Answer This Problem
Trigonometric Identities
sin(9 ± 0) = sin 0 cos 4) ± cos 0 sin 4, cos(0 f 0) = cos a cos 0 sin 0 sin 0
cos 0 cos (/’ = l[cos(0 + 0) + cos(8 – 0)] sine sin 0 = 4.[coso – coo + 4,)]
sin e cos 0 = [sin(e + 0) + sin(O – 0)]
cos26. 4[1+ cos 20] sin2 8 = [1 – cos 20] 2
0 + 0 0 — 0 +0 0 0 — 0 cos 0 + cos 0 = 2 cos cos cos 0 cos 0 = 2 sin sin
2 2 2 2
4) 0 sin 8 ± sin 0= 2 sin 0 ± cos 0 + 2 2
cost 0 + sin2 0 = 1 sect 8 – tan2 8 = 1
ei8 = cos 0 + i sin 0 [Euler’s relation]
cos 0 = + e-`0 ) sin 0 = – e-i°)
Hyperbolic Functions
cosh z = (ez + e’) = cos(iz) sinh z = (ez – e-z) = sin(iz)
tanh z = sinh z sechz = 1
cosh z cosh z
cosh2 z — sinh2 z = 1 sech2z + tanh2 z = 1
Series Expansions
f (z) = f (a) + f'(a)(z – a) + Z, f”(a)(z – a)2 + 3,f'(a)(z – a)3 + • • [Taylor’s series] ez = 1 + z + + +… ln(1 + z) = z – -1z2 + Az3 – • • [ Izi < 1]
cos z = 1 – I,z2+z— • • • sin z = z – Iz3 + -L! z5 — • 3! 5 COSh Z = 1 + -1;22 + +1,24 + • • • sinh z = z + fl,z 3 + + • • •
tan z = z + iz3 + Az 5 +•••[ lz I < ir/2] tanh z = z – + 1Z5- z5 – • • • [ I z I < 7/2]
(1 + = 1 + nz + 2!
n(n – 1) z
2 + • [ IZI < 1] [binomial series]
f dx J x2 –
= arccosh x
J x x2 -1 dx = arccos(1/x) dx
+ x2)3/2 ± x2)1/2
Some Derivatives
— tan z= sect z – d
tanh z= sech2 z dz dz
d . — sink z = cosh z —
d cosh z = sinh z
dz dz
Some Integrals
I dx arctan x
f dx = arctanh x J + x2 J – x2
dx f dx = arcsin x = arcsinh x J 1-x2 J x2
f tan x dx = — In cos x f tanh x dx = 1n cosh x dx ( f x dx
J x + x2 = +x x) J 1 + x2 = In(1 + x 2)
f x dx = + x 2 J x2
dx = arcsin(li) — (1 — x) J
f ln(x)dx = xin(x) — x
[ 1 dx
Jo -11–x 24 — mx 2 = K (m), complete elliptic integral of first kind
Classical mechanics
John r. Taylor
2005
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LIBRARY OF CONGRESS CATALOGING-IN-PUBLICATION DATA
Taylor, John R. (John Robert), 1939– Classical mechanics / John R. Taylor.
p. cm. Includes bibliographical references ISBN 1-891389-22-X (acid-free paper) 1. Mechanics. I. Title. QC125.2.T39 2004 531—dc22
2004054971
Printed in the United States of America 10 9 8 7 6 5 4 3 2 1
0,TE K OA)
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Contents
Preface xi
Ess
CHAPTER 1 Newton’s Laws of Motion 3
1.1 Classical Mechanics 3
1.2 Space and Time 4
1.3 Mass and Force 9
1.4 Newton’s First and Second Laws; Inertial Frames 13
1.5 The Third Law and Conservation of Momentum 17
1.6 Newton’s Second Law in Cartesian Coordinates 23
1.7 Two-Dimensional Polar Coordinates 26
Principal Definitions and Equations of Chapter 1 33
Problems for Chapter 1 34
CHAPTER 2 Projectiles and Charged Particles 43
2.1 Air Resistance 43
2.2 Linear Air Resistance 46
2.3 Trajectory and Range in a Linear Medium 54
2.4 Quadratic Air Resistance 57
2.5 Motion of a Charge in a Uniform Magnetic Field 65
2.6 Complex Exponentials 68
2.7 Solution for the Charge in a B Field 70
Principal Definitions and Equations of Chapter 2 71
Problems for Chapter 2 72
vi Contents
CHAPTER 3 Momentum and Angular Momentum 83
3.1 Conservation of Momentum 83
3.2 Rockets 85
3.3 The Center of Mass 87
3.4 Angular Momentum for a Single Particle 90
3.5 Angular Momentum for Several Particles 93
Principal Definitions and Equations of Chapter 3 98
Problems for Chapter 3 99
CHAPTER 4 Energy 105
4.1 Kinetic Energy and Work 105
4.2 Potential Energy and Conservative Forces 109
4.3 Force as the Gradient of Potential Energy 116
4.4 The Second Condition that F be Conservative 118
4.5 Time-Dependent Potential Energy 121
4.6 Energy for Linear One-Dimensional Systems 123
4.7 Curvilinear One-Dimensional Systems 129
4.8 Central Forces 133
4.9 Energy of Interaction of Two Particles 138
4.10 The Energy of a Multiparticle System 144
Principal Definitions and Equations of Chapter 4 148
Problems for Chapter 4 150
CHAPTER 5 Oscillations 161
5.1 Hooke’s Law 161
5.2 Simple Harmonic Motion 163
5.3 Two-Dimensional Oscillators 170
5.4 Damped Oscillations 173
5.5 Driven Damped Oscillations 179
5.6 Resonance 187
5.7 Fourier Series* 192
5.8 Fourier Series Solution for the Driven Oscillator* 197
5.9 The RMS Displacement; Parseval’s Theorem’ 203
Principal Definitions and Equations of Chapter 5 205
Problems for Chapter 5 207
* Sections marked with an asterisk could be omitted on a first reading.
Contents vii
CHAPTER 6 Calculus of Variations 215
6.1 Two Examples 216
6.2 The Euler—Lagrange Equation 218
6.3 Applications of the Euler—Lagrange Equation 221
6.4 More than Two Variables 226
Principal Definitions and Equations of Chapter 6 230
Problems for Chapter 6 230
CHAPTER 7 Lagrange’s Equations 237
7.1 Lagrange’s Equations for Unconstrained Motion 238
7.2 Constrained Systems; an Example 245
7.3 Constrained Systems in General 247
7.4 Proof of Lagrange’s Equations with Constraints 250
7.5 Examples of Lagrange’s Equations 254
7.6 Generalized Momenta and Ignorable Coordinates 266
7.7 Conclusion 267
7.8 More about Conservation Laws * 268
7.9 Lagrange’s Equations for Magnetic Forces * 272
7.10 Lagrange Multipliers and Constraint Forces * 275
Principal Definitions and Equations of Chapter 7 280
Problems for Chapter 7 281
CHAPTER 8 Two-Body Central-Force Problems 293
8.1 The Problem 293
8.2 CM and Relative Coordinates; Reduced Mass 295
8.3 The Equations of Motion 297
8.4 The Equivalent One-Dimensional Problem 300
8.5 The Equation of the Orbit 305
8.6 The Kepler Orbits 308
8.7 The Unbounded Kepler Orbits 313
8.8 Changes of Orbit 315
Principal Definitions and Equations of Chapter 8 319
Problems for Chapter 8 320
CHAPTER 9 Mechanics in Noninertial Frames 327
9.1 Acceleration without Rotation 327
9.2 The Tides 330
9.3 The Angular Velocity Vector 336
9.4 Time Derivatives in a Rotating Frame 339
viii Contents
9.5 Newton’s Second Law in a Rotating Frame 342
9.6 The Centrifugal Force 344
9.7 The Coriolis Force 348
9.8 Free Fall and the Coriolis Force 351
9.9 The Foucault Pendulum 354
9.10 Coriolis Force and Coriolis Acceleration 358
Principal Definitions and Equations of Chapter 9 359
Problems for Chapter 9 360
CHAPTER 10 Rotational Motion of Rigid Bodies 367
10.1 Properties of the Center of Mass 367
10.2 Rotation about a Fixed Axis 372
10.3 Rotation about Any Axis; the Inertia Tensor 378
10.4 Principal Axes of Inertia 387
10.5 Finding the Principal Axes; Eigenvalue Equations 389
10.6 Precession of a Top due to a Weak Torque 392
10.7 Euler’s Equations 394
10.8 Euler’s Equations with Zero Torque 397
10.9 Euler Angles * 401
10.10 Motion of a Spinning Top* 403
Principal Definitions and Equations of Chapter 10 407
Problems for Chapter 10 408
CHAPTER 11 Coupled Oscillators and Normal Modes 417
11.1 Two Masses and Three Springs 417
11.2 Identical Springs and Equal Masses 421
11.3 Two Weakly Coupled Oscillators 426
11.4 Lagrangian Approach: The Double Pendulum 430
11.5 The General Case 436
11.6 Three Coupled Pendulums 441
11.7 Normal Coordinates * 444
Principal Definitions and Equations of Chapter 11 447
Problems for Chapter 11 448
CHAPTER 12 Nonlinear Mechanics and Chaos 457
12.1 Linearity and Nonlinearity 458
12.2 The Driven Damped Pendulum DDP 462
12.3 Some Expected Features of the DDP 463
Contents ix
12.4 The DDP: Approach to Chaos 467
12.5 Chaos and Sensitivity to Initial Conditions 476
12.6 Bifurcation Diagrams 483
12.7 State-Space Orbits 487
12.8 Poincare Sections 495
12.9 The Logistic Map 498
Principal Definitions and Equations of Chapter 12 513
Problems for Chapter 12 514
CHAPTER 13 Hamiltonian Mechanics 521
13.1 The Basic Variables 522
13.2 Hamilton’s Equations for One-Dimensional Systems 524
13.3 Hamilton’s Equations in Several Dimensions 528
13.4 Ignorable Coordinates 535
13.5 Lagrange’s Equations vs. Hamilton’s Equations 536
13.6 Phase-Space Orbits 538
13.7 Liouville’s Theorem* 543
Principal Definitions and Equations of Chapter 13 550
Problems for Chapter 13 550
CHAPTER 14 Collision Theory 557
14.1 The Scattering Angle and Impact Parameter 558
14.2 The Collision Cross Section 560
14.3 Generalizations of the Cross Section 563
14.4 The Differential Scattering Cross Section 568
14.5 Calculating the Differential Cross Section 572
14.6 Rutherford Scattering 574
14.7 Cross Sections in Various Frames * 579
14.8 Relation of the CM and Lab Scattering Angles * 582
Principal Definitions and Equations of Chapter 14 586
Problems for Chapter 14 587
CHAPTER 15 Special Relativity 595
15.1 Relativity 596
15.2 Galilean Relativity 596
15.3 The Postulates of Special Relativity 601
15.4 The Relativity of Time; Time Dilation 603
15.5 Length Contraction 608
15.6 The Lorentz Transformation 610
15.7 The Relativistic Velocity-Addition Formula 615
x Contents
15.8 Four-Dimensional Space—Time; Four-Vectors 617
15.9 The Invariant Scalar Product 623
15.10 The Light Cone 625
15.11 The Quotient Rule and Doppler Effect 630
15.12 Mass, Four-Velocity, and Four-Momentum 633
15.13 Energy, the Fourth Component of Momentum 638
15.14 Collisions 644
15.15 Force in Relativity 649
15.16 Massless Particles; the Photon 652
15.17 Tensors* 656
15.18 Electrodynamics and Relativity 660
Principal Definitions and Equations of Chapter 15 664
Problems for Chapter 15 666
CHAPTER 16 Continuum Mechanics 681
16.1 Transverse Motion of a Taut String 682
16.2 The Wave Equation 685
16.3 Boundary Conditions; Waves on a Finite String * 688
16.4 The Three-Dimensional Wave Equation 694
16.5 Volume and Surface Forces 697
16.6 Stress and Strain: The Elastic Moduli 701
16.7 The Stress Tensor 704
16.8 The Strain Tensor for a Solid 709
16.9 Relation between Stress and Strain: Hooke’s Law 715
16.10 The Equation of Motion for an Elastic Solid 718
16.11 Longitudinal and Transverse Waves in a Solid 721
16.12 Fluids: Description of the Motion * 723
16.13 Waves in a Fluid* 727
Principal Definitions and Equations of Chapter 16 730
Problems for Chapter 16 732
APPENDIX Diagonalizing Real Symmetric Matrices 739
A.1 Diagonalizing a Single Matrix 739
A.2 Simultaneous Diagonalization of Two Matrices 743
Further Reading 747
Answers for Odd-Numbered Problems 749
Index 777
Preface
This book is intended for students of the physical sciences, especially physics, who have already studied some mechanics as part of an introductory physics course (“fresh- man physics” at a typical American university) and are now ready for a deeper look at the subject. The book grew out of the junior-level mechanics course which is offered by the Physics Department at Colorado and is taken mainly by physics majors, but also by some mathematicians, chemists, and engineers. Almost all of these students have taken a year of freshman physics, and so have at least a nodding acquaintance with Newton’s laws, energy and momentum, simple harmonic motion, and so on. In this book I build on this nodding acquaintance to give a deeper understanding of these basic ideas, and then go on to develop more advanced topics, such as the Lagrangian and Hamiltonian formulations, the mechanics of noninertial frames, motion of rigid bodies, coupled oscillators, chaos theory, and a few more.
Mechanics is, of course, the study of how things move — how an electron moves down your TV tube, how a baseball flies through the air, how a comet moves round the sun. Classical mechanics is the form of mechanics developed by Galileo and Newton in the seventeenth century and reformulated by Lagrange and Hamilton in the eighteenth and nineteenth centuries. For more than two hundred years, it seemed that classical mechanics was the only form of mechanics, that it could explain the motion of all conceivable systems.
Then, in two great revolutions of the early twentieth century, it was shown that classical mechanics cannot account for the motion of objects traveling close to the speed of light, nor of subatomic particles moving inside atoms. The years from about 1900 to 1930 saw the development of relativistic mechanics primarily to describe fast- moving bodies and of quantum mechanics primarily to describe subatomic systems. Faced with this competition, one might expect classical mechanics to have lost much of its interest and importance. In fact, however, classical mechanics is now, at the start of the twenty-first century, just as important and glamorous as ever. This resilience is due to three facts: First, there are just as many interesting physical systems as ever that are best described in classical terms. To understand the orbits of space vehicles and of charged particles in modern accelerators, you have to understand classical xi
xii Preface
mechanics. Second, recent developments in classical mechanics, mainly associated with the growth of chaos theory, have spawned whole new branches of physics and mathematics and have changed our understanding of the notion of causality. It is these new ideas that have attracted some of the best minds in physics back to the study of classical mechanics. Third, it is as true today as ever that a good understanding of classical mechanics is a prerequisite for the study of relativity and quantum mechanics.
Physicists tend to use the term “classical mechanics” rather loosely. Many use it for the mechanics of Newton, Lagrange, and Hamilton; for these people, “classical mechanics” excludes relativity and quantum mechanics. On the other hand, in some areas of physics, there is a tendency to include relativity as a part of “classical me- chanics”; for people of this persuasion, “classical mechanics” means “non-quantum mechanics.” Perhaps as a reflection of this second usage, some courses called “clas- sical mechanics” include an introduction to relativity, and for the same reason, I have included one chapter on relativistic mechanics, which you can use or not, as you please.
An attractive feature of a course in classical mechanics is that it is a wonderful opportunity to learn to use many of the mathematical techniques needed in so many other branches of physics — vectors, vector calculus, differential equations, complex numbers, Taylor series, Fourier series, calculus of variations, and matrices. I have tried to give at least a minimal review or introduction for each of these topics (with references to further reading) and to teach their use in the usually quite simple context of classical mechanics. I hope you will come away from this book with an increased confidence that you can really use these important tools.
Inevitably, there is more material in the book than could possibly be covered in a one-semester course. I have tried to ease the pain of choosing what to omit. The book is divided into two parts: Part I contains eleven chapters of “essential” material that should be read pretty much in sequence, while Part II contains five “further topics” that are mutually independent and any of which can be read without reference to the others. This division is naturally not very clear cut, and how you use it depends on your preparation (or that of your students). In our one-semester course at the University of Colorado, I found I needed to work steadily through most of Part I, and I only covered Part II by having students choose one of its chapters to study as a term project. (An activity they seemed to enjoy.) Some of the professors who taught from a preliminary version of the book found their students sufficiently well prepared that they could relegate the first four or five chapters to a quick review, leaving more time to cover some of Part II. At schools where the mechanics course lasts two quarters, it proved possible to cover all of Part I and much of Part II as well.
Because the chapters of Part II are mutually independent, it is possible to cover some of them before you finish Part I. For example, Chapter 12 on chaos could be covered immediately after Chapter 5 on oscillations, and Chapter 13 on Hamiltonian mechanics could be read immediately after Chapter 7 on Lagrangian mechanics. A number of sections are marked with an asterisk to indicate that they can be omitted without loss of continuity. (This is not to say that this material is unimportant. I certainly hope you’ll come back and read it later!)
As always in a physics text, it is crucial that you do lots of the exercises at the end of each chapter. I have included a large number of these to give both teacher and
Preface xiii
student plenty of choice. Some of them are simple applications of the ideas of the chapter and some are extensions of those ideas. I have listed the problems by section, so that as soon as you have read any given section you could (and probably should) try a few problems listed for that section. (Naturally, problems listed for a given section usually require knowledge of earlier sections. I promise only that you shouldn’t need material from later sections.) I have tried to grade the problems to indicate their level of difficulty, ranging from one star (*), meaning a straightforward exercise usually involving just one main concept, to three stars ( ***), meaning a challenging problem that involves several concepts and will probably take considerable time and effort. This kind of classification is quite subjective, very approximate, and surprisingly difficult to make; I would welcome suggestions for any changes you think should be made.
Several of the problems require the use of computers to plot graphs, solve differ- ential equations, and so on. None of these requires any specific software; some can be done with a relatively simple system such as MathCad or even just a spreadsheet like Excel; some require more sophisticated systems, such as Mathematica, Maple, or Matlab. (Incidentally, it is my experience that the course for which this book was written is a wonderful opportunity for the students to learn to use one of these fabu- lously useful systems.) Problems requiring the use of a computer are indicated thus: [Computer]. I have tended to grade them as *** or at least ** on the grounds that it takes a lot of time to set up the necessary code. Naturally, these problems will be easier for students who are experienced with the necessary software.
Each chapter ends with a summary called “Principal Definitions and Equations of Chapter xx.” I hope these will be useful as a check on your understanding of the chapter as you finish reading it and as a reference later on, as you try to find that formula whose details you have forgotten.
There are many people I wish to thank for their help and suggestions. At the Uni- versity of Colorado, these include Professors Larry Baggett, John Cary, Mike Dubson, Anatoli Levshin, Scott Parker, Steve Pollock, and Mike Ritzwoller. From other institu- tions, the following professors reviewed the manuscript or used a preliminary edition in their classes:
Meagan Aronson, U of Michigan Dan Bloom, Kalamazoo College Peter Blunden, U of Manitoba Andrew Cleland, UC Santa Barbara Gayle Cook, Cal Poly, San Luis Obispo Joel Fajans, UC Berkeley Richard Fell, Brandeis University Gayanath Fernando, U of Connecticut Jonathan Friedman, Amherst College David Goldhaber-Gordon, Stanford Thomas Griffy, U of Texas Elisabeth Gwinn, UC Santa Barbara Richard Hilt, Colorado College George Horton, Rutgers Lynn Knutson, U of Wisconsin
xiv Preface
Jonathan Maps, U of Minnesota, Duluth John Markert, U of Texas Michael Moloney, Rose-Hulman Institute Colin Morningstar, Carnegie Mellon Declan Mulhall, Cal Poly, San Luis Obispo Carl Mungan, US Naval Academy Robert Pompi, SUNY Binghamton Mark Semon, Bates College James Shepard, U of Colorado Richard Sonnenfeld, New Mexico Tech Edward Stern, U of Washington Michael Weinert, U of Wisconsin, Milwaukee Alma Zook, Pomona College
I am most grateful to all of these and their students for their many helpful comments. I would particularly like to thank Carl Mungan for his amazing vigilance in catching typos, obscurites, and ambiguities, and Jonathan Friedman and his student, Ben Heidenreich, who saved me from a really embarassing mistake in Chapter 10. I am especially grateful to my two friends and colleagues, Mark Semon at Bates College and Dave Goodmanson at the Boeing Aircraft Company, both of whom reviewed the manuscript with the finest of combs and gave me literally hundreds of suggestions; likewise to Christopher Taylor of the University of Wisconsin for his patient help with Mathematica and the mysteries of Latex. Bruce Armbruster and Jane Ellis of University Science Books are an author’s dream come true. My copy editor, Lee Young, is a rarity indeed, an expert in English usage and physics; he suggested many significant improvements. Finally and most of all, I want to thank my wife Debby. Being married to an author can be very trying, and she puts up with it most graciously. And, as an English teacher with the highest possible standards, she has taught me most of what I know about writing and editing. I am eternally grateful.
For all our efforts, there will surely be several errors in this book, and I would be most grateful if you could let me know of any that you find. Ancillary material, including an instructors’ manual, and other notices will be posted at the University Science Books website, www.uscibooks.com .
John R. Taylor Department of Physics University of Colorado Boulder, Colorado 80309, USA John.Taylor@Colorado.edu
PART I
Essen ia s CHAPTER 1
CHAPTER 2
CHAPTER 3
CHAPTER 4
CHAPTER 5
CHAPTER 6
CHAPTER 7
CHAPTER 8
CHAPTER 9
CHAPTER 10
CHAPTER 11
Newton’s Laws of Motion
Projectiles and Charged Particles
Momentum and Angular Momentum
Energy
Oscillations
Calculus of Variations
Lagrange’s Equations
Two-Body Central-Force Problems
Mechanics in Noninertial Frames
Rotational Motion of Rigid Bodies
Coupled Oscillators and Normal Modes
Part I of this book contains material that almost everyone would consider essential knowledge for an undergraduate physics major. Part II contains optional further topics from which you can pick according to your tastes and available time. The distinction between “essential” and “optional” is, of course, arguable, and its impact on you, the reader, depends very much on your state of preparation. For example, if you are well prepared, you might decide that the first five chapters of Part I can be treated as a quick review, or even skipped entirely. As a practical matter, the distinction is this: The eleven chapters of Part I were designed to be read in sequence, and in writing each chapter, I assumed that you would be familiar with most of the ideas of the preceding chapters— either by reading them or because you had met them elsewhere. By contrast, I tried to make the chapters of Part II independent of one another, so that you could read any of them in any order, once you knew most of the material of Part I.
CHAPTER
Newton’s Laws of. Motion
1.1 Classical Mechanics
Mechanics is the study of how things move: how planets move around the sun, how a skier moves down the slope, or how an electron moves around the nucleus of an atom. So far as we know, the Greeks were the first to think seriously about mechanics, more than two thousand years ago, and the Greeks’ mechanics represents a tremendous step in the evolution of modern science. Nevertheless, the Greek ideas were, by modern standards, seriously flawed and need not concern us here. The development of the mechanics that we know today began with the work of Galileo (1564-1642) and Newton (1642-1727), and it is the formulation of Newton, with his three laws of motion, that will be our starting point in this book.
In the late eighteenth and early nineteenth centuries, two alternative formulations of mechanics were developed, named for their inventors, the French mathematician and astronomer Lagrange (1736-1813) and the Irish mathematician Hamilton (1805- 1865). The Lagrangian and Hamiltonian formulations of mechanics are completely equivalent to that of Newton, but they provide dramatically simpler solutions to many complicated problems and are also the taking-off point for various modern developments. The term classical mechanics is somewhat vague, but it is generally understood to mean these three equivalent formulations of mechanics, and it is in this sense that the subject of this book is called classical mechanics.
Until the beginning of the twentieth century, it seemed that classical mechanics was the only kind of mechanics, correctly describing all possible kinds of motion. Then, in the twenty years from 1905 to 1925, it became clear that classical mechanics did not correctly describe the motion of objects moving at speeds close to the speed of light, nor that of the microscopic particles inside atoms and molecules. The result was the development of two completely new forms of mechanics: relativistic mechanics to describe very high-speed motions and quantum mechanics to describe the motion of microscopic particles. I have included an introduction to relativity in the “optional” Chapter 15. Quantum mechanics requires a whole separate book (or several books), and I have made no attempt to give even a brief introduction to quantum mechanics. 3
4 Chapter 1 Newton’s Laws of Motion
Although classical mechanics has been replaced by relativistic mechanics and by quantum mechanics in their respective domains, there is still a vast range of interesting and topical problems in which classical mechanics gives a complete and accurate description of the possible motions. In fact, particularly with the advent of chaos theory in the last few decades, research in classical mechanics has intensified and the subject has become one of the most fashionable areas in physics. The purpose of this book is to give a thorough grounding in the exciting field of classical mechanics. When appropriate, I shall discuss problems in the framework of the Newtonian formulation, but I shall also try to emphasize those situations where the newer formulations of Lagrange and Hamilton are preferable and to use them when this is the case. At the level of this book, the Lagrangian approach has many significant advantages over the Newtonian, and we shall be using the Lagrangian formulation repeatedly, starting in Chapter 7. By contrast, the advantages of the Hamiltonian formulation show themselves only at a more advanced level, and I shall postpone the introduction of Hamiltonian mechanics to Chapter 13 (though it can be read at any point after Chapter 7).
In writing the book, I took for granted that you have had an introduction to Newtonian mechanics of the sort included in a typical freshman course in “General Physics.” This chapter contains a brief review of the ideas that I assume you have met before.
1.2 Space and Time
Newton’s three laws of motion are formulated in terms of four crucial underlying concepts: the notions of space, time, mass, and force. This section reviews the first two of these, space and time. In addition to a brief description of the classical view of space and time, I give a quick review of the machinery of vectors, with which we label the points of space.
Space
Each point P of the three-dimensional space in which we live can be labeled by a position vector r which specifies the distance and direction of P from a chosen origin 0 as in Figure 1.1. There are many different ways to identify a vector, of which one of the most natural is to give its components (x, y, z) in the directions of three chosen perpendicular axes. One popular way to express this is to introduce three unit vectors, X, Si , i, pointing along the three axes and to write
r = ySr zi. (1.1)
In elementary work, it is probably wise to choose a single good notation, such as (1.1), and stick with it. In more advapced work, however, it is almost impossible to avoid using several different notations. Different authors have different preferences (another popular choice is to use i, j, k for what I am calling 1, ST, i) and you must get used to reading them all. Furthermore, almost every notation has its drawbacks, which can
Section 1.2 Space and Time 5
Figure 1.1 The point P is identified by its position vector r, which gives the position of P relative to a chosen origin 0. The vector r can be specified by its components (x, y, z) relative to chosen axes Oxyz.
make it unusable in some circumstances. Thus, while you may certainly choose your preferred scheme, you need to develop a tolerance for several different schemes.
It is sometimes convenient to be able to abbreviate (1.1) by writing simply
r = y, z)• (1.2)
This notation is obviously not quite consistent with (1.1), but it is usually completely unambiguous, asserting simply that r is the vector whose components are x, y, z. When the notation of (1.2) is the most convenient, I shall not hesitate to use it. For most vectors, we indicate the components by subscripts x, y, z. Thus the velocity vector v has components vx , vy , v z and the acceleration a has components ax , ay , a,.
As our equations become more complicated, it is sometimes inconvenient to write out all three terms in sums like (1.1); one would rather use the summation sign E followed by a single term. The notation of (1.1) does not lend itself to this shorthand, and for this reason I shall sometimes relabel the three components x, y, z of r as r 1 , r2 , r3 , and the three unit vectors X, ST, z as e l , e2 , e3 . That is, we define
r1 = x, r2 = y, r3 = z,
and
e l = e2 = ST, e3 = Z.
(The symbol e is commonly used for unit vectors, since e stands for the German “eins” or “one.”) With these notations, (1.1) becomes
3
r = r ie i r2e2 r3e3 = E ri ei . i=1
(1.3)
For a simple equation like this, the form (1.3) has no real advantage over (1.1), but with more complicated equations (1.3) is significantly more convenient, and I shall use this notation when appropriate.
6 Chapter 1 Newton’s Laws of Motion
Vector Operations
In our study of mechanics, we shall make repeated use of the various operations that can be performed with vectors. If r and s are vectors with components
r = (r 1 , r2 , r3) and s = (s 1 , s2 , s3),
then their sum (or resultant) r + s is found by adding corresponding components, so that
r + s = (r i + s i , r2 + S2, 7’3 S3). (1.4)
(You can convince yourself that this rule is equivalent to the familiar triangle and parallelogram rules for vector addition.) An important example of a vector sum is the resultant force on an object: When two forces Fa and Fb act on an object, the effect is the same as a single force, the resultant force, which is just the vector sum
F = Fa + Fb
as given by the vector addition law (1.4). If c is a scalar (that is, an ordinary number) and r is a vector, the product cr is
given by
cr = (cr1 , cr2 , cr3). (1.5)
This means that cr is a vector in the same direction ) as r with magnitude equal to c times the magnitude of r. For example, if an object of mass m (a scalar) has an acceleration a (a vector), Newton’s second law asserts that the resultant force F on the object will always equal the product ma as given by (1.5).
There are two important kinds of product that can be formed from any pair of vectors. First, the scalar product (or dot product) of two vectors r and s is given by either of the equivalent formulas
r • s = rs cos 0 3
r2S2 r3s3 = E rns, n=1
where r and s denote the magnitudes of the vectors r and s, and 6 is the angle between them. (For a proof that these two definitions are the same, see Problem 1.7.) For example, if a force F acts on an object that moves through a small displacement dr, the work done by the force is the scalar product F • dr, as given by either (1.6) or (1.7). Another important use of the scalar product is to define the magnitude of a vector: The magnitude (or length) of any vector r is denoted by r I or r and, by Pythagoras’s
theorem is equal to /r? r22 + r32 . By (1.7) this is the same as
r == Ill == .1% (1.8)
The scalar product r • r is often abbreviated as r 2 .
(1.6)
(1.7)
Although this is what people usually say, one should actually be careful: If c is negative, cr is in the opposite direction to r.
Section 1.2 Space and Time 7
The second kind of product of two vectors r and s is the vector product (or cross product), which is defined as the vector p = r x s with components
px = ryS, – rz Sy py = r,S, – rx s,
pz = r,Sy – ry Sx
or, equivalently
[X ST i r x s = det rx ry r, ,
s, s s
y z
where “det” stands for the determinant. Either of these definitions implies that r x s is a vector perpendicular to both r and s, with direction given by the familiar right-hand rule and magnitude rs sin B (Problem 1.15). The vector product plays an important role in the discussion of rotational motion. For example, the tendency of a force F (acting at a point r) to cause a body to rotate about the origin is given by the torque of F about 0, defined as the vector product 1″ = r x F.
Differentiation of Vectors
Many (maybe most) of the laws of physics involve vectors, and most of these involve derivatives of vectors. There are so many ways to differentiate a vector that there is a whole subject called vector calculus, much of which we shall be developing in the course of this book. For now, I shall mention just the simplest kind of vector derivative, the time derivative of a vector that depends on time. For example, the velocity v(t) of a particle is the time derivative of the particle’s position r(t); that is, v = dr/dt. Similarly the acceleration is the time derivative of the velocity, a = dv/dt.
The definition of the derivative of a vector is closely analogous to that of a scalar. Recall that if x (t) is a scalar function of t, then we define its derivative as
dx. Ax — = lim — dt At-0 At
where Ax = x (t + At) — x(t) is the change in x as the time advances from t to t + At. In exactly the same way, if r(t) is any vector that depends on t, we define its derivative as
where
dr,. Ar = um —
dt At->0 At (1.10)
Ar = r(t + At) — r(t) (1.11)
is the corresponding change in r(t). There are, of course, many delicate questions about the existence of this limit. Fortunately, none of these need concern us here: All of the vectors we shall encounter will be differentiable, and you can take for granted that the required limits exist. From the definition (1.10), one can prove that the derivative has all of the properties one would expect. For example, if r(t) and s(t)
(1.9)
8 Chapter 1 Newton’s Laws of Motion
are two vectors that depend on t, then the derivative of their sum is just what you would expect:
d dr ds
dt = —dt + dt . (1.12)
Similarly, if r(t) is a vector and f (t) is a scalar, then the derivative of the product f (t)r (t) is given by the appropriate version of the product rule,
dr df
dt (f r) = f
dt +
dt r. (1.13)
If you are the sort of person who enjoys proving these kinds of proposition, you might want to show that they follow from the definition (1.10). Fortunately, if you do not enjoy this kind of activity, you don’t need to worry, and you can safely take these results for granted.
One more result that deserves mention concerns the components of the derivative of a vector. Suppose that r, with components x, y, z, is the position of a moving particle, and suppose that we want to know the particle’s velocity v = dr Idt. When we differentiate the sum
r = + + zi, (1.14)
the rule (1.12) gives us the sum of the three separate derivatives, and, by the product rule (1.13), each of these contains two terms. Thus, in principle, the derivative of (1.14) involves six terms in all. However, the unit vectors Sr’, and i do not depend on time, so their time derivatives are zero. Therefore, three of these six terms are zero, and we are left with just three terms:
dr dx dy „ dz — = —x + —y + —z. dt dt dt dt
Comparing this with the standard expansion
V = v y ST
we see that
(1.15)
dx v = dY vx -= dt dt
and V, = dz
dt (1.16)
In words, the rectangular components of v are just the derivatives of the corresponding components of r. This is a result that we use all the time (usually without even think- ing about it) in solving elementary mechanics problems. What makes it especially noteworthy is this: It is true only because the unit vectors X, jr ‘, and i are constant, so that their derivatives are absent from (1.15). We shall find that in most coordinate systems, such as polar coordinates, the basic unit vectors are not constant, and the result corresponding to (1.16) is appreciably less transparent. In problems where we need to work in nonrectangular coordinates, it is considerably harder to write down velocities and accelerations in terms of the coordinates of r, as we shall see.
Section 1.3 Mass and Force 9
Time
The classical view is that time is a single universal parameter t on which all observers agree. That is, if all observers are equipped with accurate clocks, all properly syn- chronized, then they will all agree as to the time at which any given event occurred. We know, of course, that this view is not exactly correct: According to the theory of relativity, two observers in relative motion do not agree on all times. Nevertheless, in the domain of classical mechanics, with all speeds much much less than the speed of light, the differences among the measured times are entirely negligible, and I shall adopt the classical assumption of a single universal time (except, of course, in Chap- ter 15 on relativity). Apart from the obvious ambiguity in the choice of the origin of time (the time that we choose to label t = 0), all observers agree on the times of all events.
Reference Frames
Almost every problem in classical mechanics involves a choice (explicit or implicit) of a reference frame, that is, a choice of spatial origin and axes to label positions as in Figure 1.1 and a choice of temporal origin to measure times. The difference between two frames may be quite minor. For instance, they may differ only in their choice of the origin of time — what one frame labels t = 0 the other may label t’ = to 0. Or the two frames may have the same origins of space and time, but have different orientations of the three spatial axes. By carefully choosing your reference frame, taking advantage of these different possibilities, you can sometimes simplify your work. For example, in problems involving blocks sliding down inclines, it often helps to choose one axis pointing down the slope.
A more important difference arises when two frames are in relative motion; that is, when one origin is moving relative to the other. In Section 1.4 we shall find that not all such frames are physically equivalent? In certain special frames, called inertial frames, the basic laws hold true in their standard, simple form. (It is because one of these basic laws is Newton’s first law, the law of inertia, that these frames are called inertial.) If a second frame is accelerating or rotating relative to an inertial frame, then this second frame is noninertial, and the basic laws — in particular, Newton’s laws — do not hold in their standard form in this second frame. We shall find that the distinction between inertial and noninertial frames is central to our discussion of classical mechanics. It plays an even more explicit role in the theory of relativity.
1.3 Mass and Force
The concepts of mass and force are central to the formulation of classical mechanics. The proper definitions of these concepts have occupied many philosophers of science and are the subject of learned treatises. Fortunately we don’t need to worry much about
2 This statement is correct even in the theory of relativity.
10 Chapter 1 Newton’s Laws of Motion
force
Figure 1.2 An inertial balance compares the masses m 1 and m 2
of two objects that are attached to the opposite ends of a rigid rod.
The masses are equal if and only if a force applied at the rod’s
midpoint causes them to accelerate at the same rate, so that the
rod does not rotate.
these delicate questions here. Based on your introductory course in general physics, you have a reasonably good idea what mass and force mean, and it is easy to describe how these parameters are defined and measured in many realistic situations.
Mass
The mass of an object characterizes the object’s inertia—its resistance to being accelerated: A big boulder is hard to accelerate, and its mass is large. A little stone is easy to accelerate, and its mass is small. To make these natural ideas quantitative we have to define a unit of mass and then give a prescription for measuring the mass of any object in terms of the chosen unit. The internationally agreed unit of mass is the kilogram and is defined arbitrarily to be the mass of a chunk of platinum—iridium stored at the International Bureau of Weights and Measures outside Paris. To measure the mass of any other object, we need a means of comparing masses. In principle, this can be done with an inertial balance as shown in Figure 1.2. The two objects to be compared are fastened to the opposite ends of a light, rigid rod, which is then given a sharp pull at its midpoint. If the masses are equal, they will accelerate equally and the rod will move off without rotating; if the masses are unequal, the more massive one will accelerate less, and the rod will rotate as it moves off.
The beauty of the inertial balance is that it gives us a method of mass comparison that is based directly on the notion of mass as resistance to being accelerated. In practice, an inertial balance would be very awkward to use, and it is fortunate that there are much easier ways to compare masses, of which the easiest is to weigh the objects. As you certainly recall from your introductory physics course, an object’s mass is found to be exactly proportional to the object’s weight 3 (the gravitational force on the object) provided all measurements are made in the same location. Thus two
3 This observation goes back to Galileo’s famous experiments showing that all objects are accelerated at the same rate by gravity. The first modern experiments were conducted by the Hungarian physicist Eiityos (1848-1919), who showed that weight is proportional to mass to within
Section 1.3 Mass and Force 11
objects have the same mass if and only if they have the same weight (when weighed at the same place), and a simple, practical way to check whether two masses are equal is simply to weigh them and see if their weights are equal.
Armed with methods for comparing masses, we can easily set up a scheme to mea- sure arbitrary masses. First, we can build a large number of standard kilograms, each one checked against the original 1-kg mass using either the inertial or gravitational balance. Next, we can build multiples and fractions of the kilogram, again checking them with our balance. (We check a 2-kg mass on one end of the balance against two 1-kg masses placed together on the other end; we check two half-kg masses by verifying that their masses are equal and that together they balance a 1-kg mass; and so on.) Finally, we can measure an unknown mass by putting it on one end of the balance and loading known masses on the other end until they balance to any desired precision.
Force
The informal notion of force as a push or pull is a surprisingly good starting point for our discussion of forces. We are certainly conscious of the forces that we exert ourselves. When I hold up a sack of cement, I am very aware that I am exerting an upward force on the sack; when I push a heavy crate across a rough floor, I am aware of the horizontal force that I have to exert in the direction of motion. Forces exerted by inanimate objects are a little harder to pin down, and we must, in fact, understand something of Newton’s laws to identify such forces. If I let go of the sack of cement, it accelerates toward the ground; therefore, I conclude that there must be another force — the sack’s weight, the gravitational force of the earth — pulling it downward. As I push the crate across the floor, I observe that it does not accelerate, and I conclude that there must be another force — friction — pushing the crate in the opposite direction. One of the most important skills for the student of elementary mechanics is to learn to examine an object’s environment and identify all the forces on the object: What are the things touching the object and possibly exerting contact forces, such as friction or air pressure? And what are the nearby objects possibly exerting action-at-a-distance forces, such as the gravitational pull of the earth or the electrostatic force of some charged body?
If we accept that we know how to identify forces, it remains to decide how to measure them. As the unit of force we naturally adopt the newton (abbreviated N) defined as the magnitude of any single force that accelerates a standard kilogram mass with an acceleration of 1 m/s 2 . Having agreed what we mean by one newton, we can proceed in several ways, all of which come to the same final conclusion, of course. The route that is probably preferred by most philosophers of science is to use Newton’s second law to define the general force: A given force is 2 N if, by itself, it accelerates a standard kilogram with an acceleration of 2 m/s 2 , and so
a few parts in 109. Experiments in the last few decades have narrowed this to around one part in 1012.
12 Chapter 1 Newton’s Laws of Motion
= 2 N
Figure 1.3 One of many possible ways to define forces of any
magnitude. The lower spring balance has been calibrated to read
1 N. If the balance arm on the left is adjusted so that the lever arms
above and below the pivot are in the ratio 1 : 2 and if the force F 1 is
1 N, then the force F2 required to balance the arm is 2 N. This lets
us calibrate the upper spring balance for 2 N. By readjusting the two
lever arms, we can, in principle, calibrate the second spring balance
to read any force.
on. This approach is not much like the way we usually measure forces in practice, 4 and for our present discussion a simpler procedure is to use some spring balances.
Using our definition of the newton, we can calibrate a first spring balance to read 1 N. Then by matching a second spring balance against the first, using a balance arm as shown in Figure 1.3, we can define multiples and fractions of a newton. Once we have a fully calibrated spring balance we can, in principle, measure any unknown force, by matching it against the calibrated balance and reading off its value.
So far we have defined only the magnitude of a force. As you are certainly aware, forces are vectors, and we must also define their directions. This is easily done. If we apply a given force F (and no other forces) to any object at rest, the direction of F is defined as the direction of the resulting acceleration, that is, the direction in which the body moves off.
Now that we know, at least in principle, what we mean by positions, times, masses, and forces, we can proceed to discuss the cornerstone of our subject — Newton’s three laws of motion.
4 The approach also creates the confusing appearance that Newton’s second law is just a conse- quence of the definition of force. This is not really true: Whatever definition we choose for force, a large part of the second law is experimental. One advantage of defining forces with spring bal- ances is that it separates out the definition of force from the experimental basis of the second law. Of course, all commonly accepted definitions give the same final result for the value of any given force.
Newton’s Second La For;in y particle of mass m, the net force F on the partici mass m times the particle’s acceleration:
Section 1.4 Newton’s First and Second Laws; Inertial Frames 13
1.4 Newton’s First and Second Laws; Inertial Frames
In this chapter, I am going to discuss Newton’s laws as they apply to a point mass. A point mass, or particle, is a convenient fiction, an object with mass, but no size, that can move through space but has no internal degrees of freedom. It can have “translational” kinetic energy (energy of its motion through space) but no energy of rotation or of internal vibrations or deformations. Naturally, the laws of motion are simpler for point particles than for extended bodies, and this is the main reason that we start with the former. Later on, I shall build up the mechanics of extended bodies from our mechanics of point particles by considering the extended body as a collection of many separate particles.
Nevertheless, it is worth recognizing that there are many important problems where the objects of interest can be realistically approximated as point masses. Atomic and subatomic particles can often be considered to be point masses, and even macroscopic objects can frequently be approximated in this way. A stone thrown off the top of a cliff is, for almost all purposes, a point particle. Even a planet orbiting around the sun can usually be approximated in the same way. Thus the mechanics of point masses is more than just the starting point for the mechanics of extended bodies; it is a subject with wide application itself.
Newton’s first two laws are well known and easily stated:
Newton’s First Law (the Law of Inertia of forces, a particle moves with constant velocity v.
and
In this equation F denotes the vector sum of all the forces on the particle and a is the particle’s acceleration,
dv a = — v
dt-
d2r = dt 2
Here v denotes the particle’s velocity, and I have introduced the convenient notation of dots to denote differentiation with respect to t, as in v =r and a = v = f.
14 Chapter 1 Newton’s Laws of Motion
Both laws can be stated in various equivalent ways. For instance (the first law): In the absence of forces, a stationary particle remains stationary and a moving particle continues to move with unchanging speed in the same direction. This is, of course, exactly the same as saying that the velocity is always constant. Again, v is constant if and only if the acceleration a is zero, so an even more compact statement is this: In the absence of forces a particle has zero acceleration.
The second law can be rephrased in terms of the particle’s momentum, defined as
p = mv. (1.18)
In classical mechanics, we take for granted that the mass m of a particle never changes, so that
p =my-,– ma.
Thus the second law (1.17) can be rephrased to say that
In classical mechanics, the two forms (1.17) and (1.19) of the second law are com- pletely equivalent. 5
Differential Equations
When written in the form ml = F, Newton’s second law is a differential equation for the particle’s position r(t). That is, it is an equation for the unknown function r(t) that involves derivatives of the unknown function. Almost all the laws of physics are, or can be cast as, differential equations, and a huge proportion of a physicist’s time is spent solving these equations. In particular, most of the problems in this book involve differential equations — either Newton’s second law or its counterparts in the Lagrangian and Hamiltonian forms of mechanics. These vary widely in their difficulty. Some are so easy to solve that one scarcely notices them. For example, consider Newton’s second law for a particle confined to move along the x axis and subject to a constant force Fo,
. •i(t)
F
m .
This is a second-order differential equation for x (t) as a function of t. (Second-order because it involves derivatives of second order, but none of higher order.) To solve it
5 In relativity, the two forms are not equivalent, as we’ll see in Chapter 15. Which form is correct depends on the definitions we use for force, mass, and momentum in relativity. If we adopt the most popular definitions of these three quantities, then it is the form (1.19) that holds in relativity.
Section 1.4 Newton’s First and Second Laws; Inertial Frames 15
one has only to integrate it twice. The first integration gives the velocity
i(t)= I 1(0 dt = vo + fn
where the constant of integration is the particle’s initial velocity, and a second inte- gration gives the position
x(t) = f (t) dt = x, v ot + —Fo t2 2m
where the second constant of integration is the particle’s initial position. Solving this differential equation was so easy that we certainly needed no knowledge of the theory of differential equations. On the other hand, we shall meet lots of differential equations that do require knowledge of this theory, and I shall present the necessary theory as we need it. Obviously, it will be an advantage if you have already studied some of the theory of differential equations, but you should have no difficulty picking it up as we go along. Indeed, many of us find that the best way to learn this kind of mathematical theory is in the context of its physical applications.
Inertial Frames
On the face of it, Newton’s second law includes his first: If there are no forces on an object, then F = 0 and the second law (1.17) implies that a = 0, which is the first law. There is, however, an important subtlety, and the first law has an important role to play. Newton’s laws cannot be true in all conceivable reference frames. To see this, consider just the first law and imagine a reference frame — we’ll call it 8 — in which the first law is true. For example, if the frame 8 has its origin and axes fixed relative to the earth’s surface, then, to an excellent approximation, the first law (the law of inertia) holds with respect to the frame 8: A frictionless puck placed on a smooth horizontal surface is subject to zero force and, in accordance with the first law, it moves with constant velocity. Because the law of inertia holds, we call 8 an inertial frame. If we consider a second frame 8′ which is moving relative to S with constant velocity and is not rotating, then the same puck will also be observed to move with constant velocity relative to 8′. That is, the frame 8′ is also inertial.
If, however, we consider a third frame 8″ that is accelerating relative to 8, then, as viewed from 8″, the puck will be seen to be accelerating (in the opposite direction). Relative to the accelerating frame 8″ the law of inertia does not hold, and we say that 8″ is noninertial. I should emphasize that there is nothing mysterious about this result. Indeed it is a matter of experience. The frame 8′ could be a frame attached to a high-speed train traveling smoothly at constant speed along a straight track, and the frictionless puck, an ice cube placed on the floor of the train, as in Figure 1.4. As seen from the train (frame 8′), the ice cube is at rest and remains at rest, in accord with the first law. As seen from the ground (frame 8), the ice cube is moving with the same velocity as the train and continues to do so, again in obedience to the first law. But now consider conducting the same experiment on a second train (frame 8″) that is accelerating forward. As this train accelerates forward, the ice cube is left behind, and, relative to 8″, the ice cube accelerates backward, even though subject to no net
16 Chapter 1 Newton’s Laws of Motion
Figure 1.4 The frame 8 is fixed to the ground, while 8′ is fixed to a
train traveling at constant velocity v’ relative to 8. An ice cube placed
on the floor of the train obeys Newton’s first law as seen from both
and 8′. If the train to which 8″ is attached is accelerating forward, then,
as seen in 8″, an ice cube placed on the floor will accelerate backward,
and the first law does not hold in 8″.
force. Clearly the frame S” is noninertial, and neither of the first two laws can hold in
S”. A similar conclusion would hold if the frame 8″ had been attached to a rotating
merry-go-round. A frictionless puck, subject to zero net force, would not move in a
straight line as seen in 8″, and Newton’s laws would not hold. Evidently Newton’s two laws hold only in the special, inertial (nonaccelerating
and nonrotating) reference frames. Most philosophers of science take the view that
the first law should be used to identify these inertial frames — a reference frame 8 is
inertial if objects that are clearly subject to no forces are seen to move with constant
velocity relative to 8.6 Having identified the inertial frames by means of Newton’s
first law, we can then claim as an experimental fact that the second law holds in these
same inertial frames?
Since the laws of motion hold only in inertial frames, you might imagine that
we would confine our attention exclusively to inertial frames, and, for a while,
we shall do just that. Nevertheless, you should be aware that there are situations
where it is necessary, or at least very convenient, to work in noninertial frames.
The most important example of a noninertial frame is in fact the earth itself. To an
excellent approximation, a reference frame fixed to the earth is inertial — a fortunate
circumstance for students of physics! Nevertheless, the earth rotates on its axis once
a day and circles around the sun once a year, and the sun orbits slowly around the
center of the Milky Way galaxy. For all of these reasons, a reference frame fixed to
the earth is not exactly inertial. Although these effects are very small, there are several
phenomena — the tides and the trajectories of long-range projectiles are examples —
6 There is some danger of going in a circle here: How do we know that the object is subject to
no forces? We’d better not answer, “Because it’s traveling at constant velocity”! Fortunately, we can
argue that it is possible to identify all sources of force, such as people pushing and pulling or nearby
massive bodies exerting gravitational forces. If there are no such things around, we can reasonably
say that the object is free of forces.
