Intermediate Dynamics for Engineers

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mastercup cuus273/Oliver M. O’Reilly 978 0 521 87483 0 June 9, 2008 21:29

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INTERMEDIATE DYNAMICS FOR ENGINEERS

This book has sufficient material for two full-length semester coursesin intermediate engineering dynamics. For the first course a Newton–Euler approach is used, followed by a Lagrangian approach in the sec-ond. Using some ideas from differential geometry, the equivalence ofthese two approaches is illuminated throughout the text. In addition,this book contains comprehensive treatments of the kinematics and dy-namics of particles and rigid bodies. The subject matter is illuminatedby numerous highly structured examples and exercises featuring a widerange of applications and numerical simulations.

Oliver M. O’Reilly is a professor of mechanical engineering at theUniversity of California, Berkeley. His research interests lie in contin-uum mechanics and nonlinear dynamics, specifically in the dynamicsof rigid bodies and particles, Cosserat and directed continuua, dynam-ics of rods, history of mechanics, and vehicle dynamics. O’Reilly is theauthor of more than 50 archival publications and Engineering Dynam-ics: A Primer. He is also the recipient of the University of Californiaat Berkeley’s Distinguished Teaching Award and three departmentalteaching awards.

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Intermediate Dynamics for Engineers

A UNIFIED TREATMENT OFNEWTON–EULER AND LAGRANGIANMECHANICS

Oliver M. O’ReillyUniversity of California, Berkeley

iii

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo

Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK

First published in print format

ISBN-13 978-0-521-87483-0

ISBN-13 978-0-511-42435-9

© Oliver M. O’Reilly 2008

2008

Information on this title: www.cambridge.org/9780521874830

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Published in the United States of America by Cambridge University Press, New York

www.cambridge.org

eBook (NetLibrary)

hardback

http://www.cambridge.org/9780521874830

http://www.cambridge.org

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This book is dedicated to my adventurous daughter, Anna

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Contents

Preface page xi

PART ONE DYNAMICS OF A SINGLE PARTICLE 1

1 Kinematics of a Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1 Introduction 31.2 Reference Frames 31.3 Kinematics of a Particle 51.4 Frequently Used Coordinate Systems 61.5 Curvilinear Coordinates 91.6 Representations of Particle Kinematics 141.7 Constraints 151.8 Classification of Constraints 201.9 Closing Comments 27

Exercises 27

2 Kinetics of a Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.1 Introduction 332.2 The Balance Law for a Single Particle 332.3 Work and Power 352.4 Conservative Forces 362.5 Examples of Conservative Forces 372.6 Constraint Forces 392.7 Conservations 452.8 Dynamics of a Particle in a Gravitational Field 472.9 Dynamics of a Particle on a Spinning Cone 552.10 A Shocking Constraint 592.11 A Simple Model for a Roller Coaster 602.12 Closing Comments 64

Exercises 66

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viii Contents

3 Lagrange’s Equations of Motion for a Single Particle . . . . . . . . . . . . 70

3.1 Introduction 703.2 Lagrange’s Equations of Motion 713.3 Equations of Motion for an Unconstrained Particle 733.4 Lagrange’s Equations in the Presence of Constraints 743.5 A Particle Moving on a Sphere 783.6 Some Elements of Geometry and Particle Kinematics 803.7 The Geometry of Lagrange’s Equations of Motion 833.8 A Particle Moving on a Helix 873.9 Summary 91

Exercises 92

PART TWO DYNAMICS OF A SYSTEM OF PARTICLES 101

4 The Equations of Motion for a System of Particles . . . . . . . . . . . . . 103

4.1 Introduction 1034.2 A System of N Particles 1044.3 Coordinates 1054.4 Constraints and Constraint Forces 1074.5 Conservative Forces and Potential Energies 1104.6 Lagrange’s Equations of Motion 1114.7 Construction and Use of a Single Representative Particle 1134.8 The Lagrangian 1184.9 A Constrained System of Particles 1194.10 A Canonical Form of Lagrange’s Equations 1224.11 Alternative Principles of Mechanics 1284.12 Closing Remarks 131

Exercises 131

5 Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . . 134

5.1 Introduction 1345.2 Harmonic Oscillators 1345.3 A Dumbbell Satellite 1405.4 A Pendulum and a Cart 1435.5 Two Particles Tethered by an Inextensible String 1475.6 Closing Comments 151

Exercises 153

PART THREE DYNAMICS OF A SINGLE RIGID BODY 161

6 Rotation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

6.1 Introduction 1636.2 The Simplest Rotation 1646.3 Proper-Orthogonal Tensors 1666.4 Derivatives of a Proper-Orthogonal Tensor 168

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Contents ix

6.5 Euler’s Representation of a Rotation Tensor 1716.6 Euler’s Theorem: Rotation Tensors and Proper-Orthogonal

Tensors 1766.7 Relative Angular Velocity Vectors 1786.8 Euler Angles 1816.9 Further Representations of a Rotation Tensor 1916.10 Derivatives of Scalar Functions of Rotation Tensors 195

Exercises 198

7 Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

7.1 Introduction 2067.2 The Motion of a Rigid Body 2067.3 The Angular Velocity and Angular Acceleration Vectors 2117.4 A Corotational Basis 2127.5 Three Distinct Axes of Rotation 2137.6 The Center of Mass and Linear Momentum 2157.7 Angular Momenta 2187.8 Euler Tensors and Inertia Tensors 2197.9 Angular Momentum and an Inertia Tensor 2237.10 Kinetic Energy 2247.11 Concluding Remarks 226

Exercises 226

8 Constraints on and Potentials for Rigid Bodies . . . . . . . . . . . . . . . 237

8.1 Introduction 2378.2 Constraints 2378.3 A Canonical Function 2418.4 Integrability Criteria 2438.5 Forces and Moments Acting on a Rigid Body 2478.6 Constraint Forces and Constraint Moments 2488.7 Potential Energies and Conservative Forces and Moments 2568.8 Concluding Comments 262

Exercises 263

9 Kinetics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272

9.1 Introduction 2729.2 Balance Laws for a Rigid Body 2729.3 Work and Energy Conservation 2749.4 Additional Forms of the Balance of Angular Momentum 2769.5 Moment-Free Motion of a Rigid Body 2799.6 The Baseball and the Football 2859.7 Motion of a Rigid Body with a Fixed Point 2899.8 Motions of Rolling Spheres and Sliding Spheres 2949.9 Closing Comments 297

Exercises 299

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x Contents

10 Lagrange’s Equations of Motion for a Single Rigid Body . . . . . . . . . 307

10.1 Introduction 30710.2 Configuration Manifold of an Unconstrained Rigid Body 30810.3 Lagrange’s Equations of Motion: A First Form 31110.4 A Satellite Problem 31510.5 Lagrange’s Equations of Motion: A Second Form 31810.6 Lagrange’s Equations of Motion: Approach II 32410.7 Rolling Disks and Sliding Disks 32510.8 Lagrange and Poisson Tops 33110.9 Closing Comments 336

Exercises 336

PART FOUR SYSTEMS OF RIGID BODIES 345

11 Introduction to Multibody Systems . . . . . . . . . . . . . . . . . . . . . . 347

11.1 Introduction 34711.2 Balance Laws and Lagrange’s Equations of Motion 34711.3 Two Pin-Jointed Rigid Bodies 34911.4 A Single-Axis Rate Gyroscope 35111.5 Closing Comments 355

Exercises 355

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

A.1 Introduction 362A.2 Preliminaries: Bases, Alternators, and Kronecker Deltas 362A.3 The Tensor Product of Two Vectors 363A.4 Second-Order Tensors 364A.5 A Representation Theorem for Second-Order Tensors 364A.6 Functions of Second-Order Tensors 367A.7 Third-Order Tensors 370A.8 Special Types of Second-Order Tensors 372A.9 Derivatives of Tensors 373

Exercises 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 giventhe assignment of teaching two courses on rigid body dynamics. One of thesecourses featured Lagrange’s equations of motion, and the other featured theNewton–Euler equations. I had long struggled to resolve these two approaches toformulating 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 tothe equivalence of the Lagrangian and Newton–Euler approaches. He then wentfurther and showed how the governing equations for a rigid body formulated by useof both approaches were equivalent [27, 28]. Shades of this result could be seen inan earlier work by Greenwood [79], but Casey’s work established the equivalencein an unequivocal fashion. As is evident from this book, I subsequently adaptedand expanded on Casey’s treatment in my courses. My treatment of dynamicspresented 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 dynamicalsystems 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 equationsgoverning the motion of (often simple) mechanical systems featuring particles andrigid bodies could display surprisingly rich behavior.

There are several manners in which this book differs from a traditional text onengineering dynamics. First, I demonstrate explicitly how the equations of motionobtained by using Lagrange’s equations and the Newton–Euler equations are equiv-alent. To achieve this, my discussion of geometry and curvilinear coordinates is farmore detailed than is normally found in textbooks at this level. The second differ-ence 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 believethat this enables a far clearer derivation of many of the fundamental results in thekinematics of rigid bodies.

I have distributed as many examples as possible throughout this book and haveattempted to cite up-to-date references to them and related systems as far as fea-sible. However, I have not approached the exhaustive treatments by Papastavridis

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xii Preface

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

Using This Text

This book has been written so that it provides sufficient material for two full-lengthsemester courses in engineering dynamics. As such it contains two tracks (whichoverlap 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 Tensors6. Rotation Tensors7. Kinematics of Rigid Bodies8. Constraints on and Potentials for Rigid Bodies9. Kinetics of a Rigid Body

11. Multibody Systems

The second course, in which a Lagrangian approach is used, could be based on thefollowing chapters:

1. Kinematics of a Particle2. Kinetics of a Particle3. Lagrange’s Equations of Motion for a Single Particle4. Lagrange’s Equations of Motion for a System of Particles5. Dynamics of Systems of Particles

Appendix on Tensors6. Rotation Tensors (with particular emphasis on Section 6.8)7. Kinematics of Rigid Bodies8. Constraints on and Potentials for Rigid Bodies9. Kinetics of a Rigid Body

10. Lagrange’s Equations of Motion for a Single Rigid Body11. Multibody Systems

In discussing rotations for the second course, time constraints permit a detaileddiscussion of only the Euler angle parameterization of a rotation tensor fromChapter 6 and a brief mention of the examples on rigid body dynamics discussed inChapter 9.

Most of the exercises at the end of each chapter are highly structured and areintended as a self-study aid. As I don’t intend to publish or distribute a solutionsmanual, I have tailored the problems to provide answers that can be validated.Some of the exercises feature numerical simulations that can be performed withMatlab or Mathematica. Completing these exercises is invaluable both in terms of

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Preface xiii

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

Image Credit

The portrait of William R. Hamilton in Figure 4.6 in Subsection 4.11.3 is from theRoyal Irish Academy in Dublin, Ireland. I am grateful to Pauric Dempsey, the Headof 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, whichI have taught at the Department of Mechanical Engineering at the University ofCalifornia at Berkeley over the past decade. Some of the aims of these courses areto give senior undergraduate and first-year graduate students in mechanical engi-neering requisite skills in the area of dynamics of rigid bodies. The book is alsointended to be a sequel to my book Engineering Dynamics: A Primer, which waspublished by Springer-Verlag in 2001.

I have been blessed with the insights and questions of many remarkable studentsand the help of several dedicated teaching assistants. Space precludes mention of allof these students and assistants, but it is nice to have the opportunity to acknowl-edge some of them here: Joshua P. Coaplen, Nur Adila Faruk Senan, David Gulick,Moneer Helu, Eva Kanso, Patch Kessler, Nathan Kinkaid, Todd Lauderdale, HenryLopez, David Moody, Tom Nordenholz, Jeun Jye Ong, Sebastien Payen, BrianSpears, Philip J. Stephanou, Meng How Tan, Peter C. Varadi, and Stephane Ver-guet. I am also grateful to Chet Vignes for his careful reading of an earlier draft ofthe 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 withthe roller-coaster model and simulations of its equations of motion. Professor ChrisHall of Virginia Tech pointed out reference [118] on Lagrange’s solution of a satel-lite dynamics problem. Professor Richard Montgomery of the University of Califor-nia at Santa Cruz discussed the remarkable figure-eight solutions to the three-bodyproblem with me, Professor Glen Niebur of the University of Notre Dame providedvaluable references on Codman’s paradox, Professor Harold Soodak of the CityCollege of New York provided valuable comments on the tippe top, and Profes-sors Donald Greenwood and John Papastavridis carefully read a penultimate draft

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xiv Preface

of this book and generously provided many constructive comments and correctionsfor which I am most grateful.

Most of this book was written during the past 10 years at the University of Cali-fornia at Berkeley. The remarkable library of this institution has been an invaluableresource in my quest to distill more than 300 years of work on the subject matter inthis book. I am most grateful to the library staff for their assistance and the taxpay-ers for their support of the University of California.

Throughout this book, several references to my own research on rigid bodydynamics can be found. In addition to the students mentioned earlier, I have hadthe good fortune to work with Jim Casey and Arun Srinivasa on several aspects ofthe equations of motion for rigid bodies. The numerous citations to their works area reflection of my gratitude to them.

This book would not have been published without the help and encouragementof Peter Gordon at Cambridge University Press and would contain far more er-rors were it not for the editorial help of Victoria Danahy. Despite the assistance ofseveral other proofreaders, it is unavoidable that some typographical and technicalerrors have crept into this book, and they are my unpleasant responsibility alone. Ifyou find some on your journey through these pages, I would be pleased if you couldbring them to my attention.

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PART ONE

DYNAMICS OF A SINGLE PARTICLE

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1 Kinematics of a Particle

1.1 Introduction

One of the main goals of this book is to enable the reader to take a physical sys-tem, model it by using particles or rigid bodies, and then interpret the results of themodel. For this to happen, the reader needs to be equipped with an array of toolsand techniques, the cornerstone of which is to be able to precisely formulate thekinematics of a particle. Without this foundation in place, the future conclusions onwhich they are based either do not hold up or lack conviction.

Much of the material presented in this chapter will be repeatedly used through-out the book. We start the chapter with a discussion of coordinate systems for aparticle moving in a three-dimensional space. This naturally leads us to a discussionof curvilinear coordinate systems. These systems encompass all of the familiar co-ordinate systems, and the material presented is useful in many other contexts. Atthe conclusion of our discussion of coordinate systems and its application to particlemechanics, you should be able to establish expressions for gradient and accelerationvectors in any coordinate system.

The other major topics of this chapter pertain to constraints on the motion ofparticles. In earlier dynamics courses, these topics are intimately related to judi-cious 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 alsodiscuss time-varying constraints on the velocity vector of the particle. Along withcurvilinear coordinates, the topic of constraints is one most readers will not haveseen before and for many they will hopefully constitute an interesting thread thatwinds its way through this book.

1.2 Reference Frames

To describe the kinematics of particles and rigid bodies, we presume on the ex-istence of a space with a set of three mutually perpendicular axes that meet at acommon point P. The set of axes and the point P constitute a reference frame. InNewtonian mechanics, we also assume the existence of an inertial reference frame.In this frame, the point P moves at a constant speed.

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Path of the particle

m

O

A

v

r(t)

r(t + �t)

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

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

Depending on the application, it is often convenient to idealize the inertialreference frame. For example, for ballistics problems, the Earth’s rotation andthe 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 threeorthonormal vectors that are fixed to it (and the Earth), is then taken to approximatean inertial reference frame. This approximate inertial reference frame, however,is insufficient if we wish to explain the behavior of Foucault’s famous pendulumexperiment. In this experiment from 1851, Leon Foucault (1819–1868) ingeniouslydemonstrated the rotation of the Earth by using the motion of a pendulum.∗ Toexplain this experiment, it is sufficient to assume the existence of an inertial framewhose point P is at the fixed center of the rotating Earth and whose axes do notrotate with the Earth. As another example, when the motion of the Earth about theSun is explained, it is standard to assume that the center S of the Sun is fixed and tochoose P to be this point. The point S is then used to construct an inertial referenceframe. Other applications in celestial mechanics might need to consider the locationof the point P for the inertial reference frame as the center of mass of the solar sys-tem with the three fixed mutually perpendicular axes defined by use of certain fixedstars [80].

For the purposes of this text, we assume the existence of a fixed point O anda 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 thatthis reference frame occupies is a three-dimensional space. Vectors can be definedin this space, and an inner product for these vectors is easy to construct with the dotproduct. As such, we refer to this space as a three-dimensional Euclidean space andwe denote it by E

3.

∗ Discussions of his experiment and their interpretation can be found in [62, 138, 207]. Among hisother 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 therespective first and second time derivatives of the position vector:

v = drdt

, a = dvdt

= d2rdt2

.

It is crucial to note that, because r is measured relative to a fixed origin, v and a arethe absolute velocity and acceleration vectors. By definition, the velocity vector canbe calculated from the following limit:

v(t) = lim�t→0

r (t + �t) − r(t)�t

.

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 fixedpoint O (see, e.g., Moulton [150]). To establish an expression for this vector, weconsider the position vector of the particle at time t and t + �t. Then, the area of theparallelogram defined by these vectors is ‖r(t) × r (t + �t)‖ (see Figure 1.1). This istwice the area swept out by the particle during the interval �t. Taking the limit ofthe vector r(t)×r(t+�t)

2�t as �t → 0 and using the fact that r(t) × r(t) = 0, we arrive atan expression for the areal velocity vector A (t):

A (t) = lim�t→0

r(t) × r (t + �t)2�t

= 12

r(t) ×(

lim�t→0

r (t + �t)�t

)

= 12

r(t) ×(

lim�t→0

r (t + �t) − r (t)�t

).

That is,

A = 12

r × v. (1.1)

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The vector A plays an important role in several mechanics problems in which eitherthe angular momentum HO is constant or a component of HO is constant. Severalother 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:

T = 12

mv · v.

The definitions of the kinematical quantities that have been introduced are inde-pendent of the coordinate system that is used for E

3. In solving most problems, it iscrucial to have expressions for momenta and energies in terms of the chosen coor-dinate system. It is to this issue that we now turn.

1.4 Frequently Used Coordinate Systems

Depending on the problem of interest, there are several suitable coordinate sys-tems for E

3. 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 ofa curvilinear coordinate system {q1, q2, q3} for E

3, which we will discuss later on inthis chapter.

Cartesian Coordinate SystemFor the Cartesian coordinate system, a set of right–handed orthonormal vectors aredefined: {E1, E2, E3}. Given any vector b in E

3, this vector has the representation

b =3∑

i=1

biEi.

For the position vector r, we also have

r =3∑

i=1

xiEi,

where {x1, x2, x3} are the Cartesian coordinates of the particle. Because Ei are fixedin both magnitude and direction, their time derivatives are zero: Ei = 0.

Cylindrical Polar CoordinatesA cylindrical polar coordinate system {r, θ, z} can be defined by a Cartesian coordi-nate system as follows:

r =√

x21 + x2

2, θ = tan−1(

x2

x1

), z = x3,

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

r

r

θ

z

O

er

eθ

E1

E2

E3

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

Given a position vector r, we can write

r = x1E1 + x2E2 + x3E3

= 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.

We also notice that er = θeθ, whereas eθ = −θer. We should also verify that{er, eθ, Ez} is a right-handed orthonormal basis for E

3.∗

Spherical Polar CoordinatesA spherical polar coordinate system {R, φ, θ} can be defined by a Cartesian coordi-nate system as follows:

R =√

x21 + x2

2 + x23, θ = tan−1

(x2

x1

), φ = tan−1

⎛⎝√

x21 + x2

2

x3

⎞⎠ ,

where θ ∈ [0, 2π) and φ ∈ (0, π). Provided φ �= 0 or π, we can invert these relationsto 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 = x1E1 + x2E2 + x3E3

= 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 eachof the vectors pi is 1 and they are mutually perpendicular: p1 · p2 = 0, p2 · p3 = 0, and p1 · p3 = 0.

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r

r

θ

φ

O eθ

eR

eφ

E1

E2

E3

Figure 1.3. The spherical polar coordinates φ and θ.

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

⎢⎣eR

eφ

eθ

⎤⎥⎦ =

⎡⎢⎣

cos(θ) sin(φ) sin(θ) sin(φ) cos(φ)

cos(θ) cos(φ) sin(θ) cos(φ) − sin(φ)

− sin(θ) cos(θ) 0

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ .

To establish the relations between these vectors and those defined earlier, we firstcalculate the intermediate relations⎡

⎢⎣er

eθ

E3

⎤⎥⎦ =

⎡⎢⎣

cos(θ) sin(θ) 0


0 0 1

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ ,

⎡⎢⎣

eR

eφ

eθ

⎤⎥⎦ =

⎡⎢⎣

sin(φ) 0 cos(φ)

cos(φ) 0 − sin(φ)

0 1 0

⎤⎥⎦⎡⎢⎣

er

eθ

E3

⎤⎥⎦ . (1.2)

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, assumingthat θ and φ are functions of time, a series of long calculations using (1.2) revealsthat ⎡

⎢⎣eR

eφ

eθ

⎤⎥⎦ =

⎡⎢⎣

0 φ θ sin(φ)

−φ 0 θ cos(φ)

−θ sin(φ) −θ cos(φ) 0

⎤⎥⎦⎡⎢⎣

eR

eφ

eθ

⎤⎥⎦ . (1.3)

These relations have an interesting form: Notice that the matrix in (1.3) is skew-symmetric. We shall see numerous examples of this later on when we discuss rota-tions 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 ex-amples of a curvilinear coordinate system. The development of the vector calculusassociated 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 mechan-ics, and the material presented here has a wide range of applications. Most of ourdiscussion is based on classical works and can be found in various textbooks on ten-sor calculus. Of these books, the one closest in spirit (and notation) to our treatmenthere is that of Simmonds [198]; [139, 201] are also recommended.

Consider a curvilinear coordinate system {q1, q2, q3} that is defined by thefunctions

q1 = q1 (x1, x2, x3) ,

q2 = q2 (x1, x2, x3) ,

q3 = q3 (x1, x2, x3) . (1.4)

We assume that the functions qi are locally invertible:

x1 = x1(q1, q2, q3) ,

x2 = x2(q1, q2, q3) ,

x3 = x3(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 E

3. For instance, the cylindricalpolar coordinate θ is not uniquely defined when x2

1 + x22 = 0. This set of points

corresponds to the x3 axis.Assuming invertibility, and fixing the value of one of the curvilinear coordi-

nates, q1 say, to equal q10, we can determine the values of x1, x2, and x3 such that the

equation

q10 = q1 (x1, x2, x3)

is satisfied. The union of all the points represented by these Cartesian coordinatesdefines a surface that is known as the q1 coordinate surface (cf. Figure 1.4). If wemove on this surface we find that the coordinates q2 and q3 will vary. Indeed, thecurves 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

qj is known as a qj coordinate surface. Similarly, the curve we obtain by varying thecoordinate qk while fixing the remaining two curvilinear coordinates is known as aqk coordinate curve.

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O

q2 coordinate curveq3 coordinate curve

q1 coordinate surface

a2

a3

a1

S

Figure 1.4. An example of a q1 coordinate surface S. At a point on this surface, a1 is normalto the surface, and a2 and a3 are tangent to the surface. The q1 coordinate surface S is foliatedby curves of constant q2 and q3.

Covariant Basis Vectors

Again assuming invertibility, we can express the position vector r of any point as afunction of the curvilinear coordinates:

r =3∑

i=1

xi(q1, q2, q3)Ei.

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

ai = ∂r∂qi

=3∑

k=1

∂xk

∂qiEk.

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 toa q2 coordinate curve. In general, ai is tangent to a qi coordinate curve.

You should notice that we can express the relationship between the covariantbasis vectors and the Cartesian basis vectors in a matrix form:⎡

⎢⎣a1

a2

a3

⎤⎥⎦ =

⎡⎢⎢⎣

∂x1∂q1

∂x2∂q1

∂x3∂q1

∂x1∂q2

∂x2∂q2

∂x3∂q2

∂x1∂q3

∂x2∂q3

∂x3∂q3

⎤⎥⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ .

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

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

a1 =3∑

i=1

∂q1

∂xiEi, a2 =

3∑i=1

∂q2

∂xiEi, a3 =

3∑i=1

∂q3

∂xiEi.

That is,

ak = ∇qk.

Geometrically, ai is normal to a qi coordinate surface. However, as in the case of thecovariant basis vectors, the contravariant basis vectors are not necessarily unit vec-tors, nor do they form an orthonormal basis for E

3. Using the chain rule of calculus,we can show that

ai · aj = δij,

where δij is the Kronecker delta. As discussed in the Appendix, δi

j = 1 if i = j and is0 otherwise. It is left as an exercise for the reader to show this result.∗

Covariant and Contravariant ComponentsAs {a1, a2, a3} and {a1, a2, a3} form bases for E

3, any vector b can be described aslinear combinations of either sets of vectors:

b =3∑

i=1

biai =3∑

k=1

bkak.

The components bi are known as the contravariant components, and the compo-nents bk are known as the covariant components:

b · ai =(

3∑k=1

bkak

)· ai =

3∑k=1

bkδki = bi,

b · ai =(

3∑k=1

bkak

)· ai =

3∑k=1

bkδik = bi.

It is very important to note that bk �= b · ak in general because ai · ak is not necessar-ily equal to δi

k.The trivial case in which xi = qi deserves particular mention. For this case,

r =∑3k=1 xiEi. 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∂xj

= δjk.

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y

x6

2

−4

−4

q1 = −3

q1 = −1

q1 = 1

q1 = 3q2 = −4

q2 = 2

q2 = 4

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

An ExampleAlthough we have met three examples of curvilinear coordinate systems previously,it is useful to introduce an example that features nonorthogonal basis vectors. Con-sider 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 projec-tions 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 =q2 + (q1

)2and y = q1. Thus,

r =(

q2 + (q1)2)E1 + q1E2 + q3E3.

By taking the derivatives of this representation for r with respect to q1, q2, and q3,we see that

a1 = 2q1E1 + E2, a2 = E1, a3 = E3.

This set of vectors comprises the covariant basis vectors. By taking the gradient ofqi, we find the contravariant basis vectors:

a1 = E2, a2 = E1 − 2q1E2, a3 = E3.

It is interesting to note that a1 · a2 = −2q1 �= 0. Further, a1 and a2 do not necessarilyhave unit magnitudes. By way of illustration, a q1 coordinate curve is shown inFigure 1.6. The vector a1 is tangent to this curve, and a2 and a3 are normal tothis curve. To emphasize that a1 is not necessarily parallel to a1, a q1 coordinatesurface is also shown in the figure. It is left as an exercise for the reader toillustrate the tangent vectors a2 and a3 to the q1 coordinate surface shown in thefigure.

Some Comments on DerivativesSeveral partial derivatives of functions �(q1, q2, q3, q1, q2, q3, t) play a prominentrole in this book. When taking the partial derivative of this function with respect

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q1 coordinate curve

q1 coordinate surface

a1

a1

a1

a2

a3

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

to q2 say, we assume that t, q1, q3, and qk are constant. A related remark holdsfor the partial derivatives with respect to the velocities qj and time t. That is,

∂qk

∂qj= δk

j ,∂qk

∂qj= 0,

∂t∂qj

= 0, (1.7)

and

∂qk

∂qj= 0,

∂qk

∂qj= δk

j ,∂t∂qj

= 0. (1.8)

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 d

dt and thederivative ∂

∂t . The former derivative assumes that qi and qi are functions of time,whereas the latter assumes that they are constant:

� = d�

dt=

3∑i=1

∂�

∂qi

dqi

dt+

3∑k=1

∂�

∂qk

d2qk

dt2+ ∂�

∂t.

For example, consider the function

� = q1 + (q3)2 + 10t.

Then,

∂�

∂q1= 1,

∂�

∂q3= 2q3,

∂�

∂t= 10, � = q1 + 2q3q3 + 10.

It should be clear from this example that � �= ∂�∂t .

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1.6 Representations of Particle Kinematics

We now turn to establishing expressions for the position, velocity, and accelerationvectors of a particle in terms of the coordinate systems just mentioned. First, for theposition vector we have∗

r = x1E1 + x2E2 + x3E3

= rer + zE3

= ReR

=3∑

i=1

xi(q1, q2, q3)Ei.

Differentiating these expressions, we find

v = x1E1 + x2E2 + x3E3

= rer + rθeθ + zE3

= ReR + Rφeφ + R sin(φ)θeθ

=3∑

i=1

qiai. (1.9)

Notice the simplicity of the expression for v when expressed in terms of the covariantbasis vectors. For any given curvilinear coordinate system, if we write the positionvector as a function of the coordinates q1, q2, and q3, and then differentiate andcompare the result with v =∑3

i=1 qiai, we can read off the covariant basis vectors.For instance, (1.9)2 implies that a1 = er, a2 = reθ, and a3 = E3 for the cylindricalpolar coordinate system.

A further differentiation yields

a = x1E1 + x2E2 + x3E3

= (r − rθ2)er + (rθ + 2rθ)eθ + zE3

= (R − Rφ2 − R sin2(φ)θ2)eR + (Rφ + 2Rφ − R sin(φ) cos(φ)θ2)eφ

+ (R sin(φ)θ + 2Rθ sin(φ) + 2Rθφ cos(φ))eθ

=3∑

i=1

qiai +3∑

i=1

3∑j=1

qiqj ∂ai

∂qj.

We obtain the final representation for r after noting that ai depend on the curvilinearcoordinates, which in turn are functions of time: ak =∑3

i=1∂ak∂qi qi.

It is left as an exercise for the reader to establish expressions, using variouscoordinate systems, for the linear momentum G and the angular momentum HO.

∗ Notice that it is a mistake to assume that r =∑3i=1 qiai .

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1.7 Constraints 15

The kinetic energy T of the particle has a rather elegant representation using thecurvilinear coordinates:

T = m2

v · v

= m2

(3∑

i=1

qiai

)·(

3∑k=1

qkak

)

=3∑

i=1

3∑k=1

m2

aikqiqk, (1.10)

where

aik = aki = ak · ai.

It is also a good exercise to compute aik for a spherical polar coordinate system, andthen, with the help of the representation T =∑3

i=1

∑3k=1

m2 aikqiqk, show that

T = m2

(R2 + R2φ2 + R2 sin2(φ)θ2

).

The exercises at the end of this chapter feature this result for other coordinatesystems.

1.7 Constraints

A constraint is a kinematical restriction on the motion of the particle. They areintroduced in problems involving a particle in three manners: either as simplifyingassumptions, prescribed motions, or because of rigid connections. The constraintson the motion of a particle dictate, to a large extent, the coordinate system usedto solve the problem of determining the motion of the particle. In this section, weexamine the simplest class of constraints on the motion of a particle. Later, theseconstraints will be classified as integrable.

Classical ExamplesConsider the four mechanical systems shown in Figure 1.7. The first system is knownas the spherical pendulum. Here, a particle of mass m is attached by a rigid rod oflength L0 to a fixed point O. The constraint on the motion of the particle in thissystem can be written as

r · eR = L0.

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

r · er = L0, r · E3 = 0.

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(a) (b)

(c) (d)

spinning conical surface

moving plane

m

m

m

m

α

O O

OO

L0

L0

er

E1

E1

E1

E2

E2

E2

E3

E3

E3

Figure 1.7. Four mechanical systems featuring constraints on the motion of a particle: (a) thespherical pendulum, (b) the planar pendulum, (c) a particle moving on a plane, and (d) aparticle moving on a spinning cone.

After differentiating these equations with respect to time, we observe that the ve-locity vector of the particle has a component only in the eθ direction: v · er = 0 andv · E3 = 0. The third system involves a particle moving on a horizontal surface thatis moving with a velocity vector f (t)E3. The constraint on the motion of the particleis

r · E3 = f (t).

The final system of interest consists of a particle moving on a spinning cone. Theconstraint on the motion of the particle can be most easily described with the helpof a spherical polar coordinate system:

φ + α(t) − π

2= 0.

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 SurfaceTurning to the more general case, consider a particle constrained to move on a sur-face. With the help of a single smooth function (r, t), we assume that the constraint

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1.7 Constraints 17

O

Surface = 0

∇

m

r

Figure 1.8. A particle moving on a surface = 0. The particle in this case is subject to a singleconstraint.

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 thethree independent Cartesian coordinates of the particle. Thus the condition = 0defines a two-dimensional surface (see Figure 1.8).

The unit normal vector n to this surface is parallel to ∇ = grad() (seeFigure 1.8). Depending on the coordinate system used, this vector ∇ hasnumerous representations:

∇ = ∂

∂r=

3∑i=1

∂

∂xiEi

=3∑

i=1

∂

∂qiai

= ∂

∂rer + 1

r∂

∂θeθ + ∂

∂zE3

= ∂

∂ReR + 1

R∂

∂φeφ + 1

R sin(φ)∂

∂θeθ. (1.11)

You should notice how simple the expression for the gradient is in curvilinearcoordinates.∗

A simple differentiation of the function helps to provide the restriction itimposes on the velocity vector:

= ∂

∂r· v + ∂

∂t.

∗ To establish this result, we note that (q1, q2, q3

) = ∇ · v and (q1, q2, q3

) =∑3i=1

∂

∂qi qi . Substi-

tuting for v and comparing both expressions, we arrive at (1.11)2.

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O

Surface 1 = 0

Surface 2 = 0

∇1

∇2

m

rt

Figure 1.9. A particle subject to two constraints. The dotted curve in this figure correspondsto the curve of intersection of the surfaces 1 = 0 and 2 = 0, and the vector t is the unittangent vector to this curve.

However, if r satisfies the constraint, then (r, t) = 0 and = 0. Consequently, theconstraint = 0 implies that the velocity vector satisfies the restriction

∂

∂r· v + ∂

∂t= 0.

This result will be important in our discussion of the mechanical power of the con-straint forces.

A Particle Moving on a CurveWe now consider the more complex case of a particle moving on a curve. A curvecan 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 totwo (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 apoint 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 thecurve can be readily established. We also note that deriving the restrictions theseconstraints impose on the velocity vector follows from the corresponding results fora single constraint:

∂1

∂r· v + ∂1

∂t= 0,

∂2

∂r· v + ∂2

∂t= 0. (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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1.7 Constraints 19

A Particle Whose Motion is PrescribedThe case in which the motion is prescribed can be interpreted as a particle lying atthe intersection of three known surfaces. In other words, the particle is subject tothree 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 attheir intersection point form a basis for E

3:

∇3 · (∇1 × ∇2) �= 0.

The three conditions i = 0 can also be interpreted as three equations for the threecomponents of r.

The two primary situations in which a particle is subject to three constraintsarise when either the motion of the particle is completely controlled or the particleis subject to static friction and is therefore in a state of rest relative to a curve orsurface.

Coordinates and ConstraintsThe constraints we have considered on the motion of the particle have been de-scribed in terms of surfaces that the motion of the particle is restricted to. Thesesurfaces can be described in terms of the coordinate system used for E

3. The de-scription is greatly facilitated by a judicious choice of coordinates. For instance, ifa particle is constrained to move on a fixed plane, then we can always choose theorigin O and the Cartesian coordinates such that the constraint is easily describedby the equation x3 = constant. Similarly, if a particle is constrained to move on asphere, then spherical polar coordinates are an obvious choice.

The more sophisticated the surfaces that the particle is constrained to moveon, 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 inan appropriate curvilinear coordinate system by a simple equation, q3 = constant.Furthermore, a moving surface (r, t) = 0 can, in principle, be described by theequation q3 = f (t), where f is a function of time t. For example, suppose a particleis moving on a sphere whose radius is a known function R0(t). Then the constraintthat 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 coordinatesystem.

The Classical Examples RevisitedReturning to the four mechanical systems shown in Figure 1.7, you should convinceyourself that the constraint(s) on the motions of the particle in these systems are

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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 cylinderof radius L0 and a horizontal plane. This intersection defines a circle. If the rod’slength L0 changes with time, then the circle’s radius also changes. For the particlemoving 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

(r) = φ + α − π

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

2+ α0 + A sin(ωt).

You should verify that = 0, but ∂∂t = Aω cos(ωt). The spinning of the cone has

purposefully not been mentioned. This motion will feature in any formulation of thefriction forces acting on the particle. Further, in the event that the particle is stuckto the cone, then the particle will be subject to three constraints. This situation isdiscussed 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 typeof 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 ofthe particle – it restricts only its velocity vector. Consequently, the constraint π = 0is often known as a velocity constraint.

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1.8 Classification of Constraints 21

As we demonstrated earlier, every constraint of the form (r, t) = 0 can be dif-ferentiated to yield a restriction on the velocity vector:

∂

∂r· v + ∂

∂t= 0.

This restriction is of the form (1.13). Thus every constraint (r, t) = 0 provides aconstraint 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 func-tion (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 toHeinrich Hertz [92] (1857–1894). As noted by Lanczos [124], integrable constraintswere further classified by Ludwig Boltzmann (1844–1906) as rheonomic when =(r, t) and scleronomic when = (r) (i.e., when is not an explicit function oftime t).

The distinction between integrable and nonintegrable constraints becomes par-ticularily important when rigid bodies are concerned. However, for pedagogical pur-poses, it is desirable to introduce them when discussing single particles. We shallshortly discuss the forces needed to enforce the constraints: Such forces are knownas constraint forces. To explore the differences between integrable and noninte-grable constraints, it is best to first consider some examples. Following such anexploration, we shall discuss known criteria to determine whether or not a set ofconstraints is integrable.

Three ExamplesAs 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 (seeFigure 1.10). Two points A and B are also shown in this figure, and it is importantto notice that it is not possible for the particle to move between A and B withoutviolating the constraint xy − c = 0. The constraints xy − c = 0 and z = 0 imply thevelocity 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 ordifferential forms, see, e.g., [61, 64, 114]. It is well known that integrating factors are not unique.

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A

B

x = x1

y = x2

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

A

B

x = x1

y = x2

Figure 1.11. The motion of a particle subject to theconstraints yx = 0 and z = 0. The arrows indicate thepossible directions of motion of the particle.

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 planewithout violating the constraint yx = 0. The restriction this constraint places is thatit restricts how one can go from any A to any B. This is in marked contrast to the con-straint xy − c = 0. By multiplying yx = 0 by 1

x , for example, we see that, away fromthe y axis, this constraint is integrable.∗ Considering the possible motions shownin 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 non-

integrable constraint.

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Instead, we classify the constraint yx = 0 as a piecewise-integrable constraint.∗ Weshall discuss further unusual aspects of this constraint in Section 2.10.

Our third example is the simplest possible nonintegrable constraint on the mo-tion 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, wechoose two points A and B and use the x coordinate to parameterize a path betweenthem. Choosing

y = f (x), z = dfdx

,

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 pointsA to B, f (x) is subject to the restrictions

yA = f (xA), zA = dfdx

(xA), yB = f (xB), zB = dfdx

(xB),

where rA = xAE1 + yAE2 + zAE3 and rB = xBE1 + yBE2 + zBE3. Graphically con-structing a function f (x) that meets these restrictions is not difficult, and some ex-amples 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:

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

x − xA

xB − xA

)2

− (2 (yB − yA) − (xB − xA) (zB + zA))(

x − xA

xB − xA

)3

+ c (x − xA)2 (xB − x)2 + d sin2(

π (x − xA)xB − xA

)

+ zA (x − xA)(

xB − xxB − xA

)+ yA, (1.17)

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

∗ With the exception of that of Papastavridis [169], this classification is not typically mentioned inthe textbooks on classical and analytical mechanics. Further discussion of piecewise-integrable con-straints can be found in the interesting paper by Ruina [186]. As discussed in his paper, constraintsof this type also arise in many locomotive systems such as passive walking machines that featureimpact.

† 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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A

B

f

v

O

E1

E2

E3

Figure 1.12. Three possible motions be-tween two given points A and B of a par-ticle subject to the constraint y − zx =0. The arrows indicate the directions ofmotion of the particle and the vectorf = −zE1 + E2. The motions presentedin this figure were constructed with theassistance of (1.17).

Integrability Criteria

Suppose a constraint π = 0 is imposed on the motion of the particle. As mentionedearlier, this constraint is integrable if we can find a function (r, t) and an integrat-ing 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 aconstraint 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 q3 + e = 0. The lat-ter constraint in turn is equivalent to the constraint q3 = g, where g = e. Using thiscoordinate system, the dynamics of the particle is easier to analyze. With this inmind, several classical integrability criteria are now presented for single and multi-ple constraints.∗

A SINGLE SCLERONOMIC CONSTRAINT. The first criterion we examine pertains to con-straints of the form f · v = 0, where f is not an explicit function of time. Using acoordinate system {q1, q2, q3}, we can write the constraint π = 0 in the form

f1q1 + f2q2 + f3q3 = 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 onthe 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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for all possible choices of qi. Here,

Ic = f1

(∂f3

∂q2− ∂f2

∂q3

)+ f2

(∂f1

∂q3− ∂f3

∂q1

)+ f3

(∂f2

∂q1− ∂f1

∂q2

).

It is convenient to recall at this point the expression for the curl of a vector field Pin Cartesian coordinates:

curl(P) =(

3∑i=1

∂

∂xiEi

)× P

=(

∂P3

∂x2− ∂P2

∂x3

)E1 +

(∂P1

∂x3− ∂P3

∂x1

)E2 +

(∂P2

∂x1− ∂P1

∂x2

)E3, (1.20)

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 thatthese functions exist. Further, this criterion is local – it does not tell us if these twofunctions are the same for each point in space. For instance, although the constraintsxy + yx = 0 and xy = 0 trivially satisfy integrability criterion (1.19), only the formerhas a continuously defined (r). The function (r) for the latter constraint can bedefined in only a piecewise manner (see Figure 1.11). If we use (1.19) to examinethe 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.

A SINGLE RHEONOMIC CONSTRAINT. It is clearly of interest to present the general-ization of Jacobi’s criterion to rheonomic constraints: f · v + e = 0. The result is verysimilar 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

f1q1 + f2q2 + f3q3 + f4 = 0,

and define the variables

U1 = q1, U2 = q2, U3 = q3, U4 = t.

Clearly, f4 = e. We next form the functions

IJKL = fJ

(∂fL

∂UK− ∂fK

∂UL

)+ fK

(∂fJ

∂UL− ∂fL

∂UJ

)+ fL

(∂fK

∂UJ− ∂fJ

∂UK

).

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

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Here, the integer indices J , K, and L range from 1 to 4. A necessary and sufficientcondition for the constraint π = 0 to be integrable is that the following four equa-tions 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] orto Flanders [61].

SYSTEMS OF CONSTRAINTS. When particles are subject to several constraints, theirindependence needs to be examined. For the case of two constraints, we first expressthem both in the form

f1 · v + e1 = 0,

f2 · v + e2 = 0.

If f1 × f2 �= 0, then the constraints are said to be independent. For integrable con-straints, this is equivalent to the condition ∇1 × ∇2 �= 0. That is, the normal vec-tors to surfaces 1 = 0 and 2 = 0 are not parallel. The case of three constraints issimilar. 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 pointof 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 con-straints that are individually nonintegrable can become integrable. The most well-known 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 sys-tem of constraints is integrable. The proof of this result, which is presented in Sec-tion 8.4, uses a criterion that is due to Ferdinand G. Frobenius (1849–1917), whichwe postpone discussion of until Chapter 8. This criterion can also be used to showthat if the three constraints (1.22) are imposed on a particle, then the system of con-straints 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 theexercises at the end of this chapter.

1.9 Closing Comments

In this chapter, we have assembled many of the needed kinematical concepts andtools needed to solve problems in particle dynamics. For most readers, the novel as-pects of the chapter will have been the discussion of curvilinear coordinates andkinematical constraints. These two topics are intimately related and will featureprominently 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, andshow that the magnitude of this vector corresponds to the rate at which the particlesweeps out a particular area. Does the particle sweep out equal areas during equalperiods 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 coordinatesas

r(t) = 10er, θ(t) = ωt,

where ω �= 0. Determine the areal velocity vector A of the particle. Under whichconditions 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 Cartesiancoordinates {x = x1, y = x2, z = x3} by the relations

r =√

x21 + x2

2, θ = tan−1(

x2

x1

), z = x3.

Show that the covariant basis vectors associated with the curvilinear coordinate sys-tem, 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 = 1r

eθ, a3 = E3.

It is a good exercise to convince yourself with an illustration that a2 is tangent to aθ coordinate curve, whereas a2 is normal to a θ coordinate surface. Finally, for thiscoordinate system, show that

T = m2

(r2 + r2θ2 + z2) .

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28 Exercises 1.4–1.5

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

R =√

x21 + x2

2 + x23,

θ = tan−1(

x2

x1

),

φ = tan−1

⎛⎝√

x21 + x2

2

x3

⎞⎠ .

Show that the covariant basis vectors associated with the curvilinear coordinate sys-tem, 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 = 1R

eφ, a3 = 1R sin(φ)

eθ.

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

u = ±√

x3 +√

x23 + (x2

1 + x22

),

v = ±√

−x3 +√

x23 + (x2

1 + x22

),

θ = tan−1(

x2

x1

).

In addition, the inverse relations can be defined:

x1 = uv cos(θ), x2 = uv sin(θ), x3 = 12

(u2 − v2).

(a) In the r–x3 plane, where r is the cylindrical polar coordinate r =√

x21 + x2

2,draw several representative examples of the projections of the u and v co-ordinate surfaces. You should give a sufficient number of examples to con-vince 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 andθ coordinate curves foliate this surface.

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(c) Show that the covariant basis vectors for the parabolic coordinate systemare

a1 = ∂r∂u

= ver + uE3,

a2 = ∂r∂v

= uer − vE3,

a3 = ∂r∂θ

= uveθ.

Illuminate your results from (a) and (b) by drawing representative exam-ples of these vectors.

(d) Show that the contravariant basis vectors for the parabolic coordinate sys-tem are

a1 = grad(u) = 1u2 + v2

a1,

a2 = grad(v) = 1u2 + v2

a2,

a3 = grad(θ) = 1uv

eθ.

Again, illuminate your results from (a), (b), and (c) by drawing representa-tive 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 timederivatives.

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 equationof the helix is

r = R0, z = αR0θ,

where R0 and α are constants. Here, we use another curvilinear coordinate systemto 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 systemare

a1 = r(eθ + αE3), a2 = er + αθE3, a3 = E3.

Verify that the covariant basis vectors are not orthonormal.

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Bead of mass m

g

E1

E2

E3

Figure 1.13. A particle moving on a circular helix.

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

T = m2

((1 + α2θ2)r2 + (1 + α2)r2θ2 + ν2)

+ m2

(2νrαθ + 2νθαr + 2rθα2rθ

).

Calculate the contravariant basis vectors associated with the curvilinear co-ordinate 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 instancesof a nonorthogonal coordinate system. Referring to (1.5), calculate the functionsxk(q1, q2, q3). Draw the coordinate curves and surfaces for the curvilinear coordi-nate 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 pre-viously 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 andcontravariant basis vectors are defined by (1.24) and (1.25), respectively. In addition,

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verify that

b =3∑

i=1

biai =3∑

i=1

biai.

Furthermore, show that b �=∑3i=1 biai and b �=∑3

i=1 biai. For which values of α is{a1, a2, a3

}not a basis?

1.9. This exercise illustrates how the covariant and contravariant components of avector are related. To start, we define the following scalars by using the covariantand 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 prob-lem range from 1 to 3.

(a) For any vector b, show that the covariant and contravariant components arerelated:

bi =3∑

k=1

aikbk, bi =3∑

k=1

aikbk.

In other words, the covariant components are linear combinations of thecontravariant components and vice versa. Using a matrix notation, theseresults can be expressed as⎡

⎢⎣b1

b2

b3

⎤⎥⎦ =

⎡⎢⎣a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎥⎦⎡⎢⎣b1

b2

b3

⎤⎥⎦ ,

⎡⎢⎣b1

b2

b3

⎤⎥⎦ =

⎡⎢⎣a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎥⎦⎡⎢⎣b1

b2

b3

⎤⎥⎦ .

