on your equation, explain why it is not possible to get h = 0, and give a physical interpretation. 5 Your hand presses a block of mass m against a wall with a force FH acting at an angle θ, as shown in the figure. Find the minimum and maximum possible values of |FH | that can keep the block stationary, in terms of m, g, θ, and μs , the coefficient of static friction between the block and the wall. Check both your answers in the case of θ = 90◦ , and interpret the case where the maximum √ force is infinite.
Problem 5.
6 A skier of mass m is coasting down a slope inclined at an angle θ compared to horizontal. Assume for simplicity that the treatment of kinetic friction given in chapter 5 is appropriate here, although a soft and wet surface actually behaves a little differently. The coefficient of kinetic friction acting between the skis and the snow is μk , and in addition the skier experiences an air friction force of magnitude bv 2 , where b is a constant. (a) Find the maximum speed that the skier will attain, in terms of √ the variables m, g, θ, μk , and b. (b) For angles below a certain minimum angle θmin , the equation gives a result that is not mathematically meaningful. Find an equation for θmin , and give a physical explanation of what is happening √ for θ < θmin . 7 A gun is aimed horizontally to the west. The gun is fired, and the bullet leaves the muzzle at t = 0. The bullet’s position vector ˆ, where b, c, and d are as a function of time is r = bˆ x + ctˆ y + dt2 z positive constants. (a) What units would b, c, and d need to have for the equation to make sense? (b) Find the bullet’s velocity and acceleration as functions of time. ˆ, y ˆ , and z ˆ. (c) Give physical interpretations of b, c, d, x
Problem 9.
8 Annie Oakley, riding north on horseback at 30 mi/hr, shoots her rifle, aiming horizontally and to the northeast. The muzzle speed of the rifle is 140 mi/hr. When the bullet hits a defenseless fuzzy animal, what is its speed of impact? Neglect air resistance, and ignore the vertical motion of the bullet. Solution, p. 532 9 A cargo plane has taken off from a tiny airstrip in the Andes, and is climbing at constant speed, at an angle of θ=17◦ with respect to horizontal. Its engines supply a thrust of Fthrust = 200 kN, and the lift from its wings is Flif t = 654 kN. Assume that air resistance (drag) is negligible, so the only forces acting are thrust, lift, and weight. What is its mass, in kg? Solution, p. 533
Problem 10.
Problems
221
10 A wagon is being pulled at constant speed up a slope θ by a rope that makes an angle φ with the vertical. (a) Assuming negligible friction, show that the tension in the rope is given by the equation
FT =
sin θ FW sin(θ + φ)
,
where FW is the weight force acting on the wagon. (b) Interpret this equation in the special cases of φ = 0 and φ = Solution, p. 533 180◦ − θ. 11 The angle of repose is the maximum slope on which an object will not slide. On airless, geologically inert bodies like the moon or an asteroid, the only thing that determines whether dust or rubble will stay on a slope is whether the slope is less steep than the angle of repose. (See figure n, p. 256.) (a) Find an equation for the angle of repose, deciding for yourself what are the relevant variables. (b) On an asteroid, where g can be thousands of times lower than on Earth, would rubble be able to lie at a steeper angle of repose? Solution, p. 534 12 The figure shows an experiment in which a cart is released from rest at A, and accelerates down the slope through a distance x until it passes through a sensor’s light beam. The point of the experiment is to determine the cart’s acceleration. At B, a cardboard vane mounted on the cart enters the light beam, blocking the light beam, and starts an electronic timer running. At C, the vane emerges from the beam, and the timer stops. (a) Find the final velocity of the cart in terms of the width w of the vane and the time tb for which the sensor’s light beam was √ blocked. (b) Find the magnitude of the cart’s acceleration in terms of the √ measurable quantities x, tb , and w. (c) Analyze the forces in which the cart participates, using a table in the format introduced in section 5.3. Assume friction is negligible. (d) Find a theoretical value for the acceleration of the cart, which could be compared with the experimentally observed value extracted in part b. Express the theoretical value in terms of the angle √θ of the slope, and the strength g of the gravitational field.
Problem 12.
Problem 13 (Millikan and Gale, 1920).
222
Chapter 8
13 The figure shows a boy hanging in three positions: (1) with his arms straight up, (2) with his arms at 45 degrees, and (3) with his arms at 60 degrees with respect to the vertical. Compare the tension in his arms in the three cases.
Vectors and motion
14 Driving down a hill inclined at an angle θ with respect to horizontal, you slam on the brakes to keep from hitting a deer. Your antilock brakes kick in, and you don’t skid. (a) Analyze the forces. (Ignore rolling resistance and air friction.) (b) Find the car’s maximum possible deceleration, a (expressed as a positive number), in terms of g, θ, and the relevant coefficient of √ friction. (c) Explain physically why the car’s mass has no effect on your answer. (d) Discuss the mathematical behavior and physical interpretation of your result for negative values of θ. (e) Do the same for very large positive values of θ. 15 The figure shows the path followed by Hurricane Irene in 2005 as it moved north. The dots show the location of the center of the storm at six-hour intervals, with lighter dots at the time when the storm reached its greatest intensity. Find the time when the storm’s center had a velocity vector to the northeast and an acceleration vector to the southeast. Explain. 16 For safety, mountain climbers often wear a climbing harness and tie in to other climbers on a rope team or to anchors such as pitons or snow anchors. When using anchors, the climber usually wants to tie in to more than one, both for extra strength and for redundancy in case one fails. The figure shows such an arrangement, with the climber hanging from a pair of anchors forming a “Y” at an angle θ. The usual advice is to make θ < 90◦ ; for large values of θ, the stress placed on the anchors can be many times greater than the actual load L, so that two anchors are actually less safe than one. √ (a) Find the force S at each anchor in terms of L and θ. (b) Verify that your answer makes sense in the case of θ = 0. (c) Interpret your answer in the case of θ = 180◦ . (d) What is the smallest value of θ for which S equals or exceeds L, so that for larger angles a failure of at least one anchor is more √ likely than it would have been with a single anchor? 17 (a) The person with mass m hangs from the rope, hauling the box of mass M up a slope inclined at an angle θ. There is friction between the box and the slope, described by the usual coefficients of friction. The pulley, however, is frictionless. Find the magnitude √ of the box’s acceleration. (b) Show that the units of your answer make sense. (c) Check the physical behavior of your answer in the special cases of M = 0 and θ = −90◦ .
Problem 15.
Problem 16.
Problem 17.
Problems
223
18 Complete example 6 on p. 214 by expressing the remaining nine x and y components of the forces in terms of the five magnitudes and the small, positive angle θ ≈ 9◦ by which the crack overhangs. √ 19 Problem 16 discussed a possible correct way of setting up a redundant anchor for mountaineering. The figure for this problem shows an incorrect way of doing it, by arranging the rope in a triangle (which we’ll take to be isoceles). One of the bad things about the triangular arrangement is that it requires more force from the anchors, making them more likely to fail. (a) Using the same √ notation as in problem 16, find S in terms of L and θ. (b) Verify that your answer makes sense in the case of θ = 0, and compare with the correct setup.
Problem 19.
224
Chapter 8
Vectors and motion
Problems
225
Exercise 8: Vectors and motion Each diagram on page 227 shows the motion of an object in an x − y plane. Each dot is one location of the object at one moment in time. The time interval from one dot to the next is always the same, so you can think of the vector that connects one dot to the next as a v vector, and subtract to find Δv vectors. 1. Suppose the object in diagram 1 is moving from the top left to the bottom right. Deduce whatever you can about the force acting on it. Does the force always have the same magnitude? The same direction? Invent a physical situation that this diagram could represent. What if you reinterpret the diagram by reversing the object’s direction of motion? Redo the construction of one of the Δv vectors and see what happens. 2. What can you deduce about the force that is acting in diagram 2? Invent a physical situation that diagram 2 could represent. 3. What can you deduce about the force that is acting in diagram 3? Invent a physical situation.
226
Chapter 8
Vectors and motion
Exercise 8: Vectors and motion
227
228
Chapter 8
Vectors and motion
Chapter 9
Circular motion 9.1 Conceptual framework I now live fifteen minutes from Disneyland, so my friends and family in my native Northern California think it’s a little strange that I’ve never visited the Magic Kingdom again since a childhood trip to the south. The truth is that for me as a preschooler, Disneyland was not the Happiest Place on Earth. My mother took me on a ride in which little cars shaped like rocket ships circled rapidly around a central pillar. I knew I was going to die. There was a force trying to throw me outward, and the safety features of the ride would surely have been inadequate if I hadn’t screamed the whole time to make sure Mom would hold on to me. Afterward, she seemed surprisingly indifferent to the extreme danger we had experienced. Circular motion does not produce an outward force My younger self’s understanding of circular motion was partly right and partly wrong. I was wrong in believing that there was a force pulling me outward, away from the center of the circle. The easiest way to understand this is to bring back the parable of the bowling ball in the pickup truck from chapter 4. As the truck makes a left turn, the driver looks in the rearview mirror and thinks that some mysterious force is pulling the ball outward, but the truck is accelerating, so the driver’s frame of reference is not an inertial frame. Newton’s laws are violated in a noninertial frame, so the ball appears to accelerate without any actual force acting on it. Because we are used to inertial frames, in which accelerations are caused by
229
forces, the ball’s acceleration creates a vivid illusion that there must be an outward force.
a / 1. In the turning truck’s frame of reference, the ball appears to violate Newton’s laws, displaying a sideways acceleration that is not the result of a forceinteraction with any other object. 2. In an inertial frame of reference, such as the frame fixed to the earth’s surface, the ball obeys Newton’s first law. No forces are acting on it, and it continues moving in a straight line. It is the truck that is participating in an interaction with the asphalt, the truck that accelerates as it should according to Newton’s second law.
b / This crane fly’s halteres help it to maintain its orientation in flight.
In an inertial frame everything makes more sense. The ball has no force on it, and goes straight as required by Newton’s first law. The truck has a force on it from the asphalt, and responds to it by accelerating (changing the direction of its velocity vector) as Newton’s second law says it should.
The halteres example 1 Another interesting example is an insect organ called the halteres, a pair of small knobbed limbs behind the wings, which vibrate up and down and help the insect to maintain its orientation in flight. The halteres evolved from a second pair of wings possessed by earlier insects. Suppose, for example, that the halteres are on their upward stroke, and at that moment an air current causes the fly to pitch its nose down. The halteres follow Newton’s first law, continuing to rise vertically, but in the fly’s rotating frame of reference, it seems as though they have been subjected to a backward force. The fly has special sensory organs that perceive this twist, and help it to correct itself by raising its nose.
Circular motion does not persist without a force I was correct, however, on a different point about the Disneyland ride. To make me curve around with the car, I really did need some force such as a force from my mother, friction from the seat, or a normal force from the side of the car. (In fact, all three forces were probably adding together.) One of the reasons why Galileo failed to
230
Chapter 9
Circular motion
c / 1. An overhead view of a person swinging a rock on a rope. A force from the string is required to make the rock’s velocity vector keep changing direction. 2. If the string breaks, the rock will follow Newton’s first law and go straight instead of continuing around the circle.
refine the principle of inertia into a quantitative statement like Newton’s first law is that he was not sure whether motion without a force would naturally be circular or linear. In fact, the most impressive examples he knew of the persistence of motion were mostly circular: the spinning of a top or the rotation of the earth, for example. Newton realized that in examples such as these, there really were forces at work. Atoms on the surface of the top are prevented from flying off straight by the ordinary force that keeps atoms stuck together in solid matter. The earth is nearly all liquid, but gravitational forces pull all its parts inward. Uniform and nonuniform circular motion Circular motion always involves a change in the direction of the velocity vector, but it is also possible for the magnitude of the velocity to change at the same time. Circular motion is referred to as uniform if |v| is constant, and nonuniform if it is changing.
d / Sparks fly away along tangents to a grinding wheel.
Your speedometer tells you the magnitude of your car’s velocity vector, so when you go around a curve while keeping your speedometer needle steady, you are executing uniform circular motion. If your speedometer reading is changing as you turn, your circular motion is nonuniform. Uniform circular motion is simpler to analyze mathematically, so we will attack it first and then pass to the nonuniform case. self-check A Which of these are examples of uniform circular motion and which are nonuniform? (1) the clothes in a clothes dryer (assuming they remain against the inside of the drum, even at the top) (2) a rock on the end of a string being whirled in a vertical circle Answer, p. 543
Section 9.1
Conceptual framework
231
Only an inward force is required for uniform circular motion.
e / To make the brick go in a circle, I had to exert an inward force on the rope.
g / When a car is going straight at constant speed, the forward and backward forces on it are canceling out, producing a total force of zero. When it moves in a circle at constant speed, there are three forces on it, but the forward and backward forces cancel out, so the vector sum is an inward force.
232
Chapter 9
Figure c showed the string pulling in straight along a radius of the circle, but many people believe that when they are doing this they must be “leading” the rock a little to keep it moving along. That is, they believe that the force required to produce uniform circular motion is not directly inward but at a slight angle to the radius of the circle. This intuition is incorrect, which you can easily verify for yourself now if you have some string handy. It is only while you are getting the object going that your force needs to be at an angle to the radius. During this initial period of speeding up, the motion is not uniform. Once you settle down into uniform circular motion, you only apply an inward force. If you have not done the experiment for yourself, here is a theoretical argument to convince you of this fact. We have discussed in chapter 6 the principle that forces have no perpendicular effects. To keep the rock from speeding up or slowing down, we only need to make sure that our force is perpendicular to its direction of motion. We are then guaranteed that its forward motion will remain unaffected: our force can have no perpendicular effect, and there is no other force acting on the rock which could slow it down. The rock requires no forward force to maintain its forward motion, any more than a projectile needs a horizontal force to “help it over the top” of its arc.
f / A series of three hammer taps makes the rolling ball trace a triangle, seven hammers a heptagon. If the number of hammers was large enough, the ball would essentially be experiencing a steady inward force, and it would go in a circle. In no case is any forward force necessary.
Circular motion
Why, then, does a car driving in circles in a parking lot stop executing uniform circular motion if you take your foot off the gas? The source of confusion here is that Newton’s laws predict an object’s motion based on the total force acting on it. A car driving in circles has three forces on it (1) an inward force from the asphalt, controlled with the steering wheel; (2) a forward force from the asphalt, controlled with the gas pedal; and (3) backward forces from air resistance and rolling resistance. You need to make sure there is a forward force on the car so that the backward forces will be exactly canceled out, creating a vector sum that points directly inward.
h / Example 2.
A motorcycle making a turn example 2 The motorcyclist in figure h is moving along an arc of a circle. It looks like he’s chosen to ride the slanted surface of the dirt at a place where it makes just the angle he wants, allowing him to get the force he needs on the tires as a normal force, without needing any frictional force. The dirt’s normal force on the tires points up and to our left. The vertical component of that force is canceled by gravity, while its horizontal component causes him to curve. In uniform circular motion, the acceleration vector is inward. Since experiments show that the force vector points directly inward, Newton’s second law implies that the acceleration vector points inward as well. This fact can also be proven on purely kinematical grounds, and we will do so in the next section. Clock-comparison tests of Newton’s first law example 3 Immediately after his original statement of the first law in the Principia Mathematica, Newton offers the supporting example of a spinning top, which only slows down because of friction. He describes the different parts of the top as being held together by “cohesion,” i.e., internal forces. Because these forces act toward the center, they don’t speed up or slow down the motion. The applicability of the first law, which only describes linear motion, may be more clear if we simply take figure f as a model of rotation. Between hammer taps, the ball experiences no force, so by the first law it doesn’t speed up or slow down. Suppose that we want to subject the first law to a stringent experimental test.1 The law predicts that if we use a clock to measure the rate of rotation of an object spinning frictionlessly, it won’t “naturally” slow down as Aristotle would have expected. But what is a clock but something with hands that rotate at a fixed rate? In 1
Page 80 lists places in this book where we describe experimental tests of Newton’s first law.
Section 9.1
Conceptual framework
233
other words, we are comparing one clock with another. This is called a clock-comparison experiment. Suppose that the laws of physics weren’t purely Newtonian, and there really was a very slight Aristotelian tendency for motion to slow down in the absence of friction. If we compare two clocks, they should both slow down, but if they aren’t the same type of clock, then it seems unlikely that they would slow down at exactly the same rate, and over time they should drift further and further apart. High-precision clock-comparison experiments have been done using a variety of clocks. In atomic clocks, the thing spinning is an atom. Astronomers can observe the rotation of collapsed stars called pulars, which, unlike the earth, can rotate with almost no disturbance due to geological activity or friction induced by the tides. In these experiments, the pulsars are observed to match the rates of the atomic clocks with a drift of less than about 10−6 seconds over a period of 10 years.2 Atomic clocks using atoms of different elements drift relative to one another by no more than about 10−16 per year.3 It is not presently possible to do experiments with a similar level of precision using human-scale rotating objects. However, a set of gyroscopes aboard the Gravity Probe B satellite were allowed to spin weightlessly in a vacuum, without any physical contact that would have caused kinetic friction. Their rotation was extremely accurately monitored for the purposes of another experiment (a test of Einstein’s theory of general relativity, which was the purpose of the mission), and they were found to be spinning down so gradually that they would have taken about 10,000 years to slow down by a factor of two. This rate was consistent with estimates of the amount of friction to be expected from the small amount of residual gas present in the vacuum chambers. A subtle point in the interpretation of these experiments is that if there was a slight tendency for motion to slow down, we would have to decide what it was supposed to slow down relative to. A straight-line motion that is slowing down in some frame of reference can always be described as speeding up in some other appropriately chosen frame (problem 12, p. 89). If the laws of physics did have this slight Aristotelianism mixed in, we could wait for the anomalous acceleration or deceleration to stop. The object we were observing would then define a special or “preferred” frame of reference. Standard theories of physics do not have such a preferred frame, and clock-comparison experiments can be viewed as tests of the existence of such a frame. Another test for the existence of a preferred frame is described on p. 261. 2
Matsakis et al., Astronomy and Astrophysics 326 (1997) 924. Freely available online at adsabs.harvard.edu. 3 Gu´ena et al., arxiv.org/abs/1205.4235
234
Chapter 9
Circular motion
Discussion questions A In the game of crack the whip, a line of people stand holding hands, and then they start sweeping out a circle. One person is at the center, and rotates without changing location. At the opposite end is the person who is running the fastest, in a wide circle. In this game, someone always ends up losing their grip and flying off. Suppose the person on the end loses her grip. What path does she follow as she goes flying off? (Assume she is going so fast that she is really just trying to put one foot in front of the other fast enough to keep from falling; she is not able to get any significant horizontal force between her feet and the ground.) B Suppose the person on the outside is still holding on, but feels that she may loose her grip at any moment. What force or forces are acting on her, and in what directions are they? (We are not interested in the vertical forces, which are the earth’s gravitational force pulling down, and the ground’s normal force pushing up.) Make a table in the format shown in section 5.3. C Suppose the person on the outside is still holding on, but feels that she may loose her grip at any moment. What is wrong with the following analysis of the situation? “The person whose hand she’s holding exerts an inward force on her, and because of Newton’s third law, there’s an equal and opposite force acting outward. That outward force is the one she feels throwing her outward, and the outward force is what might make her go flying off, if it’s strong enough.”
Discussion
questions
A-D
Discussion question E.
D If the only force felt by the person on the outside is an inward force, why doesn’t she go straight in? E In the amusement park ride shown in the figure, the cylinder spins faster and faster until the customer can pick her feet up off the floor without falling. In the old Coney Island version of the ride, the floor actually dropped out like a trap door, showing the ocean below. (There is also a version in which the whole thing tilts up diagonally, but we’re discussing the version that stays flat.) If there is no outward force acting on her, why does she stick to the wall? Analyze all the forces on her. F What is an example of circular motion where the inward force is a normal force? What is an example of circular motion where the inward force is friction? What is an example of circular motion where the inward force is the sum of more than one force? G Does the acceleration vector always change continuously in circular motion? The velocity vector?
Section 9.1
Conceptual framework
235
9.2 Uniform circular motion In this section I derive a simple and very useful equation for the magnitude of the acceleration of an object undergoing constant acceleration. The law of sines is involved, so I’ve recapped it in figure i. i / The law of sines.
j / Deriving |a| = |v|2 /r uniform circular motion.
The derivation is brief, but the method requires some explanation and justification. The idea is to calculate a Δv vector describing the change in the velocity vector as the object passes through an angle θ. We then calculate the acceleration, a = Δv/Δt. The astute reader will recall, however, that this equation is only valid for motion with constant acceleration. Although the magnitude of the acceleration is constant for uniform circular motion, the acceleration vector changes its direction, so it is not a constant vector, and the equation a = Δv/Δt does not apply. The justification for using it is that we will then examine its behavior when we make the time interval very short, which means making the angle θ very small. For smaller and smaller time intervals, the Δv/Δt expression becomes a better and better approximation, so that the final result of the derivation is exact.
for
In figure j/1, the object sweeps out an angle θ. Its direction of motion also twists around by an angle θ, from the vertical dashed line to the tilted one. Figure j/2 shows the initial and final velocity vectors, which have equal magnitude, but directions differing by θ. In j/3, I’ve reassembled the vectors in the proper positions for vector subtraction. They form an isosceles triangle with interior angles θ, η, and η. (Eta, η, is my favorite Greek letter.) The law of sines gives |v| |Δv| = . sin θ sin η This tells us the magnitude of Δv, which is one of the two ingredients we need for calculating the magnitude of a = Δv/Δt. The other ingredient is Δt. The time required for the object to move through the angle θ is length of arc . Δt = |v| Now if we measure our angles in radians we can use the definition of radian measure, which is (angle) = (length of arc)/(radius), giving Δt = θr/|v|. Combining this with the first expression involving |Δv| gives |a| = |Δv|/Δt =
|v|2 sin θ 1 · · r θ sin η
.
When θ becomes very small, the small-angle approximation sin θ ≈ θ applies, and also η becomes close to 90◦ , so sin η ≈ 1, and we have
236
Chapter 9
Circular motion
an equation for |a|: |v|2 |a| = r
.
[uniform circular motion]
Force required to turn on a bike example 4 A bicyclist is making a turn along an arc of a circle with radius 20 m, at a speed of 5 m/s. If the combined mass of the cyclist plus the bike is 60 kg, how great a static friction force must the road be able to exert on the tires? Taking the magnitudes of both sides of Newton’s second law gives |F| = |ma| = m|a|
.
Substituting |a| = |v|2 /r gives |F| = m|v|2 /r ≈ 80 N (rounded off to one sig fig). Don’t hug the center line on a curve! example 5 You’re driving on a mountain road with a steep drop on your right. When making a left turn, is it safer to hug the center line or to stay closer to the outside of the road? You want whichever choice involves the least acceleration, because that will require the least force and entail the least risk of exceeding the maximum force of static friction. Assuming the curve is an arc of a circle and your speed is constant, your car is performing uniform circular motion, with |a| = |v|2 /r . The dependence on the square of the speed shows that driving slowly is the main safety measure you can take, but for any given speed you also want to have the largest possible value of r . Even though your instinct is to keep away from that scary precipice, you are actually less likely to skid if you keep toward the outside, because then you are describing a larger circle. Acceleration related to radius and period of rotation example 6 How can the equation for the acceleration in uniform circular motion be rewritten in terms of the radius of the circle and the period, T , of the motion, i.e., the time required to go around once? The period can be related to the speed as follows: circumference |v| = T = 2πr /T . Substituting into the equation |a| = |v|2 /r gives |a| =
4π2 r T2
.
k / Example 7.
Section 9.2
Uniform circular motion
237
A clothes dryer example 7 My clothes dryer has a drum with an inside radius of 35 cm, and it spins at 48 revolutions per minute. What is the acceleration of the clothes inside? We can solve this by finding the period and plugging in to the result of the previous example. If it makes 48 revolutions in one minute, then the period is 1/48 of a minute, or 1.25 s. To get an acceleration in mks units, we must convert the radius to 0.35 m. Plugging in, the result is 8.8 m/s2 . More about clothes dryers! example 8 In a discussion question in the previous section, we made the assumption that the clothes remain against the inside of the drum as they go over the top. In light of the previous example, is this a correct assumption? No. We know that there must be some minimum speed at which the motor can run that will result in the clothes just barely staying against the inside of the drum as they go over the top. If the clothes dryer ran at just this minimum speed, then there would be no normal force on the clothes at the top: they would be on the verge of losing contact. The only force acting on them at the top would be the force of gravity, which would give them an acceleration of g = 9.8 m/s2 . The actual dryer must be running slower than this minimum speed, because it produces an acceleration of only 8.8 m/s2 . My theory is that this is done intentionally, to make the clothes mix and tumble. Solved problem: The tilt-a-whirl
page 242, problem 6
Solved problem: An off-ramp
page 242, problem 7
Discussion questions A A certain amount of force is needed to provide the acceleration of circular motion. What if were are exerting a force perpendicular to the direction of motion in an attempt to make an object trace a circle of radius r , but the force isn’t as big as m|v|2 /r ? B Suppose a rotating space station, as in figure l on page 239, is built. It gives its occupants the illusion of ordinary gravity. What happens when a person in the station lets go of a ball? What happens when she throws a ball straight “up” in the air (i.e., towards the center)?
238
Chapter 9
Circular motion
l / Discussion question B. An artist’s conception of a rotating space colony in the form of a giant wheel. A person living in this noninertial frame of reference has an illusion of a force pulling her outward, toward the deck, for the same reason that a person in the pickup truck has the illusion of a force pulling the bowling ball. By adjusting the speed of rotation, the designers can make an acceleration |v|2 /r equal to the usual acceleration of gravity on earth. On earth, your acceleration standing on the ground is zero, and a falling rock heads for your feet with an acceleration of 9.8 m/s2 . A person standing on the deck of the space colony has an upward acceleration of 9.8 m/s2 , and when she lets go of a rock, her feet head up at the nonaccelerating rock. To her, it seems the same as true gravity.
9.3 Nonuniform circular motion What about nonuniform circular motion? Although so far we have been discussing components of vectors along fixed x and y axes, it now becomes convenient to discuss components of the acceleration vector along the radial line (in-out) and the tangential line (along the direction of motion). For nonuniform circular motion, the radial component of the acceleration obeys the same equation as for uniform circular motion, ar = v 2 /r
,
where v = |v|, but the acceleration vector also has a tangential component, at = slope of the graph of v versus t
.
The latter quantity has a simple interpretation. If you are going around a curve in your car, and the speedometer needle is moving, the tangential component of the acceleration vector is simply what you would have thought the acceleration was if you saw the speedometer and didn’t know you were going around a curve. Slow down before a turn, not during it. example 9 When you’re making a turn in your car and you’re afraid you may skid, isn’t it a good idea to slow down? If the turn is an arc of a circle, and you’ve already completed part of the turn at constant speed without skidding, then the road and tires are apparently capable of enough static friction to supply an acceleration of |v|2 /r . There is no reason why you would skid out now if you haven’t already. If you get nervous and brake, however, then you need to have a tangential acceleration component in addition to the radial one you were already able to produce successfully. This would require an acceleration vector with a greater magnitude, which in turn would require a larger force. Static friction might not be able to supply that much force, and you might skid out. The safer thing to do is to approach the turn at a comfortably low speed. Solved problem: A bike race
m / 1. Moving in a circle while speeding up. 2. Uniform circular motion. 3. Slowing down.
page 241, problem 5
Section 9.3
Nonuniform circular motion
239
Summary Selected vocabulary uniform circular circular motion in which the magnitude of the motion . . . . . . velocity vector remains constant nonuniform circu- circular motion in which the magnitude of the lar motion . . . . velocity vector changes radial . . . . . . . parallel to the radius of a circle; the in-out direction tangential . . . . tangent to the circle, perpendicular to the radial direction Notation ar . . . . . . . . . at . . . . . . . . .
radial acceleration; the component of the acceleration vector along the in-out direction tangential acceleration; the component of the acceleration vector tangent to the circle
Summary If an object is to have circular motion, a force must be exerted on it toward the center of the circle. There is no outward force on the object; the illusion of an outward force comes from our experiences in which our point of view was rotating, so that we were viewing things in a noninertial frame. An object undergoing uniform circular motion has an inward acceleration vector of magnitude |a| = v 2 /r
,
where v = |v|. In nonuniform circular motion, the radial and tangential components of the acceleration vector are ar = v 2 /r at = slope of the graph of v versus t
240
Chapter 9
Circular motion
.
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 When you’re done using an electric mixer, you can get most of the batter off of the beaters by lifting them out of the batter with the motor running at a high enough speed. Let’s imagine, to make things easier to visualize, that we instead have a piece of tape stuck to one of the beaters. (a) Explain why static friction has no effect on whether or not the tape flies off. (b) Analyze the forces in which the tape participates, using a table the format shown in section 5.3. (c) Suppose you find that the tape doesn’t fly off when the motor is on a low speed, but at a greater speed, the tape won’t stay on. Why would the greater speed change things? [Hint: If you don’t invoke any law of physics, you haven’t explained it.] 2
Problem 1.
Show that the expression |v|2 /r has the units of acceleration.
3 A plane is flown in a loop-the-loop of radius 1.00 km. The plane starts out flying upside-down, straight and level, then begins curving up along the circular loop, and is right-side up when it reaches the top. (The plane may slow down somewhat on the way up.) How fast must the plane be going at the top if the pilot is to experience no force from the seat or the seatbelt while at the top of √ the loop? 4 In this problem, you’ll derive the equation |a| = |v|2 /r using calculus. Instead of comparing velocities at two points in the particle’s motion and then taking a limit where the points are close together, you’ll just take derivatives. The particle’s position vector ˆ and y ˆ are the unit vectors is r = (r cos θ)ˆ x + (r sin θ)ˆ y, where x along the x and y axes. By the definition of radians, the distance traveled since t = 0 is rθ, so if the particle is traveling at constant speed v = |v|, we have v = rθ/t. (a) Eliminate θ to get the particle’s position vector as a function of time. (b) Find the particle’s acceleration vector. (c) Show that the magnitude of the acceleration vector equals v2 /r. 5 Three cyclists in a race are rounding a semicircular curve. At the moment depicted, cyclist A is using her brakes to apply a force of 375 N to her bike. Cyclist B is coasting. Cyclist C is pedaling, resulting in a force of 375 N on her bike Each cyclist, with her bike, has a mass of 75 kg. At the instant shown, the
Problem 5.
Problems
241
instantaneous speed of all three cyclists is 10 m/s. On the diagram, draw each cyclist’s acceleration vector with its tail on top of her present position, indicating the directions and lengths reasonably accurately. Indicate approximately the consistent scale you are using for all three acceleration vectors. Extreme precision is not necessary as long as the directions are approximately right, and lengths of vectors that should be equal appear roughly equal, etc. Assume all three cyclists are traveling along the road all the time, not wandering across their lane or wiping out and going off the road. Solution, p. 534 6 The amusement park ride shown in the figure consists of a cylindrical room that rotates about its vertical axis. When the rotation is fast enough, a person against the wall can pick his or her feet up off the floor and remain “stuck” to the wall without falling. (a) Suppose the rotation results in the person having a speed v. The radius of the cylinder is r, the person’s mass is m, the downward acceleration of gravity is g, and the coefficient of static friction between the person and the wall is μs . Find an equation for the speed, v, required, in terms of the other variables. (You will find that one of the variables cancels out.) (b) Now suppose two people are riding the ride. Huy is wearing denim, and Gina is wearing polyester, so Huy’s coefficient of static friction is three times greater. The ride starts from rest, and as it begins rotating faster and faster, Gina must wait longer before being able to lift her feet without sliding to the floor. Based on your equation from part a, how many times greater must the speed be before Gina can lift her feet without sliding down? Solution, p. 534
Problem 6.
7 An engineer is designing a curved off-ramp for a freeway. Since the off-ramp is curved, she wants to bank it to make it less likely that motorists going too fast will wipe out. If the radius of the curve is r, how great should the banking angle, θ, be so that for a car going at a speed v, no static friction force whatsoever is required to allow the car to make the curve? State your answer in terms of v, r, and g, and show that the mass of the car is irrelevant. Solution, p. 535
Problem 7.
8 Lionel brand toy trains come with sections of track in standard lengths and shapes. For circular arcs, the most commonly used sections have diameters of 662 and 1067 mm at the inside of the outer rail. The maximum speed at which a train can take the broader curve without flying off the tracks is 0.95 m/s. At what speed must √ the train be operated to avoid derailing on the tighter curve? 9 The figure shows a ball on the end of a string of length L attached to a vertical rod which is spun about its vertical axis by a motor. The period (time for one rotation) is P . (a) Analyze the forces in which the ball participates. (b) Find how the angle θ depends on P , g, and L. [Hints: (1)
Problem 9.
242
Chapter 9
Circular motion
Write down Newton’s second law for the vertical and horizontal components of force and acceleration. This gives two equations, which can be solved for the two unknowns, θ and the tension in the string. (2) If you introduce variables like v and r, relate them to the variables your solution is supposed to contain, and eliminate √ them.] (c) What happens mathematically to your solution if the motor is run very slowly (very large values of P )? Physically, what do you think would actually happen in this case? 10 Psychology professor R.O. Dent requests funding for an experiment on compulsive thrill-seeking behavior in guinea pigs, in which the subject is to be attached to the end of a spring and whirled around in a horizontal circle. The spring has relaxed length b, and obeys Hooke’s law with spring constant k. It is stiff enough to keep from bending significantly under the guinea pig’s weight. (a) Calculate the length of the spring when it is undergoing steady circular motion in which one rotation takes a time T . Express your √ result in terms of k, b, T , and the guinea pig’s mass m. (b) The ethics committee somehow fails to veto the experiment, but the safety committee expresses concern. Why? Does your equation do anything unusual, or even spectacular, for any particular value of T ? What do you think is the physical significance of this mathematical behavior?
Problem 10.
11 The figure shows an old-fashioned device called a flyball governor, used for keeping an engine running at the correct speed. The whole thing rotates about the vertical shaft, and the mass M is free to slide up and down. This mass would have a connection (not shown) to a valve that controlled the engine. If, for instance, the engine ran too fast, the mass would rise, causing the engine to slow back down. (a) Show that in the special case of a = 0, the angle θ is given by
θ = cos
−1
g(m + M )P 2 4π 2 mL
, Problem 11.
where P is the period of rotation (time required for one complete rotation). (b) There is no closed-form solution for θ in the general case where a is not zero. However, explain how the undesirable low-speed behavior of the a = 0 device would be improved by making a nonzero. 12 The figure shows two blocks of masses m1 and m2 sliding in circles on a frictionless table. Find the tension in the strings if√the period of rotation (time required for one rotation) is P .
Problem 12.
Problems
243
13 The acceleration of an object in uniform circular motion can be given either by |a| = |v|2 /r or, equivalently, by |a| = 4π 2 r/T 2 , where T is the time required for one cycle (example 6 on page 237). Person A says based on the first equation that the acceleration in circular motion is greater when the circle is smaller. Person B, arguing from the second equation, says that the acceleration is smaller when the circle is smaller. Rewrite the two statements so that they are less misleading, eliminating the supposed paradox. [Based on a problem by Arnold Arons.] 14 The bright star Sirius has a mass of 4.02 × 1030 kg and lies at a distance of 8.1 × 1016 m from our solar system. Suppose you’re standing on a merry-go-round carousel rotating with a period of 10 seconds, and Sirius is on the horizon. You adopt a rotating, noninertial frame of reference, in which the carousel is at rest, and the universe is spinning around it. If you drop a corndog, you see it accelerate horizontally away from the axis, and you interpret this is the result of some horizontal force. This force does not actually exist; it only seems to exist because you’re insisting on using a noninertial frame. Similarly, calculate the force that seems to act on Sirius in this frame of reference. Comment on the physical plausibility of√this force, and on what object could be exerting it. 15 In a well known stunt from circuses and carnivals, a motorcyclist rides around inside a big bowl, gradually speeding up and rising higher. Eventually the cyclist can get up to where the walls of the bowl are vertical. Let’s estimate the conditions under which a running human could do the same thing. (a) If the runner can run at speed v, and her shoes have a coefficient √ of static friction μs , what is the maximum radius of the circle? (b) Show that the units of your answer make sense. (c) Check that its dependence on the variables makes sense. (d) Evaluate your result numerically for v = 10 m/s (the speed of an olympic sprinter) and μs = 5. (This is roughly the highest coefficient of static friction ever achieved for surfaces that are not sticky. The surface has an array of microscopic fibers like a hair brush, and is inspired by the hairs on the feet of a gecko. These assumptions are not necessarily realistic, since the person would have to run at √ an angle, which would be physically awkward.)
Problem 15.
244
Chapter 9
Circular motion
Gravity is the only really important force on the cosmic scale. This falsecolor representation of saturn’s rings was made from an image sent back by the Voyager 2 space probe. The rings are composed of innumerable tiny ice particles orbiting in circles under the influence of saturn’s gravity.
Chapter 10
Gravity Cruise your radio dial today and try to find any popular song that would have been imaginable without Louis Armstrong. By introducing solo improvisation into jazz, Armstrong took apart the jigsaw puzzle of popular music and fit the pieces back together in a different way. In the same way, Newton reassembled our view of the universe. Consider the titles of some recent physics books written for the general reader: The God Particle, Dreams of a Final Theory. Without Newton, such attempts at universal understanding would not merely have seemed a little pretentious, they simply would not have occurred to anyone.
a / Johannes Kepler found a mathematical description of the motion of the planets, which led to Newton’s theory of gravity.
This chapter is about Newton’s theory of gravity, which he used to explain the motion of the planets as they orbited the sun. Whereas this book has concentrated on Newton’s laws of motion, leaving gravity as a dessert, Newton tosses off the laws of motion in the first 20 pages of the Principia Mathematica and then spends the next 130 discussing the motion of the planets. Clearly he saw this as the crucial scientific focus of his work. Why? Because in it he
245
showed that the same laws of motion applied to the heavens as to the earth, and that the gravitational force that made an apple fall was the same as the force that kept the earth’s motion from carrying it away from the sun. What was radical about Newton was not his laws of motion but his concept of a universal science of physics.
10.1 Kepler’s laws
b / Tycho Brahe made his name as an astronomer by showing that the bright new star, today called a supernova, that appeared in the skies in 1572 was far beyond the Earth’s atmosphere. This, along with Galileo’s discovery of sunspots, showed that contrary to Aristotle, the heavens were not perfect and unchanging. Brahe’s fame as an astronomer brought him patronage from King Frederick II, allowing him to carry out his historic high-precision measurements of the planets’ motions. A contradictory character, Brahe enjoyed lecturing other nobles about the evils of dueling, but had lost his own nose in a youthful duel and had it replaced with a prosthesis made of an alloy of gold and silver. Willing to endure scandal in order to marry a peasant, he nevertheless used the feudal powers given to him by the king to impose harsh forced labor on the inhabitants of his parishes. The result of their work, an Italian-style palace with an observatory on top, surely ranks as one of the most luxurious science labs ever built. Kepler described Brahe as dying of a ruptured bladder after falling from a wagon on the way home from a party, but other contemporary accounts and modern medical analysis suggest mercury poisoning, possibly as a result of court intrigue.
246
Chapter 10
Newton wouldn’t have been able to figure out why the planets move the way they do if it hadn’t been for the astronomer Tycho Brahe (1546-1601) and his protege Johannes Kepler (1571-1630), who together came up with the first simple and accurate description of how the planets actually do move. The difficulty of their task is suggested by figure c, which shows how the relatively simple orbital motions of the earth and Mars combine so that as seen from earth Mars appears to be staggering in loops like a drunken sailor.
c / As the Earth and Mars revolve around the sun at different rates, the combined effect of their motions makes Mars appear to trace a strange, looped path across the background of the distant stars.
Brahe, the last of the great naked-eye astronomers, collected extensive data on the motions of the planets over a period of many years, taking the giant step from the previous observations’ accuracy of about 10 minutes of arc (10/60 of a degree) to an unprecedented 1 minute. The quality of his work is all the more remarkable consid-
Gravity
ering that his observatory consisted of four giant brass protractors mounted upright in his castle in Denmark. Four different observers would simultaneously measure the position of a planet in order to check for mistakes and reduce random errors. With Brahe’s death, it fell to his former assistant Kepler to try to make some sense out of the volumes of data. Kepler, in contradiction to his late boss, had formed a prejudice, a correct one as it turned out, in favor of the theory that the earth and planets revolved around the sun, rather than the earth staying fixed and everything rotating about it. Although motion is relative, it is not just a matter of opinion what circles what. The earth’s rotation and revolution about the sun make it a noninertial reference frame, which causes detectable violations of Newton’s laws when one attempts to describe sufficiently precise experiments in the earth-fixed frame. Although such direct experiments were not carried out until the 19th century, what convinced everyone of the sun-centered system in the 17th century was that Kepler was able to come up with a surprisingly simple set of mathematical and geometrical rules for describing the planets’ motion using the sun-centered assumption. After 900 pages of calculations and many false starts and dead-end ideas, Kepler finally synthesized the data into the following three laws: Kepler’s elliptical orbit law The planets orbit the sun in elliptical orbits with the sun at one focus. Kepler’s equal-area law The line connecting a planet to the sun sweeps out equal areas in equal amounts of time. Kepler’s law of periods The time required for a planet to orbit the sun, called its period, is proportional to the long axis of the ellipse raised to the 3/2 power. The constant of proportionality is the same for all the planets. Although the planets’ orbits are ellipses rather than circles, most are very close to being circular. The earth’s orbit, for instance, is only flattened by 1.7% relative to a circle. In the special case of a planet in a circular orbit, the two foci (plural of “focus”) coincide at the center of the circle, and Kepler’s elliptical orbit law thus says that the circle is centered on the sun. The equal-area law implies that a planet in a circular orbit moves around the sun with constant speed. For a circular orbit, the law of periods then amounts to a statement that the time for one orbit is proportional to r3/2 , where r is the radius. If all the planets were moving in their orbits at the same speed, then the time for one orbit would simply depend on the circumference of the circle, so it would only be proportional to r to the first power. The more drastic dependence on r3/2 means
Section 10.1
Kepler’s laws
247
that the outer planets must be moving more slowly than the inner planets.
d / An ellipse is a circle that has been distorted by shrinking and stretching along perpendicular axes.
e / An ellipse can be constructed by tying a string to two pins and drawing like this with the pencil stretching the string taut. Each pin constitutes one focus of the ellipse.
10.2 Newton’s law of gravity The sun’s force on the planets obeys an inverse square law. Kepler’s laws were a beautifully simple explanation of what the planets did, but they didn’t address why they moved as they did. Did the sun exert a force that pulled a planet toward the center of its orbit, or, as suggested by Descartes, were the planets circulating in a whirlpool of some unknown liquid? Kepler, working in the Aristotelian tradition, hypothesized not just an inward force exerted by the sun on the planet, but also a second force in the direction of motion to keep the planet from slowing down. Some speculated that the sun attracted the planets magnetically. Once Newton had formulated his laws of motion and taught them to some of his friends, they began trying to connect them to Kepler’s laws. It was clear now that an inward force would be needed to bend the planets’ paths. This force was presumably an attraction between the sun and each planet. (Although the sun does accelerate in response to the attractions of the planets, its mass is so great that the effect had never been detected by the prenewtonian astronomers.) Since the outer planets were moving slowly along more gently curving paths than the inner planets, their accelerations were apparently less. This could be explained if the sun’s force was determined by distance, becoming weaker for the farther planets. Physicists were also familiar with the noncontact forces of electricity and magnetism, and knew that they fell off rapidly with distance, so this made sense. In the approximation of a circular orbit, the magnitude of the sun’s force on the planet would have to be [1]
f / If the time interval taken by the planet to move from P to Q is equal to the time interval from R to S, then according to Kepler’s equal-area law, the two shaded areas are equal. The planet is moving faster during interval RS than it did during PQ, which Newton later determined was due to the sun’s gravitational force accelerating it. The equal-area law predicts exactly how much it will speed up.
248
Chapter 10
F = ma = mv 2 /r
.
Now although this equation has the magnitude, v, of the velocity vector in it, what Newton expected was that there would be a more fundamental underlying equation for the force of the sun on a planet, and that that equation would involve the distance, r, from the sun to the object, but not the object’s speed, v — motion doesn’t make objects lighter or heavier. self-check A If eq. [1] really was generally applicable, what would happen to an object released at rest in some empty region of the solar system? Answer, p. 543
Equation [1] was thus a useful piece of information which could be related to the data on the planets simply because the planets happened to be going in nearly circular orbits, but Newton wanted
Gravity
to combine it with other equations and eliminate v algebraically in order to find a deeper truth. To eliminate v, Newton used the equation [2]
v=
circumference 2πr = T T
.
Of course this equation would also only be valid for planets in nearly circular orbits. Plugging this into eq. [1] to eliminate v gives [3]
F =
4π 2 mr T2
.
This unfortunately has the side-effect of bringing in the period, T , which we expect on similar physical grounds will not occur in the final answer. That’s where the circular-orbit case, T ∝ r3/2 , of Kepler’s law of periods comes in. Using it to eliminate T gives a result that depends only on the mass of the planet and its distance from the sun: F ∝ m/r2
.
[force of the sun on a planet of mass m at a distance r from the sun; same proportionality constant for all the planets]
(Since Kepler’s law of periods is only a proportionality, the final result is a proportionality rather than an equation, so there is no point in hanging on to the factor of 4π 2 .) As an example, the “twin planets” Uranus and Neptune have nearly the same mass, but Neptune is about twice as far from the sun as Uranus, so the sun’s gravitational force on Neptune is about four times smaller. self-check B Fill in the steps leading from equation [3] to F ∝ m/r 2 . 543
Answer, p.
The forces between heavenly bodies are the same type of force as terrestrial gravity. OK, but what kind of force was it? It probably wasn’t magnetic, since magnetic forces have nothing to do with mass. Then came Newton’s great insight. Lying under an apple tree and looking up at the moon in the sky, he saw an apple fall. Might not the earth also attract the moon with the same kind of gravitational force? The moon orbits the earth in the same way that the planets orbit the sun, so maybe the earth’s force on the falling apple, the earth’s force on the moon, and the sun’s force on a planet were all the same type of force.
g / The moon’s acceleration is 602 = 3600 times smaller than the apple’s.
There was an easy way to test this hypothesis numerically. If it was true, then we would expect the gravitational forces exerted by
Section 10.2
Newton’s law of gravity
249
the earth to follow the same F ∝ m/r2 rule as the forces exerted by the sun, but with a different constant of proportionality appropriate to the earth’s gravitational strength. The issue arises now of how to define the distance, r, between the earth and the apple. An apple in England is closer to some parts of the earth than to others, but suppose we take r to be the distance from the center of the earth to the apple, i.e., the radius of the earth. (The issue of how to measure r did not arise in the analysis of the planets’ motions because the sun and planets are so small compared to the distances separating them.) Calling the proportionality constant k, we have 2 Fearth on apple = k mapple /rearth
Fearth on moon = k mmoon /d2earth-moon
.
Newton’s second law says a = F/m, so 2 aapple = k / rearth
amoon = k / d2earth-moon
.
The Greek astronomer Hipparchus had already found 2000 years before that the distance from the earth to the moon was about 60 times the radius of the earth, so if Newton’s hypothesis was right, the acceleration of the moon would have to be 602 = 3600 times less than the acceleration of the falling apple. Applying a = v 2 /r to the acceleration of the moon yielded an acceleration that was indeed 3600 times smaller than 9.8 m/s2 , and Newton was convinced he had unlocked the secret of the mysterious force that kept the moon and planets in their orbits. Newton’s law of gravity The proportionality F ∝ m/r2 for the gravitational force on an object of mass m only has a consistent proportionality constant for various objects if they are being acted on by the gravity of the same object. Clearly the sun’s gravitational strength is far greater than the earth’s, since the planets all orbit the sun and do not exhibit any very large accelerations caused by the earth (or by one another). What property of the sun gives it its great gravitational strength? Its great volume? Its great mass? Its great temperature? Newton reasoned that if the force was proportional to the mass of the object being acted on, then it would also make sense if the determining factor in the gravitational strength of the object exerting the force was its own mass. Assuming there were no other factors affecting the gravitational force, then the only other thing needed to make quantitative predictions of gravitational forces would be a proportionality constant. Newton called that proportionality constant G, so here is the complete form of the law of gravity he hypothesized.
250
Chapter 10
Gravity
Newton’s law of gravity
F =
Gm1 m2 r2
[gravitational force between objects of mass m1 and m2 , separated by a distance r; r is not the radius of anything ]
Newton conceived of gravity as an attraction between any two masses in the universe. The constant G tells us how many newtons the attractive force is for two 1-kg masses separated by a distance of 1 m. The experimental determination of G in ordinary units (as opposed to the special, nonmetric, units used in astronomy) is described in section 10.5. This difficult measurement was not accomplished until long after Newton’s death. The units of G What are the units of G?
h / Students often have a hard time understanding the physical meaning of G. It’s just a proportionality constant that tells you how strong gravitational forces are. If you could change it, all the gravitational forces all over the universe would get stronger or weaker. Numerically, the gravitational attraction between two 1-kg masses separated by a distance of 1 m is 6.67 × 10−11 N, and this is what G is in SI units.
example 1
Solving for G in Newton’s law of gravity gives
G=
Fr 2 m1 m2
,
so the units of G must be N·m2 /kg2 . Fully adorned with units, the value of G is 6.67 × 10−11 N·m2 /kg2 . Newton’s third law example 2 Is Newton’s law of gravity consistent with Newton’s third law? The third law requires two things. First, m1 ’s force on m2 should be the same as m2 ’s force on m1 . This works out, because the product m1 m2 gives the same result if we interchange the labels 1 and 2. Second, the forces should be in opposite directions. This condition is also satisfied, because Newton’s law of gravity refers to an attraction: each mass pulls the other toward itself. Pluto and Charon example 3 Pluto’s moon Charon is unusually large considering Pluto’s size, giving them the character of a double planet. Their masses are 1.25×1022 and 1.9x1021 kg, and their average distance from one another is 1.96 × 104 km. What is the gravitational force between them? If we want to use the value of G expressed in SI (meter-kilogramsecond) units, we first have to convert the distance to 1.96 ×
Section 10.2
i / Example 3. Computerenhanced images of Pluto and Charon, taken by the Hubble Space Telescope.
Newton’s law of gravity
251
107 m. The force is
6.67 × 10−11 N·m2 /kg2 1.25 × 1022 kg 1.9 × 1021 kg 2 1.96 × 107 m = 4.1 × 1018 N The proportionality to 1/r2 in Newton’s law of gravity was not entirely unexpected. Proportionalities to 1/r2 are found in many other phenomena in which some effect spreads out from a point. For instance, the intensity of the light from a candle is proportional to 1/r2 , because at a distance r from the candle, the light has to be spread out over the surface of an imaginary sphere of area 4πr2 . The same is true for the intensity of sound from a firecracker, or the intensity of gamma radiation emitted by the Chernobyl reactor. It’s important, however, to realize that this is only an analogy. Force does not travel through space as sound or light does, and force is not a substance that can be spread thicker or thinner like butter on toast.
j / The conic sections are the curves made by cutting the surface of an infinite cone with a plane.
k / An imaginary cannon able to shoot cannonballs at very high speeds is placed on top of an imaginary, very tall mountain that reaches up above the atmosphere. Depending on the speed at which the ball is fired, it may end up in a tightly curved elliptical orbit, 1, a circular orbit, 2, a bigger elliptical orbit, 3, or a nearly straight hyperbolic orbit, 4.
252
Chapter 10
Although several of Newton’s contemporaries had speculated that the force of gravity might be proportional to 1/r2 , none of them, even the ones who had learned Newton’s laws of motion, had had any luck proving that the resulting orbits would be ellipses, as Kepler had found empirically. Newton did succeed in proving that elliptical orbits would result from a 1/r2 force, but we postpone the proof until the chapter 15 because it can be accomplished much more easily using the concepts of energy and angular momentum. Newton also predicted that orbits in the shape of hyperbolas should be possible, and he was right. Some comets, for instance, orbit the sun in very elongated ellipses, but others pass through the solar system on hyperbolic paths, never to return. Just as the trajectory of a faster baseball pitch is flatter than that of a more slowly thrown ball, so the curvature of a planet’s orbit depends on its speed. A spacecraft can be launched at relatively low speed, resulting in a circular orbit about the earth, or it can be launched at a higher speed, giving a more gently curved ellipse that reaches farther from the earth, or it can be launched at a very high speed which puts it in an even less curved hyperbolic orbit. As you go very far out on a hyperbola, it approaches a straight line, i.e., its curvature eventually becomes nearly zero. Newton also was able to prove that Kepler’s second law (sweeping out equal areas in equal time intervals) was a logical consequence of his law of gravity. Newton’s version of the proof is moderately complicated, but the proof becomes trivial once you understand the concept of angular momentum, which will be covered later in the course. The proof will therefore be deferred until section 15.7.
Gravity
self-check C Which of Kepler’s laws would it make sense to apply to hyperbolic orbits? Answer, p. 543 Solved problem: Visiting Ceres
page 266, problem 10
Solved problem: Geosynchronous orbit
page 268, problem 16
Solved problem: Why a equals g
page 267, problem 11
Solved problem: Ida and Dactyl
page 267, problem 12
Solved problem: Another solar system
page 267, problem 15
Solved problem: Weight loss
page 269, problem 19
Solved problem: The receding moon
page 268, problem 17
Discussion questions A How could Newton find the speed of the moon to plug in to a = v 2 /r ? B Two projectiles of different mass shot out of guns on the surface of the earth at the same speed and angle will follow the same trajectories, assuming that air friction is negligible. (You can verify this by throwing two objects together from your hand and seeing if they separate or stay side by side.) What corresponding fact would be true for satellites of the earth having different masses? C What is wrong with the following statement? “A comet in an elliptical orbit speeds up as it approaches the sun, because the sun’s force on it is increasing.” D Why would it not make sense to expect the earth’s gravitational force on a bowling ball to be inversely proportional to the square of the distance between their surfaces rather than their centers? E Does the earth accelerate as a result of the moon’s gravitational force on it? Suppose two planets were bound to each other gravitationally the way the earth and moon are, but the two planets had equal masses. What would their motion be like? F Spacecraft normally operate by firing their engines only for a few minutes at a time, and an interplanetary probe will spend months or years on its way to its destination without thrust. Suppose a spacecraft is in a circular orbit around Mars, and it then briefly fires its engines in reverse, causing a sudden decrease in speed. What will this do to its orbit? What about a forward thrust?
Section 10.2
Newton’s law of gravity
253
10.3 Apparent weightlessness If you ask somebody at the bus stop why astronauts are weightless, you’ll probably get one of the following two incorrect answers: (1) They’re weightless because they’re so far from the earth. (2) They’re weightless because they’re moving so fast. The first answer is wrong, because the vast majority of astronauts never get more than a thousand miles from the earth’s surface. The reduction in gravity caused by their altitude is significant, but not 100%. The second answer is wrong because Newton’s law of gravity only depends on distance, not speed. The correct answer is that astronauts in orbit around the earth are not really weightless at all. Their weightlessness is only apparent. If there was no gravitational force on the spaceship, it would obey Newton’s first law and move off on a straight line, rather than orbiting the earth. Likewise, the astronauts inside the spaceship are in orbit just like the spaceship itself, with the earth’s gravitational force continually twisting their velocity vectors around. The reason they appear to be weightless is that they are in the same orbit as the spaceship, so although the earth’s gravity curves their trajectory down toward the deck, the deck drops out from under them at the same rate. Apparent weightlessness can also be experienced on earth. Any time you jump up in the air, you experience the same kind of apparent weightlessness that the astronauts do. While in the air, you can lift your arms more easily than normal, because gravity does not make them fall any faster than the rest of your body, which is falling out from under them. The Russian air force now takes rich foreign tourists up in a big cargo plane and gives them the feeling of weightlessness for a short period of time while the plane is nose-down and dropping like a rock.
10.4 Vector addition of gravitational forces Pick a flower on earth and you move the farthest star. Paul Dirac When you stand on the ground, which part of the earth is pulling down on you with its gravitational force? Most people are tempted to say that the effect only comes from the part directly under you, since gravity always pulls straight down. Here are three observations that might help to change your mind: • If you jump up in the air, gravity does not stop affecting you just because you are not touching the earth: gravity is a noncontact force. That means you are not immune from the grav-
254
Chapter 10
Gravity
ity of distant parts of our planet just because you are not touching them. • Gravitational effects are not blocked by intervening matter. For instance, in an eclipse of the moon, the earth is lined up directly between the sun and the moon, but only the sun’s light is blocked from reaching the moon, not its gravitational force — if the sun’s gravitational force on the moon was blocked in this situation, astronomers would be able to tell because the moon’s acceleration would change suddenly. A more subtle but more easily observable example is that the tides are caused by the moon’s gravity, and tidal effects can occur on the side of the earth facing away from the moon. Thus, far-off parts of the earth are not prevented from attracting you with their gravity just because there is other stuff between you and them. • Prospectors sometimes search for underground deposits of dense minerals by measuring the direction of the local gravitational forces, i.e., the direction things fall or the direction a plumb bob hangs. For instance, the gravitational forces in the region to the west of such a deposit would point along a line slightly to the east of the earth’s center. Just because the total gravitational force on you points down, that doesn’t mean that only the parts of the earth directly below you are attracting you. It’s just that the sideways components of all the force vectors acting on you come very close to canceling out. A cubic centimeter of lava in the earth’s mantle, a grain of silica inside Mt. Kilimanjaro, and a flea on a cat in Paris are all attracting you with their gravity. What you feel is the vector sum of all the gravitational forces exerted by all the atoms of our planet, and for that matter by all the atoms in the universe. When Newton tested his theory of gravity by comparing the orbital acceleration of the moon to the acceleration of a falling apple on earth, he assumed he could compute the earth’s force on the apple using the distance from the apple to the earth’s center. Was he wrong? After all, it isn’t just the earth’s center attracting the apple, it’s the whole earth. A kilogram of dirt a few feet under his backyard in England would have a much greater force on the apple than a kilogram of molten rock deep under Australia, thousands of miles away. There’s really no obvious reason why the force should come out right if you just pretend that the earth’s whole mass is concentrated at its center. Also, we know that the earth has some parts that are more dense, and some parts that are less dense. The solid crust, on which we live, is considerably less dense than the molten rock on which it floats. By all rights, the computation of the vector sum of all the forces exerted by all the earth’s parts should be a horrendous mess.
Section 10.4
l / Gravity only appears to pull straight down because the near perfect symmetry of the earth makes the sideways components of the total force on an object cancel almost exactly. If the symmetry is broken, e.g., by a dense mineral deposit, the total force is a little off to the side.
Vector addition of gravitational forces
255
m / A, who is outside a spherical shell of mass, feels gravitational forces from every part of the shell — stronger forces from the closer parts, and weaker ones from the parts farther away. The shell theorem states that the vector sum of all the forces is the same as if all the mass had been concentrated at the center of the shell. B, at the center, is clearly weightless, because the shell’s gravitational forces cancel out. Surprisingly, C also feels exactly zero gravitational force.
n / The asteroid Toutatis, imaged by the space probe Chang’e-2 in 2012, is shaped like a bowling pin.
Actually, Newton had sound reasons for treating the earth’s mass as if it was concentrated at its center. First, although Newton no doubt suspected the earth’s density was nonuniform, he knew that the direction of its total gravitational force was very nearly toward the earth’s center. That was strong evidence that the distribution of mass was very symmetric, so that we can think of the earth as being made of layers, like an onion, with each layer having constant density throughout. (Today there is further evidence for symmetry based on measurements of how the vibrations from earthquakes and nuclear explosions travel through the earth.) He then considered the gravitational forces exerted by a single such thin shell, and proved the following theorem, known as the shell theorem: If an object lies outside a thin, spherical shell of mass, then the vector sum of all the gravitational forces exerted by all the parts of the shell is the same as if the shell’s mass had been concentrated at its center. If the object lies inside the shell, then all the gravitational forces cancel out exactly. For terrestrial gravity, each shell acts as though its mass was at the center, so the result is the same as if the whole mass was there. The second part of the shell theorem, about the gravitational forces canceling inside the shell, is a little surprising. Obviously the forces would all cancel out if you were at the exact center of a shell, but it’s not at all obvious that they should still cancel out perfectly if you are inside the shell but off-center. The whole idea might seem academic, since we don’t know of any hollow planets in our solar system that astronauts could hope to visit, but actually it’s a useful result for understanding gravity within the earth, which is an important issue in geology. It doesn’t matter that the earth is not actually hollow. In a mine shaft at a depth of, say, 2 km, we can use the shell theorem to tell us that the outermost 2 km of the earth has no net gravitational effect, and the gravitational force is the same as what would be produced if the remaining, deeper, parts of the earth were all concentrated at its center. The shell theorem doesn’t apply to things that aren’t spherical. At the point marked with a dot in figure n, we might imagine that gravity was in the direction shown by the dashed arrow, pointing toward the asteroid’s center of mass, so that the surface would be a vertical cliff almost a kilometer tall. In reality, calculations based on the assumption of uniform density show that the direction of the gravitational field is approximately as shown by the solid arrow, making the slope only about 60◦ .1 This happens because gravity at this location is more strongly affected by the nearby “neck” than by the more distant “belly.” This slope is still believed to be too steep to keep dirt and rocks from sliding off (see problem 11, p. 222). 1
256
Chapter 10
Gravity
Hudson et al., Icarus 161 (2003) 346
self-check D Suppose you’re at the bottom of a deep mineshaft, which means you’re still quite far from the center of the earth. The shell theorem says that the shell of mass you’ve gone inside exerts zero total force on you. Discuss which parts of the shell are attracting you in which directions, and how strong these forces are. Explain why it’s at least plausible that they cancel. Answer, p. 543
Discussion questions A If you hold an apple, does the apple exert a gravitational force on the earth? Is it much weaker than the earth’s gravitational force on the apple? Why doesn’t the earth seem to accelerate upward when you drop the apple? B When astronauts travel from the earth to the moon, how does the gravitational force on them change as they progress? C How would the gravity in the first-floor lobby of a massive skyscraper compare with the gravity in an open field outside of the city? D In a few billion years, the sun will start undergoing changes that will eventually result in its puffing up into a red giant star. (Near the beginning of this process, the earth’s oceans will boil off, and by the end, the sun will probably swallow the earth completely.) As the sun’s surface starts to get closer and close to the earth, how will the earth’s orbit be affected?
10.5 Weighing the earth Let’s look more closely at the application of Newton’s law of gravity to objects on the earth’s surface. Since the earth’s gravitational force is the same as if its mass was all concentrated at its center, the force on a falling object of mass m is given by 2 F = G Mearth m / rearth
.
The object’s acceleration equals F/m, so the object’s mass cancels out and we get the same acceleration for all falling objects, as we knew we should: 2 g = G Mearth / rearth
.
Newton knew neither the mass of the earth nor a numerical value for the constant G. But if someone could measure G, then it would be possible for the first time in history to determine the mass of the earth! The only way to measure G is to measure the gravitational force between two objects of known mass, but that’s an exceedingly difficult task, because the force between any two objects of ordinary
Section 10.5
Weighing the earth
257
o / Cavendish’s apparatus. The two large balls are fixed in place, but the rod from which the two small balls hang is free to twist under the influence of the gravitational forces.
p/A simplified version Cavendish’s apparatus.
of
size is extremely small. The English physicist Henry Cavendish was the first to succeed, using the apparatus shown in figures o and p. The two larger balls were lead spheres 8 inches in diameter, and each one attracted the small ball near it. The two small balls hung from the ends of a horizontal rod, which itself hung by a thin thread. The frame from which the larger balls hung could be rotated by hand about a vertical axis, so that for instance the large ball on the right would pull its neighboring small ball toward us and while the small ball on the left would be pulled away from us. The thread from which the small balls hung would thus be twisted through a small angle, and by calibrating the twist of the thread with known forces, the actual gravitational force could be determined. Cavendish set up the whole apparatus in a room of his house, nailing all the doors shut to keep air currents from disturbing the delicate apparatus. The results had to be observed through telescopes stuck through holes drilled in the walls. Cavendish’s experiment provided the first numerical values for G and for the mass of the earth. The presently accepted value of G is 6.67 × 10−11 N·m2 /kg2 . Knowing G not only allowed the determination of the earth’s mass but also those of the sun and the other planets. For instance, by observing the acceleration of one of Jupiter’s moons, we can infer the mass of Jupiter. The following table gives the distances of the planets from the sun and the masses of the sun and planets. (Other data are given in the back of the book.)
258
Chapter 10
Gravity
sun mercury venus earth mars jupiter saturn uranus neptune pluto
average distance from the sun, in units of the earth’s average distance from the sun — 0.38 0.72 1 1.5 5.2 9.5 19 30 39
mass, in units of the earth’s mass
330,000 0.056 0.82 1 0.11 320 95 14 17 0.002
Discussion questions A It would have been difficult for Cavendish to start designing an experiment without at least some idea of the order of magnitude of G. How could he estimate it in advance to within a factor of 10? B Fill in the details of how one would determine Jupiter’s mass by observing the acceleration of one of its moons. Why is it only necessary to know the acceleration of the moon, not the actual force acting on it? Why don’t we need to know the mass of the moon? What about a planet that has no moons, such as Venus — how could its mass be found?
10.6 Dark energy Until recently, physicists thought they understood gravity fairly well. Einstein had modified Newton’s theory, but certain characteristrics of gravitational forces were firmly established. For one thing, they were always attractive. If gravity always attracts, then it is logical to ask why the universe doesn’t collapse. Newton had answered this question by saying that if the universe was infinite in all directions, then it would have no geometric center toward which it would collapse; the forces on any particular star or planet exerted by distant parts of the universe would tend to cancel out by symmetry. More careful calculations, however, show that Newton’s universe would have a tendency to collapse on smaller scales: any part of the universe that happened to be slightly more dense than average would contract further, and this contraction would result in stronger gravitational forces, which would cause even more rapid contraction, and so on. When Einstein overhauled gravity, the same problem reared its ugly head. Like Newton, Einstein was predisposed to believe in a universe that was static, so he added a special repulsive term to his equations, intended to prevent a collapse. This term was not associated with any interaction of mass with mass, but represented merely an overall tendency for space itself to expand unless restrained by
Section 10.6
Dark energy
259
the matter that inhabited it. It turns out that Einstein’s solution, like Newton’s, is unstable. Furthermore, it was soon discovered observationally that the universe was expanding, and this was interpreted by creating the Big Bang model, in which the universe’s current expansion is the aftermath of a fantastically hot explosion.2 An expanding universe, unlike a static one, was capable of being explained with Einstein’s equations, without any repulsion term. The universe’s expansion would simply slow down over time due to the attractive gravitational forces. After these developments, Einstein said woefully that adding the repulsive term, known as the cosmological constant, had been the greatest blunder of his life. q / The WMAP probe’s map of the cosmic microwave background is like a “baby picture” of the universe.
This was the state of things until 1999, when evidence began to turn up that the universe’s expansion has been speeding up rather than slowing down! The first evidence came from using a telescope as a sort of time machine: light from a distant galaxy may have taken billions of years to reach us, so we are seeing it as it was far in the past. Looking back in time, astronomers saw the universe expanding at speeds that were lower, rather than higher. At first they were mortified, since this was exactly the opposite of what had been expected. The statistical quality of the data was also not good enough to constitute ironclad proof, and there were worries about systematic errors. The case for an accelerating expansion has however been supported by high-precision mapping of the dim, skywide afterglow of the Big Bang, known as the cosmic microwave background. This is discussed in more detail in section 27.4. So now Einstein’s “greatest blunder” has been resurrected. Since we don’t actually know whether or not this self-repulsion of space has a constant strength, the term “cosmological constant” has lost currency. Nowadays physicists usually refer to the phenomenon as “dark energy.” Picking an impressive-sounding name for it should not obscure the fact that we know absolutely nothing about the nature of the effect or why it exists. Dark energy is discussed in more detail on p. 27.4.4. 2
260
Chapter 10
Gravity
Section section 19.5 presents some of the evidence for the Big Bang.
10.7 A gravitational test of Newton’s first law This section describes a high-precision test of Newton’s first law. The left panel of figure r shows a mirror on the moon. By reflecting laser pulses from the mirror, the distance from the earth to the moon has been measured to the phenomenal precision of a few centimeters, or about one part in 1010 . This distance changes for a variety of known reasons. The biggest effect is that the moon’s orbit is not a circle but an ellipse, with its long axis about 11% longer than its short one. A variety of other effects can also be accounted for, including such exotic phenomena as the slightly nonspherical shape of the earth, and the gravitational forces of bodies as small and distant as Pluto. Suppose for simplicity that all these effects had never existed, so that the moon was initially placed in a perfectly circular orbit around the earth, and the earth in a perfectly circular orbit around the sun.
r / Left: The Apollo 11 mission left behind a mirror, which in this photo shows the reflection of the black sky. Right: A highly exaggerated example of an observation that would disprove Newton’s first law. The radius of the moon’s orbit gets bigger and smaller over the course of a year.