7 As I mentioned earlier, the extent to which the second law is an experimental statement depends
on how we choose to define force. If we define force by means of the second law, then to some extent
(though certainly not entirely) the law becomes a matter of definition. If we define forces by means
of spring balances, then the second law is clearly an experimentally testable proposition.
Section 1.5 The Third Law and Conservation of Momentum 17
that are most simply explained by taking into account the noninertial character of a frame fixed to the earth. In Chapter 9 we shall examine how the laws of motion must be modified for use in noninertial frames. For the moment, however, we shall confine our discussion to inertial frames.
Validity of the First Two Laws
Since the advent of relativity and quantum mechanics, we have known that Newton’s laws are not universally valid. Nevertheless, there is an immense range of phenom- ena — the phenomena of classical physics — where the first two laws are for all practical purposes exact. Even as the speeds of interest approach c, the speed of light, and relativity becomes important, the first law remains exactly true. (In relativity, just as in classical mechanics, an inertial frame is defined as one where the first law holds.) 8 As we shall see in Chapter 15, the two forms of the second law, F = ma and F = it, are no longer equivalent in relativity, although with F and p suitably defined the second law in the form F = p is still valid. In any case, the important point is this: In the classical domain, we can and shall assume that the first two laws (the second in either form) are universally and precisely valid. You can, if you wish, regard this assumption as defining a model — the classical model — of the natural world. The model is logically consistent and is such a good representation of many phenomena that it is amply worthy of our study.
1.5 The Third Law and Conservation of Momentum
Newton’s first two laws concern the response of a single object to applied forces. The third law addresses a quite different issue: Every force on an object inevitably involves a second object — the object that exerts the force. The nail is hit by the hammer, the cart is pulled by the horse, and so on. While this much is no doubt a matter of common sense, the third law goes considerably beyond our everyday experience. Newton realized that if an object 1 exerts a force on another object 2, then object 2 always exerts a force (the “reaction” force) back on object 1. This seems quite natural: If you push hard against a wall, it is fairly easy to convince yourself that the wall is exerting a force back on you, without which you would undoubtedly fall over. The aspect of the third law which certainly goes beyond our normal perceptions is this: According to the third law, the reaction force of object 2 on object 1 is always equal and opposite to the original force of 1 on 2. If we introduce the notation F21 to denote the force exerted on object 2 by object 1, Newton’s third law can be stated very compactly:
8 However, in relativity the relationship between different inertial frames — the so-called Lorentz transformation — is different from that of classical mechanics. See Section 15.6.
LITO DIDLIOJO .1NWERMETO MB! IOTEK4 n M !J %.
object 1 exert orce F2 force F l ct 1 given
s reaction
18 Chapter 1 Newton’s Laws of Motion
F21
Figure 1.5 Newton’s third law asserts that the reaction force exerted on object 1 by object 2 is equal and opposite to the force exerted on 2 by 1, that is, F12 = —F21.
This statement is illustrated in Figure 1.5, which you could think of as showing the force of the earth on the moon and the reaction force of the moon on the earth (or a proton on an electron and the electron on the proton). Notice that this figure actually goes a little beyond the usual statement (1.20) of the third law: Not only have I shown the two forces as equal and opposite; I have also shown them acting along the line joining 1 and 2. Forces with this extra property are called central forces. (They act along the line of centers.) The third law does not actually require that the forces be central, but, as I shall discuss later, most of the forces we encounter (gravity, the electrostatic force between two charges, etc.) do have this property.
As Newton himself was well aware, the third law is intimately related to the law of conservation of momentum. Let us focus, at first, on just two objects as shown in Figure 1.6, which might show the earth and the moon or two skaters on the ice. In addition to the force of each object on the other, there may be “external” forces exerted by other bodies. The earth and moon both experience forces exerted by the sun, and both skaters could experience the external force of the wind. I have shown the net external forces on the two objects as Pi’ and Fe2xt . The total force on object 1 is then
(net force on 1) F 1 F = – 12 Feixt
and similarly
(net force on 2) F2 = F21 + Fe2xt .
We can compute the rates of change of the particles’ momenta using Newton’s second law:
= F1 = Fi2 + Feixt (1.21)
Section 1.5 The Third Law and Conservation of Momentum 19
21
F 12
Figure 1.6 Two objects exert forces on each other and
may also be subject to additional “external” forces from other objects not shown.
and
P2 =F2=F21-f-F2xt . (1.22)
If we now define the total momentum of our two objects as
P = Pi ± P2 ,
then the rate of change of the total momentum is just
1.3 = 1.3 1 +
To evaluate this, we have only to add Equations (1.21) and (1.22). When we do this, the two internal forces, F 12 and F21, cancel out because of Newton’s third law, and we are left with
P = Feixt Fext Fext , (1.23)
where I have introduced the notation Fe’ denote the total external force on our two-particle system.
The result (1.23) is the first in a series of important results that let us construct a theory of many-particle systems from the basic laws for a single particle. It asserts that as far as the total momentum of a system is concerned, the internal forces have no effect. A special case of this result is that if there are no external forces (Fe’ = 0) then P = 0. Thus we have the important result:
If Fext = 0 , then P = const. (1.24)
In the absence of external forces, the total momentum of our two-particle system is
constant — a result called the principle of conservation of momentum.
Multiparticle Systems
We have proved the conservation of momentum, Equation (1.24), for a system of two particles. The extension of the result to any number of particles is straightforward in principle, but I would like to go through it in detail, because it lets me introduce some
20 Chapter 1 Newton’s Laws of Motion
oP
0 0
Figure 1.7 A five-particle system with particles labelled by a or fi = 1, 2, • • , 5. The particle a is subject to four internal forces, shown by solid arrows and denoted F 0 (the force on a by p). In addition particle a may be subject to a net external force, shown by the dashed arrow and denoted F aext.
important notation and will give you some practice using the summation notation. Let us consider then a system of N particles. I shall label the typical particle with a Greek index a or /3, either of which can take any of the values 1, 2, – • • , N. The mass of particle a is m a and its momentum is pa . The force on particle a is quite complicated: Each of the other (N — 1) particles can exert a force which I shall call Fan , the force
on a by p, as illustrated in Figure 1.7. In addition there may be a net external force on particle a, which I shall call Fr. Thus the net force on particle a is
(net force on particle a) = Fa = E Fan + Faext . (1.25) SO«
Here the sum runs over all values of /3 not equal to a. (Remember there is no force Faa because particle a cannot exert a force on itself.) According to Newton’s second law, this is the same as the rate of change of pa :
oc, E Fas 17t . (1.26) fiOcf
This result holds for each a = 1, – , N. Let us now consider the total momentum of our N-particle system,
P = I Pa a
where, of course, this sum runs over all N particles, a = 1, 2, • • • , N. If we differen- tiate this equation with respect to time, we find
or, substituting for Oa from (1.26),
= E E Fap E Faext . (1.27) a 00ce a
Section 1.5 The Third Law and Conservation of Momentum 21
The double sum here contains N (N — 1) terms in all. Each term F ap in this sum can be paired with a second term F p, (that is, F12 paired with F21, and so on), so that
E E F „p = E (Fa, + F sa). (1.28) a f 3 0a a /3 >a
The double sum on the right includes only values of a and /3 with a < p and has half as many terms as that on the left. But each term is the sum of two forces, (F aa Foa ), and, by the third law, each such sum is zero. Therefore the whole double sum in (1.28) is zero, and returning to (1.27) we conclude that
P = E a Fext _ Fext (1.29)
The result (1.29) corresponds exactly to the two-particle result (1.23). Like the latter, it says that the internal forces have no effect on the evolution of the total momentum P — the rate of change of P is determined by the net external force on the system. In particular, if the net external force is zero, we have the
Principle of Conservation of Momentum
he net x orce F”‘ on an N-particle’system is hero, t e total momentum P is
As you are certainly aware, this is one of the most important results in classical physics and is, in fact, also true in relativity and quantum mechanics. If you are not very familiar with the sorts of manipulations of sums that we used, it would be a good idea to go over the argument leading from (1.25) to (1.29) for the case of three or four particles, writing out all the sums explicitly (Problems 1.28 or 1.29). You should also convince yourself that, conversely, if the principle of conservation of momentum is true for all multiparticle systems, then Newton’s third law must be true (Problem 1.31). In other words, conservation of momentum and Newton’s third law are equivalent to one another.
Validity of Newton’s Third Law
Within the domain of classical physics, the third law, like the second, is valid with such accuracy that it can be taken to be exact. As speeds approach the speed of light, it is easy to see that the third law cannot hold: The point is that the law asserts that the action and reaction forces, F 12 (t) and F21 (t), measured at the same time t, are equal and opposite. As you certainly know, once relativity becomes important the concept of a single universal time has to be abandoned — two events that are seen as simultaneous by one observer are, in general, not simultaneous as seen by a second observer. Thus, even if the equality F 12 (t) = —F21(t) (with both times the same) were true for one observer, it would generally be false for another. Therefore, the third law cannot be valid once relativity becomes important.
22 Chapter 1 Newton’s Laws of Motion
Figure 1.8 Each of the positive charges q 1 and q2 produces
a magnetic field that exerts a force on the other charge. The
resulting magnetic forces F 12 and F21 do not obey Newton’s third law.
Rather surprisingly, there is a simple example of a well-known force — the mag- netic force between two moving charges — for which the third law is not exactly true, even at slow speeds. To see this, consider the two positive charges of Figure 1.8, with q 1 moving in the x direction and q2 moving in the y direction, as shown. The exact calculation of the magnetic field produced by each charge is complicated, but a simple argument gives the correct directions of the two fields, and this is all we need. The moving charge q 1 is equivalent to a current in the x direction. By the right-hand rule for fields, this produces a magnetic field which points in the z direction in the vicinity of q2 . By the right-hand rule for forces, this field produces a force F21 on q2 that is in the x direction. An exactly analogous argument (check it yourself) shows that the force F12 on q 1 is in the y direction, as shown. Clearly these two forces do not obey Newton’s third law!
This conclusion is especially startling since we have just seen that Newton’s third law is equivalent to the conservation of momentum. Apparently the total momentum m iv i + m 2v2 of the two charges in Figure 1.8 is not conserved. This conclusion, which is correct, serves to remind us that the “mechanical” momentum my of particles is not the only kind of momentum. Electromagnetic fields can also carry momentum, and in the situation of Figure 1.8 the mechanical momentum being lost by the two particles is going to the electromagnetic momentum of the fields.
Fortunately, if both speeds in Figure 1.8 are much less than the speed of light (v << c), the loss of mechanical momentum and the concomitant failure of the third law are completely negligible. To see this, note that in addition to the magnetic force between q i and q2 there is the electrostatic Coulomb force 9 km 21 r2 , which does obey Newton’s third law. It is a straightforward exercise (Problem 1.32) to show that the magnetic force is of order v 2/c2 times the Coulomb force. Thus only as v approaches c — and classical mechanics must give way to relativity anyway — is the violation of
9 Here k is the Coulomb force constant, often written as k = 1/(47r E0).
Section 1.6 Newton’s Second Law in Cartesian Coordinates 23
the third law by the magnetic force important. 1° We see that the unexpected situation of Figure 1.8 does not contradict our claim that in the classical domain Newton’s third law is valid, and this is what we shall assume in our discussions of nonrelativistic mechanics.
1.6 Newton’s Second Law in Cartesian Coordinates
Of Newton’s three laws, the one that we actually use the most is the second, which is often described as the equation of motion. As we have seen, the first is theoretically important to define what we mean by inertial frames but is usually of no practical use beyond this. The third law is crucially important in sorting out the internal forces in a multiparticle system, but, once we know the forces involved, the second law is what we actually use to calculate the motion of the object or objects of interest. In particular, in many simple problems the forces are known or easily found, and, in this case, the second law is all we need for solving the problem.
As we have already noted, the second law,
F (1.30)
is a second-order, differential equation” for the position vector r as a function of the time t. In the prototypical problem, the forces that comprise F are given, and our job is to solve the differential equation (1.30) for r (t) . Sometimes we are told about r (t) , and we have to use (1.30) to find some of the forces. In any case, the equation (1.30) is a vector differential equation. And the simplest way to solve such equations is almost always to resolve the vectors into their components relative to a chosen coordinate system.
Conceptually the simplest coordinate system is the Cartesian (or rectangular), with unit vectors X, ST, and z , in terms of which the net force F can then be written as
F= Fx X 1 F3,S7‘ (1.31)
and the position vector r as
(1.32)
As we noted in Section 1.2, this expansion of r in terms of its Cartesian components is especially easy to differentiate because the unit vectors X, S7 , are constant. Thus we can differentiate (1.32) twice to get the simple result
F=Ii±j3ST+Ei. (1.33)
10 The magnetic force between two steady currents is not necessarily small, even in the classical domain, but it can be shown that this force does obey the third law. See Problem 1.33.
” The force F can sometimes involve derivatives of r. (For instance the magnetic force on a moving charge involves the velocity v = 1-.) Very occasionally the force F involves a higher derivative of r, of order n > 2, in which case the second law is an nth-order differential equation.
24 Chapter 1 Newton’s Laws of Motion
That is, the three Cartesian components ofr are just the appropriate derivatives of the three coordinates x, y, z of r, and the second law (1.30) becomes
FX x f Fy Fz i = + m53 ST mE (1.34)
Resolving this equation into its three separate components, we see that Fx has to equal mz and similarly for the y and z components. That is, in Cartesian coordinates, the single vector equation (1.30) is equivalent to the three separate equations:
Fx =
F = mr < > Fy, = m5; (1.35) Fz =
This beautiful result, that, in Cartesian coordinates, Newton’s second law in three dimensions is equivalent to three one-dimensional versions of the same law, is the basis of the solution of almost all simple mechanics problems in Cartesian coordinates. Here is an example to remind you of how such problems go.
EXAMPLE 1.1 A Block Sliding down an Incline
A block of mass in is observed accelerating from rest down an incline that has coefficient of friction ,u and is at angle 8 from the horizontal. How far will it travel in time t?
Our first task is to choose our frame of reference. Naturally, we choose our spatial origin at the block’s starting position and the origin of time (t = 0) at the moment of release. As you no doubt remember from your introductory physics course, the best choice of axes is to have one axis (x say) point down the slope, one (y) normal to the slope, and the third (z) across it, as shown in Figure 1.9. This choice has two advantages: First, because the block slides straight down the slope, the motion is entirely in the x direction, and only x varies. (If we had chosen the x axis horizontal and the y axis vertical, then both x and y would vary.) Second, two of the three forces on the block are unknown (the normal force N and friction f; the weight, w = mg, we treat as known), and with our choice of axes, each of the unknowns has only one nonzero component, since N is in the y direction and f is in the (negative) x direction.
We are now ready to apply Newton’s second law. The result (1.35) means that we can analyse the three components separately, as follows:
There are no forces in the z direction, so Fz = 0. Since Fz = mE, it follows that E = 0, which implies that z (or v z ) is constant. Since the block starts from rest, this means that z is actually zero for all t. With z = 0, it follows that z is constant, and, since it too starts from zero, we conclude that z = 0 for all t. As we would certainly have guessed, the motion remains in the xy plane.
Since the block does not jump off the incline, we know that there is no motion in the y direction. In particular, ji = 0. Therefore, Newton’s second law implies that the y component of the net force is zero; that is, Fy = 0. From Figure 1.9 we see that this implies that
F = N — mg cos 0 = 0.
Section 1.6 Newton’s Second Law in Cartesian Coordinates 25
Figure 1.9 A block slides down a slope of incline O. The three forces on the block are its weight, w = mg, the normal force of the incline, N, and the frictional force f, whose magnitude is f = ,u,N . The z axis is not shown but points out of the page, that is, across the slope.
Thus the y component of the second law has told us that the unknown normal force is N = mg cos 9. Since f = ,u,N, this tells us the frictional force, f = pung cos 9, and all the forces are now known. All that remains is to use the remaining component (the y component) of the second law to solve for the actual motion.
The x component of the second law, Fx = mx , implies (see Figure 1.9) that
wx – f =Ira
Or
mg sin 0 — ,umg cos° =
The m’s cancel, and we find for the acceleration down the slope
= g (sin 0 — µ cos 0). (1.36)
Having found x, and found it to be constant, we have only to integrate it twice to find x as a function of t. First
= g (sin 0 — µ cos 6)t
(Remember that x = 0 initially, so the constant of integration is zero.) Finally,
x(t) = 0 — cos t9)t 2
(again the constant of integration is zero) and our solution is complete.
26 Chapter 1 Newton’s Laws of Motion
Figure 1.10 The definition of the polar coordinates r and 0.
1.7 Two-Dimensional Polar Coordinates
While Cartesian coordinates have the merit of simplicity, we are going to find that it is almost impossible to solve certain problems without the use of various non-Cartesian coordinate systems. To illustrate the complexities of non-Cartesian coordinates, let us consider the form of Newton’s second law in a two-dimensional problem using polar coordinates. These coordinates are defined in Figure 1.10. Instead of using the two rectangular coordinates x, y, we label the position of a particle with its distance r from 0 and the angle 0 measured up from the x axis. Given the rectangular coordinates x and y, you can calculate the polar coordinates r and 0, or vice versa, using the following relations. (Make sure you understand all four equations.’ 2 )
x = r cos y = r sin 0
r = x2 + y2 = arctan(y/x)
(1.37)
Just as with rectangular coordinates, it is convenient to introduce two unit vectors, which I shall denote by ii- and 0. To understand their definitions, notice that we can define the unit vector x as the unit vector that points in the direction of increasing x when y is fixed, as shown in Figure 1.11(a). In the same way we shall define i s- as the unit vector that points in the direction we move when r increases with 0 fixed; likewise, S is the unit vector that points in the direction we move when 0 increases with r fixed. Figure 1.11 makes clear a most important difference between the unit vectors x and Sr of rectangular coordinates and our new unit vectors r and 4. The vectors X and Sr are the same at all points in the plane, whereas the new vectors r and ofi change their directions as the position vector r moves around. We shall see that this complicates the use of Newton’s second law in polar coordinates.
Figure 1.11 suggests another way to write the unit vector r. Since r is in the same direction as r, but has magnitude 1, you can see that
r
r (1.38)
This result suggests a second role for the “hat” notation. For any vector a, we can define fi as the unit vector in the direction of a, namely a = a/lal.
12 There is a small subtlety concerning the equation for 4): You need to make sure lands in the proper quadrant, since the first and third quadrants give the same values for y/x (and likewise the second and fourth). See Problem 1.42.
Section 1.7 Two-Dimensional Polar Coordinates 27
y
(a)
(b)
Figure 1.11 (a) The unit vector x points in the direction of increas- ing x with y fixed. (b) The unit vector r points in the direction of increasing r with 0 fixed; 3 points in the direction of increasing 0 with r fixed. Unlike x, the vectors r and 4) change as the position vector r moves.
Since the two unit vectors 1- and 4) are perpendicular vectors in our two-dimensional space, any vector can be expanded in terms of them. For instance, the net force F on an object can be written
F = Foi‘b. (1.39)
If, for example, the object in question is a stone that I am twirling in a circle on the end of a string (with my hand at the origin), then Fr would be the tension in the string and Fo the force of air resistance retarding the stone in the tangential direction. The expansion of the position vector itself is especially simple in polar coordinates. From Figure 1.11(b) it is clear that
r = . (1.40)
We are now ready to ask about the form of Newton’s second law, F = mr, in polar coordinates. In rectangular coordinates, we saw that the x component of F is just z , and this is what led to the very simple result (1.35). We must now find the components of r in polar coordinates; that is, we must differentiate (1.40) with respect to t. Although (1.40) is very simple, the vector r changes as r moves. Thus when we differentiate (1.40), we shall pick up a term involving the derivative of 1′. Our first task is to find this derivative of ii-.
Figure 1.12(a) shows the position of the particle of interest at two successive times, t 1 and t2 = t 1 + At. If the corresponding angles 0 (t 1) and 0 (t2) are different, then the two unit vectors i(t 1) and 1- (t2) point in different directions. The change in i s- is shown in Figure 1.12(b), and (provided At is small) is approximately
Or AO 4)
At 4). (1.41)
(Notice that the direction of Ai is perpendicular to r , namely the direction of 4).) If we divide both sides by At and take the limit as At -÷ 0, then Ail At —> I dt and we find that
dt (1.42)
28 Chapter 1 Newton’s Laws of Motion
(a)
(b)
Figure 1.12 (a) The positions of a particle at two successive times, t j and t2 . Unless the particle is moving exactly radially, the corresponding unit vectors 1-(t 1 ) and 1.(t2 ) point in different directions. (b) The change Ai- in r is given by the triangle shown.
(For an alternative proof of this important result, see Problem 1.43.) Notice that di/dt is in the direction of 0 and is proportional to the rate of change of the angle 0 — both of which properties we would expect based on Figure 1.12.
Now that we know the derivative of r, we are ready to differentiate Equation (1.40). Using the product rule, we get two terms:
dr = rr — ,
dt
and, substituting (1.42), we find for the velocity t, or v,
v = r (1.43)
From this we can read off the polar components of the velocity:
vr = r and v0 = rc = r w (1.44)
where in the second equation I have introduced the traditional notation w for the an- gular velocity (p. While the results in (1.44) should be familiar from your introductory physics course, they are undeniably more complicated than the corresponding results in Cartesian coordinates (v x = z and vy = 57).
Before we can write down Newton’s second law, we have to differentiate a second time to find the acceleration:
d . d ., • a = —r = — (rr rrP 0),
dt dt (1.45)
where the final expression comes from substituting (1.43) for r. To complete the differentiation in (1.45), we must calculate the derivative of 0. This calculation is completely analogous to the argument leading to (1.42) and is illustrated in Figure 1.13. By inspecting this figure, you should be able to convince yourself that
d4) dt
(1.46)
Section 1.7 Two -Dimensional Polar Coordinates 29
(a)
(b)
Figure 1.13 (a) The unit vector at two successive times t 1 and t2 . (b) The change AO.
Returning to Equation (1.45), we can now carry out the differentiation to give the following five terms:
a = + i + (i4 + r.)—d4)
dt dt
or, if we use (1.42) and (1.46) to replace the derivatives of the two unit vectors,
a = — r .152 )1- +(r ± 24) isfi (1.47)
This horrible result is a little easier to understand if we consider the special case that r is constant, as is the case for a stone that I twirl on the end of a string of fixed length. With r constant, both derivatives of r are zero, and (1.47) has just two terms:
a ,— ri4
or
a = —rw2i +
where w = denotes the angular velocity and a = is the angular acceleration. This is the familiar result from elementary physics that when a particle moves around a fixed circle, it has an inward “centripetal” acceleration r co2 (or v 2 I r) and a tangential acceleration, ra. Nevertheless, when r is not constant, the acceleration includes all four of the terms in (1.47). The first term, F in the radial direction is what you would probably expect when r varies, but the final term, 2i in the direction, is harder to understand. It is called the Coriolis acceleration, and I shall discuss it in detail in Chapter 9.
Having calculated the acceleration as in (1.47), we can finally write down Newton’s second law in terms of polar coordinates:
F = ma {
Fr = m(F — 42) (1.48)
1 = m (rs;6 21-
These equations in polar coordinates are a far cry from the beautifully simple equa- tions (1.35) for rectangular coordinates. In fact, one of the main reasons for taking the
30 Chapter 1 Newton’s Laws of Motion
trouble to recast Newtonian mechanics in the Lagrangian formulation (Chapter 7) is that the latter is able to handle nonrectangular coordinates just as easily as rectangular.
You may justifiably be feeling that the second law in polar coordinates is so complicated that there could be no occasion to use it. In fact, however, there are many problems which are most easily solved using polar coordinates, and I conclude this section with an elementary example.
EXAMPLE 1.2 An Oscillating Skateboard
A “half-pipe” at a skateboard park consists of a concrete trough with a semicircu- lar cross section of radius R = 5 m, as shown in Figure 1.14. I hold a frictionless skateboard on the side of the trough pointing down toward the bottom and release it. Discuss the subsequent motion using Newton’s second law. In particular, if I release the board just a short way from the bottom, how long will it take to come back to the point of release?
Because the skateboard is constrained to move on a circular path, this prob- lem is most easily solved using polar coordinates with origin 0 at the center of the pipe as shown. (At some point in the following calculation, try writing the second law in rectangular coordinates and observe what a tangle you get.) With this choice of polar coordinates, the coordinate r of the skateboard is constant, r = R, and the position of the skateboard is completely specified by the angle 0. With r constant, the second law (1.48) takes the relatively simple form
Fr = —mR(P 2 (1.49)
Fo = m6. (1.50)
The two forces on the skateboard are its weight w = mg and the normal force N of the wall, as shown in Figure 1.14. The components of the net force F = w + N are easily seen to be
Fr = mg cos 4) — N and Fo = —mg sin 4).
w = mg
Figure 1.14 A skateboard in a semicircular trough of radius R. The board’s position is specified by the angle measured up from the bottom. The two forces on the skateboard are its weight w = mg and the normal force N.
Section 1.7 Two -Dimensional Polar Coordinates 31
Substituting for Fr into (1.49) we get an equation involving N, 0, and Fortunately, we are not really interested in N, and — even more fortunately — when we substitute for Fo into (1.50), we get an equation that does not involve
N at all:
—mg sin 0 = m /?
or, canceling the m’s and rearranging,
= — 8: sin 0. (1.51)
Equation (1.51) is the differential equation for 0(t) that determines the motion of the skateboard. Qualitatively, we can easily see the kind of motion that it implies. First, if 0 = 0, (1.51) says that = 0. Therefore, if we place the board at rest ( = 0) at the point 0 = 0, the board will never move (unless someone pushes it); that is, 0 = 0 is an equilibrium position, as you would certainly have guessed. Next, suppose that at some time, 0 is not zero and, to be definite, suppose that 0 > 0; that is, the skateboard is on the right-hand side of the half-pipe. In this case, (1.51) implies that < 0, so the acceleration is directed to the left. If the board is moving to the right it must slow down and eventually start moving to the left. 13 Once it is moving toward the left, it speeds up and returns to the bottom, where it moves over to the left. As soon as the board is on the left, the argument reverses (0 < 0, so 4 > 0) and the board must eventually return to the bottom and move over to the right again. In other words, the differential equation (1.51) implies that the skateboard oscillates back and forth, from right to left and back to the right.
The equation of motion (1.51) cannot be solved in terms of elementary func- tions, such as polynomials, trigonometric functions, or logs and exponentials. 14
Thus, if we want more quantitative information about the motion, the simplest course is to use a computer to solve it numerically (see Problem 1.50). However, if the initial angle 0, is small, we can use the small angle approximation
sin 0
(1.52)
and, within this approximation, (1.51) becomes
= (1.53)
which can be solved using elementary functions. [By this stage, you have al- most certainly recognized that our discussion of the skateboard problem closely parallels the analysis of the simple pendulum. In particular, the small-angle
13 I am taking for granted that it doesn’t reach the top and jump out of the trough. Since it was released from rest inside the trough, this is correct. Much the easiest way to prove this claim is to invoke conservation of energy, which we shan’t be discussing for a while. Perhaps, for now, you could agree to accept it as a matter of common sense.
14 Actually the solution of (1.51) is a Jacobi elliptic function. However, I shall take the point of view that for most of us the Jacobi function is not “elementary.”
32 Chapter 1 Newton’s Laws of Motion
approximation (1.52) is what let you solve the simple pendulum in your intro- ductory physics course. This parallel is, of course, no accident. Mathematically the two problems are exactly equivalent.] If we define the parameter
= R, (1.54) then (1.53) becomes
(1.55)
This is the equation of motion for our skateboard in the small-angle approxima- tion. I would like to discuss its solution in some detail to introduce several ideas that we’ll be using again and again in what follows. (If you’ve studied differential equations before, just see the next three paragraphs as a quick review.)
We first observe that it is easy to find two solutions of the equation (1.55) by inspection (that is, by inspired guessing). The function 0 (t) = A sin(cot) is clearly a solution for any value of the constant A. [Differentiating sin (cot) brings out a factor of w and changes the sin to a cos; differentiating it again brings out another w and changes the cos back to —sin. Thus the proposed solution does satisfy 4 = —co20.] Similarly, the function OW = B cos(wt) is another solution for any constant B. Furthermore, as you can easily check, the sum of these two solutions is itself a solution. Thus we have now found a whole family of solutions:
(t) = A sin (cot) B cos(wt) (1.56)
is a solution for any values of the two constants A and B. I now want to argue that every solution of the equation of motion (1.55)
has the form (1.56). In other words, (1.56) is the general solution—we have found all solutions, and we need seek no further. To get some idea of why this is, note that the differential equation (1.55) is a statement about the second derivative 4) of the unknown 0. Now, if we had actually been told what is, then we know from elementary calculus that we could find 0 by two integrations, and the result would contain two unknown constants — the two constants of integration — that would have to be determined by looking (for example) at the initial values of 0 and In other words, knowledge of would tell us that 0 itself is one of a family of functions containing precisely two undetermined constants. Of course, the differential equation (1.55) does not actually tell us
— it is an equation for in terms of 0. Nevetheless, it is plausible that such an equation would imply that 0 is one of a family of functions that contain precisely two undetermined constants. If you have studied differential equations, you know that this is the case; if you have not, then I must ask you to accept it as a plausible fact: For any given second-order differential equation [in a large class of “reasonable” equations, including (1.55) and all of the equations we shall encounter in this book], the solutions all belong to a family of functions
Principal Definitions and Equations of Chapter 1 33
containing precisely two independent constants — like the constants A and B in (1.56). (More generally, the solutions of an nth-order equation contain precisely n independent constants.)
This theorem sheds a new light on our solution (1.56). We already knew that any function of the form (1.56) is a solution of the equation of motion. Our theorem now guarantees that every solution of the equation of motion is of this form. This same argument applies to all the second-order differential equations we shall encounter. If, by hook or by crook, we can find a solution like (1.56) involving two arbitrary constants, then we are guaranteed that we have found the general solution of our equation.
All that remains is to pin down the two constants A and B for our skateboard. To do so, we must look at the initial conditions. At t = 0, Equation (1.56) implies that 0 = B. Therefore B is just the initial value of 0, which we are calling 0 o , so B = O.. At t = 0, Equation (1.56) implies that = coA . Since I released the board from rest, this means that A = 0, and our solution is
(t) = cos(cot). (1.57)
The first thing to note about this solution is that, as we anticipated on general grounds, 0 (t) oscillates, moving from positive to negative and back to positive periodically and indefinitely. In particular, the board first returns to its initial position 00 when wt = 27r. The time that this takes is called the period of the motion and is denoted z. Thus our conclusion is that the period of the skateboard’s oscillations is
27r = — = 27T
w
R
g (1.58)
We were given that R = 5 m, and g = 9.8 m/s2 . Substituting these numbers, we conclude that the skateboard returns to its starting point in a time r = 4.5 seconds.
Principal Definitions and Equations of Chapter 1
Dot and Cross Products
r•s= rs cos 9 = rx sx ry sy r,s, [Eqs. (1.6) & (1.7)]
1 rxs= (ry sz — rz sy , rz sx — rx sz , rx sy —ry sx )=det rx y r,
sx sY sz [Eq. (1.9)]
34 Chapter 1 Newton’s Laws of Motion
Inertial Frames
An inertial frame is any reference frame in which Newton’s first law holds, that is, a
nonaccelerating, nonrotating frame.
Unit Vectors of a Coordinate System
If 17, 0 are an orthogonal system of coordinates, then
= unit vector in direction of increasing c with ri and fixed
and so on, and any vector s can be expanded as s =
Newton’s Second Law in Various Coordinate Systems
Vector Form Cartesian 2D Polar
Cylindrical Polar
(x, y, z) (r, (p, 0, z)
{Fx = mi F=mr F = mj;
Fz = mE
IF, = m(F — r 2) (F = — m + 24)
{
Fr = M(16 142 ) F,, = m(pq5 + ZOO)
= mE
Problem 1.47 or 1.48 Eq. (1.35) Eq. (1.48)
Problems for Chapter 1
The problems for each chapter are arranged according to section number. A problem listed for a given
section requires an understanding of that section and earlier sections, but not of later sections. Within each section problems are listed in approximate order of difficulty. A single star (*) indicates straightforward
problems involving just one main concept. Two stars (**) identify problems that are slightly more challenging
and usually involve more than one concept. Three stars (***) indicate problems that are distinctly more
challenging, either because they are intrinsically difficult or involve lengthy calculations. Needless to say,
these distinctions are hard to draw and are only approximate. Problems that need the use of a computer are flagged thus: [Computer]. These are mostly classified as
*** on the grounds that it usually takes a long time to set up the necessary code — especially if you’re just learning the language.
SECTION 1.2 Space and Time
1.1* Given the two vectors b = + Sr and c = + find b c, 5b + 2c, b • c, and b x c.
1.2 * Two vectors are given as b = (1, 2, 3) and c = (3, 2, 1). (Remember that these statements are just a compact way of giving you the components of the vectors.) Find b c, 5b — 2c, b • c, and b x c.
1.3 * By applying Pythagoras’s theorem (the usual two-dimensional version) twice over, prove that the length r of a three-dimensional vector r = (x, y, z) satisfies r2 = x2 + y2 + z2.
Problems for Chapter 1 35
1.4 * One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors b = (1, 2, 4) and c = (4, 2, 1) by evaluating their scalar product.
1.5 * Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at 0 and the opposite corner at the point (1, 1, 1). Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.]
1.6 * By evaluating their dot product, find the values of the scalar s for which the two vectors b = x + s’Sr and c = x — sS7 are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.
1.7 * Prove that the two definitions of the scalar product r • s as rs cos 9 (1.6) and E ri s, (1.7) are equal. One way to do this is to choose your x axis along the direction of r. [Strictly speaking you should first make sure that the definition (1.7) is independent of the choice of axes. If you like to worry about such niceties, see Problem 1.16.]
1.8 * (a) Use the definition (1.7) to prove that the scalar product is distributive, that is, r • (u + v) = r • u + r • v. (b) If r and s are vectors that depend on time, prove that the product rule for differentiating products applies to r • s, that is, that
ds dr —
dt (r • s) r •
dt +
dt s .
1.9 * In elementary trigonometry, you probably learned the law of cosines for a triangle of sides a, b, and c, that c2 = a2 b2 — 2ab cos 9, where 9 is the angle between the sides a and b. Show that the law of cosines is an immediate consequence of the identity (a + b) 2 = a2 b2 + 2a • b.
1.10 * A particle moves in a circle (center 0 and radius R) with constant angular velocity w counter- clockwise. The circle lies in the xy plane and the particle is on the x axis at time t = 0. Show that the particle’s position is given by
r(t) = xR cos(wt) + y R sin(wt).
Find the particle’s velocity and acceleration. What are the magnitude and direction of the acceleration? Relate your results to well-known properties of uniform circular motion.
1.11 * The position of a moving particle is given as a function of time t to be
r(t) = xb cos(wt) + ST‘c sin(wt),
where b, c, and w are constants. Describe the particle’s orbit.
1.12 * The position of a moving particle is given as a function of time t to be
r(t) = xb cos(wt) + S7c sin(wt) + iv ot
where b, c, v o and co are constants. Describe the particle’s orbit.
1.13 * Let u be an arbitrary fixed unit vector and show that any vector b satisfies
b2 (u •b) 2 + (u x b) 2 .
Explain this result in words, with the help of a picture.
36 Chapter 1 Newton’s Laws of Motion
1.14 * Prove that for any two vectors a and b,
la + bi < (a + b).
[Hint: Work out la 131 2 and compare it with (a + b)2 .] Explain why this is called the triangle inequality.
1.15 * Show that the definition (1.9) of the cross product is equivalent to the elementary definition that r x s is perpendicular to both r and s, with magnitude rs sin 0 and direction given by the right-hand rule. [Hint: It is a fact (though quite hard to prove) that the definition (1.9) is independent of your choice of axes. Therefore you can choose axes so that r points along the x axis and s lies in the xy plane.]
1.16 ** (a) Defining the scalar product r • s by Equation (1.7), r • s = E ri si , show that Pythagoras’s theorem implies that the magnitude of any vector r is r = • r. (b) It is clear that the length of a vector does not depend on our choice of coordinate axes. Thus the result of part (a) guarantees that the scalar product r • r, as defined by (1.7), is the same for any choice of orthogonal axes. Use this to prove that r • s, as defined by (1.7), is the same for any choice of orthogonal axes. [Hint: Consider the length of the vector r
1.17 ** (a) Prove that the vector product r x s as defined by (1.9) is distributive; that is, that r x (u v) =— (r x u) (r x v). (b) Prove the product rule
d ds dr
dt —(rxs)=rx
dt at xs.
Be careful with the order of the factors.
1.18 ** The three vectors a, b, c are the three sides of the triangle ABC with angles a, ,8, y as shown in Figure 1.15. (a) Prove that the area of the triangle is given by any one of these three expressions:
area = -2 1a x bl = -2 lb X CI = -2 x al.
(b) Use the equality of these three expressions to prove the so-called law of sines, that
a
sin a sin /3 sin y
Figure 1.15 Triangle for Problem 1.18.
Problems for Chapter 1 37
1.19 ** If r, v, a denote the position, velocity, and acceleration of a particle, prove that
— d
[a • (v x r)] = a • (v x r). dt
1.20 ** The three vectors A, B, C point from the origin 0 to the three corners of a triangle. Use the result of Problem 1.18 to show that the area of the triangle is given by
(area of triangle) = 2I(B x C) + (C x A) + (A x B)1.
1.21 ** A parallelepiped (a six-faced solid with opposite faces parallel) has one corner at the origin 0 and the three edges that emanate from 0 defined by vectors a, b, c. Show that the volume of the parallelepiped is la • (b x
1.22 ** The two vectors a and b lie in the xy plane and make angles a and with the x axis. (a) By evaluating a • b in two ways [namely using (1.6) and (1.7)] prove the well-known trig identity
cos(a — 0) = cos a cos 0 + sin a sin ,8.
(b) By similarly evaluating a x b prove that
sin(a — 0) = sin a cos /3 — cos a sin 8.
1.23 ** The unknown vector v satisfies b • v = A and b x v = c, where A, b, and c are fixed and known. Find v in terms of A, b, and c.
SECTION 1.4 Newton’s First and Second Laws; Inertial Frames
1.24 * In case you haven’t studied any differential equations before, I shall be introducing the necessary ideas as needed. Here is a simple excercise to get you started: Find the general solution of the first- order equation df/dt = f for an unknown function f(t). [There are several ways to do this. One is to rewrite the equation as df/f = dt and then integrate both sides.] How many arbitrary constants does the general solution contain? [Your answer should illustrate the important general theorem that the solution to any nth-order differential equation (in a very large class of “reasonable” equations) contains n arbitrary constants.]
1.25 * Answer the same questions as in Problem 1.24, but for the differential equation dfldt = —3f .
1.26 ** The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following: I am standing on a level floor at the origin of an inertial frame S and kick a frictionless puck due north across the floor. (a) Write down the x and y coordinates of the puck as functions of time as seen from my inertial frame. (Use x and y axes pointing east and north respectively.) Now consider two more observers, the first at rest in a frame 8′ that travels with constant velocity v due east relative to 8, the second at rest in a frame 8″ that travels with constant acceleration due east relative to S. (All three frames coincide at the moment when I kick the puck, and 8″ is at rest relative to 8 at that same moment.) (b) Find the coordinates x’, y’ of the puck and describe the puck’s path as seen from 8′. (c) Do the same for 8″. Which of the frames is inertial?
1.27 ** The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following experiment: I am
38 Chapter 1 Newton’s Laws of Motion
standing on the ground (which we shall take to be an inertial frame) beside a perfectly flat horizontal turntable, rotating with constant angular velocity co. I lean over and shove a frictionless puck so that it slides across the turntable, straight through the center. The puck is subject to zero net force and, as seen from my inertial frame, travels in a straight line. Describe the puck’s path as observed by someone sitting at rest on the turntable. This requires careful thought, but you should be able to get a qualitative picture. For a quantitative picture, it helps to use polar coordinates; see Problem 1.46.
SECTION 1.5 The Third Law and Conservation of Momentum
1.28 * Go over the steps from Equation (1.25) to (1.29) in the proof of conservation of momentum, but treat the case that N = 3 and write out all the summations explicitly to be sure you understand the various manipulations.
1.29 * Do the same tasks as in Problem 1.28 but for the case of four particles (N = 4).
1.30 * Conservation laws, such as conservation of momentum, often give a surprising amount of information about the possible outcome of an experiment. Here is perhaps the simplest example: Two objects of masses m 1 and m 2 are subject to no external forces. Object 1 is traveling with velocity v when it collides with the stationary object 2. The two objects stick together and move off with common velocity v’. Use conservation of momentum to find v’ in terms of v, m 1 , and m 2 .
1.31 * In Section 1.5 we proved that Newton’s third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that F12 = — F21.]
1.32 ** If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure 1.8. The electric and magnetic fields at a point r 1
due to a charge q2 at r2 moving with constant velocity v 2 (with v2 << c) are l5
1 q2 \ [to q2 v 2 E(r i) = s and B(r 1, = — •
47r s2 47r s 2
where s = r 1 — r2 is the vector pointing from r 2 to r 1 . (The first of these you should recognize as Coulomb’s law.) If F712 and Fmi2ag denote the electric and magnetic forces on a charge q 1 at r 1 with velocity v 1 , show that Fr2ag < (v v2 /c2)Fie21 . This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges.
1.33 *** If you have some experience in electromagnetism and with vector calculus, prove that the magnetic forces, F12 and F21, between two steady current loops obey Newton’s third law. [Hints: Let the two currents be I I and /2 and let typical points on the two loops be r 1 and r2 . If dr 1 and dr2 are short segments of the loops, then according to the Biot—Savart law, the force on dr 1 due to dr2 is
it. /1 /2 — — di% x (dr2 x g) 47r s 2
where s = r 1 — r2 . The force F12 is found by integrating this around both loops. You will need to use the “B AC — CAB” rule to simplify the triple product.]
15 See, for example, David J. Griffiths, Introduction to Electrodynamics, 3rd ed., Prentice Hall, (1999), p. 440.
Problems for Chapter 1 39
1.34 *** Prove that in the absence of external forces, the total angular momentum (defined as L = ED, ra x pa ) of an N-particle system is conserved. [Hints: You need to mimic the argument from (1.25) to (1.29). In this case you need more than Newton’s third law: In addition you need to assume that the interparticle forces are central; that is, Fap acts along the line joining particles ce and $. A full discussion of angular momentum is given in Chapter 3.]
SECTION 1.6 Newton’s Second Law in Cartesian Coordinates
1.35 * A golf ball is hit from ground level with speed v o in a direction that is due east and at an angle 6 above the horizontal. Neglecting air resistance, use Newton’s second law (1.35) to find the position as a function of time, using coordinates with x measured east, y north, and z vertically up. Find the
time for the golf ball to return to the ground and how far it travels in that time.
1.36 * A plane, which is flying horizontally at a constant speed v o and at a height h above the sea, must drop a bundle of supplies to a castaway on a small raft. (a) Write down Newton’s second law for the bundle as it falls from the plane, assuming you can neglect air resistance. Solve your equations to give the bundle’s position in flight as a function of time t. (b) How far before the raft (measured horizontally) must the pilot drop the bundle if it is to hit the raft? What is this distance if v o = 50 m/s,
h = 100 m, and g ti 10 m/s2 ? (c) Within what interval of time (+At) must the pilot drop the bundle if it is to land within +10 m of the raft?
1.37 * A student kicks a frictionless puck with initial speed v o , so that it slides straight up a plane that is inclined at an angle 0 above the horizontal. (a) Write down Newton’s second law for the puck and solve to give its position as a function of time. (b) How long will the puck take to return to its starting point?
1.38 * You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle 0 with the horizontal. Choose your origin at one of the two corners that touch the floor, the x axis pointing along the bottom edge of the board, the y axis pointing up the slope, and
the z axis normal to the board. You now kick a frictionless puck that is resting at 0 so that it slides across the board with initial velocity (v ox , voy , 0). Write down Newton’s second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from 0
when it does so.
1.39 ** A ball is thrown with initial speed v o up an inclined plane. The plane is inclined at an angle above the horizontal, and the ball’s initial velocity is at an angle 0 above the plane. Choose axes
with x measured up the slope, y normal to the slope, and z across it. Write down Newton’s second law using these axes and find the ball’s position as a function of time. Show that the ball lands a distance R = 2v 2 sin 0 cos(0 0)1(g cos t 0) from its launch point. Show that for given v o and 0, the maximum 0 possible range up the inclined plane is R. = vo /[g(1 + sin OA
1.40 *** A cannon shoots a ball at an angle 6 above the horizontal ground. (a) Neglecting air resistance, use Newton’s second law to find the ball’s position as a function of time. (Use axes with x measured
horizontally and y vertically.) (b) Let r(t) denote the ball’s distance from the cannon. What is the largest possible value of 0 if r (t) is to increase throughout the ball’s flight? [Hint: Using your solution
to part (a) you can write down r 2 as x 2 + y2 , and then find the condition that r 2 is always increasing.]
40 Chapter 1 Newton’s Laws of Motion
SECTION 1.7 Two-Dimensional Polar Coordinates
1.41 * An astronaut in gravity-free space is twirling a mass m on the end of a string of length R in a circle, with constant angular velocity co. Write down Newton’s second law (1.48) in polar coordinates and find the tension in the string.
1.42 * Prove that the transformations from rectangular to polar coordinates and vice versa are given by the four equations (1.37). Explain why the equation for 0 is not quite complete and give a complete version.
1.43 * (a) Prove that the unit vector r of two-dimensional polar coordinates is equal to
= X cos 0 + 5r sin 0 (1.59)
and find a corresponding expression for 4. (b) Assuming that 0 depends on the time t, differentiate your answers in part (a) to give an alternative proof of the results (1.42) and (1.46) for the time derivatives
and cb.
1.44 * Verify by direct substitution that the function 0 (t) = A sin(wt) + B cos(wt) of (1.56) is a solution of the second-order differential equation (1.55), if, = —co20. (Since this solution involves two arbitrary constants — the coefficients of the sine and cosine functions it is in fact the general solution.)
1.45 ** Prove that if v(t) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then v(t) is orthogonal to v(t). Prove the converse that if ‘(t) is orthogonal to v (t), then Iv (t) I is constant. [Hint: Consider the derivative of v2 .] This is a very handy result. It explains why, in two-dimensional polars, di I dt has to be in the direction of 0 and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity.
1.46 ** Consider the experiment of Problem 1.27, in which a frictionless puck is slid straight across a rotating turntable through the center 0. (a) Write down the polar coordinates r, 0 of the puck as functions of time, as measured in the inertial frame S of an observer on the ground. (Assume that the puck was launched along the axis 0 = 0 at t = 0.) (b) Now write down the polar coordinates r’, 0′ of the puck as measured by an observer (frame S’) at rest on the turntable. (Choose these coordinates so that 0 and 0′ coincide at t = 0.) Describe and sketch the path seen by this second observer. Is the frame S’ inertial?
1.47 ** Let the position of a point P in three dimensions be given by the vector r = (x, y, z) in rectangular (or Cartesian) coordinates. The same position can be specified by cylindrical polar coordinates, p, 0, z, which are defined as follows: Let P’ denote the projection of P onto the xy plane; that is, P’ has Cartesian coordinates (x, y, 0). Then p and 0 are defined as the two-dimensional polar coordinates of P’ in the xy plane, while z is the third Cartesian coordinate, unchanged. (a) Make a sketch to illustrate the three cylindrical coordinates. Give expressions for p, z in terms of the Cartesian coordinates x, y, z. Explain in words what p is (“p is the distance of P from “). There are many variants in notation. For instance, some people use r instead of p. Explain why this use of r is unfortunate. (b) Describe the three unit vectors p, 0, i and write the expansion of the position vector r in terms of these unit vectors. (c) Differentiate your last answer twice to find the cylindrical components of the acceleration a =r of the particle. To do this, you will need to know the time derivatives of j) and 0. You could get these from the corresponding two-dimensional results (1.42) and (1.46), or you could derive them directly as in Problem 1.48.
Problems for Chapter 1 41
1.48 ** Find expressions for the unit vectors j), 4, and i of cylindrical polar coordinates (Problem 1.47) in terms of the Cartesian X, y , i. Differentiate these expressions with respect to time to find 01dt, clikldt, and dildt.
1.49 ** Imagine two concentric cylinders, centered on the vertical z axis, with radii R ± E, where E is very small. A small frictionless puck of thickness 2E is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates (p, 0, z) for its position (Problem 1.47), then p is fixed at p = R. while 0 and z can vary at will. Write down and solve Newton’s second law for the general motion of the puck, including the effects of gravity. Describe the puck’s motion.
1.50 *** [Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from 0 ° = 20 degrees, using the
values R = 5 m and g = 9.8 m/s2 . Make a plot of 0 against time for two or three periods. (b) On the same picture, plot the approximate solution (1.57) with the same 00 = 20°. Comment on your two graphs. Note: If you haven’t used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for “NDSolve” and how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.
1.51 *** [Computer] Repeat all of Problem 1.50 but using the initial value 0 0 = r/2.
CHAPTER
Projectiles and Charged Particles
In this chapter, I present two topics: the motion of projectiles subject to the forces of gravity and air resistance, and the motion of charged particles in uniform magnetic fields. Both problems lend themselves to solution using Newton’s laws in Cartesian coordinates, and both allow us to review and introduce some important mathematics. Above all, both are problems of great practical interest.
2.1 Air Resistance
Most introductory physics courses spend some time studying the motion of projectiles, but they almost always ignore air resistance. In many problems this is an excellent approximation; in others, air resistance is obviously important, and we need to know how to account for it. More generally, whether or not air resistance is significant, we need some way to estimate how important it really is.
