Intermediate Dynamics for Engineers: A Unified Treatment of Newton-Euler and Lagrangian Mechanics

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Intermediate Dynamics for Engineers: A Unified Treatment of Newton-Euler and Lagrangian Mechanics

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INTERMEDIATE DYNAMICS FOR ENGINEERS This book has sufficient material for two full-length semester courses in intermediate engineering dynamics. For the first course a Newton– Euler approach is used, followed by a Lagrangian approach in the second. Using some ideas from differential geometry, the equivalence of these two approaches is illuminated throughout the text. In addition, this book contains comprehensive treatments of the kinematics and dynamics of particles and rigid bodies. The subject matter is illuminated by numerous highly structured examples and exercises featuring a wide range of applications and numerical simulations. Oliver M. O’Reilly is a professor of mechanical engineering at the University of California, Berkeley. His research interests lie in continuum mechanics and nonlinear dynamics, specifically in the dynamics of rigid bodies and particles, Cosserat and directed continuua, dynamics of rods, history of mechanics, and vehicle dynamics. O’Reilly is the author of more than 50 archival publications and Engineering Dynamics: A Primer. He is also the recipient of the University of California at Berkeley’s Distinguished Teaching Award and three departmental teaching awards.

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Intermediate Dynamics for Engineers A UNIFIED TREATMENT OF NEWTON–EULER AND LAGRANGIAN MECHANICS Oliver M. O’Reilly University of California, Berkeley

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CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521874830 © Oliver M. O’Reilly 2008 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2008

ISBN-13 978-0-511-42336-9

eBook (EBL)

ISBN-13

hardback

978-0-521-87483-0

Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

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This book is dedicated to my adventurous daughter, Anna

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Contents

Preface

page xi

PART ONE

DYNAM ICS OF A SINGLE PARTICLE

1

1 Kinematics of a Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

Introduction Reference Frames Kinematics of a Particle Frequently Used Coordinate Systems Curvilinear Coordinates Representations of Particle Kinematics Constraints Classification of Constraints Closing Comments Exercises

3 3 5 6 9 14 15 20 27 27

2 Kinetics of a Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

Introduction The Balance Law for a Single Particle Work and Power Conservative Forces Examples of Conservative Forces Constraint Forces Conservations Dynamics of a Particle in a Gravitational Field Dynamics of a Particle on a Spinning Cone A Shocking Constraint A Simple Model for a Roller Coaster Closing Comments Exercises

33 33 35 36 37 39 45 47 55 59 60 64 66

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Contents

3 Lagrange’s Equations of Motion for a Single Particle . . . . . . . . . . . . 70 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Introduction Lagrange’s Equations of Motion Equations of Motion for an Unconstrained Particle Lagrange’s Equations in the Presence of Constraints A Particle Moving on a Sphere Some Elements of Geometry and Particle Kinematics The Geometry of Lagrange’s Equations of Motion A Particle Moving on a Helix Summary Exercises

PART TWO DYNAM ICS OF A SYSTEM OF PARTICLES

70 71 73 74 78 80 83 87 91 92 101

4 The Equations of Motion for a System of Particles . . . . . . . . . . . . . 103 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12

Introduction A System of N Particles Coordinates Constraints and Constraint Forces Conservative Forces and Potential Energies Lagrange’s Equations of Motion Construction and Use of a Single Representative Particle The Lagrangian A Constrained System of Particles A Canonical Form of Lagrange’s Equations Alternative Principles of Mechanics Closing Remarks Exercises

103 104 105 107 110 111 113 118 119 122 128 131 131

5 Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . . 134 5.1 5.2 5.3 5.4 5.5 5.6

Introduction Harmonic Oscillators A Dumbbell Satellite A Pendulum and a Cart Two Particles Tethered by an Inextensible String Closing Comments Exercises

PART THREE

DYNAM ICS OF A SINGLE RIGID BODY

134 134 140 143 147 151 153 161

6 Rotation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 6.1 6.2 6.3 6.4

Introduction The Simplest Rotation Proper-Orthogonal Tensors Derivatives of a Proper-Orthogonal Tensor

163 164 166 168

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Contents

ix

6.5 6.6

Euler’s Representation of a Rotation Tensor Euler’s Theorem: Rotation Tensors and Proper-Orthogonal Tensors 6.7 Relative Angular Velocity Vectors 6.8 Euler Angles 6.9 Further Representations of a Rotation Tensor 6.10 Derivatives of Scalar Functions of Rotation Tensors Exercises

171 176 178 181 191 195 198

7 Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11

Introduction The Motion of a Rigid Body The Angular Velocity and Angular Acceleration Vectors A Corotational Basis Three Distinct Axes of Rotation The Center of Mass and Linear Momentum Angular Momenta Euler Tensors and Inertia Tensors Angular Momentum and an Inertia Tensor Kinetic Energy Concluding Remarks Exercises

206 206 211 212 213 215 218 219 223 224 226 226

8 Constraints on and Potentials for Rigid Bodies . . . . . . . . . . . . . . . 237 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

Introduction Constraints A Canonical Function Integrability Criteria Forces and Moments Acting on a Rigid Body Constraint Forces and Constraint Moments Potential Energies and Conservative Forces and Moments Concluding Comments Exercises

237 237 241 243 247 248 256 262 263

9 Kinetics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9

Introduction Balance Laws for a Rigid Body Work and Energy Conservation Additional Forms of the Balance of Angular Momentum Moment-Free Motion of a Rigid Body The Baseball and the Football Motion of a Rigid Body with a Fixed Point Motions of Rolling Spheres and Sliding Spheres Closing Comments Exercises

272 272 274 276 279 285 289 294 297 299

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Contents

10 Lagrange’s Equations of Motion for a Single Rigid Body . . . . . . . . . 307 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9

Introduction Configuration Manifold of an Unconstrained Rigid Body Lagrange’s Equations of Motion: A First Form A Satellite Problem Lagrange’s Equations of Motion: A Second Form Lagrange’s Equations of Motion: Approach II Rolling Disks and Sliding Disks Lagrange and Poisson Tops Closing Comments Exercises

PART FOUR SYSTEM S OF RIGID BODIES

307 308 311 315 318 324 325 331 336 336 345

11 Introduction to Multibody Systems . . . . . . . . . . . . . . . . . . . . . . 347 11.1 11.2 11.3 11.4 11.5

Introduction Balance Laws and Lagrange’s Equations of Motion Two Pin-Jointed Rigid Bodies A Single-Axis Rate Gyroscope Closing Comments Exercises

347 347 349 351 355 355

APPENDIX: BACKGROUND ON TENSORS . . . . . . . . . . . . . . . . . . . . . . 362

A.1 A.2 A.3 A.4 A.5 A.6 A.7 A.8 A.9

Introduction Preliminaries: Bases, Alternators, and Kronecker Deltas The Tensor Product of Two Vectors Second-Order Tensors A Representation Theorem for Second-Order Tensors Functions of Second-Order Tensors Third-Order Tensors Special Types of Second-Order Tensors Derivatives of Tensors Exercises

362 362 363 364 364 367 370 372 373 374

Bibliography

377

Index

389

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Preface

The writing of this book started more than a decade ago when I was first given the assignment of teaching two courses on rigid body dynamics. One of these courses featured Lagrange’s equations of motion, and the other featured the Newton–Euler equations. I had long struggled to resolve these two approaches to formulating the equations of motion of mechanical systems. Luckily, at this time, one of my colleagues, Jim Casey, was examining the elegant works [205, 207, 208] of Synge and his co-workers on this topic. There, he found a partial resolution to the equivalence of the Lagrangian and Newton–Euler approaches. He then went further and showed how the governing equations for a rigid body formulated by use of both approaches were equivalent [27, 28]. Shades of this result could be seen in an earlier work by Greenwood [79], but Casey’s work established the equivalence in an unequivocal fashion. As is evident from this book, I subsequently adapted and expanded on Casey’s treatment in my courses. My treatment of dynamics presented in this book is also heavily influenced by the texts of Papastavridis [169] and Rosenberg [182]. It has also benefited from my graduate studies in dynamical systems at Cornell in the late 1980s. There, under the guidance of Philip Holmes, Frank Moon, Richard Rand, and Andy Ruina, I was shown how the equations governing the motion of (often simple) mechanical systems featuring particles and rigid bodies could display surprisingly rich behavior. There are several manners in which this book differs from a traditional text on engineering dynamics. First, I demonstrate explicitly how the equations of motion obtained by using Lagrange’s equations and the Newton–Euler equations are equivalent. To achieve this, my discussion of geometry and curvilinear coordinates is far more detailed than is normally found in textbooks at this level. The second difference is that I use tensors extensively when discussing the rotation of a rigid body. Here, I am following related developments in continuum mechanics, and I believe that this enables a far clearer derivation of many of the fundamental results in the kinematics of rigid bodies. I have distributed as many examples as possible throughout this book and have attempted to cite up-to-date references to them and related systems as far as feasible. However, I have not approached the exhaustive treatments by Papastavridis xi

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[169] nor its classical counterpart by Routh [184, 185]. I hope that sufficient citations to these and several other wonderful texts on dynamics have been placed throughout the text so that the interested reader has ample opportunity to explore this rewarding subject.

Using This Text This book has been written so that it provides sufficient material for two full-length semester courses in engineering dynamics. As such it contains two tracks (which overlap in places). For the first course, in which a Newton–Euler approach is used, the following chapters can be covered: 1. Kinematics of a Particle (Section 1.5 can be omitted) 2. Kinetics of a Particle Appendix on Tensors 6. Rotation Tensors 7. Kinematics of Rigid Bodies 8. Constraints on and Potentials for Rigid Bodies 9. Kinetics of a Rigid Body 11. Multibody Systems The second course, in which a Lagrangian approach is used, could be based on the following chapters: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Kinematics of a Particle Kinetics of a Particle Lagrange’s Equations of Motion for a Single Particle Lagrange’s Equations of Motion for a System of Particles Dynamics of Systems of Particles Appendix on Tensors Rotation Tensors (with particular emphasis on Section 6.8) Kinematics of Rigid Bodies Constraints on and Potentials for Rigid Bodies Kinetics of a Rigid Body Lagrange’s Equations of Motion for a Single Rigid Body Multibody Systems

In discussing rotations for the second course, time constraints permit a detailed discussion of only the Euler angle parameterization of a rotation tensor from Chapter 6 and a brief mention of the examples on rigid body dynamics discussed in Chapter 9. Most of the exercises at the end of each chapter are highly structured and are intended as a self-study aid. As I don’t intend to publish or distribute a solutions manual, I have tailored the problems to provide answers that can be validated. Some of the exercises feature numerical simulations that can be performed with Matlab or Mathematica. Completing these exercises is invaluable both in terms of

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comprehending why obtaining a set of differential equations for a system is important and for visualizing the behavior of the system predicted by the model. I also strongly recommend semester projects for the students during which they can delve into a specific problem, such as the dynamics of a wobblestone, the flight of a Frisbee, or the reorientation of a dual-spin satellite, in considerable detail. In my courses, these projects feature simulations and animations and are usually performed by students working in pairs who start working together after 7 weeks of a 15-week semester.

Image Credit The portrait of William R. Hamilton in Figure 4.6 in Subsection 4.11.3 is from the Royal Irish Academy in Dublin, Ireland. I am grateful to Pauric Dempsey, the Head of Communications and Public Affairs of this institution, for providing the image.

Acknowledgments This book is based on my class notes and exercises for two courses on dynamics, ME170, Engineering Mechanics III, and ME175, Intermediate Dynamics, which I have taught at the Department of Mechanical Engineering at the University of California at Berkeley over the past decade. Some of the aims of these courses are to give senior undergraduate and first-year graduate students in mechanical engineering requisite skills in the area of dynamics of rigid bodies. The book is also intended to be a sequel to my book Engineering Dynamics: A Primer, which was published by Springer-Verlag in 2001. I have been blessed with the insights and questions of many remarkable students and the help of several dedicated teaching assistants. Space precludes mention of all of these students and assistants, but it is nice to have the opportunity to acknowledge some of them here: Joshua P. Coaplen, Nur Adila Faruk Senan, David Gulick, Moneer Helu, Eva Kanso, Patch Kessler, Nathan Kinkaid, Todd Lauderdale, Henry ´ Payen, Brian Lopez, David Moody, Tom Nordenholz, Jeun Jye Ong, Sebastien ´ Spears, Philip J. Stephanou, Meng How Tan, Peter C. Varadi, and Stephane Verguet. I am also grateful to Chet Vignes for his careful reading of an earlier draft of the book. Many other scholars helped me with specific aspects of and topics in this book. Figure 9.1 was composed by Patch Kessler. Henry Lopez (B.E. 2006) helped me with the roller-coaster model and simulations of its equations of motion. Professor Chris Hall of Virginia Tech pointed out reference [118] on Lagrange’s solution of a satellite dynamics problem. Professor Richard Montgomery of the University of California at Santa Cruz discussed the remarkable figure-eight solutions to the three-body problem with me, Professor Glen Niebur of the University of Notre Dame provided valuable references on Codman’s paradox, Professor Harold Soodak of the City College of New York provided valuable comments on the tippe top, and Professors Donald Greenwood and John Papastavridis carefully read a penultimate draft

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of this book and generously provided many constructive comments and corrections for which I am most grateful. Most of this book was written during the past 10 years at the University of California at Berkeley. The remarkable library of this institution has been an invaluable resource in my quest to distill more than 300 years of work on the subject matter in this book. I am most grateful to the library staff for their assistance and the taxpayers for their support of the University of California. Throughout this book, several references to my own research on rigid body dynamics can be found. In addition to the students mentioned earlier, I have had the good fortune to work with Jim Casey and Arun Srinivasa on several aspects of the equations of motion for rigid bodies. The numerous citations to their works are a reflection of my gratitude to them. This book would not have been published without the help and encouragement of Peter Gordon at Cambridge University Press and would contain far more errors were it not for the editorial help of Victoria Danahy. Despite the assistance of several other proofreaders, it is unavoidable that some typographical and technical errors have crept into this book, and they are my unpleasant responsibility alone. If you find some on your journey through these pages, I would be pleased if you could bring them to my attention.

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PART ONE

DYNAMICS OF A SINGLE PARTICLE

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Kinematics of a Particle

1.1 Introduction One of the main goals of this book is to enable the reader to take a physical system, model it by using particles or rigid bodies, and then interpret the results of the model. For this to happen, the reader needs to be equipped with an array of tools and techniques, the cornerstone of which is to be able to precisely formulate the kinematics of a particle. Without this foundation in place, the future conclusions on which they are based either do not hold up or lack conviction. Much of the material presented in this chapter will be repeatedly used throughout the book. We start the chapter with a discussion of coordinate systems for a particle moving in a three-dimensional space. This naturally leads us to a discussion of curvilinear coordinate systems. These systems encompass all of the familiar coordinate systems, and the material presented is useful in many other contexts. At the conclusion of our discussion of coordinate systems and its application to particle mechanics, you should be able to establish expressions for gradient and acceleration vectors in any coordinate system. The other major topics of this chapter pertain to constraints on the motion of particles. In earlier dynamics courses, these topics are intimately related to judicious choices of coordinate systems to solve particle problems. For such problems, a constraint was usually imposed on the position vector of a particle. Here, we also discuss time-varying constraints on the velocity vector of the particle. Along with curvilinear coordinates, the topic of constraints is one most readers will not have seen before and for many they will hopefully constitute an interesting thread that winds its way through this book.

1.2 Reference Frames To describe the kinematics of particles and rigid bodies, we presume on the existence of a space with a set of three mutually perpendicular axes that meet at a common point P. The set of axes and the point P constitute a reference frame. In Newtonian mechanics, we also assume the existence of an inertial reference frame. In this frame, the point P moves at a constant speed. 3

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Kinematics of a Particle

Path of the particle m

A v Figure 1.1. The path of a particle moving in E3 . The position

vector, velocity vector, and areal velocity vector of this particle at time t and the position vector of the particle at time t + t are shown.

r(t) r(t + t)

O

Depending on the application, it is often convenient to idealize the inertial reference frame. For example, for ballistics problems, the Earth’s rotation and the translation of its center are ignored and one assumes that a point, say E, on the Earth’s surface can be considered as fixed. The point E, along with three orthonormal vectors that are fixed to it (and the Earth), is then taken to approximate an inertial reference frame. This approximate inertial reference frame, however, is insufficient if we wish to explain the behavior of Foucault’s famous pendulum ´ experiment. In this experiment from 1851, Leon Foucault (1819–1868) ingeniously demonstrated the rotation of the Earth by using the motion of a pendulum.∗ To explain this experiment, it is sufficient to assume the existence of an inertial frame whose point P is at the fixed center of the rotating Earth and whose axes do not rotate with the Earth. As another example, when the motion of the Earth about the Sun is explained, it is standard to assume that the center S of the Sun is fixed and to choose P to be this point. The point S is then used to construct an inertial reference frame. Other applications in celestial mechanics might need to consider the location of the point P for the inertial reference frame as the center of mass of the solar system with the three fixed mutually perpendicular axes defined by use of certain fixed stars [80]. For the purposes of this text, we assume the existence of a fixed point O and a set of three mutually perpendicular axes that meet at this point (see Figure 1.1). The set of axes is chosen to be the basis vectors for a Cartesian coordinate system. Clearly, the axes and the point O are an inertial reference frame. The space that this reference frame occupies is a three-dimensional space. Vectors can be defined in this space, and an inner product for these vectors is easy to construct with the dot product. As such, we refer to this space as a three-dimensional Euclidean space and we denote it by E3 . ∗

Discussions of his experiment and their interpretation can be found in [62, 138, 207]. Among his other contributions [215], Foucault is also credited with introducing the term “gyroscope.”

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1.3 Kinematics of a Particle

5

1.3 Kinematics of a Particle Suppose a single particle of mass m is in motion in E3 . The position vector of the particle relative to a fixed origin O is denoted by r (see Figure 1.1). In mechanics, this vector is usually considered to be a function of time t: r = r(t). The velocity v and acceleration a vectors of the particle are defined to be the respective first and second time derivatives of the position vector: v=

dr , dt

a=

dv d2 r = 2. dt dt

It is crucial to note that, because r is measured relative to a fixed origin, v and a are the absolute velocity and acceleration vectors. By definition, the velocity vector can be calculated from the following limit: r (t + t) − r(t) . t→0 t

v(t) = lim

We also use an overdot to denote the time derivative: v = r˙ and a = r¨ . Supplementary to the aforementioned kinematical quantities, we also have the linear momentum G of the particle: G = mv. Further, the angular momentum HO of the particle relative to O is HO = r × mv. As we now show, this vector is related to the areal velocity vector A. As used in celestial mechanics, the magnitude of the areal velocity vector is the rate at which the position vector r of the particle sweeps out an area about the fixed point O (see, e.g., Moulton [150]). To establish an expression for this vector, we consider the position vector of the particle at time t and t + t. Then, the area of the parallelogram defined by these vectors is r(t) × r (t + t) (see Figure 1.1). This is twice the area swept out by the particle during the interval t. Taking the limit of the vector r(t)×r(t+t) as t → 0 and using the fact that r(t) × r(t) = 0, we arrive at 2t an expression for the areal velocity vector A (t): r(t) × r (t + t) t→0 2t   r (t + t) 1 = r(t) × lim t→0 2 t   r (t + t) − r (t) 1 . = r(t) × lim t→0 2 t

A (t) = lim

That is, A=

1 r × v. 2

(1.1)

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Kinematics of a Particle

The vector A plays an important role in several mechanics problems in which either the angular momentum HO is constant or a component of HO is constant. Several other examples of its use are discussed in the exercises at the end of this chapter. Finally, we recall the definition of the kinetic energy T of the particle: 1 mv · v. 2 The definitions of the kinematical quantities that have been introduced are independent of the coordinate system that is used for E3 . In solving most problems, it is crucial to have expressions for momenta and energies in terms of the chosen coordinate system. It is to this issue that we now turn. T=

1.4 Frequently Used Coordinate Systems Depending on the problem of interest, there are several suitable coordinate systems for E3 . The most commonly used systems are Cartesian coordinates {x = x1 , y = x2 , z = x3 }, cylindrical polar coordinates {r, θ, z}, and spherical polar coordinates {R, φ, θ}. All of these coordinate systems can be considered as specific examples of a curvilinear coordinate system {q1 , q2 , q3 } for E3 , which we will discuss later on in this chapter. Cartesian Coordinate System For the Cartesian coordinate system, a set of right–handed orthonormal vectors are defined: {E1 , E2 , E3 }. Given any vector b in E3 , this vector has the representation

b=

3 

bi Ei .

i=1

For the position vector r, we also have r=

3 

xi Ei ,

i=1

where {x1 , x2 , x3 } are the Cartesian coordinates of the particle. Because Ei are fixed ˙ i = 0. in both magnitude and direction, their time derivatives are zero: E Cylindrical Polar Coordinates A cylindrical polar coordinate system {r, θ, z} can be defined by a Cartesian coordinate system as follows:    x2 , z = x3 , r = x21 + x22 , θ = tan−1 x1

where θ ∈ [0, 2π). Provided r = 0, then we can invert these relations to find that x1 = r cos(θ),

x2 = r sin(θ),

x3 = z.

In other words, given (x1 , x2 , x3 ), a unique (r, θ, z) exists provided (x1 , x2 ) = (0, 0). Otherwise, when r = 0, the coordinate θ is ambiguous.

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1.4 Frequently Used Coordinate Systems

7

E3 r z O

Figure 1.2. Cylindrical polar coordinates r, θ, and z.

r

E2



θ E1 er

Given a position vector r, we can write r = x1 E1 + x2 E2 + x3 E3 = r(cos(θ)E1 + sin(θ)E2 ) + zE3 = rer + zE3 , where, as shown in Figure 1.2, er = cos(θ)E1 + sin(θ)E2 . It is convenient to define the set of unit vectors {er , eθ , Ez}: er = cos(θ)E1 + sin(θ)E2 ,

eθ = cos(θ)E2 − sin(θ)E1 ,

ez = E3 .

˙ θ , whereas e˙ θ = −θe ˙ r . We should also verify that We also notice that e˙ r = θe {er , eθ , Ez} is a right-handed orthonormal basis for E3 .∗ Spherical Polar Coordinates A spherical polar coordinate system {R, φ, θ} can be defined by a Cartesian coordinate system as follows: ⎛ ⎞   2 2  + x x x2 2 1 ⎠, R = x21 + x22 + x23 , , φ = tan−1 ⎝ θ = tan−1 x1 x3

where θ ∈ [0, 2π) and φ ∈ (0, π). Provided φ = 0 or π, we can invert these relations to find x1 = R cos(θ) sin(φ),

x2 = R sin(θ) sin(φ),

x3 = R cos(φ).

Given a position vector r, we can now write r = x1 E1 + x2 E2 + x3 E3 = R sin(φ)(cos(θ)E1 + sin(θ)E2 ) + R cos(φ)E3 = ReR , where, as shown in Figure 1.3, eR = sin(φ) cos(θ)E1 + sin(φ) sin(θ)E2 + cos(φ)E3 . ∗

A basis {p1 , p2 , p3 } is right-handed if p3 · (p1 × p2 ) > 0 and is orthonormal if the magnitude of each of the vectors pi is 1 and they are mutually perpendicular: p1 · p2 = 0, p2 · p3 = 0, and p1 · p3 = 0.

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Kinematics of a Particle

eR

E3 r

Figure 1.3. The spherical polar coordinates φ and θ.