If we then observed something like what is shown in the right panel of figure r, Newton’s first law would be disproved. If space itself is symmetrical in all directions, then there is no reason for the moon’s orbit to poof up near the top of the diagram and contract near the bottom. The only possible explanation would be that there was some preferred frame of reference of the type envisioned
Section 10.7
A gravitational test of Newton’s first law
261
by Aristotle, and that our solar system was moving relative to it. Another test for a preferred frame was described in example 3 on p. 233. One could then imagine that the gravitational force of the earth on the moon could be affected by the moon’s motion relative to this frame. The lunar laser ranging data3 contain no measurable effect of the type shown in figure r, so that if the moon’s orbit is distorted in this way (or in a variety of other ways), the distortion must be less than a few centimeters. This constitutes a very strict upper limit on violation of Newton’s first law by gravitational forces. If the first law is violated, and the violation causes a fractional change in gravity that is proportional to the velocity relative to the hypothetical preferred frame, then the change is no more than about one part in 107 , even if the velocity is comparable to the speed of light.
3
262
Chapter 10
Gravity
Battat, Chandler, and Stubbs, http://arxiv.org/abs/0710.0702
Summary Selected vocabulary ellipse . . . . . . . a flattened circle; one of the conic sections conic section . . . a curve formed by the intersection of a plane and an infinite cone hyperbola . . . . another conic section; it does not close back on itself period . . . . . . . the time required for a planet to complete one orbit; more generally, the time for one repetition of some repeating motion focus . . . . . . . one of two special points inside an ellipse: the ellipse consists of all points such that the sum of the distances to the two foci equals a certain number; a hyperbola also has a focus Notation G . . . . . . . . .
the constant of proportionality in Newton’s law of gravity; the gravitational force of attraction between two 1-kg spheres at a centerto-center distance of 1 m
Summary Kepler deduced three empirical laws from data on the motion of the planets: Kepler’s elliptical orbit law: The planets orbit the sun in elliptical orbits with the sun at one focus. Kepler’s equal-area law: The line connecting a planet to the sun sweeps out equal areas in equal amounts of time. Kepler’s law of periods: The time required for a planet to orbit the sun is proportional to the long axis of the ellipse raised to the 3/2 power. The constant of proportionality is the same for all the planets. Newton was able to find a more fundamental explanation for these laws. Newton’s law of gravity states that the magnitude of the attractive force between any two objects in the universe is given by F = Gm1 m2 /r2
.
Weightlessness of objects in orbit around the earth is only apparent. An astronaut inside a spaceship is simply falling along with the spaceship. Since the spaceship is falling out from under the astronaut, it appears as though there was no gravity accelerating the astronaut down toward the deck. Gravitational forces, like all other forces, add like vectors. A gravitational force such as we ordinarily feel is the vector sum of all
Summary
263
the forces exerted by all the parts of the earth. As a consequence of this, Newton proved the shell theorem for gravitational forces: If an object lies outside a thin, uniform shell of mass, then the vector sum of all the gravitational forces exerted by all the parts of the shell is the same as if all the shell’s mass was concentrated at its center. If the object lies inside the shell, then all the gravitational forces cancel out exactly.
264
Chapter 10
Gravity
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 Roy has a mass of 60 kg. Laurie has a mass of 65 kg. They are 1.5 m apart. (a) What is the magnitude of the gravitational force of the earth on Roy? (b) What is the magnitude of Roy’s gravitational force on the earth? (c) What is the magnitude of the gravitational force between Roy and Laurie? (d) What is the magnitude of the gravitational force between Laurie √ and the sun? 2 During a solar eclipse, the moon, earth and sun all lie on the same line, with the moon between the earth and sun. Define your coordinates so that the earth and moon lie at greater x values than the sun. For each force, give the correct sign as well as the magnitude. (a) What force is exerted on the moon by the sun? (b) On the moon by the earth? (c) On the earth by the sun? (d) What total force is exerted on the sun? (e) On the moon? (f) On√the earth? 3 Suppose that on a certain day there is a crescent moon, and you can tell by the shape of the crescent that the earth, sun and moon form a triangle with a 135◦ interior angle at the moon’s corner. What is the magnitude of the total gravitational force of the earth and the sun on the moon? (If you haven’t done problem 2 already, you might want to try it first, since it’s easier, and some of its results √ can be recycled in this problem.)
Problem 3.
4 How high above the Earth’s surface must a rocket be in order to have 1/100 the weight it would have at the surface? Express √ your answer in units of the radius of the Earth. 5 The star Lalande 21185 was found in 1996 to have two planets in roughly circular orbits, with periods of 6 and 30 years. What is √ the ratio of the two planets’ orbital radii? 6 In a Star Trek episode, the Enterprise is in a circular orbit around a planet when something happens to the engines. Spock then tells Kirk that the ship will spiral into the planet’s surface unless they can fix the engines. Is this scientifically correct? Why?
Problems
265
7 (a) Suppose a rotating spherical body such as a planet has a radius r and a uniform density ρ, and the time required for one rotation is T . At the surface of the planet, the apparent acceleration of a falling object is reduced by the acceleration of the ground out from under it. Derive an equation for the apparent acceleration of √ gravity, g, at the equator in terms of r, ρ, T , and G. (b) Applying your equation from a, by what fraction is your apparent weight reduced at the equator compared to the poles, due to the √ Earth’s rotation? (c) Using your equation from a, derive an equation giving the value of T for which the apparent acceleration of gravity becomes zero, i.e., objects can spontaneously drift off the surface of the planet. √ Show that T only depends on ρ, and not on r. (d) Applying your equation from c, how long would a day have to be in order to reduce the apparent weight of objects at the equator of the Earth to zero? [Answer: 1.4 hours] (e) Astronomers have discovered objects they called pulsars, which emit bursts of radiation at regular intervals of less than a second. If a pulsar is to be interpreted as a rotating sphere beaming out a natural “searchlight” that sweeps past the earth with each rotation, use your equation from c to show that its density would have to be much greater than that of ordinary matter. (f) Astrophysicists predicted decades ago that certain stars that used up their sources of energy could collapse, forming a ball of neutrons with the fantastic density of ∼ 1017 kg/m3 . If this is what pulsars really are, use your equation from c to explain why no pulsar has ever been observed that flashes with a period of less than 1 ms or so. 8 You are considering going on a space voyage to Mars, in which your route would be half an ellipse, tangent to the Earth’s orbit at one end and tangent to Mars’ orbit at the other. Your spacecraft’s engines will only be used at the beginning and end, not during the voyage. How long would the outward leg of your trip last? (Assume √ the orbits of Earth and Mars are circular.) 9 (a) If the earth was of uniform density, would your weight be increased or decreased at the bottom of a mine shaft? Explain. (b) In real life, objects weigh slightly more at the bottom of a mine shaft. What does that allow us to infer about the Earth? 10 Ceres, the largest asteroid in our solar system, is a spherical body with a mass 6000 times less than the earth’s, and a radius which is 13 times smaller. If an astronaut who weighs 400 N on earth is visiting the surface of Ceres, what is her weight? Solution, p. 535
Problem 8.
266
Chapter 10
Gravity
11 Prove, based on Newton’s laws of motion and Newton’s law of gravity, that all falling objects have the same acceleration if they are dropped at the same location on the earth and if other forces such as friction are unimportant. Do not just say, “g = 9.8 m/s2 – it’s constant.” You are supposed to be proving that g should be the same number for all objects. Solution, p. 535 12 The figure shows an image from the Galileo space probe taken during its August 1993 flyby of the asteroid Ida. Astronomers were surprised when Galileo detected a smaller object orbiting Ida. This smaller object, the only known satellite of an asteroid in our solar system, was christened Dactyl, after the mythical creatures who lived on Mount Ida, and who protected the infant Zeus. For scale, Ida is about the size and shape of Orange County, and Dactyl the size of a college campus. Galileo was unfortunately unable to measure the time, T , required for Dactyl to orbit Ida. If it had, astronomers would have been able to make the first accurate determination of the mass and density of an asteroid. Find an equation for the density, ρ, of Ida in terms of Ida’s known volume, V , the known radius, r, of Dactyl’s orbit, and the lamentably unknown variable T . (This is the same technique that was used successfully for determining the masses and densities of the planets that have moons.) Solution, p. 535
Problem 12.
13 If a bullet is shot straight up at a high enough velocity, it will never return to the earth. This is known as the escape velocity. We will discuss escape velocity using the concept of energy later in the course, but it can also be gotten at using straightforward calculus. In this problem, you will analyze the motion of an object of mass m whose initial velocity is exactly equal to escape velocity. We assume that it is starting from the surface of a spherically symmetric planet of mass M and radius b. The trick is to guess at the general form of the solution, and then determine the solution in more detail. Assume (as is true) that the solution is of the form r = ktp , where r is the object’s distance from the center of the planet at time t, and k and p are constants. (a) Find the acceleration, and use Newton’s second law and Newton’s law of gravity to determine k and p. You should find that the √ result is independent of m. (b) What happens to the velocity as t approaches infinity? √ (c) Determine escape velocity from the Earth’s surface. 14 Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. For stars orbiting at distances of about 1014 m from the object, the orbital velocities are about 106 m/s. Assuming the orbits are circular, estimate the mass of the object, in units of the mass of the sun, 2 × 1030 kg. If the object was a tightly packed cluster of normal stars, it should be a very bright source of light. Since no visible light is detected coming from it, it is instead believed to be √ a supermassive black hole. Problems
267
15 Astronomers have detected a solar system consisting of three planets orbiting the star Upsilon Andromedae. The planets have been named b, c, and d. Planet b’s average distance from the star is 0.059 A.U., and planet c’s average distance is 0.83 A.U., where an astronomical unit or A.U. is defined as the distance from the Earth to the sun. For technical reasons, it is possible to determine the ratios of the planets’ masses, but their masses cannot presently be determined in absolute units. Planet c’s mass is 3.0 times that of planet b. Compare the star’s average gravitational force on planet c with its average force on planet b. [Based on a problem by Arnold Arons.] Solution, p. 536 16 Some communications satellites are in orbits called geosynchronous: the satellite takes one day to orbit the earth from west to east, so that as the earth spins, the satellite remains above the same point on the equator. What is such a satellite’s altitude above the surface of the earth? Solution, p. 536 17 As is discussed in more detail in example 3 on p. 380, tidal interactions with the earth are causing the moon’s orbit to grow gradually larger. Laser beams bounced off of a mirror left on the moon by astronauts have allowed a measurement of the moon’s rate of recession, which is about 1 cm per year. This means that the gravitational force acting between earth and moon is decreasing. By what fraction does the force decrease with each 27-day orbit? [Based on a problem by Arnold Arons.] Hint, p. 526 Solution, p. 536 18 Suppose that we inhabited a universe in which, instead of √ Newton’s law of gravity, we had F = k m1 m2 /r2 , where k is some constant with different units than G. (The force is still attractive.) However, we assume that a = F/m and the rest of Newtonian physics remains true, and we use a = F/m to define our mass scale, so that, e.g., a mass of 2 kg is one which exhibits half the acceleration when the same force is applied to it as to a 1 kg mass. (a) Is this new law of gravity consistent with Newton’s third law? (b) Suppose you lived in such a universe, and you dropped two unequal masses side by side. What would happen? (c) Numerically, suppose a 1.0-kg object falls with an acceleration of 10 m/s2 . What would be the acceleration of a rain drop with a mass of 0.1 g? Would you want to go out in the rain? (d) If a falling object broke into two unequal pieces while it fell, what would happen? (e) Invent a law of gravity that results in behavior that is the opposite of what you found in part b. [Based on a problem by Arnold Arons.]
268
Chapter 10
Gravity
19 (a) A certain vile alien gangster lives on the surface of an asteroid, where his weight is 0.20 N. He decides he needs to lose weight without reducing his consumption of princesses, so he’s going to move to a different asteroid where his weight will be 0.10 N. The real estate agent’s database has asteroids listed by mass, however, not by surface gravity. Assuming that all asteroids are spherical and have the same density, how should the mass of his new asteroid compare with that of his old one? (b) Jupiter’s mass is 318 times the Earth’s, and its gravity is about twice Earth’s. Is this consistent with the results of part a? If not, how do you explain the discrepancy? Solution, p. 536
20 Where would an object have to be located so that it would experience zero total gravitational force from the earth and moon? √
21 The planet Uranus has a mass of 8.68 × 1025 kg and a radius of 2.56 × 104 km. The figure shows the relative sizes of Uranus and Earth. (a) Compute the ratio gU /gE , where gU is the strength of the gravitational field at the surface of Uranus and gE is the corresponding √ quantity at the surface of the Earth. (b) What is surprising about this result? How do you explain it?
Problem 21.
22 The International Space Station orbits at an average altitude of about 370 km above sea level. Compute the value of g at √that altitude.
Problems
269
Problem 23: New Horizons at its closest approach to Jupiter. (Jupiter’s four largest moons are shown for illustrative purposes.) The masses are: sun: 1.9891 × 1030 kg Jupiter: 1.8986 × 1027 kg New Horizons: 465.0 kg
23 On Feb. 28, 2007, the New Horizons space probe, on its way to a 2015 flyby of Pluto, passed by the planet Jupiter for a gravityassisted maneuver that increased its speed and changed its course. The dashed line in the figure shows the spacecraft’s trajectory, which is curved because of three forces: the force of the exhaust gases from the probe’s own engines, the sun’s gravitational force, and Jupiter’s gravitational force. Find the magnitude of the total gravitational force acting on the probe. You will find that the sun’s force is much smaller than Jupiter’s, so that the magnitude of the total force is determined almost entirely by Jupiter’s force. However, this is a high-precision problem, and you will find that the total force is √ slightly different from Jupiter’s force.
24 On an airless body such as the moon, there is no atmospheric friction, so it should be possible for a satellite to orbit at a very low altitude, just high enough to keep from hitting the mountains. (a) Suppose that such a body is a smooth sphere of uniform density ρ and radius r. Find the velocity required for a ground-skimming √ orbit. (b) A typical asteroid has a density of about 2 g/cm3 , i.e., twice that of water. (This is a lot lower than the density of the earth’s crust, probably indicating that the low gravity is not enough to compact the material very tightly, leaving lots of empty space inside.) Suppose that it is possible for an astronaut in a spacesuit to jump at 2 m/s. Find the radius of the largest asteroid on which it would be √ possible to jump into a ground-skimming orbit.
270
Chapter 10
Gravity
25 The figure shows a region of outer space in which two stars have exploded, leaving behind two overlapping spherical shells of gas, which we assume to remain at rest. The figure is a crosssection in a plane containing the shells’ centers. A space probe is released with a very small initial speed at the point indicated by the arrow, initially moving in the direction indicated by the dashed line. Without any further information, predict as much as possible about the path followed by the probe and its changes in speed along that path.
Problem 25.
26 Approximate the earth’s density as being constant. (a) Find the gravitational field at a point P inside the earth and half-way between the center and the surface. Express your result as a ratio gP /gS relative to the field we experience at the surface. (b) As a check on your answer, make sure that the same reasoning leads to a reasonable result when the fraction 1/2 is replaced by the value 0 (P being the earth’s center) or the value 1 (P being a point on the surface).
Problems
271
27 The earth is divided into solid inner core, a liquid outer core, and a plastic mantle. Physical properties such as density change discontinuously at the boundaries between one layer and the next. Although the density is not completely constant within each region, we will approximate it as being so for the purposes of this problem. (We neglect the crust as well.) Let R be the radius of the earth as a whole and M its mass. The following table gives a model of some properties of the three layers, as determined by methods such as the observation of earthquake waves that have propagated from one side of the planet to the other. region mantle outer core inner core
outer radius/R 1 0.55 0.19
mass/M 0.69 0.29 0.017
The boundary between the mantle and the outer core is referred to as the Gutenberg discontinuity. Let gs be the strength of the earth’s gravitational field at its surface and gG its value at the Gutenberg √ discontinuity. Find gG /gs .
272
Chapter 10
Gravity
Exercise 10: The shell theorem This exercise is an approximate numerical test of the shell theorem. There are seven masses A-G, each being one kilogram. Masses A-F, each one meter from the center, form a shape like two Egyptian pyramids joined at their bases; this is a rough approximation to a six-kilogram spherical shell of mass. Mass G is five meters from the center of the main group. The class will divide into six groups and split up the work required in order to calculate the vector sum of the six gravitational forces exerted on mass G. Depending on the size of the class, more than one group may be assigned to deal with the contribution of the same mass to the total force, and the redundant groups can check each other’s results.
1. Discuss as a class what can be done to simplify the task of calculating the vector sum, and how to organize things so that each group can work in parallel with the others. 2. Each group should write its results on the board in units of piconewtons, retaining five significant figures of precision. Everyone will need to use the same value for the gravitational constant, G = 6.6743 × 10−11 N·m2 /kg2 . 3. The class will determine the vector sum and compare with the result that would be obtained with the shell theorem.
Exercise 10: The shell theorem
273
Conservation laws
274
276
In July of 1994, Comet Shoemaker-Levy struck the planet Jupiter, depositing 7 × 1022 joules of energy, and incidentally giving rise to a series of Hollywood movies in which our own planet is threatened by an impact by a comet or asteroid. There is evidence that such an impact caused the extinction of the dinosaurs. Left: Jupiter’s gravitational force on the near side of the comet was greater than on the far side, and this difference in force tore up the comet into a string of fragments. Two separate telescope images have been combined to create the illusion of a point of view just behind the comet. (The colored fringes at the edges of Jupiter are artifacts of the imaging system.) Top: A series of images of the plume of superheated gas kicked up by the impact of one of the fragments. The plume is about the size of North America. Bottom: An image after all the impacts were over, showing the damage done.
Chapter 11
Conservation of energy 11.1 The search for a perpetual motion Machine Don’t underestimate greed and laziness as forces for progress. Modern chemistry was born from the collision of lust for gold with distaste for the hard work of finding it and digging it up. Failed efforts by generations of alchemists to turn lead into gold led finally to the conclusion that it could not be done: certain substances, the chem-
277
ical elements, are fundamental, and chemical reactions can neither increase nor decrease the amount of an element such as gold. Now flash forward to the early industrial age. Greed and laziness have created the factory, the train, and the ocean liner, but in each of these is a boiler room where someone gets sweaty shoveling the coal to fuel the steam engine. Generations of inventors have tried to create a machine, called a perpetual motion machine, that would run forever without fuel. Such a machine is not forbidden by Newton’s laws of motion, which are built around the concepts of force and inertia. Force is free, and can be multiplied indefinitely with pulleys, gears, or levers. The principle of inertia seems even to encourage the belief that a cleverly constructed machine might not ever run down.
a / The magnet draws the ball to the top of the ramp, where it falls through the hole and rolls back to the bottom.
Figures a and b show two of the innumerable perpetual motion machines that have been proposed. The reason these two examples don’t work is not much different from the reason all the others have failed. Consider machine a. Even if we assume that a properly shaped ramp would keep the ball rolling smoothly through each cycle, friction would always be at work. The designer imagined that the machine would repeat the same motion over and over again, so that every time it reached a given point its speed would be exactly the same as the last time. But because of friction, the speed would actually be reduced a little with each cycle, until finally the ball would no longer be able to make it over the top. Friction has a way of creeping into all moving systems. The rotating earth might seem like a perfect perpetual motion machine, since it is isolated in the vacuum of outer space with nothing to exert frictional forces on it. But in fact our planet’s rotation has slowed drastically since it first formed, and the earth continues to slow its rotation, making today just a little longer than yesterday. The very subtle source of friction is the tides. The moon’s gravity raises bulges in the earth’s oceans, and as the earth rotates the bulges progress around the planet. Where the bulges encounter land, there is friction, which slows the earth’s rotation very gradually.
b / As the wheel spins clockwise, the flexible arms sweep around and bend and unbend. By dropping off its ball on the ramp, the arm is supposed to make itself lighter and easier to lift over the top. Picking its own ball back up again on the right, it helps to pull the right side down.
278
Chapter 11
11.2 Energy The analysis based on friction is somewhat superficial, however. One could understand friction perfectly well and yet imagine the following situation. Astronauts bring back a piece of magnetic ore from the moon which does not behave like ordinary magnets. A normal bar magnet, c/1, attracts a piece of iron essentially directly toward it, and has no left- or right-handedness. The moon rock, however, exerts forces that form a whirlpool pattern around it, 2. NASA goes to a machine shop and has the moon rock put in a lathe and machined down to a smooth cylinder, 3. If we now release a ball bearing on the surface of the cylinder, the magnetic force whips it
Conservation of energy
around and around at ever higher speeds. Of course there is some friction, but there is a net gain in speed with each revolution. Physicists would lay long odds against the discovery of such a moon rock, not just because it breaks the rules that magnets normally obey but because, like the alchemists, they have discovered a very deep and fundamental principle of nature which forbids certain things from happening. The first alchemist who deserved to be called a chemist was the one who realized one day, “In all these attempts to create gold where there was none before, all I’ve been doing is shuffling the same atoms back and forth among different test tubes. The only way to increase the amount of gold in my laboratory is to bring some in through the door.” It was like having some of your money in a checking account and some in a savings account. Transferring money from one account into the other doesn’t change the total amount. We say that the number of grams of gold is a conserved quantity. In this context, the word “conserve” does not have its usual meaning of trying not to waste something. In physics, a conserved quantity is something that you wouldn’t be able to get rid of even if you wanted to. Conservation laws in physics always refer to a closed system, meaning a region of space with boundaries through which the quantity in question is not passing. In our example, the alchemist’s laboratory is a closed system because no gold is coming in or out through the doors. Conservation of mass example 1 In figure d, the stream of water is fatter near the mouth of the faucet, and skinnier lower down. This is because the water speeds up as it falls. If the cross-sectional area of the stream was equal all along its length, then the rate of flow through a lower crosssection would be greater than the rate of flow through a crosssection higher up. Since the flow is steady, the amount of water between the two cross-sections stays constant. The crosssectional area of the stream must therefore shrink in inverse proportion to the increasing speed of the falling water. This is an example of conservation of mass.
c / A mysterious moon rock makes a perpetual motion machine.
In general, the amount of any particular substance is not conserved. Chemical reactions can change one substance into another, and nuclear reactions can even change one element into another. The total mass of all substances is however conserved: the law of conservation of mass The total mass of a closed system always remains constant. Mass cannot be created or destroyed, but only transferred from one system to another. d / Example 1.
A similar lightbulb eventually lit up in the heads of the people
Section 11.2
Energy
279
who had been frustrated trying to build a perpetual motion machine. In perpetual motion machine a, consider the motion of one of the balls. It performs a cycle of rising and falling. On the way down it gains speed, and coming up it slows back down. Having a greater speed is like having more money in your checking account, and being high up is like having more in your savings account. The device is simply shuffling funds back and forth between the two. Having more balls doesn’t change anything fundamentally. Not only that, but friction is always draining off money into a third “bank account:” heat. The reason we rub our hands together when we’re cold is that kinetic friction heats things up. The continual buildup in the “heat account” leaves less and less for the “motion account” and “height account,” causing the machine eventually to run down. These insights can be distilled into the following basic principle of physics: the law of conservation of energy It is possible to give a numerical rating, called energy, to the state of a physical system. The total energy is found by adding up contributions from characteristics of the system such as motion of objects in it, heating of the objects, and the relative positions of objects that interact via forces. The total energy of a closed system always remains constant. Energy cannot be created or destroyed, but only transferred from one system to another.
The moon rock story violates conservation of energy because the rock-cylinder and the ball together constitute a closed system. Once the ball has made one revolution around the cylinder, its position relative to the cylinder is exactly the same as before, so the numerical energy rating associated with its position is the same as before. Since the total amount of energy must remain constant, it is impossible for the ball to have a greater speed after one revolution. If it had picked up speed, it would have more energy associated with motion, the same amount of energy associated with position, and a little more energy associated with heating through friction. There cannot be a net increase in energy. Converting one form of energy to another example 2 Dropping a rock: The rock loses energy because of its changing position with respect to the earth. Nearly all that energy is transformed into energy of motion, except for a small amount lost to heat created by air friction. Sliding in to home base: The runner’s energy of motion is nearly all converted into heat via friction with the ground. Accelerating a car: The gasoline has energy stored in it, which is released as heat by burning it inside the engine. Perhaps 10%
280
Chapter 11
Conservation of energy
of this heat energy is converted into the car’s energy of motion. The rest remains in the form of heat, which is carried away by the exhaust. Cruising in a car: As you cruise at constant speed in your car, all the energy of the burning gas is being converted into heat. The tires and engine get hot, and heat is also dissipated into the air through the radiator and the exhaust. Stepping on the brakes: All the energy of the car’s motion is converted into heat in the brake shoes. Stevin’s machine example 3 The Dutch mathematician and engineer Simon Stevin proposed the imaginary machine shown in figure e, which he had inscribed on his tombstone. This is an interesting example, because it shows a link between the force concept used earlier in this course, and the energy concept being developed now. The point of the imaginary machine is to show the mechanical advantage of an inclined plane. In this example, the triangle has the proportions 3-4-5, but the argument works for any right triangle. We imagine that the chain of balls slides without friction, so that no energy is ever converted into heat. If we were to slide the chain clockwise by one step, then each ball would take the place of the one in front of it, and the over all configuration would be exactly the same. Since energy is something that only depends on the state of the system, the energy would have to be the same. Similarly for a counterclockwise rotation, no energy of position would be released by gravity. This means that if we place the chain on the triangle, and release it at rest, it can’t start moving, because there would be no way for it to convert energy of position into energy of motion. Thus the chain must be perfectly balanced. Now by symmetry, the arc of the chain hanging underneath the triangle has equal tension at both ends, so removing this arc wouldn’t affect the balance of the rest of the chain. This means that a weight of three units hanging vertically balances a weight of five units hanging diagonally along the hypotenuse.
e / Example 3.
The mechanical advantage of the inclined plane is therefore 5/3, which is exactly the same as the result, 1/ sin θ, that we got on p. 213 by analyzing force vectors. What this shows is that Newton’s laws and conservation laws are not logically separate, but rather are very closely related descriptions of nature. In the cases where Newton’s laws are true, they give the same answers as the conservation laws. This is an example of a more general idea, called the correspondence principle, about how science progresses over time. When a newer, more general theory is proposed to replace an older theory, the new theory must agree with the old one in the realm of applicability of the old theory, since the old theory only became accepted as a valid theory by being ver-
Section 11.2
Energy
281
ified experimentally in a variety of experiments. In other words, the new theory must be backward-compatible with the old one. Even though conservation laws can prove things that Newton’s laws can’t (that perpetual motion is impossible, for example), they aren’t going to disprove Newton’s laws when applied to mechanical systems where we already knew Newton’s laws were valid. Discussion question Discussion question A. The water behind the Hoover Dam has energy because of its position relative to the planet earth, which is attracting it with a gravitational force. Letting water down to the bottom of the dam converts that energy into energy of motion. When the water reaches the bottom of the dam, it hits turbine blades that drive generators, and its energy of motion is converted into electrical energy.
A Hydroelectric power (water flowing over a dam to spin turbines) appears to be completely free. Does this violate conservation of energy? If not, then what is the ultimate source of the electrical energy produced by a hydroelectric plant? B How does the proof in example 3 fail if the assumption of a frictionless surface doesn’t hold?
11.3 A numerical scale of energy Energy comes in a variety of forms, and physicists didn’t discover all of them right away. They had to start somewhere, so they picked one form of energy to use as a standard for creating a numerical energy scale. (In fact the history is complicated, and several different energy units were defined before it was realized that there was a single general energy concept that deserved a single consistent unit of measurement.) One practical approach is to define an energy unit based on heating water. The SI unit of energy is the joule, J, (rhymes with “cool”), named after the British physicist James Joule. One Joule is the amount of energy required in order to heat 0.24 g of water by 1◦ C. The number 0.24 is not worth memorizing. Note that heat, which is a form of energy, is completely different from temperature, which is not. Twice as much heat energy is required to prepare two cups of coffee as to make one, but two cups of coffee mixed together don’t have double the temperature. In other words, the temperature of an object tells us how hot it is, but the heat energy contained in an object also takes into account the object’s mass and what it is made of.1 Later we will encounter other quantities that are conserved in physics, such as momentum and angular momentum, and the method for defining them will be similar to the one we have used for energy: pick some standard form of it, and then measure other forms by comparison with this standard. The flexible and adaptable nature of this procedure is part of what has made conservation laws such a durable basis for the evolution of physics. 1 In standard, formal terminology, there is another, finer distinction. The word “heat” is used only to indicate an amount of energy that is transferred, whereas “thermal energy” indicates an amount of energy contained in an object. I’m informal on this point, and refer to both as heat, but you should be aware of the distinction.
282
Chapter 11
Conservation of energy
Heating a swimming pool example 4 If electricity costs 3.9 cents per MJ (1 MJ = 1 megajoule = 106 J), how much does it cost to heat a 26000-gallon swimming pool from 10◦ C to 18◦ C? Converting gallons to cm3 gives 26000 gallons ×
3780 cm3 = 9.8 × 107 cm3 1 gallon
.
Water has a density of 1 gram per cubic centimeter, so the mass of the water is 9.8 × 107 g. One joule is sufficient to heat 0.24 g by 1◦ C, so the energy needed to heat the swimming pool is 1J×
9.8 × 107 g 8◦ C × ◦ = 3.3 × 109 J 0.24 g 1 C = 3.3 × 103 MJ
.
The cost of the electricity is (3.3 × 103 MJ)($0.039/MJ)=$130. Irish coffee example 5 You make a cup of Irish coffee out of 300 g of coffee at 100◦ C and 30 g of pure ethyl alcohol at 20◦ C. One Joule is enough energy to produce a change of 1◦ C in 0.42 g of ethyl alcohol (i.e., alcohol is easier to heat than water). What temperature is the final mixture? Adding up all the energy after mixing has to give the same result as the total before mixing. We let the subscript i stand for the initial situation, before mixing, and f for the final situation, and use subscripts c for the coffee and a for the alcohol. In this notation, we have total initial energy = total final energy Eci + Eai = Ecf + Eaf
.
We assume coffee has the same heat-carrying properties as water. Our information about the heat-carrying properties of the two substances is stated in terms of the change in energy required for a certain change in temperature, so we rearrange the equation to express everything in terms of energy differences: Eaf − Eai = Eci − Ecf
.
Using the given ratios of temperature change to energy change, we have Eci − Ecf = (Tci − Tcf )(mc )/(0.24 g) Eaf − Eai = (Taf − Tai )(ma )/(0.42 g) Setting these two quantities to be equal, we have (Taf − Tai )(ma )/(0.42 g) = (Tci − Tcf )(mc )/(0.24 g)
Section 11.3
.
A numerical scale of energy
283
In the final mixture the two substances must be at the same temperature, so we can use a single symbol Tf = Tcf = Taf for the two quantities previously represented by two different symbols, (Tf − Tai )(ma )/(0.42 g) = (Tci − Tf )(mc )/(0.24 g)
.
Solving for Tf gives Tf =
mc ma Tci 0.24 + Tai 0.42 mc ma 0.24 + 0.42
= 96◦ C
.
Once a numerical scale of energy has been established for some form of energy such as heat, it can easily be extended to other types of energy. For instance, the energy stored in one gallon of gasoline can be determined by putting some gasoline and some water in an insulated chamber, igniting the gas, and measuring the rise in the water’s temperature. (The fact that the apparatus is known as a “bomb calorimeter” will give you some idea of how dangerous these experiments are if you don’t take the right safety precautions.) Here are some examples of other types of energy that can be measured using the same units of joules: type of energy chemical energy released by burning energy required to break an object
energy required to melt a solid substance chemical energy released by digesting food raising a mass against the force of gravity nuclear energy released in fission
example About 50 MJ are released by burning a kg of gasoline. When a person suffers a spiral fracture of the thighbone (a common type in skiing accidents), about 2 J of energy go into breaking the bone. 7 MJ are required to melt 1 kg of tin. A bowl of Cheeries with milk provides us with about 800 kJ of usable energy. Lifting 1.0 kg through a height of 1.0 m requires 9.8 J. 1 kg of uranium oxide fuel consumed by a reactor releases 2 × 1012 J of stored nuclear energy.
It is interesting to note the disproportion between the megajoule energies we consume as food and the joule-sized energies we expend in physical activities. If we could perceive the flow of energy around us the way we perceive the flow of water, eating a bowl of cereal would be like swallowing a bathtub’s worth of energy, the continual loss of body heat to one’s environment would be like an energy-hose left on all day, and lifting a bag of cement would be like flicking it with a few tiny energy-drops. The human body is tremendously
284
Chapter 11
Conservation of energy
inefficient. The calories we “burn” in heavy exercise are almost all dissipated directly as body heat. You take the high road and I’ll take the low road. example 6 Figure f shows two ramps which two balls will roll down. Compare their final speeds, when they reach point B. Assume friction is negligible. Each ball loses some energy because of its decreasing height above the earth, and conservation of energy says that it must gain an equal amount of energy of motion (minus a little heat created by friction). The balls lose the same amount of height, so their final speeds must be equal.
f / Example 6.
It’s impressive to note the complete impossibility of solving this problem using only Newton’s laws. Even if the shape of the track had been given mathematically, it would have been a formidable task to compute the balls’ final speed based on vector addition of the normal force and gravitational force at each point along the way. How new forms of energy are discovered Textbooks often give the impression that a sophisticated physics concept was created by one person who had an inspiration one day, but in reality it is more in the nature of science to rough out an idea and then gradually refine it over many years. The idea of energy was tinkered with from the early 1800’s on, and new types of energy kept getting added to the list. To establish the existence of a new form of energy, a physicist has to (1) show that it could be converted to and from other forms of energy; and (2) show that it related to some definite measurable property of the object, for example its temperature, motion, position relative to another object, or being in a solid or liquid state. For example, energy is released when a piece of iron is soaked in water, so apparently there is some form of energy already stored in the iron. The release of this energy can also be related to a definite measurable property of the chunk of metal: it turns reddish-orange. There has been a chemical change in its physical state, which we call rusting. Although the list of types of energy kept getting longer and longer, it was clear that many of the types were just variations on a theme. There is an obvious similarity between the energy needed to melt ice and to melt butter, or between the rusting of iron and many other chemical reactions. The topic of the next chapter is how this process of simplification reduced all the types of energy to a very small number (four, according to the way I’ve chosen to
Section 11.3
A numerical scale of energy
285
count them). It might seem that if the principle of conservation of energy ever appeared to be violated, we could fix it up simply by inventing some new type of energy to compensate for the discrepancy. This would be like balancing your checkbook by adding in an imaginary deposit or withdrawal to make your figures agree with the bank’s statements. Step (2) above guards against this kind of chicanery. In the 1920s there were experiments that suggested energy was not conserved in radioactive processes. Precise measurements of the energy released in the radioactive decay of a given type of atom showed inconsistent results. One atom might decay and release, say, 1.1 × 10−10 J of energy, which had presumably been stored in some mysterious form in the nucleus. But in a later measurement, an atom of exactly the same type might release 1.2 × 10−10 J. Atoms of the same type are supposed to be identical, so both atoms were thought to have started out with the same energy. If the amount released was random, then apparently the total amount of energy was not the same after the decay as before, i.e., energy was not conserved. Only later was it found that a previously unknown particle, which is very hard to detect, was being spewed out in the decay. The particle, now called a neutrino, was carrying off some energy, and if this previously unsuspected form of energy was added in, energy was found to be conserved after all. The discovery of the energy discrepancies is seen with hindsight as being step (1) in the establishment of a new form of energy, and the discovery of the neutrino was step (2). But during the decade or so between step (1) and step (2) (the accumulation of evidence was gradual), physicists had the admirable honesty to admit that the cherished principle of conservation of energy might have to be discarded. self-check A How would you carry out the two steps given above in order to establish that some form of energy was stored in a stretched or compressed spring? Answer, p. 543 Mass Into Energy Einstein showed that mass itself could be converted to and from energy, according to his celebrated equation E = mc 2 , in which c is the speed of light. We thus speak of mass as simply another form of energy, and it is valid to measure it in units of joules. The mass of a 15-gram pencil corresponds to about 1.3 × 1015 J. The issue is largely academic in the case of the pencil, because very violent processes such as nuclear reactions are required in order to convert any significant fraction of an object’s mass into energy. Cosmic rays, however, are continually striking you and your surroundings and converting part of their energy of motion into the mass of newly created particles. A single high-energy cosmic ray can create a “shower” of millions of previously nonexistent particles when it strikes the atmosphere. Einstein’s theories are discussed later in this book.
286
Chapter 11
Conservation of energy
Even today, when the energy concept is relatively mature and stable, a new form of energy has been proposed based on observations of distant galaxies whose light began its voyage to us billions of years ago. Astronomers have found that the universe’s continuing expansion, resulting from the Big Bang, has not been decelerating as rapidly in the last few billion years as would have been expected from gravitational forces. They suggest that a new form of energy may be at work.
Discussion question A I’m not making this up. XS Energy Drink has ads that read like this: All the “Energy” ... Without the Sugar! Only 8 Calories! Comment on this.
11.4 Kinetic energy The technical term for the energy associated with motion is kinetic energy, from the Greek word for motion. (The root is the same as the root of the word “cinema” for a motion picture, and in French the term for kinetic energy is “´energie cin´etique.”) To find how much kinetic energy is possessed by a given moving object, we must convert all its kinetic energy into heat energy, which we have chosen as the standard reference type of energy. We could do this, for example, by firing projectiles into a tank of water and measuring the increase in temperature of the water as a function of the projectile’s mass and velocity. Consider the following data from a series of three such experiments: m (kg) 1.00 1.00 2.00
v (m/s) 1.00 2.00 1.00
energy (J) 0.50 2.00 1.00
Comparing the first experiment with the second, we see that doubling the object’s velocity doesn’t just double its energy, it quadruples it. If we compare the first and third lines, however, we find that doubling the mass only doubles the energy. This suggests that kinetic energy is proportional to mass and to the square of velocity, KE ∝ mv 2 , and further experiments of this type would indeed establish such a general rule. The proportionality factor equals 0.5 because of the design of the metric system, so the kinetic energy of a moving object is given by 1 . KE = mv 2 2 The metric system is based on the meter, kilogram, and second, with other units being derived from those. Comparing the units on the left and right sides of the equation shows that the joule can be reexpressed in terms of the basic units as kg·m2 /s2 . Students are often mystified by the occurrence of the factor of 1/2, but it is less obscure than it looks. The metric system was
Section 11.4
Kinetic energy
287
designed so that some of the equations relating to energy would come out looking simple, at the expense of some others, which had to have inconvenient conversion factors in front. If we were using the old British Engineering System of units in this course, then we’d have the British Thermal Unit (BTU) as our unit of energy. In that system, the equation you’d learn for kinetic energy would have an inconvenient proportionality constant, KE = 1.29 × 10−3 mv 2 , with KE measured in units of BTUs, v measured in feet per second, and so on. At the expense of this inconvenient equation for kinetic energy, the designers of the British Engineering System got a simple rule for calculating the energy required to heat water: one BTU per degree Fahrenheit per pound. The inventor of kinetic energy, Thomas Young, actually defined it as KE = mv 2 , which meant that all his other equations had to be different from ours by a factor of two. All these systems of units work just fine as long as they are not combined with one another in an inconsistent way. Energy released by a comet impact example 7 Comet Shoemaker-Levy, which struck the planet Jupiter in 1994, had a mass of roughly 4 × 1013 kg, and was moving at a speed of 60 km/s. Compare the kinetic energy released in the impact to the total energy in the world’s nuclear arsenals, which is 2 × 1019 J. Assume for the sake of simplicity that Jupiter was at rest. Since we assume Jupiter was at rest, we can imagine that the comet stopped completely on impact, and 100% of its kinetic energy was converted to heat and sound. We first convert the speed to mks units, v = 6 × 104 m/s, and then plug in to the equation to find that the comet’s kinetic energy was roughly 7 × 1022 J, or about 3000 times the energy in the world’s nuclear arsenals. Is there any way to derive the equation KE = (1/2)mv 2 mathematically from first principles? No, it is purely empirical. The factor of 1/2 in front is definitely not derivable, since it is different in different systems of units. The proportionality to v 2 is not even quite correct; experiments have shown deviations from the v 2 rule at high speeds, an effect that is related to Einstein’s theory of relativity. Only the proportionality to m is inevitable. The whole energy concept is based on the idea that we add up energy contributions from all the objects within a system. Based on this philosophy, it is logically necessary that a 2-kg object moving at 1 m/s have the same kinetic energy as two 1-kg objects moving side-by-side at the same speed. Energy and relative motion Although I mentioned Einstein’s theory of relativity above, it’s more relevant right now to consider how conservation of energy relates to the simpler Galilean idea, which we’ve already studied, that motion is relative. Galileo’s Aristotelian enemies (and it is no exaggeration to call them enemies!) would probably have objected to
288
Chapter 11
Conservation of energy
conservation of energy. After all, the Galilean idea that an object in motion will continue in motion indefinitely in the absence of a force is not so different from the idea that an object’s kinetic energy stays the same unless there is a mechanism like frictional heating for converting that energy into some other form. More subtly, however, it’s not immediately obvious that what we’ve learned so far about energy is strictly mathematically consistent with the principle that motion is relative. Suppose we verify that a certain process, say the collision of two pool balls, conserves energy as measured in a certain frame of reference: the sum of the balls’ kinetic energies before the collision is equal to their sum after the collision. (In reality we’d need to add in other forms of energy, like heat and sound, that are liberated by the collision, but let’s keep it simple.) But what if we were to measure everything in a frame of reference that was in a different state of motion? A particular pool ball might have less kinetic energy in this new frame; for example, if the new frame of reference was moving right along with it, its kinetic energy in that frame would be zero. On the other hand, some other balls might have a greater kinetic energy in the new frame. It’s not immediately obvious that the total energy before the collision will still equal the total energy after the collision. After all, the equation for kinetic energy is fairly complicated, since it involves the square of the velocity, so it would be surprising if everything still worked out in the new frame of reference. It does still work out. Homework problem 13 in this chapter gives a simple numerical example, and the general proof is taken up in problem 15 on p. 372 (with the solution given in the back of the book). Discussion questions A Suppose that, like Young or Einstein, you were trying out different equations for kinetic energy to see if they agreed with the experimental data. Based on the meaning of positive and negative signs of velocity, why would you suspect that a proportionality to mv would be less likely than mv 2 ?
Discussion question B
B The figure shows a pendulum that is released at A and caught by a peg as it passes through the vertical, B. To what height will the bob rise on the right?
11.5 Power A car may have plenty of energy in its gas tank, but still may not be able to increase its kinetic energy rapidly. A Porsche doesn’t necessarily have more energy in its gas tank than a Hyundai, it is just able to transfer it more quickly. The rate of transferring energy from one form to another is called power. The definition can be written as an equation, P =
ΔE Δt
,
Section 11.5
Power
289
where the use of the delta notation in the symbol ΔE has the usual interpretation: the final amount of energy in a certain form minus the initial amount that was present in that form. Power has units of J/s, which are abbreviated as watts, W (rhymes with “lots”). If the rate of energy transfer is not constant, the power at any instant can be defined as the slope of the tangent line on a graph of E versus t. Likewise ΔE can be extracted from the area under the P -versus-t curve. Converting kilowatt-hours to joules example 8 The electric company bills you for energy in units of kilowatthours (kilowatts multiplied by hours) rather than in SI units of joules. How many joules is a kilowatt-hour? 1 kilowatt-hour = (1 kW)(1 hour) = (1000 J/s)(3600 s) = 3.6 MJ. Human wattage example 9 A typical person consumes 2000 kcal of food in a day, and converts nearly all of that directly to heat. Compare the person’s heat output to the rate of energy consumption of a 100-watt lightbulb. Looking up the conversion factor from calories to joules, we find
ΔE = 2000 kcal ×
1000 cal 4.18 J × = 8 × 106 J 1 kcal 1 cal
for our daily energy consumption. Converting the time interval likewise into mks,
Δt = 1 day ×
24 hours 60 min 60 s × × = 9 × 104 s 1 day 1 hour 1 min
.
Dividing, we find that our power dissipated as heat is 90 J/s = 90 W, about the same as a lightbulb. It is easy to confuse the concepts of force, energy, and power, especially since they are synonyms in ordinary speech. The table on the following page may help to clear this up:
290
Chapter 11
Conservation of energy
force A force is an interaction between two objects that causes a push or a pull. A force can be defined as anything that is capable of changing an object’s state of motion.
conceptual definition
operational definition
scalar vector?
or
unit Can it run out? Does it cost money?
Can it be a property of an object?
energy Heating an object, making it move faster, or increasing its distance from another object that is attracting it are all examples of things that would require fuel or physical effort. All these things can be quantified using a single scale of measurement, and we describe them all as forms of energy. If we define a unit of enA spring scale can be used ergy as the amount reto measure force. quired to heat a certain amount of water by a 1◦ C, then we can measure any other quantity of energy by transferring it into heat in water and measuring the temperature increase. vector — has a direction scalar — has no direction in space which is the di- in space rection in which it pulls or pushes newtons (N) joules (J) No. I don’t have to Yes. We pay money for pay a monthly bill for gasoline, electrical energy, the meganewtons of force batteries, etc., because required to hold up my they contain energy. house. No. A force is a rela- Yes. What a home-run tionship between two baseball has is kinetic eninteracting objects. ergy, not force. A home-run baseball doesn’t “have” force.
Section 11.5
power Power is the rate at which energy is transformed from one form to another or transferred from one object to another.
Measure the change in the amount of some form of energy possessed by an object, and divide by the amount of time required for the change to occur.
scalar — has no direction in space
watts (W) = joules/s More power means you are paying money at a higher rate. A 100-W lightbulb costs a certain number of cents per hour. Not really. A 100-W lightbulb doesn’t “have” 100 W. 100 J/s is the rate at which it converts electrical energy into light.
Power
291
Summary Selected vocabulary energy . . . . . . A numerical scale used to measure the heat, motion, or other properties that would require fuel or physical effort to put into an object; a scalar quantity with units of joules (J). power . . . . . . . The rate of transferring energy; a scalar quantity with units of watts (W). kinetic energy . . The energy an object possesses because of its motion. heat . . . . . . . . A form of energy that relates to temperature. Heat is different from temperature because an object with twice as much mass requires twice as much heat to increase its temperature by the same amount. Heat is measured in joules, temperature in degrees. (In standard terminology, there is another, finer distinction between heat and thermal energy, which is discussed below. In this book, I informally refer to both as heat.) temperature . . . What a thermometer measures. Objects left in contact with each other tend to reach the same temperature. Cf. heat. As discussed in more detail in chapter 2, temperature is essentially a measure of the average kinetic energy per molecule. Notation E . . . . J . . . . . KE . . . P . . . . W . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
energy joules, the SI unit of energy kinetic energy power watts, the SI unit of power; equivalent to J/s
Other terminology and notation Q or ΔQ . . . . . the amount of heat transferred into or out of an object K or T . . . . . . alternative symbols for kinetic energy, used in the scientific literature and in most advanced textbooks thermal energy . Careful writers make a distinction between heat and thermal energy, but the distinction is often ignored in casual speech, even among physicists. Properly, thermal energy is used to mean the total amount of energy possessed by an object, while heat indicates the amount of thermal energy transferred in or out. The term heat is used in this book to include both meanings.
292
Chapter 11
Conservation of energy
Summary Heating an object, making it move faster, or increasing its distance from another object that is attracting it are all examples of things that would require fuel or physical effort. All these things can be quantified using a single scale of measurement, and we describe them all as forms of energy. The SI unit of energy is the Joule. The reason why energy is a useful and important quantity is that it is always conserved. That is, it cannot be created or destroyed but only transferred between objects or changed from one form to another. Conservation of energy is the most important and broadly applicable of all the laws of physics, more fundamental and general even than Newton’s laws of motion. Heating an object requires a certain amount of energy per degree of temperature and per unit mass, which depends on the substance of which the object consists. Heat and temperature are completely different things. Heat is a form of energy, and its SI unit is the joule (J). Temperature is not a measure of energy. Heating twice as much of something requires twice as much heat, but double the amount of a substance does not have double the temperature. The energy that an object possesses because of its motion is called kinetic energy. Kinetic energy is related to the mass of the object and the magnitude of its velocity vector by the equation 1 KE = mv 2 2
.
Power is the rate at which energy is transformed from one form to another or transferred from one object to another, P =
ΔE Δt
.
[only for constant power]
The SI unit of power is the watt (W).
Summary
293
Problems Key √ 1
A computerized answer check is available online. A problem that requires calculus. A difficult problem. This problem is now problem 14 in chapter 12, on page 311.
2 Can kinetic energy ever be less than zero? Explain. [Based on a problem by Serway and Faughn.] 3
Estimate the kinetic energy of an Olympic sprinter.
4 You are driving your car, and you hit a brick wall head on, at full speed. The car has a mass of 1500 kg. The kinetic energy released is a measure of how much destruction will be done to the car and to your body. Calculate the energy released if you are traveling at (a) 40 mi/hr, and again (b) if you’re going 80 mi/hr. What is counterintuitive about this, and what implication does this have for √ driving at high speeds? 5 A closed system can be a bad thing — for an astronaut sealed inside a space suit, getting rid of body heat can be difficult. Suppose a 60-kg astronaut is performing vigorous physical activity, expending 200 W of power. If none of the heat can escape from her space suit, how long will it take before her body temperature rises by 6◦ C (11◦ F), an amount sufficient to kill her? Assume that the amount of heat required to raise her body temperature by 1◦ C is the same as it would be for an equal mass of water. Express √ your answer in units of minutes. 6 All stars, including our sun, show variations in their light output to some degree. Some stars vary their brightness by a factor of two or even more, but our sun has remained relatively steady during the hundred years or so that accurate data have been collected. Nevertheless, it is possible that climate variations such as ice ages are related to long-term irregularities in the sun’s light output. If the sun was to increase its light output even slightly, it could melt enough Antarctic ice to flood all the world’s coastal cities. The total sunlight that falls on Antarctica amounts to about 1 × 1016 watts. In the absence of natural or human-caused climate change, this heat input to the poles is balanced by the loss of heat via winds, ocean currents, and emission of infrared light, so that there is no net melting or freezing of ice at the poles from year to year. Suppose that the sun changes its light output by some small percentage, but there is no change in the rate of heat loss by the polar caps. Estimate the percentage by which the sun’s light output would have to increase in order to melt enough ice to raise the level of the oceans by 10 meters over a period of 10 years. (This would be enough to flood New York, London, and many other cities.) Melting 1 kg of ice requires 3 × 103 J. 294
Chapter 11
Conservation of energy
7 A bullet flies through the air, passes through a paperback book, and then continues to fly through the air beyond the book. When is there a force? When is there energy? Solution, p. 536
8 Experiments show that the power consumed by a boat’s engine is approximately proportional to third power of its speed. (We assume that it is moving at constant speed.) (a) When a boat is crusing at constant speed, what type of energy transformation do you think is being performed? (b) If you upgrade to a motor with double the power, by what factor is your boat’s crusing speed increased? [Based on a problem by Arnold Arons.] Solution, p. 537
9 Object A has a kinetic energy of 13.4 J. Object B has a mass that is greater by a factor of 3.77, but is moving more slowly by a factor of 2.34. What is object B’s kinetic energy? [Based on a problem by Arnold Arons.] Solution, p. 537
10 The moon doesn’t really just orbit the Earth. By Newton’s third law, the moon’s gravitational force on the earth is the same as the earth’s force on the moon, and the earth must respond to the moon’s force by accelerating. If we consider the earth and moon in isolation and ignore outside forces, then Newton’s first law says their common center of mass doesn’t accelerate, i.e., the earth wobbles around the center of mass of the earth-moon system once per month, and the moon also orbits around this point. The moon’s mass is 81 times smaller than the earth’s. Compare the kinetic energies of the earth and moon. (We know that the center of mass is a kind of balance point, so it must be closer to the earth than to the moon. In fact, the distance from the earth to the center of mass is 1/81 of the distance from the moon to the center of mass, which makes sense intuitively, and can be proved rigorously using the equation on page 356.)
11 My 1.25 kW microwave oven takes 126 seconds to bring 250 g of water from room temperature to a boil. What percentage of the power is being wasted? Where might the rest of the energy be going? Solution, p. 537
Problems
295
12 The multiflash photograph shows a collision between two pool balls. The ball that was initially at rest shows up as a dark image in its initial position, because its image was exposed several times before it was struck and began moving. By making measurements on the figure, determine numerically whether or not energy appears to have been conserved in the collision. What systematic effects would limit the accuracy of your test? [From an example in PSSC Physics.]
Problem 12.
13 This problem is a numerical example of the imaginary experiment discussed on p. 288 regarding the relationship between energy and relative motion. Let’s say that the pool balls both have masses of 1.00 kg. Suppose that in the frame of reference of the pool table, the cue ball moves at a speed of 1.00 m/s toward the eight ball, which is initially at rest. The collision is head-on, and as you can verify for yourself the next time you’re playing pool, the result of such a collision is that the incoming ball stops dead and the ball that was struck takes off with the same speed originally possessed by the incoming ball. (This is actually a bit of an idealization. To keep things simple, we’re ignoring the spin of the balls, and we assume that no energy is liberated by the collision as heat or sound.) (a) Calculate the total initial kinetic energy and the total final kinetic energy, and verify that they are equal. (b) Now carry out the whole calculation again in the frame of reference that is moving in the same direction that the cue ball was initially moving, but at a speed of 0.50 m/s. In this frame of reference, both balls have nonzero initial and final velocities, which are different from what they were in the table’s frame. [See also problem 15 on p. 372.]
296
Chapter 11
Conservation of energy
14 One theory about the destruction of the space shuttle Columbia in 2003 is that one of its wings had been damaged on liftoff by a chunk of foam insulation that fell off of one of its external fuel tanks. The New York Times reported on June 5, 2003, that NASA engineers had recreated the impact to see if it would damage a mock-up of the shuttle’s wing. “Before last week’s test, many engineers at NASA said they thought lightweight foam could not harm the seemingly tough composite panels, and privately predicted that the foam would bounce off harmlessly, like a Nerf ball.” In fact, the 1.7-pound piece of foam, moving at 531 miles per hour, did serious damage. A member of the board investigating the disaster said this demonstrated that “people’s intuitive sense of physics is sometimes way off.” (a) Compute the kinetic energy of the foam, and (b) compare with the energy of a 170-pound boulder moving at 5.3 miles per hour (the speed it would have if you dropped it from about knee√ level). (c) The boulder is a hundred times more massive, but its speed is a hundred times smaller, so what’s counterintuitive about your results?
15 The figure above is from a classic 1920 physics textbook by Millikan and Gale. It represents a method for raising the water from the pond up to the water tower, at a higher level, without using a pump. Water is allowed into the drive pipe, and once it is flowing fast enough, it forces the valve at the bottom closed. Explain how this works in terms of conservation of mass and energy. (Cf. example 1 on page 279.)
Problems
297
16 The following table gives the amount of energy required in order to heat, melt, or boil a gram of water. 2.05 J heat 1 g of ice by 1◦ C melt 1 g of ice 333 J heat 1 g of liquid by 1◦ C 4.19 J boil 1 g of water 2500 J heat 1 g of steam by 1◦ C 2.01 J (a) How much energy is required in order to convert 1.00 g of ice at √ -20 ◦ C into steam at 137 ◦ C? (b) What is the minimum amount of hot water that could melt √1.00 g of ice?
298
Chapter 11
Conservation of energy
Do these forms of energy have anything in common?
Chapter 12
Simplifying the energy zoo Variety is the spice of life, not of science. The figure shows a few examples from the bewildering array of forms of energy that surrounds us. The physicist’s psyche rebels against the prospect of a long laundry list of types of energy, each of which would require its own equations, concepts, notation, and terminology. The point at which we’ve arrived in the study of energy is analogous to the period in the 1960’s when a half a dozen new subatomic particles were being discovered every year in particle accelerators. It was an embarrassment. Physicists began to speak of the “particle zoo,” and it seemed that the subatomic world was distressingly complex. The particle zoo was simplified by the realization that most of the
299
new particles being whipped up were simply clusters of a previously unsuspected set of more fundamental particles (which were whimsically dubbed quarks, a made-up word from a line of poetry by James Joyce, “Three quarks for Master Mark.”) The energy zoo can also be simplified, and it is the purpose of this chapter to demonstrate the hidden similarities between forms of energy as seemingly different as heat and motion.
a / A vivid demonstration that heat is a form of motion. A small amount of boiling water is poured into the empty can, which rapidly fills up with hot steam. The can is then sealed tightly, and soon crumples. This can be explained as follows. The high temperature of the steam is interpreted as a high average speed of random motions of its molecules. Before the lid was put on the can, the rapidly moving steam molecules pushed their way out of the can, forcing the slower air molecules out of the way. As the steam inside the can thinned out, a stable situation was soon achieved, in which the force from the less dense steam molecules moving at high speed balanced against the force from the more dense but slower air molecules outside. The cap was put on, and after a while the steam inside the can reached the same temperature as the air outside. The force from the cool, thin steam no longer matched the force from the cool, dense air outside, and the imbalance of forces crushed the can.
12.1 Heat is kinetic energy What is heat really? Is it an invisible fluid that your bare feet soak up from a hot sidewalk? Can one ever remove all the heat from an object? Is there a maximum to the temperature scale? The theory of heat as a fluid seemed to explain why colder objects absorbed heat from hotter ones, but once it became clear that heat was a form of energy, it began to seem unlikely that a material substance could transform itself into and out of all those other forms of energy like motion or light. For instance, a compost pile gets hot, and we describe this as a case where, through the action of bacteria, chemical energy stored in the plant cuttings is transformed into heat energy. The heating occurs even if there is no nearby warmer object that could have been leaking “heat fluid” into the pile. An alternative interpretation of heat was suggested by the theory that matter is made of atoms. Since gases are thousands of times less dense than solids or liquids, the atoms (or clusters of atoms called molecules) in a gas must be far apart. In that case, what is keeping all the air molecules from settling into a thin film on the floor of the room in which you are reading this book? The simplest explanation is that they are moving very rapidly, continually ricocheting off of
300
Chapter 12
Simplifying the energy zoo
the floor, walls, and ceiling. Though bizarre, the cloud-of-bullets image of a gas did give a natural explanation for the surprising ability of something as tenuous as a gas to exert huge forces. Your car’s tires can hold it up because you have pumped extra molecules into them. The inside of the tire gets hit by molecules more often than the outside, forcing it to stretch and stiffen. The outward forces of the air in your car’s tires increase even further when you drive on the freeway for a while, heating up the rubber and the air inside. This type of observation leads naturally to the conclusion that hotter matter differs from colder in that its atoms’ random motion is more rapid. In a liquid, the motion could be visualized as people in a milling crowd shoving past each other more quickly. In a solid, where the atoms are packed together, the motion is a random vibration of each atom as it knocks against its neighbors. We thus achieve a great simplification in the theory of heat. Heat is simply a form of kinetic energy, the total kinetic energy of random motion of all the atoms in an object. With this new understanding, it becomes possible to answer at one stroke the questions posed at the beginning of the section. Yes, it is at least theoretically possible to remove all the heat from an object. The coldest possible temperature, known as absolute zero, is that at which all the atoms have zero velocity, so that their kinetic energies, (1/2)mv 2 , are all zero. No, there is no maximum amount of heat that a certain quantity of matter can have, and no maximum to the temperature scale, since arbitrarily large values of v can create arbitrarily large amounts of kinetic energy per atom. The kinetic theory of heat also provides a simple explanation of the true nature of temperature. Temperature is a measure of the amount of energy per molecule, whereas heat is the total amount of energy possessed by all the molecules in an object. There is an entire branch of physics, called thermodynamics, that deals with heat and temperature and forms the basis for technologies such as refrigeration. Thermodynamics is discussed in more detail in optional chapter 16, and I have provided here only a brief overview of the thermodynamic concepts that relate directly to energy, glossing over at least one point that would be dealt with more carefully in a thermodynamics course: it is really only true for a gas that all the heat is in the form of kinetic energy. In solids and liquids, the atoms are close enough to each other to exert intense electrical forces on each other, and there is therefore another type of energy involved, the energy associated with the atoms’ distances from each other. Strictly speaking, heat energy is defined not as energy associated with random motion of molecules but as any form of energy that can be conducted between objects in contact, without any force.
Section 12.1
b / Random motion of atoms in a gas, a liquid, and a solid.
Heat is kinetic energy
301
12.2 Potential energy: energy of distance or closeness We have already seen many examples of energy related to the distance between interacting objects. When two objects participate in an attractive noncontact force, energy is required to bring them farther apart. In both of the perpetual motion machines that started off the previous chapter, one of the types of energy involved was the energy associated with the distance between the balls and the earth, which attract each other gravitationally. In the perpetual motion machine with the magnet on the pedestal, there was also energy associated with the distance between the magnet and the iron ball, which were attracting each other. The opposite happens with repulsive forces: two socks with the same type of static electric charge will repel each other, and cannot be pushed closer together without supplying energy. In general, the term potential energy, with algebra symbol PE, is used for the energy associated with the distance between two objects that attract or repel each other via a force that depends on the distance between them. Forces that are not determined by distance do not have potential energy associated with them. For instance, the normal force acts only between objects that have zero distance between them, and depends on other factors besides the fact that the distance is zero. There is no potential energy associated with the normal force. The following are some commonplace examples of potential energy:
c / The skater has converted all his kinetic energy into potential energy on the way up the side of the pool.
gravitational potential energy: The skateboarder in the photo has risen from the bottom of the pool, converting kinetic energy into gravitational potential energy. After being at rest for an instant, he will go back down, converting PE back into KE. magnetic potential energy: When a magnetic compass needle is allowed to rotate, the poles of the compass change their distances from the earth’s north and south magnetic poles, converting magnetic potential energy into kinetic energy. (Eventually the kinetic energy is all changed into heat by friction, and the needle settles down in the position that minimizes its potential energy.) electrical potential energy: Socks coming out of the dryer cling together because of attractive electrical forces. Energy is required in order to separate them. potential energy of bending or stretching: The force between the two ends of a spring depends on the distance between
302
Chapter 12
Simplifying the energy zoo
them, i.e., on the length of the spring. If a car is pressed down on its shock absorbers and then released, the potential energy stored in the spring is transformed into kinetic and gravitational potential energy as the car bounces back up. I have deliberately avoided introducing the term potential energy up until this point, because it tends to produce unfortunate connotations in the minds of students who have not yet been inoculated with a careful description of the construction of a numerical energy scale. Specifically, there is a tendency to generalize the term inappropriately to apply to any situation where there is the “potential” for something to happen: “I took a break from digging, but I had potential energy because I knew I’d be ready to work hard again in a few minutes.” An equation for gravitational potential energy All the vital points about potential energy can be made by focusing on the example of gravitational potential energy. For simplicity, we treat only vertical motion, and motion close to the surface of the earth, where the gravitational force is nearly constant. (The generalization to the three dimensions and varying forces is more easily accomplished using the concept of work, which is the subject of the next chapter.) To find an equation for gravitational PE, we examine the case of free fall, in which energy is transformed between kinetic energy and gravitational PE. Whatever energy is lost in one form is gained in an equal amount in the other form, so using the notation ΔKE to stand for KEf − KEi and a similar notation for PE, we have [1]
ΔKE = −ΔP Egrav
.
It will be convenient to refer to the object as falling, so that PE is being changed into KE, but the math applies equally well to an object slowing down on its way up. We know an equation for kinetic energy, [2]
1 KE = mv 2 2
,
so if we can relate v to height, y, we will be able to relate ΔP E to y, which would tell us what we want to know about potential energy. The y component of the velocity can be connected to the height via the constant acceleration equation [3]
vf2 = vi2 + 2aΔy
d / As the skater free-falls, his PE is converted into KE. (The numbers would be equally valid as a description of his motion on the way up.)
,
and Newton’s second law provides the acceleration, [4]
a = F/m
Section 12.2
,
Potential energy: energy of distance or closeness
303
in terms of the gravitational force. The algebra is simple because both equation [2] and equation [3] have velocity to the second power. Equation [2] can be solved for v 2 to give v 2 = 2KE/m, and substituting this into equation [3], we find KEf KEi 2 =2 + 2aΔy . m m Making use of equations [1] and [4] gives the simple result ΔP Egrav = −F Δy
.
[change in gravitational PE resulting from a change in height Δy;
F is the gravitational force on the object, i.e., its weight; valid only near the surface of the earth, where F is constant]
Dropping a rock example 1 If you drop a 1-kg rock from a height of 1 m, how many joules of KE does it have on impact with the ground? (Assume that any energy transformed into heat by air friction is negligible.) If we choose the y axis to point up, then Fy is negative, and equals −(1 kg)(g) = −9.8 N. A decrease in y is represented by a negative value of Δy , Δy = −1 m, so the change in potential energy is −(−9.8 N)(−1 m) ≈ −10 J. (The proof that newtons multiplied by meters give units of joules is left as a homework problem.) Conservation of energy says that the loss of this amount of PE must be accompanied by a corresponding increase in KE of 10 J. It may be dismaying to note how many minus signs had to be handled correctly even in this relatively simple example: a total of four. Rather than depending on yourself to avoid any mistakes with signs, it is better to check whether the final result make sense physically. If it doesn’t, just reverse the sign. Although the equation for gravitational potential energy was derived by imagining a situation where it was transformed into kinetic energy, the equation can be used in any context, because all the types of energy are freely convertible into each other. Gravitational PE converted directly into heat example 2 A 50-kg firefighter slides down a 5-m pole at constant velocity. How much heat is produced? Since she slides down at constant velocity, there is no change in KE. Heat and gravitational PE are the only forms of energy that change. Ignoring plus and minus signs, the gravitational force on
304
Chapter 12
Simplifying the energy zoo
her body equals mg, and the amount of energy transformed is (mg)(5 m) = 2500 J
.
On physical grounds, we know that there must have been an increase (positive change) in the heat energy in her hands and in the flagpole. Here are some questions and answers about the interpretation of the equation ΔP Egrav = −F Δy for gravitational potential energy. Question: In a nutshell, why is there a minus sign in the equation? Answer: It is because we increase the PE by moving the object in the opposite direction compared to the gravitational force. Question: Why do we only get an equation for the change in potential energy? Don’t I really want an equation for the potential energy itself? Answer: No, you really don’t. This relates to a basic fact about potential energy, which is that it is not a well defined quantity in the absolute sense. Only changes in potential energy are unambiguously defined. If you and I both observe a rock falling, and agree that it deposits 10 J of energy in the dirt when it hits, then we will be forced to agree that the 10 J of KE must have come from a loss of 10 joules of PE. But I might claim that it started with 37 J of PE and ended with 27, while you might swear just as truthfully that it had 109 J initially and 99 at the end. It is possible to pick some specific height as a reference level and say that the PE is zero there, but it’s easier and safer just to work with changes in PE and avoid absolute PE altogether. Question: You referred to potential energy as the energy that two objects have because of their distance from each other. If a rock falls, the object is the rock. Where’s the other object? Answer: Newton’s third law guarantees that there will always be two objects. The other object is the planet earth. Question: If the other object is the earth, are we talking about the distance from the rock to the center of the earth or the distance from the rock to the surface of the earth? Answer: It doesn’t matter. All that matters is the change in distance, Δy, not y. Measuring from the earth’s center or its surface are just two equally valid choices of a reference point for defining absolute PE. Question: Which object contains the PE, the rock or the earth? Answer: We may refer casually to the PE of the rock, but technically the PE is a relationship between the earth and the rock, and we should refer to the earth and the rock together as possessing the PE. Question: How would this be any different for a force other than gravity?