Let us begin by surveying some of the basic properties of the resistive force, or drag, f of the air, or other medium, through which an object is moving. (I shall generally speak of “air resistance” since air is the medium through which most projectiles move, but the same considerations apply to other gases and often to liquids as well.) The most obvious fact about air resistance, well known to anyone who rides a bicycle, is that it depends on the speed, v, of the object concerned. In addition, for many objects, the direction of the force due to motion through the air is opposite to the velocity v. For certain objects, such as a nonrotating sphere, this is exactly true, and for many it is a good approximation. You should, however, be aware that there are situations where it is certainly not true: The force of the air on an airplane wing has a large sideways component, called the lift, without which no airplanes could fly. Nevertheless, I shall assume that f and v point in opposite directions; that is, I shall consider only objects for which the sideways force is zero, or at least small enough
43
44 Chapter 2 Projectiles and Charged Particles
w = mg
Figure 2.1 A projectile is subject to two forces, the force
of gravity, w = mg, and the drag force of air resistance,
f = — f (v)ir.
to be neglected. The situation is illustrated in Figure 2.1 and is summed up in the equation
f = – f (v)v, (2.1)
where V = v/ Iv I denotes the unit vector in the direction of v, and f (v) is the magnitude of f.
The function f (v) that gives the magnitude of the air resistance varies with v in a complicated way, especially as the object’s speed approaches the speed of sound. However, at lower speeds it is often a good approximation to write’
f (v) = by + cv 2 = f; + f
J J quad (2.2)
where flip and fquad stand for the linear and quadratic terms respectively, J
.flip = by and fquad = CV
2 (2.3)
The physical origins of these two terms are quite different: The linear term, f lip, arises from the viscous drag of the medium and is generally proportional to the viscosity of the medium and the linear size of the projectile (Problem 2.2). The quadratic term,
fquad’ arises from the projectile’s having to accelerate the mass of air with which it is continually colliding; fquad is proportional to the density of the medium and the cross- sectional area of the projectile (Problem 2.4). In particular, for a spherical projectile (a cannonball, a baseball, or a drop of rain), the coefficients b and c in (2.2) have the form
b = ,8D and c = y D2 (2.4)
where D denotes the diameter of the sphere and the coefficients 6 and y depend on the nature of the medium. For a spherical projectile in air at STP, they have the approximate values
= 1.6 x 10-4 N•s/m2 (2.5)
I Mathematically, Equation (2.2) is, in a sense, obvious. Any reasonable function is expected to have a Taylor series expansion, f = a + by + cv2 For low enough v, the first three terms should give a good approximation, and, since f = 0 when v = 0 the constant term, a, has to be zero.
Section 2.1 Air Resistance 45
and
y = 0.25 N•s 2/m4 . (2.6)
(For calculation of these two constants, see Problems 2.2 and 2.4.) You need to remember that these values are valid only for a sphere moving through air at STP. Nevertheless, they give at least a rough idea of the importance of the drag force even for nonspherical bodies moving through different gases at any normal temperatures and pressures.
It often happens that we can neglect one of the terms in (2.2) compared to the other, and this simplifies the task of solving Newton’s second law. To decide whether this does happen in a given problem, and which term to neglect, we need to compare the sizes of the two terms:
fquad C V 2 yD
.flip by ,8 v = (1 .6 x 10 3 S ) Dv
M2 (2.7)
if we use the values (2.5) and (2.6) for a sphere in air. In a given problem, we have only to substitute the values of D and v into this equation to find out if one of the terms can be neglected, as the following example illustrates.
EXAMPLE 2.1 A Baseball and Some Drops of Liquid
Assess the relative importance of the linear and quadratic drags on a baseball of diameter D =7 cm, traveling at a modest v = 5 m/s. Do the same for a drop of rain (D =1 mm and v = 0.6 m/s) and for a tiny droplet of oil used in the Millikan oildrop experiment (D = 1.5 ,um and v = 5 x 10 -5 m/s).
When we substitute the numbers for the baseball into (2.7) (remembering to convert the diameter to meters), we get
fquad 600 [baseball]. (2.8)
For this baseball, the linear term is clearly negligible and we need consider only the quadratic drag. If the ball is traveling faster, the ratio fquad/fain is even greater. At slower speeds the ratio is less dramatic, but even at 1 m/s the ratio is 100. In fact if v is small enough that the linear term is comparable to the quadratic, both terms are so small as to be negligible. Thus, for baseballs and similar objects, it is almost always safe to neglect A ir, and take the drag force to be
f = (2.9)
For the raindrop, the numbers give
fquad 1 [raindrop].
fain (2.10)
flin
Thus for this raindrop the two terms are comparable and neither can be ne- glected — which makes solving for the motion more difficult. If the drop were
46 Chapter 2 Projectiles and Charged Particles
a lot larger or were traveling much faster, then the linear term would be negligi- ble; and if the drop were much smaller or were traveling much slower, then the quadratic term would be negligible. But in general, with raindrops and similar objects, we are going to have to take both A in and fquad into account.
For the oildrop in the Millikan experiment the numbers give
fquad ti 10 [Millikan oildrop]. (2.11) fun
In this case, the quadratic term is totally negligible, and we can take
f = = (2.12)
where the second, very compact form follows because, of course, vi = v.
The moral of this example is clear: First, there are objects for which the drag force is dominantly linear, and the quadratic force can be neglected — notably, very small liquid drops in air, but also slightly larger objects in a very viscous fluid, such as a ball bearing moving through molasses. On the other hand, for most projectiles, such as golf balls, cannonballs, and even a human in free fall, the dominant drag force is quadratic, and we can neglect the linear term. This situation is a little unlucky because the linear problem is much easier to solve than the quadratic. In the following two sections, I shall discuss the linear case, precisely because it is the easier one. Nevertheless, it does have practical applications, and the mathematics used to solve it is widely used in many fields. In Section 2.4, I shall take up the harder but more usual case of quadratic drag.
To conclude this introductory section, I should mention the Reynolds number, an important parameter that features prominently in more advanced treatments of motion in fluids. As already mentioned, the linear drag f lip can be related to the viscosity of the fluid through which our projectile is moving, and the quadratic term fquad is similarly related to the inertia (and hence density) of the fluid. Thus one can relate the ratio
fquad/flin to the fundamental parameters 77, the viscosity, and 0, the density, of the fluid (see Problem 2.3). The result is that the ratio f quad/flin is of roughly the same order of magnitude as the dimensionless number R = DvQ/r), called the Reynolds number. Thus a compact and general way to summarize the foregoing discussion is to say that the quadratic drag fquad is dominant when the Reynolds number R is large, whereas the linear drag dominates when R is small.
2.2 Linear Air Resistance
Let us consider first a projectile for which the quadratic drag force is negligible, so that the force of air resistance is given by (2.12). We shall see directly that, because the drag force is linear in v, the equations of motion are very simple to solve. The two forces on the projectile are the weight w = mg and the drag force f = —by, as shown in Figure 2.2. Thus the second law, ml = F, reads
ml = mg — by. (2.13)
Section 2.2 Linear Air Resistance 47
Figure 2.2 The two forces on a projectile for which the force of air resistance is linear in the velocity, f = —bv.
An interesting feature of this form is that, because neither of the forces depends on r, the equation of motion does not involve r itself (only the first and second derivatives of r). In fact, we can rewriter as Y, and (2.13) becomes
my = mg — bv, (2.14)
a first-order differential equation for v. This simplification comes about because the forces depend only on v and not r. It means we have to solve only a first-order differential equation for v and then integrate v to find r.
Perhaps the most important simplifying feature of linear drag is that the equation of motion separates into components especially easily. For instance, with x measured to the right and y vertically downward, (2.14) resolves into
mi)x = —by, (2.15)
and
mi)y = mg — bv y . (2.16)
That is, we have two separate equations, one for v x and one for vy ; the equation for vx does not involve v y and vice versa. It is important to recognize that this happened only
because the drag force was linear in v. For instance, if the drag force were quadratic,
f = — cv 2i = —cvv = -C, / V 2 + vy 2 V
(2.17)
then in (2.14) we would have to replace the term —by with (2.17). In place of the two equations (2.15) and (2.16), we would have
MI)X X = — C,1V 2 vy V X
i)y = mg — c\/vx vx vy . Here, each equation involves both of the variables v x and vy . These two coupled differential equations are much harder to solve than the uncoupled equations of the linear case.
Because they are uncoupled, we can solve each equation for linear drag separately and then put the two solutions together. Further, each equation defines a problem that is interesting in its own right. Equation (2.15) is the equation of motion for an object
(2.18)
48 Chapter 2 Projectiles and Charged Particles
Figure 2.3 A cart moves on a horizontal frictionless track in a medium that produces a linear drag force.
(a cart with frictionless wheels, for instance) coasting horizontally in a medium that causes linear drag. Equation (2.16) describes an object (a tiny oil droplet for instance) that is falling vertically with linear air resistance. I shall solve these two separate problems in turn.
Horizontal Motion with Linear Drag
Consider an object such as the cart in Figure 2.3 coasting horizontally in a linearly resistive medium. I shall assume that at t = 0 the cart is at x = 0 with velocity vx = vxo. The only force on the cart is the drag f = —by, thus the cart inevitably slows down. The rate of slowing is determined by (2.15), which has the general form
vx = —kv x , (2.19)
where k is my temporary abbreviation for k = blm. This is a first-order differential equation for vx , whose general solution must contain exactly one arbitrary constant. The equation states that the derivative of v x is equal to —k times vx itself, and the only function with this property is the exponential function
vx (t) = Ae -kt (2.20)
which satisfies (2.19) for any value of the constant A (Problems 1.24 and 1.25). Since this solution contains one arbitrary constant, it is the general solution of our first- order equation; that is, any solution must have this form. In our case, we know that vx (0) = vxo , so that A = v xo , and we conclude that
1),(0 = Vxoe —kt = Vxoe —Ur (2.21)
where I have introduced the convenient parameter
= 1/ k = mlb [for linear drag]. (2.22)
We see that our cart slows down exponentially, as shown in Figure 2.4(a). The parameter r has the dimensions of time (as you should check), and you can see from (2.21) that when t = r, the velocity is 1/e of its initial value; that is, r is the “1/e” time for the exponentially decreasing velocity. As t oo, the velocity approaches zero.
To find the position as a function of time, we have only to integrate the velocity (2.21). Integrations of this kind can be done using the definite or indefinite integral. The definite integral has the advantage that it automatically takes care of the constant
Section 2.2 Linear Air Resistance 49
0 xc
t
(a)
(b)
Figure 2.4 (a) The velocity vx as a function of time, t, for a cart moving horizontally with a linear resistive force. As t -+ oo, v, approaches zero exponentially. (b) The position x as a function of t for the same cart. As t -+ oc, x x o„, vxo t.
of integration: Since I), = dx/dt,
fo r
1 vx (t)dt = x(t) — x(0).
(Notice that I have named the “dummy” variable of integration t’ to avoid confusion with the upper limit t.) Therefore
x(t)= x(0) + f
t
= 0 +[— vx0Te -t’
(2.23)
In the second line, I have used our assumption that x = 0 when t = 0. And in the last, I have introduced the parameter
xoo = vxor, (2.24)
which is the limit of x(t) as t —> co. We conclude that, as the cart slows down, its position approaches x,„0 asymptotically, as shown in Figure 2.4(b).
Vertical Motion with Linear Drag
Let us next consider a projectile that is subject to linear air resistance and is thrown vertically downward. The two forces on the projectile are gravity and air resistance, as shown in Figure 2.5. If we measure y vertically down, the only interesting component of the equation of motion is the y component, which reads
mi) y = mg — bvy . (2.25)
With the velocity downward (v y > 0), the retarding force is upward, while the force of gravity is downward. If v y is small, the force of gravity is more important than the drag force, and the falling object accelerates in its downward motion. This will
Q – D 2 g Vter = 6 16
[for linear drag]. (2.27)
50 Chapter 2 Projectiles and Charged Particles
f.-bv
w = mg
y
Figure 2.5 The forces on a projectile that is thrown ver- tically down, subject to linear air resistance.
continue until the drag force balances the weight. The speed at which this balance occurs is easily found by setting (2.25) equal to zero, to give v y = mg/b or
Vy = Vter
where I have defined the terminal speed
[for linear drag]. (2.26)
The terminal speed is the speed at which our projectile will eventually fall, if given the time to do so. Since it depends on m and b, it is different for different bodies. For example, if two objects have the same shape and size (b the same for both), the heavier object (m larger) will have the higher terminal speed, just as you would expect. Since
Vter is inversely proportional to the coefficient b of air resistance, we can view titer as an inverse measure of the importance of air resistance — the larger the air resistance, the smaller ti ter , again just as you would expect.
EXAMPLE 2.2 Terminal Speed of Small Liquid Drops
Find the terminal speed of a tiny oildrop in the Millikan oildrop experiment (diameter D = 1.5 gm and density Q = 840 kg/m3). Do the same for a small drop of mist with diameter D = 0.2 mm.
From Example 2.1 we know that the linear drag is dominant for these objects, so the terminal speed is given by (2.26). According to (2.4), b = j9D where
= 1.6 x 10-4 (in SI units). The mass of the drop is m = Q rc D3/6. Thus (2.26) becomes
mg titer
b
Section 2.2 Linear Air Resistance 51
This interesting result shows that, for a given density, the terminal speed is proportional to D 2. This implies that, once air resistance has become important, a large sphere will fall faster than a small sphere of the same density. 2
Putting in the numbers, we find for the oildrop
Vter = (840) x x (1.5 x 10_6)2 x (9.8) =
6.1 x 10-5 m/s [oildrop]. 6 x (1.6 x 10 -4)
In the Millikan oildrop experiment, the oildrops fall exceedingly slowly, so their speed can be measured by simply watching them through a microscope.
Putting in the numbers for the drop of mist, we find similarly that
Vter = 1.3 m/s
[drop of mist]. (2.28)
This speed is representative for a fine drizzle. For a larger raindrop, the terminal speed would be appreciably larger, but with a larger (and hence also faster) drop, the quadratic drag would need to be included in the calculation to get a reliable value for titer.
So far, we have discussed the terminal speed of a projectile (moving vertically), but we must now discuss how the projectile approaches that speed. This is determined by the equation of motion (2.25) which we can rewrite as
mi) = —b(v y titer ) . (2.29)
(Remember that titer = mg/b.) This differential equation can be solved in several ways. (For one alternative see Problem 2.9.) Perhaps the simplest is to note that it is almost the same as Equation (2.15) for the horizontal motion, except that on the right we now have (v y — titer) instead of v x . The solution for the horizontal case was the exponential function (2.20). The trick to solving our new vertical equation (2.29) is to introduce the new variable u = (vy — titer), which satisfies mu = —bu (because titeris constant).
Since this is exactly the same as Equation (2.15) for the horizontal motion, the solution for u is the same exponential, u = Ae-t ly . [Remember that the constant k in (2.20)
became k = 1/t.] Therefore,
V y — V ter = Ae -tai .
When t = 0, v y = vyo , so A = vyo — titer and our final solution for v as a function of t is
V (t) = V ter + (vyo — V ter)e-t
= -tir yoe Vter ( 1 — IT ) –
(2.30)
(2.31)
2 We are here assuming that the drag force is linear, but the same qualitative conclusion follows
for a quadratic drag force. (Problem 2.24.)
52 Chapter 2 Projectiles and Charged Particles
11 Y
V ter
t
T 2T 3T
Figure 2.6 When an object is dropped in a medium with
linear resistance, vy approaches its terminal value Vter as
shown.
This second expression gives v y (t) as the sum of two terms: The first is equal to v yo
when t = 0, but fades away to zero as t increases; the second is equal to zero when t = 0, but approaches vter as t oo. In particular, as t oo,
V (t) V ter (2.32)
just as we anticipated. Let us examine the result (2.31) in a little more detail for the case that v yo = 0;
that is, the projectile is dropped from rest. In this case (2.31) reads
V Y (t) = V ter (1 — e —t it ) . (2.33)
This result is plotted in Figure 2.6, where we see that v y starts out from 0 and approaches the terminal speed, v y ± vter, asymptotically as t oo. The significance of the time r for a falling body is easily read off from (2.33). When t = r , we see that
V y = V ter ( 1 — e -1) = 0 . 63vter •
That is, in a time r, the object reaches 63% of the terminal speed. Similar calculations give the following results:
time percent t of vt,
0 0 r 63%
2r 86% 3r 95%
Of course, the object’s speed never actually reaches v t„, but r is a good measure of how fast the speed approaches v t„. In particular, when t = 3r the speed is 95% of Net., and for many purposes we can say that after a time 3r the speed is essentially equal to vter•
Section 2.2 Linear Air Resistance 53
,M=OMMM,22231faraAVV,Z
EXAMPLE 2.3 Characteristic Time for Two Liquid Drops
Find the characteristic times, r, for the oildrop and drop of mist in Example 2.2. The characteristic time r was defined in (2.22) as r = mlb, and vter was
defined in (2.26) as vter = mg/b. Thus we have the useful relation
Vter = gt • (2.34)
Notice that this relation lets us interpret vter as the speed a falling object would acquire in a time r, if it had a constant acceleration equal to g. Also note that,
like vter , the time t is an inverse indicator of the importance of air resistance: When the coefficient b of air resistance is small, both v ter and r are large; when
b is large, both v ter and r are small. For our present purposes, the importance of (2.34) is that, since we have
already found the terminal velocities of the two drops, we can immediately find the values of r. For the Millikan oildrop, we found that v ter = 6.1 x 10-5 m/s, therefore
Vter 6.1 x 10-5 – =- g 9.8
= 6.2 x 10-6 s [oildrop].
After falling for just 20 microseconds, this oildrop will have acquired 95% of its terminal speed. For almost every purpose, the oildrop always travels at its terminal speed.
For the drop of mist of Example 2.2, the terminal speed was v ter = 1.3 m/s and so r = vt„Ig ti 0.13 s. After about 0.4 s, the drop will have acquired 95% of its terminal speed.
Whether or not our falling object starts from rest, we can find its position y as a function of time by integrating the known form (2.30) of v y ,
V y (t) = Vter + (1/y0 — V ter)e —th
Assuming that the projectile’s initial position is y = 0, it immediately follows that
y(t) f v y (t 1)dt’
vtert (v y 0 vter)T ( 1 — e- tai) – (2.35)
This equation for y(t) can now be combined with Equation (2.23) for x (t) to give us the orbit of any projectile, moving both horizontally and vertically, in a linear medium.
54 Chapter 2 Projectiles and Charged Particles
2.3 Trajectory and Range in a Linear Medium
We saw at the begining of the last section that the equation of motion for a projectile moving in any direction resolves into two separate equations, one for the horizontal and one for the vertical motion [Equations (2.15) and (2.16)]. We have solved each of these separate equations in (2.23) and (2.35), and we can now put these solutions together to give the trajectory of an arbitrary projectile moving in any direction. In this discussion it is marginally more convenient to measure y vertically upward, in which case we must reverse the sign of v t„. (Make sure you understand this point.) Thus the two equations of the orbit become
x(t) = V xo r (1 — e — t/t) y(t) (v yo + v (1— e — tit) — ter, T Vtert •
(2.36)
You can eliminate t from these two equations by solving the first for t and then substituting into the second. (See Problem 2.17.) The result is the equation for the trajectory:
y Vyo Vter X + Nei.”( in (1
x
vxo vxot (2.37)
This equation is probably too complicated to be especially illuminating, but I have plotted it as the solid curve in Figure 2.7, with the help of which you can understand some of the features of (2.37). For example, if you look at the second term on the right of (2.37), you will see that as x –> vxo r the argument of the log function approaches zero; therefore, the log term and hence y both approach —oo. That is, the trajectory has a vertical asymptote at x = vxor, as you can see in the picture. I leave it as an exercise (Problem 2.19) for you to check that if air resistance is switched off (v t, and r both approach infinity), the trajectory defined by (2.37) does indeed approach the dashed trajectory corresponding to zero air resistance.
Horizontal Range
A standard (and quite interesting) problem in elementary physics courses is to show that the horizontal range R of a projectile (subject to no air resistance of course) is
Rvac =
2vxo vyo [no air resistance] (2.38)
g
where Rvac stands for the range in a vacuum. Let us see how this result is modified by air resistance.
The range R is the value of x when y as given by (2.37) is zero. Thus R is the solution of the equation
Vyo Vter R v tert in (1
R = O.
vxot
(2.39) vxo
Section 2.3 Trajectory and Range in a Linear Medium 55
Figure 2.7 The trajectory of a projectile subject to a linear drag
force (solid curve) and the corresponding trajectory in a vacuum
(dashed curve). At first the two curves are very similar, but as t
increases, air resistance slows the projectile and pulls its trajec-
tory down, with a vertical asymptote at x = V xo t. The horizontal range of the projectile is labeled R, and the corresponding range
in vacuum Rvac.
This is a transcendental equation and cannot be solved analytically, that is, in terms of well known, elementary functions such as logs, or sines and cosines. For a given choice of parameters, it can be solved numerically with a computer (Problem 2.22), but this approach usually gives one little sense of how the solution depends on the parameters. Often a good alternative is to find some approximation that allows an approximate analytic solution. (Before the advent of computers, this was often the only way to find out what happens.) In the present case, it is often clear that the effects of air resistance should be small. This means that both vt„ and r are large and the second term in the argument of the log function is small (since it has r in its denominator). This suggests that we expand the log in a Taylor series (see Problem 2.18):
‘no – = – + 1E 2 + 1E 3 + • • .)
(2.40)
We can use this expansion for the log term in (2.39), and, provided t is large enough, we can surely neglect the terms beyond E 3 . This gives the equation
vterl R vterr [ R 1 R ) 2 + 1 R ) 31 = 0. (2.41) vxo vxo r 2 1).„,,T ) 3 vxor
This equation can be quickly tidied up. First, the second term in the first bracket cancels the first term in the second. Next, every term contains a factor of R. This implies that one solution is R = 0, which is correct — the height y is zero when x = 0. Nevertheless, this is not the solution we are interested in, and we can divide out the
56 Chapter 2 Projectiles and Charged Particles
common factor of R. A little rearrangement (and replacement of v t„/r by g) lets us rewite the equation as
2vxo vyo 2 R = R 2 .
g 3vxo r (2.42)
This may seem a perverse way to write a quadratic equation for R, but it leads us quickly to the desired approximate solution. The point is that the second term on the right is very small. (In the numerator R is certainly no more than Rvac and we are assuming that r in the denominator is very large.) Therefore, as a first approximation we get
2vxo v y0 R — Rvac
g (2.43)
This is just what we expected: For low air resistance, the range is close to Rvac . But with the help of (2.42) we can now get a second, better approximation. The last term of (2.42) is the required correction to Rvac ; because it is already small, we would certainly be satisfied with an approximate value for this correction. Thus, in evaluating the last term of (2.42), we can replace R with the approximate value R ti R vac, and we find as our second approximation [remember that the first term in (2.42) is just Rvac ]
2 R Rvac
3v t (Rvac)2
xo
Rvac ( 1 4 ,, UY° ) Uter
(2.44)
(To get the second line, I replaced the second Rvac in the previous line by 2viov yo / g and -cg by vt„.) Notice that the correction for air resistance always makes R smaller than Rvac, as one would expect. Notice also that the correction depends only on the ratio vy,,/vt„. More generally, it is easy to see (Problem 2.32) that the importance of air resistance is indicated by the ratio 04, of the projectile’s speed to the terminal speed. If v /v t, << 1 throughout the flight, the effect of air resistance is very small; if vi vtor is around 1 or more, air resistance is almost certainly important [and the approximation (2.44) is certainly no good].
EXAMPLE 2.4 Range of Small Metal Pellets
I flick a tiny metal pellet with diameter d = 0.2 mm and v = 1 m/s at 45°. Find its horizontal range assuming the pellet is gold (density Q. = 16 g/cm 3). What if it is aluminum (density Q ti 2.7 g/cm3)?
In the absence of air resistance, both pellets would have the same range,
2vxo vyo Rvac = = 10.2 cm.
Section 2.4 Quadratic Air Resistance 57
For gold, Equation (2.27) gives (as you can check) v t, ^ 21 m/s. Thus the correction term in (2.44) is
4 v y. 4 0.71 0.05.
3 Vter 3 x 21
That is, air resistance reduces the range by 5% to about 9.7 cm. The density of aluminum is about 1/6 times that of gold. Therefore the terminal speed is one sixth as big, and the correction for aluminum is 6 times greater or about 30%, giving a range of about 7 cm. For the gold pellet the correction for air resistance is quite small and could perhaps be neglected; for the aluminum pellet, the correction is still small, but is certainly not negligible.
2.4 Quadratic Air Resistance
In the last two sections we have developed a rather complete theory of projectiles subject to a linear drag force, f = —by. While we can find examples of projectiles for which the drag is linear (notably very small objects, such as the Millikan oildrop), for most of the more obvious examples of projectiles (baseballs, footballs, cannonballs, and the like) it is a far better approximation to say that the drag is pure quadratic, f = —cv4. We must, therefore, develop a corresponding theory for a quadratic drag force. On the face of it, the two theories are not so very different. In either case we have to solve the differential equation
= mg f, (2.45)
and in both cases this is a first-order differential equation for the velocity v, with f depending in a relatively simple way on v. There is, however, an important difference. In the linear case (f = —by), Equation (2.45) is a linear differential equation, inas- much as the terms that involve v are all linear in v or its derivatives. In the quadratic case, Equation (2.45) is, of course, nonlinear. And it turns out that the mathemati- cal theory of nonlinear differential equations is significantly more complicated than the linear theory. As a practical matter, we shall find that for the case of a general projectile, moving in both the x and y directions, Equation (2.45) cannot be solved in terms of elementary functions when the drag is quadratic. More generally, we shall see in Chapter 12 that for more complicated systems, nonlinearity can lead to the astonishing phenomenon of chaos, although this does not happen in the present case.
In this section, I shall start with the same two special cases discussed in Section 2.2, a body that is constrained to move horizontally, such as a railroad car on a horizontal track, and a body that moves vertically, such as a stone dropped from a window (both now with quadratic drag forces). We shall find that in these two especially simple cases the differential equation (2.45) can be solved by elementary means, and the solutions
58 Chapter 2 Projectiles and Charged Particles
introduce some important techniques and interesting results. I shall then discuss briefly the general case (motion in both the horizontal and vertical directions), which can be solved only numerically.
Horizontal Motion with Quadratic Drag
Let us consider a body moving horizontally (in the positive x direction), subject to a quadratic drag and no other forces. For example, you could imagine a cycle racer, who has crossed the finishing line and is coasting to a stop under the influence of air resistance. To the extent that the cycle is well lubricated and tires well inflated, we can ignore ordinary friction, 3 and, except at very low speeds, air resistance is purely quadratic. The x component of the equation of motion is therefore (I’ll abbreviate v x
to v)
dv 2 m – = – CV . dt
(2.46)
If we divide by v 2 and multiply by dt, we get an equation in which only the variable v appears on the left and only t on the right:4
dv m— = —c dt.
v 2 (2.47)
This trick — of rearranging a differential equation so that only one variable appears on the left and only the other on the right — is called separation of variables. When it is possible, separation of variables is often the simplest way to solve a first-order differential equation, since the solution can be found by simple integration of both sides.
Integrating Equation (2.47) we find
= —c f dt’ m f
v dv’ vo v /2 0
where vo is the initial velocity at t = 0. Notice that I have written both sides as definite integrals, with the appropriate limits, so that I shan’t have to worry about any constants of integration. I have also renamed the variables of integration as v’ and t’ to avoid
3 As I shall discuss shortly, when the cyclist slows down to a stop, air resistance becomes smaller, and eventually friction becomes the dominant force. Nevertheless, at speeds around 10 mph or more, it is a fair approximation to ignore everything but the quadratic air resistance.
4 In passing from (2.46) to (2.47), I have treated the derivative dv/dt as if it were the quotient of two separate numbers, dv and dt. As you are certainly aware this cavalier proceeding is not strictly correct. Nevertheless, it can be justified in two ways. First, in the theory of differentials, it is in fact true that dv and dt are defined as separate numbers (differentials), such that their quotient is the derivative dv/dt. Fortunately, it is quite unnecessary to know about this theory. As physicists we know that dv/dt is the limit of A v / At, as both Av and At become small, and I shall take the view that dv is just shorthand for Av (and likewise dt for At), with the understanding that it has been taken small enough that the quotient dv/dt is within my desired accuracy of the true derivative. With this understanding, (2.47), with dv on one side and dt on the other, makes perfectly good sense.
Section 2.4 Quadratic Air Resistance 59
confusion with the upper limits v and t. Both of these integrals are easily evaluated,
and we find
1 1 ml — — = —ct
vo v
or, solving for v,
v(t) = vo 0
1 ± cv.tIm 1 +tit
where I have introduced the abbreviation r for the combination of constants
(2.48)
(2.49)
T = m
[for quadratic drag]. (2.50) C Vo
As you can easily check, t is a time, with the significance that when t = t the velocity
is v = v0 /2. Notice that this parameter t is different from the t introduced in (2.22) for motion subject to linear air resistance; nevertheless, both parameters have the same general significance as indicators of the time for air resistance to slow the motion appreciably.
To find the bicycle’s position x, we have only to integrate v to give (as you should
check)
x(t) = x. + f v(t’) dt’
= vor ln (1 ± t/r), (2.51)
if we take the initial position xo to be zero. Figure 2.8 shows our results for v and x as functions of t. It is interesting to compare these graphs with the corresponding graphs of Figure 2.4 for a body coasting horizontally but subject to a linear resistance. Superficially, the two graphs for the velocity look similar. In particular, both go to zero as t oo. But in the linear case v goes to zero exponentially, whereas in the quadratic case it does so only very slowly, like 1/ t . This difference in the behavior of v manifests itself quite dramatically in the behavior of x. In the linear case, we
t t
(a)
(b)
Figure 2.8 The motion of a body, such as a bicycle, coasting horizontally and subject to a quadratic air resistance. (a) The velocity is given by (2.49) and goes to zero like 1/ t as t –>- co. (b) The position is given by (2.51) and goes to infinity as t –>- oo.
dv =
1— V2/V te2 r g dt. (2.55)
60 Chapter 2 Projectiles and Charged Particles
saw that x approaches a finite limit as t oo, but it is clear from (2.51) that in the quadratic case x increases without limit as t oo.
The striking difference in the behavior of x for quadratic and linear drags is easy to understand qualitatively. In the quadratic case, the drag is proportional to v 2 . Thus as v gets small, the drag gets very small — so small that it fails to bring the bicyle to rest at any finite value of x. This unexpected behavior serves to highlight that a drag force that is proportional to v 2 at all speeds is unrealistic. Although the linear drag and ordinary friction are very small, nevertheless as v 0 they must eventually become more important than the v2 term and cannot be ignored. In particular, one or another of these two teams (friction in the case of a bicycle) ensures that no real body can coast on to infinity!
Vertical Motion with Quadratic Drag
The case that an object moves vertically with a quadratic drag force can be solved in much the same way as the horizontal case. Consider a baseball that is dropped from a window in a high tower. If we measure the coordinate y vertically down, the equation of motion is (I’ll abbreviate v y to v now)
my = mg — cv 2 . (2.52)
Before we solve this equation, let us consider the ball’s terminal speed, the speed at which the two terms on the right of (2.52) just balance. Evidently this must satisfy c v 2 = mg, whose solution is
vter = (2.53)
For any given object (given in, g, and c), this lets us calculate the terminal speed. For example, for a baseball it gives (as we shall see in a moment) v t„ c=, 35 m/s, or nearly 80 miles per hour.
We can tidy the equation of motion (2.52) a little by using (2.53) to replace c by mg I v t2e, and canceling the factors of m:
V 2 g (1
vter (2.54)
This can be solved by separation of variables, just as in the case of horizontal motion: First we can rewrite it as
This is the desired separated form (only v on the left and only t on the right) and we can simply integrate both sides. 5 Assuming the ball starts from rest, the limits of
5 Notice that in fact any one-dimensional problem where the net force depends only on the velocity can be solved by separation of variables, since the equation my = F(v) can always be
Section 2.4 Quadratic Air Resistance 61
integration are 0 and v on the left and 0 and t on the right, and we find (as you should verify — Problem 2.35)
vter arctanh (— v
= t g V ter
(2.56)
where “arctanh” denotes the inverse hyperbolic tangent. This particular integral can be evaluated alternatively in terms of the natural log function (Problem 2.37). However, the hyperbolic functions, sinh, cosh, and tanh, and their inverses arcsinh, arccosh, and arctanh, come up so often in all branches of physics that you really should learn to use them. If you have not had much exposure to them, you might want to look at Problems 2.33 and 2.34, and study graphs of these functions.
Equation (2.56) can be solved for v to give
v = vtertanh (— gt
. V ter
(2.57)
To find the position y, we just integrate v to give
y (vte r )2
In cosh.
g L \ vter
While both of these two formulas can be cleaned up a little (see Problem 2.35), they are already sufficient to work the following example.
EXAMPLE 2.5 A Baseball Dropped from a High Tower
Find the terminal speed of a baseball (mass m = 0.15 kg and diameter D = 7 cm). Make plots of its velocity and position for the first six seconds after it is dropped from a tall tower.
The terminal speed is given by (2.53), with the coefficient of air resistance c given by (2.4) as c = y D2 where y = 0.25 N•s 2/m4 . Therefore
(0.15 kg) x (9.8 m/s 2 ) = 35 m/s (2.59)
(0.251\1-s 2/m4) x (0.07 m) 2 vter =
or nearly 80 miles per hour. It is interesting to note that fast baseball pitchers can pitch a ball considerably faster than vter . Under these conditions, the drag force is actually greater than the ball’s weight!
The plots of v and y can be made by hand, but are, of course, much easier with the help of computer software such as Mathcad or Mathematica that can make the plots for you. Whatever method we choose, the results are as shown in Figure 2.9, where the solid curves show the actual velocity and position while the dashed curves are the corresponding values in a vacuum. The actual velocity levels out,
written as m dv I F(v) = dt. Of course there is no assurance that this can be integrated analytically if F(v) is too complicated, but it does guarantee a straightforward numerical solution at worst. See Problem 2.7.
(2.58)
62 Chapter 2 Projectiles and Charged Particles
Figure 2.9 The motion of a baseball dropped from the top of a high tower (solid curves). The corresponding motion in a vacuum is shown with long dashes. (a) The actual velocity approaches the ball’s terminal velocity v ter = 35 m/s as t oo. (b) The graph of position against time falls further and further behind the corresponding vacuum graph. When t = 6 s, the baseball has dropped about 130 meters; in a vacuum, it would have dropped about 180 meters.
approaching the terminal value v t, = 35 m/s as t oo, whereas the velocity in a vacuum would increase without limit Initially, the position increases just as it would in a vacuum (that is, y = z gt 2), but falls behind as v increases and the air resistance becomes more important. Eventually, y approaches a straight line of the form y = vtert + const. (See Problem 2.35.)
Quadratic Drag with Horizontal and Vertical Motion
The equation of motion for a projectile subject to quadratic drag,
mr = mg — cv 2i
= mg — cvv, (2.60)
resolves into its horizontal and vertical components (with y measured vertically upward) to give
mix = —C V 2 2+ VV
X y X
(2.61) m 1) 3, = —mg — vx2 v; v y .
These are two differential equations for the two unknown functions v x (t) and v y (t), but each equation involves both vx and vy . In particular, neither equation is the same as for an object that moves only in the x direction or only in the y direction. This means that we cannot solve these two equations by simply pasting together our two separate solutions for horizontal and vertical motion. Worse still, it turns out that the two equations (2.61) cannot be solved analytically at all. The only way to solve them is numerically, which we can only do for specified numerical initial conditions (that is, specified values of the initial position and velocity). This means that we cannot find the general solution; all we can do numerically is to find the particular solution corresponding to any chosen initial conditions. Before I discuss some general properties of the solutions of (2.61), let us work out one such numerical solution.
Section 2.4 Quadratic Air Resistance 63
EXAMPLE 2.6 Trajectory of a Baseball
The baseball of Example 2.5 is now thrown with velocity 30 m/s (about 70 mi/h) at 50° above the horizontal from a high cliff. Find its trajectory for the first eight seconds of flight and compare with the corresponding trajectory in a vacuum. If the same baseball was thrown with the same initial velocity on horizontal ground how far would it travel before landing? That is, what is its horizontal range?
We have to solve the two coupled differential equations (2.61) with the initial conditions
vxo = vo cos 0 = 19.3 m/s and vy0 = vo sin 0 = 23.0 m/s
and xo = yo = 0 (if we put the origin at the point from which the ball is thrown). This can be done with systems such as Mathematica, Matlab, or Maple, or with programming languages such as “C” or Fortran. Figure 2.10 shows the resulting trajectory, found using the function “NDSolve” in Mathematica.
Several features of Figure 2.10 deserve comment. Obviously the effect of air resistance is to lower the trajectory, as compared to the vacuum trajectory (shown dashed). For example, we see that in a vacuum the high point of the trajectory occurs at t ti 2.3 s and is about 27 m above the starting point; with air resistance, the high point comes just before t = 2.0 s and is at about 21 m. In a vacuum, the ball would continue to move indefinitely in the x direction. The
y (m)
Figure 2.10 Trajectory of a baseball thrown off a cliff and subject to quadratic air resistance (solid curve). The initial velocity is 30 m/s at 50° above the horizontal; the terminal speed is 35 m/s. The dashed curve shows the corresponding trajectory in a vacuum. The dots show the ball’s position at one-second intervals. Air resistance slows the horizontal motion, so that the ball approaches a vertical asymptote just beyond x = 100 meters.
64 Chapter 2 Projectiles and Charged Particles
effect of air resistance is to slow the horizontal motion so that x never moves to the right of a vertical asymptote near x = 100 m.
The horizontal range of the baseball is easily read off the figure as the value of x when y returns to zero. We see that R ti 59 m, as opposed to the range in vacuum, Rvac ti 90 m. The effect of air resistance is quite large in this example, as we might have anticipated: The ball was thrown with a speed only a little less than the terminal speed (30 vs 35 m/s), and this means that the force of air resistance is only a little less than that of gravity. This being the case, we should expect air resistance to change the trajectory appreciably.
This example illustrates several of the general features of projectile motion with a quadratic drag force. Although we cannot solve analytically the equations of motion (2.61) for this problem, we can use the equations to prove various general properties of the trajectory. For example, we noticed that the baseball reached a lower maximum height, and did so sooner, than it would have in a vacuum. It is easy to prove that this will always be the case: As long as the projectile is moving upward (v y > 0), the force
of air resistance has a downward y component. Thus the downward acceleration is greater than g (its value in vacuum). Therefore a graph of v y against t slopes down from
vy,, more quickly than it would in vacuum, as shown in Figure 2.11. This guarantees that vy reaches zero sooner than it would in vacuum, and that the ball travels less distance (in the y direction) before reaching the high point. That is, the ball’s high point occurs sooner, and is lower, than it would be in a vacuum.
Figure 2.11 Graph of v y against t for a projectile that is thrown upward (vyo > 0) and is subject to a quadratic resistance (solid curve). The dashed line (slope = —g) is the corresponding graph when there is no air resistance. The projectile moves upward until it reaches its maximum height when vy = 0. During this time, the drag force is downward and the downward acceleration is always greater than g. Therefore, the curve slopes more steeply than the dashed line, and the projectile reaches its high point sooner than it would in a vacuum. Since the area under the curve is less than that under the dashed line, the projectile’s maximum height is less than it would be in a vacuum.
Section 2.5 Motion of a Charge in a Uniform Magnetic Field 65
I claimed that the baseball of Example 2.6 approaches a vertical asymptote as t —> oo, and we can now prove that this is always the case. First, it is easy to convince yourself that once the ball starts moving downward, it continues to accelerate downward, with v y approaching — vt„ as 1 —> oo. At the same time vx continues to decrease and approaches zero. Thus the square root in both of the equations (2.61) approaches v t„. In particular, when t is large, the equation for v, can be approximated
by
vx CVter vx = —kv x
m
say. The solution of this equation is, of course, an exponential function, v x = Ae —kt , and we see that vx approaches zero very rapidly (exponentially) as t oo. This guarantees that x, which is the integral of vx ,
t
x(t) = f v x (t f)dt’,
approaches a finite limit as t —> oo, and the trajectory has a finite vertical asymptote as claimed.
2.5 Motion of a Charge in a Uniform Magnetic Field
Another interesting application of Newton’s laws, and (like projectile motion) an application that lets me introduce some important mathematical methods, is the motion of a charged particle in a magnetic field. I shall consider here a particle of charge q (which I shall usually take to be positive), moving in a uniform magnetic field B that points in the z direction as shown in Figure 2.12. The net force on the particle is just the magnetic force
F = qv x B, (2.62)
z
B
Figure 2.12 A charged particle moving in a uniform mag- netic field that points in the z direction.
66 Chapter 2 Projectiles and Charged Particles
so the equation of motion can be written as
mir = qv x B. (2.63)
[As with projectiles, the force depends only on the velocity (not the position), so the second law reduces to a first-order differential equation for v.]
As is so often the case, the simplest way to solve the equation of motion is to resolve it into components. The components of v and B are
v = (vx , vy , v z )
and
B= (0, 0, B),
from which we can read off the components of v x B:
v x B = (v y B, —v x B, 0).
Thus the three components of (2.63) are
mi)x = q B v y (2.64)
mi)y = —qBv x (2.65)
= 0. (2.66)
The last of these says simply that v z , the component of the particle’s velocity in the direction of B, is constant:
vZ = const,
a result we could have anticipated since the magnetic force is always perpendicular to B. Because vz is constant, we shall focus most of our attention on v, and v y . In fact, we can even think of them as comprising a two-dimensional vector (v x , vy ), which is just the projection of v onto the xy plane and can be called the transverse velocity,
(vx , v y ) = transverse velocity.
To simplify the equations (2.64) and (2.65) for v x and vy , I shall define the parameter
(.0 = qB
(2.67) m
which has the dimensions of inverse time and is called the cyclotron frequency. With this notation, Equations (2.64) and (2.65) become
vx
—
cov y 1 v y = —coy x •
These two coupled differential equations can be solved in a host of different ways. I would like to describe one that makes use of complex numbers. Though perhaps
(2.68)
Section 2.5 Motion of a Charge in a Uniform Magnetic Field 67
Figure 2.13 The complex number ii = v x + i vy is repre- sented as a point in the complex plane. The arrow pointing from 0 to /7 is literally a picture of the transverse velocity vector (v x , v y ).
not the easiest solution, this method has surprisingly wide application in many ar- eas of physics. (For an alternative solution that avoids complex numbers, see Prob- lem 2.54.)
The two variables vx and vy are, of course, real numbers. However, there is nothing to prevent us from defining a complex number
g = v, + ivy , (2.69)
where i denotes the square root of —1 (called j by most engineers), i = ,N,/ —1 (and
ti is the Greek letter eta). If we draw the complex number ri in the complex plane, or Argand diagram, then its two components are v x and vy as shown in Figure 2.13; in other words, the representation of ri in the complex plane is a picture of the two- dimensional transverse velocity (v i , v y ).
The advantages of introducing the complex number ri appear when we evaluate its derivative. Using (2.68), we find that
r) = V, + ay = wvy — icor), = —iw(v x + iv y )
or
ii = —icon. (2.70)
We see that the two coupled equations for v x and vy have become a single equation for the complex number r7. Furthermore, it is an equation of the now familiar form U = ku, whose solution we know to be the exponential u = Ae kt . Thus we can immediately write down the solution for 77:
II = Ae -i” (2.71)
Before we discuss the significance of this solution, I would like to review a few properties of complex exponentials in the next section. If you are very familiar with these ideas, by all means skip this material.
68 Chapter 2 Projectiles and Charged Particles
2.6 Complex Exponentials
While you are certainly familiar with the exponential function ex for a real variable x, you may not be so at home with ez when z is complex.6 For the real case there are several possible definitions of ex (for instance, as the function that is equal to its own derivative). The definition that extends most easily to the complex case is the Taylor series (see Problem 2.18)
Z 2
Z 3
ez = 1 Z — — • • . 2! 3!
(2.72)
For any value of z, real or complex, large or small, this series converges to give a well- defined value for ez . By differentiating it, you can easily convince yourself that it has the expected property that it equals its own derivative. And one can show (not always so easily) that it has all the other familiar properties of the exponential function — for instance, that ezew = e (z+w ) . (See Problems 2.50 and 2.51.) In particular, the function Ae k z (with A and k any constants, real or complex) has the property that
d (Ae kz) = k (Ae kz) . dz
(2.73)
Since it satisfies this same equation whatever the value of A, it is, in fact, the general solution of the first-order equation dfldz=kf. At the end of the last section, I introduced the complex number 1)(0 and showed that it satisfied the equation ri = —icor). We are now justified in saying that this guarantees that ri must be the exponential function anticipated in (2.71).
We shall be particularly concerned with the exponential of a pure imaginary number, that is, ei 8 where 0 is a real number. The Taylor series (2.72) for this function reads
(0)2 (0) 3 (0)4 e it9 = 1+ i0 +
2! 3! 4! (2.74)
Noting that i 2 = —1, i 3 = —i, and so on, you can see that all of the even powers in this series are real, while all of the odd powers are pure imaginary. Regrouping accordingly, we can rewrite (2.74) to read
9 — +.. .— +… +i 3!
(2.75)
The series in the first brackets is the Taylor series for cos 0, and that in the second brackets is sin 0 (Problem 2.18). Thus we have proved the important relation:
ei0 = cos 0 + i sin 0. (2.76)
6 For a review of some elementary properties of complex numbers, see Problems 2.45 to 2.49.
Section 2.6 Complex Exponentials 69
(a)
(b)
Figure 2.14 (a) Euler’s formula, (2.76), implies that the complex num-
ber eta lies on the unit circle (the circle of radius 1, centered on the origin 0) with polar angle 0. (b) The complex constant A = aei’ lies on a cir- cle of radius a with polar angle S. The function 77 (t) = Ae-i cut lies on the same circle but with polar angle (3 — cot) and moves clockwise around the circle as t advances.
This result, known as Euler’s formula, is illustrated in Figure 2.14(a). Note especially that the complex number ei° has polar angle 0 and, since cos t 0 + sin2 0 = 1, the magnitude of ei0 is 1; that is, ei9 lies on the unit circle, the circle with radius 1 centered at O.
Our main concern is with a complex number of the form 17 = Ae -i”. The co- efficient A is a fixed complex number, which can be expressed as A = aei8 , where a = IAI is the magnitude, and 3 is the polar angle of A, as shown in Figure 2.14(b). (See Problem 2.45.) The number 77 can therefore be written as
77 = Ae-i” = aei8 e-iwt = aei(6-wt) • (2.77)
Thus i has the same magnitude as A (namely a), but has polar angle equal to (6 — wt), as shown in Figure 2.14(b). As a function of t, the number 11 moves clockwise around the circle of radius a with angular velocity co.
It is important that you get a good feel for the role of the complex constant A = aei6 in (2.77): If A happened to equal 1, then 17 would be just >7 = e -i”, which lies on the unit circle, moving clockwise with angular velocity w and starting from the real axis (r) = 1) when t = 0. If A = a is real but not equal to 1, then it simply magnifies the unit circle to a circle of radius a, around which ri moves with the same angular speed and starting from the real axis, at 17 = a when t = 0. Finally if A = aei5 , then the effect of the angle 8 is to rotate ri through the fixed angle 8, so that ri starts out at t = 0 with polar angle S.
Armed with these mathematical results, we can now return to the charged particle in a magnetic field.
70 Chapter 2 Projectiles and Charged Particles
2.7 Solution for the Charge in a B Field
Mathematically, the solution for the velocity v of our charged particle in a B field is complete, and all that remains is to interpret it physically. We already know that v z , the component along B, is constant. The components (v x , vy ) transverse to B we have represented by the complex number 77 = vx + ivy , and we have seen that Newton’s second law implies that 77 has the time dependence ri = Ae -iw t , moving uniformly around the circle of Figure 2.14(b). Now, the arrow shown in that figure, pointing from 0 to 77, is in fact a pictorial representation of the transverse velocity (v x , v y ). Therefore this transverse velocity changes direction, turning clockwise, with constant angular velocity w = q Blm and with constant magnitude. Because v z is constant, this suggests that the particle undergoes a spiralling, or helical, motion. To verify this, we have only to integrate v to find r as a function of t.
That vz is constant implies that
z(t) = zo vzo t. (2.78)
The motion of x and y is most easily found by introducing another complex number
= x iy
where is the Greek letter xi. In the complex plane, is a picture of the transverse position (x, y). Clearly, the derivative of is 77, that is, = 77. Therefore,
=J 77 dt f Ae -iwt dt i A —e + constant. (2.79)
If we rename the coefficient i A /w as C and the constant of integration as X + i Y, this implies that
x + iy = Ce-c wt ± (X ± iY).
By redefining our origin so that the z axis goes through the point (X, Y), we can eliminate the constant term on the right to give
x iy = Ce-iwt
(2.80)
and, by setting t = 0, we can identify the remaining constant C as
C = xo iyo .
This result is illustrated in Figure 2.15. We see there that the transverse position (x, y) moves clockwise round a circle with angular velocity w = q B 1m. Meanwhile z as given by (2.78) increases steadily, so the particle actually describes a uniform helix whose axis is parallel to the magnetic field.
7 1 am assuming the charge q is positive; if q is negative, then w = qB/m is negative, meaning
that the transverse velocity rotates counterclockwise.
Principal Definitions and Equations of Chapter 2 71
Figure 2.15 Motion of a charge in a uniform magnetic field in the z direction. The transverse position (x, y) moves around a circle as shown, while the coordinate z moves with constant velocity into or out of the page.