(b) By choosing b = ar and as, and using the symmetries of akm and ars, showthat

3∑k=1

aikakj =3∑

k=1

akiakj =3∑

k=1

akiajk = δji .

It might be helpful to realize that, by using matrices, one of the precedingresults has the following representation:⎡

⎢⎣a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎥⎦⎡⎢⎣a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎥⎦ =

⎡⎢⎣1 0 0

0 1 00 0 1

⎤⎥⎦ .

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1.10. With the help of integrability criteria (1.19) and (1.21), show that only one ofthe following constraints is integrable:

xx + yy = −e(t), zy + x = 0, cos(z)y − sin(z)x = 0.

In addition, show that the integrable constraint corresponds to a particle moving ona cylinder whose radius varies with time.

1.11. With the help of integrability criterion (1.21), show that one of the followingconstraints is nonintegrable:

zx + y = −e(t), z = 0.

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

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2 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 fromplanetary motion to a model for a roller coaster. Our discussion of the behaviorof these systems predicted by the models relies heavily on numerical integration ofthe equations of motion provided by F = ma, and it is presumed that the reader isfamiliar with the numerical integration of ordinary differential equations.

Two of the most important types of forces featured in many applications areconservative forces and constraint forces. For the former, the gravitational forcebetween two particles is the prototypical example, whereas the most common con-straint force in particle mechanics is the normal force. It is crucial to be able to prop-erly formulate and represent conservative and constraint forces, and we will spenda considerable amount of time discussing them in this chapter. In contrast to mosttexts in dynamics, here we consider friction forces to be types of constraint forces.

For most applications, exact (or analytical) solutions are not available and re-course to numerical methods is often the only course of action. In validating thesesolutions, any conservations that might be present are crucial. To this end, conser-vations of momentum and energy are discussed at length and we also show (withthe help of two examples) how angular momentum conservation can often be ex-ploited. The examples discussed in this chapter are far from exhaustive. Althoughseveral other examples are included in the exercises, they too do not come close toencompassing 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 textsby 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 forthis particle is known as the balance of linear momentum, Newton’s second law, or

33

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Euler’s first law. The integral (or impulse momentum) form of this law is

G(t) − G (t0) =∫ t

t0F(τ)dτ, (2.1)

where F is the resultant force acting on the particle and G = mv is the linear momen-tum. Notice that this form of the balance law does not assume that v is differentiablewith respect to time t. As a result, it is valid in impact problems, among others. Anexample of this is discussed in Section 2.10.

If we assume that G is differentiable with respect to time, then we can differ-entiate both sides of the integral form of the balance of linear momentum to obtainthe local form:

F = G.

Assuming that the mass of the particle is constant, we can write

F = mr. (2.2)

This law represents three (scalar) equations that relate F and the rate of changeof linear momentum of the particle. We refer to F = ma as the balance of linearmomentum.∗

Our emphasis in this chapter is on the principle F = ma, but it is misleading tobelieve that this is the sole accepted principle in dynamics. Indeed, since the creationof this principle by Newton over 300 years ago, several alternative (and often equiv-alent) principles of dynamics have been proposed, and we shall postpone discussionof several of them until Section 4.11.

It is convenient to write the balance law as a set of first-order ordinary differen-tial equations:

v = r, F = mv.

In the absence of constraints, these represent six scalar (differential) equationsfor the six unknowns r(t) and v(t). To solve these equations, six initial conditionsr(t0) and v(t0) must be specified. Alternatively, if the problem is formulated as aboundary-value problem, then a combination of six initial and final conditions onr(t) and v(t) must be prescribed.

If we write F = ma using a Cartesian coordinate system, then we find thefollowing three equations:

mx1 = F · E1,

mx2 = F · E2,

mx3 = F · E3.

∗ Discussions of the historical threads from Newton’s Principia [152] that lead to Euler’s explicitformulation 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 − rθ2) = F · er,

m(rθ + 2rθ

) = F · eθ,

mz = F · E3. (2.3)

Finally, if we use a spherical polar coordinate system, we find that

m(R − Rφ2 − R sin2(φ)θ2) = F · eR,

m(Rφ + 2Rφ − R sin(φ) cos(φ)θ2) = F · eφ,

m(R sin(φ)θ + 2Rθ sin(φ) + 2Rθφ cos(φ)) = F · eθ. (2.4)

Notice that these equations are different projections of F = ma onto a set of basisvectors for E

3.Establishing the component representations of F = ma for various coordinate

systems can be a laborious task. However, Lagrange’s equations of motion allow usto 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, whent = tB, the particle is at B: r(tB) = rB. During the interval of time that the particlemoves 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

m

O

A B

P

v

rrA rB

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

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with respect to time, of the mechanical power:

WAB =∫ tB

tAP · vdt.

Notice that this is a line integral, and we are using t to parameterize the path of theparticle. Depending on the choice of coordinate system, the integral in this expres-sion has several equivalent representations.

As an example, suppose that a force P = Peθ acts on a particle, and the mo-tion 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 forceis

P · v = ωPLeαt,

and the work performed by the force is

WAB =∫ tB

tAωPLeαtdt = ωPL

α(eαtB − eαtA) ,

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 duringany motion of the particle is independent of the path of particle. When a result fromvector calculus is used, the path independence implies that P is the gradient of ascalar function U = U (r):

P = −∇U.

The function U is known as the potential energy associated with the force P, and theminus 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−U. To see this, we simply examine U and use the definition of a conservative force:

−U = −∂U∂r

· v = −(−P) · v = P · v.

This result holds for all motions of the particle and is very useful when wewish to establish expressions for the rate of change of the total energy E of aparticle.

To check if a given force P is conservative, one approach is to find a potentialfunction U such that

P · v = −U

holds for all motions of the particle. This approach reduces to solving a set of cou-pled 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 forU (r, θ, z):

Pr = −∂U∂r

, Pθ = −1r

∂U∂θ

, Pz = −∂U∂z

, (2.5)

where P = Prer + Pθeθ + PzE3. You might notice that the solution to (2.5) will yielda potential energy U(r, θ, z) modulo an additive constant. This constant is usually setby 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 itscurl. The idea here is based on the identity curl(grad(V)) = 0, where V = V(r) is anyscalar 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 thefollowing conditions:

∂P3

∂x2= ∂P2

∂x3,

∂P1

∂x3= ∂P3

∂x1,

∂P2

∂x1= ∂P1

∂x2,

where Pi = P · Ei.

2.5 Examples of Conservative Forces

The three main types of conservative forces in engineering dynamics are constantforces, spring forces, and gravitational force fields.

Constant ForcesAll constant forces are conservative. To see this, let C denote a constant force andlet Uc = −C · r. Now, ∇Uc = −C, and, consequently, Uc is the potential energy asso-ciated with C. The most common examples of constant forces are the gravitationalforces −mgE2 and −mgE3, and their associated potentials are mgE2 · r and mgE3 · r,respectively.

Spring ForcesConsider the spring shown in Figure 2.2. One end of the spring is attached to a fixedpoint A, and the other end is attached to a particle of mass m. When the spring isunstretched, it has a length L0. Clearly, the stretched length of the spring is ||r − rA||,and the extension/compression of the spring is

ε = ||r − rA|| − L0.

The potential energy Us associated with the spring is

Us = f (ε)

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

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spring

m

O

A

r

rA

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

where f is a function of the change in length of the spring. Evaluating the gradientof Us, we find the spring force Fs:

Fs = −∂Us

∂r= −∂f

∂ε

r − rA

||r − rA|| .

To establish this result, we use the identity

∂ε

∂r= ∂

∂r(||r − rA|| − L0) = r − rA

||r − rA|| .

This identity is left as an exercise to establish.∗

The most common spring in engineering dynamics is a linear spring. For thisspring,

Us = K2

(||r − rA|| − L0)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 changein length. Examples of nonlinear springs include those in which f is a polynomialfunction in ε. For instance,

f (ε) = Aε2 + Bε4,

where A > 0 and B are constants. Such a spring is known as hardening if B > 0 andsoftening if B < 0.

Newton’s Gravitational ForceDating to Newton in the late 1600s, the force exerted by a body of mass M on an-other 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 ∂√

x·x∂x = x√

x·x = x||x|| . This result is equiv-

alent to showing that the gradient of ||r|| is eR.

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2.6 Constraint Forces 39

The body of mass M is assumed to be located at the origin O, and G is the universalgravitation constant. In this model for the gravitational force, both bodies aremodeled as mass particles. Later on, in Sections 4.5 and 8.7, we shall examine gen-eralizations of this force field to systems of particles and rigid bodies, respectively.Because the magnitude of Fn depends on the distance squared between the twobodies, 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 rathersimple 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 isthis 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. Examplesof these forces include reaction forces, normal forces, and tension forces in inexten-sible strings. Given a constraint (r, t) = 0 on the motion of the particle, there is nouniversal prescription for the associated constraint force. Choosing the correct pre-scription 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 thatwe need to solve the six equations

r = v, v = 1m

F,

subject to the restrictions on r and v,

(r, t) = 0,∂

∂r· v + ∂

∂t= 0.

To close this system of equations, an additional unknown is introduced in the formof 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 mostcommonly used, which dates to Joseph-Louis Lagrange (1736–1813), is referred toin 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 particlesatisfies the constraint. However, freedom is available to include other arbitrarynonnormal components. This is the reason why prescriptions of constraint forcesfeaturing frictional components are valid.

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Surface = 0

m

O

s1

s2

∇

n

r

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

A Single ConstraintConsider 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 right-handed orthonormal basis {s1, s2, n} for E

3. This is not in general a constant set ofvectors; rather, it changes as we move from point to point along the surface. Thefinal 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 con-straint 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 unde-termined 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 thatis sometimes known as the Lagrange principle or Lagrange–D’Alembert principle.On physical grounds, Lagrange’s prescription is justified if the surface on which theparticle is constrained to move is smooth.

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s1

s2

N = ∇

nvrel

Ff

Fc = N + Ff

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

In the event that the surface is rough, an alternative prescription, which is dueto Charles Augustin Coulomb (1736–1806), can be used∗:

Fc = λ∇ + Ff ,

where the normal force N = λ∇ and the friction force is

Ff = −µd ||λ∇|| v − vs

||v − vs|| .

Here, µd is known as the coefficient of dynamic friction. Notice that the tangentialcomponents of Fc are governed by the behavior of vrel = vrel1 s1 + vrel2 s2 and opposethe 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∂

∂t+ Ff · v,

where we used the identity

= ∇ · v + ∂

∂t= 0.

For the Lagrange prescription, Ff = 0, we can now see that, if the surface that theparticle is moving on is fixed, i.e., = (r), then Fc does no work. Otherwise, thisconstraint force is expected to do work because its normal component ensures thatpart of the velocity vector of the particle is vs. For the Coulomb prescription, ex-cept when vs = 0, it is not possible to predict if work is done on the particle by theconstraint 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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eR − L0 = 0. Consequently, ∇ = eR. In addition, v − vs = L0φeφ + L0 sin(φ)θeθ. Inconclusion,

Fc = λeR − µd|λ| φeφ + sin(φ)θeθ√φ2 + sin2(φ)θ2

.

In this expression |λ| is the magnitude of the normal force exerted by the sphereon the particle. If we now consider the spherical pendulum, then Fc is given by theLagrange prescription: Fc = λeR. In the spherical pendulum, −λ is the tension in therod connecting the particle to the fixed point O.

Two ConstraintsWhen a particle is subject to two constraints, 1(r, t) = 0 and 2(r, t) = 0, then it canbe considered as constrained to move on a curve. The curve in question is formedby 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 thisvector by first observing that ∇1 and ∇2 are both normal to the surfaces that thecurve 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 hasa velocity. We denote this velocity by vc. The velocity vector of the particle relativeto the curve is v − vc = vt.

We now turn to prescriptions for the constraint force. The first prescription isthe 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 thecase of a single constraint, this prescription is valid when the curve that the particlemoves 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 + Ff ,

where the friction force is

Ff = −µd ||λ1∇1 + λ2∇2|| v − vc

||v − vc|| .

The friction force opposes the motion of the particle relative to the curve and thenormal 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 + Ff · v

= −λ1∂1

∂t− λ2

∂2

∂t+ Ff · v,

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where we again use the identities

1 = ∇1 · v + ∂1

∂t= 0, 2 = ∇ψ 2 · v + ∂2

∂t= 0.

For the Lagrange prescription, Ff = 0, and we can now see that, if the curve thatthe particle is moving on is fixed, i.e., 1 = 1(r) and 2 = 2(r), then Fc does nowork. Otherwise, this constraint force is expected to do work because its normalcomponents force part of the velocity vector of the particle to be vc. As in the caseof a single constraint, for the Coulomb prescription, except when vc = 0, it is notpossible to predict if work is done on the particle by this force.

We now consider some examples. Recall that the planar pendulum consists ofa particle of mass m that is attached by a rod of length L0 to a fixed point O. Theparticle is also constrained to move on a vertical plane. In short, 1 = r · er − L0 = 0and 2 = r · E3 = 0. With a little work, we find that ∇1 = er and ∇2 = E3. Forthis mechanical system, Lagrange’s prescription is appropriate:

Fc = λ1er + λ2E3.

For this system, λ1er can be interpreted as the tension force in the rod and λ2E3 canbe interpreted as the normal force exerted by the plane on the particle. If we letL0 = 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 particlemoving on a rough circle whose radius L0 = L0(t). The particle is subject to the sameconstraints as it is in the planar pendulum; however, Lagrange’s prescription is notvalid. Instead, we now have

Fc = λ1er + λ2E3 − µd

√λ2

1 + λ22

θ∣∣θ∣∣eθ,

where we used the fact that v − vc = L0θeθ.

Three ConstraintsThe reader may have noticed that our expressions for the constraint force when weemployed Coulomb’s prescription were not valid when the particle was stationaryrelative to the surface or curve that it was constrained to move on. This is becausewe view this case as corresponding to the motion of the particle subject to threeconstraints: i (r, t) = 0, i = 1, 2, 3. As mentioned earlier, when a particle is subjectto three constraints, the three equations i(r, t) = 0 can in principle be solved todetermine 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, thesole purpose of F = ma is to determine the constraint force Fc.

For the case in which the particle is subject to three constraints, the Lagrangeprescription and a prescription based on static Coulomb friction are equivalent.This equivalence holds in spite of the distinct physical situations these prescriptionspertain 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 tacitlyassumed to be independent, {∇1,∇2,∇3} forms a basis for E

3. 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 thecurve or surface on which it lies is

Fc = N + Ff ,

where the magnitude of Ff is restricted by the static friction criterion:

‖Ff ‖ ≤ µs ||N|| ,where µs is the coefficient of static friction. Again, the Coulomb prescription pro-vides a vector Fc with three independent components. In other words, both prescrip-tions 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 pre-scription or Coulomb’s prescription.

Nonintegrable ConstraintsOur discussion of constraint forces has focused entirely on the case of integrableconstraints. If a nonintegrable constraint,

f · v + e = 0,

is imposed on the particle, we need to discuss a prescription for the associated con-straint 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 thenonintegrable constraint turns out to be integrable, then the prescription we employwill agree with Lagrange’s prescription we discussed earlier.

As a further example, suppose the motion of the particle is subject to two con-straints, 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 parti-cle is

Fc = λ1∇ + λ2f.

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2.7 Conservations 45

Suppose that the applied force acting on the particle is Fa; then the equations gov-erning the motion of the particle are

(r, t) = 0, f · v + e = 0,

r = v, v = 1m

(Fa + λ1∇ + λ2f).

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 quan-tity is said to be conserved if it is constant during the motion of the particle. Theconserved quantities are often known as integrals of motion. The solutions of manyproblems in particle mechanics are based on the observation that either a mo-mentum or an energy (or both) is conserved. At this stage in the development ofthe field, most of these conservations are obvious and are deduced by inspection.However, for future purposes it is useful to understand the conditions for suchconservations. We shall consider numerous examples of these conservations lateron.

Conservation of Linear MomentumThe linear momentum G of a particle is defined as G = mv. Recalling the integralform of the balance of linear momentum,

G(t) − G (t0) =∫ t

t0F(τ)dτ,

we see that G(t) is conserved during an interval of time (t0, t) if∫ t

t0F(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 di-

rection of a given vector b(t) being conserved. That is, ddt (G · b) = 0. For this to

happen,

˙G · b = G · b + G · b = F · b + G · b = 0.

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 exampleis to consider a particle impacting a smooth vertical wall. Then the components of Gin the two tangential directions are conserved. For these two examples, the vector bis constant.

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Conservation of Angular MomentumThe angular momentum of a particle relative to a fixed point O is HO = r × G. Toestablish how HO changes during the motion of a particle, a simple calculation isneeded:

HO = v × G + r × G = v × mv + r × F = r × F.

It is important to note that we used F = ma during this calculation. The final resultis known as the angular momentum theorem for a particle:

HO = r × F.

In words, the rate of change of angular momentum is equal to the moment of theresultant force.

Conservation of angular momentum usually arises in two forms. First, the en-tire vector is conserved, and, second, a component, say c(t), is conserved. Forthe first case, we see from the angular momentum theorem that HO is conservedif F is parallel to r. Problems in which this arises are known as central forceproblems. Dating to Newton, they occupy an important place in the history ofdynamics. When the angular momentum theorem is used, it is easy to see thatthe second form of conservation, HO · c is constant, arises when r × F · c + HO · c= 0.

Conservation of EnergyAs a prelude to discussing the conservation of energy, we first need to discuss thework–energy theorem. This theorem is a result that is established by use of F = maand relates the time rate of change of kinetic energy to the mechanical power ofF:

T = F · v.

This theorem is the basis for establishing conservation of energy results for a singleparticle.

The proof of the work–energy theorem is very straightforward. First, recall thatT = 1

2 mv · v. Differentiating T, we find

T = ddt

(12

mv · v)

= 12

(mv · v + mv · v) = mv · v.

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, wefirst divide the forces acting on the particle into the sum of a resultant conservativeforce P = − ∂U

∂r and a nonconservative force Pncon: F = P + Pncon. Here, U is the sum

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2.8 Dynamics of a Particle in a Gravitational Field 47

of the potential energies of the conservative forces acting on the particle. From thework–energy theorem, we find

T = F · v

= P · v + Pncon · v

= − ∂U∂r

· v + Pncon · v

= − 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 forcesdo no work, then the total energy of the particle is conserved.

To examine whether energy is conserved, it usually suffices to check whetherPncon · v = 0. To see this, let us consider the example of the spherical pendulumwhose length L0 = L0(t). For this particle,

P = −mgE3, Pncon = λeR.

Consequently,

E = T + mgE3 · r

and

Pncon · v = λeR · v = L0λ.

As a result, when the length of the pendulum is constant, L0 = 0, then E is con-served. On the other hand, if L0 �= 0, then the constraint force λeR does work bygiving 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 center-pieces in Isaac Newton’s Principia.∗ Over 100 years later, it was also discussed inwonderful detail in Lagrange’s famous text Mecanique Analytique.† Newton waspartially motivated to study this problem because of Johannes Kepler’s (1571–1630) famous three laws of motion for the (then known) planets in the solarsystem:

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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x

y

m

O

a

b

a(1 − e)

rθ P

A

h < 0 h > 0

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

orbit. One of the foci of the ellipse is at O, and the eccentricity e =√

1 − b2

a2 of the ellipse isless 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 equaltimes.

III. If a denotes the semimajor axis of the elliptical orbit and T denotes the period

of the orbit, then, for any of two planets, a31

a32

= T21

T22

.

Concise discussions of Kepler’s laws can be found in [150, 175, 188, 220]. Some ofthese authors note that the laws were based on astronomical data taken with thenaked eye.

In our analysis we assume that the body of mass M is fixed. As can be seen inExercise 4.6 from Chapter 4, this restriction is easily removed and the results pre-sented can be readily applied to deduce the motions of m and M. Many of the resultspresented feature terminology associated with ellipses, and, for convenience, a sum-marization of many of the terms associated with an ellipse, such as its eccentricity eand axes a and b, is given in Figure 2.5. The area swept out by the particle can bedetermined by integrating the areal vector (1.1): A = 1

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 AUand 11.862 years, and Mars’ are 1.524 AU and 1.881 years, respectively. We notethat

0.3873

0.2412= 1.00,

5.2033

11.8622= 1.00,

1.5243

1.8812= 1.00.

All of these results are in agreement with Kepler’s third law.Here, we start by setting up the coordinates for this problem and establish-

ing the equations of motion. Our analysis of the equations of motion then ex-ploits conservation of angular momentum to show that the motion must be planar.

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Following this, we reduce the equations of motion to a single second-order differen-tial equation that we nondimensionalize and integrate numerically.∗ An alternativeapproach, which is used in most textbooks, will also be discussed. These analysesenable us to classify all five possible types of trajectories of the particle.

KinematicsWe pick as the origin O the fixed particle of mass M. Then the position vector of theparticle of mass m is r, and it is convenient to pick cylindrical polar coordinates forthis position vector:

r = rer + zE3.

Representations for the velocity and acceleration vectors in terms of cylindrical po-lar coordinates were established earlier and we do not rewrite them here.

Equations of MotionThe equations of motion for the particle can be obtained from F = ma, where F issolely due to Newton’s gravitational force:

Fn = −GMm

||r||3 r.

You should recall that this force is conservative, and its potential energy is denotedby Un.

Using (2.3), we can write out the component forms of Fn = ma. The result willbe three differential equations:

m(r − rθ2) = − GMmr(√

r2 + z2)3 ,

m(rθ + 2rθ

) = 0,

mz = − GMmz(√r2 + z2

)3 . (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.

ConservationsThe solutions of differential equations (2.6) conserve two important kinematicalquantities. 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 reductionprocedure that is used to incorporate momentum conservation in a variety of mechanical systemsranging from the problem at hand to Lagrange and Poisson tops. For further details on this proce-dure, 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 readerto demonstrate these results by using the work–energy and angular momentumtheorems.

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 theorbital plane). We also observe that the normal to this plane is parallel to HO. Ifwe allow ourselves the freedom to choose E3, then we can pick this vector such thatHO = hE3, where h = mr2θ. Consequently, r and v will have components in onlythe E1 and E2 directions: z(t) = 0 and z(t) = 0. We henceforth exploit the fact thatangular momentum is conserved and the motion is planar. Because HO and the arealvelocity vector are synonymous, angular momentum conservation for this problemis often known as the “integrals of area.”∗

We have tacitly ignored the case in which r (t0) ‖ v (t0). In this case, HO is zeroand must remain so. Consequently, the motion of the particle is a straight line. Forconvenience, we can choose this line to lie on the E1 − E2 plane. It can be shownthat the motion of the particle will eventually lead to a collision with the particle ofmass M at the origin. Thus, without an initial angular momentum, a collision for thissystem would be unavoidable.

Determining the Motion of the ParticleBecause the angular momentum is conserved, we choose E3 such that z(t) = 0 andz(t) = 0 for the particle. That is, the direction of HO is E3. As a result, the equationsof motion reduce to

m(r − rθ2) = −GMm

r2,

m(rθ + 2rθ

) = 0. (2.7)

Now the second of these equations can be expressed as

ddt

(mr2θ

) = 0. (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:

mr − h2

mr3= −GMm

r2. (2.9)

Here, h is determined from the initial position and velocity of the particle:

h = HO · E3 = (mr (t0) × v (t0)) · E3.

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

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For a particle with no angular momentum, we see from (2.8) that θ = 0 and hencer = −GM

r2 . It is not difficult to see that this equation implies that r(t) → 0 as t in-creases. This is the collision we discussed earlier.

Given r(t0) and v(t0), we can determine h and then integrate (2.9) to determiner(t).∗ We can then compute the coordinate θ(t) by integrating

θ = hmr2

. (2.10)

We can then find expressions for x(t) = r(t) cos (θ(t)) and y(t) = r(t) sin (θ(t)) andconstruct 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 = h2

GMm2. (2.11)

We also show, using (2.10), that θ is constant:

θ(t) = ωK = h

mr20

=√

GM

r30

. (2.12)

The frequency ωK is known as the Kepler frequency. Physically, the particle is mov-ing in a circular orbit of radius r0 at constant speed r0ωK about the fixed body ofmass 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, wechoose the dimensionless variable w = r

r0and time τ = ωKt. Now using identities of

the form r = drdt = dτ

dtdrdτ

= ωKdrdτ

, we can simplify (2.9) and (2.10) to

d2w

dτ2= 1

w3− 1

w2,

dθ

dτ= 1

w2. (2.13)

Notice that we have reduced the problem of determining the motion of the particleto the integration of two differential equations.

Differential equation (2.13)1 is a second-order differential equation for w(τ). Asopposed to exhaustive displays of w(τ) for several sets of initial conditions, a quali-tative method of representing the solutions of (2.13)1 is to construct what is knownas a phase portrait. In this portrait, dw

dτis plotted as a function of w(τ).† Equilibria

of the second-order differential equation correspond to points in the phase portraitwhere dw

dτ= 0 and w(τ) is a constant. To find such points, we set dw

dτ= 0 and d2w

dτ2 = 0in the governing second-order differential equation for w(τ) and solve for the result-ing constant values of w(τ). Later examples in this chapter will feature differentialequations with multiple equilibria.

∗ Differential equation (2.9) has an analytical solution that can be expressed in terms of Jacobi’selliptic 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 ordinarydifferential equations (see, for example, the texts of [10, 18, 81, 229]).

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w

dwdτ

cee e

hh

p

1

1

2

4

−1

−2

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

Returning to the problem at hand, the phase portrait of (2.13)1 is shown inFigure 2.6. There, we see an equilibrium point at

(w, dw

dτ

) = (1, 0) that correspondsto a circular orbit of the particle. The closed orbits enclosing this point correspondto elliptical orbits of the form shown in Figure 2.7.∗ The remaining orbits shownin this figure correspond to hyperbolic orbits of the particle. For these orbits, theparticle circles around the equilibrium once and never returns. The interesting casein which the particle describes a parabolic orbit is also shown in this figure. Thetrajectory on the phase portrait corresponding to this orbit separates the ellipticaland hyperbolic trajectories.†

The Orbital MotionsAn alternative approach to the one outlined in the previous section is followed inmost textbooks on dynamics. This approach involves solving for the motion of theparticle 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) toshow that

r = − hm

ddθ

(1r

), r = − h2

m2r2

d2

dθ2

(1r

).

∗ A second integration involving dθdτ

= 1w2 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, dwdτ

)= (0.5, 0) connects the fixed point at

(w, dw

dτ

)= (∞, 0) and corresponds to the parabolic

orbit of the particle.

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x = x1

x = x1x = x1

x = x1x = x1

y = x2

y = x2y = x2

y = x2y = x2

O

OO

OO

h < 0

h > 0

h > 0

h > 0

rr

r

r

(a) (b)

(c)

(d) (e)

θ

θ θP

θP

θP

Figure 2.7. Schematic of the four types of orbits of a particle of mass m moving about a fixedpoint 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 θp

are considered.

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

d2

dθ2

(1r

)+ 1

r= 1

r0. (2.14)

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

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Equation (2.14) is a linear ordinary differential equation for 1r as a function of θ

that has the exact solution

r = r (θ) = r0 (1 + e cos (θ − θp))−1, (2.15)

where e and θp are constants (which are determined from the initial conditions for

the position and velocity of the particle). We can then integrate mr2(θ)h

dθ = dt to de-termine 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; whene = 1, the orbit is parabolic; and when e > 1, the orbit is hyperbolic. Referring toFigure 2.5 for the elliptical orbit, it is easy to see that b = h2

GMm2 .To solve for e and θp, let us assume that r (t0) and v (t0) are given. Then we can

determine HO and specify E3 and h. To compute θp, we first calculate r by using(2.15) and the chain rule:

r(t) = −(

GMm

h

)e sin (θ(t) − θp) . (2.16)

We also compute the value E0 of total energy of the particle

E0 = m2

v (t0) · v (t0) − GMm∣∣∣∣r (t0)∣∣∣∣ . (2.17)

Now as the total energy is conserved, the value of this kinematical quantity whenθ = θp is also equal to E0. With some manipulations of (2.15) and (2.16), it can beshown that

E0 = G2M2m3

2h2

(e2 − 1

). (2.18)

We now have a method of determining e and θp 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 θp. With these values, r(θ) isspecified and θ(t) can be calculated. Depending on the value of e, the orbits will beone 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):

d2udθ2

+ u = 1, (2.19)

where u = 1w

= r0r . The phase portrait of this equation is shown in Figure 2.8. In

contrast to the earlier phase portraint, here the trajectory corresponding to theparabolic orbit is easily distinguished. However, as in the previous case, to gaina physical interpretation of the trajectories shown in Figure 2.8, it is necessary toreconstruct a position vector of the particle corresponding to the particular orbit.

CommentsFor this problem, the rare event occurs that a complete classification of the motionsof 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 55

u

dudθ

c

ee

p

hhh

h

1

1

4

2

-1

-2

Figure 2.8. The phase portrait of (2.19). The trajectories labeled e and h correspond to ellip-tical and hyperbolic orbits of the particle, the point c corresponds to a circular orbit, and thetrajectory labeled p corresponds to the parabolic orbits. In this figure, the arrow indicates thedirection 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 relatedproblems involving escape velocities and transfer orbits can be found in several text-books; 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 theproblem for now, we wish to show that the elliptical (and circular) orbits are inagreement with Kepler’s laws. We first note that satisfaction of the first law is triv-ial, and the second law is a consequence of angular momentum conservation. To seethat the third law is satisfied, we need to compute the period T of the particle exe-cuting an elliptical orbit. We leave it as an exercise to show that T = 2π√

GMa

32 , and,

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 andunstretched length L0. We assume that the surface of the cone is rough and that thecone is spinning about its axis of symmetry with an angular speed �0. Our goal isto establish the equations of motion for the particle and discuss some features of itsdynamics.

Coordinates, Constraints, and VelocitiesAs discussed in Section 1.7, when the particle is moving on the surface of the cone, itis 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 ∇ = 1Reφ. Because the cone

is rotating with a speed �0, the velocity of the particle relative to the cone is

vrel = ReR + R cos(α)(θ − �0

)eθ.

We defer discussion of the case in which the particle is stuck to the cone.

ForcesThe 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 thatthe particle is moving relative to the surface of the cone, we find that the constraintforce Fc acting on the particle has the representation

Fc = N + Ff ,

where the normal force N is parallel to ∇:

N = λ

Reφ, Ff = −µd ||N|| vrel

||vrel|| .

Thus the total force on the particle is F = Fc + Fs − mgE3.

The Equations of MotionTo 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)θ2) = −K (R − L0) + Ff · eR − mg sin(α),

m(R cos(α)θ + 2Rθ cos(α)

) = Ff · eθ,

−mR sin(α) cos(α)θ2 = λ

R+ mg cos(α). (2.20)

The first two of these equations are ordinary differential equations for R and θ, andthe third equation can be solved for λ (and hence the normal force) as a function ofthe motion of the particle.

To integrate these equations, it is desirable to nondimensionalize them. We use

L0 as a measure of length and√

L0g as a measure of time:

τ = t√

gL0

, w = RL0

.

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2.9 Dynamics of a Particle on a Spinning Cone 57

With the help of identities of the type R = dRdt = dτ

dtdRdτ

=√

gL0

dRdτ

, we can rewrite

(2.20) in the form

d2w

dτ2= w cos2(α)

(dθ

dτ

)2

− ω2(w − 1) − sin(α)

−µkndwdτ√

w2( dθ

dτ− ω0

)2 + ( dwdτ

)2 ,

ddτ

(w2 cos2(α)

dθ

dτ

)= −µkn (w cos(α))

w( dθ

dτ− ω0

)√w2( dθ

dτ− ω0

)2 + ( dwdτ

)2 .

(2.21)

In these equations, the constants and dimensionless normal force are

ω2 = KL0

mg, ω0 = �

√L0

g, n = ||N||

mg= cos(α) + h2

wtan(α),

and we can also show that dimensionless versions of total energy E and angularmomentum HO · E3 are

EmgL0

= 12

((dw

dτ

)2

+(

dθ

dτ

)2

w2 cos2(α)

)+ ω2

2(w − 1)2 + w sin(α),

h = HO · E3

m√

gL30

= w2 cos2(α)dθ

dτ.

Notice that, by nondimensionalizing the equations of motion, we have reduced thenumber of parameters by two.

The Static Friction CaseWhen the particle is stuck to the cone, its velocity vector is v = R0�0 cos(α)eθ. In ad-dition, the particle is subject to three constraints and the friction and normal forcesconstitute three undetermined forces that enforce these constraints:

Fc = λ

R0eφ + λ1

R0 cos(α)eθ + λ2eR.

To determine λ, λ1, and λ2, we examine F = ma. With some manipulations, we con-clude that

Ff = (K (R0 − L0) + mg sin(α) − mR0 cos2(α)�20

)eR,

+ mR0 cos(α)�0eθ,

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 tocheck this sufficiency we need to examine the static friction criterion ‖Ff ‖ ≤ µs ||N||.

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RL0

1L0

dRdτ

x1

x1

x1

x1 x2

x2

x2

x2

1 8

20

−20

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

If this criterion holds, then a particle that is stuck on the surface of the cone willremain stuck on the surface. Otherwise it slips, and the initial direction in which itslips is parallel to Ff .∗

The Smooth ConeWhen 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 angularmomentum to write a single equation for w:

d2w

dτ2= h2

w3 cos2(α)− ω2(w − 1) − sin(α). (2.22)

Integrating this equation, we can find w(τ). Another integration usingw2 cos2(α) dθ

dτ= h provides θ(τ). The equilibrium point at

(w, dθ

dτ

) = (w0, 0), wherew0 is the solution of

h2

w30 cos2(α)

− ω2(w0 − 1) − sin(α) = 0,

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

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2.10 A Shocking Constraint 59

corresponds to a circular orbit of the particle that has a radius r = L0w0 cos(α). Someof the other trajectories of w in the w − dw

dτplane are shown in Figure 2.9. In this

figure, we have also constructed possible trajectories of the particle correspondingto w(τ). Unlike the problem of the particle subject to Fn that we discussed in Section2.8, here the classifications of the trajectories for the particle on the cone defy asimple classification.

2.10 A Shocking Constraint

We now return to the constraint yx = 0, discussed earlier [see (1.15)]. Our interestis to determine the equations of motion of a particle that is subject to this constraintand that is also under the influence of an applied force Fa = P1E1 + P2E2.

First, we assume that the constraint force that enforces yx = 0 has the standardprescription

Fc = λxEy, (2.23)

where λ is a Lagrange multiplier. We note that, when x = 0, Fc = 0. From a balanceof 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 isplanar.

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 rec-tilinear:

y(t) = y0, x(t) = x0 + x0t +∫ t

0

∫ τ

0

P1

mdudτ, λ = Fy

x(t)(2.25)

and

x = 0, y(t) = y0 + y0t +∫ t

0

∫ τ

0

P2

mdudτ. (2.26)

Here x0 = x(0), x0 = x(0), y0 = y(0), and y0 = y(0) are initial conditions. From thesesolutions to the equations of motion, we observe that λ is not defined when x = 0.

The ShockReferring to Figure 1.11, we recall that this constraint has the unusual feature thatit can be decomposed into two piecewise integrable constraints:

y = 0 when x �= 0, and x = 0.

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x = x1x = x1

y = x2y = x2

Ic1

Ic2

A AB

B

(a) (b)

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

At the points where the particle makes a transition from one of these integrableconstraints to the other, its velocity vector v will be discontinuous and thereforeits acceleration vector cannot be defined. At such a transition, the prescription forconstraint force (2.23) does not hold, and instead we can calculate only the impulseIc that is due to this force by using (2.1). Supposing that the transition occurs at timet = T, we will have

Ic = limσ→0

(mv (T + σ) − mv (T − σ) −

∫ T+σ

T−σ

Fadτ

).

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 discontinuousvelocity 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 tothe situation shown in Figure 2.10(a), where there is no discontinuity in the motionof the particle. That is, the shock is absent and consequently Ic = 0.

If the constraint cannot supply the impulse Ic, then the particle is effectivelysubject 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 exercisefor the reader to imagine a rigid wall placed to the left of the y axis in Figure 1.11 asa 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 theslightest nudge will set the cart in motion. The presence of Coulomb friction withstick–slip changes this scenario. It will eventually bring the cart to a halt, and it maybring the cart to a halt near the top of the roller coaster. Indeed, if there is sufficientstatic 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 61

x1

yA

B C

E1

E2

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 bethe dynamic friction coefficient.

Here, a very simple model is presented for the roller coaster that captures itsstick–slip behavior.∗ First, we establish a differential equation governing the mo-tion of the roller coaster, and then we use numerical integrations to investigate thedynamics of the roller coaster.

The Equations of MotionOne model for the dynamics of a cart on a roller coaster is to model the cart as aparticle 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 + Ff . It is a standard exercise to calcu-late the unit tangent et, the unit normal en, and the binormal eb vectors to this curve[159]:

et = 1√1 + f ′ 2

(E1 + f

′E2

),

en = sgn(

f′′)√

1 + f ′ 2

(E1 + f

′E2

),

eb = et × en,

where the prime denotes the derivative with respect to x1. These three vectors con-stitute the Frenet triad, and in calculating this triad we assume that x1 > 0. A nor-mal force N = Nen + λ2E3, a friction force Ff 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 accelerationvector of the particle is

a = vet + κv2en,

∗ 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:

κ =∣∣f ′′ ∣∣(√

1 + f ′ 2)3 .

Taking the et and en components of F = ma, we can easily calculate the equationsgoverning the motion of the cart and the normal force:

m((

1 + f′ 2)

x1 + f′′f

′x2

1

)= −mgf ′ − µd

√1 + f ′ 2 ||N|| x1

|x1| , (2.27)

where

N = 1√1 + f ′ 2

(sgn(

f′′)

mg +∣∣∣f ′′∣∣∣mx2

1

)en, if f

′′ �= 0,

= mg√1 + f ′ 2

en, when f′′ = 0. (2.28)

These equations apply when the cart is moving and the friction is dynamic. In theevent that the cart is stationary, static friction acts and, provided the static frictioncriterion is satisfied, ∣∣∣∣∣∣

mgf′√

1 + f ′ 2

∣∣∣∣∣∣ ≤ µs ||N|| , (2.29)

the cart remains stationary. With the help of (2.28), (2.29) can be expressed in thesimple form ∣∣∣ f ′

∣∣∣ ≤ µs. (2.30)

This equation can be viewed as the basis for the classical experiment to measurethe coefficient of static friction: We place a block on an inclined plane and slowlyincrease the angle of inclination until slipping occurs. The tangent of the angle ofinclination is equal to µs. We shall shortly use (2.30) to establish a continuum ofpoints at which the roller coaster can remain in a state of rest.

We now choose a cosinusoidal track,

f (x1) = Acos(

πx1

L0

). (2.31)

In addition, we employ the following nondimensionalizations:

x = x1

L0, τ =

√g

L0t.

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 interestingchoice would be Euler’s spiral (clothoid) that features in “loop-the-loop” rollercoasters.

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2.11 A Simple Model for a Roller Coaster 63

x

dxdτ

−2 2

−1.5

1.5

Figure 2.12. The phase portrait of (2.27)–(2.29) when friction is absent. The points on thex = x1

L0axis labeled by · correspond to equilibria of the cart. For this figure, A

L0= 0.25, and

µs = µd = 0.0.

States of RestFor a cart on a smooth roller coaster, the motion of the cart will be perpetual, anda portion of its phase portrait is shown in Figure 2.12. We note the presence of anequilibrium at

(x = 0, dx

dτ= 0). This point corresponds to the cart’s being stationary

at the top of the roller coaster. Referring to Figure 2.13(a), we refer to equilibriaof this type as saddles. The two equilibria at

(x = ±1, dx

dτ= 0)

represent a stationarycart at the bottom of one of the valleys of the roller coaster. Examining the phaseportrait in Figure 2.13(c), we easily see why equilibria of this type are known ascenters. The equilibria at

(x = (−2, 0, 2), dx

dτ= 0)

correspond to a stationary cart atone of the crests of the roller coaster.

When stick–slip friction is present, the phase portrait changes dramatically (seeFigure 2.14). First, all of the saddles have split, and between their two split halveswe have what we call a sticking region [see Figure 2.13(b)]. Depending on the valueof x1, this region contains either x1 > 0 or x1 < 0. If the cart’s state enters this regionthen the cart will stop. That is, the cart will come to rest near a crest of the rollercoaster. Similarly, the equilibria of the smooth roller coaster at the floor of its valleyshave now transformed from discrete points to regions surrounding these points [seeFigure 2.13(d)]. These regions are also sticking regions, and if the cart’s state entersthis region, then the cart will stop.

The size of the sticking region is easy to compute by use of (2.30), and a graphi-cal 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 eventuallyany point (x, 0) on the dx

dτaxis will become an equilibrium, and thus a state of rest

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xx

x x

dxdτ

dxdτ

dxdτ

dxdτ(a) (b)

(c) (d)

−0.2

−0.2

−0.2 −0.2

−0.2

−0.20.2

0.2

0.2 0.2

0.2

0.2

0.80.8 1.21.2 s

s

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, whereasfor (b) and (d), µd = 0.1. For the latter cases, the sticking regions s are shown for the case inwhich µs = 0.3.

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

µs ≥ πAL0

,

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


A vast amount of material has been covered in this chapter, starting with descrip-tions of various forces, discussions of the balance laws, and analyses of various ap-plications. The analyses we employed invariably featured the numerical integrationof an ordinary differential equation and an interpretation of its solutions. Develop-ing physical interpretations of the results provided by the model is one of the mostrewarding 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

x

dxdτ

−2 2

−1.5

1.5

Figure 2.14. The phase portrait of (2.27)–(2.29). Although it is not evident from the figure, thediscrete equilibria of the frictionless case shown in Figure 2.12 are now replaced with familiesof equilibria that correspond to possible resting (sticking) states for the cart. For this figure,AL0

= 0.25 and µd = 0.1.

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

x = x1L0

x = x1L0

∣∣ f ′ ∣∣

µs = 0.3

−2

−2 2

2

0.2

ss ss

sssss

Figure 2.15. A graphical method to compute the possi-ble sticking regions s of the cart on a smooth rollercoaster when µs = 0.3. The method is based on ex-amining (2.30) for the choice (2.31). That is,

∣∣ f ′ ∣∣ =∣∣∣Aπ

L0sin(

πx1L0

)∣∣∣. Examples of the sticking regions can be

seen in Figure 2.13.

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EXERCISES

2.1. Which of the following force fields are conservative/nonconservative?

P = x1E1 + x3E2,

P = x2E1 + x1E2,

P = x1x2E1,

P = −L0 sin(θ)E1 + L0 cos(θ)E2,

where L0 is a constant. For the conservative force fields, what are the associatedpotential 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, theposition r (t1) at some later time t1 is known. Argue that the conservation of energycan be used to determine the speed ‖v‖ of the particle. Give three distinct physicalexamples of applications of this result.

2.3. In contrast to Exercise 2.2, here consider a particle that is moving on asmooth fixed surface. The constraint force acting on the particle is prescribed byuse of Lagrange’s prescription, and the applied forces acting on the particle areconservative. Prove that E is again conserved. In addition, show that the speed ofthe particle can be determined at a known position r(t1) if the initial position andvelocity vectors are known. Finally, give three distinct physical examples of theapplication of this result.

2.4. A particle is free to move on a smooth horizontal surface x3 = 0. At the sametime, a rough plane propels the particle in the E1 direction. That is, the constraintson 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 avelocity vector v.

(a) Show that the areal velocity vector A is conserved if the resultant force Facting on the particle is a central force.∗

(b) Show that conservation of angular momentum HO is synonymous withconservation of the areal velocity vector.

(c) Suppose a particle is moving on a horizontal table under the action of aspring force, a normal force, and a vertical gravitational force −mgE3. Oneend 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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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 periodof time?

(d) Derive the equations of motion for the particle in (c). Using the nondimen-sionalizations

τ =√

Km

t, x = rL0

,

and conservation of angular momentum, show that the motion of theparticle can be found by integrating the following differential equations:

d2xdτ2

− β2

x3= − (x − 1) ,

dθ

dτ= β

x2, (2.32)

where

β = h

L20

√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, constructthe phase portraits of (2.32)1. For a selection of the orbits on each of thesephase portraits, construct images of the motion of the particle.∗

(e) Verify that the areal velocity vector is conserved for the motions of theparticle you found in (d).

2.6. A particle of mass m is in motion about a fixed planet of mass M. The externalforce acting on the body is assumed to be a conservative force P. The potentialenergy UP associated with this force is a function of ||r||, where r is the positionvector 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 thatthis conservation implies that the motion of the particle is planar. Thisplane, which is known as the orbital plane, also contains O. Show that theparticle sweeps out equal areas on its orbital plane in equal times.

(c) Write out the equations of motion of the particle using a spherical polarcoordinate system.

(d) Using the conservation of HO, show that the equations of (c) can besimplified to

m(r − rθ2) = −∂UP

∂r,

mr2θ = 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 ona smooth cone.

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(e) Show that the solutions to (2.33) conserve the total energy E of theparticle.

2.7. A particle of mass m is free to move on the inner surface of a rough sphereof constant radius R0. The center of the sphere is located at the origin O, and theparticle is attached to a fixed point A whose position vector is aEx + bEy by alinear 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 = R0eR, derive expres-sions for the acceleration vector a and angular momentum HO of theparticle.

(b) What is the velocity vector of the particle relative to a point on the surfaceof 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 equationsgoverning the motion of the particle are

mR0(φ − sin(φ) cos(φ)θ2) = mg sin(φ) − K (||x|| − L0)x · eφ

||x||

−µd ||N|| φ√φ2 + sin2(φ)θ2

,

1R0 sin(φ)

ddt

(mR2

0 sin2(φ)θ)

= −K (||x|| − L0)x · eθ

||x||

−µd ||N|| sin(φ)θ√φ2 + sin2(φ)θ2

,

where x = R0eR − aEx − bEy.

(e) Show that the normal force exerted by the surface on the particle is

N =(

mgE3 · eR + K (||x|| − L0)x · eR

||x||)

eR

− mR0(φ2 + sin2(φ)θ2)eR.

(f) For the case in which the particle is not moving relative to the surface,show that

Fc = mgE3 + K (||x|| − L0)x

||x|| .

What is the static friction criterion for this case?

(g) Show that the total energy of the particle decreases with time if the particlemoves relative to the surface.

(h) If the spring is removed and the surface is assumed to be smooth, provethat 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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simplify to

dθ

dτ= h

sin2(φ),

d2φ

dτ2= h2 cos(φ)

sin3(φ)+ sin(φ).

In these equations, h =(

1mR2

0

√R0g

)H0 · E3 and τ =

√g

R0t.

(i) Suppose that the sphere is smooth. Using the fact that the total energyE of the particle is conserved, show that the criterion for the particle toremain on the surface of the sphere is

mg (3 cos(φ) − 2 cos (φ0)) − mR0

(φ2

0 + sin2 (φ0) θ20

)> 0,

where φ0 is the value of the initial φ coordinate of the particle and θ0 andφ0 are the initial velocities. Suppose the particle is placed on top of thesphere. 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 followingconstraints:

(xE3 + E2) · v = 0, (E2) · v + e(t) = 0. (2.34)

(a) Show that one of the constraints is integrable whereas the other is non-integrable. 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) + µ2E2,

a gravitational force −mgE3 acts on the particle. With the help of thebalance of linear momentum F = ma, specify the equations governing themotion of the particle and the constraint forces.