φ O

E2

r



θ E1

For future purposes, it is convenient to define the right-handed orthonormal set of vectors {eR , eφ , eθ }: ⎡

eR





cos(θ) sin(φ)

sin(θ) sin(φ)

cos(φ)

− sin(θ)

cos(θ)

0

⎤⎡

E1



⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ eφ ⎦ = ⎣cos(θ) cos(φ) sin(θ) cos(φ) − sin(φ)⎦ ⎣E2 ⎦ . eθ

E3

To establish the relations between these vectors and those defined earlier, we first calculate the intermediate relations ⎡ ⎤ ⎡ ⎤⎡ ⎤ er cos(θ) sin(θ) 0 E1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ eθ ⎦ = ⎣− sin(θ) cos(θ) 0⎦ ⎣E2 ⎦ , E3 0 ⎡ ⎤ ⎡ sin(φ) eR ⎢ ⎥ ⎢ ⎣ eφ ⎦ = ⎣cos(φ) eθ

0

0 0 0 1

E3 ⎤⎡ ⎤ cos(φ) er ⎥⎢ ⎥ − sin(φ)⎦ ⎣ eθ ⎦ . 1

(1.2)

E3

0

These results enable us to transform among the three distinct sets of basis vectors. As with the cylindrical polar coordinate system, the basis vectors we defined for the spherical polar coordinate system vary with the coordinates. Indeed, assuming that θ and φ are functions of time, a series of long calculations using (1.2) reveals that ⎡ ⎤ ⎡ ⎤⎡ ⎤ e˙ R 0 φ˙ θ˙ sin(φ) eR ⎢ ⎥ ⎢ ⎥⎢ ⎥ ˙ ˙ (1.3) 0 θ cos(φ)⎦ ⎣ eφ ⎦ . ⎣ e˙ φ ⎦ = ⎣ −φ e˙ θ

−θ˙ sin(φ) −θ˙ cos(φ)

0



These relations have an interesting form: Notice that the matrix in (1.3) is skewsymmetric. We shall see numerous examples of this later on when we discuss rotations and their time derivatives. Our later discussion should allow us to verify (1.3) rather easily.

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1.5 Curvilinear Coordinates

9

1.5 Curvilinear Coordinates The preceeding examples of coordinate systems can be considered as specific examples of a curvilinear coordinate system. The development of the vector calculus associated with such a system will be the focal point of this section of the book. Curvilinear coordinate systems have featured prominently in all areas of mechanics, and the material presented here has a wide range of applications. Most of our discussion is based on classical works and can be found in various textbooks on tensor calculus. Of these books, the one closest in spirit (and notation) to our treatment here is that of Simmonds [198]; [139, 201] are also recommended. Consider a curvilinear coordinate system {q1 , q2 , q3 } that is defined by the functions q1 = qˆ 1 (x1 , x2 , x3 ) , q2 = qˆ 2 (x1 , x2 , x3 ) , q3 = qˆ 3 (x1 , x2 , x3 ) .

(1.4)

We assume that the functions qˆ i are locally invertible:   x1 = xˆ 1 q1 , q2 , q3 ,   x2 = xˆ 2 q1 , q2 , q3 ,   x3 = xˆ 3 q1 , q2 , q3 .

(1.5)

This invertibility implies that, given the curvilinear coordinates of any point in E3 , there is a unique set of Cartesian coordinates for this point and vice versa. Usually, the invertibility breaks down at several points in E3 . For instance, the cylindrical polar coordinate θ is not uniquely defined when x21 + x22 = 0. This set of points corresponds to the x3 axis. Assuming invertibility, and fixing the value of one of the curvilinear coordinates, q1 say, to equal q10 , we can determine the values of x1 , x2 , and x3 such that the equation q10 = qˆ 1 (x1 , x2 , x3 ) is satisfied. The union of all the points represented by these Cartesian coordinates defines a surface that is known as the q1 coordinate surface (cf. Figure 1.4). If we move on this surface we find that the coordinates q2 and q3 will vary. Indeed, the curves on the q1 coordinate surface that we find by varying q2 while keeping q3 fixed are known as q2 coordinate curves. More generally, the surface corresponding to a constant value of a coordinate j q is known as a qj coordinate surface. Similarly, the curve we obtain by varying the coordinate qk while fixing the remaining two curvilinear coordinates is known as a qk coordinate curve.

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q3 coordinate curve q2 coordinate curve

a1

a3

S

a2

q1 coordinate surface O Figure 1.4. An example of a q1 coordinate surface S. At a point on this surface, a1 is normal

to the surface, and a2 and a3 are tangent to the surface. The q1 coordinate surface S is foliated by curves of constant q2 and q3 .

Covariant Basis Vectors Again assuming invertibility, we can express the position vector r of any point as a function of the curvilinear coordinates: r=

3 

  xˆ i q1 , q2 , q3 Ei .

i=1

It is also convenient to define the covariant basis vectors a1 , a2 , and a3 : ai = =

∂r ∂qi 3  ∂ xˆ k k=1

∂qi

Ek.

Mathematically, when we take the derivative with respect to q2 we fix q1 and q3 ; consequently, a2 points in the direction of increasing q2 . As a result, a2 is tangent to a q2 coordinate curve. In general, ai is tangent to a qi coordinate curve. You should notice that we can express the relationship between the covariant basis vectors and the Cartesian basis vectors in a matrix form: ⎡ ⎤ ⎡ ∂ xˆ 1 ∂ xˆ 2 ∂ xˆ 3 ⎤ ⎡ ⎤ a1 E1 ∂q1 ∂q1 ∂q1 ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ∂ x ˆ ∂ x ˆ ∂ x ˆ 3⎥ 1 2 ⎣a2 ⎦ = ⎢ ⎣ ∂q2 ∂q2 ∂q2 ⎦ ⎣E2 ⎦ . a3

∂ xˆ 1 ∂q3

∂ xˆ 2 ∂q3

∂ xˆ 3 ∂q3

E3

It is a good exercise to write out the matrix in the preceding equation for various examples of curvilinear coordinate systems, for instance, cylindrical polar coordinates.

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Contravariant Basis Vectors Curvilinear coordinate systems also have a second set of associated basis vectors: {a1 , a2 , a3 }. These vectors are known as the contravariant basis vectors. One method of defining them is as follows:

a1 =

3  ∂ qˆ 1 i=1

∂xi

a2 =

Ei ,

3  ∂ qˆ 2 i=1

∂xi

a3 =

Ei ,

3  ∂ qˆ 3 i=1

∂xi

Ei .

That is, ak = ∇qk. Geometrically, ai is normal to a qi coordinate surface. However, as in the case of the covariant basis vectors, the contravariant basis vectors are not necessarily unit vectors, nor do they form an orthonormal basis for E3 . Using the chain rule of calculus, we can show that ai · a j = δij , where δij is the Kronecker delta. As discussed in the Appendix, δij = 1 if i = j and is 0 otherwise. It is left as an exercise for the reader to show this result.∗ Covariant and Contravariant Components As {a1 , a2 , a3 } and {a1 , a2 , a3 } form bases for E3 , any vector b can be described as linear combinations of either sets of vectors:

b=

3 

bi ai =

i=1

3 

bkak.

k=1

The components bi are known as the contravariant components, and the components bk are known as the covariant components: b · ai =

 3  k=1

b · ai =

 3 

 bkak · ai =

3 

bkδki = bi ,

k=1

 bkak · ai =

k=1

3 

bkδik = bi .

k=1

It is very important to note that bk = b · ak in general because ai · ak is not necessarily equal to δik. The trivial case in which xi = qi deserves particular mention. For this case, 3 r = k=1 xi Ei . Consequently, ai = Ei . In addition, ai = Ei , and the covariant and contravariant basis vectors are equal. ∗

The starting point for this exercise is to note that

∂xk ∂x j

j

= δk .

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Kinematics of a Particle

y q =3

q2 = −4 q2 = 2

1

2

q =1 1

q = −1 1

−4

q1 = −3

6

x q2 = 4

−4

Figure 1.5. Projections of the q1 and q2 coordinate surfaces for coordinate system (1.6) on the

x–y plane.

An Example Although we have met three examples of curvilinear coordinate systems previously, it is useful to introduce an example that features nonorthogonal basis vectors. Consider the following coordinate system for Euclidean three-space:

q1 = y,

q2 = x − y2 ,

q3 = z.

(1.6)

Here, x = x1 , y = x2 , and z = x3 are Cartesian coordinates. Representative projections of the coordinate surfaces for q1 and q2 are shown in Figure 1.5. For this coordinate system, it is straightforward to invert (1.6) to see that x =  2 q2 + q1 and y = q1 . Thus,   2  r = q2 + q1 E1 + q1 E2 + q3 E3 . By taking the derivatives of this representation for r with respect to q1 , q2 , and q3 , we see that a1 = 2q1 E1 + E2 ,

a2 = E1 ,

a3 = E3 .

This set of vectors comprises the covariant basis vectors. By taking the gradient of qi , we find the contravariant basis vectors: a1 = E2 ,

a2 = E1 − 2q1 E2 ,

a3 = E3 .

It is interesting to note that a1 · a2 = −2q1 = 0. Further, a1 and a2 do not necessarily have unit magnitudes. By way of illustration, a q1 coordinate curve is shown in Figure 1.6. The vector a1 is tangent to this curve, and a2 and a3 are normal to this curve. To emphasize that a1 is not necessarily parallel to a1 , a q1 coordinate surface is also shown in the figure. It is left as an exercise for the reader to illustrate the tangent vectors a2 and a3 to the q1 coordinate surface shown in the figure. Some Comments on Derivatives Several partial derivatives of functions (q1 , q2 , q3 , q˙1 , q˙2 , q˙3 , t) play a prominent role in this book. When taking the partial derivative of this function with respect

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a1 a1

q1 coordinate surface a2

a1

q1 coordinate curve

a3

Figure 1.6. A q1 coordinate curve and a q1 coordinate surface showing representative examples of normal (a2 and a3 ) vectors and tangent (a1 ) vectors to the curve. Note that a1 is normal to the q1 coordinate surface and a1 is not parallel to a1 .

to q2 say, we assume that t, q1 , q3 , and q˙ k are constant. A related remark holds for the partial derivatives with respect to the velocities q˙ j and time t. That is, ∂qk = δkj, ∂qj

∂ q˙ k = 0, ∂qj

∂t = 0, ∂qj

(1.7)

∂qk = 0, ∂ q˙ j

∂ q˙ k = δkj, ∂ q˙ j

∂t = 0. ∂ q˙ j

(1.8)

and

In all these equations, j and k range from 1 to 3. You may have noticed that (1.7)1 was used in our calculations of ai . It is easy to be confused about the distinction between the derivative dtd and the derivative ∂t∂ . The former derivative assumes that qi and q˙ i are functions of time, whereas the latter assumes that they are constant: ˙ = 

d  ∂ dqi  ∂ d2 qk ∂ = + . + dt ∂qi dt ∂ q˙ k dt2 ∂t 3

3

i=1

k=1

For example, consider the function  = q1 + (q˙ 3 )2 + 10t. Then, ∂ = 1, ∂q1

∂ = 2q˙ 3 , ∂ q˙ 3

∂ = 10, ∂t

˙ = It should be clear from this example that 

∂ . ∂t

˙ = q˙ 1 + 2q˙ 3 q¨ 3 + 10. 

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Kinematics of a Particle

1.6 Representations of Particle Kinematics We now turn to establishing expressions for the position, velocity, and acceleration vectors of a particle in terms of the coordinate systems just mentioned. First, for the position vector we have∗ r = x1 E1 + x2 E2 + x3 E3 = rer + zE3 = ReR =

3 

  xˆ i q1 , q2 , q3 Ei .

i=1

Differentiating these expressions, we find v = x˙1 E1 + x˙2 E2 + x˙3 E3 ˙ θ + z˙ E3 = re ˙ r + rθe ˙ R + Rφe ˙ θ ˙ φ + R sin(φ)θe = Re =

3 

q˙ i ai .

(1.9)

i=1

Notice the simplicity of the expression for v when expressed in terms of the covariant basis vectors. For any given curvilinear coordinate system, if we write the position vector as a function of the coordinates q1 , q2 , and q3 , and then differentiate and  compare the result with v = 3i=1 q˙ i ai , we can read off the covariant basis vectors. For instance, (1.9)2 implies that a1 = er , a2 = reθ , and a3 = E3 for the cylindrical polar coordinate system. A further differentiation yields a = x¨1 E1 + x¨2 E2 + x¨3 E3 ˙ θ + zE = (r¨ − rθ˙2 )er + (rθ¨ + 2r˙θ)e ¨ 3 = (R¨ − Rφ˙ 2 − R sin2 (φ)θ˙2 )eR + (Rφ¨ + 2R˙ φ˙ − R sin(φ) cos(φ)θ˙2 )eφ + (R sin(φ)θ¨ + 2R˙ θ˙ sin(φ) + 2Rθ˙φ˙ cos(φ))eθ =

3  i=1

q¨ i ai +

3  3  i=1 j=1

q˙ i q˙ j

∂ai . ∂qj

We obtain the final representation for r¨ after noting that ai depend on the curvilinear  k i coordinates, which in turn are functions of time: a˙ k = 3i=1 ∂a q˙ . ∂qi It is left as an exercise for the reader to establish expressions, using various coordinate systems, for the linear momentum G and the angular momentum HO. ∗

Notice that it is a mistake to assume that r =

3 i=1

qi ai .

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15

The kinetic energy T of the particle has a rather elegant representation using the curvilinear coordinates: m T = v·v 2   3   3  m  i k = q˙ ai · q˙ ak 2 i=1

=

k=1

3  3  m i=1 k=1

2

aikq˙ i q˙ k,

(1.10)

where aik = aki = ak · ai . It is also a good exercise to compute aik for a spherical polar coordinate system, and   then, with the help of the representation T = 3i=1 3k=1 m2 aikq˙ i q˙ k, show that  m  ˙2 T= R + R2 φ˙ 2 + R2 sin2 (φ)θ˙2 . 2 The exercises at the end of this chapter feature this result for other coordinate systems.

1.7 Constraints A constraint is a kinematical restriction on the motion of the particle. They are introduced in problems involving a particle in three manners: either as simplifying assumptions, prescribed motions, or because of rigid connections. The constraints on the motion of a particle dictate, to a large extent, the coordinate system used to solve the problem of determining the motion of the particle. In this section, we examine the simplest class of constraints on the motion of a particle. Later, these constraints will be classified as integrable. Classical Examples Consider the four mechanical systems shown in Figure 1.7. The first system is known as the spherical pendulum. Here, a particle of mass m is attached by a rigid rod of length L0 to a fixed point O. The constraint on the motion of the particle in this system can be written as

r · eR = L0 . By differentiating this equation, we see that the velocity vector satisfies the relation v · eR = 0. The second system we consider is the planar pendulum. Again, the particle is attached by a rigid rod of length L0 to a fixed point O, but it is also assumed to move on a vertical plane. The constraints on the motion of the particle are r · er = L0 ,

r · E3 = 0.

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Kinematics of a Particle

(b)

(a) E3

E2

m L0

O

O

E2

E1

L0

E1 m (c)

E3

(d) m E3

moving plane

O

m

α

E2

O

er

E1 spinning conical surface Figure 1.7. Four mechanical systems featuring constraints on the motion of a particle: (a) the

spherical pendulum, (b) the planar pendulum, (c) a particle moving on a plane, and (d) a particle moving on a spinning cone.

After differentiating these equations with respect to time, we observe that the velocity vector of the particle has a component only in the eθ direction: v · er = 0 and v · E3 = 0. The third system involves a particle moving on a horizontal surface that is moving with a velocity vector f˙ (t)E3 . The constraint on the motion of the particle is r · E3 = f (t). The final system of interest consists of a particle moving on a spinning cone. The constraint on the motion of the particle can be most easily described with the help of a spherical polar coordinate system: φ + α(t) −

π = 0. 2

For all four systems, we have selected a coordinate system in which the constraint(s) on the motion of the particle is easily described. A Particle Moving on a Surface Turning to the more general case, consider a particle constrained to move on a surface. With the help of a single smooth function (r, t), we assume that the constraint

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17



m

r Surface = 0 O Figure 1.8. A particle moving on a surface = 0. The particle in this case is subject to a single

constraint.

can be described in a standard (canonical) form: (r, t) = 0. At each instant in time, this equation can be interpreted as a single condition on the three independent Cartesian coordinates of the particle. Thus the condition = 0 defines a two-dimensional surface (see Figure 1.8). The unit normal vector n to this surface is parallel to ∇ = grad( ) (see Figure 1.8). Depending on the coordinate system used, this vector ∇ has numerous representations:  ∂ ∂ = Ei ∂r ∂xi 3

∇ =

i=1

=

3  ∂ i a ∂qi i=1

=

∂ 1 ∂ ∂ er + eθ + E3 ∂r r ∂θ ∂z

=

∂ ∂ 1 ∂ 1 eR + eφ + eθ . ∂R R ∂φ R sin(φ) ∂θ

(1.11)

You should notice how simple the expression for the gradient is in curvilinear coordinates.∗ A simple differentiation of the function helps to provide the restriction it imposes on the velocity vector: ˙ = ∗

∂ ∂ ·v+ . ∂r ∂t

     ˙ q1 , q2 , q3 = 3i=1 ˙ q1 , q2 , q3 = ∇ · v and To establish this result, we note that tuting for v and comparing both expressions, we arrive at (1.11)2 .

∂ i q˙ . ∂qi

Substi-

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Kinematics of a Particle

∇ 2 ∇ 1 m t

r

Surface 1 = 0 O

Surface 2 = 0

Figure 1.9. A particle subject to two constraints. The dotted curve in this figure corresponds

to the curve of intersection of the surfaces 1 = 0 and 2 = 0, and the vector t is the unit tangent vector to this curve.

˙ = 0. Consequently, the However, if r satisfies the constraint, then (r, t) = 0 and constraint = 0 implies that the velocity vector satisfies the restriction ∂ ∂ ·v+ = 0. ∂r ∂t This result will be important in our discussion of the mechanical power of the constraint forces. A Particle Moving on a Curve We now consider the more complex case of a particle moving on a curve. A curve can be defined by the intersection of two surfaces. Using the previous developments, we consider the condition that the particle move on the curve to be equivalent to two (simultaneous) constraints:

1 (r, t) = 0,

2 (r, t) = 0.

This situation is shown in Figure 1.9. The normal vectors to the two surfaces at a point of their intersection are assumed not to be parallel: ∇ 1 × ∇ 2 = 0. That is, the two constraints 1 = 0 and 2 = 0 are assumed to be independent. Once 1 and 2 are given, then expressions for the two normal vectors to the curve can be readily established. We also note that deriving the restrictions these constraints impose on the velocity vector follows from the corresponding results for a single constraint: ∂ 1 ∂ 1 ·v+ = 0, ∂r ∂t

∂ 2 ∂ 2 ·v+ = 0. ∂r ∂t

(1.12)

If the curve is fixed, then 1 and 2 are not explicit functions of time. In this case, (1.12) can be used to show the expected result that v is tangent to the curve.

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A Particle Whose Motion is Prescribed The case in which the motion is prescribed can be interpreted as a particle lying at the intersection of three known surfaces. In other words, the particle is subject to three constraints:

1 (r, t) = 0,

2 (r, t) = 0,

3 (r, t) = 0.

We assume that these constraints are independent and thus their normal vectors at their intersection point form a basis for E3 : ∇ 3 · (∇ 1 × ∇ 2 ) = 0. The three conditions i = 0 can also be interpreted as three equations for the three components of r. The two primary situations in which a particle is subject to three constraints arise when either the motion of the particle is completely controlled or the particle is subject to static friction and is therefore in a state of rest relative to a curve or surface. Coordinates and Constraints The constraints we have considered on the motion of the particle have been described in terms of surfaces that the motion of the particle is restricted to. These surfaces can be described in terms of the coordinate system used for E3 . The description is greatly facilitated by a judicious choice of coordinates. For instance, if a particle is constrained to move on a fixed plane, then we can always choose the origin O and the Cartesian coordinates such that the constraint is easily described by the equation x3 = constant. Similarly, if a particle is constrained to move on a sphere, then spherical polar coordinates are an obvious choice. The more sophisticated the surfaces that the particle is constrained to move on, then the more difficult it becomes to choose an appropriate coordinate system. Help is at hand: The surfaces (r) = 0 of interest in this book can be described in an appropriate curvilinear coordinate system by a simple equation, q3 = constant. Furthermore, a moving surface (r, t) = 0 can, in principle, be described by the equation q3 = f (t), where f is a function of time t. For example, suppose a particle is moving on a sphere whose radius is a known function R0 (t). Then the constraint that the particle move on the sphere is simply described by

R = R0 (t). Here, we are choosing the spherical polar coordinate system to be our coordinate system. The Classical Examples Revisited Returning to the four mechanical systems shown in Figure 1.7, you should convince yourself that the constraint(s) on the motions of the particle in these systems are

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Kinematics of a Particle

individually of the form = 0. Specifically, for the spherical pendulum, = r · eR − L0 . That is, we may imagine the particle in a spherical pendulum as moving on a sphere. For the planar pendulum, we have 1 = r · er − L0 ,

2 = r · E3 = 0.

In this case, the particle can be visualized as moving on the intersection of a cylinder of radius L0 and a horizontal plane. This intersection defines a circle. If the rod’s length L0 changes with time, then the circle’s radius also changes. For the particle moving on the horizontal surface, = r · E3 − f (t). Notice that in this example = (r, t). For the final system, the particle moving on a cone, the constraint on the motion of the particle can be represented by = 0, where π (r) = φ + α − . 2 If the cone were moving in a manner such that α = α(t), then the function = (r, t). For example, suppose α = α0 + A sin(ωt); then (r, t) = φ −

π + α0 + A sin(ωt). 2

˙ = 0, but ∂ = Aω cos(ωt). The spinning of the cone has You should verify that ∂t purposefully not been mentioned. This motion will feature in any formulation of the friction forces acting on the particle. Further, in the event that the particle is stuck to the cone, then the particle will be subject to three constraints. This situation is discussed in Section 2.9.

1.8 Classification of Constraints All of the constraints discussed so far can be individually written in the form (r, t) = 0. Thus they are often known as positional constraints. We now define a further type of constraint: π = 0,

(1.13)

where π = f · v + e, and f = f(r, t) and e = e(r, t). The constraint π = 0 does not restrict the position of the particle – it restricts only its velocity vector. Consequently, the constraint π = 0 is often known as a velocity constraint.

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21

As we demonstrated earlier, every constraint of the form (r, t) = 0 can be differentiated to yield a restriction on the velocity vector: ∂ ∂ ·v+ = 0. ∂r ∂t This restriction is of the form (1.13). Thus every constraint (r, t) = 0 provides a constraint f · v + e = 0. However, the converse is not true. A constraint π = 0 that can be integrated to yield a constraint of the form (r, t) = 0 is said to be an integrable (or holonomic) constraint. More precisely, given a constraint π = 0, if we can find an integrating factor k = k (r, t) and a function (r, t), such that∗ k (f · v + e) = ∇ · v +

∂ , ∂t

then the constraint π = 0 is said to be integrable. Otherwise, the constraint π = 0 is said to be nonintegrable (or nonholonomic). The terminology here dates to Heinrich Hertz [92] (1857–1894). As noted by Lanczos [124], integrable constraints were further classified by Ludwig Boltzmann (1844–1906) as rheonomic when = (r, t) and scleronomic when = (r) (i.e., when is not an explicit function of time t). The distinction between integrable and nonintegrable constraints becomes particularily important when rigid bodies are concerned. However, for pedagogical purposes, it is desirable to introduce them when discussing single particles. We shall shortly discuss the forces needed to enforce the constraints: Such forces are known as constraint forces. To explore the differences between integrable and nonintegrable constraints, it is best to first consider some examples. Following such an exploration, we shall discuss known criteria to determine whether or not a set of constraints is integrable. Three Examples As a first example, we suppose that the particle is subject to the constraints

xy − c = 0,

z = 0.

That is, the particle is constrained to move on a hyperbola in the x − y plane (see Figure 1.10). Two points A and B are also shown in this figure, and it is important to notice that it is not possible for the particle to move between A and B without violating the constraint xy − c = 0. The constraints xy − c = 0 and z = 0 imply the velocity constraints: (xE2 + yE1 ) · v = 0,

E3 · v = 0.

(1.14)

These conditions imply that v has no component normal to the hyperbola xy = c. Constraints (1.14) are both clearly of the form f · v + e = 0, where e = 0 and ∗

Further background on integrating factors can be found in most texts on differential equations or differential forms, see, e.g., [61, 64, 114]. It is well known that integrating factors are not unique.

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Kinematics of a Particle

y = x2

B

x = x1

Figure 1.10. The motion of a particle subject to the constraints xy = c and z = 0, where c is a positive constant. The arrows shown on the hyperbolae indicate the possible directions of motion of the particle.

A

y = x2 B

x = x1

Figure 1.11. The motion of a particle subject to the constraints yx ˙ = 0 and z = 0. The arrows indicate the possible directions of motion of the particle.

A

f = xE2 + yE1 , and e = 0 and f = E3 , respectively. By construction, constraints (1.14) are both integrable. As a second example, let us examine the following constraints: (xE2 ) · v = 0,

z = 0.