Section 12.2
Potential energy: energy of distance or closeness
305
Answer: It wouldn’t. The result was derived under the assumption of constant force, but the result would be valid for any other situation where two objects interacted through a constant force. Gravity is unusual, however, in that the gravitational force on an object is so nearly constant under ordinary conditions. The magnetic force between a magnet and a refrigerator, on the other hand, changes drastically with distance. The math is a little more complex for a varying force, but the concepts are the same. Question: Suppose a pencil is balanced on its tip and then falls over. The pencil is simultaneously changing its height and rotating, so the height change is different for different parts of the object. The bottom of the pencil doesn’t lose any height at all. What do you do in this situation? Answer: The general philosophy of energy is that an object’s energy is found by adding up the energy of every little part of it. You could thus add up the changes in potential energy of all the little parts of the pencil to find the total change in potential energy. Luckily there’s an easier way! The derivation of the equation for gravitational potential energy used Newton’s second law, which deals with the acceleration of the object’s center of mass (i.e., its balance point). If you just define Δy as the height change of the center of mass, everything works out. A huge Ferris wheel can be rotated without putting in or taking out any PE, because its center of mass is staying at the same height. self-check A A ball thrown straight up will have the same speed on impact with the ground as a ball thrown straight down at the same speed. How can this be explained using potential energy? Answer, p. 543
Discussion question A You throw a steel ball up in the air. How can you prove based on conservation of energy that it has the same speed when it falls back into your hand? What if you throw a feather up — is energy not conserved in this case?
12.3 All energy is potential or kinetic
e / All these energy transformations turn out at the atomic level to be changes in potential energy resulting from changes in the distances between atoms.
306
Chapter 12
In the same way that we found that a change in temperature is really only a change in kinetic energy at the atomic level, we now find that every other form of energy turns out to be a form of potential energy. Boiling, for instance, means knocking some of the atoms (or molecules) out of the liquid and into the space above, where they constitute a gas. There is a net attractive force between essentially any two atoms that are next to each other, which is why matter always prefers to be packed tightly in the solid or liquid state unless we supply enough potential energy to pull it apart into a gas. This explains why water stops getting hotter when it reaches the
Simplifying the energy zoo
boiling point: the power being pumped into the water by your stove begins going into potential energy rather than kinetic energy. As shown in figure e, every stored form of energy that we encounter in everyday life turns out to be a form of potential energy at the atomic level. The forces between atoms are electrical and magnetic in nature, so these are actually electrical and magnetic potential energies. Although light is a topic of the second half of this course, it is useful to have a preview of how it fits in here. Light is a wave composed of oscillating electric and magnetic fields, so we can include it under the category of electrical and magnetic potential energy. Even if we wish to include nuclear reactions in the picture, there still turn out to be only four fundamental types of energy: kinetic energy (including heat) gravitational potential energy electrical and magnetic potential energy nuclear potential energy Discussion question A Referring back to the pictures at the beginning of the chapter, how do all these forms of energy fit into the shortened list of categories given above?
f / This figure looks similar to the previous ones, but the scale is a million times smaller. The little balls are the neutrons and protons that make up the tiny nucleus at the center of the uranium atom. When the nucleus splits (fissions), the potential energy change is partly electrical and partly a change in the potential energy derived from the force that holds atomic nuclei together (known as the strong nuclear force).
g / A pellet of plutonium-238 glows with its own heat. Its nuclear potential energy is being converted into heat, a form of kinetic energy. Pellets of this type are used as power supplies on some space probes.
Section 12.3
All energy is potential or kinetic
307
Summary Selected vocabulary potential energy the energy having to do with the distance between two objects that interact via a noncontact force Notation PE . . . . . . . . .
potential energy
Other terminology and notation U or V . . . . . . symbols used for potential energy in the scientific literature and in most advanced textbooks Summary Historically, the energy concept was only invented to include a few phenomena, but it was later generalized more and more to apply to new situations, for example nuclear reactions. This generalizing process resulted in an undesirably long list of types of energy, each of which apparently behaved according to its own rules. The first step in simplifying the picture came with the realization that heat was a form of random motion on the atomic level, i.e., heat was nothing more than the kinetic energy of atoms. A second and even greater simplification was achieved with the realization that all the other apparently mysterious forms of energy actually had to do with changing the distances between atoms (or similar processes in nuclei). This type of energy, which relates to the distance between objects that interact via a force, is therefore of great importance. We call it potential energy. Most of the important ideas about potential energy can be understood by studying the example of gravitational potential energy. The change in an object’s gravitational potential energy is given by ΔP Egrav = −Fgrav Δy
,
[if Fgrav is constant, i.e., the
the motion is all near the Earth’s surface]
The most important thing to understand about potential energy is that there is no unambiguous way to define it in an absolute sense. The only thing that everyone can agree on is how much the potential energy has changed from one moment in time to some later moment in time.
308
Chapter 12
Simplifying the energy zoo
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 Can gravitational potential energy ever be negative? Note that the question refers to P E, not ΔP E, so that you must think about how the choice of a reference level comes into play. [Based on a problem by Serway and Faughn.]
2 A ball rolls up a ramp, turns around, and comes back down. When does it have the greatest gravitational potential energy? The greatest kinetic energy? [Based on a problem by Serway and Faughn.] 3 (a) You release a magnet on a tabletop near a big piece of iron, and the magnet leaps across the table to the iron. Does the magnetic potential energy increase, or decrease? Explain. (b) Suppose instead that you have two repelling magnets. You give them an initial push towards each other, so they decelerate while approaching each other. Does the magnetic potential energy increase, or decrease? Explain. 4 Let Eb be the energy required to boil one kg of water. (a) Find an equation for the minimum height from which a bucket of water must be dropped if the energy released on impact is to vaporize it. Assume that all the heat goes into the water, not into the dirt it strikes, and ignore the relatively small amount of energy required to heat the water from room temperature to 100◦ C. [Numerical check, not for credit: Plugging in Eb = 2.3 MJ/kg should give a result of √ 230 km.] (b) Show that the units of your answer in part a come out right based on the units given for Eb . 5 A grasshopper with a mass of 110 mg falls from rest from a height of 310 cm. On the way down, it dissipates 1.1 mJ of heat due to air resistance. At what speed, in m/s, does it hit the ground? Solution, p. 537 6 A person on a bicycle is to coast down a ramp of height h and then pass through a circular loop of radius r. What is the smallest value of h for which the cyclist will complete the loop without √ falling? (Ignore the kinetic energy of the spinning wheels.)
Problems
309
7 A skateboarder starts at rest nearly at the top of a giant cylinder, and begins rolling down its side. (If he started exactly at rest and exactly at the top, he would never get going!) Show that his board loses contact with the pipe after he has dropped by a height equal to one third the radius of the pipe. Solution, p. 537 8 (a) A circular hoop of mass m and radius r spins like a wheel while its center remains at rest. Its period (time required for one revolution) is T . Show that its kinetic energy equals 2π 2 mr2 /T 2 . (b) If such a hoop rolls with its center moving at velocity v, its kinetic energy equals (1/2)mv 2 , plus the amount of kinetic energy found in the first part of this problem. Show that a hoop rolls down an inclined plane with half the acceleration that a frictionless sliding block would have. 9 Students are often tempted to think of potential energy and kinetic energy as if they were always related to each other, like yin and yang. To show this is incorrect, give examples of physical situations in which (a) PE is converted to another form of PE, and (b) KE is converted to another form of KE. Solution, p. 538
Problem 7.
10 Lord Kelvin, a physicist, told the story of how he encountered James Joule when Joule was on his honeymoon. As he traveled, Joule would stop with his wife at various waterfalls, and measure the difference in temperature between the top of the waterfall and the still water at the bottom. (a) It would surprise most people to learn that the temperature increased. Why should there be any such effect, and why would Joule care? How would this relate to the energy concept, of which he was the principal inventor? (b) How much of a gain in temperature should there be between the top and bottom of a 50-meter waterfall? (c) What assumptions did you have to make in order to calculate your answer to part b? In reality, would the temperature change be more than or less than what√you calculated? [Based on a problem by Arnold Arons.] 11 Make an order-of-magnitude estimate of the power represented by the loss of gravitational energy of the water going over Niagara Falls. If the hydroelectric plant at the bottom of the falls could convert 100% of this to electrical power, roughly how many households could be powered? Solution, p. 538 12 When you buy a helium-filled balloon, the seller has to inflate it from a large metal cylinder of the compressed gas. The helium inside the cylinder has energy, as can be demonstrated for example by releasing a little of it into the air: you hear a hissing sound, and that sound energy must have come from somewhere. The total amount of energy in the cylinder is very large, and if the valve is inadvertently damaged or broken off, the cylinder can behave like bomb or a rocket. Suppose the company that puts the gas in the cylinders prepares
310
Chapter 12
Simplifying the energy zoo
cylinder A with half the normal amount of pure helium, and cylinder B with the normal amount. Cylinder B has twice as much energy, and yet the temperatures of both cylinders are the same. Explain, at the atomic level, what form of energy is involved, and why cylinder B has twice as much. 13 At a given temperature, the average kinetic energy per molecule is a fixed value, so for instance in air, the more massive oxygen molecules are moving more slowly on the average than the nitrogen molecules. The ratio of the masses of oxygen and nitrogen molecules is 16.00 to 14.01. Now suppose a vessel containing some air is surrounded by a vacuum, and the vessel has a tiny hole in it, which allows the air to slowly leak out. The molecules are bouncing around randomly, so a given molecule will have to “try” many times before it gets lucky enough to head out through the hole. Find the rate at which oxygen leaks divided by the rate at which nitrogen leaks. (Define this rate according to the fraction of the gas that leaks out in a given time, not the mass or number of √ molecules leaked per unit time.) 14 Explain in terms of conservation of energy why sweating cools your body, even though the sweat is at the same temperature as your body. Describe the forms of energy involved in this energy transformation. Why don’t you get the same cooling effect if you wipe the sweat off with a towel? Hint: The sweat is evaporating. 15 Anya and Ivan lean over a balcony side by side. Anya throws a penny downward with an initial speed of 5 m/s. Ivan throws a penny upward with the same speed. Both pennies end up on the ground below. Compare their kinetic energies and velocities on impact. 16 (a) A circular hoop of mass m and radius r spins like a wheel while its center remains at rest. Let ω (Greek letter omega) be the number of radians it covers per unit time, i.e., ω = 2π/T , where the period, T , is the time for one revolution. Show that its kinetic energy equals (1/2)mω 2 r2 . (b) Show that the answer to part a has the right units. (Note that radians aren’t really units, since the definition of a radian is a unitless ratio of two lengths.) (c) If such a hoop rolls with its center moving at velocity v, its kinetic energy equals (1/2)mv 2 , plus the amount of kinetic energy found in part a. Show that a hoop rolls down an inclined plane with half the acceleration that a frictionless sliding block would have.
Problems
311
17 The figure shows two unequal masses, M and m, connected by a string running over a pulley. This system was analyzed previously in problem 10 on p. 172, using Newton’s laws. (a) Analyze the system using conservation of energy instead. Find the speed the weights gain after being released from rest and trav√ eling a distance h. (b) Use your result from part a to find the acceleration, reproducing √ the result of the earlier problem.
Problem 17.
312
Chapter 12
Simplifying the energy zoo
Chapter 13
Work: the transfer of mechanical energy 13.1 Work: the transfer of mechanical energy The concept of work The mass contained in a closed system is a conserved quantity, but if the system is not closed, we also have ways of measuring the amount of mass that goes in or out. The water company does this with a meter that records your water use. Likewise, we often have a system that is not closed, and would like to know how much energy comes in or out. Energy, however, is not a physical substance like water, so energy transfer cannot be measured with the same kind of meter. How can we tell, for instance, how much useful energy a tractor can “put out” on one tank of gas? The law of conservation of energy guarantees that all the chem-
313
ical energy in the gasoline will reappear in some form, but not necessarily in a form that is useful for doing farm work. Tractors, like cars, are extremely inefficient, and typically 90% of the energy they consume is converted directly into heat, which is carried away by the exhaust and the air flowing over the radiator. We wish to distinguish the energy that comes out directly as heat from the energy that serves to accelerate a trailer or to plow a field, so we define a technical meaning of the ordinary word “work” to express the distinction: definition of work Work is the amount of energy transferred into or out of a system, not counting energy transferred by heat conduction. a / Work is a transfer of energy.
self-check A Based on this definition, is work a vector, or a scalar? What are its units? Answer, p. 543
The conduction of heat is to be distinguished from heating by friction. When a hot potato heats up your hands by conduction, the energy transfer occurs without any force, but when friction heats your car’s brake shoes, there is a force involved. The transfer of energy with and without a force are measured by completely different methods, so we wish to include heat transfer by frictional heating under the definition of work, but not heat transfer by conduction. The definition of work could thus be restated as the amount of energy transferred by forces. b / The tractor raises the weight over the pulley, increasing its gravitational potential energy.
Calculating work as force multiplied by distance The examples in figures b-d show that there are many different ways in which energy can be transferred. Even so, all these examples have two things in common: 1. A force is involved.
c / The tractor accelerates the trailer, increasing its kinetic energy.
d / The tractor pulls a plow. Energy is expended in frictional heating of the plow and the dirt, and in breaking dirt clods and lifting dirt up to the sides of the furrow.
2. The tractor travels some distance as it does the work. In b, the increase in the height of the weight, Δy, is the same as the distance the tractor travels, which we’ll call d. For simplicity, we discuss the case where the tractor raises the weight at constant speed, so that there is no change in the kinetic energy of the weight, and we assume that there is negligible friction in the pulley, so that the force the tractor applies to the rope is the same as the rope’s upward force on the weight. By Newton’s first law, these forces are also of the same magnitude as the earth’s gravitational force on the weight. The increase in the weight’s potential energy is given by F Δy, so the work done by the tractor on the weight equals F d, the product of the force and the distance moved: W = Fd
314
Chapter 13
Work: the transfer of mechanical energy
.
In example c, the tractor’s force on the trailer accelerates it, increasing its kinetic energy. If frictional forces on the trailer are negligible, then the increase in the trailer’s kinetic energy can be found using the same algebra that was used on page 303 to find the potential energy due to gravity. Just as in example b, we have W = Fd
.
Does this equation always give the right answer? Well, sort of. In example d, there are two quantities of work you might want to calculate: the work done by the tractor on the plow and the work done by the plow on the dirt. These two quantities can’t both equal F d. Most of the energy transmitted through the cable goes into frictional heating of the plow and the dirt. The work done by the plow on the dirt is less than the work done by the tractor on the plow, by an amount equal to the heat absorbed by the plow. It turns out that the equation W = F d gives the work done by the tractor, not the work done by the plow. How are you supposed to know when the equation will work and when it won’t? The somewhat complex answer is postponed until section 13.6. Until then, we will restrict ourselves to examples in which W = F d gives the right answer; essentially the reason the ambiguities come up is that when one surface is slipping past another, d may be hard to define, because the two surfaces move different distances.
e / The baseball pitcher put kinetic energy into the ball, so he did work on it. To do the greatest possible amount of work, he applied the greatest possible force over the greatest possible distance.
We have also been using examples in which the force is in the same direction as the motion, and the force is constant. (If the force was not constant, we would have to represent it with a function, not a symbol that stands for a number.) To summarize, we have: rule for calculating work (simplest version) The work done by a force can be calculated as W = Fd
,
if the force is constant and in the same direction as the motion. Some ambiguities are encountered in cases such as kinetic friction.
Section 13.1
Work: the transfer of mechanical energy
315
Mechanical work done in an earthquake example 1 In 1998, geologists discovered evidence for a big prehistoric earthquake in Pasadena, between 10,000 and 15,000 years ago. They found that the two sides of the fault moved 6.7 m relative to one another, and estimated that the force between them was 1.3 × 1017 N. How much energy was released?
f / Example 1.
Multiplying the force by the distance gives 9 × 1017 J. For comparison, the Northridge earthquake of 1994, which killed 57 people and did 40 billion dollars of damage, released 22 times less energy. The fall factor example 2 Counterintuitively, the rock climber may be in more danger in figure g/1 than later when she gets up to position g/2. Along her route, the climber has placed removable rock anchors (not shown) and carabiners attached to the anchors. She clips the rope into each carabiner so that it can travel but can’t pop out. In both 1 and 2, she has ascended a certain distance above her last anchor, so that if she falls, she will drop through a height h that is about twice this distance, and this fall height is about the same in both cases. In fact, h is somewhat larger than twice her height above the last anchor, because the rope is intentionally designed to stretch under the big force of a falling climber who suddenly brings it taut.
g / Example 2. Surprisingly, the climber is in more danger at 1 than at 2. The distance d is the amount by which the rope will stretch while work is done to transfer the kinetic energy of a fall out of her body.
To see why we want a stretchy rope, consider the equation F = W /d in the case where d is zero; F would theoretically become infinite. In a fall, the climber loses a fixed amount of gravitational energy mgh. This is transformed into an equal amount of kinetic energy as she falls, and eventually this kinetic energy has to be transferred out of her body when the rope comes up taut. If the rope was not stretchy, then the distance traveled at the point where the rope attaches to her harness would be zero, and the force exerted would theoretically be infinite. Before the rope reached the theoretically infinite tension F it would break (or her back would break, or her anchors would be pulled out of the rock). We want the rope to be stretchy enough to make d fairly big, so that dividing W by d gives a small force.1 In g/1 and g/2, the fall h is about the same. What is different is the length L of rope that has been paid out. A longer rope can stretch more, so the distance d traveled after the “catch” is proportional to L. Combining F = W /d, W ∝ h, and d ∝ L, we have F ∝ h/L. For these reasons, rock climbers define a fall factor f = h/L. The larger fall factor in g/1 is more dangerous.
1
Actually F isn’t constant, because the tension in the rope increases steadily as it stretches, but this is irrelevant to the present analysis.
316
Chapter 13
Work: the transfer of mechanical energy
Machines can increase force, but not work. Figure h shows a pulley arrangement for doubling the force supplied by the tractor (book 1, section 5.6). The tension in the lefthand rope is equal throughout, assuming negligible friction, so there are two forces pulling the pulley to the left, each equal to the original force exerted by the tractor on the rope. This doubled force is transmitted through the right-hand rope to the stump.
h / The pulley doubles the force the tractor can exert on the stump.
It might seem as though this arrangement would also double the work done by the tractor, but look again. As the tractor moves forward 2 meters, 1 meter of rope comes around the pulley, and the pulley moves 1 m to the left. Although the pulley exerts double the force on the stump, the pulley and stump only move half as far, so the work done on the stump is no greater that it would have been without the pulley. The same is true for any mechanical arrangement that increases or decreases force, such as the gears on a ten-speed bike. You can’t get out more work than you put in, because that would violate conservation of energy. If you shift gears so that your force on the pedals is amplified, the result is that you just have to spin the pedals more times. No work is done without motion. It strikes most students as nonsensical when they are told that if they stand still and hold a heavy bag of cement, they are doing no work on the bag. Even if it makes sense mathematically that W = F d gives zero when d is zero, it seems to violate common sense. You would certainly become tired! The solution is simple. Physicists have taken over the common word “work” and given it a new technical meaning, which is the transfer of energy. The energy of the bag of cement is not changing, and that is what the physicist means by saying no work is done on the bag. There is a transformation of energy, but it is taking place entirely within your own muscles, which are converting chemical energy into heat. Physiologically, a human muscle is not like a tree limb, which can support a weight indefinitely without the expenditure of energy. Each muscle cell’s contraction is generated by zillions of little molecular machines, which take turns supporting the tension. When a
Section 13.1
Work: the transfer of mechanical energy
317
particular molecule goes on or off duty, it moves, and since it moves while exerting a force, it is doing work. There is work, but it is work done by one molecule in a muscle cell on another. Positive and negative work When object A transfers energy to object B, we say that A does positive work on B. B is said to do negative work on A. In other words, a machine like a tractor is defined as doing positive work. This use of the plus and minus signs relates in a logical and consistent way to their use in indicating the directions of force and motion in one dimension. In figure i, suppose we choose a coordinate system with the x axis pointing to the right. Then the force the spring exerts on the ball is always a positive number. The ball’s motion, however, changes directions. The symbol d is really just a shorter way of writing the familiar quantity Δx, whose positive and negative signs indicate direction.
i / Whenever energy is transferred out of the spring, the same amount has to be transferred into the ball, and vice versa. As the spring compresses, the ball is doing positive work on the spring (giving up its KE and transferring energy into the spring as PE), and as it decompresses the ball is doing negative work (extracting energy).
While the ball is moving to the left, we use d < 0 to represent its direction of motion, and the work done by the spring, F d, comes out negative. This indicates that the spring is taking kinetic energy out of the ball, and accepting it in the form of its own potential energy. As the ball is reaccelerated to the right, it has d > 0, F d is positive, and the spring does positive work on the ball. Potential energy is transferred out of the spring and deposited in the ball as kinetic energy. In summary: rule for calculating work (including cases of negative work) The work done by a force can be calculated as W = Fd
,
if the force is constant and along the same line as the motion. The quantity d is to be interpreted as a synonym for Δx, i.e., positive and negative signs are used to indicate the direction of motion. Some ambiguities are encountered in cases such as kinetic friction. self-check B In figure i, what about the work done by the ball on the spring? Answer, p. 543
There are many examples where the transfer of energy out of an object cancels out the transfer of energy in. When the tractor pulls the plow with a rope, the rope does negative work on the tractor and positive work on the plow. The total work done by the rope is zero, which makes sense, since it is not changing its energy. It may seem that when your arms do negative work by lowering
318
Chapter 13
Work: the transfer of mechanical energy
a bag of cement, the cement is not really transferring energy into your body. If your body was storing potential energy like a compressed spring, you would be able to raise and lower a weight all day, recycling the same energy. The bag of cement does transfer energy into your body, but your body accepts it as heat, not as potential energy. The tension in the muscles that control the speed of the motion also results in the conversion of chemical energy to heat, for the same physiological reasons discussed previously in the case where you just hold the bag still. One of the advantages of electric cars over gasoline-powered cars is that it is just as easy to put energy back in a battery as it is to take energy out. When you step on the brakes in a gas car, the brake shoes do negative work on the rest of the car. The kinetic energy of the car is transmitted through the brakes and accepted by the brake shoes in the form of heat. The energy cannot be recovered. Electric cars, however, are designed to use regenerative braking. The brakes don’t use friction at all. They are electrical, and when you step on the brake, the negative work done by the brakes means they accept the energy and put it in the battery for later use. This is one of the reasons why an electric car is far better for the environment than a gas car, even if the ultimate source of the electrical energy happens to be the burning of oil in the electric company’s plant. The electric car recycles the same energy over and over, and only dissipates heat due to air friction and rolling resistance, not braking. (The electric company’s power plant can also be fitted with expensive pollutionreduction equipment that would be prohibitively expensive or bulky for a passenger car.)
j / Left: No mechanical work occurs in the man’s body while he holds himself motionless. There is a transformation of chemical energy into heat, but this happens at the microscopic level inside the tensed muscles. Right: When the woman lifts herself, her arms do positive work on her body, transforming chemical energy into gravitational potential energy and heat. On the way back down, the arms’ work is negative; gravitational potential energy is transformed into heat. (In exercise physiology, the man is said to be doing isometric exercise, while the woman’s is concentric and then eccentric.)
k / Because the force is in the opposite direction compared to the motion, the brake shoe does negative work on the drum, i.e., accepts energy from it in the form of heat.
Section 13.1
Work: the transfer of mechanical energy
319
Discussion questions A Besides the presence of a force, what other things differentiate the processes of frictional heating and heat conduction? B Criticize the following incorrect statement: “A force doesn’t do any work unless it’s causing the object to move.” C To stop your car, you must first have time to react, and then it takes some time for the car to slow down. Both of these times contribute to the distance you will travel before you can stop. The figure shows how the average stopping distance increases with speed. Because the stopping distance increases more and more rapidly as you go faster, the rule of one car length per 10 m.p.h. of speed is not conservative enough at high speeds. In terms of work and kinetic energy, what is the reason for the more rapid increase at high speeds?
Discussion question C.
13.2 Work in three dimensions A force perpendicular to the motion does no work. Suppose work is being done to change an object’s kinetic energy. A force in the same direction as its motion will speed it up, and a force in the opposite direction will slow it down. As we have already seen, this is described as doing positive work or doing negative work on the object. All the examples discussed up until now have been of motion in one dimension, but in three dimensions the force can be at any angle θ with respect to the direction of motion.
m / A force can do positive, negative, or zero work, depending on its direction relative to the direction of the motion.
What if the force is perpendicular to the direction of motion? We have already seen that a force perpendicular to the motion results in circular motion at constant speed. The kinetic energy does not change, and we conclude that no work is done when the force is perpendicular to the motion. So far we have been reasoning about the case of a single force acting on an object, and changing only its kinetic energy. The result is more generally true, however. For instance, imagine a hockey puck sliding across the ice. The ice makes an upward normal force, but does not transfer energy to or from the puck.
320
Chapter 13
Work: the transfer of mechanical energy
Forces at other angles Suppose the force is at some other angle with respect to the motion, say θ = 45◦ . Such a force could be broken down into two components, one along the direction of the motion and the other perpendicular to it. The force vector equals the vector sum of its two components, and the principle of vector addition of forces thus tells us that the work done by the total force cannot be any different than the sum of the works that would be done by the two forces by themselves. Since the component perpendicular to the motion does no work, the work done by the force must be W = F |d|
,
[work done by a constant force]
where the vector d is simply a less cumbersome version of the notation Δr. This result can be rewritten via trigonometry as W = |F||d| cos θ
.
[work done by a constant force]
Even though this equation has vectors in it, it depends only on their magnitudes, and the magnitude of a vector is a scalar. Work is therefore still a scalar quantity, which only makes sense if it is defined as the transfer of energy. Ten gallons of gasoline have the ability to do a certain amount of mechanical work, and when you pull in to a full-service gas station you don’t have to say “Fill ’er up with 10 gallons of south-going gas.”
n / Work is only done by the component of the force parallel to the motion.
Students often wonder why this equation involves a cosine rather than a sine, or ask if it would ever be a sine. In vector addition, the treatment of sines and cosines seemed more equal and democratic, so why is the cosine so special now? The answer is that if we are going to describe, say, a velocity vector, we must give both the component parallel to the x axis and the component perpendicular to the x axis (i.e., the y component). In calculating work, however, the force component perpendicular to the motion is irrelevant — it changes the direction of motion without increasing or decreasing the energy of the object on which it acts. In this context, it is only the parallel force component that matters, so only the cosine occurs. self-check C (a) Work is the transfer of energy. According to this definition, is the horse in the picture doing work on the pack? (b) If you calculate work by the method described in this section, is the horse in figure o doing work on the pack? Answer, p. 544
Pushing a broom example 3 If you exert a force of 21 N on a push broom, at an angle 35 degrees below horizontal, and walk for 5.0 m, how much work do you do? What is the physical significance of this quantity of work?
o / Self-check. (Breaking Trail, by Walter E. Bohl.)
Using the second equation above, the work done equals (21 N)(5.0 m)(cos 35◦ ) = 86 J
.
Section 13.2
Work in three dimensions
321
The form of energy being transferred is heat in the floor and the broom’s bristles. This comes from the chemical energy stored in your body. (The majority of the calories you burn are dissipated directly as heat inside your body rather than doing any work on the broom. The 86 J is only the amount of energy transferred through the broom’s handle.) A violin example 4 As a violinist draws the bow across a string, the bow hairs exert both a normal force and a kinetic frictional force on the string. The normal force is perpendicular to the direction of motion, and does no work. However, the frictional force is in the same direction as the motion of the bow, so it does work: energy is transferred to the string, causing it to vibrate. One way of playing a violin more loudly is to use longer strokes. Since W = F d, the greater distance results in more work. A second way of getting a louder sound is to press the bow more firmly against the strings. This increases the normal force, and although the normal force itself does no work, an increase in the normal force has the side effect of increasing the frictional force, thereby increasing W = F d. The violinist moves the bow back and forth, and sound is produced on both the “up-bow” (the stroke toward the player’s left) and the “down-bow” (to the right). One may, for example, play a series of notes in alternation between up-bows and down-bows. However, if the notes are of unequal length, the up and down motions tend to be unequal, and if the player is not careful, she can run out of bow in the middle of a note! To keep this from happening, one can move the bow more quickly on the shorter notes, but the resulting increase in d will make the shorter notes louder than they should be. A skilled player compensates by reducing the force. Up until now, we have not found any physically useful way to define the multiplication of two vectors. It would be possible, for instance, to multiply two vectors component by component to form a third vector, but there are no physical situations where such a multiplication would be useful. The equation W = |F||d| cos θ is an example of a sort of multiplication of vectors that is useful. The result is a scalar, not a vector, and this is therefore often referred to as the scalar product of the vectors F and d. There is a standard shorthand notation for this operation, A · B = |A||B| cos θ
,
[definition of the notation A · B;
θ is the angle between vectors A and B] and because of this notation, a more common term for this operation
322
Chapter 13
Work: the transfer of mechanical energy
is the dot product. In dot product notation, the equation for work is simply W =F·d . The dot product has the following geometric interpretation: A · B = |A|(component of B parallel to A) = |B|(component of A parallel to B) The dot product has some of the properties possessed by ordinary multiplication of numbers, A·B=B·A A · (B + C) = A · B + A · C (cA) · B = c (A · B)
,
but it lacks one other: the ability to undo multiplication by dividing. If you know the components of two vectors, you can easily calculate their dot product as follows: A · B = Ax Bx + Ay By + Az Bz
.
(This can be proved by first analyzing the special case where each vector has only an x component, and the similar cases for y and z. We can then use the rule A · (B + C) = A · B + A · C to make a generalization by writing each vector as the sum of its x, y, and z components. See homework problem 17.) Magnitude expressed with a dot product example 5 If we take the dot product of any vector b with itself, we find b · b = bx xˆ + by yˆ + bz zˆ · bx xˆ + by yˆ + bz zˆ = bx2 + by2 + bz2
,
so its magnitude can be expressed as |b| =
√
b·b
.
We will often write b2 to mean b · b, when the context makes it clear what is intended. For example, we could express kinetic energy as (1/2)m|v|2 , (1/2)mv·v, or (1/2)mv 2 . In the third version, nothing but context tells us that v really stands for the magnitude of some vector v. Towing a barge example 6 A mule pulls a barge with a force F=(1100 N)ˆx + (400 N)ˆy, and the total distance it travels is (1000 m)ˆx. How much work does it do? The dot product is 1.1 × 106 N·m = 1.1 × 106 J.
Section 13.2
Work in three dimensions
323
13.3 Varying force Up until now we have done no actual calculations of work in cases where the force was not constant. The question of how to treat such cases is mathematically analogous to the issue of how to generalize the equation (distance) = (velocity)(time) to cases where the velocity was not constant. There, we found that the correct generalization was to find the area under the graph of velocity versus time. The equivalent thing can be done with work: general rule for calculating work The work done by a force F equals the area under the curve on a graph of F versus x. (Some ambiguities are encountered in cases such as kinetic friction.) The examples in this section are ones in which the force is varying, but is always along the same line as the motion, so F is the same as F . self-check D In which of the following examples would it be OK to calculate work using F d , and in which ones would you have to use the area under the F − x graph? (a) A fishing boat cruises with a net dragging behind it. (b) A magnet leaps onto a refrigerator from a distance. (c) Earth’s gravity does work on an outward-bound space probe. Answer, p. 544 p / The spring does work on the cart. (Unlike the ball in section 13.1, the cart is attached to the spring.)
An important and straightforward example is the calculation of the work done by a spring that obeys Hooke’s law, F ≈ −k (x − xo )
.
The minus sign is because this is the force being exerted by the spring, not the force that would have to act on the spring to keep it at this position. That is, if the position of the cart in figure p is to the right of equilibrium, the spring pulls back to the left, and vice-versa.
q / The area of the shaded triangle gives the work done by the spring as the cart moves from the equilibrium position to position x .
We calculate the work done when the spring is initially at equilibrium and then decelerates the car as the car moves to the right. The work done by the spring on the cart equals the minus area of the shaded triangle, because the triangle hangs below the x axis. The area of a triangle is half its base multiplied by its height, so 1 W = − k (x − xo )2 2
.
This is the amount of kinetic energy lost by the cart as the spring decelerates it. It was straightforward to calculate the work done by the spring in this case because the graph of F versus x was a straight line, giving
324
Chapter 13
Work: the transfer of mechanical energy
a triangular area. But if the curve had not been so geometrically simple, it might not have been possible to find a simple equation for the work done, or an equation might have been derivable only using calculus. Optional section 13.4 gives an important example of such an application of calculus.
r / Example 7.
Energy production in the sun example 7 The sun produces energy through nuclear reactions in which nuclei collide and stick together. The figure depicts one such reaction, in which a single proton (hydrogen nucleus) collides with a carbon nucleus, consisting of six protons and six neutrons. Neutrons and protons attract other neutrons and protons via the strong nuclear force, so as the proton approaches the carbon nucleus it is accelerated. In the language of energy, we say that it loses nuclear potential energy and gains kinetic energy. Together, the seven protons and six neutrons make a nitrogen nucleus. Within the newly put-together nucleus, the neutrons and
Section 13.3
Varying force
325
protons are continually colliding, and the new proton’s extra kinetic energy is rapidly shared out among all the neutrons and protons. Soon afterward, the nucleus calms down by releasing some energy in the form of a gamma ray, which helps to heat the sun. The graph shows the force between the carbon nucleus and the proton as the proton is on its way in, with the distance in units of femtometers (1 fm=10−15 m). Amusingly, the force turns out to be a few newtons: on the same order of magnitude as the forces we encounter ordinarily on the human scale. Keep in mind, however, that a force this big exerted on a single subatomic particle such as a proton will produce a truly fantastic acceleration (on the order of 1027 m/s2 !). Why does the force have a peak around x = 3 fm, and become smaller once the proton has actually merged with the nucleus? At x = 3 fm, the proton is at the edge of the crowd of protons and neutrons. It feels many attractive forces from the left, and none from the right. The forces add up to a large value. However if it later finds itself at the center of the nucleus, x = 0, there are forces pulling it from all directions, and these force vectors cancel out. We can now calculate the energy released in this reaction by using the area under the graph to determine the amount of mechanical work done by the carbon nucleus on the proton. (For simplicity, we assume that the proton came in “aimed” at the center of the nucleus, and we ignore the fact that it has to shove some neutrons and protons out of the way in order to get there.) The area under the curve is about 17 squares, and the work represented by each square is (1 N)(10−15 m) = 10−15 J
,
so the total energy released is about (10−15 J/square)(17 squares) = 1.7 × 10−14 J
.
This may not seem like much, but remember that this is only a reaction between the nuclei of two out of the zillions of atoms in the sun. For comparison, a typical chemical reaction between two atoms might transform on the order of 10−19 J of electrical potential energy into heat — 100,000 times less energy! As a final note, you may wonder why reactions such as these only occur in the sun. The reason is that there is a repulsive electrical force between nuclei. When two nuclei are close together, the electrical forces are typically about a million times weaker than the nuclear forces, but the nuclear forces fall off much more quickly with distance than the electrical forces, so the electrical force is the dominant one at longer ranges. The sun is a very hot gas, so
326
Chapter 13
Work: the transfer of mechanical energy
the random motion of its atoms is extremely rapid, and a collision between two atoms is sometimes violent enough to overcome this initial electrical repulsion.
13.4
Applications of calculus
The student who has studied integral calculus will recognize that the graphical rule given in the previous section can be reexpressed as an integral, x2
F dx
W =
.
x1
We can then immediately find by the fundamental theorem of calculus that force is the derivative of work with respect to position, F =
dW dx
.
For example, a crane raising a one-ton block on the moon would be transferring potential energy into the block at only one sixth the rate that would be required on Earth, and this corresponds to one sixth the force. Although the work done by the spring could be calculated without calculus using the area of a triangle, there are many cases where the methods of calculus are needed in order to find an answer in closed form. The most important example is the work done by gravity when the change in height is not small enough to assume a constant force. Newton’s law of gravity is F =
GM m r2
,
which can be integrated to give
r2
GM m dr r2 r1 1 1 = −GM m − r2 r1
W =
.
Section 13.4
Applications of calculus
327
13.5 Work and potential energy The techniques for calculating work can also be applied to the calculation of potential energy. If a certain force depends only on the distance between the two participating objects, then the energy released by changing the distance between them is defined as the potential energy, and the amount of potential energy lost equals minus the work done by the force, ΔP E = −W
.
The minus sign occurs because positive work indicates that the potential energy is being expended and converted to some other form. It is sometimes convenient to pick some arbitrary position as a reference position, and derive an equation for once and for all that gives the potential energy relative to this position P Ex = −Wref→x
.
[potential energy at a point x]
To find the energy transferred into or out of potential energy, one then subtracts two different values of this equation. These equations might almost make it look as though work and energy were the same thing, but they are not. First, potential energy measures the energy that a system has stored in it, while work measures how much energy is transferred in or out. Second, the techniques for calculating work can be used to find the amount of energy transferred in many situations where there is no potential energy involved, as when we calculate the amount of kinetic energy transformed into heat by a car’s brake shoes. A toy gun example 8 A toy gun uses a spring with a spring constant of 10 N/m to shoot a ping-pong ball of mass 5 g. The spring is compressed to 10 cm shorter than its equilibrium length when the gun is loaded. At what speed is the ball released? The equilibrium point is the natural choice for a reference point. Using the equation found previously for the work, we have PEx =
1 k (x − xo )2 2
.
The spring loses contact with the ball at the equilibrium point, so the final potential energy is PEf = 0
.
The initial potential energy is 1 (10 N/m)(0.10 m)2 2 = 0.05 J.
PEi =
328
Chapter 13
Work: the transfer of mechanical energy
.
The loss in potential energy of 0.05 J means an increase in kinetic energy of the same amount. The velocity of the ball is found by solving the equation K E = (1/2)mv 2 for v , 2K E v= m (2)(0.05 J) = 0.005 kg = 4 m/s
.
Gravitational potential energy example 9 We have already found the equation ΔPE = −F Δy for the gravitational potential energy when the change in height is not enough to cause a significant change in the gravitational force F . What if the change in height is enough so that this assumption is no longer valid? Use the equation W = GMm(1/r2 − 1/r1 ) derived in section 13.4 to find the potential energy, using r = ∞ as a reference point. The potential energy equals minus the work that would have to be done to bring the object from r1 = ∞ to r = r2 , which is PE = −
GMm r
.
This is simpler than the equation for the work, which is an example of why it is advantageous to record an equation for potential energy relative to some reference point, rather than an equation for work. Although the equations derived in the previous two examples may seem arcane and not particularly useful except for toy designers and rocket scientists, their usefulness is actually greater than it appears. The equation for the potential energy of a spring can be adapted to any other case in which an object is compressed, stretched, twisted, or bent. While you are not likely to use the equation for gravitational potential energy for anything practical, it is directly analogous to an equation that is extremely useful in chemistry, which is the equation for the potential energy of an electron at a distance r from the nucleus of its atom. As discussed in more detail later in the course, the electrical force between the electron and the nucleus is proportional to 1/r2 , just like the gravitational force between two masses. Since the equation for the force is of the same form, so is the equation for the potential energy.
Section 13.5
Work and potential energy
329
s / The twin Voyager space probes were perhaps the greatest scientific successes of the space program. Over a period of decades, they flew by all the planets of the outer solar system, probably accomplishing more of scientific interest than the entire space shuttle program at a tiny fraction of the cost. Both Voyager probes completed their final planetary flybys with speeds greater than the escape velocity at that distance from the sun, and so headed on out of the solar system on hyperbolic orbits, never to return. Radio contact has been lost, and they are now likely to travel interstellar space for billions of years without colliding with anything or being detected by any intelligent species.
Discussion questions A What does the graph of PE = (1/2)k (x − xo )2 look like as a function of x ? Discuss the physical significance of its features. B What does the graph of PE = −GMm/r look like as a function of r ? Discuss the physical significance of its features. How would the equation and graph change if some other reference point was chosen rather than r = ∞? C Starting at a distance r from a planet of mass M , how fast must an object be moving in order to have a hyperbolic orbit, i.e., one that never comes back to the planet? This velocity is called the escape velocity. Interpreting the result, does it matter in what direction the velocity is? Does it matter what mass the object has? Does the object escape because it is moving too fast for gravity to act on it? D
Does a spring have an “escape velocity?”
E Calculus-based question: If the form of energy being transferred is F = dW /dx and W = F dx become potential energy, then the equations F = −dPE /dx and PE = − F dx . How would you then apply the following calculus concepts: zero derivative at minima and maxima, and the second derivative test for concavity up or down.
13.6 When does work equal force times distance? In the example of the tractor pulling the plow discussed on page 315, the work did not equal F d. The purpose of this section is to explain more fully how the quantity F d can and cannot be used. To simplify things, I write F d throughout this section, but more generally everything said here would be true for the area under the
330
Chapter 13
Work: the transfer of mechanical energy
graph of F versus d. The following two theorems allow most of the ambiguity to be cleared up. the work-kinetic-energy theorem The change in kinetic energy associated with the motion of an object’s center of mass is related to the total force acting on it and to the distance traveled by its center of mass according to the equation ΔKEcm = Ftotal dcm . This can be proved based on Newton’s second law and the equation KE = (1/2)mv 2 . Note that despite the traditional name, it does not necessarily tell the amount of work done, since the forces acting on the object could be changing other types of energy besides the KE associated with its center of mass motion. The second theorem does relate directly to work: When a contact force acts between two objects and the two surfaces do not slip past each other, the work done equals F d, where d is the distance traveled by the point of contact. This one has no generally accepted name, so we refer to it simply as the second theorem. A great number of physical situations can be analyzed with these two theorems, and often it is advantageous to apply both of them to the same situation. An ice skater pushing off from a wall example 10 The work-kinetic energy theorem tells us how to calculate the skater’s kinetic energy if we know the amount of force and the distance her center of mass travels while she is pushing off. The second theorem tells us that the wall does no work on the skater. This makes sense, since the wall does not have any source of energy. Absorbing an impact without recoiling? example 11 Is it possible to absorb an impact without recoiling? For instance, would a brick wall “give” at all if hit by a ping-pong ball? There will always be a recoil. In the example proposed, the wall will surely have some energy transferred to it in the form of heat and vibration. The second theorem tells us that we can only have nonzero work if the distance traveled by the point of contact is nonzero.
Section 13.6
When does work equal force times distance?
331
Dragging a refrigerator at constant velocity example 12 Newton’s first law tells us that the total force on the refrigerator must be zero: your force is canceling the floor’s kinetic frictional force. The work-kinetic energy theorem is therefore true but useless. It tells us that there is zero total force on the refrigerator, and that the refrigerator’s kinetic energy doesn’t change. The second theorem tells us that the work you do equals your hand’s force on the refrigerator multiplied by the distance traveled. Since we know the floor has no source of energy, the only way for the floor and refrigerator to gain energy is from the work you do. We can thus calculate the total heat dissipated by friction in the refrigerator and the floor. Note that there is no way to find how much of the heat is dissipated in the floor and how much in the refrigerator. Accelerating a cart example 13 If you push on a cart and accelerate it, there are two forces acting on the cart: your hand’s force, and the static frictional force of the ground pushing on the wheels in the opposite direction. Applying the second theorem to your force tells us how to calculate the work you do. Applying the second theorem to the floor’s force tells us that the floor does no work on the cart. There is no motion at the point of contact, because the atoms in the floor are not moving. (The atoms in the surface of the wheel are also momentarily at rest when they touch the floor.) This makes sense, since the floor does not have any source of energy. The work-kinetic energy theorem refers to the total force, and because the floor’s backward force cancels part of your force, the total force is less than your force. This tells us that only part of your work goes into the kinetic energy associated with the forward motion of the cart’s center of mass. The rest goes into rotation of the wheels.
13.7 The dot product Up until now, we have not found any physically useful way to define the multiplication of two vectors. It would be possible, for instance, to multiply two vectors component by component to form a third vector, but there are no physical situations where such a multiplication would be useful. The equation W = |F||d| cos θ is an example of a sort of multiplication of vectors that is useful. The result is a scalar, not a vector, and this is therefore often referred to as the scalar product of the vectors F and d. There is a standard shorthand notation for
332
Chapter 13
Work: the transfer of mechanical energy
this operation, A · B = |A||B| cos θ
,
[definition of the notation A · B;
θ is the angle between vectors A and B] and because of this notation, a more common term for this operation is the dot product. In dot product notation, the equation for work is simply W =F·d . The dot product has the following geometric interpretation: A · B = |A|(component of B parallel to A) = |B|(component of A parallel to B) The dot product has some of the properties possessed by ordinary multiplication of numbers, A·B=B·A A · (B + C) = A · B + A · C (cA) · B = c (A · B)
,
but it lacks one other: the ability to undo multiplication by dividing. If you know the components of two vectors, you can easily calculate their dot product as follows: A · B = Ax Bx + Ay By + Az Bz
.
(This can be proved by first analyzing the special case where each vector has only an x component, and the similar cases for y and z. We can then use the rule A · (B + C) = A · B + A · C to make a generalization by writing each vector as the sum of its x, y, and z components. See homework problem 17.) Magnitude expressed with a dot product example 14 If we take the dot product of any vector b with itself, we find b · b = bx xˆ + by yˆ + bz zˆ · bx xˆ + by yˆ + bz zˆ = bx2 + by2 + bz2
,
so its magnitude can be expressed as √ . |b| = b · b We will often write b2 to mean b · b, when the context makes it clear what is intended. For example, we could express kinetic energy as (1/2)m|v|2 , (1/2)mv·v, or (1/2)mv 2 . In the third version, nothing but context tells us that v really stands for the magnitude of some vector v. Towing a barge example 15 A mule pulls a barge with a force F=(1100 N)ˆx + (400 N)ˆy, and the total distance it travels is (1000 m)ˆx. How much work does it do? The dot product is 1.1 × 106 N·m = 1.1 × 106 J.
Section 13.7
The dot product
333
Summary Selected vocabulary work . . . . . . . . the amount of energy transferred into or out of a system, excluding energy transferred by heat conduction Notation W . . . . . . . . .
work
Summary Work is a measure of the transfer of mechanical energy, i.e., the transfer of energy by a force rather than by heat conduction. When the force is constant, work can usually be calculated as W = F |d|
,
[only if the force is constant]
where d is simply a less cumbersome notation for Δr, the vector from the initial position to the final position. Thus, • A force in the same direction as the motion does positive work, i.e., transfers energy into the object on which it acts. • A force in the opposite direction compared to the motion does negative work, i.e., transfers energy out of the object on which it acts. • When there is no motion, no mechanical work is done. The human body burns calories when it exerts a force without moving, but this is an internal energy transfer of energy within the body, and thus does not fall within the scientific definition of work. • A force perpendicular to the motion does no work. When the force is not constant, the above equation should be generalized as the area under the graph of F versus d. Machines such as pulleys, levers, and gears may increase or decrease a force, but they can never increase or decrease the amount of work done. That would violate conservation of energy unless the machine had some source of stored energy or some way to accept and store up energy. There are some situations in which the equation W = F |d| is ambiguous or not true, and these issues are discussed rigorously in section 13.6. However, problems can usually be avoided by analyzing the types of energy being transferred before plunging into the math. In any case there is no substitute for a physical understanding of the processes involved. The techniques developed for calculating work can also be applied to the calculation of potential energy. We fix some position
334
Chapter 13
Work: the transfer of mechanical energy
as a reference position, and calculate the potential energy for some other position, x, as P Ex = −Wref→x
.
The following two equations for potential energy have broader significance than might be suspected based on the limited situations in which they were derived: 1 P E = k (x − xo )2 2
.
[potential energy of a spring having spring constant k, when stretched or compressed from the equilibrium position xo ; analogous equations apply for the twisting, bending, compression, or stretching of any object.] GM m r [gravitational potential energy of objects of masses M and m, separated by a distance r; an analogous equation applies to the electrical potential energy of an electron in an atom.] PE = −
Summary
335
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 Two speedboats are identical, but one has more people aboard than the other. Although the total masses of the two boats are unequal, suppose that they happen to have the same kinetic energy. In a boat, as in a car, it’s important to be able to stop in time to avoid hitting things. (a) If the frictional force from the water is the same in both cases, how will the boats’ stopping distances compare? Explain. (b) Compare the times required for the boats to stop. 2 In each of the following situations, is the work being done positive, negative, or zero? (a) a bull paws the ground; (b) a fishing boat pulls a net through the water behind it; (c) the water resists the motion of the net through it; (d) you stand behind a pickup truck and lower a bale of hay from the truck’s bed to the ground. Explain. [Based on a problem by Serway and Faughn.] 3 In the earth’s atmosphere, the molecules are constantly moving around. Because temperature is a measure of kinetic energy per molecule, the average kinetic energy of each type of molecule is the same, e.g., the average KE of the O2 molecules is the same as the average KE of the N2 molecules. (a) If the mass of an O2 molecule is eight times greater than that of a He atom, what is the ratio of their average speeds? Which way is the ratio, i.e., which is typically moving faster? (b) Use your result from part a to explain why any helium occurring naturally in the atmosphere has long since escaped into outer space, never to return. (Helium is obtained commercially by extracting it from rocks.) You may want to do problem 21 first, for insight. 4 Weiping lifts a rock with a weight of 1.0 N through a height of 1.0 m, and then lowers it back down to the starting point. Bubba pushes a table 1.0 m across the floor at constant speed, requiring a force of 1.0 N, and then pushes it back to where it started. (a) Compare the total work done by Weiping and Bubba. (b) Check that your answers to part a make sense, using the definition of work: work is the transfer of energy. In your answer, you’ll need to discuss what specific type of energy is involved in each case. 5 In one of his more flamboyant moments, Galileo wrote “Who does not know that a horse falling from a height of three or four cubits will break his bones, while a dog falling from the same height or a cat from a height of eight or ten cubits will suffer no injury? Equally harmless would be the fall of a grasshopper from a tower or the fall of an ant from the distance of the moon.” Find the speed
336
Chapter 13
Work: the transfer of mechanical energy
of an ant that falls to earth from the distance of the moon at the moment when it is about to enter the atmosphere. Assume it is released from a point that is not actually near the moon, so the moon’s gravity is negligible. You will need the result of example 9 √ on p. 329. 6
[Problem 6 has been deleted.]
7 (a) The crew of an 18th century warship is raising the anchor. The anchor has a mass of 5000 kg. The water is 30 m deep. The chain to which the anchor is attached has a mass per unit length of 150 kg/m. Before they start raising the anchor, what is the total weight of the anchor plus the portion of the chain hanging out of the ship? (Assume that the buoyancy of the anchor and is negligible.) (b) After they have raised the anchor by 1 m, what is the weight they are raising? (c) Define y = 0 when the anchor is resting on the bottom, and y = +30 m when it has been raised up to the ship. Draw a graph of the force the crew has to exert to raise the anchor and chain, as a function of y. (Assume that they are raising it slowly, so water resistance is negligible.) It will not be a constant! Now find the area under the graph, and determine the work done by the crew in raising the anchor, in joules. √ (d) Convert your answer from (c) into units of kcal. 8 In the power stroke of a car’s gasoline engine, the fuel-air mixture is ignited by the spark plug, explodes, and pushes the piston out. The exploding mixture’s force on the piston head is greatest at the beginning of the explosion, and decreases as the mixture expands. It can be approximated by F = a/x, where x is the distance from the cylinder to the piston head, and a is a constant with units of N·m. (Actually a/x1.4 would be more accurate, but the problem works out more nicely with a/x!) The piston begins its stroke at x = x1 , and ends at x = x2 . The 1965 Rambler had six cylinders, each with a = 220 N·m, x1 = 1.2 cm, and x2 = 10.2 cm. (a) Draw a neat, accurate graph of F vs x, on graph paper. (b) From the area under the curve, derive the amount of work done √ in one stroke by one cylinder. (c) Assume the engine is running at 4800 r.p.m., so that during one minute, each of the six cylinders performs 2400 power strokes. (Power strokes only happen every other revolution.) Find the en√ gine’s power, in units of horsepower (1 hp=746 W). (d) The compression ratio of an engine is defined as x2 /x1 . Explain in words why the car’s power would be exactly the same if x1 and x2 were, say, halved or tripled, maintaining the same compression ratio of 8.5. Explain why this would not quite be true with the more realistic force equation F = a/x1.4 .
Problem 8: A cylinder from the 1965 Rambler’s engine. The piston is shown in its pushed out position. The two bulges at the top are for the valves that let fresh air-gas mixture in. Based on a figure from Motor Service’s Automotive Encyclopedia, Toboldt and Purvis.
Problems
337
9 The magnitude of the force between two magnets separated by a distance r can be approximated as kr−3 for large values of r. The constant k depends on the strengths of the magnets and the relative orientations of their north and south poles. Two magnets are released on a slippery surface at an initial distance ri , and begin sliding towards each other. What will be the total kinetic energy of the two magnets when they reach a final distance rf ?√(Ignore friction.) 10 A car starts from rest at t = 0, and starts speeding up with constant acceleration. (a) Find the car’s kinetic energy in terms of its mass, m, acceleration, a, and the time, t. (b) Your answer in the previous part also equals the amount of work, W , done from t = 0 until time t. Take the derivative of the previous expression to find the power expended by the car at time t. (c) Suppose two cars with the same mass both start from rest at the same time, but one has twice as much acceleration as the other. At any moment, how many times more power is being dissipated by the more√quickly accelerating car? (The answer is not 2.) 11 A space probe of mass m is dropped into a previously unexplored spherical cloud of gas and dust, and accelerates toward the center of the cloud under the influence of the cloud’s gravity. Measurements of its velocity allow its potential energy, P E, to be determined as a function of the distance r from the cloud’s center. The mass in the cloud is distributed in a spherically symmetric way, so its density, ρ(r), depends only on r and not on the angular coordinates. Show that by finding P E, one can infer ρ(r) as follows: d 1 ρ(r) = 2 4πGmr dr
r
2 dP E
dr
.
12 A rail gun is a device like a train on a track, with the train propelled by a powerful electrical pulse. Very high speeds have been demonstrated in test models, and rail guns have been proposed as an alternative to rockets for sending into outer space any object that would be strong enough to survive the extreme accelerations. Suppose that the rail gun capsule is launched straight up, and that the force of air friction acting on it is given by F = be−cx , where x is the altitude, b and c are constants, and e is the base of natural logarithms. The exponential decay occurs because the atmosphere gets thinner with increasing altitude. (In reality, the force would probably drop off even faster than an exponential, because the capsule would be slowing down somewhat.) Find the amount of kinetic energy lost by the capsule due to air friction between when it is launched and when it is completely beyond the atmosphere. (Gravity is negligible, since the air friction force is much greater than the √ gravitational force.) 338
Chapter 13
Work: the transfer of mechanical energy
13 A certain binary star system consists of two stars with masses m1 and m2 , separated by a distance b. A comet, originally nearly at rest in deep space, drops into the system and at a certain point in time arrives at the midpoint between the two stars. For that moment in time, find its velocity, v, symbolically in terms of b,√m1 , m2 , and fundamental constants. 14 An airplane flies in the positive direction along the x axis, through crosswinds that exert a force F = (a + bx)ˆ x + (c + dx)ˆ y. Find the work done by the wind on the plane, and by the plane on the wind, in traveling from the origin to position x. 15 In 1935, Yukawa proposed an early theory of the force that held the neutrons and protons together in the nucleus. His equation for the potential energy of two such particles, at a center-tocenter distance r, was P E(r) = gr−1 e−r/a , where g parametrizes the strength of the interaction, e is the base of natural logarithms, and would a is about 10−15 m. Find the force between two nucleons that √ be consistent with this equation for the potential energy. 16 Prove that the dot product defined in section 13.7 is rotationally invariant in the sense of section 7.5. 17 Fill in the details of the proof of A·B = Ax Bx +Ay By +Az Bz on page 333. 18
Does it make sense to say that work is conserved? Solution, p. 538
19 (a) Suppose work is done in one-dimensional motion. What happens to the work if you reverse the direction of the positive coordinate axis? Base your answer directly on the definition of work. (b) Now answer the question based on the W = F d rule. 20 A microwave oven works by twisting molecules one way and then the other, counterclockwise and then clockwise about their own centers, millions of times a second. If you put an ice cube or a stick of butter in a microwave, you’ll observe that the solid doesn’t heat very quickly, although eventually melting begins in one small spot. Once this spot forms, it grows rapidly, while the rest of the solid remains solid; it appears that a microwave oven heats a liquid much more rapidly than a solid. Explain why this should happen, based on the atomic-level description of heat, solids, and liquids. (See, e.g., figure b on page 301.) Don’t repeat the following common mistakes: In a solid, the atoms are packed more tightly and have less space between them. Not true. Ice floats because it’s less dense than water. In a liquid, the atoms are moving much faster. No, the difference in average speed between ice at −1◦ C and water at 1◦ C is only 0.4%.
Problems
339
21 Starting at a distance r from a planet of mass M , how fast must an object be moving in order to have a hyperbolic orbit, i.e., one that never comes back to the planet? This velocity is called the escape velocity. Interpreting the result, does it matter in what direction the velocity is? Does it matter what mass the object has? Does the object escape because it is moving too fast for gravity to √ act on it? 22 The figure, redrawn from Gray’s Anatomy, shows the tension of which a muscle is capable. The variable x is defined as the contraction of the muscle from its maximum length L, so that at x = 0 the muscle has length L, and at x = L the muscle would theoretically have zero length. In reality, the muscle can only contract to x = cL, where c is less than 1. When the muscle is extended to its maximum length, at x = 0, it is capable of the greatest tension, To . As the muscle contracts, however, it becomes weaker. Gray suggests approximating this function as a linear decrease, which would theoretically extrapolate to zero at x = L. (a) Find the maximum work the muscle can do in one contraction, in terms of c, L, and √ To . (b) Show that your answer to part a has the right units. (c) Show that your answer to part a has the right behavior when c = 0 and when c = 1. (d) Gray also states that the absolute maximum tension To has been found to be approximately proportional to the muscle’s crosssectional area A (which is presumably measured at x = 0), with proportionality constant k. Approximating the muscle as a cylinder, show that your answer from part a can be reexpressed in terms √ of the volume, V , eliminating L and A. (e) Evaluate your result numerically for a biceps muscle with a volume of 200 cm3 , with c = 0.8 and k = 100 N/cm2 as estimated by √ Gray.
Problem 22.
23 A car accelerates from rest. At low speeds, its acceleration is limited by static friction, so that if we press too hard on the gas, we will “burn rubber” (or, for many newer cars, a computerized traction-control system will override the gas pedal). At higher speeds, the limit on acceleration comes from the power of the engine, which puts a limit on how fast kinetic energy can be developed. (a) Show that if a force F is applied to an object moving at speed v, the power required is given by P = vF . (b) Find the speed v at which we cross over from the first regime described above to the second. At speeds higher than this, the engine does not have enough power to burn rubber. Express your result in terms of the car’s power P , its mass m, the coefficient of static √ friction μs , and g. (c) Show that your answer to part b has units that make sense.
340
Chapter 13
Work: the transfer of mechanical energy
(d) Show that the dependence of your answer on each of the four variables makes sense physically. (e) The 2010 Maserati Gran Turismo Convertible has a maximum power of 3.23 × 105 W (433 horsepower) and a mass (including a 50kg driver) of 2.03 × 103 kg. (This power is the maximum the engine can supply at its optimum frequency of 7600 r.p.m. Presumably the automatic transmission is designed so a gear is available in which the engine will be running at very nearly this frequency when the car is at moving at v.) Rubber on asphalt has μs ≈ 0.9. Find v for this car. Answer: 18 m/s, or about 40 miles per hour. (f) Our analysis has neglected air friction, which can probably be approximated as a force proportional to v 2 . The existence of this force is the reason that the car has a maximum speed, which is 176 miles per hour. To get a feeling for how good an approximation it is to ignore air friction, find what fraction of the engine’s maximum power is being used to overcome air resistance when the car is moving at the speed v found in part e. Answer: 1% 24 Most modern bow hunters in the U.S. use a fancy mechanical bow called a compound bow, which looks nothing like what most people imagine when they think of a bow and arrow. It has a system of pulleys designed to produce the force curve shown in the figure, where F is the force required to pull the string back, and x is the distance between the string and the center of the bow’s body. It is not a linear Hooke’s-law graph, as it would be for an old-fashioned bow. The big advantage of the design is that relatively little force is required to hold the bow stretched to point B on the graph. This is the force required from the hunter in order to hold the bow ready while waiting for a shot. Since it may be necessary to wait a long time, this force can’t be too big. An old-fashioned bow, designed to require the same amount of force when fully drawn, would shoot arrows at much lower speeds, since its graph would be a straight line from A to B. For the graph shown in the figure (taken from realistic data), find the speed at which a 26 g arrow is released, assuming that 70% of the mechanical work done by the hand is actually transmitted to the arrow. (The other 30% is lost to frictional heating inside√the bow and kinetic energy of the recoiling and vibrating bow.)
Problem 24.
Problems
341
342
Chapter 13
Work: the transfer of mechanical energy
Pool balls exchange momentum.
Chapter 14
Conservation of momentum In many subfields of physics these days, it is possible to read an entire issue of a journal without ever encountering an equation involving force or a reference to Newton’s laws of motion. In the last hundred and fifty years, an entirely different framework has been developed for physics, based on conservation laws. The new approach is not just preferred because it is in fashion. It applies inside an atom or near a black hole, where Newton’s laws do not. Even in everyday situations the new approach can be superior. We have already seen how perpetual motion machines could be designed that were too complex to be easily debunked by Newton’s laws. The beauty of conservation laws is that they tell us something must remain the same, regardless of the complexity of the process. So far we have discussed only two conservation laws, the laws of conservation of mass and energy. Is there any reason to believe that further conservation laws are needed in order to replace Newton’s laws as a complete description of nature? Yes. Conservation of mass and energy do not relate in any way to the three dimensions of space, because both are scalars. Conservation of energy, for instance, does not prevent the planet earth from abruptly making a 90-degree turn and heading straight into the sun, because kinetic energy does not depend on direction. In this chapter, we develop a new conserved quantity, called momentum, which is a vector.
343
14.1 Momentum A conserved quantity of motion Your first encounter with conservation of momentum may have come as a small child unjustly confined to a shopping cart. You spot something interesting to play with, like the display case of imported wine down at the end of the aisle, and decide to push the cart over there. But being imprisoned by Dad in the cart was not the only injustice that day. There was a far greater conspiracy to thwart your young id, one that originated in the laws of nature. Pushing forward did nudge the cart forward, but it pushed you backward. If the wheels of the cart were well lubricated, it wouldn’t matter how you jerked, yanked, or kicked off from the back of the cart. You could not cause any overall forward motion of the entire system consisting of the cart with you inside. In the Newtonian framework, we describe this as arising from Newton’s third law. The cart made a force on you that was equal and opposite to your force on it. In the framework of conservation laws, we cannot attribute your frustration to conservation of energy. It would have been perfectly possible for you to transform some of the internal chemical energy stored in your body to kinetic energy of the cart and your body. The following characteristics of the situation suggest that there may be a new conservation law involved: A closed system is involved. All conservation laws deal with closed systems. You and the cart are a closed system, since the well-oiled wheels prevent the floor from making any forward force on you. Something remains unchanged. The overall velocity of the system started out being zero, and you cannot change it. This vague reference to “overall velocity” can be made more precise: it is the velocity of the system’s center of mass that cannot be changed. Something can be transferred back and forth without changing the total amount. If we define forward as positive and backward as negative, then one part of the system can gain positive motion if another part acquires negative motion. If we don’t want to worry about positive and negative signs, we can imagine that the whole cart was initially gliding forward on its well-oiled wheels. By kicking off from the back of the cart, you could increase your own velocity, but this inevitably causes the cart to slow down.
344
Chapter 14
Conservation of momentum
It thus appears that there is some numerical measure of an object’s quantity of motion that is conserved when you add up all the objects within a system. Momentum Although velocity has been referred to, it is not the total velocity of a closed system that remains constant. If it was, then firing a gun would cause the gun to recoil at the same velocity as the bullet! The gun does recoil, but at a much lower velocity than the bullet. Newton’s third law tells us Fgun
on bullet
= −Fbullet
on gun
,
and assuming a constant force for simplicity, Newton’s second law allows us to change this to mbullet
Δvgun Δvbullet = −mgun Δt Δt
.
Thus if the gun has 100 times more mass than the bullet, it will recoil at a velocity that is 100 times smaller and in the opposite direction, represented by the opposite sign. The quantity mv is therefore apparently a useful measure of motion, and we give it a name, momentum, and a symbol, p. (As far as I know, the letter “p” was just chosen at random, since “m” was already being used for mass.) The situations discussed so far have been one-dimensional, but in three-dimensional situations it is treated as a vector. definition of momentum for material objects The momentum of a material object, i.e., a piece of matter, is defined as p = mv , the product of the object’s mass and its velocity vector.
The units of momentum are kg·m/s, and there is unfortunately no abbreviation for this clumsy combination of units. The reasoning leading up to the definition of momentum was all based on the search for a conservation law, and the only reason why we bother to define such a quantity is that experiments show it is conserved: the law of conservation of momentum In any closed system, the vector sum of all the momenta remains constant, , p1i + p2i + . . . = p1f + p2f + . . . where i labels the initial and f the final momenta. (A closed system is one on which no external forces act.)
Section 14.1
Momentum
345
This chapter first addresses the one-dimensional case, in which the direction of the momentum can be taken into account by using plus and minus signs. We then pass to three dimensions, necessitating the use of vector addition. A subtle point about conservation laws is that they all refer to “closed systems,” but “closed” means different things in different cases. When discussing conservation of mass, “closed” means a system that doesn’t have matter moving in or out of it. With energy, we mean that there is no work or heat transfer occurring across the boundary of the system. For momentum conservation, “closed” means there are no external forces reaching into the system. A cannon example 1 A cannon of mass 1000 kg fires a 10-kg shell at a velocity of 200 m/s. At what speed does the cannon recoil? The law of conservation of momentum tells us that pcannon,i + pshell,i = pcannon,f + pshell,f
.
Choosing a coordinate system in which the cannon points in the positive direction, the given information is pcannon,i = 0 pshell,i = 0 pshell,f = 2000 kg·m/s
.
We must have pcannon,f = −2000 kg·m/s, so the recoil velocity of the cannon is −2 m/s.
Ion drive for propelling spacecraft example 2 The experimental solar-powered ion drive of the Deep Space 1 space probe expels its xenon gas exhaust at a speed of 30,000 m/s, ten times faster than the exhaust velocity for a typical chemical-fuel rocket engine. Roughly how many times greater is the maximum speed this spacecraft can reach, compared with a chemical-fueled probe with the same mass of fuel (“reaction mass”) available for pushing out the back as exhaust? Momentum equals mass multiplied by velocity. Both spacecraft are assumed to have the same amount of reaction mass, and the ion drive’s exhaust has a velocity ten times greater, so the momentum of its exhaust is ten times greater. Before the engine starts firing, neither the probe nor the exhaust has any momentum, so the total momentum of the system is zero. By conservation of momentum, the total momentum must also be zero after
346
Chapter 14
Conservation of momentum
a / The ion drive engine of the NASA Deep Space 1 probe, shown under construction (left) and being tested in a vacuum chamber (right) prior to its October 1998 launch. Intended mainly as a test vehicle for new technologies, the craft nevertheless carried out a successful scientific program that included a flyby of a comet.
all the exhaust has been expelled. If we define the positive direction as the direction the spacecraft is going, then the negative momentum of the exhaust is canceled by the positive momentum of the spacecraft. The ion drive allows a final speed that is ten times greater. (This simplified analysis ignores the fact that the reaction mass expelled later in the burn is not moving backward as fast, because of the forward speed of the already-moving spacecraft.) Generalization of the momentum concept As with all the conservation laws, the law of conservation of momentum has evolved over time. In the 1800’s it was found that a beam of light striking an object would give it some momentum, even though light has no mass, and would therefore have no momentum
Section 14.1
Momentum
347
according to the above definition. Rather than discarding the principle of conservation of momentum, the physicists of the time decided to see if the definition of momentum could be extended to include momentum carried by light. The process is analogous to the process outlined on page 285 for identifying new forms of energy. The first step was the discovery that light could impart momentum to matter, and the second step was to show that the momentum possessed by light could be related in a definite way to observable properties of the light. They found that conservation of momentum could be successfully generalized by attributing to a beam of light a momentum vector in the direction of the light’s motion and having a magnitude proportional to the amount of energy the light possessed. The momentum of light is negligible under ordinary circumstances, e.g., a flashlight left on for an hour would only absorb about 10−5 kg·m/s of momentum as it recoiled.
b / Steam and other gases boiling off of the nucleus of Halley’s comet. This close-up photo was taken by the European Giotto space probe, which passed within 596 km of the nucleus on March 13, 1986.