(c) With the help of the work–energy theorem T = F · v, prove that the totalenergy of the particle is not conserved. Give a physical interpretation forthis lack of conservation.

(d) Using the results from (b), determine the motion of the particle and theconstraint force Fc.

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3 Lagrange’s Equations of Motion for aSingle Particle

3.1 Introduction

The balance of linear momentum F = ma for a particle can be traced to Newtonin the late 17th century. As we have seen, this vector-valued equation yields threedifferential equations from which the motion of a particle can be determined. In thecenturies that followed, alternative principles of mechanics were proposed. Someof them, such as the principle of least action, also yielded equations of motion thatwere equivalent to those obtained with F = ma. Others did not, and the equivalenceof, and interrelationships between, the principles of mechanics remains one of thecentral issues for any student of dynamics.

At the end of the 18th century, a formulation of the equations of motion for asingle particle appeared in a famous text by Lagrange [121].∗ Among their attrac-tive features, Lagrange’s equations of motion could easily accommodate integrableconstraints, and they (remarkably) have the same canonical form both for singleparticles 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 dis-cussed and several forms of these equations are established. For example [see (3.2)],

ddt

(∂L∂qi

)− ∂L

∂qi= Fncon · ai.

Many of the forms presented can be used with dynamic Coulomb friction and non-conservative forces. One of the most important features of our discussion is the em-phasis on the equivalence of Lagrange’s equations of motion to F = ma. Althoughthis equivalence is not sufficiently discussed in most textbooks, it can be found inmany 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 atransparent manner, and we follow many aspects of his exposition in this chapter.

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

70

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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 rigidbodies.

3.2 Lagrange’s Equations of Motion

There are several approaches to deriving Lagrange’s equations of motion that ap-pear in the literature. Among them, a variational principle known as Hamilton’sprinciple (or the principle of least action) is arguably the most popular, whereas anapproach based on D’Alembert’s principle was used by Lagrange [121]. Lagrange’soriginal developments were in the context of mechanical systems subject to holo-nomic constraints, and his equations were subsequently extended to systems withnonholonomic constraints by Edward J. Routh (1831–1907) (see Section 24 in Chap-ter IV of [183]) and Aurel Voss (1845–1931) [221].∗

Here, an approach is used that is rooted in differential geometry and is con-tained in some texts on this subject (see, for example, Synge and Schild [208]). Itprobably migrated from there to the classic text by Synge and Griffith [207] and hasbeen recently revived by Casey [27].

Two IdentitiesWe assume that a curvilinear coordinate system has been chosen for E

3. The velocityvector v consequently has the representation

v =3∑

i=1

qiai.

In addition, the kinetic energy has the representations

T = m2

v · v = m2

3∑i=1

3∑k=1

ai · akqiqk.

It is crucial to notice that T = T(q1, q2, q3, q1, q2, q3

).

We now consider in succession the partial derivatives of T with respect to thecoordinates and their velocities. We wish to establish the following results:

∂T∂qi

= mv · ai,∂T∂qi

= mv · ai.

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∂qi

= ∂

∂qi

(m2

v · v)

= mv ·(

∂v∂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–Vossequations of motion.

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72 Lagrange’s Equations of Motion for a Single Particle

To proceed with the goal of concluding that ∂T∂qi = mv · ai, we first note that v =∑3

k=1 qkak and that ∂qk

∂qi = 0. As a consequence,

∂T∂qi

= mv ·(

3∑k=1

qk ∂ak

∂qi

).

The remaining steps use the fact that ak = ∂r∂qk :

∂T∂qi

= mv ·(

3∑k=1

qk ∂ak

∂qi

)

= mv ·(

3∑k=1

qk ∂2r∂qi∂qk

=3∑

k=1

qk ∂2r∂qk∂qi

)

= mv ·(

3∑k=1

qk ∂

∂qk

(∂r∂qi

= ai

))

= mv · ai.

We achieve the last step by noting that f =∑3k=1

∂f∂qk qk for any function f =

f(q1, q2, q3

).

The next result, which is far easier to establish, involves the partial derivative ofT with respect to a velocity. The reason this result is easier to establish is becausethe basis vectors ai do not depend on qk. Getting on with the proof, we have

∂T∂qi

= mv ·(

3∑k=1

∂qk

∂qiak

)

= mv ·(

3∑k=1

δki ak

)

= mv · ai.

This completes the proofs of both identities.

A Covariant Form of Lagrange’s EquationsIt is crucial to note that Lagrange’s equations are equivalent to F = ma. The formof Lagrange’s equations of motion discussed here is derived from this balance lawby taking its covariant components, i.e., dotting it with ai.

To start, we consider

ddt

(∂T∂qi

)− ∂T

∂qi= d

dt(mv · ai) − mv · ai

= ma · ai + mv · ai − mv · ai

= ma · ai

= F · ai.

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3.3 Equations of Motion for an Unconstrained Particle 73

In conclusion, we have a covariant form of Lagrange’s equations of motion:

ddt

(∂T∂qi

)− ∂T

∂qi= F · ai. (3.1)

We can appreciate some of the beauty of this equation by using it to establish com-ponent forms of F = ma for various curvilinear coordinate systems.

The LagrangianAnother form of Lagrange’s equations arises when we decompose the force F intoits conservative and nonconservative parts:

F = −∇U + Fncon,

where the potential energy U = U(q1, q2, q3). As

∇U =3∑

k=1

∂U∂qk

ak,∂U∂qk

= 0,

we find that Lagrange’s equations can be rewritten in the form

ddt

(∂T∂qi

− ∂U∂qi

)−(

∂T∂qi

− ∂U∂qi

)= Fncon · ai.

Introducing the Lagrangian L = T − U, we find an alternative form of Lagrange’sequations:

ddt

(∂L∂qi

)− ∂L

∂qi= Fncon · ai. (3.2)

If there are no nonconservative forces acting on the particle, then the right-handside of these equations vanishes. In addition, to calculate the equations of motion aminimal 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 thecurvilinear 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 = m2

(R2 + R2 sin2(φ)θ2 + R2φ2).

Notice that T does not depend on θ.

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We obtain Lagrange’s equations of motion for the spherical polar coordinatesystem by first calculating the six partial derivatives of T:

∂T∂R

= mR sin2(φ)θ2 + mRφ2,∂T∂φ

= mR2 sin(φ) cos(φ)θ2,

∂T∂θ

= 0,∂T

∂R= mR,

∂T

∂φ= mR2φ,

∂T

∂θ= mR2 sin2(φ)θ.

Using these results, we find the covariant form of Lagrange’s equations:

ddt

(∂T

∂R= mR

)−(

∂T∂R

= mR sin2(φ)θ2 + mRφ2)

= F · eR,

ddt

(∂T

∂φ= mR2φ

)−(

∂T∂φ

= mR2 sin(φ) cos(φ)θ2)

= F · Reφ,

ddt

(∂T

∂θ= mR2 sin2(φ)θ

)−(

∂T∂θ

= 0)

= F · R sin(φ)eθ. (3.3)

Clearly, these equations were far easier to calculate than an alternative approachthat 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

= m2

(R2 + R2 sin2(φ)θ2 + R2φ2) − mgR cos(φ).

We can calculate Lagrange’s equations of motion using L,

ddt

(∂L∂qk

)− ∂L

∂qk= 0,

or by substituting for F in (3.3). It is left as an exercise to show that both approachesare equivalent.

3.4 Lagrange’s Equations in the Presence of Constraints

The previous discussion of Lagrange’s equations did not address situations in whichconstraints on the motion of the particle were present. It is to this matter that wenow turn our attention. With integrable constraints, whose constraint forces are pre-scribed by use of Lagrange’s prescription, the beauty and power of Lagrange’s equa-tions are manifested. In this case, it is possible to choose the curvilinear coordinatesqi such that the equations of motion decouple into two sets. The first set describesthe unconstrained motion of the particle, and the second set yields the constraintforces 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 themthroughout these sections as Approach I and Approach II. For the novice, we highlyrecommended the first approach. As in the previous section, our exposition followsthat of Casey [27].

PreliminariesWe 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 inte-grable constraint has the form

(r, t) = q3 − d(t) = 0,

and that the constraint forces are prescribed by use of Lagrange’s prescription:

Fc = λ1a3 + λ2

(3∑

i=1

f iai

).

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

F = λ1a3 + λ2

(3∑

i=1

f iai

)− ∇U + Fancon.

The total nonconservative force acting on the particle is Fncon = Fc + Fancon.The kinetic energy of the particle is

T = m2

v · v = m2

3∑i=1

3∑k=1

aik(q1, q2, q3) qiqk,

where aik = ai · ak. Imposing the integrable constraint on T, we find the constrainedkinetic energy:

T = T2 + T1 + T0, (3.4)

where

T2 = m2

2∑i=1

2∑k=1

aikqiqk, T1 = m2∑

i=1

ai3qid, T0 = m2

a33d2. (3.5)

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In these expressions,

aik = aik(q1, q2, t) = aik(q1, q2, q3 = d(t))

is the constrained metric tensor. Notice that we use a tilde (˜) to denote impositionof the integrable constraint(s) and that the subscripts on T2, T1 and T0 refer to thepowers of qi.

A direct calculation shows that∗

∂T∂q1

∣∣∣q3=d,q3=d

= ∂T∂q1

,∂T∂q2

∣∣∣q3=d,q3=d

= ∂T∂q2

,

∂T∂q1

∣∣∣q3=d,q3=d

= ∂T∂q1

,∂T∂q2

∣∣∣q3=d,q3=d

= ∂T∂q2

,

∂T∂q3

∣∣∣q3=d,q3=d

�= ∂T∂q3

= 0,∂T∂q3

∣∣∣q3=d,q3=d

�= ∂T∂q3

= 0.

(3.6)

In these relations, the partial derivative of T is evaluated prior to imposing the con-straint q3 = d(t). These relations imply that we can use T to obtain the first twoLagrange’s equations of motion, but not the third. Results that are identical in formto (3.6) pertain to the partial derivatives of L and L and U and U.

Notice that we did not impose the nonintegrable constraint on the kinetic en-ergy T and the Lagrangian L. It is possible to do this, but the result is not useful tous here.

Approach IIn the first approach, we evaluate the partial derivatives in Lagrange’s equations ofmotion (3.2) in the absence of any constraints:

ddt

(∂L∂qk

)− ∂L

∂qk= Fancon · ak + (Fc = 0) · ak.

Explicitly, these equations are

ddt

(m

3∑i=1

ai1qi

)−

3∑i=1

3∑r=1

m2

∂air

∂q1qiqr + ∂U

∂q1= Fancon · a1,

ddt

(m

3∑i=1

ai2qi

)−

3∑i=1

3∑r=1

m2

∂air

∂q2qiqr + ∂U


ddt

(m

3∑i=1

ai3qi

)−

3∑i=1

3∑r=1

m2

∂air

∂q3qiqr + ∂U

∂q3= Fancon · a3.

Notice that we have not introduced the constraint forces on the right-hand side ofthese equations. That is, in the preceding equations F = −∇U + Fancon.

We now impose the integrable constraint q3 = d(t) and introduce the nonin-tegrable 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,

q3 = d,

f1q1 + f2q2 + f3d + e = 0,

ddt

(3∑

i=1

mai1qi

)−

3∑i=1

3∑r=1

m2

∂air

∂q1qiqr + ∂U

∂q1= λ2 f1 + Fancon · a1,

ddt

(m

3∑i=1

ai2qi

)−

3∑i=1

3∑r=1

m2

∂air

∂q2qiqr + ∂U


ddt

(m

3∑i=1

ai3qi

)−

3∑i=1

3∑r=1

m2

∂air

∂q3qiqr + ∂U

∂q3= λ1 + λ2 f3 + Fancon · a3. (3.7)

We have refrained from ornamenting U, f i, ai, and aik with a tilde in the last four ofthese 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 yielddifferential 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 = λ1a3 actingon the particle.

Approach IIIn Approach II, we work directly with L = T − U. Here, as

L = L(q1, q2, q1, q2, t

),

the partial derivatives of L with respect to q3 and q3 are zero. Consequently, using(3.6) and (3.2), we find that there are only two Lagrange’s equations:

ddt

(∂L∂q1

)− ∂L

∂q1= Fc · a1 + Fancon · a1,

ddt

(∂L∂q2

)− ∂L

∂q2= Fc · a2 + Fancon · a2.

Introducing the expression for the constraint force Fc and the nonintegrable con-straint, we find the equations governing λ2, q1(t), and q2(t) are

f1q1 + f2q2 + f3d + e = 0,

ddt

(∂L∂q1

)− ∂L


ddt

(∂L∂q2

)− ∂L

∂q2= λ2 f2 + Fancon · a2. (3.8)

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Notice that λ1 does not feature in these equations. In addition, if no nonintegrableconstraint were present, then the differential equations provided by Approach IIare all that are needed to determine q1(t) and q2(t). Equations (3.8)2,3 are examplesof 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 aparticle 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 nonconserva-tive force DReθ, where D is a constant. Later on, we shall impose a nonintegrableconstraint on the motion of the particle.

For the problem at hand, it is convenient to use a spherical polar coordinatesystem:

q1 = θ, q2 = φ, q3 = R.

Using this coordinate system, we can write the integrable constraint R = d(t) in theform

(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

T = m2

(R2 + R2 sin2(φ)θ2 + R2φ2

), U = mgR cos(φ).

The constrained kinetic and potential energies are

T = m2

(d2 + d2 sin2(φ)θ2 + d2φ2

), U = mgd cos(φ). (3.9)

Finally, the covariant basis vectors are

a1 = R sin(φ)eθ, a2 = Reφ, a3 = eR.

Their constrained counterparts ai 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:

ddt

(∂L

∂θ

)− ∂L

∂θ= F · a1

= Fc · a1 + Ddeθ · a1

= Dd2 sin(φ),

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3.5 A Particle Moving on a Sphere 79

ddt

(∂L

∂φ

)− ∂L

∂φ= F · a2

= Fc · a2 + Ddeθ · a2

= 0.

Evaluating the partial derivatives of the constrained Lagrangian, we find that theseequations become

ddt

(md2 sin2(φ)θ) = Dd2 sin(φ),

ddt

(md2φ

)− md2 sin(φ) cos(φ)θ2 − mgd sin(φ) = 0. (3.10)

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)]:

ddt

(∂L

∂θ= mR2 sin2(φ)θ

)−(

∂L∂θ

= 0)

= Fancon · R sin(φ)eθ,

ddt

(∂L

∂φ= mR2φ

)−(

∂L∂φ

= mR2 sin(φ) cos(φ)θ2 + mgR sin(φ))

= Fancon · Reφ,

ddt

(∂L

∂R= mR

)−(

∂L∂R

= mR sin2(φ)θ2 + mRφ2 − mg cos(φ))

= Fancon · eR.

Next, we impose the integrable constraint and introduce the constraint force Fc tofind the equations of motion:

ddt

(md2 sin2(φ)θ) = Dd2 sin(φ),

ddt

(md2φ) − md2 sin(φ) cos(φ)θ2 − mgd sin(φ) = 0,

ddt

(md) − (md sin2(φ)θ2 + mdφ2 − mg cos(φ)) = λ. (3.11)

Notice that the first two of these equations are identical to (3.10), whereas the thirdequation is an equation for the constraint force Fc.

We could now introduce an additional constraint:

f1θ + f2φ + f3R + e = 0.

Using Lagrange’s prescription, we find that the total constraint force on the particleis

Fc = λeR + λ2

(f1

R sin(φ)eθ + f2

Reφ + f3eR

).

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To obtain the equations of motion for the case in which the nonintegrable constraintis active, we need to introduce only the constraint force associated with the nonin-tegrable constraint on the right-hand side of (3.11) and to append the nonintegrableconstraint to the resulting equations:

f1θ + f2φ + f3d + e = 0,

ddt

(md2 sin2(φ)θ) = Dd2 sin(φ) + λ2 f1,

ddt

(md2φ) − md2 sin(φ) cos(φ)θ2 − mgd sin(φ) = λ2 f2,

ddt

(md) − (md sin2(φ)θ2 + mdφ2 − mg cos(φ)) = λ + λ2 f3.

It is left as an exercise to show what additional simplifications to these equationsarise if the nonintegrable constraint were integrable with f1 = 0, f2 = 1, f3 = 0, ande = 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 signif-icance, some material from differential geometry needs to be presented. Our treat-ment is limited to the ingredients we shall shortly need and, as such, it cannot do jus-tice to this wonderful subject. Mercifully, there are several excellent texts that canbe 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 paperby Lutzen [132] for a historical overview of the interaction between geometry anddynamics 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, andwe call them manifolds. In the case of a curve, a single coordinate is needed tolocally parameterize the points P on this manifold, and so it is considered to bea one-dimensional manifold. For a surface, two coordinates are needed to locallyparameterize the points on the surface and so the surface is considered to be atwo-dimensional manifold. In an obvious generalization, subsets of E

3 such as solidspheres and solid ellipsoids are considered to be three-dimensional manifolds.

Previously, in Section 1.5, curvilinear coordinates were introduced. For a givensurface (or curve) we used these coordinates both to label points on the manifoldand to define the manifold. For example, for a sphere of radius R0, the sphericalpolar coordinates φ and θ label points on the sphere and the coordinate R canbe used to define the sphere: R = R0. Similarly, for a circle, the cylindrical polarcoordinate θ can be used to label points on the circle, and the coordinates r and z canbe used to define the circle. The curvilinear coordinate system we use to label pointson the manifold is known as a chart. For some manifolds, such as a plane, a straightline, 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 81

(a) (b)

P

PC

TPC

S

TPS

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

manifold. For surfaces such as spheres, for which a set of spherical polar coordinateswill not be defined at the poles, at least two charts are needed. With terminologyborrowed 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 thisspace by TPM. If the manifold is n-dimensional, then TPM is also n-dimensional.For example, the tangent space TPC is a line for the curve shown in Figure 3.1(a), andthe tangent space TPS is a plane for the sphere S shown in Figure 3.1(b). Continuingwith the sphere as an example, if we fix a point P on the sphere, this is equivalent tofixing 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 TPS at P, and any tangent vector to the sphereat this point can be expressed in terms of these vectors. Related remarks apply at apoint 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 = E

3, the dimension ofTPM is 3.

Returning to the example of a sphere, we choose two points P1 and P2 on thesphere of radius R and consider a path V between them (see Figure 3.2). We wishto measure the distance one would travel along V . To do this, we first parameterizethe curve with a parameter u, where u = uα at Pα. The curve can then be uniquelydescribed by the functions θ(u) and φ(u).∗ To determine the distance �s traveledalong the curve, one method would be to evaluate the following integral:

�s =∫ u2

u1

√R2 sin2 (φ)

dθ

dudθ

du+ R2

dφ

dudφ

dudu. (3.12)

Referring to (3.4), (3.5), and (3.9), we can express the integrand on the right-handside 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

P2

VVVVV

Svrel

Figure 3.2. A curve V connecting two points on asphere S. The velocity vector vrel of a particle movingon this curve would lie in the tangent plane TPS at eachpoint P of the curve.

sphere:

�s =∫ u2

u1

√2T2

m

(dtdu

)du

=∫ t2

t1

√2T2

mdt =

∫ t2

t1

||vrel|| dt.

For the second integral, we changed variables and parameterized the path by usingt rather than u. Turning to the example of a circle, the reader is invited to showthat a measure of distance corresponding to (3.12) can be established by using thesingle polar coordinate θ. For M = E

3, the measure of distance can be defined in astandard manner by use of Cartesian coordinates:

�s =∫ u2

u1

√√√√ 3∑k=1

dxk

dudxk

dudu.

It is easy to see that this expression can be rewritten as∫ t2

t1

√2Tm dt.

From the previous discussion, for an n-dimensional manifold M, a measure ofdistance similar to that provided by (3.12) can be established∗:

�s =∫ u2

u1

n∑i=1

n∑k=1

aik∂qi

∂u∂qk

∂udu.

Parameterizing the path by using time t instead of a variable u, we find

�s =∫ t2

t1

n∑i=1

n∑k=1

aikqiqkdt. (3.13)

∗ The index n here is either 1, 2, or 3. In later chapters, we shall see that n can range from 1 to 3N fora 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 (represen-tative) particle of mass m moving on a manifold M. The manifold is known asthe configuration manifold. The velocity of the particle relative to M is simplyvrel =∑n

k=1 qkak. Thus (3.13) can be expressed as

�s =∫ t2

t1

||vrel|| dt. (3.14)

We see from this expression that, in defining �s, we have also defined a measure ofthe magnitude of a vector vrel ∈ TPM. Such a measure is known as a metric, and amanifold that is equipped with a metric is known as a Riemannian manifold. The ter-minology here pays tribute to Georg F. B. Riemann (1826–1866) and his remarkablework [179].

Distance measure (3.13) is clearly intimately related to the kinetic energy of aparticle, and, following Synge [205],

ds =n∑

i=1

n∑k=1

aikqiqkdt (3.15)

is known as the kinematical line-element. This measure of distance has a long his-tory 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 distancesuch as meters and feet are possible and is intimately related to Einstein’s theory ofrelativity.

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, fora single particle the configuration manifold corresponds to the physical surface orcurve that the particle moves on. However, for a system of particles or rigid bodiessubject to constraints, this is not the case, and the construction of a single represen-tative particle is nontrivial. Indeed, for a system of particles, such a construction wasexplicitly recorded only rather recently by Casey [27].∗ Subsequently, he extendedthis 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 mo-tion obtained by use of Approach II are projections of F = ma onto the covariantbasis vectors for the unconstrained coordinates. That is, we are projecting F = maonto the basis for TPM.

For those who have not yet found this perspective, let us recall the example ofthe particle moving on the sphere of radius R = d(t). There, we obtained the twoLagrange’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 spaceTPS to a point P of the sphere S. Furthermore, because the constraint force associ-ated with the integrable constraint λeR is perpendicular to the sphere, this force didnot appear in the two Lagrange’s equations.

An important feature of nonintegrable constraints is that the constraint forceassociated with these constraints is not decoupled from the equations governingthe unconstrained motion. This deficiency in Lagrange’s equations of motion canbe removed by use of alternative forms of F = ma that are suited to nonintegrablyconstrained systems, but we do not approach this vast subject here. We now delvea 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 ConstraintFirst, let us consider the case in which the particle is subject to a single integrableconstraint:

(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 (seeFigure 3.3). The velocity of the particle relative to this surface has the representation

vrel = q1a1 + q2a2.

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 theparticle. Thus an unconstrained particle has three degrees-of-freedom, whereas theparticle 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 tangentplane TPM to M at P. We can also define a relative kinetic energy Trel = T2:

Trel = m2

vrel · vrel =2∑

i=1

2∑k=1

m2

ai · akqiqk.

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

s(t1) − s(t0) =∫ t1

t0

√vrel · vreldt

=∫ t1

t0

√2Trel

mdt.

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3.7 The Geometry of Lagrange’s Equations of Motion 85

O

q1 coordinate curveq2 coordinate curve

q3 = d(t) coordinate surface

a1

a2

a3

M

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) coordinatesurface.

We can differentiate this result with respect to t to find the kinematical line-elementds:

ds =√

2Trel

mdt =

√√√√ 2∑i=1

2∑k=1

ai · akdqidqk. (3.16)

As emphasized previously in Section 3.6, notice that the measure of distance is de-fined 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 constraintforces associated with the integrable constraint are prescribed by use of Lagrange’sprescription: Fc = λa3. Then Fc · a1 = Fc · a2 = 0, and the constraint force does notappear in the first two Lagrange’s equations. The forces

Q1 = F · a1, Q2 = F · a2

are known as the generalized forces, and the expressions we use for them are equiv-alent 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) andq2(t):

ddt

(∂L∂q1

)− ∂L


ddt

(∂L∂q2

)− ∂L

∂q2= Fancon · a2. (3.17)

Imposing a nonintegrable constraint on the motion of the particle will not changeM. Furthermore, this constraint will, in general, introduce constraint forces into

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Equations (3.17). These forces will destroy the decoupling that Lagrange’s equa-tions achieve for integrable constraints.

A Particle Subject to Two Integrable ConstraintsWe 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 areeasily represented. At each instant of time, the intersection of the two surfaces1 = 0 and 2 = 0 in E

3 defines a curve – in this case a q1 coordinate curve (seeFigure 3.4). The configuration manifold M in this case corresponds to the q1 coor-dinate curve. This coordinate is the generalized coordinate for the system.

We can easily represent the velocity vector of the particle relative to M, whichwe again denote by vrel:

vrel = q1a1.

It should be clear that a1 is tangent to M. Indeed, a1 evaluated at P is a basis vectorfor the one-dimensional space TPM. In addition, we can associate with vrel a relativekinetic energy:

Trel = m2

vrel · vrel = m2

a1 · a1q1q1.

Paralleling previous developments [see (3.15)], the kinematical line-element for Mis

ds =√

2Trel

mdt =

√a1 · a1dq1dq1.

O q2 = d2(t) coordinate surface

q3 = d3(t) coordinate surface

a1

r

m

Figure 3.4. A particle moving on a curve. In this figure, a1 is tangent to the q1 coordinate curvecorresponding to q2 = d2(t) and q3 = d3(t). That is, the q1 coordinate curve is the configura-tion manifold M. The vectors a2 and a3, which are not shown, are normal to M.

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3.8 A Particle Moving on a Helix 87

O

increasing s & θ

E1

E2

E3

et

et

en

en

eb

eb

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.

With regard to Lagrange’s equations of motion, if the constraint forces are pre-scribed by use of Lagrange’s prescription, Fc = λ1a3 + λ1a2, then we can easily findthe differential equation governing q1(t) by using Approach II:

ddt

(∂L∂q1

)− ∂L

∂q1= Fancon · a1.

Here, L = L(q1, q1, t

)and F · a1 is the generalized force for this problem.

3.8 A Particle Moving on a Helix

As an illustrative example, we turn our attention to establishing results for a particlethat 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 sev-eral reasons. First, it is a prototypical problem to illustrate how the Serret–Frenetformulae and the Frenet triad {et, en, eb} help to determine the motion of a parti-cle on a space curve. Second, we can use this example to illustrate a nonorthogonalcurvilinear coordinate system.∗

Curvilinear Coordinates, Basis Vectors, and Other KinematicsA helix is defined by the intersection of two surfaces: a cylinder r = R and a helicoidz = cθ, where c and R are constants. To conveniently define these surfaces, we definea curvilinear coordinate system:

q1 = θ = tan−1(

x2

x1

), q2 = r =

√x2

1 + x22,

q3 = η = z − αrθ = x3 − α

√x2

1 + x22 tan−1

(x2

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 = x1E1 + x2E2 + x3E3

= r cos(θ)E1 + r sin(θ)E2 + (η + αrθ) E3.

Our labeling of the coordinates minimizes subsequent manipulations. You shouldnote that the curvilinear coordinate system is not defined when r = 0. That is, it hasthe 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 = 1r

eθ, a2 = er, a3 = E3 − αθer − αeθ.

You should notice that ai · aj = δji , 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) , en = −er, eb = 1√1 + α2

(E3 − αeθ) .

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. Youshould 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 E

3, we have the general representation v =∑3i=1 qiai. From this result, we can immediately write

v = θ(reθ + αrE3) + r(er + αθE3) + ηE3.

Furthermore, the kinetic energy of the particle is

T = m2

(r2 + r2θ2 + (η + αrθ + αrθ)2) .

When the particle is in motion on the helix, it is subject to two constraints 1 = 0and 2 = 0:

1(r, t) = q2 − R,

2(r, t) = q3.

In preparation for writing expressions for the constraint forces acting on a particlemoving on the helix, you should calculate the gradient of these two functions. Youmight also notice that θ is the generalized coordinate for a particle moving on thehelix that we will be using.

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3.8 A Particle Moving on a Helix 89

ForcesWe assume that an applied force Fa acts on the particle. In addition, we assume thatthe friction is of the Coulomb type. Consequently, if the particle is moving relativeto the helix,

F = Fa + λ1a2 + λ2a3 + Ff ,

where

Ff = −µd∣∣∣∣λ1a2 + λ2a3

∣∣∣∣ θa1∣∣∣∣θa1∣∣∣∣ .

On the other hand, if the particle is not moving relative to the helix, i.e., θ is constant,then

F = Fa + λ1a2 + λ2a3 + λ3a1.

The friction force in this case is subject to the static friction criterion:∣∣∣∣Ff∣∣∣∣ ≤ µs ||N|| ,

where

Ff =(

(λ1a2 + λ2a3 + λ3a1) · a1

||a1||)

a1

||a1|| ,

N = λ1a2 + λ2a3 + λ3a1 − Ff .

Balance of Linear Momentum and Lagrange’s EquationsFor an unconstrained particle moving in E

3, we have the three Lagrange’s equations:

ddt

(∂T∂qi

)− ∂T

∂qi= F · ai.

For the present coordinate system θ, r, η, these equations read

ddt

(∂T

∂θ= mr2θ + mαr(η + αrθ + αrθ)

)

−(

∂T∂θ

= mαr(η + αrθ + αrθ))

= F · (a1 = reθ + αrE3),

ddt

(∂T∂r

= mr + mαθ(η + αrθ + αrθ))

−(

∂T∂r

= mrθ2 + mαθ(η + αrθ + αrθ))

= F · (a2 = er + αθE3),

ddt

(∂T∂η

= m(η + αrθ + αrθ))

−(

∂T∂η

= 0)

= F · (a3 = E3) .

Equations of Motion for the Particle on the HelixWe obtain the equations of motion for the particle on the helix from the precedingequations 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, wefind three equations:

ddt

(m(1 + α2)R2θ

) = Fa · a1 − µd∣∣∣∣λ1a2 + λ2a3

∣∣∣∣ ||a1|| θ

|θ| ,

ddt

(mα2Rθθ

)− m(1 + α2)Rθ2 = Fa · a2 + λ1 − µd∣∣∣∣λ1a2 + λ2a3

∣∣∣∣ θa1 · a2∣∣∣∣θa1∣∣∣∣ ,

ddt

(mαRθ

) = Fa · a3 + λ2 − µd∣∣∣∣λ1a2 + λ2a3

∣∣∣∣ θa1 · a3∣∣∣∣θa1∣∣∣∣ ,

where

a1 = Reθ + αRE3, a2 = er + αθE3, a3 = E3.

These three equations provide a differential equation for the unconstrained motionof 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 particleis not moving relative to the helix), we find, from F = ma, three equations for thethree unknowns:

λ3 = −Fa · a1, λ2 = −Fa · a2, λ1 = −Fa · a3.

It remains to use the static friction criterion, but this is left as an easy exercise.

The Particle on a Smooth HelixIn this case,

F = Fa + λ1a2 + λ2a3,

and because Fc · a1 = 0, Lagrange’s equations of motion decouple:

ddt

(m(1 + α2)R2θ

) = Fa · a1,

ddt

(mα2Rθθ

)− m(1 + α2)Rθ2 = Fa · a2 + λ1,

ddt

(mαRθ

) = Fa · a3 + λ2.

Consequently, the desired differential equation is

m(1 + α2)R2θ = Fa · (Reθ + αRE3),

and the constraint force is

Fc = λ1a2 + λ2a3

= (mα2Rθθ − mRθ2 − Fa · a2)

a2 + (mαRθ − Fa · a3)

a3.

Once θ as a function of time has been calculated from the ordinary differential equa-tion, 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 forceis gravitational, Fa = −mgE3. Then, from the preceding equations,

m(1 + α2)R2θ = −mgαR. (3.18)

Subject to the initial conditions θ(t0) = θ0 and θ(t0) = ω0, this equation has the solu-tion

θ(t) = θ0 + ω0(t − t0) − gα

2R(1 + α2)(t − t0)2.

Using this result, we find that the constraint force is

Fc =(

mgαθ(t)1 + α2

− mRθ2(t))

a2 + mg1 + α2

(E3 − α(θer + eθ)) ,

where θ(t) is as previously given.

Some ObservationsSuppose one is interested in determining only the differential equation governingthe 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 byimposing the constraints on the expression for T:

T = m2

(R2θ2 + (αRθ)2) = m

2

((1 + α2)R2θ2) .

Furthermore,

ddt

(∂T

∂θ

)− ∂T

∂θ= F · a1 = F · (Reθ + αRE3) = Fa · (Reθ + αRE3) .

A quick calculation shows that the resulting differential equation is identical to thatobtained previously [(3.18)].

Clearly, Lagrange’s equations calculated with Approach II (i.e., with T) havetheir advantages, but they cannot accommodate dynamic friction forces. It is, how-ever, the standard approach to Lagrange’s equations in the literature and textbooks.You should note that ∂T

∂r = ∂T∂η

= ∂T∂r = ∂T

∂η= 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 werepresented. The most fundamental of these forms is [see (3.1)]

ddt

(∂T∂qi

)− ∂T

∂qi= F · ai.

In one of the following exercises, we establish two other forms of these equations byexpanding 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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92 Exercise 3.1

decompose the forces acting on the particle into conservative and nonconservativeforces, then we can transform (3.1) to (3.2):

ddt

(∂L∂qi

)− ∂L

∂qi= Fncon · ai.

Now suppose that an integrable constraint is imposed on the particle, that this con-straint can be written as q3 − f (t) = 0, and that the constraint force associated withthis constraint is Fc = λa3. In this case, Lagrange’s equations of motion can be usedto readily provide a set of differential equations for the generalized coordinates q1

and q2:

ddt

(∂L∂qα

)− ∂L

∂qα= Fncon · aα, α = 1, 2. (3.19)

These equations feature the constrained Lagrangian L that we obtain from L byimposing the integrable constraint q3 = f (t), and, most important, do not feature λ.That is, equations of motion (3.19) are reactionless. This case, in which all the con-straints are integrable and the constraint forces are prescribed by use of Lagrange’sprescription, is an example of a mechanical system subject to “ideal constraints.”

We also discussed the situation in which nonintegrable constraints were im-posed on the system and outlined how the equations of motion could be obtained inthese circumstances. The imposition of nonintegrable constraints will not affect thenumber of generalized coordinates, the configuration manifold, or the kinematicalline-element.

The summary just presented will be identical for systems of particles, rigid bod-ies, and systems of both particles and rigid bodies. The only major differences arethat the calculation of the kinetic energy becomes significantly more complicatedfor these systems and that the right-hand sides of Lagrange’s equations feature sev-eral forces and moments. Despite these differences, the decoupling of the equationsof motion into a set of reactionless equations governing the generalized coordinateswill hold if Lagrange’s prescription for the constraint forces is used. This is one ofthe most remarkable features of Lagrange’s equations for systems subject to inte-grable constraints.

EXERCISES

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

a1 = ∂r∂u

= ver + uE3, a2 = ∂r∂v

= uer − vE3,

a3 = ∂r∂θ

= uveθ,

and

a1 = 1u2 + v2

a1, a2 = 1u2 + v2

a2, a3 = 1uv

eθ.

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(a) Consider a particle of mass m that is acted on by a force F and is free tomove in E

3. Show that the equations of motion of the particle are

ddt

(m(u2 + v2)u

)− m(u2 + v2)u − mv2uθ2 = F · a1,

ddt

(m(u2 + v2)v

)− m(u2 + v2)v − mu2vθ2 = F · a2,

ddt

(mu2v2θ

) = F · a3.

(b) Next, we are interested in a particle that is moving on the parabolic surfaceof revolution:

c2 = −z +√

z2 + r2,

where c is a constant. A vertical gravitational force −mgE3 acts on the par-ticle. Using the results of (a), derive the equations governing the uncon-strained motion of the particle and show that the normal force acting onthe particle is

N = − (mu2c + mu2cθ2 + mgc)

a2.

Show that the two second-order differential equations governing thegeneralized coordinates can be written as a single second-order differentialequation:

(m(u2 + c2)

)u + mu2u − h2

mu3c2= −mgu, (3.20)

where h is a constant. (This constant is none other than HO · E3, which isan integral of motion). Noting that the units of u and c are meters1/2, whatis a dimensionless form of equations of motion (3.20)?

(c) Show that the solutions of (3.20) conserve the energy

E = m2

(u2 + c2)u2 + h2

2mu2c2+ mg

2(u2 − c2).

How does one arrive at this expression for E?

3.2. For many mechanical systems a canonical form of Lagrange’s equations canbe established that is suited to numerical integrations. Here, we establish one suchform [see (3.23)].∗ This problem is adapted from the texts of McConnell [139] andSynge and Schild [208]. We take this opportunity to note that (3.23) can be found inan early paper by Ricci and Levi-Civita [178].

We start by recalling the covariant component forms of Lagrange’s equations ofmotion 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 establishedfor any mechanical system that features scleronomic integrable constraints and constraint forcesand moments that are prescribed by use of Lagrange’s prescription.

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94 Exercise 3.2

force F =∑3

i=1 Fiai =∑3

i=1 Fiai∗:

ddt

(∂T∂qk

)− ∂T

∂qk= F · ak,

where

T = T(qr, qs) = m2

3∑i=1

3∑k=1

aikqiqk, (3.21)

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:

m3∑

i=1

akiqi + m3∑

i=1

3∑s=1

[si, k]qiqs = G · ak = Fk, (3.22)

where a Christoffel symbol of the first kind is defined by

[si, k] = ∂as

∂qi· ak.

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

mqk + m3∑

i=1

3∑s=1

�ksiq

iqs = G · ak = Fk, (3.23)

where a Christoffel symbol of the second kind is defined by

�kij = ∂ai

∂qj· ak.

Notice that �kij = �k

ji. This form of Lagrange’s equations is used in numerical simu-lations of mechanical systems.†

(a) Show that the Christoffel symbols have the representations

[si, k] = 12

(∂aki

∂qs+ ∂ask

∂qi− ∂asi

∂qk

),

and

�kij =

3∑r=1

akr[i j, r].

∗ 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 = q1, . . . , x6 =q3, the contravariant form of Lagrange’s equations can be easily placed in the form x = f(x, t).

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(b) Starting from Lagrange’s equations,

ddt

(∂T∂qk

)− ∂T

∂qk= F · ak,

derive the following representation for the covariant component form∗:

m3∑

i=1

akiqi + m3∑

i=1

3∑s=1

[si, k]qiqs = G · ak = Fk.

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

m3∑

i=1

akiqi + m3∑

s=1

3∑i=1

[si, k]qiqs = G · ak = Fk,

derive the following representation for the contravariant component form†:

mqk + m3∑

s=1

3∑i=1

�ksiq

iqs = G · ak = Fk.

(d) For which coordinate system do the Christoffel symbols vanish?

3.3. Recall that for spherical polar coordinates, {R, φ, θ}, the covariant basis vectorsare

a1 = eR, a2 = Reφ, a3 = R sin(φ)eθ,

and the contravariant basis vectors are

a1 = eR, a2 = 1R

eφ, a3 = 1R sin(φ)

eθ.

Furthermore, the linear momentum and kinetic energy of a particle of mass m are

G = mRa1 + mφa2 + mθa3, T = m2

(R2 + R2φ2 + R2 sin2(φ)θ2

).

(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’sequations of motion. In your solution, avoid explicitly calculating the 27Christoffel symbols of the first kind.

(b) For a particle of mass m that is in motion in E3 under the influence of a re-

sultant force F, establish the three contravariant components of Lagrange’sequations of motion. In your solution, avoid explicitly calculating the 27Christoffel symbols of the second kind.

∗ Hint : Expand the partial derivatives of T using the representation (3.21). Then, take the appropriatetime derivative and reorganize the resulting equation by using the aforementioned symmetries. Youmay 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 relabelingof the indices, you should get the final desired result. Notice that the covariant component andcontravariant component forms of these equations can be viewed as linear combinations of eachother.

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3.4. Consider a particle that is in motion on a rough surface. A curvilinear coordi-nate system q1, q2, q3 is chosen such that the surface can be described by the equa-tion

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 = q1a1 + q2a2.

(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, twoequivalent prescriptions for the constraint force are

Fc = N + Ff =3∑

i=1

λiai,

where the tildes are dropped for convenience.

(i) Show that ⎡⎢⎣

λ1

λ2

λ3

⎤⎥⎦ =

⎡⎢⎣

a11 a12 0

a12 a22 0

a13 a23 1

⎤⎥⎦⎡⎢⎣

F 1f

F2f

N

⎤⎥⎦ ,

where N, and F1f and F2

f uniquely define the normal force N and frictionforce Ff , respectively, and aik = ai · ak with i, k = 1, 2, 3.

(ii) For which coordinate systems do F1f = λ1, F2

f = λ2, and N = λ3? Givean 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 thefriction force is static.

3.5. Consider a particle of mass m that is in motion on a helicoid. In terms of cylin-drical 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 = 1r

eθ, a2 = er, a3 = E3 − α

reθ.

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(b) Consider a particle moving on the smooth helicoid:

(i) What is the constraint on the motion of the particle, and what is a pre-scription for the constraint force Fc enforcing this constraint?

(ii) Show that the equations governing the unconstrained motion of theparticle are

ddt

(m(r2 + α2) θ) = −mgα,

ddt

(mr) − mrθ2 = 0. (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 equationfor r(t), a differential equation for θ(t), and an equation for the constraintforce enforcing the nonintegrable constraint. Indicate how you would solvethese equations to determine the motion of the particle and the constraintforces acting on it.

3.6. Consider a particle of mass m that is free to move on the smooth inner surfaceof a hemisphere of radius R0 (cf. Figure 3.6). The particle is under the influence of agravitational force −mgE3.

m

O

E1

E2

E3

rg

Figure 3.6. Schematic of a particle of mass m moving on the inside of a hemisphere of radiusR0.

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

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(b) Using Lagrange’s equations, establish the equations of motion for the par-ticle and an expression for the constraint force.

(c) Prove that the total energy E and the angular momentum HO · E3 of theparticle are conserved.

(d) Show that the normal force acting on the particle can be expressed as afunction of the position of the particle and its initial energy E0:

N =(

−2E0

R0+ 2mg sin (φ) + mg cos (φ)

)eR.

(e) Numerically integrate the equations of motion of the particle and show thatthere are instances for which it will always remain on the surface of thehemisphere.

3.7. As shown in Figure 3.7, consider a bead of mass m that is free to move ona smooth semicircular wire of radius R0. The wire has a constant angular velocity�0E3 and whirls about the configuration shown in the figure. The particle is also

m

O

E1

E2

E3

r

g

�(t)

�(t)

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

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under the influence of a gravitational force −mgE3. This is a classical problem thatis discussed in several textbooks (see, for example, [78]).

(a) Using a spherical polar coordinate system, what are the two constraints onthe 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 theparticle:

φ =(

�20 cos (φ) + g

R0

)sin (φ) . (3.25)

After nondimensionalizing (3.25), numerically integrate the resulting dif-ferential equation and construct its phase portrait for values of 0.5, 1.0, 1.5of the parameter g

R0�20.

(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:

φ0 = 0, φ0 = π, φ0 = cos−1

(− g

R0�20

).

Give physical interpretations for these equilibria and show that the thirdone is possible if, and only if, �2

0 is sufficiently large. How do these resultscorrelate to your phase portraits?

(d) Starting from the work–energy theorem T = F · v, prove that the totalenergy of the particle is not conserved:

E = NθR0�0 sin (φ) ,

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 andthe kinetic energy T of the particle.

(a) Suppose a particle is constrained to move on a rough parabolic surface de-scribed by the equation

x − y2 = −4.

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

(b) As illustrated in Figure 3.8, suppose a particle is constrained to move on thesmooth parabola

x − y2 = −4, z = 0.

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

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100 Exercise 3.9

my

x6

2

−4

−4

q2 = −4

Figure 3.8. Schematic of a particle movingon a parabola in the x − y plane.

3.9. Consider a particle of mass m moving on a smooth ellipsoid:

x2

a2+ y2

b2+ z2

c2= 1.

(a) With the help of a suitable curvilinear coordinate system, establish expres-sions for the constrained kinetic energy of the particle.∗

(b) Assuming that a gravitational force −mgE3 acts on the particle, establishthe two second-order differential equations governing the motion of theparticle.

(c) Numerically integrate the equations you found in (b) and discuss featuresof 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 OFPARTICLES

101

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102

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4 The Equations of Motion for a Systemof Particles

4.1 Introduction

In this chapter, we establish Lagrange’s equations for a system of particles by start-ing with the balances of linear momentum for each of the particles. Our derivationis based on the results presented in Chapter 15 of Synge and Griffith [207].∗ Wesupplement their work with a discussion of constraints and potential energies. Toexamine the geometry inherent in Lagrange’s equations of motion for the system ofparticles, we use the construction of a representative single particle by Casey [27].All the work presented in this chapter emphasizes the equivalence of Lagrange’sequations 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’sprinciple, Gauss’ principle of least constraint, and Hamilton’s principle are also pre-sented in Section 4.11. The chapter closes with a discussion of a canonical form ofLagrange’s equations of motion in which time-independent integrable constraintsare present.

For many specific problems, we can obtain Lagrange’s equations by merelycalculating the kinetic and potential energies of the system. This approach isused in most dynamics textbooks, and neither the construction of a single particlenor the components of force vectors are mentioned.† Indeed, once we establishLagrange’s equations we can also ignore the explicit construction of the singleparticle. However, for many cases – which are not possible to treat using theapproach adopted in most dynamics textbooks – we find that the use of Synge’s andGriffith’s representation of Lagrange’s equations of motion allows us to tremen-dously increase the range of application of Lagrange’s equations. For instance, aswill be shown in some of the examples in Chapter 5, dynamic Coulomb friction isaccommodated.

∗ Related derivations for systems of particles can be found in several texts. For example, Section 6-6of Greenwood [79] and Section 21 of Whittaker [228].

† These texts use either the principle of virtual velocities or Hamilton’s principle (also known as theprinciple of least action) to derive Lagrange’s equations.

103

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104 The Equations of Motion for a System of Particles

mi

O

ri

Fi

E1

E2

E3

Figure 4.1. A single particle of mass mi in E3. The position

vector of this particle is ri, and the resultant external forceacting on the particle is Fi .

4.2 A System of N Particles

Here, we are interested in establishing the equations of motion for a system of Nparticles. The first step in this development is to discuss the individual elements inthe system of particles.

We consider a system of N particles, each of which is in motion in three-dimensional Euclidean space E

3. For the particle of mass mi (see Figure 4.1), theposition vector is

ri =3∑

j=1

xjiEj.

We also recall that the kinetic energy of the particle is

Ti = 12

mivi · vi.

It is also convenient to recall that the linear momentum of the particle of mass mi isGi = mi ri = mivi.

The resultant force acting on the particle of mass mi has the representation

Fi =3∑

j=1

F ji Ej.

The balance of linear momentum for the particle of mass mi is

Fi = mivi.

As a consequence of the balance of linear momentum for the particle, we have theangular momentum theorem:

HOi = ri × Fi,

where HOi = ri × mivi is the angular momentum of the particle relative to the fixedpoint O. A second consequence of the linear momentum balance is the work–energytheorem

Ti = Fi · vi

for the particle of mass mi.

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4.3 Coordinates 105

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

T = T1 + · · · + TN.

With the help of the work–energy theorem for each particle, we can determine thecorresponding 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 ofparticles.

The Center of MassFor the system of particles, we can define a center of mass C. This point, which liesin E

3, has the position vector r, where

r = 1m1 + · · · + mN

(m1r1 + · · · + mNrN) .

It is easy to show from this result that the linear momentum of a system of particleshas the representation G = (m1 + · · · + mN) ˙r. Next we examine a particle of massm moving in E

3N, 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. Indeedfor many systems of two particles, we use one coordinate system to describe r1 andanother to describe the relative position vector r2 − r1. For instance, for the systemshown in Figure 4.2, we might use Cartesian coordinates for r1 and spherical polarcoordinates for r2 − r1:

r1 = xE1 + yE2 + zE3, r2 = xE1 + yE2 + zE3 + R2eR2 . (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

m2O

g

E1

E2

E3

Figure 4.2. A particle of mass m1 attachedby a rigid rod of length L0 to a particle ofmass m2.