(1.15)

The motions of the particle that satisfy these constraints are shown in Figure 1.11. Notice that it is possible to move between any two points A and B on the x–y plane without violating the constraint yx ˙ = 0. The restriction this constraint places is that it restricts how one can go from any A to any B. This is in marked contrast to the constraint xy − c = 0. By multiplying yx ˙ = 0 by 1x , for example, we see that, away from the y axis, this constraint is integrable.∗ Considering the possible motions shown in Figure 1.11, it is not surprising to note that we cannot find a smooth function to conclude that the constraint yx ˙ = 0 is integrable throughout the entirety of E3 . ∗

Here, 1x is an example of the integrating factor k (r, t) mentioned earlier in the definition of a nonintegrable constraint.

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1.8 Classification of Constraints

23

Instead, we classify the constraint yx ˙ = 0 as a piecewise-integrable constraint.∗ We shall discuss further unusual aspects of this constraint in Section 2.10. Our third example is the simplest possible nonintegrable constraint on the motion of a particle.† The constraint is (−zE1 + E2 ) · v = 0.

(1.16)

That is, y˙ − zx˙ = 0. To demonstrate the type of restrictions y˙ − zx˙ = 0 imposes, we choose two points A and B and use the x coordinate to parameterize a path between them. Choosing y = f (x),

z=

df , dx

where f (x) is any sufficiently smooth function, we observe that the constraint −zx˙ + y˙ = 0 is satisfied. In order that the particle be able to move between any two points A to B, f (x) is subject to the restrictions yA = f (xA),

zA =

df (xA), dx

yB = f (xB),

zB =

df (xB), dx

where rA = xAE1 + yAE2 + zAE3 and rB = xBE1 + yBE2 + zBE3 . Graphically constructing a function f (x) that meets these restrictions is not difficult, and some examples are presented in Figure 1.12. A specific example of f (x) is discussed in Pars [170], and a slightly modified version of it is presented here: 

 x − xA 2 xB − xA   x − xA 3 − (2 (yB − yA) − (xB − xA) (zB + zA)) xB − xA   2 2 2 π (x − xA) + c (x − xA) (xB − x) + d sin xB − xA   xB − x + yA, + zA (x − xA) xB − xA

f (x) = (3 (yB − yA) − (xB − xA) (zB + zA))

(1.17)

where c and d are arbitrary constants. It is left as an exercise for the reader to verify that an infinite number of paths between A and B are possible without violating the constraint y˙ − zx˙ = 0. Shortly, we shall verify that this constraint is indeed nonintegrable. ∗



With the exception of that of Papastavridis [169], this classification is not typically mentioned in the textbooks on classical and analytical mechanics. Further discussion of piecewise-integrable constraints can be found in the interesting paper by Ruina [186]. As discussed in his paper, constraints of this type also arise in many locomotive systems such as passive walking machines that feature impact. A proof of this statement can be found in Section 163 of Forsyth [64]. Our discussion of constraint (1.16) is based on the treatments presented in Goursat [75] and Pars [170].

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Kinematics of a Particle

f v Figure 1.12. Three possible motions between two given points A and B of a particle subject to the constraint y˙ − zx˙ = 0. The arrows indicate the directions of motion of the particle and the vector f = −zE1 + E2 . The motions presented in this figure were constructed with the assistance of (1.17).

E3 E2 A O E1

B

Integrability Criteria Suppose a constraint π = 0 is imposed on the motion of the particle. As mentioned earlier, this constraint is integrable if we can find a function (r, t) and an integrating factor k such that ˙ = k (f · v + e) .

(1.18)

Otherwise, the constraint f · v + e = 0 is nonintegrable. It is desirable to know if a constraint is integrable, because we can then, in principle, find a coordinate system {q1 , q2 , q3 } such that f · v + e = 0 is equivalent to the constraint q˙ 3 + e = 0. The latter constraint in turn is equivalent to the constraint q3 = g, where g˙ = e. Using this coordinate system, the dynamics of the particle is easier to analyze. With this in mind, several classical integrability criteria are now presented for single and multiple constraints.∗ The first criterion we examine pertains to constraints of the form f · v = 0, where f is not an explicit function of time. Using a coordinate system {q1 , q2 , q3 }, we can write the constraint π = 0 in the form

A SINGLE SCLERONOMIC CONSTRAINT.

f 1 q˙ 1 + f 2 q˙ 2 + f 3 q˙ 3 = 0. A necessary and sufficient condition for f · v = 0 to be integrable is that† Ic = 0 ∗



(1.19)

For additional discussion and illustrative examples from mechanics, the texts of Papastavridis [169] and Rosenberg [182] are recommended. Here, the historical remarks on these criteria are based on the paper by Hawkins [91]. Classical proofs of this result can be found in Section 151 of Forsyth [64], Section 442 of Goursat [74], and in Papastavridis [169]. A proof featuring differential forms can be found in Flanders [61], who refers to this result as Frobenius’ integration theorem.

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1.8 Classification of Constraints

25

for all possible choices of qi . Here,       ∂f 3 ∂f 1 ∂f 2 ∂f 2 ∂f 3 ∂f 1 − 3 + f2 − 1 + f3 − 2 . Ic = f 1 ∂q2 ∂q ∂q3 ∂q ∂q1 ∂q It is convenient to recall at this point the expression for the curl of a vector field P in Cartesian coordinates:  3   ∂ curl(P) = Ei × P ∂xi i=1       ∂P3 ∂P2 ∂P1 ∂P3 ∂P2 ∂P1 E1 + E2 + E3 , (1.20) = − − − ∂x2 ∂x3 ∂x3 ∂x1 ∂x1 ∂x2 where Pi = P · Ei . With the help of this expression, it is easy to see that criterion (1.19) can also be expressed in the compact form f · (curl (f)) = 0. We refer to (1.19) as Jacobi’s criterion after its discoverer Carl G. J. Jacobi (1804– 1851). Satisfaction of (1.19) does not tell us what (r) or k(r) are; it indicates only that these functions exist. Further, this criterion is local – it does not tell us if these two functions are the same for each point in space. For instance, although the constraints xy˙ + yx˙ = 0 and xy˙ = 0 trivially satisfy integrability criterion (1.19), only the former has a continuously defined (r). The function (r) for the latter constraint can be defined in only a piecewise manner (see Figure 1.11). If we use (1.19) to examine the constraint y˙ − zx˙ = 0, then we find that∗ Ic = −z (0 − 0) + 1 (−1 − 0) + 0 (0 − 0) . As Ic = −1 = 0, the constraint y˙ − zx˙ = 0 is nonintegrable. It is clearly of interest to present the generalization of Jacobi’s criterion to rheonomic constraints: f · v + e = 0. The result is very similar in form to that for a scleronomic constraint, but it is more tedious to evaluate. To proceed, we express the constraint π = 0 in the form

A SINGLE RHEONOMIC CONSTRAINT.

f 1 q˙ 1 + f 2 q˙ 2 + f 3 q˙ 3 + f 4 = 0, and define the variables U 1 = q1 ,

U2 = q2,

U3 = q3,

U 4 = t.

Clearly, f 4 = e. We next form the functions       ∂f L ∂f J ∂f K ∂f K ∂f L ∂f J + f + f . − − − IJKL = f J K L ∂UK ∂UL ∂UL ∂UJ ∂UJ ∂UK ∗

That is, we choose q1 = x, q2 = y, and q3 = z.

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Kinematics of a Particle

Here, the integer indices J , K, and L range from 1 to 4. A necessary and sufficient condition for the constraint π = 0 to be integrable is that the following four equations hold for all q1 , q2 , q3 , t: IJKL = 0,

for all J, K, L ∈ {1, 2, 3, 4}, L = J = K, K = L.

(1.21)

For a proof of this theorem, the reader is referred to Section 161 of Forsyth [64] or to Flanders [61]. When particles are subject to several constraints, their independence needs to be examined. For the case of two constraints, we first express them both in the form

SYSTEMS OF CONSTRAINTS.

f1 · v + e1 = 0, f2 · v + e2 = 0. If f1 × f2 = 0, then the constraints are said to be independent. For integrable constraints, this is equivalent to the condition ∇ 1 × ∇ 2 = 0. That is, the normal vectors to surfaces 1 = 0 and 2 = 0 are not parallel. The case of three constraints is similar. We first express each of them in the form f1 · v + e1 = 0, f2 · v + e2 = 0, f3 · v + e3 = 0.

(1.22)

Then, the condition for their independence is that f1 · (f2 × f3 ) = 0. If the constraints are integrable, then this condition is equivalent to ∇ 1 · (∇ 2 × ∇ 3 ) = 0. Geometrically, this means that the normal vectors at the point of intersection of the surfaces 1 = 0, 2 = 0, and 3 = 0 form a basis. The presence of more than one constraint can also imply that a system of constraints that are individually nonintegrable can become integrable. The most wellknown instance occurs when two scleronomic constraints are imposed on a particle [169]: f1 · v = 0, f2 · v = 0,

(1.23)

where the functions f1 and f2 are functions of r, and f1 × f2 = 0. In this case, the system of constraints is integrable. The proof of this result, which is presented in Section 8.4, uses a criterion that is due to Ferdinand G. Frobenius (1849–1917), which we postpone discussion of until Chapter 8. This criterion can also be used to show that if the three constraints (1.22) are imposed on a particle, then the system of constraints is integrable. Consequently, the motion of the particle is prescribed. Other

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Exercises 1.1–1.3

27

instances of multiple constraints on the motion of a particle are discussed in the exercises at the end of this chapter.

1.9 Closing Comments In this chapter, we have assembled many of the needed kinematical concepts and tools needed to solve problems in particle dynamics. For most readers, the novel aspects of the chapter will have been the discussion of curvilinear coordinates and kinematical constraints. These two topics are intimately related and will feature prominently in the forthcoming chapters.

EXERCISES

1.1. Consider a particle whose motion is described in Cartesian coordinates as r(t) = cE2 + 10tE1 , where c is a constant. Determine the areal velocity vector A of the particle, and show that the magnitude of this vector corresponds to the rate at which the particle sweeps out a particular area. Does the particle sweep out equal areas during equal periods of time? In your solution, you should also consider the case in which c = 0. 1.2. Consider a particle whose motion is described in cylindrical polar coordinates as r(t) = 10er ,

θ(t) = ωt,

where ω = 0. Determine the areal velocity vector A of the particle. Under which conditions does the particle sweep out equal areas during equal periods of time? 1.3. Recall that the cylindrical polar coordinates {r, θ, z} are defined in Cartesian coordinates {x = x1 , y = x2 , z = x3 } by the relations    −1 x2 2 2 , z = x3 . r = x1 + x2 , θ = tan x1 Show that the covariant basis vectors associated with the curvilinear coordinate system, q1 = r, q2 = θ, and q3 = z, are a1 = er ,

a2 = reθ ,

a3 = E3 .

In addition, show that the contravariant basis vectors are a1 = er ,

a2 =

1 eθ , r

a3 = E3 .

It is a good exercise to convince yourself with an illustration that a2 is tangent to a θ coordinate curve, whereas a2 is normal to a θ coordinate surface. Finally, for this coordinate system, show that  m 2 r˙ + r2 θ˙2 + z˙ 2 . T= 2

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Exercises 1.4–1.5

1.4. Recall that the spherical polar coordinates {R, φ, θ} are defined in Cartesian coordinates {x = x1 , y = x2 , z = x3 } by the relations 

R=

x21 + x22 + x23 ,   −1 x2 , θ = tan x1 ⎛ ⎞ x21 + x22 ⎠. φ = tan−1 ⎝ x3

Show that the covariant basis vectors associated with the curvilinear coordinate system, q1 = R, q2 = φ, and q3 = θ, are a1 = eR ,

a2 = Reφ ,

a3 = R sin(φ)eθ .

In addition, show that the contravariant basis vectors are a1 = eR ,

a2 =

1 eφ , R

a3 =

1 eθ . R sin(φ)

1.5. In the parabolic coordinate system, the coordinates {u, v, θ} can be defined in Cartesian coordinates {x = x1 , y = x2 , z = x3 } by the relations  u = ± x3 +



 v = ± −x3 + θ = tan

−1



x2 x1

  x23 + x21 + x22 ,

   x23 + x21 + x22 ,  .

In addition, the inverse relations can be defined: x1 = uv cos(θ),

x2 = uv sin(θ),

x3 =

1 2 (u − v2 ). 2

 (a) In the r–x3 plane, where r is the cylindrical polar coordinate r = x21 + x22 , draw several representative examples of the projections of the u and v coordinate surfaces. You should give a sufficient number of examples to convince yourself that u, v, and θ can be used as a coordinate system. (b) In the x1 –x2 –x3 space, draw a u coordinate surface. Illustrate how the v and θ coordinate curves foliate this surface.

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Exercises 1.5–1.6

29

(c) Show that the covariant basis vectors for the parabolic coordinate system are ∂r a1 = = ver + uE3 , ∂u ∂r a2 = = uer − vE3 , ∂v ∂r = uveθ . a3 = ∂θ Illuminate your results from (a) and (b) by drawing representative examples of these vectors. (d) Show that the contravariant basis vectors for the parabolic coordinate system are a1 = grad(u) =

1 a1 , u2 + v2

a2 = grad(v) =

1 a2 , u2 + v2

a3 = grad(θ) =

1 eθ . uv

Again, illuminate your results from (a), (b), and (c) by drawing representative examples of these vectors. (e) Where are the singularities of the parabolic coordinate system? Verify that, at these singularities, the contravariant basis vectors are not defined. (f) For a particle of mass m that is moving in E3 , establish expressions for the kinetic energy T and linear momentum G in terms of {u, v, θ} and their time derivatives. 1.6. A classical problem is to determine the motion of a particle on a circular helix (see Figure 1.13). In terms of the cylindrical polar coordinates r, θ, z, the equation of the helix is r = R0 ,

z = αR0 θ,

where R0 and α are constants. Here, we use another curvilinear coordinate system to define the motion of particle: q1 = θ,

q2 = r,

q3 = ν = z − αrθ.

A q3 coordinate surface is known as a right helicoid. (a) Show that the covariant basis vectors associated with this coordinate system are a1 = r(eθ + αE3 ),

a2 = er + αθE3 ,

a3 = E3 .

Verify that the covariant basis vectors are not orthonormal.

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Exercises 1.6–1.8

Bead of mass m

g

E3 E2 E1 Figure 1.13. A particle moving on a circular helix.

(b) Show that the kinetic energy of the particle has the representation T=

 m (1 + α2 θ2 )r˙2 + (1 + α2 )r2 θ˙2 + ν˙ 2 2  m ˙ + 2r˙θα ˙ 2 rθ . 2˙νrαθ ˙ + 2˙νθαr + 2

Calculate the contravariant basis vectors associated with the curvilinear coordinate system. 1.7. Consider the following curvilinear coordinate system: q1 = x1 sec(α),

q2 = x2 − x1 tan(α),

q3 = x3 ,

where α is a constant. This coordinate system is one of the simplest instances of a nonorthogonal coordinate system. Referring to (1.5), calculate the functions xˆ k(q1 , q2 , q3 ). Draw the coordinate curves and surfaces for the curvilinear coordinate system and then show that the covariant and contravariant basis vectors are a1 = cos(α)E1 + sin(α)E2 ,

a2 = E2 ,

a3 = E3 ,

(1.24)

and a1 = sec(α)E1 ,

a2 = − tan(α)E1 + E2 ,

a3 = E3 ,

(1.25)

respectively. Illustrate these vectors on the coordinate curves and surfaces you previously drew. For which values of α is {a1 , a2 , a3 } not a basis? 1.8. Given a vector b = 10E1 + 5E2 + 6E3 , calculate its covariant bi and contravariant bi components when the covariant and contravariant basis vectors are defined by (1.24) and (1.25), respectively. In addition,

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Exercises 1.8–1.9

31

verify that 3 

b=

bi ai =

i=1

Furthermore, show that b =  1 2 3 a , a , a not a basis?

3 i=1

3 

bi ai .

i=1

bi ai and b =

3

bi ai . For which values of α is

i=1

1.9. This exercise illustrates how the covariant and contravariant components of a vector are related. To start, we define the following scalars by using the covariant and contravariant basis vectors: aik = aik(qr ) = ai · ak,

aik = aik(qr ) = ai · ak.

You should notice that aik = aki and aik = aki . The indices i, k, r, and s in this problem range from 1 to 3. (a) For any vector b, show that the covariant and contravariant components are related: bi =

3 

bi =

aikbk,

k=1

3 

aikbk.

k=1

In other words, the covariant components are linear combinations of the contravariant components and vice versa. Using a matrix notation, these results can be expressed as ⎡ ⎤ ⎡ ⎤ ⎡ 1⎤ b1 a11 a12 a13 b ⎢ ⎥ ⎢ ⎥ ⎢ 2⎥ ⎣b2 ⎦ = ⎣a21 a22 a23 ⎦ ⎣b ⎦ , b3

a31

⎡ ⎤ ⎡ 11 a b1 ⎢ 2 ⎥ ⎢ 21 = ⎣b ⎦ ⎣ a b3 a31

a32

a33

a12

a13

a22 a32

b3

⎤⎡ ⎤ b1 ⎥⎢ ⎥ 23 ⎦ ⎣ ⎦ . a b2 a33 b3

(b) By choosing b = ar and as , and using the symmetries of akm and ars , show that 3  k=1

aikakj =

3  k=1

aki akj =

3 

j

aki a jk = δi .

k=1

It might be helpful to realize that, by using matrices, one of the preceding results has the following representation: ⎡ ⎤ ⎡ 11 ⎤ ⎡ ⎤ a11 a12 a13 a12 a13 a 1 0 0 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣a21 a22 a23 ⎦ ⎣a21 a22 a23 ⎦ = ⎣0 1 0⎦ . 0 0 1 a31 a32 a33 a31 a32 a33

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Exercises 1.10–1.11

1.10. With the help of integrability criteria (1.19) and (1.21), show that only one of the following constraints is integrable: xx˙ + yy˙ = −e(t),

z˙y + x˙ = 0,

cos(z)y˙ − sin(z)x˙ = 0.

In addition, show that the integrable constraint corresponds to a particle moving on a cylinder whose radius varies with time. 1.11. With the help of integrability criterion (1.21), show that one of the following constraints is nonintegrable: zx˙ + y˙ = −e(t),

z˙ = 0.

If both constraints are imposed on the particle simultaneously, then show that the system of constraints is integrable. Give a geometric description of the line that the particle is constrained to move on.

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Kinetics of a Particle

2.1 Introduction In this chapter, the balance law F = ma for a single particle plays a central role. This law is then used to examine models for several physical systems ranging from planetary motion to a model for a roller coaster. Our discussion of the behavior of these systems predicted by the models relies heavily on numerical integration of the equations of motion provided by F = ma, and it is presumed that the reader is familiar with the numerical integration of ordinary differential equations. Two of the most important types of forces featured in many applications are conservative forces and constraint forces. For the former, the gravitational force between two particles is the prototypical example, whereas the most common constraint force in particle mechanics is the normal force. It is crucial to be able to properly formulate and represent conservative and constraint forces, and we will spend a considerable amount of time discussing them in this chapter. In contrast to most texts in dynamics, here we consider friction forces to be types of constraint forces. For most applications, exact (or analytical) solutions are not available and recourse to numerical methods is often the only course of action. In validating these solutions, any conservations that might be present are crucial. To this end, conservations of momentum and energy are discussed at length and we also show (with the help of two examples) how angular momentum conservation can often be exploited. The examples discussed in this chapter are far from exhaustive. Although several other examples are included in the exercises, they too do not come close to encompassing the vast array of solved problems in the mechanics of a single particle. To this end, it is recommended that the interested reader consult the classical texts by Routh [185] and Whittaker [228] and the more recent texts by Baruh [14], Moon [146], and Sheck [190].

2.2 The Balance Law for a Single Particle Consider a single particle of mass m that is moving in E3 . As usual, the position vector of the particle relative to a fixed origin O is denoted by r. The balance law for this particle is known as the balance of linear momentum, Newton’s second law, or 33

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Kinetics of a Particle

Euler’s first law. The integral (or impulse momentum) form of this law is  t G(t) − G (t0 ) = F(τ)dτ,

(2.1)

t0

where F is the resultant force acting on the particle and G = mv is the linear momentum. Notice that this form of the balance law does not assume that v is differentiable with respect to time t. As a result, it is valid in impact problems, among others. An example of this is discussed in Section 2.10. If we assume that G is differentiable with respect to time, then we can differentiate both sides of the integral form of the balance of linear momentum to obtain the local form: ˙ F = G. Assuming that the mass of the particle is constant, we can write F = m¨r.

(2.2)

This law represents three (scalar) equations that relate F and the rate of change of linear momentum of the particle. We refer to F = ma as the balance of linear momentum.∗ Our emphasis in this chapter is on the principle F = ma, but it is misleading to believe that this is the sole accepted principle in dynamics. Indeed, since the creation of this principle by Newton over 300 years ago, several alternative (and often equivalent) principles of dynamics have been proposed, and we shall postpone discussion of several of them until Section 4.11. It is convenient to write the balance law as a set of first-order ordinary differential equations: v = r˙ ,

F = mv. ˙

In the absence of constraints, these represent six scalar (differential) equations for the six unknowns r(t) and v(t). To solve these equations, six initial conditions r(t0 ) and v(t0 ) must be specified. Alternatively, if the problem is formulated as a boundary-value problem, then a combination of six initial and final conditions on r(t) and v(t) must be prescribed. If we write F = ma using a Cartesian coordinate system, then we find the following three equations: mx¨ 1 = F · E1 , mx¨ 2 = F · E2 , mx¨ 3 = F · E3 . ∗

Discussions of the historical threads from Newton’s Principia [152] that lead to Euler’s explicit formulation of (2.2) in [51] can be found in several works by Clifford A. Truesdell (1919–2000); see, for example, [216, 217].

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2.3 Work and Power

35

On the other hand, if a cylindrical polar coordinate system is used, we have   m r¨ − rθ˙2 = F · er ,   m rθ¨ + 2r˙θ˙ = F · eθ , mz¨ = F · E3 .

(2.3)

Finally, if we use a spherical polar coordinate system, we find that   m R¨ − Rφ˙ 2 − R sin2 (φ)θ˙2 = F · eR , m(Rφ¨ + 2R˙ φ˙ − R sin(φ) cos(φ)θ˙2 ) = F · eφ , m(R sin(φ)θ¨ + 2R˙ θ˙ sin(φ) + 2Rθ˙ φ˙ cos(φ)) = F · eθ .

(2.4)

Notice that these equations are different projections of F = ma onto a set of basis vectors for E3 . Establishing the component representations of F = ma for various coordinate systems can be a laborious task. However, Lagrange’s equations of motion allow us to do this in a very easy manner. We will examine these equations in Section 3.2.

2.3 Work and Power The mechanical power P of the force P acting on a particle of mass m is defined to be P = P · v. Clearly, if P is perpendicular to v, then the power of the force is zero. As shown in Figure 2.1, consider a motion of the particle between two points: A and B. We suppose that at time t = tA the particle is at A: r(tA) = rA. Similarly, when t = tB, the particle is at B: r(tB) = rB. During the interval of time that the particle moves from A to B, we suppose that a force P, among others, acts on the particle. The work WAB performed by P during this time interval is defined to be the integral,

Path of the particle P

m v A

B

Figure 2.1. A force P acting on a particle as it moves from A to B.

r rA

rB

O

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Kinetics of a Particle

with respect to time, of the mechanical power:  tB WAB = P · vdt. tA

Notice that this is a line integral, and we are using t to parameterize the path of the particle. Depending on the choice of coordinate system, the integral in this expression has several equivalent representations. As an example, suppose that a force P = Peθ acts on a particle, and the motion of the particle is r(t) = Leαt (cos(ωt)Ex + sin(ωt)Ey ), where L, α, and ω = θ˙ are constant. A straightforward calculation shows that the power of this force is P · v = ωPLeαt , and the work performed by the force is  tB ωPL αtB WAB = ωPLeαt dt = (e − eαtA ) , α tA where, in evaluating the integral, we have assumed that α = 0.