The tail of a comet example 3 Momentum is not always equal to mv . Like many comets, Halley’s comet has a very elongated elliptical orbit. About once per century, its orbit brings it close to the sun. The comet’s head, or nucleus, is composed of dirty ice, so the energy deposited by the intense sunlight boils off steam and dust, b. The sunlight does not just carry energy, however — it also carries momentum. The momentum of the sunlight impacting on the smaller dust particles pushes them away from the sun, forming a tail, c. By analogy with matter, for which momentum equals mv , you would expect that massless light would have zero momentum, but the equation p = mv is not the correct one for light, and light does have momentum. (The gases typically form a second, distinct tail whose motion is controlled by the sun’s magnetic field.) The reason for bringing this up is not so that you can plug numbers into a formulas in these exotic situations. The point is that the conservation laws have proven so sturdy exactly because they can easily be amended to fit new circumstances. Newton’s laws are no longer at the center of the stage of physics because they did not have the same adaptability. More generally, the moral of this story is the provisional nature of scientific truth. It should also be noted that conservation of momentum is not a consequence of Newton’s laws, as is often asserted in textbooks. Newton’s laws do not apply to light, and therefore could not possibly be used to prove anything about a concept as general as the conservation of momentum in its modern form. Momentum compared to kinetic energy
c / Halley’s comet, in a much less magnified view from a ground-based telescope.
348
Chapter 14
Momentum and kinetic energy are both measures of the quantity of motion, and a sideshow in the Newton-Leibnitz controversy over who invented calculus was an argument over whether mv (i.e., momentum) or mv 2 (i.e., kinetic energy without the 1/2 in front)
Conservation of momentum
was the “true” measure of motion. The modern student can certainly be excused for wondering why we need both quantities, when their complementary nature was not evident to the greatest minds of the 1700’s. The following table highlights their differences.
kinetic energy . . . is a scalar. is not changed by a force perpendicular to the motion, which changes only the direction of the velocity vector.
momentum . . . is a vector is changed by any force, since a change in either the magnitude or the direction of the velocity vector will result in a change in the momentum vector. is always positive, and cannot cancel cancels with momentum in the opout. posite direction. can be traded for other forms of en- is always conserved in a closed sysergy that do not involve motion. KE tem. is not a conserved quantity by itself. is quadrupled if the velocity is dou- is doubled if the velocity is doubled. bled.
A spinning top example 4 A spinning top has zero total momentum, because for every moving point, there is another point on the opposite side that cancels its momentum. It does, however, have kinetic energy. Why a tuning fork has two prongs example 5 A tuning fork is made with two prongs so that they can vibrate in opposite directions, canceling their momenta. In a hypothetical version with only one prong, the momentum would have to oscillate, and this momentum would have to come from somewhere, such as the hand holding the fork. The result would be that vibrations would be transmitted to the hand and rapidly die out. In a two-prong fork, the two momenta cancel, but the energies don’t. Momentum and kinetic energy in firing a rifle example 6 The rifle and bullet have zero momentum and zero kinetic energy to start with. When the trigger is pulled, the bullet gains some momentum in the forward direction, but this is canceled by the rifle’s backward momentum, so the total momentum is still zero. The kinetic energies of the gun and bullet are both positive scalars, however, and do not cancel. The total kinetic energy is allowed to increase, because kinetic energy is being traded for other forms of energy. Initially there is chemical energy in the gunpowder. This chemical energy is converted into heat, sound, and kinetic energy. The gun’s “backward” kinetic energy does not refrigerate the shooter’s shoulder!
Section 14.1
d / Examples 4 and 5. The momenta cancel, but the energies don’t.
Momentum
349
The wobbly earth example 7 As the moon completes half a circle around the earth, its motion reverses direction. This does not involve any change in kinetic energy, and the earth’s gravitational force does not do any work on the moon. The reversed velocity vector does, however, imply a reversed momentum vector, so conservation of momentum in the closed earth-moon system tells us that the earth must also change its momentum. In fact, the earth wobbles in a little “orbit” about a point below its surface on the line connecting it and the moon. The two bodies’ momentum vectors always point in opposite directions and cancel each other out. The earth and moon get a divorce example 8 Why can’t the moon suddenly decide to fly off one way and the earth the other way? It is not forbidden by conservation of momentum, because the moon’s newly acquired momentum in one direction could be canceled out by the change in the momentum of the earth, supposing the earth headed the opposite direction at the appropriate, slower speed. The catastrophe is forbidden by conservation of energy, because both their energies would have to increase greatly. Momentum and kinetic energy of a glacier example 9 A cubic-kilometer glacier would have a mass of about 1012 kg. If it moves at a speed of 10−5 m/s, then its momentum is 107 kg · m/s. This is the kind of heroic-scale result we expect, perhaps the equivalent of the space shuttle taking off, or all the cars in LA driving in the same direction at freeway speed. Its kinetic energy, however, is only 50 J, the equivalent of the calories contained in a poppy seed or the energy in a drop of gasoline too small to be seen without a microscope. The surprisingly small kinetic energy is because kinetic energy is proportional to the square of the velocity, and the square of a small number is an even smaller number. Discussion questions A If all the air molecules in the room settled down in a thin film on the floor, would that violate conservation of momentum as well as conservation of energy? B A refrigerator has coils in back that get hot, and heat is molecular motion. These moving molecules have both energy and momentum. Why doesn’t the refrigerator need to be tied to the wall to keep it from recoiling from the momentum it loses out the back?
350
Chapter 14
Conservation of momentum
14.2 Collisions in one dimension Physicists employ the term “collision” in a broader sense than ordinary usage, applying it to any situation where objects interact for a certain period of time. A bat hitting a baseball, a radioactively emitted particle damaging DNA, and a gun and a bullet going their separate ways are all examples of collisions in this sense. Physical contact is not even required. A comet swinging past the sun on a hyperbolic orbit is considered to undergo a collision, even though it never touches the sun. All that matters is that the comet and the sun exerted gravitational forces on each other. The reason for broadening the term “collision” in this way is that all of these situations can be attacked mathematically using the same conservation laws in similar ways. In the first example, conservation of momentum is all that is required. Getting rear-ended example 10 Ms. Chang is rear-ended at a stop light by Mr. Nelson, and sues to make him pay her medical bills. He testifies that he was only going 35 miles per hour when he hit Ms. Chang. She thinks he was going much faster than that. The cars skidded together after the impact, and measurements of the length of the skid marks and the coefficient of friction show that their joint velocity immediately after the impact was 19 miles per hour. Mr. Nelson’s Nissan weighs 3100 pounds, and Ms. Chang ’s Cadillac weighs 5200 pounds. Is Mr. Nelson telling the truth?
e / This Hubble Space Telescope photo shows a small galaxy (yellow blob in the lower right) that has collided with a larger galaxy (spiral near the center), producing a wave of star formation (blue track) due to the shock waves passing through the galaxies’ clouds of gas. This is considered a collision in the physics sense, even though it is statistically certain that no star in either galaxy ever struck a star in the other. (This is because the stars are very small compared to the distances between them.)
Since the cars skidded together, we can write down the equation for conservation of momentum using only two velocities, v for Mr. Nelson’s velocity before the crash, and v for their joint velocity afterward: . mN v = mN v + mC v Solving for the unknown, v , we find mC v = 1+ v mN
.
Although we are given the weights in pounds, a unit of force, the ratio of the masses is the same as the ratio of the weights, and we find v = 51 miles per hour. He is lying. The above example was simple because both cars had the same velocity afterward. In many one-dimensional collisions, however, the two objects do not stick. If we wish to predict the result of such a collision, conservation of momentum does not suffice, because both velocities after the collision are unknown, so we have one equation in two unknowns. Conservation of energy can provide a second equation, but its application is not as straightforward, because kinetic energy is only the particular form of energy that has to do with motion. In many
Section 14.2
Collisions in one dimension
351
collisions, part of the kinetic energy that was present before the collision is used to create heat or sound, or to break the objects or permanently bend them. Cars, in fact, are carefully designed to crumple in a collision. Crumpling the car uses up energy, and that’s good because the goal is to get rid of all that kinetic energy in a relatively safe and controlled way. At the opposite extreme, a superball is “super” because it emerges from a collision with almost all its original kinetic energy, having only stored it briefly as potential energy while it was being squashed by the impact. Collisions of the superball type, in which almost no kinetic energy is converted to other forms of energy, can thus be analyzed more thoroughly, because they have KEf = KEi , as opposed to the less useful inequality KEf < KEi for a case like a tennis ball bouncing on grass. Pool balls colliding head-on example 11 Two pool balls collide head-on, so that the collision is restricted to one dimension. Pool balls are constructed so as to lose as little kinetic energy as possible in a collision, so under the assumption that no kinetic energy is converted to any other form of energy, what can we predict about the results of such a collision? Pool balls have identical masses, so we use the same symbol m for both. Conservation of momentum and no loss of kinetic energy give us the two equations
Gory Details of the Proof in Example 11 The equation A + B = C + D says that the change in one ball’s velocity is equal and opposite to the change in the other’s. We invent a symbol x = C − A for the change in ball 1’s velocity. The second equation can then be rewritten as A2 + B 2 = (A + x )2 +(B − x )2 . Squaring out the quantities in parentheses and then simplifying, we get 0 = Ax − Bx + x 2 . The equation has the trivial solution x = 0, i.e., neither ball’s velocity is changed, but this is physically impossible because the balls can’t travel through each other like ghosts. Assuming x = 0, we can divide by x and solve for x = B − A. This means that ball 1 has gained an amount of velocity exactly right to match ball 2’s initial velocity, and viceversa. The balls must have swapped velocities.
352
Chapter 14
mv1i + mv2i = mv1f + mv2f 1 1 1 1 2 2 + mv2f mv1i2 + mv2i2 = mv1f 2 2 2 2 The masses and the factors of 1/2 can be divided out, and we eliminate the cumbersome subscripts by replacing the symbols v1i ,... with the symbols A, B, C, and D: A+B =C+D 2
A + B2 = C 2 + D2
.
A little experimentation with numbers shows that given values of A and B, it is impossible to find C and D that satisfy these equations unless C and D equal A and B, or C and D are the same as A and B but swapped around. A formal proof of this fact is given in the sidebar. In the special case where ball 2 is initially at rest, this tells us that ball 1 is stopped dead by the collision, and ball 2 heads off at the velocity originally possessed by ball 1. This behavior will be familiar to players of pool. Often, as in the example above, the details of the algebra are the least interesting part of the problem, and considerable physical insight can be gained simply by counting the number of unknowns and comparing to the number of equations. Suppose a beginner at
Conservation of momentum
pool notices a case where her cue ball hits an initially stationary ball and stops dead. “Wow, what a good trick,” she thinks. “I bet I could never do that again in a million years.” But she tries again, and finds that she can’t help doing it even if she doesn’t want to. Luckily she has just learned about collisions in her physics course. Once she has written down the equations for conservation of energy and no loss of kinetic energy, she really doesn’t have to complete the algebra. She knows that she has two equations in two unknowns, so there must be a well-defined solution. Once she has seen the result of one such collision, she knows that the same thing must happen every time. The same thing would happen with colliding marbles or croquet balls. It doesn’t matter if the masses or velocities are different, because that just multiplies both equations by some constant factor. The discovery of the neutron This was the type of reasoning employed by James Chadwick in his 1932 discovery of the neutron. At the time, the atom was imagined to be made out of two types of fundamental particles, protons and electrons. The protons were far more massive, and clustered together in the atom’s core, or nucleus. Attractive electrical forces caused the electrons to orbit the nucleus in circles, in much the same way that gravitational forces kept the planets from cruising out of the solar system. Experiments showed that the helium nucleus, for instance, exerted exactly twice as much electrical force on an electron as a nucleus of hydrogen, the smallest atom, and this was explained by saying that helium had two protons to hydrogen’s one. The trouble was that according to this model, helium would have two electrons and two protons, giving it precisely twice the mass of a hydrogen atom with one of each. In fact, helium has about four times the mass of hydrogen. Chadwick suspected that the helium nucleus possessed two additional particles of a new type, which did not participate in electrical forces at all, i.e., were electrically neutral. If these particles had very nearly the same mass as protons, then the four-to-one mass ratio of helium and hydrogen could be explained. In 1930, a new type of radiation was discovered that seemed to fit this description. It was electrically neutral, and seemed to be coming from the nuclei of light elements that had been exposed to other types of radiation. At this time, however, reports of new types of particles were a dime a dozen, and most of them turned out to be either clusters made of previously known particles or else previously known particles with higher energies. Many physicists believed that the “new” particle that had attracted Chadwick’s interest was really a previously known particle called a gamma ray, which was electrically neutral. Since gamma rays have no mass, Chadwick decided to try to determine the new particle’s mass and see if it was nonzero and approximately equal
Section 14.2
Collisions in one dimension
353
to the mass of a proton. Unfortunately a subatomic particle is not something you can just put on a scale and weigh. Chadwick came up with an ingenious solution. The masses of the nuclei of the various chemical elements were already known, and techniques had already been developed for measuring the speed of a rapidly moving nucleus. He therefore set out to bombard samples of selected elements with the mysterious new particles. When a direct, head-on collision occurred between a mystery particle and the nucleus of one of the target atoms, the nucleus would be knocked out of the atom, and he would measure its velocity.
f / Chadwick’s subatomic pool table. A disk of the naturally occurring metal polonium provides a source of radiation capable of kicking neutrons out of the beryllium nuclei. The type of radiation emitted by the polonium is easily absorbed by a few mm of air, so the air has to be pumped out of the left-hand chamber. The neutrons, Chadwick’s mystery particles, penetrate matter far more readily, and fly out through the wall and into the chamber on the right, which is filled with nitrogen or hydrogen gas. When a neutron collides with a nitrogen or hydrogen nucleus, it kicks it out of its atom at high speed, and this recoiling nucleus then rips apart thousands of other atoms of the gas. The result is an electrical pulse that can be detected in the wire on the right. Physicists had already calibrated this type of apparatus so that they could translate the strength of the electrical pulse into the velocity of the recoiling nucleus. The whole apparatus shown in the figure would fit in the palm of your hand, in dramatic contrast to today’s giant particle accelerators.
Suppose, for instance, that we bombard a sample of hydrogen atoms with the mystery particles. Since the participants in the collision are fundamental particles, there is no way for kinetic energy to be converted into heat or any other form of energy, and Chadwick thus had two equations in three unknowns: equation #1: conservation of momentum
354
Chapter 14
Conservation of momentum
equation #2: no loss of kinetic energy unknown #1: mass of the mystery particle unknown #2: initial velocity of the mystery particle unknown #3: final velocity of the mystery particle The number of unknowns is greater than the number of equations, so there is no unique solution. But by creating collisions with nuclei of another element, nitrogen, he gained two more equations at the expense of only one more unknown: equation #3: conservation of momentum in the new collision equation #4: no loss of kinetic energy in the new collision unknown #4: final velocity of the mystery particle in the new collision He was thus able to solve for all the unknowns, including the mass of the mystery particle, which was indeed within 1% of the mass of a proton. He named the new particle the neutron, since it is electrically neutral. Discussion question A Good pool players learn to make the cue ball spin, which can cause it not to stop dead in a head-on collision with a stationary ball. If this does not violate the laws of physics, what hidden assumption was there in the example above?
14.3 Relationship of momentum to the center of mass g / In this multiple-flash photograph, we see the wrench from above as it flies through the air, rotating as it goes. Its center of mass, marked with the black cross, travels along a straight line, unlike the other points on the wrench, which execute loops.
We have already discussed the idea of the center of mass on p. 67, but using the concept of momentum we can now find a mathematical method for defining the center of mass, explain why the motion of an object’s center of mass usually exhibits simpler motion than any other point, and gain a very simple and powerful way of understanding collisions. The first step is to realize that the center of mass concept can be applied to systems containing more than one object. Even something like a wrench, which we think of as one object, is really made
Section 14.3
Relationship of momentum to the center of mass
355
of many atoms. The center of mass is particularly easy to visualize in the case shown on the left, where two identical hockey pucks collide. It is clear on grounds of symmetry that their center of mass must be at the midpoint between them. After all, we previously defined the center of mass as the balance point, and if the two hockey pucks were joined with a very lightweight rod whose own mass was negligible, they would obviously balance at the midpoint. It doesn’t matter that the hockey pucks are two separate objects. It is still true that the motion of their center of mass is exceptionally simple, just like that of the wrench’s center of mass. The x coordinate of the hockey pucks’ center of mass is thus given by xcm = (x1 + x2 )/2, i.e., the arithmetic average of their x coordinates. Why is its motion so simple? It has to do with conservation of momentum. Since the hockey pucks are not being acted on by any net external force, they constitute a closed system, and their total momentum is conserved. Their total momentum is h / Two hockey pucks collide. Their mutual center of mass traces the straight path shown by the dashed line.
mv1 + mv2 = m(v1 + v2 ) Δx1 Δx2 =m + Δt Δt m = Δ (x1 + x2 ) Δt 2Δxcm =m Δt = mtotal vcm In other words, the total momentum of the system is the same as if all its mass was concentrated at the center of mass point. Since the total momentum is conserved, the x component of the center of mass’s velocity vector cannot change. The same is also true for the other components, so the center of mass must move along a straight line at constant speed. The above relationship between the total momentum and the motion of the center of mass applies to any system, even if it is not closed. total momentum related to center of mass motion The total momentum of any system is related to its total mass and the velocity of its center of mass by the equation ptotal = mtotal vcm
.
What about a system containing objects with unequal masses, or containing more than two objects? The reasoning above can be generalized to a weighted average xcm =
m1 x1 + m2 x2 + . . . m1 + m2 + . . .
,
with similar equations for the y and z coordinates.
356
Chapter 14
Conservation of momentum
Momentum in different frames of reference Absolute motion is supposed to be undetectable, i.e., the laws of physics are supposed to be equally valid in all inertial frames of reference. If we first calculate some momenta in one frame of reference and find that momentum is conserved, and then rework the whole problem in some other frame of reference that is moving with respect to the first, the numerical values of the momenta will all be different. Even so, momentum will still be conserved. All that matters is that we work a single problem in one consistent frame of reference. One way of proving this is to apply the equation ptotal = mtotal vcm . If the velocity of frame B relative to frame A is vBA , then the only effect of changing frames of reference is to change vcm from its original value to vcm + vBA . This adds a constant onto the momentum vector, which has no effect on conservation of momentum. The center of mass frame of reference A particularly useful frame of reference in many cases is the frame that moves along with the center of mass, called the center of mass (c.m.) frame. In this frame, the total momentum is zero. The following examples show how the center of mass frame can be a powerful tool for simplifying our understanding of collisions. A collision of pool balls viewed in the c.m. frame example 12 If you move your head so that your eye is always above the point halfway in between the two pool balls, you are viewing things in the center of mass frame. In this frame, the balls come toward the center of mass at equal speeds. By symmetry, they must therefore recoil at equal speeds along the lines on which they entered. Since the balls have essentially swapped paths in the center of mass frame, the same must also be true in any other frame. This is the same result that required laborious algebra to prove previously without the concept of the center of mass frame. The slingshot effect example 13 It is a counterintuitive fact that a spacecraft can pick up speed by swinging around a planet, if it arrives in the opposite direction compared to the planet’s motion. Although there is no physical contact, we treat the encounter as a one-dimensional collision, and analyze it in the center of mass frame. Figure j shows such a “collision,” with a space probe whipping around Jupiter. In the sun’s frame of reference, Jupiter is moving. What about the center of mass frame? Since Jupiter is so much more massive than the spacecraft, the center of mass is essentially fixed at Jupiter’s center, and Jupiter has zero velocity in the center of mass frame, as shown in figure k. The c.m. frame is moving to the left compared to the sun-fixed frame used in j, so
Section 14.3
i / Moving your head so that you are always looking down from right above the center of mass, you observe the collision of the two hockey pucks in the center of mass frame.
j / The slingshot effect viewed in the sun’s frame of reference. Jupiter is moving to the left, and the collision is head-on.
k / The slingshot viewed in the frame of the center of mass of the Jupiter-spacecraft system.
Relationship of momentum to the center of mass
357
the spacecraft’s initial velocity is greater in this frame. Things are simpler in the center of mass frame, because it is more symmetric. In the complicated sun-fixed frame, the incoming leg of the encounter is rapid, because the two bodies are rushing toward each other, while their separation on the outbound leg is more gradual, because Jupiter is trying to catch up. In the c.m. frame, Jupiter is sitting still, and there is perfect symmetry between the incoming and outgoing legs, so by symmetry we have v1f = −v1i . Going back to the sun-fixed frame, the spacecraft’s final velocity is increased by the frames’ motion relative to each other. In the sun-fixed frame, the spacecraft’s velocity has increased greatly. The result can also be understood in terms of work and energy. In Jupiter’s frame, Jupiter is not doing any work on the spacecraft as it rounds the back of the planet, because the motion is perpendicular to the force. But in the sun’s frame, the spacecraft’s velocity vector at the same moment has a large component to the left, so Jupiter is doing work on it. Discussion questions A Make up a numerical example of two unequal masses moving in one dimension at constant velocity, and verify the equation ptotal = mtotal vcm over a time interval of one second. B A more massive tennis racquet or baseball bat makes the ball fly off faster. Explain why this is true, using the center of mass frame. For simplicity, assume that the racquet or bat is simply sitting still before the collision, and that the hitter’s hands do not make any force large enough to have a significant effect over the short duration of the impact.
14.4 Momentum transfer The rate of change of momentum As with conservation of energy, we need a way to measure and calculate the transfer of momentum into or out of a system when the system is not closed. In the case of energy, the answer was rather complicated, and entirely different techniques had to be used for measuring the transfer of mechanical energy (work) and the transfer of heat by conduction. For momentum, the situation is far simpler. In the simplest case, the system consists of a single object acted on by a constant external force. Since it is only the object’s velocity that can change, not its mass, the momentum transferred is Δp = mΔv
,
which with the help of a = F/m and the constant-acceleration equation a = Δv/Δt becomes Δp = maΔt = FΔt
358
Chapter 14
Conservation of momentum
.
Thus the rate of transfer of momentum, i.e., the number of kg·m/s absorbed per second, is simply the external force, F=
Δp Δt
.
[relationship between the force on an object and the rate of change of its momentum; valid only if the force is constant] This is just a restatement of Newton’s second law, and in fact Newton originally stated it this way. As shown in figure l, the relationship between force and momentum is directly analogous to that between power and energy. The situation is not materially altered for a system composed of many objects. There may be forces between the objects, but the internal forces cannot change the system’s momentum. (If they did, then removing the external forces would result in a closed system that could change its own momentum, like the mythical man who could pull himself up by his own bootstraps. That would violate conservation of momentum.) The equation above becomes Ftotal =
Δptotal Δt
l / Power and force are the rates at which energy and momentum are transferred.
.
[relationship between the total external force on a system and the rate of change of its total momentum; valid only if the force is constant] Walking into a lamppost example 14 Starting from rest, you begin walking, bringing your momentum up to 100 kg·m/s. You walk straight into a lamppost. Why is the momentum change of −100 kg·m/s caused by the lamppost so much more painful than the change of +100 kg · m/s when you started walking? The situation is one-dimensional, so we can dispense with the vector notation. It probably takes you about 1 s to speed up initially, so the ground’s force on you is F = Δp/Δt ≈ 100 N. Your impact with the lamppost, however, is over in the blink of an eye, say 1/10 s or less. Dividing by this much smaller Δt gives a much larger force, perhaps thousands of newtons. (The negative sign simply indicates that the force is in the opposite direction.) This is also the principle of airbags in cars. The time required for the airbag to decelerate your head is fairly long, the time required for your face to travel 20 or 30 cm. Without an airbag, your face would hit the dashboard, and the time interval would be the much shorter time taken by your skull to move a couple of centimeters while your face compressed. Note that either way, the same amount of mechanical work has to be done on your head: enough to eliminate all its kinetic energy.
Section 14.4
m / The airbag increases so as to reduce F = Δp/Δt .
Momentum transfer
Δt
359
Ion drive for spacecraft example 15 The ion drive of the Deep Space 1 spacecraft, pictured on page 347 and discussed in example 2, produces a thrust of 90 mN (millinewtons). It carries about 80 kg of reaction mass, which it ejects at a speed of 30,000 m/s. For how long can the engine continue supplying this amount of thrust before running out of reaction mass to shove out the back? Solving the equation F = Δp/Δt for the unknown Δt, and treating force and momentum as scalars since the problem is onedimensional, we find Δp F mexhaust Δvexhaust = F (80 kg)(30, 000 m/s) = 0.090 N = 2.7 × 107 s
Δt =
= 300 days
A toppling box example 16 If you place a box on a frictionless surface, it will fall over with a very complicated motion that is hard to predict in detail. We know, however, that its center of mass moves in the same direction as its momentum vector points. There are two forces, a normal force and a gravitational force, both of which are vertical. (The gravitational force is actually many gravitational forces acting on all the atoms in the box.) The total force must be vertical, so the momentum vector must be purely vertical too, and the center of mass travels vertically. This is true even if the box bounces and tumbles. [Based on an example by Kleppner and Kolenkow.]
n / Example 16.
The area under the force-time graph
o / The F − t graph for a tennis racquet hitting a ball might look like this. The amount of momentum transferred equals the area under the curve.
360
Chapter 14
Few real collisions involve a constant force. For example, when a tennis ball hits a racquet, the strings stretch and the ball flattens dramatically. They are both acting like springs that obey Hooke’s law, which says that the force is proportional to the amount of stretching or flattening. The force is therefore small at first, ramps up to a maximum when the ball is about to reverse directions, and ramps back down again as the ball is on its way back out. The equation F = Δp/Δt, derived under the assumption of constant acceleration, does not apply here, and the force does not even have a single well-defined numerical value that could be plugged in to the equation. As with similar-looking equations such as v = Δp/Δt, the equation F = Δp/Δt is correctly generalized by saying that the force is the slope of the p − t graph.
Conservation of momentum
Conversely, if we wish to find Δp from a graph such as the one in figure o, one approach would be to divide the force by the mass of the ball, rescaling the F axis to create a graph of acceleration versus time. The area under the acceleration-versus-time graph gives the change in velocity, which can then be multiplied by the mass to find the change in momentum. An unnecessary complication was introduced, however, because we began by dividing by the mass and ended by multiplying by it. It would have made just as much sense to find the area under the original F − t graph, which would have given us the momentum change directly. Discussion question A Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities: 1. The collision is instantaneous. 2. The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact. 3. The collision takes a finite amount of time, during which the ball and bat are bending or being compressed. How can two of these be ruled out based on energy or momentum considerations?
14.5 Momentum in three dimensions In this section we discuss how the concepts applied previously to one-dimensional situations can be used as well in three dimensions. Often vector addition is all that is needed to solve a problem: An explosion
example 17
Astronomers observe the planet Mars as the Martians fight a nuclear war. The Martian bombs are so powerful that they rip the planet into three separate pieces of liquified rock, all having the same mass. If one fragment flies off with velocity components v1x = 0 v1y = 1.0 × 104 km/hr
,
and the second with v2x = 1.0 × 104 km/hr v2y = 0
p / Example 17.
,
(all in the center of mass frame) what is the magnitude of the third one’s velocity?
Section 14.5
Momentum in three dimensions
361
In the center of mass frame, the planet initially had zero momentum. After the explosion, the vector sum of the momenta must still be zero. Vector addition can be done by adding components, so mv1x + mv2x + mv3x = 0
,
mv1y + mv2y + mv3y = 0
,
and
where we have used the same symbol m for all the terms, because the fragments all have the same mass. The masses can be eliminated by dividing each equation by m, and we find v3x = −1.0 × 104 km/hr v3y = −1.0 × 104 km/hr which gives a magnitude of
2 + v2 |v3 | = v3x 3y = 1.4 × 104 km/hr
The center of mass In three dimensions, we have the vector equations Ftotal =
Δptotal Δt
and ptotal = mtotal vcm
.
The following is an example of their use. The bola example 18 The bola, similar to the North American lasso, is used by South American gauchos to catch small animals by tangling up their legs in the three leather thongs. The motion of the whirling bola through the air is extremely complicated, and would be a challenge to analyze mathematically. The motion of its center of mass, however, is much simpler. The only forces on it are gravitational, so Ftotal = mtotal g . Using the equation Ftotal = Δptotal /Δt, we find Δptotal /Δt = mtotal g
,
and since the mass is constant, the equation ptotal = mtotal vcm allows us to change this to mtotal Δvcm /Δt = mtotal g
362
Chapter 14
Conservation of momentum
.
The mass cancels, and Δvcm /Δt is simply the acceleration of the center of mass, so acm = g
.
In other words, the motion of the system is the same as if all its mass was concentrated at and moving with the center of mass. The bola has a constant downward acceleration equal to g, and flies along the same parabola as any other projectile thrown with the same initial center of mass velocity. Throwing a bola with the correct rotation is presumably a difficult skill, but making it hit its target is no harder than it is with a ball or a single rock.
q / Example 18.
[Based on an example by Kleppner and Kolenkow.] Counting equations and unknowns Counting equations and unknowns is just as useful as in one dimension, but every object’s momentum vector has three components, so an unknown momentum vector counts as three unknowns. Conservation of momentum is a single vector equation, but it says that all three components of the total momentum vector stay constant, so we count it as three equations. Of course if the motion happens to be confined to two dimensions, then we need only count vectors as having two components. A two-car crash with sticking example 19 Suppose two cars collide, stick together, and skid off together. If we know the cars’ initial momentum vectors, we can count equations and unknowns as follows: unknown #1: x component of cars’ final, total momentum unknown #2: y component of cars’ final, total momentum equation #1: conservation of the total px equation #2: conservation of the total py Since the number of equations equals the number of unknowns, there must be one unique solution for their total momentum vector after the crash. In other words, the speed and direction at which their common center of mass moves off together is unaffected by factors such as whether the cars collide center-to-center or catch each other a little off-center. Shooting pool example 20 Two pool balls collide, and as before we assume there is no decrease in the total kinetic energy, i.e., no energy converted from KE into other forms. As in the previous example, we assume we are given the initial velocities and want to find the final velocities. The equations and unknowns are: unknown #1: x component of ball #1’s final momentum unknown #2: y component of ball #1’s final momentum
Section 14.5
Momentum in three dimensions
363
unknown #3: x component of ball #2’s final momentum unknown #4: y component of ball #2’s final momentum equation #1: conservation of the total px equation #2: conservation of the total py equation #3: no decrease in total KE Note that we do not count the balls’ final kinetic energies as unknowns, because knowing the momentum vector, one can always find the velocity and thus the kinetic energy. The number of equations is less than the number of unknowns, so no unique result is guaranteed. This is what makes pool an interesting game. By aiming the cue ball to one side of the target ball you can have some control over the balls’ speeds and directions of motion after the collision. It is not possible, however, to choose any combination of final speeds and directions. For instance, a certain shot may give the correct direction of motion for the target ball, making it go into a pocket, but may also have the undesired side-effect of making the cue ball go in a pocket. Calculations with the momentum vector The following example illustrates how a force is required to change the direction of the momentum vector, just as one would be required to change its magnitude. A turbine example 21 In a hydroelectric plant, water flowing over a dam drives a turbine, which runs a generator to make electric power. The figure shows a simplified physical model of the water hitting the turbine, in which it is assumed that the stream of water comes in at a 45◦ angle with respect to the turbine blade, and bounces off at a 90◦ angle at nearly the same speed. The water flows at a rate R, in units of kg/s, and the speed of the water is v . What are the magnitude and direction of the water’s force on the turbine?
r / Example 21.
In a time interval Δt, the mass of water that strikes the blade is RΔt, and the magnitude of its initial momentum is mv = v RΔt. The water’s final momentum vector is of the same magnitude, but in the perpendicular direction. By Newton’s third law, the water’s force on the blade is equal and opposite to the blade’s force on the water. Since the force is constant, we can use the equation
Fblade on water =
Δpwater Δt
.
Choosing the x axis to be to the right and the y axis to be up, this
364
Chapter 14
Conservation of momentum
can be broken down into components as Δpwater,x Δt −v RΔt − 0 = Δt = −v R
Fblade on water,x =
and Δpwater,y Δt 0 − (−v RΔt) = Δt = vR .
Fblade on water,y =
The water’s force on the blade thus has components Fwater on blade,x = v R Fwater on blade,y = −v R
.
In situations like this, it is always a good idea to check that the result makes sense physically. The x component of the water’s force on the blade is positive, which is correct since we know the blade will be pushed to the right. The y component is negative, which also makes sense because the water must push the blade down. The magnitude of the water’s force on the blade is |Fwater on blade | =
√
2v R
and its direction is at a 45-degree angle down and to the right. Discussion questions A The figures show a jet of water striking two different objects. How does the total downward force compare in the two cases? How could this fact be used to create a better waterwheel? (Such a waterwheel is known as a Pelton wheel.)
Discussion question A.
Section 14.5
Momentum in three dimensions
365
14.6
Applications of calculus
By now you will have learned to recognize the circumlocutions I use in the sections without calculus in order to introduce calculus-like concepts without using the notation, terminology, or techniques of calculus. It will therefore come as no surprise to you that the rate of change of momentum can be represented with a derivative, Ftotal =
dptotal dt
.
And of course the business about the area under the F − t curve is really an integral, Δptotal = Ftotal dt, which can be made into an integral of a vector in the more general three-dimensional case: Δptotal = Ftotal dt . In the case of a material object that is neither losing nor picking up mass, these are just trivially rearranged versions of familiar equations, e.g., F = mdv/dt rewritten as F = d(mv)/dt. The following is a less trivial example, where F = ma alone would not have been very easy to work with. Rain falling into a moving cart example 22 If 1 kg/s of rain falls vertically into a 10-kg cart that is rolling without friction at an initial speed of 1.0 m/s, what is the effect on the speed of the cart when the rain first starts falling? The rain and the cart make horizontal forces on each other, but there is no external horizontal force on the rain-plus-cart system, so the horizontal motion obeys F =
d(mv ) =0 dt
We use the product rule to find 0=
dv dm v +m dt dt
.
We are trying to find how v changes, so we solve for dv /dt, dv v dm =− dt m dt 1 m/s =− (1 kg/s) 10 kg = −0.1 m/s2
.
(This is only at the moment when the rain starts to fall.) Finally we note that there are cases where F = ma is not just less convenient than F = dp/dt but in fact F = ma is wrong and F = dp/dt is right. A good example is the formation of a comet’s
366
Chapter 14
Conservation of momentum
tail by sunlight. We cannot use F = ma to describe this process, since we are dealing with a collision of light with matter, whereas Newton’s laws only apply to matter. The equation F = dp/dt, on the other hand, allows us to find the force experienced by an atom of gas in the comet’s tail if we know the rate at which the momentum vectors of light rays are being turned around by reflection from the atom.
Section 14.6
Applications of calculus
367
Summary Selected vocabulary momentum . . . a measure of motion, equal to mv for material objects collision . . . . . an interaction between moving objects that lasts for a certain time center of mass . . the balance point or average position of the mass in a system Notation p. . . . . . . . . . cm . . . . . . . . .
the momentum vector center of mass, as in xcm , acm , etc.
Other terminology and notation impulse, I, J . . the amount of momentum transferred, Δp elastic collision . one in which no KE is converted into other forms of energy inelastic collision one in which some KE is converted to other forms of energy Summary If two objects interact via a force, Newton’s third law guarantees that any change in one’s velocity vector will be accompanied by a change in the other’s which is in the opposite direction. Intuitively, this means that if the two objects are not acted on by any external force, they cannot cooperate to change their overall state of motion. This can be made quantitative by saying that the quantity m1 v1 + m2 v2 must remain constant as long as the only forces are the internal ones between the two objects. This is a conservation law, called the conservation of momentum, and like the conservation of energy, it has evolved over time to include more and more phenomena unknown at the time the concept was invented. The momentum of a material object is p = mv
,
but this is more like a standard for comparison of momenta rather than a definition. For instance, light has momentum, but has no mass, and the above equation is not the right equation for light. The law of conservation of momentum says that the total momentum of any closed system, i.e., the vector sum of the momentum vectors of all the things in the system, is a constant. An important application of the momentum concept is to collisions, i.e., interactions between moving objects that last for a certain amount of time while the objects are in contact or near each other. Conservation of momentum tells us that certain outcomes of a collision are impossible, and in some cases may even be sufficient to predict the motion after the collision. In other cases, conservation of momentum does not provide enough equations to find all the unknowns. In some collisions, such as the collision of a superball with
368
Chapter 14
Conservation of momentum
the floor, very little kinetic energy is converted into other forms of energy, and this provides one more equation, which may suffice to predict the outcome. The total momentum of a system can be related to its total mass and the velocity of its center of mass by the equation ptotal = mtotal vcm
.
The center of mass, introduced on an intuitive basis in book 1 as the “balance point” of an object, can be generalized to any system containing any number of objects, and is defined mathematically as the weighted average of the positions of all the parts of all the objects, m1 x 1 + m2 x 2 + . . . , xcm = m1 + m2 + . . . with similar equations for the y and z coordinates. The frame of reference moving with the center of mass of a closed system is always a valid inertial frame, and many problems can be greatly simplified by working them in the inertial frame. For example, any collision between two objects appears in the c.m. frame as a head-on one-dimensional collision. When a system is not closed, the rate at which momentum is transferred in or out is simply the total force being exerted externally on the system. If the force is constant, Ftotal =
Δptotal Δt
.
When the force is not constant, the force equals the slope of the tangent line on a graph of p versus t, and the change in momentum equals the area under the F − t graph.
Summary
369
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 Derive a formula expressing the kinetic energy of an object in √ terms of its momentum and mass. 2 Two people in a rowboat wish to move around without causing the boat to move. What should be true about their total momentum? Explain. 3 A learjet traveling due east at 300 mi/hr collides with a jumbo jet which was heading southwest at 150 mi/hr. The jumbo jet’s mass is five times greater than that of the learjet. When they collide, the learjet sticks into the fuselage of the jumbo jet, and they fall to earth together. Their engines stop functioning immediately after the collision. On a map, what will be the direction from the location of the collision to the place where the wreckage hits√the ground? (Give an angle.) 4 A bullet leaves the barrel of a gun with a kinetic energy of 90 J. The gun barrel is 50 cm long. The gun has a mass of 4 kg, the bullet 10 g. √ (a) Find the bullet’s final velocity. √ (b) Find the bullet’s final momentum. (c) Find the momentum of the recoiling gun. (d) Find the kinetic energy of the recoiling gun, and explain √why the recoiling gun does not kill the shooter. Problem 5
5 The graph shows the force, in meganewtons, exerted by a rocket engine on the rocket as a function of time. If the rocket’s mass is 4000 kg, at what speed is the rocket moving when the engine
370
Chapter 14
Conservation of momentum
stops firing? Assume it goes straight up, and neglect the force of √ gravity, which is much less than a meganewton. 6 Cosmic rays are particles from outer space, mostly protons and atomic nuclei, that are continually bombarding the earth. Most of them, although they are moving extremely fast, have no discernible effect even if they hit your body, because their masses are so small. Their energies vary, however, and a very small minority of them have extremely large energies. In some cases the energy is as much as several Joules, which is comparable to the KE of a well thrown rock! If you are in a plane at a high altitude and are so incredibly unlucky as to be hit by one of these rare ultra-high-energy cosmic rays, what would you notice, the momentum imparted to your body, the energy dissipated in your body as heat, or both? Base your conclusions on numerical estimates, not just random speculation. (At these high speeds, one should really take into account the deviations from Newtonian physics described by Einstein’s special theory of relativity. Don’t worry about that, though.) 7 Show that for a body made up of many equal masses, the equation for the center of mass becomes a simple average of all the positions of the masses. 8 The figure shows a view from above of a collision about to happen between two air hockey pucks sliding without friction. They have the same speed, vi , before the collision, but the big puck is 2.3 times more massive than the small one. Their sides have sticky stuff on them, so when they collide, they will stick together. At what angle will they emerge from the collision? In addition to giving a numerical answer, please indicate by drawing on the figure how your angle is defined. Solution, p. 538 9 A flexible rope of mass m and length L slides without friction over the edge of a table. Let x be the length of the rope that is hanging over the edge at a given moment in time. (a) Show that x satisfies the equation of motion d2 x/dt2 = gx/L. [Hint: Use F = dp/dt, which allows you to handle the two parts of the rope separately even though mass is moving out of one part and into the other.] (b) Give a physical explanation for the fact that a larger value of x on the right-hand side of the equation leads to a greater value of the acceleration on the left side. (c) When we take the second derivative of the function x(t) we are supposed to get essentially the same function back again, except for a constant out in front. The function ex has the property that it is unchanged by differentiation, so it is reasonable to look for solutions to this problem that are of the form x = bect , where b and c are constants. Show that this does indeed provide a solution for two specific values of c (and for any value of b). (d) Show that the sum of any two solutions to the equation of motion
Problem 8
Problems
371
is also a solution. (e) Find the solution for the case where the rope starts at rest at t = 0 with some nonzero value of x. 10 A very massive object with velocity v collides head-on with an object at rest whose mass is very small. No kinetic energy is converted into other forms. Prove that the low-mass object recoils with velocity 2v. [Hint: Use the center-of-mass frame of reference.] 11 When the contents of a refrigerator cool down, the changed molecular speeds imply changes in both momentum and energy. Why, then, does a fridge transfer power through its radiator coils, but not force? Solution, p. 538 12 A 10-kg bowling ball moving at 2.0 m/s hits a 1.0-kg bowling pin, which is initially at rest. The other pins are all gone already, and the collision is head-on, so that the motion is one-dimensional. Assume that negligible amounts of heat and sound are produced. Find the velocity of the pin immediately after the collision. 13 A rocket ejects exhaust with an exhaust velocity u. The rate at which the exhaust mass is used (mass per unit time) is b. We assume that the rocket accelerates in a straight line starting from rest, and that no external forces act on it. Let the rocket’s initial mass (fuel plus the body and payload) be mi , and mf be its final mass, after all the fuel is used up. (a) Find the rocket’s final velocity, v, in terms of u, mi , and mf . Neglect the effects of special relativity. (b) A typical exhaust velocity for chemical rocket engines is 4000 m/s. Estimate the initial mass of a rocket that could accelerate a one-ton payload to 10% of the speed of light, and show that this design won’t work. (For the sake of the estimate, ignore the mass of the fuel tanks. The speed is fairly small compared to c, so it’s not √ an unreasonable approximation to ignore relativity.) 14 A firework shoots up into the air, and just before it explodes it has a certain momentum and kinetic energy. What can you say about the momenta and kinetic energies of the pieces immediately after the explosion? [Based on a problem from PSSC Physics.] Solution, p. 538 15 Suppose a system consisting of pointlike particles has a total kinetic energy Kcm measured in the center-of-mass frame of reference. Since they are pointlike, they cannot have any energy due to internal motion. (a) Prove that in a different frame of reference, moving with velocity u relative to the center-of-mass frame, the total kinetic energy equals Kcm + M |u|2 /2, where M is the total mass. [Hint: You can save yourself a lot of writing if you express the total kinetic energy using the dot product.] Solution, p. 539 (b) Use this to prove that if energy is conserved in one frame of
372
Chapter 14
Conservation of momentum
reference, then it is conserved in every frame of reference. The total energy equals the total kinetic energy plus the sum of the potential energies due to the particles’ interactions with each other, which we assume depends only on the distance between particles. [For a simpler numerical example, see problem 13 on p. 296.] 16 The big difference between the equations for momentum and kinetic energy is that one is proportional to v and one to v 2 . Both, however, are proportional to m. Suppose someone tells you that there’s a third quantity, funkosity, defined as f = m2 v, and that funkosity is conserved. How do you know your leg is being pulled? Solution, p. 539 17 A mass m moving at velocity v collides with a stationary target having the same mass m. Find the maximum amount of √ energy that can be released as heat and sound. 18 Two blobs of putty collide head-on. The collision is completely symmetric: the blobs are of equal mass, and they collide at equal speeds. What becomes of the energy the blobs had before the collision? The momentum?
Problems
373
374
Chapter 14
Conservation of momentum
A tornado touches down in Spring Hill, Kansas, May 20, 1957.
Chapter 15
Conservation of angular momentum “Sure, and maybe the sun won’t come up tomorrow.” Of course, the sun only appears to go up and down because the earth spins, so the cliche should really refer to the unlikelihood of the earth’s stopping its rotation abruptly during the night. Why can’t it stop? It wouldn’t violate conservation of momentum, because the earth’s rotation doesn’t add anything to its momentum. While California spins in one direction, some equally massive part of India goes the opposite way, canceling its momentum. A halt to Earth’s rotation would entail a drop in kinetic energy, but that energy could simply be converted into some other form, such as heat. Other examples along these lines are not hard to find. A hydrogen atom spins at the same rate for billions of years. A high-diver who is rotating when he comes off the board does not need to make
375
any physical effort to continue rotating, and indeed would be unable to stop rotating before he hit the water. These observations have the hallmarks of a conservation law: A closed system is involved. Nothing is making an effort to twist the earth, the hydrogen atom, or the high-diver. They are isolated from rotation-changing influences, i.e., they are closed systems. Something remains unchanged. There appears to be a numerical quantity for measuring rotational motion such that the total amount of that quantity remains constant in a closed system. Something can be transferred back and forth without changing the total amount. In figure a, the jumper wants to get his feet out in front of him so he can keep from doing a “face plant” when he lands. Bringing his feet forward would involve a certain quantity of counterclockwise rotation, but he didn’t start out with any rotation when he left the ground. Suppose we consider counterclockwise as positive and clockwise as negative. The only way his legs can acquire some positive rotation is if some other part of his body picks up an equal amount of negative rotation. This is why he swings his arms up behind him, clockwise.
a / An early photograph of an old-fashioned long-jump.
What numerical measure of rotational motion is conserved? Car engines and old-fashioned LP records have speeds of rotation measured in rotations per minute (r.p.m.), but the number of rotations per minute (or per second) is not a conserved quantity. A twirling figure skater, for instance, can pull her arms in to increase her r.p.m.’s. The first section of this chapter deals with the numerical definition of the quantity of rotation that results in a valid conservation law.
376
Chapter 15
Conservation of angular momentum
15.1 Conservation of angular momentum When most people think of rotation, they think of a solid object like a wheel rotating in a circle around a fixed point. Examples of this type of rotation, called rigid rotation or rigid-body rotation, include a spinning top, a seated child’s swinging leg, and a helicopter’s spinning propeller. Rotation, however, is a much more general phenomenon, and includes noncircular examples such as a comet in an elliptical orbit around the sun, or a cyclone, in which the core completes a circle more quickly than the outer parts. If there is a numerical measure of rotational motion that is a conserved quantity, then it must include nonrigid cases like these, since nonrigid rotation can be traded back and forth with rigid rotation. For instance, there is a trick for finding out if an egg is raw or hardboiled. If you spin a hardboiled egg and then stop it briefly with your finger, it stops dead. But if you do the same with a raw egg, it springs back into rotation because the soft interior was still swirling around within the momentarily motionless shell. The pattern of flow of the liquid part is presumably very complex and nonuniform due to the asymmetric shape of the egg and the different consistencies of the yolk and the white, but there is apparently some way to describe the liquid’s total amount of rotation with a single number, of which some percentage is given back to the shell when you release it. The best strategy is to devise a way of defining the amount of rotation of a single small part of a system. The amount of rotation of a system such as a cyclone will then be defined as the total of all the contributions from its many small parts. The quest for a conserved quantity of rotation even requires us to broaden the rotation concept to include cases where the motion doesn’t repeat or even curve around. If you throw a piece of putty at a door, the door will recoil and start rotating. The putty was traveling straight, not in a circle, but if there is to be a general conservation law that can cover this situation, it appears that we must describe the putty as having had some “rotation,” which it then gave up to the door. The best way of thinking about it is to attribute rotation to any moving object or part of an object that changes its angle in relation to the axis of rotation. In the puttyand-door example, the hinge of the door is the natural point to think of as an axis, and the putty changes its angle as seen by someone standing at the hinge. For this reason, the conserved quantity we are investigating is called angular momentum. The symbol for angular momentum can’t be a or m, since those are used for acceleration and mass, so the symbol L is arbitrarily chosen instead.
b / An overhead view of a piece of putty being thrown at a door. Even though the putty is neither spinning nor traveling along a curve, we must define it as having some kind of “rotation” because it is able to make the door rotate.
c / As seen by someone standing at the axis, the putty changes its angular position. We therefore define it as having angular momentum.
Imagine a 1-kg blob of putty, thrown at the door at a speed of 1 m/s, which hits the door at a distance of 1 m from the hinge. We define this blob to have 1 unit of angular momentum. When
Section 15.1
Conservation of angular momentum
377
it hits the door, the door will recoil and start rotating. We can use the speed at which the door recoils as a measure of the angular momentum the blob brought in.1 Experiments show, not surprisingly, that a 2-kg blob thrown in the same way makes the door rotate twice as fast, so the angular momentum of the putty blob must be proportional to mass, L∝m
.
Similarly, experiments show that doubling the velocity of the blob will have a doubling effect on the result, so its angular momentum must be proportional to its velocity as well, L ∝ mv
d / A putty blob thrown directly at the axis has no angular motion, and therefore no angular momentum. It will not cause the door to rotate.
You have undoubtedly had the experience of approaching a closed door with one of those bar-shaped handles on it and pushing on the wrong side, the side close to the hinges. You feel like an idiot, because you have so little leverage that you can hardly budge the door. The same would be true with the putty blob. Experiments would show that the amount of rotation the blob can give to the door is proportional to the distance, r, from the axis of rotation, so angular momentum must also be proportional to r, L ∝ mvr
e / Only the component of the velocity vector perpendicular to the dashed line should be counted into the definition of angular momentum.
.
.
We are almost done, but there is one missing ingredient. We know on grounds of symmetry that a putty ball thrown directly inward toward the hinge will have no angular momentum to give to the door. After all, there would not even be any way to decide whether the ball’s rotation was clockwise or counterclockwise in this situation. It is therefore only the component of the blob’s velocity vector perpendicular to the door that should be counted in its angular momentum, L = mv⊥ r
.
More generally, v⊥ should be thought of as the component of the object’s velocity vector that is perpendicular to the line joining the object to the axis of rotation. We find that this equation agrees with the definition of the original putty blob as having one unit of angular momentum, and we can now see that the units of angular momentum are (kg·m/s)·m, i.e., kg·m2 /s. This gives us a way of calculating the angular momentum of any material object or any system consisting of material objects: 1
We assume that the door is much more massive than the blob. Under this assumption, the speed at which the door recoils is much less than the original speed of the blob, so the blob has lost essentially all its angular momentum, and given it to the door.
378
Chapter 15
Conservation of angular momentum
angular momentum of a material object The angular momentum of a moving particle is L = mv⊥ r
,
where m is its mass, v⊥ is the component of its velocity vector perpendicular to the line joining it to the axis of rotation, and r is its distance from the axis. Positive and negative signs are used to describe opposite directions of rotation. The angular momentum of a finite-sized object or a system of many objects is found by dividing it up into many small parts, applying the equation to each part, and adding to find the total amount of angular momentum.
Note that r is not necessarily the radius of a circle. (As implied by the qualifiers, matter isn’t the only thing that can have angular momentum. Light can also have angular momentum, and the above equation would not apply to light.) Conservation of angular momentum has been verified over and over again by experiment, and is now believed to be one of the three most fundamental principles of physics, along with conservation of energy and momentum. A figure skater pulls her arms in example 1 When a figure skater is twirling, there is very little friction between her and the ice, so she is essentially a closed system, and her angular momentum is conserved. If she pulls her arms in, she is decreasing r for all the atoms in her arms. It would violate conservation of angular momentum if she then continued rotating at the same speed, i.e., taking the same amount of time for each revolution, because her arms’ contributions to her angular momentum would have decreased, and no other part of her would have increased its angular momentum. This is impossible because it would violate conservation of angular momentum. If her total angular momentum is to remain constant, the decrease in r for her arms must be compensated for by an overall increase in her rate of rotation. That is, by pulling her arms in, she substantially reduces the time for each rotation.
Section 15.1
f / A figure skater pulls in her arms so that she can execute a spin more rapidly.
Conservation of angular momentum
379
g / Example 2.
h / Example 3. A view of the earth-moon system from above the north pole. All distances have been highly distorted for legibility. The earth’s rotation is counterclockwise from this point of view (arrow). The moon’s gravity creates a bulge on the side near it, because its gravitational pull is stronger there, and an “anti-bulge” on the far side, since its gravity there is weaker. For simplicity, let’s focus on the tidal bulge closer to the moon. Its frictional force is trying to slow down the earth’s rotation, so its force on the earth’s solid crust is toward the bottom of the figure. By Newton’s third law, the crust must thus make a force on the bulge which is toward the top of the figure. This causes the bulge to be pulled forward at a slight angle, and the bulge’s gravity therefore pulls the moon forward, accelerating its orbital motion about the earth and flinging it outward.
380
Chapter 15
Changing the axis example 2 An object’s angular momentum can be different depending on the axis about which it rotates. Figure g shows shows two doubleexposure photographs a viola player tipping the bow in order to cross from one string to another. Much more angular momentum is required when playing near the bow’s handle, called the frog, as in the panel on the right; not only are most of the atoms in the bow at greater distances, r , from the axis of rotation, but the ones in the tip also have more momentum, p. It is difficult for the player to quickly transfer a large angular momentum into the bow, and then transfer it back out just as quickly. (In the language of section 15.4, large torques are required.) This is one of the reasons that string players tend to stay near the middle of the bow as much as possible. Earth’s slowing rotation and the receding moon example 3 As noted in chapter 1, the earth’s rotation is actually slowing down very gradually, with the kinetic energy being dissipated as heat by friction between the land and the tidal bulges raised in the seas by the earth’s gravity. Does this mean that angular momentum is not really perfectly conserved? No, it just means that the earth is not quite a closed system by itself. If we consider the earth and moon as a system, then the angular momentum lost by the earth must be gained by the moon somehow. In fact very precise measurements of the distance between the earth and the moon have been carried out by bouncing laser beams off of a mirror left there by astronauts, and these measurements show that the moon is receding from the earth at a rate of 4 centimeters per year! The moon’s greater value of r means that it has a greater
Conservation of angular momentum
angular momentum, and the increase turns out to be exactly the amount lost by the earth. In the days of the dinosaurs, the days were significantly shorter, and the moon was closer and appeared bigger in the sky. But what force is causing the moon to speed up, drawing it out into a larger orbit? It is the gravitational forces of the earth’s tidal bulges. The effect is described qualitatively in the caption of the figure. The result would obviously be extremely difficult to calculate directly, and this is one of those situations where a conservation law allows us to make precise quantitative statements about the outcome of a process when the calculation of the process itself would be prohibitively complex. Restriction to rotation in a plane Is angular momentum a vector, or a scalar? It does have a direction in space, but it’s a direction of rotation, not a straight-line direction like the directions of vectors such as velocity or force. It turns out (see problem 29) that there is a way of defining angular momentum as a vector, but in this book the examples will be confined to a single plane of rotation, i.e., effectively two-dimensional situations. In this special case, we can choose to visualize the plane of rotation from one side or the other, and to define clockwise and counterclockwise rotation as having opposite signs of angular momentum. Discussion question A Conservation of plain old momentum, p, can be thought of as the greatly expanded and modified descendant of Galileo’s original principle of inertia, that no force is required to keep an object in motion. The principle of inertia is counterintuitive, and there are many situations in which it appears superficially that a force is needed to maintain motion, as maintained by Aristotle. Think of a situation in which conservation of angular momentum, L, also seems to be violated, making it seem incorrectly that something external must act on a closed system to keep its angular momentum from “running down.”
15.2 Angular momentum in planetary motion We now discuss the application of conservation of angular momentum to planetary motion, both because of its intrinsic importance and because it is a good way to develop a visual intuition for angular momentum. Kepler’s law of equal areas states that the area swept out by a planet in a certain length of time is always the same. Angular momentum had not been invented in Kepler’s time, and he did not even know the most basic physical facts about the forces at work. He thought of this law as an entirely empirical and unexpectedly simple way of summarizing his data, a rule that succeeded in describing
Section 15.2
Angular momentum in planetary motion
381
and predicting how the planets sped up and slowed down in their elliptical paths. It is now fairly simple, however, to show that the equal area law amounts to a statement that the planet’s angular momentum stays constant. There is no simple geometrical rule for the area of a pie wedge cut out of an ellipse, but if we consider a very short time interval, as shown in figure i, the shaded shape swept out by the planet is very nearly a triangle. We do know how to compute the area of a triangle. It is one half the product of the base and the height: 1 area = bh 2
.
We wish to relate this to angular momentum, which contains the variables r and v⊥ . If we consider the sun to be the axis of rotation, then the variable r is identical to the base of the triangle, r = b. Referring to the magnified portion of the figure, v⊥ can be related to h, because the two right triangles are similar: i / The planet’s angular momentum is related to the rate at which it sweeps out area.
h v⊥ = distance traveled |v| The area can thus be rewritten as 1 v⊥ (distance traveled) area = r 2 |v|
.
The distance traveled equals |v|Δt, so this simplifies to 1 area = rv⊥ Δt 2
.
We have found the following relationship between angular momentum and the rate at which area is swept out: L = 2m
area Δt
.
The factor of 2 in front is simply a matter of convention, since any conserved quantity would be an equally valid conserved quantity if you multiplied it by a constant. The factor of m was not relevant to Kepler, who did not know the planets’ masses, and who was only describing the motion of one planet at a time. We thus find that Kepler’s equal-area law is equivalent to a statement that the planet’s angular momentum remains constant. But wait, why should it remain constant? — the planet is not a closed system, since it is being acted on by the sun’s gravitational force. There are two valid answers. The first is that it is actually the total angular momentum of the sun plus the planet that is conserved. The sun, however, is millions of times more massive than the typical planet, so it accelerates very little in response to the planet’s gravitational force. It is thus a good approximation to say that the sun
382
Chapter 15
Conservation of angular momentum
doesn’t move at all, so that no angular momentum is transferred between it and the planet. The second answer is that to change the planet’s angular momentum requires not just a force but a force applied in a certain way. In section 15.4 we discuss the transfer of angular momentum by a force, but the basic idea here is that a force directly in toward the axis does not change the angular momentum. Discussion questions A Suppose an object is simply traveling in a straight line at constant speed. If we pick some point not on the line and call it the axis of rotation, is area swept out by the object at a constant rate? Would it matter if we chose a different axis? B The figure is a strobe photo of a pendulum bob, taken from underneath the pendulum looking straight up. The black string can’t be seen in the photograph. The bob was given a slight sideways push when it was released, so it did not swing in a plane. The bright spot marks the center, i.e., the position the bob would have if it hung straight down at us. Does the bob’s angular momentum appear to remain constant if we consider the center to be the axis of rotation? What if we choose a different axis?
Discussion question A.
Discussion question B.
15.3 Two theorems about angular momentum With plain old momentum, p, we had the freedom to work in any inertial frame of reference we liked. The same object could have different values of momentum in two different frames, if the frames were not at rest with respect to each other. Conservation of momentum, however, would be true in either frame. As long as we employed a single frame consistently throughout a calculation, everything would work. The same is true for angular momentum, and in addition there is an ambiguity that arises from the definition of an axis of rotation. For a wheel, the natural choice of an axis of rotation is obviously the axle, but what about an egg rotating on its side? The egg
Section 15.3
Two theorems about angular momentum
383
has an asymmetric shape, and thus no clearly defined geometric center. A similar issue arises for a cyclone, which does not even have a sharply defined shape, or for a complicated machine with many gears. The following theorem, the first of two presented in this section without proof, explains how to deal with this issue. Although I have put descriptive titles above both theorems, they have no generally accepted names. the choice of axis theorem It is entirely arbitrary what point one defines as the axis for purposes of calculating angular momentum. If a closed system’s angular momentum is conserved when calculated with one choice of axis, then it will also be conserved for any other choice. Likewise, any inertial frame of reference may be used.
j / Example 4.
Colliding asteroids described with different axes example 4 Observers on planets A and B both see the two asteroids colliding. The asteroids are of equal mass and their impact speeds are the same. Astronomers on each planet decide to define their own planet as the axis of rotation. Planet A is twice as far from the collision as planet B. The asteroids collide and stick. For simplicity, assume planets A and B are both at rest.
k / Everyone has a strong tendency to think of the diver as rotating about his own center of mass. However, he is flying in an arc, and he also has angular momentum because of this motion.
With planet A as the axis, the two asteroids have the same amount of angular momentum, but one has positive angular momentum and the other has negative. Before the collision, the total angular momentum is therefore zero. After the collision, the two asteroids will have stopped moving, and again the total angular momentum is zero. The total angular momentum both before and after the collision is zero, so angular momentum is conserved if you choose planet A as the axis. The only difference with planet B as axis is that r is smaller by a factor of two, so all the angular momenta are halved. Even though the angular momenta are different than the ones calculated by planet A, angular momentum is still conserved. The earth spins on its own axis once a day, but simultaneously travels in its circular one-year orbit around the sun, so any given part of it traces out a complicated loopy path. It would seem difficult to calculate the earth’s angular momentum, but it turns out that there is an intuitively appealing shortcut: we can simply add up the angular momentum due to its spin plus that arising from its center of mass’s circular motion around the sun. This is a special case of the following general theorem:
l / This rigid object has angular momentum both because it is spinning about its center of mass and because it is moving through space.
384
Chapter 15
the spin theorem An object’s angular momentum with respect to some outside axis A can be found by adding up two parts: (1) The first part is the object’s angular momentum found by using its own center of mass as the axis, i.e., the angular
Conservation of angular momentum
momentum the object has because it is spinning. (2) The other part equals the angular momentum that the object would have with respect to the axis A if it had all its mass concentrated at and moving with its center of mass. A system with its center of mass at rest example 5 In the special case of an object whose center of mass is at rest, the spin theorem implies that the object’s angular momentum is the same regardless of what axis we choose. (This is an even stronger statement than the choice of axis theorem, which only guarantees that angular momentum is conserved for any given choice of axis, without specifying that it is the same for all such choices.) Angular momentum of a rigid object example 6 A motorcycle wheel has almost all its mass concentrated at the outside. If the wheel has mass m and radius r , and the time required for one revolution is T , what is the spin part of its angular momentum? This is an example of the commonly encountered special case of rigid motion, as opposed to the rotation of a system like a hurricane in which the different parts take different amounts of time to go around. We don’t really have to go through a laborious process of adding up contributions from all the many parts of a wheel, because they are all at about the same distance from the axis, and are all moving around the axis at about the same speed. The velocity is all perpendicular to the spokes, v⊥ = v = (circumference)/T = 2πr /T
,
and the angular momentum of the wheel about its center is L = mv⊥ r = m(2πr /T )r = 2πmr 2 /T
.
Note that although the factors of 2π in this expression is peculiar to a wheel with its mass concentrated on the rim, the proportionality to m/T would have been the same for any other rigidly rotating object. Although an object with a noncircular shape does not have a radius, it is also true in general that angular momentum is proportional to the square of the object’s size for fixed values of m and T . For instance doubling an object’s size doubles both the v⊥ and r factors in the contribution of each of its parts to the total angular momentum, resulting in an overall factor of four increase.
Section 15.3
Two theorems about angular momentum
385
The figure shows some examples of angular momenta of various shapes rotating about their centers of mass. The equations for their angular momenta were derived using calculus, as described in my calculus-based book Simple Nature. Do not memorize these equations!
The hammer throw example 7 In the men’s Olympic hammer throw, a steel ball of radius 6.1 cm is swung on the end of a wire of length 1.22 m. What fraction of the ball’s angular momentum comes from its rotation, as opposed to its motion through space? It’s always important to solve problems symbolically first, and plug in numbers only at the end, so let the radius of the ball be b, and the length of the wire . If the time the ball takes to go once around the circle is T , then this is also the time it takes to revolve once around its own axis. Its speed is v = 2π/T , so its angular momentum due to its motion through space is mv = 2πm2 /T . Its angular momentum due to its rotation around its own center is (4π/5)mb2 /T . The ratio of these two angular momenta is (2/5)(b/)2 = 1.0×10−3 . The angular momentum due to the ball’s spin is extremely small.
386
Chapter 15
Conservation of angular momentum
Toppling a rod example 8 A rod of length b and mass m stands upright. We want to strike the rod at the bottom, causing it to fall and land flat. Find the momentum, p, that should be delivered, in terms of m, b, and g. Can this really be done without having the rod scrape on the floor? This is a nice example of a question that can very nearly be answered based only on units. Since the three variables, m, b, and g, all have different units, they can’t be added or subtracted. The only way to combine them mathematically is by multiplication or division. Multiplying one of them by itself is exponentiation, so in general we expect that the answer must be of the form p = Amj bk g l
m / Example 8.
,
where A, j, k, and l are unitless constants. The result has to have units of kg·m/s. To get kilograms to the first power, we need j =1
,
meters to the first power requires k +l =1
,
and seconds to the power −1 implies l = 1/2
.
We find j = 1, k = 1/2, and l = 1/2, so the solution must be of the form p = Am bg . Note that no physics was required! Consideration of units, however, won’t help us to find the unitless constant A. Let t be the time the rod takes to fall, so that (1/2)gt 2 = b/2. If the rod is going to land exactly on its side, then the number of revolutions it completes while in the air must be 1/4, or 3/4, or 5/4, . . . , but all the possibilities greater than 1/4 would cause the head of the rod to collide with the floor prematurely. The rod must therefore rotate at a rate that would cause it to complete a full rotation in a time T = 4t, and it has angular momentum L = (π/6)mb2 /T . The momentum lost by the object striking the rod is p, and by conservation of momentum, this is the amount of momentum, in the horizontal direction, that the rod acquires. In other words, the rod will fly forward a little. However, this has no effect on the solution to the problem. More importantly, the object striking the rod loses angular momentum bp/2, which is also transferred to the rod. Equating this to the expression above for L, we find p = (π/12)m bg.
Section 15.3
Two theorems about angular momentum
387
Finally, we need to know whether this can really be done without having the foot of the rod scrape on the floor. The figure shows that the answer is no for this rod of finite width, but it appears that the answer would be yes for a sufficiently thin rod. This is analyzed further in homework problem 28 on page 409. Discussion question A In the example of the colliding asteroids, suppose planet A was moving toward the top of the page, at the same speed as the bottom asteroid. How would planet A’s astronomers describe the angular momenta of the asteroids? Would angular momentum still be conserved?
15.4 Torque: the rate of transfer of angular momentum Force can be interpreted as the rate of transfer of momentum. The equivalent in the case of angular momentum is called torque (rhymes with “fork”). Where force tells us how hard we are pushing or pulling on something, torque indicates how hard we are twisting on it. Torque is represented by the Greek letter tau, τ , and the rate of change of an object’s angular momentum equals the total torque acting on it, ΔL τtotal = . Δt (If the angular momentum does not change at a constant rate, the total torque equals the slope of the tangent line on a graph of L versus t.) As with force and momentum, it often happens that angular momentum recedes into the background and we focus our interest on the torques. The torque-focused point of view is exemplified by the fact that many scientifically untrained but mechanically apt people know all about torque, but none of them have heard of angular momentum. Car enthusiasts eagerly compare engines’ torques, and there is a tool called a torque wrench which allows one to apply a desired amount of torque to a screw and avoid overtightening it. n / Energy, momentum, and angular momentum can be transferred. The rates of transfer are called power, force, and torque.
Torque distinguished from force Of course a force is necessary in order to create a torque — you can’t twist a screw without pushing on the wrench — but force and torque are two different things. One distinction between them is direction. We use positive and negative signs to represent forces in the two possible directions along a line. The direction of a torque, however, is clockwise or counterclockwise, not a linear direction. The other difference between torque and force is a matter of leverage. A given force applied at a door’s knob will change the door’s angular momentum twice as rapidly as the same force applied halfway between the knob and the hinge. The same amount of force produces different amounts of torque in these two cases.
388
Chapter 15
Conservation of angular momentum
It is possible to have a zero total torque with a nonzero total force. An airplane with four jet engines, o, would be designed so that their forces are balanced on the left and right. Their forces are all in the same direction, but the clockwise torques of two of the engines are canceled by the counterclockwise torques of the other two, giving zero total torque. Conversely we can have zero total force and nonzero total torque. A merry-go-round’s engine needs to supply a nonzero torque on it to bring it up to speed, but there is zero total force on it. If there was not zero total force on it, its center of mass would accelerate! Relationship between force and torque How do we calculate the amount of torque produced by a given force? Since it depends on leverage, we should expect it to depend on the distance between the axis and the point of application of the force. We’ll derive an equation relating torque to force for a particular very simple situation, and state without proof that the equation actually applies to all situations.
o / The plane’s four engines produce zero total torque but not zero total force.