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with related definitions for eφ2 and eθ2 . Notice that eR2 points from m1 to m2. Thecoordinate system most suited to a given system of particles will, as in the case of asingle particle, depend on the constraints and applied forces.

In general, for the system of N particles we will use a coordinate system de-noted by q1, . . . , q3N. As in the case of a single particle, we assume that we can de-termine the unique Cartesian coordinates for a system of particles once q1, . . . , q3N

are known, and vice versa:

qK = qK(x11, x2

1, x31, . . . , x1

N, x2N, x3

N

)(K = 1, . . . , 3N),

xji = xj

i(q1, . . . , q3N) (i = 1, . . . , N and j = 1, 2, 3).

To calculate Lagrange’s equations of motion, we need to calculate the following 3N2

vectors:

∂ri

∂qK(K = 1, . . . , 3N and i = 1, . . . , N).

These vectors play the role of the basis vectors ak that we used with the single par-ticle earlier.

As an example, let us return to (4.2). In these equations, we gave examples ofthe coordinates for a two-particle system:

q1 = x = x11, q2 = y = x2

1, q3 = z = x31, q4 = R2, q5 = θ2, q6 = φ2.

For this coordinate system, and with the help of (4.2),

∂r1

∂qj= Ej,

∂r1

∂q(j+3)= 0,

and

∂r2

∂qj= Ej,

∂r2

∂q4= eR2 ,

∂r2

∂q5= R2 sin(φ2)eθ2 ,

∂r2

∂q6= R2eφ2 ,

where j = 1, 2, 3 in these two sets of equations. It is left to the student to realize howcoordinate system (4.2) can also be used to describe the kinematics of the particlesystem shown in Fig. 4.3.

m1

m2

Olinear spring

E1

E2

E3

Figure 4.3. A particle of mass m1 attached by a linear spring of stiffness K and unstretchedlength L0 to a particle of mass m2.

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4.4 Constraints and Constraint Forces 107

4.4 Constraints and Constraint Forces

For a single particle, one of the key notions we encountered earlier was kinematicconstraints. We now turn to examining this topic for a system of particles. As weshall see, the extension to this case is not as difficult as it first may seem.

Let us start with a physical system and use it to establish expressions for con-straint forces. We will later show that these constraint forces are compatible withLagrange’s prescription. Suppose we have two particles connected by a rigid rod oflength L0 (see Figure 4.2). Then the constraint imposed by the rod on their motionis ||r1 − r2|| − L0 = 0. To enforce this constraint, the particles will be subject to equaland opposite constraint forces:

Fc1 = −µt,

Fc2 = µt, (4.3)

where µ is the tension force in the rod and the unit vector t points from m1 to m2:

t = r2 − r1

||r2 − r1|| .

If we describe r2 − r1 = L0eR2 , which we did earlier, then t = eR2 . Writing the con-straint as

= 0,

where

= ‖r2 − r1‖ − L0,

then, as

∂

∂r1= − r2 − r1

‖r1 − r2‖ ,∂

∂r2= r2 − r1

‖r1 − r2‖ ,

we observe that (4.3) can be expressed as

Fc1 = µ∂

∂r1, Fc2 = µ

∂

∂r2. (4.4)

This result will help motivate Lagrange’s prescription for the constraint forces actingon a system of particles.

As a second example, consider the three-particle system shown in Figure 4.4.This model is a prototypical model for a braking vehicle and is used to explain thevehicle instability that often occurs when the front wheels lock during braking.∗ Weassume that the motion of this cart is planar. As a result the motion of the centerof mass C of the system is planar r = xE1 + yE2. Further, we need to supplement xand y only by an angle φ to determine the position of any point on the cart. Thus,

∗ For further details on this matter, and references to the vast literature on this prototypical noninte-grably constrained system, see O’Reilly and Tongue [165] and Ruina [186].

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m1

m2

m3

OQ

Cφ

Vehicle

E1E1

E2

e1

e1

e2

Figure 4.4. A model for a braking three-wheeled vehicle. The distributed mass of the vehicleis modeled as three mass particles of masses m1, m2, and m3. The particles of mass m1 andm2 model wheels that are rolling without slipping, and the particle of m3 models a slippingwheel.

for this system, we make the following choices for the nine coordinates q1, . . . , q9:

q1 = x, q2 = y, q3 = φ, q3+k = r2 · Ek, q6−k = r3 · Ek.

Here, k = 1, 2, 3 and r = xE1 + yE2. The unit vector e1 = cos(φ)E1 + sin(φ)E2 inFigure 4.4 is perpendicular to the line connecting m1 and m2 and e2 is tangent to thisline. The constraint that the velocity of the particle of mass m1 is always normal toe2 can be written in several equivalent forms. For example,

e2 · v1 = 0.

This constraint is nonintegrable. By inspection, the force that enforces this con-straint is parallel to e2:

Fc1 = µe2. (4.5)

This force ensures that m1 moves only in the e1 direction.

Lagrange’s PrescriptionWe now have sufficient motivation to motivate Lagrange’s prescription for the con-straint forces acting on a system of particles. We suppose that the system of particlesis subject to an integrable constraint:

= 0, (4.6)

where

= (r1, r2, . . . , rN, t).

Constraint (4.6) can be differentiated to yield a constraint of the form

N∑i=1

fi · vi + e = 0,

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4.4 Constraints and Constraint Forces 109

where

fi = ∂

∂ri,

e = ∂

∂t.

When evaluating ∂∂r1

, for example, we assume that r2, . . . , rN are fixed. Lagrange’sprescription for the constraint force Fci acting on the particle of mass mi is

Fci = µfi = µ∂

∂ri.

Result (4.4) discussed earlier is an example of this prescription. As with a single par-ticle in the presence of dynamic friction, Lagrange’s prescription is not universallyapplicable. Indeed, we shall encounter some examples later on when this prescrip-tion cannot be used because of the presence of friction.∗

Clearly Lagrange’s prescription can also be applied to nonintegrable con-straints. Such constraints have the functional form

f1 · v1 + · · · + fN · vN + e = 0, (4.7)

where fi = fi (r1, . . . , rN, t) and e = e (r1, . . . , rN, t). Lagrange’s prescription for sucha constraint is

Fci = µfi (i = 1, . . . , N).

This prescription is equivalent to one we discussed earlier for the example of anonintegrably constrained system of particles [see (4.5)]. It should be obvioushow Lagrange’s prescription can be extended to systems of (integrable andnonintegrable) constraints on the system of particles.

The Power of the Constraint ForcesThe constraints we consider can all be written in the form (4.7). Further, if the con-straint forces are consistent with Lagrange’s prescription, then the power expendedby these forces is easily calculated:

P =N∑

i=1

Fci · vi = µ

N∑i=1

fi · vi = −µe.

Thus, when e = 0, the constraint forces do no work. For an integrable constraint thisarises when ψ is not an explict function of time: = (r1, . . . , rN). In a previousexample, shown in Figure 4.2, we note that depended on r1 and r2. If, in addition,L0 were a function of time, then = (r1, r2, t).

∗ See, for example, Subsection 8.6.2 in Chapter 8.

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Newton’s Third LawFor the constraint ||r1 − r2|| − L0 = 0 mentioned earlier, Lagrange’s prescriptionyields

Fc1 = µr1 − r2

||r1 − r2|| ,

Fc2 = µr2 − r1

||r1 − r2|| .

Notice that these constraint forces point along the rod – which is what we wouldexpect from physical grounds. Furthermore, Fc2 = −Fc1 – which is none other thanNewton’s third law.∗

4.5 Conservative Forces and Potential Energies

Conservative forces in the dynamics of systems of particles commonly occur. Forinstance, they arise when two particles are connected by a spring (see Figure 4.3) orwhen each particle is attracted to a central body by a gravitational force field. In thissection, we discuss how to prescribe the conservative forces Fcon1, . . . , FconN actingon their respective particle of a system of particles. The system is assumed to have apotential energy

U = U (r1, r2, . . . , rN) . (4.8)

Notice that we are presuming that this is the most general form of the potentialenergy of conservative forces in a system of particles.

General form (4.8) encompasses the inverse gravitational law between two par-ticles of mass m1 and m2,

Un = − Gm1m2

||r2 − r1|| ,

and the potential energy of a spring force between two particles of mass m1 and m2:

Us = K2

(||r2 − r1|| − L0)2.

Here, G is the universal gravitational constant and K is the spring constant.To calculate the conservative forces, we equate the time derivative of U to the

negative power of the conservative forces. After some rearraging, we find that

N∑i=1

(Fconi + ∂U

∂ri

)· vi = 0. (4.9)

∗ For more details on this interesting result, see Noll [153], O’Reilly and Srinivasa [163], and Subsec-tion 8.6.4 in this book.

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4.6 Lagrange’s Equations of Motion 111

Assuming that, for each i, Fconi + ∂U∂ri

is independent of v1, . . . , vN and that (4.9) istrue for all possible v1, . . . , vN, then we conclude that

Fconi = −∂U∂ri

(i = 1, . . . , N). (4.10)

This is the prescription for the conservative forces.For the spring force and gravitational forces associated with Un and Us that we

defined earlier, prescription (4.10) yields

Fcon1 = − Gm1m2

||r2 − r1||2r1 − r2

||r2 − r1|| , Fcon2 = − Gm1m2

||r2 − r1||2r2 − r1

||r2 − r1|| ,

and

Fcon1 = −K2

(||r2 − r1|| − L0)r1 − r2

||r2 − r1|| ,

Fcon2 = −K2

(||r2 − r1|| − L0)r2 − r1

||r2 − r1|| ,

respectively. You might again notice that the pairs of forces obey Newton’s thirdlaw identically. This interesting result was first pointed out by Lanczos [124] andNoll [153].

4.6 Lagrange’s Equations of Motion

For a system of unconstrained particles, the equations of motion consist of Nbalances of linear momenta:

mi ri = Fi (i = 1, . . . , N).

We now show that these equations are equivalent to Lagrange’s equations ofmotion:

ddt

(∂T∂qK

)− ∂T

∂qK= �K (K = 1, . . . , 3N) , (4.11)

where

�K =N∑

i=1

Fi · ∂ri

∂qK.

As there are no constraints on the system, all of the coordinates q1, . . . , q3N aregeneralized coordinates and �K are the generalized forces. Our results here areadapted from the work of Synge and Griffith. Their derivation is presented inSection 15.1 of [207]. Shortly, an alternative derivation of (4.11) that is due to Casey[27] will be presented.∗

∗ See Equation (4.20) in Section 4.7.

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Lagrange’s equations presume that each individual particle’s position vector isexpressed as a function of the coordinates q1, . . . , q3N:

ri = ri(q1, . . . , q3N) (i = 1, . . . , N).

With these coordinates, we then compute vi and construct T.

Derivation of Lagrange’s Equations of MotionBefore proceeding with the derivation of Lagrange’s equations of motion, we estab-lish an important result:

∂vi

∂qK= ∂ri

∂qK. (4.12)

This identity is often (fondly) referred to as “canceling the dots.” To verify the iden-tity, we use the chain rule to calculate vi:

vi = ri =3N∑

J=1

qJ ∂ri

∂qJ.

Taking the partial derivative of both sides of this expression with respect to qK thenyields (4.12).

We are now in a position to show that

∂T∂qK

=N∑

i=1

mivi ·(

∂ri

∂qK

),

∂T∂qK

=N∑

i=1

mivi · ddt

(∂ri

∂qK

). (4.13)

First,

∂T∂qK

= ∂

∂qK

(N∑

i=1

mi

2vi · vi

)

=N∑

i=1

mivi · ∂vi

∂qK

=N∑

i=1

mivi · ∂ri

∂qK.

Notice that we used (4.12) in the final stages of this calculation. The other resultfollows similarily and so we skip some of the intermediate stages:

∂T∂qK

=N∑

i=1

mivi · ∂vi

∂qK

=N∑

i=1

mivi ·(

∂

∂qK

(3N∑

J=1

qJ ∂ri

∂qJ

))

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4.7 Construction and Use of a Single Representative Particle 113

=N∑

i=1

mivi ·(

3N∑J=1

qJ ∂

∂qJ

(∂ri

∂qK

))

=N∑

i=1

mivi · ddt

(∂ri

∂qK

).

In the last stages of this calculation, we used the identities

∂2ri

∂qK∂qJ= ∂2ri

∂qJ ∂qK, f =

3N∑J=1

qJ ∂f∂qJ

,

where f = f(q1, . . . , q3N

).

To derive Lagrange’s equations of motion, we now follow a familiar series ofsteps with the help of (4.13):

ddt

(∂T∂qK

)− ∂T

∂qK= d

dt

(N∑

i=1

mivi · ∂ri

∂qK

)−

N∑i=1

mivi · ddt

(∂ri

∂qK

)

=N∑

i=1

(mivi · ∂ri

∂qK

)

=N∑

i=1

Fi · ∂ri

∂qK

= �K, (4.14)

where the force �K is

�K =N∑

i=1

Fi · ∂ri

∂qK.

We have now shown how Lagrange’s equations of motion for a system of particlescan be established by using the balances of linear momenta for each of the particles.

4.7 Construction and Use of a Single Representative Particle

An alternative derivation of Lagrange’s equations was presented recently byCasey [27]. In this work, he constructs a single representative particle moving ina 3N-dimensional space subject to a force. With the help of this construction, thederivation of Lagrange’s equations of motion follows easily, and tremendous insightcan be gained into the geometry of Lagrange’s equations of motion.

In this section, we follow Casey [27], and construct a single particle of mass mthat is moving in E

3N. The position vector of this particle is r. The kinetic energyT of this particle is the same as the kinetic energy of the system of particles that itrepresents, and the force on the particle is such that

T = � · r, � = mr.

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In Casey [27], m is chosen (without loss in generality) to be the sum of the masses:m = m1 + · · · + mN. Here, we assume only that m > 0. As we noted earlier in Sec-tion 3.6, the interested reader is also referred to Chapter 1 of Lanczos [124] for arelated discussion of the geometry of a mechanical system.

The Configuration SpaceThe space E

3N, which is known as the configuration space, is equipped with a Carte-sian coordinate system. Consequently, for any vector b,

b =3N∑

K=1

bKeK =N∑

i=1

3∑j=1

b3i+j−3e3i+j−3.

Here, {e1, . . . , e3N} is a fixed orthonormal basis for E3N. We also define two other

sets of basis vectors:

e3i+j−3 =√

mi

me3i+j−3,

e3i+j−3 =√

mmi

e3i+j−3.

Here i = 1, . . . , N and j = 1, 2, 3. Notice that

eK · eJ = δJK.

The particle has a position vector r and a force vector �. Both of these vectors haverepresentations similar to that for b.

Prescription for the Position Vector rAs mentioned earlier, the position vector r is defined by the criterion that the kineticenergy of the particle of mass m is identical to the kinetic energy of the system ofparticles:

T = 12

mv · v = 12

N∑i=1

mivi · vi.

Substituting,

v = r =N∑

i=1

3∑j=1

r3i+j−3e3i+j−3, vi = ri =3∑

j=1

xjiEj,

and equating kinetic energies, we find one solution for r:

r =N∑

i=1

3∑j=1

(ri · Ej)

√mi

me3i+j−3 =

N∑i=1

3∑j=1

(ri · Ej) e3i+j−3. (4.15)

It is a good exercise to show that the other solutions for r can be shown to be equiv-alent to (4.15) modulo a translation of the origin and a relabeling of axes of E

3N.Notice how the mass ratios mi

m are subsumed into the basis vectors e3i+j−3 in (4.15).

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Prescription for the Force Vector �

The force vector � is prescribed by the requirement that

mv = �.

Substituting (4.15) and using the balance of linear momentum for each particle, oneis led to the prescription

� =N∑

i=1

3∑j=1

(Fi · Ej)√

mmi

e3i+j−3 =N∑

i=1

3∑j=1

(Fi · Ej) e3i+j−3. (4.16)

Again its interesting to note how the mass ratios mim are subsumed into the basis

vectors e3i+j−3 in (4.15).Next, it is often convenient to note that

� · e3i+j−3 = Fi · Ej,

which follows from (4.16). It is also an easy exercise to show that the work–energytheorems for the individual particles, Ti = Fi · vi (no sum on i), give a work–energytheorem for the particle of mass m:

T = � · v.

This theorem can be used to establish energy conservation results for the system ofparticles.

Curvilinear CoordinatesWe then use the curvilinear coordinates q1, . . . , q3N to define basis vectors for E

3N:

aK = ∂r∂qK

= ∂

∂qK

⎛⎝ N∑

i=1

3∑j=1

(ri · Ej = xj

i

)e3i+j−3

⎞⎠

and

aJ = grad(qJ ) =N∑

i=1

3∑j=1

(∂qJ

∂xji

)e3i+j−3.

It can be shown that aK · aJ = δJK. For most problems, it is not necessary to explicitly

calculate the contravariant basis vectors aJ .

EquivalencesFor future reference, it is important to note that

� · aK =N∑

i=1

Fi · ∂ri

∂qK. (4.17)

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To establish this result a series of identities and the definition xps = rs · Ep are

invoked:

� · aK = � · ∂r∂qK

=⎛⎝ N∑

i=1

3∑j=1

(Fi · Ej) e3i+j−3

⎞⎠ · ∂

∂qK

⎛⎝ N∑

s=1

3∑p=1

(rs · Ep) e3s+p−3

⎞⎠

=N∑

i=1

3∑j=1

N∑s=1

3∑p=1

(Fi · Ej)(

∂xps

∂qK

)e3i+j−3 · e3s+p−3

=N∑

i=1

3∑j=1

N∑s=1

3∑p=1

(Fi · Ej)(

∂xps

∂qK

)δi

sδjp

=N∑

i=1

3∑j=1

(Fi · Ej)

(∂xj

i

∂qK

)

=N∑

i=1

Fi · ∂

∂qK

⎛⎝ 3∑

j=1

xjiEj

⎞⎠

=N∑

i=1

Fi · ∂ri

∂qK.

We conclude that the desired identity has been established.Identity (4.17) is similar to several other results pertaining to the system of par-

ticles and the single particle of mass m. For instance, consider a function

� = � (r, t) = � (r1, . . . , rN, t) .

With some minor manipulations, we find several representations for the derivativeof � with respect to a curvilinear coordinate:

∂�

∂qK= ∂�

∂r· ∂r∂qK

=N∑

i=1

∂�

∂ri· ∂ri

∂qK. (4.18)

These results can be used to establish equivalences between conservative forces andconstraint forces acting on a system of particles and those acting on the single parti-cle of mass m.

Constraints and Constraint ForcesFor the single particle, we can express the constraint = 0, where [as in (4.6)]

(r1, r2, . . . , rN, t) = 0,

as a constraint on the motion of the single particle of mass m:

= (r, t) = 0.

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This implies that the motion of the single particle is subject to the constraint and, byuse of Lagrange’s prescription, a constraint force �c, respectively,

f · v + e = 0, �c = µf,

where

f = ∂

∂r, e = ∂

∂t.

Furthermore, the integrable constraint implies that the particle moves on a configu-ration manifold M that is a (3N − 1)-dimensional subset of the configuration spaceE

3N.It is important to note that the prescriptions of �c and Fci are consistent with

each other. Indeed, the equivalence of the constraint forces Fci acting on the systemof particles and the constraint force �c acting on the single particle of mass m canbe inferred from (4.17) and (4.18).

Conservative Forces and Potential EnergiesFor the single particle, we can express the potential energy U as a function of themotion of the single particle of mass m:

U(r1, r2, . . . , rN) = U(r).

Calculating ˙U and equating it to −�con · v, we find that∗

�con = −∂U∂r

. (4.19)

Again, it is important to note that the prescriptions of �con and Fconi [see (4.10)]are consistent with each other. As with constraint forces, the equivalence can beinferred from (4.17) and (4.18).

Lagrange’s Equations: An Unconstrained System of ParticlesWe are now in a position to establish Lagrange’s equations of motion for the singleparticle of mass m. Because⎧⎪⎨

⎪⎩m1v1 = F1

. . .

mNvN = FN

⎫⎪⎬⎪⎭ is equivalent to mv = �,

these equations will be none other than Lagrange’s equations for the system of Nparticles.

First, for the single particle of mass m, it is easy to see that

v =3N∑i=1

qKaK.

∗ Essentially, we are solving the equation(�con + ∂U

∂r

)· v = 0 for all possible motions of the system.

For this to hold, the terms in the parentheses, assuming they are independent of v, must vanish, andwe arrive at (4.19).

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Consequently, we have the two intermediate results

∂T∂qK

= mv · aK,∂T∂qK

= mv · aK.

With the assistance of these results we find that

ddt

(∂T∂qK

)− ∂T

∂qK= d

dt(mv · aK) − mv · aK

= ddt

(mv) · aK

= � · aK.

We have just established Lagrange’s equations for a system of particles:

ddt

(∂T∂qK

)− ∂T

∂qK= � · aK. (4.20)

These equations give the motion r = r(t) of the particle of mass m and hence themotion of the system of particles.

We notice that, because of (4.17), Equations (4.20) are identical to those weestablished earlier. In particular,

�K =N∑

i=1

Fi · ∂ri

∂qK= � · aK.

For many problems it is more convenient to use F1, . . . , FN instead of �. However,(4.20) provides wonderful insight into the geometry of Lagrange’s equations for asystem of particles.

4.8 The Lagrangian

In many mechanical systems, the sole forces that act on the system are conservative.In this case, we can use the potential energy to define a Lagrangian for the systemof particles:

L = T − U.

Using the Lagrangian, and with the help of the identity∗

∂U∂qK

=N∑

i=1

∂U∂ri

· ∂ri

∂qK,

it is easy to see that Lagrange’s equations for a system of particles is

ddt

(∂L∂qK

)− ∂L

∂qK= QK. (4.21)

∗ Those readers who read Section 4.7 will have seen this identity earlier as (4.18).

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4.9 A Constrained System of Particles 119

Here, QK are the nonconservative forces acting on the system of particles. The non-conservative forces have the equivalent representations

QK = �K + ∂U∂qK

=N∑

i=1

Fnconi · ∂ri

∂qK

= �ncon · aK.

Here,

Fnconi = Fi + ∂U∂ri

, �ncon = � + ∂U∂r

.

Notice that if QK = 0 – as it is in many celestial mechanics problems – one can writethe equations of motion for a system of particles without ever having to calculate anacceleration vector.

When constraints are present, a form of Lagrange’s equations similar to (4.21)can also be obtained provided the constraints are integrable and the constraintforces are prescribed by use of Lagrange’s prescription. In this case, the equationsare identical to (4.21) with the kinetic and potential energies being replaced withtheir constrained counterparts.∗

4.9 A Constrained System of Particles

Consider a system comprising particles that are subject to one integrable and onenonintegrable constraint. We choose the curvilinear coordinates to express theseconstraints as

q3N − r(t) = 0,

N∑i=1

fi · vi + e = 0.

These constraints are equivalent to the following constraints on the motion of thesingle (representative) particle:

q3N − r(t) = 0, f · v + e = 0.

Thus the generalized coordinates for this system are q1, . . . , q3N. These coordinatesparameterize the (3N − 1)-dimensional configuration manifold M. This manifoldlies in the 3N-dimensional configuration space that was discussed in Section 4.7 inconjunction with Casey’s construction of the representative particle. A representa-tive example is sketched in Figure 4.5.

∗ That is, in the notation subsequently presented, T is replaced with T and U is replaced with U. As aresult, L is replaced with L = T − U.

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Configuration manifold M

m v

r

�c

O

Figure 4.5. The representative particle moving on a configuration manifold. For the exampleshown in this figure, the configuration manifold M is fixed and �c is normal to M.

Assuming that the constraint forces associated with these constraints are com-patible with Lagrange’s prescription, the forces acting on the system of particleshave the decompositions

Fi = −∂U∂ri

+ µ1∂q3N

∂ri+ µ2fi + Pi,

where Pi is the resultant of the nonconservative and nonconstraint forces acting onthe particle of mass mi. It is important to note that

�K =N∑

i=1

Fi · ∂ri

∂qK

= − ∂U∂qK

+ µ1δ3NK + µ2 fK + �K, (4.22)

where

�K =N∑

i=1

Pi · ∂ri

∂qK, fK =

N∑i=1

fi · ∂ri

∂qK.

From these results, it is easy to construct � =∑3NK=1 �KaK.

Approach IFor the first approach, we expand the partial derivatives of T and U and then imposethe integrable constraint q3N = r(t) on the resulting equations. It can be seen thatthe resulting Lagrange’s equations of motion decouple into two sets:[

ddt

(∂T∂qS

)− ∂T

∂qS= − ∂U

∂qS+ µ2 fS + �S

]q3N=r,q3N=r

,

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4.9 A Constrained System of Particles 121

and

[ddt

(∂T

∂q3N

)− ∂T

∂q3N= − ∂U

∂q3N+ µ1 + µ2 f3N + �3N

]q3N=r,q3N=r

,

where S = 1, . . . , 3N − 1, and we emphasize that the constraint q3N = r(t) is im-posed after the partial derivatives of T and U have been calculated.

Approach IIFor comparison, we now use the second approach. In this case we first impose theintegrable constraint on the kinetic and potential energies, in addition to the basisvectors and constraints. For instance,

T = T(q1, . . . , q3N−1, q1, . . . , q3N−1, t)

= T(q1, . . . , q3N−1, q3N = r(t), q1, . . . , q3N−1, q3N = r(t)).

As in the case of a single particle, T can be used to construct a line-element for theconfiguration manifold M.

We note that, because

∂T∂q3N

= 0,∂T

∂q3N= 0,

we are unable to obtain an expression for µ1. In other words, because we have elim-inated the coordinate associated with the integrable constraint, we can obtain only3N − 1 Lagrange’s equations:

ddt

(∂T∂qS

)− ∂T

∂qS= − ∂U

∂qS+ µ2 f S + �S.

Further, no information on the constraint force enforcing the integrable constraintis obtained when this approach is used.

Geometric ConsiderationsOnce the integrable constraint has been imposed, the single particle of mass mmoves on the (3N − 1)-dimensional submanifold of the configuration space E

3N. Asbefore, this submanifold is known as the configuration manifold M. A measure ofthe distance traveled by the single particle on this manifold can be found by use ofT. Indeed, this energy can be decomposed as

T = T0 + T1 + T2,

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where

T0 = m2

a3N3Nr2,

T1 = m3N−1∑K=1

aK3NqKr,

Trel = T2 = m2

3N−1∑K=1

3N−1∑J=1

aKJ qKqJ . (4.23)

In the event that the single particle of mass m is determined, we also have thedecomposition:

aJK = aK · aJ (J, K = 1, . . . , 3N) .

The mass m in (4.23) is usually chosen to be m1 + · · · + mN, although otherselections such as m = 1 are equally admissible.

The kinematical line-element ds for M is

ds =(√

2Trel

m

)dt.

This quantity may also be written as

ds =√√√√3N−1∑

K=1

3N−1∑J=1

aKJ dqKdqJ .

As was the case previously, the imposition of a nonintegrable constraint will notchange M or ds.

4.10 A Canonical Form of Lagrange’s Equations

We now consider a system of particles in which the constraints are integrable andtime independent and the constraint forces are prescribed by use of Lagrange’s pre-scription. In this case, it suffices to examine those equations associated with thegeneralized coordinates in order to find the equations of motion. In what follows,we establish two alternative forms of Lagrange’s equations of motion [see (4.30) and(4.32)].

PreliminariesWe are interested in a system of N particles subject to a set of C (scleronomic)integrable constraints of the form∗

1 (r1, . . . , rN) = 0, . . . , C (r1, . . . , rN) = 0. (4.24)

∗ The generalization of our developments to instances in which the constraints are rheonomic can befound elsewhere, for example, Ginsberg [71].

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4.10 A Canonical Form of Lagrange’s Equations 123

We assume that the coordinates q1, . . . , q3N are chosen such that the generalized co-ordinates for the system are q1, . . . , qM, where M = 3N − C. That is, the constraints1 = 0, . . . , C = 0 are expressed in terms of the coordinates qM+1, . . . , q3N.

The constrained kinetic energy for the system has the representation

T = m2

M∑K=1

M∑J=1

aKJ qJ qK,

where we have chosen m = m1 + · · · + mN. We note that aKJ = aJK and assume thatthis matrix is invertible. The inverse is aJK. For example, if M = 3, then⎡

⎢⎣a11 a12 a13

a12 a22 a23

a13 a23 a33

⎤⎥⎦⎡⎢⎣a11 a12 a13

a12 a22 a23

a13 a23 a33

⎤⎥⎦ =

⎡⎢⎣1 0 0

0 1 00 0 1

⎤⎥⎦ .

We also define the Christoffel symbols of the first kind as

[SJ, K] = 12

(∂aKJ

∂qS+ ∂aKS

∂qJ− ∂aSJ

∂qK

)(J, K, S = 1, . . . , 3N) . (4.25)

There are (3N)3 of these symbols, but many of them are not distinct: [SJ, K] =[JS, K]. The Christoffel symbols of the second kind are

�KIJ =

3N∑R=1

aKR [IJ, R] (I, J, K = 1, . . . , 3N) . (4.26)

To gain further insight into these symbols, it is very useful to examine the singleparticle of mass m.

The Representative ParticleCasey’s construction of the representative particle is very useful for exploring theChristoffel symbols. Using the particle and its configuration space, we have a set ofcovariant, aK, and contravariant, aJ , basis vectors for E

3N. These vectors can be usedto define aJK, aJK, and the Christoffel symbols:

aJK = aJ · aK, aJK = aJ · aK,

[SI, K] = ∂aS

∂qI· aK, �J

KS = ∂aK

∂qS· aJ . (4.27)

Here, all the indices range from 1 to 3N. Using (4.27), the symmetries aJK = aKJ ,aJK = aKJ , [SI, K] = [IS, K], and �J

KS = �JSK should be transparent. We also observe

that the Christoffel symbols are none other than the covariant and contravariantcomponents of ∂aK

∂qS . It is a good exercise to show how (4.25) and (4.26) can be usedto establish (4.27)3,4 when (4.27)1,2 apply.

Derivatives of the Kinetic EnergyPreparatory to establishing the covariant and contravariant forms, we first recordthe derivatives of the kinetic energy. What follows is based entirely on the

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constrained kinetic energy, and we now drop the tildes ornamenting the variouskinematical quantities.

First, we note that

∂T∂qR

= ∂

∂qR

(m2

M∑K=1

M∑J=1

aKJ qJ qK

)

= m2

M∑K=1

M∑J=1

aKJ

⎛⎜⎜⎜⎝�

��

δJR

∂qJ

∂qRqK + qJ

��

��δK

R

∂qK

∂qR

⎞⎟⎟⎟⎠

= mM∑

K=1

aRKqK.

To arrive at this result, we used the symmetries aRK = aKR and the fact that aRK areindependent of qS. Similarly,

∂T∂qR

= ∂

∂qR

(m2

M∑K=1

M∑J=1

aKJ qJ qK

)

= m2

M∑K=1

M∑J=1

∂aKJ

∂qRqJ qK.

We next differentiate m∑M

K=1 aRKqK with respect to time. With some rearrangingand relabeling of the indices,∗ we find

ddt

(∂T∂qR

)= m

M∑K=1

aRKqK + mM∑

K=1

M∑S=1

∂aRK

∂qSqSqK

= mM∑

K=1

aRKqK + m2

M∑K=1

M∑S=1

∂aRK

∂qSqSqK

+ m2

M∑J=1

M∑I=1

∂aIR

∂qJqI qJ

= mM∑

K=1

aRKqK + m2

M∑K=1

M∑S=1

(∂aRK

∂qS+ ∂aSR

∂qK

)qSqK.

∗ Although ∂aRK∂qS �= ∂aRS

∂qK , because we are summing over the indices S and K, we always have that∑MK=1

∑MS=1

∂aRK∂qS qSqK =∑M

K=1∑M

S=1∂aRS∂qK qSqK.

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Summarizing, the derivatives of T have the representations

∂T∂qR

= m2

M∑K=1

M∑J=1

∂aKJ

∂qRqJ qK,

ddt

(∂T∂qR

)= m

M∑K=1

aRKqK + m2

M∑K=1

M∑S=1

(∂aRK

∂qS+ ∂aSR

∂qK

)qSqK. (4.28)

A Covariant Form of Lagrange’s Equations of MotionA covariant form of Lagrange’s equations of motion was previously established:

ddt

(∂T∂qR

)− ∂T

∂qR= �R (R = 1, . . . , M) . (4.29)

Here �1, . . . , �M are the generalized forces. We now expand the derivatives of T toestablish another form of this equation. With the help of (4.28) and the definition ofthe Christoffel symbol of the first kind, it is straightforward to show that Lagrange’sequations of motion can be written in the form

mM∑

K=1

aRKqK + mM∑

K=1

M∑S=1

[SK, R] qKqS = �R (R = 1, . . . , M) . (4.30)

As mentioned in the exercises at the end of Chapter 3, this form of Lagrange’s equa-tions appears in several texts on differential geometry,∗ and, in the case of a singleparticle, was an exercise at the end of Chapter 3.

From (4.30), it should be apparent that, if we know aJK, then we can immedi-ately write the left-hand side of Lagrange’s equations of motion. Further, for thesystem at hand the kinematical line-element ds is

ds =√√√√ M∑

R=1

M∑K=1

aRKqRqKdt.

Knowledge of ds (which also implies knowledge of aJK) enables us to write the left-hand side of Lagrange’s equations.

A Contravariant Form of Lagrange’s Equations of MotionIf we multiply both sides of the covariant form of Lagrange’s equations by the in-verse of aRK, then we will find the contravariant form of Lagrange’s equations ofmotion. Because we have imposed the integrable constraints and are interested in

∗ See, for example, McConnell [139] or Synge and Schild [208].

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only the first M equations of motion, we can find only the M contravariant equationsfrom the M covariant equations provided∗:

⎡⎢⎢⎣

a11 · · · a1M

.... . .

...a1M · · · aMM

⎤⎥⎥⎦⎡⎢⎢⎣

a11 · · · a1M

.... . .

...a1M · · · aMM

⎤⎥⎥⎦ =

⎡⎢⎢⎣

1 · · · 0...

. . ....

0 · · · 1

⎤⎥⎥⎦ . (4.31)

That is, we require several aKJ ’s to be zero:

⎡⎢⎢⎢⎢⎣

a1(M+1) a1(M+2) · · · a1(3N)

a2(M+1) a2(M+2) · · · a2(3N)...

.... . .

...a3N(M+1) a3N(M+2) · · · a3N(3N)

⎤⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎣

0 0 · · · 0

0 0 · · · 0...

.... . .

...0 0 · · · 0

⎤⎥⎥⎥⎥⎥⎦ .

When (4.31) holds, the right-hand side of Lagrange’s equations of motion trans-forms from the covariant components of � to the contravariant components:

�J =M∑

R=1

aJR�R.

In summary, the contravariant form of Lagrange’s equations of motion is

mqJ + mM∑

K=1

M∑S=1

�JKSqKqS = �J (J = 1, . . . , M) . (4.32)

This form of Lagrange’s equations is very useful in numerical simulations becausewe have explicit expressions for q1, . . . , qM.

An ExampleA simple example with which to explore Equations (4.30) and (4.32) is to considera single particle moving in E

3 under the action of a force F. We describe the motionof this particle by using a spherical polar coordinate system: q1 = R, q2 = φ, andq3 = θ. That is, M = 3. We can also consider our discussion here to be a solution toExercise 3.3.

It should by now be trivial to establish that

T = m2

(R2 + R2 sin2(φ)θ2 + R2φ2

).

∗ In terms of the representative particle, these restrictions are equivalent to aJ · aK = 0 for all J =1, . . . , M and all K = M + 1, . . . , 3N.

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From the expression for T, we can immediately deduce that

⎡⎢⎣

a11 a12 a13

a12 a22 a23

a13 a23 a33

⎤⎥⎦ =

⎡⎢⎣

1 0 0

0 R2 0

0 0 R2 sin2(φ)

⎤⎥⎦ ,

⎡⎢⎣

a11 a12 a13

a12 a22 a23

a13 a23 a33

⎤⎥⎦ =

⎡⎢⎣

1 0 0

0 R−2 0

0 0 R−2 sin−2(φ)

⎤⎥⎦ .

We readily deduce the covariant form of Lagrange’s equations of motion fromddt

(∂T∂qi

)− ∂T

∂qi = Fi by expanding the time derivative:

m

⎡⎢⎣

1 0 0

0 R2 0

0 0 R2 sin2(φ)

⎤⎥⎦⎡⎢⎣

R

φ

θ

⎤⎥⎦+

⎡⎢⎣

C1

C2

C3

⎤⎥⎦ =

⎡⎢⎣

F1

F2

F3

⎤⎥⎦ , (4.33)

where

⎡⎢⎣

C1

C2

C3

⎤⎥⎦ =

⎡⎢⎢⎣

−mR(

sin2(φ)θ2 + φ2)

−mR sin(φ) cos(φ)θ2 + 2mRRφ

2mR sin2(φ)Rθ + 2mR2 sin(φ) cos(φ)φθ

⎤⎥⎥⎦ ,

⎡⎢⎣

F1

F2

F3

⎤⎥⎦ =

⎡⎢⎣

F · eR

F · Reφ

F · R sin(φ)eθ

⎤⎥⎦ ,

⎡⎢⎣

F1

F2

F3

⎤⎥⎦ =

⎡⎢⎢⎣

F · eR

F · 1Reφ

F · 1R sin(φ) eθ

⎤⎥⎥⎦ =

⎡⎢⎣

a11 a12 a13

a12 a22 a23

a13 a23 a33

⎤⎥⎦⎡⎢⎣

F1

F2

F3

⎤⎥⎦ .

From (4.33), we can read off the Christoffel symbols of the first kind.For example, [22, 1] = −R, [33, 1] = −R sin2(φ), [12, 2] = R, and [13, 3] =R sin2(φ).

We can determine the contravariant form of (4.33) by multiplying both sides of(4.33) by the inverse of

[aij]:

m

⎡⎢⎣

R

φ

θ

⎤⎥⎦+

⎡⎢⎢⎣

−mR(sin2(φ)θ2 + φ2)

−m sin(φ) cos(φ)R θ2 + 2m

R Rφ

2mR Rθ + 2m cot(φ)φθ

⎤⎥⎥⎦ =

⎡⎢⎣

F1

F2

F3

⎤⎥⎦ . (4.34)

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Again, we can read off the Christoffel symbols of the second kind from the right-hand side of these equations. For example, �1

22 = −R, �233 = − sin(φ) cos(φ), and

�323 = cot(φ).

4.11 Alternative Principles of Mechanics

Newton’s tremendous contributions to mechanics in the 17th century still left manyunanswered questions. Among these questions were the formulation of equations ofmotion for rigid bodies and deformable media. To this end, several principles of me-chanics were postulated in the subsequent centuries: among them, Jean Bernoulli’sprinciple of virtual work from 1717, D’Alembert’s principle from 1743 [44], Gauss’principle of least constraint in 1829 [68], and Hamilton’s principle in 1835 [89].∗ Thepurpose of this section is to briefly outline how some of these principles are relatedto the balances of linear momenta used to establish the equations of motion for asystem of particles.

4.11.1 Principle of Virtual Work and D’Alembert’s Principle

We first consider the principle of virtual work and D’Alembert’s principle appliedto a system of N particles. These principles are often the basis for treatments ofLagrange’s equations of motion in many texts.† We assume that the particles aresubject to a single constraint:

f1 · v1 + · · · + fN · vN + e = 0. (4.35)

The principle of virtual work and D’Alembert’s principle collectively state that themotion of the system of particles is such that the following equation is satisfied:

(Fa1 − m1r1) · d1 + · · · + (FaN − mN rN) · dN = 0 (4.36)

for all possible choices of the vectors d1, . . . , dN that satisfy the condition

f1 · d1 + · · · + fN · dN = 0. (4.37)

Notice that the e present in (4.35) is notably absent from (4.37). The vectors dK areknown as virtual displacements and are usually denoted by δrK. The virtual workperformed by the applied force FaK is defined as FaK · dK; thus (4.36) states that thecombined virtual work of the applied forces FaK and the inertial forces −mKrK iszero.

Our aim is to obtain the equations of motion of the system of particles from(4.36). To this end, we can introduce Lagrange multipliers to accommodate the

∗ Further background on these (and many other principles) can be found in Dugas [48], Szabo [209],and Truesdell [216, 217].

† See, for example, Section 4.9 of Baruh [14], Section 2.1 of Greenwood [78], or Section 46 of Synge[206]. As noted by many authors, the principle of virtual work pertains to static problems, and itsextension to dynamics requires the application of D’Alembert’s principle; hence the combinationof these principles in this section.

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4.11 Alternative Principles of Mechanics 129

constraint on the vectors dK∗:

N∑K=1

(FaK − mKrN) · dK + µ

N∑K=1

fK · dK = 0. (4.38)

As a consequence of the Lagrange multiplier µ, the vectors dK can be varied inde-pendently. For (4.38) to hold for all such displacements, it is necessary and sufficientthat

mKrK = FaK + µfK (K = 1, . . . , N) . (4.39)

However, (4.39) are none other than the balances of linear momenta for a system ofparticles subject to constraint (4.35), for which the constraint forces are prescribedby use of Lagrange’s prescription:

FcK = µfK (K = 1, . . . , N) . (4.40)

We can now easily proceed to establish Lagrange’s equations of motion for the sys-tem of particles. Thus the principle of virtual work and D’Alembert’s principle com-bined with Lagrange’s prescription for the constraint forces collectively lead to thebalance laws for a constrained system of particles.

When constraint (4.35) is integrable, we can interpret (4.37) as a normality con-dition for the vectors dK. Further, it is not too difficult to see that constraint forceprescription (4.40) implies that this force is normal to the configuration manifold Min this case.

4.11.2 Gauss’ Principle of Least Constraint

Gauss’ principle of least constraint was published in 1829 [68]. It is a remarkableinterpretation of the role played by constraint forces in a mechanical system. Re-stricting our attention to particles, suppose that we have a system of N particlesthat are subject to constraint (4.35) and suppose that the constraint forces are pre-scribed by use of Lagrange’s prescription [i.e., (4.40) holds]. Then, in any motion ofthe system that satisfies the constraints, the constraint forces FcK = µfK are the leastneeded to ensure that the constraint is satisfied. That is, Lagrange’s prescription isin a sense optimal!

Since its introduction in 1829, the principle of least constraint has played a keyrole in several seminal developments in mechanics (see, for example, Hertz [92]). Alucid discussion of the principle can be found in Udwadia and Kalaba [218], and itsextension to cases for which Lagrange’s prescription does not hold can be found inO’Reilly and Srinivasa [162].

∗ When constraint (4.35) is integrable, this is precisely the approach taken by Lagrange in [121, 122](cf. Section 4 in Chapter 11 of Dugas [48]).

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t

q (t0)

q

q (t1)

t0 t1

Figure 4.6. Some of the possible paths q connecting two configurations, q (t0) and q (t1), of aparticle. The actual path is the one that satisfies the differential equations F = ma. Hamilton’sprinciple states that this path extremizes I = ∫ t1

t0Ldt [see (4.41)]. The portrait of Hamilton is

from the Royal Irish Academy in Dublin, Ireland.

4.11.3 Hamilton’s Principle

The remaining principle of interest was discovered by Sir William R. Hamilton(1805–1865) and first published in [89] for a system of unconstrained particles sub-ject solely to conservative forces. For ease of exposition, we restrict our discussionto a single particle of mass m and suppose that three coordinates q = (q1, q2, q3

)are

used to parameterize its position vector. Then Hamilton’s principle states that themotion of the system between a given initial configuration q (t0) and a given finalconfiguration q (t1) is such that it extremizes the action integral∗:

I =∫ t1

t0Ldt, (4.41)

where the Lagrangian L = T(qi, qi

)− U(qi)

and t0 and t1 are fixed instances of time.As illustrated in Figure 4.6, there are an infinite number of paths q(t) that can con-nect two possible configurations, and so finding the one that extremizes I appears tobe a daunting task.

However, it was known long before (4.41) appeared in print in 1835 that thenecessary conditions for q1(t), q2(t), q3(t) to extremize I was that q1(t), q2(t), q3(t)satisfy the following differential equations:

ddt

(∂L∂qk

)− ∂L

∂qk= 0 (k = 1, 2, 3) . (4.42)

In fact, in the context of extremizing I, (4.42) are known as the Euler–Lagrangeequations.† Thus Hamilton’s principle implies that the motion of the system satisfies

∗ To extremize is to minimize or maximize.† Problems featuring the extremization of I can be solved by use of the calculus of variations. In the

middle of the 18th century, Euler and Lagrange played seminal roles in the development of thiscalculus [72]. It was on the subject of this calculus that a 19-year-old Lagrange first wrote to Eulerin 1755.

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Exercise 4.1 131

(4.42). At this stage, it should be transparent that (4.42) are equivalent to F = mafor a particle. In conclusion, for an unconstrained particle, Hamilton’s principle isequivalent to the balance of linear momentum. Thus the motion of the system pro-vided by the solution to F = ma extremizes I!

Subsequent extensions to Hamilton’s principle were made by several authors,and these included the case in which the system of particles was subject to integrableconstraints. Jacobi [102] in particular used this principle to great effect when exam-ining the motion of particles on smooth surfaces and the shortest distance betweentwo points on a surface. Although Hamilton’s principle cannot readily be applied tosystems of particles subject to nonintegrable constraints, it has helped form severalpillars of modern physics.

4.12 Closing Remarks

In this chapter, a derivation of Lagrange’s equations of motion was presented fora system of particles, and it was shown how they can be modified when constraintsare introduced. Our developments emphasized that these equations are equivalentto the Newtonian balance laws of linear momenta for each particle. This importantfeature enables us to confidently calculate the forces �K =∑N

i=1 Fi · ∂ri∂qK that appear

on the right-hand side of Lagrange’s equations of motion:

ddt

(∂T∂qK

)− ∂T

∂qK= �K (R = 1, . . . , 3N) .

We are also able to introduce integrable and nonintegrable constraints and the con-straint forces associated with them.

On a deeper level, if we wish to consider the system of particles as a single par-ticle moving on a configuration manifold M that is embedded in a 3N-dimensionalEuclidean space, then Casey’s construction of the representative particle enables usto do this in a straightforward manner. In the next chapter, we turn to examples inwhich Lagrange’s equations of motion for several systems of particle are establishedand discussed.

EXERCISES

4.1. What are the kinematical line-elements ds, generalized coordinates, and con-figuration manifolds M of the following systems:

(a) a particle attached to a fixed point by a spring;

(b) a particle attached to a fixed point by a rod of length L(t);

(c) a harmonic oscillator consisting of one particle;

(d) a planar double pendulum;

(e) a spherical double pendulum.

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4.2. Under which circumstances do constraint forces perform no work?

4.3. Many authors assume that the kinetic energy of a system of particles is apositive-definite function of the velocities qK. Letting T =

∑3NI=1

∑3NJ=1

m2 aIJ qI qJ is

equivalent to saying that the matrix whose components are m2 aIJ is positive-definite.∗

Using the spherical pendulum as an example, show that certain representationsof T are not always positive-definite. Specifically, if one uses spherical polar coordi-nates, then the positive-definiteness breaks down at the singularities of this coordi-nate system.

4.4. In Casey’s construction of the single particle, what are the distinctions amongthe bases {eK}, {eK}, and {eK}? For a given system of two particles, how does oneconstruct these bases for E

6?