2.4 Conservative Forces A force P acting on a particle is said to be conservative if the work done by P during any motion of the particle is independent of the path of particle. When a result from vector calculus is used, the path independence implies that P is the gradient of a scalar function U = U (r): P = −∇U. The function U is known as the potential energy associated with the force P, and the minus sign in the equation relating P to the gradient of U is a historical convention. Various representations of the gradient can be found in (1.11). It is important to notice that, if P is conservative, then its mechanical power is ˙ To see this, we simply examine U˙ and use the definition of a conservative force: −U. −U˙ = −

∂U · v = −(−P) · v = P · v. ∂r

This result holds for all motions of the particle and is very useful when we wish to establish expressions for the rate of change of the total energy E of a particle. To check if a given force P is conservative, one approach is to find a potential function U such that P · v = −U˙ holds for all motions of the particle. This approach reduces to solving a set of coupled partial differential equations for U. For example, if cylindrical coordinates

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2.5 Examples of Conservative Forces

37

are used, one needs to solve the following three partial differential equations for U (r, θ, z): Pr = −

∂U , ∂r

Pθ = −

1 ∂U , r ∂θ

Pz = −

∂U , ∂z

(2.5)

where P = Pr er + Pθ eθ + PzE3 . You might notice that the solution to (2.5) will yield a potential energy U(r, θ, z) modulo an additive constant. This constant is usually set by the condition that U = 0 when the coordinates have a certain set of values. Another approach to ascertain if a given force P is conservative is to examine its curl. The idea here is based on the identity curl(grad(V)) = 0, where V = V(r) is any scalar function of r. Clearly, if the given force P is conservative, then curl(P) = 0.∗ Consequently, if curl(P) = 0, then the Cartesian components of P must satisfy the following conditions: ∂P3 ∂P2 = , ∂x2 ∂x3

∂P1 ∂P3 = , ∂x3 ∂x1

∂P2 ∂P1 = , ∂x1 ∂x2

where Pi = P · Ei .

2.5 Examples of Conservative Forces The three main types of conservative forces in engineering dynamics are constant forces, spring forces, and gravitational force fields. Constant Forces All constant forces are conservative. To see this, let C denote a constant force and let Uc = −C · r. Now, ∇Uc = −C, and, consequently, Uc is the potential energy associated with C. The most common examples of constant forces are the gravitational forces −mgE2 and −mgE3 , and their associated potentials are mgE2 · r and mgE3 · r, respectively. Spring Forces Consider the spring shown in Figure 2.2. One end of the spring is attached to a fixed point A, and the other end is attached to a particle of mass m. When the spring is unstretched, it has a length L0 . Clearly, the stretched length of the spring is ||r − rA||, and the extension/compression of the spring is

= ||r − rA|| − L0 . The potential energy Us associated with the spring is Us = f ( ) ∗

An expression for the curl of a vector field was presented earlier in (1.20).

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m

r spring

Figure 2.2. A spring whose ends are attached to a particle and a fixed point A.

O rA A where f is a function of the change in length of the spring. Evaluating the gradient of Us , we find the spring force Fs : Fs = −

∂Us ∂f r − rA . =− ∂r ∂ ||r − rA||

To establish this result, we use the identity ∂ ∂ r − rA = . (||r − rA|| − L0 ) = ||r − rA|| ∂r ∂r This identity is left as an exercise to establish.∗ The most common spring in engineering dynamics is a linear spring. For this spring, Us =

K (||r − rA|| − L0 )2 , 2

Fs = −K (||r − rA|| − L0 )

r − rA . ||r − rA||

In words, the potential energy of a linear spring is a quadratic function of its change in length. Examples of nonlinear springs include those in which f is a polynomial function in . For instance, f ( ) = A 2 + B 4 , where A > 0 and B are constants. Such a spring is known as hardening if B > 0 and softening if B < 0. Newton’s Gravitational Force Dating to Newton in the late 1600s, the force exerted by a body of mass M on another body of mass m is modeled as

Fn = mg, where g=− ∗

GM ||r||3

r.

To help you with this, it is convenient to first establish that alent to showing that the gradient of ||r|| is eR .

√ ∂ x·x ∂x

=

√x x·x

=

x ||x|| .

This result is equiv-

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The body of mass M is assumed to be located at the origin O, and G is the universal gravitation constant. In this model for the gravitational force, both bodies are modeled as mass particles. Later on, in Sections 4.5 and 8.7, we shall examine generalizations of this force field to systems of particles and rigid bodies, respectively. Because the magnitude of Fn depends on the distance squared between the two bodies, Newton’s force field is often known as the inverse-square law. The force Fn is conservative, and its potential energy is Un = −

GMm . ||r||

It should be transparent from the expressions for Un and Fn that they have rather simple representations when a spherical polar coordinate system is used. Newton’s gravitational force field is attractive: It tends to pull m toward M. Thus, we have the interesting question of what keeps the two bodies from colliding. The answer, as you know from other courses, is the change of momentum of m. It is this delicate balance that allows m to steadily orbit M in a circular orbit.

2.6 Constraint Forces A constraint force Fc is a force that ensures that a constraint is enforced. Examples of these forces include reaction forces, normal forces, and tension forces in inextensible strings. Given a constraint (r, t) = 0 on the motion of the particle, there is no universal prescription for the associated constraint force. Choosing the correct prescription depends on the physical situation that the constraint represents. However, when we turn to solving for the motion of the particle by using F = ma, we see that we need to solve the six equations r˙ = v,

v˙ =

1 F, m

subject to the restrictions on r and v, (r, t) = 0,

∂ ∂ ·v+ = 0. ∂r ∂t

To close this system of equations, an additional unknown is introduced in the form of the constraint force Fc . The prescription of Fc must be such that F = ma and (r, t) can be used to determine Fc and r(t). There are no unique prescriptions for constraint forces. The prescription most commonly used, which dates to Joseph-Louis Lagrange (1736–1813), is referred to in this book as the Lagrange prescription. As shown in O’Reilly and Srinivasa [162], this prescription is necessary and sufficient to ensure that the motion of the particle satisfies the constraint. However, freedom is available to include other arbitrary nonnormal components. This is the reason why prescriptions of constraint forces featuring frictional components are valid.

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n

s2

m r s1

Surface = 0 O Figure 2.3. A particle moving on a surface = 0. The vectors s1 and s2 are unit tangent vectors to this surface at the point of contact of the particle and the surface.

A Single Constraint Consider the case of a particle subject to a single constraint:

(r, t) = 0. Referring to Figure 2.3, we recall that the unit normal vector n to this surface is n=

∇ . ||∇ ||

Knowing n, we can construct a unit tangent vector s1 to the surface. In addition, by defining another unit tangent vector, s2 = n × s1 , we have constructed a righthanded orthonormal basis {s1 , s2 , n} for E3 . This is not in general a constant set of vectors; rather, it changes as we move from point to point along the surface. The final ingredient we need is to denote the velocity vector of the point of the surface (which the particle is in contact with) as vs . With this background in mind, we now consider two prescriptions for the constraint force Fc . The first prescription is known as the Lagrange prescription: Fc = λ∇ , where we must determine λ = λ(t) by using F = ma. In other words, λ is an undetermined Lagrange multiplier. As discussed in Casey [27], the constraint force Fc is normal to the surface = 0 and is identical to a virtual work prescription that is sometimes known as the Lagrange principle or Lagrange–D’Alembert principle. On physical grounds, Lagrange’s prescription is justified if the surface on which the particle is constrained to move is smooth.

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41

Ff N = ∇ Figure 2.4. The constraint force Fc acting on a particle moving on a rough surface. The velocity vector vrel = v − vs is the velocity vector of the particle relative to the point of its contact with the surface.

vrel

n

Fc = N + F f s2

s1

In the event that the surface is rough, an alternative prescription, which is due to Charles Augustin Coulomb (1736–1806), can be used∗ : Fc = λ∇ + F f , where the normal force N = λ∇ and the friction force is F f = −µd ||λ∇ ||

v − vs . ||v − vs ||

Here, µd is known as the coefficient of dynamic friction. Notice that the tangential components of Fc are governed by the behavior of vrel = vrel1 s1 + vrel2 s2 and oppose the motion of the particle relative to the surface (cf. Figure 2.4). The velocity ||v − vs || is sometimes known as the slip speed. The mechanical power of the constraint force Fc is Fc · v = λ∇ · v + F f · v = −λ

∂ + F f · v, ∂t

where we used the identity ˙ = ∇ · v +

∂ = 0. ∂t

For the Lagrange prescription, F f = 0, we can now see that, if the surface that the particle is moving on is fixed, i.e., = (r), then Fc does no work. Otherwise, this constraint force is expected to do work because its normal component ensures that part of the velocity vector of the particle is vs . For the Coulomb prescription, except when vs = 0, it is not possible to predict if work is done on the particle by the constraint force. As a first example, consider a particle moving on a rough sphere of radius L0 whose center is fixed at the origin O. For this surface, the constraint is (r) = r · ∗

This prescription is often known as Amontons–Coulomb friction, after Guillaume Amontons (1663– 1705).

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˙ θ . In ˙ φ + L0 sin(φ)θe eR − L0 = 0. Consequently, ∇ = eR . In addition, v − vs = L0 φe conclusion, ˙ θ ˙ φ + sin(φ)θe φe Fc = λeR − µd |λ|  . φ˙ 2 + sin2 (φ)θ˙2 In this expression |λ| is the magnitude of the normal force exerted by the sphere on the particle. If we now consider the spherical pendulum, then Fc is given by the Lagrange prescription: Fc = λeR . In the spherical pendulum, −λ is the tension in the rod connecting the particle to the fixed point O. Two Constraints When a particle is subject to two constraints, 1 (r, t) = 0 and 2 (r, t) = 0, then it can be considered as constrained to move on a curve. The curve in question is formed by the instantaneous intersection of the surfaces defined by the constraints. At each point on the curve there is a unit tangent vector t. We can define this vector by first observing that ∇ 1 and ∇ 2 are both normal to the surfaces that the curve lies on (cf. Figure 1.9). Consequently,

t=

∇ 1 × ∇ 2 . ||∇ 1 × ∇ 2 ||

For each instant in time, the point of the curve that is in contact with the particle has a velocity. We denote this velocity by vc . The velocity vector of the particle relative to the curve is v − vc = vt. We now turn to prescriptions for the constraint force. The first prescription is the Lagrange prescription: Fc = λ1 ∇ 1 + λ2 ∇ 2 , where λ1 = λ1 (t) and λ2 = λ2 (t) are both determined by use of F = ma. As in the case of a single constraint, this prescription is valid when the curve that the particle moves on is smooth, and it provides a constraint force that is normal to the curve. For the rough case, we use Coulomb’s prescription: Fc = λ1 ∇ 1 + λ2 ∇ 2 + F f , where the friction force is F f = −µd ||λ1 ∇ 1 + λ2 ∇ 2 ||

v − vc . ||v − vc ||

The friction force opposes the motion of the particle relative to the curve and the normal force N with λ1 ∇ 1 + λ2 ∇ 2 . The mechanical power of the constraint force Fc for this case is Fc · v = λ1 ∇ 1 · v + λ2 ∇ 2 · v + F f · v = −λ1

∂ 1 ∂ 2 − λ2 + F f · v, ∂t ∂t

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43

where we again use the identities ˙ 1 = ∇ 1 · v +

∂ 1 = 0, ∂t

˙ 2 = ∇ψ

2

·v+

∂ 2 = 0. ∂t

For the Lagrange prescription, F f = 0, and we can now see that, if the curve that the particle is moving on is fixed, i.e., 1 = 1 (r) and 2 = 2 (r), then Fc does no work. Otherwise, this constraint force is expected to do work because its normal components force part of the velocity vector of the particle to be vc . As in the case of a single constraint, for the Coulomb prescription, except when vc = 0, it is not possible to predict if work is done on the particle by this force. We now consider some examples. Recall that the planar pendulum consists of a particle of mass m that is attached by a rod of length L0 to a fixed point O. The particle is also constrained to move on a vertical plane. In short, 1 = r · er − L0 = 0 and 2 = r · E3 = 0. With a little work, we find that ∇ 1 = er and ∇ 2 = E3 . For this mechanical system, Lagrange’s prescription is appropriate: Fc = λ1 er + λ2 E3 . For this system, λ1 er can be interpreted as the tension force in the rod and λ2 E3 can be interpreted as the normal force exerted by the plane on the particle. If we let L0 = L0 (t), the prescription for the constraint force will not change. A system that is related to the planar pendulum can be imagined as a particle moving on a rough circle whose radius L0 = L0 (t). The particle is subject to the same constraints as it is in the planar pendulum; however, Lagrange’s prescription is not valid. Instead, we now have  θ˙ Fc = λ1 er + λ2 E3 − µd λ21 + λ22   eθ , θ˙ ˙ θ. where we used the fact that v − vc = L0 θe Three Constraints The reader may have noticed that our expressions for the constraint force when we employed Coulomb’s prescription were not valid when the particle was stationary relative to the surface or curve that it was constrained to move on. This is because we view this case as corresponding to the motion of the particle subject to three constraints: i (r, t) = 0, i = 1, 2, 3. As mentioned earlier, when a particle is subject to three constraints, the three equations i (r, t) = 0 can in principle be solved to determine the motion r(t) of the particle. We denote the resulting solution by f(t), i.e., r(t) = f(t). In other words, the motion is completely prescribed. In this case, the sole purpose of F = ma is to determine the constraint force Fc . For the case in which the particle is subject to three constraints, the Lagrange prescription and a prescription based on static Coulomb friction are equivalent. This equivalence holds in spite of the distinct physical situations these prescriptions pertain to.

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To examine the equivalence, let us first use Lagrange’s prescription: Fc = λ1 ∇ 1 + λ2 ∇ 2 + λ3 ∇ 3 . Here, λ1 , λ2 , and λ3 are functions of time. Because the three constraints are tacitly assumed to be independent, {∇ 1 , ∇ 2 , ∇ 3 } forms a basis for E3 . Consequently, Lagrange’s prescription provides a vector Fc with three independent components. Coulomb’s static friction prescription for a particle that is not moving relative to the curve or surface on which it lies is Fc = N + F f , where the magnitude of F f is restricted by the static friction criterion: F f  ≤ µs ||N|| , where µs is the coefficient of static friction. Again, the Coulomb prescription provides a vector Fc with three independent components. In other words, both prescriptions state that Fc consists of three independent unknown functions of time. If we now assume that the resultant force F has the decomposition F = Fc + Fa , where Fa are the nonconstraint forces, then we see how Fc is determined from F = ma: ¨ Fc = −Fa + ma = −Fa + mf. This solution Fc will be the same regardless of whether one uses Lagrange’s prescription or Coulomb’s prescription. Nonintegrable Constraints Our discussion of constraint forces has focused entirely on the case of integrable constraints. If a nonintegrable constraint,

f · v + e = 0, is imposed on the particle, we need to discuss a prescription for the associated constraint force. To this end, we adopt a conservative approach and use the prescription Fc = λf. The main reason for adopting this prescription is as follows: In the event that the nonintegrable constraint turns out to be integrable, then the prescription we employ will agree with Lagrange’s prescription we discussed earlier. As a further example, suppose the motion of the particle is subject to two constraints, one of which is integrable: (r, t) = 0,

f · v + e = 0.

Using Lagrange’s prescription, we find that the constraint force acting on the particle is Fc = λ1 ∇ + λ2 f.

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45

Suppose that the applied force acting on the particle is Fa ; then the equations governing the motion of the particle are (r, t) = 0, r˙ = v,

f · v + e = 0, v˙ =

1 (Fa + λ1 ∇ + λ2 f). m

This set of equations constitutes eight equations for the eight unknowns: λ1 , λ2 , r, and v.

2.7 Conservations For a given particle and system of forces acting on the particle, a kinematical quantity is said to be conserved if it is constant during the motion of the particle. The conserved quantities are often known as integrals of motion. The solutions of many problems in particle mechanics are based on the observation that either a momentum or an energy (or both) is conserved. At this stage in the development of the field, most of these conservations are obvious and are deduced by inspection. However, for future purposes it is useful to understand the conditions for such conservations. We shall consider numerous examples of these conservations later on. Conservation of Linear Momentum The linear momentum G of a particle is defined as G = mv. Recalling the integral form of the balance of linear momentum,

 G(t) − G (t0 ) =

t

F(τ)dτ, t0

t we see that G(t) is conserved during an interval of time (t0 , t) if t0 F(τ)dτ = 0. The simplest case of this conservation arises when F(τ) = 0. Another form of this conservation pertains to a component of G in the direction of a given vector b(t) being conserved. That is, dtd (G · b) = 0. For this to happen, ˙ · b + G · b˙ = F · b + G · b˙ = 0. G ˙· b = G In words, if F · b + G · b˙ = 0, then G · b is conserved. Examples of conservation of linear momentum arise in many problems. For example, consider a particle under the influence of a gravitational force F = −mgE3 . For this problem, the E1 and E2 components of G are conserved. Another example is to consider a particle impacting a smooth vertical wall. Then the components of G in the two tangential directions are conserved. For these two examples, the vector b is constant.

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Conservation of Angular Momentum The angular momentum of a particle relative to a fixed point O is HO = r × G. To establish how HO changes during the motion of a particle, a simple calculation is needed:

˙ = v × mv + r × F = r × F. ˙O =v×G+r×G H It is important to note that we used F = ma during this calculation. The final result is known as the angular momentum theorem for a particle: ˙ O = r × F. H In words, the rate of change of angular momentum is equal to the moment of the resultant force. Conservation of angular momentum usually arises in two forms. First, the entire vector is conserved, and, second, a component, say c(t), is conserved. For the first case, we see from the angular momentum theorem that HO is conserved if F is parallel to r. Problems in which this arises are known as central force problems. Dating to Newton, they occupy an important place in the history of dynamics. When the angular momentum theorem is used, it is easy to see that the second form of conservation, HO · c is constant, arises when r × F · c + HO · c˙ = 0.

Conservation of Energy As a prelude to discussing the conservation of energy, we first need to discuss the work–energy theorem. This theorem is a result that is established by use of F = ma and relates the time rate of change of kinetic energy to the mechanical power of F:

T˙ = F · v. This theorem is the basis for establishing conservation of energy results for a single particle. The proof of the work–energy theorem is very straightforward. First, recall that T = 12 mv · v. Differentiating T, we find d T˙ = dt



 1 1 mv · v = (mv˙ · v + mv · v) ˙ = mv˙ · v. 2 2

However, we know that mv˙ = F, and so substituting for mv, ˙ we find that T˙ = F · v, as required. To examine situations in which the total energy of a particle is conserved, we first divide the forces acting on the particle into the sum of a resultant conservative force P = − ∂U and a nonconservative force Pncon : F = P + Pncon . Here, U is the sum ∂r

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2.8 Dynamics of a Particle in a Gravitational Field

of the potential energies of the conservative forces acting on the particle. From the work–energy theorem, we find T˙ = F · v = P · v + Pncon · v ∂U · v + Pncon · v ∂r = − U˙ + Pncon · v.

=−

Defining the total energy E of the particle by E = T + U, we see that E˙ = Pncon · v. This result states that if, during a motion of the particle, the nonconservative forces do no work, then the total energy of the particle is conserved. To examine whether energy is conserved, it usually suffices to check whether Pncon · v = 0. To see this, let us consider the example of the spherical pendulum whose length L0 = L0 (t). For this particle, P = −mgE3 ,

Pncon = λeR .

Consequently, E = T + mgE3 · r and ˙ 0 λ. Pncon · v = λeR · v = L ˙ 0 = 0, then E is conAs a result, when the length of the pendulum is constant, L ˙ served. On the other hand, if L0 = 0, then the constraint force λeR does work by giving the particle a velocity in the eR direction.

2.8 Dynamics of a Particle in a Gravitational Field The problem of a body of mass m orbiting a body of mass M is one of the centerpieces in Isaac Newton’s Principia.∗ Over 100 years later, it was also discussed in wonderful detail in Lagrange’s famous text M´ecanique Analytique.† Newton was partially motivated to study this problem because of Johannes Kepler’s (1571– 1630) famous three laws of motion for the (then known) planets in the solar system: I. The planets move in elliptical paths with the Sun at one of the foci. ∗ †

See Section III of Book 1 of [152]. See Section VII of the Second Part of [121].

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y a m b

r θ

O A

P x

a(1 − e)

hˆ < 0

hˆ > 0

Figure 2.5. Schematic of a particle of mass m moving about a fixed  point O in an elliptical 2

orbit. One of the foci of the ellipse is at O, and the eccentricity e = 1 − ba2 of the ellipse is less than 1, where a and b are the lengths of the semimajor and semiminor axes of the ellipse, respectively. Point A is known as the apocenter, and P is known as the pericenter.

II. The vector connecting the Sun to the planet sweeps out equal areas in equal times. III. If a denotes the semimajor axis of the elliptical orbit and T denotes the period a3 T2 of the orbit, then, for any of two planets, a13 = T12 . 2

2

Concise discussions of Kepler’s laws can be found in [150, 175, 188, 220]. Some of these authors note that the laws were based on astronomical data taken with the naked eye. In our analysis we assume that the body of mass M is fixed. As can be seen in Exercise 4.6 from Chapter 4, this restriction is easily removed and the results presented can be readily applied to deduce the motions of m and M. Many of the results presented feature terminology associated with ellipses, and, for convenience, a summarization of many of the terms associated with an ellipse, such as its eccentricity e and axes a and b, is given in Figure 2.5. The area swept out by the particle can be 1 determined by integrating the areal vector (1.1): A = 2m HO. The third law is remarkable when we note from [188] that the semimajor axis a and orbital period T for the planet Mercury are 0.387 astronomical units (AU) and 0.241 years, the Earth’s are 1 AU and 1 year, Jupiter’s are 5.203 AU and 11.862 years, and Mars’ are 1.524 AU and 1.881 years, respectively. We note that 0.3873 = 1.00, 0.2412

5.2033 = 1.00, 11.8622

1.5243 = 1.00. 1.8812

All of these results are in agreement with Kepler’s third law. Here, we start by setting up the coordinates for this problem and establishing the equations of motion. Our analysis of the equations of motion then exploits conservation of angular momentum to show that the motion must be planar.

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49

Following this, we reduce the equations of motion to a single second-order differential equation that we nondimensionalize and integrate numerically.∗ An alternative approach, which is used in most textbooks, will also be discussed. These analyses enable us to classify all five possible types of trajectories of the particle. Kinematics We pick as the origin O the fixed particle of mass M. Then the position vector of the particle of mass m is r, and it is convenient to pick cylindrical polar coordinates for this position vector:

r = rer + zE3 . Representations for the velocity and acceleration vectors in terms of cylindrical polar coordinates were established earlier and we do not rewrite them here. Equations of Motion The equations of motion for the particle can be obtained from F = ma, where F is solely due to Newton’s gravitational force:

Fn = −

GMm ||r||3

r.

You should recall that this force is conservative, and its potential energy is denoted by Un . Using (2.3), we can write out the component forms of Fn = ma. The result will be three differential equations:   GMmr m r¨ − rθ˙2 = − √ 3 , r2 + z2   m rθ¨ + 2r˙θ˙ = 0, GMmz m¨z = − √ 3 . r2 + z2

(2.6)

For a given set of six initial conditions,† these equations provide r(t), θ(t), and z(t), and hence can be used to predict the position of the particle of mass m. Conservations The solutions of differential equations (2.6) conserve two important kinematical quantities. First, they conserve the total energy E = T + U, where U = Un . Second, ∗



The reduction procedure we use is equivalent to the so-called Routhian or Lagrangian reduction procedure that is used to incorporate momentum conservation in a variety of mechanical systems ranging from the problem at hand to Lagrange and Poisson tops. For further details on this procedure, the reader is referred to Gantmacher [67], Chapter 2 of Karapetyan and Rumyantsev [108], and Marsden and Ratiu [138]. The six initial conditions needed are r (t0 ), θ (t0 ), z (t0 ), r˙ (t0 ), θ˙ (t0 ), and z˙ (t0 ).