Consider a pointlike object which is initially at rest at a distance r from the axis we have chosen for defining angular momentum. We first observe that a force directly inward or outward, along the line connecting the axis to the object, does not impart any angular momentum to the object. A force perpendicular to the line connecting the axis and the object does, however, make the object pick up angular momentum. Newton’s second law gives F , a= m and assuming for simplicity that the force is constant, the constant acceleration equation a = Δv/Δt allows us to find the velocity the object acquires after a time Δt, F Δt . Δv = m We are trying to relate force to a change in angular momentum, so we multiply both sides of the equation by mr to give
p / The simple physical situation we use to derive an equation for torque. A force that points directly in at or out away from the axis produces neither clockwise nor counterclockwise angular momentum. A force in the perpendicular direction does transfer angular momentum.
mΔvr = F Δtr ΔL = F Δtr Dividing by Δt gives the torque: ΔL = Fr Δt τ = Fr
Section 15.4
.
.
Torque: the rate of transfer of angular momentum
389
If a force acts at an angle other than 0 or 90◦ with respect to the line joining the object and the axis, it would be only the component of the force perpendicular to the line that would produce a torque,
τ = F⊥ r
.
Although this result was proved under a simplified set of circumstances, it is more generally valid: relationship between force and torque The rate at which a force transfers angular momentum to an object, i.e., the torque produced by the force, is given by
|τ | = r|F⊥ |
,
where r is the distance from the axis to the point of application of the force, and F⊥ is the component of the force that is perpendicular to the line joining the axis to the point of application.
q / The geometric relationships referred to in the relationship between force and torque.
The equation is stated with absolute value signs because the positive and negative signs of force and torque indicate different things, so there is no useful relationship between them. The sign of the torque must be found by physical inspection of the case at hand. From the equation, we see that the units of torque can be written as newtons multiplied by meters. Metric torque wrenches are calibrated in N·m, but American ones use foot-pounds, which is also a unit of distance multiplied by a unit of force. We know from our study of mechanical work that newtons multiplied by meters equal joules, but torque is a completely different quantity from work, and nobody writes torques with units of joules, even though it would be technically correct. self-check A Compare the magnitudes and signs of the four torques shown in the figure. Answer, p. 544
390
Chapter 15
Conservation of angular momentum
How torque depends on the direction of the force example 9 How can the torque applied to the wrench in the figure be expressed in terms of r , |F |, and the angle θ between these two vectors? The force vector and its F⊥ component form the hypotenuse and one leg of a right triangle,
and the interior angle opposite to F⊥ equals θ. The absolute value of F⊥ can thus be expressed as F⊥ = |F| sin θ
,
|τ| = r |F| sin θ
.
leading to
Sometimes torque can be more neatly visualized in terms of the quantity r⊥ shown in figure r, which gives us a third way of expressing the relationship between torque and force: |τ | = r⊥ |F|
.
Of course you would not want to go and memorize all three equations for torque. Starting from any one of them you could easily derive the other two using trigonometry. Familiarizing yourself with them can however clue you in to easier avenues of attack on certain problems.
r / The quantity r⊥ .
The torque due to gravity Up until now we’ve been thinking in terms of a force that acts at a single point on an object, such as the force of your hand on the wrench. This is of course an approximation, and for an extremely realistic calculation of your hand’s torque on the wrench you might need to add up the torques exerted by each square millimeter where your skin touches the wrench. This is seldom necessary. But in the case of a gravitational force, there is never any single point at which the force is applied. Our planet is exerting a separate tug on every brick in the Leaning Tower of Pisa, and the total gravitational torque on the tower is the sum of the torques contributed by all the little forces. Luckily there is a trick that allows us to avoid such a massive calculation. It turns out that for purposes of computing
Section 15.4
Torque: the rate of transfer of angular momentum
391
the total gravitational torque on an object, you can get the right answer by just pretending that the whole gravitational force acts at the object’s center of mass. Gravitational torque on an outstretched arm example 10 Your arm has a mass of 3.0 kg, and its center of mass is 30 cm from your shoulder. What is the gravitational torque on your arm when it is stretched out horizontally to one side, taking the shoulder to be the axis?
s / Example 10.
The total gravitational force acting on your arm is |F | = (3.0 kg)(9.8 m/s2 ) = 29 N
.
For the purpose of calculating the gravitational torque, we can treat the force as if it acted at the arm’s center of mass. The force is straight down, which is perpendicular to the line connecting the shoulder to the center of mass, so F⊥ = |F | = 29 N
.
Continuing to pretend that the force acts at the center of the arm, r equals 30 cm = 0.30 m, so the torque is τ = r F⊥ = 9 N·m
.
Cow tipping example 11 In 2005, Dr. Margo Lillie and her graduate student Tracy Boechler published a study claiming to debunk cow tipping. Their claim was based on an analysis of the torques that would be required to tip a cow, which showed that one person wouldn’t be able to make enough torque to do it. A lively discussion ensued on the popular web site slashdot.org (“news for nerds, stuff that matters”) concerning the validity of the study. Personally, I had always assumed that cow-tipping was a group sport anyway, but as a physicist, I also had some quibbles with their calculation. Here’s my own analysis. There are three forces on the cow: the force of gravity FW , the ground’s normal force FN , and the tippers’ force FA . As soon as the cow’s left hooves (on the right from our point of view) break contact with the ground, the ground’s force is being applied only to hooves on the other side. We don’t know the ground’s force, and we don’t want to find it. Therefore we take the axis to be at its point of application, so that its torque is zero. For the purpose of computing torques, we can pretend that gravity acts at the cow’s center of mass, which I’ve placed a little lower than the center of its torso, since its legs and head also have some mass, and the legs are more massive than the head and
t / Example 11.
392
Chapter 15
Conservation of angular momentum
stick out farther, so they lower the c.m. more than the head raises it. The angle θW between the vertical gravitational force and the line rW is about 14◦ . (An estimate by Matt Semke at the University of Nebraska-Lincoln gives 20◦ , which is in the same ballpark.) To generate the maximum possible torque with the least possible force, the tippers want to push at a point as far as possible from the axis, which will be the shoulder on the other side, and they want to push at a 90 degree angle with respect to the radius line rA . When the tippers are just barely applying enough force to raise the cow’s hooves on one side, the total torque has to be just slightly more than zero. (In reality, they want to push a lot harder than this — hard enough to impart a lot of angular momentum to the cow fair in a short time, before it gets mad and hurts them. We’re just trying to calculate the bare minimum force they can possibly use, which is the question that science can answer.) Setting the total torque equal to zero, τN + τW + τA = 0
,
and letting counterclockwise torques be positive, we have 0 − mgrW sin θW + FA rA sin 90◦ = 0
rW mg sin θW rA 1 ≈ (680 kg)(9.8 m/s2 ) sin 14◦ 1.5 = 1100 N .
FA =
The 680 kg figure for the typical mass of a cow is due to Lillie and Boechler, who are veterinarians, so I assume it’s fairly accurate. My estimate of 1100 N comes out significantly lower than their 1400 N figure, mainly because their incorrect placement of the center of mass gives θW = 24◦ . I don’t think 1100 N is an impossible amount of force to require of one big, strong person (it’s equivalent to lifting about 110 kg, or 240 pounds), but given that the tippers need to impart a large angular momentum fairly quickly, it’s probably true that several people would be required. The main practical issue with cow tipping is that cows generally sleep lying down. Falling on its side can also seriously injure a cow.
Section 15.4
Torque: the rate of transfer of angular momentum
393
Discussion questions A This series of discussion questions deals with past students’ incorrect reasoning about the following problem. Suppose a comet is at the point in its orbit shown in the figure. The only force on the comet is the sun’s gravitational force. Discussion question B.
Throughout the question, define all torques and angular momenta using the sun as the axis. (1) Is the sun producing a nonzero torque on the comet? Explain. (2) Is the comet’s angular momentum increasing, decreasing, or staying the same? Explain. Explain what is wrong with the following answers. In some cases, the answer is correct, but the reasoning leading up to it is wrong. (a) Incorrect answer to part (1): “Yes, because the sun is exerting a force on the comet, and the comet is a certain distance from the sun.” (b) Incorrect answer to part (1): “No, because the torques cancel out.” (c) Incorrect answer to part (2): “Increasing, because the comet is speeding up.” B Which claw hammer would make it easier to get the nail out of the wood if the same force was applied in the same direction? C You whirl a rock over your head on the end of a string, and gradually pull in the string, eventually cutting the radius in half. What happens to the rock’s angular momentum? What changes occur in its speed, the time required for one revolution, and its acceleration? Why might the string break?
Discussion question E.
D A helicopter has, in addition to the huge fan blades on top, a smaller propeller mounted on the tail that rotates in a vertical plane. Why? E The photo shows an amusement park ride whose two cars rotate in opposite directions. Why is this a good design?
394
Chapter 15
Conservation of angular momentum
15.5 Statics Equilibrium There are many cases where a system is not closed but maintains constant angular momentum. When a merry-go-round is running at constant angular momentum, the engine’s torque is being canceled by the torque due to friction. When an object has constant momentum and constant angular momentum, we say that it is in equilibrium. This is a scientific redefinition of the common English word, since in ordinary speech nobody would describe a car spinning out on an icy road as being in equilibrium. Very commonly, however, we are interested in cases where an object is not only in equilibrium but also at rest, and this corresponds more closely to the usual meaning of the word. Trees and bridges have been designed by evolution and engineers to stay at rest, and to do so they must have not just zero total force acting on them but zero total torque. It is not enough that they don’t fall down, they also must not tip over. Statics is the branch of physics concerned with problems such as these. Solving statics problems is now simply a matter of applying and combining some things you already know: • You know the behaviors of the various types of forces, for example that a frictional force is always parallel to the surface of contact.
u / The windmills are not closed systems, but angular momentum is being transferred out of them at the same rate it is transferred in, resulting in constant angular momentum. To get an idea of the huge scale of the modern windmill farm, note the sizes of the trucks and trailers.
• You know about vector addition of forces. It is the vector sum of the forces that must equal zero to produce equilibrium. • You know about torque. The total torque acting on an object must be zero if it is to be in equilibrium. • You know that the choice of axis is arbitrary, so you can make a choice of axis that makes the problem easy to solve. In general, this type of problem could involve four equations in four unknowns: three equations that say the force components add up to zero, and one equation that says the total torque is zero. Most cases you’ll encounter will not be this complicated. In the following example, only the equation for zero total torque is required in order to get an answer.
Section 15.5
Statics
395
A flagpole example 12 A 10-kg flagpole is being held up by a lightweight horizontal cable, and is propped against the foot of a wall as shown in the figure. If the cable is only capable of supporting a tension of 70 N, how great can the angle α be without breaking the cable? All three objects in the figure are supposed to be in equilibrium: the pole, the cable, and the wall. Whichever of the three objects we pick to investigate, all the forces and torques on it have to cancel out. It is not particularly helpful to analyze the forces and torques on the wall, since it has forces on it from the ground that are not given and that we don’t want to find. We could study the forces and torques on the cable, but that doesn’t let us use the given information about the pole. The object we need to analyze is the pole.
v / Example 12.
The pole has three forces on it, each of which may also result in a torque: (1) the gravitational force, (2) the cable’s force, and (3) the wall’s force. We are free to define an axis of rotation at any point we wish, and it is helpful to define it to lie at the bottom end of the pole, since by that definition the wall’s force on the pole is applied at r = 0 and thus makes no torque on the pole. This is good, because we don’t know what the wall’s force on the pole is, and we are not trying to find it. With this choice of axis, there are two nonzero torques on the pole, a counterclockwise torque from the cable and a clockwise torque from gravity. Choosing to represent counterclockwise torques as positive numbers, and using the equation |τ| = r |F | sin θ, we have rcable |Fcable | sin θcable − rgr av |Fgr av | sin θgr av = 0
.
A little geometry gives θcable = 90◦ − α and θgr av = α, so rcable |Fcable | sin(90◦ − α) − rgr av |Fgr av | sin α = 0
.
The gravitational force can be considered as acting at the pole’s center of mass, i.e., at its geometrical center, so rcable is twice rgr av , and we can simplify the equation to read 2|Fcable | sin(90◦ − α) − |Fgr av | sin α = 0
.
These are all quantities we were given, except for α, which is the angle we want to find. To solve for α we need to use the trig identity sin(90◦ − x) = cos x, 2|Fcable | cos α − |Fgr av | sin α = 0
396
Chapter 15
Conservation of angular momentum
,
which allows us to find |Fcable | |Fgr av | |Fcable | −1 2 α = tan |Fgr av | 70 N −1 = tan 2× 98 N ◦ = 55 .
tan α = 2
Art! example 13 The abstract sculpture shown in figure w contains a cube of mass m and sides of length b. The cube rests on top of a cylinder, which is off-center by a distance a. Find the tension in the cable. There are four forces on the cube: a gravitational force mg, the force FT from the cable, the upward normal force from the cylinder, FN , and the horizontal static frictional force from the cylinder, Fs . The total force on the cube in the vertical direction is zero: FN − mg = 0
w / Example 13.
.
As our axis for defining torques, it’s convenient to choose the point of contact between the cube and the cylinder, because then neither Fs nor FN makes any torque. The cable’s torque is counterclockwise, the torque due to gravity is clockwise. Letting counterclockwise torques be positive, and using the convenient equation τ = r⊥ F , we find the equation for the total torque: bFT − mga = 0
.
We could also write down the equation saying that the total horizontal force is zero, but that would bring in the cylinder’s frictional force on the cube, which we don’t know and don’t need to find. We already have two equations in the two unknowns FT and FN , so there’s no need to make it into three equations in three unknowns. Solving the first equation for FN = mg, we then substitute into the second equation to eliminate FN , and solve for FT = (a/b)mg. As a check, our result makes sense when a = 0; the cube is balanced on the cylinder, so the cable goes slack.
Section 15.5
Statics
397
Stable and unstable equilibria
x / Stable libria.
and
unstable
equi-
A pencil balanced upright on its tip could theoretically be in equilibrium, but even if it was initially perfectly balanced, it would topple in response to the first air current or vibration from a passing truck. The pencil can be put in equilibrium, but not in stable equilibrium. The things around us that we really do see staying still are all in stable equilibrium. Why is one equilibrium stable and another unstable? Try pushing your own nose to the left or the right. If you push it a millimeter to the left, your head responds with a gentle force to the right, which keeps your nose from flying off of your face. If you push your nose a centimeter to the left, your face’s force on your nose becomes much stronger. The defining characteristic of a stable equilibrium is that the farther the object is moved away from equilibrium, the stronger the force is that tries to bring it back. The opposite is true for an unstable equilibrium. In the top figure, the ball resting on the round hill theoretically has zero total force on it when it is exactly at the top. But in reality the total force will not be exactly zero, and the ball will begin to move off to one side. Once it has moved, the net force on the ball is greater than it was, and it accelerates more rapidly. In an unstable equilibrium, the farther the object gets from equilibrium, the stronger the force that pushes it farther from equilibrium.
y / The dancer’s equilibrium is unstable. If she didn’t constantly make tiny adjustments, she would tip over.
This idea can be rephrased in terms of energy. The difference between the stable and unstable equilibria shown in figure x is that in the stable equilibrium, the potential energy is at a minimum, and moving to either side of equilibrium will increase it, whereas the unstable equilibrium represents a maximum. Note that we are using the term “stable” in a weaker sense than in ordinary speech. A domino standing upright is stable in the sense we are using, since it will not spontaneously fall over in response to a sneeze from across the room or the vibration from a passing truck. We would only call it unstable in the technical sense if it could be toppled by any force, no matter how small. In everyday usage, of course, it would be considered unstable, since the force required to topple it is so small. An application of calculus example 14 Nancy Neutron is living in a uranium nucleus that is undergoing fission. Nancy’s potential energy as a function of position can be approximated by PE = x 4 − x 2 , where all the units and numerical constants have been suppressed for simplicity. Use calculus to locate the equilibrium points, and determine whether they are stable or unstable.
z / Example 14.
The equilibrium points occur where the PE is at a minimum or maximum, and minima and maxima occur where the derivative
398
Chapter 15
Conservation of angular momentum
(which equals minus the force on Nancy) is zero. This deriva3 tive is dPE/dx √ = 4x − 2x, and setting it equal to zero, we have x = 0, ±1/ 2. Minima occur where the second derivative is positive, and maxima where it is negative. The second derivative is 12x 2 − √ 2, which is negative at x = 0 (unstable) and positive at x = ±1/ 2 (stable). Interpretation: the graph of the PE is shaped like a rounded letter ‘W,’ with the two troughs representing the two halves of the splitting nucleus. Nancy is going to have to decide which half she wants to go with.
15.6 Simple Machines: the lever Although we have discussed some simple machines such as the pulley, without the concept of torque we were not yet ready to address the lever, which is the machine nature used in designing living things, almost to the exclusion of all others. (We can speculate what life on our planet might have been like if living things had evolved wheels, gears, pulleys, and screws.) The figures show two examples of levers within your arm. Different muscles are used to flex and extend the arm, because muscles work only by contraction. Analyzing example aa physically, there are two forces that do work. When we lift a load with our biceps muscle, the muscle does positive work, because it brings the bone in the forearm in the direction it is moving. The load’s force on the arm does negative work, because the arm moves in the direction opposite to the load’s force. This makes sense, because we expect our arm to do positive work on the load, so the load must do an equal amount of negative work on the arm. (If the biceps was lowering a load, the signs of the works would be reversed. Any muscle is capable of doing either positive or negative work.) There is also a third force on the forearm: the force of the upper arm’s bone exerted on the forearm at the elbow joint (not shown with an arrow in the figure). This force does no work, because the elbow joint is not moving.
aa / The biceps muscle flexes the arm.
Because the elbow joint is motionless, it is natural to define our torques using the joint as the axis. The situation now becomes quite simple, because the upper arm bone’s force exerted at the elbow neither does work nor creates a torque. We can ignore it completely. In any lever there is such a point, called the fulcrum. If we restrict ourselves to the case in which the forearm rotates with constant angular momentum, then we know that the total torque on the forearm is zero, τmuscle + τload = 0
ab / The arm.
triceps
extends
the
.
If we choose to represent counterclockwise torques as positive, then the muscle’s torque is positive, and the load’s is negative. In terms
Section 15.6
Simple Machines: the lever
399
of their absolute values, |τmuscle | = |τload |
.
Assuming for simplicity that both forces act at angles of 90◦ with respect to the lines connecting the axis to the points at which they act, the absolute values of the torques are rmuscle Fmuscle = rload Farm
,
where rmuscle , the distance from the elbow joint to the biceps’ point of insertion on the forearm, is only a few cm, while rload might be 30 cm or so. The force exerted by the muscle must therefore be about ten times the force exerted by the load. We thus see that this lever is a force reducer. In general, a lever may be used either to increase or to reduce a force. Why did our arms evolve so as to reduce force? In general, your body is built for compactness and maximum speed of motion rather than maximum force. This is the main anatomical difference between us and the Neanderthals (their brains covered the same range of sizes as those of modern humans), and it seems to have worked for us. As with all machines, the lever is incapable of changing the amount of mechanical work we can do. A lever that increases force will always reduce motion, and vice versa, leaving the amount of work unchanged. It is worth noting how simple and yet how powerful this analysis was. It was simple because we were well prepared with the concepts of torque and mechanical work. In anatomy textbooks, whose readers are assumed not to know physics, there is usually a long and complicated discussion of the different types of levers. For example, the biceps lever, aa, would be classified as a class III lever, since it has the fulcrum and load on the ends and the muscle’s force acting in the middle. The triceps, ab, is called a class I lever, because the load and muscle’s force are on the ends and the fulcrum is in the middle. How tiresome! With a firm grasp of the concept of torque, we realize that all such examples can be analyzed in much the same way. Physics is at its best when it lets us understand many apparently complicated phenomena in terms of a few simple yet powerful concepts.
400
Chapter 15
Conservation of angular momentum
15.7 Proof of Kepler’s elliptical orbit law Kepler determined purely empirically that the planets’ orbits were ellipses, without understanding the underlying reason in terms of physical law. Newton’s proof of this fact based on his laws of motion and law of gravity was considered his crowning achievement both by him and by his contemporaries, because it showed that the same physical laws could be used to analyze both the heavens and the earth. Newton’s proof was very lengthy, but by applying the more recent concepts of conservation of energy and angular momentum we can carry out the proof quite simply and succinctly, and without calculus. The basic idea of the proof is that we want to describe the shape of the planet’s orbit with an equation, and then show that this equation is exactly the one that represents an ellipse. Newton’s original proof had to be very complicated because it was based directly on his laws of motion, which include time as a variable. To make any statement about the shape of the orbit, he had to eliminate time from his equations, leaving only space variables. But conservation laws tell us that certain things don’t change over time, so they have already had time eliminated from them. There are many ways of representing a curve by an equation, of which the most familiar is y = ax + b for a line in two dimensions. It would be perfectly possible to describe a planet’s orbit using an x − y equation like this, but remember that we are applying conservation of angular momentum, and the space variables that occur in the equation for angular momentum are the distance from the axis, r, and the angle between the velocity vector and the r vector, which we will call φ. The planet will have φ=90◦ when it is moving perpendicular to the r vector, i.e., at the moments when it is at its smallest or greatest distances from the sun. When φ is less than 90◦ the planet is approaching the sun, and when it is greater than 90◦ it is receding from it. Describing a curve with an r − φ equation is like telling a driver in a parking lot a certain rule for what direction to steer based on the distance from a certain streetlight in the middle of the lot. The proof is broken into the three parts for easier digestion. The first part is a simple and intuitively reasonable geometrical fact about ellipses, whose proof we relegate to the caption of figure ad; you will not be missing much if you merely absorb the result without reading the proof.
ac / The r − φ representation of a curve.
ad / Proof that the two angles labeled φ are in fact equal: The definition of an ellipse is that the sum of the distances from the two foci stays constant. If we move a small distance along the ellipse, then one distance shrinks by an amount cos φ1 , while the other grows by cos φ2 . These are equal, so φ1 = φ2 .
(1) If we use one of the two foci of an ellipse as an axis for defining the variables r and φ, then the angle between the tangent line and the line drawn to the other focus is the same as φ, i.e., the two angles labeled φ in figure ad are in fact equal. The other two parts form the meat of our proof. We state the
Section 15.7
Proof of Kepler’s elliptical orbit law
401
results first and then prove them. (2) A planet, moving under the influence of the sun’s gravity with less than the energy required to escape, obeys an equation of the form 1 sin φ = , −pr2 + qr where p and q are positive constants that depend on the planet’s energy and angular momentum. (3) A curve is an ellipse if and only if its r − φ equation is of the form 1 sin φ = , −pr2 + qr where p and q are constants that depend on the size and shape of the ellipse and p is greater than zero. Proof of part (2) The component of the planet’s velocity vector that is perpendicular to the r vector is v⊥ = v sin φ, so conservation of angular momentum tells us that L = mrv sin φ is a constant. Since the planet’s mass is a constant, this is the same as the condition rv sin φ = constant
.
Conservation of energy gives 1 GM m mv 2 − = constant . 2 r We solve the first equation for v and plug into the second equation to eliminate v. Straightforward algebra then leads to the equation claimed above, with the constant p being positive because of our assumption that the planet’s energy is insufficient to escape from the sun, i.e., its total energy is negative. Proof of part (3) We define the quantities α, d, and s as shown in the figure. The law of cosines gives d2 = r2 + s2 − 2rs cos α
ae / Proof of part (3).
.
Using α = 180◦ −2φ and the trigonometric identities cos(180◦ −x) = − cos x and cos 2x = 1 − 2 sin2 x, we can rewrite this as d2 = r2 + s2 − 2rs 2 sin2 φ − 1 . Straightforward algebra transforms this into (r + s)2 − d2 sin φ = . 4rs Since r + s is constant, the top of the fraction is constant, and the denominator can be rewritten as 4rs = 4r(constant − r), which is equivalent to the desired form.
402
Chapter 15
Conservation of angular momentum
Summary Selected vocabulary angular momen- a measure of rotational motion; a conserved tum . . . . . . . . quantity for a closed system axis . . . . . . . . An arbitrarily chosen point used in the definition of angular momentum. Any object whose direction changes relative to the axis is considered to have angular momentum. No matter what axis is chosen, the angular momentum of a closed system is conserved. torque . . . . . . the rate of change of angular momentum; a numerical measure of a force’s ability to twist on an object equilibrium . . . a state in which an object’s momentum and angular momentum are constant stable equilibrium one in which a force always acts to bring the object back to a certain point unstable equilib- one in which any deviation of the object from rium . . . . . . . . its equilibrium position results in a force pushing it even farther away Notation L. . . . . . . . . . t . . . . . . . . . . T . . . . . . . . .
angular momentum torque the time required for a rigidly rotating body to complete one rotation
Other terminology and notation period . . . . . . . a name for the variable T defined above moment of iner- the proportionality constant in the equation tia, I . . . . . . . L = 2πI/T Summary Angular momentum is a measure of rotational motion which is conserved for a closed system. This book only discusses angular momentum for rotation of material objects in two dimensions. Not all rotation is rigid like that of a wheel or a spinning top. An example of nonrigid rotation is a cyclone, in which the inner parts take less time to complete a revolution than the outer parts. In order to define a measure of rotational motion general enough to include nonrigid rotation, we define the angular momentum of a system by dividing it up into small parts, and adding up all the angular momenta of the small parts, which we think of as tiny particles. We arbitrarily choose some point in space, the axis, and we say that anything that changes its direction relative to that point possesses angular momentum. The angular momentum of a single particle is L = mv⊥ r
,
where v⊥ is the component of its velocity perpendicular to the line
Summary
403
joining it to the axis, and r is its distance from the axis. Positive and negative signs of angular momentum are used to indicate clockwise and counterclockwise rotation. The choice of axis theorem states that any axis may be used for defining angular momentum. If a system’s angular momentum is constant for one choice of axis, then it is also constant for any other choice of axis. The spin theorem states that an object’s angular momentum with respect to some outside axis A can be found by adding up two parts: (1) The first part is the object’s angular momentum found by using its own center of mass as the axis, i.e., the angular momentum the object has because it is spinning. (2) The other part equals the angular momentum that the object would have with respect to the axis A if it had all its mass concentrated at and moving with its center of mass. Torque is the rate of change of angular momentum. The torque a force can produce is a measure of its ability to twist on an object. The relationship between force and torque is |τ | = r|F⊥ |
,
where r is the distance from the axis to the point where the force is applied, and F⊥ is the component of the force perpendicular to the line connecting the axis to the point of application. Statics problems can be solved by setting the total force and total torque on an object equal to zero and solving for the unknowns.
404
Chapter 15
Conservation of angular momentum
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 You are trying to loosen a stuck bolt on your RV using a big wrench that is 50 cm long. If you hang from the wrench, and your mass is 55 kg, what is the maximum torque you can exert on√the bolt? 2 A physical therapist wants her patient to rehabilitate his injured elbow by laying his arm flat on a table, and then lifting a 2.1 kg mass by bending his elbow. In this situation, the weight is 33 cm from his elbow. He calls her back, complaining that it hurts him to grasp the weight. He asks if he can strap a bigger weight onto his arm, only 17 cm from his elbow. How much mass should she tell him to use so that he will be exerting the same torque? (He is √ raising his forearm itself, as well as the weight.) 3 An object thrown straight up in the air is momentarily at rest when it reaches the top of its motion. Does that mean that it is in equilibrium at that point? Explain. 4 An object is observed to have constant angular momentum. Can you conclude that no torques are acting on it? Explain. [Based on a problem by Serway and Faughn.] 5 A person of weight W stands on the ball of one foot. Find the tension in the calf muscle and the force exerted by the shinbones on the bones of the foot, in terms of W , a, and b. For simplicity, assume that all the forces are at 90-degree angles to the foot, i.e., neglect the angle between the foot and the floor. 6 Two objects have the same momentum vector. Assume that they are not spinning; they only have angular momentum due to their motion through space. Can you conclude that their angular momenta are the same? Explain. [Based on a problem by Serway and Faughn.]
Problem 5.
Problems
405
7 The sun turns on its axis once every 26.0 days. Its mass is 2.0 × 1030 kg and its radius is 7.0 × 108 m. Assume it is a rigid sphere of uniform density. √ (a) What is the sun’s angular momentum? In a few billion years, astrophysicists predict that the sun will use up all its sources of nuclear energy, and will collapse into a ball of exotic, dense matter known as a white dwarf. Assume that its radius becomes 5.8 × 106 m (similar to the size of the Earth.) Assume it does not lose any mass between now and then. (Don’t be fooled by the photo, which makes it look like nearly all of the star was thrown off by the explosion. The visually prominent gas cloud is actually thinner than the best laboratory vacuum ever produced on earth. Certainly a little bit of mass is actually lost, but it is not at all unreasonable to make an approximation of zero loss of mass as we are doing.) (b) What will its angular momentum be? √ (c) How long will it take to turn once on its axis?
Problem 7.
8 A uniform ladder of mass m and length L leans against a smooth wall, making an angle q with respect to the ground. The dirt exerts a normal force and a frictional force on the ladder, producing a force vector with magnitude F1 at an angle φ with respect to the ground. Since the wall is smooth, it exerts only a normal force on the ladder; let its magnitude be F2 . (a) Explain why φ must be greater than θ. No math is needed. (b) Choose any numerical values you like for m and L, and show that the ladder can be in equilibrium (zero torque and zero total force vector) for θ = 45.00◦ and φ = 63.43◦ . Problems 8 and 9.
10 (a) Find the minimum horizontal force which, applied at the axle, will pull a wheel over a step. Invent algebra symbols for whatever quantities you find to be relevant, and give your answer in symbolic form. [Hints: There are four forces on the wheel at first, but only three when it lifts off. Normal forces are always perpendicular to the surface of contact. Note that the corner of the step cannot be perfectly sharp, so the surface of contact for this force really coincides with the surface of the wheel.] (b) Under what circumstances does your result become infinite? Give a physical interpretation.
Problem 10.
406
9 Continuing the previous problem, find an equation for φ in terms of θ, and show that m and L do not enter into the equation. Do not assume any numerical values for any of the variables. You will need the trig identity sin(a − b) = sin a cos b − sin b cos a. (As a numerical check on your result, you may wish to check that the angles given in part b of the previous problem satisfy your equation.) √
Chapter 15
Conservation of angular momentum
11 A yo-yo of total mass m consists of two solid cylinders of radius R, connected by a small spindle of negligible mass and radius r. The top of the string is held motionless while the string unrolls from the spindle. Show that the acceleration of the yo-yo is g/(1 + R2 /2r2 ). [Hint: The acceleration and the tension in the string are unknown. Use τ = ΔL/Δt and F = ma to determine these two unknowns.] 12 A ball is connected by a string to a vertical post. The ball is set in horizontal motion so that it starts winding the string around the post. Assume that the motion is confined to a horizontal plane, i.e., ignore gravity. Michelle and Astrid are trying to predict the final velocity of the ball when it reaches the post. Michelle says that according to conservation of angular momentum, the ball has to speed up as it approaches the post. Astrid says that according to conservation of energy, the ball has to keep a constant speed. Who is right? [Hint: How is this different from the case where you whirl a rock in a circle on a string and gradually reel in the string?] 13 In the 1950’s, serious articles began appearing in magazines like Life predicting that world domination would be achieved by the nation that could put nuclear bombs in orbiting space stations, from which they could be dropped at will. In fact it can be quite difficult to get an orbiting object to come down. Let the object have energy E = KE + P E and angular momentum L. Assume that the energy is negative, i.e., the object is moving at less than escape velocity. Show that it can never reach a radius less than GM m 2EL2 −1 + 1 + 2 2 3 . rmin = 2E G M m [Note that both factors are negative, giving a positive result.] 14
[Problem 14 has been deleted.]
15
[Problem 15 has been deleted.]
16 Two bars of length L are connected with a hinge and placed on a frictionless cylinder of radius r. (a) Show that the angle θ shown in the figure is related to the unitless ratio r/L by the equation r cos2 θ = L 2 tan θ
.
Problem 16.
(b) Discuss the physical behavior of this equation for very large and very small values of r/L. 17 You wish to determine the mass of a ship in a bottle without taking it out. Show that this can be done with the setup shown in the figure, with a scale supporting the bottle at one end, provided that it is possible to take readings with the ship slid to several different locations. Note that you can’t determine the position of
Problem 17.
Problems
407
the ship’s center of mass just by looking at it, and likewise for the bottle. In particular, you can’t just say, “position the ship right on top of the fulcrum” or “position it right on top of the balance.” 18 Two atoms will interact via electrical forces between their protons and electrons. One fairly good approximation to the potential energy is the Lennard-Jones formula,
a 6 a 12 , −2 P E(r) = k r r where r is the center-to-center distance between the atoms and k is a positive constant. Show that (a) there is an equilibrium point at r = a, (b) the equilibrium is stable, and (c) the energy required to bring the atoms from their equilibrium separation to infinity is k. Hint, p. 526 19 Suppose that we lived in a universe in which Newton’s law of gravity gave forces proportional to r−7 rather than r−2 . Which, if any, of Kepler’s laws would still be true? Which would be completely false? Which would be different, but in a way that could be calculated with straightforward algebra? 20 The figure shows scale drawing of a pair of pliers being used to crack a nut, with an appropriately reduced centimeter grid. Warning: do not attempt this at home; it is bad manners. If the force required to crack the nut is 300 N, estimate the force required of the person’s hand. Solution, p. 539 21 Show that a sphere of radius R that is rolling without slipping has angular momentum and momentum in the ratio L/p = (2/5)R. Problem 20.
22 Suppose a bowling ball is initially thrown so that it has no angular momentum at all, i.e., it is initially just sliding down the lane. Eventually kinetic friction will get it spinning fast enough so that it is rolling without slipping. Show that the final velocity of the ball equals 5/7 of its initial velocity. [Hint: You’ll need the result of problem 21.] 23 The rod in the figure is supported by the finger and the string. (a) Find the tension, T , in the string, and the force, F , from the √ finger, in terms of m, b, L, and g. (b) Comment on the cases b = L and b = L/2. (c) Are any values of b unphysical? 24 Two horizontal tree branches on the same tree have equal diameters, but one branch is twice as long as the other. Give a quantitative comparison of the torques where the branches join the trunk. [Thanks to Bong Kang.]
Problem 23.
408
Chapter 15
Conservation of angular momentum
25 (a) Alice says Cathy’s body has zero momentum, but Bob says Cathy’s momentum is nonzero. Nobody is lying or making a mistake. How is this possible? Give a concrete example. (b) Alice and Bob agree that Dong’s body has nonzero momentum, but disagree about Dong’s angular momentum, which Alice says is zero, and Bob says is nonzero. Explain. 26 Penguins are playful animals. Tux the Penguin invents a new game using a natural circular depression in the ice. He waddles at top speed toward the crater, aiming off to the side, and then hops into the air and lands on his belly just inside its lip. He then bellysurfs, moving in a circle around the rim. The ice is frictionless, so his speed is constant. Is Tux’s angular momentum zero, or nonzero? What about the total torque acting on him? Take the center of the crater to be the axis. Explain your answers. 27 Make a rough estimate of the mechanical advantage of the lever shown in the figure. In other words, for a given amount of force applied on the handle, how many times greater is the resulting force on the cork? 28 In example 8 on page 387, prove that if the rod is sufficiently thin, it can be toppled without scraping on the floor. Solution, p. 539
Problem 27.
29 A massless rod of length has weights, each of mass m, attached to its ends. The rod is initially put in a horizontal position, and laid on an off-center fulcrum located at a distance b from the rod’s center. The rod will topple. (a) Calculate the total gravitational torque on the rod directly, by adding the two torques. (b) Verify that this gives the same result as would have been obtained by taking the entire gravitational force as acting at the center of mass. 30 A skilled motorcyclist can ride up a ramp, fly through the air, and land on another ramp. Why would it be useful for the rider to speed up or slow down the back wheel while in the air?
Problems
409
Exercise 15: Torque Equipment: • rulers with holes in them • spring scales (two per group)
While one person holds the pencil which forms the axle for the ruler, the other members of the group pull on the scale and take readings. In each case, calculate the total torque on the ruler, and find out whether it equals zero to roughly within the accuracy of the experiment. Finish the calculations for each part before moving on to the next one.
410
Chapter 15
Conservation of angular momentum
Chapter 16
Thermodynamics This chapter is optional, and should probably be omitted from a twosemester survey course. It can be covered at any time after chapter 13. In a developing country like China, a refrigerator is the mark of a family that has arrived in the middle class, and a car is the ultimate symbol of wealth. Both of these are heat engines: devices for converting between heat and other forms of energy. Unfortunately for the Chinese, neither is a very efficient device. Burning fossil fuels has made China’s big cities the most polluted on the planet, and the country’s total energy supply isn’t sufficient to support American levels of energy consumption by more than a small fraction of China’s population. Could we somehow manipulate energy in a more efficient way? Conservation of energy is a statement that the total amount of energy is constant at all times, which encourages us to believe that any energy transformation can be undone — indeed, the laws of physics you’ve learned so far don’t even distinguish the past from the future. If you get in a car and drive around the block, the net effect is to consume some of the energy you paid for at the gas station, using it to heat the neighborhood. There would not seem to be any fundamental physical principle to prevent you from
411
recapturing all that heat and using it again the next time you want to go for a drive. More modestly, why don’t engineers design a car engine so that it recaptures the heat energy that would otherwise be wasted via the radiator and the exhaust? Hard experience, however, has shown that designers of more and more efficient engines run into a brick wall at a certain point. The generators that the electric company uses to produce energy at an oil-fueled plant are indeed much more efficient than a car engine, but even if one is willing to accept a device that is very large, expensive, and complex, it turns out to be impossible to make a perfectly efficient heat engine — not just impossible with present-day technology, but impossible due to a set of fundamental physical principles known as the science of thermodynamics. And thermodynamics isn’t just a pesky set of constraints on heat engines. Without thermodynamics, there is no way to explain the direction of time’s arrow — why we can remember the past but not the future, and why it’s easier to break Humpty Dumpty than to put him back together again.
16.1 Pressure and temperature When we heat an object, we speed up the mind-bogglingly complex random motion of its molecules. One method for taming complexity is the conservation laws, since they tell us that certain things must remain constant regardless of what process is going on. Indeed, the law of conservation of energy is also known as the first law of thermodynamics. But as alluded to in the introduction to this chapter, conservation of energy by itself is not powerful enough to explain certain empirical facts about heat. A second way to sidestep the complexity of heat is to ignore heat’s atomic nature and concentrate on quantities like temperature and pressure that tell us about a system’s properties as a whole. This approach is called macroscopic in contrast to the microscopic method of attack. Pressure and temperature were fairly well understood in the age of Newton and Galileo, hundreds of years before there was any firm evidence that atoms and molecules even existed. Unlike the conserved quantities such as mass, energy, momentum, and angular momentum, neither pressure nor temperature is additive. Two cups of coffee have twice the heat energy of a single cup, but they do not have twice the temperature. Likewise, the painful pressure on your eardrums at the bottom of a pool is not affected if you insert or remove a partition between the two halves of the pool. Pressure We restrict ourselves to a discussion of pressure in fluids at rest and in equilibrium. In physics, the term “fluid” is used to mean
412
Chapter 16
Thermodynamics
either a gas or a liquid. The important feature of a fluid can be demonstrated by comparing with a cube of jello on a plate. The jello is a solid. If you shake the plate from side to side, the jello will respond by shearing, i.e., by slanting its sides, but it will tend to spring back into its original shape. A solid can sustain shear forces, but a fluid cannot. A fluid does not resist a change in shape unless it involves a change in volume. If you’re at the bottom of a pool, you can’t relieve the pain in your ears by turning your head. The water’s force on your eardrum is always the same, and is always perpendicular to the surface where the eardrum contacts the water. If your ear is on the east side of your head, the water’s force is to the west. If you keep your head in the same spot while turning around so your ear is on the north, the force will still be the same in magnitude, and it will change its direction so that it is still perpendicular to the eardrum: south. This shows that pressure has no direction in space, i.e., it is a scalar. The direction of the force is determined by the orientation of the surface on which the pressure acts, not by the pressure itself. A fluid flowing over a surface can also exert frictional forces, which are parallel to the surface, but the present discussion is restricted to fluids at rest. Experiments also show that a fluid’s force on a surface is proportional to the surface area. The vast force of the water behind a dam, for example, in proportion to the dam’s great surface area. (The bottom of the dam experiences a higher proportion of its force.) Based on these experimental results, it appears that the useful way to define pressure is as follows. The pressure of a fluid at a given point is defined as F⊥ /A, where A is the area of a small surface inserted in the fluid at that point, and F⊥ is the component of the fluid’s force on the surface which is perpendicular to the surface. This is essentially how a pressure gauge works. The reason that the surface must be small is so that there will not be any significant difference in pressure between one part of it and another part. The SI units of pressure are evidently N/m2 , and this combination can be abbreviated as the pascal, 1 Pa=1 N/m2 . The pascal turns out to be an inconveniently small unit, so car tires, for example, have recommended pressures imprinted on them in units of kilopascals.
a / A simple pressure gauge consists of a cylinder open at one end, with a piston and a spring inside. The depth to which the spring is depressed is a measure of the pressure. To determine the absolute pressure, the air needs to be pumped out of the interior of the gauge, so that there is no air pressure acting outward on the piston. In many practical gauges, the back of the piston is open to the atmosphere, so the pressure the gauge registers equals the pressure of the fluid minus the pressure of the atmosphere.
Pressure in U.S. units example 1 In U.S. units, the unit of force is the pound, and the unit of distance is the inch. The unit of pressure is therefore pounds per square inch, or p.s.i. (Note that the pound is not a unit of mass.)
Section 16.1
Pressure and temperature
413
Atmospheric pressure in U.S. and metric units example 2 A figure that many people in the U.S. remember is that atmospheric pressure is about 15 pounds per square inch. What is this in metric units? 15 lb 2
1 in
=
68 N (0.0254 m)2
= 1.0 × 105 N/m2 = 100 kPa
Only pressure differences are normally significant. If you spend enough time on an airplane, the pain in your ears subsides. This is because your body has gradually been able to admit more air into the cavity behind the eardrum. Once the pressure inside is equalized with the pressure outside, the inward and outward forces on your eardrums cancel out, and there is no physical sensation to tell you that anything unusual is going on. For this reason, it is normally only pressure differences that have any physical significance. Thus deep-sea fish are perfectly healthy in their habitat because their bodies have enough internal pressure to cancel the pressure from the water in which they live; if they are caught in a net and brought to the surface rapidly, they explode because their internal pressure is so much greater than the low pressure outside. Getting killed by a pool pump example 3 My house has a pool, which I maintain myself. A pool always needs to have its water circulated through a filter for several hours a day in order to keep it clean. The filter is a large barrel with a strong clamp that holds the top and bottom halves together. My filter has a prominent warning label that warns me not to try to open the clamps while the pump is on, and it shows a cartoon of a person being struck by the top half of the pump. The crosssectional area of the filter barrel is 0.25 m2 . Like most pressure gauges, the one on my pool pump actually reads the difference in pressure between the pressure inside the pump and atmospheric pressure. The gauge reads 90 kPa. What is the force that is trying to pop open the filter? If the gauge told us the absolute pressure of the water inside, we’d have to find the force of the water pushing outward and the force of the air pushing inward, and subtract in order to find the total force. Since air surrounds us all the time, we would have to do such a subtraction every time we wanted to calculate anything useful based on the gauge’s reading. The manufacturers of the gauge decided to save us from all this work by making it read the difference in pressure between inside and outside, so all we have
414
Chapter 16
Thermodynamics
to do is multiply the gauge reading by the cross-sectional area of the filter: F = PA = (90 × 103 N/m2 )(0.25 m2 ) = 22000 N That’s a lot of force! The word “suction” and other related words contain a hidden misunderstanding related to this point about pressure differences. When you suck water up through a straw, there is nothing in your mouth that is attracting the water upward. The force that lifts the water is from the pressure of the water in the cup. By creating a partial vacuum in your mouth, you decreased the air’s downward force on the water so that it no longer exactly canceled the upward force. Variation of pressure with depth The pressure within a fluid in equilibrium can only depend on depth, due to gravity. If the pressure could vary from side to side, then a piece of the fluid in between, b, would be subject to unequal forces from the parts of the fluid on its two sides. But fluids do not exhibit shear forces, so there would be no other force that could keep this piece of fluid from accelerating. This contradicts the assumption that the fluid was in equilibrium. self-check A How does this proof fail for solids?
b / This doesn’t happen. If pressure could vary horizontally in equilibrium, the cube of water would accelerate horizontally. This is a contradiction, since we assumed the fluid was in equilibrium.
Answer, p. 544
To find the variation with depth, we consider the vertical forces acting on a tiny, imaginary cube of the fluid having height Δy and areas dA on the top and bottom. Using positive numbers for upward forces, we have Pbottom ΔA − Ptop ΔA − Fg = 0
.
The weight of the fluid is Fg = mg = ρV g = ρ ΔAΔy g, where ρ is the density of the fluid, so the difference in pressure is ΔP = −ρgΔy
.
c / This does happen. The sum of the forces from the surrounding parts of the fluid is upward, canceling the downward force of gravity.
[variation in pressure with depth for a fluid of density ρ in equilibrium; positive y is up.]
The factor of ρ explains why we notice the difference in pressure when diving 3 m down in a pool, but not when going down 3 m of stairs. Note also that the equation only tells us the difference in pressure, not the absolute pressure. The pressure at the surface of a swimming pool equals the atmospheric pressure, not zero, even though the depth is zero at the surface. The blood in your body does not even have an upper surface.
Section 16.1
d / The pressure is the same at all the points marked with dots.
Pressure and temperature
415
Pressure of lava underneath a volcano example 4 A volcano has just finished erupting, and a pool of molten lava is lying at rest in the crater. The lava has come up through an opening inside the volcano that connects to the earth’s molten mantle. The density of the lava is 4.1 g/cm3 . What is the pressure in the lava underneath the base of the volcano, 3000 m below the surface of the pool? ΔP = ρgΔy = (4.1 g/cm3 )(9.8 m/s2 )(3000 m) = (4.1 × 106 g/m3 )(9.8 m/s2 )(3000 m) = (4.1 × 103 kg/m3 )(9.8 m/s2 )(3000 m) = 1.2 × 108 N/m2 = 1.2 × 108 Pa This is the difference between the pressure we want to find and atmospheric pressure at the surface. The latter, however, is tiny compared to the ΔP we just calculated, so what we’ve found is essentially the pressure, P. Atmospheric pressure This example uses calculus.
example 5
Gases, unlike liquids, are quite compressible, and at a given temperature, the density of a gas is approximately proportional to the pressure. The proportionality constant is discussed in section 16.2, but for now let’s just call it k , ρ = k P. Using this fact, we can find the variation of atmospheric pressure with altitude, assuming constant temperature: dP = −ρg dy dP = −k Pg dy dP = −k g dy P ln P = −k gy + constant P = (constant)e
−k gy
[integrating both sides] [exponentiating both sides]
Pressure falls off exponentially with height. There is no sharp cutoff to the atmosphere, but the exponential gets extremely small by the time you’re ten or a hundred miles up. Temperature e / We have to wait for the thermometer to equilibrate its temperature with the temperature of Irene’s armpit.
416
Chapter 16
Thermal equilibrium We use the term temperature casually, but what is it exactly? Roughly speaking, temperature is a measure of how concentrated the heat energy is in an object. A large, massive object with very little heat energy in it has a low temperature.
Thermodynamics
But physics deals with operational definitions, i.e., definitions of how to measure the thing in question. How do we measure temperature? One common feature of all temperature-measuring devices is that they must be left for a while in contact with the thing whose temperature is being measured. When you take your temperature with a fever thermometer, you wait for the mercury inside to come up to the same temperature as your body. The thermometer actually tells you the temperature of its own working fluid (in this case the mercury). In general, the idea of temperature depends on the concept of thermal equilibrium. When you mix cold eggs from the refrigerator with flour that has been at room temperature, they rapidly reach a compromise temperature. What determines this compromise temperature is conservation of energy, and the amount of energy required to heat or cool each substance by one degree. But without even having constructed a temperature scale, we can see that the important point is the phenomenon of thermal equilibrium itself: two objects left in contact will approach the same temperature. We also assume that if object A is at the same temperature as object B, and B is at the same temperature as C, then A is at the same temperature as C. This statement is sometimes known as the zeroth law of thermodynamics, so called because after the first, second, and third laws had been developed, it was realized that there was another law that was even more fundamental. Thermal expansion The familiar mercury thermometer operates on the principle that the mercury, its working fluid, expands when heated and contracts when cooled. In general, all substances expand and contract with changes in temperature. The zeroth law of thermodynamics guarantees that we can construct a comparative scale of temperatures that is independent of what type of thermometer we use. If a thermometer gives a certain reading when it’s in thermal equilibrium with object A, and also gives the same reading for object B, then A and B must be the same temperature, regardless of the details of how the thermometers works. What about constructing a temperature scale in which every degree represents an equal step in temperature? The Celsius scale has 0 as the freezing point of water and 100 as its boiling point. The hidden assumption behind all this is that since two points define a line, any two thermometers that agree at two points must agree at all other points. In reality if we calibrate a mercury thermometer and an alcohol thermometer in this way, we will find that a graph of one thermometer’s reading versus the other is not a perfectly straight y = x line. The subtle inconsistency becomes a drastic one when we try to extend the temperature scale through the points where mercury and alcohol boil or freeze. Gases, however, are much more consistent among themselves in their thermal expansion than
Section 16.1
f / Thermal equilibrium can be prevented. Otters have a coat of fur that traps air bubbles for insulation. If a swimming otter was in thermal equilibrium with cold water, it would be dead. Heat is still conducted from the otter’s body to the water, but much more slowly than it would be in a warm-blooded animal that didn’t have this special adaptation.
g / A hot air balloon is inflated. Because of thermal expansion, the hot air is less dense than the surrounding cold air, and therefore floats as the cold air drops underneath it and pushes it up out of the way.
Pressure and temperature
417
solids or liquids, and the noble gases like helium and neon are more consistent with each other than gases in general. Continuing to search for consistency, we find that noble gases are more consistent with each other when their pressure is very low.
h / A simplified version of an ideal gas thermometer. The whole instrument is allowed to come into thermal equilibrium with the substance whose temperature is to be measured, and the mouth of the cylinder is left open to standard pressure. The volume of the noble gas gives an indication of temperature.
As an idealization, we imagine a gas in which the atoms interact only with the sides of the container, not with each other. Such a gas is perfectly nonreactive (as the noble gases very nearly are), and never condenses to a liquid (as the noble gases do only at extremely low temperatures). Its atoms take up a negligible fraction of the available volume. Any gas can be made to behave very much like this if the pressure is extremely low, so that the atoms hardly ever encounter each other. Such a gas is called an ideal gas, and we define the Celsius scale in terms of the volume of the gas in a thermometer whose working substance is an ideal gas maintained at a fixed (very low) pressure, and which is calibrated at 0 and 100 degrees according to the melting and boiling points of water. The Celsius scale is not just a comparative scale but an additive one as well: every step in temperature is equal, and it makes sense to say that the difference in temperature between 18 and 28◦ C is the same as the difference between 48 and 58. Absolute zero and the kelvin scale
i / The volume of 1 kg of neon gas as a function of temperature (at standard pressure). Although neon would actually condense into a liquid at some point, extrapolating the graph to zero volume gives the same temperature as for any other gas: absolute zero.
We find that if we extrapolate a graph of volume versus temperature, the volume becomes zero at nearly the same temperature for all gases: -273◦ C. Real gases will all condense into liquids at some temperature above this, but an ideal gas would achieve zero volume at this temperature, known as absolute zero. The most useful temperature scale in scientific work is one whose zero is defined by absolute zero, rather than by some arbitrary standard like the melting point of water. The ideal temperature scale for scientific work, called the Kelvin scale, is the same as the Celsius scale, but shifted by 273 degrees to make its zero coincide with absolute zero. Scientists use the Celsius scale only for comparisons or when a change in temperature is all that is required for a calculation. Only on the Kelvin scale does it make sense to discuss ratios of temperatures, e.g., to say that one temperature is twice as hot as another. Which temperature scale to use example 6 You open an astronomy book and encounter the equation (light emitted) = (constant) × T 4 for the light emitted by a star as a function of its surface temperature. What temperature scale is implied? The equation tells us that doubling the temperature results in the emission of 16 times as much light. Such a ratio only makes sense if the Kelvin scale is used.
418
Chapter 16
Thermodynamics
16.2 Microscopic description of an ideal gas Evidence for the kinetic theory Why does matter have the thermal properties it does? The basic answer must come from the fact that matter is made of atoms. How, then, do the atoms give rise to the bulk properties we observe? Gases, whose thermal properties are so simple, offer the best chance for us to construct a simple connection between the microscopic and macroscopic worlds. A crucial observation is that although solids and liquids are nearly incompressible, gases can be compressed, as when we increase the amount of air in a car’s tire while hardly increasing its volume at all. This makes us suspect that the atoms in a solid are packed shoulder to shoulder, while a gas is mostly vacuum, with large spaces between molecules. Most liquids and solids have densities about 1000 times greater than most gases, so evidently each molecule in a gas is separated from its nearest neighbors by a space something like 10 times the size of the molecules themselves. If gas molecules have nothing but empty space between them, why don’t the molecules in the room around you just fall to the floor? The only possible answer is that they are in rapid motion, continually rebounding from the walls, floor and ceiling. In chapter 12, we have already seen some of the evidence for the kinetic theory of heat, which states that heat is the kinetic energy of randomly moving molecules. This theory was proposed by Daniel Bernoulli in 1738, and met with considerable opposition, because there was no precedent for this kind of perpetual motion. No rubber ball, however elastic, rebounds from a wall with exactly as much energy as it originally had, nor do we ever observe a collision between balls in which none of the kinetic energy at all is converted to heat and sound. The analogy is a false one, however. A rubber ball consists of atoms, and when it is heated in a collision, the heat is a form of motion of those atoms. An individual molecule, however, cannot possess heat. Likewise sound is a form of bulk motion of molecules, so colliding molecules in a gas cannot convert their kinetic energy to sound. Molecules can indeed induce vibrations such as sound waves when they strike the walls of a container, but the vibrations of the walls are just as likely to impart energy to a gas molecule as to take energy from it. Indeed, this kind of exchange of energy is the mechanism by which the temperatures of the gas and its container become equilibrated. Pressure, volume, and temperature A gas exerts pressure on the walls of its container, and in the kinetic theory we interpret this apparently constant pressure as the averaged-out result of vast numbers of collisions occurring every second between the gas molecules and the walls. The empirical
Section 16.2
Microscopic description of an ideal gas
419
facts about gases can be summarized by the relation P V ∝ nT ,
[ideal gas]
which really only holds exactly for an ideal gas. Here n is the number of molecules in the sample of gas. Volume related to temperature example 7 The proportionality of volume to temperature at fixed pressure was the basis for our definition of temperature. Pressure related to temperature example 8 Pressure is proportional to temperature when volume is held constant. An example is the increase in pressure in a car’s tires when the car has been driven on the freeway for a while and the tires and air have become hot. We now connect these empirical facts to the kinetic theory of a classical ideal gas. For simplicity, we assume that the gas is monoatomic (i.e., each molecule has only one atom), and that it is confined to a cubical box of volume V , with L being the length of each edge and A the area of any wall. An atom whose velocity has an x component vx will collide regularly with the left-hand wall, traveling a distance 2L parallel to the x axis between collisions with that wall. The time between collisions is Δt = 2L/vx , and in each collision the x component of the atom’s momentum is reversed from −mvx to mvx . The total force on the wall is F =
Δpx,1 Δpx,2 + + ... Δt1 Δt2
[monoatomic ideal gas]
,
where the indices 1, 2, . . . refer to the individual atoms. Substituting Δpx,i = 2mvx,i and Δti = 2L/vx,i , we have 2 2 mvx,1 mvx,2 F = + + ... L L
[monoatomic ideal gas]
.
2 is twice the contribution to the kinetic energy The quantity mvx,i from the part of the atom’s center of mass motion that is parallel to the x axis. Since we’re assuming a monoatomic gas, center of mass motion is the only type of motion that gives rise to kinetic energy. (A more complex molecule could rotate and vibrate as well.) If the quantity inside the sum included the y and z components, it would be twice the total kinetic energy of all the molecules. By symmetry, it must therefore equal 2/3 of the total kinetic energy, so
F =
2KEtotal 3L
[monoatomic ideal gas]
.
Dividing by A and using AL = V , we have P =
420
Chapter 16
Thermodynamics
2KEtotal 3V
[monoatomic ideal gas]
.
This can be connected to the empirical relation P V ∝ nT if we multiply by V on both sides and rewrite KEtotal as nKEav , where KEav is the average kinetic energy per molecule: 2 P V = nKEav 3
[monoatomic ideal gas]
.
For the first time we have an interpretation for the temperature based on a microscopic description of matter: in a monoatomic ideal gas, the temperature is a measure of the average kinetic energy per molecule. The proportionality between the two is KEav = (3/2)kT , where the constant of proportionality k, known as Boltzmann’s constant, has a numerical value of 1.38 × 10−23 J/K. In terms of Boltzmann’s constant, the relationship among the bulk quantities for an ideal gas becomes P V = nkT
,
[ideal gas]
which is known as the ideal gas law. Although I won’t prove it here, this equation applies to all ideal gases, even though the derivation assumed a monoatomic ideal gas in a cubical box. (You may have seen it written elsewhere as P V = N RT , where N = n/NA is the number of moles of atoms, R = kNA , and NA = 6.0 × 1023 , called Avogadro’s number, is essentially the number of hydrogen atoms in 1 g of hydrogen.) Pressure in a car tire example 9 After driving on the freeway for a while, the air in your car’s tires heats up from 10◦ C to 35◦ C. How much does the pressure increase? The tires may expand a little, but we assume this effect is small, so the volume is nearly constant. From the ideal gas law, the ratio of the pressures is the same as the ratio of the absolute temperatures, P2 /P1 = T2 /T1 = (308 K)/(283 K) = 1.09
,
or a 9% increase. Earth’s senescence example 10 Microbes were the only life on Earth up until the relatively recent advent of multicellular life, and are arguably still the dominant form of life on our planet. Furthermore, the sun has been gradually heating up ever since it first formed, and this continuing process will soon (“soon” in the sense of geological time) eliminate multicellular life again. Heat-induced decreases in the atmosphere’s CO2 content will kill off all complex plants within about
Section 16.2
Microscopic description of an ideal gas
421
500 million years, and although some animals may be able to live by eating algae, it will only be another few hundred million years at most until the planet is completely heat-sterilized. Why is the sun getting brighter? The only thing that keeps a star like our sun from collapsing due to its own gravity is the pressure of its gases. The sun’s energy comes from nuclear reactions at its core, and the net result of these reactions is to fuse hydrogen atoms into helium atoms. It takes four hydrogens to make one helium, so the number of atoms in the sun is continuously decreasing. Since PV = nk T , this causes a decrease in pressure, which makes the core contract. As the core contracts, collisions between hydrogen atoms become more frequent, and the rate of fusion reactions increases. A piston, a refrigerator, and a space suit example 11 Both sides of the equation PV = nk T have units of energy. Suppose the pressure in a cylinder of gas pushes a piston out, as in the power stroke of an automobile engine. Let the cross-sectional area of the piston and cylinder be A, and let the piston travel a small distance Δx. Then the gas’s force on the piston F = PA does an amount of mechanical work W = F Δx = PAΔx = PΔV , where ΔV is the change in volume. This energy has to come from somewhere; it comes from cooling the gas. In a car, what this means is that we’re harvesting the energy released by burning the gasoline. In a refrigerator, we use the same process to cool the gas, which then cools the food. In a space suit, the quantity PΔV represents the work the astronaut has to do because bending her limbs changes the volume of the suit. The suit inflates under pressure like a balloon, and doesn’t want to bend. This makes it very tiring to work for any significant period of time.
j / A space suit (example 11).
422
Chapter 16
Thermodynamics
16.3 Entropy Efficiency and grades of energy Some forms of energy are more convenient than others in certain situations. You can’t run a spring-powered mechanical clock on a battery, and you can’t run a battery-powered clock with mechanical energy. However, there is no fundamental physical principle that prevents you from converting 100% of the electrical energy in a battery into mechanical energy or vice-versa. More efficient motors and generators are being designed every year. In general, the laws of physics permit perfectly efficient conversion within a broad class of forms of energy. Heat is different. Friction tends to convert other forms of energy into heat even in the best lubricated machines. When we slide a book on a table, friction brings it to a stop and converts all its kinetic energy into heat, but we never observe the opposite process, in which a book spontaneously converts heat energy into mechanical energy and starts moving! Roughly speaking, heat is different because it is disorganized. Scrambling an egg is easy. Unscrambling it is harder. We summarize these observations by saying that heat is a lower grade of energy than other forms such as mechanical energy. Of course it is possible to convert heat into other forms of energy such as mechanical energy, and that is what a car engine does with the heat created by exploding the air-gasoline mixture. But a car engine is a tremendously inefficient device, and a great deal of the heat is simply wasted through the radiator and the exhaust. Engineers have never succeeded in creating a perfectly efficient device for converting heat energy into mechanical energy, and we now know that this is because of a deeper physical principle that is far more basic than the design of an engine.
k / The temperature difference between the hot and cold parts of the air can be used to extract mechanical energy, for example with a fan blade that spins because of the rising hot air currents.
Heat engines Heat may be more useful in some forms than in other, i.e., there are different grades of heat energy. In figure k, the difference in temperature can be used to extract mechanical work with a fan blade. This principle is used in power plants, where steam is heated by burning oil or by nuclear reactions, and then allowed to expand through a turbine which has cooler steam on the other side. On a smaller scale, there is a Christmas toy that consists of a small propeller spun by the hot air rising from a set of candles, very much like the setup shown in the figure. In figure l, however, no mechanical work can be extracted because there is no difference in temperature. Although the air in l has the same total amount of energy as the air in k, the heat in l is a lower grade of energy, since none of it is accessible for doing
l / If the temperature of the air is first allowed to become uniform, then no mechanical energy can be extracted. The same amount of heat energy is present, but it is no longer accessible for doing mechanical work.
Section 16.3
Entropy
423
mechanical work.
m / The beginning of the first expansion stroke, in which the working gas is kept in thermal equilibrium with the hot reservoir.
In general, we define a heat engine as any device that takes heat from a reservoir of hot matter, extracts some of the heat energy to do mechanical work, and expels a lesser amount of heat into a reservoir of cold matter. The efficiency of a heat engine equals the amount of useful work extracted, W , divided by the amount of energy we had to pay for in order to heat the hot reservoir. This latter amount of heat is the same as the amount of heat the engine extracts from the high-temperature reservoir, QH . (The letter Q is the standard notation for a transfer of heat.) By conservation of energy, we have QH = W + QL , where QL is the amount of heat expelled into the low-temperature reservoir, so the efficiency of a heat engine, W/QH , can be rewritten as efficiency = 1 −
n / The beginning of the second expansion stroke, in which the working gas is thermally insulated. The working gas cools because it is doing work on the piston and thus losing energy.
o / The beginning of the first compression stroke. The working gas begins the stroke at the same temperature as the cold reservoir, and remains in thermal contact with it the whole time. The engine does negative work.
QL QH
.
[efficiency of any heat engine]
It turns out that there is a particular type of heat engine, the Carnot engine, which, although not 100% efficient, is more efficient than any other. The grade of heat energy in a system can thus be unambiguously defined in terms of the amount of heat energy in it that cannot be extracted, even by a Carnot engine. How can we build the most efficient possible engine? Let’s start with an unnecessarily inefficient engine like a car engine and see how it could be improved. The radiator and exhaust expel hot gases, which is a waste of heat energy. These gases are cooler than the exploded air-gas mixture inside the cylinder, but hotter than the air that surrounds the car. We could thus improve the engine’s efficiency by adding an auxiliary heat engine to it, which would operate with the first engine’s exhaust as its hot reservoir and the air as its cold reservoir. In general, any heat engine that expels heat at an intermediate temperature can be made more efficient by changing it so that it expels heat only at the temperature of the cold reservoir. Similarly, any heat engine that absorbs some energy at an intermediate temperature can be made more efficient by adding an auxiliary heat engine to it which will operate between the hot reservoir and this intermediate temperature.
p / The beginning of the second compression stroke, in which mechanical work is absorbed, heating the working gas back up to TH .
Based on these arguments, we define a Carnot engine as a heat engine that absorbs heat only from the hot reservoir and expels it only into the cold reservoir. Figures m-p show a realization of a Carnot engine using a piston in a cylinder filled with a monoatomic ideal gas. This gas, known as the working fluid, is separate from, but exchanges energy with, the hot and cold reservoirs. It turns out that this particular Carnot engine has an efficiency given by efficiency = 1 −
424
Chapter 16
Thermodynamics
TL TH
,
[efficiency of a Carnot engine]
where TL is the temperature of the cold reservoir and TH is the temperature of the hot reservoir. (A proof of this fact is given in my book Simple Nature, which you can download for free.) Even if you do not wish to dig into the details of the proof, the basic reason for the temperature dependence is not so hard to understand. Useful mechanical work is done on strokes m and n, in which the gas expands. The motion of the piston is in the same direction as the gas’s force on the piston, so positive work is done on the piston. In strokes o and p, however, the gas does negative work on the piston. We would like to avoid this negative work, but we must design the engine to perform a complete cycle. Luckily the pressures during the compression strokes are lower than the ones during the expansion strokes, so the engine doesn’t undo all its work with every cycle. The ratios of the pressures are in proportion to the ratios of the temperatures, so if TL is 20% of TH , the engine is 80% efficient. We have already proved that any engine that is not a Carnot engine is less than optimally efficient, and it is also true that all Carnot engines operating between a given pair of temperatures TH and TL have the same efficiency. Thus a Carnot engine is the most efficient possible heat engine. Entropy We would like to have some numerical way of measuring the grade of energy in a system. We want this quantity, called entropy, to have the following two properties: (1) Entropy is additive. When we combine two systems and consider them as one, the entropy of the combined system equals the sum of the entropies of the two original systems. (Quantities like mass and energy also have this property.) (2) The entropy of a system is not changed by operating a Carnot engine within it. It turns out to be simpler and more useful to define changes in entropy than absolute entropies. Suppose as an example that a system contains some hot matter and some cold matter. It has a relatively high grade of energy because a heat engine could be used to extract mechanical work from it. But if we allow the hot and cold parts to equilibrate at some lukewarm temperature, the grade of energy has gotten worse. Thus putting heat into a hotter area is more useful than putting it into a cold area. Motivated by these considerations, we define a change in entropy as follows: ΔS
=
Q T
q / Entropy can be understood using the metaphor of a water wheel. Letting the water levels equalize is like letting the entropy maximize. Taking water from the high side and putting it into the low side increases the entropy. Water levels in this metaphor correspond to temperatures in the actual definition of entropy.