4.5. Consider the following function that depends on the motion of two particles:

V = V (||r2 − r1|| , t) .

This function is representative of a potential energy function and an integrableconstraint.

(a) For any vector x, establish the following result:

d‖x‖dt

= x‖x‖ · x.

The easiest way to show this result is to represent x by Cartesian coordi-nates.

(b) Show that

V = ∂V∂r1

· v1 + ∂V∂r2

· v2 + ∂V∂t

,

where

∂V∂r1

= − ∂V∂r2

= −∂V∂x

x‖x‖

and

x = ‖x‖, x = r2 − r1.

(c) Consider a system of two particles subject to a constraint = 0, where

= (‖r2 − r1‖, t) . (4.43)

Using Lagrange’s prescription, show that Fc1 = −Fc2. Notice that this isNewton’s third law of motion. Give two examples of a physical constraintfor which has the form (4.43) and for which the constraint forces thatenforce this constraint are equal and opposite.

∗ Recall that a matrix C is positive-definite if, for all nonzero x, xTCx > 0 and xTCx = 0 only whenx = 0.

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(d) Consider a system of two particles subject to a conservative force that hasa potential energy function:

U = U (‖r2 − r1‖) . (4.44)

Using the identity,

U = −Fcon1 · v1 − Fcon2 · v2,

show that Fcon1 = −Fcon2. Notice that this (again) is Newton’s third law ofmotion. Give two instances in which U has the form (4.44) and demonstratethat the conservative forces are equal and opposite.

4.6. The two-body problem consists of a system of two particles, one of massm1 = M and the other of mass m2 = m. The sole forces on the system are theresult of a Newtonian gravitational force field that has a potential energy functionUn = − GMm

||r2−r1|| .

(a) For this system, show that the linear momentum of the system is conservedand that the center of mass C moves at a constant speed in a straight line.In addition, show that

r2 − r = Mm + M

(r2 − r1) , r1 − r = − mm + M

(r2 − r1) .

Here, r is the position vector of the center of mass.

(b) Show that the angular momentum of the system of particles relative to C isconserved.

(c) Show that the total energy of the system of particles is conserved.

(d) Show that the differential equations governing the motions of m1 and m2

can be written in the following forms∗:

M(r1 − ¨r

) = −GMm(

mM + m

)2 r1 − r

||r1 − r||3 ,

m(r2 − ¨r

) = −GMm(

MM + m

)2 r2 − r

||r2 − r||3 . (4.45)

(e) Argue that the results of Section 2.8 can be applied to (4.45) to determinethe orbital motions of the particles about their center of mass C.

∗ In celestial mechanics, expressing the equations of motion in this manner is equivalent to using whatis known as a barycentric coordinate system (see, for example, [220]), that is, a coordinate systemwhose origin is at the center of mass.

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5 Dynamics of Systems of Particles

5.1 Introduction

In this chapter, several examples of systems of particles are discussed. We payparticular attention to how the equations of motion for these systems are estab-lished by use of Lagrange’s equations. The examples we discuss are classical andrange from simple harmonic oscillators to dumbbell satellites and pendula. Ourgoals are to illuminate the developments of the previous chapter and to presentrepresentative examples.

Examples that are closely related to the ones we discuss can be found in manydynamics texts. Most of these texts use alternative formulations of Lagrange’sequations of motion that do not readily accommodate nonconservative forces.Here, because we have established an equivalence between Lagrange’s equationsof motion and the balances of linear momenta, we are easily able to incorporatenonconservative forces such as dynamic Coulomb friction. This chapter closeswith a brief discussion of some recent works on the dynamics of systems ofparticles.

5.2 Harmonic Oscillators

We first consider simple examples involving a system of two particles. The systemshown in Figure 5.1 is the first of several related systems that we discuss in thissection.

Referring to the figure, we see that a particle of mass m1 is connected by a springof stiffness K1 and unstretched length L1 to a fixed support. It is also connected bya spring of stiffness K2 and unstretched length L2 to a particle of mass m2. Bothparticles are constrained to move in the E1 direction:

r1 = (L1 + x1) E1 + y1E2 + z1E3,

r2 = (L1 + L2 + x2) E1 + y2E2 + z2E3.

Notice that x1 and x2 measure the displacement of the particles from the unstretchedspring states.

134

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5.2 Harmonic Oscillators 135

m1 m2

O

Smooth surface

g

E1

E2

Figure 5.1. The system of two particles moving on a smooth horizontal surface. The particlesare connected by linear springs.

CoordinatesHere we denote the six coordinates by

q1 = x1, q2 = x2, q3 = y1, q4 = y2, q5 = z1, q6 = z2.

For the systems of interest, the particles are constrained to move on a line. Thus thesystem is subject to four constraints:

1 = 0, 2 = 0, 3 = 0, 4 = 0,

where

1 = y1, 2 = y2, 3 = z1, 4 = z2.

It should be noted that

∂1

∂r1= E2,

∂2

∂r1= 0,

∂3

∂r1= E3,

∂4

∂r1= 0,

and

∂1

∂r2= 0,

∂2

∂r2= E2,

∂3

∂r2= 0,

∂4

∂r2= E3.

These expressions will be used later to prescribe constraint forces.

Kinetic and Potential EnergiesIt is easy to see that the kinetic and potential energies of the system are

T = m1

2

(x2

1 + y21 + z2

1

)+ m2

2

(x2

2 + y22 + z2

2

),

U = K1

2x2

1 + K2

2(x2 − x1)2 + m1gy1 + m2gy2. (5.1)

Imposing the four integrable constraints on these energies, we can determine theirconstrained counterparts:

T = m1

2x2

1 + m2

2x2

2, U = 12

K1x21 + 1

2K2 (x2 − x1)2

.

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Constraint ForcesIf the surface that the particles are moving on is smooth, then Lagrange’s prescrip-tion can be invoked:

Fc1 = µ1E2 + µ3E3, Fc2 = µ2E2 + µ4E3.

Here, µ1, µ2, µ3, and µ4 are normal force components. If the surface is rough, thenthese prescriptions must be altered to include the friction forces:

Fc1 = µ1E2 + µ3E3 − µd||µ1E2 + µ3E3|| x1

|x1|E1,

Fc2 = µ2E2 + µ4E3 − µd||µ2E2 + µ4E3|| x2

|x2|E1.

The total force acting on each particle is composed of the constraint force and theconservative forces that are due to gravity and the springs.

The Representative ParticleIt is easy to construct the representative particle of mass m for this system. First,the position vector of this particle, which is moving in E

6, is defined with the helpof (4.15):

r = (x1 + L1) e1 + y1e2 + z1e3 + (x2 + L1 + L2) e4 + y2e5 + z2e6.

Using this expression, we easily calculate the six covariant vectors aJ and the sixcontravariant vectors aJ :

a1 = e1, a2 = e4, a3 = e2, a4 = e5, a5 = e3, a6 = e6,

and

a1 = e1, a2 = e4, a3 = e2, a4 = e5, a5 = e3, a6 = e6.

If the kinetic energy T = m2 r · r were calculated, it would be identical to (5.1).

The constraints on the motion of the single particle have the following repre-sentations:

1 = r · e2, 2 = r · e5, 3 = r · e3, 4 = r · e6.

Using these representations, we easily see that the constraint force �c that we wouldprescribe by using Lagrange’s prescription is

�c = µ1e2 + µ2e5 + µ3e3 + µ4e6. (5.2)

When friction is present, this prescription is inadequate.

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The force vector � can be determined by use of prescription (4.16):

� =(

−K1 (x1) + K2 (x2 − x1) − µd||µ1E2 + µ3E3|| x1

|x1|)

e1

+ (µ1 − m1g) e2 + µ3e3 + (µ2 − m2g) e5 + µ4e6

+(

−K2 (x2 − x1) − µd||µ2E2 + µ4E3|| x2

|x2|)

e4.

Notice that the constraint force contributions to this vector are consistent with ourprescription for �c in (5.2) when µd = 0. We have now completely specified all theingredients needed to establish Lagrange’s equations of motion by using the repre-sentative particle of mass m.

The Generalized Coordinates and Configuration ManifoldThere are four integrable constraints on the system, so the generalized coordinatesare x1 and x2. The configuration manifold M is simply the plane E

2 and the kine-matical line-element is

ds =√

m1

m1 + m2

(dx1

dt

)2

+ m2

m1 + m2

(dx2

dt

)2

dt.

Notice that this line-element is easily related to the standard measure of distance

traveled along a curve (x(τ), y(τ)) on a plane:

√( dxdτ

)2 +(

dydτ

)2dτ. We shall shortly

introduce a nonintegrable constraint into the system. This constraint will not affectM or ds.

Equations of Motion for the OscillatorWe first consider the case shown in Figure 5.1 in which friction is absent. For thissystem, the constraint forces are prescribed by use of Lagrange’s prescription andthe other forces acting on the system are conservative. Thus �1 = − ∂U

∂q1 and �2 =− ∂U

∂q2 . In conclusion, the equations of motion can be obtained by use of the followingform of Lagrange’s equations of motion:

ddt

(∂T∂qα

)− ∂T

∂qα= − ∂U

∂qα, (5.3)

where α = 1, 2. With a small amount of work, we find the equations of motion:

m1x1 = −K1x1 − K2 (x1 − x2) ,

m2x2 = −K2 (x2 − x1) .

These equations are classical and are easily seen to be equivalent to F1 · E1 = m1r1

and F2 · E1 = m2r2, respectively.

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m1 m2

O

Rough surface

g

PE1

E2

Figure 5.2. A system of two particles moving on a rough horizontal surface.

The Influences of Dynamic Friction and an Applied ForceWe now consider the same system for particles, but assume that dynamic Coulombfriction is present and a forcing P = P cos(ωt)E1 is applied to the particle of massm2 (see Figure 5.2). We have already determined all of the needed ingredients todetermine the equations of motion for this system. The major difference is thatLagrange’s equations of motion in the form (5.3) are inadequate. It is necessary touse the following form of these equations:

ddt

(∂T∂qK

)− ∂T

∂qK= �K. (5.4)

Here, �K = F1 · ∂r1∂qK + F2 · ∂r2

∂qK , or equivalently if we use the single representativeparticle, �K = � · aK. Clearly we need to append P = P cos(ωt)E1 to F2 andP cos(ωt)e4 to �.

From (5.4) and the expressions we have established for T and the force, we findthe following six equations:

m1x1 = −K1x1 − K2 (x1 − x2) − µdm1gx1

|x1| ,

m2x2 = −K2 (x2 − x1) − µdm2gx2

|x2| + P cos(ωt),

0 = N1y − m1g,

0 = N2y − m2g,

0 = N1z,

0 = N2z. (5.5)

The last four equations yield the constraint forces: N1y = µ1, N2y = µ2, N1z = µ3,and N2z = µ4.

Stick–Slip OscillationsOne of the most interesting features of the oscillator arises when either or bothof the velocities of the particles vanish. In this case, the number of integrableconstraints increases and the constraint forces must be altered. The resulting

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oscillations are often termed stick–slip: behavior we encountered earlier in ourmodel for the roller coaster. Because the number of differential equations govern-ing the motion is different for these two types of friction, numerical simulations ofthe equations of motion can be challenging.∗

For instance, suppose m1 is instantaneously at rest (sticks); then the additionalconstraint x1 = x10, where x10 is a constant, needs to be imposed. The constraintforce Fc1 is now

Fc1 = µ1E2 + µ3E3 + µ5E1,

where µ5E1 is the static friction force. This force is subject to the static frictioncriterion:

|µ5| ≤ µs

√µ2

1 + µ23,

where µs is the coefficient of static friction. Instead of (5.5), the equations of motionfor the oscillator in this case are

µ5 = K1x10 + K2 (x10 − x2) ,

m2x2 = −K2(x2 − x10) − µdm2gx2

|x2| + P cos(ωt), (5.6)

where ∣∣K1x10 + K2 (x10 − x2)∣∣ ≤ µsm1g.

Thus, if there is sufficient friction to match the spring forces, then x10 will remainstationary. Otherwise, it will tend to slip in the direction of the resultant spring forceand (5.5) is then used to determine the motion.

Imposing a Nonintegrable ConstraintWe now consider the introduction of a nonintegrable constraint in the system con-sidered in Section 5.2. The constraint is

x1x2 − x2 = 0.

We can also express this constraint as

(x2E1) · v1 + (−E1) · v2 = 0,

or, for the representative particle of mass m, as(x2e1 − e4) · v = 0.

It is left as an exercise to show that the constraint is nonintegrable.Using Lagrange’s prescription, we find that the additional constraint forces on

the particles are µ5x2E1 and −µ5E1. For the representative particle of mass m, �c

∗ For further details on numerical schemes for stick–slip oscillators, see [23, 50, 112]. Interesting ex-amples of such oscillators and their application to various fields, including brake squeal, can befound in [21, 111].

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m1

m2

r1

r2

r

O

C

E1

E2

E3

Figure 5.3. A particle of mass m1 attached by linear spring of stiffness K and unstretchedlength L0 to a particle of mass m2. Each particle is attracted to the fixed point O by a Newto-nian gravitational force field.

needs to be augmented by µ5(x2e1 − e4). Starting from Lagrange’s equations of mo-tion in the form (5.4), we quickly arrive at the equations of motion:

m1x1 = −K1x1 − K2(x1 − x2) − µdm1gx1

|x1| + µ5x2,

m2x2 = −K2(x2 − x1) − µdm2gx2

|x2| − µ5.

The last four equations are identical to those recorded in (5.5). The two equationsof motion are supplemented by the constraint x1x2 − x2 = 0.

5.3 A Dumbbell Satellite

In the 1960s, several simple models for deformable satellites orbiting a planet ofmass M appeared. Here we consider one of these models and discuss some featuresof the dynamics predicted by it. In particular, we will notice a coupling between themotion of the center of mass and the rotation of the satellite. This coupling, whichis elegantly explained in Beletskii’s text on satellite dynamics [16], is induced by thegravitational forces acting on the satellite.

The model we discuss here lumps the mass distribution of the satellite into twomass particles at the extremeties of a spring of stiffness K and unstretched lengthL0 (see Figure 5.3). The gravitational force exerted on the satellite by the planet ofmass M it is orbiting is modeled as a Newtonian gravitational force field exerted oneach of the particles. In what follows, we discuss how the equations of motion forthis model can be determined.

CoordinatesOur first task is to choose coordinates for the position vectors r1 and r2 of the massparticles. One reasonable choice would be to pick Cartesian coordinates for bothparticles; another would be to choose Cartesian coordinates for r1 and sphericalpolar coordinates for r2 − r1. A third alternative would be to pick Cartesian

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5.3 A Dumbbell Satellite 141

coordinates for the position vector r of the center of mass C of the system and a setof spherical polar coordinates for r2 − r and r1 − r. We elect the third choice here.

Denoting r2 − r1 by ReR, where R = ||r2 − r1||,

eR = sin(φ) (cos(θ)E1 + sin(θ)E2) + cos(φ)E3.

Some elementary algebra shows that

r1 = r −(

m2

m1 + m2

)ReR, r2 = r +

(m1

m1 + m2

)ReR.

With our choice of coordinates, r = xE1 + yE2 + zE3.For future reference, we label our coordinates

q1 = x, q2 = y, q3 = z, q4 = R, q5 = φ, q6 = θ. (5.7)

We also record the following 12 partial derivatives:

∂r1

∂x= E1,

∂r1

∂y= E2,

∂r1

∂z= E3,

∂r1

∂R= −

(m2

m1 + m2

)eR,

∂r1

∂φ= −

(m2

m1 + m2

)Reφ,

∂r1

∂θ= −

(m2

m1 + m2

)R sin(φ)eθ,

and

∂r2

∂x= E1,

∂r2

∂y= E2,

∂r2

∂z= E3,

∂r2

∂R=(

m1

m1 + m2

)eR,

∂r2

∂φ=(

m1

m1 + m2

)Reφ,

∂r2

∂θ=(

m1

m1 + m2

)R sin(φ)eθ.

Because the forces acting on the system of particles are conservative, we can deter-mine Lagrange’s equations of motion without calculating these vectors. Our moti-vation for calculating them here is that they will illuminate a certain relationship inSection 5.3.

Kinetic and Potential EnergiesThe kinetic energy of the system is the sum of the kinetic energies of the particles:

T = m1

2v1 · v1 + m2

2v2 · v2

= m1 + m2

2

(x2 + y2 + z2)+ m1m2

2(m1 + m2)

(R2 + R2φ2 + R2 sin2 (φ) θ2

).

To arrive at this expression, a substantial amount of algebraic cancellationsoccurred.

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The potential energy of the system is due to Newtonian gravitation and thespring kinetic energy:

U = − GMm1∣∣∣∣∣∣r −(

m2m1+m2

)ReR

∣∣∣∣∣∣ − GMm2∣∣∣∣∣∣r +(

m1m1+m2

)ReR

∣∣∣∣∣∣ + K2

(R − L0)2.

We could also have included the Newtonian gravitational force that m1 exerts on m2

and vice versa but this will not significantly add to the discussion.

Lagrange’s Equations of MotionFor this system of particles, there are no constraints and the applied forces are con-servative. As a result, we can use Lagrange’s equations of motion in the form

ddt

(∂L∂qS

)− ∂L

∂qS= 0 (S = 1, . . . , 6) .

Equipped with the expressions for T and U recorded in the previous subsection,we can derive the equations straightforwardly, and the details are not presentedhere.

Because the sole forces acting on this system are conservative, the solutions tothe equations of motion should preserve the total energy E = T + U. In addition,the angular momentum of this system relative to O, HO, should also be conserved.The former momentum has the following representation:

HO = r1 × m1v1 + r2 × m2v2

= r × mv + m1m2

m1 + m2R2 (φeθ − θ sin(φ)eφ

).

The conservation of HO implies that the linear speeds, x, y, and z, of the center ofmass are coupled to the angular speeds θ and φ of the satellite. We shall observe thiscoupling later in rigid body models for satellites.

The Generalized Coordinates and Configuration ManifoldThere are no constraints on the system, so the generalized coordinates are x, y, z, R,θ, and φ. The configuration manifold is simply E

6, and the kinematical line-element

is ds =√

2Tm1+m2

dt.

Comments on the Equations of MotionThe equations of motion found by use of Lagrange’s equations are equivalent tothose that would be obtained directly from F1 = m1r1 and F2 = m2r2. Indeed, if wereturn to the derivation of Lagrange’s equations in Section 4.6, then it is easy tosee that these equations are linear combinations of F1 = m1r1 and F2 = m2r2. Forinstance, Lagrange’s equation of motion for R,

ddt

(∂L

∂R

)− ∂L

∂R= 0,

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5.4 A Pendulum and a Cart 143

is equivalent to

(F1 = m1r1) · ∂r1

∂R+ (F2 = m2r2) · ∂r2

∂R= 0.

If we change our selection for the coordinates used for the system, this changesthe precise linear combinations of the components of F1 = m1r1 and F2 = m2r2 thatconstitute Lagrange’s equations of motion.

Related SystemsWe can modify the model presented in this section in a variety of manners. First, thespring can be replaced with a rigid bar. For this model, E and HO are still conserved.It is interesting to observe that, even for this simple system, the gravitational forcefield is equivalent to a force acting at the center of mass C and a moment relative tothis point.

5.4 A Pendulum and a Cart

Consider the system of two particles shown in Figure 5.4. A particle of mass m1 isfree to move on a smooth horizontal rail and is connected to a particle of mass m2

by a spring of stiffness K and unstretched length L0. The motion of both particles isassumed to be planar.

Coordinates and ConstraintsAs usual, the first task is to choose coordinates for the position vectors r1 and r2 ofthe mass particles. In anticipation of imposing the constraints, we choose Cartesiancoordinates for r1 and cylindrical polar coordinates for r2 − r1:

r1 = xE1 + yE2 + z1E3, r2 = xE1 + yE2 + rer + (z1 + z2) E3.

m1

m2

O

Smooth horizontal rail

Linear spring

g

r

θ

E1

E2

Figure 5.4. A system of two particles connected by a linear spring of stiffness K and un-stretched length L0. The particle of mass m1 is free to move on a smooth horizontal rail andthe second particle moves on the x − y plane.

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The angle θ is measured from the E1 direction and is counterclockwise positive. Thatis, when m2 is directly below m1, θ = 3π

2 .We label the coordinates as follows:

q1 = x, q2 = r, q3 = θ, q4 = y, q5 = z1, q6 = z2 = (r2 − r1) · E3. (5.8)

The constraints on this system of particles are all integrable:

1 = 0, 2 = 0, 3 = 0,

where

1 = y = r1 · E2, 2 = z1 = r1 · E3, 3 = z2 = (r2 − r1) · E3.

At this stage, it is prudent to compute that

∂1

∂r1= E2,

∂1

∂r2= 0,

∂2

∂r1= E3,

∂2

∂r2= 0,

∂3

∂r1= −E3,

∂3

∂r2= E3.

We could also have used Cartesian coordinates to parameterize r2 − r1, and thischoice would be a good exercise to pursue.

Kinetic and Potential EnergiesThe potential energy of the system is due to gravity and the spring potential energy:

U = m1gy + m2gy + m2gr sin(θ) + K2

(||r2 − r1|| − L0)2.

Imposing the integrable constraints, we find that

U = m2gr sin(θ) + K2

(r − L0)2.

The kinetic energy of the system is also easy to calculate:

T = m1 + m2

2

(x2 + y2 + z2

1

)+ m2

2

(r2 + r2θ2 + z2

2

)+ m2

(xr cos(θ) + yr sin(θ) − xrθ sin(θ) + yrθ cos(θ)

)+ m2z1z2. (5.9)

This expression simplifies when we impose the following constraints:

T = m1 + m2

2x2 + m2

2

(r2 + r2θ2)+ m2

(xr cos(θ) − xrθ sin(θ)

).

Generalized Coordinates and ConstraintsThe system has three generalized coordinates: x, r, θ. The configuration manifold Mfor this system is a three-dimensional manifold in E

6 that is parameterized by thesecoordinates. Because r ranges from 0 to ∞ and θ ranges from 0 to 2π, these two

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5.4 A Pendulum and a Cart 145

coordinates parameterize E2 in its entirety. Further, as x ranges from −∞ to ∞, we

see that this coordinate parameterizes E. We conclude that M is E3.

We can find the kinematical line-element for the configuration manifold by us-ing T:

ds =√

2Tm1 + m2

dt,

where we have chosen m = m1 + m2.

Constraint Forces and Resultant ForcesThe constraint forces on the system can be computed by use of Lagrange’s prescrip-tion:

Fc1 =3∑

i=1

µi∂i

∂r1= µ1E2 + µ2E3 − µ3E3,

Fc2 =3∑

i=1

µi∂i

∂r2= µ3E3.

Notice that the Lagrange multipliers µi are equivalent to the normal forces: N1 =µ1E2 + (µ2 − µ3) E3 and N2 = µ3E3.

For completeness, the total resultant forces F1 and F2 acting on the particles are

F1 = K (r − L0) er + (µ1 − m1g) E2 + (µ2 − µ3) E3,

F2 = −K (r − L0) er − m2gE2 + µ3E3,

respectively. We shall use these expressions later on to determine the force vector �.

Lagrange’s Equations of MotionWe can find the equations of motion for this system by using any of the variety offorms for Lagrange’s equations of motion. Arguably, the easiest approach is to usethe form involving Lagrangian (4.21):

ddt

(∂L∂qS

)− ∂L

∂qS= QS =

2∑i=1

Fnconi · ∂ri

∂qS(S = 1, . . . , 6) . (5.10)

However, as the constraint forces are compatible with Lagrange’s prescription, forthe first three of these equations the right-hand side will be zero: Q1 = 0, Q2 = 0,and Q3 = 0.

To find the equations of motion, it suffices to examine the first three ofLagrange’s equations of motion:

ddt

(∂L∂qJ

)− ∂L

∂qJ= 0 (J = 1, . . . , 3) .

Here, L = T − U. Evaluating the partial derivatives of L = T − U and then calcu-lating d

dt , we find with some rearrangement that the equations of motion can be

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expressed in the form⎡⎢⎣

m1 + m2 m2 cos(θ) −m2r sin(θ)

m2 cos(θ) m2 0

−m2r sin(θ) 0 m2r2

⎤⎥⎦⎡⎢⎣

x

r

θ

⎤⎥⎦ = −

⎡⎢⎣

f1

f2

f3

⎤⎥⎦ , (5.11)

where f1, f2, and f3 are quadratic in the velocities:

f1 = −2m2rθ sin(θ) − m2rθ2 cos(θ),

f2 = −m2rθ2 + m2g sin(θ) + K (r − L0) ,

f3 = 2m2rrθ + m2gr cos(θ). (5.12)

The form (5.11) of the equations of motion is easy to implement numerically. Itis also a canonical form for many mechanical systems featuring time-independent(scleronomic) integrable constraints. You should also notice that the matrix on theleft-hand side of (5.11) can be obtained by inspection from T.

Solving for the Constraint ForcesTo determine the constraint forces Fc1 and Fc2, we first solve for µi by using threeof Lagrange’s equations of motion:

ddt

(∂L∂qJ

)− ∂L

∂qJ= QJ (J = 4, . . . , 6) .

We leave the intermediate steps as an exercise:

Fc1 = (m1 + m2) gE2 + ddt

(m2r sin(θ) + m2rθ cos(θ)

)E2,

Fc2 = 0.

It is easy to observe from these expressions the expected result that Fc1 · v1 + Fc2 ·v2 = 0.

ConservationsThe solutions to equations of motion (5.11) conserve two kinematical quantities.The first is the total energy E of the system. To see this conservation, it suffices tonote that none of the constraint forces do work and all of the remaining forces actingon the system are conservative. Thus the work–energy theorem easily leads to theconclusion that E = 0, where E = T + U. The second conservation can be deducedfrom (5.11)1. That is, the linear momentum of the system in the horizontal direction,G · E1, is conserved.

The Single Representative ParticleWe computed Lagrange’s equations of motion for the system without explicitly cal-culating the position vector r of the single particle of mass m moving in E

6. If we

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5.5 Two Particles Tethered by an Inextensible String 147

were to use this particle to compute (5.11), then we would first need to use (4.15) todefine

r = xe1 + ye2 + z1e3 + (x + r cos(θ)) e4 + (y + r sin(θ)) e5 + (z1 + z2) e6.

Using this expression, we can easily calculate the six covariant vectors aJ and the sixcontravariant vectors aJ . For example, a1 = e1 + e4 and a6 = e6. As expected, thekinetic energy T = m

2 r · r will be identical to (5.9).The force vector � can be determined by use of prescription (4.16):

� = K (r − L0) cos(θ)e1 + (µ1 − m1g + K (r − L0) sin(θ)) e2

+ (µ2 − µ3) e3 − K (r − L0) cos(θ)e4

+ (−m2g − K (r − L0) sin(θ)) e5 + µ3e6.

We used the expressions for F1 and F2 recorded earlier to compute this vector. If wecompute � · aS and compare the results with those we obtained by using F1 · ∂r1

∂qS +F2 · ∂r2

∂qS , then the two sets of expressions for �S should be identical.

RemarksThe system discussed here is also a good candidate to explore the covariant (4.30)and contravariant (4.32) forms of Lagrange’s equations of motion that we discussedearlier. Indeed, by using (5.11), we can readily compute the

[aij]

matrix and theChristoffel symbols of the first kind.

5.5 Two Particles Tethered by an Inextensible String

As shown in Figure 5.5, a particle of mass m1 is connected by an inextensible stringof length L0, which passes through a smooth eyelet at O to a particle of mass m2.The particle of mass m1 moves on a rough horizontal plane, and the particle of massm2 is free to move in space. The goal of the following analysis is to establish theequations of motion for the system of particles and then discuss certain conservedquantities associated with their solutions. To make the problem tractible, we assumethat the string remains taut and that the particle of mass m2 does not collide withthe underside of the horizontal plane.

The Coordinates and Other Kinematical QuantitiesTo describe the kinematics of this system of particles, a cylindrical polar coordinatesystem {r1, θ1, z1} is used to describe the motion of the particle of mass m1 and aspherical polar coordinate system {R2, φ2, θ2} is used to describe the motion of theparticle of mass m2:

r1 = r1er1 + z1E3, r2 = R2eR2 .

We define six coordinates as follows:

q1 = r1, q2 = θ1, q3 = φ2, q4 = θ2, q5 = x, q6 = z1. (5.13)

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m1

m2

O

g

Inextensible string

Rough horizontal plane

E1

E2

E3

Figure 5.5. A system of two particles connected by an inextensible string of length L0.

Notice that we have introduced a new coordinate x:

R2 = L0 − r1 + x.

Using the cylindrical polar coordinate system and the spherical polar coordinatesystems, we have

r1 = r1er1 + z1E3, r2 = (L0 + x − r1) eR2 .

Hence,

∂r1

∂r1= er1 ,

∂r1

∂θ1= r1eθ1 ,

∂r1

∂φ2= 0,

∂r1

∂θ2= 0,

∂r1

∂x= 0,

∂r1

∂z1= E3,

∂r2

∂r1= −eR2 ,

∂r2

∂θ1= 0,

∂r2

∂φ2= (L0 + x − r1) eφ2 ,

∂r2

∂θ2= (L0 + x − r1) sin(φ2)eθ2 ,

∂r2

∂x= eR2 ,

∂r2

∂z1= 0.

The Potential and Kinetic EnergiesThe potential energy of the system is due to gravity:

U = m1gE3 · r1 + m2gE3 · r2 = m1gz1 + m2g (L0 + x − r1) cos (φ2) .

To calculate the kinetic energy of the system we first need expressions for the veloc-ity vectors. The expression for v1 is easy to find:

v1 = r1er1 + r1θ1eθ1 + z1E3.

To calculate v2, we recall the expression for this vector in spherical polar coordinatesand then substitute for R2 and R2:

v2 = (x − r1) eR2 + (L0 + x − r1) sin (φ2) θ2eθ2 + (L0 + x − r1) φ2eφ2 .

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5.5 Two Particles Tethered by an Inextensible String 149

The kinetic energy T of the system of particles is

T = m1

2v1 · v1 + m2

2v2 · v2

= m1

2

(r2

1 + r21θ

21 + z2

1

)+ m2

2

((x − r1)2 + (L0 + x − r1)2 sin2 (φ2) θ2

2 + (L0 + x − r1)2φ2

2

).

Notice that T is a function of q1, . . . , q6 and their time derivatives.

The Constraints and Constraint ForcesThe constraints on the motion of the system are twofold. First, the particles areconnected by an inextensible string of length L0, and, second, the motion of m1 isplanar. In terms of the coordinates q1, . . . , q6, the two constraints are

x = 0, z1 = 0.

The constraint forces associated with these constraints correspond to the tensionforce in the string and the friction and normal forces on m1:

Fc1 = µ1er1 + µ2E3 − µd||µ2E3|| vrel

||vrel|| ,

Fc2 = µ1eR2 ,

where vrel = r1er1 + r1θ1eθ1 .The constraint forces associated with the inextensible string can also be pre-

scribed by use of Lagrange’s prescription. Let

1 = ||r2|| + ||r1|| − L0.

Then the inextensibility constraint is 1 = 0. Using Lagrange’s prescription, we find

Fc1 = µ1r1

||r1|| = µ1er1 ,

Fc2 = µ1eR2 .

Notice that we imposed the constraint z1 = 0 to simplify the expression for Fc1. Theconstraint forces associated with the constraint 2 = 0, where 2 = z1, are not pre-scribed by Lagrange’s prescription because it features a dynamic friction force.

The Equations of MotionLagrange’s equations will provide four differential equations for the generalizedcoordinates and two equations for µ1 and µ2. For ease of exposition, first the

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differential equations are presented:

(m1 + m2) r1 − m1r1θ21

+ m2 (L0 − r1)(

sin2(φ2) θ22 + φ2

2

)= −µd||µ2E3||vrel · er1

||vrel|| + m2g cos(φ2) ,

ddt

(m1r2

1θ1) = −µdr1||µ2E3||vrel · eθ1

||vrel|| ,

ddt

(m1(L0 − r1)2

φ2

)− m2(L0 − r1)2

θ22 sin(φ2) cos(φ2) = m2g (L0 − r1) sin (φ2) ,

ddt

(m2 (L0 − r1)2 sin2(φ2) θ2

)= 0. (5.14)

In addition, the two equations for µ1 and µ2 are

µ1 = m2g cos(φ2) − m2r1 − m2 (L0 − r1)(

sin2(φ2) θ22 + φ2

2

),

µ2 = m1g.

To find the preceding equations, we started with Lagrange’s equations:

ddt

(∂T∂qK

)− ∂T

∂q= F1 · ∂r1

∂qK+ F2 · ∂r2

∂qK(K = 1, . . . , 6) .

We then substituted for T, F1, and F2. After the partial derivatives of T were cal-culated, we imposed the two constraints and performed some rearranging. Some ofthe details of these calculations are subsequently provided:

F1 · er1 − F2 · eR2 = −µd||µ2E3||vrel · er1

||vrel|| + m2g cos (φ2) ,

F1 · r1eθ1 + F2 · 0 = −µdr1||µ2E3||vrel · eθ1

||vrel|| ,

F1 · 0 + F2 · (L0 − r1) eφ2 = m2g (L0 − r1) sin (φ2) ,

F1 · 0 + F2 · (L0 − r1) sin (φ2) eθ2 = 0,

F1 · 0 + F2 · eR2 = µ1 − m2g cos (φ2) ,

F1 · E3 + F2 · 0 = µ2 − m1g.

The Lack of Energy ConservationTo prove that the total energy of the system of particles is not conserved, we startwith the work–energy theorem T = F1 · v1 + F2 · v2 and substitute for the appliedand constraint forces:

T = (−m1gE3 + Fc1) · v1 + (−m2gE3 + Fc2) · v2.

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Making use of the potential energy function U, we can express this equation as

E = µ1 (er1 · v1 + eR2 · v2) + µ2E3 · v1 − µd||µ2E3|| vrel

||vrel|| · v1,

where the total energy E = T + U. Thus, as v1 = vrel and is normal to E3, and µ2 =m1g, we surmise that

E = µ1 (er1 · v1 + eR2 · v2) − µdm2g||v1||.

However, er1 · v1 + eR2 · v2 = r1 + R2, and this sum is zero because r1 + R2 = L0. Inconclusion,

E = −µdm2g||v1||.

Notice that E ≤ 0 as expected because of the friction force.

Conservations of Angular MomentaIf friction is absent, then we find from (5.14)2,4 that HO1 · E3 = m1r2

1θ1 and HO2 ·E3 = m2 (L0 − r1)2 sin2 (φ2) θ2 are conserved. That is, the angular momentum ofeach particle relative to O in the E3 direction is conserved.

Configuration Manifold and Its GeometryThe configuration manifold for this system is a four-dimensional subspace of E

6 pa-rameterized by r ∈ (0, L0), θ1 ∈ [0, 2π), θ2 ∈ [0, 2π), and φ2 ∈ (0, π). The kinematicalline-element ds for this manifold is given by

ds =√

2T2

m1 + m2dt,

where we find T2 from T by imposing the constraints and collecting all those termsthat are quadratic in the generalized velocities:

T2 = m1

2

(r2

1 + r21θ

21

)+ m2

2

(r2

1 + (L0 − r1)2 sin2 (φ2) θ22 + (L0 − r1)2

φ22

).

To visualize the configuration manifold, one would give a two-dimensional pictureof a plane with the coordinates r1 cos (θ1)–r1 sin (θ1). This would be supplemented bya three-dimensional image featuring a sphere of radius 1 parameterized by φ2 and θ2.


Problems involving systems of particles have played a key role in the developmentof dynamics. Specifically, mention is made here of a model for the celestial system ofthe Sun, Earth, and Moon, known as the three-body problem. In this problem, thethree bodies are modeled as particles subject to the mutual interaction that is due to

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m1

m1

m1

m2m2

m2

m3

m3

m3

(a) (b)

(c)

Figure 5.6. Representative orbits from the three-body problem: (a), (b) examples of La-grange’s equilateral triangle solutions, and (c) the figure-eight solution. For the solutionsshown in this figure m1 = m2 = m3.

a Newtonian gravitational force field. That is, the potential energy for the system is[cf. (4.9)]

Un = − Gm1m2

||r2 − r1|| − Gm3m1

||r1 − r3|| − Gm2m3

||r3 − r2|| , (5.15)

where r1, r2, and r3 are the position vectors of the particles of masses m1, m2, andm3, respectively.

Famous exact solutions to special cases of the three-body problem range fromthe equilateral triangle solution by Lagrange [117] in 1772∗ to the figure-eight so-lution that was only recently found numerically by Moore [147] and Moore andNauenberg [148] and proven to exist by Chenciner and Montgomery [37, 145] (seeFigure 5.6).† Apart from its paucity of exact solutions, the three-body problem isalso well known because of the profound analysis of this system by Henri Poincare(1854–1912) in the late 1880s (see [4, 13, 45]). His analysis is considered to be thefirst description of chaos in mathematical models for physical systems and formedone of the cornerstones for the field of chaos in dynamical systems that achievedpopular attention some 100 years later in the late 1980s.

The three-body and two-body problems are special cases of the more generaln-body problem. In celestial mechanics, the n-body problem is synonymous withmodels for our Solar System and has attracted some of the most celebrated scien-tists in history. It was also the problem that led Hamilton to discover his famous

∗ A discussion of this famous solution can be found in many texts on celestial mechanics, for example,[93, 150, 220].

† The interested reader is referred to the on-line article by Casselman [31], where simulations ofseveral three-body problems can be found.

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equations of motion in [88] and his variational principle in [89]. Unfortunately, wedo not have the opportunity to explore the three-body and n-body problems in anydetail here; the interested reader is referred to the previously cited texts.

The problems just mentioned do not feature constraints on the motions ofthe particles, and they are often formulated without using Lagrange’s equationsof motion. However, problems featuring particles connected by rigid links oftenfeature in simple models for artificial satellites orbiting a celestial body and invarious pendulum systems. For these models, Lagrange’s equations of motion areideally suited to the task of establishing a set of governing ordinary differentialequations that are free from constraint forces. In the following exercises, problemsfeaturing systems of particles of this type are emphasized.

EXERCISES

5.1. Consider the systems of particles discussed in Section 5.2. Suppose a time-dependent force P(t)E1 acted on the particle m2.∗ What are the equations of motionfor each of these systems?

5.2. Again, consider the systems of particles discussed in Section 5.2. Suppose, inaddition to the springs, there are viscous dashpots in these systems.† Then, what arethe equations of motion?

5.3. Here, we are interested in establishing a particular representation for the equa-tions governing the motion of two unconstrained particles. In a subsequent exercise,one can impose constraints to yield the equations of motion of a pendulum system.

Consider the system of particles shown in Figure 5.7. The particles are free tomove in E

3 under the influences of resultant external forces F1 and F2, respectively.

m1m2

Og

E1

E2

E3

Figure 5.7. A system of two particles.

(a) To establish the equations of motion for the single particle, we use acylindrical polar coordinate system {r1, θ1, z1} for the particle of mass m1.For the second particle, it is convenient to describe its motion with theassistance of the relative position vector r21 = r2 − r1. We describe this

∗ This force is not conservative.† The forces from these dashpots are not conservative.

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154 Exercise 5.3

vector by using a spherical polar coordinate system {R2, φ2, θ2}. Show thatthe position vector of the single particle is

r = r1 cos(θ1)e1 + r1 sin(θ1)e2 + z1e3

+ (r1 cos(θ1) + R2 sin(φ2) cos(θ2))e4

+ (r1 sin(θ1) + R2 sin(φ2) sin(θ2))e5 + (z1 + R2 cos(φ2))e6.

(b) Using r and the curvilinear coordinate system it induces on E6,

q1 = r1, q2 = θ1, q3 = z1, q4 = R2, q5 = φ2, q6 = θ2,

show that the six covariant basis vectors aJ = ∂r∂qJ are

a1 = ∂r∂r1

= cos(θ1)e1 + sin(θ1)e2 + cos(θ1)e4 + sin(θ1)e5,

a2 = ∂r∂θ1

= −r1 sin(θ1)e1 + r1 cos(θ1)e2 − r1 sin(θ1)e4 + r1 cos(θ1)e5,

a3 = ∂r∂z1

= e3 + e6,

a4 = ∂r∂R2

= sin(φ2) cos(θ2)e4 + sin(φ2) sin(θ2)e5 + cos(φ2)e6,

a5 = ∂r∂φ2

= R2 cos(φ2) cos(θ2)e4 + R2 cos(φ2) sin(θ2)e5 − R2 sin(φ2)e6,

a6 = ∂r∂θ2

= −R2 sin(φ2) sin(θ2)e4 + R2 sin(φ2) cos(θ2)e5.

(c) Show that the six contravariant basis vectors have the following represen-tations:

a1 = cos(θ1)e1 + sin(θ1)e2,

a2 = − sin(θ1)r1

e1 + cos(θ1)r1

e2,

a3 = e3,

a4 = sin(φ2)(cos(θ2)(e4 − e1) + sin(θ2)(e5 − e2)) + cos(φ2)(e6 − e3),

a5 = cos(φ2)R2

(cos(θ2)(e4 − e1) + sin(θ2)(e5 − e2)

)− sin(φ2)R2

(e6 − e3),

a6 = − sin(θ2)R2 sin(φ2)

(e4 − e1) + cos(θ2)R2 sin(φ2)

(e5 − e2).

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(d) Show that the kinetic energy T of the particle of mass m = m1 + m2 is

T = m1 + m2

2

(r2

1 + r21θ

21 + z2

1

)+ m2

2

(R2

2 + R22φ

22 + R2

2 sin2(φ2)θ22

)+ m2 cos(φ2)

(R2z1 + φ2r1R2 cos(θ21) + φ2θ1r1R2 sin(θ21)

)− m2 sin(φ2)

(R2φ2z1 − r1R2θ1θ2 cos(θ21) − r1R2 cos(θ21)

)− m2 sin(φ2)

(−r1R2θ1 sin(θ21) + r1θ2R2 sin(θ21)),

where we have used the abbreviation θ21 = θ2 − θ1. This expression for thekinetic energy follows from the definition

T = m2

v · v = m1

2v1 · v1 + m2

2v2 · v2.

(e) If the forces acting on the particles are F1 = −m1gE3 and F2 = −m2gE3,then what are the force � and potential energy U associated with this force?

(f) What are the six Lagrange’s equations governing the motion of the particleof mass m?∗

5.4. As shown in Figure 5.8, two particles of mass m1 and m2 are connected by arigid massless rod of length L2. The rod is connected to m1 by a ball-and-socketjoint. In addition, the particle of mass m1 is connected by a rigid massless rod oflength L1 to a fixed point O. The connection between the rod and the point O isthrough a pin joint and is such that the motion of m1 is in the E1 − E2 plane.

m1

m2

O

g

E1

E2

E3

Figure 5.8. A planar double pendulum.

(a) What are the three constraints on the motion of the particle of mass m?

(b) Using Lagrange’s prescription, what is the constraint force �c acting on theparticle of mass m? You should also, if possible, verify that the componentsof this force are physically realistic.

∗ You should refrain if possible from expanding the time derivative here – it will entail a considerableamount of algebra.

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156 Exercise 5.4–5.5

(c) Starting from the final results of Exercise 5.3, establish Lagrange’sequations of motion for the pendulum system. In your solution, clearlydistinguish the equations governing the motion of the particle and theequations giving the components of �c.

(d) Let us now establish some of the equations of (c) by using an equivalentapproach. Impose the constraints on T to determine the constrained kineticenergy T. In addition, determine the constrained potential energy U. Verifythat the following equations correspond to those you obtained from (c)∗:

ddt

(∂T

∂θ1

)− ∂T

∂θ1= � · a2 = − ∂U

∂θ1,

ddt

(∂T

∂φ2

)− ∂T

∂φ2= � · a5 = − ∂U

∂φ2,

ddt

(∂T

∂θ2

)− ∂T

∂θ2= � · a6 = − ∂U

∂θ2.

(e) Suppose a nonintegrable constraint is imposed on the pendulum systemdiscussed in (d):

f1 · v1 + f2 · v2 + e = 0.

Show that this constraint can be expressed as

f · v + e = 0.

In addition, what are the equations governing the motion of the non-integrably constrained system? Illustrate your solution with an non-integrable constraint of your choice.

5.5. As shown in Figure 5.9, a model for an artificial satellite consists of twoparticles of mass m1 and m2 connected by a rigid massless rod of length L0. A thirdparticle of mass m3 is assumed to be stationary at the fixed point O. In addition tothe constraint force in the rod, the system is subject to conservative forces whosepotential energy function is given by (5.15).

(a) What are the four constraints on the motion of the system of particles?

(b) Using Lagrange’s prescription, what are the constraint forces acting on theparticles of mass m1 and m2? You should also, if possible, verify that thecomponents of these forces are physically realistic.

(c) Using a set of Cartesian coordinates to describe the location of the centerof mass C of the satellite of mass m1 + m2 and a set of spherical polarcoordinates to parameterize the position of m2 relative to C, establish anexpression for the kinetic energy of the system.

(d) Establish the equations of motion for the system.

∗ It is crucial to note that ∂T∂r1

= ∂T∂R2

= ∂T∂z1

= ∂T∂r1

= ∂T∂R2

= ∂T∂z1

= 0.

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m1

m2

m3

O

E1

E2

E3

Figure 5.9. Schematic of a model for a satelliteorbiting a fixed body of mass m3.

(e) Show that the solutions to the equations of motion for the system conservethe total energy of the system and the angular momentum of the systemrelative to O.

(f) Show that it is possible for C to execute a steady circular motion about O.What are the possible orientations of the rigid massless rod of length L0

during such motions?

5.6. As shown in Figure 5.10, a particle of mass m1 is connected by a linear springof stiffness K1 and unstretched length L0 to a fixed point O. A second particle ofmass m2 is attached by a rod of length L2 to the particle of mass m1 with a pinjoint. For this system, which is a variation on the classical system of a planar doublependulum, we assume that the motions of m1 and m2 are constrained to move onthe E1 − E2 plane.

m1

m2

O E1

E2

g

linear spring

rigid massless rod

Figure 5.10. A system of two parti-cles connected by a rigid masslessrod of length L2.

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158 Exercise 5.6

To describe the kinematics of this system, a cylindrical polar coordinate system{r1, θ1, z1} is used to parameterize the motion of the particle of mass m1 and anothercylindrical polar coordinate system {r2, θ2, z2} is used to parameterize the motion ofthe particle of mass m2 relative to m1:

r1 = r1er1 + z1E3, r2 = r1 + r2er2 + z2E3.

We define six coordinates as follows:

q1 = θ1, q2 = θ2, q3 = r1, q4 = r2, q5 = z1, q6 = z2. (5.16)

(a) With the help of (5.16), what are the 12 vectors ∂r1∂qK and ∂r2

∂qK ? Here,K = 1, . . . , 6.

(b) What are the three constraints on the motion of the system of particles?Argue that the constraint forces Fc1 and Fc2 acting on the individualparticles have the prescriptions

Fc1 = µ1er2 + µ2E3, Fc2 = −µ1er2 + µ3E3. (5.17)

Compute the following six components:

�cK = Fc1 · ∂r1

∂qK+ Fc2 · ∂r2

∂qK.

Comment on the values of the first three components.

(c) In terms of the coordinates q1, . . . , q3 and their time derivatives, what arethe kinetic energy T and potential energy U of the constrained system ofparticles?

(d) What are Lagrange’s equations of motion for the generalized coordinatesof this system of particles?