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the angular momentum HO of the particle is conserved. It is left to the reader to demonstrate these results by using the work–energy and angular momentum theorems. The conservation of angular momentum implies that r (t) × v (t) = r (t0 ) × v (t0 ) . Now the initial position r (t0 ) and velocity v (t0 ) vectors define a plane, in general, and thus the motion of the particle remains on this plane (which is known as the orbital plane). We also observe that the normal to this plane is parallel to HO. If we allow ourselves the freedom to choose E3 , then we can pick this vector such that ˆ 3 , where hˆ = mr2 θ. ˙ Consequently, r and v will have components in only HO = hE the E1 and E2 directions: z(t) = 0 and z˙ (t) = 0. We henceforth exploit the fact that angular momentum is conserved and the motion is planar. Because HO and the areal velocity vector are synonymous, angular momentum conservation for this problem is often known as the “integrals of area.”∗ We have tacitly ignored the case in which r (t0 )  v (t0 ). In this case, HO is zero and must remain so. Consequently, the motion of the particle is a straight line. For convenience, we can choose this line to lie on the E1 − E2 plane. It can be shown that the motion of the particle will eventually lead to a collision with the particle of mass M at the origin. Thus, without an initial angular momentum, a collision for this system would be unavoidable. Determining the Motion of the Particle Because the angular momentum is conserved, we choose E3 such that z(t) = 0 and z˙ (t) = 0 for the particle. That is, the direction of HO is E3 . As a result, the equations of motion reduce to

  GMm m r¨ − rθ˙2 = − 2 , r   m rθ¨ + 2r˙θ˙ = 0.

(2.7)

Now the second of these equations can be expressed as d 2  mr θ˙ = 0. dt

(2.8)

This equation is equivalent to the conservation of HO · E3 . Using this conservation, we can eliminate θ˙ from (2.7) and arrive at a single governing differential equation: m¨r −

hˆ 2 GMm =− 2 . 3 mr r

Here, hˆ is determined from the initial position and velocity of the particle: hˆ = HO · E3 = (mr (t0 ) × v (t0 )) · E3 . ∗

See, for example, Section 86 of Moulton [150].

(2.9)

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51

For a particle with no angular momentum, we see from (2.8) that θ˙ = 0 and hence r¨ = − GM . It is not difficult to see that this equation implies that r(t) → 0 as t inr2 creases. This is the collision we discussed earlier. Given r(t0 ) and v(t0 ), we can determine hˆ and then integrate (2.9) to determine ∗ r(t). We can then compute the coordinate θ(t) by integrating θ˙ =

hˆ . mr2

(2.10)

We can then find expressions for x(t) = r(t) cos (θ(t)) and y(t) = r(t) sin (θ(t)) and construct the orbit of the particle. The easiest solution of (2.9) to compute is the one for which r is constant: r(t) = r0 and r(t) ˙ = 0. In this case, (2.9) is satisfied provided r0 =

hˆ 2 . GMm2

We also show, using (2.10), that θ˙ is constant: hˆ ˙ = ωK = θ(t) = mr02

(2.11)



GM . r03

(2.12)

The frequency ωK is known as the Kepler frequency. Physically, the particle is moving in a circular orbit of radius r0 at constant speed r0 ωK about the fixed body of mass M. In numerically integrating (2.9), the time scale of the integration is very long, and it is convenient to nondimensionalize the equations of motion. To do this, we choose the dimensionless variable w = rr0 and time τ = ωK t. Now using identities of dr dr the form r˙ = dr = dτ = ωK dτ , we can simplify (2.9) and (2.10) to dt dt dτ 1 1 d2 w = 3 − 2, 2 dτ w w

dθ 1 = 2. dτ w

(2.13)

Notice that we have reduced the problem of determining the motion of the particle to the integration of two differential equations. Differential equation (2.13)1 is a second-order differential equation for w(τ). As opposed to exhaustive displays of w(τ) for several sets of initial conditions, a qualitative method of representing the solutions of (2.13)1 is to construct what is known as a phase portrait. In this portrait, dw is plotted as a function of w(τ).† Equilibria dτ of the second-order differential equation correspond to points in the phase portrait 2 where dw = 0 and w(τ) is a constant. To find such points, we set dw = 0 and ddτw2 = 0 dτ dτ in the governing second-order differential equation for w(τ) and solve for the resulting constant values of w(τ). Later examples in this chapter will feature differential equations with multiple equilibria. ∗



Differential equation (2.9) has an analytical solution that can be expressed in terms of Jacobi’s elliptic functions. However, this is beyond our scope here, and the reader is referred to Whittaker [228] for details on how such an integration can be performed. Phase portraits are a standard method for the graphical representation of solutions to ordinary differential equations (see, for example, the texts of [10, 18, 81, 229]).

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2

h

h 1

p c

dw dτ

e

e

e

−1

−2

1

4

w

Figure 2.6. The phase portrait of (2.13)1 . The trajectories labeled e and h correspond to elliptical and hyperbolic orbits of the particle, the point c corresponds to a circular orbit, and the trajectory labeled p corresponds to the parabolic orbits. The arrows in this figure correspond to the directions of increasing τ.

Returning to the problem at hand, the phase portrait of (2.13)1 is shown in   Figure 2.6. There, we see an equilibrium point at w, dw = (1, 0) that corresponds dτ to a circular orbit of the particle. The closed orbits enclosing this point correspond to elliptical orbits of the form shown in Figure 2.7.∗ The remaining orbits shown in this figure correspond to hyperbolic orbits of the particle. For these orbits, the particle circles around the equilibrium once and never returns. The interesting case in which the particle describes a parabolic orbit is also shown in this figure. The trajectory on the phase portrait corresponding to this orbit separates the elliptical and hyperbolic trajectories.†

The Orbital Motions An alternative approach to the one outlined in the previous section is followed in most textbooks on dynamics. This approach involves solving for the motion of the particle as a function of θ rather than of time t. For this approach, we first use the chain rule and (2.7)2 in the form (2.10) to show that

r˙ = − ∗ †

  hˆ d 1 , m dθ r

r¨ = −

  hˆ 2 d2 1 . m2 r2 dθ2 r

dθ A second integration involving dτ = w12 is needed to construct these orbits, and this is left as an exercise. In the parlance of dynamical systems theory, the homoclinic orbit that passes through the point  

w, dw dτ = (0.5, 0) connects the fixed point at w, orbit of the particle.

dw dτ

= (∞, 0) and corresponds to the parabolic

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2.8 Dynamics of a Particle in a Gravitational Field

y = x2

(a)

(b)

53

y = x2

r θ x = x1

O

x = x1

O

y = x2

hˆ > 0

(c)

r θ

θP x = x1

O hˆ > 0 y = x2

(d)

(e)

y = x2

r r

hˆ < 0

θP θP

O

x = x1

O

x = x1

hˆ > 0 Figure 2.7. Schematic of the four types of orbits of a particle of mass m moving about a fixed point O: (a) line, (b) circular orbit (e = 0), (c) elliptical orbit (0 < e < 1), (d) parabolic orbit (e = 1), and (e) a hyperbolic orbit (e > 1). For each of the orbits (c)–(e), distinct values of θ p are considered.

Using the second of these results to rewrite (2.7)1 , we find the differential equation d2 dθ2

  1 1 1 + = . r r r0

where r0 was defined earlier [see (2.11)].

(2.14)

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Equation (2.14) is a linear ordinary differential equation for that has the exact solution r = r (θ) = r0 (1 + e cos (θ − θ p))−1 ,

1 r

as a function of θ (2.15)

where e and θ p are constants (which are determined from the initial conditions for 2 the position and velocity of the particle). We can then integrate mrhˆ (θ) dθ = dt to determine an analytical expression for θ(t). This is left as an exercise. Solution (2.15) represents a conic section, and from the theory of conic sections, it is known that, when e = 0, the orbit (r(θ)) is circular; when 0 < e < 1, the orbit is elliptical; when e = 1, the orbit is parabolic; and when e > 1, the orbit is hyperbolic. Referring to hˆ 2 Figure 2.5 for the elliptical orbit, it is easy to see that b = GMm 2. To solve for e and θ p, let us assume that r (t0 ) and v (t0 ) are given. Then we can ˆ To compute θ p, we first calculate r˙ by using determine HO and specify E3 and h. (2.15) and the chain rule:   GMm e sin (θ(t) − θ p) . r(t) ˙ =− (2.16) hˆ We also compute the value E0 of total energy of the particle E0 =

m v (t0 ) · v (t0 ) − 2

GMm  .  r (t0 )

(2.17)

Now as the total energy is conserved, the value of this kinematical quantity when θ = θ p is also equal to E0 . With some manipulations of (2.15) and (2.16), it can be shown that E0 =

 G2 M2 m3  2 e −1 . 2hˆ 2

(2.18)

We now have a method of determining e and θ p from a given set of initial conditions. The procedure is to compute E0 by use of (2.17) and then use (2.18) to compute e ≥ 0. Once e is known, then (2.16) can be used to compute θ p. With these values, r(θ) is specified and θ(t) can be calculated. Depending on the value of e, the orbits will be one of four types: a circle, an ellipse, a parabola, and a hyperbola (see Figure 2.7). For completeness, we could also have nondimensionalized (2.14): d2 u + u = 1, dθ2

(2.19)

where u = w1 = rr0 . The phase portrait of this equation is shown in Figure 2.8. In contrast to the earlier phase portraint, here the trajectory corresponding to the parabolic orbit is easily distinguished. However, as in the previous case, to gain a physical interpretation of the trajectories shown in Figure 2.8, it is necessary to reconstruct a position vector of the particle corresponding to the particular orbit. Comments For this problem, the rare event occurs that a complete classification of the motions of the particle are possible. For many of the problems that are discussed later on in

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2.9 Dynamics of a Particle on a Spinning Cone

2

55

h

h

h h

1

p c

du dθ

e

e

-1

-2

1

u

4

Figure 2.8. The phase portrait of (2.19). The trajectories labeled e and h correspond to elliptical and hyperbolic orbits of the particle, the point c corresponds to a circular orbit, and the trajectory labeled p corresponds to the parabolic orbits. In this figure, the arrow indicates the direction of increasing θ. For the parabolic trajectory, this angle ranges from −π → π.

this book, this classification has not been performed and indeed may not be possible. As a result, our previous discussion will be a benchmark. We have not exhausted the literature on this problem, and discussions of related problems involving escape velocities and transfer orbits can be found in several textbooks; see, for example, Baruh [14]. Generalizations of the problem also abound, and we shall discuss two of them at later stages in this book. Before we leave the problem for now, we wish to show that the elliptical (and circular) orbits are in agreement with Kepler’s laws. We first note that satisfaction of the first law is trivial, and the second law is a consequence of angular momentum conservation. To see that the third law is satisfied, we need to compute the period T of the particle exe3 cuting an elliptical orbit. We leave it as an exercise to show that T = √2π a 2 , and, GM consequently, the third law is satisfied.

2.9 Dynamics of a Particle on a Spinning Cone As shown in Figure 1.7(d), a particle of mass m moves on the surface of a cone. It is attached to the fixed apex O of the cone by a linear spring of stiffness K and unstretched length L0 . We assume that the surface of the cone is rough and that the cone is spinning about its axis of symmetry with an angular speed 0 . Our goal is to establish the equations of motion for the particle and discuss some features of its dynamics. Coordinates, Constraints, and Velocities As discussed in Section 1.7, when the particle is moving on the surface of the cone, it is subject to a single constraint = 0. This constraint can be conveniently expressed

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by use of a spherical polar coordinate system: =φ+α−

π . 2

For future reference, we note that the gradient of is ∇ = R1 eφ . Because the cone is rotating with a speed 0 , the velocity of the particle relative to the cone is   ˙ R + R cos(α) θ˙ − 0 eθ . vrel = Re We defer discussion of the case in which the particle is stuck to the cone. Forces The particle is under the influence of a gravitational force −mgE3 and a spring force:

Fs = −K (R − L0 ) eR , where K is the stiffness of the spring and L0 is its unstretched length. Assuming that the particle is moving relative to the surface of the cone, we find that the constraint force Fc acting on the particle has the representation Fc = N + F f , where the normal force N is parallel to ∇ : N=

λ eφ , R

F f = −µd ||N||

vrel . ||vrel ||

Thus the total force on the particle is F = Fc + Fs − mgE3 . The Equations of Motion To obtain the equations of motion, we express F = ma in spherical polar coordinates [see (2.4)] and impose the constraint = 0 to find

  m R¨ − R cos2 (α)θ˙2 = −K (R − L0 ) + F f · eR − mg sin(α),   m R cos(α)θ¨ + 2R˙ θ˙ cos(α) = F f · eθ , −mR sin(α) cos(α)θ˙2 =

λ + mg cos(α). R

(2.20)

The first two of these equations are ordinary differential equations for R and θ, and the third equation can be solved for λ (and hence the normal force) as a function of the motion of the particle. To integrate these equations,  it is desirable to nondimensionalize them. We use L0 as a measure of length and

L0 g

as a measure of time: 

τ=t

g , L0

w=

R . L0

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2.9 Dynamics of a Particle on a Spinning Cone

With the help of identities of the type R˙ =

dR dt

57

=

dτ dR dt dτ

=



g dR , L0 dτ

we can rewrite

(2.20) in the form  2 d2 w dθ 2 = w cos (α) − ω2 (w − 1) − sin(α) dτ2 dτ − µkn  w2

 dθ dτ

dw dτ

− ω0

2

+

 dw 2 , dτ

  dθ   w dτ − ω0 d dθ 2 2 w cos (α) = −µkn (w cos(α))   2  dw 2 . dτ dτ dθ 2 w dτ − ω0 + dτ (2.21) In these equations, the constants and dimensionless normal force are  L0 KL0 ||N|| h2 2 ω = , ω0 =  , n= = cos(α) + tan(α), mg g mg w and we can also show that dimensionless versions of total energy E and angular momentum HO · E3 are     2 E dw 2 1 dθ ω2 2 2 = + w cos (α) + (w − 1)2 + w sin(α), mgL0 2 dτ dτ 2 h=

dθ HO · E3  = w2 cos2 (α) . dτ m gL30

Notice that, by nondimensionalizing the equations of motion, we have reduced the number of parameters by two. The Static Friction Case When the particle is stuck to the cone, its velocity vector is v = R0 0 cos(α)eθ . In addition, the particle is subject to three constraints and the friction and normal forces constitute three undetermined forces that enforce these constraints:

Fc =

λ λ1 eθ + λ2 eR . eφ + R0 R0 cos(α)

To determine λ, λ1 , and λ2 , we examine F = ma. With some manipulations, we conclude that   F f = K (R0 − L0 ) + mg sin(α) − mR0 cos2 (α)20 eR , ˙ 0 eθ , + mR0 cos(α)   N = − mg cos(α) + mR0 20 sin(α) cos(α) eφ . Such a state of the particle is sustained provided sufficient friction is present, and to check this sufficiency we need to examine the static friction criterion F f  ≤ µs ||N||.

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Kinetics of a Particle

x2 x2

x1

x1 20

1 dR L0 dτ

1

8

R L0

x2 −20 x1

x2 x1

Figure 2.9. The phase portrait of (2.22) and the corresponding planar projections of the trajectories of the particle. For this figure, we assumed that α = 20◦ , h = 5, and ω2 = 10. Consequently the equilbrium point corresponding to a circular trajectory of the particle has the coordinates (w0 , 0) = (1.58896, 0).

If this criterion holds, then a particle that is stuck on the surface of the cone will remain stuck on the surface. Otherwise it slips, and the initial direction in which it slips is parallel to F f .∗ The Smooth Cone When the particle is moving on a smooth cone, we can simplify (2.21) considerably. Indeed, as in the earlier particle problem, we can exploit the conservation of angular momentum to write a single equation for w:

d2 w h2 − ω2 (w − 1) − sin(α). = 3 2 dτ w cos2 (α)

(2.22)

Integrating this equation, we can find w(τ). Another integration using  dθ  dθ w2 cos2 (α) dτ = (w0 , 0), where = h provides θ(τ). The equilibrium point at w, dτ w0 is the solution of w30 ∗

h2 − ω2 (w0 − 1) − sin(α) = 0, cos2 (α)

The initial slip direction must be specified in order that the initial motion of the particle slipping on the cone can be determined. The prescription of the initial slip direction allows one to specify ||vvrel || even though vrel = 0.

rel

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2.10 A Shocking Constraint

59

corresponds to a circular orbit of the particle that has a radius r = L0 w0 cos(α). Some of the other trajectories of w in the w − dw plane are shown in Figure 2.9. In this dτ figure, we have also constructed possible trajectories of the particle corresponding to w(τ). Unlike the problem of the particle subject to Fn that we discussed in Section 2.8, here the classifications of the trajectories for the particle on the cone defy a simple classification.

2.10 A Shocking Constraint We now return to the constraint yx ˙ = 0, discussed earlier [see (1.15)]. Our interest is to determine the equations of motion of a particle that is subject to this constraint and that is also under the influence of an applied force Fa = P1 E1 + P2 E2 . First, we assume that the constraint force that enforces yx ˙ = 0 has the standard prescription Fc = λxEy ,

(2.23)

where λ is a Lagrange multiplier. We note that, when x = 0, Fc = 0. From a balance of linear momentum, we find that the equations of motion for the particle are xy˙ = 0, mx¨ = P1 , my¨ = P2 + λx, mz¨ = 0.

(2.24)

The equation for motion in the z direction is trivial to integrate and interpret, and, for convenience, we henceforth ignore this direction and assume that the motion is planar. When some modest restrictions are imposed on P1 and P2 , governing equations (2.24)1,2,3 have exact solutions that are easy to establish provided the motion is rectilinear:  t τ Fy P1 dudτ, λ= (2.25) y(t) = y0 , x(t) = x0 + x˙ 0 t + x(t) 0 0 m and x = 0,

y(t) = y0 + y˙ 0 t +

 t 0

τ 0

P2 dudτ. m

(2.26)

Here x0 = x(0), x˙ 0 = x(0), ˙ y0 = y(0), and y˙ 0 = y(0) ˙ are initial conditions. From these solutions to the equations of motion, we observe that λ is not defined when x = 0. The Shock Referring to Figure 1.11, we recall that this constraint has the unusual feature that it can be decomposed into two piecewise integrable constraints:

y˙ = 0 when x = 0,

and x = 0.

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Kinetics of a Particle

(a)

(b)

y = x2

y = x2 Ic2 B

x = x1 B

A

x = x1 A Ic1

Figure 2.10. Two possible motions of a particle subject to a constraint yx ˙ = 0. In (a), the particle moves from A to B and there is no impulse Ic when x = 0, whereas in (b) the particle experiences two instances in which v is not continuous.

At the points where the particle makes a transition from one of these integrable constraints to the other, its velocity vector v will be discontinuous and therefore its acceleration vector cannot be defined. At such a transition, the prescription for constraint force (2.23) does not hold, and instead we can calculate only the impulse Ic that is due to this force by using (2.1). Supposing that the transition occurs at time t = T, we will have    T+σ Ic = lim mv (T + σ) − mv (T − σ) − Fa dτ . σ→0

T−σ

For example, to achieve a motion that goes from point A to point B in Figure 2.10(b), the particle needs to perform a motion for which it will possess a discontinuous velocity vector in at least two locations. That is, the particle will experience a shock. Impulses Ic1 and Ic2 shown in the figure enable these shocks. This is in contrast to the situation shown in Figure 2.10(a), where there is no discontinuity in the motion of the particle. That is, the shock is absent and consequently Ic = 0. If the constraint cannot supply the impulse Ic , then the particle is effectively subject to a single holonomic constraint. This constraint is either y = y0 or x = 0, depending on the initial position and velocity of the particle. It is left as an exercise for the reader to imagine a rigid wall placed to the left of the y axis in Figure 1.11 as a method of realizing the constraint yx ˙ = 0.

2.11 A Simple Model for a Roller Coaster Imagine being in a cart at the top of a roller coaster. If there is no friction, then the slightest nudge will set the cart in motion. The presence of Coulomb friction with stick–slip changes this scenario. It will eventually bring the cart to a halt, and it may bring the cart to a halt near the top of the roller coaster. Indeed, if there is sufficient static friction, then the cart can come to a halt at any location on the track, and the

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2.11 A Simple Model for a Roller Coaster

E2

61

A y E1

x1

B

C

Figure 2.11. Schematic of a particle on a cosinusoidal path.

chief quantity that governs how fast the halting occurs in this extreme case will be the dynamic friction coefficient. Here, a very simple model is presented for the roller coaster that captures its stick–slip behavior.∗ First, we establish a differential equation governing the motion of the roller coaster, and then we use numerical integrations to investigate the dynamics of the roller coaster. The Equations of Motion One model for the dynamics of a cart on a roller coaster is to model the cart as a particle of mass m that is moving on a fixed plane curve: y = f (x1 ), z = 0. That is, the particle is subject to two constraints, 1 = 0 and 2 = 0, where

1 = y − f (x1 ) ,

2 = z.

These constraints will be enforced by Fc = N + F f . It is a standard exercise to calculate the unit tangent et , the unit normal en , and the binormal eb vectors to this curve [159]: et = 

1

 

E1 + f E2 ,

1 + f 2 

  sgn f 

E1 + f E2 , en =  1 + f 2 eb = et × en , where the prime denotes the derivative with respect to x1 . These three vectors constitute the Frenet triad, and in calculating this triad we assume that x˙ 1 > 0. A normal force N = Nen + λ2 E3 , a friction force F f et , and a vertical gravitational force −mgE2 act on the cart (see Figure 2.11). Now, for a particle moving on a curve with a velocity v = vet , the acceleration vector of the particle is a = ve ˙ t + κv2 en , ∗

The work presented on this model was performed in collaboration with Henry Lopez [130].

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where κ is the curvature of the space curve: 

 f  κ =  3 .

2 1+ f Taking the et and en components of F = ma, we can easily calculate the equations governing the motion of the cart and the normal force:     x˙ 1

2

m 1+ f x¨ 1 + f f x˙ 21 = −mgf − µd 1 + f 2 ||N|| , (2.27) |x˙ 1 | where



  

  1   sgn f mg +  f  mx˙ 21 en , N=  1 + f 2 mg =  en ,

2 1+ f



if f = 0,



when f = 0.

(2.28)

These equations apply when the cart is moving and the friction is dynamic. In the event that the cart is stationary, static friction acts and, provided the static friction criterion is satisfied,      mgf    ≤ µs ||N|| , (2.29)    1 + f 2  the cart remains stationary. With the help of (2.28), (2.29) can be expressed in the simple form     (2.30) f  ≤ µs . This equation can be viewed as the basis for the classical experiment to measure the coefficient of static friction: We place a block on an inclined plane and slowly increase the angle of inclination until slipping occurs. The tangent of the angle of inclination is equal to µs . We shall shortly use (2.30) to establish a continuum of points at which the roller coaster can remain in a state of rest. We now choose a cosinusoidal track,   πx1 f (x1 ) = A cos . (2.31) L0 In addition, we employ the following nondimensionalizations:  g x1 x= , τ= t. L0 L0 Of course, other tracks are possible, and the reader is referred to Shaw and Haddow [193], where other interesting choices of f (x) can be found. A further interesting choice would be Euler’s spiral (clothoid) that features in “loop-the-loop” roller coasters.

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2.11 A Simple Model for a Roller Coaster

63

dx dτ

1.5

−2

x

2

−1.5 Figure 2.12. The phase portrait of (2.27)–(2.29) when friction is absent. The points on the

x = Lx1 axis labeled by · correspond to equilibria of the cart. For this figure, 0 µs = µd = 0.0.

A L0

= 0.25, and

States of Rest For a cart on a smooth roller coaster, the motion of the cart will be perpetual, and a portion of its phase portrait is shown in Figure 2.12. We note the presence of an   equilibrium at x = 0, dx = 0 . This point corresponds to the cart’s being stationary dτ at the top of the roller coaster. Referring to Figure 2.13(a), we refer to equilibria   of this type as saddles. The two equilibria at x = ±1, dx = 0 represent a stationary dτ cart at the bottom of one of the valleys of the roller coaster. Examining the phase portrait in Figure 2.13(c), we easily see why equilibria of this type are known as   centers. The equilibria at x = (−2, 0, 2), dx = 0 correspond to a stationary cart at dτ one of the crests of the roller coaster. When stick–slip friction is present, the phase portrait changes dramatically (see Figure 2.14). First, all of the saddles have split, and between their two split halves we have what we call a sticking region [see Figure 2.13(b)]. Depending on the value of x1 , this region contains either x˙ 1 > 0 or x˙ 1 < 0. If the cart’s state enters this region then the cart will stop. That is, the cart will come to rest near a crest of the roller coaster. Similarly, the equilibria of the smooth roller coaster at the floor of its valleys have now transformed from discrete points to regions surrounding these points [see Figure 2.13(d)]. These regions are also sticking regions, and if the cart’s state enters this region, then the cart will stop. The size of the sticking region is easy to compute by use of (2.30), and a graphical method is shown in Figure 2.15. As µs gets larger, the size of the sticking region (or sticking states) surrounding the equilibria when µd = 0 grows, and eventually any point (x, 0) on the dx axis will become an equilibrium, and thus a state of rest dτ

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64

Kinetics of a Particle dx dτ

(a)

(b)

0.2

0.2

−0.2

0.2

x

0.2

x

−0.2

dx dτ

(d)

0.2

dx dτ

0.2

x 1.2

0.8

s

−0.2

−0.2

(c)

dx dτ

0.8

−0.2

s

x 1.2

−0.2

Figure 2.13. Expanded views of the phase portraits in the neighborhood of equilibria of (2.27) with f specified by (2.31). For (a) and (c), µd = 0 and the roller coaster is smooth, whereas for (b) and (d), µd = 0.1. For the latter cases, the sticking regions s are shown for the case in which µs = 0.3.

for the roller coaster.∗ This phenomenon is also easy to explain physically. We note for completeness that, for the present choice of f (x), if µs ≥

πA , L0

then it is possible to stick at any location on the roller coaster. This can also be inferred from the graphical technique shown in Figure 2.15.