[change in entropy when adding heat Q to matter at temperature T ; ΔS is negative if heat is taken out]
Section 16.3
Entropy
425
A system with a higher grade of energy has a lower entropy. Entropy is additive. example 12 Since changes in entropy are defined by an additive quantity (heat) divided by a non-additive one (temperature), entropy is additive. Entropy isn’t changed by a Carnot engine. The efficiency of a heat engine is defined by efficiency = 1 − QL /QH
example 13 ,
and the efficiency of a Carnot engine is efficiency = 1 − TL /TH
,
so for a Carnot engine we have QL /QH = TL /TH , which can be rewritten as QL /TL = QH /TH . The entropy lost by the hot reservoir is therefore the same as the entropy gained by the cold one. Entropy increases in heat conduction. example 14 When a hot object gives up energy to a cold one, conservation of energy tells us that the amount of heat lost by the hot object is the same as the amount of heat gained by the cold one. The change in entropy is −Q/TH + Q/TL , which is positive because TL < T H . Entropy is increased by a non-Carnot engine. example 15 The efficiency of a non-Carnot engine is less than 1 - TL /TH , so QL /QH > TL /TH and QL /TL > QH /TH . This means that the entropy increase in the cold reservoir is greater than the entropy decrease in the hot reservoir. A book sliding to a stop example 16 A book slides across a table and comes to a stop. Once it stops, all its kinetic energy has been transformed into heat. As the book and table heat up, their entropies both increase, so the total entropy increases as well. Examples 14-16 involved closed systems, and in all of them the total entropy either increased or stayed the same. It never decreased. Here are two examples of schemes for decreasing the entropy of a closed system, with explanations of why they don’t work. Using a refrigerator to decrease entropy? example 17 A refrigerator takes heat from a cold area and dumps it into a hot area. (1) Does this lead to a net decrease in the entropy of a closed system? (2) Could you make a Carnot engine more efficient by running a refrigerator to cool its low-temperature reservoir and eject heat into its high-temperature reservoir? (1) No. The heat that comes off of the radiator coils on the back of your kitchen fridge is a great deal more than the heat the fridge removes from inside; the difference is what it costs to run your fridge. The heat radiated from the coils is so much more
426
Chapter 16
Thermodynamics
than the heat removed from the inside that the increase in the entropy of the air in the room is greater than the decrease of the entropy inside the fridge. The most efficient refrigerator is actually a Carnot engine running in reverse, which leads to neither an increase nor a decrease in entropy. (2) No. The most efficient refrigerator is a reversed Carnot engine. You will not achieve anything by running one Carnot engine in reverse and another forward. They will just cancel each other out. Maxwell’s daemon example 18 Physicist James Clerk Maxwell imagined pair of neighboring rooms, their air being initially in thermal equilibrium, having a partition across the middle with a tiny door. A miniscule daemon is posted at the door with a little ping-pong paddle, and his duty is to try to build up faster-moving air molecules in room B and slower ones in room A. For instance, when a fast molecule is headed through the door, going from A to B, he lets it by, but when a slower than average molecule tries the same thing, he hits it back into room A. Would this decrease the total entropy of the pair of rooms? No. The daemon needs to eat, and we can think of his body as a little heat engine. His metabolism is less efficient than a Carnot engine, so he ends up increasing the entropy rather than decreasing it. Observation such as these lead to the following hypothesis, known as the second law of thermodynamics: The entropy of a closed system always increases, or at best stays the same: ΔS ≥ 0. At present my arguments to support this statement may seem less than convincing, since they have so much to do with obscure facts about heat engines. A more satisfying and fundamental explanation for the continual increase in entropy was achieved by Ludwig Boltzmann, and you may wish to learn more about Boltzmann’s ideas from my book Simple Nature, which you can download for free. Briefly, Boltzmann realized that entropy was a measure of randomness or disorder at the atomic level, and disorder doesn’t spontaneously change into order. To emphasize the fundamental and universal nature of the second law, here are a few examples. Entropy and evolution example 19 A favorite argument of many creationists who don’t believe in evolution is that evolution would violate the second law of thermodynamics: the death and decay of a living thing releases heat (as
Section 16.3
Entropy
427
when a compost heap gets hot) and lessens the amount of energy available for doing useful work, while the reverse process, the emergence of life from nonliving matter, would require a decrease in entropy. Their argument is faulty, since the second law only applies to closed systems, and the earth is not a closed system. The earth is continuously receiving energy from the sun. The heat death of the universe example 20 Victorian philosophers realized that living things had low entropy, as discussed in example 19, and spent a lot of time worrying about the heat death of the universe: eventually the universe would have to become a high-entropy, lukewarm soup, with no life or organized motion of any kind. Fortunately (?), we now know a great many other things that will make the universe inhospitable to life long before its entropy is maximized. Life on earth, for instance, will end when the sun evolves into a giant star and vaporizes our planet. Hawking radiation example 21 Any process that could destroy heat (or convert it into nothing but mechanical work) would lead to a reduction in entropy. Black holes are supermassive stars whose gravity is so strong that nothing, not even light, can escape from them once it gets within a boundary known as the event horizon. Black holes are commonly observed to suck hot gas into them. Does this lead to a reduction in the entropy of the universe? Of course one could argue that the entropy is still there inside the black hole, but being able to “hide” entropy there amounts to the same thing as being able to destroy entropy. The physicist Steven Hawking was bothered by this question, and finally realized that although the actual stuff that enters a black hole is lost forever, the black hole will gradually lose energy in the form of light emitted from just outside the event horizon. This light ends up reintroducing the original entropy back into the universe at large.
428
Chapter 16
Thermodynamics
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 (a) Show that under conditions of standard pressure and temperature, the volume of a sample of an ideal gas depends only on the number of molecules in it. (b) One mole is defined as 6.0 × 1023 atoms. Find the volume of one mole of an ideal gas, in units of liters, at standard temperature√and pressure (0◦ C and 101 kPa). 2 A gas in a cylinder expands its volume by an amount ΔV , pushing out a piston. Show that the work done by the gas on the piston is given by ΔW = P ΔV . 3 (a) A helium atom contains 2 protons, 2 electrons, and √ 2 neutrons. Find the mass of a helium atom. √ (b) Find the number of atoms in 1 kg of helium. (c) Helium gas is monoatomic. Find the amount of heat needed to raise the temperature of 1 kg of helium by 1 degree C. (This is √ known as helium’s heat capacity at constant volume.) 4 Refrigerators, air conditioners, and heat pumps are heat engines that work in reverse. You put in mechanical work, and it the effect is to take heat out of a cooler reservoir and deposit heat in a warmer one: QL + W = QH . As with the heat engines discussed previously, the efficiency is defined as the energy transfer you want (QL for a refrigerator or air conditioner, QH for a heat pump) divided by the energy transfer you pay for (W ). Efficiencies are supposed to be unitless, but the efficiency of an air conditioner is normally given in terms of an EER rating (or a more complex version called an SEER). The EER is defined as QL /W , but expressed in the barbaric units of of Btu/watt-hour. A typical EER rating for a residential air conditioner is about 10 Btu/watt-hour, corresponding to an efficiency of about 3. The standard temperatures used for testing an air conditioner’s efficiency are 80◦ F (27◦ C) inside and 95◦ F (35◦ C) outside. (a) What would be the EER rating of a reversed Carnot engine used √ as an air conditioner? (b) If you ran a 3-kW residential air conditioner, with an efficiency of 3, for one hour, what would be the effect on the total entropy of the universe? Is your answer consistent with the second law of √ thermodynamics? 5 (a) Estimate the pressure at the center of the Earth, assuming it is of constant density throughout. Use the technique of example 5 on page 416. Note that g is not constant with respect to depth
Problems
429
— it equals Gmr/b3 for r, the distance from the center, less than b, the earth’s radius.1 State your result in terms of G, m, and b. (b) Show that your answer from part a has the right units for pressure. √ (c) Evaluate the result numerically. (d) Given that the earth’s atmosphere is on the order of one thousandth the thickness of the earth’s radius, and that the density of the earth is several thousand times greater than the density of the lower atmosphere, check that your result is of a reasonable order of magnitude. 6 (a) Determine the ratio between the escape velocities from the √ surfaces of the earth and the moon. (b) The temperature during the lunar daytime gets up to about 130◦ C. In the extremely thin (almost nonexistent) lunar atmosphere, estimate how the typical velocity of a molecule would compare with that of the same type of molecule in the earth’s atmosphere. As√ sume that the earth’s atmosphere has a temperature of 0◦ C. (c) Suppose you were to go to the moon and release some fluorocarbon gas, with molecular formula Cn F2n+2 . Estimate what is the smallest fluorocarbon molecule (lowest n) whose typical velocity would be lower than that of an N2 molecule on earth in proportion to the moon’s lower escape velocity. The moon would be able to √ retain an atmosphere made of these molecules. 7 Most of the atoms in the universe are in the form of gas that is not part of any star or galaxy: the intergalactic medium (IGM). The IGM consists of about 10−5 atoms per cubic centimeter, with a typical temperature of about 103 K. These are, in some sense, the density and temperature of the universe (not counting light, or the exotic particles known as “dark matter”). Calculate the pressure of the universe (or, speaking more carefully, the typical pressure√due to the IGM). 8 A sample of gas is enclosed in a sealed chamber. The gas consists of molecules, which are then split in half through some process such as exposure to ultraviolet light, or passing an electric spark through the gas. The gas returns to thermal equilibrium with the surrounding room. How does its pressure now compare with its pressure before the molecules were split?
9 The figure shows a demonstration performed by Otto von Guericke for Emperor Ferdinand III, in which two teams of horses failed to pull apart a pair of hemispheres from which the air had been evacuated. (a) What object makes the force that holds the 1
Derivation: The shell theorem tells us that the gravitational field at r is the same as if all the mass existing at greater depths was concentrated at the earth’s center. Since volume scales like the third power of distance, this constitutes a fraction (r/b)3 of the earth’s mass, so the field is (Gm/r2 )(r/b)3 = Gmr/b3 .
430
Chapter 16
Thermodynamics
Problem 9.
hemispheres together? (b) The hemispheres are in a museum in Berlin, and have a diameter of 65 cm. What is the amount of force holding them together? (Hint: The answer would be the same if they were cylinders or pie plates rather then hemispheres.) 10 Even when resting, the human body needs to do a certain amount of mechanical work to keep the heart beating. This quantity is difficult to define and measure with high precision, and also depends on the individual and her level of activity, but it’s estimated to be about 1 to 5 watts. Suppose we consider the human body as nothing more than a pump. A person who is just lying in bed all day needs about 1000 kcal/day worth of food to stay alive. (a) Estimate the person’s thermodynamic efficiency as a pump, and (b) compare with the maximum possible efficiency imposed by the laws of thermodynamics for a heat engine operating across the difference between a body temperature of 37◦ C and an ambient temperature Answer, p. 545 of 22◦ C. (c) Interpret your answer.
Problems
431
Vibrations and waves
432
434
The vibrations of this electric bass string are converted to electrical vibrations, then to sound vibrations, and finally to vibrations of our eardrums.
Chapter 17
Vibrations Dandelion. Cello. Read those two words, and your brain instantly conjures a stream of associations, the most prominent of which have to do with vibrations. Our mental category of “dandelion-ness” is strongly linked to the color of light waves that vibrate about half a million billion times a second: yellow. The velvety throb of a cello has as its most obvious characteristic a relatively low musical pitch — the note you are spontaneously imagining right now might be one whose sound vibrations repeat at a rate of a hundred times a second. Evolution has designed our two most important senses around the assumption that not only will our environment be drenched with information-bearing vibrations, but in addition those vibrations will often be repetitive, so that we can judge colors and pitches by the rate of repetition. Granting that we do sometimes encounter nonrepeating waves such as the consonant “sh,” which has no recognizable pitch, why was Nature’s assumption of repetition nevertheless so right in general? Repeating phenomena occur throughout nature, from the orbits of electrons in atoms to the reappearance of Halley’s Comet every 75 years. Ancient cultures tended to attribute repetitious phenomena
435
like the seasons to the cyclical nature of time itself, but we now have a less mystical explanation. Suppose that instead of Halley’s Comet’s true, repeating elliptical orbit that closes seamlessly upon itself with each revolution, we decide to take a pen and draw a whimsical alternative path that never repeats. We will not be able to draw for very long without having the path cross itself. But at such a crossing point, the comet has returned to a place it visited once before, and since its potential energy is the same as it was on the last visit, conservation of energy proves that it must again have the same kinetic energy and therefore the same speed. Not only that, but the comet’s direction of motion cannot be randomly chosen, because angular momentum must be conserved as well. Although this falls short of being an ironclad proof that the comet’s orbit must repeat, it no longer seems surprising that it does. a / If we try to draw a nonrepeating orbit for Halley’s Comet, it will inevitably end up crossing itself.
Conservation laws, then, provide us with a good reason why repetitive motion is so prevalent in the universe. But it goes deeper than that. Up to this point in your study of physics, I have been indoctrinating you with a mechanistic vision of the universe as a giant piece of clockwork. Breaking the clockwork down into smaller and smaller bits, we end up at the atomic level, where the electrons circling the nucleus resemble — well, little clocks! From this point of view, particles of matter are the fundamental building blocks of everything, and vibrations and waves are just a couple of the tricks that groups of particles can do. But at the beginning of the 20th century, the tables were turned. A chain of discoveries initiated by Albert Einstein led to the realization that the so-called subatomic “particles” were in fact waves. In this new world-view, it is vibrations and waves that are fundamental, and the formation of matter is just one of the tricks that waves can do.
17.1 Period, frequency, and amplitude
b / A spring has an equilibrium length, 1, and can be stretched, 2, or compressed, 3. A mass attached to the spring can be set into motion initially, 4, and will then vibrate, 4-13.
Figure b shows our most basic example of a vibration. With no forces on it, the spring assumes its equilibrium length, b/1. It can be stretched, 2, or compressed, 3. We attach the spring to a wall on the left and to a mass on the right. If we now hit the mass with a hammer, 4, it oscillates as shown in the series of snapshots, 4-13. If we assume that the mass slides back and forth without friction and that the motion is one-dimensional, then conservation of energy proves that the motion must be repetitive. When the block comes back to its initial position again, 7, its potential energy is the same again, so it must have the same kinetic energy again. The motion is in the opposite direction, however. Finally, at 10, it returns to its initial position with the same kinetic energy and the same direction of motion. The motion has gone through one complete cycle, and will now repeat forever in the absence of friction. The usual physics terminology for motion that repeats itself over
436
Chapter 17
Vibrations
and over is periodic motion, and the time required for one repetition is called the period, T . (The symbol P is not used because of the possible confusion with momentum.) One complete repetition of the motion is called a cycle. We are used to referring to short-period sound vibrations as “high” in pitch, and it sounds odd to have to say that high pitches have low periods. It is therefore more common to discuss the rapidity of a vibration in terms of the number of vibrations per second, a quantity called the frequency, f . Since the period is the number of seconds per cycle and the frequency is the number of cycles per second, they are reciprocals of each other,
f = 1/T
c / Position-versus-time graphs for half a period and a full period.
.
A carnival game example 1 In the carnival game shown in figure d, the rube is supposed to push the bowling ball on the track just hard enough so that it goes over the hump and into the valley, but does not come back out again. If the only types of energy involved are kinetic and potential, this is impossible. Suppose you expect the ball to come back to a point such as the one shown with the dashed outline, then stop and turn around. It would already have passed through this point once before, going to the left on its way into the valley. It was moving then, so conservation of energy tells us that it cannot be at rest when it comes back to the same point. The motion that the customer hopes for is physically impossible. There is a physically possible periodic motion in which the ball rolls back and forth, staying confined within the valley, but there is no way to get the ball into that motion beginning from the place where we start. There is a way to beat the game, though. If you put enough spin on the ball, you can create enough kinetic friction so that a significant amount of heat is generated. Conservation of energy then allows the ball to be at rest when it comes back to a point like the outlined one, because kinetic energy has been converted into heat.
d / Example 1.
Period and frequency of a fly’s wing-beats example 2 A Victorian parlor trick was to listen to the pitch of a fly’s buzz, reproduce the musical note on the piano, and announce how many times the fly’s wings had flapped in one second. If the fly’s wings flap, say, 200 times in one second, then the frequency of their motion is f = 200/1 s = 200 s−1 . The period is one 200th of a second, T = 1/f = (1/200) s = 0.005 s.
Section 17.1
Period, frequency, and amplitude
437
Units of inverse second, s−1 , are awkward in speech, so an abbreviation has been created. One Hertz, named in honor of a pioneer of radio technology, is one cycle per second. In abbreviated form, 1 Hz = 1 s−1 . This is the familiar unit used for the frequencies on the radio dial. Frequency of a radio station example 3 KKJZ’s frequency is 88.1 MHz. What does this mean, and what period does this correspond to? The metric prefix M- is mega-, i.e., millions. The radio waves emitted by KKJZ’s transmitting antenna vibrate 88.1 million times per second. This corresponds to a period of T = 1/f = 1.14 × 10−8 s
.
This example shows a second reason why we normally speak in terms of frequency rather than period: it would be painful to have to refer to such small time intervals routinely. I could abbreviate by telling people that KKJZ’s period was 11.4 nanoseconds, but most people are more familiar with the big metric prefixes than with the small ones. Units of frequency are also commonly used to specify the speeds of computers. The idea is that all the little circuits on a computer chip are synchronized by the very fast ticks of an electronic clock, so that the circuits can all cooperate on a task without getting ahead or behind. Adding two numbers might require, say, 30 clock cycles. Microcomputers these days operate at clock frequencies of about a gigahertz.
e / 1. The amplitude of the vibrations of the mass on a spring could be defined in two different ways. It would have units of distance. 2. The amplitude of a swinging pendulum would more naturally be defined as an angle.
We have discussed how to measure how fast something vibrates, but not how big the vibrations are. The general term for this is amplitude, A. The definition of amplitude depends on the system being discussed, and two people discussing the same system may not even use the same definition. In the example of the block on the end of the spring, e/1, the amplitude will be measured in distance units such as cm. One could work in terms of the distance traveled by the block from the extreme left to the extreme right, but it would be somewhat more common in physics to use the distance from the center to one extreme. The former is usually referred to as the peak-to-peak amplitude, since the extremes of the motion look like mountain peaks or upside-down mountain peaks on a graph of position versus time. In other situations we would not even use the same units for amplitude. The amplitude of a child on a swing, or a pendulum, e/2, would most conveniently be measured as an angle, not a distance, since her feet will move a greater distance than her head. The electrical vibrations in a radio receiver would be measured in electrical units such as volts or amperes.
438
Chapter 17
Vibrations
17.2 Simple harmonic motion Why are sine-wave vibrations so common? If we actually construct the mass-on-a-spring system discussed in the previous section and measure its motion accurately, we will find that its x−t graph is nearly a perfect sine-wave shape, as shown in figure f/1. (We call it a “sine wave” or “sinusoidal” even if it is a cosine, or a sine or cosine shifted by some arbitrary horizontal amount.) It may not be surprising that it is a wiggle of this general sort, but why is it a specific mathematically perfect shape? Why is it not a sawtooth shape like 2 or some other shape like 3? The mystery deepens as we find that a vast number of apparently unrelated vibrating systems show the same mathematical feature. A tuning fork, a sapling pulled to one side and released, a car bouncing on its shock absorbers, all these systems will exhibit sine-wave motion under one condition: the amplitude of the motion must be small. It is not hard to see intuitively why extremes of amplitude would act differently. For example, a car that is bouncing lightly on its shock absorbers may behave smoothly, but if we try to double the amplitude of the vibrations the bottom of the car may begin hitting the ground, f/4. (Although we are assuming for simplicity in this chapter that energy is never dissipated, this is clearly not a very realistic assumption in this example. Each time the car hits the ground it will convert quite a bit of its potential and kinetic energy into heat and sound, so the vibrations would actually die out quite quickly, rather than repeating for many cycles as shown in the figure.)
f / Sinusoidal and non-sinusoidal vibrations.
The key to understanding how an object vibrates is to know how the force on the object depends on the object’s position. If an object is vibrating to the right and left, then it must have a leftward force on it when it is on the right side, and a rightward force when it is on the left side. In one dimension, we can represent the direction of the force using a positive or negative sign, and since the force changes from positive to negative there must be a point in the middle where the force is zero. This is the equilibrium point, where the object would stay at rest if it was released at rest. For convenience of notation throughout this chapter, we will define the origin of our coordinate system so that x equals zero at equilibrium. The simplest example is the mass on a spring, for which the force on the mass is given by Hooke’s law, F = −kx
.
We can visualize the behavior of this force using a graph of F versus x, as shown in figure g. The graph is a line, and the spring constant, k, is equal to minus its slope. A stiffer spring has a larger value of k and a steeper slope. Hooke’s law is only an approximation, but
Section 17.2
g / The force exerted by an ideal spring, which behaves exactly according to Hooke’s law.
Simple harmonic motion
439
it works very well for most springs in real life, as long as the spring isn’t compressed or stretched so much that it is permanently bent or damaged. The following important theorem, whose proof is given in optional section 17.3, relates the motion graph to the force graph. Theorem: A linear force graph makes a sinusoidal motion graph. If the total force on a vibrating object depends only on the object’s position, and is related to the objects displacement from equilibrium by an equation of the form F = −kx, then the object’s motion displays a sinusoidal graph with period T = 2π m/k. Even if you do not read the proof, it is not too hard to understand why the equation for the period makes sense. A greater mass causes a greater period, since the force will not be able to whip a massive object back and forth very rapidly. A larger value of k causes a shorter period, because a stronger force can whip the object back and forth more rapidly.
h / Seen from close up, any F − x curve looks like a line.
This may seem like only an obscure theorem about the mass-ona-spring system, but figure h shows it to be far more general than that. Figure h/1 depicts a force curve that is not a straight line. A system with this F − x curve would have large-amplitude vibrations that were complex and not sinusoidal. But the same system would exhibit sinusoidal small-amplitude vibrations. This is because any curve looks linear from very close up. If we magnify the F − x graph as shown in figure h/2, it becomes very difficult to tell that the graph is not a straight line. If the vibrations were confined to the region shown in h/2, they would be very nearly sinusoidal. This is the reason why sinusoidal vibrations are a universal feature of all vibrating systems, if we restrict ourselves to small amplitudes. The theorem is therefore of great general significance. It applies throughout the universe, to objects ranging from vibrating stars to vibrating nuclei. A sinusoidal vibration is known as simple harmonic motion. Period is approximately independent of amplitude, if the amplitude is small. Until now we have not even mentioned the most counterintu itive aspect of the equation T = 2π m/k: it does not depend on amplitude at all. Intuitively, most people would expect the mass-ona-spring system to take longer to complete a cycle if the amplitude was larger. (We are comparing amplitudes that are different from each other, but both small enough that the theorem applies.) In fact the larger-amplitude vibrations take the same amount of time as the small-amplitude ones. This is because at large amplitudes, the force is greater, and therefore accelerates the object to higher
440
Chapter 17
Vibrations
speeds. Legend has it that this fact was first noticed by Galileo during what was apparently a less than enthralling church service. A gust of wind would now and then start one of the chandeliers in the cathedral swaying back and forth, and he noticed that regardless of the amplitude of the vibrations, the period of oscillation seemed to be the same. Up until that time, he had been carrying out his physics experiments with such crude time-measuring techniques as feeling his own pulse or singing a tune to keep a musical beat. But after going home and testing a pendulum, he convinced himself that he had found a superior method of measuring time. Even without a fancy system of pulleys to keep the pendulum’s vibrations from dying down, he could get very accurate time measurements, because the gradual decrease in amplitude due to friction would have no effect on the pendulum’s period. (Galileo never produced a modernstyle pendulum clock with pulleys, a minute hand, and a second hand, but within a generation the device had taken on the form that persisted for hundreds of years after.)
The pendulum example 4 Compare the periods of pendula having bobs with different masses. From the equation T = 2π m/k , we might expect that a larger mass would lead to a longer period. However, increasing the mass also increases the forces that act on the pendulum: gravity and the tension in the string. This increases k as well as m, so the period of a pendulum is independent of m.
Discussion questions
A Suppose that a pendulum has a rigid arm mounted on a bearing, rather than a string tied at its top with a knot. The bob can then oscillate with center-to-side amplitudes greater than 90◦ . For the maximum amplitude of 180◦ , what can you say about the period?
B In the language of calculus, Newton’s second law for a simple harmonic oscillator can be written in the form d2 x /dt 2 = −(. . .)x , where “. . . ” refers to a constant, and the minus sign says that if we pull the object away from equilibrium, a restoring force tries to bring it back to equlibrium, which is the opposite direction. This is why we get motion that looks like a sine or cosine function: these are functions that, when differentiated twice, give back the original function but with an opposite sign. Now consider the example described in discussion question A, where a pendulum is upright or nearly upright. How does the analysis play out differently?
Section 17.2
Simple harmonic motion
441
17.3 Proofs In this section we prove (1) that a linear F − x graph gives sinusoidal motion, (2) that the period of the motion is 2π m/k, and (3) that the period is independent of the amplitude. You may omit this section without losing the continuity of the chapter. The basic idea of the proof can be understood by imagining that you are watching a child on a merry-go-round from far away. Because you are in the same horizontal plane as her motion, she appears to be moving from side to side along a line. Circular motion viewed edge-on doesn’t just look like any kind of back-and-forth motion, it looks like motion with a sinusoidal x−t graph, because the sine and cosine functions can be defined as the x and y coordinates of a point at angle θ on the unit circle. The idea of the proof, then, is to show that an object acted on by a force that varies as F = −kx has motion that is identical to circular motion projected down to one dimension. The v 2 /r expression will also fall out at the end. i / The object moves along the circle at constant speed, but even though its overall speed is constant, the x and y components of its velocity are continuously changing, as shown by the unequal spacing of the points when projected onto the line below. Projected onto the line, its motion is the same as that of an object experiencing a force F = −kx .
The moons of Jupiter. example 5 Before moving on to the proof, we illustrate the concept using the moons of Jupiter. Their discovery by Galileo was an epochal event in astronomy, because it proved that not everything in the universe had to revolve around the earth as had been believed. Galileo’s telescope was of poor quality by modern standards, but figure j shows a simulation of how Jupiter and its moons might appear at intervals of three hours through a large present-day instrument. Because we see the moons’ circular orbits edge-on, they appear to perform sinusoidal vibrations. Over this time period, the innermost moon, Io, completes half a cycle.
j / Example 5.
442
Chapter 17
Vibrations
For an object performing uniform circular motion, we have |a| =
v2 r
.
The x component of the acceleration is therefore ax =
v2 cos θ r
,
where θ is the angle measured counterclockwise from the x axis. Applying Newton’s second law, Fx v2 = − cos θ m r v2 Fx = −m cos θ r
,
so .
Since our goal is an equation involving the period, it is natural to eliminate the variable v = circumference/T = 2πr/T , giving Fx = −
4π 2 mr cos θ T2
.
The quantity r cos θ is the same as x, so we have Fx = −
4π 2 m x T2
.
Since everything is constant in this equation except for x, we have proved that motion with force proportional to x is the same as circular motion projected onto a line, and therefore that a force proportional to x gives sinusoidal motion. Finally, we identify the constant factor of 4π 2 m/T 2 with k, and solving for T gives the desired equation for the period, m T = 2π . k Since this equation is independent of r, T is independent of the amplitude, subject to the initial assumption of perfect F = −kx behavior, which in reality will only hold approximately for small x.
Section 17.3
Proofs
443
Summary Selected vocabulary periodic motion . motion that repeats itself over and over period . . . . . . . the time required for one cycle of a periodic motion frequency . . . . . the number of cycles per second, the inverse of the period amplitude . . . . the amount of vibration, often measured from the center to one side; may have different units depending on the nature of the vibration simple harmonic motion whose x − t graph is a sine wave motion . . . . . . Notation T . . . . f . . . . . A . . . . k. . . . .
. . . .
. . . .
. . . .
. . . .
. . . .
period frequency amplitude the slope of the graph of F versus x, where F is the total force acting on an object and x is the object’s position; for a spring, this is known as the spring constant.
Other terminology and notation ν . . . . . . . . . . The Greek letter ν, nu, is used in many books for frequency. ω . . . . . . . . . . The Greek letter ω, omega, is often used as an abbreviation for 2πf . Summary Periodic motion is common in the world around us because of conservation laws. An important example is one-dimensional motion in which the only two forms of energy involved are potential and kinetic; in such a situation, conservation of energy requires that an object repeat its motion, because otherwise when it came back to the same point, it would have to have a different kinetic energy and therefore a different total energy. Not only are periodic vibrations very common, but small-amplitude vibrations are always sinusoidal as well. That is, the x − t graph is a sine wave. This is because the graph of force versus position will always look like a straight line on a sufficiently small scale. This type of vibration is called simple harmonic motion. In simple harmonic motion, the period is independent of the amplitude, and is given by . T = 2π m/k
444
Chapter 17
Vibrations
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 Find an equation for the frequency of simple harmonic motion √ in terms of k and m. 2 Many single-celled organisms propel themselves through water with long tails, which they wiggle back and forth. (The most obvious example is the sperm cell.) The frequency of the tail’s vibration is typically about 10-15 Hz. To what range of periods does this range of frequencies correspond? 3 (a) Pendulum 2 has a string twice as long as pendulum 1. If we define x as the distance traveled by the bob along a circle away from the bottom, how does the k of pendulum 2 compare with the k of pendulum 1? Give a numerical ratio. [Hint: the total force on the bob is the same if the angles away from the bottom are the same, but equal angles do not correspond to equal values of x.] (b) Based on your answer from part (a), how does the period of pendulum 2 compare with the period of pendulum 1? Give a numerical ratio. 4 A pneumatic spring consists of a piston riding on top of the air in a cylinder. The upward force of the air on the piston is given by Fair = ax−1.4 , where a is a constant with funny units of N · m1.4 . For simplicity, assume the air only supports the weight, FW , of the piston itself, although in practice this device is used to support some other object. The equilibrium position, x0 , is where FW equals −Fair . (Note that in the main text I have assumed the equilibrium position to be at x = 0, but that is not the natural choice here.) Assume friction is negligible, and consider a case where the amplitude of the vibrations is very small. Let a = 1.0 N·m1.4 , x0 = 1.00 m, and FW = −1.00 N. The piston is released from x = 1.01 m. Draw a neat, accurate graph of the total force, F , as a function of x, on graph paper, covering the range from x = 0.98 m to 1.02 m. Over this small range, you will find that the force is very nearly proportional to x − x0 . Approximate the curve with a straight line, find its slope, and derive the approximate period of √ oscillation.
Problem 4.
5 Consider the same pneumatic piston described in problem 4, but now imagine that the oscillations are not small. Sketch a graph of the total force on the piston as it would appear over this wider range of motion. For a wider range of motion, explain why the vibration of the piston about equilibrium is not simple harmonic motion, and sketch a graph of x vs t, showing roughly how the curve is different from a sine wave. [Hint: Acceleration corresponds to the
Problems
445
curvature of the x − t graph, so if the force is greater, the graph should curve around more quickly.] 6 Archimedes’ principle states that an object partly or wholly immersed in fluid experiences a buoyant force equal to the weight of the fluid it displaces. For instance, if a boat is floating in water, the upward pressure of the water (vector sum of all the forces of the water pressing inward and upward on every square inch of its hull) must be equal to the weight of the water displaced, because if the boat was instantly removed and the hole in the water filled back in, the force of the surrounding water would be just the right amount to hold up this new “chunk” of water. (a) Show that a cube of mass m with edges of length b floating upright (not tilted) in a fluid of density ρ will have a draft (depth to which it sinks below the waterline) h given at equilibrium by h0 = m/b2 ρ. (b) Find the total force on the cube when its draft is h, and verify that plugging in h − h0 gives a total force of zero. (c) Find the cube’s period of oscillation as it bobs up and down in the water, and show that√can be expressed in terms of and g only. 7 The figure shows a see-saw with two springs at Codornices Park in Berkeley, California. Each spring has spring constant k, and a kid of mass m sits on each seat. (a) Find the period of vibration in terms of the variables k, m, a, and b. (b) Discuss the special case where a = b, rather than a > b as in the real see-saw. (c) Show√that your answer to part a also makes sense in the case of b = 0. 8 Show that the equation T = 2π m/k has units that make sense.
Problem 7.
9 A hot scientific question of the 18th century was the shape of the earth: whether its radius was greater at the equator than at the poles, or the other way around. One method used to attack this question was to measure gravity accurately in different locations on the earth using pendula. If the highest and lowest latitudes accessible to explorers were 0 and 70 degrees, then the the strength of gravity would in reality be observed to vary over a range from about 9.780 to 9.826 m/s2 . This change, amounting to 0.046 m/s2 , is greater than the 0.022 m/s2 effect to be expected if the earth had been spherical. The greater effect occurs because the equator feels a reduction due not just to the acceleration of the spinning earth out from under it, but also to the greater radius of the earth at the equator. What is the accuracy with which the period of a one-second pendulum would have to be measured in order to prove that the earth was not a sphere, and that it bulged at the equator?
446
Chapter 17
Vibrations
Exercise 17: Vibrations Equipment: • air track and carts of two different masses • springs • spring scales
Place the cart on the air track and attach springs so that it can vibrate. 1. Test whether the period of vibration depends on amplitude. Try at least one moderate amplitude, for which the springs do not go slack, at least one amplitude that is large enough so that they do go slack, and one amplitude that’s the very smallest you can possibly observe. 2. Try a cart with a different mass. Does the period change by the expected factor, based on the equation T = 2π m/k? 3. Use a spring scale to pull the cart away from equilibrium, and make a graph of force versus position. Is it linear? If so, what is its slope? 4. Test the equation T = 2π m/k numerically.
Exercise 17: Vibrations
447
448
Chapter 17
Vibrations
Top: A series of images from a film of the Tacoma Narrows Bridge vibrating on the day it was to collapse. Middle: The bridge immediately before the collapse, with the sides vibrating 8.5 meters (28 feet) up and down. Note that the bridge is over a mile long. Bottom: During and after the final collapse. The right-hand picture gives a sense of the massive scale of the construction.
Chapter 18
Resonance Soon after the mile-long Tacoma Narrows Bridge opened in July 1940, motorists began to notice its tendency to vibrate frighteningly in even a moderate wind. Nicknamed “Galloping Gertie,” the bridge collapsed in a steady 42-mile-per-hour wind on November 7 of the same year. The following is an eyewitness report from a newspaper editor who found himself on the bridge as the vibrations approached the breaking point. “Just as I drove past the towers, the bridge began to sway violently from side to side. Before I realized it, the tilt became so violent that I lost control of the car... I jammed on the brakes and
449
got out, only to be thrown onto my face against the curb. “Around me I could hear concrete cracking. I started to get my dog Tubby, but was thrown again before I could reach the car. The car itself began to slide from side to side of the roadway. “On hands and knees most of the time, I crawled 500 yards or more to the towers... My breath was coming in gasps; my knees were raw and bleeding, my hands bruised and swollen from gripping the concrete curb... Toward the last, I risked rising to my feet and running a few yards at a time... Safely back at the toll plaza, I saw the bridge in its final collapse and saw my car plunge into the Narrows.” The ruins of the bridge formed an artificial reef, one of the world’s largest. It was not replaced for ten years. The reason for its collapse was not substandard materials or construction, nor was the bridge under-designed: the piers were hundred-foot blocks of concrete, the girders massive and made of carbon steel. The bridge was destroyed because of the physical phenomenon of resonance, the same effect that allows an opera singer to break a wine glass with her voice and that lets you tune in the radio station you want. The replacement bridge, which has lasted half a century so far, was built smarter, not stronger. The engineers learned their lesson and simply included some slight modifications to avoid the resonance phenomenon that spelled the doom of the first one.
18.1 Energy in vibrations One way of describing the collapse of the bridge is that the bridge kept taking energy from the steadily blowing wind and building up more and more energetic vibrations. In this section, we discuss the energy contained in a vibration, and in the subsequent sections we will move on to the loss of energy and the adding of energy to a vibrating system, all with the goal of understanding the important phenomenon of resonance. Going back to our standard example of a mass on a spring, we find that there are two forms of energy involved: the potential energy stored in the spring and the kinetic energy of the moving mass. We may start the system in motion either by hitting the mass to put in kinetic energy or by pulling it to one side to put in potential energy. Either way, the subsequent behavior of the system is identical. It trades energy back and forth between kinetic and potential energy. (We are still assuming there is no friction, so that no energy is converted to heat, and the system never runs down.) The most important thing to understand about the energy content of vibrations is that the total energy is proportional to the
450
Chapter 18
Resonance
square of the amplitude. Although the total energy is constant, it is instructive to consider two specific moments in the motion of the mass on a spring as examples. When the mass is all the way to one side, at rest and ready to reverse directions, all its energy is potential. We have already seen that the potential energy stored in a spring equals (1/2)kx2 , so the energy is proportional to the square of the amplitude. Now consider the moment when the mass is passing through the equilibrium point at x = 0. At this point it has no potential energy, but it does have kinetic energy. The velocity is proportional to the amplitude of the motion, and the kinetic energy, (1/2)mv 2 , is proportional to the square of the velocity, so again we find that the energy is proportional to the square of the amplitude. The reason for singling out these two points is merely instructive; proving that energy is proportional to A2 at any point would suffice to prove that energy is proportional to A2 in general, since the energy is constant. Are these conclusions restricted to the mass-on-a-spring example? No. We have already seen that F = −kx is a valid approximation for any vibrating object, as long as the amplitude is small. We are thus left with a very general conclusion: the energy of any vibration is approximately proportional to the square of the amplitude, provided that the amplitude is small. Water in a U-tube example 1 If water is poured into a U-shaped tube as shown in the figure, it can undergo vibrations about equilibrium. The energy of such a vibration is most easily calculated by considering the “turnaround point” when the water has stopped and is about to reverse directions. At this point, it has only potential energy and no kinetic energy, so by calculating its potential energy we can find the energy of the vibration. This potential energy is the same as the work that would have to be done to take the water out of the righthand side down to a depth A below the equilibrium level, raise it through a height A, and place it in the left-hand side. The weight of this chunk of water is proportional to A, and so is the height through which it must be lifted, so the energy is proportional to A2 .
a / Example 1.
The range of energies of sound waves example 2 The amplitude of vibration of your eardrum at the threshold of pain is about 106 times greater than the amplitude with which it vibrates in response to the softest sound you can hear. How many times greater is the energy with which your ear has to cope for the painfully loud sound, compared to the soft sound? The amplitude is 106 times greater, and energy is proportional to the square of the amplitude, so the energy is greater by a factor
Section 18.1
Energy in vibrations
451
of 1012 . This is a phenomenally large factor! We are only studying vibrations right now, not waves, so we are not yet concerned with how a sound wave works, or how the energy gets to us through the air. Note that because of the huge range of energies that our ear can sense, it would not be reasonable to have a sense of loudness that was additive. Consider, for instance, the following three levels of sound: barely audible wind quiet conversation . . . . heavy metal concert . .
105 times more energy than the wind 1012 times more energy than the wind
In terms of addition and subtraction, the difference between the wind and the quiet conversation is nothing compared to the difference between the quiet conversation and the heavy metal concert. Evolution wanted our sense of hearing to be able to encompass all these sounds without collapsing the bottom of the scale so that anything softer than the crack of doom would sound the same. So rather than making our sense of loudness additive, mother nature made it multiplicative. We sense the difference between the wind and the quiet conversation as spanning a range of about 5/12 as much as the whole range from the wind to the heavy metal concert. Although a detailed discussion of the decibel scale is not relevant here, the basic point to note about the decibel scale is that it is logarithmic. The zero of the decibel scale is close to the lower limit of human hearing, and adding 1 unit to the decibel measurement corresponds to multiplying the energy level (or actually the power per unit area) by a certain factor.
18.2 Energy lost from vibrations
b / Friction has the effect of pinching the x − t graph of a vibrating object.
Until now, we have been making the relatively unrealistic assumption that a vibration would never die out. For a realistic mass on a spring, there will be friction, and the kinetic and potential energy of the vibrations will therefore be gradually converted into heat. Similarly, a guitar string will slowly convert its kinetic and potential energy into sound. In all cases, the effect is to “pinch” the sinusoidal x − t graph more and more with passing time. Friction is not necessarily bad in this context — a musical instrument that never got rid of any of its energy would be completely silent! The dissipation of the energy in a vibration is known as damping. self-check A Most people who try to draw graphs like those shown on the left will tend to shrink their wiggles horizontally as well as vertically. Why is this wrong? Answer, p. 544
In the graphs in figure b, I have not shown any point at which
452
Chapter 18
Resonance
the damped vibration finally stops completely. Is this realistic? Yes and no. If energy is being lost due to friction between two solid surfaces, then we expect the force of friction to be nearly independent of velocity. This constant friction force puts an upper limit on the total distance that the vibrating object can ever travel without replenishing its energy, since work equals force times distance, and the object must stop doing work when its energy is all converted into heat. (The friction force does reverse directions when the object turns around, but reversing the direction of the motion at the same time that we reverse the direction of the force makes it certain that the object is always doing positive work, not negative work.) Damping due to a constant friction force is not the only possibility however, or even the most common one. A pendulum may be damped mainly by air friction, which is approximately proportional to v 2 , while other systems may exhibit friction forces that are proportional to v. It turns out that friction proportional to v is the simplest case to analyze mathematically, and anyhow all the important physical insights can be gained by studying this case. If the friction force is proportional to v, then as the vibrations die down, the frictional forces get weaker due to the lower speeds. The less energy is left in the system, the more miserly the system becomes with giving away any more energy. Under these conditions, the vibrations theoretically never die out completely, and mathematically, the loss of energy from the system is exponential: the system loses a fixed percentage of its energy per cycle. This is referred to as exponential decay. A non-rigorous proof is as follows. The force of friction is proportional to v, and v is proportional to how far the objects travels in one cycle, so the frictional force is proportional to amplitude. The amount of work done by friction is proportional to the force and to the distance traveled, so the work done in one cycle is proportional to the square of the amplitude. Since both the work and the energy are proportional to A2 , the amount of energy taken away by friction in one cycle is a fixed percentage of the amount of energy the system has. self-check B Figure c shows an x-t graph for a strongly damped vibration, which loses half of its amplitude with every cycle. What fraction of the energy is lost in each cycle? Answer, p. 544
It is customary to describe the amount of damping with a quantity called the quality factor, Q, defined as the number of cycles required for the energy to fall off by a factor of 535. (The origin of this obscure numerical factor is e2π , where e = 2.71828 . . . is the base of natural logarithms. Choosing this particular number causes some of our later equations to come out nice and simple.) The terminology arises from the fact that friction is often considered a bad
Section 18.2
c / The amplitude with each cycle.
Energy lost from vibrations
is
halved
453
thing, so a mechanical device that can vibrate for many oscillations before it loses a significant fraction of its energy would be considered a high-quality device. Exponential decay in a trumpet example 3 The vibrations of the air column inside a trumpet have a Q of about 10. This means that even after the trumpet player stops blowing, the note will keep sounding for a short time. If the player suddenly stops blowing, how will the sound intensity 20 cycles later compare with the sound intensity while she was still blowing? The trumpet’s Q is 10, so after 10 cycles the energy will have fallen off by a factor of 535. After another 10 cycles we lose another factor of 535, so the sound intensity is reduced by a factor of 535 × 535 = 2.9 × 105 . The decay of a musical sound is part of what gives it its character, and a good musical instrument should have the right Q, but the Q that is considered desirable is different for different instruments. A guitar is meant to keep on sounding for a long time after a string has been plucked, and might have a Q of 1000 or 10000. One of the reasons why a cheap synthesizer sounds so bad is that the sound suddenly cuts off after a key is released. Q of a stereo speaker example 4 Stereo speakers are not supposed to reverberate or “ring” after an electrical signal that stops suddenly. After all, the recorded music was made by musicians who knew how to shape the decays of their notes correctly. Adding a longer “tail” on every note would make it sound wrong. We therefore expect that stereo speaker will have a very low Q, and indeed, most speakers are designed with a Q of about 1. (Low-quality speakers with larger Q values are referred to as “boomy.”) We will see later in the chapter that there are other reasons why a speaker should not have a high Q.
18.3 Putting energy into vibrations
d / 1. Pushing a child on a swing gradually puts more and more energy into her vibrations. 2. A fairly realistic graph of the driving force acting on the child. 3. A less realistic, but more mathematically simple, driving force.
454
Chapter 18
When pushing a child on a swing, you cannot just apply a constant force. A constant force will move the swing out to a certain angle, but will not allow the swing to start swinging. Nor can you give short pushes at randomly chosen times. That type of random pushing would increase the child’s kinetic energy whenever you happened to be pushing in the same direction as her motion, but it would reduce her energy when your pushing happened to be in the opposite direction compared to her motion. To make her build up her energy, you need to make your pushes rhythmic, pushing at the same point in each cycle. In other words, your force needs to form a repeating pattern with the same frequency as the normal frequency
Resonance
of vibration of the swing. Graph d/1 shows what the child’s x − t graph would look like as you gradually put more and more energy into her vibrations. A graph of your force versus time would probably look something like graph 2. It turns out, however, that it is much simpler mathematically to consider a vibration with energy being pumped into it by a driving force that is itself a sine-wave, 3. A good example of this is your eardrum being driven by the force of a sound wave. Now we know realistically that the child on the swing will not keep increasing her energy forever, nor does your eardrum end up exploding because a continuing sound wave keeps pumping more and more energy into it. In any realistic system, there is energy going out as well as in. As the vibrations increase in amplitude, there is an increase in the amount of energy taken away by damping with each cycle. This occurs for two reasons. Work equals force times distance (or, more accurately, the area under the force-distance curve). As the amplitude of the vibrations increases, the damping force is being applied over a longer distance. Furthermore, the damping force usually increases with velocity (we usually assume for simplicity that it is proportional to velocity), and this also serves to increase the rate at which damping forces remove energy as the amplitude increases. Eventually (and small children and our eardrums are thankful for this!), the amplitude approaches a maximum value, e, at which energy is removed by the damping force just as quickly as it is being put in by the driving force.
e / The amplitude approaches a maximum.
This process of approaching a maximum amplitude happens extremely quickly in many cases, e.g., the ear or a radio receiver, and we don’t even notice that it took a millisecond or a microsecond for the vibrations to “build up steam.” We are therefore mainly interested in predicting the behavior of the system once it has had enough time to reach essentially its maximum amplitude. This is known as the steady-state behavior of a vibrating system. Now comes the interesting part: what happens if the frequency of the driving force is mismatched to the frequency at which the system would naturally vibrate on its own? We all know that a radio station doesn’t have to be tuned in exactly, although there is only a small range over which a given station can be received. The designers of the radio had to make the range fairly small to make it possible to eliminate unwanted stations that happened to be nearby in frequency, but it couldn’t be too small or you wouldn’t be able to adjust the knob accurately enough. (Even a digital radio can be tuned to 88.0 MHz and still bring in a station at 88.1 MHz.) The ear also has some natural frequency of vibration, but in this case the range of frequencies to which it can respond is quite broad. Evolution has made the ear’s frequency response as broad as possible because it was to our ancestors’ advantage to be able to hear everything from a low roars to a high-pitched shriek.
Section 18.3
Putting energy into vibrations
455
The remainder of this section develops four important facts about the response of a system to a driving force whose frequency is not necessarily the same as the system’s natural frequency of vibration. The style is approximate and intuitive, but proofs are given in section 18.4. First, although we know the ear has a frequency — about 4000 Hz — at which it would vibrate naturally, it does not vibrate at 4000 Hz in response to a low-pitched 200 Hz tone. It always responds at the frequency at which it is driven. Otherwise all pitches would sound like 4000 Hz to us. This is a general fact about driven vibrations: (1) The steady-state response to a sinusoidal driving force occurs at the frequency of the force, not at the system’s own natural frequency of vibration. Now let’s think about the amplitude of the steady-state response. Imagine that a child on a swing has a natural frequency of vibration of 1 Hz, but we are going to try to make her swing back and forth at 3 Hz. We intuitively realize that quite a large force would be needed to achieve an amplitude of even 30 cm, i.e., the amplitude is less in proportion to the force. When we push at the natural frequency of 1 Hz, we are essentially just pumping energy back into the system to compensate for the loss of energy due to the damping (friction) force. At 3 Hz, however, we are not just counteracting friction. We are also providing an extra force to make the child’s momentum reverse itself more rapidly than it would if gravity and the tension in the chain were the only forces acting. It is as if we are artificially increasing the k of the swing, but this is wasted effort because we spend just as much time decelerating the child (taking energy out of the system) as accelerating her (putting energy in). Now imagine the case in which we drive the child at a very low frequency, say 0.02 Hz or about one vibration per minute. We are essentially just holding the child in position while very slowly walking back and forth. Again we intuitively recognize that the amplitude will be very small in proportion to our driving force. Imagine how hard it would be to hold the child at our own headlevel when she is at the end of her swing! As in the too-fast 3 Hz case, we are spending most of our effort in artificially changing the k of the swing, but now rather than reinforcing the gravity and tension forces we are working against them, effectively reducing k. Only a very small part of our force goes into counteracting friction, and the rest is used in repetitively putting potential energy in on the upswing and taking it back out on the downswing, without any long-term gain. We can now generalize to make the following statement, which
456
Chapter 18
Resonance
is true for all driven vibrations: (2) A vibrating system resonates at its own natural frequency.1 That is, the amplitude of the steady-state response is greatest in proportion to the amount of driving force when the driving force matches the natural frequency of vibration. An opera singer breaking a wine glass example 5 In order to break a wineglass by singing, an opera singer must first tap the glass to find its natural frequency of vibration, and then sing the same note back. Collapse of the Nimitz Freeway in an earthquake example 6 I led off the chapter with the dramatic collapse of the Tacoma Narrows Bridge, mainly because a it was well documented by a local physics professor, and an unknown person made a movie of the collapse. The collapse of a section of the Nimitz Freeway in Oakland, CA, during a 1989 earthquake is however a simpler example to analyze.
f / The collapsed section the Nimitz Freeway.
of
An earthquake consists of many low-frequency vibrations that occur simultaneously, which is why it sounds like a rumble of indeterminate pitch rather than a low hum. The frequencies that we can hear are not even the strongest ones; most of the energy is in the form of vibrations in the range of frequencies from about 1 Hz to 10 Hz. Now all the structures we build are resting on geological layers of dirt, mud, sand, or rock. When an earthquake wave comes along, the topmost layer acts like a system with a certain natural frequency of vibration, sort of like a cube of jello on a plate being shaken from side to side. The resonant frequency of the layer depends on how stiff it is and also on how deep it is. The illfated section of the Nimitz freeway was built on a layer of mud, and analysis by geologist Susan E. Hough of the U.S. Geological Survey shows that the mud layer’s resonance was centered on about 2.5 Hz, and had a width covering a range from about 1 Hz to 4 Hz. When the earthquake wave came along with its mixture of frequencies, the mud responded strongly to those that were close to its own natural 2.5 Hz frequency. Unfortunately, an engineering analysis after the quake showed that the overpass itself had a resonant frequency of 2.5 Hz as well! The mud responded strongly to the earthquake waves with frequencies close to 2.5 Hz, and the bridge responded strongly to the 2.5 Hz vibrations of the mud, causing sections of it to collapse. Collapse of the Tacoma Narrows Bridge example 7 Let’s now examine the more conceptually difficult case of the
Section 18.3
Putting energy into vibrations
457
Tacoma Narrows Bridge. The surprise here is that the wind was steady. If the wind was blowing at constant velocity, why did it shake the bridge back and forth? The answer is a little complicated. Based on film footage and after-the-fact wind tunnel experiments, it appears that two different mechanisms were involved. The first mechanism was the one responsible for the initial, relatively weak vibrations, and it involved resonance. As the wind moved over the bridge, it began acting like a kite or an airplane wing. As shown in the figure, it established swirling patterns of air flow around itself, of the kind that you can see in a moving cloud of smoke. As one of these swirls moved off of the bridge, there was an abrupt change in air pressure, which resulted in an up or down force on the bridge. We see something similar when a flag flaps in the wind, except that the flag’s surface is usually vertical. This back-and-forth sequence of forces is exactly the kind of periodic driving force that would excite a resonance. The faster the wind, the more quickly the swirls would get across the bridge, and the higher the frequency of the driving force would be. At just the right velocity, the frequency would be the right one to excite the resonance. The wind-tunnel models, however, show that the pattern of vibration of the bridge excited by this mechanism would have been a different one than the one that finally destroyed the bridge. The bridge was probably destroyed by a different mechanism, in which its vibrations at its own natural frequency of 0.2 Hz set up an alternating pattern of wind gusts in the air immediately around it, which then increased the amplitude of the bridge’s vibrations. This vicious cycle fed upon itself, increasing the amplitude of the vibrations until the bridge finally collapsed. As long as we’re on the subject of collapsing bridges, it is worth bringing up the reports of bridges falling down when soldiers marching over them happened to step in rhythm with the bridge’s natural frequency of oscillation. This is supposed to have happened in 1831 in Manchester, England, and again in 1849 in Anjou, France. Many modern engineers and scientists, however, are suspicious of the analysis of these reports. It is possible that the collapses had more to do with poor construction and overloading than with resonance. The Nimitz Freeway and Tacoma Narrows Bridge are far better documented, and occurred in an era when engineers’ abilities to analyze the vibrations of a complex structure were much more advanced. Emission and absorption of light waves by atoms example 8 In a very thin gas, the atoms are sufficiently far apart that they can act as individual vibrating systems. Although the vibrations are of a very strange and abstract type described by the theory of quantum mechanics, they nevertheless obey the same basic rules as ordinary mechanical vibrations. When a thin gas made of a cer-
458
Chapter 18
Resonance
tain element is heated, it emits light waves with certain specific frequencies, which are like a fingerprint of that element. As with all other vibrations, these atomic vibrations respond most strongly to a driving force that matches their own natural frequency. Thus if we have a relatively cold gas with light waves of various frequencies passing through it, the gas will absorb light at precisely those frequencies at which it would emit light if heated. (3) When a system is driven at resonance, the steady-state vibrations have an amplitude that is proportional to Q.
This is fairly intuitive. The steady-state behavior is an equilibrium between energy input from the driving force and energy loss due to damping. A low-Q oscillator, i.e., one with strong damping, dumps its energy faster, resulting in lower-amplitude steady-state motion. self-check C If an opera singer is shopping for a wine glass that she can impress her friends by breaking, what should she look for? Answer, p. 544
Piano strings ringing in sympathy with a sung note example 9 A sufficiently loud musical note sung near a piano with the lid raised can cause the corresponding strings in the piano to vibrate. (A piano has a set of three strings for each note, all struck by the same hammer.) Why would this trick be unlikely to work with a violin? If you have heard the sound of a violin being plucked (the pizzicato effect), you know that the note dies away very quickly. In other words, a violin’s Q is much lower than a piano’s. This means that its resonances are much weaker in amplitude. Our fourth and final fact about resonance is perhaps the most surprising. It gives us a way to determine numerically how wide a range of driving frequencies will produce a strong response. As shown in the graph, resonances do not suddenly fall off to zero outside a certain frequency range. It is usual to describe the width of a resonance by its full width at half-maximum (FWHM) as illustrated in figure g.
g / The definition of the width at half maximum.
full
(4) The FWHM of a resonance is related to its Q and its resonant frequency fres by the equation FWHM =
fres Q
.
(This equation is only a good approximation when Q is large.)
Section 18.3
Putting energy into vibrations
459
Why? It is not immediately obvious that there should be any logical relationship between Q and the FWHM. Here’s the idea. As we have seen already, the reason why the response of an oscillator is smaller away from resonance is that much of the driving force is being used to make the system act as if it had a different k. Roughly speaking, the half-maximum points on the graph correspond to the places where the amount of the driving force being wasted in this way is the same as the amount of driving force being used productively to replace the energy being dumped out by the damping force. If the damping force is strong, then a large amount of force is needed to counteract it, and we can waste quite a bit of driving force on changing k before it becomes comparable to the damping force. If, on the other hand, the damping force is weak, then even a small amount of force being wasted on changing k will become significant in proportion, and we cannot get very far from the resonant frequency before the two are comparable. Changing the pitch of a wind instrument example 10 A saxophone player normally selects which note to play by choosing a certain fingering, which gives the saxophone a certain resonant frequency. The musician can also, however, change the pitch significantly by altering the tightness of her lips. This corresponds to driving the horn slightly off of resonance. If the pitch can be altered by about 5% up or down (about one musical half-step) without too much effort, roughly what is the Q of a saxophone? Five percent is the width on one side of the resonance, so the full width is about 10%, FWHM / fr es = 0.1. This implies a Q of about 10, i.e., once the musician stops blowing, the horn will continue sounding for about 10 cycles before its energy falls off by a factor of 535. (Blues and jazz saxophone players will typically choose a mouthpiece that has a low Q, so that they can produce the bluesy pitch-slides typical of their style. “Legit,” i.e., classically oriented players, use a higher-Q setup because their style only calls for enough pitch variation to produce a vibrato.) Decay of a saxophone tone example 11 If a typical saxophone setup has a Q of about 10, how long will it take for a 100-Hz tone played on a baritone saxophone to die down by a factor of 535 in energy, after the player suddenly stops blowing? A Q of 10 means that it takes 10 cycles for the vibrations to die down in energy by a factor of 535. Ten cycles at a frequency of 100 Hz would correspond to a time of 0.1 seconds, which is not very long. This is why a saxophone note doesn’t “ring” like a note played on a piano or an electric guitar. Q of a radio receiver example 12 A radio receiver used in the FM band needs to be tuned in to
460
Chapter 18
Resonance
within about 0.1 MHz for signals at about 100 MHz. What is its Q? Q = fr es /FWHM = 1000. This is an extremely high Q compared to most mechanical systems. Q of a stereo speaker example 13 We have already given one reason why a stereo speaker should have a low Q: otherwise it would continue ringing after the end of the musical note on the recording. The second reason is that we want it to be able to respond to a large range of frequencies. Nuclear magnetic resonance example 14 If you have ever played with a magnetic compass, you have undoubtedly noticed that if you shake it, it takes some time to settle down, h/1. As it settles down, it acts like a damped oscillator of the type we have been discussing. The compass needle is simply a small magnet, and the planet earth is a big magnet. The magnetic forces between them tend to bring the needle to an equilibrium position in which it lines up with the planet-earth-magnet.
h / Example 14. 1. A compass needle vibrates about the equilibrium position under the influence of the earth’s magnetic forces. 2. The orientation of a proton’s spin vibrates around its equilibrium direction under the influence of the magnetic forces coming from the surrounding electrons and nuclei.
Essentially the same physics lies behind the technique called Nuclear Magnetic Resonance (NMR). NMR is a technique used to deduce the molecular structure of unknown chemical substances, and it is also used for making medical images of the inside of people’s bodies. If you ever have an NMR scan, they will actually tell you you are undergoing “magnetic resonance imaging” or “MRI,” because people are scared of the word “nuclear.” In fact, the nuclei being referred to are simply the non-radioactive nuclei of atoms found naturally in your body. Here’s how NMR works. Your body contains large numbers of hydrogen atoms, each consisting of a small, lightweight electron orbiting around a large, heavy proton. That is, the nucleus of a hydrogen atom is just one proton. A proton is always spinning on its own axis, and the combination of its spin and its electrical charge cause it to behave like a tiny magnet. The principle identical to that of an electromagnet, which consists of a coil of wire through which electrical charges pass; the circling motion of the charges in the coil of wire makes it magnetic, and in the same way, the circling motion of the proton’s charge makes it magnetic. Now a proton in one of your body’s hydrogen atoms finds itself surrounded by many other whirling, electrically charged particles: its own electron, plus the electrons and nuclei of the other nearby atoms. These neighbors act like magnets, and exert magnetic forces on the proton, h/2. The k of the vibrating proton is simply a measure of the total strength of these magnetic forces. Depending on the structure of the molecule in which the hydrogen atom finds itself, there will be a particular set of magnetic forces acting on the proton and a particular value of k . The NMR apparatus
Section 18.3
i / A member of the author’s family, who turned out to be healthy.
j / A three-dimensional computer reconstruction of the shape of a human brain, based on magnetic resonance data.
Putting energy into vibrations
461
bombards the sample with radio waves, and if the frequency of the radio waves matches the resonant frequency of the proton, the proton will absorb radio-wave energy strongly and oscillate wildly. Its vibrations are damped not by friction, because there is no friction inside an atom, but by the reemission of radio waves. By working backward through this chain of reasoning, one can determine the geometric arrangement of the hydrogen atom’s neighboring atoms. It is also possible to locate atoms in space, allowing medical images to be made. Finally, it should be noted that the behavior of the proton cannot be described entirely correctly by Newtonian physics. Its vibrations are of the strange and spooky kind described by the laws of quantum mechanics. It is impressive, however, that the few simple ideas we have learned about resonance can still be applied successfully to describe many aspects of this exotic system. Discussion question
k / Driving at a frequency above resonance.
A Nikola Tesla, one of the inventors of radio and an archetypical mad scientist, told a credulous reporter in 1912 the following story about an application of resonance. He built an electric vibrator that fit in his pocket, and attached it to one of the steel beams of a building that was under construction in New York. Although the article in which he was quoted didn’t say so, he presumably claimed to have tuned it to the resonant frequency of the building. “In a few minutes, I could feel the beam trembling. Gradually the trembling increased in intensity and extended throughout the whole great mass of steel. Finally, the structure began to creak and weave, and the steelworkers came to the ground panic-stricken, believing that there had been an earthquake. ... [If] I had kept on ten minutes more, I could have laid that building flat in the street.” Is this physically plausible?
18.4 Proofs l / Driving at resonance.
m / Driving at a below resonance.
462
frequency
Chapter 18
Our first goal is to predict the amplitude of the steady-state vibrations as a function of the frequency of the driving force and the amplitude of the driving force. With that equation in hand, we will then prove statements 2, 3, and 4 from section 18.3. We assume without proof statement 1, that the steady-state motion occurs at the same frequency as the driving force. As with the proof in chapter 17, we make use of the fact that a sinusoidal vibration is the same as the projection of circular motion onto a line. We visualize the system shown in figures k-m, in which the mass swings in a circle on the end of a spring. The spring does not actually change its length at all, but it appears to from the flattened perspective of a person viewing the system edgeon. The radius of the circle is the amplitude, A, of the vibrations as seen edge-on. The damping force can be imagined as a backward drag force supplied by some fluid through which the mass is
Resonance
moving. As usual, we assume that the damping is proportional to velocity, and we use the symbol b for the proportionality constant, |Fd | = bv. The driving force, represented by a hand towing the mass with a string, has a tangential component |Ft | which counteracts the damping force, |Ft | = |Fd |, and a radial component Fr which works either with or against the spring’s force, depending on whether we are driving the system above or below its resonant frequency. The speed of the rotating mass is the circumference of the circle divided by the period, v = 2πA/T , its acceleration (which is directly inward) is a = v 2 /r, and Newton’s second law gives a = F/m = 1 (kA + Fr )/m. We write fo for 2π k/m. Straightforward algebra yields Fr 2πm 2 [1] = . f − fo2 Ft bf This is the ratio of the wasted force to the useful force, and we see that it becomes zero when the system is driven at resonance. The amplitude of the vibrations can be found by attacking the equation |Ft | = bv = 2πbAf , which gives [2]
A=
|Ft | 2πbf
.(2)
However, we wish to know the amplitude in terms of |F|, not |Ft |. From now on, let’s drop the cumbersome magnitude symbols. With the Pythagorean theorem, it is easily proved that F
2 1 + FFrt
Ft =
[3]
, (3)
and equations 1-3 can then be combined to give the final result [4]
A=
F
2π 4π 2 m2 (f 2 − fo2 )2 + b2 f 2
.
Statement 2: maximum amplitude at resonance Equation [4] makes it plausible that the amplitude is maximized when the system is driven at close to its resonant frequency. At f = fo , the first term inside the square root vanishes, and this makes the denominator as small as possible, causing the amplitude to be as big as possible. (Actually this is only approximately true, because it is possible to make A a little bigger by decreasing f a little below fo , which makes the second term smaller. This technical issue is addressed in homework problem 3 on page 467.) Statement 3: amplitude at resonance proportional to Q Equation [4] shows that the amplitude at resonance is proportional to 1/b, and the Q of the system is inversely proportional to b, so the amplitude at resonance is proportional to Q.
Section 18.4
Proofs
463
Statement 4: FWHM related to Q We will satisfy ourselves by proving only the proportionality F W HM ∝ fo /Q, not the actual equation F W HM = fo /Q. The energy is proportional to A2 , i.e., to the inverse of the quantity inside the square root in equation [4]. At resonance, the first term inside the square root vanishes, and the half-maximum points occur at frequencies for which the whole quantity inside the square root is double its value at resonance, i.e., when the two terms are equal. At the half-maximum points, we have 2
f −
fo2
=
FWHM fo ± 2
2
− fo2
1 = ±fo · FWHM + FWHM2 4 If we assume that the width of the resonance is small compared to the resonant frequency, then the FWHM2 term is negligible compared to the fo · FWHM term, and setting the terms in equation 4 equal to each other gives 4π 2 m2 (fo FWHM)2 = b2 f 2
.
We are assuming that the width of the resonance is small compared to the resonant frequency, so f and fo can be taken as synonyms. Thus, b FWHM = . 2πm We wish to connect this to Q, which can be interpreted as the energy of the free (undriven) vibrations divided by the work done by damping in one cycle. The former equals kA2 /2, and the latter is proportional to the force, bv ∝ bAfo , multiplied by the distance traveled, A. (This is only a proportionality, not an equation, since the force is not constant.) We therefore find that Q is proportional to k/bfo . The equation for the FWHM can then be restated as a proportionality FWHM ∝ k/Qfo m ∝ fo /Q.
464
Chapter 18
Resonance
Summary Selected vocabulary damping . . . . . the dissipation of a vibration’s energy into heat energy, or the frictional force that causes the loss of energy quality factor . . the number of oscillations required for a system’s energy to fall off by a factor of 535 due to damping driving force . . . an external force that pumps energy into a vibrating system resonance . . . . the tendency of a vibrating system to respond most strongly to a driving force whose frequency is close to its own natural frequency of vibration steady state . . . the behavior of a vibrating system after it has had plenty of time to settle into a steady response to a driving force Notation Q . . . . . . . . . fo . . . . . . . . .
f . . . . . . . . . .
the quality factor the natural (resonant) frequency of a vibrating system, i.e., the frequency at which it would vibrate if it was simply kicked and left alone the frequency at which the system actually vibrates, which in the case of a driven system is equal to the frequency of the driving force, not the natural frequency
Summary The energy of a vibration is always proportional to the square of the amplitude, assuming the amplitude is small. Energy is lost from a vibrating system for various reasons such as the conversion to heat via friction or the emission of sound. This effect, called damping, will cause the vibrations to decay exponentially unless energy is pumped into the system to replace the loss. A driving force that pumps energy into the system may drive the system at its own natural frequency or at some other frequency. When a vibrating system is driven by an external force, we are usually interested in its steady-state behavior, i.e., its behavior after it has had time to settle into a steady response to a driving force. In the steady state, the same amount of energy is pumped into the system during each cycle as is lost to damping during the same period. The following are four important facts about a vibrating system being driven by an external force: (1) The steady-state response to a sinusoidal driving force occurs at the frequency of the force, not at the system’s own natural frequency of vibration.
Summary
465
(2) A vibrating system resonates at its own natural frequency. That is, the amplitude of the steady-state response is greatest in proportion to the amount of driving force when the driving force matches the natural frequency of vibration. (3) When a system is driven at resonance, the steady-state vibrations have an amplitude that is proportional to Q. (4) The FWHM of a resonance is related to its Q and its resonant frequency fo by the equation FWHM =
fo . Q
(This equation is only a good approximation when Q is large.)
466
Chapter 18
Resonance
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 If one stereo system is capable of producing 20 watts of sound power and another can put out 50 watts, how many times greater is the amplitude of the sound wave that can be created by the more powerful system? (Assume they are playing the same music.)
2 Many fish have an organ known as a swim bladder, an air-filled cavity whose main purpose is to control the fish’s buoyancy an allow it to keep from rising or sinking without having to use its muscles. In some fish, however, the swim bladder (or a small extension of it) is linked to the ear and serves the additional purpose of amplifying sound waves. For a typical fish having such an anatomy, the bladder has a resonant frequency of 300 Hz, the bladder’s Q is 3, and the maximum amplification is about a factor of 100 in energy. Over what range of frequencies would the amplification be at least a factor of 50?
3 As noted in section 18.4, it is only approximately true that the careamplitude has its maximum at f = (1/2π) k/m. Being more ful, we should actually define two different symbols, f0 = (1/2π) k/m and fo for the slightly different frequency at which the amplitude is a maximum, i.e., the actual resonant frequency. In this notation, the amplitude as a function of frequency is
A= 2π
F
4π 2 m2 f 2 − f02
2
. + b2 f 2
Show that the maximum occurs not at fo but rather at the frequency b2 1 2 f = f0 − 2 2 = f02 − FWHM2 8π m 2 Hint: Finding the frequency that minimizes the quantity inside the square root is equivalent to, but much easier than, finding the frequency that maximizes the amplitude.