(e) Suppose a nonintegrable constraint∗

r1θ1 + L2θ2 = 0 (5.18)

is imposed on the system of particles. After expressing this constraint inthe form f1 · v1 + f2 · v2 = 0, argue that

Fc1 = µ1er2 + µ2E3 + µ4 (eθ1 − eθ2 ) , Fc2 = −µ1er2 + µ3E3 + µ4eθ2 .

With the help of your results from (d), determine the equations of motionfor the system of particles.

(f) Starting from the work–energy theorem T = F1 · v1 + F2 · v2, show that thetotal energy E is conserved.

(g) Suppose that the spring is replaced with a rigid rod of length L1 andnonintegrable constraint (5.18) is removed. In this case, which is the classic

∗ Referring to the discussion of constraint (1.16) in Chapter 1, this constraint is arguably the simplestnonintegrable constraint that we can impose on this system.

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planar double pendulum, show that the equations governing the motion ofthe system are

(1 + α)d2θ1

dτ2+ αβ cos (θ2 − θ1)

d2θ2

dτ2− αβ

(dθ2

dτ

)2

sin (θ2 − θ1)

= − (1 + α) cos (θ1) ,

d2θ2

dτ2+ 1

βcos (θ2 − θ1)

d2θ1

dτ2+ 1

β

(dθ1

dτ

)2

sin (θ2 − θ1)

= − 1β

cos (θ2) . (5.19)

In writing (5.19), we have used the following dimensionless parameters andtime variable:

α = m2

m1, β = L2

L1, τ =

√g

L1t.

The configuration manifold M for this system is a torus. What is thekinematical line-element for M?

(h) Numerically integrate (5.19) for a variety of initial conditions and illustrateyour solutions on the configuration manifold for the planar double pendu-lum. You should verify that your solutions conserve the total energy of thesystem.

5.7. This problem is adapted from Section 156 of Whittaker [228] and the introduc-tion to [30]. Consider a system of N particles, and, following Lecture 4 from Jacobi[102], define the following function:

J = 12

N∑k=1

mk ||rk||2 .

The quantity 2J is often known as the moment of inertia of the system of particles.

(a) Assuming that the center of mass C of the system is stationary and locatedat the origin, show that J has the equivalent representation:

J = 14

N∑k=1

N∑j=1

mkmj

M

∣∣∣∣rk − rj∣∣∣∣2 , (5.20)

where M = m1 + · · · + mN.

(b) As in (a), assuming that the center of mass C of the system is station-ary, show that the kinetic energy of the system of particles has therepresentation

T = 14

N∑k=1

N∑j=1

mkmj

M

∣∣∣∣vk − vj∣∣∣∣2 . (5.21)

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160 Exercise 5.7

(c) Now suppose that the system of particles is in motion subject to aconservative Newtonian force field:

Un = −12

N∑j=1

N∑k=1,k�=j

Gmkmj∣∣∣∣rk − rj∣∣∣∣ . (5.22)

The presence of the 12 in the expression for Un should be noted: It is needed

to ensure that the summations on k and j yield the correct expression forUn. With the help of (5.20)–(5.22) and balances of linear momenta for eachparticle, establish Jacobi’s equation:

J = 2T + Un. (5.23)

This equation is also known as the Lagrange–Jacobi equation (see, forexample, [220]).

(d) For the orbits of the three-body problem shown in Figure 5.6(b), show thatT = − 1

2 Un.

(e) Show that J is a measure of the distance squared from the origin of the con-figuration space E

3N for the representative particle discussed in Section 4.7.

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PART THREE

DYNAMICS OF A SINGLE RIGIDBODY

161

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162

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6 Rotation Tensors

6.1 Introduction

One of the key features of the rigid body dynamics problems that we will shortlyexamine is the presence of a variable axis of rotation. This is one of the reasonsfor the richness of phenomena in rigid body dynamics. It is also a reason why thissubject is intimidating. To quote the mechanician Louis Poinsot (1777–1859), from[172], “. . . if we have to consider the motion of a body of sensible shape, it must beallowed that the idea which we form of it is very obscure.” In this chapter, severalrepresentations of rotations are discussed that will enable us to establish both a clearpicture of rigid body motions and straightforward proofs of several major results. Tothis end, many results on two key kinematical quantities for rigid bodies, rotationtensors and their associated angular velocity vectors, are discussed in considerabledetail.

The subject of rotations in rigid body dynamics has a wonderful history, a widerange of interesting results, and an impressive list of contributors. Here, however,space limits the presentation of only the handful of results that are most relevant toour purposes. From a historical perspective, much of what is presented was estab-lished by Leonhard Euler (1707–1783) in his great works on rigid body dynamicsthat started to appear in the 1750s. The foundations Euler established were builtupon by such notables as Cayley, Gauss, Hamilton, and Rodrigues in the early partof the 19th century. Despite the wealth of their results, the subject of rotationsremains an active area of discovery (and some unfortunate reinvention) to thisday.

Our exposition makes extensive use of tensors, and the background requiredfor these quantities is provided in the Appendix. Following related developmentsin continuum mechanics, tensors are an invaluable tool for providing concise expla-nations of many important results. They are not universally invoked in dynamicstexts, and the interested reader is invited to compare our treatment with those inthe textbooks cited in the bibliography.

We start this chapter with a discussion of a rotation in which the axis of rotationis fixed, and this example is used to introduce several of the key concepts in this

163

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164 Rotation Tensors

chapter. The example also serves to illuminate many of the generalizations that arerequired for characterizing more elaborate rotations. To this end, our expositionthen turns to proper-orthogonal tensors in Section 6.3, and many of their interestingfeatures are shown there. Then Euler’s representation of a rotation tensor ispresented, and we examine his important theorem that any proper-orthogonaltensor is a rotation tensor. This result has many consequences for rigid bodydynamics. One of the intriguing aspects of rotation tensors is the vast range ofrepresentations. Here the Euler angle representation is emphasized, but sufficientmaterial is present for the interested reader to examine many of the other represen-tations such as the Euler–Rodrigues symmetric parameters (often synonymous withquaternions).

Our principal reference for this chapter is the authoritative review by Shuster[196]. This material is supplemented with the elegant relative angular velocity vectorthat was proposed by Casey and Lam [29], the dual Euler basis [160], and a proofof Euler’s theorem by Guo [83]. In the interests of presenting a unified treatment,our notation differs from these references in several places, but the differences areeasily deciphered once the material in this chapter has been comprehended.

6.2 The Simplest Rotation

To motivate many of our later developments, we start with the simplest case of arotation about a fixed axis p3 through an angle θ = θ(t). This example should be fa-miliar to you from many different venues. To describe this rotation, we considerthe action of this rotation on this set of orthonormal right-handed basis vectors{p1, p2, p3

}. As shown in Figure 6.1, we suppose that these vectors are transformed

to the set {t1, t2, t3} by the rotation.Using a matrix notation, we can represent the transformation from the set of

basis vectors{p1, p2, p3

}to {t1, t2, t3} as

⎡⎢⎣

t1

t2

t3

⎤⎥⎦ = R

⎡⎢⎣

p1

p2

p3

⎤⎥⎦ , (6.1)

θ

θ

θ

p1

p2t1

t2

p3 = t3

Figure 6.1. The transformations of various basis vectors in-duced by a rotation about p3 through an angle θ.

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6.2 The Simplest Rotation 165

where

R =

⎡⎢⎣

cos(θ) sin(θ) 0


0 0 1

⎤⎥⎦ .

It is easy to see that the matrix R in this equation has a determinant of +1 and thatits inverse is its transpose: R−1 = RT. That is, the matrix R is proper-orthogonal.

By differentiating (6.1) with respect to time, we find that⎡⎢⎣

t1

t2

t3

⎤⎥⎦ = θ

⎡⎢⎣


− cos(θ) − sin(θ) 0

0 0 0

⎤⎥⎦⎡⎢⎣

p1

p2

p3

⎤⎥⎦ .

Using the result R−1 = RT, we can easily replace pi in this equation with ti:⎡⎢⎣

t1

t2

t3

⎤⎥⎦ = θ

⎡⎢⎣


− cos(θ) − sin(θ) 0

0 0 0

⎤⎥⎦⎡⎢⎣

cos(θ) − sin(θ) 0

sin(θ) cos(θ) 0

0 0 1

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦

= θ

⎡⎢⎣

0 1 0

−1 0 0

0 0 0

⎤⎥⎦

︸︷︷︸RRT

⎡⎢⎣

t1

t2

t3

⎤⎥⎦ . (6.2)

Notice that this is equivalent to the familiar results t1 = θ t2 and t2 = −θ t1. It shouldalso be clear from (6.2) that RRT is a skew-symmetric matrix. A vector θ p3 = θt3 canbe introduced that has the useful property that

tk = θ p3 × tk (k = 1, 2, 3) . (6.3)

You should notice how the vector θ p3 can be inferred from the components of RRT.It is convenient to use a tensor notation to describe the rotation we have been

discussing. In particular, we can write (6.1) in the form∗

tk = Rpk (k = 1, 2, 3) , (6.4)

where R is the tensor

R = cos(θ) (p1 ⊗ p1 + p2 ⊗ p2) − sin(θ) (p1 ⊗ p2 − p2 ⊗ p1) + p3 ⊗ p3.

It is left as an exercise to verify that this representation of the rotation is equivalentto (6.1).† Indeed, because

I − p3 ⊗ p3 = p1 ⊗ p1 + p2 ⊗ p2, εp3 = p1 ⊗ p2 − p2 ⊗ p1,

∗ Background on tensor notation can be found the Appendix.† The definition of the tensor product needed to establish this equivalence can be found in (A.1) in

the Appendix. The alternating tensor ε is defined in Section A.7 [see, in particular, (A.11).]

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we can express the tensor R entirely in terms of the axis of rotation p3 and the angleof rotation θ:

R = cos(θ) (I − p3 ⊗ p3) − sin(θ)εp3 + p3 ⊗ p3. (6.5)

As we shall see later, this representation naturally leads to the general form of atensor that represents a rotation about an arbitrary axis through an arbitrary angleof rotation. It is left as another exercise to show that RT = R−1 and that det(R) = 1.

Differentiating (6.4), we find that

tk = Rpk + R��0

pk

= R RTtk︸︷︷︸=pk

= (RRT) tk.

Some straightforward computations are used to show that

R = −θ sin(θ) (p1 ⊗ p1 + p2 ⊗ p2) − θ cos(θ) (p1 ⊗ p2 − p2 ⊗ p1) ,

RRT = θ (−p1 ⊗ p2 + p2 ⊗ p1) .

With the help of these results, we conclude that

tk = (RRT) tk

= θ p3 × tk (k = 1, 2, 3) .

As expected, this result is in agreement with (6.3). You might have already noticedthat θ p3 = θ t3 is the axial vector of RRT.

We have demonstrated how several results for a familiar rotation can be ex-pressed by using a tensor notation. This notation will prove to be very usefulnext when we wish to examine more complex rotations. To this end, we need toanswer several questions. The first among them is this: What is the representa-tion for a rotation about an arbitrary axis? Once this is known, the question ofwhether its time derivative leads to a vector such as θp3 then presents itself. Theanswers to these questions were first formulated by Euler in the 1750s. However,numerous alternative representations for his solutions have appeared since then,and we shall leverage several of these representations in the remainder of thischapter.

6.3 Proper-Orthogonal Tensors

To proceed with our treatment of rotations, we appear to regress somewhat anddiscuss proper-orthogonal tensors. This discussion will lay the foundations for thepossibility of three-parameter representations of rotations and subsequently the ex-istence of angular velocity vectors. It is also a starting point for several investigationson experimental measurements of rotations. We recall that a proper-orthogonal

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6.3 Proper-Orthogonal Tensors 167

second-order tensor R is a tensor that has a unit determinant and whose inverseis its transpose:

RTR = RRT = I, det(R) = 1. (6.6)

The first of these equations implies that there are six restrictions on the nine com-ponents of R. Consequently, only three components of R are independent. In otherwords, any proper-orthogonal tensor can be parameterized by three independentparameters.

The tensors of interest here are second-order tensors, any one of which has therepresentation

R =3∑

i=1

3∑k=1

Rikpi ⊗ pk.

Let us now consider the transformation induced by R on the basis vectors p1, p2,and p3. We define

t1 = Rp1 =3∑

i=1

Ri1pi,

t2 = Rp2 =3∑

i=1

Ri2pi,

t3 = Rp3 =3∑

i=1

Ri3pi.

You should notice that, by using the vectors ti, R has the representation

R = t1 ⊗ p1 + t2 ⊗ p2 + t3 ⊗ p3.

We now wish to show that {t1, t2, t3} is a right-handed orthonormal basis. First, letus verify the orthonormality:

ti · tk = Rpi · Rpk = RTRpi · pk

= pi · pk = δik.

Hence the vectors ti are orthonormal. To establish right-handedness, we use thedefinition of the determinant [see (A.6)]:

[t1, t2, t3] = [Rp1, Rp2, Rp3]

= det(R)[p1, p2, p3]

= (1)(1) = 1.

Hence {t1, t2, t3} is a right-handed orthonormal basis.∗

∗ It is left as an exercise to verify that this result holds for the example of a simple rotation presentedin (6.1).

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A proper-orthogonal tensor also has a rather unusual representation. To arriveat it, we embark on a series of manipulations:

R = RRRT = R

(3∑

i=1

3∑k=1

Rikpi ⊗ pk

)RT

=3∑

i=1

3∑k=1

RikR (pi ⊗ pk) RT

=3∑

i=1

3∑k=1

Rik (Rpi ⊗ Rpk)

=3∑

i=1

3∑k=1

Rikti ⊗ tk.

In summary, we have the following representations for R:

R =3∑

i=1

3∑k=1

Rikpi ⊗ pk =3∑

i=1

3∑k=1

Rikti ⊗ tk =3∑

i=1

ti ⊗ pi.

Notice that the components of R for the first two representations are identical, andthe handedness of {p1, p2, p3} is transferred without change by R to {t1, t2, t3}.

We observed that the components Rik of R are equal to tk · pi. As this productis equal to the cosine of the angle between tk and pi, each Rik is often referred to asa direction cosine. Consequently the matrix [Rik] is often known as the direction co-sine matrix. Clearly, the nine angles whose cosines are tk · pi are not all independent,for if they were then [Rik] would have nine independent components and this wouldcontradict the requirement RTR = I. Indeed, as we shall shortly see, it is possible toarrive at three independent angles to parameterize R, but these angles are not alleasily related to the angles between pi and tk.

6.4 Derivatives of a Proper-Orthogonal Tensor

Here we consider a proper-orthogonal tensor R that is a function of time: R = R(t).Consider the derivative of RRT:

ddt

(RRT) = RRT + RRT.

However, I = O, so the right-hand side of the preceding equation is zero. Hence,

RRT = −RRT = − (RRT)T .

In other words, RRT is a skew-symmetric second-order tensor. We define

�R = RRT,

in part because this tensor appears in numerous places later on. The tensor �R isknown as the angular velocity tensor (of R).

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6.4 Derivatives of a Proper-Orthogonal Tensor 169

The skew-symmetry of RRT allows us to define the angular velocity vector ωR:

ωR = −12ε[RRT] .

As mentioned in the Appendix, ωR × a = (RRT)a for all vectors a.∗ The most com-mon example of the calculation of an axial vector arises when we consider the mo-tion of a rigid body rotating about the E3 direction. In this case, we will see laterthat the skew-symmetric tensor

�R = � (E2 ⊗ E1 − E1 ⊗ E2) .

Consequently, we can compute that

ε[�R] = −2�E3.

We conclude that the axial vector of �R is the angular velocity vector �E3. It alsouseful to check that

(� (E2 ⊗ E1 − E1 ⊗ E2)) a = �E3 × a

for all vectors a.In a similar manner, we can also show that RTR is a skew-symmetric tensor and

define an angular velocity tensor �0R and another angular vector ω0R :

�0R = RTR, ω0R = −12ε[RTR]. (6.7)

At a later stage, the reader should be able to show that Rω0R = ω and R�0R RT =�R. A key to establishing one of these results is the identity, which holds for allorthogonal Q,

ε[QBQT] = det (Q) Q (ε [B]) .

Notice how this identity simplifies when Q is a proper-orthogonal tensor.

Corotational DerivativesRecall that, for a proper-orthogonal tensor R, we have the representation

R =3∑

i=1

ti ⊗ pi.

Now suppose that pi = 0. Then,

�R = RRT =(

3∑i=1

ti ⊗ pi +3∑

i=1

ti ⊗��0

pi

)RT =

(3∑

i=1

ti ⊗ pi

)RT

=3∑

i=1

ti ⊗ ti.

∗ In Section A.7 of the Appendix, several examples featuring the calculation of the axial vector ωR ofa given skew-symmetric tensor �R are presented.

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If we now consider �Rtk, we find a familiar result:

ti = �Rti = ωR × ti.

It is left as an exercise to show the less familiar result

ti = R(ω0R × pi).

It is important to note that if ti are defined by use of a proper-orthogonal tensorR and a fixed basis pi, then their time derivatives can be expressed in terms of theangular velocity vector of the rotation tensor and the basis vectors ti.

Given any second-order tensor A and any vector a, we have the following rep-resentations:

a =3∑

i=1

aiti, A =3∑

i=1

3∑k=1

Aikti ⊗ tk.

If we assume that a is a function of time, then

a =3∑

i=1

aiti + ai ti

=3∑

i=1

aiti + ai(ωR × ti)

=3∑

i=1

aiti + ωR × a.

Similarly, if we assume that A is a function of time, then

A =3∑

i=1

3∑k=1

Aikti ⊗ tk +3∑

i=1

3∑k=1

Aikti ⊗ tk +3∑

i=1

3∑k=1

Aikti ⊗ tk

=3∑

i=1

3∑k=1

Aikti ⊗ tk +3∑

i=1

3∑k=1

Aik(�Rti) ⊗ tk +3∑

i=1

3∑k=1

Aikti ⊗ (�Rtk)

=3∑

i=1

3∑k=1

Aikti ⊗ tk + �RA − A�R.

The derivativesoA and

oa are known as the corotational derivatives (with respect to

R) of A and a, respectively. They are the respective derivatives of A and a if thevectors ti are constant:

oA=

3∑i=1

3∑k=1

Aikti ⊗ tk,oa=

3∑i=1

aiti.

Using the recently established expression foroa, we observe that

a = oa + ωR × a.

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6.5 Euler’s Representation of a Rotation Tensor 171

Similarly, for any second-order tensor A, we have

A = oA + �RA − A�R.

The terms involving the angular velocity vectors and tensors in these expressionsare the result of the orthonormal vectors ti changing with time.

We shall subsequently use several distinct corotational derivatives, and we wishto do so without introducing a laborious notation. To this end, it will be explicitlystated which rotation tensor the corotational derivative pertains to in situations inwhich confusion is possible.

6.5 Euler’s Representation of a Rotation Tensor

Leonhard Euler defined a rotation by using an angle of rotation φ and an axis ofrotation r.∗ Using notation introduced over a century later by Gibbs,† Euler’s rep-resentation for a tensor that produces this rotation can be written as

R = L(φ, r) = cos(φ)(I − r ⊗ r) − sin(φ)(εr) + r ⊗ r, (6.8)

where r is a unit vector and φ is a counterclockwise angle of rotation. We referto (6.8) as Euler’s representation of a rotation tensor and use the function L toprescribe the rotation tensor associated with an axis and angle of rotation. The threeindependent parameters of R are the angle of rotation and the two independentcomponents of the unit vector r. The reason r is known as the axis of rotation lies inthe fact that it is invariant under the action of R: Rr = r. We shall shortly examinethe role of φ.

It is interesting to examine some of the features of representation (6.8). To thisend, we define an orthonormal basis {p1, p2, p3} with p3 = r. Using this basis, wehave

I = p1 ⊗ p1 + p2 ⊗ p2 + p3 ⊗ p3,

r ⊗ r = p3 ⊗ p3,

I − r ⊗ r = p1 ⊗ p1 + p2 ⊗ p2,

−(εr) = p2 ⊗ p1 − p1 ⊗ p2.

Consequently we can write

R = L(φ, r = p3) = cos(φ)(p1 ⊗ p1 + p2 ⊗ p2)

+ sin(φ)(p2 ⊗ p1 − p1 ⊗ p2) + p3 ⊗ p3.

∗ This representation can be seen in Section 49 in one of Euler’s great papers on rigid body dynamicsfrom 1775 [56]. There, he provides expressions for the components of the tensor R in terms of anangle of rotation φ and the direction cosines p, q, r of the axis of rotation. In our notation, r =pE1 + qE2 + rE3.

† See Section 129 of [231].

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Using the identity (a ⊗ b)T = b ⊗ a, we find

RT = L (φ, r = p3) = cos(φ)(p1 ⊗ p1 + p2 ⊗ p2)

+ sin(φ)(p1 ⊗ p2 − p2 ⊗ p1) + p3 ⊗ p3.

It is now easy to check that RRT is the identity tensor, and the details are left to thereader. Next, we examine the determinant of R:

det(R) = det

⎡⎢⎣

cos(φ) − sin(φ) 0

sin(φ) cos(φ) 0

0 0 1

⎤⎥⎦

= 1.

In conclusion, RTR = I and det(R) = 1. Thus R is a proper-orthogonal tensor. Weshall examine the converse of this statement shortly.

Composition of RotationsSuppose we have a rotation through an angle φ1 about an axis r1 and we follow thisby a rotation through an angle φ2 about an axis r2; then the tensor

Rc = L (φ2, r2) L (φ1, r1)

represents the composite rotation. It is left as an exercise to show that Rc is a proper-orthogonal tensor. As can be shown later by Euler’s theorem, this implies that Rc isa rotation tensor. Thus the composition of two rotations is also a rotation.

As tensor multiplication is noncommutative, the order in which we consider thecomposition is important. In general,

L (φ2, r2) L (φ1, r1) �= L (φ1, r1) L (φ2, r2) .

The exceptions arise when r1 ‖ r2 or one of the angles of rotation is 0. The fact thatrotations are sequence dependent will play a major role later on in the definition ofEuler angle sequences.

Euler’s FormulaWe now examine the action of R on a vector a. As shown in Figure 6.2, the part ofa that is parallel to r is unaltered by the transformation, whereas the part of a thatis perpendicular to r is rotated through an angle φ counterclockwise about r. To seethis, it is convenient to decompose a:

a = a⊥ + a‖,

where

a⊥ = a − (a · r)r = (I − r ⊗ r)a, a‖ = (a · r)r = (r ⊗ r)a.

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Ra

ra

φa‖ a⊥

Ra⊥

Figure 6.2. The transformation of a vector a by the rotation tensorR = L (φ, r).

With the assistance of this decomposition, we now compute that

Ra = L(φ, r)a

= (cos(φ)(I − r ⊗ r) − sin(φ)(εr) + r ⊗ r) a

= cos(φ)(I − r ⊗ r)a − sin(φ)(εr)a + (r ⊗ r)a

= cos(φ)(I − r ⊗ r)a + sin(φ)r × a + (r · a)r,

= cos(φ)a⊥ + sin(φ)r × a⊥ + a‖. (6.9)

Noting that cos(φ)a⊥ + sin(φ)r × a⊥ is a rotation of a⊥ about r, we obtain the desiredconclusion. The final expression for Ra in (6.9) is known as Euler’s formula.

Remarks on Euler’s RepresentationEuler’s representation (6.8) is unusual in several respects. First, you should noticethat

L(φ, r) = L(−φ,−r).

This implies that there are two different representations for the same rotation ten-sor. Second, as R is a rotation tensor, R−1 = RT, and this leads to two easy repre-sentations for these tensors:

R−1 = RT = L(−φ, r) = L(φ,−r).

In words, the inverse can be calculated by either reversing the angle of rotation orinverting the axis of rotation. Another peculiarity is that L(φ = 0, r) = I holds forall vectors r. Finally, we note that Euler’s representation is used to define otherrepresentations of rotation tensors in this book.

6.5.1 Calculating the Axis and Angle of Rotation

Given a rotation tensor R, it is a standard exercise to calculate the axis of rotationr and the angle of rotation θ associated with this tensor. First, one is normally

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presented with the matrix components of R with respect to a basis, say {p1, p2, p3}:

R =3∑

i=1

3∑k=1

Rikpi ⊗ pk.

If we compare this representation with (6.8), we find that

Rik = ((cos(θ)(I − r ⊗ r) − sin(θ)(εr) + r ⊗ r) pk) · pi

= cos(θ) (δik − rirk) + rirk −3∑

j=1

sin(θ)εjikrj,

where

r =3∑

i=1

ripi.

Expanding the expressions for the components of R, we find the matrix represen-tation⎡⎢⎢⎣

R11 R12 R13

R21 R22 R23

R31 R32 R33

⎤⎥⎥⎦ = cos (θ)

⎡⎢⎢⎣

1 − r21 −r1r2 −r1r3

−r1r2 1 − r22 −r2r3

−r1r3 −r2r3 1 − r23

⎤⎥⎥⎦+

⎡⎢⎢⎣

r21 r1r2 r1r3

r1r2 r22 r2r3

r1r3 r2r3 r23

⎤⎥⎥⎦

+ sin (θ)

⎡⎢⎣

0 −r3 r2

r3 0 −r1

−r2 r1 0

⎤⎥⎦ . (6.10)

Notice that this matrix can be expressed as the sum of a symmetric and skew-symmetric matrix. The skew-symmetric part is the only part that changes when wetranspose the tensor.

To determine the angle of rotation, we calculate the trace of R:

cos(θ) = 12

(tr(R) − 1) = 12

(R11 + R22 + R33 − 1) . (6.11)

Looking at the skew-symmetric part of R, we find that⎡⎢⎣

r1

r2

r3

⎤⎥⎦ = − 1

2 sin(θ)

⎡⎢⎣

R23 − R32

R31 − R13

R12 − R21

⎤⎥⎦ . (6.12)

To verify the calculated value of r, we could examine the eigenvectors of R: Theeigenvector corresponding to the unit eigenvalue should be parallel to the axis ofrotation r.

It is interesting to notice that, if R =∑3i=1 ti ⊗ pi, then r has the same compo-

nents with respect to both sets of basis vectors:

r =3∑

i=1

ripi =3∑

i=1

riti.

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The proof of this result is based on the observation that R has the same componentswith respect to the bases pi ⊗ pk and ti ⊗ tk. That is, R =∑3

i=1

∑3k=1 Rikpi ⊗ pk =∑3

i=1

∑3k=1 Rikti ⊗ tk.

AN EXAMPLE. As an example, suppose that the components of a rotation tensor Rare ⎡

⎢⎢⎣R11 R12 R13

R21 R22 R23

R31 R32 R33

⎤⎥⎥⎦ =

⎡⎢⎣

0.835959 −0.283542 −0.469869

0.271321 0.957764 −0.0952472

0.47703 −0.0478627 0.877583

⎤⎥⎦ .

We can compute the angle of rotation θ of this tensor by using (6.11):

cos(θ) = 12

(0.835959 + 0.877583 + 0.957764 − 1) .

That is, θ = 33.3161◦. We can calculate the axis of rotation r by using (6.12):

r = 0.043135p1 − 0.861981p2 + 0.505103p3

= 0.043135t1 − 0.861981t2 + 0.505103t3.

In writing this result, we are emphasizing that the components of r in the basis{p1, p2, p3

}and {t1, t2, t3} are identical.

The Associated Angular Velocity VectorGiven Euler’s representation (6.8) we assume that R = R(t). This implies, in gen-eral, that φ = φ(t) and r = r(t). We now seek to establish representations for ωR.

As a preliminary result, we note that, because r is a unit vector, r · r = 0. Inaddition, ε = 0 and I = 0. Now, starting from (6.8),

R = L(φ, r) = cos(φ)(I − r ⊗ r) − sin(φ)(εr) + r ⊗ r,

we differentiate to find

R = −φ sin(φ)(I − r ⊗ r) − φ cos(φ)(εr) + (1 − cos(φ))(r ⊗ r + r ⊗ r) − sin(φ)(εr).

To proceed further, we define a right-handed orthonormal basis {t1, t2, t3}, such thatt3 = r at a given instant in time. Then, at the same instant in time,

εr = (t1 ⊗ t2 − t2 ⊗ t1),

r = at1 + bt2,

r × r = at2 − bt1,

εr = a(t2 ⊗ t3 − t3 ⊗ t2) + b(t3 ⊗ t1 − t1 ⊗ t3). (6.13)

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In these expressions, a and b are the scalar components of r. Using (6.13) along withsome manipulations, we find that

RRT = −φ(t1 ⊗ t2 − t2 ⊗ t1) + (a(1 − cos(φ))

+ bsin(φ))(t1 ⊗ t3 − t3 ⊗ t1) + (−b(1 − cos(φ))

+ a sin(φ))(t3 ⊗ t2 − t2 ⊗ t3)

= −φεt3 − (a(1 − cos(φ)) + bsin(φ))εt2 − (−b(1 − cos(φ))

+a sin(φ))εt1.

With the assistance of (6.13), we can now write the desired final result:

�R = RRT = −φεr − (1 − cos(φ))ε(r × r) − sin(φ)εr.

The associated angular velocity vector is

ωR = φr + sin(φ)r + (1 − cos(φ))r × r. (6.14)

If r is constant, then the expression for the angular velocity vector simplifies con-siderably. It is interesting to note that a constant ωR does not necessarily imply aconstant r. However, as shown in [161], it is usually possible to choose a fixed ba-sis{p1, p2, p3

}where R =∑3

k=1 tk ⊗ pk so that a constant angular velocity implies aconstant φ and a constant r.

6.6 Euler’s Theorem: Rotation Tensors and Proper-Orthogonal Tensors

A tensor Q is proper-orthogonal if, and only if,

QTQ = I, det(Q) = 1.

From our discussion of rotation tensors, it follows that a rotation tensor is a proper-orthogonal tensor. However, is the converse true? In other words, is every proper-orthogonal tensor a rotation tensor? The affirmative answer to this question isknown as Euler’s theorem.

Our approach to presenting the proof of Euler’s theorem is adapted from awonderful paper by Zhong-heng Guo [83]. One of the key results needed is to re-call that we can define two of the invariants, IA = tr(A) and IIIA = det(A), of anysecond-order tensor A by using the scalar triple product:

[Aa, b, c] + [a, Ab, c] + [a, b, Ac] = IA[a, b, c],

[Aa, Ab, Ac] = IIIA[a, b, c],

where a, b, and c are any three vectors (see Section A.6).For a proper-orthogonal tensor, the fact that det(Q) = 1 implies that this tensor

has an eigenvalue λ = 1. There are several ways to see this, but first consider

QT (Q − I) = I − QT.

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6.6 Euler’s Theorem: Rotation Tensors and Proper-Orthogonal Tensors 177

Taking the determinant of both sides and using the fact that det(QT) = 1, one findsthat

det (Q − I) = (−1)3det (Q − I) .

It follows that

det (Q − I) = 0, (6.15)

and thus Q has an eigenvalue of 1.As Q has a unit eigenvalue, it has an eigenvector u such that

Qu = u.

When this equation is multiplied by QT, it follows that

u = Qu = QTu.

Thus u is a unit eigenvector of both Q and QT.Now consider a vector v ⊥ u. Some manipulations show that

Qv · Qu = Qv · u,

Qv · Qu = QTQv · u = v · u = 0.

Consequently,

v ⊥ u if, and only if, Qv ⊥ u.

We henceforth assume that u and v have unit magnitudes and define a vector wsuch that [u, v, w] = 1. Let us now calculate IQ:

IQ = [Qu, v, w] + [u, Qv, w] + [u, v, Qw]

= [u, v, w] + [u, Qv, w] + [u, v, Qw]

= 1 + v · Qv + w · Qw.

= 1 + v · Qv + (u × v) · (Qu × Qv) .

= 1 + v · Qv + (u · u) (v · Qv) − (u · Qv) (v · u) .

= 1 + 2v · Qv.

We define an angle ν such that

cos(ν) = v · Qv.

It is important to notice that this angle is an invariant of Q and that

Qv = cos(ν)v + sin(ν)w

= cos(ν)v + sin(ν) (u × v) . (6.16)

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As mentioned previously, any vector a can be decomposed into two compo-nents: a‖ and a⊥, where a⊥ · u = 0 and a = a‖ + a⊥. Now,

Qa‖ = a‖,

and, from (6.16),

Qa⊥ = cos(ν)a⊥ + sin(ν) (u × a⊥) .

It follows that

Qa = (a · u) u + cos(ν)(I − u ⊗ u)a + sin(ν)u × a.

Comparing this equation with (6.9), we conclude that Q has the form (6.8) of arotation tensor. Further, as the preceding equation holds for any proper-orthogonalQ, we conclude that every proper-orthogonal tensor is a rotation tensor. Thus weuse these terms interchangeably.

The result that every proper-orthogonal tensor is a rotation tensor is creditedto Euler and dates to 1775 [55]:

Every proper-orthogonal tensor is a rotation tensor.

We shall shortly revisit this result and see why it is also known as Euler’s theoremon the motion of a rigid body. To verify that a tensor A is a rotation tensor, weinvoke Euler’s theorem and simply show that A is proper-orthogonal: ATA = I anddet (A) = 1.

6.7 Relative Angular Velocity Vectors

Consider two rotation tensors: R1 = R1(t) and R2 = R2(t). It is straightforward toshow that the product R = R2R1 of these two tensors is also a rotation tensor. Wenow wish to calculate its angular velocity tensor and vector. To do this, we use thework of Casey and Lam [29], who defined a very useful and intuitive relative angularvelocity vector.

To discuss the relative angular velocity vector, it is convenient to define threesets of right-handed orthonormal vectors: {1t1, 1t2, 1t3}, {2t1, 2t2, 2t3}, and {p1, p2, p3}.In this section, we assume that pi = 0. For a given R1 and R2, two of these sets canbe defined by use of the representations

R1 =3∑

i=1

1ti ⊗ pi, R2 =3∑

i=1

2ti ⊗ 1ti.

Notice that

R =3∑

i=1

2ti ⊗ pi.

In words, R transforms the vector pi into the vector 2ti.

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6.7 Relative Angular Velocity Vectors 179

Following Casey and Lam [29], let us consider the following relative angularvelocity tensor:

�R2 = �R − �R1 .

Using the definition of the angular velocity tensors and the fact that R = R2R1 +R2R1, we find that

�R2 = �R − �R1 = RRT − R1RT1

= R2R1RT1 RT

2 + R2R1RT1 RT

2 − R1RT1

= R2RT2 + R2�R1 RT

2 − �R1

= R2RT2 + R2�R1 RT

2 + �TR1

= (R2 + R2�R1 + �TR1

R2)

RT2 .

However,

oR2 = R2 + R2�R1 + �T

R1R2,

where the corotational derivativeoR2 is defined to be

oR2 =

3∑i=1

3∑k=1

R2ik1ti ⊗ 1tk,

where R2ik = ((R2)1tk) · 1ti. In words,oR2 is the derivative of the tensor R2 assuming

that 1ti are constant.In conclusion, the relative angular velocity tensor �R2 is

�R2 = �R − �R1 = oR2 RT

2 .

If we denote the axis of rotation of R2 by r2 and its angle of rotation by φ2, then wecan parallel the derivation of (6.14) to find that the relative angular velocity vectorhas the representation

ωR2 = ωR − ωR1 = φ2r2 + sin (φ2)or2 + (1 − cos (φ2)) r2× o

r2 . (6.17)

In this equation,

ωR2 = −12ε[ oR2 RT

2

], (6.18)

and the corotational derivative of r2 =∑3i=1 ri (1ti) is

or2= r1 (1t1) + r2 (1t2) + r3 (1t3) .

Formula (6.17) will prove to be exceedingly useful when calculating the angular ve-locity vector associated with various representations of a rotation tensor. In partic-ular, for the Euler angle representation, we decompose R into the product of three

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rotation tensors, and we shall invoke (6.17) twice to get a representation for the an-gular velocity vector corresponding to R. The manner in which we do this is similarto the example now presented.

6.7.1 An Example

To illustrate the convenience of relative angular velocity result (6.17), let us consideran example. Suppose we have two rotations R1 and R, where

R1 = cos(ψ)(I − E3 ⊗ E3) − sin(ψ)εE3 + E3 ⊗ E3,

R = R2R1.

Here, the relative rotation tensor R2 is chosen to correspond to a rotation throughan angle θ about t2 = cos(ψ)E2 − sin(ψ)E1:

R2 = cos(θ)(I − t2 ⊗ t2) − sin(θ)εt2 + t2 ⊗ t2.

The tensor R1 defines a transformation consisting of a rotation through an angle ψ

about E3. This rotation transforms Ei to ti, where

t1 = cos(ψ)E1 + sin(ψ)E1, t2 = − sin(ψ)E1 + cos(ψ)E2, t3 = E3.

In addition, R consists of the rotation R1 followed by a rotation through an angle θ

about t2.∗

To calculate ωR1 we can appeal to (6.14) to find that

ωR1 = ψE3 + sin(ψ)E3 + (1 − cos(ψ))E3 × E3

= ψE3.

To calculate ωR we cannot appeal directly to (6.14) because we do not know the axisand angle of rotation of R.† Instead, we use the relative angular velocity vector. Todo this, we first need to write R2 with respect to an appropriate basis. As

R1 = t1 ⊗ E1 + t2 ⊗ E2 + t3 ⊗ E3,

the appropriate basis is ti ⊗ tk:

R2 = cos(θ)(t3 ⊗ t3 + t1 ⊗ t1) − sin(θ) (t3 ⊗ t1 − t1 ⊗ t3) + t2 ⊗ t2.

Calculating the corotational rate of this tensor, we take its derivative, keeping ti

fixed:oR2= −θ sin(θ)(t3 ⊗ t3 + t1 ⊗ t1) − θ cos(θ) (t3 ⊗ t1 − t1 ⊗ t3) .

Hence,

�R2 = θ (t1 ⊗ t3 − t3 ⊗ t1) ,

∗ In the notation of the previous section, pi is replaced with Ei and 1ti is replaced with ti .† The interested reader might wish to use the famed Rodrigues formula [(6.49)] that is discussed in

the exercises to compute these quantities and then one could use (6.14).

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6.8 Euler Angles 181

and, with the help of (6.18), we conclude that

ωR2 = θ t2. (6.19)

As an alternative method of calculating (6.19), we can use (6.17) directly.Thus we replace r2 and φ2 in (6.17) with t2 and θ, respectively. We then appeal

to the fact thatot2= 0. In summary,

ωR2 = θ t2 + sin(θ)ot2 + (1 − cos(θ))t2 × o

t2

= θ t2.

This alternative method is clearly equivalent to, but more attractive than, the

method that involved calculatingoR2.

Combining the expressions for ωR1 and ωR2 , we arrive at an expression for theangular velocity vector for R:

ωR = θ t2 + ψE3.

The intuitive nature of this result is often surprising. It is very easy to use (6.14) tosee that ωR2 �= ωR2 .

6.8 Euler Angles

The most popular representation of a rotation tensor is based on the use of threeEuler angles. As discussed in [38, 230], this representation dates to works by Euler[54, 57] that he first presented in 1751.∗ In these papers, he shows how three anglescan be used to parameterize a rotation, and he also establishes expressions for thecorotational components of the angular velocity vector.

One interpretation of the Euler angles involves a decomposition of the rotationtensor into a product of three fairly simple rotations:

R = R(γ1, γ2, γ3) = L(γ3, g3)L(γ2, g2)L(γ1, g1). (6.20)

Here, {γi} are the Euler angles and the set of unit vectors {gi} is known as the Eulerbasis. The function L(θ, b) is defined by use of the Euler representation:

L(θ, b) = cos(θ)(I − b ⊗ b) − sin(θ)(εb) + b ⊗ b,

∗ For discussions of these papers, and several other interesting historical facts on the developmentof representations for rotations, see Blanc’s introduction to parts of Euler’s collected works in[59, 60], Cheng and Gupta [38], and Wilson [230]. Although [57] dates to the 18th century, it wasfirst published posthumously in 1862.

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ψ

ψ

θ

θ

φ

φφ

p1

p′1

p′′1

p2

p3 = g1

t1

t2t3 = g3

g2

Figure 6.3. Schematic of the 3–2–3 set of Euler angles and the individual rotations these anglesrepresent. In this figure, the three Euler angles are denoted by ψ = γ1, θ = γ2, and φ = γ3,respectively, and the rotation tensor R that they parameterize transforms pi to ti . The imageon the left-hand side is a portrait of Leonhard Euler.

where b is a unit vector and θ is the counterclockwise angle of rotation. In general,g3 is a function of γ2 and γ1 and g2 is a function of γ1. As we shall shortly see,there are 12 possible choices of the Euler angles. For example, Figure 6.3 illustratesthese angles for a set of 3–2–3 Euler angles. Because there are three Euler angles,the parameterization of a rotation tensor by use of these angles is an example of athree-parameter representation.

If we assume that g1 is constant, then the angular velocity vector associatedwith the Euler angle representation can be established by use of the relativeangular velocity vector. In this case, there are two relative angular velocity vectors[cf. (6.17)]. For the first rotation, the angular velocity vector is γ1g1 [cf. (6.14)].The angular velocity of the second rotation relative to the first rotation is γ2g2,and the angular velocity of the third rotation relative to the second rotation isγ3g3.∗ Combining the two relative angular velocity vectors with γ1g1, we concludethat

ωR = γ3g3 + γ2g2 + γ1g1. (6.21)

If the rotation tensor R transforms the vectors pi into the set ti, then it is possible toexpress the Euler basis in terms of either set of vectors.

∗ The calculation of these angular velocity vectors is similar to the example discussed in Subsection6.7.1.

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φ

φ

ψ

ψθ

θ

p1

p2 t′1

t′1

t′2

t′3

t′′1

t′′2

t′′3

t′′3 t2

t3

Figure 6.4. The transformations of various basis vectors induced by the individual angles in aset of 3–2–1 Euler angles.

Alternative approaches to establishing (6.21) can be found in textbooks. Inone such approach, all three Euler angles are considered to be infinitesimal. Agood example of this approach can be found in Section 2.9 of Lurie [131]. Anotherapproach, which can be found in [109, 185] and dates to Euler, features sphericalgeometry. Finally, a third (lengthy) approach involves directly differentiating (6.20)and then computing �R and its axial vector.

For the Euler angles to effectively parameterize all rotations, we need toassume that we can find γk and γk such that, for any given ωR, (6.21) holds. Forthis to happen, it is necessary and sufficient that the vectors g1, g2, and g3 span E

3.When these vectors are not linearly independent, we say that the Euler angles havea singularity. This singularity is unavoidable for Euler angles and is often knownas a “gimbal lock” (cf. [196]). We shall find that this singularity occurs for certainvalues of γ2, and we shall restrict this angle to avoid these singularities.

For future purposes, it is also convenient to define the dual Euler basis {gj}:gj · gi = δ

ji , where δ

ji is the Kronecker delta.∗ Notice that

ωR · gi = γi. (6.22)

To determine this basis, one expresses gi in terms of a right-handed basis, say {ti}.Then, to determine g2, say, we write g2 = at1 + bt2 + ct3 and solve the three equa-tions, g1 · g2 = 0, g2 · g2 = 1, and g3 · g2 = 0 for the three unknowns a, b, and c. Whenthe Euler angles have singularities, one will find that the two dual Euler basis vec-tors g1 and g3 cannot be defined. Another point of interest is the observation thatg2 = g2 for all possible sets of Euler angles.

We now turn to examining two different choices of the Euler angles: the 3–2–1set and the 3–1–3 set. Both of these sets are popular in different communities. Forinstance, the aircraft and vehicle dynamics community favors the 3–2–1 Euler anglesto parameterize yaw–pitch–roll behavior, whereas problems involving spinningrigid bodies in dynamics often lend themselves to the 3–1–3 set. We shall subse-quently discuss examples of both sets. One of the exercises at the end of this chapter

∗ The dual Euler basis vectors are analogous to the contravariant basis vectors and have similar usesin dynamics. We shall use the dual Euler basis later in various contexts, and a rapid summary ofthese uses can be found in O’Reilly [160].

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involves a comprehensive investigation into the set of 3–2–3 Euler angles. It is highlyrecommended that you complete this exercise after reading this section of the book.

6.8.1 3-2-1 Euler Angles

To elaborate further on the Euler angles, we now consider the 3–2–1 set of Eulerangles (see Figure 6.4). These are arguably among the most popular sets of Eulerangles.∗ In several communities, they are known as examples of the Tait and/orBryan angles [after Peter G. Tait (1831–1901) and George H. Bryan (1864–1928)]or the Euler–Cardan angles [after Euler and Girolamo Cardano (1501–1576)].

First, suppose that the rotation tensor has the representation

R =3∑

i=1

ti ⊗ pi,

where {pi} is a fixed Cartesian basis. The first rotation is about p3 through an angleψ . This rotation transforms pi to t′i. The second rotation is about the t′2 axis throughan angle θ. This rotation transforms t′i to t′′i . The third and last rotation is through anangle φ about the axis t′′1 = t1. Thus

R = R(γ1 = ψ, γ2 = θ, γ3 = φ) = L(φ, t1)L (θ, t′2) L(ψ, p3).

Here,

ti = L(φ, t1)t′′i , t′′i = L(θ, t′′2 = t′2)t′i, t′i = L(ψ, t′3 = p3)pi.

It is not difficult to express the various basis vectors as linear combinations of eachother: ⎡

⎢⎣t′1t′2t′3

⎤⎥⎦ =

⎡⎢⎣

cos(ψ) sin(ψ) 0

− sin(ψ) cos(ψ) 0

0 0 1

⎤⎥⎦⎡⎢⎣

p1

p2

p3

⎤⎥⎦ ,

⎡⎢⎣

t′′1t′′2t′′3

⎤⎥⎦ =

⎡⎢⎣

cos(θ) 0 − sin(θ)

0 1 0

sin(θ) 0 cos(θ)

⎤⎥⎦⎡⎢⎣

t′1t′2t′3

⎤⎥⎦ ,

⎡⎢⎣

t1

t2

t3

⎤⎥⎦ =

⎡⎢⎣

1 0 0

0 cos(φ) sin(φ)

0 − sin(φ) cos(φ)

⎤⎥⎦⎡⎢⎣

t′′1t′′2t′′3

⎤⎥⎦ . (6.23)

The inverses of these relationships are easy to obtain once you realize that each ofthe three matrices in (6.23) is orthogonal. As the inverse of an orthogonal matrix is

∗ For example, they are used in Greenwood’s text [79], Rao’s text [176], and numerous texts on vehi-cle and aircraft dynamics.

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its transpose, we quickly arrive at the sought-after results:

⎡⎢⎣

p1

p2

p3

⎤⎥⎦ =

⎡⎢⎣

cos(ψ) − sin(ψ) 0

sin(ψ) cos(ψ) 0

0 0 1

⎤⎥⎦⎡⎢⎣

t′1t′2t′3

⎤⎥⎦ ,

⎡⎢⎣

t′1t′2t′3

⎤⎥⎦ =

⎡⎢⎣

cos(θ) 0 sin(θ)

0 1 0

− sin(θ) 0 cos(θ)

⎤⎥⎦⎡⎢⎣

t′′1t′′2t′′3

⎤⎥⎦ ,

⎡⎢⎣

t′′1t′′2t′′3

⎤⎥⎦ =

⎡⎢⎣

1 0 0

0 cos(φ) − sin(φ)

0 sin(φ) cos(φ)

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦ . (6.24)

Relationships (6.23) and (6.24) can be combined to express ti in terms of pk and viceversa. Later, they will also be used to obtain representations for the Euler basis interms of pk and ti.