There are many examples of the helical motion of a charged particle along a magnetic field; for example, cosmic-ray particles (charged particles hitting the earth from space) can get caught by the earth’s magnetic field and spiral north or south along the field lines. If the z component of the velocity happens to be zero, then the spiral reduces to a circle. In the cyclotron, a device for accelerating charged particles to high energies, the particles are trapped in circular orbits in this way. They are slowly accelerated by the judiciously timed application of an electric field. The angular frequency of the orbit is, of course, co = qB/m (which is why this is called the cyclotron frequency). The radius of the orbit is
v my r = — = — = B .
co q B q B (2.81)
This radius increases as the particles accelerate, so that they eventually emerge at the outer edge of the circular magnets that produce the magnetic field.
The same method that we have used here for a charge in a magnetic field can also be used for a particle in magnetic and electric fields, but since this complication adds nothing to the method of solution, I shall leave you to try it for yourself in Problems 2.53 and 2.55.
Principal Definitions and Equations of Chapter 2
Linear and Quadratic Drags
Provided the speed v is well below that of sound, the magnitude of the drag force f = —f (v)i , on an object moving through a fluid is usually well approximated as
f (v) = fii. + /quad
where
flin = by = 13Th) and fquad = cv2 = y D2V2. [Eqs. (2.2) to (2.6)]
72 Chapter 2 Projectiles and Charged Particles
Here D denotes the linear size of the object. For a sphere, D is the diameter and, for a sphere in air at STP, j3 = 1.6 x 10 -4 1\1-s/m2 and y = 0.25 I\T•s2/m4 .
The Lorentz Force on a Charged Particle
F = q (E v x B). [Eq. (2.62) & Problem 2.53]
Problems for Chapter 2
Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***).
SECTION 2.1 Air Resistance
2.1 * When a baseball flies through the air, the ratio fquad/fiin of the quadratic to the linear drag force is given by (2.7). Given that a baseball has diameter 7 cm, find the approximate speed v at which the two drag forces are equally important. For what approximate range of speeds is it safe to treat the drag force as purely quadratic? Under normal conditions is it a good approximation to ignore the linear term? Answer the same questions for a beach ball of diameter 70 cm.
2.2 * The origin of the linear drag force on a sphere in a fluid is the viscosity of the fluid. According to Stokes’s law, the viscous drag on a sphere is
flir, = 37r iiDv (2.82)
where ri is the viscosity 8 of the fluid, D the sphere’s diameter, and v its speed. Show that this expression reproduces the form (2.3) for Ain , with b given by (2.4) as b = 8D. Given that the viscosity of air at STP is 17 = 1.7 x 10-5 N•s/m2 , verify the value of ,8 given in (2.5).
2.3 * (a) The quadratic and linear drag forces on a moving sphere in a fluid are given by (2.84) and (2.82) (Problems 2.4 and 2.2). Show that the ratio of these two kinds of drag force can be written as
fquad/fiin = R/48,9 where the dimensionless Reynolds number R is
R= Dvp
(2.83)
where D is the sphere’s diameter, v its speed, and Q and 77 are the fluid’s density and viscosity. Clearly the Reynolds number is a measure of the relative importance of the two kinds of drag. 1° When R is
8 For the record, the viscosity of a fluid is defined as follows: Imagine a wide channel along which fluid is flowing (x direction) such that the velocity v is zero at the bottom (y = 0) and increases toward the top (y = h), so that successive layers of fluid slide across one another with a velocity gradient dvldy. The force F with which an area A of any one layer drags the fluid above it is proportional to A and to dv/dy, and )7 is defined as the constant of proportionality; that is, F =ri A dvldy.
9 The numerical factor 48 is for a sphere. A similar result holds for other bodies, but the numerical factor is different for different shapes.
1° The Reynolds number is usually defined by (2.83) for flow involving any object, with D defined as a typical linear dimension. One sometimes hears the claim that R is the ratio f quad,/ „du . Since f f quad, lin = R /48 for a sphere,
this claim would be better phrased as “R is roughly of the order of fquadf,,f;in.”
Problems for Chapter 2 73
very large, the quadratic drag is dominant and the linear can be neglected; vice versa when R is very
small. (b) Find the Reynolds number for a steel ball bearing (diameter 2 mm) moving at 5 cm/s through glycerin (density 1.3 g/cm 3 and viscosity 12 N•s/m2 at STP).
2.4 ** The origin of the quadratic drag force on any projectile in a fluid is the inertia of the fluid that the projectile sweeps up. (a) Assuming the projectile has a cross-sectional area A (normal to its velocity) and speed v, and that the density of the fluid is Q, show that the rate at which the projectile encounters fluid (mass/time) is QA v. (b) Making the simplifying assumption that all of this fluid is accelerated to the speed v of the projectile, show that the net drag force on the projectile is pit v 2 . It is certainly not true that all the fluid that the projectile encounters is accelerated to the full speed v, but one might guess that the actual force would have the form
fquad = KOAV2
(2.84)
where K is a number less than 1, which would depend on the shape of the projectile, with K small for a streamlined body, and larger for a body with a flat front end. This proves to be true, and for a sphere the factor K is found to be K = 1/4. (c) Show that (2.84) reproduces the form (2.3) for fquad, with c given by (2.4) as c = y D2 . Given that the density of air at STP is Q = 1.29 kg/m3 and that K = 1/4 for
a sphere, verify the value of y given in (2.6).
SECTION 22 Linear Air Resistance
2.5 * Suppose that a projectile which is subject to a linear resistive force is thrown vertically down with a speed v yo which is greater than the terminal speed v t„. Describe and explain how the velocity varies with time, and make a plot of v y against t for the case that v yo = 2vt„.
2.6 * (a) Equation (2.33) gives the velocity of an object dropped from rest. At first, when v y is small, air resistance should be unimportant and (2.33) should agree with the elementary result v y = gt for free fall in a vacuum. Prove that this is the case. [Hint• Remember the Taylor series for ex = 1+ x + x 2 /2! + x 3 /3! + • • • , for which the first two or three terms are certainly a good approximation when x is small.] (b) The position of the dropped object is given by (2.35) with v y0 = 0. Show similarly that this reduces to the familiar y = igt2 when t is small.
2.7 * There are certain simple one-dimensional problems where the equation of motion (Newton’s second law) can always be solved, or at least reduced to the problem of doing an integral. One of these (which we have met a couple of times in this chapter) is the motion of a one-dimensional particle subject to a force that depends only on the velocity v, that is, F = F(v). Write down Newton’s second law and separate the variables by rewriting it as m dv/F(v) = dt. Now integrate both sides of this equation and show that
dv’ t = m fvo F(v).
Provided you can do the integral, this gives t as a function of v. You can then solve to give v as a function of t. Use this method to solve the special case that F(v) = Fo , a constant, and comment on your result. This method of separation of variables is used again in Problems 2.8 and 2.9.
2.8 * A mass m has velocity vo at time t = 0 and coasts along the x axis in a medium where the drag force is F(v) = —cv 3/2 . Use the method of Problem 2.7 to find v in terms of the time t and the other given parameters. At what time (if any) will it come to rest?
74 Chapter 2 Projectiles and Charged Particles
2.9 * We solved the differential equation (2.29), mi, y = —b(vy — vmr), for the velocity of an object falling through air, by inspection — a most respectable way of solving differential equations. Never- theless, one would sometimes like a more systematic method, and here is one. Rewrite the equation in the “separated” form
m dv Y —b dt V y Vter
and integrate both sides from time 0 to t to find vy as a function of t. Compare with (2.30).
2.10 ** For a steel ball bearing (diameter 2 mm and density 7.8 g/cm 3) dropped in glycerin (density 1.3 g/cm3 and viscosity 12 N•s/m2 at STP), the dominant drag force is the linear drag given by (2.82) of Problem 2.2. (a) Find the characteristic time t and the terminal speed v t„. [In finding the latter, you should include the buoyant force of Archimedes. This just adds a third force on the right side of Equation (2.25).] How long after it is dropped from rest will the ball bearing have reached 95% of its terminal speed? (b) Use (2.82) and (2.84) (with K = 1/4 since the ball bearing is a sphere) to compute the ratio
J quad ? at the terminal speed. Was it a good approximation to neglect f Jequadi,
2.11 ** Consider an object that is thrown vertically up with initial speed v o in a linear medium. (a) Measuring y upward from the point of release, write expressions for the object’s velocity v y (t) and position y(t). (b) Find the time for the object to reach its highest point and its position y max at that point. (c) Show that as the drag coefficient approaches zero, your last answer reduces to the well-known result ymax = zvo2 /g for an object in the vacuum. [Hint: If the drag force is very small, the terminal speed is very big, so v o/vm, is very small. Use the Taylor series for the log function to approximate ln(1 3) by 3 — Z 3 2 . (For a little more on Taylor series see Problem 2.18.)]
2.12 ** Problem 2.7 is about a class of one-dimensional problems that can always be reduced to doing an integral. Here is another. Show that if the net force on a one-dimensional particle depends only on position, F = F(x), then Newton’s second law can be solved to find v as a function of x given by
V 2 = V 2 + —2 x
f F(x’) dx’ . 0
x, (2.85)
[Hint: Use the chain rule to prove the following handy relation, which we could call the “v dv I dx rule”: If you regard v as a function of x, then
dv 1 dv 2 = v — =
dx 2 dx (2.86)
Use this to rewrite Newton’s second law in the separated form m d(v 2) = 2F (x) dx and then integrate from x o to x.] Comment on your result for the case that F(x) is actually a constant. (You may recognise your solution as a statement about kinetic energy and work, both of which we shall be discussing in Chapter 4.)
2.13 ** Consider a mass m constrained to move on the x axis and subject to a net force F = —kx where k is a positive constant. The mass is released from rest at x = x0 at time t = 0. Use the result (2.85) in Problem 2.12 to find the mass’s speed as a function of x; that is, dx I dt = g(x) for some function g(x). Separate this as dx I g(x) = dt and integrate from time 0 to t to find x as a function of t. (You may recognize this as one way — not the easiest — to solve the simple harmonic oscillator.)
2.14 *** Use the method of Problem 2.7 to solve the following: A mass m is constrained to move along the x axis subject to a force F (v) = —Foe” I V , where Fo and V are constants. (a) Find v(t) if the initial
Problems for Chapter 2 75
velocity is v o > 0 at time t = 0. (b) At what time does it come instantaneously to rest? (c) By integrating v(t), you can find x(t). Do this and find how far the mass travels before coming instantaneously to rest.
SECTION 2.3 Trajectory and Range in a Linear Medium
2.15 * Consider a projectile launched with velocity (v, o , vyo) from horizontal ground (with x measured
horizontally and y vertically up). Assuming no air resistance, find how long the projectile is in the air and show that the distance it travels before landing (the horizontal range) is 2vxo v y „/ g.
2.16 * A golfer hits his ball with speed v o at an angle 6 above the horizontal ground. Assuming that
the angle 0 is fixed and that air resistance can be neglected, what is the minimum speed v o(min) for
which the ball will clear a wall of height h, a distanced away? Your solution should get into trouble if
the angle 0 is such that tan 0 < h 1 d. Explain. What is vo (min) if 6 = 25°, d = 50 m, and h = 2 m?
2.17 * The two equations (2.36) give a projectile’s position (x, y) as a function of t. Eliminate t to give y as a function of x. Verify Equation (2.37).
2.18 * Taylor’s theorem states that, for any reasonable function f (x), the value of f at a point (x 6) can be expressed as an infinite series involving f and its derivatives at the point x:
f + 8) = f (x) + f'(x)8 + 2
—1 ! f” (x)8 2 +
3 — 1 r(x)6 3 ± • • • (2.87)
where the primes denote successive derivatives of f (x). (Depending on the function this series may
converge for any increment 3 or only for values of 8 less than some nonzero “radius of convergence.”) This theorem is enormously useful, especially for small values of 3, when the first one or two terms of the series are often an excellent approximation.” (a) Find the Taylor series for ln(1 + 8). (b) Do the same for cos 8. (c) Likewise sin 3. (d) And e8 .
2.19 * Consider the projectile of Section 2.3. (a) Assuming there is no air resistance, write down the position (x, y) as a function of t, and eliminate t to give the trajectory y as a function of x. (b) The correct trajectory, including a linear drag force, is given by (2.37). Show that this reduces to your answer for part (a) when air resistance is switched off (r and v t, = gr both approach infinity). [Hint:
Remember the Taylor series (2.40) for ln(1 —
2.20 ** [Computer] Use suitable graph-plotting software to plot graphs of the trajectory (2.36) of a projectile thrown at 45°above the horizontal and subject to linear air resistance for four different values of the drag coefficient, ranging from a significant amount of drag down to no drag at all. Put all four trajectories on the same plot. [Hint: In the absence of any given numbers, you may as well choose convenient values. For example, why not take v xo = vyo = 1 and g = 1. (This amounts to choosing your units of length and time so that these parameters have the value 1.) With these choices, the strength of the drag is given by the one parameter v t„ = t, and you might choose to plot the trajectories for
Vter = 0-351, 3, and oo (that is, no drag at all), and for times from t = 0 to 3. For the case that v ter = oo,
you’ll probably want to write out the trajectory separately.]
2.21 *** A gun can fire shells in any direction with the same speed v o . Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and z measured vertically up, show that
11 For more details on Taylor’s series see, for example, Mary Boas, Mathematical Methods in the Physical Sci- ences (Wiley, 1983), p. 22 or Donald McQuarrie, Mathematical Methods for Scientists and Engineers (University
Science Books, 2003), p. 94.
76 Chapter 2 Projectiles and Charged Particles
the gun can hit any object inside the surface
V 2 g 2 Z =
2 2v p
0
Describe this surface and comment on its dimensions.
2.22 *** [Computer] The equation (2.39) for the range of a projectile in a linear medium cannot be solved analytically in terms of elementary functions. If you put in numbers for the several parameters, then it can be solved numerically using any of several software packages such as Mathematica, Maple, and MatLab. To practice this, do the following: Consider a projectile launched at angle 0 above the horizontal ground with initial speed v 0 in a linear medium. Choose units such that v o = 1 and g = 1. Suppose also that the terminal speed vter = 1. (With v o = vter , air resistance should be fairly important.) We know that in a vacuum, the maximum range occurs at 0 = r/4 0.75. (a) What is the maximum range in a vacuum? (b) Now solve (2.39) for the range in the given medium at the same angle 0 = 0.75. (c) Once you have your calculation working, repeat it for some selection of values of 0 within which the maximum range probably lies. (You could try 0 = 0.4, 0.5, • • • , 0.8.) (d) Based on these results, choose a smaller interval for 0 where you’re sure the maximum lies and repeat the process. Repeat it again if necessary until you know the maximum range and the corresponding angle to two significant figures. Compare with the vacuum values.
SECTION 2.4 Quadratic Air Resistance
2.23 * Find the terminal speeds in air of (a) a steel ball bearing of diameter 3 mm, (b) a 16-pound steel shot, and (c) a 200-pound parachutist in free fall in the fetal position. In all three cases, you can safely assume the drag force is purely quadratic. The density of steel is about 8 g/cm 3 and you can treat the parachutist as a sphere of density 1 g/cm 3 .
2.24 * Consider a sphere (diameter D, density psph) falling through air (density Q air) and assume that the drag force is purely quadratic. (a) Use Equation (2.84) from Problem 2.4 (with K = 1/4 for a sphere) to show that the terminal speed is
vter = — 8
D g Osph
3 Qair (2.88)
(b) Use this result to show that of two spheres of the same size, the denser one will eventually fall faster. (c) For two spheres of the same material, show that the larger will eventually fall faster.
2.25 * Consider the cyclist of Section 2.4, coasting to a halt under the influence of a quadratic drag force. Derive in detail the results (2.49) and (2.51) for her velocity and position, and verify that the constant r = m/cv o is indeed a time.
2.26 * A typical value for the coefficient of quadratic air resistance on a cyclist is around c = 0.20 N/(m/s) 2 . Assuming that the total mass (cyclist plus cycle) is m = 80 kg and that at t = 0 the cyclist has an initial speed v o = 20 m/s (about 45 mi/h) and starts to coast to a stop under the influence of air resistance, find the characteristic time r = m/cv o . How long will it take him to slow to 15 m/s? What about 10 m/s? And 5 m/s? (Below about 5 m/s, it is certainly not reasonable to ignore friction, so there is no point pursuing this calculation to lower speeds.)
Problems for Chapter 2 77
2.27 * I kick a puck of mass m up an incline (angle of slope = 0) with initial speed v o . There is no friction between the puck and the incline, but there is air resistance with magnitude f (v) = cv 2 . Write down and solve Newton’s second law for the puck’s velocity as a function of t on the upward journey. How long does the upward journey last?
2.28 * A mass m has speed vo at the origin and coasts along the x axis in a medium where the drag force is F(v) = —cv 312 . Use the “v dv/dx rule” (2.86) in Problem 2.12 to write the equation of motion in the separated form m v dvl F (v) = dx, and then integrate both sides to give x in terms of v (or vice versa). Show that it will eventually travel a distance 2m lvo/c.
2.29 * The terminal speed of a 70-kg skydiver in spread-eagle position is around 50 m/s (about 115 mi/h). Find his speed at times t = 1, 5, 10, 20, 30 seconds after he jumps from a stationary balloon. Compare with the corresponding speeds if there were no air resistance.
2.30 * Suppose we wish to approximate the skydiver of Problem 2.29 as a sphere (not a very promising approximation, but nevertheless the kind of approximation physicists sometimes like to make). Given the mass and terminal speed, what should we use for the diameter of the sphere? Does your answer seem reasonable?
2.31 ** A basketball has mass m = 600 g and diameter D = 24 cm. (a) What is its terminal speed? (b) If it is dropped from a 30-m tower, how long does it take to hit the ground and how fast is it going when it does so? Compare with the corresponding numbers in a vacuum.
2.32 ** Consider the following statement: If at all times during a projectile’s flight its speed is much less than the terminal speed, the effects of air resistance are usually very small. (a) Without reference to the explicit equations for the magnitude of v t„, explain clearly why this is so. (b) By examining the explicit formulas (2.26) and (2.53) explain why the statement above is even more useful for the case of quadratic drag than for the linear case. [Hint: Express the ratio f 1 mg of the drag to the weight in terms of the ratio v /v ter .]
2.33 ** The hyperbolic functions cosh z and sinh z are defined as follows:
ez e’ cosh z = and sinh z =
ez — e — z
2 2
for any z, real or complex. (a) Sketch the behavior of both functions over a suitable range of real values of z. (b) Show that cosh z = cos(iz). What is the corresponding relation for sinh z? (c) What are the derivatives of cosh z and sinh z? What about their integrals? (d) Show that cosh2 z — sinh2 z = 1. (e) Show that f dx 1 ,\/1± x 2 = arcsinh x. [Hint: One way to do this is to make the substitution x = sinh z.]
2.34 ** The hyperbolic function tanh z is defined as tanh z = sinh z/ cosh z, with cosh z and sinh z defined as in Problem 2.33. (a) Prove that tanh z = tan(iz). (b) What is the derivative of tanh z? (c) Show that f dz tanh z = In cosh z. (d) Prove that 1 — tanh 2 z = sech2z, where sech z = 1/ cosh z. (e) Show that f dx/(1 — x 2) = arctanh x .
2.35 ** (a) Fill in the details of the arguments leading from the equation of motion (2.52) to Equations (2.57) and (2.58) for the velocity and position of a dropped object subject to quadratic air resistance. Be sure to do the two integrals involved. (The results of Problem 2.34 will help.) (b) Tidy the two equations by introducing the parameter r = v t„Ig• Show that when t = r, v has reached 76% of its terminal value. What are the corresponding percentages when t = 2r and 3z? (c) Show that when t >> r, the position is approximately y v t ter const. [Hint: The definition of cosh x (Problem 2.33)
78 Chapter 2 Projectiles and Charged Particles
gives you a simple approximation when x is large.] (d) Show that for t small, Equation (2.58) for the position gives y ti Igt2 . [Use the Taylor series for cosh x and for ln(1 +
2.36 ** Consider the following quote from Galileo’s Dialogues Concerning Two New Sciences:
Aristotle says that “an iron ball of 100 pounds falling from a height of one hundred cubits reaches the ground before a one-pound ball has fallen a single cubit.” I say that they arrive at the same time. You find, on making the experiment, that the larger outstrips the smaller by two finger-breadths, that is, when the larger has reached the ground, the other is short of it by two finger-breadths.
We know that the statement attributed to Aristotle is totally wrong, but just how close is Galileo’s claim that the difference is just “two finger breadths”? (a) Given that the density of iron is about 8 g/cm 3, find the terminal speeds of the two iron balls. (b) Given that a cubit is about 2 feet, use Equation (2.58) to find the time for the heavier ball to land and then the position of the lighter ball at that time. How far apart are they?
2.37 ** The result (2.57) for the velocity of a falling object was found by integrating Equation (2.55) and the quickest way to do this is to use the integral f du I (1 — u 2) = arctanh u. Here is another way to do it: Integrate (2.55) using the method of “partial fractions,” writing
1 _ 1C1 1
1 — u 2
–
2 1 + u
•
1— u
which lets you do the integral in terms of natural logs. Solve the resulting equation to give v as a function of t and show that your answer agrees with (2.57).
2.38 ** A projectile that is subject to quadratic air resistance is thrown vertically up with initial speed vo . (a) Write down the equation of motion for the upward motion and solve it to give v as a function of t. (b) Show that the time to reach the top of the trajectory is
trop = (v terl g) arctan (vo / vt„) .
(c) For the baseball of Example 2.5 (with titer = 35 m/s), find t rop for the cases that v o = 1, 10, 20, 30, and 40 m/s, and compare with the corresponding values in a vacuum.
2.39 ** When a cyclist coasts to a stop, he is actually subject to two forces, the quadratic force of air resistance, f = —cv 2 (with c as given in Problem 2.26), and a constant frictional force f t., of about 3 N. The former is dominant at high and medium speeds, the latter at low speed. (The frictional force is a combination of ordinary friction in the bearings and rolling friction of the tires on the road.) (a) Write down the equation of motion while the cyclist is coasting to a stop. Solve it by separating variables to give t as a function of v. (b) Using the numbers of Problem 2.26 (and the value f f, = 3 N given above) find how long it takes the cyclist to slow from his initial 20 m/s to 15 m/s. How long to slow to 10 and 5 m/s? How long to come to a full stop? If you did Problem 2.26, compare with the answers you got there ignoring friction entirely.
2.40 ** Consider an object that is coasting horizontally (positive x direction) subject to a drag force f = —by — cv 2 . Write down Newton’s second law for this object and solve for v by separating variables. Sketch the behavior of v as a function of t. Explain the time dependence for t large. (Which force term is dominant when t is large?)
2.41 ** A baseball is thrown vertically up with speed v o and is subject to a quadratic drag with magnitude f (v) = cv 2 . Write down the equation of motion for the upward journey (measuring y vertically up) and show that it can be rewritten as i) = —g[l + (v/vt„) 2 ]. Use the “v dv dx rule”
Problems for Chapter 2 79
(2.86) to write I) as v dvldy, and then solve the equation of motion by separating variables (put all terms involving v on one side and all terms involving y on the other). Integrate both sides to give y in terms of v, and hence v as a function of y. Show that the baseball’s maximum height is
2 v vo
2 2 Vter in ter
Ymax = 2 2g
+ V ter
(2.89)
If vo = 20 m/s (about 45 mph) and the baseball has the parameters given in Example 2.5 (page 61),
what is ym„? Compare with the value in a vacuum.
2.42 ** Consider again the baseball of Problem 2.41 and write down the equation of motion for the downward journey. (Notice that with a quadratic drag the downward equation is different from the upward one, and has to be treated separately.) Find v as a function of y and, given that the downward journey starts at ym„ as given in (2.89), show that the speed when the ball returns to the ground is
vt„voilvL vo2 . Discuss this result for the cases of very much and very little air resistance. What
is the numerical value of this speed for the baseball of Problem 2.41? Compare with the value in a vacuum.
2.43 *** [Computer] The basketball of Problem 2.31 is thrown from a height of 2 m with initial velocity vo = 15 m/s at 45° above the horizontal. (a) Use appropriate software to solve the equations of motion (2.61) for the ball’s position (x, y) and plot the trajectory. Show the corresponding trajectory in the absence of air resistance. (b) Use your plot to find how far the ball travels in the horizontal direction before it hits the floor. Compare with the corresponding range in a vacuum.
2.44 *** [Computer] To get an accurate trajectory for a projectile one must often take account of several complications. For example, if a projectile goes very high then we have to allow for the reduction in air resistance as atmospheric density decreases. To illustrate this, consider an iron cannonball (diameter 15 cm, density 7.8 g/cm 3) that is fired with initial velocity 300 m/s at 50 degrees above the horizontal. The drag force is approximately quadratic, but since the drag is proportional to the atmospheric density and the density falls off exponentially with height, the drag force is f = c(y)v 2 where c(y) = y D2 exp(—y/A) with y given by (2.6) and A ti 10, 000 m. (a) Write down the equations of motion for the cannonball and use appropriate software to solve numerically for x(t) and y(t) for 0 < t < 3.5 s. Plot the ball’s trajectory and find its horizontal range. (b) Do the same calculation ignoring the variation of atmospheric density [that is, setting c(y) = c(0)], and yet again ignoring air resistance entirely. Plot all three trajectories for 0 < t < 3.5 s on the same graph. You will find that in this case air resistance makes a huge difference and that the variation of air resistance makes a small, but not negligible, difference.
SECTION 2.6 Complex Exponentials
2.45 * (a) Using Euler’s relation (2.76), prove that any complex number z = x iy can be written in the form z = rei° , where r and 0 are real. Describe the significance of r and 0 with reference to the complex plane. (b) Write z = 3 + 4i in the form z = reie (c) Write z = 2e -i7113 in the form x iy.
2.46 * For any complex number z = x iy, the real and imaginary parts are defined as the real numbers Re(z) = x and Im(z) = y. The modulus or absolute value is lz I = \/x 2 + y2 and the phase or angle is the value of 6 when z is expressed as z = re° . The complex conjugate is z* = x — iy. (This last is the notation used by most physicists; most mathematicians use -z-.) For each of the following complex numbers, find the real and imaginary parts, the modulus and phase, and the complex conjugate,
80 Chapter 2 Projectiles and Charged Particles
and sketch z and z* in the complex plane:
(a) z = 1 + i (b) z = 1 — (c) z = ,4e -in (d) z 5eiwt .
In part (d), w is a constant and t is the time.
2.47 * For each of the following two pairs of numbers compute z + w, z — w, zw, and z/w.
(a) z = 6 + 8i and w = 3 — 4i (b) z = 8e iTh/3 and w = 4e i’/ 6 .
Notice that for adding and subtracting complex numbers, the form x + iy is more convenient, but for multiplying and especially dividing, the form rei9 is more convenient. In part (a), a clever trick for finding z/w without converting to the form rei° is to multiply top and bottom by w*; try this one both ways.
2.48 * Prove that lz I = ,N/z*z for any complex number z.
2.49 * Consider the complex number z = e’9 = cos 8 + i sin 9. (a) By evaluating z 2 two different ways, prove the trig identities cos 26 = cos 2 9 — sin2 9 and sin 29 = 2 sin 8 cos 9. (b) Use the same technique to find corresponding identities for cos 39 and sin 39.
2.50 * Use the series definition (2.72) of ez to prove that 12 dez ldz = ez.
2.51 ** Use the series definition (2.72) of ez to prove that ezew = ez+w. . [Hint: If you write down the left side as a product of two series, you will have a huge sum of terms like zn Wm. If you group together all the terms for which n + m is the same (call it p) and use the binomial theorem, you will find you have the series for the right side.]
SECTION 2.7 Solution for the Charge in a B Field
2.52 * The transverse velocity of the particle in Sections 2.5 and 2.7 is contained in (2.77), since + ivy . By taking the real and imaginary parts, find expressions for v x and vy separately. Based
on these expressions describe the time dependence of the transverse velocity.
2.53 * A charged particle of mass m and positive charge q moves in uniform electric and magnetic fields, E and B, both pointing in the z direction. The net force on the particle is F = q(E + v x B). Write down the equation of motion for the particle and resolve it into its three components. Solve the equations and describe the particle’s motion.
2.54 ** In Section 2.5 we solved the equations of motion (2.68) for the transverse velocity of a charge in a magnetic field by the trick of using the complex number ri = v, + i v y . As you might imagine,
the equations can certainly be solved without this trick. Here is one way: (a) Differentiate the first of
equations (2.68) with respect to t and use the second to give you a second-order differential equation for vx . This is an equation you should recognize [if not, look at Equation (1.55)] and you can write down its general solution. Once you know v x , (2.68) tells you v y . (b) Show that the general solution you get here is the same as the general solution contained in (2.77), as disentangled in Problem 2.52.
12 1f you are the type who worries about mathematical niceties, you may be wondering if it is permissible to differentiate an infinite series. Fortunately, in the case of a power series (such as this), there is a theorem that guarantees the series can be differentiated for any z inside the “radius of convergence.” Since the radius of convergence of the series for ez is infinite, we can differentiate it for any value of z.
Problems for Chapter 2 81
2.55 *** A charged particle of mass m and positive charge q moves in uniform electric and magnetic
fields, E pointing in the y direction and B in the z direction (an arrangement called “crossed E and B fields”). Suppose the particle is initially at the origin and is given a kick at time t = 0 along the x axis with vx = vxo (positive or negative). (a) Write down the equation of motion for the particle and resolve it into its three components. Show that the motion remains in the plane z = 0. (b) Prove that there is a unique value of vxo, called the drift speed v dr , for which the particle moves undeflected through the fields. (This is the basis of velocity selectors, which select particles traveling at one chosen speed from a beam with many different speeds.) (c) Solve the equations of motion to give the particle’s velocity as a function of t, for arbitrary values of vx0 . [Hint: The equations for (v x , vy ) should look very like Equations (2.68) except for an offset of v x by a constant. If you make a change of variables of the
form U., = — vd, and u y = vy , the equations for (ux , u y ) will have exactly the form (2.68), whose general solution you know.] (d) Integrate the velocity to find the position as a function of t and sketch the trajectory for various values of vx.o.
CHAPTER
Momentum and Angular Momentum
In this and the next chapter I shall describe the great conservation laws of momentum, angular momentum, and energy. These three laws are closely related to one another and are perhaps the most important of the small number of conservation laws that are regarded as cornerstones of all modern physics. Curiously, in classical mechanics the first two laws (momentum and angular momentum) are very different from the last (energy). It is a relatively easy matter to prove the first two from Newton’s laws (indeed we already have proved conservation of momentum), whereas the proof of energy conservation is surprisingly subtle. I discuss momentum and angular momentum in this rather short chapter and energy in Chapter 4, which is appreciably longer.
3.1 Conservation of Momentum
In Chapter 1 we examined a system of N particles labeled a = 1, • • • , N. We found that as long as all the internal forces obey Newton’s third law, the rate of change of the system’s total linear momentum P = p i + • pN = E pa is determined entirely by the external forces on the system:
P = Fext (3.1)
where Fext denotes the total external force on the system. Because of the third law, the internal forces all cancel out of the rate of change of the total momentum. In particular, if the system is isolated, so that the total external force is zero, we have the
Principle of Conservation of Mo entum
e net e t r e on de system is zero, the syst total t chani ornen urn P is cougar)t.
83
84 Chapter 3 Momentum and Angular Momentum
If our system contains just one particle (N = 1), then all forces on the particle are external, and the conservation of momentum is reduced to the not very interesting statement that, in the absence of any forces, the momentum of a single particle is constant, which is just Newton’s first law. However, if our system has two or more particles (N > 2), then momentum conservation is a nontrivial and often useful property, as the following simple and well-known example will remind you.
EXAMPLE 3.1 An Inelastic Collision of Two Bodies
Two bodies (two lumps of putty, for example, or two cars at an intersection) have masses m 1 and m 2 and velocities v 1 and v2 . The two bodies collide and lock together, so they move off as a single unit, as shown in Figure 3.1. (A collision in which the bodies lock together like this is said to be perfectly inelastic.) Assuming that any external forces are negligible during the brief moment of collision, find the velocity v just after the collision.
The initial total momentum, just before the collision, is
= mivi m2v2
and the final momentum, just after the collision, is
Pfin = m + m2v = (m 1 + m2)v.
(Notice that this last equation illustrates the useful result that, once two bodies have locked together, we can find their momentum by considering them as a single body of mass m 1 + m2.) By conservation of momentum these two momenta must be equal, Pfin = Pin , and we can easily solve to give the final velocity,
M1V1 M2V2 V =
m I -I- m2
We see that the final velocity is just the weighted average of the original veloc- ities v 1 and v2 , weighted by the corresponding masses m 1 and m 2 .
(3.2)
Figure 3.1 A perfectly inelastic collision between two lumps of putty.
Section 3.2 Rockets 85
An important special case is when one of the bodies is initially at rest, as when a speeding car rams a stationary car at a stop light. With v 2 = 0, Equation
(3.2) reduces to
1111 V = V1.
m + m2 (3.3)
In this case the final velocity is always in the same direction as v 1 but is reduced by the factor m i / (m i + m 2). The result (3.3) is used by police investigating car crashes, since it lets them find the unknown velocity v 1 of a speeder who has rear-ended a stationary car, in terms of quantities that can be measured after the event. (The final velocity v can be found from the skid marks of the combined wreck.)
This sort of analysis of collisions, using conservation of momentum, is an important tool in solving many problems ranging from nuclear reactions, through car crashes, to collisions of galaxies.
3.2 Rockets
A beautiful example of the use of momentum conservation is the analysis of rocket propulsion. The basic problem that is solved by the rocket is this: With no external agent to push on or be pushed by, how does an object get itself moving? You can put yourself in the same difficulty by imagining yourself stranded on a perfectly frictionless frozen lake. The simplest way to get yourself to shore is to take off anything that is dispensible, such as a boot, and throw it as hard as possible away from the shore. By Newton’s third law, when you push one way on the boot, the boot pushes in the opposite direction on you. Thus as you throw the boot, the reaction force of the boot on you will cause you to recoil in the opposite direction and then glide across the ice to shore. A rocket does essentially the same thing. Its motor is designed to hurl the spent fuel out of the back of the rocket, and by the third law, the fuel pushes the rocket forward.
To analyse a rocket’s motion quantitatively we must examine the total momentum. Consider the rocket shown in Figure 3.2 with mass m, traveling in the positive x direction (so I can abbreviate vx as just v) and ejecting spent fuel at the exhaust speed vex relative to the rocket. Since the rocket is ejecting mass, the rocket’s mass m is steadily decreasing. At time t, the momentum is P(t) = my. A short time later at t dt, the rocket’s mass is (m + dm), where dm is negative, and its momentum is (m + dm)(v dv). The fuel ejected in the time dt has mass (—dm) and velocity
Concerning the use of the small quantities like dt and dm, I recommend again the view that they are small but nonzero increments, with dt chosen sufficiently small that dm divided by dt is (within whatever we have chosen as our desired accuracy) equal to the derivative dm/dt. For more details, see the footnote immediately before Equation (2.47).
86 Chapter 3 Momentum and Angular Momentum
v ___,….
Figure 3.2 A rocket of mass m travels to the right with speed v and ejects spent fuel with exhaust speed v ex rela- tive to the rocket.
v — vex relative to the ground. Thus the total momentum (rocket plus the fuel just ejected) at t + dt is
P(t + dt) = (m + dm)(v + dv) — dm(v — v ex ) = my + m dv + dm vex
where I have neglected the doubly small product dm d v. Therefore, the change in total momentum is
dP = P(t + dt) — P(t) = m dv + dm vex . (3.4)
If there is a net external force F’ (gravity, for instance), this change of momentum is F’ di. (See Problem 3.11.) Here I shall assume that there are no external forces, so that P is constant and dP = 0. Therefore
m dv = —dm v ex . (3.5)
Dividing both sides by dt, we can rewrite this as
Tr ) = — thvex (3.6)
where —rh is the rate at which the rocket’s engine is ejecting mass. This equation looks just like Newton’s second law (mi) = F) for an ordinary particle, except that the product —th vex on the right plays the role of the force. For this reason this product is often called the thrust:
thrust = —th Vex . (3.7)
(Since rii is negative, this defines the thrust to be positive.) Equation (3.5) can be solved by separation of variables. Dividing both sides by m
gives
dm dv = — vex — • •
If the exhaust speed v ex is constant, this equation can be integrated to give
v — v o = vex ln(m o/m)
(3.8)
where vo is the initial velocity and m o is the initial mass of the rocket (including fuel and payload). This result puts a significant restriction on the maximum speed of the rocket. The ratio m o/m is largest when all the fuel is burned and m is just the mass
Section 3.3 The Center of Mass 87
of rocket plus payload. Even if, for example, the original mass is 90% fuel, this ratio is only 10, and, since In 10 = 2.3, this says that the speed gained, v — v 0 , cannot be more than 2.3 times v ex . This means that rocket engineers try to make v ex as big as possible and also design multistage rockets, which can jettison the heavy fuel tanks of the early stages to reduce the total mass for later stages. 2
3.3 The Center of Mass
Several of the ideas of Section 3.1 can be rephrased in terms of the important notion of a system’s center of mass. Let us consider a group of N particles, a = 1, • • • , N, with masses m e, and positions ra measured relative to an origin O. The center of mass (or CM) of this system is defined to be the position (relative to the same origin 0)
R= — 1 E
N
mara = t71 ir + • • • m N r
NM a=1 (3.9)
where M denotes the total mass of all of the particles, M = E ma . The first thing to note about this definition is that it is a vector equation. The CM position is a vector R with three components (X, Y, Z), and Equation (3.9) is equivalent to three equations giving these three components,
N N , N 1 11 1
X= E 77„.„,,, Y = — E mayor, Z = — E mazes . M m , M a=1 a=i a=1 Either way, the CM position R is a weighted average of the positions r 1 , , rN , in which each position ra is weighted by the corresponding mass m a . (Equivalently, it is the sum of the ra , each multiplied by the fraction of the total mass at r a .)
To get a feeling for the CM, it may help to consider the case of just two particles (N = 2). In this case, the definition (3.9) reads
m m 2r2 R =
mi + m2 (3.10)
It is easy to verify that the CM position has several familiar properties. For example, you can show (Problem 3.18) that the CM defined by (3.10) lies on the line joining the two particles, as shown in Figure 3.3. It is also easy to show that the distances of the CM from m 1 and m 2 are in the ratio m 2 / m i , so that the CM lies closer to the more massive particle. (In Figure 3.3 this ratio is 1/3.) In particular, if m 1 is much greater than m 2, the CM will be very close to r 1 . More generally, going back to Equation (3.9) for the CM of N particles, we see that if m 1 is much greater than any of the other masses (as is the case for the sun as compared to all the planets), then m 1 ti M while ma « M for all other particles; this means that R is very close to r 1 . Thus, for example, the CM of the solar system is very close to the sun.
2 Jettisoning the fuel tanks of stage 1 reduces the inital and final masses of stage 2 by the same amount. This increases the ratio m o/m when we apply (3.8) to stage 2. See Problem 3.12.
88 Chapter 3 Momentum and Angular Momentum
Figure 3.3 The CM of two particles lies at the position (m ir i + m2r2)/M. You can prove that this lies on
the line joining m 1 to m 2 , as shown, and that the distances of the CM from m 1 and m 2 are in the ratio m 2/m 1 .
We can now write the total momentum P of any N-particle system in terms of the system’s CM as follows:
P = pa MR (3.11)
a a
where the last equality is just the derivative of the definition (3.9) of R (multiplied by M). This remarkable result says that the total momentum of the N particles is exactly
the same as that of a single particle of mass M and velocity equal to that of the CM. We get an even more striking result when we differentiate (3.11). According to
(3.1), the derivative of P is just F”t . Therefore, (3.11) implies that
Fext = MR. (3.12)
That is, the center of mass R moves exactly as if it were a single particle of mass M, subject to the net external force on the system. This result is the main reason why we can often treat extended bodies, such as baseballs and planets, as if they were point particles. Provided a body is small compared to the scale of its trajectory, its CM position R is a good representative of its overall location, and (3.12) implies that R moves just like a point particle.
Given the importance of the CM, you need to feel comfortable calculating the CM position for various systems. You may have had plenty of practice in introductory physics or in a calculus course, but, in case you didn’t, there are several exercises at the end of this chapter. One important point to bear in mind is that when the mass in a body is distributed continuously, the sum in the definition (3.9) goes over to an integral
R=— 1
m f rdm= —1 f prdV (3.13)
where p is the mass density of the body, dV denotes an element of volume, and the integral runs over the whole body (that is, everywhere p 0). We shall be using similar integrals to evaluate the moment of inertia tensor in Chapter 10. Meanwhile, here is one example:
Section 3.3 The Center of Mass 89
EXAMPLE 3.2 The CM of a Solid Cone
Find the CM position for the uniform solid cone shown in Figure 3.4. It is perhaps obvious by symmetry that the CM lies on the axis of symmetry
(the z axis), but this also follows immediately from the integral (3.13). For example, if you consider the x component of that integral, it is easy to see that the contribution from any point (x, y, z) is exactly cancelled by that from the
point (—x, y, z). That is, the integral for X is zero. Because the same argument applies to Y, the CM lies on the z axis. To find the height Z of the CM, we must evaluate the integral
Z.- 1 f pzdV = 0 — f zdxdydz
where I could take the factor p outside the integral since p is constant throughout the cone (as long as we understand the integral is limited to the inside of the cone) and I have changed the volume element dV to dx dy dz. For any given z, the integral over x and y runs over a circle of radius r = Rz I h, giving a factor of rrr2 R2z2″ /12, so that
pyr R 2 fh z 3dz =
pm R2 h4 3 Z
h = Mh 2 0 Mh 2 4 4
where in the last step I replaced the mass M by O times the volume or M =
ipn- R 2h. We conclude that the CM is on the axis of the cone at a distance 4 h from the vertex (or ih from the base).
Figure 3.4 A solid cone, centered on the z axis, with vertex at the origin and uniform mass density Q. Its height is h and its base has radius R.
90 Chapter 3 Momentum and Angular Momentum
3.4 Angular Momentum for a Single Particle
In many ways the conservation of angular momentum parallels the conservation of ordinary (or “linear”) momentum. Nevertheless, I would like to review the formalism in detail, first for a single particle and then for a multiparticle system. This will introduce several important ideas and some useful mathematics.
The angular momentum £ of a single particle is defined as the vector
r x p. (3.14)
Here r x p is the vector product of the particle’s position vector r, relative to the chosen origin 0, and its momentum p, as shown in Figure 3.5. Notice that because r depends on the choice of origin, the same is true of £: The angular momentum £ (unlike the linear momentum p) depends on the choice of origin, and we should, strictly speaking, refer to £ as the angular momentum relative to O.
The time rate of change of is easily found:
d = —
dt (r x p) =(r x p) (r x to). (3.15)
(You can easily check that the product rule can be used for differentiating vector products, as long as you are careful to keep the vectors in the right order. See Problem 1.17.) In the first term on the right, we can replace p by mi, and, because the cross product of any two parallel vectors is zero, the first term is zero. In the second term, we can replace p by the net force F on the particle, and we get
(3.16)
Here F (Greek capital gamma) denotes the net torque about 0 on the particle, defined as r x F. (Other popular symbols for torque are r and N.) In words, (3.16) says that the rate of change of a particle’s angular momentum about the origin 0 is equal to the net applied torque about 0. Equation (3.16) is the rotational analog of the equation
F for the linear momentum, and (3.16) is often described as the rotational form of Newton’s second law.
Figure 3.5 For any particle with position r relative to the origin O and momentum p, the angular momentum about 0 is defined as the vector t = r x p. For the case shown, t points into the page.
Section 3.4 Angular Momentum for a Single Particle 91
Sun
Figure 3.6 A planet (mass m) is subject to the central force
of the sun (mass M). If we choose the origin at the sun, then r x F = 0, and the planet’s angular momentum about 0 is constant.
In many one-particle problems one can choose the origin 0 so that the net torque F (about the chosen 0) is zero. In this case, the particle’s angular momentum about 0 is constant. Consider, for example, a single planet (or comet) orbiting the sun. The only force on the planet is the gravitational pull Gm M I r 2 of the sun, as shown in Figure 3.6. A crucial property of the gravitational force is that it is central, that is, directed along the line joining the two centers. This means that F is parallel (actually, antiparallel) to the position vector r measured from the sun, and hence that r x F = 0. Thus if we choose our origin at the sun, the planet’s angular momentum about 0 is constant, a fact that greatly simplifies the analysis of planetary motion. For example, because r x p is constant, r and p must remain in a fixed plane; in other words, the planet’s orbit is confined to a single plane containing the sun, and the problem is reduced to two dimensions, a result we shall exploit in Chapter 8.
Kepler’s Second Law
One of the earliest triumphs for Newton’s mechanics was that he was able to explain Kepler’s second law as a simple consequence of conservation of angular momentum. Newton’s laws of motion were published in 1687 in his famous book Principia. Nearly eighty years earlier, the German astronomer Johannes Kepler (1571-1630) had published his three laws of planetary motion. 3 These laws are quite different from Newton’s laws in that they are simply mathematical descriptions of the observed motion of the planets. For example, the first law states that the planets move around the sun in ellipses with the sun at one focus. Kepler’s laws make no attempt to explain planetary motion in terms of more fundamental ideas; they are just summaries — brilliant summaries, requiring great insight, but nonetheless just summaries — of the observed motions of the planets. All three of Kepler’s laws turn out to be consequences of Newton’s laws of motion. I shall derive the first and third of the Kepler laws in Chapter 8. The second we are ready to discuss now.
3 Kepler’s first two laws appeared in his book Astronomia Nova in 1609 and the third in another book, Harmonices Mundi, published in 1619.
92 Chapter 3 Momentum and Angular Momentum
Figure 3.7 The orbit of a planet with the sun fixed at O. Kepler’s second law asserts that if the two pairs of points P. Q and 1)’,Q’ are separated by equal time intervals, dt = dt’, then the two areas dA and dA’ are equal.
Kepler’s second law is generally stated something like this:
Keplers Second Law
ach planet incases around the sun, ne drawn rein i the planet to die s cps c zt equ a n equal times.
This rather curious statement is illustrated in Figure 3.7, which shows the path of a planet or comet — the law applies to comets as well — orbiting about the sun at the origin 0. (Throughout this discussion, I shall make the approximation that the sun is fixed; we shall see how to allow for the very small motion of the sun in Chapter 8.) The area “swept out” by the planet moving between any two points P and Q is just the area of the triangle 0 P Q. (Strictly speaking the “triangle” is the area between the two lines OP and 0 Q and the arc P Q. However, it is sufficient to consider pairs of points P and Q that are close together, in which case the difference between the arc P Q and the straight line P Q is negligible.) I shall denote the time elapsed between the planet’s visiting P and Q by dt and the corresponding area of OP Q by dA. Kepler’s second law asserts that if we choose any other pair of points P’ and Q’ separated by the same time interval (dt’ = dt), then the area 0 P’ Q’ will be the same as 0 P Q, or dA’ = dA. Equivalently we can divide both sides of this equality by dt and assert that the rate at which the planet sweeps out area, dAldt, is the same at all points on the orbit; that is, dA/dt is constant.
To prove this result, we note first that the line OP is just the position vector r, and P Q is the displacement dr = v dt. Now, it is a well-known property of the vector
Section 3.5 Angular Momentum for Several Particles 93
product that if two sides of a triangle are given by vectors a and b, then the area of the triangle is 4. 1a x bl . (See Problem 3.24.) Thus the area of the triangle OPQ is
dA = zIr x v dtl.
Replacing v by p/m and dividing both sides by dt, we find that
dA 1 1
dt =
2m Ir x =
2m (3.17)
where denotes the magnitude of the angular momentum e = r x p. Since the planet’s angular momentum about the sun is conserved, this establishes that dA/dt is constant, which, as we have seen, is the content of Kepler’s second law.
An alternative proof of the same result adds some additional insight: It is a straightforward exercise to show that (Problem 3.27)
= mr 2 co (3.18)
where w = is is the planet’s angular velocity around the sun. And it is an equally simple geometrical exercise to show that the rate of sweeping out area is
dA/dt = Zr 2w. (3.19)
Comparison of (3.18) and (3.19) shows that is constant if and only if dA/dt is constant. That is, conservation of angular momentum is exactly equivalent to Kepler’s second law. In addition, we see that as the planet (or comet) approaches closer to the sun (r decreasing) its angular velocity co necessarily increases. Specifically, co is inversely proportional to r2 ; for example, if the value of r at point P’ is half that at P, then the angular velocity co at P’ is four times that at P.
It is interesting to note that our proof of Kepler’s second law depended only on the fact that the gravitational force is central and hence that the planet’s angular momentum about the sun is constant. Thus Kepler’s second law is true for an object that moves under the influence of any central force. By contrast, we shall see in Chapter 8 that the first and third laws (in particular the first, which says that the orbits are ellipses with the sun at one focus) depend on the inverse-square nature of the gravitational force and are not true for other force laws.
3.5 Angular Momentum for Several Particles
Let us next discuss a system of N particles, a = 1, 2, , N, each with its angular momentum lc, = ro, x pa (with all of the re, measured from the same origin 0, of course). We define the total angular momentum L as
N N
L= ta =>ra x pa • (3.20) ot=i ci=i
94 Chapter 3 Momentum and Angular Momentum
Differentiating with respect to t and using the result (3.16), we find that
L=EQa = rc, x F, (3.21) a a
where, as usual, Fa denotes the net force on particle a. This result shows that the rate of change of L is just the net torque on the whole system, an important result in its own right. However, my interest now is to separate the effects of the internal and external forces. As in Equation (1.25) we write Fa as
(net force on particle a) = Fa = E Fas + raxt (3.22) 00ce
where, as before, Fa18 denotes the force exerted on particle a by particle p, and FeaKt is the net force exerted on particle a by all agents outside our N-particle system.