2.12 Closing Comments A vast amount of material has been covered in this chapter, starting with descriptions of various forces, discussions of the balance laws, and analyses of various applications. The analyses we employed invariably featured the numerical integration of an ordinary differential equation and an interpretation of its solutions. Developing physical interpretations of the results provided by the model is one of the most rewarding aspects of dynamics; however, it can also be the most time consuming. ∗

For the phase portrait shown in Figure 2.13(b), the sticking region s is the interval [−0.124755, 0.124755], and for situation shown in Figure 2.13(d), the sticking region s is the interval [1. − 0.124755, 1.124755].

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2.12 Closing Comments

65 dx dτ

1.5

−2

x

2

−1.5 Figure 2.14. The phase portrait of (2.27)–(2.29). Although it is not evident from the figure, the discrete equilibria of the frictionless case shown in Figure 2.12 are now replaced with families of equilibria that correspond to possible resting (sticking) states for the cart. For this figure, A = 0.25 and µd = 0.1. L 0

In many of the chapters to follow, several more examples of such interpretations are presented, and you are strongly encouraged to take the time to do this when completing the exercises in this book or performing your own research.   f 

µs = 0.3 0.2 Figure 2.15. A graphical method to compute the possi-

ble sticking regions s of the cart on a smooth roller coaster when µs = 0.3. The method is based on  examining (2.30) for the choice (2.31). That is,  f  =     Aπ  πx  L0 sin L01 . Examples of the sticking regions can be seen in Figure 2.13. s

−2

2

s

−2 s

s

s

2 s

s

s

x=

x1 L0

x=

x1 L0

s

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Exercises 2.1–2.5 EXERCISES

2.1. Which of the following force fields are conservative/nonconservative? P = x1 E1 + x3 E2 , P = x2 E1 + x1 E2 , P = x1 x2 E1 , P = −L0 sin(θ)E1 + L0 cos(θ)E2 , where L0 is a constant. For the conservative force fields, what are the associated potential energies? 2.2. Consider a particle of mass m that is moving in E3 . Suppose the only forces acting on the particle are conservative. Starting from the work–energy theorem, prove that the total energy E of the particle is conserved. Suppose during a motion, for which the initial conditions r0 and v0 are known, the position r (t1 ) at some later time t1 is known. Argue that the conservation of energy can be used to determine the speed v of the particle. Give three distinct physical examples of applications of this result. 2.3. In contrast to Exercise 2.2, here consider a particle that is moving on a smooth fixed surface. The constraint force acting on the particle is prescribed by use of Lagrange’s prescription, and the applied forces acting on the particle are conservative. Prove that E is again conserved. In addition, show that the speed of the particle can be determined at a known position r(t1 ) if the initial position and velocity vectors are known. Finally, give three distinct physical examples of the application of this result. 2.4. A particle is free to move on a smooth horizontal surface x3 = 0. At the same time, a rough plane propels the particle in the E1 direction. That is, the constraints on the motion of the particle are 1 = 0 and 2 = 0, where 1 = 1 (r) = x3 ,

2 = 2 (r, t) = x1 − f (t).

Give a prescription for the constraint force acting on the particle. 2.5. Suppose a particle of mass m is in motion and has a position vector r and a velocity vector v. (a) Show that the areal velocity vector A is conserved if the resultant force F acting on the particle is a central force.∗ (b) Show that conservation of angular momentum HO is synonymous with conservation of the areal velocity vector. (c) Suppose a particle is moving on a horizontal table under the action of a spring force, a normal force, and a vertical gravitational force −mgE3 . One end of the spring is attached to the fixed origin O, and the other is attached ∗

A force P is said to be central if P is parallel to r.

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Exercises 2.5–2.6

67

to the particle. The spring has a stiffness K and unstretched length L0 . What can you say about the area swept out by the particle in a given period of time? (d) Derive the equations of motion for the particle in (c). Using the nondimensionalizations  r K τ= t, x= , m L0 and conservation of angular momentum, show that the motion of the particle can be found by integrating the following differential equations: d2 x β2 − 3 = − (x − 1) , dτ2 x

dθ β = 2, dτ x

(2.32)

where β=

L20

h √ Km

and h is a constant that depends on the initial conditions of the motion. For a selection of values of β, e.g., β = −20, −2, −1, 0, 1, 2, 20, construct the phase portraits of (2.32)1 . For a selection of the orbits on each of these phase portraits, construct images of the motion of the particle.∗ (e) Verify that the areal velocity vector is conserved for the motions of the particle you found in (d). 2.6. A particle of mass m is in motion about a fixed planet of mass M. The external force acting on the body is assumed to be a conservative force P. The potential energy UP associated with this force is a function of ||r||, where r is the position vector of the particle relative to the fixed center O of the planet. (a) Prove that r is parallel to P. (b) Show that the angular momentum HO of the particle is conserved and that this conservation implies that the motion of the particle is planar. This plane, which is known as the orbital plane, also contains O. Show that the particle sweeps out equal areas on its orbital plane in equal times. (c) Write out the equations of motion of the particle using a spherical polar coordinate system. (d) Using the conservation of HO, show that the equations of (c) can be simplified to   ∂UP m r¨ − rθ˙2 = − , ∂r mr2 θ˙ = h,

(2.33)

where h is a constant. ∗

Your results will be qualitatively similar to those presented in Section 2.9 for the particle moving on a smooth cone.

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Exercises 2.6–2.7

(e) Show that the solutions to (2.33) conserve the total energy E of the particle. 2.7. A particle of mass m is free to move on the inner surface of a rough sphere of constant radius R0 . The center of the sphere is located at the origin O, and the particle is attached to a fixed point A whose position vector is aEx + bEy by a linear spring of unstretched length L0 and stiffness K. A vertical gravitational force −mgE3 also acts on the particle. (a) Using a spherical polar coordinate system, r = R0 eR , derive expressions for the acceleration vector a and angular momentum HO of the particle. (b) What is the velocity vector of the particle relative to a point on the surface of the sphere? (c) Give a prescription for the constraint force Fc acting on the particle. (d) If the particle is moving relative to the surface, show that the equations governing the motion of the particle are x · eφ mR0 (φ¨ − sin(φ) cos(φ)θ˙2 ) = mg sin(φ) − K (||x|| − L0 ) ||x|| − µd ||N|| 

φ˙

,

φ˙ 2 + sin2 (φ)θ˙2

 d 1 x · eθ mR20 sin2 (φ)θ˙ = −K (||x|| − L0 ) ||x|| R0 sin(φ) dt − µd ||N|| 

sin(φ)θ˙

,

φ˙ 2 + sin2 (φ)θ˙2

where x = R0 eR − aEx − bEy . (e) Show that the normal force exerted by the surface on the particle is   x · eR eR N = mgE3 · eR + K (||x|| − L0 ) ||x|| − mR0 (φ˙ 2 + sin2 (φ)θ˙2 )eR . (f) For the case in which the particle is not moving relative to the surface, show that x Fc = mgE3 + K (||x|| − L0 ) . ||x|| What is the static friction criterion for this case? (g) Show that the total energy of the particle decreases with time if the particle moves relative to the surface. (h) If the spring is removed and the surface is assumed to be smooth, prove that the angular momentum HO · E3 is conserved. Using this conservation, show that the dimensionless equations governing the motion of the particle

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Exercises 2.7–2.8

69

simplify to h d2 φ cos(φ) dθ = , + sin(φ). = h2 3 2 dτ dτ2 sin (φ) sin (φ)     g R0 1 In these equations, h = mR H · E and τ = t. 0 3 2 g R0 0

(i) Suppose that the sphere is smooth. Using the fact that the total energy E of the particle is conserved, show that the criterion for the particle to remain on the surface of the sphere is   mg (3 cos(φ) − 2 cos (φ0 )) − mR0 φ˙ 20 + sin2 (φ0 ) θ˙20 > 0, where φ0 is the value of the initial φ coordinate of the particle and θ˙0 and φ˙ 0 are the initial velocities. Suppose the particle is placed on top of the √ sphere. If the particle is given an initial speed v0 > gR0 , show that it will immediately lose contact with the sphere. 2.8. Consider a particle of mass m whose motion is subject to the following constraints: (xE3 + E2 ) · v = 0,

(E2 ) · v + e(t) = 0.

(2.34)

(a) Show that one of the constraints is integrable whereas the other is nonintegrable. In addition, for the integrable constraint, specify the function (r, t) = 0. (b) Suppose that, in addition to the constraint force Fc = µ1 (xE3 + E2 ) + µ2 E2 , a gravitational force −mgE3 acts on the particle. With the help of the balance of linear momentum F = ma, specify the equations governing the motion of the particle and the constraint forces. (c) With the help of the work–energy theorem T˙ = F · v, prove that the total energy of the particle is not conserved. Give a physical interpretation for this lack of conservation. (d) Using the results from (b), determine the motion of the particle and the constraint force Fc .

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Lagrange’s Equations of Motion for a Single Particle

3.1 Introduction The balance of linear momentum F = ma for a particle can be traced to Newton in the late 17th century. As we have seen, this vector-valued equation yields three differential equations from which the motion of a particle can be determined. In the centuries that followed, alternative principles of mechanics were proposed. Some of them, such as the principle of least action, also yielded equations of motion that were equivalent to those obtained with F = ma. Others did not, and the equivalence of, and interrelationships between, the principles of mechanics remains one of the central issues for any student of dynamics. At the end of the 18th century, a formulation of the equations of motion for a single particle appeared in a famous text by Lagrange [121].∗ Among their attractive features, Lagrange’s equations of motion could easily accommodate integrable constraints, and they (remarkably) have the same canonical form both for single particles and systems of particles as well as for systems of rigid bodies. In this chapter, Lagrange’s equations of motion for a single particle are discussed and several forms of these equations are established. For example [see (3.2)], d dt



∂L ∂ q˙i

 −

∂L = Fncon · ai . ∂qi

Many of the forms presented can be used with dynamic Coulomb friction and nonconservative forces. One of the most important features of our discussion is the emphasis on the equivalence of Lagrange’s equations of motion to F = ma. Although this equivalence is not sufficiently discussed in most textbooks, it can be found in many of the classical texts on dynamics, such as those of Synge and Griffith [207] and Whittaker [228]. A recent paper by Casey [27] explores this equivalence in a transparent manner, and we follow many aspects of his exposition in this chapter. ∗

70

Four editions of Lagrange’s great work, M´ecanique Analytique, appeared in the years 1789, 1811, 1853, and 1888. The last two of these editions were posthumous. An English translation of the second edition was recently published [122].

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3.2 Lagrange’s Equations of Motion

71

Casey’s work will also feature later on when we discuss systems of particles and rigid bodies.

3.2 Lagrange’s Equations of Motion There are several approaches to deriving Lagrange’s equations of motion that appear in the literature. Among them, a variational principle known as Hamilton’s principle (or the principle of least action) is arguably the most popular, whereas an approach based on D’Alembert’s principle was used by Lagrange [121]. Lagrange’s original developments were in the context of mechanical systems subject to holonomic constraints, and his equations were subsequently extended to systems with nonholonomic constraints by Edward J. Routh (1831–1907) (see Section 24 in Chapter IV of [183]) and Aurel Voss (1845–1931) [221].∗ Here, an approach is used that is rooted in differential geometry and is contained in some texts on this subject (see, for example, Synge and Schild [208]). It probably migrated from there to the classic text by Synge and Griffith [207] and has been recently revived by Casey [27]. Two Identities We assume that a curvilinear coordinate system has been chosen for E3 . The velocity vector v consequently has the representation

v=

3 

q˙i ai .

i=1

In addition, the kinetic energy has the representations 3 3 m  m v·v= ai · akq˙i q˙k. 2 2 i=1 k=1  1 2 3 1 2 3 It is crucial to notice that T = T q , q , q , q˙ , q˙ , q˙ . We now consider in succession the partial derivatives of T with respect to the coordinates and their velocities. We wish to establish the following results:

T=

∂T = mv · a˙ i , ∂qi

∂T = mv · ai . ∂ q˙i

These two elegant results form the basis for Lagrange’s equations of motion. First, we start with the derivative of T with respect to a coordinate:  ∂T ∂ m v·v = i i ∂q ∂q 2   ∂v . = mv · ∂qi ∗

For further details on the historical development of Lagrange’s equations, see Papastavridis [167, 169]. Equations of motion (3.8)2,3 are examples of what could be referred to as the Routh–Voss equations of motion.

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Lagrange’s Equations of Motion for a Single Particle ∂T ˙ i , we first note that v = To proceed with the goal of concluding that ∂q i = mv · a 3 ∂ q˙k k k=1 q˙ ak and that ∂qi = 0. As a consequence,  3   ∂ak ∂T k = mv · q˙ . ∂qi ∂qi k=1

∂r The remaining steps use the fact that ak = ∂q k:  3   ∂ak ∂T k = mv · q˙ ∂qi ∂qi k=1

= mv ·

 3  k=1

= mv ·

 3  k=1

 ∂2r ∂2r q˙k i k = q˙k k i ∂q ∂q ∂q ∂q 3

k=1

∂ q˙k k ∂q



∂r = ai ∂qi





= mv · a˙ i .  ∂f k We achieve the last step by noting that f˙ = 3k=1 ∂q ˙ for any function f = kq  1 2 3 f q ,q ,q . The next result, which is far easier to establish, involves the partial derivative of T with respect to a velocity. The reason this result is easier to establish is because the basis vectors ai do not depend on q˙k. Getting on with the proof, we have  3   ∂ q˙k ∂T = mv · ak ∂ q˙i ∂ q˙i  = mv ·

k=1

3 

 δkiak

k=1

= mv · ai . This completes the proofs of both identities. A Covariant Form of Lagrange’s Equations It is crucial to note that Lagrange’s equations are equivalent to F = ma. The form of Lagrange’s equations of motion discussed here is derived from this balance law by taking its covariant components, i.e., dotting it with ai . To start, we consider   ∂T d ∂T d − i = (mv · ai ) − mv · a˙ i dt ∂ q˙i ∂q dt

= ma · ai + mv · a˙ i − mv · a˙ i = ma · ai = F · ai .

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3.3 Equations of Motion for an Unconstrained Particle

In conclusion, we have a covariant form of Lagrange’s equations of motion:   ∂T d ∂T − i = F · ai . i dt ∂ q˙ ∂q

73

(3.1)

We can appreciate some of the beauty of this equation by using it to establish component forms of F = ma for various curvilinear coordinate systems. The Lagrangian Another form of Lagrange’s equations arises when we decompose the force F into its conservative and nonconservative parts:

F = −∇U + Fncon , where the potential energy U = U(q1 , q2 , q3 ). As ∇U =

3  ∂U k a , ∂qk k=1

∂U = 0, ∂ q˙k

we find that Lagrange’s equations can be rewritten in the form     ∂T d ∂T ∂U ∂U − = Fncon · ai . − − dt ∂ q˙i ∂ q˙i ∂qi ∂qi Introducing the Lagrangian L = T − U, we find an alternative form of Lagrange’s equations:   d ∂L ∂L − i = Fncon · ai . (3.2) i dt ∂ q˙ ∂q If there are no nonconservative forces acting on the particle, then the right-hand side of these equations vanishes. In addition, to calculate the equations of motion a minimal amount of vector calculus is required – it is sufficient to calculate v and U.

3.3 Equations of Motion for an Unconstrained Particle To illustrate the ease of Lagrange’s equations, we consider the case in which the curvilinear coordinates chosen are the spherical polar coordinates: q1 = R, q2 = φ, and q3 = θ. For these coordinates, we have a1 = eR ,

a2 = Reφ ,

a3 = R sin(φ)eθ ,

and T=

m ˙2 (R + R2 sin2 (φ)θ˙2 + R2 φ˙2 ). 2

Notice that T does not depend on θ.

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Lagrange’s Equations of Motion for a Single Particle

We obtain Lagrange’s equations of motion for the spherical polar coordinate system by first calculating the six partial derivatives of T: ∂T = mR sin2 (φ)θ˙2 + mRφ˙2 , ∂R ∂T = 0, ∂θ ∂T ˙ = mR2 φ, ∂ φ˙

∂T = mR2 sin(φ) cos(φ)θ˙2 , ∂φ ∂T ˙ = mR, ∂ R˙ ∂T ˙ = mR2 sin2 (φ)θ. ∂ θ˙

Using these results, we find the covariant form of Lagrange’s equations:     ∂T d ∂T 2 2 2 ˙ ˙ ˙ = mR sin (φ)θ + mRφ = F · eR , = mR − dt ∂ R˙ ∂R     ∂T d ∂T 2 ˙ 2 2 ˙ = mR φ − = mR sin(φ) cos(φ)θ = F · Reφ , dt ∂ φ˙ ∂φ     d ∂T ∂T = 0 = F · R sin(φ)eθ . = mR2 sin2 (φ)θ˙ − dt ∂ θ˙ ∂θ

(3.3)

Clearly, these equations were far easier to calculate than an alternative approach that involves differentiating r = ReR twice with respect to t. Let us now suppose that the only force acting on the particle is gravity: F = −mgE3 ,

U = mgE3 · r = mgR cos(φ).

For this case, the Lagrangian L is L= T−U m = (R˙ 2 + R2 sin2 (φ)θ˙2 + R2 φ˙2 ) − mgR cos(φ). 2 We can calculate Lagrange’s equations of motion using L,   ∂L d ∂L − k = 0, dt ∂ q˙k ∂q or by substituting for F in (3.3). It is left as an exercise to show that both approaches are equivalent.

3.4 Lagrange’s Equations in the Presence of Constraints The previous discussion of Lagrange’s equations did not address situations in which constraints on the motion of the particle were present. It is to this matter that we now turn our attention. With integrable constraints, whose constraint forces are prescribed by use of Lagrange’s prescription, the beauty and power of Lagrange’s equations are manifested. In this case, it is possible to choose the curvilinear coordinates qi such that the equations of motion decouple into two sets. The first set describes the unconstrained motion of the particle, and the second set yields the constraint forces as functions of the unconstrained motion.

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3.4 Lagrange’s Equations in the Presence of Constraints

75

There are two approaches to obtaining Lagrange’s equations. We refer to them throughout these sections as Approach I and Approach II. For the novice, we highly recommended the first approach. As in the previous section, our exposition follows that of Casey [27]. Preliminaries We assume that the particle is subject to an integrable constraint,

(r, t) = 0, and a nonintegrable constraint, f · v + e = 0. Further, we assume that the curvilinear coordinates are chosen such that the integrable constraint has the form (r, t) = q3 − d(t) = 0, and that the constraint forces are prescribed by use of Lagrange’s prescription:  3   3 i Fc = λ1 a + λ2 fia . i=1

Notice that f i = f · ai . Suppose that there is an applied force Fa acting on the particle. This applied force can be decomposed into conservative and nonconservative parts: Fa = −∇U + Fancon . The resultant force acting on the particle is  3   F = λ1 a3 + λ2 f i ai − ∇U + Fancon . i=1

The total nonconservative force acting on the particle is Fncon = Fc + Fancon . The kinetic energy of the particle is m m    1 2 3 i k v·v= aik q , q , q q˙ q˙ , 2 2 3

T=

3

i=1 k=1

where aik = ai · ak. Imposing the integrable constraint on T, we find the constrained kinetic energy: T˜ = T˜ 2 + T˜ 1 + T˜ 0 ,

(3.4)

where m  T˜ 2 = a˜ ikq˙i q˙k, 2 2

2

i=1 k=1

T˜ 1 = m

2  i=1

˙ a˜ i3 q˙i d,

m T˜ 0 = a˜ 33 d˙ 2 . 2

(3.5)

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Lagrange’s Equations of Motion for a Single Particle

In these expressions, a˜ ik = a˜ ik(q1 , q2 , t) = aik(q1 , q2 , q3 = d(t)) is the constrained metric tensor. Notice that we use a tilde (˜) to denote imposition of the integrable constraint(s) and that the subscripts on T˜ 2 , T˜ 1 and T˜ 0 refer to the powers of q˙i . A direct calculation shows that∗ ∂T  ∂ T˜ ∂T  ∂ T˜ = , = 2,  3 1 1 2 3 3 3 ˙ ˙ ∂ q˙ q =d,q˙ =d ∂ q˙ ∂ q˙ q =d,q˙ =d ∂ q˙   ˜ ∂T  ∂T ∂T  ∂ T˜ (3.6) = 1, = 2, 3  1 2 3 3 3 ∂q q =d,q˙ =d˙ ∂q ∂q q =d,q˙ =d˙ ∂q ∂T  ∂ T˜ ∂T  ∂ T˜ =  = 0, =  = 0. 3  3 3 3 ∂ q˙ q =d,q˙3 =d˙ ∂ q˙ ∂q q3 =d,q˙3 =d˙ ∂q3 In these relations, the partial derivative of T is evaluated prior to imposing the constraint q3 = d(t). These relations imply that we can use T˜ to obtain the first two Lagrange’s equations of motion, but not the third. Results that are identical in form ˜ and U and U. ˜ to (3.6) pertain to the partial derivatives of L and L Notice that we did not impose the nonintegrable constraint on the kinetic energy T and the Lagrangian L. It is possible to do this, but the result is not useful to us here. Approach I In the first approach, we evaluate the partial derivatives in Lagrange’s equations of motion (3.2) in the absence of any constraints:   d ∂L ∂L − k = Fancon · ak + (Fc = 0) · ak. dt ∂ q˙k ∂q

Explicitly, these equations are   3 3  3   m ∂air i r ∂U d ai1 q˙i − q˙ q˙ + 1 = Fancon · a1 , m dt 2 ∂q1 ∂q 

i=1

i=1 r=1



   m ∂air d ∂U ai2 q˙i − q˙i q˙r + 2 = Fancon · a2 , m dt 2 ∂q2 ∂q 

d m dt

3

i=1

3 

3

i=1 r=1

 i

ai3 q˙

i=1

3



3  3  m ∂air i=1 r=1

2 ∂q3

q˙i q˙r +

∂U = Fancon · a3 . ∂q3

Notice that we have not introduced the constraint forces on the right-hand side of these equations. That is, in the preceding equations F = −∇U + Fancon . We now impose the integrable constraint q3 = d(t) and introduce the nonintegrable constraint and the constraint forces. The resulting equations govern the ∗

Suppose g = 10t2 . Then



∂g  ∂t t=5

= 2(10)(5) = 100. In words, we evaluate the derivative of g with

respect to t and then substitute t = 5 in the resulting function.