By using (6.24) we can find a representation for the components Rij = (Rpj) · pi.The components are easily expressed by a matrix representation:

⎡⎢⎢⎣

R11 R12 R13

R21 R22 R23

R31 R32 R33

⎤⎥⎥⎦ =

⎡⎢⎣


sin(ψ) cos(ψ) 0

0 0 1

⎤⎥⎦⎡⎢⎣

cos(θ) 0 sin(θ)

0 1 0

− sin(θ) 0 cos(θ)

⎤⎥⎦

×

⎡⎢⎣

1 0 0


0 sin(φ) cos(φ)

⎤⎥⎦ . (6.25)

Representation (6.25) is the transpose of what one might naively expect. However,recalling that Rik = pi · tk will hopefully resolve this initial surprise. Indeed, it is use-ful to note that

⎡⎢⎣

p1

p2

p3

⎤⎥⎦ =

⎡⎢⎢⎣

R11 R12 R13

R21 R22 R23

R31 R32 R33

⎤⎥⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦ ,

⎡⎢⎣

t1

t2

t3

⎤⎥⎦ =

⎡⎢⎢⎣

R11 R21 R31

R12 R22 R32

R13 R23 R33

⎤⎥⎥⎦⎡⎢⎣

p1

p2

p3

⎤⎥⎦ .

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THE EULER BASIS. As mentioned earlier, by examining the individual rotations[cf. (6.23)], we can show that the Euler basis vectors have the representations⎡

⎢⎣g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

p3

t′2t1

⎤⎥⎦ =

⎡⎢⎣

− sin(θ) sin(φ) cos(θ) cos(φ) cos(θ)


1 0 0

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦ . (6.26)

Alternatively, we can also express the Euler basis in terms of the basis vectors{p1, p2, p3

}:⎡

⎢⎣g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

p3

t′2t1

⎤⎥⎦ =

⎡⎢⎣

0 0 1


cos(θ) cos(ψ) cos(θ) sin(ψ) − sin(θ)

⎤⎥⎦⎡⎢⎣

p1

p2

p3

⎤⎥⎦ . (6.27)

This representation is useful for establishing the components ωR · pk.

THE DUAL EULER BASIS. We can now determine the dual Euler basis vectors gk. Werecall the remarks following (6.22) and express each of the dual Euler basis vectorsin terms of their components relative to the basis {t1, t2, t3}. That is,

gk = gk1t1 + gk2t2 + gk3t3.

Combining these results for the dual Euler basis vectors, we observe that⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

g11 g12 g13

g21 g22 g23

g31 g32 g33

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦ . (6.28)

With some manipulation, the relations gi · gk = δki can be expressed as nine equa-

tions for the nine unknowns gik:⎡⎢⎣

g11 g21 g31

g12 g22 g32

g13 g23 g33

⎤⎥⎦⎡⎢⎣

− sin(θ) sin(φ) cos(θ) cos(φ) cos(θ)


1 0 0

⎤⎥⎦ =

⎡⎢⎣

1 0 0

0 1 0

0 0 1

⎤⎥⎦ .

Isolating the matrix[gik]

on the left-hand side of this equation, we find that⎡⎢⎣

g11 g12 g13

g21 g22 g23

g31 g32 g33

⎤⎥⎦ =

⎡⎢⎣

0 sin(φ) sec(θ) cos(φ) sec(θ)


1 sin(φ) tan(θ) cos(φ) tan(θ)

⎤⎥⎦ .

That is, the dual Euler basis vectors have the representations⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣




⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦ . (6.29)

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(a) (b)

ψψ

φ

θ

θθ

p1p1

p2p2

g1 = p3g1 = p3

t′1t

′1

t′′3t

′′3

g1

g2 = t′2g2 = t

′2

g3 = t1

g3

Figure 6.5. (a) The dual Euler basis vectors gk for the 3–2–1 set of Euler angles: g1 ‖ t′′3, g2 = t′2,and g3 ‖ t′1. (b) The Euler angles ψ and θ serve as coordinates for the Euler basis vector g3 = t1

in a manner that is similar to the role that spherical polar coordinates play in parameterizingeR and eφ.

If we had used (6.27) in place of (6.26) to calculate the dual Euler basis vectors,then we would have found the following representations:

⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

cos(ψ) tan(θ) sin(ψ) tan(θ) 1


cos(ψ) sec(θ) sin(ψ) sec(θ) 0

⎤⎥⎦⎡⎢⎣

p1

p2

p3

⎤⎥⎦ . (6.30)

Expressions for the dual Euler basis vectors in terms of t′k are easily inferred from(6.30) and are shown in Figure 6.5(a).

For completeness, we note that, when θ �= ±π2 , one can also express the Euler

basis vectors in terms of the dual Euler basis vectors:

⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

1 0 − sin(θ)

0 1 0

− sin(θ) 0 1

⎤⎥⎦⎡⎢⎣

g1

g2

g3

⎤⎥⎦ . (6.31)

The simplicity of this relationship (related versions of which hold for the other 11sets of Euler angles) is surprising.

SINGULARITIES. If we examine (6.26), we see that the Euler basis fails to be a ba-sis for E

3 when θ = ±π2 . One of the easiest ways to see this fact is to consider

ψ and θ + π2 to be spherical polar coordinates for t1 [see Figure 6.5(b)]. When

θ = ±π2 , g1 = p3 = ±g3, and the Euler basis does not span E

3. To avoid the afore-mentioned singularity, it is necessary to place restrictions on the second Euler angle:θ ∈ (−π

2 , π2

). The other two angles are free to range from 0 to 2π.

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ANGULAR VELOCITY VECTORS. The angular velocity vector ωR associated with the3–2–1 Euler angles has several representations:

ωR = −12ε[RRT] =

3∑i=1

γigi

= φt1 + θ t′2 + ψp3

= (−ψ sin(θ) + φ)t1 + (ψ sin(φ) cos(θ) + θ cos(φ))t2

+ (ψ cos(φ) cos(θ) − θ sin(φ))t3.

To arrive at the primitive representation ω = φt1 + θ t′2 + ψp3, we needed tocompute two relative angular velocity vectors. We calculated the first of these, θ t′2,assuming that t′i were fixed. The explicit details of this calculation are easily inferredfrom our earlier example in Subsection 6.7.1. We calculated the second relativeangular velocity vector, φt1, by using the relative rotation tensor L

(φ, t′′1

), assuming

that t′′i were fixed.It is also interesting to note that the angular velocity vector ω0R has the repre-

sentations

ω0R = −12ε[RTR

] = RTωR

= φRTt1 + θ RTt′2 + ψRTp3

= (−ψ sin(θ) + φ)p1 + (ψ sin(φ) cos(θ) + θ cos(φ))p2

+(ψ cos(φ) cos(θ) − θ sin(φ))p3.

In establishing this result, we used the fact that RTti = pi.

6.8.2 3–1–3 Euler Angles

We can parallel the developments of the previous section for another popular setof Euler angles: the 3–1–3 Euler angles (see Figure 6.6). These are the set of Eu-ler angles that Lagrange used,∗ and they are also used in Arnol’d [9], Landau andLifshitz [125], and Thomson [214], among many others. For motions of a spinningtop, the Euler angles are identified with precession, nutation, and spin, respectively.A closely related set of Euler angles, the 3–2–3 set, are discussed in Exercise 6.2 atthe end of this chapter.

Paralleling the developments for the 3–2–1 Euler angles:

R = L(φ, t3 = t′′3

)L(θ, t′1

)L(ψ, p3),

where {p3, t′1, t3} is the Euler basis and

t′i = L (ψ, p3) pi, t′′i = L(θ, t′1

)t′i, ti = L

(φ, t′′3

)t′′i .

∗ See [118] and Section IX of the Second Part of [121]: His φ, ω, ψ correspond to our φ, θ, ψ , respec-tively.

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φ

φ

ψ

ψθ

θ

p1

p2t′1

t′2

t′2

t′3

t′′1

t′′2

t′′2

t′′3

t1

t2

Figure 6.6. The transformations of various basis vectors induced by the individual angles in aset of 3–1–3 Euler angles.

Harking back to many of the celestial mechanics applications for this set of Eulerangles, the line passing through the origin that is parallel to t′1 is often known as theline of nodes [214]. The angular velocity vector has the representations

ωR = −12ε[RRT] =

3∑i=1

γigi = φt3 + θ t′1 + ψp3. (6.32)

We are using the same notation for the three Euler angles as we did for the 3–2–1set. However, it should be clear that θ and φ represent different angles of rotationfor these two set of Euler angles.

EULER BASIS. It is not difficult to show that the Euler basis {gi} has the representa-tions ⎡

⎢⎣g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

p3

t′1t3

⎤⎥⎦ =

⎡⎢⎣

sin(φ) sin(θ) cos(φ) sin(θ) cos(θ)


0 0 1

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦

=

⎡⎢⎣

0 0 1

cos(ψ) sin(ψ) 0

sin(θ) sin(ψ) − sin(θ) cos(ψ) cos(θ)

⎤⎥⎦⎡⎢⎣

p1

p2

p3

⎤⎥⎦ . (6.33)

With these results and (6.32), two other representations for ωR can be obtained, butthis is left as an exercise.

SINGULARITIES. As with all sets of Euler angles, the 3–1–3 Euler angles are subjectto restrictions. For the 3–1–3 set, we can find the restrictions by examining whenthe Euler basis fails to be a basis. Fortunately, it is easy to see from Figure 6.7(a)that this is the case when t3 = ±p3. As a result, restrictions are placed on the secondangle:

φ ∈ [0, 2π), θ ∈ (0, π), ψ ∈ [0, 2π).

The fact that the restriction needs to be placed on the second Euler angle is consis-tent with corresponding results for the other 11 sets of Euler angles.

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(a) (b)

ψψ

φφ

θθ

p1p1

p2p2

g1 = p3g1 = p3

t′2

g1

g2 = t′1

g2 = t′1

g3 = t3g3 = t3

g3

Figure 6.7. (a) The Euler basis and (b) dual Euler basis vectors for the 3–1–3 set of Eulerangles. For this set of Euler angles, g1 ‖ t′2, g2 = t′1, and g3 ‖ t′′2.

DUAL EULER BASIS. By following the procedure that led to (6.29), we find that thedual Euler basis

{gi}

has the representation∗

⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

sin(φ)cosec(θ) cos(φ)cosec(θ) 0


− sin(φ) cot(θ) − cos(φ) cot(θ) 1

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦ . (6.34)

Similarly, ⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

− sin(ψ)cot(θ) cos(ψ)cot(θ) 1

cos(ψ) sin(ψ) 0

sin(ψ)cosec(θ) − cos(ψ)cosec(θ) 0

⎤⎥⎦⎡⎢⎣

p1

p2

p3

⎤⎥⎦ .

The dual Euler basis vectors are shown in Figure 6.7. It is important to observe thatthey are not defined when this set of Euler angles has its singularities at θ = 0 andθ = π.

Using (6.34), we can show that

ωR · g1 = ψ, ωR · g2 = θ, ωR · g3 = φ.

We can also establish the following results:⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

1 0 cos(θ)

0 1 0

cos(θ) 0 1

⎤⎥⎦⎡⎢⎣

g1

g2

g3

⎤⎥⎦ . (6.35)

It is a valuable exercise to compare the results just presented for the 3–1–3 Eulerangles with those presented earlier for the 3–2–1 set. One issue that will arise in thiscomparison is the different ranges that the second Euler angle θ possesses for thetwo sets.

∗ That is, one calculates the inverse of the transpose of the 3 × 3 matrix in (6.33) that describes gi interms of tk.

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6.9 Further Representations of a Rotation Tensor 191

Other Sets of Euler AnglesFor the Euler basis, one has three choices for g1 and, because g1 �= g2, two choicesfor g2. Finally, there are two choices of g3. Consequently, there are 2 × 2 × 3 = 12choices of the vectors for the Euler basis. The easiest method to see which set ofEuler angles is being used is to specify the angular velocity vector. Here, expres-sions are given for each of the 12 sets of Euler angles for a rotation tensor R =∑3

i=1 ti ⊗ pi:

1–2–3 Set: ωR = ψp1 + θ t′2 + φt3,

3–2–3 Set: ωR = ψp3 + θ t′2 + φt3,

1–2–1 Set: ωR = ψp1 + θ t′2 + φt1,

1–3–1 Set: ωR = ψp1 + θ t′3 + φt1,

1–3–2 Set: ωR = ψp1 + θ t′3 + φt2,

2–3–1 Set: ωR = ψp2 + θ t′3 + φt1,

2–3–2 Set: ωR = ψp2 + θ t′3 + φt2,

2–1–2 Set: ωR = ψp2 + θ t′1 + φt2,

2–1–3 Set: ωR = ψp2 + θ t′1 + φt3,

3–1–3 Set: ωR = ψp3 + θ t′1 + φt3,

2–3–1 Set: ωR = ψp2 + θ t′3 + φt1,

3–1–2 Set: ωR = ψp3 + θ t′1 + φt2.

The sets of Euler angles, 121, 131, 232, 212, 313, and 323, are known as the symmetricsets, whereas the other six sets are known as asymmetric sets. The latter sets are alsoknown as the Cardan angles, Tait angles, or Bryan angles. Tait’s original discussion(of what we would refer to as 1–2–3 Euler angles) can be seen in Section 12 of his1868 paper [210]. In his seminal text [24] on aircraft stability that was published in1911, Bryan introduced what we would refer to as a 2–3–1 set of Euler angles (seeFigure 6.8). It is interesting to recall that the Wright brothers first successful flightwas in 1903.

For all sets of Euler angles, a singularity is present for certain values of thesecond angle θ. At these values g1 = ±g3, and the Euler basis fails to be a ba-sis for E

3. To avoid these singularities it is often necessary to use two differentsets of Euler angles and to switch from one set to the other as a singularity isapproached.

6.9 Further Representations of a Rotation Tensor

Apart from Euler’s representation and the Euler angle representation, there areseveral other representations of a rotation tensor. Most of them are discussed in

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φ

φ

φ θ

θ

G

x

z

Figure 6.8. Reproduction of Figure 3 in Bryan’s seminal text [24] on the stability of aircraft.The angles θ and φ in this figure are the pitch and roll angles, respectively, of the aircraft. Thepoint G is the center of mass of the aircraft, and x and z label the corotational bases for theaircraft.

Shuster’s review article [196], and we now discuss two of them. These are repre-sentations that are due to Olinde Rodrigues [181] in 1840 and the Euler parameterrepresentation.

The Rodrigues VectorThe Rodrigues representation is based on the vector:

λ = tan(

φ

2

)r.

This vector is sometimes called the Gibbs vector. Clearly the Rodrigues vector λ isnot a unit vector. Indeed, when φ = 0, λ = 0, and when φ = π, λ is undefined. Con-sequently, if φ varies through π, then we cannot use the Rodrigues representationsubsequently discussed.

With the assistance of the identities

sin(φ) = 2 tan(φ

2 )

1 + tan2(φ

2 ), cos(φ) = 1 − tan2(φ

2 )

1 + tan2(φ

2 ),

you should be able to verify that

cos(φ) = 1 − λ · λ

1 + λ · λ, sin(φ) = 2λ · r

1 + λ · λ,

Substituting for r and φ in (6.8), we find the Rodrigues representation:

R = R(λ) = 11 + λ · λ

((1 − λ · λ)I + 2λ ⊗ λ − 2(ελ)) .

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6.9 Further Representations of a Rotation Tensor 193

The angular velocity vector associated with this representation is

ωR = 21 + λ · λ

(λ − λ × λ

).

This vector can be calculated by directly substituting into the earlier result associ-ated with Euler representation (6.14).

The Euler–Rodrigues Symmetric ParametersOne of the most popular four-parameter representations uses the four Euler–Rodrigues symmetric parameters e0 and e.∗ These parameters are often known asthe Euler parameters and have very interesting historical connections.† They can bedefined as

e0 = cos(

φ

2

), e = sin

(φ

2

)r.

As a consequence of their definition, the parameters are subject to what is knownas the Euler parameter constraint‡:

e20 + e · e = 1.

We also note that

λ = ee0

.

It is possible to express r and φ in terms of the parameters e0 and e, but this is left asan exercise.

By substituting for r and φ in Euler representation (6.8), we find the Euler–Rodrigues symmetric parameter representation:

R = R(e0, e) = (e20 − e · e

)I + 2e ⊗ e − 2e0(εe).

∗ The four parameters, e0 and the three components of e, are often considered to be the four com-ponents of a quaternion q = e0 + e1i + e2 j + e3k, where ei are the components of e relative to aright-handed orthonormal basis, and i, j, and k are bases vectors for the quaternion. Consequently,Euler–Rodrigues symmetric parameters are sometimes referred to as (unit) quaternions.

† Excellent discussions can be found in Altmann [2, 3] and Gray [77].‡ The relaxation of this constraint is discussed in O’Reilly and Varadi [166] who show, among other

matters, how it can be visualized by using Hoberman’s sphere. This topic is also intimately relatedto Gauss’ mutation of space [69].

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Suppose R =∑3i=1 ti ⊗ pi; then it is easy to show that e · tk = e · pk. If one writes out

expressions for the components of Rik = tk · pi,∗⎡⎢⎢⎣

R11 R12 R13

R21 R22 R23

R31 R32 R33

⎤⎥⎥⎦ = (e2

0 − e21 − e2

2 − e23

)⎡⎢⎢⎣

1 0 0

0 1 0

0 0 1

⎤⎥⎥⎦+

⎡⎢⎢⎣

2e21 2e1e2 2e1e3

2e1e2 2e22 2e2e3

2e1e3 2e2e3 2e23

⎤⎥⎥⎦

+

⎡⎢⎣

0 −2e0e3 2e0e2

2e0e3 0 −2e0e1

−2e0e2 2e0e1 0

⎤⎥⎦ , (6.36)

then it is easy to see that the rotation tensor is a quadratic function of the four pa-rameters e0, e1, e2, e3. As a result, this representation has several computational ad-vantages over other representations. You may also notice how easy it is to establishan expression for the components of RT from (6.36).

The angular velocity vector associated with this representation is

ωR = 2 (e0e − e0e + e × e) .

Again, we can calculate this vector by directly substituting into the earlier resultassociated with Euler representation (6.14) for ωR. Notice that the angular velocityvector is a relatively simple function of the Euler–Rodrigues symmetric parametersand their derivatives.

In one of the exercises at the end of this chapter, further results pertaining tothe Euler–Rodrigues parameters for the composition of two rotation tensors arepresented. These results, which date to Olinde Rodrigues (1794–1851) in 1840, areremarkably elegant. Modern applications of these parameters arise in the estimationof the rotation tensor (attitude) of spacecraft and in computer vision and robotics.In these areas, one considers measurements of two sets of vectors that are relatedby an unknown rotation tensor R and a translation d. The estimation of R and d canbe rendered as a least-squares estimation problem that is known as the orthogonalProcrustes problem [95, 192] and in the satellite dynamics community as the Wahbaproblem after Grace Wahba [222]. That is, given the N ≥ 2 measurements a1, . . . , aN

and b1, . . . , bN, which are related by a rotation R and a translation d, determine theoptimal R and d such that the following function is minimized:

W = 12

N∑K=1

αK ||bK − RaK − d||2 , (6.37)

where αK is a scalar weight for the Kth measurement. When d = 0, Davenportshowed that, by parameterizing R in (6.37) by e0 and e, it is possible to find a

∗ The corresponding matrix representation for Euler’s representation was established earlier; see(6.10).

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6.10 Derivatives of Scalar Functions of Rotation Tensors 195

very elegant solution to the Wahba problem.∗ The corresponding elegant solutionto the Procrustes problem ( i.e., when d �= 0) was established later by Horn [95].†

Summary of the Angular Velocity VectorsIt is useful to summarize the representations we have discussed for the angular ve-locity vector corresponding to a rotation through an angle φ about an axis r:

ωR =3∑

i=1

γigi

= φr + sin(φ)r + (1 − cos(φ))r × r

= 2 (e0e − e0e + e × e)

= 21 + λ · λ

(λ − λ × λ

). (6.38)

With some minor manipulations of these equations, one can also obtain expressionsfor r, e, and λ.

6.10 Derivatives of Scalar Functions of Rotation Tensors

Consider a function U = U(R). We wish to calculate the time derivative of this func-tion. One of the complications is that R has several representations, and for each ofthem a different representation of the derivative will be found. It will be subse-quently revealed that a simple expression for the derivative of U can be found interms of a vector uR and the dual Euler basis. This result is presented in Equation(6.41).

To start, we use the simplest representation for R:

R =3∑

i=1

3∑k=1

Rikpi ⊗ pk,

where pk = 0. Then, as U = U(R) = U (Rik, pj), we find by using the chain rule that

U =3∑

i=1

3∑k=1

∂U∂Rik

Rik.

We can express this result by using the trace operator:

U = tr(

∂U∂R

RT)

,

∗ Discussions of Davenport’s solution and those of others, along with extensions to Wahba’s originalformulation, can be found in [136, 137, 195, 197].

† Additional references to other solutions to this problem can be found in [46, 49, 200].

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where

∂U∂R

=3∑

i=1

3∑k=1

∂U∂Rik

pi ⊗ pk.

Noting that �R = RRT, and, after introducing the angular velocity tensor, we canthen write

U = tr(

∂U∂R

RT�TR

).

However, as �R is skew-symmetric, tr(A�T

R

) = 0 for all A = AT. Consequently,only the skew-symmetric part of ∂U

∂R RT contributes to U:

U = tr(

∂U∂R

RT�TR

)= tr

(UR�T

R

),

where we have introduced the skew-symmetric operator:

UR = 12

(∂U∂R

RT − R(

∂U∂R

)T)

.

As UR is a skew-symmetric tensor, we can calculate a vector that corresponds totwice the axial vector of UR

∗:

uR = −ε [UR] . (6.39)

The existence of this vector allows us to establish the following representations forU:

U = tr(

∂U∂R

RT)

= tr(UR�T)

= uR · ωR. (6.40)

Because we have established numerous representations for ωR, we next invoke thefinal form of the representation for U to determine representations for uR.

As an example, suppose U is parameterized by use of Euler angles: U = U(γk).

Then, invoking (6.40)2,

U =3∑

k=1

∂U∂γk

γk = uR ·3∑

j=1

γ jgj.

∗ The reason for the absence of the factor 12 in (6.39) can be inferred from identity (A.12), which is

discussed in the Exercises at the end of the Appendix.

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6.10 Derivatives of Scalar Functions of Rotation Tensors 197

With the help of the dual Euler basis, we conclude that

uR =3∑

i=1

∂U∂γi

gi. (6.41)

This is the simplest, and most useful, representation that we know of for uR. It firstappeared in [163].

We now assume that R is parameterized by use of one of the other three meth-ods mentioned previously. By evaluating U, using identity (6.40)2 and representa-tions (6.38), and following the procedure that led to (6.41), we can find three otherrepresentations for uR. With details omitted, a summary of the representations isnow presented:

uR =3∑

i=1

∂U∂γi

gi

= ∂U∂φ

r + 12

(cot(

φ

2

)(I − r ⊗ r) − εr

)∂U∂r

= 12

((e0I − εe)

∂U∂e

− ∂U∂e0

e)

= 12

(I + λ ⊗ λ − ελ)∂U∂λ

, (6.42)

where U(R) = U(e0, e) = U(γ1, γ2, γ3) = U(φ, r) = U(λ). Representation (6.42)4

was first established by Simmonds [199]. We also note that a representation thatis closely related to (6.42)2 is discussed in Antman [5].

Several of the partial derivatives in (6.42) need to be carefully evaluated. Forexample, because r is a unit vector, the derivative ∂U

∂r must be evaluated on the sur-face r · r = 1. Related remarks pertain to ∂U

∂e , ∂U∂e0

, and ∂U∂R . In other words, these are

tangential or surface derivatives. One method of evaluating them is to parameterizeR by the Euler angles, and then transform from the parameters of interest to theEuler angles. Indeed, (6.42), the chain rule, and the identity εgi = − ∂R

∂γi RT can beused to show that

∂U∂e0

= uR · 2e,∂U∂e

= 2(e0I + εe)uR,

∂U∂R

= −(εuR)R,∂U∂φ

= uR · r,

∂U∂r

=(

sin(φ)(I − r ⊗ r) + 2 sin2(

φ

2

)εr)

uR.

As discussed in O’Reilly [160] and Simmonds [199], results (6.42) can be used toestablish moment potentials associated with conservative moments. We shall use(6.42)1 extensively.

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EXERCISES

6.1. Consider the tensor R:

R = cos(θ)E1 ⊗ E1 + sin(θ)E2 ⊗ E1 + cos(θ)E2 ⊗ E2

− sin(θ)E1 ⊗ E2 + E3 ⊗ E3.

(a) Show that R also has the representations

R = er ⊗ E1 + eθ ⊗ E2 + E3 ⊗ E3

= cos(θ)er ⊗ er + sin(θ)eθ ⊗ er + cos(θ)eθ ⊗ eθ

− sin(θ)er ⊗ eθ + E3 ⊗ E3.

(b) Show that

�R = θ (E2 ⊗ E1 − E1 ⊗ E2)

= θ (eθ ⊗ er − er ⊗ eθ) ,

ωR = θE3.

6.2. Recall that three Euler angles can be used to parameterize a rotation tensor R.In this exercise, we consider the 3–2–3 set of Euler angles:

R = L(φ, t3 = t′′3

)L(θ, t′2

)L (ψ, E3) .

This set of Euler angles is used in several texts, for example, Section 4.2 of Ginsberg[71], Kelvin and Tait [109], Routh [184, 185], and Whittaker [228],∗ and is illustratedin Figure 6.3.

(a) Draw figures illustrating the relationships between (i) Ei and t′i, (ii) t′i andt′′i , and (iii) t′′i and ti.

(b) Explain why the second angle of rotation θ is restricted to lie between 0 andπ

(c) For this set of Euler angles, show that the Euler basis has the representa-tions ⎡

⎢⎣g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

− sin(θ) cos(φ) sin(φ) sin(θ) cos(θ)

sin(φ) cos(φ) 0

0 0 1

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦

=

⎡⎢⎣

0 0 1


cos(ψ) sin(θ) sin(ψ) sin(θ) cos(θ)

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ .

∗ Whittaker’s notation is similar to ours except his ψ corresponds to our φ and vice versa. If you arecomparing his expression for the components of ωR with ours, you will see that there is a typograph-ical error in his expression for ω2 in Section 16 of [228]. Routh’s notation is similar to ours, but hiscoordinate axes are left-handed.

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Exercise 6.2 199

In addition, show that the Euler angles are such that⎡⎢⎣

t1

t2

t3

⎤⎥⎦ =

⎡⎢⎣

cos(φ) sin(φ) 0

− sin(φ) cos(φ) 0

0 0 1

⎤⎥⎦⎡⎢⎣

cos(θ) 0 − sin(θ)

0 1 0

sin(θ) 0 cos(θ)

⎤⎥⎦

×

⎡⎢⎣

cos(ψ) sin(ψ) 0


0 0 1

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ . (6.43)

(d) Using (6.43), derive expressions for the components Rik of R. These com-ponents have a variety of representations

Rik = (Rtk) · ti = (REk) · Ei = tk · Ei.

(e) Recall that every rotation tensor R has an axis of rotation r and angle ofrotation. With the help of the results presented in Subsection 6.5.1, selectfour different values of the set (φ, θ, ψ) and determine the correspondingaxis of rotation and the angle of rotation. Give physical interpretations forthe four sets of values of the Euler angles that you have selected.

(f) For this set of Euler angles, show that the dual Euler basis has the repre-sentation ⎡

⎢⎣g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

− cos(φ)cosec(θ) sin(φ)cosec(θ) 0

sin(φ) cos(φ) 0

cos(φ) cot(θ) − sin(φ) cot(θ) 1

⎤⎥⎦⎡⎢⎣

t1

t2

t3

⎤⎥⎦ .

With the help of these results, verify the following expressions for the dualEuler basis in terms of the bases {E1, E2, E3} and

{g1, g2, g3

}:

⎡⎢⎣

g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

− cos(ψ)cot(θ) − sin(ψ)cot(θ) 1


cos(ψ)cosec(θ) sin(ψ)cosec(θ) 0

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦

=

⎡⎢⎣

cosec2(θ) 0 −cot(θ)cosec(θ)

0 1 0

−cot(θ)cosec(θ) 0 cosec2(θ)

⎤⎥⎦⎡⎢⎣

g1

g2

g3

⎤⎥⎦ .

(g) For this set of angles, show that the angular velocity vector has the repre-sentation

ωR = φt3 + θ t′2 + ψE3. (6.44)

You will need to use two distinct corotational derivatives to find this repre-sentation. The first of these fixes t′i whereas the second fixes t′′i .

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(h) For this set of angles, show that the components of ωR relative to the basis{E1, E2, E3} have the representations

�1 = φ sin (θ) cos (ψ) − θ sin (ψ) ,

�2 = φ sin (θ) sin (ψ) + θ cos (ψ) ,

�3 = φ cos (θ) + ψ,

where �i = ωR · Ei. These representations can be found in Section 257 ofRouth [185].

(i) Suppose that ωk(t) = ωR · tk are known functions. Show that⎡⎢⎣

ψ

θ

φ

⎤⎥⎦ =

⎡⎢⎣

− cos(φ) cosec(θ) sin(φ) cosec(θ) 0

sin(φ) cos(φ) 0

cos(φ) cot(θ) − sin(φ) cot(θ) 1

⎤⎥⎦⎡⎢⎣

ω1

ω2

ω3

⎤⎥⎦ . (6.45)

(j) This problem involves the numerical integration of (6.45). Given

ω1(t) = 0.2 sin(0.5t), ω2(t) = 0.2 sin(0.05t), ω3(t) = 10ω1(t),

and initial values for the Euler angles of your choice, determine φ(t), θ(t),and ψ(t). How can these results be used to determine tk(t)?

6.3. Recall that a rotation tensor L representing a counterclockwise rotation aboutan axis p though an angle ν has the representation

L = L(ν, p) = cos(ν)(I − p ⊗ p) − sin(ν)εp + p ⊗ p,

and its associated angular velocity vector has the representation

ωL = νp + sin(ν)p + (1 − cos(ν))p × p.

Consider two rotation tensors:

Q1 = L(θ, E3), Q2 = L(φ, e1)L(θ, E3),

where

e1 = cos(θ)E1 + sin(θ)E2, e2 = cos(θ)E2 − sin(θ)E1, e3 = E3.

(a) Show that Q1 has the representation

Q1 = e1 ⊗ E1 + e2 ⊗ E2 + e3 ⊗ E3.

(b) Give an example of a system of two rigid bodies for which the rotationtensor of one body is Q1 and the rotation tensor of the second body is Q2.

(c) Given that the relative rotation tensor R2 = Q2QT1 , show that

ωR2 = φe1 + sin(φ)θ e2 + (1 − cos(φ))θ e3.

(d) Explain why

ωR2 �= ωR2 = ωQ2 − ωQ1 .

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Exercise 6.4 201

6.4. The parameterization of the rotation tensor by use of Euler parameters (unitquaternions or symmetric Euler–Rodriques parameters) has the beautiful conse-quence that the formula for the composition of two rotations is very elegant. Indeed,the same results for two tensors described by Euler angles are very unwieldy. In thisproblem, we use Euler parameters to explore some results pertaining to rotationtensors.

Consider two rotation tensors A and B:

A = (e20 − e · e)I + 2e ⊗ e − 2e0(εe),

B = ( f 20 − f · f)I + 2f ⊗ f − 2f0(εf). (6.46)

Here, {e0, e} and { f0, f} represent two sets of Euler parameters:

e0 = cos(

φ

2

), e = sin

(φ

2

)a, f0 = cos

(θ

2

), f = sin

(θ

2

)b, (6.47)

where a is the axis of rotation of A and b is the axis of rotation of B, φ is the angle ofrotation for A, and θ is the angle of rotation for B. That is, the tensor A correspondsto a counterclockwise rotation of φ about a.

(a) Recall the representation for a rotation tensor A in terms of the angle ofrotation φ and the axis of rotation a:

A(φ, a) = cos(φ)(I − a ⊗ a) − sin(φ)εa + a ⊗ a.

Verify that A has the representation (6.46)1.

(b) Letting e =∑3i=1 eiEi, what are Aik = (AEk) · Ei?

(c) Show that

C = BA = (g20 − g · g)I + 2g ⊗ g − 2g0(εg), (6.48)

where

g0 = e0 f0 − e · f, g = e0f + f0e + f × e. (6.49)

This result was first established by Rodrigues [181] in 1840. The earliestEnglish commentary on it is by Cayley [32] in 1845.

(d) In terms of e0, f0, f, and e, what are the Euler parameters of the rotationtensors AB and BTAT?

(e) Using the results of (d), show that the compositions of rotations is not, ingeneral, commutative: i.e., AB �= BA.

(f) Recall that the angular velocity vector associated with A has the represen-tation

ωA = 2 (e0e − e0e + e × e) ,

where ωA = − 12ε[AAT

].

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(i) What are ωB and ωC?

(ii) Give an explanation for the following result:

ωC �= ωB + ωA.

(g) If e0 = f0 = 1√2, a = E3, and b = E2, then what does the rotation tensor C

represent? Illustrate your solution by showing how C transforms the basis{E1, E2, E3}.

6.5. The latitude (λ) and longitude (θ) of a point on the Earth’s surface are illus-trated schematically in Figure 6.9. In navigation systems, one uses these angles todefine the downward direction ez, the northerly direction ex, and the easterly direc-tion ey: ⎡

⎢⎣ex

ey

ez

⎤⎥⎦ =

⎡⎢⎣

− cos(θ) sin(λ) − sin(θ) sin(λ) cos(λ)


− cos(θ) cos(λ) − sin(θ) cos(λ) − sin(λ)

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ . (6.50)

Here, the triad {E1, E2, E3} is a set of fixed right-handed Cartesian basis vectors.

λ

θ

E1

E2

E3

ex ex

ey

ey

ez

−ez

North

East

Down

Figure 6.9. The angles of longitude θ and latitude λ.

(a) Suppose that R = ex ⊗ E1 + ey ⊗ E2 + ez ⊗ E3. Verify that this rotationtensor can be composed of two rotation tensors R1 and R2, where R1 cor-responds to a rotation about E3 through an angle θ and R2 corresponds toa rotation about ey through an angle −π

2 − λ.

(b) Given a vector x,

x = xxex + xyey + xzez = X1E1 + X2E2 + X3E3.

Show that

xx =3∑

i=1

Ri1Xi, xy =3∑

i=1

Ri2Xi, xz =3∑

i=1

Ri3Xi,

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Exercise 6.12 203

where

R =3∑

i=1

3∑k=1

RikEi ⊗ Ek.

How are Rik’s related to the matrix in Equation (6.50)?

6.6. Consider two rotation tensors A and B, where

A =3∑

i=1

ti ⊗ Ei, B =3∑

i=1

ei ⊗ ti,

and {E1, E2, E3} is a fixed, right-handed orthonormal basis for E3.

(a) Show that

B = oB +�AB − B�A,

whereoB is the corotational derivative of B assuming that ti are fixed, and

�A = AAT.

(b) Consider the rotation tensor C = BA. Using the results of (a), show that

�B = oB BT,

where the relative angular velocity tensor �B and the angular velocity ten-sor �C are, respectively,

�B = �C − �A, �C = CCT.

Why is �B skew-symmetric, and what does this imply for the product �Bb,where b is any vector?

(c) Consider the following examples of two tensors:

A = cos(ψ)(I − E3 ⊗ E3) − sin(ψ)εE3 + E3 ⊗ E3,

B = cos(θ)(I − t1 ⊗ t1) − sin(θ)εt1 + t1 ⊗ t1,

where

t1 = cos(ψ)E1 + sin(ψ)E2.

(i) What are A andoB?

(ii) Using Equation (6.14), what are ωA and ωB?

(iii) With the help of (6.14), verify that

ωB = θ t1 − ψ (cos(θ)E3 − sin(θ)t2) + ψE3.

Here, ωB is the angular velocity vector associated with B.

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6.7. Show that the rotation tensor L(π2 , E3)L(π

2 , E1) is equivalent to a rotation ten-sor L( 2π

3 , p), where

p = 1√3

(E1 + E2 + E3) .

This result is very useful in analyzing material symmetry groups of crystals.

6.8. Show that∗

L(π

2, E2

)L(π

2, E3

)L(−π

2, E1

)= L

(π

2, E3

).

This result has a interpretation that is sometimes used to show that successive rota-tions about three perpendicular axes can be reproduced by a single rotation aboutone of the axes. It is also the source for one explanation of a phenomenon in biome-chanics that is known as Codman’s paradox [40]. †

6.9. Examine Sections 1 and 24–30 of Euler [54], in which he introduces the Eulerangles (p, q, r). Show that a set of 1–3–1 Euler angles is being used, where

ψ = p, θ = q, φ = π − r.

You may wish to examine his expression for the components R = e1 · ω, P = e2 · ω,and Q = e3 · ω on page 205 of [54] to help with this. Related results, but with adifferent notation, can be found in [57]. In both of these papers you will find, amongother matters, Euler’s discussion of the components of Euler and inertia tensors.

6.10. If p1, p2 and p3 are any right-handed set of orthonormal basis vectors, thenestablish the Rodrigues–Hamilton theorem:

L (π, p3) L (π, p2) L (π, p1) = I. (6.51)

This result is discussed in Section 3 of Whittaker [228], and he credits it toRodrigues [181] and Hamilton [90]. Whittaker presents a proof that involves purelygeometric arguments, and it is a good exercise to compare your proof with his. TheRodrigues–Hamilton theorem will be used later to show that a constant moment isnot conservative.

6.11. Suppose a set of 3–2–1 Euler angles is used to parameterize a rotation tensorR. If we denote the values of these angles by α1, α2, and α3, respectively, then whatare the corresponding values of the 3–2–1 Euler angles for the inverse RT of thisrotation?

∗ For the rotation tensors discussed in this problem, you will need to compute εEk. If you need assis-tance with this, then please see (A.11) in the Appendix.

† As quoted in Politti et al. [174], Codman’s paradox occurs if you first place your right arm hang-ing down along your side with your thumb pointing forward and your fingers pointing toward theground. Now elevate your arm horizontally so that your fingers point to the right, then rotate yourarm in the horizontal plane so that your fingers now point forward, and finally rotate your armdownward so that your fingers eventually point toward the ground. After these three rotations, youwill notice that your thumb points to the left. That is, your arm has rotated by 90◦. The fact thatyou get this rotation without having performed a rotation about the longitudinal axis of your arm isknown as Codman’s paradox.

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Exercise 6.12 205

6.12. In Section 1.7 of Kane et al. [106] two distinct sets of Euler angles aredefined. The first corresponds to “body-angles” and the second corresponds to“space-angles.” Thus they define 24 sets of Euler angles. Verify that the 3–2–1Euler angles discussed in Subsection 6.8.1 are equivalent to the “body-three: 3–2–1”angles in [106]. In addition, show how a set of “space-three: 3–2–1” angles can beused to parameterize R. Denoting the “space-three: 3–2–1” angles by β1, β2, and β3,respectively, then, in your solution to this problem, you will find the representation⎡⎢⎢⎣

R11 R12 R13

R21 R22 R23

R31 R32 R33

⎤⎥⎥⎦ =

⎡⎢⎣1 0 0

0 cos (β3) − sin (β3)0 sin (β3) cos (β3)

⎤⎥⎦×

⎡⎢⎣

cos (β2) 0 sin (β2)

0 1 0

− sin (β2) 0 cos (β2)

⎤⎥⎦

×

⎡⎢⎣

cos (β1) − sin (β1) 0

sin (β1) cos (β1) 0

0 0 1

⎤⎥⎦ .

This result should be compared with corresponding representation (6.25) for the“body-three: 3–2–1” angles. For assistance with this problem, the discussion of“body-fixed” and “space-fixed” rotations in Section 3.2 of Ginsberg [71] and arelated discussion in Section 7.14 of Pars [170] might be helpful.

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7 Kinematics of Rigid Bodies

7.1 Introduction

This chapter contains results on the three-dimensional kinematics of rigid bodies.We discuss several useful classical representations for the velocity and accelerationvectors of any material point of a rigid body. We also discuss the angular velocityvector ω, the linear momentum G, angular momenta H, HO, and HA, and kinetic en-ergy T of rigid bodies and the inertias that accompany them. The chapter concludeswith a discussion of the configuration manifold for a rigid body.

The origin of most of the material in this chapter can be traced to Euler’s sem-inal work on rigid body dynamics in the 1750s. Since that time, his theory has beenused to develop models for a wide range of mechanical systems and various treat-ments of his work have appeared. Recently, a tensor-based notation has been usedby Beatty [15], Casey [26, 28], Fox [65], Greenwood [80], and Gurtin [84]. In thisbook, we follow their work as it leads to transparent developments particularly withregards to inertias and angular velocities.

7.2 The Motion of a Rigid Body

To discuss the kinematics of rigid bodies, it is convenient to follow some develop-ments in continuum mechanics and define the reference and present configurationof a rigid body. First, a body B is considered to be a collection of material points(mass particles or particles). We denote a material point of B by X. The positionof the material point X, relative to a fixed origin, at time t is denoted by x (seeFigure 7.1). The present (or current) configuration κt of the body is a smooth, one-to-one, onto function that has a continuous inverse. It maps material points X of Bto points in three-dimensional Euclidean space: x = κt(X). As the location x of theparticle X changes with time, this function depends on time, hence the subscript t. Itis important to note that κt defines the state of the body at time t.

We also define a fixed reference configuration κ0 of the body. This configura-tion is defined by the invertible function X = κ0(X). Using the invertibility of thisfunction, we can use the position vector X of a material point X in the referenceconfiguration to uniquely define the material point of interest. Later, we will use the

206

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7.2 The Motion of a Rigid Body 207

κ0

κt

O

X

X

Y

Y

X

X

X

YX

x

y

x

π

Figure 7.1. The reference κ0 and present κt configurations of a body B.

reference configuration to determine many of the properties of a body, such as itsmass m and inertia tensor J0. Using the reference configuration, we can define themotion of the body as a function of X and t:

x = χ(X, t).

Notice that the motion of a material point of B depends on the instant of time andthe material point of interest.

Euler’s TheoremFor rigid bodies, the nature of the function χ(X, t) can be simplified dramatically.First, for rigid bodies the distance between any two mass particles, say X1 and X2,remains constant for all motions. Mathematically, this is equivalent to saying that

||x1 − x2|| = ||X1 − X2|| . (7.1)

Second, the motion of the rigid body preserves orientations. In 1775, Euler [55, 56]showed that the motion of a body that satisfies (7.1) is such that

x1 − x2 = Q (X1 − X2) , (7.2)

where Q is a rotation tensor. As you may recall, this rotation tensor has an associ-ated axis and angle of rotation.

We can use (7.2) to tell if the rotation of a rigid body is nontrivial. That is, ifQ �= I, we can pick two points X1 and X2 of the rigid body and examine how therelative position vector x1 − x2 varies as a function of time. If, for all choices of X1

and X2, the relative position vector is unaltered in direction, then Q = I; otherwisethe rigid body is rotating. This is, perhaps, the most useful interpretation of Q: thetransformation that takes vectors between two points in the body and transformsthem into their present state.

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If we assume that one point of the body is fixed then we can simplify (7.2) bychoosing the fixed point to be the origin:

x(t) = Q(t)X. (7.3)

We can then infer Euler’s theorem on the motion of a rigid body:

Every motion of a rigid body about a fixed point is a rotation about an axis through thefixed point.

The axis here is the axis of rotation of Q(t). Because the motion of the body in ques-tion is from the configuration κ0 to κt, this axis depends on the choice of referenceconfiguration.

We can arrive at an alternative, and more common, interpretation of Euler’stheorem that does not feature the reference configuration κ0. To do this, we againconsider the motion of the body with a fixed point during the interval t ∈ [t0, t]. Wefind from (7.3) that

x (t) = Q (t) QT (t0) x (t0) .

Thus the motion of the body at the end of the time interval is characterized bythe rotation tensor Q (t) QT (t0).∗ Invoking Euler’s theorem, the axis of rotation ofQ (t) QT (t0) is the axis of rotation for the motion of the rigid body. It is emphasizedthat, during the motion in question, the body’s present configuration changes fromκt0 to κt.

Representations for the General MotionA general motion of a rigid body is one in which the body may not have a fixedpoint. In this case, it is easy to argue that a uniform translation of the reference con-figuration can be imposed on the rigid body so that an arbitrary one of its materialpoints XP is placed at its location xP(t) in the present configuration. The rigid bodyis then rotated about XP so as to occupy its present configuration κt:

x(t) − xP(t) = Q(t) (X − XP) .

That is,

x = Q(t)X + d(t), (7.4)

where d(t) = xP(t) − Q(t)XP. In words, (7.4) states that the most general motion ofa rigid body is a translation and a rotation.† It is one of the general representationsof the rigid body motion that we will often use in our subsequent discussions.

∗ In general, the tensor Q (t) QT (t0) has an axis of rotation and an angle of rotation that differ fromthose associated with Q (t).

† It should not come as a surprise that this result was also known to Euler. See his remarks in theintroductory sections to [52, 54].

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7.2 The Motion of a Rigid Body 209

κ0

κt

κt0

O

X

X

X

X

x (t)x (t0)

Q(t)

Q(t0)

Q (t) QT (t0)

Figure 7.2. Schematic of the configurations κt0 and κt of a rigid body. The reference configu-ration κ0 and illustrations of the roles played by several rotation tensors are also shown.

To discuss an alternative to (7.4) that does not feature the reference configura-tion, we consider the motion of the body during a time interval [t0, t] (see Figure 7.2).With the help of (7.4), we find that

x (t0) = Q (t0) X + d (t0) , x (t) = Q (t) X + d (t) .

Combining these equations, we arrive at an alternative representation of (7.4):

x (t) = Q (t) QT (t0) x (t0) + z (t) , (7.5)

where z (t) = d (t) − QT (t0) d (t0).∗

Result (7.5) is a convenient departure point to discuss a third alternative repre-sentation of rigid body motion. This representation of rigid body motion is synony-mous with a famous theorem, credited to Michel Chasles (1793–1880), on this topic(see Section 5 of [228] and references therein). Here, the motion of a rigid bodyis decomposed into a screw motion (see Figure 7.3). That is, the motion is consid-ered to be a rotation through an angle θ about an axis s(t) followed by a translationσ(t)s(t) along that axis. The screw axis s(t) is the axis of rotation of Q (t) QT (t0), andφ(t) is this tensor’s angle of rotation. The translational component σs(t) and the lo-cation ρ of the intercept of the screw axis can in principle be determined from thethree components of z(t) by use of three equations:

(I − Q(t)) ρ(t) + σ(t)s(t) = z(t). (7.6)

∗ You may notice that we can choose the reference configuration κ0 to be identical to κt0 . Then,Q (t0) = I, and representations (7.4) and (7.5) will be identical.

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κt

κt0

φ(t)

s(t)

O

X

X

x (t)

x (t0)

ρ1

ρ2 Q (t) QT (t0)

Figure 7.3. Schematic of the configurations κt0 and κt of a rigid body showing the screw axiss(t) and angle of rotation φ(t). Here, the rigid body is translated along the screw axis by anamount σs and rotated about s(t) through an angle φ. Two possible choices of ρ(t) are alsoshown. For the first, ρ(t) = ρ1 = ρ1E1 + ρ2E2 is chosen to be the intercept of the screw axiswith the E1 − E2 plane, and, for the second, ρ(t) = ρ2, where ρ2 is the vector from the originthat intersects the screw axis at a right angle: ρ2 · s = 0.