Substituting into (3.21), we find that
= E ra x Fai3 + E ra x Faext • (3.23) a 0Aa a
Equation (3.23) corresponds to Equation (1.27) in our discussion of linear momen- tum back in Chapter 1, and we can rework it in much the same way as there, with one interesting additional twist. We can regroup the terms of the double sum, pairing each term aP with the corresponding term Pa, to give4
E E ra x Fa18 = E E(r a x Fa18 + r18 x Fsa ). (3.24) a /3 0a a 13>a
If we assume that all the internal forces obey the third law (F as = —Fsa ), then we can rewrite the sum on the right as
E E (ra — rp ) x Fa18 . (3.25) a 13>a
To understand this sum, we must examine the vector (ra — r18 ) = ra18 , say. This is illustrated in Figure 3.8, where we see that ra18 is the vector pointing toward particle a from particle /3. If, in addition to satisfying the third law, the forces Fa18 are all central, then the two vectors r a18 and Fat point along the same line, and their cross product is zero.
Returning to Equation (3.23), we conclude that, provided our various assumptions are valid, the double sum in (3.23) is zero. The remaining single sum is just the net external torque, and we conclude that
L = (3.26)
In particular, if the net external torque is zero, we have the
4 Be sure you understand what has happened here. For example, I have paired the term r 1 x F12 with the term r2 x F21.
Section 3.5 Angular Momentum for Several Particles 95
Figure 3.8 The vector rah =(ra — rd points to particle a from particle /3. If the force F,, ,6 is central (points along the line joining a and p), then rap and Fad are collinear and their cross product is zero.
Principle of Conservation of Angular Momentum the net external torque on an N-particle system is /cm the system
annular momentum i> =` Er x pa is constant,
The validity of this principle depends on our two assumptions that all internal forces Foo are central and satisfy the third law. Since these assumptions are almost always valid, the principle (as stated) is likewise. It is of the greatest utility in solving many problems, as I shall illustrate shortly with a couple of simple examples.
The Moment of Inertia
Before discussing an example, it is worth noting that the calculation of angular momenta does not always require one to go back to the basic definition (3.20). As you probably recall from your introductory physics course, for a rigid body rotating about a fixed axis (for example, a wheel rotating on its fixed axle), the rather complicated sum (3.20) can be expressed in terms of the moment of inertia and the angular velocity of rotation. Specifically, if we take the axis of rotation to be the z axis, then L z , the z component of angular momentum, is just L z = Iw, where I is the moment of inertia of the body for the given axis, and w is the angular velocity of rotation. We shall prove and generalize this result in Chapter 10, or you can prove it yourself with the guidance of Problem 3.30. For now, I shall ask you to carry it over from introductory physics. In particular, as you may recall, the moments of inertia of various standard bodies are known. For example, for a uniform disk (mass M, radius R) rotating about its axis, I = 1M R 2 . For a uniform solid sphere rotating about a diameter, I = s M R 2 . In general, for any multiparticle system, I = E m apa2 , where pa is the distance of the mass m a from the axis of rotation.
96 Chapter 3 Momentum and Angular Momentum
EXAMPLE 3.3 Collision of a Lump of Putty with a Turntable
A uniform circular turntable (mass M, radius R, center 0) is at rest in the xy plane and is mounted on a frictionless axle, which lies along the vertical z axis. I throw a lump of putty (mass m) with speed v toward the edge of the turntable, so it approaches along a line that passes within a distance s b of 0, as shown in Figure 3.9. When the putty hits the turntable, it sticks to the edge, and the two rotate together with angular velocity w. Find co.
This problem is easily solved using conservation of angular momentum. Because the turntable is mounted on a frictionless axle, there is no torque on the table in the z direction. Therefore the z component of the external torque on the system is zero, and 1,, is conserved. (This is true even if we include gravity, which acts in the z direction and contributes nothing to the torque in the z direction.) Before the collision, the turntable has zero angular momentum, while the putty has t = r x p, which points in the z direction. Thus the initial total angular momentum has z component
L in = tz r(mv) sin 0 = mvb.
After the collision, the putty and turntable rotate together about the z axis with total moment of inertia 6 I = (m M I2)R 2 , and the z component of the final angular momentum is L zfin = ICO. Therefore, conservation of angular momentum in the form L 1 = Lfin tells us that z z
mvb = (m M/2)R 2w,
Figure 3.9 A lump of putty of mass m is thrown with velocity v at a stationary turntable. The putty’s line of approach passes within the distance b of the table’s center 0.
5 In collision theory — the theory of collisions, usually between atomic or subatomic particles—
the distance b is called the impact parameter. 6 This is m R2 for the putty stuck at radius R plus 4 mR 2 for the uniform turntable.
Section 3.5 Angular Momentum for Several Particles 97
or, solving for co,
m vb = (rn + M12) R2.
(3.27)
This answer is not especially interesting. What is interesting is that we were able to find it with so comparatively little effort. This is typical of the conservation laws, that they can answer many questions so simply. The kind of analysis used here can be used in many situations (such as nuclear reactions) where an incident projectile is absorbed by a stationary target and its angular momentum is shared between the two bodies.
Angular Momentum about the CM
The conservation of angular momentum and the more general result (3.26), L = rext,
were derived on the assumption that all quantities were measured in an inertial frame, so that Newton’s second law could be invoked. This required that both L and Fe’ be measured about an origin 0 fixed in some inertial frame. Remarkably, the same two results also hold if L and feat are measured about the center of mass — even if the CM is being accelerated and so is not fixed in an inertial frame. That is,
—L(about CM) = T”t (about CM) (3.28) dt
and hence, if fex t (about CM) = 0, then L(about CM) is conserved. We shall prove this result in Chapter 10, or you can prove it yourself with the guidance of Problem 3.37. I mention it now, because it allows a very simple solution to various problems, as the following example illustrates.
EXAMPLE 3.4 A Sliding and Spinning Dumbbell
A dumbbell consisting of two equal masses m mounted on the ends of a rigid massless rod of length 2b is at rest on a frictionless horizontal table, lying on the x axis and centered on the origin, as shown in Figure 3.10. At time t = 0, the left mass is given a sharp tap, in the shape of a horizontal force F in the y direction, lasting for a short time At. Describe the subsequent motion.
There are actually two parts to this problem: We must find the initial motion immediately after the impulse, and then the subsequent, force-free motion. The initial motion is not hard to guess, but let us derive it using the tools of this chapter. The only external force is the force F acting in the y direction for the brief time At. Since P F= ext , the total momentum just after the impulse is P = F At. Since P = Mit. (with M = 2m), we conclude that the CM starts moving directly up the y axis with velocity
= = F At/2m.
While the force F is acting, there is a torque Fex t = Fb about the CM, and so, according to (3.28), the initial angular momentum (just after the impulse has
m + • • • m N rN m ares = [Eq. (3.9)] R =
M
98 Chapter 3 Momentum and Angular Momentum
X
Figure 3.10 The left mass of the dumbbell is given a sharp tap in the y direction.
ceased) is L = Fb At. Since L = ho, with I = 2mb2 , we conclude that the dumbbell is spinning clockwise, with initial angular velocity
w= F At I2mb.
The clockwise rotation of the dumbbell means that the left mass is moving up relative to the CM with speed cob, and its total initial velocity is
vieft= v crn + cob = F At /m.
By the same token the right mass is moving down relative to the CM, and its total initial velocity is
Vright = Van COb = 0.
That is, the right mass is initially stationary, while the left one carries all the momentum F At of the system.
The subsequent motion is very straightforward. Once the impulse has ceased, there are no external forces or torques. Thus the CM continues to move straight up the y axis with constant speed, and the dumbbell continues to rotate with constant angular momentum about the CM and hence constant angular velocity.
Principal Definitions and Equations of Chapter 3
Equation of Motion for a Rocket
= _thvex Fext . [Eqs. (3.6) & (3.29)]
The Center of Mass of Several Particles
where M is the total mass of all particles, M = E m a .
Problems for Chapter 3 99
Angular Momentum
For a single particle with position r (relative to an origin 0) and momentum p, the angular momentum about 0 is
= r x p. [Eq. (3.14)]
For several particles, the total angular momentum is
L .E.ea = ro, x p,• [Eq. (3.20)] a=1 a=1
Provided all the internal forces are central,
L = rext
[Eq. (3.26)]
where Text is the net external torque.
Problems for Chapter 3
Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***).
SECTION 3.1 Conservation of Momentum
3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell’s speed relative to the gun is v.) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell’s speed relative to the ground is v /(1 m/M).
3.2 * A shell traveling with speed v o exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed vo . What is the velocity of the other fragment?
3.3 * A shell traveling with velocity vo explodes into three pieces of equal masses. Just after the explosion, one piece has velocity v 1 = vo and the other two have velocities v2 and v3 that are equal in magnitude (v2 = v3) but mutually perpendicular. Find v2 and v3 and sketch the three velocities.
3.4 ** Two hobos, each of mass m h , are standing at one end of a stationary railroad flatcar with frictionless wheels and mass m t-c . Either hobo can run to the other end of the flatcar and jump off with the same speed u (relative to the car). (a) Use conservation of momentum to find the speed of the recoiling car if the two men run and jump simultaneously. (b) What is it if the second man starts running only after the first has already jumped? Which procedure gives the greater speed to the car? [Hint: The speed u is the speed of either hobo, relative to the car just after he has jumped; it has the same value for either man and is the same in parts (a) and (b).]
3.5 ** Many applications of conservation of momentum involve conservation of energy as well, and we haven’t yet begun our discussion of energy. Nevertheless, you know enough about energy from your introductory physics course to handle some problems of this type. Here is one elegant example: An elastic collision between two bodies is defined as a collision in which the total kinetic energy of the two bodies after the collision is the same as that before. (A familiar example is the collision
100 Chapter 3 Momentum and Angular Momentum
between two billiard balls, which generally lose extremely little of their total kinetic energy.) Consider an elastic collision between two equal mass bodies, one of which is initially at rest. Let their velocities be v i and v2 = 0 before the collision, and v’1 and v; after. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that the angle between vi and v; is 90°. This result was important in the history of atomic and nuclear physics: That two bodies emerged from a collision traveling on perpendicular paths was strongly suggestive that they had equal mass and had undergone an elastic collision.
SECTION 3.2 Rockets
3.6 * In the early stages of the Saturn V rocket’s launch, mass was ejected at about 15,000 kg/s, with a speed vex ti 2500 m/s relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton ti 9000 newtons) and compare with the rocket’s initial weight (about 3000 tons).
3.7 * The first couple of minutes of the launch of a space shuttle can be described very roughly as follows: The initial mass is 2 x 10 6 kg, the final mass (after 2 minutes) is about 1 x 10 6 kg, the average exhaust speed v ex is about 3000 m/s, and the initial velocity is, of course, zero. If all this were taking place in outer space, with negligible gravity, what would be the shuttle’s speed at the end of this stage? What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)?
3.8 * A rocket (initial mass m 0) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass Am o of its fuel, for how long can it hover? [Hint: Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables t and m. Take v„ to be constant.] (b) If vex ti 3000 m/s and A 10%, for how long could the rocket hover just above the earth’s surface?
3.9 * From the data in Problem 3.7 you can find the space shuttle’s initial mass and the rate of ejecting mass —tic (which you may assume is constant). What is the minimum exhaust speed v ex for which the shuttle would just begin to lift as soon as burn is fully underway? [Hint: The thrust must at least balance the shuttle’s weight.]
3.10 * Consider a rocket (initial mass m o) accelerating from rest in free space. At first, as it speeds up, its momentum p increases, but as its mass m decreases p eventually begins to decrease. For what value of m is p maximum?
3.11 ** (a) Consider a rocket traveling in a straight line subject to an external force F” t acting along the same line. Show that the equation of motion is
m U = vex + Fext. (3.29)
[Review the derivation of Equation (3.6) but keep the external force term.] (b) Specialize to the case of a rocket taking off vertically (from rest) in a gravitational field g, so the equation of motion becomes
my = — Mv ex — mg. (3.30)
Assume that the rocket ejects mass at a constant rate, tit = — k (where k is a positive constant), so that m = mo — kt. Solve equation (3.30) for v as a function of t, using separation of variables (that is, rewriting the equation so that all terms involving v are on the left and all terms involving t on the right). (c) Using the rough data from Problem 3.7, find the space shuttle’s speed two minutes into flight, assuming (what is nearly true) that it travels vertically up during this period and that g doesn’t change appreciably. Compare with the corresponding result if there were no gravity. (d) Describe what would
Problems for Chapter 3 101
happen to a rocket that was designed so that the first term on the right of Equation (3.30) was smaller than the initial value of the second.
3.12 ** To illustrate the use of a multistage rocket consider the following: (a) A certain rocket carries 60% of its initial mass as fuel. (That is, the mass of fuel is 0.6m 0 .) What is the rocket’s final speed, accelerating from rest in free space, if it burns all its fuel in a single stage? Express your answer as a multiple of vex . (b) Suppose instead it burns the fuel in two stages as follows: In the first stage it burns a mass 0.3m o of fuel. It then jettisons the first-stage fuel tank, which has a mass of 0.1m o , and then burns the remaining 0.3m 0 of fuel. Find the final speed in this case, assuming the same value of v ex
throughout, and compare.
3.13 ** If you have not already done it, do Problem 3.11(b) and find the speed v(t) of a rocket accelerating vertically from rest in a gravitational field g. Now integrate v(t) and show that the rocket’s height as a function of t is
1 M y (t) = V ex t
2 gt
2 MP
k ex in (
m /
Using the numbers given in Problem 3.7, estimate the space shuttle’s height after two minutes.
3.14 ** Consider a rocket subject to a linear resistive force, f = —by, but no other external forces. Use Equation (3.29) in Problem 3.11 to show that if the rocket starts from rest and ejects mass at a constant rate k = —th, then its speed is given by
k b I k v
b vex[l ( m 1 .
mo
SECTION 3.3 The Center of Mass
3.15 * Find the position of the center of mass of three particles lying in the xy plane at r 1 = (1, 1, 0), r2 = (1, —1, 0), and r 3 = (0, 0, 0), if m 1 = m2 and m 3 = 10m 1 . Illustrate your answer with a sketch and comment.
3.16 * The masses of the earth and sun are Me = 6.0 x 1024 and Ms ti 2.0 x 1030 (both in kg) and their center-to-center distance is 1.5 x 10 8 km. Find the position of their CM and comment. (The radius of the sun is R s 7.0 x 105 km.)
3.17 * The masses of the earth and moon are M,==, 6.0 x 1024 and Mm 7.4 x 1022 (both in kg) and their center to center distance is 3.8 x 10 5 km. Find the position of their CM and comment. (The radius of the earth is Re ti 6.4 x 103 km.)
3.18 ** (a) Prove that the CM of any two particles always lies on the line joining them, as illustrated in Figure 3.3. [Write down the vector that points from m 1 to the CM and show that it has the same direction as the vector from m l to m 2 .] (b) Prove that the distances from the CM to m 1 and m 2 are in the ratio m 2/m l . Explain why if m 1 is much greater than m 2 , the CM lies very close to the position of m 1 .
3.19 ** (a) We know that the path of a projectile thrown from the ground is a parabola (if we ignore air resistance). In the light of the result (3.12), what would be the subsequent path of the CM of the pieces if the projectile exploded in midair? (b) A shell is fired from level ground so as to hit a target 100 m away. Unluckily the shell explodes prematurely and breaks into two equal pieces. The two pieces land
102 Chapter 3 Momentum and Angular Momentum
at the same time, and one lands 100 m beyond the target. Where does the other piece land? (c) Is the same result true if they land at different times (with one piece still landing 100 m beyond the target)?
3.20 ** Consider a system comprising two extended bodies, which have masses M 1 and M2 and centers of mass at R 1 and R2. Prove that the CM of the whole system is at
R= MiRI M2R2
• + M2
This beautiful result means that in finding the CM of a complicated system, you can treat its component parts just like point masses positioned at their separate centers of mass — even when the component parts are themselves extended bodies.
3.21 ** A uniform thin sheet of metal is cut in the shape of a semicircle of radius R and lies in the xy plane with its center at the origin and diameter lying along the x axis. Find the position of the CM using polar coordinates. [In this case the sum (3.9) that defines the CM position becomes a two-dimensional integral of the form f r a dA where a denotes the surface mass density (mass/area) of the sheet and dA is the element of area dA = r dr d0.]
3.22 ** Use spherical polar coordinates r, 9, 4) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the xy plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is dV = r2dr sin 9 d0 dO. (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)
3.23 *** [Computer] A grenade is thrown with initial velocity v o from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: v. = (4, 4), g = 1, and 0 < t < 4 (and with x measured horizontally and y vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at t = 1, 2, 3, 4. (b) At t = 4, when the grenade’s velocity is v, it explodes into two equal pieces, one of which moves off with velocity v -I- Av. What is the velocity of the other piece? (c) Assuming that Av = (1, 3), add to your original plot the paths of the two pieces for 4 < t < 9. Insert marks to show their positions at t = 5, 6, 7, 8, 9. Find some way to show clearly that the CM of the two pieces continues to follow the original parabolic path.
SECTION 3.4 Angular Momentum for a Single Particle
3.24 * If the vectors a and b form two of the sides of a triangle, prove that z a x b I is equal to the area of the triangle.
3.25 * A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius ro with angular velocity wo, but I now pull the string down through the hole until a length r remains between the hole and the particle. What is the particle’s angular velocity now?
3.26 * A particle moves under the influence of a central force directed toward a fixed origin 0. (a) Explain why the particle’s angular momentum about 0 is constant. (b) Give in detail the argument that the particle’s orbit must lie in a single plane containing 0.
3.27 ** Consider a planet orbiting the fixed sun. Take the plane of the planet’s orbit to be the xy plane, with the sun at the origin, and label the planet’s position by polar coordinates (r, 4)). (a) Show that the
Problems for Chapter 3 103
planet’s angular momentum has magnitude = mr2w, where co = 4 is the planet’s angular velocity about the sun. (b) Show that the rate at which the planet “sweeps out area” (as in Kepler’s second law) is d A Idt = 4r2w, and hence that d A Idt = £/2m. Deduce Kepler’s second law.
SECTION 3.5 Angular Momentum for Several Particles
3.28 * For a system of just three particles, go through in detail the argument leading from (3.20) to (3.26), L = rext, writing out all the summations explicitly.
3.29 * A uniform spherical asteroid of radius R o is spinning with angular velocity co o . As the aeons go by, it picks up more matter until its radius is R. Assuming that its density remains the same and that the additional matter was originally at rest relative to the asteroid (anyway on average), find the asteroid’s new angular velocity. (You know from elementary physics that the moment of inertia is s MR 2 .) What is the final angular velocity if the radius doubles?
3.30 ** Consider a rigid body rotating with angular velocity co about a fixed axis. (You could think of a door rotating about the axis defined by its hinges.) Take the axis of rotation to be the z axis and use cylindrical polar coordinates poi , Oa , z cy to specify the positions of the particles a = 1, , N that make up the body. (a) Show that the velocity of the particle a is paw in the 0 direction. (b) Hence show that the z component of the angular momentum f a of particle a is m apa2w. (c) Show that the z component L z of the total angular momentum can be written as L z = I co where I is the moment of inertia (for the axis in question),
N
,L ma pa2 . (3.31) a=1
3.31 ** Find the moment of inertia of a uniform disc of mass M and radius R rotating about its axis, by replacing the sum (3.31) by the appropriate integral and doing the integral in polar coordinates.
3.32 ** Show that the moment of inertia of a uniform solid sphere rotating about a diameter is s MR 2 . The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the z axis. The element of volume is dV = r2dr sin 6) dB d/. (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)
3.33 ** Starting from the sum (3.31) and replacing it by the appropriate integral, find the moment of inertia of a uniform thin square of side 2b, rotating about an axis perpendicular to the square and passing through its center.
3.34 ** A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its CM is traveling vertically up at speed v o and it is rotating with angular velocity co o . To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should vo be, if the rod is to have made exactly n rotations when it returns to his hand?
3.35 ** Consider a uniform solid disk of mass M and radius R, rolling without slipping down an incline which is at angle y to the horizontal. The instantaneous point of contact between the disk and the incline is called P. (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration i) of the disk by applying the result L = tex t for rotation about P. (Remember that L = Ico and the moment of inertia for rotation about a point on the circumference is 2 MR 2 . The condition that the disk not slip is that v = Rco and hence i) = Rt.) (c) Derive the same result by applying rext
104 Chapter 3 Momentum and Angular Momentum
to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton’s second law to the motion of the CM. The moment of inertia for rotation about the CM is 4 MR2 .)
3.36 ** Repeat the calculations of Example 3.4 (page 97) for the case that the force F acts in a “northeasterly” direction at angle y from the x axis. What are the velocities of the two masses just after the impulse has been applied? Check your answers for the cases that y = 0 and y = 90°.
3.37 *** A system consists of N masses m a at positions ra relative to a fixed origin 0. Let ra denote the position of m a relative to the CM; that is, r at = ra — R. (a) Make a sketch to illustrate this last equation. (b) Prove the useful relation that > m ares = 0. Can you explain why this relation is nearly obvious? (c) Use this relation to prove the result (3.28) that the rate of change of the angular momentum about the CM is equal to the total external torque about the CM. (This result is surprising since the CM may be accelerating, so that it is not necessarily a fixed point in any inertial frame.)
CHAPTER
Energy
This chapter takes up the conservation of energy. You will see that the analysis of energy conservation is surprisingly more complicated than the corresponding discus- sions of linear and angular momenta in Chapter 3. The main reason for the difference is this: In almost all problems of classical mechanics there is only one kind of linear momentum (p = my for each particle), and one kind of angular momentum (f = r x p for each particle). By contrast, energy comes in many different and important forms: kinetic, several kinds of potential, thermal, and more. It is the processes that trans- form energy from one kind to another that complicate the use of energy conservation. We shall see that conservation of energy is a quite subtle business, even for a system consisting of just a single particle.
One manifestation of the relative difficulty of the discussion of energy is that we shall need some new tools from vector calculus, namely, the concepts of the gradient and the curl. I shall introduce these important ideas as we need them.
4.1 Kinetic Energy and Work
As I have said, there are many different kinds of energy. Perhaps the most basic is kinetic energy (or KE), which for a single particle of mass in traveling with speed v is defined to be
T = 1 mv 2 2 •
(4.1)
Let us imagine the particle moving through space and examine the change in its kinetic energy as it moves between two neighboring points r 1 and r 1 + dr on its path as shown in Figure 4.1. The time derivative of T is easily evaluated if we note that v 2 = v • v, so that
d T d • — = -,m — (v • v) = m(v • v + v – V) = tn .@•v .
dt dt (4.2) 105
106 Chapter 4 Energy
Figure 4.1 Three points on the path of a particle:
r 1 , r 1 + dr (with dr infinitesimal) and r2 .
By the second law, the factor my is equal to the net force F on the particle, so that
dT = F • v.
dt (4.3)
If we multiply both sides by dt, then since v dt is the displacement dr we find
dT = F • dr. (4.4)
The expression on the right, F • dr, is defined to be the work done by the force F in the displacement dr. Thus we have proved the Work—KE theorem, that the change in the particle’s kinetic energy between two neighboring points on its path is equal to the work done by the net force as it moves between the two points.’
So far we have proved the Work—KE theorem only for an infinitesimal displace- ment dr, but it generalizes easily to larger displacements. Consider the two points shown as r 1 and r2 in Figure 4.1. We can divide the path between these points 1 and 2 into a large number of very small segments, to each of which we can apply the in- finitesimal result (4.4). Adding all of these results, we find that the total change in T going from 1 to 2 is the sum > F • dr of all the infinitesimal works done in all the infinitesimal displacements between points 1 and 2:
AT T2 — = F • dr. (4.5)
In the limit that all the displacements dr go to zero, this sum becomes an integral:
2 EF • dr —> f F • dr. (4.6)
1 Two points that can be puzzling at first: The work F • dr can be negative, if for example F and dr point in opposite directions. While the notion of a force doing negative work conflicts with our everyday notion of work, it is perfectly consisent with the physicist’s definition: A force in the opposite direction to the displacement reduces the KE, so, by the work—KE theorem, the corresponding work has to be negative. Second, if F and dr are perpendicular, then the work F • dr is zero. Again this conflicts with our everyday sense of work, but is consistent with the physicist’s usage: A force that is perpendicular to the displacement does not change the KE.
Section 4.1 Kinetic Energy and Work 107
This integral, called a line integral, 2 is a generalization of the integral f f (x) dx over a single variable x, and its definition as the limit of the sum of many small pieces is closely analogous. If you feel any doubt about the symbol fl F • dr on the right of (4.6), think of it as being just the sum on the left (with all the displacements infinitesimally small). In evaluating a line integral, it is usually possible to convert it into an ordinary integral over a single variable, as the following examples show. Notice that, as the name implies, the line integral depends (in general) on the path that the particle followed from point 1 to point 2. The particular line integral on the right of (4.6) is called the work done by the force F moving between points 1 and 2 along the path concerned.
EXAMPLE 4.1 Three Line Integrals
Evaluate the line integral for the work done by the two-dimensional force F = (y, 2x) going from the origin 0 to the point P = (1, 1) along each of the three paths shown in Figure 4.2. Path a goes from 0 to Q = (1, 0) along the x axis and then from Q straight up to P, path b goes straight from 0 to P along the line y = x, and path c goes round a quarter circle centered on Q.
The integral along path a is easily evaluated in two parts, if we note that on OQ the displacements have the form dr = (dx, 0), while on QP they are dr = (0, dy). Thus
1 1 Wa = f F • dr = f F • dr ± f F • dr = f Fx (x, 0) dx + f Fy (1, y) dy
a 0 0
=0 + 2 J dy = 2.
Figure 4.2 Three different paths, a, b, and c, from the
origin to the point P = (1, 1).
2 Not an especially happy name for those of us who think of a line as something straight. However, there are curved lines as well as straight lines, and in general a line integral can involve a curved line, such as the path shown in Figure 4.1.
108 Chapter 4 Energy
On the path b, x = y, so that dx = dy, and
1
Wb = f F • dr = f (F x dx Fy dy) = f (x 2x)dx = 1.5. 0
Path c is conveniently expressed parametrically as
r (x, y) = (1 — cos 0, sin 0)
where 8 is the angle between 0 Q and the line from Q to the point (x, y), with 0 < 8 < 7/2. Thus on path c
dr = (dx, dy) = (sin 0, cos 0) d0 and
= f F • dr = f (Fx dx Fy dy)
7 f
/2 = [sin 2 0 + 2(1 — cos 0) cos 6] d8 = 2 — 7/4 = 1.21.
Some more examples can be found in Problems 4.2 and 4.3 and, if you have never studied line integrals, you may want to try some of these.
With the notation of the line integral, we can rewrite the result (4.5) as
where I have introduced the notation W (1 -± 2) for the work done by F moving from point 1 to point 2. The result is the Work—KE theorem for arbitrary displacements, large or small: The change in a particle’s KE as it moves between points 1 and 2 is the work done by the net force.
It is important to remember that the work that appears on the right of (4.7) is the work done by the net force F on the particle. In general, F is the vector sum of various separate forces
F = ± • • • ± F, Fi . i=i
(For example, the net force on a projectile is the sum of two forces, the weight and air resistance.) It is a most convenient fact that to evaluate the work done by the net force F, we can simply add up the works done by the separate forces F 1 , • • • , Fn . This claim is easily proved as follows:
2 2
W(1 —›- 2) = f F • dr = f E Fi • dr 2
_> Fi • dr = w, ( 1_, 2). . (4.8)
Section 4.2 Potential Energy and Conservative Forces 109
The crucial step, from the first line to the second, is justified because the integral of a sum of n terms is the same as the sum of the n individual integrals. The Work–KE theorem can therefore be rewritten as
T2 – = (1 -> 2). (4.9) i =
In practice, one almost always uses the theorem in this way: Calculate the work W i done by each of the n separate forces on the particle and then set A T equal to the sum
of all the Wi . If the net force on a particle is zero, then the Work–KE theorem tells us that the
particle’s kinetic energy is constant. This simply says that the speed v is constant, which, though true, is not very interesting, since it already follows from Newton’s first law.
4.2 Potential Energy and Conservative Forces
The next step in the development of the energy formalism is to introduce the notion of potential energy (or PE) corresponding to the forces on an object. As you probably recall, not every force lends itself to the definition of a corresponding potential energy. Those special forces that do have a corresponding potential energy (with the required properties) are called conservative forces, and we must discuss the properties that distinguish conservative from nonconservative forces. Specifically, we shall find that there are two conditions that a force must satisfy to be considered conservative.
To simplify our discussion, let us assume at first that there is only one force acting on the object of interest — the gravitational force on a planet by its sun, or the electric force qE on a charge in an electric field (with no other forces present). The force F may depend on many different variables: It may depend on the object’s position r. (The farther the planet is from the sun, the weaker the gravitational pull.) It may depend on the object’s velocity, as is the case with air resistance; and it may depend on the time t, as would be the case for a charge in a time-varying electric field. Finally, if the force is exerted by humans, it will depend on a host of imponderables hoW tired they are feeling, how conveniently they are situated to push, and so on.
The first condition for a force F to be conservative is that F depends only on the position r of the object on which it acts; it must not depend on the velocity, the time, or any variables other than r. This sounds, and is, quite restrictive, but there are plenty of forces that have this property: The gravitational force of the sun on a planet (position r relative to the sun) can be written as
F(r) = GmM
r 2
which evidently depends only on the variable r. (The parameters G, m, M are constant for a given planet and given sun.) Similarly, the electrostatic force F(r) = qE(r) on a charge q by a static electric field E(r) has this property. Forces that do not satisfy this condition include the force of air resistance (which depends on the velocity), friction
110 Chapter 4 Energy
Figure 4.3 Three different paths, a, b, and c, joining the same two points 1 and 2.
(which depends on the direction of motion), the magnetic force (which depends on the velocity), and the force of a time-varying electric field E(r, t) (which obviously depends on time).
The second condition that a force must satisfy to be called conservative concerns the work done by the force as the object on which it acts moves between two points r 1 and r2 (or just 1 and 2 for short),
2
W (1—* 2) = F -dr. (4.10)
Figure 4.3 shows two points, 1 and 2, and three different paths connecting them. It is entirely possible that the work done between points 1 and 2, as defined by the integral (4.10), has different values depending on which of the three paths, a, b, or c, the particle happens to follow. For example, consider the force of sliding friction as I push a heavy crate across the floor. This force has a constant magnitude, Ffric say, and is always opposite to the direction of motion. Thus the work done by friction as the crate moves from 1 to 2 is given by (4.10) to be
Wfric (1 2) = — Ffii,L,
where L denotes the length of the path followed. The three paths of Figure 4.3 have different lengths, and Whic (1 —> 2) will have a different value for each of the three paths.
On the other hand, there are forces with the property that the work W (1 –> 2) is the same for any path connecting the same two points 1 and 2. An example of a force with this property is the gravitational force, Fgrav = mg, of the earth on an object close to the earth’s surface. It is easy to show (Problem 4.5) that, because g is a constant vector pointing vertically down, the work done in this case is
Wgrav (1 2) = —mgh, (4.11)
where h is just the vertical height gained between points 1 and 2. This work is the same for any two paths between the given points 1 and 2. This property, the path
Section 4.2 Potential Energy and Conservative Forces 111
independence of the work it does, is the second condition that a force must satisfy to be considered conservative, and we are now ready to state the two conditions:
Conditions for a Force to be Conservative
A fOCe F act: on a par isle is co se conservativeative if and only if it sa eS: MO’ damns:
i) F depends only on t p on r and not on the ve loci ty Or the tiff t car any other variable); thatF F(r). For any two mints I and 2, the work 2) don by F is the same for all paths between l and 2.
The reason for the name “conservative” and for the importance of the concept is this: If all forces on an object are conservative, we can define a quantity called the potential energy (or just PE), denoted U(r), a function only of position, with the property that the total mechanical energy
E = KE + PE = T U(r) (4.12)
is constant; that is, E is conserved. To define the potential energy U (r) corresponding to a given conservative force,
we first choose a reference point ro at which U is defined to be zero. (For example, in the case of gravity near the earth’s surface, we often define U to be zero at ground level.) We then define U(r), the potential energy at an arbitrary point r, to be3
In words, U (r) is minus the work done by F if the particle moves from the reference point ro to the point of interest r, as in Figure 4.4. (We shall see the reason for the minus sign shortly.) Notice that the definition (4.13) only makes sense because of the property (ii) of conservative forces. If the work integral in (4.13) were different for different paths, then (4.13) would not define a unique function 4 U(r).
3 Notice that I have called the variable of integration r’ to avoid confusion with the upper limit r 4 The definition (4.13) also depends on property (i) of conservative forces, but in a slightly subtler
way. If F depended on another variable besides r (for instance, t or v), then the right side of (4.13) would depend on when or how the particle moved from r o to r, and again there would be no uniquely defined U(r).
112 Chapter 4 Energy
Figure 4.4 The potential energy U(r) at any point r is de- fined as minus the work done by F if the particle moves from the reference point ro to r. This gives a well-defined function U (r) only if this work is independent of the path followed — that is, the force is conservative.
EXAMPLE 4.2 Potential Energy of a Charge in a Uniform Electric Field
A charge q is placed in a uniform electric field pointing in the x direction with strength Eo , so that the force on q is F = qE = qEoX. Show that this force is conservative and find the corresponding potential energy.
The work done by F going between any two points 1 and 2 along any path is
2 2 2
W(1 —> 2) = f F -dr = qEo f • dr = qE0 f dx = qEo (x 2 — x i). (4.14) 1 1
This depends only on the two end points 1 and 2. (In fact it depends only on their x coordinates x 1 and x2 .) Certainly, it is independent of the path, and the force is conservative. To define the corresponding potential energy U (r), we must first pick a reference point r o at which U will be zero. A natural choice is the origin, ro = 0, in which case the potential energy is U (r) = — W (0 —k r) or, according to (4.14),
U(r) = —qE ox.
We can now derive a crucial expression for the work done by F in terms of the potential energy U(r). Let r 1 and r2 be any two points as in Figure 4.5. If r o is the reference point at which U is zero, then it is clear from Figure 4.5 that
W(ro r2) = W(ro —> r 1) W(r —> r2 )
and hence
W (r i r2) = W(ro —> r2) — W (ro —> r 1 ). (4.15)
Each of the two terms on the right is (minus) the potential energy at the corresponding point. Thus we have proved that the work on the left is just the difference of these two potential energies:
W(ri r2) = — [U(r2) — U(ri)] = — AU. (4.16)
Section 4.2 Potential Energy and Conservative Forces 113
Figure 4.5 The work W (r i —> r2) going from r 1 to r2 is the same as W (ro —> r2) minus W(rc, –> r i ). This result is inde- pendent of what path we use for either limb of the journey, provided the force concerned is conservative.
The usefulness of this result emerges when we combine it with the Work—KE theorem (4.7):
AT = W(r i —> r2 ).
Comparing this with (4.16), we see that
AT = —AU
or, moving the right side across to the left, 5
A(T + U) -= 0.
That is, the mechanical energy
(4.17)
(4.18)
(4.19)
E = T U (4.20)
does not change as the particle moves from r 1 to r2 . Since the points r 1 and r2 were any two points on the particle’s trajectory, we have the important conclusion: If the force on a particle is conservative, then the particle’s mechanical energy never changes; that is, the particle’s energy is conserved, which explains the use of the adjective “conservative.”
Several Forces
So far we have established the conservation of energy for a particle subject to a single conservative force. If the particle is subject to several forces, all of them conservative, our result generalizes easily. For instance, imagine a mass suspended from the ceiling by a spring. This mass is subject to two forces, the forces of gravity (Fgray 1 and the , spring (Fspr). The force of gravity is certainly conservative (as I’ve already argued), and, provided the spring obeys Hooke’s law, Fspr is likewise (see Problem 4.42). We
5 We now see the reason for the minus sign in the definition of U. It gives the minus sign on the right of (4.18), which in turn gives the desired plus sign on the left of (4.19).
114 Chapter 4 Energy
can define separate potential energies for each force, Ugrav for Fgrav and U for F U. Fspr , each with the crucial property (4.16) that the change in U gives (minus) the work done
by the corresponding force. According to the Work—KE theorem, the change in the mass’s kinetic energy is
A T = Wgrav W spr
= Ugrav A Uspr) , (4.21)
where the second line follows from the properties of the two separate potential energies. Rearranging this equation, we see that A (T + Ugrav Uspr) = 0. That is, the total mechanical energy, defined as E =T Ugrav Uspr , is conserved.
The argument just given extends immediately to the case of n forces on a particle, so long as they are all conservative. If for each force F i we define a corresponding potential energy Ui , then we have the
Principle of Con ervation of Energy One Particle of>the n forces F n) acting on a Particle are conservative, each
corresnding pot a energy U,(r), cmec hanical ene with its po as
E U T Ut ) U, (4.22)
n time.
Nonconservative Forces
If some of the forces on our particle are nonconservative, then we cannot define corre- sponding potential energies; nor can we define a conserved mechanical energy. Nev- ertheless, we can define potential energies for all of the forces that are conservative, and then recast the Work—KE theorem in a form that shows how the nonconservative forces change the particle’s mechanical energy. First, we divide the net force on the particle into two parts, the conservative part F., and the nonconservative part Fnc . For Fcons we can define a potential energy, which we’ll call just U. By the Work—KE theorem, the change in kinetic energy between any two times is
A T = W = Wcons Wnc • (4.23)
The first term on the right is just — A U and can be moved to the left side to give A (T + U) = Wne . If we define the mechanical energy as E =T U, then we see that
AE A(T + U) = Wnc- (4.24)
Mechanical energy is no longer conserved, but we have the next best thing. The mechanical energy changes to precisely the extent that the nonconservative forces do work on our particle. In many problems the only nonconservative force is the force
defin
Section 4.2 Potential Energy and Conservative Forces 115
of sliding friction, which usually does negative work. (The frictional force f is in the direction opposite to the motion, so the work f • dr is negative.) In this case ke is negative and (4.24) tells us that the object loses mechanical energy in the amount “stolen” by friction. All of these ideas are illustrated by the following simple example.
EXAMPLE 4.3 Block Sliding Down an Incline
Consider again the block of Example 1.1 and find its speed v when it reaches the bottom of the slope, a distance d from its starting point.
The setup and the forces on the block are shown in Figure 4.6. The three forces on the block are its weight, w = mg, the normal force of the incline, N, and the frictional force f, whose magnitude we found in Example 1.1 to be f = pcmg cos 0. The weight mg is conservative, and the corresponding potential energy is (as you certainly recall from introductory physics, but see Problem 4.5)
U = mgy
where y is the block’s vertical height above the bottom of the slope (if we choose the zero of PE at the bottom). The normal force does no work, since it is perpendicular to the direction of motion, so will not contribute to the energy balance. The frictional force does work Wfric fd = — Amgd cos 0. The change in kinetic energy is A T = Tf Ti = lmv2 and the change in potential energy is A U = Uf Ui = —mgh = —mgd sin 0. Thus (4.24) reads
AT + AU = Wfric
Or
im v2 — mgd sin 0 = mumgd cos 9.
Solving for v we find
v = \/2gd (sin 0 — ,u, cos 9).
Figure 4.6 A block on an incline of angle 0. The length of the slope is d, and the height is h = d sin O.
116 Chapter 4 Energy
As usual, you should check that this answer agrees with common sense. For example, does it give the expected answer when 9 = 900 ? What about 0 = 0? (The case 9 = 0 is a bit subtler.)
4.3 Force as the Gradient of Potential Energy
We have seen that the potential energy U(r) corresponding to a force F(r) can be expressed as an integral of F(r) as in (4.13). This suggests that we should be able to write F(r) as some kind of derivative of U(r). This suggestion proves correct, though to implement it we shall need some mathematics that you may not have met before. Specifically, since F(r) is a vector [while U (r) is a scalar] we shall be involved in some vector calculus.
Let us consider a particle acted on by a conservative force F (r), with corresponding potential energy U (r), and examine the work done by F(r) in a small displacement from r to r + dr. We can evaluate this work in two ways. On the one hand, it is, by definition,
W(r r+dr) = F(r) • dr
= Fx dx + Fy dy + dz, (4.25)
for any small displacement dr with components (dx, dy, dz). On the other hand, we have seen that the work W (r r+dr) is the same as
(minus) the change in PE in the displacement:
W(r r+dr) = —dU = —[U(r + dr) — U(r)]
= —[U(x + dx,y + dy,z + dz) — U(x,y,z)]. (4.26)
In the second line, I have replaced the position vector r by its components to emphasize that U is really a function of the three variables (x, y, z). Now, for functions of one variable, a difference like that in (4.26) can be expressed in terms of the derivative:
df = f (x + dx) — f (x) = df
dx dx. (4.27)
This is really no more than the definition of the derivative. 6 For a function of three variables, such as U(x, y, z), the corresponding result is
dU = U(x + dx, y + dy, z + dz) — U(x, y, z)
= — au
dx + — au
dy + — au
dz ax ay az
(4.28)
where the three derivatives are the partial derivatives with respect to the three in- dependent variables (x, y, z). [For example, au/ax is the rate of change of U as x
6 Strictly speaking, this equation is exact only in the limit that dx 0. As usual, I take the view that dx is small enough (though nonzero) that the two sides are equal within our chosen accuracy target.
Section 4.3 Force as the Gradient of Potential Energy 117
changes, with y and z fixed, and is found by differentiating U(x, y, z) with respect to x treating y and z as constants. See Problems 4.10 and 4.11 for some examples.] Substituting (4.28) into (4.26), we find that the work done in the small displacement from r to r + dr is
au , au W(r —> r+dr) = —
au [—ax + — dy + —dz (4.29)
ax B y az
The two expressions (4.25) and (4.29) are both valid for any small displacement dr. In particular, we can choose dr to point in the x direction, in which case dy = dz = 0 and the last two terms in both (4.25) and (4.29) are zero. Equating the remaining terms, we see that Fx = —au/ax. By choosing dr to point in the y or z directions, we get corresponding results for Fy and F2 , and we conclude that
F =— au au aU
F , F = — x Y ay z az (4.30)
That is, F is the vector whose three components are minus the three partial derivatives of U with respect to x, y, and z. A slightly more compact way to write this result is this:
,s yau ,au aU F = —x— — — — z—. ax ay az
(4.31)
Relationships like (4.31) between a vector (F) and a scalar (U) come up over and over again in physics. For example, the electric field E is related to the electrostatic potential V in exactly the same way. More generally, given any scalar f(r), the vector whose three components are the partial derivatives of f(r) is called the gradient of f, denoted Vf:
,af ,af ,af Vf x— + y— + z—
ax ay a .
z (4.32)
The symbol Vf is pronounced “grad f” The symbol V by itself is called “grad,” or “del,” or “nabla.” With this notation, (4.31) is abbreviated to
This important relation gives us the force F in terms of derivatives of U, just as the definition (4.13) gave U as an integral of F. When a force F can be expressed in the form (4.33), we say that F is derivable from a potential energy. Thus, we have shown that any conservative force is derivable from a potential energy. 7
7 1 am following standard terminology here. Notice that we have defined “conservative” so that a conservative force conserves energy and is derivable from a potential energy. This is occasionally confusing, since there are forces (such as the magnetic force on a charge or the normal force on a sliding object) that do no work and hence conserve energy, but are not “conservative” in the sense defined here, since they are not derivable from a potential energy. This unfortunate confusion seldom causes trouble, but you may want to register it somewhere in the back of your mind.
118 Chapter 4 Energy
EXAMPLE 4.4 Finding F from U
The potential energy of a certain particle is U = Axy 2 B sin Cz, where A, B and C are constants. What is the corresponding force?
To find F we have only to evaluate the three partial derivatives in (4.31). In doing this, you must remember that au/ax is found by differentiating with respect to x, treating y and z as constant, and so on. Thus au/ax = Ay2 , and so on, and the final result is
F = —(x Ay 2 Sr 2Axy i BC cos Cz).
It is sometimes convenient to remove the f from (4.32) and to write
a a a V=x—
ax +y—
a +z—
az .
y (4.34)
In this view, V is a vector differential operator that can be applied to any scalar f and produces the vector given in (4.32).
A very useful application of the gradient is given by (4.28), whose right-hand side you will recognize as V U • dr. Thus, if we replace U by an arbitrary scalar f , we see that the change in f resulting from a small displacement dr is just
This useful relation is the three-dimensional analog of Equation (4.27) for a function of one variable. It shows the sense in which the gradient is the three-dimensional equivalent of the ordinary derivative in one dimension.
If you have never met the V notation before, it will take a little getting used to. Meanwhile, you can just think of (4.33) as a convenient shorthand for the three equations (4.30). For practice using the gradient, you could look at Problems 4.12 through 4.19.
4.4 The Second Condition that F be Conservative
We have seen that one of the two conditions that a force F be conservative is that the work fi2 F • dr which it does moving between any two points 1 and 2 must be independent of the path followed. You are certainly to be excused if you don’t see how we could test whether a given force has this property. Checking the value of the integral for every pair of points and every path joining those points is indeed a formidable prospect! Fortunately, we never need to do this. There is a simple test, which can be quickly applied to any force that is given in analytic form. This test involves another of the basic concepts of vector calculus, this time the so-called curl of a vector.
Section 4.4 The Second Condition that F be Conservative 119
It can be shown (though I shall not do so here 8) that a force F has the desired property, that the work it does is independent of path, if and only if
V xF= 0 (4.36)
everywhere. The quantity V x F is called the curl of F, or just “curl F,” or “del cross F.” It is defined by taking the cross product of V and F just as if the components of V, namely (a/ax, a/ay, a/az), were ordinary numbers. To see what this means, consider first the cross product of two ordinary vectors A and B. In the table below, I have listed the components of A, B, and A x B:
vector x component y component z component
A A x A Y A z B Bx B Y Bz
A x B A y Bz — A z By A z Bx — A x B, A x B y — A y Bx
(4.37)
The components of V x F are found in exactly the same way, except that the entries in the first row are differential operators. Thus,
vector x component y component z component V a/ax a/ay a/az F Fx Fy Fz
V X F — — — ay z y az x ax z ax Y a y x
(4.38)
No one would claim that (4.36) is obviously equivalent to the condition that fi2 F • dr is path-independent, but it is, and it provides an easily applied test for the path-independence property, as the following example shows.
EXAMPLE 4.5 Is the Coulomb Force Conservative?
Consider the force F on a charge q due to a fixed charge Q at the origin. Show that it is conservative and find the corresponding potential energy U. Check that — V U = F.
The force in question is the Coulomb force, as shown in Figure 4.7(a),
kq F = r =—r (4.39) r 2 r 3
where k denotes the Coulomb force constant, often written as 1/(470, and y is just an abbreviation for the constant kq Q. From the last expression we can read off the components of F, and using (4.38) we can calculate the components of V x F. For example, the x component is
a a _ a ( yz a ( yy) (V x F) x = Fz — — , (4.40) y —
ay az ay r 3 az r 3
8 The condition (4.36) follows from a result called Stokes’s theorem. If you would like to explore this a little, see Problem 4.25. For more details, see any text on vector calculus or mathematical methods. I particularly like Mathematical Methods in the Physical Sciences by Mary Boas (Wiley,
1983), p. 260.
120 Chapter 4 Energy
F
Q (a)
(b)
Figure 4.7 (a) The Coulomb force F = yi./r 2 of the fixed charge Q on the charge q. (b) The work done by F as q moves from ro to r can be evaluated following a path that
goes radially outward to P and then around a circle to r.
The two derivatives here are easily evaluated: First, since az/ay = ay/az = 0, we can rewrite (4.40) as
(V X F)„ = yz (1-r-3) — yy . (4.41) ay az
Next recall that
r (x 2 + y2 ± z2 )
1/2
so that, for example,
ar y — ay r
(4.42)
(Check this one using the chain rule.) We can now evaluate the two remaining derivatives in (4.41) to give (remember the chain rule again)
(V x F)x = yz (-4 . y —) — yy (-4 •
r = 0.
r r
The other two components work in exactly the same way (check it, if you don’t believe me), and we conclude that V x F = 0. According to the result (4.36), this guarantees that F satisfies the second condition to be conservative. Since it certainly satisfies the first condition (it depends only on the variable r), we have proved that F is conservative. (The proof that V x F = 0 is considerably quicker in spherical polar coordinates. See Problem 4.22.)
The potential energy is defined by the work integral (4.13),
U(r) = — f F(e) • dr’ (4.43) ro
where ro is the (as yet unspecified) reference point where U (ro) = 0. Fortu- nately, we know that this integral is independent of path, so we can choose whatever path is most convenient. One possibility is shown in Figure 4.7(b), where I have chosen a path that goes radially outward to the point labeled P and then around a circle (centered on Q) to r. On the first segment, F(e) and
Section 4.5 Time-Dependent Potential Energy 121
dr’ are collinear, and F(r’) • dr’ = (y I r’2)dr’ . On the second, F(r’) and dr’ are perpendicular, so no work is done along this segment, and the total work is just that of the first segment,
r v
U(r) = — dr’ = 3± — Y. (4.44) f r 0 r’2 r ro
Finally, it is usual in this problem to choose the reference point r o at infinity, so that the second term here is zero. With this choice (and replacing y by kqQ) we arrive at the well-known formula for the potential energy of the charge q due to Q,
U(r) = U(r) = kqQ
(4.45) r
Notice that the answer depends only on the magnitude r of the position vector r and not on the direction.
To check V U let us evaluate the x component:
a ( kqQ) kqQ ar (VU), =
= (4.46)
ax r ) r2 ax where the last expression follows from the chain rule. The derivative ar/ax is x/r [compare Equation (4.42)], so
(VU)„ = —kqQ— x
= —Ft , r3
as given by (4.39). The other two components work in exactly the same way, and we have shown that
V U = —F
(4.47)
as required.