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3.4 Lagrange’s Equations in the Presence of Constraints

77

motion of the particle and the constraint forces: q3 = d, ˙ q˙3 = d, f 1 q˙1 + f 2 q˙2 + f 3 d˙ + e = 0,

  3 3  3  d  m ∂air i r ∂U i mai1 q˙ − q˙ q˙ + 1 = λ2 f 1 + Fancon · a1 , dt 2 ∂q1 ∂q 

i=1



i=1 r=1

   m ∂air d ∂U ai2 q˙i − q˙i q˙r + 2 = λ2 f 2 + Fancon · a2 , m 2 dt 2 ∂q ∂q 

3

i=1



3

3

i=1 r=1

   m ∂air d ∂U ai3 q˙i − q˙i q˙r + 3 = λ1 + λ2 f 3 + Fancon · a3 . m 3 dt 2 ∂q ∂q 3

i=1

3

3

(3.7)

i=1 r=1

We have refrained from ornamenting U, f i , ai , and aik with a tilde in the last four of these equations. It is crucial to notice that if the nonintegrable constraint were absent, then (3.7) would reduce to two sets of equations. The first of these sets, (3.7)4,5 , would yield differential equations for the unconstrained motion, q1 (t), and q2 (t), of the particle, whereas the second set, (3.7)6 , would provide the constraint force Fc = λ1 a3 acting on the particle. Approach II ˜ = T˜ − U. ˜ Here, as In Approach II, we work directly with L

  ˜ =L ˜ q1 , q2 , q˙1 , q˙2 , t , L ˜ with respect to q3 and q˙3 are zero. Consequently, using the partial derivatives of L (3.6) and (3.2), we find that there are only two Lagrange’s equations:  ˜  ∂L − ∂ q˙1  ˜  d ∂L − dt ∂ q˙2 d dt

˜ ∂L = Fc · a˜ 1 + Fancon · a˜ 1 , ∂q1 ˜ ∂L = Fc · a˜ 2 + Fancon · a˜ 2 . ∂q2

Introducing the expression for the constraint force Fc and the nonintegrable constraint, we find the equations governing λ2 , q1 (t), and q2 (t) are f 1 q˙1 + f 2 q˙2 + f 3 d˙ + e = 0,  ˜  ˜ d ∂L ∂L − 1 = λ2 f 1 + Fancon · a˜ 1 , 1 dt ∂ q˙ ∂q  ˜  ˜ ∂L d ∂L − 2 = λ2 f 2 + Fancon · a˜ 2 . 2 dt ∂ q˙ ∂q

(3.8)

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Lagrange’s Equations of Motion for a Single Particle

Notice that λ1 does not feature in these equations. In addition, if no nonintegrable constraint were present, then the differential equations provided by Approach II are all that are needed to determine q1 (t) and q2 (t). Equations (3.8)2,3 are examples of the Routh–Voss equations of motion.

3.5 A Particle Moving on a Sphere To clarify the two approaches just discussed, we now consider the example of a particle moving on a smooth sphere whose radius R is a known function of time: R = d(t). The particle is subject to a conservative force −mgE3 and a nonconservative force DReθ , where D is a constant. Later on, we shall impose a nonintegrable constraint on the motion of the particle. For the problem at hand, it is convenient to use a spherical polar coordinate system: q1 = θ,

q2 = φ,

q3 = R.

Using this coordinate system, we can write the integrable constraint R = d(t) in the form (r, t) = R − d(t) = 0. As the sphere is smooth, we can use Lagrange’s prescription, Fc = λeR . The kinetic and potential energies of a particle in the chosen coordinate system are  m  ˙2 R + R2 sin2 (φ)θ˙2 + R2 φ˙2 , U = mgR cos(φ). T= 2 The constrained kinetic and potential energies are  m  ˙2 T˜ = d + d2 sin2 (φ)θ˙2 + d2 φ˙2 , 2

U˜ = mgd cos(φ).

(3.9)

Finally, the covariant basis vectors are a1 = R sin(φ)eθ ,

a2 = Reφ ,

a3 = eR .

Their constrained counterparts a˜ i are easily inferred from these expressions. First, we use Approach II to obtain the equations governing θ(t) and φ(t). There are two equations:  ˜ ˜ d ∂L ∂L − = F · a˜ 1 dt ∂ θ˙ ∂θ = Fc · a˜ 1 + Ddeθ · a˜ 1 = Dd2 sin(φ),

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3.5 A Particle Moving on a Sphere

d dt

79

 ˜ ˜ ∂L ∂L = F · a˜ 2 − ˙ ∂φ ∂φ = Fc · a˜ 2 + Ddeθ · a˜ 2 = 0.

Evaluating the partial derivatives of the constrained Lagrangian, we find that these equations become d ˙ = Dd2 sin(φ), (md2 sin2 (φ)θ) dt d 2  md φ˙ − md2 sin(φ) cos(φ)θ˙2 − mgd sin(φ) = 0. (3.10) dt Notice that the constraint force λeR is absent from these equations. Alternatively, using Approach I, we start with the unconstrained Lagrangian L and establish three equations of motion [cf. (3.3)]:     d ∂L ∂L 2 2 ˙ =0 = mR sin (φ)θ − dt ∂ θ˙ ∂θ

d dt



d dt

= Fancon · R sin(φ)eθ ,   ∂L ∂L = mR2 sin(φ) cos(φ)θ˙2 + mgR sin(φ) = mR2 φ˙ − ∂φ ∂ φ˙ 



= Fancon · Reφ ,    ∂L ∂L = mR˙ − = mR sin2 (φ)θ˙2 + mRφ˙2 − mg cos(φ) ∂R ∂ R˙ = Fancon · eR .

Next, we impose the integrable constraint and introduce the constraint force Fc to find the equations of motion: d ˙ = Dd2 sin(φ), (md2 sin2 (φ)θ) dt d ˙ − md2 sin(φ) cos(φ)θ˙2 − mgd sin(φ) = 0, (md2 φ) dt d ˙ − (md sin2 (φ)θ˙2 + mdφ˙2 − mg cos(φ)) = λ. (md) dt

(3.11)

Notice that the first two of these equations are identical to (3.10), whereas the third equation is an equation for the constraint force Fc . We could now introduce an additional constraint: f 1 θ˙ + f 2 φ˙ + f 3 R˙ + e = 0. Using Lagrange’s prescription, we find that the total constraint force on the particle is   f1 f2 Fc = λeR + λ2 eθ + eφ + f 3 eR . R sin(φ) R

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To obtain the equations of motion for the case in which the nonintegrable constraint is active, we need to introduce only the constraint force associated with the nonintegrable constraint on the right-hand side of (3.11) and to append the nonintegrable constraint to the resulting equations: f 1 θ˙ + f 2 φ˙ + f 3 d˙ + e = 0, d ˙ = Dd2 sin(φ) + λ2 f 1 , (md2 sin2 (φ)θ) dt d ˙ − md2 sin(φ) cos(φ)θ˙2 − mgd sin(φ) = λ2 f 2 , (md2 φ) dt d ˙ − (md sin2 (φ)θ˙2 + mdφ˙2 − mg cos(φ)) = λ + λ2 f 3 . (md) dt It is left as an exercise to show what additional simplifications to these equations arise if the nonintegrable constraint were integrable with f 1 = 0, f 2 = 1, f 3 = 0, and e = 0. In this case, one will see that the particle moves on a circle of radius d sin(φ0 ).

3.6 Some Elements of Geometry and Particle Kinematics As a prelude to our discussion of Lagrange’s equations and their geometrical significance, some material from differential geometry needs to be presented. Our treatment is limited to the ingredients we shall shortly need and, as such, it cannot do justice to this wonderful subject. Mercifully, there are several excellent texts that can be recommended to remedy this: [47, 149, 155, 201]. Reading Chapter 1 of Lanczos [124] for a related discussion on kinetic energy and geometry and the recent paper ¨ by Lutzen [132] for a historical overview of the interaction between geometry and dynamics in the 19th century is highly recommended. Here, we are interested in surfaces and curves that are in E3 . We assume that these entities are smooth. That is, they are without edges and sharp corners, and we call them manifolds. In the case of a curve, a single coordinate is needed to locally parameterize the points P on this manifold, and so it is considered to be a one-dimensional manifold. For a surface, two coordinates are needed to locally parameterize the points on the surface and so the surface is considered to be a two-dimensional manifold. In an obvious generalization, subsets of E3 such as solid spheres and solid ellipsoids are considered to be three-dimensional manifolds. Previously, in Section 1.5, curvilinear coordinates were introduced. For a given surface (or curve) we used these coordinates both to label points on the manifold and to define the manifold. For example, for a sphere of radius R0 , the spherical polar coordinates φ and θ label points on the sphere and the coordinate R can be used to define the sphere: R = R0 . Similarly, for a circle, the cylindrical polar coordinate θ can be used to label points on the circle, and the coordinates r and z can be used to define the circle. The curvilinear coordinate system we use to label points on the manifold is known as a chart. For some manifolds, such as a plane, a straight line, and a circle, a single chart suffices to enable the labeling of each point on the

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3.6 Some Elements of Geometry and Particle Kinematics

(b)

(a)

81

TP S P

TP C C

P S

Figure 3.1. Two examples of manifolds and the tangent spaces to points on them: (a) a curve C and (b) a sphere S.

manifold. For surfaces such as spheres, for which a set of spherical polar coordinates will not be defined at the poles, at least two charts are needed. With terminology borrowed from cartography, the set of all charts for a manifold is known as an atlas. At each point P of a manifold M we define a tangent space, and we denote this space by TP M. If the manifold is n-dimensional, then TP M is also n-dimensional. For example, the tangent space TP C is a line for the curve shown in Figure 3.1(a), and the tangent space TP S is a plane for the sphere S shown in Figure 3.1(b). Continuing with the sphere as an example, if we fix a point P on the sphere, this is equivalent to fixing the polar coordinates φ = φ0 and θ = θ0 . The vectors eφ = eφ (φ0 , θ0 ) = sin (φ0 ) (cos (θ0 ) E1 + sin (θ0 ) E2 ) + cos (φ0 ) E3 , eθ = eθ (θ0 ) = − sin (θ0 ) E1 + cos (θ0 ) E2 , form a basis for the tangent space TP S at P, and any tangent vector to the sphere at this point can be expressed in terms of these vectors. Related remarks apply at a point on a curve, but now only a single vector is needed to span the tangent space. As a final example, for a particle that is free to move in M = E3 , the dimension of TP M is 3. Returning to the example of a sphere, we choose two points P1 and P2 on the sphere of radius R and consider a path V between them (see Figure 3.2). We wish to measure the distance one would travel along V. To do this, we first parameterize the curve with a parameter u, where u = uα at Pα . The curve can then be uniquely described by the functions θ(u) and φ(u).∗ To determine the distance s traveled along the curve, one method would be to evaluate the following integral:  u2  dφ dφ dθ dθ + R2 du. (3.12) s = R2 sin2 (φ) du du du du u1 Referring to (3.4), (3.5), and (3.9), we can express the integrand on the right-hand side of this equation in terms of the kinetic energy of a particle moving on the ∗

For example, if the curve were a segment of the equator, then θ (u) = θ (u1 ) +

(u2 −u1 ) 2πR

and φ(u) =

π 2.

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P1 Figure 3.2. A curve V connecting two points on a

V

sphere S. The velocity vector vrel of a particle moving on this curve would lie in the tangent plane TP S at each point P of the curve.

P2 vrel S

sphere: 



u2

s =

u1







 dt du du

2T˜ 2 dt = m

t2

=

2T˜ 2 m

t1



t2

||vrel || dt.

t1

For the second integral, we changed variables and parameterized the path by using t rather than u. Turning to the example of a circle, the reader is invited to show that a measure of distance corresponding to (3.12) can be established by using the single polar coordinate θ. For M = E3 , the measure of distance can be defined in a standard manner by use of Cartesian coordinates:   u2  3  dxk dxk  s = du. du du u1 k=1

t  It is easy to see that this expression can be rewritten as t12 2T dt. m From the previous discussion, for an n-dimensional manifold M, a measure of distance similar to that provided by (3.12) can be established∗ :  s =

u2

n  n 

u1

a˜ ik

i=1 k=1

∂qi ∂qk du. ∂u ∂u

Parameterizing the path by using time t instead of a variable u, we find  t2  n  n s = a˜ ikq˙i q˙kdt. t1



(3.13)

i=1 k=1

The index n here is either 1, 2, or 3. In later chapters, we shall see that n can range from 1 to 3N for a system of N particles and from 1 to 6 for a single rigid body.

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3.7 The Geometry of Lagrange’s Equations of Motion

83

Following the advocacy of Hertz [92], we can conveniently imagine a (representative) particle of mass m moving on a manifold M. The manifold is known as the configuration manifold. The velocity of the particle relative to M is simply  vrel = nk=1 q˙ka˜ k. Thus (3.13) can be expressed as  t2 ||vrel || dt. (3.14) s = t1

We see from this expression that, in defining s, we have also defined a measure of the magnitude of a vector vrel ∈ TP M. Such a measure is known as a metric, and a manifold that is equipped with a metric is known as a Riemannian manifold. The terminology here pays tribute to Georg F. B. Riemann (1826–1866) and his remarkable work [179]. Distance measure (3.13) is clearly intimately related to the kinetic energy of a particle, and, following Synge [205], ds =

n  n 

a˜ ikq˙i q˙kdt

(3.15)

i=1 k=1

is known as the kinematical line-element. This measure of distance has a long history with important contributions by several esteemed figures such as Jacobi [103] and Ricci and Levi-Civita [178]. Other choices of ds are available (see, for instance, [123, 124, 149, 205]), and some of them feature prominently in relativistic mechanics. The freedom of selection is similar to the notion that different measures of distance such as meters and feet are possible and is intimately related to Einstein’s theory of relativity. We remarked earlier on a (representative) particle of mass m moving on M. Clearly, for a single particle, such a construction can be easily achieved. Indeed, for a single particle the configuration manifold corresponds to the physical surface or curve that the particle moves on. However, for a system of particles or rigid bodies subject to constraints, this is not the case, and the construction of a single representative particle is nontrivial. Indeed, for a system of particles, such a construction was explicitly recorded only rather recently by Casey [27].∗ Subsequently, he extended this construction to a single rigid body [28] and a system of rigid bodies [30].

3.7 The Geometry of Lagrange’s Equations of Motion Some readers will have gained the perspective that the Lagrange’s equations of motion obtained by use of Approach II are projections of F = ma onto the covariant basis vectors for the unconstrained coordinates. That is, we are projecting F = ma onto the basis for TP M. For those who have not yet found this perspective, let us recall the example of the particle moving on the sphere of radius R = d(t). There, we obtained the two Lagrange’s equations for the θ and φ by taking the d sin(φ)eθ and deφ components ∗

We shall examine his construction in Section 4.7.

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of F = ma. These two vectors, d sin(φ)eθ and deφ , form a basis for the tangent space TP S to a point P of the sphere S. Furthermore, because the constraint force associated with the integrable constraint λeR is perpendicular to the sphere, this force did not appear in the two Lagrange’s equations. An important feature of nonintegrable constraints is that the constraint force associated with these constraints is not decoupled from the equations governing the unconstrained motion. This deficiency in Lagrange’s equations of motion can be removed by use of alternative forms of F = ma that are suited to nonintegrably constrained systems, but we do not approach this vast subject here. We now delve a little more deeply into the geometry inherent in Lagrange’s equations of motion. Our discussion is based on Casey [27], Lanczos [124], and Synge [205]. A Particle Subject to a Single Integrable Constraint First, let us consider the case in which the particle is subject to a single integrable constraint:

(r, t) = q3 − d(t) = 0, where d3 (t) is a known function. When considered in E3 , the constraint = 0 represents a moving two-dimensional surface: In this case, a q3 coordinate surface. As mentioned earlier, this surface is known as the configuration manifold M (see Figure 3.3). The velocity of the particle relative to this surface has the representation vrel = q˙1 a˜ 1 + q˙2 a˜ 2 . The coordinates q1 and q2 in this case are known as the generalized coordinates, and the number of these coordinates is the number of degrees-of-freedom of the particle. Thus an unconstrained particle has three degrees-of-freedom, whereas the particle constrained to move on the surface has only two. Recall that at each point P of M, {a1 , a2 } evaluated at P is a basis for the tangent plane TP M to M at P. We can also define a relative kinetic energy Trel = T˜ 2 :  m m vrel · vrel = a˜ i · a˜ kq˙i q˙k. 2 2 2

Trel =

2

i=1 k=1

We now consider a particle moving on M. To calculate the distance traveled by the particle on the surface in a given interval t1 − t0 of time t, we integrate the magnitude of its velocity vrel with respect to time:  s(t1 ) − s(t0 ) =

t1

t0

 = t0

t1

√ vrel · vrel dt 

2Trel dt. m

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3.7 The Geometry of Lagrange’s Equations of Motion

q2 coordinate curve a3

q1 coordinate curve a2

M

a1

q3 = d(t) coordinate surface O Figure 3.3. The configuration manifold M of a particle moving on a surface. Here, the coor-

dinates q1 and q2 are known as the generalized coordinates and M is the q3 = d(t) coordinate surface.

We can differentiate this result with respect to t to find the kinematical line-element ds:    2 2   2Trel ds = (3.16) a˜ i · a˜ kdqi dqk. dt =  m i=1 k=1

As emphasized previously in Section 3.6, notice that the measure of distance is defined by the kinetic energy Trel . It is left as an exercise to show that the integral of (3.16) is none other than (3.12). With regard to Lagrange’s equations of motion, suppose that the constraint forces associated with the integrable constraint are prescribed by use of Lagrange’s prescription: Fc = λa˜ 3 . Then Fc · a˜ 1 = Fc · a˜ 2 = 0, and the constraint force does not appear in the first two Lagrange’s equations. The forces Q1 = F · a˜ 1 ,

Q2 = F · a˜ 2

are known as the generalized forces, and the expressions we use for them are equivalent to those used in other texts on dynamics (for example, [14, 80]). In addition, Approach II can be used to obtain the differential equations governing q1 (t) and q2 (t):  ˜  ˜ ∂L d ∂L − = Fancon · a˜ 1 , dt ∂ q˙1 ∂q1  ˜  ˜ ∂L d ∂L − 2 = Fancon · a˜ 2 . (3.17) 2 dt ∂ q˙ ∂q Imposing a nonintegrable constraint on the motion of the particle will not change M. Furthermore, this constraint will, in general, introduce constraint forces into

85

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Equations (3.17). These forces will destroy the decoupling that Lagrange’s equations achieve for integrable constraints. A Particle Subject to Two Integrable Constraints We now turn to the case in which a particle is subject to two integrable constraints:

1 (r, t) = q3 − d3 (t) = 0,

2 (r, t) = q2 − d2 (t) = 0.

Notice that we have chosen the curvilinear coordinates so that the constraints are easily represented. At each instant of time, the intersection of the two surfaces 1 = 0 and 2 = 0 in E3 defines a curve – in this case a q1 coordinate curve (see Figure 3.4). The configuration manifold M in this case corresponds to the q1 coordinate curve. This coordinate is the generalized coordinate for the system. We can easily represent the velocity vector of the particle relative to M, which we again denote by vrel : vrel = q˙1 a˜ 1 . It should be clear that a˜ 1 is tangent to M. Indeed, a1 evaluated at P is a basis vector for the one-dimensional space TP M. In addition, we can associate with vrel a relative kinetic energy: Trel =

m m vrel · vrel = a˜ 1 · a˜ 1 q˙1 q˙1 . 2 2

Paralleling previous developments [see (3.15)], the kinematical line-element for M is   2Trel ds = dt = a˜ 1 · a˜ 1 dq1 dq1 . m

m a˜ 1 r

O

q3 = d3 (t) coordinate surface q2 = d2 (t) coordinate surface

Figure 3.4. A particle moving on a curve. In this figure, a˜ 1 is tangent to the q1 coordinate curve corresponding to q2 = d2 (t) and q3 = d3 (t). That is, the q1 coordinate curve is the configuration manifold M. The vectors a˜ 2 and a˜ 3 , which are not shown, are normal to M.

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3.8 A Particle Moving on a Helix

87

increasing s & θ eb en et Figure 3.5. A helix and its associated Frenet triad {et , en , eb}. Here, et is the unit tangent vector, en is the unit principal normal vector, and eb = et × en is the binormal vector.

eb en et E3 E1

O

E2

With regard to Lagrange’s equations of motion, if the constraint forces are prescribed by use of Lagrange’s prescription, Fc = λ1 a˜ 3 + λ1 a˜ 2 , then we can easily find the differential equation governing q1 (t) by using Approach II:  ˜  ˜ ∂L d ∂L − = Fancon · a˜ 1 . dt ∂ q˙1 ∂q1   ˜ =L ˜ q1 , q˙1 , t and F · a˜ 1 is the generalized force for this problem. Here, L

3.8 A Particle Moving on a Helix As an illustrative example, we turn our attention to establishing results for a particle that is in motion on a helix (see Figure 3.5). The helix can be either rough or smooth, and a variety of applied forces are considered. This example is interesting for several reasons. First, it is a prototypical problem to illustrate how the Serret–Frenet formulae and the Frenet triad {et , en , eb} help to determine the motion of a particle on a space curve. Second, we can use this example to illustrate a nonorthogonal curvilinear coordinate system.∗ Curvilinear Coordinates, Basis Vectors, and Other Kinematics A helix is defined by the intersection of two surfaces: a cylinder r = R and a helicoid z = cθ, where c and R are constants. To conveniently define these surfaces, we define a curvilinear coordinate system:    x2 q1 = θ = tan−1 , q2 = r = x21 + x22 , x1    x2 . q3 = η = z − αrθ = x3 − α x21 + x22 tan−1 x1 ∗

This is a coordinate system in which ai are not necessarily parallel to ai .

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It is appropriate to notice that r = x1 E1 + x2 E2 + x3 E3 = r cos(θ)E1 + r sin(θ)E2 + (η + αrθ) E3 . Our labeling of the coordinates minimizes subsequent manipulations. You should note that the curvilinear coordinate system is not defined when r = 0. That is, it has the same singularities as the cylindrical and spherical polar coordinate systems. The coordinates θ, r, and η can be used to define bases for E3 : a1 = reθ + αrE3 ,

a2 = er + αθE3 ,

a3 = E3 .

In addition, using the representation of the gradient in cylindrical polar coordinates, we find that the contravariant basis vectors are a1 =

1 eθ , r

a2 = er ,

a3 = E3 − αθer − αeθ .

j

You should notice that ai · a j = δi , as expected. Further, neither the covariant basis nor the contravariant basis is orthogonal. You may recall that the Frenet triad for the helix of radius R is (from [159]) et = √

1 1 + α2

(eθ + αE3 ) ,

1 eb = √ (E3 − αeθ ) . 1 + α2

en = −er ,

Furthermore, the torsion τ, curvature κ, and arc-length parameter s of the helix are τ=

α , R(1 + α2 )

κ=

1 , R(1 + α2 )

s = R 1 + α2 (θ − θ0 ) − s0 .

These results also apply to a helix for which α and R are functions of time. You should verify that a1 is parallel to et and that a2 and a3 are in the plane formed by en and eb. For a particle moving freely in E3 , we have the general representation v = 3 i i=1 q˙ ai . From this result, we can immediately write ˙ θ + αrE3 ) + r(e v = θ(re ˙ r + αθE3 ) + ηE ˙ 3. Furthermore, the kinetic energy of the particle is  m 2 ˙2 . r˙ + r2 θ˙2 + (η˙ + αrθ T= ˙ + αrθ) 2 When the particle is in motion on the helix, it is subject to two constraints 1 = 0 and 2 = 0: 1 (r, t) = q2 − R, 2 (r, t) = q3 . In preparation for writing expressions for the constraint forces acting on a particle moving on the helix, you should calculate the gradient of these two functions. You might also notice that θ is the generalized coordinate for a particle moving on the helix that we will be using.