Problems
467
4 (a) Let W be the amount of work done by friction in the first cycle of oscillation, i.e., the amount of energy lost to heat. Find the fraction of the original energy E that remains in the oscillations after n cycles of motion. (b) From this, prove the equation
W 1− E
Q
= e−2π
(recalling that the number 535 in the definition of Q is e2π ). (c) Use this to prove the approximation 1/Q ≈ (1/2π)W/E. (Hint: Use the approximation ln(1 + x) ≈ x, which is valid for small values of x.) 5 The goal of this problem is to refine the proportionality FWHM ∝ fres /Q into the equation FWHM = fres /Q, i.e., to prove that the constant of proportionality equals 1. (a) Show that the work done by a damping force F = −bv over one cycle of steady-state motion equals Wdamp = −2π 2 bf A2 . Hint: It is less confusing to calculate the work done over half a cycle, from x = −A to x = +A, and then double it. (b) Show that the fraction of the undriven oscillator’s energy lost to damping over one cycle is |Wdamp |/E = 4π 2 bf /k. (c) Use the previous result, combined with the result of problem 4, to prove that Q equals k/2πbf . (d) Combine the preceding result for Q with the equation FWHM = b/2πm from section 18.4 to prove the equation FWHM = fres /Q. 6 (a) We observe that the amplitude of a certain free oscillation √ decreases from Ao to Ao /Z after n oscillations. Find its Q. (b) The figure is from Shape memory in Spider draglines, Emile, Le Floch, and Vollrath, Nature 440:621 (2006). Panel 1 shows an electron microscope’s image of a thread of spider silk. In 2, a spider is hanging from such a thread. From an evolutionary point of view, it’s probably a bad thing for the spider if it twists back and forth while hanging like this. (We’re referring to a back-and-forth rotation about the axis of the thread, not a swinging motion like a pendulum.) The authors speculate that such a vibration could make the spider easier for predators to see, and it also seems to me that it would be a bad thing just because the spider wouldn’t be able to control its orientation and do what it was trying to do. Panel 3 shows a graph of such an oscillation, which the authors measured using a video camera and a computer, with a 0.1 g mass hung from it in place of a spider. Compared to human-made fibers such as kevlar or copper wire, the spider thread has an unusual set of properties:
468
Chapter 18
Resonance
1. It has a low Q, so the vibrations damp out quickly. 2. It doesn’t become brittle with repeated twisting as a copper wire would. 3. When twisted, it tends to settle in to a new equilibrium angle, rather than insisting on returning to its original angle. You can see this in panel 2, because although the experimenters initially twisted the wire by 35 degrees, the thread only performed oscillations with an amplitude much smaller than ±35 degrees, settling down to a new equilibrium at 27 degrees. 4. Over much longer time scales (hours), the thread eventually resets itself to its original equilbrium angle (shown as zero degrees on the graph). (The graph reproduced here only shows the motion over a much shorter time scale.) Some humanmade materials have this “memory” property as well, but they typically need to be heated in order to make them go back to their original shapes. Focusing on property number 1, estimate the Q of spider silk from √ the graph.
Problems
469
Problem 6.
470
Chapter 18
Resonance
Exercise 18: Resonance 1. Compare the oscillator’s energies at A, B, C, and D.
2. Compare the Q values of the two oscillators.
3. Match the x-t graphs in #2 with the amplitude-frequency graphs below.
Exercise 18: Resonance
471
472
Chapter 18
Resonance
“The Great Wave Off Kanagawa,” by Katsushika Hokusai (1760-1849).
Chapter 19
Free waves Your vocal cords or a saxophone reed can vibrate, but being able to vibrate wouldn’t be of much use unless the vibrations could be transmitted to the listener’s ear by sound waves. What are waves and why do they exist? Put your fingertip in the middle of a cup of water and then remove it suddenly. You will have noticed two results that are surprising to most people. First, the flat surface of the water does not simply sink uniformly to fill in the volume vacated by your finger. Instead, ripples spread out, and the process of flattening out occurs over a long period of time, during which the water at the center vibrates above and below the normal water level. This type of wave motion is the topic of the present chapter. Second, you have found that the ripples bounce off of the walls of the cup, in much the same way that a ball would bounce off of a wall. In the next chapter we discuss what happens to waves that have a boundary around them. Until then, we confine ourselves to wave phenomena that can be analyzed as if the medium (e.g., the water) was infinite and the same everywhere. It isn’t hard to understand why removing your fingertip creates ripples rather than simply allowing the water to sink back down uniformly. The initial crater, (a), left behind by your finger has sloping sides, and the water next to the crater flows downhill to fill in the hole. The water far away, on the other hand, initially has
a / Dipping a finger in some water, 1, causes a disturbance that spreads outward, 2.
473
no way of knowing what has happened, because there is no slope for it to flow down. As the hole fills up, the rising water at the center gains upward momentum, and overshoots, creating a little hill where there had been a hole originally. The area just outside of this region has been robbed of some of its water in order to build the hill, so a depressed “moat” is formed, (b). This effect cascades outward, producing ripples.
474
Chapter 19
Free waves
b / The two circular patterns of ripples pass through each other. Unlike material objects, wave patterns can overlap in space, and when this happens they combine by addition.
19.1 Wave motion There are three main ways in which wave motion differs from the motion of objects made of matter. 1. Superposition The most profound difference is that waves do not display have anything analogous to the normal forces between objects that come in contact. Two wave patterns can therefore overlap in the same region of space, as shown in figure b. Where the two waves coincide, they add together. For instance, suppose that at a certain location in at a certain moment in time, each wave would have had a crest 3 cm above the normal water level. The waves combine at this point to make a 6-cm crest. We use negative numbers to represent depressions in the water. If both waves would have had a troughs measuring -3 cm, then they combine to make an extra-deep -6 cm trough. A +3 cm crest and a -3 cm trough result in a height of zero, i.e., the waves momentarily cancel each other out at that point. This additive rule is referred to as the principle of superposition, “superposition” being merely a fancy word for “adding.” Superposition can occur not just with sinusoidal waves like the ones in the figure above but with waves of any shape. The figures on the following page show superposition of wave pulses. A pulse is simply a wave of very short duration. These pulses consist only of a single hump or trough. If you hit a clothesline sharply, you will observe pulses heading off in both directions. This is analogous to
Section 19.1
Wave motion
475
the way ripples spread out in all directions when you make a disturbance at one point on water. The same occurs when the hammer on a piano comes up and hits a string. Experiments to date have not shown any deviation from the principle of superposition in the case of light waves. For other types of waves, it is typically a very good approximation for low-energy waves. Discussion question A In figure c/3, the fifth frame shows the spring just about perfectly flat. If the two pulses have essentially canceled each other out perfectly, then why does the motion pick up again? Why doesn’t the spring just stay flat?
c / These pictures show the motion of wave pulses along a spring. To make a pulse, one end of the spring was shaken by hand. Movies were filmed, and a series of frame chosen to show the motion. 1. A pulse travels to the left. 2. Superposition of two colliding positive pulses. 3. Superposition of two colliding pulses, one positive and one negative.
476
Chapter 19
Free waves
d / As the wave pattern passes the rubber duck, the duck stays put. The water isn’t moving forward with the wave.
2. The medium is not transported with the wave. Figure d shows a series of water waves before it has reached a rubber duck (left), having just passed the duck (middle) and having progressed about a meter beyond the duck (right). The duck bobs around its initial position, but is not carried along with the wave. This shows that the water itself does not flow outward with the wave. If it did, we could empty one end of a swimming pool simply by kicking up waves! We must distinguish between the motion of the medium (water in this case) and the motion of the wave pattern through the medium. The medium vibrates; the wave progresses through space. self-check A In figure e, you can detect the side-to-side motion of the spring because the spring appears blurry. At a certain instant, represented by a single photo, how would you describe the motion of the different parts of the spring? Other than the flat parts, do any parts of the spring have zero velocity? Answer, p. 544
A worm example 1 The worm in the figure is moving to the right. The wave pattern, a pulse consisting of a compressed area of its body, moves to the left. In other words, the motion of the wave pattern is in the opposite direction compared to the motion of the medium. e / As the wave pulse goes by, the ribbon tied to the spring is not carried along. The motion of the wave pattern is to the right, but the medium (spring) is moving up and down, not to the right.
Section 19.1
Wave motion
477
f / Example 2. The surfer is dragging his hand in the water.
Surfing example 2 The incorrect belief that the medium moves with the wave is often reinforced by garbled secondhand knowledge of surfing. Anyone who has actually surfed knows that the front of the board pushes the water to the sides, creating a wake — the surfer can even drag his hand through the water, as in in figure f. If the water was moving along with the wave and the surfer, this wouldn’t happen. The surfer is carried forward because forward is downhill, not because of any forward flow of the water. If the water was flowing forward, then a person floating in the water up to her neck would be carried along just as quickly as someone on a surfboard. In fact, it is even possible to surf down the back side of a wave, although the ride wouldn’t last very long because the surfer and the wave would quickly part company. 3. A wave’s velocity depends on the medium.
g / Example wave.
3:
a
breaking
A material object can move with any velocity, and can be sped up or slowed down by a force that increases or decreases its kinetic energy. Not so with waves. The magnitude of a wave’s velocity depends on the properties of the medium (and perhaps also on the shape of the wave, for certain types of waves). Sound waves travel at about 340 m/s in air, 1000 m/s in helium. If you kick up water waves in a pool, you will find that kicking harder makes waves that are taller (and therefore carry more energy), not faster. The sound waves from an exploding stick of dynamite carry a lot of energy, but are no faster than any other waves. Thus although both waves and physical objects carry energy as they move through space, the energy of the wave relates to its amplitude, not to its speed. In the following section we will give an example of the physical relationship between the wave speed and the properties of the medium. Breaking waves example 3 The velocity of water waves increases with depth. The crest of a wave travels faster than the trough, and this can cause the wave to break.
h / Example 4. The boat has run up against a limit on its speed because it can’t climb over its own wave. Dolphins get around the problem by leaping out of the water.
478
Chapter 19
Once a wave is created, the only reason its speed will change is if it enters a different medium or if the properties of the medium change. It is not so surprising that a change in medium can slow down a wave, but the reverse can also happen. A sound wave traveling through a helium balloon will slow down when it emerges into the air, but if it enters another balloon it will speed back up again! Similarly, water waves travel more quickly over deeper water, so a wave will slow down as it passes over an underwater ridge, but speed
Free waves
up again as it emerges into deeper water. Hull speed example 4 The speeds of most boats, and of some surface-swimming animals, are limited by the fact that they make a wave due to their motion through the water. The boat in figure h is going at the same speed as its own waves, and can’t go any faster. No matter how hard the boat pushes against the water, it can’t make the wave move ahead faster and get out of the way. The wave’s speed depends only on the medium. Adding energy to the wave doesn’t speed it up, it just increases its amplitude. A water wave, unlike many other types of wave, has a speed that depends on its shape: a broader wave moves faster. The shape of the wave made by a boat tends to mold itself to the shape of the boat’s hull, so a boat with a longer hull makes a broader wave that moves faster. The maximum speed of a boat whose speed is limited by this effect is therefore closely related to the length of its hull, and the maximum speed is called the hull speed. Sailboats designed for racing are not just long and skinny to make them more streamlined — they are also long so that their hull speeds will be high. Wave patterns
i / Circular patterns.
and
linear
wave
spherical
wave
If the magnitude of a wave’s velocity vector is preordained, what about its direction? Waves spread out in all directions from every point on the disturbance that created them. If the disturbance is small, we may consider it as a single point, and in the case of water waves the resulting wave pattern is the familiar circular ripple, i/1. If, on the other hand, we lay a pole on the surface of the water and wiggle it up and down, we create a linear wave pattern, i/2. For a three-dimensional wave such as a sound wave, the analogous patterns would be spherical waves and plane waves, j. Infinitely many patterns are possible, but linear or plane waves are often the simplest to analyze, because the velocity vector is in the same direction no matter what part of the wave we look at. Since all the velocity vectors are parallel to one another, the problem is effectively one-dimensional. Throughout this chapter and the next, we will restrict ourselves mainly to wave motion in one dimension, while not hesitating to broaden our horizons when it can be done without too much complication.
j / Plane and patterns.
Section 19.1
Wave motion
479
Discussion questions A
[see above]
B Sketch two positive wave pulses on a string that are overlapping but not right on top of each other, and draw their superposition. Do the same for a positive pulse running into a negative pulse. C A traveling wave pulse is moving to the right on a string. Sketch the velocity vectors of the various parts of the string. Now do the same for a pulse moving to the left. D In a spherical sound wave spreading out from a point, how would the energy of the wave fall off with distance? k / Hitting a key on a piano causes a hammer to come up from underneath and hit a string (actually a set of three strings). The result is a pair of pulses moving away from the point of impact.
19.2 Waves on a string
l / A string is struck with a hammer, 1, and two pulses fly off, 2.
After the qualitative discussion, we will use simple approximations to investigate the speed of a wave pulse on a string. This quick and dirty treatment is then followed by a rigorous attack using the methods of calculus, which may be skipped by the student who has not studied calculus. How far you penetrate in this section is up to you, and depends on your mathematical self-confidence. If you skip some of the math, you should nevertheless absorb the significance of the result, discussed on p. 484.
So far you have learned some counterintuitive things about the behavior of waves, but intuition can be trained. The first half of this section aims to build your intuition by investigating a simple, onedimensional type of wave: a wave on a string. If you have ever stretched a string between the bottoms of two open-mouthed cans to talk to a friend, you were putting this type of wave to work. Stringed instruments are another good example. Although we usually think of a piano wire simply as vibrating, the hammer actually strikes it quickly and makes a dent in it, which then ripples out in both directions. Since this chapter is about free waves, not bounded ones, we pretend that our string is infinitely long.
Intuitive ideas Consider a string that has been struck, l/1, resulting in the creation of two wave pulses, 2, one traveling to the left and one to the right. This is analogous to the way ripples spread out in all directions from a splash in water, but on a one-dimensional string, “all directions” becomes “both directions.”
m / A continuous string can be modeled as a series of discrete masses connected by springs.
480
Chapter 19
We can gain insight by modeling the string as a series of masses connected by springs. (In the actual string the mass and the springiness are both contributed by the molecules themselves.) If we look at various microscopic portions of the string, there will be some areas that are flat, m/1, some that are sloping but not curved, 2, and some that are curved, 3 and 4. In example 1 it is clear that both the forces on the central mass cancel out, so it will not accelerate. The
Free waves
same is true of 2, however. Only in curved regions such as 3 and 4 is an acceleration produced. In these examples, the vector sum of the two forces acting on the central mass is not zero. The important concept is that curvature makes force: the curved areas of a wave tend to experience forces resulting in an acceleration toward the mouth of the curve. Note, however, that an uncurved portion of the string need not remain motionless. It may move at constant velocity to either side. Approximate treatment We now carry out an approximate treatment of the speed at which two pulses will spread out from an initial indentation on a string. For simplicity, we imagine a hammer blow that creates a triangular dent, n/1. We will estimate the amount of time, t, required until each of the pulses has traveled a distance equal to the width of the pulse itself. The velocity of the pulses is then ±w/t.
n / A triangular pulse spreads out.
As always, the velocity of a wave depends on the properties of the medium, in this case the string. The properties of the string can be summarized by two variables: the tension, T , and the mass per unit length, μ (Greek letter mu). If we consider the part of the string encompassed by the initial dent as a single object, then this object has a mass of approximately μw (mass/length × length = mass). (Here, and throughout the derivation, we assume that h is much less than w, so that we can ignore the fact that this segment of the string has a length slightly greater than w.) Although the downward acceleration of this segment of the string will be neither constant over time nor uniform across the string, we will pretend that it is constant for the sake of our simple estimate. Roughly speaking, the time interval between n/1 and 2 is the amount of time required for the initial dent to accelerate from rest and reach its normal, flattened position. Of course the tip of the triangle has a longer distance to travel than the edges, but again we ignore the complications and simply assume that the segment as a whole must travel a distance h. Indeed, it might seem surprising that the triangle would so neatly spring back to a perfectly flat shape. It is an experimental fact that it does, but our analysis is too crude to address such details. The string is kinked, i.e., tightly curved, at the edges of the triangle, so it is here that there will be large forces that do not cancel out to zero. There are two forces acting on the triangular hump, one of magnitude T acting down and to the right, and one of the same magnitude acting down and to the left. If the angle of the sloping sides is θ, then the total force on the segment equals 2T sin θ. Dividing the triangle into two right triangles, we see that sin θ equals h divided by the length of one of the sloping sides. Since h is much less than w, the length of the sloping side is essentially
Section 19.2
Waves on a string
481
the same as w/2, so we have sin θ = 2h/w, and F = 4T h/w. The acceleration of the segment (actually the acceleration of its center of mass) is a = F/m = 4T h/μw2
.
The time required to move a distance h under constant acceleration a is found by solving h = 12 at2 to yield 2h t= a μ =w . 2T Our final result for the velocity of the pulses is |v| = =
w t 2T μ
.
The remarkable feature of this result is that the velocity of the pulses does not depend at all on w or h, i.e., any triangular pulse has the same speed. It is an experimental fact (and we will also prove rigorously in the following subsection) that any pulse of any kind, triangular or otherwise, travels along the string at the same speed. Of course, after so many approximations we cannot expect to have gotten all the numerical factors right. The correct result for the velocity of the pulses is T v= . μ The importance of the above derivation lies in the insight it brings —that all pulses move with the same speed — rather than in the details of the numerical result. The reason for our too-high value for the velocity is not hard to guess. It comes from the assumption that the acceleration was constant, when actually the total force on the segment would diminish as it flattened out. Rigorous derivation using calculus (optional) After expending considerable effort for an approximate solution, we now display the power of calculus with a rigorous and completely general treatment that is nevertheless much shorter and easier. Let the flat position of the string define the x axis, so that y measures how far a point on the string is from equilibrium. The motion of the string is characterized by y(x, t), a function of two variables. Knowing that the force on any small segment of string depends
482
Chapter 19
Free waves
on the curvature of the string in that area, and that the second derivative is a measure of curvature, it is not surprising to find that the infinitesimal force dF acting on an infinitesimal segment dx is given by
dF = T
d2 y dx dx2
.
(This can be proved by vector addition of the two infinitesimal forces acting on either side.) The acceleration is then a = dF/dm, or, substituting dm = μdx, d2 y T d2 y = dt2 μ dx2
.
The second derivative with respect to time is related to the second derivative with respect to position. This is no more than a fancy mathematical statement of the intuitive fact developed above, that the string accelerates so as to flatten out its curves. Before even bothering to look for solutions to this equation, we note that it already proves the principle of superposition, because the derivative of a sum is the sum of the derivatives. Therefore the sum of any two solutions will also be a solution. Based on experiment, we expect that this equation will be satisfied by any function y(x, t) that describes a pulse or wave pattern moving to the left or right at the correct speed v. In general, such a function will be of the form y = f (x − vt) or y = f (x + vt), where f is any function of one variable. Because of the chain rule, each derivative with respect to time brings out a factor of ±v. Evaluating the second derivatives on both sides of the equation gives
(±v)2 f =
T f μ
.
Squaring gets rid of the sign, and we find that we have a valid solution for any function f , provided that v is given by v=
T μ
.
Section 19.2
Waves on a string
483
Significance of the result
This specific result for the speed of waves on a string, v = T /μ, is utterly unimportant. Don’t memorize it. Don’t take notes on it. Try to erase it from your memory. What is important about this result is that it is an example of two things that are usually true, at least approximately, for mechanical waves in general: 1. The speed at which a wave moves does not depend on the size or shape of the wave. 2. The speed of a mechanical wave depends on a combination of two properties of the medium: some measure of its inertia and some measure of its tightness, i.e., the strength of the force trying to bring the medium back toward equilibrium. self-check B (a) What is it about the equation v = T /μ that relates to fact 1 above? (b) In the equation v = T /μ, which variable is a measure of inertia, and which is a measure of tightness? (c) Now suppose that we produce compressional wave pulses in a metal rod by tapping the end of the rod with a hammer. What physical properties of the rod would play the roles of inertia and tightness? How would you expect the speed of compressional waves in lead to compare with their speed in aluminum? Answer, p. 544
19.3 Sound and light waves Sound waves The phenomenon of sound is easily found to have all the characteristics we expect from a wave phenomenon: • Sound waves obey superposition. Sounds do not knock other sounds out of the way when they collide, and we can hear more than one sound at once if they both reach our ear simultaneously. • The medium does not move with the sound. Even standing in front of a titanic speaker playing earsplitting music, we do not feel the slightest breeze. • The velocity of sound depends on the medium. Sound travels faster in helium than in air, and faster in water than in helium. Putting more energy into the wave makes it more intense, not faster. For example, you can easily detect an echo when you clap your hands a short distance from a large, flat wall, and the delay of the echo is no shorter for a louder clap.
484
Chapter 19
Free waves
Although not all waves have a speed that is independent of the shape of the wave, and this property therefore is irrelevant to our collection of evidence that sound is a wave phenomenon, sound does nevertheless have this property. For instance, the music in a large concert hall or stadium may take on the order of a second to reach someone seated in the nosebleed section, but we do not notice or care, because the delay is the same for every sound. Bass, drums, and vocals all head outward from the stage at 340 m/s, regardless of their differing wave shapes. If sound has all the properties we expect from a wave, then what type of wave is it? It must be a vibration of a physical medium such as air, since the speed of sound is different in different media, such as helium or water. Further evidence is that we don’t receive sound signals that have come to our planet through outer space. The roars and whooshes of Hollywood’s space ships are fun, but scientifically wrong.1 We can also tell that sound waves consist of compressions and expansions, rather than sideways vibrations like the shimmying of a snake. Only compressional vibrations would be able to cause your eardrums to vibrate in and out. Even for a very loud sound, the compression is extremely weak; the increase or decrease compared to normal atmospheric pressure is no more than a part per million. Our ears are apparently very sensitive receivers! Unlike a wave on a string, which vibrates in the direction perpendicular to the direction in which the wave pattern moves, a sound wave is a longitudinal wave, i.e., one in which the vibration is forward and backward along the direction of motion. Light waves Entirely similar observations lead us to believe that light is a wave, although the concept of light as a wave had a long and tortuous history. It is interesting to note that Isaac Newton very influentially advocated a contrary idea about light. The belief that matter was made of atoms was stylish at the time among radical thinkers (although there was no experimental evidence for their existence), and it seemed logical to Newton that light as well should be made of tiny particles, which he called corpuscles (Latin for “small objects”). Newton’s triumphs in the science of mechanics, i.e., the study of matter, brought him such great prestige that nobody bothered to 1 Outer space is not a perfect vacuum, so it is possible for sounds waves to travel through it. However, if we want to create a sound wave, we typically do it by creating vibrations of a physical object, such as the sounding board of a guitar, the reed of a saxophone, or a speaker cone. The lower the density of the surrounding medium, the less efficiently the energy can be converted into sound and carried away. An isolated tuning fork, left to vibrate in interstellar space, would dissipate the energy of its vibration into internal heat at a rate many orders of magnitude greater than the rate of sound emission into the nearly perfect vacuum around it.
Section 19.3
Sound and light waves
485
question his incorrect theory of light for 150 years. One persuasive proof that light is a wave is that according to Newton’s theory, two intersecting beams of light should experience at least some disruption because of collisions between their corpuscles. Even if the corpuscles were extremely small, and collisions therefore very infrequent, at least some dimming should have been measurable. In fact, very delicate experiments have shown that there is no dimming. The wave theory of light was entirely successful up until the 20th century, when it was discovered that not all the phenomena of light could be explained with a pure wave theory. It is now believed that both light and matter are made out of tiny chunks which have both wave and particle properties. For now, we will content ourselves with the wave theory of light, which is capable of explaining a great many things, from cameras to rainbows. If light is a wave, what is waving? What is the medium that wiggles when a light wave goes by? It isn’t air. A vacuum is impenetrable to sound, but light from the stars travels happily through zillions of miles of empty space. Light bulbs have no air inside them, but that doesn’t prevent the light waves from leaving the filament. For a long time, physicists assumed that there must be a mysterious medium for light waves, and they called it the aether (not to be confused with the chemical). Supposedly the aether existed everywhere in space, and was immune to vacuum pumps. The details of the story are more fittingly reserved for later in this course, but the end result was that a long series of experiments failed to detect any evidence for the aether, and it is no longer believed to exist. Instead, light can be explained as a wave pattern made up of electrical and magnetic fields.
486
Chapter 19
Free waves
19.4 Periodic waves Period and frequency of a periodic wave You choose a radio station by selecting a certain frequency. We have already defined period and frequency for vibrations, but what do they signify in the case of a wave? We can recycle our previous definition simply by stating it in terms of the vibrations that the wave causes as it passes a receiving instrument at a certain point in space. For a sound wave, this receiver could be an eardrum or a microphone. If the vibrations of the eardrum repeat themselves over and over, i.e., are periodic, then we describe the sound wave that caused them as periodic. Likewise we can define the period and frequency of a wave in terms of the period and frequency of the vibrations it causes. As another example, a periodic water wave would be one that caused a rubber duck to bob in a periodic manner as they passed by it. The period of a sound wave correlates with our sensory impression of musical pitch. A high frequency (short period) is a high note. The sounds that really define the musical notes of a song are only the ones that are periodic. It is not possible to sing a non-periodic sound like “sh” with a definite pitch.
o / A graph of pressure versus time for a periodic sound wave, the vowel “ah.”
p / A similar graph for a nonperiodic wave, “sh.”
The frequency of a light wave corresponds to color. Violet is the high-frequency end of the rainbow, red the low-frequency end. A color like brown that does not occur in a rainbow is not a periodic light wave. Many phenomena that we do not normally think of as light are actually just forms of light that are invisible because they fall outside the range of frequencies our eyes can detect. Beyond the red end of the visible rainbow, there are infrared and radio waves. Past the violet end, we have ultraviolet, x-rays, and gamma rays. Graphs of waves as a function of position Some waves, light sound waves, are easy to study by placing a detector at a certain location in space and studying the motion as a function of time. The result is a graph whose horizontal axis is time. With a water wave, on the other hand, it is simpler just to look at the wave directly. This visual snapshot amounts to a graph of the height of the water wave as a function of position. Any wave can be represented in either way.
q / A strip chart recorder.
An easy way to visualize this is in terms of a strip chart recorder, an obsolescing device consisting of a pen that wiggles back and forth as a roll of paper is fed under it. It can be used to record a person’s electrocardiogram, or seismic waves too small to be felt as a noticeable earthquake but detectable by a seismometer. Taking the seismometer as an example, the chart is essentially a record of the ground’s wave motion as a function of time, but if the paper was set to feed at the same velocity as the motion of an earthquake wave, it
Section 19.4
Periodic waves
487
would also be a full-scale representation of the profile of the actual wave pattern itself. Assuming, as is usually the case, that the wave velocity is a constant number regardless of the wave’s shape, knowing the wave motion as a function of time is equivalent to knowing it as a function of position. Wavelength
r / A water wave profile created by a series of repeating pulses.
Any wave that is periodic will also display a repeating pattern when graphed as a function of position. The distance spanned by one repetition is referred to as one wavelength. The usual notation for wavelength is λ, the Greek letter lambda. Wavelength is to space as period is to time.
s / Wavelengths of linear and circular water waves.
Wave velocity related to frequency and wavelength Suppose that we create a repetitive disturbance by kicking the surface of a swimming pool. We are essentially making a series of wave pulses. The wavelength is simply the distance a pulse is able to travel before we make the next pulse. The distance between pulses is λ, and the time between pulses is the period, T , so the speed of the wave is the distance divided by the time, v = λ/T . This important and useful relationship is more commonly written in terms of the frequency, v = fλ
488
Chapter 19
Free waves
.
Wavelength of radio waves example 5 8 The speed of light is 3.0 × 10 m/s. What is the wavelength of the radio waves emitted by KMHD, a station whose frequency is 89.1 MHz? Solving for wavelength, we have λ = v /f = (3.0 × 108 m/s)/(89.1 × 106 s−1 ) = 3.4 m The size of a radio antenna is closely related to the wavelength of the waves it is intended to receive. The match need not be exact (since after all one antenna can receive more than one wavelength!), but the ordinary “whip” antenna such as a car’s is 1/4 of a wavelength. An antenna optimized to receive KMHD’s signal would have a length of 3.4 m/4 = 0.85 m.
t / Ultrasound, i.e., sound with frequencies higher than the range of human hearing, was used to make this image of a fetus. The resolution of the image is related to the wavelength, since details smaller than about one wavelength cannot be resolved. High resolution therefore requires a short wavelength, corresponding to a high frequency.
The equation v = f λ defines a fixed relationship between any two of the variables if the other is held fixed. The speed of radio waves in air is almost exactly the same for all wavelengths and frequencies (it is exactly the same if they are in a vacuum), so there is a fixed relationship between their frequency and wavelength. Thus we can say either “Are we on the same wavelength?” or “Are we on the same frequency?” A different example is the behavior of a wave that travels from a region where the medium has one set of properties to an area where the medium behaves differently. The frequency is now fixed,
Section 19.4
u / A water wave traveling into a region with a different depth changes its wavelength.
Periodic waves
489
because otherwise the two portions of the wave would otherwise get out of step, causing a kink or discontinuity at the boundary, which would be unphysical. (A more careful argument is that a kink or discontinuity would have infinite curvature, and waves tend to flatten out their curvature. An infinite curvature would flatten out infinitely fast, i.e., it could never occur in the first place.) Since the frequency must stay the same, any change in the velocity that results from the new medium must cause a change in wavelength. The velocity of water waves depends on the depth of the water, so based on λ = v/f , we see that water waves that move into a region of different depth must change their wavelength, as shown in the figure on the left. This effect can be observed when ocean waves come up to the shore. If the deceleration of the wave pattern is sudden enough, the tip of the wave can curl over, resulting in a breaking wave. A note on dispersive waves The discussion of wave velocity given here is actually an oversimplification for a wave whose velocity depends on its frequency and wavelength. Such a wave is called a dispersive wave. Nearly all the waves we deal with in this course are non-dispersive, but the issue becomes important in quantum physics, as discussed in more detail in optional section 35.2.
Sinusoidal waves Sinusoidal waves are the most important special case of periodic waves. In fact, many scientists and engineers would be uncomfortable with defining a waveform like the “ah” vowel sound as having a definite frequency and wavelength, because they consider only sine waves to be pure examples of a certain frequency and wavelengths. Their bias is not unreasonable, since the French mathematician Fourier showed that any periodic wave with frequency f can be constructed as a superposition of sine waves with frequencies f , 2f , 3f , ... In this sense, sine waves are the basic, pure building blocks of all waves. (Fourier’s result so surprised the mathematical community of France that he was ridiculed the first time he publicly presented his theorem.) However, what definition to use is a matter of utility. Our sense of hearing perceives any two sounds having the same period as possessing the same pitch, regardless of whether they are sine waves or not. This is undoubtedly because our ear-brain system evolved to be able to interpret human speech and animal noises, which are periodic but not sinusoidal. Our eyes, on the other hand, judge a color as pure (belonging to the rainbow set of colors) only if it is a sine wave.
490
Chapter 19
Free waves
Discussion question A Suppose we superimpose two sine waves with equal amplitudes but slightly different frequencies, as shown in the figure. What will the superposition look like? What would this sound like if they were sound waves?
19.5 The Doppler effect Figure v shows the wave pattern made by the tip of a vibrating rod which is moving across the water. If the rod had been vibrating in one place, we would have seen the familiar pattern of concentric circles, all centered on the same point. But since the source of the waves is moving, the wavelength is shortened on one side and lengthened on the other. This is known as the Doppler effect. Note that the velocity of the waves is a fixed property of the medium, so for example the forward-going waves do not get an extra boost in speed as would a material object like a bullet being shot forward from an airplane. We can also infer a change in frequency. Since the velocity is constant, the equation v = f λ tells us that the change in wavelength must be matched by an opposite change in frequency: higher frequency for the waves emitted forward, and lower for the ones emitted backward. The frequency Doppler effect is the reason for the familiar dropping-pitch sound of a race car going by. As the car approaches us, we hear a higher pitch, but after it passes us we hear a frequency that is lower than normal.
v / The pattern of waves made by a point source moving to the right across the water. Note the shorter wavelength of the forward-emitted waves and the longer wavelength of the backward-going ones.
The Doppler effect will also occur if the observer is moving but the source is stationary. For instance, an observer moving toward a stationary source will perceive one crest of the wave, and will then be surrounded by the next crest sooner than she otherwise would have, because she has moved toward it and hastened her encounter with it. Roughly speaking, the Doppler effect depends only the relative motion of the source and the observer, not on their absolute state of motion (which is not a well-defined notion in physics) or on their velocity relative to the medium. Restricting ourselves to the case of a moving source, and to waves emitted either directly along or directly against the direction of motion, we can easily calculate the wavelength, or equivalently the frequency, of the Doppler-shifted waves. Let v be the velocity of the waves, and vs the velocity of the source. The wavelength of the
Section 19.5
The Doppler effect
491
forward-emitted waves is shortened by an amount vs T equal to the distance traveled by the source over the course of one period. Using the definition f = 1/T and the equation v = f λ, we find for the wavelength of the Doppler-shifted wave the equation
vs λ . λ = 1 − v A similar equation can be used for the backward-emitted waves, but with a plus sign rather than a minus sign. Doppler-shifted sound from a race car example 6 If a race car moves at a velocity of 50 m/s, and the velocity of sound is 340 m/s, by what percentage are the wavelength and frequency of its sound waves shifted for an observer lying along its line of motion? For an observer whom the car is approaching, we find 1−
vs = 0.85 v
,
so the shift in wavelength is 15%. Since the frequency is inversely proportional to the wavelength for a fixed value of the speed of sound, the frequency is shifted upward by 1/0.85 = 1.18, i.e., a change of 18%. (For velocities that are small compared to the wave velocities, the Doppler shifts of the wavelength and frequency are about the same.) Doppler shift of the light emitted by a race car example 7 What is the percent shift in the wavelength of the light waves emitted by a race car’s headlights? Looking up the speed of light in the front of the book, v = 3.0 × 108 m/s, we find 1−
vs = 0.99999983 v
,
i.e., the percentage shift is only 0.000017%. The second example shows that under ordinary earthbound circumstances, Doppler shifts of light are negligible because ordinary things go so much slower than the speed of light. It’s a different story, however, when it comes to stars and galaxies, and this leads us to a story that has profound implications for our understanding of the origin of the universe. w / Example 8. A Doppler radar image of Hurricane Katrina, in 2005.
492
Chapter 19
Doppler radar example 8 The first use of radar was by Britain during World War II: antennas on the ground sent radio waves up into the sky, and detected the echoes when the waves were reflected from German planes.
Free waves
Later, air forces wanted to mount radar antennas on airplanes, but then there was a problem, because if an airplane wanted to detect another airplane at a lower altitude, it would have to aim its radio waves downward, and then it would get echoes from the ground. The solution was the invention of Doppler radar, in which echoes from the ground were differentiated from echoes from other aircraft according to their Doppler shifts. A similar technology is used by meteorologists to map out rainclouds without being swamped by reflections from the ground, trees, and buildings. Optional topic: Doppler shifts of light If Doppler shifts depend only on the relative motion of the source and receiver, then there is no way for a person moving with the source and another person moving with the receiver to determine who is moving and who isn’t. Either can blame the Doppler shift entirely on the other’s motion and claim to be at rest herself. This is entirely in agreement with the principle stated originally by Galileo that all motion is relative. On the other hand, a careful analysis of the Doppler shifts of water or sound waves shows that it is only approximately true, at low speeds, that the shifts just depend on the relative motion of the source and observer. For instance, it is possible for a jet plane to keep up with its own sound waves, so that the sound waves appear to stand still to the pilot of the plane. The pilot then knows she is moving at exactly the speed of sound. The reason this doesn’t disprove the relativity of motion is that the pilot is not really determining her absolute motion but rather her motion relative to the air, which is the medium of the sound waves. Einstein realized that this solved the problem for sound or water waves, but would not salvage the principle of relative motion in the case of light waves, since light is not a vibration of any physical medium such as water or air. Beginning by imagining what a beam of light would look like to a person riding a motorcycle alongside it, Einstein eventually came up with a radical new way of describing the universe, in which space and time are distorted as measured by observers in different states of motion. As a consequence of this theory of relativity, he showed that light waves would have Doppler shifts that would exactly, not just approximately, depend only on the relative motion of the source and receiver. The resolution of the motorcycle paradox is given in example refeg:einstein-motorcycle on p. 681, and a quantitative discussion of Doppler shifts of light is given on p. 684.
The Big Bang As soon as astronomers began looking at the sky through telescopes, they began noticing certain objects that looked like clouds in deep space. The fact that they looked the same night after night meant that they were beyond the earth’s atmosphere. Not knowing what they really were, but wanting to sound official, they called them “nebulae,” a Latin word meaning “clouds” but sounding more impressive. In the early 20th century, astronomers realized that although some really were clouds of gas (e.g., the middle “star” of Orion’s sword, which is visibly fuzzy even to the naked eye when
Section 19.5
x / The galaxy M51. Under high magnification, the milky clouds reveal themselves to be composed of trillions of stars.
The Doppler effect
493
y / How do astronomers know what mixture of wavelengths a star emitted originally, so that they can tell how much the Doppler shift was? This image (obtained by the author with equipment costing about $5, and no telescope) shows the mixture of colors emitted by the star Sirius. (If you have the book in black and white, blue is on the left and red on the right.) The star appears white or bluish-white to the eye, but any light looks white if it contains roughly an equal mixture of the rainbow colors, i.e., of all the pure sinusoidal waves with wavelengths lying in the visible range. Note the black “gap teeth.” These are the fingerprint of hydrogen in the outer atmosphere of Sirius. These wavelengths are selectively absorbed by hydrogen. Sirius is in our own galaxy, but similar stars in other galaxies would have the whole pattern shifted toward the red end, indicating they are moving away from us.
z / The telescope at Wilson used by Hubble.
494
Mount
Chapter 19
conditions are good), others were what we now call galaxies: virtual island universes consisting of trillions of stars (for example the Andromeda Galaxy, which is visible as a fuzzy patch through binoculars). Three hundred years after Galileo had resolved the Milky Way into individual stars through his telescope, astronomers realized that the universe is made of galaxies of stars, and the Milky Way is simply the visible part of the flat disk of our own galaxy, seen from inside. This opened up the scientific study of cosmology, the structure and history of the universe as a whole, a field that had not been seriously attacked since the days of Newton. Newton had realized that if gravity was always attractive, never repulsive, the universe would have a tendency to collapse. His solution to the problem was to posit a universe that was infinite and uniformly populated with matter, so that it would have no geometrical center. The gravitational forces in such a universe would always tend to cancel out by symmetry, so there would be no collapse. By the 20th century, the belief in an unchanging and infinite universe had become conventional wisdom in science, partly as a reaction against the time that had been wasted trying to find explanations of ancient geological phenomena based on catastrophes suggested by biblical events like Noah’s flood. In the 1920’s astronomer Edwin Hubble began studying the Doppler shifts of the light emitted by galaxies. A former college football player with a serious nicotine addiction, Hubble did not set out to change our image of the beginning of the universe. His autobiography seldom even mentions the cosmological discovery for which he is now remembered. When astronomers began to study the Doppler shifts of galaxies, they expected that each galaxy’s direction and velocity of motion would be essentially random. Some would be approaching us, and their light would therefore be Doppler-shifted to the blue end of the spectrum, while an equal number would be expected to have red shifts. What Hubble discovered instead was that except for a few very nearby ones, all the galaxies had red shifts, indicating that they were receding from us at a hefty fraction of the speed of light. Not only that, but the ones farther away were receding more quickly. The speeds were directly proportional to their distance from us. Did this mean that the earth (or at least our galaxy) was the center of the universe? No, because Doppler shifts of light only depend on the relative motion of the source and the observer. If we see a distant galaxy moving away from us at 10% of the speed of light, we can be assured that the astronomers who live in that galaxy will see ours receding from them at the same speed in the opposite direction. The whole universe can be envisioned as a rising loaf of raisin bread. As the bread expands, there is more and more space between the raisins. The farther apart two raisins are, the
Free waves
greater the speed with which they move apart. Extrapolating backward in time using the known laws of physics, the universe must have been denser and denser at earlier and earlier times. At some point, it must have been extremely dense and hot, and we can even detect the radiation from this early fireball, in the form of microwave radiation that permeates space. The phrase Big Bang was originally coined by the doubters of the theory to make it sound ridiculous, but it stuck, and today essentially all astronomers accept the Big Bang theory based on the very direct evidence of the red shifts and the cosmic microwave background radiation. What the Big Bang is not Finally it should be noted what the Big Bang theory is not. It is not an explanation of why the universe exists. Such questions belong to the realm of religion, not science. Science can find ever simpler and ever more fundamental explanations for a variety of phenomena, but ultimately science takes the universe as it is according to observations. Furthermore, there is an unfortunate tendency, even among many scientists, to speak of the Big Bang theory as a description of the very first event in the universe, which caused everything after it. Although it is true that time may have had a beginning (Einstein’s theory of general relativity admits such a possibility), the methods of science can only work within a certain range of conditions such as temperature and density. Beyond a temperature of about 109 degrees C, the random thermal motion of subatomic particles becomes so rapid that its velocity is comparable to the speed of light. Early enough in the history of the universe, when these temperatures existed, Newtonian physics becomes less accurate, and we must describe nature using the more general description given by Einstein’s theory of relativity, which encompasses Newtonian physics as a special case. At even higher temperatures, beyond about 1033 degrees, physicists know that Einstein’s theory as well begins to fall apart, but we don’t know how to construct the even more general theory of nature that would work at those temperatures. No matter how far physics progresses, we will never be able to describe nature at infinitely high temperatures, since there is a limit to the temperatures we can explore by experiment and observation in order to guide us to the right theory. We are confident that we understand the basic physics involved in the evolution of the universe starting a few minutes after the Big Bang, and we may be able to push back to milliseconds or microseconds after it, but we cannot use the methods of science to deal with the beginning of time itself.
Section 19.5
The Doppler effect
495
Discussion questions A If an airplane travels at exactly the speed of sound, what would be the wavelength of the forward-emitted part of the sound waves it emitted? How should this be interpreted, and what would actually happen? What happens if it’s going faster than the speed of sound? Can you use this to explain what you see in figure aa? B If bullets go slower than the speed of sound, why can a supersonic fighter plane catch up to its own sound, but not to its own bullets? aa / Shock waves from by the X-15 rocket plane, flying at 3.5 times the speed of sound.
C If someone inside a plane is talking to you, should their speech be Doppler shifted? D The plane in figure ab was photographed when it was traveling at a speed close to the speed of sound. Comparing figures aa and ab, how can we tell from the angles of the cones that the speed is much lower in figure ab?
ab / As in figure aa, this plane shows a shock wave. The sudden decompression of the air causes water droplets to condense, forming a cloud.
496
Chapter 19
Free waves
Summary Selected vocabulary superposition . . the adding together of waves that overlap with each other medium . . . . . a physical substance whose vibrations constitute a wave wavelength . . . . the distance in space between repetitions of a periodic wave Doppler effect . . the change in a wave’s frequency and wavelength due to the motion of the source or the observer or both Notation λ. . . . . . . . . .
wavelength (Greek letter lambda)
Summary Wave motion differs in three important ways from the motion of material objects: (1) Waves obey the principle of superposition. When two waves collide, they simply add together. (2) The medium is not transported along with the wave. The motion of any given point in the medium is a vibration about its equilibrium location, not a steady forward motion. (3) The velocity of a wave depends on the medium, not on the amount of energy in the wave. (For some types of waves, notably water waves, the velocity may also depend on the shape of the wave.) Sound waves consist of increases and decreases (typically very small ones) in the density of the air. Light is a wave, but it is a vibration of electric and magnetic fields, not of any physical medium. Light can travel through a vacuum. A periodic wave is one that creates a periodic motion in a receiver as it passes it. Such a wave has a well-defined period and frequency, and it will also have a wavelength, which is the distance in space between repetitions of the wave pattern. The velocity, frequency, and wavelength of a periodic wave are related by the equation v = f λ. A wave emitted by a moving source will be shifted in wavelength and frequency. The shifted wavelength is given by the equation
vs λ = 1 − λ v
,
where v is the velocity of the waves and vs is the velocity of the source, taken to be positive or negative so as to produce a Dopplerlengthened wavelength if the source is receding and a Dopplershortened one if it approaches. A similar shift occurs if the observer
Summary
497
is moving, and in general the Doppler shift depends approximately only on the relative motion of the source and observer if their velocities are both small compared to the waves’ velocity. (This is not just approximately but exactly true for light waves, and as required by Einstein’s Theory of Relativity.)
498
Chapter 19
Free waves
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 The following is a graph of the height of a water wave as a function of position, at a certain moment in time.
Trace this graph onto another piece of paper, and then sketch below it the corresponding graphs that would be obtained if (a) the amplitude and frequency were doubled while the velocity remained the same; (b) the frequency and velocity were both doubled while the amplitude remained unchanged; (c) the wavelength and amplitude were reduced by a factor of three while the velocity was doubled. Explain all your answers. [Problem by Arnold Arons.] 2 (a) The graph shows the height of a water wave pulse as a function of position. Draw a graph of height as a function of time for a specific point on the water. Assume the pulse is traveling to the right. (b) Repeat part a, but assume the pulse is traveling to the left. (c) Now assume the original graph was of height as a function of time, and draw a graph of height as a function of position, assuming the pulse is traveling to the right. (d) Repeat part c, but assume the pulse is traveling to the left. Explain all your answers. [Problem by Arnold Arons.]
Problem 2.
Problems
499
3 The figure shows one wavelength of a steady sinusoidal wave traveling to the right along a string. Define a coordinate system in which the positive x axis points to the right and the positive y axis up, such that the flattened string would have y = 0. Copy the figure, and label with y = 0 all the appropriate parts of the string. Similarly, label with v = 0 all parts of the string whose velocities are zero, and with a = 0 all parts whose accelerations are zero. There is more than one point whose velocity is of the greatest magnitude. Pick one of these, and indicate the direction of its velocity vector. Do the same for a point having the maximum magnitude of acceleration. Explain all your answers.
Problem 3.
[Problem by Arnold Arons.] 4 Find an equation for the relationship between the Dopplershifted frequency of a wave and the frequency of the original wave, for the case of a stationary observer and a source moving directly toward or away from the observer. 5 Suggest a quantitative experiment to look for any deviation from the principle of superposition for surface waves in water. Make it simple and practical. 6 The musical note middle C has a frequency of 262 Hz. What √ are its period and wavelength? 7 Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to a the rest of a band to make this much of a change in pitch due to the Doppler effect? 8 In section 19.2, we saw that the speed of waves on a string depends on the ratio of T /μ, i.e., the speed of the wave is greater if the string is under more tension, and less if it has more inertia. This is true in general: the speed of a mechanical wave always depends on the medium’s inertia in relation to the restoring force (tension, stiffness, resistance to compression,...) Based on these ideas, explain why the speed of sound in air is significantly greater on a hot day, while the speed of sound in liquids and solids shows almost no variation with temperature.
500
Chapter 19
Free waves
A cross-sectional view of a human body, showing the vocal tract.
Chapter 20
Bounded waves Speech is what separates humans most decisively from animals. No other species can master syntax, and even though chimpanzees can learn a vocabulary of hand signs, there is an unmistakable difference between a human infant and a baby chimp: starting from birth, the human experiments with the production of complex speech sounds. Since speech sounds are instinctive for us, we seldom think about them consciously. How do we do control sound waves so skillfully? Mostly we do it by changing the shape of a connected set of hollow cavities in our chest, throat, and head. Somehow by moving the boundaries of this space in and out, we can produce all the vowel sounds. Up until now, we have been studying only those properties of waves that can be understood as if they existed in an infinite, open space. In this chapter we address what happens when a wave is confined within a certain space, or when a wave pattern encounters the boundary between two different media, as when a light wave moving through air encounters a glass windowpane.
501
a / A diver photographed this fish, and its reflection, from underwater. The reflection is the one on top, and is formed by light waves that went up to the surface of the water, but were then reflected back down into the water.
20.1 Reflection, transmission, and absorption Reflection and transmission Sound waves can echo back from a cliff, and light waves are reflected from the surface of a pond. We use the word reflection, normally applied only to light waves in ordinary speech, to describe any such case of a wave rebounding from a barrier. Figure b shows a circular water wave being reflected from a straight wall. In this chapter, we will concentrate mainly on reflection of waves that move in one dimension, as in figure c. Wave reflection does not surprise us. After all, a material object such as a rubber ball would bounce back in the same way. But waves are not objects, and there are some surprises in store. First, only part of the wave is usually reflected. Looking out through a window, we see light waves that passed through it, but a person standing outside would also be able to see her reflection in the glass. A light wave that strikes the glass is partly reflected and partly transmitted (passed) by the glass. The energy of the original wave is split between the two. This is different from the behavior of the rubber ball, which must go one way or the other, not both. Second, consider what you see if you are swimming underwater and you look up at the surface. You see your own reflection. This is utterly counterintuitive, since we would expect the light waves to burst forth to freedom in the wide-open air. A material projectile shot up toward the surface would never rebound from the water-air
502
Chapter 20
Bounded waves
boundary! Figure a shows a similar example. What is it about the difference between two media that causes waves to be partly reflected at the boundary between them? Is it their density? Their chemical composition? Ultimately all that matters is the speed of the wave in the two media. A wave is partially reflected and partially transmitted at the boundary between media in which it has different speeds. For example, the speed of light waves in window glass is about 30% less than in air, which explains why windows always make reflections. Figures d/1 and 2 show examples of wave pulses being reflected at the boundary between two coil springs of different weights, in which the wave speed is different.
b / Circular water waves are reflected from a boundary on the left.
Reflections such as b and c, where a wave encounters a massive fixed object, can usually be understood on the same basis as cases like d/1 and 2 later in his section, where two media meet. Example c, for instance, is like a more extreme version of example d/1. If the heavy coil spring in d/1 was made heavier and heavier, it would end up acting like the fixed wall to which the light spring in c has been attached. self-check A In figure c, the reflected pulse is upside-down, but its depth is just as big as the original pulse’s height. How does the energy of the reflected pulse compare with that of the original? Answer, p. 545
Fish have internal ears. example 1 Why don’t fish have ear-holes? The speed of sound waves in a fish’s body is not much different from their speed in water, so sound waves are not strongly reflected from a fish’s skin. They pass right through its body, so fish can have internal ears. Whale songs traveling long distances example 2 Sound waves travel at drastically different speeds through rock, water, and air. Whale songs are thus strongly reflected at both the bottom and the surface. The sound waves can travel hundreds of miles, bouncing repeatedly between the bottom and the surface, and still be detectable. Sadly, noise pollution from ships has nearly shut down this cetacean version of the internet. Long-distance radio communication. example 3 Radio communication can occur between stations on opposite sides of the planet. The mechanism is similar to the one explained in example 2, but the three media involved are the earth, the atmosphere, and the ionosphere. self-check B Sonar is a method for ships and submarines to detect each other by producing sound waves and listening for echoes. What properties would an underwater object have to have in order to be invisible to sonar? Answer, p. 545
Section 20.1
c / A wave on a spring, initially traveling to the left, is reflected from the fixed end.
Reflection, transmission, and absorption
503
The use of the word “reflection” naturally brings to mind the creation of an image by a mirror, but this might be confusing, because we do not normally refer to “reflection” when we look at surfaces that are not shiny. Nevertheless, reflection is how we see the surfaces of all objects, not just polished ones. When we look at a sidewalk, for example, we are actually seeing the reflecting of the sun from the concrete. The reason we don’t see an image of the sun at our feet is simply that the rough surface blurs the image so drastically.
d / 1. A wave in the lighter spring, where the wave speed is greater, travels to the left and is then partly reflected and partly transmitted at the boundary with the heavier coil spring, which has a lower wave speed. The reflection is inverted. 2. A wave moving to the right in the heavier spring is partly reflected at the boundary with the lighter spring. The reflection is uninverted.
504
Chapter 20
Bounded waves
Inverted and uninverted reflections Notice how the pulse reflected back to the right in example d/1 comes back upside-down, whereas the one reflected back to the left in 2 returns in its original upright form. This is true for other waves as well. In general, there are two possible types of reflections, a reflection back into a faster medium and a reflection back into a slower medium. One type will always be an inverting reflection and one noninverting. It’s important to realize that when we discuss inverted and uninverted reflections on a string, we are talking about whether the wave is flipped across the direction of motion (i.e., upside-down in these drawings). The reflected pulse will always be reversed front to back, as shown in figure e. This is because it is traveling in the other direction. The leading edge of the pulse is what gets reflected first, so it is still ahead when it starts back to the left — it’s just that “ahead” is now in the opposite direction. Absorption
e / 1. An uninverted reflection. The reflected pulse is reversed front to back, but is not upside-down. 2. An inverted reflection. The reflected pulse is reversed both front to back and top to bottom.
So far we have tacitly assumed that wave energy remains as wave energy, and is not converted to any other form. If this was true, then the world would become more and more full of sound waves, which could never escape into the vacuum of outer space. In reality, any mechanical wave consists of a traveling pattern of vibrations of some physical medium, and vibrations of matter always produce heat, as when you bend a coat-hangar back and forth and it becomes hot. We can thus expect that in mechanical waves such as water waves, sound waves, or waves on a string, the wave energy will gradually be converted into heat. This is referred to as absorption. The wave suffers a decrease in amplitude, as shown in figure f. The decrease in amplitude amounts to the same fractional change for each unit of distance covered. For example, if a wave decreases from amplitude 2 to amplitude 1 over a distance of 1 meter, then after traveling another meter it will have an amplitude of 1/2. That is, the reduction in amplitude is exponential. This can be proven as follows. By the principle of superposition, we know that a wave of amplitude 2 must behave like the superposition of two identical waves of amplitude 1. If a single amplitude-1 wave would die down to amplitude 1/2 over a certain distance, then two amplitude-1 waves superposed on top of one another to make amplitude 1 + 1 = 2 must die down to amplitude 1/2 + 1/2 = 1 over the same distance.
f / A pulse traveling through a highly absorptive medium.
self-check C As a wave undergoes absorption, it loses energy. Does this mean that it slows down? Answer, p. 545
In many cases, this frictional heating effect is quite weak. Sound waves in air, for instance, dissipate into heat extremely slowly, and the sound of church music in a cathedral may reverberate for as much
Section 20.1
Reflection, transmission, and absorption
505
as 3 or 4 seconds before it becomes inaudible. During this time it has traveled over a kilometer! Even this very gradual dissipation of energy occurs mostly as heating of the church’s walls and by the leaking of sound to the outside (where it will eventually end up as heat). Under the right conditions (humid air and low frequency), a sound wave in a straight pipe could theoretically travel hundreds of kilometers before being noticeably attenuated.
g / X-rays are light waves with a very high frequency. They are absorbed strongly by bones, but weakly by flesh.
In general, the absorption of mechanical waves depends a great deal on the chemical composition and microscopic structure of the medium. Ripples on the surface of antifreeze, for instance, die out extremely rapidly compared to ripples on water. For sound waves and surface waves in liquids and gases, what matters is the viscosity of the substance, i.e., whether it flows easily like water or mercury or more sluggishly like molasses or antifreeze. This explains why our intuitive expectation of strong absorption of sound in water is incorrect. Water is a very weak absorber of sound (viz. whale songs and sonar), and our incorrect intuition arises from focusing on the wrong property of the substance: water’s high density, which is irrelevant, rather than its low viscosity, which is what matters. Light is an interesting case, since although it can travel through matter, it is not itself a vibration of any material substance. Thus we can look at the star Sirius, 1014 km away from us, and be assured that none of its light was absorbed in the vacuum of outer space during its 9-year journey to us. The Hubble Space Telescope routinely observes light that has been on its way to us since the early history of the universe, billions of years ago. Of course the energy of light can be dissipated if it does pass through matter (and the light from distant galaxies is often absorbed if there happen to be clouds of gas or dust in between). Soundproofing example 4 Typical amateur musicians setting out to soundproof their garages tend to think that they should simply cover the walls with the densest possible substance. In fact, sound is not absorbed very strongly even by passing through several inches of wood. A better strategy for soundproofing is to create a sandwich of alternating layers of materials in which the speed of sound is very different, to encourage reflection. The classic design is alternating layers of fiberglass and plywood. The speed of sound in plywood is very high, due to its stiffness, while its speed in fiberglass is essentially the same as its speed in air. Both materials are fairly good sound absorbers, but sound waves passing through a few inches of them are still not going to be absorbed sufficiently. The point of combining them is that a sound wave that tries to get out will be strongly reflected at each of the fiberglass-plywood boundaries, and will bounce back and forth many times like a ping pong ball. Due to all the back-
506
Chapter 20
Bounded waves
and-forth motion, the sound may end up traveling a total distance equal to ten times the actual thickness of the soundproofing before it escapes. This is the equivalent of having ten times the thickness of sound-absorbing material. The swim bladder example 5 The swim bladder of a fish, which was first discussed in homework problem 2 in chapter 18, is often located right next to the fish’s ear. As discussed in example 1 on page 503, the fish’s body is nearly transparent to sound, so it’s actually difficult to get any of the sound wave energy to deposit itself in the fish so that the fish can hear it! The physics here is almost exactly the same as the physics of example 4 above, with the gas-filled swim bladder playing the role of the low-density material. Radio transmission example 6 A radio transmitting station, such as a commercial station or an amateur “ham” radio station, must have a length of wire or cable connecting the amplifier to the antenna. The cable and the antenna act as two different media for radio waves, and there will therefore be partial reflection of the waves as they come from the cable to the antenna. If the waves bounce back and forth many times between the amplifier and the antenna, a great deal of their energy will be absorbed. There are two ways to attack the problem. One possibility is to design the antenna so that the speed of the waves in it is as close as possible to the speed of the waves in the cable; this minimizes the amount of reflection. The other method is to connect the amplifier to the antenna using a type of wire or cable that does not strongly absorb the waves. Partial reflection then becomes irrelevant, since all the wave energy will eventually exit through the antenna. The tympanogram example 7 The tympanogram is a medical procedure used to diagnose problems with the middle ear. The middle ear is a chamber, normally filled with air, lying between the eardrum (tympanic membrane) and the inner ear. It contains a tiny set of bones that act as a system of levers to amplify the motion of the eardrum and transmit it to the inner ear. The air pressure in the inner ear is normally equalized via the Eustachian tube, which connects to the throat; when you feel uncomfortable pressure in your ear while flying, it’s because the pressure has not yet equalized. Ear infections or allergies can cause the middle ear to become filled with fluid, and the Eustachian tube can also become blocked, so that the pressure in the inner ear cannot become equalized.
h/A 7.
tympanometer,
example
The tympanometer has a probe that is inserted into the ear, with several holes. One hole is used to send a 226 Hz sound wave into
Section 20.1
Reflection, transmission, and absorption
507
the ear canal. The ear has evolved so as to transmit a maximum amount of wave motion to the inner ear. Any change in its physical properties will change its behavior from its normal optimum, so that more sound energy than normal is reflected back. A second hole in the probe senses the reflected wave. If the reflection is stronger than normal, there is probably something wrong with the inner ear. The full physical analysis is fairly complex. The middle ear has some of the characteristics of a mass oscillating on a spring, but it also has some of the characteristics of a medium that carries waves. Crudely, we could imagine that an infected, fluid-filled middle ear would act as a medium that differed greatly from the air in the outer ear, causing a large amount of reflection. Equally crudely, we could forget about the wave ideas and think of the middle ear purely as as a mass on a spring. We expect resonant behavior, and there is in fact such a resonance, which is typically at a frequency of about 600 Hz in adults, so the 226 Hz frequency emitted by the probe is actually quite far from resonance. If the mechanisms of the middle ear are jammed and cannot vibrate, then it is not possible for energy of the incoming sound wave to be turned into energy of vibration in the middle ear, and therefore by conservation of energy we would expect all of the sound to be reflected. Sometimes the middle ear’s mechanisms can get jammed because of abnormally high or low pressure, because the Eustachian tube is blocked and cannot equalize the pressure with the outside environment. Diagnosing such a condition is the purpose of the third hole in the probe, which is used to vary the pressure in the ear canal. The amount of reflection is measured as a function of this pressure. If the reflection is minimized for some value of the pressure that is different than atmospheric pressure, it indicates that that is the value of the pressure in the middle ear; when the pressures are equalized, the forces on the eardrum cancel out, and it can relax to its normal position, unjamming the middle ear’s mechanisms. Discussion question A A sound wave that underwent a pressure-inverting reflection would have its compressions converted to expansions and vice versa. How would its energy and frequency compare with those of the original sound? Would it sound any different? What happens if you swap the two wires where they connect to a stereo speaker, resulting in waves that vibrate in the opposite way?
508
Chapter 20
Bounded waves
20.2 Quantitative treatment of reflection In this optional section we analyze the reasons why reflections occur at a speed-changing boundary, predict quantitatively the intensities of reflection and transmission, and discuss how to predict for any type of wave which reflections are inverting and which are noninverting. The gory details are likely to be of interest mainly to students with concentrations in the physical sciences, but all readers are encouraged at least to skim the first two subsections for physical insight. Why reflection occurs To understand the fundamental reasons for what does occur at the boundary between media, let’s first discuss what doesn’t happen. For the sake of concreteness, consider a sinusoidal wave on a string. If the wave progresses from a heavier portion of the string, in which its velocity is low, to a lighter-weight part, in which it is high, then the equation v = f λ tells us that it must change its frequency, or its wavelength, or both. If only the frequency changed, then the parts of the wave in the two different portions of the string would quickly get out of step with each other, producing a discontinuity in the wave, i/1. This is unphysical, so we know that the wavelength must change while the frequency remains constant, 2. But there is still something unphysical about figure 2. The sudden change in the shape of the wave has resulted in a sharp kink at the boundary. This can’t really happen, because the medium tends to accelerate in such a way as to eliminate curvature. A sharp kink corresponds to an infinite curvature at one point, which would produce an infinite acceleration, which would not be consistent with the smooth pattern of wave motion envisioned in figure 2. Waves can have kinks, but not stationary kinks.
i / 1. A change in frequency without a change in wavelength would produce a discontinuity in the wave. 2. A simple change in wavelength without a reflection would result in a sharp kink in the wave.
We conclude that without positing partial reflection of the wave, we cannot simultaneously satisfy the requirements of (1) continuity of the wave, and (2) no sudden changes in the slope of the wave. (The student who has studied calculus will recognize this as amounting to an assumption that both the wave and its derivative are continuous functions.) Does this amount to a proof that reflection occurs? Not quite. We have only proven that certain types of wave motion are not valid solutions. In the following subsection, we prove that a valid solution can always be found in which a reflection occurs. Now in physics, we normally assume (but seldom prove formally) that the equations of motion have a unique solution, since otherwise a given set of initial conditions could lead to different behavior later on, but the Newtonian universe is supposed to be deterministic. Since the solution must be unique, and we derive below a valid solution involving a reflected pulse, we will have ended up with what amounts
Section 20.2
Quantitative treatment of reflection
509
to a proof of reflection. Intensity of reflection We will now show, in the case of waves on a string, that it is possible to satisfy the physical requirements given above by constructing a reflected wave, and as a bonus this will produce an equation for the proportions of reflection and transmission and a prediction as to which conditions will lead to inverted and which to uninverted reflection. We assume only that the principle of superposition holds, which is a good approximations for waves on a string of sufficiently small amplitude. Let the unknown amplitudes of the reflected and transmitted waves be R and T , respectively. An inverted reflection would be represented by a negative value of R. We can without loss of generality take the incident (original) wave to have unit amplitude. Superposition tells us that if, for instance, the incident wave had double this amplitude, we could immediately find a corresponding solution simply by doubling R and T . Just to the left of the boundary, the height of the wave is given by the height 1 of the incident wave, plus the height R of the part of the reflected wave that has just been created and begun heading back, for a total height of 1 + R. On the right side immediately next to the boundary, the transmitted wave has a height T . To avoid a discontinuity, we must have 1+R=T j / A pulse being partially reflected and partially transmitted at the boundary between two strings in which the speed of waves is different. The top drawing shows the pulse heading to the right, toward the heavier string. For clarity, all but the first and last drawings are schematic. Once the reflected pulse begins to emerge from the boundary, it adds together with the trailing parts of the incident pulse. Their sum, shown as a wider line, is what is actually observed.
.
Next we turn to the requirement of equal slopes on both sides of the boundary. Let the slope of the incoming wave be s immediately to the left of the junction. If the wave was 100% reflected, and without inversion, then the slope of the reflected wave would be −s, since the wave has been reversed in direction. In general, the slope of the reflected wave equals −sR, and the slopes of the superposed waves on the left side add up to s − sR. On the right, the slope depends on the amplitude, T , but is also changed by the stretching or compression of the wave due to the change in speed. If, for example, the wave speed is twice as great on the right side, then the slope is cut in half by this effect. The slope on the right is therefore s(v1 /v2 )T , where v1 is the velocity in the original medium and v2 the velocity in the new medium. Equality of slopes gives s − sR = s(v1 /v2 )T , or 1−R=
v1 T v2
.
Solving the two equations for the unknowns R and T gives R=
510
Chapter 20
Bounded waves
v2 − v1 v2 + v1
and
T =
2v2 v2 + v1
.
The first equation shows that there is no reflection unless the two wave speeds are different, and that the reflection is inverted in reflection back into a fast medium. The energies of the transmitted and reflected waves always add up to the same as the energy of the original wave. There is never any abrupt loss (or gain) in energy when a wave crosses a boundary. (Conversion of wave energy to heat occurs for many types of waves, but it occurs throughout the medium.) The equation for T , surprisingly, allows the amplitude of the transmitted wave to be greater than 1, i.e., greater than that of the incident wave. This does not violate conservation of energy, because this occurs when the second string is less massive, reducing its kinetic energy, and the transmitted pulse is broader and less strongly curved, which lessens its potential energy. Inverted and uninverted reflections in general For waves on a string, reflections back into a faster medium are inverted, while those back into a slower medium are uninverted. Is this true for all types of waves? The rather subtle answer is that it depends on what property of the wave you are discussing. Let’s start by considering wave disturbances of freeway traffic. Anyone who has driven frequently on crowded freeways has observed the phenomenon in which one driver taps the brakes, starting a chain reaction that travels backward down the freeway as each person in turn exercises caution in order to avoid rear-ending anyone. The reason why this type of wave is relevant is that it gives a simple, easily visualized example of our description of a wave depends on which aspect of the wave we have in mind. In steadily flowing freeway traffic, both the density of cars and their velocity are constant all along the road. Since there is no disturbance in this pattern of constant velocity and density, we say that there is no wave. Now if a wave is touched off by a person tapping the brakes, we can either describe it as a region of high density or as a region of decreasing velocity.
k / A disturbance traffic.
in
freeway
The freeway traffic wave is in fact a good model of a sound wave, and a sound wave can likewise be described either by the density (or pressure) of the air or by its speed. Likewise many other types of waves can be described by either of two functions, one of which is often the derivative of the other with respect to position. Now let’s consider reflections. If we observe the freeway wave in a mirror, the high-density area will still appear high in density, but velocity in the opposite direction will now be described by a negative number. A person observing the mirror image will draw the same density graph, but the velocity graph will be flipped across the x axis, and its original region of negative slope will now have positive slope. Although I don’t know any physical situation that would
Section 20.2
l / In the mirror image, the areas of positive excess traffic density are still positive, but the velocities of the cars have all been reversed, so areas of positive excess velocity have been turned into negative ones.
Quantitative treatment of reflection
511
correspond to the reflection of a traffic wave, we can immediately apply the same reasoning to sound waves, which often do get reflected, and determine that a reflection can either be density-inverting and velocity-noninverting or density-noninverting and velocity-inverting. This same type of situation will occur over and over as one encounters new types of waves, and to apply the analogy we need only determine which quantities, like velocity, become negated in a mirror image and which, like density, stay the same. A light wave, for instance consists of a traveling pattern of electric and magnetic fields. All you need to know in order to analyze the reflection of light waves is how electric and magnetic fields behave under reflection; you don’t need to know any of the detailed physics of electricity and magnetism. An electric field can be detected, for example, by the way one’s hair stands on end. The direction of the hair indicates the direction of the electric field. In a mirror image, the hair points the other way, so the electric field is apparently reversed in a mirror image. The behavior of magnetic fields, however, is a little tricky. The magnetic properties of a bar magnet, for instance, are caused by the aligned rotation of the outermost orbiting electrons of the atoms. In a mirror image, the direction of rotation is reversed, say from clockwise to counterclockwise, and so the magnetic field is reversed twice: once simply because the whole picture is flipped and once because of the reversed rotation of the electrons. In other words, magnetic fields do not reverse themselves in a mirror image. We can thus predict that there will be two possible types of reflection of light waves. In one, the electric field is inverted and the magnetic field uninverted. In the other, the electric field is uninverted and the magnetic field inverted.
20.3 Interference effects
m / Seen from this angle, the optical coating on the lenses of these binoculars appears purple and green. (The color varies depending on the angle from which the coating is viewed, and the angle varies across the faces of the lenses because of their curvature.)