4.5 Time-Dependent Potential Energy
We sometimes have occasion to study a force F(r,t) that satisfies the second condition to be conservative (V x F = 0), but, because it is time-dependent, does not satisfy the first condition. In this case, we can still define a potential energy U (r, t) with the property that F = — V U, but it is no longer the case that total mechanical energy, E = T U, is conserved. Before I justify these claims, let me give an example of this situation. Figure 4.8 shows a small charge q in the vicinity of a charged conducting sphere (for example, a Van de Graaff generator) with a charge Q (t) that is slowly leaking away through the moist air to ground. Because Q(t) changes with time, the force that it exerts on the small charge q is explicitly time-dependent. Nevertheless, the spatial dependence of the force is the same as for the time-independent Coulomb force of Example 4.5 (page 119). Exactly the same analysis as in that example shows that V x F = 0.
Q (t) F =
kqQ(t) r2
q
122 Chapter 4 Energy
Figure 4.8 The charge Q (t) on the conducting sphere is slowly leaking away, so the force on the small charge q varies with time, even if its position r is constant.
Let me now justify the claims made above. First, since V x F(r, t) = 0, the same mathematical theorem quoted in connection with Equation (4.36) guarantees that the work integral fi2 F(r, t) • dr (evaluated at any one time t) is path independent. This means we can define a function U(r, t) by an integral exactly analogous to (4.13),
U(r, t) = — f F (r/ , t) • dr’, ro
(4.48)
and, for the same reasons as before, F(r,t) = —VU(r, t). (See Problem 4.27.) In this case, we can say the force F is derivable from the time-dependent potential energy U(r, t).
So far everything has gone through just as before, but now the story changes. We can define the mechanical energy as E = T + U, but it is no longer true that E is conserved. If you review carefully the argument leading to Equation (4.19), you may be able to see what goes wrong, but we can in any case show directly that E =T + U changes as the particle moves on its path. As before, consider any two neighboring points on the particle’s path at times t and t dt. Exactly as in (4.4), the change in kinetic energy is
dT = — dT
dt = (mir •v)dt = F • dr. dt
(4.49)
Meanwhile, U(r, t) = U (x , y, z, t) is a function of four variables (x, y, z, t) and
aU au au au dU = —dx + —dy + —dz + —dt.
ax ay az at (4.50)
You will recognize the first three terms on the right as VU • dr = —F • dr. Thus
au dU = —F • dr —dt.
at (4.51)
When we add this to Equation (4.49) the first two terms cancel, and we are left with
u d(T + U) = a —dt.
at (4.52)
Clearly it is only when U is independent of t (that is, au/at = 0) that the mechanical energy E = T + U is conserved.
Section 4.6 Energy for Linear One-Dimensional Systems 123
Returning to the example of Figure 4.8, we can understand this conclusion and see what has happened to conservation of energy. Imagine that I hold the charge q stationary at the position of Figure 4.8, while the charge on the sphere leaks away. Under these conditions, the KE of q doesn’t change, but the potential energy kq Q (t) / r slowly diminishes to zero. Clearly T U is not constant. However, while mechanical energy is not conserved, total energy is conserved: The loss of mechanical energy is exactly balanced by the gain of thermal energy as the discharge current heats up the surrounding air. This example suggests, what is true, that the potential energy depends explicitly on time in precisely those situations where mechanical energy gets transformed to some other form of energy or to mechanical energy of other bodies external to the system of interest.
4.6 Energy for Linear One-Dimensional Systems
So far we have discussed the energy of a particle that is free to move in all three dimensions. Many interesting problems involve an object that is constrained to move in just one dimension, and the analysis of such problems is remarkably simpler than the general case. Oddly enough, there is some ambiguity in what a physicist means by a “one-dimensional system.” Many introductory physics texts start out discussing the motion of a one-dimensional system, by which they mean an object (a railroad car, for instance) that is confined to move on a perfectly straight, or linear, track. In discussing such linear systems, we naturally take the x axis to coincide with the track, and the position of the object is then specified by the single coordinate x. In this section I shall focus on linear one-dimensional systems. However, there are much more complicated systems, such as a roller coaster on its curving track, that are also one-dimensional, inasmuch as their position can be specified by a single parameter (such as the distance of the roller coaster along its track). As I shall discuss in the next section, energy conservation for such curvilinear one-dimensional systems is just as straightforward as for a perfectly straight track.
To begin, let us consider an object constrained to move along a perfectly straight track, which we take to be the x axis. The only component of any force F that can do work is the x component, and we can simply ignore the other two components. Therefore the work done by F is the one-dimensional integral
x2 W (x i x 2) = f Fx (x) dx. (4.53)
If the force is to be conservative. Fx must satisfy the two usual conditions: (i) It must depend only on the position x [as I have already implied in writing the integral (4.53)]. (ii) The work (4.53) must be independent of path. The remarkable feature of one- dimensional systems is that the first condition already guarantees the second, so the latter is superfluous. To understand this property, you have only to recognize that in one dimension there is only a small choice of paths connecting any two points. Consider, for example, the two points A and B shown in Figure 4.9. The obvious path between points A and B is the path that goes from A directly to B (let’s call this path
124 Chapter 4 Energy
x A B<
Figure 4.9 The path called ABC B goes from A past B and on to C, then back to B.
“AB”). Another possibility, shown in the figure, is to go from A past B to C and then
back to B (let’s call this one “ABCB”). The work done along this path can be broken up as follows:
W(ABCB) = W(AB) W(BC) + W(CB).
Now, provided the force depends only on the position x [condition (i)] each
increment of work going from B to C is exactly equal (but of opposite sign) to the corresponding contribution going from C to B. That is, the last two terms on the right cancel, and we conclude that
W(ABC B) = W(AB),
as required. One can of course concoct a path from A to B that doubles back and forth many times, but a little thought should convince you that any such path can be broken into a number of segments some of which together traverse the direct path AB exactly once, and all the rest of which cancel in pairs. Thus the work done on any
path between A and B is the same as that on the direct path AB, and we have proved that in one dimension the first condition for a force to be conservative guarantees the second.
Graphs of the Potential Energy
A second useful feature of one-dimensional systems is that with only one independent variable (x) we can plot the potential energy U(x), and, as we shall see, this makes it easy to visualize the behavior of the system. Assuming all forces on the object are conservative, we define the potential energy as
U (x) = — f F x (x) dx’ xo
(4.54)
where Fx is the x component of the net force on the particle. For example, for a mass on the end of a spring obeying Hooke’s law, the force is Fx = —kx , and, if we choose
the reference point x. = 0, Equation (4.54) gives the celebrated result
U = kx 2 2
for any spring obeying Hooke’s law. Corresponding to the three-dimensional result F = — V U, we have the simpler
result in one dimension
dU F
dx • (4.55)
Section 4.6 Energy for Linear One-Dimensional Systems 125
Figure 4.10 The graph of potential energy U(x) against x for any one-dimensional system can be thought of as a picture of a roller coaster track. The force Fx = — dU ldx tends to push the object “downhill” as at x 1 and x2 . At the points x3 and x4, where U(x) is minimum or maximum, dUldx= 0 and the force is zero; such points are therefore points of equilibrium.
If we plot the potential energy against x as in Figure 4.10, we can easily see qualita- tively how the object has to behave. The direction of the net force is given by (4.55) as “downhill” on the graph of U(x) — to the left at x i and to the right at x 2 . It follows that the object always accelerates in the “downhill” direction — a property that reminds one of the motion of a roller coaster, which also always accelerates downhill. This analogy is not an accident: For a roller coaster, U (x) is mgh (where h is the height
above ground) and the graph of U(x) against x has the same shape as a graph of h against x, which is just a picture of the track. For any one -dimensional system, we can always think about the graph of U(x) as a picture of a roller coaster, and common sense will generally tell us the kind of motion that is possible at different places, as I now describe.
At points, such as x 3 and x4 , where dU I dx = 0 and U (x) is minimum or maximum, the net force is zero, and the object can remain in equilibrium. That is, the condition dU ldx = 0 characterizes points of equilibrium. At x 3 , where d2 Uldx 2 > 0 and U(x) is minimum, a small displacement from equilibrium causes a force which pushes the object back to equilibrium (back to the left on the right of x 3 , back to the right on the left of x3). In other words, equilibrium points where d2 U/dx 2 > 0 and U(x) is minimum are points of stable equilibrium. At equilibrium points like x4 where d2 U/dx 2 < 0 and U(x) is maximum, a small displacement leads to a force away from equilibrium, and the equilibrium is unstable.
If the object is moving then its kinetic energy is positive and its total energy is necessarily greater than U(x). For example, suppose the object is moving some- where near the equilibrium point x = b in Figure 4.11. Its total energy has to be greater than U (b) and could, for example, equal the value shown as E in that fig- ure. If the object happens to be on the right of b and moving toward the right, its PE will increase and its KE must therefore decrease until the object reaches the turning point labeled c, where U (c) = E and the KE is zero. At x = c the object stops and, with the force back to the left, it accelerates back toward x = b. It can- not now stop until once again the KE is zero, and this occurs at the turning point a, where U (a) = E and the object accelerates back to the right. Since the whole
126 Chapter 4 Energy
E
Figure 4.11 If an object starts out near x = b with the
energy E shown, it is trapped in the valley or “well”
between the two hills and oscillates between the turning
points at x = a and c where U(x) = E and the kinetic
energy is zero.
cycle now repeats itself, we see that if the object starts out between two hills and
its energy is lower than the crest of both hills, then the object is trapped in the
valley or “well” and oscillates indefinitely between the two turning points where
U(x) = E.
Suppose the cart again starts out between the two hills but with energy higher than
the crest of the right hill though still lower than the left. In this case, it will escape
to the right since E > U(x) everywhere on the right, and it can never stop once it is
moving in that direction. Finally, if the energy is higher than both hills, the cart can
escape in either direction.
These considerations play an important role in many fields. An example from
molecular physics is illustrated in Figure 4.12, which shows the potential energy of a
typical diatomic molecule, such as HC1, as a function of the distance between the two
atoms. This potential energy function governs the radial motion of the hydrogen atom
(in the case of HC1) as it vibrates in and out from the much heavier chlorine atom.
The zero of energy has been chosen where the two atoms are far apart (at infinity)
and at rest. Notice that the independent variable is the interatomic distance r which,
by its definition, is always positive, 0 < r < oo. As r -f 0, the potential energy gets
very large, indicating that the two atoms repel one another when very close together
(because of the Coulomb repulsion of the nuclei). If the energy is positive (E > 0)
the H atom can escape to infinity, since there is no “hill” to trap it; the H atom can
come in from infinity, but it will stop at the turning point r = a and (in the absence of
any mechanism to take up some of its energy) it will move away to infinity again. On
the other hand, if E < 0, the H atom is trapped and will oscillate in and out between
the two turning points shown at r = b and r = d. The equilibrium separation of the
molecule is at the point shown as r = c. It is the states with E < 0 that correspond to what we normally regard as the HC1 molecule. To form such a molecule, two separate
atoms (with E > 0) must come together to a separation somewhere near r = c, and
some process, such as emission of light, must remove enough energy to leave the two
atoms trapped with E < 0.
Section 4.6 Energy for Linear One-Dimensional Systems 127
Figure 4.12 The potential energy for a typical diatomic molecule such as HC1, plotted as a function of the distance r between the two atoms. If E > 0, the two atoms cannot approach closer than the turning point r = a, but they can move apart to infinity. If E < 0, they are trapped between the turning points at b and d and form a bound molecule. The equilibrium separation is r = c.
Complete Solution of the Motion
A third remarkable feature of one-dimensional conservative systems is that we can — at least in principle — use the conservation of energy to obtain a complete solution of the motion, that is, to find the position x as a function of time t. Since E = T U(x) is conserved, with U(x) a known function (in the context of a given problem) and E
determined by the initial conditions, we can solve for T = 2ini 2 = E — U(x) and hence for the velocity x as a function of x:
= — U(x). (4.56)
(Notice that there is an ambiguity in the sign since energy considerations cannot determine the direction of the velocity. For this reason, the method described here usually does not work in a truly three-dimensional problem. In one dimension, you can almost always decide the sign of •i by inspection, though you must remember to do so.)
Knowing the velocity as a function of x, we can now find x as a function of t,
using separation of variables, as follows: We first rewrite the definition •i = dx/dt as
dt = dx x
[Since z = .i(x), this separates the variables t and x.] Next, we can integrate between any initial and final points to give
tf = xf dx
(4.57) x 1 1 X
128 Chapter 4 Energy
This gives the time for travel between any initial and final positions of interest. If we substitute for •i from (4.56) (and assume, to be definite, that z is positive) then the time to go from the initial x o at time 0 to an arbitrary x at time t is
t = r dx’ Im r dx’ Jxo •i(x’) V 2 ix. E — U(x’)
(4.58)
(As usual, I’ve renamed the variable of integration as x’ to avoid confusion with the upper limit x.) The integral (4.58) depends on the particular form of U(x) in the problem at hand. Assuming we can do the integral [and we can at least do it numerically for any given U(x)], it gives us t as a function of x. Finally we can solve to give x as a function of t, and our solution is complete, as the following simple example illustrates.
EXAMPLE 4.6 Free Fall
I drop a stone from the top of a tower at time t = 0. Use conservation of energy to find the stone’s position x (measured down from the top of the tower, where x = 0) as a function of t. Neglect air resistance.
The only force on the stone is gravity, which is, of course, conservative. The corresponding potential energy is
U(x) = —mgx.
(Remember x is measured downward.) Since the stone is at rest when x = 0, the total energy is E = 0, and according to (4.56) the velocity is
— 2
N/E — U(x) = V2gx
(a result that is well known from elementary kinematics). Thus
t = f x dx’ f x dx’ Jo 1(x’) = Jo V2gx’
As anticipated, this gives t as a function of x, and we can solve to give the familiar result
x = gt2 . This simple example, involving the gravitational potential energy U(x) = —mgx, can be solved many different (and some simpler) ways, but the energy method used here can be used for any potential energy function U(x). In some cases, the integral (4.58) can be evaluated in terms of elementary functions, and we obtain an analytic solution of the problem; for example, if U(x) = 4kx 2 (as for a mass on the end of a spring), the integral turns out to be an inverse sine function, which implies that x oscillates sinusoidally with time, as we should expect (see Problem 4.28). For some potential energies, the integral cannot be
2x
g
Section 4.7 Curvilinear One-Dimensional Systems 129
done in terms of elementary functions, but can nonetheless be related to func- tions that are tabulated (see Problem 4.38). For some problems, the only way to do the integral (4.58) is to do it numerically.
4.7 Curvilinear One-Dimensional Systems
So far the only one-dimensional system I have discussed is an object constrained to move along a linear path, with position specified by the coordinate x. There are other, more general, systems that can equally be said to be one-dimensional, inasmuch as their position is specified by a single number. An example of such a one-dimensional system is a bead threaded on a curved rigid wire as illustrated in Figure 4.13. (Another is a roller coaster confined to a curved track.) The position of the bead can be specified by a single parameter, which we can choose as the distance s, measured along the wire, from a chosen origin 0. With this choice of coordinate, the discussion of the curved one-dimensional track parallels closely that of the straight track, as I now show.
The coordinate s of our bead corresponds, of course, to x for a cart on a straight track. The speed of the bead is easily seen to be s, and the kinetic energy is therefore just
T = 2
as compared to the familiar 4mi 2 for the straight track. The force is a little more complicated. As our bead moves on the curved wire the net normal force is not zero; on the contrary, the normal force is what constrains the bead to follow its assigned curving path. (For this reason, the normal force is called the force of constraint.) On the other hand, the normal force does no work, and it is the tangential component Ftang of the net force that is our chief concern. In particular, it is fairly easy to show (Problem 4.32) that
Ftang = ms
Figure 4.13 An object constrained to move on a curved
track can be considered to be a one-dimensional system, with the position specified by the distance s (measured
along the track) of the object from an origin 0. The system
shown is a bead threaded on a stiff wire, bent into a double loop-the-loop.
130 Chapter 4 Energy
(just as Fx = na on a straight track). Further, if all the forces on the bead that have a tangential component are conservative, we can define a corresponding potential energy U(s) such that Ftang = -LclU Ids, and the total mechanical energy E = T U(s) is constant. The whole discussion of Section 4.6 can now be applied to the bead on a curved wire (or any other object constrained to move on a one-dimensional path). In particular, those points where U(s) is a minimum are points of stable equilibrium, and those where U(s) is maximum are points of unstable equilibrium.
There are many systems that appear to be much more complicated than the bead on a wire, but are nonetheless one-dimensional and can be treated in much the same way. Here is an example.
EXAMPLE 4.7 Stability of a Cube Balanced on a Cylinder
A hard rubber cylinder of radius r is held fixed with its axis horizontal, and a wooden cube of side 2b is balanced on top of the cylinder, with its center vertically above the cylinder’s axis and four of its sides parallel to the axis. The cube cannot slip on the rubber of the cylinder, but it can of course rock from side to side, as shown in Figure 4.14. By examining the cube’s potential energy, find out if the equilibrium with the cube centered above the cylinder is stable or unstable.
Let us first note that the system is one-dimensional, since its position as it rocks from side to side can be specified by a single coordinate, for instance the angle 9 through which it has turned. (We could also specify it by the distance s of the cube’s center from equilibrium, but the angle is a little more convenient. Either way the system’s position is specified by a single coordinate, and our problem is definitely one-dimensional.) The constraining forces are the normal
Figure 4.14 A cube, of side 2b and center C, is placed on a fixed horizontal cylinder of radius r and center 0. It is originally put so that C is centered above 0, but it can roll from side to side without slipping.
Section 4.7 Curvilinear One-Dimensional Systems 131
and frictional forces of the cylinder on the cube; that is, these two forces constrain the cube to move only as shown in Figure 4.14. Since neither of these does any work we need not consider them explicitly. The only other force on the cube is gravity, and we know from elementary physics that this is conservative and that the gravitational potential energy is the same as for a point mass at the center of the cube; that is, U = mgh, where h is the height of C above the origin, as shown in Figure 4.14. (See Problem 4.6.) The length of the line shown as OB
is just r b, while the length BC is the distance the cube has rolled around the cylinder, namely re. Therefore h = (r b) cos 0 + r0 sin 0 and the potential energy is
U(0) = mgh = mg[(r b) cos 0 + r0 sin 0]. (4.59)
To find the equilibrium position (or positions) we must find the points where
dU I d0 vanishes. (Strictly speaking I haven’t proved this very plausible claim yet for this kind of constrained system; I’ll discuss it shortly.) The derivative is easily seen to be (check this for yourself )
dU = mg [re cos 0 — b sin 0].
de
This vanishes at 0 = 0, confirming the obvious — that 0 = 0 is a point of equilibrium. To decide whether this equilibrium is stable, we have only to differentiate again and find the value of d2 U/d0 2 at the equilibrium position. This gives (as you should check)
d02
(at 0 = 0). If the cube is smaller than the cylinder (that is, b < r), this second derivative is positive, which means that U(0) has a minimum at 0 = 0 and the equilibrium is stable; if the cube is balanced on the cylinder, it will remain there indefinitely. On the other hand, if the cube is larger than the cylinder (b > r), the second derivative (4.60) is negative, the equilibrium is unstable, and the smallest disturbance will cause the cube to roll and fall off the cylinder.
Further Generalizations
There are many other, more complicated systems that are still legitimately described as one dimensional. Such systems may comprise several bodies, but the bodies are joined by struts or strings in such a way that just one parameter is needed to describe the system’s position. An example of such a system is the Atwood machine shown in Figure 4.15, which consists of two masses, m 1 and m 2, suspended from opposite ends of a massless, inextensible string that passes over a frictionless pulley. (To simplify the discussion, I shall assume the pulley is massless, although it is easy to allow for a mass of the pulley.) The two masses can move up and down, but the forces of the pulley on the string and the string on the masses constrain matters so that the mass m2 can move up only to the extent that m 1 moves down by exactly the same distance.
d2U = mg (r — b) (4.60)
132 Chapter 4 Energy
x
m2g
Figure 4.15 An Atwood machine consisting of two masses,
m 1 and m 2 , suspended by a massless inextensible string
that passes over a massless, frictionless pulley. Because the
string’s length is fixed, the position of the whole system is
specified by the distance x of m i below any convenient fixed level. The forces on the two masses are their weights m 1 g
and m 2 g, and the tension forces FT (which are equal since the pulley and string are massless).
Thus the position of the whole system can be specified by a single parameter, for example the height x of m 1 below the pulley’s center as shown, and the system is again one-dimensional.9
Let us consider the energies of the masses m 1 and m 2 . The forces acting on them are gravity and the tension in the string. Since gravity is conservative, we can introduce potential energies U 1 and U2 for the gravitational forces, and our previous considerations imply that in any displacement of the system,
A Ti + A Ul = witen (4.61) and
A 72 + A u2 Wien (4.62)
where the terms W’ denote the work done by the tension on m 1 and m2 . Now, in the absence of friction, the tension is the same all along the string. Thus, although the tension certainly does work on the two individual masses, the work done on m 1 is equal and opposite to that done on m 2, when m 1 moves down and m 2 moves an equal distance up (or vice versa). That is,
wr w2ten . (4.63)
9 You may object, correctly, that the masses can also move sideways. If this worries you, we can thread each mass over a vertical frictionless rod, but these rods are actually unnecessary: As long as we refrain from pushing the masses sideways, each will remain in a vertical line of its own accord.
Section 4.8 Central Forces 133
Thus, if we add the two energy equations (4.61) and (4.62), the terms involving the tension in the string cancel and we are left with
A (Ti Ui T2 + U2) = O.
That is, the total mechanical energy
E = + Ui + T2 + U2 (4.64)
is conserved. The beauty of this result is that all reference to the constraining forces of the string and pulley has disappeared.
It turns out that many systems which contain several particles that are constrained in some way (by strings, struts, or a track on which they must move, etc.) can be treated in this same way: The constraining forces are crucially important in determining how the system moves, but they do no work on the system as a whole. Thus in considering the total energy of the system, we can simply ignore the constraining forces. In particular, if all other forces are conservative (as with our example of the Atwood machine), we can define a potential energy U oi for each particle a, and the total energy
N
E_ (Ta + U„) 01=1
is constant. If the system is also one-dimensional (position specified by just one parameter, as with the Atwood machine), then all of the considerations of Section 4.6 apply.
A careful discussion of constrained systems is far easier in the Lagrangian formula- tion of mechanics than in the Newtonian. Thus I shall postpone any further discussion to Chapter 7. In particular, the proof that a stable equilibrium normally corresponds to a minimum of the potential energy (for a large class of constrained systems) is sketched in Problem 7.47.
4.8 Central Forces
A three-dimensional situation that has some of the simplicity of one-dimensional problems is a particle that is subject to a central force, that is, a force that is everywhere directed toward or away from a fixed “force center.” If we take the force center to be the origin, a central force has the form
F(r) = (4.65)
where the function f(r) gives the magnitude of the force (and is positive if the force is outward and negative if it is inward). An example of a central force is the Coulomb force on a charge q due to a second charge Q at the origin; this has the familiar form
F(r) = kqQ r, r2 (4.66)
134 Chapter 4 Energy
which is obviously an example of (4.65), with the magnitude function given by f(r) = kqQ1r 2 . The Coulomb force has two additional properties not shared by all central forces: First, as we have proved, it is conservative. Second, it is spherically symmetric or rotationally invariant; that is, the magnitude function f(r) in (4.65) is independent of the direction of r and, hence, has the same value at all points at the same distance from the origin. A compact way to express this second property of spherical symmetry is to observe that the magnitude function f(r) depends only on the magnitude of the vector r and not its direCtion, so can be written as
f(r) = f(r)• (4.67)
A remarkable feature of central forces is that the two properties just mentioned always go together: A central force that is conservative is automatically spherically symmetric, and, conversely, a central force that is spherically symmetric is automati- cally conservative. These two results can be proved in several ways, but the most direct proofs involve the use of spherical polar coordinates. Therefore, before offering any proofs, I shall briefly review the definition of these coordinates.
Spherical Polar Coordinates
The position of any point P is, of course, identified by the vector r pointing from the origin 0 to P. The vector r can be specified by its Cartesian coordinates (x, y, z), but in problems involving spherical symmetry it is almost always more convenient to specify r by its spherical polar coordinates (r, 9, 0), as defined in Figure 4.16. The first coordinate r is just the distance of P from the origin; that is, r = jr I, as usual. The angle 9 is the angle between r and the z axis. The angle 0, often called the azimuth, is the angle from the x axis to the projection of r on the xy plane, as shown. 10 It is a simple exercise (Problem 4.40) to relate the Cartesian coordinates (x, y, z) to the polar coordinates (r, 9, q5) and vice versa. For example, by inspecting Figure 4.16 you should be able to convince yourself that
x = r sin 0 cos 0, y = r sin 9 sin 0, and z = r cos 9. (4.68)
A beautiful use of spherical coordinates, which may help you to visualize them, is to specify positions on the surface of the earth. If we choose the origin at the center of the earth, then all points on the surface have the same value of r, namely the radius of the earth.” Thus positions on the surface can be specified by giving just the two angles (9, 0). If we choose our z axis to coincide with the north polar axis, then it is easy to see from Figure 4.16 that 9 gives the latitude of the point P, measured down from the north pole. (Since latitude is traditionally measured up from the equator, our angle 0 is often called the colatitude.) Similarly, 0 is the longitude measured east from the meridian of the x axis.
10 You should be aware that, while the definitions given here are those always used by physicists, most mathematics texts reverse the roles of 0 and 0.
” Actually the earth isn’t perfectly spherical, so r isn’t quite constant, but this doesn’t change the conclusion that any position on the surface can be specified by giving 0 and 0.
Section 4.8 Central Forces 135
Figure 4.16 The spherical polar coordinates (r, 0, 0) of a
point P are defined so that r is the distance of P from the
origin, 0 is the angle between the line OP and the z axis, and 0 is the angle of the line OQ from the x axis, where
Q is the projection of P onto the xy plane.
The statement that a function f (r) is spherically symmetric is simply the statement that, with r expressed in spherical polars, f is independent of 0 and 0. This is what we mean when we write f (r) = f (r), and the test for spherical symmetry is simply that the two partial derivatives of /a9 and of/a¢ are both zero everywhere.
The unit vectors r, 9, and 0 are defined in the usual way: First, r is the unit vector pointing in the direction of movement if r increases with 0 and 0 fixed. Thus, as shown in Figure 4.17, the vector r points radially outward, and is just the unit vector in the direction of r as usual. (On the surface of the earth, r points upward, in the direction of the local vertical.) Similarly, 9 points in the direction of increasing 0 with r and 0 fixed, that is, southward along a line of longitude. Finally, 0 points in the direction of increasing 0 with r and 0 fixed, that is, eastward along a circle of latitude.
Since the three unit vectors 1′, 9, and 4; are mutually perpendicular, we can evaluate dot products in spherical polars in just the same way as in Cartesians. Thus, if
a = cio e + (10 and
b bo b + bo ;/)
then (make sure you see this)
a • b = ark, ao be aobo . (4.69)
Like the unit vectors of two-dimensional polar coordinates, the unit vectors and 0 vary with position, and, as was the case in two dimensions, this variability complicates many calculations involving differentiation, as we shall now see.
136 Chapter 4 Energy
The Gradient in Spherical Polar Coordinates
In Cartesian coordinates, we have seen that the components of V f are precisely the partial derivatives of f with respect to x, y, and z,
,af ,af ,af Vf=x—+y—+z—. (4.70)
ax ay az
The corresponding expression for V f in polar coordinates is not so straightforward. To find it, recall from (4.35) that, in a small displacement dr, the change in any function f (r) is
df =V f • dr. (4.71)
To evaluate the small vector dr in polar coordinates, we must examine carefully what happens to the point r when we change r, 0, and 0: A small change dr in r moves the point a distance dr radially out, in the direction of I. As you can see from Figure 4.17, a small change dO in 6 moves the point around a circle of longitude (radius r) through a distance r dO in the direction of 0. (Note well the factor of r — the distance is not just d0.) Similarly, a small change d0 in 0 moves the point around a circle of latitude (radius r sin 6) through a distance r sin 0 d4. Putting all this together, we see that
dr = drid-rd00±rsin0d¢4.
Knowing the components of dr, we can now evaluate the dot product in (4.71) in terms of the unknown components of Vf ,
df = (V f), dr + (V n o r dO ± (V Do r sin 0 dO. (4.72)
Figure 4.17 The three unit vectors of spherical polar co- ordinates at the point P. The vector r points radially out, 6 points “south” along a line of longitude, and 0 points “east” around a circle of latitude.
(Vf), = a f
(VA =
r ae’
or, a little more compactly,
„af ,slaf of ar
V f = r 0— ± r ae r sin e ao
(4.75)
and (Vf) 0 = 1 of
(4.74) r sin e ao
Section 4.8 Central Forces 137
Meanwhile, since f is a function of the three variables r, 6, 4), the change in f is, of course,
df = — af
dr + — af a
dO + — f
dO. ar ae ao
(4.73)
Comparing (4.72) and (4.73), we conclude that the components of Vf in spherical
polars are
Similar considerations apply to the curl and other operators of vector calculus, all of which are markedly more complicated in spherical polar coordinates (and all other non-Cartesian coordinates) than in Cartesian coordinates. Since the formulas for these operators are very hard to remember, I have listed the more important ones inside the back cover. Proofs can be found in any textbook of vector calculus.’ 2 Armed with these ideas, let us return to central forces.
Conservative and Spherically Symmetric, Central Forces
I claimed earlier that a central force is conservative if and only if it is spherically symmetric. This claim can be proved several different ways. The quickest proofs (though not necessarily the most insightful) use spherical polar coordinates. Let us assume first that the central force F(r) is conservative and try to prove that it must be spherically symmetric. Since it is conservative, it can be expressed in the form —VU , which according to (4.75), has the form
,au -lay 1 au F(r) = —VU = —r— — 0— (4.76)
ar r ae r sin0 Bo
Since F(r) is central, only its radial component can be nonzero, and the last two terms in (4.76) must be zero. This requires that au/a6 = au lao = 0; that is, U(r) is spherically symmetric, and (4.76) reduces to
au F(r) = —r .
Br
Since U is spherically symmetric (depends only on r), the same is true of a( for, and we see that the central force F(r) is indeed spherically symmetric. I shall leave the proof of the converse result, that a central force which is spherically symmetric is necessarily conservative, to the problems at the end of this chapter. (See Problems 4.43
12 See, for example, Mary L. Boas, Mathematical Methods in the Physical Sciences, John Wiley, 1983, p. 431.
138 Chapter 4 Energy
and 4.44, but the simplest proof mimics almost exactly the analysis of the Coulomb force in Example 4.5.)
The importance of these results is this: First, because a force F(r) that is central and spherically symmetric has a magnitude that depends only on r, it is nearly as simple as a one-dimensional force. Second, although F(r) is certainly not actually a one- dimensional force (its direction still depends on 0 and 0), we shall see in Chapter 8 that any problem involving this kind of force is mathematically equivalent to a certain related one-dimensional problem.
4.9 Energy of Interaction of Two Particles
Almost all of our discussion of energy has focused on the energy of a single particle (or any larger object that can be approximated as a particle). It is now time to extend the discussion to systems of several particles, and I shall naturally start with just two particles. In this section, I shall suppose that the two particles interact via forces F12 (on particle 1 by particle 2) and F21 (on particle 2 by particle 1), but that there are no other, external, forces. In general, the force F12 could depend on the positions of both particles, so can be written as
F12 = F12(r11 r2) ,
and by Newton’s third law
F12 = — F21.
As an example of such a two-particle system we could consider an isolated binary star, in which case the only two forces are the gravitational attraction of each star for the other. If we denote the vector pointing to star 1 from star 2 by r, as in Figure 4.18, the force F12 is just the familiar
Grnim 2 , F12 = r =
Gin im2 r.
r2 r3
r r1 — r2
Figure 4.18 The vector r pointing to particle 1 from particle 2 is just r = (r 1 — r2).
Section 4.9 Energy of Interaction of Two Particles 139
The vector r can be written in terms of the two positions r 1 and r2 . In fact, as can be seen in Figure 4.18,
r = r 1 — r2 .
Thus the force F12, expressed as a function of r 1 and r2 , is
G m 1 111 2
3 F12 =
1ri (r 1 — r2). (4.77)
— r2 1
A striking property of the force (4.77) is that it depends on the two positions r 1 and r2 only through the particular combination r 1 — r2 . This property is not an accident, and is in fact true of any isolated two-particle system. The reason is that any isolated system must be translationally invariant: If we bodily translate the system to a new position, without changing the relative positions of the particles, the interparticle forces should remain the same. This is illustrated in Figure 4.19, which shows a pair of points r 1 and r2 and a second pair of points s 1 and s2 , with s 1 — s2 = r 1 — r2 . Since the two points r 1 and r2 could be simultaneously translated to s 1 and s2 , the force
F12(r1, r2) must be the same as F 12 (s 1 , s2) for any points satisfying r 1 — r2 = s 1 — s2 . In other words, F 12 (r 1 , r2) depends only on r 1 — r2 , as claimed, and we can write
F12 = F12(rl r2)• (4.78)
The result (4.78) greatly simplifies our discussion. We can learn almost everything about the force F 12 by fixing r2 at any convenient point. In particular, let us temporarily fix r2 at the origin, in which case (4.78) reduces to just F 12 (r 1 ). (This maneuver amounts to translating both particles until particle 2 is at the origin, and we know that the force is unaffected by any such translation.) With r 2 fixed, our discussion of the force on a single particle now applies. For example, if the force F12 on particle 1 is to be conservative, then it must satisfy
V1 x F12 = 0 (4.79)
r 1
1 2
r2 S1
s2
Figure 4.19 If r 1 — r2 = s i — s2 , then two particles at r 1 and r2 could be bodily translated to s 1 and s2 without affecting their relative positions. This means that the force between the particles at r 1 and r2 must be the same as that at s 1 and s2 .
140 Chapter 4 Energy
where V1 is the differential operator
„ a a ayi
a + z az1
with respect to the coordinates (x 1 , y i , z 1) of particle 1. If (4.79) is satisfied, we can define a potential energy U(r 1) such that the force on particle 1 is
F12 =
This gives the force F 12 for the case that particle 2 is at the origin. To find it for particle 2 anywhere else we have only to translate back to an arbitrary position by replacing r 1 with r 1 — r2 to give
F12 = -V1U(r1 (4.80)
Notice that I don’t have to change the operator V 1 , since an operator like a/axi is unchanged by addition of a constant to x 1 .
To find the reaction force F21 on particle 2, we have only to invoke Newton’s third law, which says that F21 = -F12. That is, we have only to change the sign of (4.80). We can re-express this by noticing that
V1 U(r1 — r2) = —V2 U(r 1 — r2 ), (4.81)
where V2 denotes the gradient with respect to the coordinates of particle 2. (To prove this, invoke the chain rule. See Problem 4.50.) So, instead of changing the sign of (4.80) to find F21, we can simply replace V 1 by V2 to give
F21 = -V2U(T1 – r2). (4.82)
Equations (4.80) and (4.82) are a beautiful result that generalizes to multiparticle systems. To emphasize what they say, let me rewrite them as
(Force on particle 1) = (Force on particle 2) = — V 2 U.
There is a single potential energy function U, from which we can derive both forces. To find the force on particle 1, we just take the gradient of U with respect to the coordinates of particle 1; to find the force on particle 2, we take the gradient with respect to the coordinates of particle 2.
Before generalizing this result to multiparticle systems, let us consider the conser- vation of energy for our two-particle system. Figure 4.20 shows the orbits of the two particles. During a short time interval dt , particle 1 moves through dr 1 and particle 2 through dr2 , and work is done on both particles by the corresponding forces. By the work—KE theorem
dT1 = (work on 1) = dr ‘ • F12
and similarly
d T2 = (work on 2) = dr2 • F21.
(4.83)
Section 4.9 Energy of Interaction of Two Particles 141
2 r2
Figure 4.20 Motion of two interacting particles. During a short time interval dt, particle 1 moves from r 1 to r 1 + dr 1
and particle 2 from r2 to r2 + dr2 .
Adding these, we find for the change in the total kinetic energy T = T1 + T2,
dT = dT1 + dT2 = (work on 1) + (work on 2)
= Wtot
where
Wot = dr i • F12 dr2 • F21
(4.84)
denotes the total work done on both particles. Replacing F21 by —F 12 and then replacing F12 with (4.80), we can rewrite W tot as
Wtot = (dr1 — dr2) • F12 = d(r 1 r2) • [—V1 U(r1 — r2)]. (4.85)
If we rename (r 1 — r2) as r, then the right side of this equation can be seen to be just (minus) the change in the potential energy, and we find that 13
Wtot = dr • VU(r) = —dU
(4.86)
where the last step follows from the property (4.35) of the gradient operator. It is worth pausing to appreciate this important result. The total work W tot is the sum of two terms, the work done by F12 as particle 1 moves through dr 1 plus the work done by F21 as particle 2 moves through dr 2 . According to (4.86), the potential energy U takes both of these terms into account and W tot is simply —dU.
Returning to the total kinetic energy, we now see that according to (4.84) the change dT is just —dU. Moving the term dU to the other side, we conclude that
d(T + U) = O.
That is, the total energy,
E = T + U = + T2 U, (4.87)
” If you invoke the chain rule for differentiation, you can see that it makes no difference whether we write V1 U(r) or V U (r).
142 Chapter 4 Energy
of our two-particle system is conserved. Note well that the total energy of our two particles contains two kinetic terms (of course), but only one potential term, since U accounts for the work done by both of the forces F12 and F21.
Elastic Collisions
Elastic collisions give a simple application of these ideas. An elastic collision is a collision between two particles (or bodies that can be treated as particles) that interact via a conservative force that goes to zero as their separation r 1 — r2 increases. Since the force goes to zero as I r 1 — r2 I oo, the potential energy U (r — r2) approaches a constant, which we may as well take to be zero. For example, the two particles could be an electron and a proton, or they could be two billiard balls. That the force between two billiard balls is conservative is not obvious, but it is a fact that billiard balls are manufactured so that they behave like almost perfect (that is, conservative) springs when they are forced together. It is certainly easy to think of other objects (such as lumps of putty), for which the interobject force is nonconservative, and the collisions of such objects are not elastic.
In a collision, the two particles start out far apart, approach one another, and then move apart again. Because the forces are conservative, the total energy is conserved; that is, T U = constant (where, of course, T = T1 + T2). But when the particles are far apart, U is zero. Thus if we use the subscripts “in” and “fin” to label the situations well before and well after the particles come together, then conservation of energy implies that
Tin = Tfin (4.88)
In other words, an elastic collision can be characterized as a collision in which two particles come together and re-emerge with their total kinetic energy unchanged. However, it is important to remember that there is no principle of conservation of kinetic energy. On the contrary, while the particles are close together their PE is nonzero and their KE certainly is changing. It is only when they are well separated that the PE is negligible and conservation of energy leads to the result (4.88).
The foregoing discussion may suggest that elastic collisions should be a very com- mon occurence. All that is needed is two particles whose interaction is conservative. In practice, elastic collisions are not as widespread as this seems to imply. The trouble comes from the requirement that it be two particles that enter and leave the collision. For example, if we fire one billiard ball at a second with sufficient energy, the two balls may shatter. Similarly, if we fire an electron with sufficient energy at an atom, the atom may fall apart or, at least, change the internal motion of its constituents. Even in the collision of two genuine particles, such as an electron and a proton, relativity tells us that, with sufficient energy, new particles can be created. Clearly, at high enough energy, the assumption that the two objects entering a collision can be approximated as indivisible particles eventually breaks down, and we cannot assume that collisions will be elastic, even if all the underlying forces are conservative. Nevertheless, at rea- sonably low energies there are many situations where collisions are perfectly elastic:
1 2 •
v i
Section 4.9 Energy of Interaction of Two Particles 143
At sufficiently low energy, collisions of an electron with an atom always are, and to a good approximation, the same is true of billiard balls.
Elastic collisionS- provide several simple illustrations of the uses of conservation of energy and momentum, of which the following is one.
EXAMPLE 4.8 An Equal-Mass, Elastic Collision
Consider an elastic collision between two particles of equal mass, m 1 = m 2 = m (for example, two electrons, or two billiard balls), as shown in Figure 4.21. Prove that if particle 2 is initially at rest then the angle between the two outgoing velocities is 0 = 90°.
Conservation of momentum implies that mv i vI2 or
vi = v l + V2 .
That the collision is elastic implies that zmvi = zmvi 2 + 4mAT22 or
2 r2 r 2 V 1 -= V 1 + V2
Squaring (4.89), we find that
2 /2 r 2 V 1 = V 1 + 2v’ • v’ v 2 2 ‘
and comparing the last two equations we see that
V I1 • V2i =- 0’ ‘
that is, vl and v12 are perpendicular (unless one of them is zero, in which case the angle between them is undefined). This result was useful in atomic and nuclear physics; when an unknown projectile hit a stationary target particle, the fact that the two emerged traveling at 90° was taken as evidence that the collision was elastic and the two particles had equal masses.
Figure 4.21 Elastic collision between two equal- mass particles. Particle 1 enters with velocity v 1 and collides with the stationary particle 2. The angle between the two final velocities vi and v 12 is 8.
144 Chapter 4 Energy
4.10 The Energy of a Multiparticle System
We can extend our discussion of two particles to N particles fairly easily. The main complication is notational: The large number of E signs can make it hard to see clearly what is going on. For this reason, I shall start by considering the case of four particles (N = 4) and write out all of the various sums explicitly.
Four Particles
Let us consider, then, four particles, as shown in Figure 4.22. The particles can interact with each other (for example, they could be charged, so that each particle experiences the Coulomb force from the three others), and they may be subject to external forces, such as gravity or the Coulomb force of nearby charged bodies. In defining the energy of this system, the easy part is the kinetic energy T, which is, of course, the sum of four terms,
T = Ti + T2 + T3 + T4 , (4.90)
one term TCY 2 = m a 0/2 V for each particle. i To define the potential energy, we must examine the forces on the particles. First,
there are the internal forces of the four particles interacting with each other. For each pair of particles there is an action—reaction pair of forces; for example, particles 3 and 4 produce the forces F34 and F43 shown in Figure 4.22. I shall take for granted that each of these interparticle forces Fad is unaffected by the presence of the other particles and any external bodies. For example, F34 is just the same as if particles 1
4 F43
F341.16—–• 3
1 •
• 2
Figure 4.22 A system of four particles a = 1, 2, 3, 4. For each pair of particles, ap, there is an action—reaction pair of forces, Fap and Fi5a , such as the pair F34 and F43 shown. In addition, each particle a may be subject to an external net force Fec,xt . The four particles could be charged dust motes floating in the air, with the forces Fad being electrostatic and Feaxt being gravity plus buoyancy of the air.
Section 4.10 The Energy of a Multiparticle System 145
and 2 and all external bodies were removed. 14 Thus, we can treat the two forces F34
and F43 exactly as in Section 4.9. Provided the forces are conservative, we can define a potential energy
U34 = U34 (r3 — r4)
(4.91)
and the corresponding forces are the appropriate gradients as in (4.83)
F34 = — V3 U34 and
F43 = — V4 U34. (4.92)
There are in all six distinct pairs of particles, 12, 13, 14, 23, 24, 34, and for each pair we can define a corresponding potential energy U12, • • • , U34 from which the corresponding forces are obtained in the same way.
Each of the external forces Feuxt depends only on the corresponding position r a . (The force F1′, for instance, depends on the position r 1 , but not on r2 , r3 , r4 .) Therefore, we can handle F aext exactly as we did the force on a single particle. In particular, if Fat is conservative, we can introduce a potential energy Ur t (ra ) and the corresponding force is given by
Faext = Va Urt (ra )
(4.93)
where, of course, V a denotes differentiation with respect to the coordinates of parti- cle a.
We can now put all the potential energies together and define the total potential energy as the sum
= Uint uext irf T T I TT I TT I TT i u\
kt-)12, -r 13 m Li2,3 1-1 24 L-1 34/
(urct uxt uxt u4ext) . (4.94)
In this definition, U 1 nt is the sum over all six pairs of particles of the pairwise potential energies, U12, • • • , U34, and Uext is the sum of the four potential energies, ulext , • rirt• u arising from the external forces.
It is a fairly straightforward matter to show (see Problem 4.51 for more details) that the force on particle a is just (minus) the gradient of U with respect to the coordinates (x„, y,, z„). Consider, for instance, the gradient —V, U. When — Vl acts on the first line of (4.94), its action on the first three terms, U12 + U,3 + U14 gives precisely the three internal forces, F,2 + F,3 + F14. Acting on the last three terms,
U23 + U24 + U34, it produces zero, since none of these depend on r 1 . When —V1 acts on the second line of (4.94), its action on the first term, Ur, produces the external
14 This is quite a subtle point. I am, of course, not denying that the extra particles exert extra forces on particle 3. I claim only that the force of particle 4 on particle 3 is independent of the presence or absence of particles 1 and 2 and any external bodies. One could imagine a world where this claim was false (the presence of particle 1 could somehow change the force of 4 on 3), but experiment seems to confirm that in our world my claim is true.
146 Chapter 4 Energy
force Fr t . Acting on the last three terms it produces zero, since none of them depend on r 1 . Accordingly,
= F12 ± F13 + F14 + Fei xt
= (net force on particle 1). (4.95)
In exactly the same way, we can prove that in general
as expected. The second crucial property of our definition of potential energy U is that (provided
all the forces concerned are conservative, so we can define U), the total energy, defined as E =T + U, is conserved. We prove this in the now familiar way (for more details, see Problem 4.52): Apply the work—KE theorem to each of the four particles and add the results to show that, in any short time interval, dT = Wtot where Wtot denotes the total work done by all forces on all particles. Next show that W tot = —dU, and conclude that dT = —dU, and hence
dE = dT + dU =O.
That is, energy is conserved.
N Particles
The extension of these ideas to an arbitrary number of particles is now quite straight- forward, and I shall just write down the principal formulas. For N particles, labeled a = 1, • • , N, the total kinetic energy is just the sum of the N separate kinetic energies
T = E Tc, =E -2m. V:* a a
Assuming that all forces are conservative, for each pair of particles, a,8, we introduce the potential energy Uap that describes their interaction, and for each particle a we introduce the potential energy Ua”t that describes the net external force on that particle. The total potential energy is then
U = uint u ext uap uaext . (4.97) a /3>a a
(Here the condition 13 > a in the double sum makes sure we don’t double count the internal interactions Uo . For instance, we include U12 but not U21 .)
With the potential energy U defined in this way, the net force on any particle a is given by — Voi U, as in Equation (4.96), and total energy E = T + U is conserved. Finally, if any forces are nonconservative, we can define U as the potential energy pertaining to the conservative forces and then show that, in this case, dE = Wne where Wric is the work done by the nonconservative forces.
Section 4.10 The Energy of a Multiparticle System 147
Rigid Bodies
While the formalism of the last two sections is fairly general and complicated, you can perhaps take some comfort that most applications of the formalism are much simpler than the formalism itself. As one simple example, consider a rigid body, such as a golf ball or a meteorite, made up of N atoms. The number N is typically very large, but the energy formalism just developed usually turns out to be very simple. As you probably recall from elementary physics, the total kinetic energy of the N particles rigidly bound together is just the kinetic energy of the center-of-mass motion plus the kinetic energy of rotation. (I’ll be proving this in Chapter 10, but I hope you’ll accept it for now.) The potential energy of the internal, interatomic forces as given by
(4.97) is
u int = E E (las (ra — rs ). a p>a
(4.98)
If the interatomic forces are central (as is usually the case), then, as we saw in Section 4.8, the potential energy Uai3 actually depends on just the magnitude of ra — r,6 (not its direction). Thus we can rewrite (4.98) as
E E (las (Ira — rB I ). (4.99) a )3>a
Now, as a rigid body moves, the positions ra of its constituent atoms can, of course, move, but the distance Ira — ri3 I between any two atoms cannot change. (This is, in fact, the definition of a rigid body.) Therefore, if the body concerned is truly rigid, none of the terms in (4.99) can change. That is, the potential energy U int of the internal forces is a constant and can, therefore, be ignored. Thus, in applying energy considerations to a rigid body we can entirely ignore U int and have to worry only about the energy 1./’ corresponding to the external forces. Since this latter energy is often a very simple function (see the following example), energy considerations as applied to a rigid body are usually very straightforward.
EXAMPLE 4.9 A Cylinder Rolling down an Incline
A uniform rigid cylinder of radius R rolls without slipping down a sloping track as shown in Figure 4.23. Use energy conservation to find its speed v when it reaches a vertical height h below its point of release.
In accordance with the preceding discussion we can ignore the internal forces that hold the cylinder together. The external forces on the cylinder are the normal and frictional forces of the track and gravity. The first two do no work, and gravity is conservative. As you certainly recall from introductory physics, the gravitational potential energy of an extended body is the same as if all the mass were concentrated at the center of mass. (See Problem 4.6.) Therefore,
uext m gy
where Y is the height of the cylinder’s CM measured up from any convenient reference level. The kinetic energy of the cylinder is T = Z M v 2 + 11 cot , where
148 Chapter 4 Energy
Figure 4.23 A uniform cylinder starts from rest and rolls without slipping down a slope through a total vertical drop h= Yin — Yfin (with the CM coordinate Y measured vertically up).
I is its moment of inertia, I = 4mR2 , and w is its angular velocity of rolling, w = v/R. Thus the final kinetic energy is
T = -3-MV2 4
and the initial KE is zero. Therefore, conservation of energy in the form A T = —AU’ implies that
2m v2 = _mg (yfin — Y in) = Mgh 4
and hence that the final speed is
v =
Principal Definitions and Equations of Chapter 4
Work—KE Theorem
The change in KE of a particle as it moves from point 1 to point 2 is
2 AT -T2 —= f F • dr W(1 2) [Eq. (4.7)]
where T =lrn v2 and W(1 2) is the work which is done by the total force F on the particle and is defined by the preceding integral.