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89

Forces We assume that an applied force Fa acts on the particle. In addition, we assume that the friction is of the Coulomb type. Consequently, if the particle is moving relative to the helix,

F = Fa + λ1 a˜ 2 + λ2 a˜ 3 + F f , where   θ˙a˜ 1 F f = −µd λ1 a˜ 2 + λ2 a˜ 3    . θ˙a˜ 1  On the other hand, if the particle is not moving relative to the helix, i.e., θ is constant, then F = Fa + λ1 a˜ 2 + λ2 a˜ 3 + λ3 a˜ 1 . The friction force in this case is subject to the static friction criterion:   F f  ≤ µs ||N|| , where



a˜ 1 F f = (λ1 a˜ + λ2 a˜ + λ3 a˜ ) · ||a˜ 1 || 2

3

1



a˜ 1 , ||a˜ 1 ||

N = λ1 a˜ 2 + λ2 a˜ 3 + λ3 a˜ 1 − F f . Balance of Linear Momentum and Lagrange’s Equations For an unconstrained particle moving in E3 , we have the three Lagrange’s equations:   ∂T d ∂T − i = F · ai . i dt ∂ q˙ ∂q

For the present coordinate system θ, r, η, these equations read   d ∂T ˙ ˙ + αrθ) = mr2 θ˙ + mαr(η˙ + αrθ dt ∂ θ˙   ∂T ˙ = F · (a1 = reθ + αrE3 ), = mαr( ˙ η˙ + αrθ ˙ + αrθ) − ∂θ   d ∂T ˙ = mr˙ + mαθ(η˙ + αrθ ˙ + αrθ) dt ∂ r˙   ∂T ˙ η˙ + αrθ ˙ = F · (a2 = er + αθE3 ), − ˙ + αrθ) = mrθ˙2 + mαθ( ∂r     d ∂T ∂T ˙ = m(η˙ + αrθ ˙ + αrθ) − = 0 = F · (a3 = E3 ) . dt ∂ η˙ ∂η Equations of Motion for the Particle on the Helix We obtain the equations of motion for the particle on the helix from the preceding equations by substituting for the resultant force and imposing the constraints. With

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some algebra, for the case in which the particle is moving relative to the helix, we find three equations:    d θ˙ m(1 + α2 )R2 θ˙ = Fa · a˜ 1 − µd λ1 a˜ 2 + λ2 a˜ 3  ||a˜ 1 || , ˙ dt |θ|   θ˙a˜ 1 · a˜ 2  d 2 mα Rθθ˙ − m(1 + α2 )Rθ˙2 = Fa · a˜ 2 + λ1 − µd λ1 a˜ 2 + λ2 a˜ 3    , θ˙a˜ 1  dt   θ˙a˜ 1 · a˜ 3  d mαRθ˙ = Fa · a˜ 3 + λ2 − µd λ1 a˜ 2 + λ2 a3    , θ˙a˜ 1  dt where a˜ 1 = Reθ + αRE3 ,

a˜ 2 = er + αθE3 ,

a˜ 3 = E3 .

These three equations provide a differential equation for the unconstrained motion of the particle and two equations for the unknowns λ1 and λ2 . For the case in which the motion of the particle is specified (i.e., the particle is not moving relative to the helix), we find, from F = ma, three equations for the three unknowns: λ3 = −Fa · a˜ 1 ,

λ2 = −Fa · a˜ 2 ,

λ1 = −Fa · a˜ 3 .

It remains to use the static friction criterion, but this is left as an easy exercise. The Particle on a Smooth Helix In this case,

F = Fa + λ1 a˜ 2 + λ2 a˜ 3 , and because Fc · a˜ 1 = 0, Lagrange’s equations of motion decouple:  d m(1 + α2 )R2 θ˙ = Fa · a˜ 1 , dt  d 2 mα Rθθ˙ − m(1 + α2 )Rθ˙2 = Fa · a˜ 2 + λ1 , dt  d mαRθ˙ = Fa · a˜ 3 + λ2 . dt Consequently, the desired differential equation is m(1 + α2 )R2 θ¨ = Fa · (Reθ + αRE3 ), and the constraint force is Fc = λ1 a˜ 2 + λ2 a˜ 3     ¨ − mRθ˙2 − Fa · a˜ 2 a˜ 2 + mαRθ¨ − Fa · a˜ 3 a˜ 3 . = mα2 Rθθ Once θ as a function of time has been calculated from the ordinary differential equation, then Fc as a function of time can be determined.

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3.9 Summary

91

To illustrate the previous equations, consider the case in which the applied force is gravitational, Fa = −mgE3 . Then, from the preceding equations, m(1 + α2 )R2 θ¨ = −mgαR.

(3.18)

˙ 0 ) = ω0 , this equation has the soluSubject to the initial conditions θ(t0 ) = θ0 and θ(t tion gα θ(t) = θ0 + ω0 (t − t0 ) − (t − t0 )2 . 2R(1 + α2 ) Using this result, we find that the constraint force is   mgαθ(t) mg 2 ˙ Fc = − mRθ (t) a˜ 2 + (E3 − α(θer + eθ )) , 2 1+α 1 + α2 where θ(t) is as previously given. Some Observations Suppose one is interested in determining only the differential equation governing the unconstrained motion of the particle moving on a smooth helix. In other words, the constraint forces are of no concern. One can obtain this differential equation by imposing the constraints on the expression for T:

 m  m 2 2 ˙2 = T˜ = R θ˙ + (αRθ) (1 + α2 )R2 θ˙2 . 2 2 Furthermore,   ∂ T˜ d ∂ T˜ − = F · a˜ 1 = F · (Reθ + αRE3 ) = Fa · (Reθ + αRE3 ) . ˙ dt ∂ θ ∂θ A quick calculation shows that the resulting differential equation is identical to that obtained previously [(3.18)]. ˜ have Clearly, Lagrange’s equations calculated with Approach II (i.e., with T) their advantages, but they cannot accommodate dynamic friction forces. It is, however, the standard approach to Lagrange’s equations in the literature and textbooks. ˜ ˜ ˜ ˜ You should note that ∂∂Tr˙ = ∂∂Tη˙ = ∂∂rT = ∂∂ηT = 0. Consequently, we cannot recover the other two Lagrange’s equations once we have imposed the constraints.

3.9 Summary In this chapter, several forms of Lagrange’s equations of motion for a particle were presented. The most fundamental of these forms is [see (3.1)]   d ∂T ∂T − i = F · ai . dt ∂ q˙i ∂q In one of the following exercises, we establish two other forms of these equations by expanding the partial derivatives with respect to the coordinates and their velocities. These two forms are a covariant form (3.22) and a contravariant form (3.23). If we

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Exercise 3.1

decompose the forces acting on the particle into conservative and nonconservative forces, then we can transform (3.1) to (3.2):   ∂L d ∂L − i = Fncon · ai . dt ∂ q˙i ∂q Now suppose that an integrable constraint is imposed on the particle, that this constraint can be written as q3 − f (t) = 0, and that the constraint force associated with this constraint is Fc = λa3 . In this case, Lagrange’s equations of motion can be used to readily provide a set of differential equations for the generalized coordinates q1 and q2 :  ˜  ˜ d ∂L ∂L α = 1, 2. (3.19) − α = Fncon · a˜ α , α dt ∂ q˙ ∂q ˜ that we obtain from L by These equations feature the constrained Lagrangian L 3 imposing the integrable constraint q = f (t), and, most important, do not feature λ. That is, equations of motion (3.19) are reactionless. This case, in which all the constraints are integrable and the constraint forces are prescribed by use of Lagrange’s prescription, is an example of a mechanical system subject to “ideal constraints.” We also discussed the situation in which nonintegrable constraints were imposed on the system and outlined how the equations of motion could be obtained in these circumstances. The imposition of nonintegrable constraints will not affect the number of generalized coordinates, the configuration manifold, or the kinematical line-element. The summary just presented will be identical for systems of particles, rigid bodies, and systems of both particles and rigid bodies. The only major differences are that the calculation of the kinetic energy becomes significantly more complicated for these systems and that the right-hand sides of Lagrange’s equations feature several forces and moments. Despite these differences, the decoupling of the equations of motion into a set of reactionless equations governing the generalized coordinates will hold if Lagrange’s prescription for the constraint forces is used. This is one of the most remarkable features of Lagrange’s equations for systems subject to integrable constraints. EXERCISES

3.1. Recall from Exercise 1.5 in Chapter 1 that, for a parabolic coordinate system {u, v, θ}, a1 =

∂r = ver + uE3 , ∂u a3 =

a2 =

∂r = uer − vE3 , ∂v

∂r = uveθ , ∂θ

and a1 =

1 a1 , u2 + v2

a2 =

1 a2 , u2 + v2

a3 =

1 eθ . uv

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Exercises 3.1–3.2

93

(a) Consider a particle of mass m that is acted on by a force F and is free to move in E3 . Show that the equations of motion of the particle are  d m(u2 + v2 )u˙ − m(u˙ 2 + v˙ 2 )u − mv2 uθ˙2 = F · a1 , dt  d m(u2 + v2 )v˙ − m(u˙ 2 + v˙ 2 )v − mu2 vθ˙2 = F · a2 , dt d 2 2  mu v θ˙ = F · a3 . dt (b) Next, we are interested in a particle that is moving on the parabolic surface of revolution: c2 = −z +

z2 + r2 ,

where c is a constant. A vertical gravitational force −mgE3 acts on the particle. Using the results of (a), derive the equations governing the unconstrained motion of the particle and show that the normal force acting on the particle is   N = − mu˙ 2 c + mu2 cθ˙2 + mgc a2 . Show that the two second-order differential equations governing the generalized coordinates can be written as a single second-order differential equation:   m(u2 + c2 ) u¨ + mu˙ 2 u −

h2 = −mgu, mu3 c2

(3.20)

where h is a constant. (This constant is none other than HO · E3 , which is an integral of motion). Noting that the units of u and c are meters1/2 , what is a dimensionless form of equations of motion (3.20)? (c) Show that the solutions of (3.20) conserve the energy E=

h2 mg 2 m 2 (u + c2 )u˙ 2 + (u − c2 ). + 2 2 2 2mu c 2

How does one arrive at this expression for E? 3.2. For many mechanical systems a canonical form of Lagrange’s equations can be established that is suited to numerical integrations. Here, we establish one such form [see (3.23)].∗ This problem is adapted from the texts of McConnell [139] and Synge and Schild [208]. We take this opportunity to note that (3.23) can be found in an early paper by Ricci and Levi-Civita [178]. We start by recalling the covariant component forms of Lagrange’s equations of motion for a particle that is in motion under the influence of a resultant external ∗

As will become evident from the developments of later chapters, a related form can be established for any mechanical system that features scleronomic integrable constraints and constraint forces and moments that are prescribed by use of Lagrange’s prescription.

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Exercise 3.2

force F =

3 i=1

Fi ai =

3 i=1

F i ai ∗ :   d ∂T ∂T − k = F · ak, k dt ∂ q˙ ∂q

where m  aikq˙i q˙k, 2 3

T = T(qr , q˙s ) =

3

(3.21)

i=1 k=1

and aik = aik(qr ) = ai · ak,

aik = aik(qr ) = ai · ak.

You should notice that aik = aki and aik = aki . Here, we wish to show that Lagrange’s equations can be written in two other equivalent forms. The first one is the covariant form: m

3 

aki q¨ + m i

i=1

3  3 

˙ · ak = Fk, [si, k]q˙i q˙s = G

(3.22)

i=1 s=1

where a Christoffel symbol of the first kind is defined by [si, k] =

∂as · ak. ∂qi

It is important to note that [si, k] = [is, k]. The second form of Lagrange’s equations is known as the contravariant form: mq¨ k + m

3  3 

˙ · ak = F k, ksi q˙i q˙s = G

(3.23)

i=1 s=1

where a Christoffel symbol of the second kind is defined by kij =

∂ai k ·a . ∂qj

Notice that kij = kji . This form of Lagrange’s equations is used in numerical simulations of mechanical systems.† (a) Show that the Christoffel symbols have the representations   1 ∂aki ∂ask ∂asi [si, k] = + − k , 2 ∂qs ∂qi ∂q and kij =

3 

akr [ij, r].

r=1 ∗ †

The indices i, j, k, r, and s range from 1 to 3. Most numerical integration packages assume that the differential equations to be integrated are of the form x˙ = f(x, t). By defining the set of variables (states) x1 = q1 , . . . , x3 = q3 , x4 = q˙1 , . . . , x6 = q˙3 , the contravariant form of Lagrange’s equations can be easily placed in the form x˙ = f(x, t).

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Exercises 3.2–3.3

95

(b) Starting from Lagrange’s equations,   ∂T d ∂T − k = F · ak, k dt ∂ q˙ ∂q derive the following representation for the covariant component form∗ : m

3 

aki q¨ i + m

i=1

3  3  ˙ · ak = Fk. [si, k]q˙i q˙s = G i=1 s=1

(c) Starting from Lagrange’s equations in the form m

3 

aki q¨ i + m

i=1

3  3  ˙ · ak = Fk, [si, k]q˙i q˙s = G s=1 i=1

derive the following representation for the contravariant component form† : mq¨ k + m

3  3 

˙ · ak = F k. ksi q˙i q˙s = G

s=1 i=1

(d) For which coordinate system do the Christoffel symbols vanish? 3.3. Recall that for spherical polar coordinates, {R, φ, θ}, the covariant basis vectors are a1 = eR ,

a2 = Reφ ,

a3 = R sin(φ)eθ ,

and the contravariant basis vectors are a1 = eR ,

a2 =

1 eφ , R

a3 =

1 eθ . R sin(φ)

Furthermore, the linear momentum and kinetic energy of a particle of mass m are  m  ˙2 ˙ 1 + mφa ˙ 3, ˙ 2 + mθa G = mRa R + R2 φ˙2 + R2 sin2 (φ)θ˙2 . T= 2 (a) For a particle of mass m that is in motion in E3 under the influence of a resultant force F, establish the three covariant components of Lagrange’s equations of motion. In your solution, avoid explicitly calculating the 27 Christoffel symbols of the first kind. (b) For a particle of mass m that is in motion in E3 under the influence of a resultant force F, establish the three contravariant components of Lagrange’s equations of motion. In your solution, avoid explicitly calculating the 27 Christoffel symbols of the second kind. ∗



Hint : Expand the partial derivatives of T using the representation (3.21). Then, take the appropriate time derivative and reorganize the resulting equation by using the aforementioned symmetries. You may need to relabel certain indices to obtain the desired results. Hint : Multiply the covariant form by ask and sum over k. After some rearranging and relabeling of the indices, you should get the final desired result. Notice that the covariant component and contravariant component forms of these equations can be viewed as linear combinations of each other.

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Exercises 3.4–3.5

3.4. Consider a particle that is in motion on a rough surface. A curvilinear coordinate system q1 , q2 , q3 is chosen such that the surface can be described by the equation q3 = d(t), where d(t) is a known function of time t. (a) Suppose that the particle is moving on the rough surface. (i) Argue that vrel = q˙1 a˜ 1 + q˙2 a˜ 2 . (ii) Give a prescription for the constraint force acting on the particle. (b) Suppose that the particle is stationary on the rough surface. In this case, two equivalent prescriptions for the constraint force are Fc = N + F f =

3 

λi ai ,

i=1

where the tildes are dropped for convenience. (i) Show that ⎡ ⎤ ⎡ a11 λ1 ⎢ ⎥ ⎢ ⎣λ2 ⎦ = ⎣a12

a12 a22

⎥⎢ ⎥ 0⎦ ⎣F f2 ⎦ ,

a13

a23

1

λ3

0

⎤⎡

F f1



N

where N, and F f1 and F f2 uniquely define the normal force N and friction force F f , respectively, and aik = ai · ak with i, k = 1, 2, 3. (ii) For which coordinate systems do F f1 = λ1 , F f2 = λ2 , and N = λ3 ? Give an example to illustrate your answer. (c) Suppose that a spring force and a gravitational force also act on the particle. Prove that the total energy of the particle is not conserved, even when the friction force is static. 3.5. Consider a particle of mass m that is in motion on a helicoid. In terms of cylindrical polar coordinates r, θ, z, the equation of the right helicoid is z = αθ, where α is a constant. A gravitational force −mgE3 acts on the particle. (a) Consider the following curvilinear coordinate system for E3 : q1 = θ,

q2 = r,

q3 = ν = z − αθ.

Show that a1 = reθ + αE3 ,

a2 = er ,

a3 = E3 ,

and that a1 =

1 eθ , r

a2 = er ,

a3 = E3 −

α eθ . r

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Exercises 3.5–3.6

97

(b) Consider a particle moving on the smooth helicoid: (i) What is the constraint on the motion of the particle, and what is a prescription for the constraint force Fc enforcing this constraint? (ii) Show that the equations governing the unconstrained motion of the particle are   d 2 m r + α2 θ˙ = −mgα, dt d ˙ − mrθ˙ 2 = 0. (mr) dt

(3.24)

(iii) Prove that the angular momentum HO · E3 is not conserved. (c) Suppose the nonintegrable constraint rθ˙ + h(t) = 0 is imposed on the particle. Establish a second-order differential equation for r(t), a differential equation for θ(t), and an equation for the constraint force enforcing the nonintegrable constraint. Indicate how you would solve these equations to determine the motion of the particle and the constraint forces acting on it. 3.6. Consider a particle of mass m that is free to move on the smooth inner surface of a hemisphere of radius R0 (cf. Figure 3.6). The particle is under the influence of a gravitational force −mgE3 .

E3

O

E2

g E1

r

m

Figure 3.6. Schematic of a particle of mass m moving on the inside of a hemisphere of radius

R0 .

(a) Using a spherical polar coordinate system, what is the constraint on the motion of the particle? Give a prescription for the constraint force acting on the particle.

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Exercises 3.6–3.7

(b) Using Lagrange’s equations, establish the equations of motion for the particle and an expression for the constraint force. (c) Prove that the total energy E and the angular momentum HO · E3 of the particle are conserved. (d) Show that the normal force acting on the particle can be expressed as a function of the position of the particle and its initial energy E0 :   2E0 + 2mg sin (φ) + mg cos (φ) eR . N= − R0 (e) Numerically integrate the equations of motion of the particle and show that there are instances for which it will always remain on the surface of the hemisphere. 3.7. As shown in Figure 3.7, consider a bead of mass m that is free to move on a smooth semicircular wire of radius R0 . The wire has a constant angular velocity 0 E3 and whirls about the configuration shown in the figure. The particle is also

(t)

E3

g

O

E2 r

E1

m (t)

Figure 3.7. Schematic of a particle of mass m moving on a semicircular path that is being whirled about the vertical at a speed (t) = 0 .

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Exercises 3.7–3.8

99

under the influence of a gravitational force −mgE3 . This is a classical problem that is discussed in several textbooks (see, for example, [78]). (a) Using a spherical polar coordinate system, what are the two constraints on the motion of the particle? Give a prescription for the constraint force Fc acting on the particle. (b) Using Lagrange’s equations, establish the equation of motion for the particle:   g sin (φ) . (3.25) φ¨ = 20 cos (φ) + R0 After nondimensionalizing (3.25), numerically integrate the resulting differential equation and construct its phase portrait for values of 0.5, 1.0, 1.5 of the parameter R g2 . 0

0

(c) Recall that an equilibrium point x = x0 of the differential equation x¨ = f (x) is such that x˙ = 0 and f (x0 ) = 0. Show that (3.25) has three equilibria:   g −1 φ0 = 0, φ0 = π, φ0 = cos − . R0 20 Give physical interpretations for these equilibria and show that the third one is possible if, and only if, 20 is sufficiently large. How do these results correlate to your phase portraits? (d) Starting from the work–energy theorem T˙ = F · v, prove that the total energy of the particle is not conserved: ˙ = Nθ R0 0 sin (φ) , E where Nθ is the eθ component of the normal force acting on the particle. 3.8. Consider a particle of mass m moving in E3 . If coordinate system (1.6) is used to describe its kinematics, then establish expressions for the velocity vector v and the kinetic energy T of the particle. (a) Suppose a particle is constrained to move on a rough parabolic surface described by the equation x − y2 = −4. Give a prescription for the constraint force Fc acting on the particle, and establish the equations of motion for the particle. (b) As illustrated in Figure 3.8, suppose a particle is constrained to move on the smooth parabola x − y2 = −4,

z = 0.

Give a prescription for the constraint force Fc acting on the particle, and establish the equations of motion for the particle.

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Exercise 3.9

y

m

q2 = −4

2 −4

6

x

Figure 3.8. Schematic of a particle moving

on a parabola in the x − y plane.

−4 3.9. Consider a particle of mass m moving on a smooth ellipsoid: y2 z2 x2 + + = 1. a2 b2 c2 (a) With the help of a suitable curvilinear coordinate system, establish expressions for the constrained kinetic energy of the particle.∗ (b) Assuming that a gravitational force −mgE3 acts on the particle, establish the two second-order differential equations governing the motion of the particle. (c) Numerically integrate the equations you found in (b) and discuss features of the motion of the particle.



The coordinate systems you need are not discussed in this text but are readily found in the literature. They are often known as “confocal ellipsoidal coordinates.”

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PART TWO

DYNAMICS OF A SYSTEM OF PARTICLES

101

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The Equations of Motion for a System of Particles

4.1 Introduction In this chapter, we establish Lagrange’s equations for a system of particles by starting with the balances of linear momentum for each of the particles. Our derivation is based on the results presented in Chapter 15 of Synge and Griffith [207].∗ We supplement their work with a discussion of constraints and potential energies. To examine the geometry inherent in Lagrange’s equations of motion for the system of particles, we use the construction of a representative single particle by Casey [27]. All the work presented in this chapter emphasizes the equivalence of Lagrange’s equations of motion for a system of particles and the balances of linear momenta. For completeness, a brief discussion of the principle of virtual work, D’Alembert’s principle, Gauss’ principle of least constraint, and Hamilton’s principle are also presented in Section 4.11. The chapter closes with a discussion of a canonical form of Lagrange’s equations of motion in which time-independent integrable constraints are present. For many specific problems, we can obtain Lagrange’s equations by merely calculating the kinetic and potential energies of the system. This approach is used in most dynamics textbooks, and neither the construction of a single particle nor the components of force vectors are mentioned.† Indeed, once we establish Lagrange’s equations we can also ignore the explicit construction of the single particle. However, for many cases – which are not possible to treat using the approach adopted in most dynamics textbooks – we find that the use of Synge’s and Griffith’s representation of Lagrange’s equations of motion allows us to tremendously increase the range of application of Lagrange’s equations. For instance, as will be shown in some of the examples in Chapter 5, dynamic Coulomb friction is accommodated. ∗ †

Related derivations for systems of particles can be found in several texts. For example, Section 6-6 of Greenwood [79] and Section 21 of Whittaker [228]. These texts use either the principle of virtual velocities or Hamilton’s principle (also known as the principle of least action) to derive Lagrange’s equations.

103

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The Equations of Motion for a System of Particles

Fi E3

mi Figure 4.1. A single particle of mass mi in E3 . The position vector of this particle is ri , and the resultant external force acting on the particle is Fi .

ri O

E2

E1

4.2 A System of N Particles Here, we are interested in establishing the equations of motion for a system of N particles. The first step in this development is to discuss the individual elements in the system of particles. We consider a system of N particles, each of which is in motion in threedimensional Euclidean space E3 . For the particle of mass mi (see Figure 4.1), the position vector is ri =

3 

j

xi E j .

j=1

We also recall that the kinetic energy of the particle is 1 mi vi · vi . 2 It is also convenient to recall that the linear momentum of the particle of mass mi is Gi = mi r˙i = mi vi . The resultant force acting on the particle of mass mi has the representation Ti =

Fi =

3 

j

Fi E j .

j=1

The balance of linear momentum for the particle of mass mi is Fi = mi v˙i . As a consequence of the balance of linear momentum for the particle, we have the angular momentum theorem: H˙ Oi = ri × Fi , where HOi = ri × mi vi is the angular momentum of the particle relative to the fixed point O. A second consequence of the linear momentum balance is the work–energy theorem T˙ i = Fi · vi for the particle of mass mi .

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4.3 Coordinates

105

The Kinetic Energy For the system of particles, the combined (total) kinetic energy T is the sum of the kinetic energies:

T = T1 + · · · + TN . With the help of the work–energy theorem for each particle, we can determine the corresponding result for the system of particles: T˙ = F1 · v1 + · · · + FN · vN .

(4.1)

This result is used to establish energy conservation (or lack thereof) in a system of particles. The Center of Mass For the system of particles, we can define a center of mass C. This point, which lies in E3 , has the position vector r¯ , where

r¯ =

1 (m1 r1 + · · · + mN rN ) . m1 + · · · + mN

It is easy to show from this result that the linear momentum of a system of particles has the representation G = (m1 + · · · + mN ) ¯r˙. Next we examine a particle of mass m moving in E3N , and it is important not to confuse this particle with C.

4.3 Coordinates In many problems, Cartesian coordinates for ri are not a convenient choice. Indeed for many systems of two particles, we use one coordinate system to describe r1 and another to describe the relative position vector r2 − r1 . For instance, for the system shown in Figure 4.2, we might use Cartesian coordinates for r1 and spherical polar coordinates for r2 − r1 : r1 = xE1 + yE2 + zE3 ,

r2 = xE1 + yE2 + zE3 + R2 eR2 .

(4.2)

In this equation, R2 = L0 is the length of the rod connecting the particles, and eR2 = sin(φ2 ) (cos(θ2 )E1 + sin(θ2 )E2 ) + cos(φ2 )E3 . m1

E3 Figure 4.2. A particle of mass m1 attached by a rigid rod of length L0 to a particle of mass m2 .

g O

E1

m2 E2

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