512
Chapter 20
If you look at the front of a pair of high-quality binoculars, you will notice a greenish-blue coating on the lenses. This is advertised as a coating to prevent reflection. Now reflection is clearly undesirable — we want the light to go in the binoculars — but so far I’ve described reflection as an unalterable fact of nature, depending only on the properties of the two wave media. The coating can’t change the speed of light in air or in glass, so how can it work? The key is that the coating itself is a wave medium. In other words, we have a three-layer sandwich of materials: air, coating, and glass. We will analyze the way the coating works, not because optical coatings are an important part of your education but because it provides a good example of the general phenomenon of wave interference effects. There are two different interfaces between media: an air-coating boundary and a coating-glass boundary. Partial reflection and partial transmission will occur at each boundary. For ease of visual-
Bounded waves
ization let’s start by considering an equivalent system consisting of three dissimilar pieces of string tied together, and a wave pattern consisting initially of a single pulse. Figure n/1 shows the incident pulse moving through the heavy rope, in which its velocity is low. When it encounters the lighter-weight rope in the middle, a faster medium, it is partially reflected and partially transmitted. (The transmitted pulse is bigger, but nevertheless has only part of the original energy.) The pulse transmitted by the first interface is then partially reflected and partially transmitted by the second boundary, 3. In figure 4, two pulses are on the way back out to the left, and a single pulse is heading off to the right. (There is still a weak pulse caught between the two boundaries, and this will rattle back and forth, rapidly getting too weak to detect as it leaks energy to the outside with each partial reflection.) Note how, of the two reflected pulses in 4, one is inverted and one uninverted. One underwent reflection at the first boundary (a reflection back into a slower medium is uninverted), but the other was reflected at the second boundary (reflection back into a faster medium is inverted). Now let’s imagine what would have happened if the incoming wave pattern had been a long sinusoidal wave train instead of a single pulse. The first two waves to reemerge on the left could be in phase, o/1, or out of phase, 2, or anywhere in between. The amount of lag between them depends entirely on the width of the middle segment of string. If we choose the width of the middle string segment correctly, then we can arrange for destructive interference to occur, 2, with cancellation resulting in a very weak reflected wave. This whole analysis applies directly to our original case of optical coatings. Visible light from most sources does consist of a stream of short sinusoidal wave-trains such as the ones drawn above. The only real difference between the waves-on-a-rope example and the case of an optical coating is that the first and third media are air and glass, in which light does not have the same speed. However, the general result is the same as long as the air and the glass have light-wave speeds that either both greater than the coating’s or both less than the coating’s.
n / A rope consisting of three sections, the middle one being lighter.
o / Two reflections, are superimposed. One reflection is inverted.
The business of optical coatings turns out to be a very arcane one, with a plethora of trade secrets and “black magic” techniques handed down from master to apprentice. Nevertheless, the ideas you have learned about waves in general are sufficient to allow you to come to some definite conclusions without any further technical knowledge. The self-check and discussion questions will direct you along these lines of thought. The example of an optical coating was typical of a wide variety of wave interference effects. With a little guidance, you are now ready to figure out for yourself other examples such as the rainbow
Section 20.3
p / A soap bubble interference effects.
Interference effects
displays
513
pattern made by a compact disc, a layer of oil on a puddle, or a soap bubble. self-check D 1. Color corresponds to wavelength of light waves. Is it possible to choose a thickness for an optical coating that will produce destructive interference for all colors of light? 2. How can you explain the rainbow colors on the soap bubble in figure p? Answer, p. 545
Discussion questions A Is it possible to get complete destructive interference in an optical coating, at least for light of one specific wavelength? B Sunlight consists of sinusoidal wave-trains containing on the order of a hundred cycles back-to-back, for a length of something like a tenth of a millimeter. What happens if you try to make an optical coating thicker than this? C Suppose you take two microscope slides and lay one on top of the other so that one of its edges is resting on the corresponding edge of the bottom one. If you insert a sliver of paper or a hair at the opposite end, a wedge-shaped layer of air will exist in the middle, with a thickness that changes gradually from one end to the other. What would you expect to see if the slides were illuminated from above by light of a single color? How would this change if you gradually lifted the lower edge of the top slide until the two slides were finally parallel? D An observation like the one described in discussion question C was used by Newton as evidence against the wave theory of light! If Newton didn’t know about inverting and noninverting reflections, what would have seemed inexplicable to him about the region where the air layer had zero or nearly zero thickness?
514
Chapter 20
Bounded waves
20.4 Waves bounded on both sides In the examples discussed in section 20.3, it was theoretically true that a pulse would be trapped permanently in the middle medium, but that pulse was not central to our discussion, and in any case it was weakening severely with each partial reflection. Now consider a guitar string. At its ends it is tied to the body of the instrument itself, and since the body is very massive, the behavior of the waves when they reach the end of the string can be understood in the same way as if the actual guitar string was attached on the ends to strings that were extremely massive, q. Reflections are most intense when the two media are very dissimilar. Because the wave speed in the body is so radically different from the speed in the string, we should expect nearly 100% reflection. Although this may seem like a rather bizarre physical model of the actual guitar string, it already tells us something interesting about the behavior of a guitar that we would not otherwise have understood. The body, far from being a passive frame for attaching the strings to, is actually the exit path for the wave energy in the strings. With every reflection, the wave pattern on the string loses a tiny fraction of its energy, which is then conducted through the body and out into the air. (The string has too little cross-section to make sound waves efficiently by itself.) By changing the properties of the body, moreover, we should expect to have an effect on the manner in which sound escapes from the instrument. This is clearly demonstrated by the electric guitar, which has an extremely massive, solid wooden body. Here the dissimilarity between the two wave media is even more pronounced, with the result that wave energy leaks out of the string even more slowly. This is why an electric guitar with no electric pickup can hardly be heard at all, and it is also the reason why notes on an electric guitar can be sustained for longer than notes on an acoustic guitar. If we initially create a disturbance on a guitar string, how will the reflections behave? In reality, the finger or pick will give the string a triangular shape before letting it go, and we may think of this triangular shape as a very broad “dent” in the string which will spread out in both directions. For simplicity, however, let’s just imagine a wave pattern that initially consists of a single, narrow pulse traveling up the neck, r/1. After reflection from the top end, it is inverted, 3. Now something interesting happens: figure 5 is identical to figure 1. After two reflections, the pulse has been inverted twice and has changed direction twice. It is now back where it started. The motion is periodic. This is why a guitar produces sounds that have a definite sensation of pitch. self-check E Notice that from r/1 to r/5, the pulse has passed by every point on the string exactly twice. This means that the total distance it has traveled
Section 20.4
q / A model of a guitar string.
r / The motion of a pulse on the string.
s / A tricky way to double the frequency.
Waves bounded on both sides
515
equals 2L, where L is the length of the string. Given this fact, what are the period and frequency of the sound it produces, expressed in terms Answer, p. 545 of L and v , the velocity of the wave?
Note that if the waves on the string obey the principle of superposition, then the velocity must be independent of amplitude, and the guitar will produce the same pitch regardless of whether it is played loudly or softly. In reality, waves on a string obey the principle of superposition approximately, but not exactly. The guitar, like just about any acoustic instrument, is a little out of tune when played loudly. (The effect is more pronounced for wind instruments than for strings, but wind players are able to compensate for it.) Now there is only one hole in our reasoning. Suppose we somehow arrange to have an initial setup consisting of two identical pulses heading toward each other, as in figure s. They will pass through each other, undergo a single inverting reflection, and come back to a configuration in which their positions have been exactly interchanged. This means that the period of vibration is half as long. The frequency is twice as high. t / Using the sum of four sine waves to approximate the triangular initial shape of a plucked guitar string.
This might seem like a purely academic possibility, since nobody actually plays the guitar with two picks at once! But in fact it is an example of a very general fact about waves that are bounded on both sides. A mathematical theorem called Fourier’s theorem states that any wave can be created by superposing sine waves. Figure t shows how even by using only four sine waves with appropriately chosen amplitudes, we can arrive at a sum which is a decent approximation to the realistic triangular shape of a guitar string being plucked. The one-hump wave, in which half a wavelength fits on the string, will behave like the single pulse we originally discussed. We call its frequency fo . The two-hump wave, with one whole wavelength, is very much like the two-pulse example. For the reasons discussed above, its frequency is 2fo . Similarly, the three-hump and four-hump waves have frequencies of 3fo and 4fo . Theoretically we would need to add together infinitely many such wave patterns to describe the initial triangular shape of the string exactly, although the amplitudes required for the very high frequency parts would be very small, and an excellent approximation could be achieved with as few as ten waves. We thus arrive at the following very general conclusion. Whenever a wave pattern exists in a medium bounded on both sides by media in which the wave speed is very different, the motion can be broken down into the motion of a (theoretically infinite) series of sine waves, with frequencies fo , 2fo , 3fo , ... Except for some technical details, to be discussed below, this analysis applies to a vast range of sound-producing systems, including the air column within the human vocal tract. Because sounds composed of this kind of pattern of frequencies are so common, our ear-brain system has evolved so
516
Chapter 20
Bounded waves
as to perceive them as a single, fused sensation of tone. Musical applications Many musicians claim to be able to pick out by ear several of the frequencies 2fo , 3fo , ..., called overtones or harmonics of the fundamental fo , but they are kidding themselves. In reality, the overtone series has two important roles in music, neither of which depends on this fictitious ability to “hear out” the individual overtones. First, the relative strengths of the overtones is an important part of the personality of a sound, called its timbre (rhymes with “amber”). The characteristic tone of the brass instruments, for example, is a sound that starts out with a very strong harmonic series extending up to very high frequencies, but whose higher harmonics die down drastically as the attack changes to the sustained portion of the note. Second, although the ear cannot separate the individual harmonics of a single musical tone, it is very sensitive to clashes between the overtones of notes played simultaneously, i.e., in harmony. We tend to perceive a combination of notes as being dissonant if they have overtones that are close but not the same. Roughly speaking, strong overtones whose frequencies differ by more than 1% and less than 10% cause the notes to sound dissonant. It is important to realize that the term “dissonance” is not a negative one in music. No matter how long you search the radio dial, you will never hear more than three seconds of music without at least one dissonant combination of notes. Dissonance is a necessary ingredient in the creation of a musical cycle of tension and release. Musically knowledgeable people don’t use the word “dissonant” as a criticism of music, although dissonance can be used in a clumsy way, or without providing any contrast between dissonance and consonance.
u / Graphs of loudness versus frequency for the vowel “ah,” sung as three different musical notes. G is consonant with D, since every overtone of G that is close to an overtone of D (*) is at exactly the same frequency. G and C# are dissonant together, since some of the overtones of G (x) are close to, but not right on top of, those of C#.
Standing waves Figure v shows sinusoidal wave patterns made by shaking a rope. I used to enjoy doing this at the bank with the pens on chains, back in the days when people actually went to the bank. You might think that I and the person in the photos had to practice for a long time in order to get such nice sine waves. In fact, a sine wave is the only shape that can create this kind of wave pattern, called a standing wave, which simply vibrates back and forth in one place without moving. The sine wave just creates itself automatically when you find the right frequency, because no other shape is possible. If you think about it, it’s not even obvious that sine waves should be able to do this trick. After all, waves are supposed to travel at a set speed, aren’t they? The speed isn’t supposed to be zero! Well, we can actually think of a standing wave as a superposition of a moving
Section 20.4
Waves bounded on both sides
517
v / Standing waves on a spring.
w / Sine waves add to make sine waves. Other functions don’t have this property.
sine wave with its own reflection, which is moving the opposite way. Sine waves have the unique mathematical property, w, that the sum of sine waves of equal wavelength is simply a new sine wave with the same wavelength. As the two sine waves go back and forth, they always cancel perfectly at the ends, and their sum appears to stand still. Standing wave patterns are rather important, since atoms are really standing-wave patterns of electron waves. You are a standing wave! Harmonics on string instruments example 8 Figure x shows a violist playing what string players refer to as a natural harmonic. The term “harmonic” is used here in a somewhat different sense than in physics. The musician’s pinkie is pressing very lightly against the string — not hard enough to make it touch the fingerboard — at a point precisely at the center of the string’s length. As shown in the diagram, this allows the string to vibrate at frequencies 2fo , 4fo , 6fo , . . ., which have stationary points at the center of the string, but not at the odd multiples fo , 3fo , . . .. Since all the overtones are multiples of 2fo , the ear perceives 2fo as the basic frequency of the note. In musical terms, doubling the frequency corresponds to raising the pitch by an octave. The technique can be used in order to make it easier to play high notes in rapid passages, or for its own sake, because of the change in timbre.
x / Example 8.
518
Chapter 20
Bounded waves
Standing-wave patterns of air columns The air column inside a wind instrument behaves very much like the wave-on-a-string example we’ve been concentrating on so far, the main difference being that we may have either inverting or noninverting reflections at the ends. Some organ pipes are closed at both ends. The speed of sound is different in metal than in air, so there is a strong reflection at the closed ends, and we can have standing waves. These reflections are both density-noninverting, so we get symmetric standing-wave patterns, such as the one shown in figure z/1. Figure y shows the sound waves in and around a bamboo Japanese flute called a shakuhachi, which is open at both ends of the air column. We can only have a standing wave pattern if there are reflections at the ends, but that is very counterintuitive — why is there any reflection at all, if the sound wave is free to emerge into open space, and there is no change in medium? Recall the reason why we got reflections at a change in medium: because the wavelength changes, so the wave has to readjust itself from one pattern to another, and the only way it can do that without developing a kink is if there is a reflection. Something similar is happening here. The only difference is that the wave is adjusting from being a plane wave to being a spherical wave. The reflections at the open ends are density-inverting, z/2, so the wave pattern is pinched off at the ends. Comparing panels 1 and 2 of the figure, we see that although the wave pattens are different, in both cases the wavelength is the same: in the lowest-frequency standing wave, half a wavelength fits inside the tube. Thus, it isn’t necessary to memorize which type of reflection is inverting and which is inverting. It’s only necessary to know that the tubes are symmetric. Finally, we can have an asymmetric tube: closed at one end and open at the other. A common example is the pan pipes, aa, which are closed at the bottom and open at the top. The standing wave with the lowest frequency is therefore one in which 1/4 of a wavelength fits along the length of the tube, as shown in figure z/3.
y / Surprisingly, sound waves undergo partial reflection at the open ends of tubes as well as closed ones.
z / Graphs of excess density versus position for the lowestfrequency standing waves of three types of air columns. Points on the axis have normal air density.
Sometimes an instrument’s physical appearance can be misleading. A concert flute, ab, is closed at the mouth end and open at the other, so we would expect it to behave like an asymmetric air column; in reality, it behaves like a symmetric air column open at both ends, because the embouchure hole (the hole the player blows over) acts like an open end. The clarinet and the saxophone look similar, having a mouthpiece and reed at one end and an open end at the other, but they act different. In fact the clarinet’s air column has patterns of vibration that are asymmetric, the saxophone symmetric. The discrepancy comes from the difference between the conical tube of the sax and the cylindrical tube of the clarinet. The adjustment of the wave pattern from a plane wave to a spherical
Section 20.4
Waves bounded on both sides
519
wave is more gradual at the flaring bell of the saxophone. self-check F Draw a graph of density versus position for the first overtone of the air column in a tube open at one end and closed at the other. This will be the next-to-longest possible wavelength that allows for a point of maximum vibration at one end and a point of no vibration at the other. How many times shorter will its wavelength be compared to the wavelength of the lowest-frequency standing wave, shown in the figure? Based on this, how many times greater will its frequency be? Answer, p. 545
aa / A pan pipe is an asymmetric air column, open at the top and closed at the bottom.
ab / A concert flute looks like an asymmetric air column, open at the mouth end and closed at the other. However, its patterns of vibration are symmetric, because the embouchure hole acts like an open end.
520
Chapter 20
Bounded waves
Summary Selected vocabulary reflection . . . . . the bouncing back of part of a wave from a boundary transmission . . . the continuation of part of a wave through a boundary absorption . . . . the gradual conversion of wave energy into heating of the medium standing wave . . a wave pattern that stays in one place Notation λ. . . . . . . . . .
wavelength (Greek letter lambda)
Summary Whenever a wave encounters the boundary between two media in which its speeds are different, part of the wave is reflected and part is transmitted. The reflection is always reversed front-to-back, but may also be inverted in amplitude. Whether the reflection is inverted depends on how the wave speeds in the two media compare, e.g., a wave on a string is uninverted when it is reflected back into a segment of string where its speed is lower. The greater the difference in wave speed between the two media, the greater the fraction of the wave energy that is reflected. Surprisingly, a wave in a dense material like wood will be strongly reflected back into the wood at a wood-air boundary. A one-dimensional wave confined by highly reflective boundaries on two sides will display motion which is periodic. For example, if both reflections are inverting, the wave will have a period equal to twice the time required to traverse the region, or to that time divided by an integer. An important special case is a sinusoidal wave; in this case, the wave forms a stationary pattern composed of a superposition of sine waves moving in opposite direction.
Summary
521
Problems Key √
A computerized answer check is available online. A problem that requires calculus. A difficult problem.
1 Light travels faster in warmer air. Use this fact to explain the formation of a mirage appearing like the shiny surface of a pool of water when there is a layer of hot air above a road. (For simplicity, pretend that there is actually a sharp boundary between the hot layer and the cooler layer above it.) 2 (a) Using the equations from optional section 20.2, compute the amplitude of light that is reflected back into air at an air-water interface, relative to the amplitude of the incident wave. The speeds of light in air and water are 3.0× 108 and 2.2× 108 m/s, respectively. (b) Find the energy of the reflected wave as a fraction of the incident energy. [Hint: The answers to the two parts are not the same.] √ 3 A concert flute produces its lowest note, at about 262 Hz, when half of a wavelength fits inside its tube. Compute the length of the flute. Answer, p. 546
C D E F G A B
4 (a) A good tenor saxophone player can play all of the following notes without changing her fingering, simply by altering the tightness of her lips: E (150 Hz), E (300 Hz), B (450 Hz), and E (600 Hz). How is this possible? (I’m not asking you to analyze the coupling between the lips, the reed, the mouthpiece, and the air column, which is very complicated.) (b) Some saxophone players are known for their ability to use this technique to play “freak notes,” i.e., notes above the normal range of the instrument. Why isn’t it possible to play notes below the normal range using this technique?
261.6 Hz 293.7 329.6 349.2 392.0 440.0 466.2
5 The table gives the frequencies of the notes that make up the key of F major, starting from middle C and going up through all seven notes. (a) Calculate the first four or five harmonics of C and G, and determine whether these two notes will be consonant or dissonant. (b) Do the same for C and B. [Hint: Remember that harmonics that differ by about 1-10% cause dissonance.]
Problem 5.
522
6 Brass and wind instruments go up in pitch as the musician warms up. As a typical empirical example, a trumpet’s frequency might go up by about 1%. Let’s consider possible physical reasons for the change in pitch. (a) Solids generally expand with increasing temperature, because the stronger random motion of the atoms tends to bump them apart. Brass expands by 1.88×10−5 per degree C. Would this tend to raise the pitch, or lower it? Estimate the size of the effect in comparison with the observed change in frequency.
Chapter 20
Bounded waves
(b) The speed of sound in a gas is proportional to the square root of the absolute temperature, where zero absolute temperature is -273 degrees C. As in part a, analyze the size and direction of the effect. 7 Your exhaled breath contains about 4.5% carbon dioxide, and is therefore more dense than fresh air by about 2.3%. By analogy with the treatment of waves on a string in section 19.2, we expect that the speed of sound will be inversely proportional to the square root of the density of the gas. Calculate the effect on the frequency produced by a wind instrument.
Problems
523
Three essential mathematical skills More often than not when a search-and-rescue team finds a hiker dead in the wilderness, it turns out that the person was not carrying some item from a short list of essentials, such as water and a map. There are three mathematical essentials in this course.
1. Converting units basic technique: section 0.9, p. 30; conversion of area, volume, etc.: section 1.1, p. 41 Examples: 0.7 kg ×
103 g = 700 g 1 kg
.
To check that we have the conversion factor the right way up (103 rather then 1/103 ), we note that the smaller unit of grams has been compensated for by making the number larger. For units like m2 , kg/m3 , etc., we have to raise the conversion factor to the appropriate power: 3 3 10 mm mm3 3 3 4m × = 4 × 109 m × = 4 × 109 mm3 3 1m m Examples with solutions — p. 37, #6; p. 59, #10 Problems you can check at lightandmatter.com/area1checker.html — p. 37, #5; p. 37, #4; p. 37, #7; p. 59, #1; p. 60, #19
2. Reasoning about ratios and proportionalities The technique is introduced in section 1.2, p. 43, in the context of area and volume, but it applies more generally to any relationship in which one variable depends on another raised to some power. Example: When a car or truck travels over a road, there is wear and tear on the road surface, which incurs a cost. Studies show that the cost per kilometer of travel C is given by C = kw4
,
where w is the weight per axle and k is a constant. The weight per axle is abot 13 times higher for a semi-trailer than for my Honda Fit. How many times greater is the cost imposed on the federal government when the semi travels a given distance on an interstate freeway? First we convert the equation into a proportionality by throwing out k, which is the same for both vehicles: C ∝ w4 Next we convert this proportionality to a statement about ratios: 4 w1 C1 = ≈ 29, 000 C2 w2 Since the gas taxes paid by the trucker are nowhere near 29,000 times more than those I pay to drive my Fit the same distance, the federal government is effectively awarding a massive subsidy to the trucking company. Plus my Fit is cuter.
524
Chapter 20
Bounded waves
Examples with solutions — p. 59, #11; p. 59, #12; p. 60, #17; p. 115, #16; p. 116, #22; p. 242, #6; p. 266, #10; p. 267, #15; p. 269, #19; p. 295, #8; p. 295, #9 Problems you can check at lightandmatter.com/area1checker.html — p. 60, #16; p. 60, #18; p. 61, #23; p. 62, #24; p. 62, #25; p. 189, #7; p. 242, #8; p. 265, #5; p. 266, #8; p. 269, #21; p. 294, #4; p. 405, #2
3. Vector addition section 7.3, p. 198 Example: The Δr vector from San Diego to Los Angeles has magnitude 190 km and direction 129◦ counterclockwise from east. The one from LA to Las Vegas is 370 km at 38◦ counterclockwise from east. Find the distance and direction from San Diego to Las Vegas. Graphical addition is discussed on p. 198. Here we concentrate on analytic addition, which involves adding the x components to find the total x component, and similarly for y. The trig needed in order to find the components of the second leg (LA to Vegas) is laid out in figure e on p. 195 and explained in detail in example 3 on p. 195: Δx2 = (370 km) cos 38◦ = 292 km Δy2 = (370 km) sin 38◦ = 228 km (Since these are intermediate results, we keep an extra sig fig to avoid accumulating too much rounding error.) Once we understand the trig for one example, we don’t need to reinvent the wheel every time. The pattern is completely universal, provided that we first make sure to get the angle expressed according to the usual trig convention, counterclockwise from the x axis. Applying the pattern to the first leg, we have: Δx1 = (190 km) cos 129◦ = −120 km Δy1 = (190 km) sin 129◦ = 148 km For the vector directly from San Diego to Las Vegas, we have Δx = Δx1 + Δx2 = 172 km Δy = Δy1 + Δy2 = 376 km
.
The distance from San Diego to Las Vegas is found using the Pythagorean theorem, (172 km)2 + (376 km)2 = 410 km (rounded to two sig figs because it’s one of our final results). The direction is one of the two possible values of the inverse tangent tan−1 (Δy/Δx) = {65◦ , 245◦ }
.
Consulting a sketch shows that the first of these values is the correct one. Examples with solutions — p. 221, #8; p. 221, #9; p. 371, #8 Problems you can check at lightandmatter.com/area1checker.html — p. 204, #3; p. 204, #4; p. 220, #1; p. 220, #3; p. 223, #16; p. 265, #3; p. 270, #23; p. 370, #3
Problems
525
Hints for volume 1 Hints for chapter 10 Page 268, problem 17: If you try to calculate the two forces and subtract, your calculator will probably give a result of zero due to rounding. Instead, reason about the fractional amount by which the quantity 1/r2 will change. As a warm-up, you may wish to observe the percentage change in 1/r2 that results from changing r from 1 to 1.01. Hints for chapter 15 Page 408, problem 18: The first two parts can be done more easily by setting a = 1, since the value of a only changes the distance scale. One way to do part b is by graphing.
Solutions to selected problems for volume 1 Solutions for chapter 0 Page 37, problem 6:
134 mg ×
10−3 g 10−3 kg × = 1.34 × 10−4 kg 1 mg 1g
Page 38, problem 8: (a) Let’s do 10.0 g and 1000 g. The arithmetic mean is 505 grams. It comes out to be 0.505 kg, which is consistent. (b) The geometric mean comes out to be 100 g or 0.1 kg, which is consistent. (c) If we multiply meters by meters, we get square meters. Multiplying grams by grams should give square grams! This sounds strange, but it makes sense. Taking the square root of square grams (g2 ) gives grams again. (d) No. The superduper mean of two quantities with units of grams wouldn’t even be something with units of grams! Related to this shortcoming is the fact that the superduper mean would fail the kind of consistency test carried out in the first two parts of the problem. Page 38, problem 12: (a) They’re all defined in terms of the ratio of side of a triangle to another. For instance, the tangent is the length of the opposite side over the length of the adjacent side. Dividing meters by meters gives a unitless result, so the tangent, as well as the other trig functions, is unitless. (b) The tangent function gives a unitless result, so the units on the right-hand side had better cancel out. They do, because the top of the fraction has units of meters squared, and so does the bottom. Solutions for chapter 1 Page 59, problem 10: 2
1 mm ×
1 cm 10 mm
2
= 10−2 cm2
Page 59, problem 11: The bigger scope has a diameter that’s ten times greater. Area scales as the square of the linear
526
dimensions, so A ∝ d2 , or in the language of ratios A1 /A2 = (d1 /d2 )2 = 100. Its light-gathering power is a hundred times greater. Page 59, problem 12: Since they differ by two steps on the Richter scale, the energy of the bigger quake is 104 times greater. The wave forms a hemisphere, and the surface area of the hemisphere √ over which the 2 energy isspread is proportional to the square of its radius, A ∝ r , or r ∝ A, which means r1 /r2 = A1 /A2 . If the amount of vibration was the same, then the surface areas must be in the ratio A1 /A2 = 104 , which means that the ratio of the radii is 102 . Page 60, problem 17: The cone of mixed gin and vermouth is the same shape as the cone of vermouth, but its linear dimensions are doubled. Translating the proportionality V ∝ L3 into an equation about ratios, we have V1 /V2 = (L1 /L2 )3 = 8. Since the ratio of the whole thing to the vermouth is 8, the ratio of gin to vermouth is 7. Page 60, problem 19: The proportionality V ∝ L3 can be restated in terms of ratios as V1 /V2 = (L1 /L2 )3 = (1/10)3 = 1/1000, so scaling down the linear dimensions by a factor of 1/10 reduces the volume by 1/1000, to a milliliter. Page 61, problem 21: Let’s estimate the Great Wall’s volume, and then figure out how many bricks that would represent. The wall is famous because it covers pretty much all of China’s northern border, so let’s say it’s 1000 km long. From pictures, it looks like it’s about 10 m high and 10 m wide, so the total volume would be 106 m × 10 m × 10 m = 108 m3 . If a single brick has a volume of 1 liter, or 10−3 m3 , then this represents about 1011 bricks. If one person can lay 10 bricks in an hour (taking into account all the preparation, etc.), then this would be 1010 man-hours. Page 62, problem 26: Directly guessing the number of jelly beans would be like guessing volume directly. That would be a mistake. Instead, we start by estimating the linear dimensions, in units of beans. The contents of the jar look like they’re about 10 beans deep. Although the jar is a cylinder, its exact geometrical shape doesn’t really matter for the purposes of our order-of-magnitude estimate. Let’s pretend it’s a rectangular jar. The horizontal dimensions are also something like 10 beans, so it looks like the jar has about 10 × 10 × 10 or ∼ 103 beans inside. Solutions for chapter 2 Page 88, problem 4:
1 light-year = vΔt = 3 × 108 m/s (1 year) 365 days 24 hours 3600 s × × = 3 × 108 m/s (1 year) × 1 year 1 day 1 hour = 9.5 × 1015 m Page 88, problem 5: Velocity is relative, so having to lean tells you nothing about the train’s velocity. Fullerton is moving at a huge speed relative to Beijing, but that doesn’t produce any noticeable effect in
527
either city. The fact that you have to lean tells you that the train is changing its speed, but it doesn’t tell you what the train’s current speed is. Page 88, problem 7: To the person riding the moving bike, bug A is simply going in circles. The only difference between the motions of the two wheels is that one is traveling through space, but motion is relative, so this doesn’t have any effect on the bugs. It’s equally hard for each of them. Page 89, problem 10: In one second, the ship moves v meters to the east, and the person moves v meters north relative to the deck. Relative to the water, he traces the √ diagonal of a triangle whose length is given 2 2 1/2 by the Pythagorean theorem, (v + v ) = 2v. Relative to the water, he is moving at a 45-degree angle between north and east. Solutions for chapter 3 Page 115, problem 14:
Page 115, problem 15: Taking g to be 10 m/s2 , the bullet loses 10 m/s of speed every second, so it will take 10 s to come to a stop, and then another 10 s to come back down, for a total of 20 s. Page 115, problem 16: √ Δx = 21 at2 , so for a fixed value of Δx, we have t ∝ 1/ a. Translating this into the language of √ ratios gives tM /tE = aE /aM = 3 = 1.7. Page 115, problem 17: dx dt = 10 − 3t2 dv a= dt = −6t v=
= −18 m/s2 Page 116, problem 18: √ (a) Solving for Δx = 12 at2 for a, we find a = 2Δx/t2 = 5.51 m/s2 . (b) v = 2aΔx = 66.6 m/s. (c) The actual car’s final velocity is less than that of the idealized constant-acceleration car. If the real car and the idealized car covered the quarter mile in the same time but the real car was moving more slowly at the end than the idealized one, the real car must have been going faster than the idealized car at the beginning of the race. The real car apparently has a greater acceleration at the beginning, and less acceleration at the end. This make sense, because every car has some maximum speed, which is the speed beyond which it cannot accelerate.
528
Page 116, problem 19: Since the lines are at intervals of one m/s and one second, each box represents one meter. From t = 0 to t = 2 s, the area under the curve represents a positive Δx of 6 m. (The triangle has half the area of the 2 × 6 rectangle it fits inside.) After t = 2 s, the area above the curve represents negative Δx. To get −6 m worth of area, we need to go out to t = 6 s, at which point the triangle under the axis has a width of 4 s and a height of 3 m/s, for an area of 6 m (half of 3 × 4). Page 116, problem 20: (a) We choose a coordinate system with positive pointing to the right. Some people might expect that the ball would slow down once it was on the more gentle ramp. This may be true if there is significant friction, but Galileo’s experiments with inclined planes showed that when friction is negligible, a ball rolling on a ramp has constant acceleration, not constant speed. The speed stops increasing as quickly once the ball is on the more gentle slope, but it still keeps on increasing. The a-t graph can be drawn by inspecting the slope of the v-t graph.
(b) The ball will roll back down, so the second half of the motion is the same as in part a. In the first (rising) half of the motion, the velocity is negative, since the motion is in the opposite direction compared to the positive x axis. The acceleration is again found by inspecting the slope of the v-t graph.
Page 116, problem 21: This is a case where it’s probably easiest to draw the acceleration graph first. While the ball is in the air (bc, de, etc.), the only force acting on it is gravity, so it must have the same, constant acceleration during each hop. Choosing a coordinate system where the positive x axis points up, this becomes a negative acceleration (force in the opposite direction compared to the axis). During the short times between hops when the ball is in contact with the ground (cd, ef, etc.), it experiences a large acceleration, which turns around its velocity very rapidly. These short positive accelerations probably aren’t constant, but it’s hard to know how they’d really look. We just idealize them as constant accelerations. Similarly, the hand’s force on the ball during the time ab is probably not constant, but we can draw it that way, since we don’t know
529
how to draw it more realistically. Since our acceleration graph consists of constant-acceleration segments, the velocity graph must consist of line segments, and the position graph must consist of parabolas. On the x graph, I chose zero to be the height of the center of the ball above the floor when the ball is just lying on the floor. When the ball is touching the floor and compressed, as in interval cd, its center is below this level, so its x is negative.
Page 116, problem 22: We have vf2 = 2aΔx, so the distance is proportional to the square of the velocity. To get up to half the speed, the ball needs 1/4 the distance, i.e., L/4. Solutions for chapter 4 Page 143, problem 7: a = Δv/Δt, and also a = F/m, so Δv a mΔv = F (1000 kg)(50 m/s − 20 m/s) = 3000 N = 10 s
Δt =
Page 144, problem 10: (a) This is a measure of the box’s resistance to a change in its state of motion, so it measures the box’s mass. The experiment would come out the same in lunar gravity. (b) This is a measure of how much gravitational force it feels, so it’s a measure of weight. In lunar gravity, the box would make a softer sound when it hit. (c) As in part a, this is a measure of its resistance to a change in its state of motion: its mass. Gravity isn’t involved at all. Solutions for chapter 5 Page 173, problem 14: (a)
530
top spring’s rightward force on connector ...connector’s leftward force on top spring bottom spring’s rightward force on connector ...connector’s leftward force on bottom spring hand’s leftward force on connector ...connector’s rightward force on hand Looking at the three forces on the connector, we see that the hand’s force must be double the force of either spring. The value of x − xo is the same for both springs and for the arrangement as a whole, so the spring constant must be 2k. This corresponds to a stiffer spring (more force to produce the same extension). (b) Forces in which the left spring participates: hand’s leftward force on left spring ...left spring’s rightward force on hand right spring’s rightward force on left spring ...left spring’s leftward force on right spring Forces in which the right spring participates: left spring’s leftward force on right spring ...right spring’s rightward force on left spring wall’s rightward force on right spring ...right spring’s leftward force on wall Since the left spring isn’t accelerating, the total force on it must be zero, so the two forces acting on it must be equal in magnitude. The same applies to the two forces acting on the right spring. The forces between the two springs are connected by Newton’s third law, so all eight of these forces must be equal in magnitude. Since the value of x − xo for the whole setup is double what it is for either spring individually, the spring constant of the whole setup must be k/2, which corresponds to a less stiff spring. Page 173, problem 16: (a) Spring constants in parallel add, so the spring constant has to be proportional to the crosssectional area. Two springs in series give half the spring constant, three springs in series give 1/3, and so on, so the spring constant has to be inversely proportional to the length. Summarizing, we have k ∝ A/L. (b) With the Young’s modulus, we have k = (A/L)E.The spring constant has units of N/m, so the units of E would have to be N/m2 . Page 174, problem 18: (a) The swimmer’s acceleration is caused by the water’s force on the swimmer, and the swimmer makes a backward force on the water, which accelerates the water backward. (b) The club’s normal force on the ball accelerates the ball, and the ball makes a backward normal force on the club, which decelerates the club. (c) The bowstring’s normal force accelerates the arrow, and the arrow also makes a backward normal force on the string. This force on the string causes the string to accelerate less rapidly than it would if the bow’s force was the only one acting on it. (d) The tracks’ backward frictional force slows the locomotive down. The locomotive’s forward frictional force causes the whole planet earth to accelerate by a tiny amount, which is too small to measure because the earth’s mass is so great. Page 174, problem 20: The person’s normal force on the box is paired with the box’s normal force on the person. The
531
dirt’s frictional force on the box pairs with the box’s frictional force on the dirt. The earth’s gravitational force on the box matches the box’s gravitational force on the earth. Page 175, problem 26: (a) A liter of water has a mass of 1.0 kg. The mass is the same in all three locations. Mass indicates how much an object resists a change in its motion. It has nothing to do with gravity. (b) The term “weight” refers to the force of gravity on an object. The bottle’s weight on earth is FW = mg = 9.8 N. Its weight on the moon is about one sixth that value, and its weight in interstellar space is zero. Solutions for chapter 6 Page 188, problem 5: (a) The easiest strategy is to find the time spent aloft, and then find the range. The vertical motion and the horizontal motion are independent. The vertical motion has acceleration −g, and the cannonball spends enough time in the air to reverse its vertical velocity component completely, so we have Δvy = vyf − vyo = −2v sin θ
.
The time spent aloft is therefore Δt = Δvy /ay = 2v sin θ/g
.
During this time, the horizontal distance traveled is R = vx Δt = 2v 2 sin θ cos θ/g
.
(b) The range becomes zero at both θ = 0 and at θ = 90◦ . The θ = 0 case gives zero range because the ball hits the ground as soon as it leaves the mouth of the cannon. A 90-degree angle gives zero range because the cannonball has no horizontal motion. Solutions for chapter 8 Page 221, problem 8: We want to find out about the velocity vector vBG of the bullet relative to the ground, so we need to add Annie’s velocity relative to the ground vAG to the bullet’s velocity vector vBA relative to her. Letting the positive x axis be east and y north, we have vBA,x = (140 mi/hr) cos 45◦ = 100 mi/hr vBA,y = (140 mi/hr) sin 45◦ = 100 mi/hr and vAG,x = 0 vAG,y = 30 mi/hr
532
.
The bullet’s velocity relative to the ground therefore has components vBG,x = 100 mi/hrand vBG,y = 130 mi/hr
.
Its speed on impact with the animal is the magnitude of this vector |vBG | = (100 mi/hr)2 + (130 mi/hr)2 = 160 mi/hr (rounded off to 2 significant figures). Page 221, problem 9: Since its velocity vector is constant, it has zero acceleration, and the sum of the force vectors acting on it must be zero. There are three forces acting on the plane: thrust, lift, and gravity. We are given the first two, and if we can find the third we can infer its mass. The sum of the y components of the forces is zero, so 0 = Fthrust,y + Flif t,y + FW ,y = |Fthrust | sin θ + |Flif t | cos θ − mg
.
The mass is m = (|Fthrust | sin θ + |Flif t | cos θ)/g = 6.9 × 104 kg Page 221, problem 10: (a) Since the wagon has no acceleration, the total forces in both the x and y directions must be zero. There are three forces acting on the wagon: FT , FW , and the normal force from the ground, FN . If we pick a coordinate system with x being horizontal and y vertical, then the angles of these forces measured counterclockwise from the x axis are 90◦ − φ, 270◦ , and 90◦ + θ, respectively. We have Fx,total = |FT | cos(90◦ − φ) + |FW | cos(270◦ ) + |FN | cos(90◦ + θ) Fy,total = |FT | sin(90◦ − φ) + |FW | sin(270◦ ) + |FN | sin(90◦ + θ)
,
which simplifies to 0 = |FT | sin φ − |FN | sin θ 0 = |FT | cos φ − |FW | + |FN | cos θ. The normal force is a quantity that we are not given and do not wish to find, so we should choose it to eliminate. Solving the first equation for |FN | = (sin φ/ sin θ)|FT |, we eliminate |FN | from the second equation, 0 = |FT | cos φ − |FW | + |FT | sin φ cos θ/ sin θ and solve for |FT |, finding |FT | =
|FW | cos φ + sin φ cos θ/ sin θ
.
533
Multiplying both the top and the bottom of the fraction by sin θ, and using the trig identity for sin(θ + φ) gives the desired result, |FT | =
sin θ |FW | sin(θ + φ)
.
(b) The case of φ = 0, i.e., pulling straight up on the wagon, results in |FT | = |FW |: we simply support the wagon and it glides up the slope like a chair-lift on a ski slope. In the case of φ = 180◦ − θ, |FT | becomes infinite. Physically this is because we are pulling directly into the ground, so no amount of force will suffice. Page 222, problem 11: (a) If there was no friction, the angle of repose would be zero, so the coefficient of static friction, μs , will definitely matter. We also make up symbols θ, m and g for the angle of the slope, the mass of the object, and the acceleration of gravity. The forces form a triangle just like the one in example 5 on p. 213, but instead of a force applied by an external object, we have static friction, which is less than μs |FN |. As in that example, |Fs | = mg sin θ, and |Fs | < μs |FN |, so mg sin θ < μs |FN |
.
From the same triangle, we have |FN | = mg cos θ, so mg sin θ < μs mg cos θ
.
Rearranging, θ < tan−1 μs
.
(b) Both m and g canceled out, so the angle of repose would be the same on an asteroid. Solutions for chapter 9 Page 241, problem 5: Each cyclist has a radial acceleration of v 2 /r = 5 m/s2 . The tangential accelerations of cyclists A and B are 375 N/75 kg = 5 m/s2 .
Page 242, problem 6: (a) The inward normal force must be sufficient to produce circular motion, so |FN | = mv 2 /r
.
We are searching for the minimum speed, which is the speed at which the static friction force is just barely able to cancel out the downward gravitational force. The maximum force of static friction is |Fs | = μs |FN | ,
534
and this cancels the gravitational force, so |Fs | = mg
.
Solving these three equations for v gives v= (b) Greater by a factor of
√
gr μs
.
3.
Page 242, problem 7: The inward force must be supplied by the inward component of the normal force, |FN | sin θ = mv 2 /r
.
The upward component of the normal force must cancel the downward force of gravity, |FN | cos θ = mg. Eliminating |FN | and solving for θ, we find θ = tan
−1
v2 gr
.
Solutions for chapter 10 Page 266, problem 10: Newton’s law of gravity tells us that her weight will be 6000 times smaller because of the asteroid’s smaller mass, but 132 = 169 times greater because of its smaller radius. Putting these two factors together gives a reduction in weight by a factor of 6000/169, so her weight will be (400 N)(169)/(6000) = 11 N. Page 267, problem 11: Newton’s law of gravity says F = Gm1 m2 /r2 , and Newton’s second law says F = m2 a, so Gm1 m2 /r2 = m2 a. Since m2 cancels, a is independent of m2 . Page 267, problem 12: Newton’s second law gives F = mD aD
,
where F is Ida’s force on Dactyl. Using Newton’s universal law of gravity, F= GmI mD /r2 ,and the equation a = v 2 /r for circular motion, we find GmI mD /r2 = mD v 2 /r. Dactyl’s mass cancels out, giving
GmI /r2 = v 2 /r.
Dactyl’s velocity equals the circumference of its orbit divided by the time for one orbit: v = 2πr/T . Inserting this in the above equation and solving for mI , we find mI =
4π 2 r3 GT 2
,
535
so Ida’s density is ρ = mI /V =
4π 2 r3 GV T 2
.
Page 267, problem 15: Newton’s law of gravity depends on the inverse square of the distance, so if the two planets’ masses had been equal, then the factor of 0.83/0.059 = 14 in distance would have caused the force on planet c to be 142 = 2.0 × 102 times weaker. However, planet c’s mass is 3.0 times greater, so the force on it is only smaller by a factor of 2.0 × 102 /3.0 = 65. Page 268, problem 16: The reasoning is reminiscent of section 10.2. From Newton’s second law we have F = ma = mv 2 /r = m(2πr/T )2 /r = 4π 2 mr/T 2
,
and Newton’s law of gravity gives F = GM m/r2 , where M is the mass of the earth. Setting these expressions equal to each other, we have 4π 2 mr/T 2 = GM m/r2
,
which gives r=
GM T 2 4π 2
1/3
= 4.22 × 104 km
.
This is the distance from the center of the earth, so to find the altitude, we need to subtract the radius of the earth. The altitude is 3.58 × 104 km. Page 268, problem 17: Any fractional change in r results in double that amount of fractional change in 1/r2 . For example, raising r by 1% causes 1/r2 to go down by very nearly 2%. A 27-day orbit is 1/13.5 of a year, so the fractional change in 1/r2 is 2×
(1/13.5) cm 1 km × 5 = 4 × 10−12 5 3.84 × 10 km 10 cm
Page 269, problem 19: (a) The asteroid’s mass depends on the cube of its radius, and for a given mass the surface gravity depends on r−2 . The result is that surface gravity is directly proportional to radius. Half the gravity means half the radius, or one eighth the mass. (b) To agree with a, Earth’s mass would have to be 1/8 Jupiter’s. We assumed spherical shapes and equal density. Both planets are at least roughly spherical, so the only way out of the contradiction is if Jupiter’s density is significantly less than Earth’s. Solutions for chapter 11 Page 294, problem 7: A force is an interaction between two objects, so while the bullet is in the air, there is no force. There is only a force while the bullet is in contact with the book. There is energy the whole
536
time, and the total amount doesn’t change. The bullet has some kinetic energy, and transfers some of it to the book as heat, sound, and the energy required to tear a hole through the book. Page 295, problem 8: (a) The energy stored in the gasoline is being changed into heat via frictional heating, and also probably into sound and into energy of water waves. Note that the kinetic energy of the propeller and the boat are not changing, so they are not involved in the energy transformation. (b) The crusing speed would be greater by a factor of the cube root of 2, or about a 26% increase. Page 295, problem 9: We don’t have actual masses and velocities to plug in to the equation, but that’s OK. We just have to reason in terms of ratios and proportionalities. Kinetic energy is proportional to mass and to the square of velocity, so B’s kinetic energy equals (13.4 J)(3.77)/(2.34)2 = 9.23 J Page 295, problem 11: Room temperature is about 20◦ C. The fraction of the energy that actually goes into heating the water is (250 g)/(0.24 g· ◦ C/J) × (100◦ C − 20◦ C) = 0.53 (1.25 × 103 J/s) (126 s) So roughly half of the energy is wasted. The wasted energy might be in several forms: heating of the cup, heating of the oven itself, or leakage of microwaves from the oven. Solutions for chapter 12 Page 309, problem 5: Etotal,i = Etotal,f P Ei + heati = P Ef + KEf + heatf 1 mv 2 = P Ei − P Ef + heati − heatf 2 = −ΔP E − Δheat −ΔP E − Δheat v= 2 m = 6.4 m/s Page 310, problem 7: Let θ be the angle by which he has progressed around the pipe. Conservation of energy gives Etotal,i = Etotal,f P Ei = P Ef + KEf Let’s make P E = 0 at the top, so 1 . 0 = mgr(cos θ − 1) + mv 2 2 While he is still in contact with the pipe, the radial component of his acceleration is ar =
v2 r
,
537
and making use of the previous equation we find ar = 2g(1 − cos θ)
.
There are two forces on him, a normal force from the pipe and a downward gravitation force from the earth. At the moment when he loses contact with the pipe, the normal force is zero, so the radial component, mg cos θ, of the gravitational force must equal mar , mg cos θ = 2mg(1 − cos θ)
,
which gives cos θ =
2 3
.
The amount by which he has dropped is r(1 − cos θ), which equals r/3 at this moment. Page 310, problem 9: (a) Example: As one child goes up on one side of a see-saw, another child on the other side comes down. (b) Example: A pool ball hits another pool ball, and transfers some KE. Page 310, problem 11: Suppose the river is 1 m deep, 100 m wide, and flows at a speed of 10 m/s, and that the falls are 100 m tall. In 1 second, the volume of water flowing over the falls is 103 m3 , with a mass of 106 kg. The potential energy released in one second is (106 kg)(g)(100 m) = 109 J, so the power is 109 W. A typical household might have 10 hundred-watt applicances turned on at any given time, so it consumes about 103 watts on the average. The plant could supply a about million households with electricity. Solutions for chapter 13 Page 339, problem 18: No. Work describes how energy was transferred by some process. It isn’t a measurable property of a system. Solutions for chapter 14 Page 371, problem 8: Let m be the mass of the little puck and M = 2.3m be the mass of the big one. All we need to do is find the direction of the total momentum vector before the collision, because the total momentum vector is the same after the collision. Given the two components of the momentum vector px = M v and py = mv, the direction of the vector is tan−1 (py /px ) = 23◦ counterclockwise from the big puck’s original direction of motion. Page 372, problem 11: Momentum is a vector. The total momentum of the molecules is always zero, since the momenta in different directions cancal out on the average. Cooling changes individual molecular momenta, but not the total. Page 372, problem 14: By conservation of momentum, the total momenta of the pieces after the explosion is the same as the momentum of the firework before the explosion. However, there is no law of conservation of kinetic energy, only a law of conservation of energy. The chemical energy in the gunpowder is converted into heat and kinetic energy when it explodes. All we can say about the kinetic energy of the pieces is that their total is greater than the kinetic energy before the explosion.
538
Page 372, problem 15: (a) Particle i had velocity vi in the center-of-mass frame, and has velocity vi + u in the new frame. The total kinetic energy is 1 m1 (v1 + u)2 + . . . 2
,
where “. . . ” indicates that the sum continues for all the particles. Rewriting this in terms of the vector dot product, we have 1 1 m1 (v1 + u) · (v1 + u) + . . . = m1 (v1 · v1 + 2u · v1 + u · u) + . . . 2 2
.
When we add up all the terms like the first one, we get Kcm . Adding up all the terms like the third one, we get M |u|2 /2. The terms like the second term cancel out: m1 u · v1 + . . . = u · (m1 v1 + . . .)
,
where the sum in brackets equals the total momentum in the center-of-mass frame, which is zero by definition. (b) Changing frames of reference doesn’t change the distances between the particles, so the potential energies are all unaffected by the change of frames of reference. Suppose that in a given frame of reference, frame 1, energy is conserved in some process: the initial and final energies add up to be the same. First let’s transform to the center-of-mass frame. The potential energies are unaffected by the transformation, and the total kinetic energy is simply reduced by the quantity M |u1 |2 /2, where u1 is the velocity of frame 1 relative to the center of mass. Subtracting the same constant from the initial and final energies still leaves them equal. Now we transform to frame 2. Again, the effect is simply to change the initial and final energies by adding the same constant. Page 373, problem 16: A conservation law is about addition: it says that when you add up a certain thing, the total always stays the same. Funkosity would violate the additive nature of conservation laws, because a two-kilogram mass would have twice as much funkosity as a pair of one-kilogram masses moving at the same speed. Solutions for chapter 15 Page 408, problem 20: The pliers are not moving, so their angular momentum remains constant at zero, and the total torque on them must be zero. Not only that, but each half of the pliers must have zero total torque on it. This tells us that the magnitude of the torque at one end must be the same as that at the other end. The distance from the axis to the nut is about 2.5 cm, and the distance from the axis to the centers of the palm and fingers are about 8 cm. The angles are close enough to 90◦ that we can pretend they’re 90 degrees, considering the rough nature of the other assumptions and measurements. The result is (300 N)(2.5 cm) = (F )(8 cm), or F = 90 N. Page 409, problem 28: The foot of the rod is moving in a circle relative to the center of the rod, with speed v = πb/T , and acceleration v 2 /(b/2) = (π 2 /8)g. This acceleration is initially upward, and is greater in magnitude than g, so the foot of the rod will lift off without dragging. We could also worry about whether the foot of the rod would make contact with the floor again before the rod finishes up flat on its back. This is a question that can be settled by graphing, or simply by
539
inspection of figure m on page 387. The key here is that the two parts of the acceleration are both independent of m and b, so the result is univeral, and it does suffice to check a graph in a single example. In practical terms, this tells us something about how difficult the trick is to do. Because π 2 /8 = 1.23 isn’t much greater than unity, a hit that is just a little too weak (by a factor of 1.231/2 = 1.11) will cause a fairly obvious qualitative change in the results. This is easily observed if you try it a few times with a pencil.
Answers to self-checks for volume 1 Answers to self-checks for chapter 0 Page 17, self-check A: If only he has the special powers, then his results can never be reproduced. Page 19, self-check B: They would have had to weigh the rays, or check for a loss of weight in the object from which they were have emitted. (For technical reasons, this was not a measurement they could actually do, hence the opportunity for disagreement.) Page 25, self-check C: A dictionary might define “strong” as “possessing powerful muscles,” but that’s not an operational definition, because it doesn’t say how to measure strength numerically. One possible operational definition would be the number of pounds a person can bench press. Page 29, self-check D: A microsecond is 1000 times longer than a nanosecond, so it would seem like 1000 seconds, or about 20 minutes. Page 30, self-check E: Exponents have to do with multiplication, not addition. The first line should be 100 times longer than the second, not just twice as long. Page 33, self-check F: The various estimates differ by 5 to 10 million. The CIA’s estimate includes a ridiculous number of gratuitous significant figures. Does the CIA understand that every day, people in are born in, die in, immigrate to, and emigrate from Nigeria? Page 33, self-check G: (1) 4; (2) 2; (3) 2 Answers to self-checks for chapter 1 Page 42, self-check A: 1 yd2 × (3 ft/1 yd)2 = 9 ft2 1 yd3 × (3 ft/1 yd)3 = 27 ft3 Page 48, self-check B: C1 /C2 = (w1 /w2 )4 Answers to self-checks for chapter 2 Page 71, self-check A: Coasting on a bike and coasting on skates give one-dimensional center-of-mass motion, but running and pedaling require moving body parts up and down, which makes the center of mass move up and down. The only example of rigid-body motion is coasting on skates. (Coasting on
540
a bike is not rigid-body motion, because the wheels twist.) Page 71, self-check B: By shifting his weight around, he can cause the center of mass not to coincide with the geometric center of the wheel. Page 72, self-check C: (1) a point in time; (2) time in the abstract sense; (3) a time interval Page 73, self-check D: Zero, because the “after” and “before” values of x are the same. Page 80, self-check E: (1) The effect only occurs during blastoff, when their velocity is changing. Once the rocket engines stop firing, their velocity stops changing, and they no longer feel any effect. (2) It is only an observable effect of your motion relative to the air. Answers to self-checks for chapter 3 Page 93, self-check A: Its speed increases at a steady rate, so in the next second it will travel 19 cm. Answers to self-checks for chapter 4 Page 134, self-check A: (1) The case of ρ = 0 represents an object falling in a vacuum, i.e., there is no density of air. The terminal velocity would be infinite. Physically, we know that an object falling in a vacuum would never stop speeding up, since there would be no force of air friction to cancel the force of gravity. (2) The 4-cm ball would have a mass that was greater by a factor of 4 × 4 × 4, but its cross-sectional area would be greater by a factor of 4 × 4. Its terminal velocity would be greater by a factor of 43 /42 = 2. (3) It isn’t of any general importance. It’s just an example of one physical situation. You should not memorize it. Page 137, self-check B: (1) This is motion, not force. (2) This is a description of how the sub is able to get the water to produce a forward force on it. (3) The sub runs out of energy, not force. Answers to self-checks for chapter 5 Page 149, self-check A: The sprinter pushes backward against the ground, and by Newton’s third law, the ground pushes forward on her. (Later in the race, she is no longer accelerating, but the ground’s forward force is needed in order to cancel out the backward forces, such as air friction.) Page 157, self-check B: (1) It’s kinetic friction, because her uniform is sliding over the dirt. (2) It’s static friction, because even though the two surfaces are moving relative to the landscape, they’re not slipping over each other. (3) Only kinetic friction creates heat, as when you rub your hands together. If you move your hands up and down together without sliding them across each other, no heat is produced by the static friction. Page 157, self-check C: By the POFOSTITO mnemonic, we know that each of the bird’s forces on the trunk will be of the same type as the corresponding force of the tree on the bird, but in the opposite direction. The bird’s feet make a normal force on the tree that is to the right and a static frictional force
541
that is downward. Page 158, self-check D: Frictionless ice can certainly make a normal force, since otherwise a hockey puck would sink into the ice. Friction is not possible without a normal force, however: we can see this from the equation, or from common sense, e.g., while sliding down a rope you do not get any friction unless you grip the rope. Page 159, self-check E: (1) Normal forces are always perpendicular to the surface of contact, which means right or left in this figure. Normal forces are repulsive, so the cliff’s force on the feet is to the right, i.e., away from the cliff. (2) Frictional forces are always parallel to the surface of contact, which means right or left in this figure. Static frictional forces are in the direction that would tend to keep the surfaces from slipping over each other. If the wheel was going to slip, its surface would be moving to the left, so the static frictional force on the wheel must be in the direction that would prevent this, i.e., to the right. This makes sense, because it is the static frictional force that accelerates the dragster. (3) Normal forces are always perpendicular to the surface of contact. In this diagram, that means either up and to the left or down and to the right. Normal forces are repulsive, so the ball is pushing the bat away from itself. Therefore the ball’s force is down and to the right on this diagram. Answers to self-checks for chapter 6 Page 181, self-check A: The wind increases the ball’s overall speed. If you think about it in terms of overall speed, it’s not so obvious that the increased speed is exactly sufficient to compensate for the greater distance. However, it becomes much simpler if you think about the forward motion and the sideways motion as two separate things. Suppose the ball is initially moving at one meter per second. Even if it picks up some sideways motion from the wind, it’s still getting closer to the wall by one meter every second. Answers to self-checks for chapter 7 Page 193, self-check A: v = Δr/Δt Page 194, self-check B:
Page 199, self-check C: A − B is equivalent to A + (−B), which can be calculated graphically by reversing B to form −B, and then adding it to A. Answers to self-checks for chapter 8 Page 210, self-check A: (1) It is speeding up, because the final velocity vector has the greater magnitude. (2) The result would be zero, which would make sense. (3) Speeding up produced a Δv vector in the same direction as the motion. Slowing down would have given a Δv that pointed backward. Page 211, self-check B:
542
As we have already seen, the projectile has ax = 0 and ay = −g, so the acceleration vector is pointing straight down. Answers to self-checks for chapter 9 Page 231, self-check A: (1) Uniform. They have the same motion as the drum itself, which is rotating as one solid piece. No part of the drum can be rotating at a different speed from any other part. (2) Nonuniform. Gravity speeds it up on the way down and slows it down on the way up. Answers to self-checks for chapter 10 Page 248, self-check A: It would just stay where it was. Plugging v = 0 into eq. [1] would give F = 0, so it would not accelerate from rest, and would never fall into the sun. No astronomer had ever observed an object that did that! Page 249, self-check B: F ∝ mr/T 2 ∝ mr/(r3/2 )2 ∝ mr/r3 = m/r2 Page 252, self-check C: The equal-area law makes equally good sense in the case of a hyperbolic orbit (and observations verify it). The elliptical orbit law had to be generalized by Newton to include hyperbolas. The law of periods doesn’t make sense in the case of a hyperbolic orbit, because a hyperbola never closes back on itself, so the motion never repeats. Page 257, self-check D: Above you there is a small part of the shell, comprising only a tiny fraction of the earth’s mass. This part pulls you up, while the whole remainder of the shell pulls you down. However, the part above you is extremely close, so it makes sense that its force on you would be far out of proportion to its small mass. Answers to self-checks for chapter 11 Page 286, self-check A: (1) A spring-loaded toy gun can cause a bullet to move, so the spring is capable of storing energy and then converting it into kinetic energy. (2) The amount of energy stored in the spring relates to the amount of compression, which can be measured with a ruler. Answers to self-checks for chapter 12 Page 306, self-check A: Both balls start from the same height and end at the same height, so they have the same Δy. This implies that their losses in potential energy are the same, so they must both have gained the same amount of kinetic energy. Answers to self-checks for chapter 13 Page 314, self-check A: Work is defined as the transfer of energy, so like energy it is a scalar with units of joules. Page 318, self-check B: Whenever energy is transferred out of the spring, the same amount has to be transferred into the ball, and vice versa. As the spring compresses, the ball is doing positive work on the spring (giving up its KE and transferring energy into the spring as PE), and as it decompresses the
543
ball is doing negative work (extracting energy). Page 321, self-check C: (a) No. The pack is moving at constant velocity, so its kinetic energy is staying the same. It is only moving horizontally, so its gravitational potential energy is also staying the same. No energy transfer is occurring. (b) No. The horse’s upward force on the pack forms a 90-degree angle with the direction of motion, so cos θ = 0, and no work is done. Page 324, self-check D: Only in (a) can we use F d to calculate work. In (b) and (c), the force is changing as the distance changes. Answers to self-checks for chapter 15 Page 390, self-check A: 1, 2, and 4 all have the same sigm, because they are trying to twist the wrench clockwise. The sign of torque 3 is opposite to the signs of the others. The magnitude of torque 3 is the greatest, since it has a large r, and the force is nearly all perpendicular to the wrench. Torques 1 and 2 are the same because they have the same values of r and F⊥ . Torque 4 is the smallest, due to its small r. Answers to self-checks for chapter 16 Page 415, self-check A: Solids can exert shear forces. A solid could be in an equilibrium in which the shear forces were canceling the forces due to unequal pressures on the sides of the cube. Answers to self-checks for chapter 18 Page 452, self-check A: The horizontal axis is a time axis, and the period of the vibrations is independent of amplitude. Shrinking the amplitude does not make the cyles and faster. Page 453, self-check B: Energy is proportional to the square of the amplitude, so its energy is four times smaller after every cycle. It loses three quarters of its energy with each cycle. Page 459, self-check C: She should tap the wine glasses she finds in the store and look for one with a high Q, i.e., one whose vibrations die out very slowly. The one with the highest Q will have the highest-amplitude response to her driving force, making it more likely to break. Answers to self-checks for chapter 19 Page 477, self-check A: The leading edge is moving up, the trailing edge is moving down, and the top of the hump is motionless for one instant. Page 484, self-check B: (a) It doesn’t have w or h in it. (b) Inertia is measured by μ, tightness by T . (c) Inertia would be measured by the density of the metal, tightness by its resistance to compression. Lead is more dense than aluminum, and this would tend to make the speed of the waves lower in lead. Lead is also softer, so it probably has less resistance to compression, and we would expect this to provide an additional effect in the same direction. Compressional waves will definitely be slower in lead than in aluminum.
544
Answers to self-checks for chapter 20 Page 503, self-check A: The energy of a wave is usually proportional to the square of its amplitude. Squaring a negative number gives a positive result, so the energy is the same. Page 503, self-check B: A substance is invisible to sonar if the speed of sound waves in it is the same as in water. Reflections only occur at boundaries between media in which the wave speed is different. Page 505, self-check C: No. A material object that loses kinetic energy slows down, but a wave is not a material object. The velocity of a wave ordinarily only depends on the medium, not the amplitude. The speed of a soft sound, for example, is the same as the speed of a loud sound. Page 514, self-check D: 1. No. To get the best possible interference, the thickness of the coating must be such that the second reflected wave train lags behind the first by an integer number of wavelengths. Optimal performance can therefore only be produced for one specific color of light. The typical greenish color of the coatings shows that they do the worst job for green light. 2. Light can be reflected either from the outer surface of the film or from the inner surface, and there can be either constructive or destructive interference between the two reflections. We see a pattern that varies across the surface because its thickness isn’t constant. We see rainbow colors because the condition for destructive or constructive interference depends on wavelength. White light is a mixture of all the colors of the rainbow, and at a particular place on the soap bubble, part of that mixture, say red, may be reflected strongly, while another part, blue for example, is almost entirely transmitted. Page 515, self-check E: The period is the time required to travel a distance 2L at speed v, T = 2L/v. The frequency is f = 1/T = v/2L. Page 520, self-check F: . Three quarters of a wavelength fit in the tube, so the The wave pattern will look like this: wavelength is three times shorter than that of the lowest-frequency mode, in which one quarter of a wave fits. Since the wavelength is smaller by a factor of three, the frequency is three times higher. Instead of fo , 2fo , 3fo , 4fo , . . ., the pattern of wave frequencies of this air column goes fo , 3fo , 5fo , 7fo , . . .
Answers for volume 1 Answers for chapter 1 Page 61, problem 23: Check: The actual number of species of lupine occurring in the San Gabriels is 22. You should find that your answer comes out in the same ballpark as this figure, but not exactly the same, of course, because the scaling rule is only a generalization. Answers for chapter 16 Page 431, problem 10: (a) ∼ 2 − 10% (b) 5% (c) The high end for the body’s actual efficiency is higher than the limit imposed by the laws of thermodynamics. However, the high end of the 1-5 watt range quoted in
545
the problem probably includes large people who aren’t just lying around. Still, it’s impressive that the human body comes so close to the thermodynamic limit. Answers for chapter 20 Page 522, problem 3: Check: The actual length of a flute is about 66 cm.
546
Photo credits for volume 1 Except as specifically noted below or in a parenthetical credit in the caption of a figure, all the illustrations in this book are under my own copyright, and are copyleft licensed under the same license as the rest of the book. In some cases it’s clear from the date that the figure is public domain, but I don’t know the name of the artist or photographer; I would be grateful to anyone who could help me to give proper credit. I have assumed that images that come from U.S. government web pages are copyright-free, since products of federal agencies fall into the public domain. I’ve included some public-domain paintings; photographic reproductions of them are not copyrightable in the U.S. (Bridgeman Art Library, Ltd. v. Corel Corp., 36 F. Supp. 2d 191, S.D.N.Y. 1999). When “PSSC Physics” is given as a credit, it indicates that the figure is from the first edition of the textbook entitled Physics, by the Physical Science Study Committee. The early editions of these books never had their copyrights renewed, and are now therefore in the public domain. There is also a blanket permission given in the later PSSC College Physics edition, which states on the copyright page that “The materials taken from the original and second editions and the Advanced Topics of PSSC PHYSICS included in this text will be available to all publishers for use in English after December 31, 1970, and in translations after December 31, 1975.” Credits to Millikan and Gale refer to the textbooks Practical Physics (1920) and Elements of Physics (1927). Both are public domain. (The 1927 version did not have its copyright renewed.) Since it is possible that some of the illustrations in the 1927 version had their copyrights renewed and are still under copyright, I have only used them when it was clear that they were originally taken from public domain sources. In a few cases, I have made use of images under the fair use doctrine. However, I am not a lawyer, and the laws on fair use are vague, so you should not assume that it’s legal for you to use these images. In particular, fair use law may give you less leeway than it gives me, because I’m using the images for educational purposes, and giving the book away for free. Likewise, if the photo credit says “courtesy of ...,” that means the copyright owner gave me permission to use it, but that doesn’t mean you have permission to use it. Cover Spider: Wikimedia Commons user Opoterser, CC-BY-SA. Contents See photo credits below, referring to the places where the images appear in the main text. 15 Mars Climate Orbiter: NASA/JPL/CIT. 26 Standard kilogram: Bo Bengtsen, GFDL licensed. Further retouching by Wikipedia user Greg L and by B. Crowell. 41 Bee: Wikipedia user Fir0002, CC-BY-SA licensed. 56 Jar of jellybeans: Flickr user cbgrfx123, CC-BY-SA licensed. 57 Amphicoelias: Wikimedia commons users Dinoguy2, Niczar, ArthurWeasley, Steveoc 86, Dropzink, and Piotr Jaworski, CC-BY-SA licensed. 61 E. Coli bacteria: Eric Erbe, digital colorization by Christopher Pooley, both of USDA, ARS, EMU. A public-domain product of the Agricultural Research Service.. 62 Galaxy: ESO, CC-BY license. 62 Stacked oranges: Wikimedia Commons user J.J. Harrison, CC-BY-SA license. 68 Trapeze: Calvert Litho. Co., Detroit, ca. 1890. 71 High jumper: Dunia Young. 71 Gymnastics wheel: Copyright Hans Genten, Aachen, Germany. “The copyright holder of this file allows anyone to use it for any purpose, provided that this remark is referenced or copied.”. 79 Jets over New York: U.S. Air Force, Tech. Sgt. Sean Mateo White, public domain work of the U.S. Government. 79 Aristotle: Francesco Hayez, 1811. 79 Angel Stadium: U.S. Marine Corps, Staff Sgt. Chad McMeen, public domain work of the U.S. Government. 79 Shanghai: Agnieszka Bojczuk, CC-BY-SA. 79 Rocket sled: U.S. Air Force, public domain work of the U.S. Government. 90 Tuna’s migration: Modified from a figure in Block et al. 91 Galileo’s trial: Cristiano Banti (1857). 97 Gravity map: US Navy, European Space Agency, D. Sandwell, and W. Smith. 118 Astronaut jumping: NASA. 118 Flea jumping: Burrows and Sutton, “Biomechanics of jumping in the flea,” J. Exp. Biology 214 (2011) 836. Used under the U.S. fair use exception to copyright. 119 Sprinter: Drawn from a photo provided by the German Federal Archives under a CC-BY-SA license. 119 Dinosaur: Redrawn from art by Wikimedia Commons user Dinoguy2, CCBY-SA license. 123 Newton: Godfrey Kneller, 1702. 148 Space shuttle launch: NASA. 149 Swimmer: Karen Blaha, CC-BY-SA licensed. 157 Partridge: Redrawn from K.P. Dial, “Wing-Assisted Incline Running and the Evolution of Flight,” Science 299 (2003) 402. 159 Locomotive: Locomotive Cyclopedia of American Practice, 1922, public domain. 160 Crop duster: NASA Langley Research Center, public domain. 160 Wind tunnel: Jeff Caplan/NASA Langley, public domain. 160 Turbulence: C. Fukushima and J. Westerweel, Technical University of Delft, The Netherlands, CC-BY license. 160 Series of vortices: Wikimedia Commons user Onera, CC-BY license. 161 Hummer: Wikimedia commons user Bull-Doser, public domain. 161 Golf ball: Wikimedia Commons user Paolo Neo, CC-BY-SA license. 161 Shark: Wikimedia Commons user Pterantula, CC-BY-SA license. 161 Prius: Wikimedia commons user IFCAR, public domain. 162 Dog: From a photo by Wikimedia Commons user Ron Armstrong, CC-BY licensed. 165 Golden Gate Bridge:
547
View more...
Comments