Principal Definitions and Equations of Chapter 4 149
Conservative Forces and Potential Energy
A force F on a particle is conservative if (i) it depends only on the particle’s position, F = F(r), and (ii) for any two points 1 and 2, the work W(1 —> 2) done by F is the same for all paths joining 1 and 2 (or equivalently, V x F = 0). [Sections 4.2 & 4.4]
If F is conservative, we can define a corresponding potential energy so that
and
U(r) — W (ro –> r) — f F(r’) • dr’ ro
[Eq. (4.13)]
F = —VU. [Eq. (4.33)]
If all the forces on a particle are conservative with corresponding potential energies U1 , , Uri , then the total mechanical energy
E=T Ul • • + [Eq. (4.22)]
is constant. More generally if there are also nonconservative forces, 0E = Wnc , the work done by the nonconservative forces.
Central Forces
A force F(r) is central if it is everywhere directed toward or away from a “force center.” If we take the latter to be the origin,
F(r) = f (r)i. [Eq. (4.65)]
A central force is spherically symmetric [f (r) = f (r)] if and only if it is conservative. [Sec. (4.8)]
Energy of a Multiparticle System
If all forces (internal and external) on a multiparticle system are conservative, the total potential energy,
U = utnt + uext E E ua, + E uaext [Eq. (4.97)]
satisfies
and
a p>a
(net force on particle a) = —Va U [Eq. (4.96)]
T U = constant. [Problem 4.52]
150 Chapter 4 Energy
Problems for Chapter 4
Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***).
SECTION 4.1 Kinetic Energy and Work
4.1 * By writing a • b in terms of components prove that the product rule for differentiation applies to the dot product of two vectors; that is,
—
d
(a • b) —
da
• b + a • —
db
.
dt dt dt
4.2 ** Evaluate the work done
P
W = f F • dr = f (Fx dx Fydy) 0 0
(4.100)
by the two-dimensional force F = (x 2 , 2xy) along the three paths joining the origin to the point P = (1, 1) as shown in Figure 4.24(a) and defined as follows: (a) This path goes along the x axis to Q = (1, 0) and then straight up to P. (Divide the integral into two pieces, fic; = f g + (b) On this path y = x 2 , and you can replace the term dy in (4.100) by dy = 2x dx and convert the whole integral into an integral over x. (c) This path is given parametrically as x = t 3 , y = t 2 . In this case rewrite x, y, dx, and dy in (4.100) in terms of t and dt, and convert the integral into an integral over t.
4.3 ** Do the same as in Problem 4.2, but for the force F = (—y, x) and for the three paths joining P and Q shown in Figure 4.24(b) and defined as follows: (a) This path goes straight from P = (1, 0) to
the origin and then straight to Q = (0, 1). (b) This is a straight line from P to Q. (Write y as a function of x and rewrite the integral as an integral over x.) (c) This is a quarter-circle centered on the origin. (Write x and y in polar coordinates and rewrite the integral as an integral over 0.)
4.4 ** A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius ro with angular velocity wo , but I now pull the string down through the hole until a length r remains between the hole and the particle. (a) What is the particle’s angular velocity now? (b) Assuming that I pull the string so slowly that we can approximate the particle’s path by a
(a)
(b)
Figure 4.24 (a) Problem 4.2. (b) Problem 4.3
Problems for Chapter 4 151
circle of slowly shrinking radius, calculate the work I did pulling the string. (c) Compare your answer to part (b) with the particle’s gain in kinetic energy.
SECTION 4.2 Potential Energy and Conservative Forces
4.5 * (a) Consider a mass m in a uniform gravitational field g, so that the force on m is mg, where g is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point 2, show that the work done by gravity is Wgray (1 —> 2) = —mgh where h is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that g can be considered constant). (b) Show that, if we choose axes with y measured vertically up, the gravitational potential energy is U = mgy (if we choose U = 0 at the
origin).
4.6 * For a system of N particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is,
MgY a
where M = E ma is the total mass and R = (X, Y, Z) is the position of the CM, with the y coordinate measured vertically up. [Hint: We know from Problem 4.5 that U, =
4.7 * Near to the point where I am standing on the surface of Planet X, the gravitational force on a mass m is vertically down but has magnitude my y 2 where y is a constant and y is the mass’s height above the horizontal ground. (a) Find the work done by gravity on a mass m moving from r 1 to r2 , and use your answer to show that gravity on Planet X, although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height h above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it’s clear what they are; don’t worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height h, how fast will it be going when it reaches the ground?
4.8 ** Consider a small frictionless puck perched at the top of a fixed sphere of radius R. If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck’s speed as a function of its height, then use Newton’s second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]
4.9 ** (a) The force exerted by a one-dimensional spring, fixed at one end, is F = —kx, where x is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is U = 4kx2 , if we choose U to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass m suspended from the other end and constrained to move in the vertical direction only. Find the extension x o of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form i ky 2 if we use the coordinate y equal to the displacement measured from the new equilibrium position at x = xo (and redefine our reference point so that U = 0 at y = 0).
152 Chapter 4 Energy
SECTION 4.3 Force as the Gradient of Potential Energy
4.10 * Find the partial derivatives with respect to x, y, and z of the following functions: (a) f (x, y, z) = ax e bxy cy 2 , (b)g(x, y, z) = sin (axyz 2), (c) h(x, y, z) = aexYlz
2 , where a, b, and c are constants.
Remember that to evaluate a flax you differentiate with respect to x treating y and z as constants. 4.11 * Find the partial derivatives with respect to x, y, and z of the following functions: (a) f (x, y, z) = aye 2byz cz 2 , (b) g(x, y, z) = cos(axy2z 3), (c) h(x, y, z) = ar, where a, b, and c are constants and r = \/x 2 + y2 + z2 . Remember that to evaluate of/ax you differentiate with respect to x treating y and z as constants.
4.12 * Calculate the gradient Vf of the following functions, f (x, y, z): (a) f = x 2 + z3 . (b) f = ky, where k is a constant. (c) f = r = -/x2 + y2 + z2 . [Hint: Use the chain rule.] (d) f = 1/ r .
4.13 * Calculate the gradient Vf of the following functions, f (x, y, z): (a) f = ln(r ), (b) f = rn, (c) f = g(r), where r = ,./x 2 + y2 + z2 and g(r) is some unspecified function of r. [Hint: Use the chain rule.]
4.14 * Prove that if f (r) and g(r) are any two scalar functions of r, then
V (fg) , fVg+ gVf
4.15 * For f (r) = x 2 + 2y2 3z2 , use the approximation (4.35) to estimate the change in f if we move from the point r = (1, 1, 1) to (1.01, 1.03, 1.05). Compare with the exact result.
4.16 * If a particle’s potential energy is U(r) = k(x2 + y2 + z 2), ) where k is a constant, what is the force on the particle?
4.17 * A charge q in a uniform electric field E o experiences a constant force F = q Eo . (a) Show that this force is conservative and verify that the potential energy of the charge at position r is U (r) = —qE0 • r. (b) By doing the necessary derivatives, check that F = — V U .
4.18 ** Use the property (4.35) of the gradient to prove the following important results: (a) The vector Vf at any point r is perpendicular to the surface of constant f through r. (Choose a small displacement dr that lies in a surface of constant f . What is df for such a displacement?) (b) The direction of Vf at any point r is the direction in which f increases fastest as we move away from r. (Choose a small displacement dr = eu, where u is a unit vector and E is fixed and small. Find the direction of u for which the corresponding df is maximum, bearing in mind that a • b = ab cos 9.)
4.19 ** (a) Describe the surfaces defined by the equation f = const, where f = x 2 + 4y2 . (b) Using the results of Problem 4.18, find a unit normal to the surface f = 5 at the point (1, 1, 1). In what direction should one move from this point to maximize the rate of change of f?
SECTION 4.4 The Second Condition that F be Conservative
4.20 * Find the curl, V x F, for the following forces: (a) F = kr; (b) F = (Ax, By e , Cz 3); (c) F = (Ay 2 , Bx, Cz), where A, B, C and k are constants.
4.21 * Verify that the gravitational force —GMini 1r 2 on a point mass m at r, due to a fixed point mass M at the origin, is conservative and calculate the corresponding potential energy.
4.22 * The proof in Example 4.5 (page 119) that the Coulomb force is conservative is considerably simplified if we evaluate V x F using spherical polar coordinates. Unfortunately, the expression for V x F in spherical polar coordinates is quite messy and hard to derive. However, the answer is given
Problems for Chapter 4 153
inside the back cover, and the proof can be found in any book on vector calculus or mathematical
methods. 15 Taking the expression inside the back cover on faith, prove that the Coulomb force F =
y r 2 is conservative.
4.23 ** Which of the following forces is conservative? (a) F = k(x , 2y, 3z) where k is a constant.
(b) F = k(y, x, 0). (c) F = k(—y, x, 0). For those which are conservative, find the corresponding
potential energy U, and verify by direct differentiation that F = — V U.
4.24*** An infinitely long, uniform rod of mass p, per unit length is situated on the z axis. (a) Calculate the gravitational force F on a point mass m at a distance p from the z axis. (The gravitational force between two point masses is given in Problem 4.21.) (b) Rewrite F in terms of the rectangular
coordinates (x, y, z) of the point and verify that V x F = 0. (c) Show that V x F = 0 using the expression for V x F in cylindrical polar coordinates given inside the back cover. (d) Find the corresponding potential energy U.
4.25 *** The proof that the condition V x F = 0 guarantees the path independence of the work
fi2 F • dr done by F is unfortunately too lengthy to be included here. However, the following three
exercises capture the main points: 16 (a) Show that the path independence of f i2 F • dr is equivalent to the statement that the integral fr, F • dr around any closed path F is zero. (By tradition, the symbol f is used for integrals around a closed path — a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2, consider the work done by F going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction.] (b) Stokes’s theorem asserts that A, F • dr = f (V x F) dA, where the integral on the right is a surface integral over a surface for which the path F is the boundary, and ii. and dA are a unit normal to the surface and an element of area. Show that Stokes’s theorem implies that if V x F = 0 everywhere, then A, F • dr = 0. (c) While the general proof of Stokes’s theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let F denote a rectangular closed path lying in a plane perpendicular to the z direction and bounded by the lines x = B, x = B + b, y = C and y = C c. For this simple path (traced counterclockwise as seen from above), prove Stokes’s theorem that
F • dr = f (V x F) • ill d A
where it = z and the integral on the right runs over the flat, rectangular area inside F. [Hint: The integral on the left contains four terms, two of which are integrals over x and two over y. If you pair them in this way, you can combine each pair into a single integral with an integrand of the form Fx (x, C c, z) — F x (x, C, z) (or a similar term with the roles of x and y exchanged). You can rewrite this integrand as an integral over y of a Fx (x, y, z)/ay (and similarly with the other term), and you’re home.]
SECTION 4.5 Time-Dependent Potential Energy
4.26 * A mass m is in a uniform gravitational field, which exerts the usual force F = mg vertically down, but with g varying with time, g = g (t). Choosing axes with y measured vertically up and defining U = mgy as usual, show that F = — V U as usual, but, by differentiating E = 4mv2 + U with respect to t, show that E is not conserved.
15 See, for example, Mathematical Methods in the Physical Sciences by Mary Boas (Wiley, 1983), p. 435. 16 For a complete discussion see, for example, Mathematical Methods, Boas, Ch. 6, Sections 8-11.
154 Chapter 4 Energy
Figure 4.25 Problem 4.30
4.27 ** Suppose that the force F(r, t) depends on the time t but still satisfies V x F = 0. It is a mathematical fact (related to Stokes’s theorem as discussed in Problem 4.25) that the work integral fi2 F(r, t) • dr (evaluated at any one time t) is independent of the path taken between the points 1 and 2. Use this to show that the time-dependent PE defined by (4.48), for any fixed time t, has the claimed property that F(r, t) = — V U (r, t). Can you see what goes wrong with the argument leading to Equation (4.19), that is, conservation of energy?
SECTION 4.6 Energy for Linear One-Dimensional Systems
4.28 ** Consider a mass m on the end of a spring of force constant k and constrained to move along the horizontal x axis. If we place the origin at the spring’s equilibrium position, the potential energy is 2 kx 2 . At time t = 0 the mass is sitting at the origin and is given a sudden kick to the right so that it moves out to a maximum displacement at x max = A and then continues to oscillate about the origin. (a) Write down the equation for conservation of energy and solve it to give the mass’s velocity x in terms of the position x and the total energy E. (b) Show that E = 2kA 2, and use this to eliminate E from your expression for Use the result (4.58), t = f dxf li(x’), to find the time for the mass to move from the origin out to a position x. (c) Solve the result of part (b) to give x as a function of t and show that the mass executes simple harmonic motion with period 27,/m1k.
4.29 ** [Computer] A mass m confined to the x axis has potential energy U = kx4 with k > 0. (a) Sketch this potential energy and qualitatively describe the motion if the mass is initially stationary at x = 0 and is given a sharp kick to the right at t = 0. (b) Use (4.58) to find the time for the mass to reach its maximum displacement x max = A. Give your answer as an integral over x in terms of m, A, and k. Hence find the period r of oscillations of amplitude A as an integral. (c) By making a suitable change of variables in the integral, show that the period r is inversely proportional to the amplitude A. (d) The integral of part (b) cannot be evaluated in terms of elementary functions, but it can be done numerically. Find the period for the case that m = k = A = 1.
SECTION 4.7 Curvilinear One-Dimensional Systems
4.30 * Figure 4.25 shows a child’s toy, which has the shape of a cylinder mounted on top of a hemisphere. The radius of the hemisphere is R and the CM of the whole toy is at a height h above the floor. (a) Write down the gravitational potential energy when the toy is tipped to an angle 0 from the vertical. [You need to find the height of the CM as a function of 0. It helps to think first about the height of the hemisphere’s center 0 as the toy tilts.] (b) For what values of R and h is the equilibrium at 0 = 0 stable?
Problems for Chapter 4 155
Figure 4.26 Problem 4.34
4.31 * (a) Write down the total energy E of the two masses in the Atwood machine of Figure 4.15 in terms of the coordinate x and x. (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate x by differentiating the equation E = const. Check that the equation of motion is the same as you would obtain by applying Newton’s second law to each mass and eliminating the unknown tension from the two resulting equations.
4.32 ** Consider the bead of Figure 4.13 threaded on a curved rigid wire. The bead’s position is specified by its distance s, measured along the wire from the origin. (a) Prove that the bead’s speed v is just v (Write v in terms of its components, dx/dt, etc., and find its magnitude using Pythagoras’s theorem.) (b) Prove that m’s’ = Ftang , the tangential component of the net force on the bead. (One way to do this is to take the time derivative of the equation v 2 = v • v. The left side should lead you to :5′ and the right to Ftang .) (c) One force on the bead is the normal force N of the wire (which constrains the bead to stay on the wire). If we assume that all other forces (gravity, etc.) are conservative, then their resultant can be derived from a potential energy U. Prove that Ftang = —dU Ids. This shows that one- dimensional systems of this type can be treated just like linear systems, with x replaced by s and Fx
by Ftang•
4.33 ** [Computer] (a) Verify the expression (4.59) for the potential energy of the cube balanced on a cylinder in Example 4.7 (page 130). (b) Make plots of U(8) for b = 0.9r and b = 1.1r. (You may as well choose units such that r, m, and g are all equal to 1.) (c) Use your plots to confirm the findings of Example 4.7 concerning the stability of the equilibrium at 0 = 0. Are there any other equilibrium points and are they stable?
4.34 ** An interesting one-dimensional system is the simple pendulum, consisting of a point mass m, fixed to the end of a massless rod (length 1), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure 4.26. The pendulum’s position can be specified by its angle 0 from the equilibrium position. (It could equally be specified by its distance s from equilibrium — indeed s = 10—but the angle is a little more convenient.) (a) Prove that the pendulum’s potential energy (measured from the equilibrium level) is
U(0) = mgl(1 — cos 0). (4.101)
Write down the total energy E as a function of 0 and 0. (b) Show that by differentiating your expression for E with respect to t you can get the equation of motion for 0 and that the equation of motion is just the familiar F = /a (where F is the torque, I is the moment of inertia, and a is the angular acceleration 0). (c) Assuming that the angle 0 remains small throughout the motion, solve for 0 (t) and show that the motion is periodic with period
to = 270 / g. (4.102)
156 Chapter 4 Energy
Figure 4.27 Problem 4.36
(The subscript “o” is to emphasize that this is the period for small oscillations.)
4.35 ** Consider the Atwood machine of Figure 4.15, but suppose that the pulley has radius R and moment of inertia I. (a) Write down the total energy of the two masses and the pulley in terms of the coordinate x and I. (Remember that the kinetic energy of a spinning wheel is i ho2 .) (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate x by differentiating the equation E = const. Check that the equation of motion is the same as you would obtain by applying Newton’s second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations.
4.36 ** A metal ball (mass m) with a hole through it is threaded on a frictionless vertical rod. A massless string (length 1) attached to the ball runs over a massless, frictionless pulley and supports a block of mass M, as shown in Figure 4.27. The positions of the two masses can be specified by the one angle 9. (a) Write down the potential energy U(9). (The PE is given easily in terms of the heights shown as h and H. Eliminate these two variables in favor of 9 and the constants b and 1. Assume that the pulley and ball have negligible size.) (b) By differentiating U(0) find whether the system has an equilibrium position, and for what values of m and M equilibrium can occur. Discuss the stability of any equilibrium positions.
4.37 *** [Computer] Figure 4.28 shows a massless wheel of radius R, mounted on a frictionless, horizontal axle. A point mass M is glued to the edge of the wheel, and a mass m hangs from a string wrapped around the perimeter of the wheel. (a) Write down the total PE of the two masses as a function of the angle 0. (b) Use this to find the values of m and M for which there are any positions of equilibrium Describe the equilibrium positions, discuss their stability, and explain your answers in terms of torques. (c) Plot U (0) for the cases that m = 0.7M and m = 0.8M, and use your graphs to describe the behavior of the system if I release it from rest at 0 = 0. (d) Find the critical value of m/M on one side of which the system oscillates and on the other side of which it does not (if released from rest at 0 = 0).
4.38 *** [Computer] Consider the simple pendulum of Problem 4.34. You can get an expression for the pendulum’s period (good for large oscillations as well as small) using the method discussed in connection with (4.57), as follows: (a) Using (4.101) for the PE, find 0 as a function of 0. Next use (4.57), in the form t = f d04, to write the time for the pendulum to travel from 0 = 0 to its maximum value (the amplitude) O. Because this time is a quarter of the period t, you can now write down the period. Show that
Problems for Chapter 4 157
Figure 4.28 Problem 4.37
1 dO 2 f 1 du = to —7r f = r 0
° -isin2 (e13/2) — sin2(4)/2) 7 . ,,/1-.2,/1- A2.2′ (4.103)
where ro is the period (4.102) (Problem 4.34) for small oscillations and A = sin(eD/2). [To get the first expression you will need to use the trig identity for 1 — cos 4) in terms of sin 2 (4)/2). To get the second you need to make the substitution sin(4)/2) = Au.] These integrals cannot be evaluated in terms of elementary functions. However, the second integral is a standard integral called the complete elliptic integral of the first kind, sometimes denoted K (A 2), whose values are tabulated 17 and are known to computer software such as Mathematica [which calls it EllipticK(A 2)]. (b) If you have access to computer software that knows this function, make a plot of r/r o for amplitudes 0 < < 3 rad. Comment. What becomes of r as the amplitude of oscillation approaches 7r? Explain.
4.39 *** (a) If you have not already done so, do Problem 4.38(a). (b) If the amplitude eD is small then so is A = sin el3/2. If the amplitude is very small, we can simply ignore the last square root in (4.103). Show that this gives the familiar result for the small-amplitude period, r = To = 2701g. (c) If the amplitude is small but not very small, we can improve on the approximation of part (b). Use the binomial expansion to give the approximation 1/- ‘/1 — A 2u2 ti 1 + 1A 2 u 2 and show that, in this approximation, (4.103) gives
r = ro [l sin2 (43/2)].
What percentage correction does the second term represent for an amplitude of 45°? (The exact answer for elp = 45° is 1.040 ro to four significant figures.)
SECTION 4.8 Central Forces
4.40 * (a) Verify the three equations (4.68) that give x, y, z in terms of the spherical polar coordinates r, 9, (b) Find expressions for r, in terms of x, y, z.
17 See, for example, M.Abramowitz and I.Stegun, Handbook of Mathematical Functions, Dover, New York, 1965. Be warned that different authors use different notations. In particular, some authors call the exact same integral K (A) .
158 Chapter 4 Energy
Figure 4.29 Problem 4.44
4.41 * A mass m moves in a circular orbit (centered on the origin) in the field of an attractive central force with potential energy U = krn . Prove the virial theorem that T = nU 12.
4.42 * In one dimension, it is obvious that a force obeying Hooke’s law is conservative (since F = —kx depends only on the position x, and this is sufficient to guarantee that F is conservative in one dimension). Consider instead a spring that obeys Hooke’s law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass m at its other end, for instance.) Write down the force F(r) exerted by the spring in terms of its length r and its equilibrium length ro . Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]
4.43 ** In Section 4.8, I claimed that a force F(r) that is central and spherically symmetric is automati- cally conservative. Here are two ways to prove it: (a) Since F(r) is central and spherically symmetric, it must have the form F(r) = f (r)r. Using Cartesian coordinates, show that this implies that V x F = 0. (b) Even quicker, using the expression given inside the back cover for V x F in spherical polars, show that V x F = 0.
4.44 ** Problem 4.43 suggests two proofs that a central, spherically symmetric force is automatically conservative, but neither proof makes really clear why this is so. Here is a proof that is less complete but more insightful: Consider any two points A and B and two different paths ACB and ADB connecting them as shown in Figure 4.29. Path ACB goes radially out from A until it reaches the radius rB of B, and then around a sphere (center 0) to B. Path ADB goes around a sphere of radius rA until it reaches the line OB, and then radially out to B. Explain clearly why the work done by a central, spherically symmetric force F is the same along both paths. (This doesn’t prove that the work is the same along any two paths from A to B. If you want you can complete the proof by showing that any path can be approximated by a series of paths moving radially in or out and paths of constant r.)
4.45 ** In Section 4.8, I proved that a force F(r) = f (r)i that is central and conservative is automat- ically spherically symmetric. Here is an alternative proof: Consider the two paths ACB and ADB of
Figure 4.29, but with rB = rA + dr where dr is infinitesimal. Write down the work done by F(r) going around both paths, and use the fact that they must be equal to prove that the magnitude function f (r) must be the same at points A and D; that is, f (r) = f (r) and the force is spherically symmetric.
Problems for Chapter 4 159
SECTION 4.9 Energy of Interaction of Two Particles
4.46 * Consider an elastic collision of two particles as in Example 4.8 (page 143), but with unequal masses, m 1 0 m2 . Show that the angle 0 between the two outgoing velocities satisfies 9 < 7r/2 if m i > m2 , but 0 > r/2 if m i < m2 .
4.47 * Consider a head-on elastic collision between two particles. (Since the collision is head-on, the motion is confined to a single straight line and is therefore one-dimensional.) Prove that the relative velocity after the collision is equal and opposite to that before. That is, v 1 — v2 = —(1/1 — v;), where
v 1 and v2 are the initial velocities and vi and v’2 the corresponding final velocities.
4.48* A particle of mass m 1 and speed v 1 collides with a second particle of mass m 2 at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that m 1 <-< m 2 and that
M2 << M 1′
4.49 ** Both the Coulomb and gravitational forces lead to potential energies of the form U = y — r2 1, where y denotes ka 1 q2 in the case of the Coulomb force and —Gm 1 m 2 for gravity, and r 1 and r2
are the positions of the two particles. Show in detail that —V1 U is the force on particle 1 and —V 2 U that on particle 2.
4.50 ** The formalism of the potential energy of two particles depends on the claim in (4.81) that
V1 U(r 1 — r2) = —V 2 U(r 1 — r2).
Prove this. (Use the chain rule for differentiation. The proof in three dimensions is notationally awkward, so prove the one-dimensional result that
a a
ax f (x i — x 2) = — — f (x i — x 2)
, ax2
and then convince yourself that it extends to three dimensions.)
SECTION 4.10 The Energy of a Multiparticle System
4.51 ** Write out the arguments of all the potential energies of the four-particle system in (4.94). For instance U = U(r 1 , r2 , • • , r4), whereas U34 = U34 (r3 – r4). Show in detail that the net force on particle 3 (for instance) is given by —V3 U. [You know that the separate forces, internal and external, are given by (4.92) and (4.93).]
4.52 ** Consider the four-particle system of Section 4.10. (a) Write down the work—KE theorem for each of the four particles separately and, by adding these four equations, show that the change in the total KE in a short time interval dt is dT = Wtot where Wtot is the total work done on all particles by all forces. [This shouldn’t take more than two or three lines.] (b) Next show that Wtot = —dU where dU is the change in total PE during the same time interval. Deduce that the total mechanical energy E = T U is conserved.
4.53 ** (a) Consider an electron (charge —e and mass m) in a circular orbit of radius r around a fixed proton (charge +e). Remembering that the inward Coulomb force ke g /r2 is what gives the electron its centripetal acceleration, prove that the electron’s KE is equal to —4 times its PE; that is, T = —1U and hence E = z U. (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius r around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches
160 Chapter 4 Energy
from afar with kinetic energy T2. When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius r’. (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy E long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables T2, r, and r’.
CHAPTER
Oscillations
Almost any system that is displaced from a position of stable equilibrium exhibits oscillations. If the displacement is small, the oscillations are almost always of the type called simple harmonic. Oscillations, and particularly simple harmonic oscillations, are therefore extremely widespread. They are also extremely useful. For example, all good clocks depend on an oscillator to regulate their time keeping: The first reliable clocks used a pendulum; the first accurate watches (historically crucial in navigation) used an oscillating balance wheel; modern watches use the oscillations of a quartz crystal; and today’s most accurate clocks, such as the atomic clock at the National Institute for Standards and Technology in Boulder, Colorado, use the oscillations of an atom. In this chapter, we shall explore the physics and mathematics of oscillations. I shall begin with simple harmonic oscillations and then go on to damped oscillations (oscillations that die out because of resistive forces) and driven oscillations (oscillations that are maintained by an outside driving force, as in all clocks). The last three sections of this chapter describe the use of Fourier series in finding the motion of an oscillator driven by an arbitrary periodic driving force.
5.1 Hooke’s Law
As you are certainly aware, a mass on the end of a spring that obeys Hooke’s law executes oscillations of the type that we call simple harmonic. Before we review the proof of this claim, let us first ask why Hooke’s law is so important and appears so frequently. Hooke’s law asserts that the force exerted by a spring has the form (for now we’ll restrict ourselves to a spring confined to the x axis)
Fx (x) = —kx (5.1)
where x is the displacement of the spring from its equilibrium length and k is a positive number called the force constant. That k is positive means that the equilibrium at x = 0 is stable: When x = 0 there is no force, when x > 0 (displacement to the right) the 161
162 Chapter 5 Oscillations
force is negative (back to the left), and when x < 0 (displacement to the left) the force is positive (back to the right); either way, the force is a restoring force, and the equilibrium is stable. (If k were negative, the force would be away from the origin, and the equilibrium would be unstable, in which case we do not expect to see oscillations.) An exactly equivalent way to state Hooke’s law is that the potential energy is
U(x) = 2 kx 2 .
Consider now an arbitrary conservative one-dimensional system which is specified by a coordinate x and has potential energy U (x). Suppose that the system has a stable equilibrium position x = x0 , which we may as well take to be the origin (x 0 = 0). Now consider the behavior of U(x) in the vicinity of the equilibrium position. Since any reasonable function can be expanded in a Taylor series, we can safely write
U(x) = U(0) + c(o)x + 2u”(0)x2+ • • • . (5.2)
As long as x remains small, the first three terms in this series should be a good approximation. The first term is a constant, and, since we can always subtract a constant from U(x) without affecting any physics, we may as well redefine U (0) to be zero. Because x = 0 is an equilibrium point, U'(0) = 0 and the second term in the series (5.2) is automatically zero. Because the equilibrium is stable, U”(0) is positive. Renaming U”(0) as k, we conclude that for small displacements it is always a good approximation to take’
U(x) = Ikx 2 . (5.3)
That is, for sufficiently small displacements from stable equilibrium, Hooke’s law is always valid. Notice that if U”(0) were negative, then k would also be negative, and the equilibrium would be unstable — a case we’re not interested in just now. Hooke’s law in the form (5.3) crops up in many situations, although it is certainly not necessary that the coordinate be the rectangular coordinate x, as the following example illustrates.
EXAMPLE 5.1 The Cube Balanced on a Cylinder
Consider again the cube of Example 4.7 (page 130) and show that for small angles B the potential energy takes the Hooke’s-law form U(9) = 40 2 .
We saw in that example that
U(0) = mg[(r b) cos 0 ± r0 sing].
If 0 is small we can make the approximations cos 9 ti 1 — 0 2/2 and sin 0 ti 0 , so that
U(0) mg[(r b)(1 — 0 2) + r02] = mg (r b) m g (r — b)0 2 ,
1 The only exception is if Un(0) happens to be zero, but I shall not worry about this exceptional case here.
Section 5.2 Simple Harmonic Motion 163
which, apart from the uninteresting constant, has the form 4/(0 2 with “spring
constant” k = mg(r — b). Notice that the equilibrium is stable (k positive) only when r > b, a condition we had already found in Example 4.7.
Figure 5.1 A mass m with potential energy U(x) = zkx 2 and total energy E oscillates between the two turning points at x = +A, where U(x) = E and the kinetic energy is zero.
As discussed in Section 4.6, the general features of the motion of any one- dimensional system can be understood from a graph of U(x) against x. For the Hooke’s-law potential energy (5.3), this graph is a parabola, as shown in Figure 5.1. If a mass in has potential energy of this form and has any total energy E > 0, it is trapped and oscillates between the two turning points where U(x) = E, so that the kinetic energy is zero and the mass is instantaneously at rest. Because U(x) is symmetric about x = 0, the two turning points are equidistant on opposite sides of the origin and are traditionally denoted x = ±A, where A is called the amplitude of the oscillations.
5.2 Simple Harmonic Motion
We are now ready to examine the equation of motion (that is, Newton’s second law) for a mass m that is displaced from a position of stable equilibrium. To be definite, let us consider a cart on a frictionless track attached to a fixed spring as sketched in Figure 5.2. We have seen that we can approximate the potential energy by (5.3) or, equivalently, the force by Fx (x) = —kx. Thus the equation of motion is mx= Fx = —kx or
= –x = —co2X
(5.4) m
where I have introduced the constant
w =- 1,1 — , rn
which we shall see is the angular frequency with which the cart will oscillate. Although we have arrived at Equation (5.4) in the context of a cart on a spring moving along
164 Chapter 5 Oscillations
x = 0 (equilibrium)
Figure 5.2 A cart of mass m oscillating on the end of a spring.
the x axis, we shall see eventually that it applies to many different oscillating systems in many different coordinate systems. For example, we have already seen in Equation (1.55) that the angle 4) that gives the position of a pendulum (or a skateboard in a half-pipe) is governed by the same equation, 4 = —0) 20, at least for small values of 0. In this section I am going to review the properties of the solutions of (5.4). Unfortunately, there are many different ways to write the same solution, all of which have their advantages, and you should be comfortable with them all.
The Exponential Solutions
Equation (5.4) is a second-order, linear, homogeneous differential equation 2 and so has two independent solutions. These two independent solutions can be chosen in several different ways, but perhaps the most convenient is this:
x(t) = e iwt and x(t) =
As you can easily check, both of these functions do satisfy (5.4). Further, any constant multiple of either solution is also a solution, and likewise any sum of such multiples. Thus the function
x(t) = Cie iwt (5.5)
is also a solution for any two constants C 1 and C2. (That any linear combination of solutions like this is itself a solution is called the superposition principle and plays a crucial role in many branches of physics.) Since this solution (5.5) contains two arbitrary constants, it is the general solution of our second-order equation (5.4). 3
Therefore, any solution can be expressed in the form (5.5) by suitable choice of the coefficients C 1 and C2.
2 Linear because it contains no higher powers of x or its derivatives than the first power, and homogeneous because every term is a first power (that is, there is no term independent of x and its derivatives).
3 Recall the result, discussed below Equation (1.56), that the general solution of a second-order differential equation contains precisely two arbitrary constants.
Section 5.2 Simple Harmonic Motion 165
The Sine and Cosine Solutions
The exponential functions in (5.5) are so convenient to handle that (5.5) is often the best form of the solution. Nevertheless, this form does have one disadvantage. We know, of course, that x (t) is real, whereas the two exponentials in (5.5) are complex. This means the coefficients C 1 and C2 must be chosen carefully to ensure that x (t) itself is real. I shall return to this point shortly, but first I shall rewrite (5.5) in another useful way. From Euler’s formula (2.76) we know that the two exponentials in (5.5)
can be written as
e±iwt = cos(cot) i sin(cot).
Substituting into (5.5) and regrouping we find that
x(t) = (C 1 + C2 ) cos(cot) i (CI — C2) sin(wt)
= Bi cos(wt) + B2 sin(wt) (5.6)
where B 1 and B2 are simply new names for the coefficients in the previous line,
B 1 – ± C2 and B2 -= i (C1 C2). (5.7)
The form (5.6) can be taken as the definition of simple harmonic motion (or SHM): Any motion that is a combination of a sine and cosine of this form is called simple harmonic. Because the functions cos(wt) and sin(cot) are real, the requirement that x (t) be real means simply that the coefficients B 1 and B2 must be real.
We can easily identify the coefficients B 1 and B2 in terms of the initial conditions of the problem. Clearly at t = 0, (5.6) implies that x (0) = B 1 . That is, B 1 is just the initial position x (0) = x 0 . Similarly, by differentiating (5.6), we can identify coB 2 as the initial velocity v o .
If I start the oscillations by pulling the cart aside to x = x0 and releasing it from rest (vo = 0), then B2 = 0 in (5.6) and only the cosine term survives, so that
x(t) = cos(cot). (5.8)
If I launch the cart from the origin (x o = 0) by giving it a kick at t = 0, only the sine term survives, and
x(t) = — v.
sin(wt).
These two simple cases are illustrated in Figure 5.3. Notice that both solutions, like the general solution (5.6), are periodic because both the sine and cosine are. Since the argument of both sine and cosine is cot, the function x(t) repeats itself after the time r for which cot = 27. That is, the period is
27r = – = 2 (5.9)
166 Chapter 5 Oscillations
(a)
(b)
Figure 5.3 (a) Oscillations in which the cart is released from x o at t = 0 follow a cosine curve. (b) If the cart is kicked from the origin at t = 0, the oscillations follow a sine curve with initial slope v o . In either case the period of the oscilla- tions is r = 2jr/w = k and is the same whatever the values of x o or vo .
The Phase-Shifted Cosine Solution
The general solution (5.6) is harder to visualize than the two special cases of Fig- ure 5.3, and it can usefully be rewritten as follows: First, we define yet another constant
A = ,\/B12 + B22 . (5.10)
Notice that A is the hypotenuse to a right triangle whose other two sides are B I and B2. I have indicated this in Figure 5.4, where I have also defined 3 as the lower angle of that triangle. We can now rewrite (5.6) as
[ x(t) = A — B1
cos(wt) + sin(wt) ] A
= A[ cos 8 cos(cot) + sin 3 sin (wt)]
= A cos(wt — O. (5.11)
From this form it is clear that the cart is oscillating with amplitude A, but instead of being a simple cosine as in (5.8), it is a cosine which is shifted in phase: When t = 0 the argument of the cosine is —8, and the oscillations lag behind the simple cosine by the phase shift 6. We have derived the result (5.11) from Newton’s second law, but, as so often happens, one can derive the same result in more than one way. In particular,
Figure 5.4 The constants A and 6 are defined in terms of B l and B2 as shown.
Section 5.2 Simple Harmonic Motion 167
(5.11) can also be derived using the energy approach discussed in Section 4.6. (See
Problem 4.28.)
Solution as the Real Part of a Complex Exponential
There is still another useful way to write our solution, in terms of the complex exponentials of (5.5). The coefficients C 1 and C2 there are related to the coefficients
B 1 and B2 of the sine-cosine form by Equation (5.7), which we can solve to give
C 1 = z(B i – i B2 ) and C2 = i (B1 i B2). (5.12) Since B 1 and B2 are real, this shows that both C 1 and C2 are generally complex and
that C2 is the complex conjugate of C 1 ,
C2 = C
(Recall that for any complex number z = x iy, the complex conjugate z* is defined
as4 z* = x – iy.) Thus the solution (5.5) can be written as
x(t) = C Cie-iwt (5.13)
where the whole second term on the right is just the complex conjugate of the first. (See Problem 5.35 if this isn’t clear to you.) Now, for any complex number z = x + iy,
z z* = (x iy) (x – iy) = 2x = 2 Re z
where Re z denotes the real part of z (namely x). Thus (5.13) can be written as
x (t) = 2 Re C i ei”.
If we define a final constant C = 2C 1 , we see from Equation (5.12) and Figure 5.4 that
C = B 1 – iB 2 = Ae -i8 (5.14)
and
x(t) = Re Cei wt = Re Ae i(w”) .
This beautiful result is illustrated in Figure 5.5. The complex number Ae i (w”) moves counterclockwise with angular velocity co around a circle of radius A. Its real part [namely x(t)] is the projection of the complex number onto the real axis. While the complex number goes around the circle, this projection oscillates back and forth on the x axis, with angular frequency w and amplitude A. Specifically, x(t) = A cos(wt – 6), in agreement with (5.11).
4 While most physicists use the notation z*, mathematicians almost always use E for the complex conjugate of z.
168 Chapter 5 Oscillations
Figure 5.5 The position x (t) of the oscillating cart is the real part of the complex number Aei (t -S) As the latter moves around the circle of radius A, the former oscillates back and forth on the x axis with amplitude A.
EXAMPLE 5.2 A Bottle in a Bucket
A bottle is floating upright in a large bucket of water as shown in Figure 5.6. In equilibrium it is submerged to a depth do below the surface of the water. Show that if it is pushed down to a depth d and released, it will execute harmonic motion, and find the frequency of its oscillations. If do = 20 cm, what is the period of the oscillations?
The two forces on the bottle are its weight mg downward and the upward buoyant force pg Ad, where Q is the density of water and A is the cross-sectional area of the bottle. (Remember that Archidemes’ principle says that the buoyant force is pg times the volume submerged, which is just Ad.) The equilibrium depth do is determined by the condition
mg = pgAdo . (5.15)
Figure 5.6 The bottle shown has been loaded with sand so that it floats upright in a bucket of water. Its equilibrium depth is d = do.
Section 5.2 Simple Harmonic Motion 169
Suppose now the bottle is at a depth d x. (This defines x as the distance from equilibrium, always the best coordinate to use.) Newton’s second law now reads
mx = mg — pgA(d o x).
By (5.15), the first and second terms on the right cancel, and we’re left with —pgAxlm. But again by (5.15), pgAlm= gldo , so the equation of motion
becomes
= –x, 0
which is exactly the equation for simple harmonic motion. We conclude that the bottle moves up and down in SHM with angular frequency co = .\/gIdo . A remarkable feature of this result is that the frequency of oscillations is indepen- dent of m, Q, and A; also, the frequency is the same as that of a simple pendulum of length 1 = do . If do = 20 cm, then the period is
= — = 27T = / 0.20 m
g V 9.8 m/s 2 = 0.9 sec.
Try this experiment yourself! But be aware that the details of the flow of water around the bottle complicate the situation considerably. The calculation here is a very simplified version of the truth.
Energy Considerations
To conclude this section on simple harmonic motion, let us consider briefly the energy of the oscillator (the cart on a spring or whatever it is) as it oscillates back and forth. Since x (t) = A cos(wt — 8), the potential energy is just
U = 2kx 2 = IkA 2 cos2 (wt — 8).
Differentiating x (t) to give the velocity, we find for the kinetic energy
T = 2 imi2 = 2 l inco2A2 sin2(wt — 8)
1 = -2 kA
2 sine (cot — 8)
where the second line results from replacing w2 by k I m. We see that both U and T oscillate between 0 and 4kA 2, with their oscillations perfectly out of step — when U is maximum T is zero and vice versa. In particular, since sin e 0 + cos2 B = 1, the total energy is constant,
E = T U = kA2 , (5.16)
as it has to be for any conservative force.
170 Chapter 5 Oscillations
5.3 Two-Dimensional Oscillators
In two or three dimensions, the possibilities for oscillation are considerably richer than in one dimension. The simplest possibility is the so-called isotropic harmonic oscillator, for which the restoring force is proportional to the displacement from equilibrium, with the same constant of proportionality in all directions:
F = —kr. (5.17)
That is, F.„ = —kx, F y = —ky (and F, = —kz in three dimensions), all with the same constant k. This force is a central force directed toward the equilibrium position, which we may as well take to be the origin, as sketched in Figure 5.7(a). Figure 5.7(b) shows an arrangement of four identical springs that would produce a force of this form; it is easy to see that if the mass at the center is moved away from its equilibrium position it will experience a net inward force, and it is not too hard to show (Problem 5.19) that this inward force has the form (5.17) for small displacements r. 5 Another example of a two-dimensional isotropic oscillator is (at least approximately) a ball bearing rolling near the bottom of a large spherical bowl. Two important three-dimensional examples are an atom vibrating near its equilibrium in a symmetric crystal, and a proton (or neutron) as it moves inside a nucleus.
Let us consider a particle that is subject to this type of force and suppose, for simplicity, that it is confined to two dimensions. The equation of motion, = F/m, separates into two independent equations:
_ co2 ••
—602 Y (5.18)
where I have introduced the familiar angular frequency w = fig m (which is the same in both x and y equations because the same is true of the force constants). Each of
Y
71 r4,_ kr
X >
0
(a) (b)
Figure 5.7 (a) A restoring force that is proportional to r defines the isotropic harmonic oscillator. (b) The mass at the center of this arrangement of springs would experience a net force of the form F = —kr as it moves in the plane of the four springs.
5 1t is perhaps worth pointing out that one does not get a force of the form (5.17) by simply attaching a mass to a spring whose other end is anchored to the origin.
(c) 5 = 7/4
Section 5.3 Two – Dimensional Oscillators 171
these two equations has exactly the form of the one-dimensional equation discussed in the last section, and the solutions are [as in (5.11)1
x(t) = Ax cos(cot — 3s ) 1 y(t) = A y cos(wt y)
(5.19)
where the four constants A x , A y , 3x, and 8 are determined by the initial conditions of the problem. By redefining the origin of time, we can dispose of the phase shift 3 s , but, in general, we cannot also dispose of the corresponding phase in the y solution. Thus the simplest form for the general solution is
x(t) = A s cos(cot) y(t) = A y cos(cot — 6)
(5.20)
where 8 = 6y — 6 x is the relative phase of the y and x oscillations. (See Problem 5.15.) The behavior of the solution (5.20) depends on the values of the three constants
A x , Ay , and 8. If either A s or Ay is zero, the particle executes simple harmonic motion along one of the axes. (The ball bearing in the bowl rolls back and forth through the origin, moving in the x direction only or the y direction only.) If neither A x nor A y
is zero, the motion depends critically on the relative phase 6. If 6 = 0, then x(t) and y(t) rise and fall in step, and the point (x, y) moves back and forth on the slanting line that joins (A s , A y) to (—A s , —A y ), as shown in Figure 5.8(a). If 8 = 7r/2, then x and y oscillate out of step, with x at an extreme when y is zero, and vice versa; the point (x, y) describes an ellipse with semimajor and semiminor axes A s and Ay , as in Figure 5.8(b). For other values of 6, the point (x, y) moves around a slanting ellipse, as shown for the case 3 = 7r/4 in Figure 5.8(c). (For a proof that the path really is an ellipse, see Problem 8.11.)
In the anisotropic oscillator, the components of the restoring force are propor- tional to the components of the displacement, but with different constants of propor- tionality:
Fx = —kx x, Fy = —ky y, and = —k z z. (5.21)
(a) S =0 (b) 8 = 71-12
Figure 5.8 Motion of a two-dimensional isotropic oscillator as given by (5.20). (a) If 8 = 0, then x and y execute simple harmonic motion in step, and the point (x, y) moves back and forth along a slanting line as shown. (b) If 8 = 7r/2, then (x, y) moves around an ellipse with axes along the x and y axes. (c) In general (for example, 8 = 7/4), the point (x, y) moves around a slanted ellipse as shown.
172 Chapter 5 Oscillations
An example of such a force is the force felt by an atom displaced from its equilibrium position in a crystal of low symmetry, where it experiences different force constants along the different axes. For simplicity, I shall again consider a particle in two dimensions, for which Newton’s second law separates into two separate equations just as in (5.18):
_ co 2x
Y = — 60 2
Y• (5.22)
The only difference between this and (5.18) is that there are now different frequencies for the different axes, co, = .1k,Im and so on. The solution of these two equations is just like (5.20):
x(t) = Ax cos(cox t) y (t) = Ay COS(Wyt — 6).
(5.23)
Because of the two different frequencies, there is a much richer variety of possible motions. If cox /coy is a rational number, it is fairly easy to see (Problem 5.17) that the motion is periodic, and the resulting path is called a Lissajous figure (after the French physicist Jules Lissajous, 1822-1880). For example, Figure 5.9(a) shows an orbit of a particle for which cox /coy = 2 and the x motion repeats itself twice as often as the y motion. In the case shown, the result is a figure eight. If cox /coy is irrational, the motion is more complicated and never repeats itself. This case is illustrated, with cox /coy = -s/2, in Figure 5.9(b). This kind of motion is called quasiperiodic: The motion of the separate coordinates x and y is periodic, but because the two periods are incompatible, the motion of r = (x, y) is not.
(a) 60, = 2 coy (b) co, = V G coy
Figure 5.9 (a) One possible path for an anisotropic oscillator with co, = 2 and coy = 1. You can see that x goes back and forth twice in the time that y does so once, and the motion then repeats itself exactly. (b) A path for the case cox = Vi and co = 1 from t = 0 to t = 24. In this case the path never repeats itself, although, if we wait long enough, it will come arbitrarily close to any point in the rectangle bounded by x = ±Ax and y = ±Ay.
Section 5.4 Damped Oscillations 173
5.4 Damped Oscillations
I shall now return to the one-dimensional oscillator, and take up the possibility that there are resistive forces that will damp the oscillations. There are several possibilities for the resistive force. Ordinary sliding friction is approximately constant in magni- tude, but always directed opposite to the velocity. The resistance offered by a fluid, such as air or water, depends on the velocity in a complicated way. However, as we saw in Chapter 2, it is sometimes a reasonable approximation to assume that the re- sistive force is proportional to v or (under different circumstances) to v 2 . Here I shall assume that the resistive force is proportional to v; specifically, f -= —by. One of my main reasons is that this case leads to an especially simple equation to solve, and the equation is itself a very important equation that appears in several other contexts and is therefore well worth studying. 6
Consider, then, an object in one dimension, such as a cart attached to a spring, that is subject to a Hooke’s law force, —kx, and a resistive force, —b. . The net force on the object is —b.i kx, and Newton’s second law reads (if I move the two force terms over to the left side)
± kx = 0. (5.24)
One of the beautiful things about physics is the way the same mathematical equation can arise in totally different physical contexts, so that our understanding of the equation in one situation carries over immediately to the other. Before we set about solving Equation (5.24), I would like to show how the same equation appears in the study of LRC circuits. An LRC circuit is a circuit containing an inductor (inductance L), a capacitor (capacitance C), and a resistor (resistance R), as sketched in Figure 5.10. I have chosen the positive direction for the current to be counterclockwise, and the charge q(t) to be the charge on the left-hand plate of the capacitor [with —q (t) on the right], so that I (t) = (t). If we follow around the circuit in the positive direction, the electric potential drops by Li = L4 across the inductor, by RI = R4 across the resistor, and by q /C across the capacitor. Applying Kirchoff’s second rule for circuits, we conclude that
. 1 Rq —q = O. (5.25)
This has exactly the form of Equation (5.24) for the damped oscillator, and anything that we learn about the equation for the oscillator will be immediately applicable to the LRC circuit. Notice that the inductance L of the electric circuit plays the role of the mass of the oscillator, the resistance term R4 corresponds to the resistive force, and 1/ C to the spring constant k.
6 You should be aware, however, that although the case I am considering — that the resistive force f is linear in v — is very important, it is nevertheless a very special case. I shall describe some of the startling complications that can occur when f is not linear in v in Chapter 12 on nonlinear mechanics and chaos.
174 Chapter 5 Oscillations
Figure 5.10 An LRC circuit.
Let us now return to mechanics and the differential equation (5.24). To solve this equation it is convenient to divide by in and then introduce two other constants. I shall rename the constant b I m as 2p,
— b
. 2P. m
(5.26)
This parameter 8, which can be called the damping constant, is simply a convenient way to characterize the strength of the damping force — as with b, large 8 corresponds to a large damping force and conversely. I shall rename the constant k I m as 6002 , that is,
coo = (5.27)
Notice that too is precisely what I was calling to in the previous two sections. I have added the subscript because, once we admit resistive forces, various other frequencies become important. From now on, I shall use the notation w o to denote the system’s natural frequency, the frequency at which it would oscillate if there were no resistive force present, as given by (5.27). With these notations, the equation of motion (5.24) for the damped oscillator becomes
Notice that both of the parameters f and too have the dimensions of inverse time, that is, frequency.
Equation (5.28) is another second-order, linear, homogeneous equation [the last was (5.4)]. Therefore, if by any means we can spot two independent 7 solutions, x i (t)
and x2 (t) say, then any solution must have the form Cix i (t) + C2x2(t). What this
7 It is about time I gave you a definition of “independent.” In general this is a little complicated, but for two functions it is easy: Two functions are independent if neither is a constant multiple of the other. Thus the two functions sin(x) and cos(x) are independent; likewise the two functions x and x 2 ; but the two functions x and 3x are not.
Section 5.4 Damped Oscillations 175
means is that we are free to play a game of inspired guessing to find ourselves two independent solutions; if by hook or by crook we can spot two solutions, then we have the general solution.
In particular, there is nothing to stop us trying to find a solution of the form
for which
and
x(t) = ert
I = re’
= r2 ert
(5.29)
Substituting into (5.28) we see that our guess (5.29) satisfies (5.28) if and only if
2 r z i6r + co2 =0 0 (5.30)
[an equation sometimes called the auxiliary equation for the differential equation
(5.28)]. The solutions of this equation are, of course, r = -p ± . ✓
,52 – (002. Thus if we define the two constants
r =— ,8 2 — o_) 2 0 (5.31)
r2 _p \A82 2
then the two functions eri t and er2t are two independent solutions of (5.28) and the general solution is
x(t) = Cleat
C2er2t
(5.32)
= e-Pt ( C ie -1132’4 t + C2e- N/ P2-
(5.33)
This solution is rather too messy to be especially illuminating, but, by examining various ranges of the damping constant /3, we can begin to see what (5.33) entails.
Undamped Oscillation
If there is no damping then the damping constant 13 is zero, the square root in the exponents of (5.33) is just i coo , and our solution reduces to
x(t) = C ie 01 + C2e -iwc’t , (5.34)
the familiar solution for the undamped harmonic oscillator.
Weak Damping
Suppose next that the damping constant f3 is small. Specifically, suppose that
fi < wo (5.35)
176 Chapter 5 Oscillations
a condition sometimes called underdamping. In this case, the square root in the exponents of (5.33) is again imaginary, and we can write
2 (0 2 = i l/(0 2 /82 = o o
where
CO1 = -1(002 — /32 • (5.36)
The parameter co l is a frequency, which is less than the natural frequency co o . In the important case of very weak damping (/3 << coo), co l is very close to coo . With this notation, the solution (5.33) becomes
x(t) = e (C i e iw’ t c2e -iwit)
(5.37)
This solution is the product of two factors: The first, e-ft , is a decaying exponen- tial, which steadily decreases toward zero. The second factor has exactly the form (5.34) of undamped oscillations, except that the natural frequency co o is replaced by the somewhat lower frequency co l . We can rewrite the second factor, as in Equation (5.11), in the form A cos(co i t — 8) and our solution becomes
x(t) = Ae – fi t cos(w i t — 8). (5.38)
This solution clearly describes simple harmonic motion of frequency co l with an exponentially decreasing amplitude Ae – fi t , as shown in Figure 5.11. The result (5.38) suggests another interpretation of the damping constant /3. Since ,8 has the dimensions of inverse time, 1//3 is a time, and we now see that it is the time in which the amplitude function Ae – fi t falls to 1/e of its initial value. Thus, at least for underdamped oscillations, /3 can be seen as the decay parameter, a measure of the rate at which the motion dies out,
(decay parameter) = /3
x(t)
[underdamped motion].
Figure 5.11 Underdamped oscillations can be thought of as sim- ple harmonic oscillations with an exponentially decreasing amplitude Ae – fit . The dashed curves are the envelopes, ±Ae – i3t .
Section 5.4 Damped Oscillations 177
The larger the more rapidly the oscillations die out, at least for the case /3 < co o that we are discussing here.
Strong Damping
Suppose instead that the damping constant 13 is large. Specifically suppose that
/3 > (00, (5.39)
a condition sometimes called overdamping. In this case, the square root in the exponents of (5.33) is real and our solution is
x(t) = C i e- (16 2-a)°) t C2e- (13+ N/ P2-4) t (5.40) Here we have two real exponential functions, both of which decrease as time goes by (since the coefficients of t in both exponents are negative). In this case, the motion is so damped that it completes no bona fide oscillations. Figure 5.12 shows a typical case in which the oscillator was given a kick from 0 at t = 0; it slid out to a maximum displacement and then slid ever more slowly back again, returning to the origin only in the limit that t co. The first term on the right of (5.40) decreases more slowly than the second, since the coefficient in its exponent is the smaller of the two. Thus the long-term motion is dominated by this first term. In particular, the rate at which the motion dies out can be characterized by the coefficient in the first exponent,
(decay parameter) = — \/132 602 [overdamped motion]. (5.41)
Careful inspection of (5.41) shows that — contrary to what one might expect — the rate of decay of overdamped motion gets smaller if the damping constant is made bigger. (See Problem 5.20.)
0
Figure 5.12 Overdamped motion in which the oscillator is kicked from the origin at t = 0. It moves out to a maximum displacement and then moves back toward 0 asymptotically as t —> oo.
Critical Damping
The boundary between underdamping and overdamping is called critical damping and occurs when the damping constant is equal to the natural frequency, /3 = coo . This
178 Chapter 5 Oscillations
case has some interesting features, especially from a mathematical point of view. When /3 = wo the two solutions that we found in (5.33) are the same solution, namely
x(t) = e-)51′ (5.42)
[This happened because the two solutions of the auxiliary equation (5.30) happen to coincide when /3 = coo l This is the one case where our inspired guess, to seek a solution of the form x (t) = ert , fails to find us two solutions of the equation of motion, and we have to find a second solution by some other method. Fortunately, in this case, it is not hard to spot a second solution: As you can easily check, the function
x(t) = to- i3t (5.43)
is also a solution of the equation of motion (5.28) in the special case that /3 = co,. (See Problems 5.21 and 5.24.) Thus the general solution for the case of critical damping is
x(t) = Cl e- fi t + C2 t e –
fi t . (5.44)
Notice that both terms contain the same exponential factor e-Pt . Since this factor is what dominates the decay of the oscillations as t Do, we can say that both terms decay at about the same rate, with decay parameter
(decay parameter) = /3 = co o [critical damping].
It is interesting to compare the rates at which the various types of damped oscil- lation die out. We have seen that in each case, this rate is determined by a “decay parameter,” which is just the coefficient of t in the exponent of the dominant expo- nential factor in x(t). Our findings can be summarized as follows:
damping /3 decay parameter
none /3 = 0 0 under /3 < wo 18 critical /3 = 0)0 ,8 over /3 > (D. 13 — //32 — (02 o
Figure 5.13 is a plot of the decay parameter as a function of /3 and shows clearly that the motion dies out most quickly when /3 = w o ; that is, when the damping is critical. There are situations where one wants any oscillations to die out as quickly as possible. For example, one wants the needle of an analog meter (a voltmeter or pressure gauge, for instance) to settle down rapidly on the correct reading. Similarly, in a car, one wants the oscillations caused by a bumpy road to decay quickly. In such cases one must arrange for the oscillations to be damped (by the shock absorbers in a car), and for the quickest results the damping should be reasonably close to critical.
Section 5.5 Driven Damped Oscillations 179
co,
Figure 5.13 The decay parameter for damped oscillations as a function of the damping constant /3. The decay parameter is biggest, and the motion dies out most quickly, for critical damping, with p = wo .
5.5 Driven Damped Oscillations
Any natural oscillator, left to itself, eventually comes to rest, as the inevitable damping forces drain its energy. Thus if one wants the oscillations to continue, one must arrange for some external “driving” force to maintain them. For example, the motion of the pendulum in a grandfather clock is driven by periodic pushes caused by the clock’s weights; the motion of a young child on a swing is maintained by periodic pushes from a parent. If we denote the external driving force by F(t) and if we assume as before that the damping force has the form —by, then the net force on the oscillator is —by — kx + F(t) and the equation of motion can be written as
mz + bx + kx = F(t). (5.45)
Like its counterpart for undriven oscillations, this differential equation crops up in several other areas of physics. A prominent example is the LRC circuit of Figure 5.10. If we want the oscillating current in that circuit to persist, we must apply a driving EMF, E(t), in which case the equation of motion for the circuit becomes
1,4 + + — 1
q = E(t)
(5.46)
in perfect correspondence with (5.45). As before, we can tidy Equation (5.45) if we divide the equation by rn and replace
blm by 2/3 and klm by w02 . In addition, I shall denote F(t)/m by
F(t) f (t) ,
m
the force per unit mass. With this notation, (5.45) becomes
(5.47)
180 Chapter 5 Oscillations
Linear Differential Operators
Before we discuss how to solve this equation, I would like to streamline our notation. It turns out that it is very helpful to think of the left side of (5.48) as the result of a certain operator acting on the function x(t). Specifically, we define the differential operator
2 D = —
d
dt2 213—
dt ° w 2 . (5.49)
The meaning of this definition is simply that when D acts on x it gives the left side of (5.48):
Dx ± 2/3X + coo2x.
This definition is obviously a notational convenience — the equation (5.48) becomes just
Dx = f (5.50)
— but it is much more: The notion of an operator like (5.49) proves to be a powerful mathematical tool, with applications throughout physics. For the moment the impor- tant thing is that the operator is linear: We know from elementary calculus that the derivative of ax (where a is a constant) is just ax and that the derivative of x 1 + x2 is just x1 + 12. Since this also applies to second derivatives, it applies to the operator D:
D(ax) = aDx and D(x 1 + x2) = Dx 1 + Dx 2 .
(Make sure you understand what these two equations mean.) We can combine these into a single equation:
D(ax i + bx2) = aDx] bDx 2 (5.51)
for any two constants a and b and any two functions x l (t) and x2(t). Any operator that satisfies this equation is called a linear operator.
We have actually used the property (5.51) of linear operators before. The equation (5.28) for a damped oscillator (not driven) can be written as
Dx = 0. (5.52)
The superposition principle asserts that if x i and x2 are solutions of this equation, then so is ax 1 + bx2 for any constants a and b. In our new operator notation, the proof is very simple: We are given that Dx 1 = 0 and Dx 2 = 0, and using (5.51) it immediately follows that
D(ax i + bx2) = aDxi bDx 2 = 0 + = 0;
that is, ax i + bx2 is also a solution. The equation (5.52), Dx = 0, for the undriven oscillator is called a homogeneous
equation, since every term involves either x or one of its derivatives exactly once. The equation (5.50), Dx = f , is called an inhomogeneous equation, since it contains the
Section 5.5 Driven Damped Oscillations 181
inhomogeneous term f , which does not involve x at all. Our job now is to solve this inhomogeneous equation.
Particular and Homogeneous Solutions
Using our new operator notation, we can find the general solution of Equation (5.48) surprisingly easily; in fact, we’ve already done most of the work. Suppose first that we have somehow spotted a solution; that is, we’ve found a function xp (t) that satisfies
Dx P = f • (5.53)
We call this function x p (t) a particular solution of the equation, and the subscript “p” stands for “particular.” Next let us consider for a moment the homogeneous equation Dx = 0 and suppose we have a solution xh (t), satisfying
Dx h = 0. (5.54)
We’ll call this function a homogeneous solution. 8 We already know all about the solutions of the homogeneous equation, and we know from (5.32) that x h must have the form
xh (t) = Cier’t C2er2t , (5.55)
where both exponentials die out as t co. We’re now ready to prove the crucial result. First, if x p is a particular solution
satisfying (5.53), then x p + xh is another solution, for
D(xp x h) = Dxp Dxh = f +0= f.
Given the one particular solution xp , this gives us a large number of other solutions xp + xh . And we have in fact found all the solutions, since the function x h contains two arbitrary constants, and we know that the general solution of any second-order equation contains exactly two arbitrary constants. Therefore x p + xh , with xh given by (5.55) is the general solution.
This result means that all we have to do is somehow to find a single particular solution xp (t) of the equation of motion (5.48), and we shall have every solution in the form x (t) = xp (t) x h (t).
Complex Solutions for a Sinusoidal Driving Force
I shall now specialize to the case that the driving force f (t) is a sinusoidal function of time,
f (t) = fo cos(wt), (5.56)
8 Another common name is the complementary function. This has the disadvantage that it’s hard
to remember which is “particular” and which “complementary.”
182 Chapter 5 Oscillations
where fo denotes the amplitude of the driving force [actually the amplitude divided by the oscillator’s mass, since f (t) = F(t)/ m] and co is the angular frequency of the driving force. (Be careful to distinguish the driving frequency w from the natural
frequency co o of the oscillator. These are entirely independent frequencies, although we shall see that the oscillator responds most when w w o.) The driving force for many driven oscillators is at least approximately sinusoidal. For example, even the parent pushing the child on a swing can be crudely approximated by (5.56); the driving EMF induced in your radio’s circuits by a broadcast signal is almost perfectly of this form. Probably the chief importance of sinusoidal driving forces is that, according to Fourier’s theorem, 9 essentially any driving force can be built up as a series of sinusoidal forces.
Let us therefore assume the driving force is given by (5.56), so that the equation of motion (5.48) takes the form
+ 213.i + coo2x = fo cos(cot). (5.57)
Solving this equation is greatly simplified by the following trick: For any solution of (5.57), there must be a solution of the same equation, but with the cosine on the right replaced by a sine function. (After all, these two differ only by a shift in the origin of time.) Accordingly, there must also be a function y(t) that satisfies
± 2,85, + wo2y = fo sin(cot). (5.58)
Suppose now we define the complex function
z(t) = x(t) iy(t), (5.59)
with x(t) as its real part and y(t) as its imaginary part. If we multiply (5.58) by i and add it to (5.57), we find that
+2& coo2z foeiwt . (5.60)
Although it may not yet look it, Equation (5.60) is a tremendous advance. Because of the simple properties of the exponential function, (5.60) is much easier to solve than either (5.57) or (5.58). And as soon as we find a solution z(t) of (5.60), we have only to take its real part to have a solution of the equation (5.57) whose solutions we actually want.
In seeking a solution of (5.60), we are obviously free to try any function we like. In particular, let’s see if there is a solution of the form
z(t) = Ceiwt , (5.61)
where C is an as yet undetermined constant. If we substitute this guess into the left side of (5.60), we get
(-602 + 20a) coo2)Ceiwt = Lei t
9 Named for the French mathematician, the Baron Jean Baptiste Joseph Fourier, 1768-1830. See Sections 5.7-5.9.
or
2 A2 = (0)2 (02
f. )2 ± 4162(02
. (5.64)
Section 5.5 Driven Damped Oscillations 183
In other words, our guess (5.61) is a solution of (5.60) if and only if
C = fo (5.62) CO – CO2 ± 2i,8w 0
and we have succeeded in finding a particular solution of the equation of motion. Before we take the real part of z(t) = Ce” t , it is convenient to write the complex
coefficient C in the form
C = Ae —i6 (5.63)
where A and 8 are real. [Any complex number can be written in this form; the particular notation is chosen to match (5.14).] To identify A and 6, we must compare (5.62) and (5.63). First, taking the absolute value squared of both equations we find that
A 2 = cc* = 1’0 fo (002 w2 + 20(0 (002 — (02 — 2i p a)
(Make sure you understand this derivation. See Problem 5.35 for some guidance.) We are going to see in a moment that A is just the amplitude of the oscillations caused by the driving force f (t). Thus the result (5.64) is the most important result of this discussion. It shows how the amplitude of oscillations depends on the various parameters. In particular, we see that the amplitude is biggest when co ° ti co, so that the denominator is small; in other words, the oscillator responds best when driven at a frequency w that is close to its natural frequency co., as you would probably have guessed.
Before we continue to discuss the properties of our solution, we need to identify the phase angle 8 in (5.63). Comparing (5.63) and (5.62) and rearranging, we see that
foe is = A(0)02 w2 + 200.
Since fo and A are real, this says that the phase angle 8 is the same as the phase angle of the complex number (0)02 — w2) 2ifico. This relationship is illustrated in Figure 5.14, from which we conclude that
2136° 8 = arctan (0 2 (02 (5.65)
Our quest for a particular solution is now complete. The “fictitious” complex solution introduced in (5.59) is
z(t) = = Aei(‘-6)
and the real part of this is the solution we are seeking,
x(t) = A cos(cot — 8)
(5.66)
where the real constants A and 8 are given by (5.64) and (5.65).
184 Chapter 5 Oscillations
2
Figure 5.14 The phase angle 8 is the angle of this triangle.
The solution (5.66) is just one particular solution of the equation of motion. The general solution is found by adding any solution of the corresponding homogeneous equation, as given by (5.55); that is, the general solution is
x(t) = A cos(wt — S) + C2e’ 2t . (5.67)
Because the two extra terms in this general solution both die out exponentially as time passes, they are called transients. They depend on the initial conditions of the problem but are eventually irrelevant: The long-term behavior of our solution is dominated by the cosine term. Thus the particular solution (5.66) is the solution with which we are usually concerned, and we shall explore its properties in the next section.
Before we discuss an example of the motion (5.67), it is important that you be very clear as to the type of system to which (5.67) applies, namely, any oscillator for which both the restoring force (—kx) and the resistive force (—b1) are linear — a driven, damped linear oscillator, whose equation of motion (5.45) is a linear differential equation. Because nonlinear differential equations are often hard to solve, most mechanics texts have until recently focussed on linear equations. This created the false impression that linear equations were in some sense the norm, and that the solution (5.67) was the only (or, at least, the only important) way for an oscillator to behave. As we shall see in Chapter 12 on nonlinear mechanics and chaos, an oscillator whose equation of motion is nonlinear can behave in ways that are astonishingly different from (5.67). One important reason for studying the linear oscillator here is to give you a backdrop against which to study the nonlinear oscillator later on.
The details of the motion (5.67) depend on the strength of the damping param- eter /3. To be specific, let us assume that our oscillator is weakly damped, with /3 less than the natural frequency co o (underdamping). In this case, we know that the two transient terms of (5.67) can be rewritten as in (5.38), so that
You need to think very carefully about this potentially confusing formula. The second term on the right is the homogeneous or transient term, and I have added the subscript “tr” to distinguish the constants A tr and S t, from the A and S of the first term. The two constants A tr and St, are arbitrary constants; (5.68) is a possible motion of our system for any values of A tr and kr , which are determined by the initial conditions.
Section 5.5 Driven Damped Oscillations 185
The factor e — fi t makes clear that this transient term decays exponentially and is indeed irrelevant to the long-term behavior. As it decays, the transient term oscillates with the angular frequency co l of the undriven (but still damped) oscillator, as in (5.36). The first term is our particular solution and its two constants A and 8 are certainly not arbitrary; they are determined by (5.64) and (5.65) in terms of the parameters of the system. This term oscillates with the frequency w of the driving force and with unchanging amplitude, for as long as the driving force is maintained.
EXAMPLE 5.3 Graphing a Driven Damped Linear Oscillator
Make a plot of x (t) as given by (5.68) for a driven damped linear oscillator which is released from rest at the origin at time t = 0, with the following parameters: Drive freqency w = 27. , natural frequency coo = 5co, decay constant /3 = co o/20, and driving amplitude fo = 1000. Show the first five drive cycles.
The choice of drive frequency equal to 27 means that the drive period is T = 27r/C0 = 1; this means simply that we have chosen to measure time in units of the drive period — often a convenient choice. That w o = 5w means that our oscillator has a natural frequency five times the drive frequency; this will let us distinguish easily between the two on a graph. That ,f3 = wo/20 means that the oscillator is rather weakly damped. Finally the choice of fo = 1000 is just a choice of our unit of force; the reason for this odd-seeming choice is that it leads to a conveniently sized amplitude of oscillation (namely A close to 1).
Our first task is to determine the various constants in (5.68) in terms of the given parameters. In fact this is easier if we rewrite the transient term of (5.68) in the “cosine plus sine” form, so that
x(t) = A cos(cot — 6) + e— $t [B i cos(w it) + B2 sin(w it)]. (5.69)
The constants A and 8 are determined by (5.64) and (5.65), which, for the given parmeters, yield
A = 1.06 and 8 = 0.0208.
The frequency co l is
(0,= – p2 = 9.987m,
which is very close to wo, as we would expect for a weakly damped oscillator. To find B 1 and B2, we must equate x(0) as given by (5.69) to its given initial value xo , and likewise the corresponding expression for l o to the initial value vo . This gives two simultaneous equations for B 1 and B2, which are easily solved (Problem 5.33) to give
B 1 = x o — A cos 8 and B2 = 1 (210 — (OA sin 8 + pB 1) (Di
or, with the numbers, including the initial conditions x o = vo = 0,
B 1 = —1.05 and B2 = —0.0572.
(5.70)
186 Chapter 5 Oscillations
(a)
(b)
Figure 5.15 The response of a damped, linear oscillator to a si- nusoidal driving force, with the time t shown in units of the drive period. (a) The driving force is a pure cosine as a function of time. (b) The resulting motion for the initial conditons x o = vo = 0. For the first two or three drive cycles, the transient motion is clearly vis- ible, but after that only the long-term motion remains, oscillating sinusoidally at exactly the drive frequency. As explained in the text, the sinusoidal motion after t ti 3 is called an attractor.
Putting all of these numbers into (5.69), we can now plot the motion, as shown in Figure 5.15, where part (a) shows the driving force f(t) = fo cos(wt) and part (b) the resulting motion x (t) of the oscillator. The driving force is, of course, perfectly sinusoidal with period 1. The resulting motion is much more interesting. After about three drive cycles (t 3), the motion is indistinguishable from a pure cosine, oscillating at exactly the drive frequency; that is, the transients have died out and only the long-term motion remains. Before t 3, however, the effects of the transients are clearly visible. Since they oscillate at the faster natural frequency wo , they show up as a rapid succession of bumps and dips. In fact, you can easily see that there are five such bumps within the first drive cycle, indicating that wo = 5w.
Because the transient motion depends on the initial values xo and vo , different values of xo and vo would lead to quite different initial motion. (See Problem 5.36.) After a short time however (a couple of cycles in this example), the initial differences disappear and the motion settles down to the same sinusoidal motion of the particular solution (5.66), irrespective of the initial conditions. For this reason, the motion of (5.66) is sometimes called an attractor — the motions corresponding to several different initial conditions are “attracted” to the particular motion (5.66). For the linear oscillator discussed here, there is a unique attractor (for a given driving force): Every possible motion of the system, whatever its initial conditions, is attracted to the same motion (5.66). We shall see in Chapter 12 that for nonlinear oscillators there can be several different attractors and that for some values of the parameters the motion of
Section 5.6 Resonance 187
an attractor can be far more complicated than simple harmonic oscillation at the drive
frequency. The amplitude and phase of the attractor seen in Figure 5.15(b) depend on the
parameters of the driving force (but not, of course, on the initial conditions). The dependence of the amplitude and phase on these parameters is the subject of the next
section.
5.6 Resonance
In the previous section we considered a damped oscillator that is being driven by a sinusoidal driving force (actually force divided by mass) f (t) = f0 cos(wt) with
angular frequency co. We saw that, apart from transient motions that die out quickly, the system’s response is to oscillate sinusoidally at the same frequency, w:
x(t) = A cos(cot — 6) ,
with amplitude A given by (5.64),
A2 = fo2 (0) 2 0)2)2 4fi2w2
(5.71)
and phase shift 8 given by (5.65). The most obvious feature of (5.71) is that the amplitude A of the response is
proportional to the amplitude of the driving force, A a f0 , a result you might have guessed. More interesting is the dependence of A on the frequencies co. (the natural frequency of the oscillator) and w (the frequency of the driver), and on the damping constant /3. The most interesting case is when the damping constant /3 is very small, and this is the case I shall discuss. With p small, the second term in the denominator of (5.71) is small. If w0 and w are very different, then the first term in the denominator of (5.71) is large, and the amplitude of the driven oscillations is small. On the other hand, if wo is very close to w, both terms in the denominator are small, and the amplitude A is large. This means that if we vary either w 0 or w, there can be quite dramatic changes in the amplitude of the oscillator’s motion. This is illustrated in Figure 5.16, which shows A 2 as a function of w 0 with w fixed, for a rather weakly damped system (/3 = 0.1w). (Note that, because the energy of the system is proportional to A 2 , it is usual to make plots of A 2 rather than A.)
Although the behavior illustrated in Figure 5.16 is startlingly dramatic, the qualita- tive features are what you might have expected. Left to its own devices, the oscillator vibrates at its natural frequency w 0 (or at the slightly lower frequency co l if we allow for the damping). If we try to force it to vibrate at a frequency w, then for values of w close to wo the oscillator responds very well, but if w is far from w0 , it hardly re- sponds at all. We refer to this phenomenon — the dramatically greater response of an oscillator when driven at the right frequency — as resonance.
An everyday application of resonance is the reception of radio signals by the LRC circuit in your radio. As we saw, the equation of motion of an LRC circuit is exactly
188 Chapter 5 Oscillations
0
Figure 5.16 The amplitude squared, A 2 , of a driven oscillator,
shown as a function of the natural frequency coo , with the driving frequency co fixed. The response is dramatically largest when coo
and co are close.
the same as that of a driven oscillator, and LRC circuits show the same phenomenon of resonance. When you tune your radio to receive station KVOD at 90.1 MHz, you are adjusting an LRC circuit in the radio so that its natural frequency is 90.1 MHz. The many radio stations in your neighborhood are all sending out signals, each at its own frequency and each inducing a tiny EMF in the circuit of your radio. But only the signal with the right frequency actually succeeds in driving an appreciable current, which mimics the signal sent out by your favorite KVOD and reproduces its broadcast sounds.
An example of a mechanical resonance of the kind discussed here is the behavior of a car driving on a “washboard” road that has worn into a series of regularly spaced bumps. Each time a wheel crosses a bump, it is given an upward impulse, and the frequency of these impulses depends on the car’s speed. There is one speed at which the frequency of these impulses equals the natural frequency of the wheels’ vibration on the springs l° and the wheels resonate, causing an uncomfortable ride. If the car drives slower or faster than this speed, it goes “off resonance” and the ride is much smoother.
Another example occurs when a platoon of soldiers marches across a bridge. A bridge, like almost any mechanical system, has certain natural frequencies, and if the soldiers happened to march with a frequency equal to one of these natural frequencies, the bridge could conceivably resonate sufficiently violently to break. For this reason, soldiers break step when marching across a bridge.
The details of the resonance phenomenon are a bit complicated. For example, the exact location of the maximum response depends on whether we vary co o with co fixed or vice versa. The amplitude A is a maximum when the denominator,
denominator = (0)02 — 0)2) 2 4p20)2 (5.72)
10 It is the wheels (plus axles) that exhibit the resonant oscillations; the much heavier body of the car is relatively unaffected.
Section 5.6 Resonance 189
of (5.71) is a minimum. If we vary wo with w fixed (as when you tune a radio to pick up your favorite station) this minimum obviously occurs when wo = w, making the first term zero. On the other hand, if we vary w with we, fixed (which is what happens in many applications), then the second term in (5.72) also varies and a straightforward differentiation shows that the maximum occurs when
co = (02 = ‘0)02 2,82 . (5.73)
However, when ,8 << wo (as is usually the most interesting case), the difference between (5.73) and w = wo is negligible.
We have met so many different frequencies in this chapter that it may be worth pausing to review them. First there is w o the natural frequency of the oscillator (in the absence of any damping). Next when we added in a little damping, we found
that the same system oscillated sinusoidally with frequency w 1 = 1/w02 — ,8 2 under an
exponentially decaying envelope. Then we added a driving force with frequency w, which can, in principle, take on any value independently of the previous two. However, the response of the driven oscillator is biggest when w w o ; specifically, if we vary w with wo fixed, the maximum response comes when w = w 2 as defined by (5.73). To summarize:
wo = ./k/m = natural frequency of undamped oscillator
0) 1 = 2 ,82 = frequency of damped oscillator
w = frequency of driving force
(02 = \/(0.2 — 2 ,8 2 p = value of w at which response is maximum.
In any case, the maximum amplitude of the driven oscillations is found by putting wo ti w in (5.71), to give
fo A max • 2Pwo
(5.74)
This shows that smaller values of the damping constant /3 lead to larger values of the maximum amplitude of oscillation, as illustrated in Figure 5.17.”
Width of the Resonance; the Q Factor
You can see clearly from Figure 5.17 that if we make the damping constant p smaller, the resonance peak not only gets higher, but also gets narrower. We can make this idea more precise by defining the width (or full width at half maximum or FWHM) as the interval between the two points where A 2 is equal to half its maximum height. It
11 In this figure I chose to plot A 2 against w with wo fixed, rather than the other way around as in Figure 5.16. Note that the curves have very similar shapes either way.
190 Chapter 5 Oscillations
Figure 5.17 The amplitude for driven oscillations as a function of the driving frequency (t) for three different values of the damp- ing constant p. Note how as ,8 decreases the resonance peak gets higher and sharper.
is a simple exercise (Problem 5.41) to show that the two half-maximum points are at co coo + ,B, as in Figure 5.18. Thus the full width at half maximum is
FWHM 2p (5.75)
or, equivalently, the half width at half maximum is
HWHM (5.76)
The sharpness of the resonance peak is indicated by the ratio of its width 2fi to its position, coo . For many purposes, we want a very sharp resonance, so it is common practice to define a quality factor Q as the reciprocal of this ratio:
(5.77)
coo — i 3/ \
wo+
Figure 5.18 The full width at half maximum (FWHM) is the distance between the points where A 2 is half its maximum value.
Section 5.6 Resonance 191
A large Q indicates a narrow resonance, and vice versa. For example, clocks depend on the resonance in an oscillator (a pendulum, or a quartz crystal, for instance) to regulate the mechanism to move with very well-defined frequency. This requires that the width 2/3 be very small compared to the natural frequency co o . In other words, a good clock needs a high Q. The Q for a typical pendulum may be around 100; that for a quartz crystal around 10,000. Therefore quartz watches keep much better time than a typical grandfather clock. I2
There is another way to look at the quality factor Q. We saw that, in the absence of any driving force, the oscillations die out in a time of order 11p,
(decay time) = 1//3.
(This was actually the time for the amplitude to drop to 1/e of its initial value.) The period of a single oscillation is, of course,
period = 27r/coo .
(Remember we’re assuming that ,8 << wo, so we don’t need to distinguish between coo and coi .) Thus we can rewrite the definition of Q as
wo 1//3 decay time — = = 2/3 27r/co. period
(5.78)
The ratio on the right is just the number of periods in the decay time. Thus, the quality factor Q is 7/- times the number of cycles our oscillator makes in one decay time. 13
The Phase at Resonance
The phase difference 6 by which the oscillator’s motion lags behind the driving force is given by (5.65) as
8 2/3w = arctan CO 2 – 0)2
0
(5.79)
Let us follow this phase as we vary w, starting well below a narrow resonance small). With w << coo , (5.79) implies that 8 is very small; that is, while w << coo the oscillations are almost perfectly in step with the driving force. (This was the case in Figure 5.15.) As w is increased toward co o , so 6 slowly increases. At resonance, where w = coo , the argument of the arctangent in (5.79) is infinite, so 8 = 7r/2 and the oscillations are 90° behind the driving force. Once co > co o, the argument of the
12 Actually, both quartz watches and grandfather clocks keep much better time than this simple discussion would suggest. A good chronometer keeps the frequency very close to the center of the resonance. Thus the variability of the frequency is actually much smaller than the width of the resonance. Nevertheless, the stated conclusion is correct.
13 Yet another definition (and perhaps the most fundamental) is that Q = 27r times the ratio of the energy stored in the oscillator to the energy dissipated in one cycle. See Problem 5.44.
0
192 Chapter 5 Oscillations
6
= 0.03 coo
0.3w,
7F –
7/2 –
Figure 5.19 The phase shift S increases from 0 through 7/2
to Jr as the driving frequency co passes through resonance. The narrower the resonance, the more suddenly this increase occurs.
The solid curve is for a relatively narrow resonance (/3 = 0.03(00 or Q = 16.7), and the dashed curve is for a wider resonance
(0= 0.3(00 or Q = 1.67).
arctangent is negative and approaches 0 as co increases; thus 6 increases beyond 7/2 and eventually approaches 7. In particular, once co >> co., the oscillations are almost perfectly out of step with the driving force. All of this behavior is illustrated for two different values of )6 in Figure 5.19. Notice, in particular, that the narrower the resonance, the more quickly 6 jumps from 0 to 7.
In the resonances of classical mechanics, the behavior of the phase (as in Fig- ure 5.19) is usually less important than that of the amplitude (as in Figure 5.18). 14 In atomic and nuclear collisions, the phase shift is often the quantity of primary inter- est. Such collisions are governed by quantum mechanics, but there is a corresponding phenomenon of resonance. A beam of neutrons, for example, can “drive” a target nucleus. When the energy of the beam equals a resonant energy of the system (in quantum mechanics energy plays the role of frequency) a resonance occurs and the phase shift increases rapidly from 0 to 7.
5.7 Fourier Series*
* Fourier series have broad application in almost every area of modern physics. Nevertheless, we shall not be using them again until Chapter 16. Thus, if you are pressed for time you could omit the last three sections of this chapter on a first reading.
In the last two sections, we have discussed an oscillator that is driven by a sinusoidal driving force f (t) = fo cos(cot). There are two main reasons for the importance of
14 The behavior of 8 can, nevertheless, be observed. Make a simple pendulum from a piece of string and a metal nut, and drive it by holding it at the top and moving your hand from side to side. The most obvious thing is that you will be most successful at driving it when your frequency equals the natural frequency, but you can also see that when you drive more slowly the pendulum moves in step with your hand, whereas when you move more quickly the pendulum moves oppositely to your hand.
Section 5.7 Fourier Series* 193
f(t)
T-p
t
(a) (b)
Figure 5.20 Two examples of periodic functions with period T. (a) A rectangular pulse, which could represent a hammer hitting a nail with a constant force at
intervals of r, or a digital signal in a telephone line. (b) A smooth periodic signal, which could be the pressure variation of a musical instrument.
sinusoidal driving forces: The first is simply that there are many important systems in which the driving force is sinusoidal — the electrical circuit in a radio is a good example. The second is somewhat subtler. It turns out that any periodic driving force can be built up from sinusoidal forces using the powerful technique of Fourier series. Thus, in a sense that I shall try to describe, by solving the motion with a sinusoidal driver we have already solved the motion with any periodic driver. Before we can appreciate this wonderful result, we need to review some aspects of Fourier series. In this section I sketch the needed properties of Fourier series; 15 in the next we can apply them to the driven oscillator.
Let us consider a function f (t) that is periodic with period r; that is, the function repeats itself every time t advances by the period 1″:
f (t + -r) = f (t)
whatever the value of t. We can describe a function with this property as being r – periodic. A simple example of a r -periodic function would be the force exerted on a nail by a hammer that is being swung at intervals of r, as sketched in Figure 5.20(a). Another could be the pressure exerted on your ear drum by a note played by a musical instrument, as sketched in 5.20(b). It is easy to think up many more periodic functions. In particular, there are lots of sinusoidal functions that are periodic with any given period: The functions
cos(27rt/r), cos(47rt/r), cos(67rt/r), • • • (5.80)
are all r -periodic, as are the corresponding sine functions. (If t is increased by the amount T, each of these functions returns to its original value — see Figure 5.21.) We can write these sinusoidal functions a little more compactly if we introduce the angular frequency w = 27r/r, , in which case all of the functions of (5.80) and the corresponding sines can be written as
cos(ncot) and sin(ncot) [n = 0, 1, 2, • • •]. (5.81)
15 As usual, I shall try to describe all of the theory that we shall be needing. For more details, see, for example, Mathematical Methods in the Physical Sciences by Mary Boas (Wiley, 1983), Ch. 7.
194 Chapter 5 Oscillations
f(t) f(t)
cos(277117) cos (47a/r)
Figure 5.21 Any function of the form cos(2n7rt/r) (or the corresponding sine) is periodic with period r if n is an integer. Notice that cos(4rrt/r) also has the smaller period r/2, but this doesn’t change the fact that it has period r as well.
(If n = 0 the cosine function is just the constant 1 — which is certainly periodic — while the sine is 0 and not at all interesting.)
That the sine and cosine functions (5.81) are all t -periodic is reasonably obvious. (Be sure you can see this.) What is truly amazing is that, in a sense, these sine and cosine functions define all possible r -periodic functions: In 1807 the French math- ematician Jean Baptiste Fourier (1768-1830) realized that every r -periodic function can be written as a linear combination of the sines and cosines of (5.81). That is, if f (t) is any 16 periodic function with period T then it can be expressed as the sum
where the constants an and bn depend on the function f (t). This extraordinarily useful result is called Fourier’s theorem, and the sum (5.82) is called the Fourier series for f (t)
It is not hard to see why Fourier’s theorem met with considerable surprise, and even skepticism, when he first published it. It claims that a discontinous function, such as the rectangular pulse of Figure 5.20(a), can be built up with sine and cosine functions that are continuous and perfectly smooth. Surprising or not, this turns out to be true, as we shall see by example shortly. Perhaps even more surprising, it is often the case that one gets an excellent approximation by retaining just the first few terms of a Fourier series. Thus, instead of having to handle a fairly disagreeable and possibly discontinuous function, we have only to handle a reasonably small number of sines and cosines. Before we discuss the application of Fourier’s theorem to the driven oscillator, we need to look at a few properties of Fourier series.
The proof of Fourier’s theorem is difficult — indeed it was many years after Fourier’s discovery before a satisfactory proof was found — and I shall simply ask you to accept it. However, once the result is accepted it is easy to learn to use it. In
16 As always with theorems of this kind, there are certain restrictions on the “reasonableness” of the function f (t), but certainly Fourier’s theorem is valid for all of the functions we shall have occasion to use.
r/2 ao = — f (t) dt
fr/2
1Ar/2 f max Az = — f fmax dt –
-Ar/2
Section 5.7 Fourier Series* 195
particular, for any given periodic function f (t) it is easy to find the coefficients an and bn . Problem 5.48 gives you the opportunity to show that these coefficients are given by
and
(asjti(i; 44) di.` .. (5 .j34)
Unfortunately the coefficients for n = 0 require separate attention. Since the term sin ncot in (5.82) is identically zero for n = 0, the coefficient b0 is irrelevant and we can simply define it to be zero. It is very easy to show (Problem 5.46) that
Armed with these formulas for the Fourier coefficients, it is easy to find the Fourier series for any given periodic function. In the following example, we do this for the rectangular pulses of Figure 5.20(a).
1
EXAMPLE 5.4 Fourier Series for the Rectangular Pulse
Find the Fourier series for the periodic rectangular pulse f (t) shown in Figure 5.22 in terms of the period z, the pulse height fmax’ and the duration of the pulse A . Using the values r = 1, fmax = 1, and AZ = 0.25, plot f (t), as well as the sum of the first three terms of its Fourier series, and the sum of the firs _ t eleven terms.
Our first task is to calculate the Fourier coefficients an and bn for the given function. First according to (5.85) the constant term ac, is
(5.86)
Chapter 5 Oscillations
T -> fmax —4-
—112
7;2 AT
AT/2
Figure 5.22 A periodic rectangular pulse. The period is t, the duration of the pulse is AT, and the pulse height is fmax .
where the change in the limits of integration was allowed because the integrand f (t) is zero outside ± A T/2. Next, according to (5.83), all the other a coefficients (n > 1) are given by
2 r/2 an = — f f (t) cos(nwt) dt
—r/2
2fmax At/2 cos(nwt) dt
t J—At/2
4 fmax Ar/2 (
2t 2f n Ot cos dt = sin
T 0
7/-n
n -c ) (5.87)
Notice that in passing from the second to the third line, I used a trick that is often useful in evaluating Fourier coefficients. The integrand on the second line, cos(ncot), is an even function; that is, it has the same value at any point t as at —t. Therefore we could replace any integral from —T to T by twice the integral from 0 to T.
Finally the b coefficients are all exactly zero, for if you examine the integral (5.84) you will see that (in this case) the integrand is an odd function; that is, its value at any point t is the negative of its value at —t. [Moving from t to —t leaves f (t) unchanged, but reverses the sign of sin(nwt).] Therefore any integral from —T to T is zero, since the left half exactly cancels the right.
The required Fourier series is therefore
00
f (t) = ao an cos(nwt) (5.88) n=1
with the constant term a 0 given by (5.86) and all the remaining a coefficients (n > 1) by (5.87). If we put in the given numbers, these coefficients can all be evaluated and the resulting Fourier series is
f (t) = f max [o .25 + 0.24 cos(27t) + 0.23 cos(47tt) + 0.20 cos(67t)
+ 0.16 cos(87t) + 0.12 cos(107rt) + 0.08 cos(127rt) + • • • ] (5.89)
Section 5.8 Fourier Series Solution for the Driven Oscillator* 197
f f(t) f f(t)
(a) 3 terms (b) 11 terms
Figure 5.23 (a) The sum of the first three terms of the Fourier series for the rectangular pulse of Figure 5.22. (b) The sum of the first 11 terms.
The practical value of a Fourier series is usually greatest if the series con- verges rapidly, so that we get a reliable approximation by retaining just the first few terms of the series. Figure 5.23(a) shows the sum of the first three terms of the series (5.89) and the rectangular pulse itself. As you might expect, with just three smooth terms we do not get a sensationally accurate approximation to the original discontinuous function. Nevertheless, the three terms do a remarkable job of imitating the general shape. By the time we have included 11 terms, as in Figure 5.23(b), the fit is amazing. 17 In the next section, we shall use the method of Fourier series to solve for the motion of an oscillator driven by the periodic pulses of this example. We shall find the solution as a Fourier series, which con- verges so quickly that just the first 3 or 4 terms tell us most of what is interesting to know.
5.8 Fourier Series Solution for the Driven Oscillator*
* This section contains a beautiful application of the method of Fourier series. Important as it is to understand this method, you can nevertheless omit the section without loss of continuity.
In this section we shall combine our knowledge of Fourier series (Section 5.7) with our solution of the sinusoidally driven oscillator (Section 5.5) to solve for the motion of an oscillator that is driven by an arbitrary periodic driving force. To see how this works, let us return to the equation of motion (5.48)
+ 2,6.Z + coo2x = f
where x = x (t) is the position of the oscillator, /3 is the damping constant, w o is the natural frequency, and f = f (t) is any periodic driving force (actually force/mass) with period r. As before it is convenient to rewrite this in the compact form
Dx = f
17 Notice, however, that the Fourier series still has a little difficulty in the immediate neigh- borhood of the discontinuities in f (t). This tendency for the Fourier series to overshoot near a discontinuity is called the Gibbs phenomenon.
198 Chapter 5 Oscillations
where D stands for the linear differential operator
2 „ d 2 D = z p — co .
dt2 dt °
The use of Fourier series to solve this problem hinges on the following observation: Suppose that the force f (t) is the sum of two forces, f (t) = f1 (t) f2 (t), for each of which we have already solved the equation of motion. That is, we already know functions x i (t) and x2 (t) that satisfy
Dx t = fi and Dx 2 = f2 .
Then the solution 18 to the problem of interest is just the sum x(t) = x l(t) x 2 (t), as we can easily show:
Dx = D(x l + x2) = Dxi + Dx2 = + f2 = f
where the crucial second step is valid because D is linear. This argument would work equally well however many terms were in the sum for f (t), so we have the conclusion: If the driving force f (t) can be expressed as the sum of any number of terms
f (t) = E fn(t) and if we know the solutions x,, (t) for each of the individual forces f,i (t), then the solution for the total driving force f (t) is just the sum
x(t) = xn (t). n
This result is ideally suited for use in combination with Fourier’s theorem. Any periodic driving force f (t) can be expanded in a Fourier series of sines and cosines, and we already know the solutions for sinusoidal driving forces. Thus, by adding these sinusoidal solutions together, we can find the solution for any periodic driving force. To simplify our writing, let us suppose that the driving force f (t) contains only cosine terms in its Fourier series. [This was the case for the rectangular pulse of Example 5.4, and is true for any even function — satisfying f (—t) = f (t) — because this condition guarantees that the coefficients of the sine terms are all zero.] In this case, the driving force can be written as
18 Strictly speaking we should not speak of the solution, since our second-order differential equation has many solutions. However, we know that the difference between any two solutions is transient — decays to zero — and our main interest is in the long-term behavior, which is therefore essentially unique.
Section 5.8 Fourier Series Solution for the Driven Oscillator* 199
where f, denotes the nth Fourier coefficient of f (t), and w = 2n- / -c as usual. Now, each individual term f„cos(ncot) has the same form (5.56) that we assumed for the sinusoidal driving force in Section 5.5 (except that the amplitude fc, has become fn and the frequency co has become nco). The corresponding solution was given in (5.66), 19
xn (t) = An cos(nwt — 8n ) (5.91)
where
fn A n = (5.92)
from (5.64), and
l 2 — n2w2)2 4182n2c02
6, = arctan 2,8nco
co 2 n 2602 0
(5.93)
from (5.65). Since (5.91) is the solution for the driving force fn cos(ncot), the solution for the complete force (5.90) is the sum
This completes the solution for the long-term motion of an oscillator driven by a periodic driving force f (t). To summarize, the steps are:
1. Find the coefficients fn in the Fourier series (5.90) for the given driving force f (t)•
2. Calculate the quantities A n and 8,„ as given by (5.92) and (5.93). 3. Write down the solution x (t) as the Fourier series (5.94).
In practice, one usually needs to include surprisingly few terms of the solution (5.94) to get a satisfactory approximation, as the following example illustrates 2 0
EXAMPLE 5.5 An Oscillator Driven by a Rectangular Pulse
Consider a weakly damped oscillator that is being driven by the periodic rectan- gular pulses of Example 5.4 (Figure 5.22). Let the natural period of the oscillator be r, = 1, so that the natural frequency is co o = 27r, and let the damping constant
19 The n = 0, constant term needs separate consideration. It is easy to see that for a constant force fo the solution is xo = fo/w02 . This is actually exactly what you get if you just set n = 0 in (5.92) and (5.93).
20 The solution contained in Equations (5.92) to (5.94) can be written more compactly if you don’t mind using complex notation. See Problem 5.51.
200 Chapter 5 Oscillations
be /3 = 0.2. Let the pulse last for a time Ar = 0.25 and have a height f max = 1. Calculate the first six Fourier coefficients A n for the long-term motion x (t) of the oscillator, assuming first that the drive period is the same as the natural period, t = to = 1. Plot the resulting motion for several complete oscillations. Repeat these exercises for r = 1.5t0 , 2.0r0, and 2.5ro .
Before we look at any of these exercises, it is worth thinking of a real system this problem might represent. One simple possibility is a mass hanging from the end of a spring, to which a professor is applying regularly spaced upward taps at intervals r. An even more familiar example would be a child in a swing, to whom a parent is giving regularly spaced pushes — though in this case we need to be careful to keep the amplitude small to justify the use of Hooke’s law. We are told to start by taking r = r o = 1; that is, the parent is pushing the child at exactly the natural frequency.
The Fourier coefficients fn of the driving force were already calculated in Equations (5.86) and (5.87) of Example 5.4 (where they were called an ). If we substitute these into (5.92) for the coefficients A n and put in the given numbers (including r = r0 = 1), we find for the first six Fourier coefficients A o , • , A5:
Ao Al A2 A3 A4 A5
63 1791 27 5 0 1
(Since the numbers are rather small, I have quoted the values multiplied by 10 4 ; that is, A o = 63 x 10-4 , A l = 1791 x 10 -4 , and so on.) Two things stand out about these numbers: First, after A 1, they get rapidly smaller, and for almost all purposes it would be an excellent approximation to ignore all but the first three terms in the Fourier series for x(t). Second, the coefficient A l is vastly bigger than all the rest. This is easy to understand if you look at (5.92) for the coefficient A 1 : Since co = coo (remember the parent is pushing the child at the swing’s natural frequency) and n = 1, the first term in the denominator is exactly zero, the denominator is anomalously small, and A l is anomalously big compared to all the other coefficients. In other words, when the driving frequency is the same as the natural frequency, the n = 1 term in the Fourier series for x (t) is at resonance, and the oscillator responds especially strongly with frequency coo .
Before we can plot x (t) as given by (5.94), we need to calculate the phase shifts Sn using (5.93). This is easily done, though I shan’t waste space displaying the results. We cannot actually plot the infinite series (5.94); instead, we must pick some finite number of terms with which to approximate x (t). In the present case, it seems clear that three terms would be plenty, but to be on the safe side I’ll use six. Figure 5.24 shows x(t) as approximated by the sum of the first six terms in (5.94). At the scale shown, this approximate graph is completely indistinguishable from the exact result, which in turn is indistinguishable from a pure cosine21 with frequency equal to the natural frequency of the oscillator. The strong response at the natural frequency is just what we would expect. For
21 Actually it’s a pure sine, but this is really cos(wt — 8 1 ) with S i = 7r/2 as we should have expected because we’re exactly on resonance.
Section 5.8 Fourier Series Solution for the Driven Oscillator* 201
x(t) T
0.2
0.1
1 2 3 4 5 6
—0.1
Figure 5.24 The motion of a linear oscillator, driven by periodic rectangular pulses, with the drive period r equal to the natural period ro of the oscillator (and hence to = coo). The horizontal axis shows time in units of the natural period to . As expected the motion is almost perfectly sinusoidal, with period equal to the natural period.
instance, anyone who has pushed a child on a swing knows that the most efficient way to get the child swinging high is to administer regularly spaced pushes at intervals of the natural period — that is, r = r o — and that the swing will then oscillate vigorously at its natural frequency.
A driving force with any other period r can be treated in exactly the same way. The Fourier coefficients A o , • • • , A5 for all of the values of r requested above are shown in Table 5.1.
Table 5.1 The first six Fourier coefficients A, for the motion x (t) of a linear oscillator driven by periodic rectangular pulses, for four different drive periods r = to , 1.5ro , 2.0r0 , and 2.5r0 . All values have been multiplied by 104 .
A0 At A2 A3 A4 A5
1.0 ro 63 1791 27 5 0 1 1.5 ro 42 145 89 18 6 2 2.0 ro 32 82 896 40 13 6 2.5 ro 25 59 130 97 25 11
The entries in the four rows of this table deserve careful examination. The first row ( -r = ro) shows the coefficients already discussed, the most prominent feature of which is that the n = 1 coefficient is far the largest, because it is exactly on resonance. In the next row (r = 1.5r.), the n = 1 Fourier component has moved well away from resonance, and A l has dropped by a factor of 12 or so. Some of the other coefficients have increased a bit, but the net effect is that the oscillator moves much less than when r = r o. This is clearly visible in Figures 5.25(a) and (b), which show x(t) (as approximated by the first six terms of its Fourier series) for these two values of the drive period.
r = r = r = r =
0.1 0.1
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