However, (I − Q(t)) is noninvertible,∗ and so ρ(t) is not uniquely defined. As aresult, several choices of ρ can be found in the literature (see, for example, the twochoices shown in Figure 7.3). Choosing ρ(t) to be normal to s leads to the followingsolutions for σ(t) and ρ(t):

ρ(t) = 12

(z⊥(t) + cot

(φ(t)

2

)s(t) × z(t)

), σ(t) = s(t) · z(t), (7.7)

where

z⊥(t) = z(t) − (z(t) · s(t)) s(t).

It is left as an exercise for the reader to verify (7.7). You should notice that a specialcase of the solution occurs when Q = I and ρ is indeterminate.

∗ There are many ways to see this result. The first is to refer the reader to the derivation of (6.15).Alternatively, one can use Euler representation (6.8) for a rotation tensor and observe that s is thezero eigenvalue of (I − Q(t)).

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7.3 The Angular Velocity and Angular Acceleration Vectors 211

7.3 The Angular Velocity and Angular Acceleration Vectors

The motion of a rigid body is defined by (7.4) or, equivalently, by (7.5). Because Q isa rotation tensor, we can define an angular velocity tensor � and an angular velocityvector ω∗:

� = QQT, ω = −12ε[QQT] .

The vector ω is the angular velocity vector of the rigid body, and � is the angular ve-locity tensor of the rigid body. As we know, the rotation tensor Q can be representedin a variety of manners, for instance, Euler angles or the Euler representation, andso too can its angular velocity vector. However, here it is convenient to omit explicitmention of these representations.

By differentiating the angular velocity vector, we find the angular accelerationvector of the rigid body:

α = ω.

You should notice that

α = −12ε[QQT + QQT] = −1

2ε[�],

where we used the fact that ε = O.We can use result (7.2) to determine the relative velocity and acceleration vec-

tors of any two points X1 and X2 of the rigid body:

v1 − v2 = x1 − x2

= Q (X1 − X2)

= QQTQ (X1 − X2)

= �Q (X1 − X2)

= ω × (x1 − x2) .

A further differentiation and some manipulations give the relative accelerationvectors:

a1 − a2 = v1 − v2

= ω × (x1 − x2) + ω × (x1 − x2)

= α × (x1 − x2) + ω × (v1 − v2) .

The final forms of the relative velocity and acceleration vectors are expressed asfunctions of t, x1, and x2. They can also be expressed as functions of t, X1, and X2.

∗ As QT (t0) = 0, ω is also the angular velocity vector associated with the rotation tensor Q(t)QT (t0).Thus representations (7.4) and (7.5) have the same angular velocity vectors.

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κ0 κtO

X1

X1

X2

X2

X3

X3

E1

E2

E3

e1

e2

e3

Figure 7.4. The corotational basis {e1, e2, e3} and the fixed Cartesian basis {E1, E2, E3}.

7.4 A Corotational Basis

It is convenient, when discussing the dynamics of rigid bodies, to introduce anotherbasis {e1, e2, e3}, which is known as a corotational basis.∗ Here, we define such abasis and point out some features of its use. Our discussion of the corotational basisfollows Casey [26] with some minor changes.

As is well known in the kinematics of rigid bodies, knowledge of the positionvectors of three material points suffices to determine the motion of the rigid body.Indeed, this is the premise for many navigation schemes and is the motivationfor our construction of a corotational basis. Referring to Figure 7.4, we start bypicking three material points X1, X2, and X3 of the body. These points are chosensuch that the orthonormal vectors E1 and E2 point from X3 toward X1 and X2,respectively:

E1 ‖ X1 − X3, E2 ‖ X2 − X3.

We then complete the (fixed) right-handed Cartesian basis by defining

E3 = E1 × E2.

Now consider the present locations of the three material points. Because Qpreserves lengths and orientations, the two vectors x1 − x3, x2 − x3 will retain theirrelative orientation. As a result, using (7.2), we define two orthonormal members ofa corotational basis by choosing them to point from x3 toward x1 and x2, respectively:

e1 ‖ x1 − x3, e2 ‖ x2 − x3.

We then define e3 = e1 × e2. As mentioned earlier, the basis {e1, e2, e3} is known asthe corotational basis.

It should be transparent that

Q = e1 ⊗ E1 + e2 ⊗ E2 + e3 ⊗ E3.

∗ This basis is often referred to as a body-fixed frame or an embedded frame.

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7.5 Three Distinct Axes of Rotation 213

This result follows from (7.2) and our previous discussions on representations ofrotation tensors. Because the corotational basis moves with the body, we can useour previous results for relative velocities and accelerations to see that

ei = ω × ei, ei = α × ei + ω × (ω × ei) ,

where i = 1, 2, 3.As the corotational basis is a basis for E

3, for any vector r we have the represen-tation

r =3∑

i=1

riei.

When the components ri are constant, then the vector is known as a corotationalvector. The most trivial examples of corotational vectors are e1, e2, and e3. The timederivative r of r has the representations

v = r

=3∑

i=1

riei +3∑

i=1

riei

= or + ω × r,

whereor is the corotational derivative (with respect to Q) of the vector r. A related

expression can be obtained for r:

a = v

=3∑

i=1

riei + 23∑

i=1

riei +3∑

i=1

riei

=3∑

i=1

riei + 2ω × or + ω × (ω × r) + α × r.

The presence of the Coriolis acceleration 2ω × or in the expression for a arises

because we have chosen to express r in a basis that is not fixed. You should alsoobserve that, if r is a corotational vector, then

or = 0, and the Coriolis acceleration

vanishes.

7.5 Three Distinct Axes of Rotation

It is possible to define three distinct axes of rotation for a rigid body. These axes arecommonly used in mechanics and navigation, and to discuss them it is convenient torecall the representations of rigid body motion (7.4) and (7.5). The rotation tensorQ(t) associated with the former has an axis of rotation q(t) and an angle of rotationθ(t), and the screw axis s(t) and angle φ(t) are the axis and angle of rotation, respec-tively, of Q (t) QT (t0). A third axis of rotation, which is known as the instantaneous

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ω

r

qr‖r⊥ Figure 7.5. An example of r(t) for a motion of a rigid body for which ω

is constant and r is a corotational vector. In this figure, r⊥ · ω = 0 andr⊥ + r‖ = r. The evolution of a possible axis of rotation q(t) for Q(t) isalso shown.

axis of rotation i, can also be defined. This axis is a unit vector parallel to the angularvelocity vector ω∗:

i(t) = ω

||ω|| .

Except in the simple case in which ω is constant, i does not have an associated angleof rotation. The terminology “instantaneous axis” can be appreciated from the ob-servation that r = ω × r for any corotational vector. Thus, if ω is constant, then, asshown in Figure 7.5, r(t) will appear to rotate about i.

Our definition of the instantaneous axis of rotation is identical to that used inclassical works on rigid body dynamics, for example, Poinsot [172] and Sections 405–406 of Poisson [173]. It is not universally adapted. For instance, in the literature onkinematics of anatomical joints the terminology instantaneous axis of rotation oftenrefers to s and not i.†

In general, the axes q, s, and i are not identical. However, (6.14) can be used torelate the axes and two angles of rotation:

ω = ||ω|| i

= φs + sin(φ)s + (1 − cos(φ))s × s

= θq + sin(θ)q + (1 − cos(θ))q × q. (7.8)

In addition, Rodrigues formula (6.49) can be used to relate q, s, φ, and θ, but this isleft as an exercise.

In the course of examining the rotation tensors from various problems in rigidbody dynamics, you can numerically compute s, q, and i. There you will easily findexamples in which these axes are distinct. It is, however, also of interest to con-sider examples in which some of these axes are equal. Two such examples are nowpresented. First, suppose that the body’s rotation tensor describes a steady rotation

∗ You may wish to recall that the angular velocity vector associated with Q (t) and Q (t) QT (t0) areidentical.

† The interested reader is referred to Woltring [232] and Woltring et al. [233] for further discussionon this matter.

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7.6 The Center of Mass and Linear Momentum 215

00

1

π

2π

tT

θ(t)

ν(t) − ν0

i = s = E3

E1

E2

e1

e2

e3

q

Figure 7.6. Plots of the loci of the extremities of the corotational vectors ei(t) and the axesof rotation s, q, and i for the rotation tensor (7.9). A plot of ν(t) is also shown in this figurewhere T = 2π

ν0.

about E3 through an angle ν, where ν = ν0 is constant: Q(t) = L (ν0t, E3). In thiscase, it is easy to compute that s(t) = i(t) = q(t) = E3 and ω = ν0E3.

Now consider an example in which Q describes a rotation about a time-varyingaxis of rotation at constant speed∗:

Q(t) = 2q(t) ⊗ q(t) − I, (7.9)

where the axis of rotation of Q(t) is

q(t) = cos(

ν

2

)E1 + sin

(ν

2

)E2, ν(t) = ν0 (t − t0) + ν0,

and ν0 and ν0 are constants. You should notice that, with the possible help of (6.14),ω = ν0E3, and that the angle of rotation θ for this rotation tensor is π. A standardcalculation also reveals that

Q(t)QT (t0) = L (−ν(t) + ν0, e3) .

That is, the rotation tensor Q(t)QT (t0) corresponds to the familiar rotation aboutE3 through an angle ν(t) − ν0. It now follows that s �= q. Indeed, (7.9) is the simplestexample of a rotation where ω is constant but the axis q is not parallel to ω that weknow of. The temporal behavior of ei(t) = QEi, the axes of rotation, and the anglesof rotation θ(t) and ν(t) are shown in Figure 7.6.

7.6 The Center of Mass and Linear Momentum

It is convenient (and traditional) to define a special material point of the rigid body,which we refer to as the center of mass X. We then use the velocity vector v of thispoint in the present configuration to define a very useful expression for the linearmomentum G of the rigid body.

∗ This example is adapted from [161]. Other examples of rotations with constant angular velocityvectors but distinct axes s and q can be found in [161].

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The Center of MassThe position vectors of the center of mass of the body in its reference and presentconfigurations are defined by

X =∫R0

Xρ0dV∫R0

ρ0dV, x =

∫R xρdv∫R ρdv

, (7.10)

where ρ0 = ρ0(X) and ρ = ρ(x, t) are the mass densities per unit volume of the bodyin κ0 and κt, respectively. The regions R0 and R denote the regions of E

3 occupiedby the body in κ0 and κt, respectively. If a body is homogeneous, then ρ0 is a constantthat is independent of X.

The principle of mass conservation states that the mass of the body is conserved.That is,

dm = ρ0dV = ρdv,

or, equivalently,

m =∫R0

ρ0dV =∫R

ρdv.

It follows immediately from (7.10) that

mX =∫R0

Xρ0dV, mx =∫R

xρdv.

We can also find from these results that

0 =∫R0

(X − X)ρ0dV, 0 =∫R

(x − x)ρdv.

These identities play key roles in establishing expressions for momenta and energiesof a rigid body.

A special feature of rigid bodies is that the center of mass behaves as if itwere a material point, which we denote by X. To see this we follow [26] andconsider

mx =∫R

xρdv

=∫R

(QX + d) ρdv

=∫R0

(QX + d) ρ0dV

= Q(∫

R0

Xρ0dV)

+(∫

R0

ρ0dV)

d

= mQX + md.

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7.6 The Center of Mass and Linear Momentum 217

Consequently,

x = QX + d. (7.11)

Recalling (7.4), this implies that the center of mass of the rigid body behaves as if itwere a material point of the rigid body. For many bodies, such as a rigid homoge-neous sphere, the center of mass corresponds to the geometric center of the sphere,whereas for others, such as a rigid circular ring, it does not correspond to a materialpoint.

It is convenient in many treatments of rigid bodies to define a reference frameconsisting of the center of mass X and the corotational basis {e1, e2, e3}. Such aframe is known as the corotational reference frame. In the treatment presented inthis book, we often express the relative position vectors of particles x − x in thisframe.

Linear MomentumBy definition, the linear momentum G of a rigid body is

G =∫R

vρdv.

That is, the linear momentum of a rigid body is the sum of the linear momenta ofits constituents. Using the center of mass, we can establish an alternative expressionfor G with the help of the definitions of m and x:

G = mv,

where v = ˙x is the velocity vector of the center of mass. You may recall that a relatedresult holds for a (finite) system of particles.

Relative Position VectorsFor a material point X of a rigid body, it is convenient to define the relative positionvectors π and :

π = x − x, = X − X.

Representative examples of these vectors are displayed in Figure 7.1. With the as-sistance of (7.2) and (7.11), we see that

π = Q. (7.12)

Using the corotational basis, we also easily see that

π · ei = · Ei.

This implies that the relative position vectors have the representations

=3∑

i=1

�iEi, π =3∑

i=1

�iei. (7.13)

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Furthermore, the corotational derivative (with respect to Q) of π is zero: π = ω × π.That is, π is a corotational vector.

7.7 Angular Momenta

Angular momenta of a rigid body are its most important distinctive feature whencompared with a particle. In particular, the angular momentum relative to twopoints, the center of mass X and a fixed point O, is of considerable importance. Forconvenience, we assume that the fixed point O is also the origin (see Figure 7.1).

By definition, the angular momenta of a rigid body relative to its center of massX, H, a fixed point O, HO, and a point A, HA, are

H =∫R

(x − x) × vρdv,

HO =∫R

x × vρdv,

HA =∫R

(x − xA) × vρdv.

The position vectors in these expressions are relative to the fixed point O, and xA

is the position vector of the point A. You should notice that the velocity vector inthese expressions is the absolute velocity vector.

The aforementioned angular momenta are related by simple and important for-mulae. To find one of these formulae, we perform some manipulations on HO:

HO =∫R

x × vρdv

=∫R

(π + x) × vρdv

=∫R

π × vρdv +∫R

x × vρdv

= H + x ×∫R

vρdv.

That is,

HO = H + x × G. (7.14)

This relation states that the angular momentum of a rigid body relative to a fixedpoint O is the sum of the angular momentum of the rigid body about its center ofmass and the angular momentum of its center of mass relative to O. Paralleling theestablishment of (7.14), it can also be shown that

HA = H + (x − xA) × G, HO = HA + xA × G. (7.15)

These results have obvious similarities to (7.14).

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7.8 Euler Tensors and Inertia Tensors 219

7.8 Euler Tensors and Inertia Tensors

To use the balance laws for a rigid body it is convenient to consider some furtherdevelopments of the angular momentum H. These developments are considerablyaided by use of the Euler tensors E0 and E and the inertia tensors J0 and J.

Euler TensorsWe next define the Euler tensors (relative to the center of mass of the rigid body):

E0 =∫R0

⊗ ρ0dV, E =∫R

π ⊗ πρdv.

You should notice that E and E0 are symmetric.Using mass conservation, (7.12), and the identity (Aa) ⊗ (Bb) = A(a ⊗ b)BT,

we easily see that

E = QE0QT.

Furthermore,

(Eei) · ek = (E0Ei) · Ek.

This implies that E has the representation

E =3∑

i=1

3∑k=1

Eikei ⊗ ek,

where Eik = (E0Ek) · Ei are the constant components of E0. In other words, al-though E is a function of time, its components, relative to the corotational basis,are constant. Furthermore, these constants are identical to the components of theconstant Euler tensor E0.

Inertia TensorsThe inertia tensors J0 and J can be defined by use of the Euler tensors:

J0 = tr(E0)I − E0, J = tr(E)I − E.

Using the definitions of the Euler tensors and the identity tr(a ⊗ b) = a · b, we canrestate these definitions as

J0 =∫R0

(( · )I − ⊗ ) ρ0dV, J =∫R

((π · π)I − π ⊗ π) ρdv. (7.16)

It is easy to see that

J = QJ0QT, JT0 = J0, JT = J.

The first of these results implies that

J =3∑

i=1

3∑k=1

Jikei ⊗ ek,

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where Jki = (J0Ei) · Ek are the constant components of J0.∗ The symmetry of theinertia tensors also implies that Jki = Jik.

Additional RelationshipsAnother set of interesting results follows by inverting the relationships between theEuler and inertia tensors. To do this, we use the definition of the inertia tensor interms of the Euler tensor and the fact that tr(I) = 3:

E0 = 12

tr(J0)I − J0, E = 12

tr(J)I − J. (7.17)

These results are useful for obtaining the Euler tensors from tabulations of the in-ertia tensor that are found in numerous undergraduate textbooks on dynamics.

It is a good exercise to substitute (7.13) into (7.16) to see that the components ofJ0 relative to the basis {E1, E2, E3} are volume integrals involving quadratic powersof the components of . For example,

J011 = (J0E1) · E1 =∫R0

(y2 + z2)ρ0dV,

J012 = (J0E2) · E1 = −∫R0

xyρ0dV,

J013 = (J0E3) · E1 = −∫R0

xzρ0dV, (7.18)

where

x = π · e1 = · E1, y = π · e2 = · E2, z = π · e3 = · E3.

A similar exercise with the components of E0 shows that

E011 = (E0E1) · E1 =∫R0

x2ρ0dV,

E012 = (E0E2) · E1 =∫R0

xyρ0dV,

E013 = (E0E3) · E1 =∫R0

xzρ0dV.

You should notice the simple relationship between the off-diagonal components ofE0 and J0. Again, one can then use tables of inertias found in textbooks to determinethe components of J0 and J.

Both inertia tensors J0 and J are symmetric. It can also be shown that they arepositive-definite. This allows us to choose {E1, E2, E3} such that these vectors arethe eigenvectors of J0, and, consequently,

J0 = λ1E1 ⊗ E1 + λ2E2 ⊗ E2 + λ3E3 ⊗ E3.

∗ Not surprisingly, this is similar to the situation we encountered earlier with the Euler tensors E andE0.

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7.8 Euler Tensors and Inertia Tensors 221

Here, λi are known as the principal moments of inertia. As J = QJ0QT and ei =QEi, we also have

J = λ1e1 ⊗ e1 + λ2e2 ⊗ e2 + λ3e3 ⊗ e3.

It is common to refer to ei as the principal axes of the rigid body. Now because ofthe definition of the components J0ik of the inertia tensor, the principal momentsneed to satisfy certain inequalities:

λ1 + λ2 > λ3, λ2 + λ3 > λ1, λ3 + λ1 > λ2.

These inequalities are easy to establish by use of definitions (7.18) and are useful inselecting representative examples of inertias. For instance, (7.19) imply that a rigidbody with λ1 = 1, λ2 = 2, and λ3 = 3 is not physically realizable.

It can also be shown that the eigenvectors of J0 are the eigenvectors of E0.Consequently, if we choose {E1, E2, E3} to be the eigenvectors of J0, then

E0 = e1E1 ⊗ E1 + e2E2 ⊗ E2 + e3E3 ⊗ E3.

By use of identity (7.17)1, the constants ei can be related to the principal momentsof inertia:

e1 = 12

(λ2 + λ3 − λ1) , e2 = 12

(λ1 + λ3 − λ2) ,

e3 = 12

(λ1 + λ2 − λ3) . (7.19)

As E = QE0QT and ei = QEi, we also have

E = e1e1 ⊗ e1 + e2e2 ⊗ e2 + e3e3 ⊗ e3.

Notice that the principal axis ej corresponding to the maximum value of λi corre-sponds to the minimum value of ei.

Some ExamplesIn many studies of satellites, it is convenient to model the satellite as a set of con-nected dumbbells. For such systems, the integrals in the definition of the Euler andinertia tensors degenerate into summations over a discrete number of particles. Forinstance, an example of such an arrangement is shown in Figure 7.7. For the bodyshown in this figure, it is easy to calculate E0:

E0 =3∑

k=1

2mkL2kEk ⊗ Ek.

Consequently, the inertia tensor is

J0 = 2(m2L2

2 + m3L23

)E1 ⊗ E1 + 2

(m1L2

1 + m3L23

)E2 ⊗ E2

+ 2(m1L2

1 + m2L22

)E3 ⊗ E3.

It is left as an exercise to write E and J.

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O

X

m1

m1

m2m2

m3

m3

L1

L2

L3

E1

E2

E3

Figure 7.7. The reference configuration of a dumbell satellite consisting of six masses joinedto a center by three massless, rigid rods of lengths 2L1, 2L2, 2L3.

The simplest inertia tensor arises when the body is a homogeneous sphere ofradius R or a homogeneous cube of length a:

J = J0 = 2mR2

5I, J = J0 = ma2

6I,

respectively. For these bodies, any three mutually perpendicular unit vectors areprincipal axes.

The next class of bodies is for those with an axis of symmetry. For instance,a homogeneous circular rod of length L and radius R has a moment of inertiatensor

J0 = mR2

2E3 ⊗ E3 +

(mR2

4+ mL2

12

)(I − E3 ⊗ E3) ,

where E3 is the axis of symmetry of the circular rod in its reference configuration.Most bodies, however, do not have an axis of symmetry. Consider the homo-

geneous ellipsoid shown in Figure 7.8. The equation for the lateral surface of theellipsoid is

x2

a2+ y2

b2+ z2

c2= 1.

X

Ellipsoid

E1

E2

E3

Figure 7.8. An ellipsoid of mass m.

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7.9 Angular Momentum and an Inertia Tensor 223

The inertia tensor of the ellipsoid is

J0 = m5

(b2 + c2)E1 ⊗ E1 + m

5

(a2 + c2)E2 ⊗ E2 + m

5

(a2 + b2)E3 ⊗ E3.

It is left as an exercise to write the Euler tensor E0 for the ellipsoid.∗

It is crucial to note that for all of the preceding examples we have used theproperty that any body has three principal axes. Writing the inertia and Euler ten-sors with respect to these axes provides their simplest possible representations.

7.9 Angular Momentum and an Inertia Tensor

The result we now wish to establish is that H = Jω. This is arguably one of the mostimportant results in rigid body dynamics. In particular, with the assistance of J0, fora particular rigid body it allows us to write a tractable expression for H.

We now reconsider the angular momentum H,

H =∫R

(x − x) × vρdv

=∫R

π × vρdv

=∫R

π × (v + ω × π)ρdv

=∫R

π × vρdv +∫R

π × (ω × π)ρdv.

However, because X is the center of mass and the velocity vector v of X is indepen-dent of the region of integration, we can take v outside the integral:

H =∫R

π × vρdv +∫R

π × (ω × π)ρdv

=∫R

πρdv × v +∫R


= 0 × v +∫R


=∫R

π × (ω × π)ρdv.

Notice that we also used the identity∫R πρdv = 0 in the next-to-last step of this

calculation. Summarizing, we have

H =∫R

π × (ω × π)ρdv =∫R

((π · π)ω − (π · ω)π) ρdv. (7.20)

∗ For assistance, see (7.19).

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In writing this equation, we used the identity a × (b × c) = (a · c)b − (a · b)c. By useof the definition of the inertia tensor, J, it should now be apparent that

H = Jω.

As mentioned earlier, this is one of the most important results in the kinematics ofrigid bodies.

You should notice that H = Jω implies that there is a linear transformation be-tween angular velocity and angular momentum. Furthermore, unless ω is an eigen-vector of J, H and ω will not be parallel.

7.10 Kinetic Energy

The kinetic energy of a rigid body has a very convenient representation. This repre-sentation, which was first established by a contemporary of Euler, Johann S. Koenig(1712–1757), is known as the Koenig decomposition:

T = 12

mv · v + 12

(Jω) · ω. (7.21)

Here, a derivation of this result is given.The kinetic energy T of a rigid body is defined to be

T = 12

∫R

v · vρdv.

We can simplify this expression for the energy by expressing the velocity vector v as

v = v + ω × π.

Substituting this expression into T, we have

T = 12

∫R

v · vρdv

= 12

∫R

v · vρdv +∫R

(ω × π) · vρdv + 12

∫R

(ω × π) · (ω × π)ρdv.

However, we have the following identities∫R

v · vρdv = v · v∫R

ρdv = mv · v,

∫R

(ω × π) · vρdv =(

ω ×(∫

Rπρdv

))· v = (ω × 0) · v = 0,

∫R

(ω × π) · (ω × π)ρdv =∫R

(ω · ω)(π · π) − (ω · π)2ρdv

= ω ·((∫

R(π · π)I − π ⊗ πρdv

)ω

)= ω · (Jω) .

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7.10 Kinetic Energy 225

Substituting these results into the previous expression for T, we find the desiredresult:

T = 12

mv · v + 12

(Jω) · ω.

This result is known as the Koenig decomposition of the kinetic energy of a rigidbody. In words, the kinetic energy of a rigid body is equal to the sum of the kineticenergy of its center of mass and the rotational kinetic energy Trot = 1

2ω · (Jω) of therigid body.

Comments on the Rotational Kinetic EnergyThe previous development of the Koenig decomposition showed that 2Trot = ω · Jω.This representation is used in the vast majority of works on rigid body dynamics. Anequivalent, complementary representation using J0 can also be found,∗ and we nowdiscuss this representation.

Consider the angular velocity vector ω0:

ω0 = QTω.

If ω =∑3i=1 ωiei, then it follows that

ω0 =3∑

i=1

ωiEi.

In addition, it can also be shown that ω0 is the axial vector of QTQ:

ω0 = −12ε[QTQ] = −1

2ε[QT�Q].

It is a good exercise to see what the representation of ω0 is when the 3–2–1 or 3–1–3Euler angles are used.

We now use the relationship between J and J0 to see that

2Trot = ω · (Jω)

= Qω0 · (JQω0)

= ω0 · (QTJQω0)

= ω0 · (J0ω0) .

This is the final desired result:

Trot = 12ω · Jω = 1

2ω0 · J0ω0.

The advantage of the representation involving ω0 is that J0 is a constant. Thus, whentaking derivatives of Trot with respect to the parameters used for Q, we need toconsider only the derivatives of ω0.

∗ See, for example, [28] and Chapter 15 of [138].

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226 Exercise 7.1

7.11 Concluding Remarks

We have now assembled most of the kinematical quantities required for character-izing the motion of a rigid body. The precise representations for these quantitiesthat are used generally depend on the problem at hand. For instance, we shall lateruse a set of Euler angles for a spinning-top problem that differs from a set used for aproblem featuring a satellite. Another example arises when we examine the dynam-ics of a rolling sphere. There, we will choose to use a fixed basis representation ofits angular velocity vector, ω =∑3

i=1 �iEi, whereas we will use the Euler basis forthe satellite problem. These choices are guided by experience and are often not in-tuitively obvious. It is hoped that the examples in the exercises and chapters aheadwill help you gain this needed experience. You might also have noticed that we haveyet to discuss constraints on the motions of rigid bodies. The next chapter is devotedto this topic.

EXERCISES

7.1. Recall that the rotation tensor Q of a rigid body has the representations

Q =3∑

i=1

ei ⊗ Ei =3∑

i=1

3∑k=1

Qikei ⊗ ek =3∑

i=1

3∑k=1

QikEi ⊗ Ek.

We recall that the corotational rates of a vector a and a tensor A relative to therotation tensor Q are defined as

oa = a − ω × a,

oA = A − �A + A�.

In these expressions, � = QQT and ω = − 12ε[QQT].

(a) If ai = a · ei and Aik = (Aek) · ei, then show that

oa =

3∑i=1

aiei,oA =

3∑i=1

3∑k=1

Aikei ⊗ ek.

Give physical interpretations of these results.

(b) A vector z is said to be corotational if z = QZ, where Z is constant. Giveexamples of such vectors from rigid body dynamics and show that the coro-tational rate relative to Q of a corotational vector is 0.

(c) Construct a definition of a corotational tensor and give two prominent ex-amples of such tensors from rigid body dynamics.

(d) Establish the following identities:

ω = oω, Q = QT

oQQ.

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(e) Suppose a rigid body is in motion and the corotational rate of ω is 0. Whatis the angular acceleration of the rigid body and what is the rotation tensorof the rigid body?∗

7.2. As shown in Figure 7.9, a robotic arm can be used to move a payload. The robotconsists of

1. a drive shaft that rotates about the E3 axis through an angle ψ ,

2. an axle A that rotates about g2 through an angle θ relative to the drive shaft,

3. an arm that rotates about g3 through an angle φ relative to axle A.

The payload is rigidly attached to the robotic arm. In this question, the drive shaft,the axle A, the robotic arm, and the payload are assumed to be rigid.

P

O

Axle

Drive shaft

Robotic arm

Payload

ψ

θ

φ

E1

E2

E3

g2

g3

Figure 7.9. Schematic of a robot consistingof a robotic arm, drive shaft, and axle. Themotors used to actuate the robot are notshown.

(a) What are the angular velocity vectors of the drive shaft, the axle A, therobotic arm, and the payload?

(b) Which set of Euler angles is being used to parameterize the rotation tensorQ of the payload?

(c) If the position vector r of a point X on the payload is

r = HE3 + Lg3,

then establish an expression for the velocity vector v of the point X.

(d) Suppose after a time interval t1 − t0 the point X of has returned to its origi-nal location in space:

r(t1) = r(t0).

Show that the payload will have rotated through an angle φ(t1) − φ(t0)about the axis g3(t0) during this interval of time. In other words, the

∗ A solution to this problem can be found in O’Reilly and Payen [161].

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rotation tensor Q(t1)QT(t0) corresponds to a rotation of φ(t1) − φ(t0) aboutg3(t0).

7.3. Consider the circular disk shown in Figure 7.10. The motion of the disk is givenby the position vector y of an arbitrary material point Y of the disk and the rotationtensor Q of the rigid disk.

g

P

O

X

ψ

φ

E1

E2

E3

e′′1

e′′1

e′′2

e1

e2

Figure 7.10. The present configuration of a circular disk moving with one point in contact witha fixed horizontal plane.

(a) Starting from the results that the motions of any points X and Y of the diskhave the representations

x = QX + d, y = QY + d,

show that their relative velocity vector and relative acceleration vectorsatisfy

x − y = ω × (x − y) ,

x − y = ω × (x − y) + ω × (ω × (x − y)) .

(b) To parameterize the rotation tensor of the disk, a set of 3–1–3 Euler anglesis used. With the assistance of Figure 7.10, prescribe a reference configura-tion for the disk. For which orientations of the disk in its present configura-tion do the singularities of the 3–1–3 Euler angles occur?

(c) A sensor is mounted onto the disk and is aligned with the e3 axis so that itmeasures ω · e3 = ω3(t). Show that, in general,∫ t1

t0ω3(t)dt �= φ(t1) − φ(t0). (7.22)

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Give a physical interpretation of the case in which the integral of ω3(t) doesyield the angle φ.∗

7.4. Consider a rigid body of mass m whose inertia tensors are defined as

J = QJ0QT, J0 =∫R0

( · )I − ⊗ ρodV.

(a) Explain why J0 and J are symmetric.

(b) Show that Ei · (J0Ek) = ei · (Jek), where Q =∑3i=1 ei ⊗ Ei.

(c) Show that J has the representation

J =3∑

i=1

λiei ⊗ ei, (7.23)

where λi are the principal values of J0 and Ei = QTei are the principal di-rections of J0.

(d) Establish the following identities:

J = �J − J�, H = Jω + ω × (Jω), ˙Jω · ω = 2H · ω.

How do these results simplify if J0 = µI, where µ is a constant? Note that,in the first of these results, you are showing that the corotational rate of J

relative to Q is zero:oJ = 0.

(e) Suppose one used the representation

J =3∑

p=1

3∑n=1

J pnEp ⊗ En.

Starting from (7.23), show that

J pn =3∑

i=1

QpiλiQni,

where Qpi = (QEi) · Ep. Why are the components J pn of J not constant?

7.5. Recall that a tensor that is intimately related to the familiar inertia tensor is theEuler tensor:

E = QE0QT, E0 =∫R0

⊗ ρodV.

(a) Show that

J0 = tr(E0)I − E0, J = tr(E)I − E.

(b) Verify that Ei · (E0Ek) = ei · (Eek), where Q =∑3i=1 ei ⊗ Ei.

(c) Establish the following results:oE = 0, E = �E − E�,

whereoE denotes the corotational derivative of E with respect to Q.

∗ An application of (7.22) to the navigation of motorcycles can be found in Coaplen et al. [39].

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(d) What are the Euler tensors for a sphere of mass m and radius R, and acylinder of mass m, radius R, and length L? You might find it convenientto use the relationships of the form E0 = 1

2 tr (J0) I − J0 that were discussedearlier in this chapter in Section 7.8.

7.6. Recall that the angular momentum H and rotational kinetic energy Trot of arigid body have the representations

H = Jω, Trot = 12ω · Jω.

Here, J is the moment of inertia tensor of the rigid body relative to its center ofmass. Choosing {Ei} to be the principal directions of J0, then

J0 =3∑

i=1

λiEi ⊗ Ei, J = QJ0QT =3∑

i=1

λiei ⊗ ei,

where Q =∑3i=1 ei ⊗ Ei.

(a) Using a set of 3–2–3 Euler angles to parameterize Q,∗ show that

ω = (θ sin(φ) − ψsin(θ) cos(φ))e1

+ (θ cos(φ) + ψsin(φ) sin(θ))e2

+ (ψcos(θ) + φ)e3.

(b) Using a set of 3–2–3 Euler angles, establish an expression for H and Trot asfunctions of the Euler angles and their time derivatives.

7.7. Recall the definitions of the angular momenta of a rigid body relative to itscenter of mass and a point A:

H =∫R

π × vρdv, HA =∫R

πA × vρdv. (7.24)

Here, π = x − x and πA = x − xA, where xA is the position vector of the point A andx is the position vector of the center of mass.

(a) Starting from (7.24)2, and with the assistance of (7.24)1, show that

HA = H + (x − xA) × G,

where G is the linear momentum of the rigid body.

(b) Suppose that A is a point of the rigid body. Show that

πA = QA,

where A = X − XA and XA is the position vector of A in the referenceconfiguration. If vA = 0, show in addition that

v = ω × (x − xA).

∗ This set of Euler angles is discussed in Exercise 6.2. In particular, the student is referred to (6.44) inExercise 6.2(g).

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(c) Again supposing that A is a point of the rigid body and that vA = 0, showthat

HA = JAω,

where JA is the inertia tensor of the rigid body relative to A:

JA =∫R

(πA · πA)I − πA ⊗ πAρdv.

Verify that

JA = QJA0 QT,

where

JA0 =

∫R0

((A · A)I − A ⊗ A) ρ0dV.

(d) Using the previous results, prove the parallel axis theorem∗:

JA0 = J0 + m((X − XA) · (X − XA)I − (X − XA) ⊗ (X − XA)). (7.25)

You should be able to see from this result how a parallel axis theorem re-lating two Euler tensors could be established.

(e) Suppose that the rigid body is a circular cylinder of mass m, length L, andradius R, where

X = 0, XA = −xE1 − zE3,

J0 = mL2

12(I − E3 ⊗ E3) + mR2

4(I + E3 ⊗ E3) .

Then, what is JA0 ?

7.8. Consider the rectangular parallelepiped of mass m and dimensions a, b, and cshown in Figure 7.11. Relative to the principal axes {A1, A2, A3}, the inertia tensorof this rigid body has the representation

JO =3∑

i=1

λiAi ⊗ Ai,

where

λ1 = m12

(b2 + c2), λ2 = m12

(a2 + c2), λ3 = m12

(a2 + b2).

The right-handed set of basis vectors {E1, E2, E3} is oriented so that E3 passes di-agonally through the parallelepiped. That is, E3 is parallel to the line connecting Xand the point P:

E3 = 1√a2 + b2 + c2

(aA1 − bA2 + cA3) .

∗ This representation of the parallel axis theorem is due to Fox [65].

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232 Exercise 7.8

In addition,

E1 = cos(α)A1 − sin(α)A3,

where

α = tan−1(

ac

).

P

ab

cX

E1

E2E3

A1

A2

A3

Figure 7.11. Schematic of a parallelepiped.

(a) After calculating E2, verify that the transformation from the basis{A1, A2, A3} to the basis {E1, E2, E3} can be written in the form

E1

E2

E3

= R

A1

A2

A3

, (7.26)

where

R =

R11 R12 R13

R21 R22 R23

R31 R32 R33

=

1 0 0

0 cos(β) sin(β)

0 − sin(β) cos(β)

cos(α) 0 − sin(α)

0 1 0

sin(α) 0 cos(α)

,

and

cos(β) =√

a2 + c2√

a2 + b2 + c2, sin(β) = b√

a2 + b2 + c2.

(b) Show that the components J0ik = (J0Ek) · Ei are

J0ik =3∑

r=1

RirλrRkr.

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Using a matrix notation, these equations are equivalent to

J011 J012 J013

J012 J022 J023

J013 J023 J033

= R

λ1 0 0

0 λ2 0

0 0 λ3

RT.

(c) If the body is given an angular velocity ω = ωe3, where ei = QEi, then es-tablish expressions for the angular momentum H of the body relative to itscenter of mass and the rotational kinetic energy Trot of the body. How dothese expressions simplify if a = b = c?

7.9. In this question, you will establish results for a set of Euler angles: the 3–1–2set. This set of angles is similar to the 3–2–1 set discussed in Subsection 6.8.1. Wethen apply these results to solve a navigation problem.

We shall decompose the rotation tensor Q into the product of three rotation ten-sors. The first rotation is about E3 through an angle ψ : L (ψ, E3). The second ro-tation is an angle θ about e′

1 = cos(ψ)E1 + sin(ψ)E2: L(θ, e′1). The third rotation is

through an angle φ about

e2 = e′′2 = cos(θ)e′

2 + sin(θ)e′3.

(a) Show that

e1

e2

e3

=

Q11 Q21 Q31

Q12 Q22 Q32

Q13 Q23 Q33

E1

E2

E3

,

where Qij = (QEj) · Ei.

(b) Show that the Euler basis for the 3–1–2 set of Euler angles has the repre-sentation

g1

g2

g3

=

0 0 1

cos(ψ) sin(ψ) 0

− cos(θ) sin(ψ) cos(θ) cos(ψ) sin(θ)

E1

E2

E3

.

Using the relations gk · gi = δki , show that the dual Euler basis for the 3–1–2

set of Euler angles has the representation

g1

g2

g3

=

sin(ψ) tan(θ) − cos(ψ) tan(θ) 1

cos(ψ) sin(ψ) 0

− sec(θ) sin(ψ) sec(θ) cos(ψ) 0

E1

E2

E3

.

You should notice that the second Euler angle needs to be restricted toθ ∈ (−π

2 , π2

).

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234 Exercise 7.9

(c) By examining the three rotations that compose Q and with the help of theresults in (a), show that

Q11 Q12 Q13

Q21 Q22 Q23

Q31 Q32 Q33

= CBA, (7.27)

where

A =

cos(φ) 0 sin(φ)

0 1 0

− sin(φ) 0 cos(φ)

,

B =

1 0 0

0 cos(θ) − sin(θ)

0 sin(θ) cos(θ)

,

C =


sin(ψ) cos(ψ) 0

0 0 1

.

(d) Select four different values of the set (φ, θ, ψ) and determine the axis ofrotation q and the angle of rotation of Q.

(e) The angular velocity vector associated with the 3–1–2 Euler angles has therepresentations

ω = ψE3 + θe′1 + φe2 = ω1e1 + ω2e2 + ω3e3.

Show that

ω1

ω2

ω3

=

− sin(φ) cos(θ) 0 cos(φ)

sin(θ) 1 0

cos(φ) cos(θ) 0 sin(φ)

ψ

φ

θ

,

and

ψ

φ

θ

=

− sec(θ) sin(φ) 0 sec(θ) cos(φ)

tan(θ) sin(φ) 1 − tan(θ) cos(φ)

cos(φ) 0 sin(φ)

ω1

ω2

ω3

. (7.28)

These sets of differential equations are very important in strap-down navi-gation systems.

(f) Suppose that a set of 3–1–2 Euler angles is used to parameterize the rota-tional motion of a vehicle: Q =∑3

i=1 ei ⊗ Ei. Here, e1 is taken to point inthe forward direction along the vehicle and E3 is taken to point verticallydownward. The first Euler angle ψ is known as the yaw angle, the secondangle θ is the roll angle, and the third angle φ is called the pitch angle.

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(i) A set of three gyroscopes is mounted on the vehicle and provides threesignals ωi(t) (see, for example, the signals shown in Figure 7.12).∗ Arguethat, with knowledge of the initial orientation of the vehicle, integrating(7.28), the orientation of the vehicle can be determined.

(ii) Suppose the vehicle is initially level. As time progresses the gyroscopesregister the following signals:

ω1(t) = 0.02 sin(0.5t),

ω2(t) = 0.02 sin(0.05t),

ω3(t) = 2.0 sin(0.5t).

Determine ek(t) for the vehicle.

(iii) How does the axis of rotation q of the vehicle change and is this axisparallel to the instantaneous axis of rotation i = ω

||ω||?

0

100

100−100

time (s)

o/s

ω1 + 40

ω2 − 40

ω3

Figure 7.12. Time traces of ω3(t), ω2(t) − 40, and ω1(t) + 40 from a set of three gyroscopesmounted on a motorcycle. The motorcycle is moving in a rectangular path and the offsets ±40are imposed so that the three signals can be distinguished. It is also interesting to observe theamount of noise in the signals for ωi(t).

7.10. As shown in Figure 7.13, a Poisson top is an axisymmetric body with a sharpapex that is free to move on a flat surface. The material point of the top in con-tact with the surface is labeled P. The position vectors of this point relative to thefixed point O in the present and reference configurations are denoted by xP and XP,respectively. The material point corresponding to the center of mass of the top is de-noted by X, and the position vectors of X relative to the fixed point O in the presentand reference configurations are denoted by x and X, respectively.

(a) A set of 3–2–1 Euler angles is used to describe the orientation of the top.Denoting the first angle by ψ , the second by θ, and the third by φ, givesketches showing how these angles describe the orientation of the top.

∗ The principle of operation of a class of gyroscopes is discussed in Section 11.4.

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236 Exercise 7.10

g

O

X

X

P

P

E1

E1

E2

E2

E3

E3

e1

e2

e3

κ0

κt

Figure 7.13. The fixed reference κ0 and present κt configurations of a Poisson top. The surfacethat the top moves on is taken to be the E1 − E2 plane.

(b) For which orientations of the top does the set of 3–2–1 Euler angles havesingularities? Now, suppose a set of 3–1–3 Euler angles were used to pa-rameterize the rotation tensor of the top. For which orientations of the topdoes the 3–1–3 set have singularities?

(c) Starting from the result that for any point X on a rigid body,

x = QX + d,

show that

x = Q (X − XP) + xP, x = ω × (x − xP) + xP.

(d) The moment of inertia tensor of the top in its reference configuration hasthe representation

JO = λaE3 ⊗ E3 + λt (I − E3 ⊗ E3) .

If

P = XP − X = −L3E3, ω = ω1e1 + ω2e2 + ω3e3,

then show that the angular momentum vector of the top relative to the pointP is

HP = λaω3e3 + (λt + mL23

)(ω1e1 + ω2e2) + L3e3 × mxP.

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8 Constraints on and Potentialsfor Rigid Bodies

8.1 Introduction

In this chapter, constraints and the forces and moments that enforce them inthe dynamics of rigid bodies are discussed. In particular, the constraints associ-ated with interconnected rigid bodies, rolling rigid bodies, and sliding rigid bod-ies along with prescriptions for the associated constraint forces and moments arepresented. We also discuss Lagrange’s prescription for constraint forces and con-straint moments and outline its limitations. It also proves convenient to discusspotential energies and their associated conservative forces and moments. Our dis-cussion includes as examples springs and central gravitational fields. It has ob-vious parallels to the treatment of constraints and their associated forces andmoments.

8.2 Constraints

Constraints in the motions of rigid bodies usually arise in two manners. First, therigid body is connected to another rigid body in such a way that its relative motionis constrained. The connections in question are usually in the form of joints. Thesecond manner in which constraints arise occurs when one rigid body is rolling orsliding on the other. As in particles, the constraints we discuss can be classified asintegrable or nonintegrable, and this classification is important in dynamics becauseit may lead to considerable simplification in the formulation of the equations gov-erning the motion.

In our discussion, we consider two rigid bodies, B1 and B2 (see Figure 8.1). Wecan consider the case of a rigid body interacting with the ground by prescribing thevelocity vector of the center of mass and the angular velocity vector of either oneof the bodies as functions of time. If more than two rigid bodies are involved inthe constraint, then it is not too difficult to generalize the discussion subsequentlypresented. It should also be noted that the discussion presented here of the varioustypes of joints and contacts is not exhaustive.

237

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238 Constraints on and Potentials for Rigid Bodies

O

X1

X2

x1x2

Figure 8.1. The present configurations and centers of mass of two rigid bodies B1 and B2.

Connected Rigid BodiesAs shown in Figure 8.2, consider two rigid bodies B1 and B2 that are connected atthe point P by a joint. The position vector of P on Bα relative to Xα is denoted byπPα. Because the point P occupies the same location for both bodies, we have thefollowing three constraints∗:

x1 + πP1 = x2 + πP2. (8.1)

These constraints also imply that

v1 + ω1 × πP1 = v2 + ω2 × πP2. (8.2)

Notice that by integrating (8.2) with respect to time and setting the constants of in-tegration to 0, we will arrive at (8.1). Consequently, constraints (8.2) are integrable.

The joint at P may have the ability to restrict the rotation of B2 relative to B1.There are two types of joints to consider: the pin (or revolute) joint and the ball-and-socket joint. The pin joint arises in gyroscopes where it serves to connect theinner and outer gimbals. On the other hand, ball-and-socket joints do not place anyrestriction on the relative angular velocity vector ω2 − ω1.

Consider the case of a pin joint. Let s3 be a unit vector that is parallel to theaxis of relative rotation permitted by the joint, and let {s1, s2, s3} be an orthonormalbasis. The pin joint ensures that the relative rotation Q2QT

1 is a rotation about s3.These restrictions are most easily expressed in terms of the relative angular velocityvector. Specifically, a pin joint imposes the constraints

(ω1 − ω2) · s1 = 0, (ω1 − ω2) · s2 = 0.

It should be clear that these two (integrable) constraints are supplemented by (8.2).In addition, it is sufficient to use a single angle to parameterize Q2QT

1 .

∗ By taking the components of this vector equation with respect to a basis, one arrives at three inde-pendent scalar equations. Hence the vector equation represents three constraints.

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8.2 Constraints 239

O

P

X1

X2

x1

x2πP1

πP2

Figure 8.2. The present configurations of two rigidbodies B1 and B2 that are connected by a joint at P.

Clearly each of the individual constraints mentioned to this point can be writtenin the form

π = 0, (8.3)

where

π = f1 · v1 + f2 · v2 + h1 · ω1 + h2 · ω2 + e.

Here, f1, f2, h1, h2, and e are functions of x1, x2, Q1, Q2, and t. Furthermore, (8.3) canbe integrated with respect to time to yield a function �:

� = � (x1, x2, Q1, Q2, t) .

Here,

� = π.

In other words, for this case, the constraint π = 0 is said to be integrable or holo-nomic. In general, the constraints associated with connections are usually integrable.This is in marked contrast to the next set of situations we consider.

Rolling and Sliding Rigid BodiesConsider the situation shown in Figure 8.3. Here, two rigid bodies are instanta-neously in contact at the point P. It is assumed that there is a well-defined unitnormal n to the surfaces of both bodies at the point P. In addition, we use n to de-fine an orthonormal basis {s1, s2, n}, where s1 and s2 are tangent to the surfaces ofboth bodies at the point of contact P.

As the point of contact P coincides with a material point of each body, we have

xp1 = x1 + πp1, vp1 = v1 + ω1 × πp1, (8.4)

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O

P

X1

X2

x1

x2πP1

πP2

n

s1

s2

Figure 8.3. The present configurations of two rigid bodies B1 and B2 that are in contact at thepoint P.

and

xp2 = x2 + πp2, vp2 = v2 + ω2 × πp2. (8.5)

It is important to note that it is not generally possible to differentiate (8.4)1 to arriveat (8.4)2, nor (8.5)1 to arrive at (8.5)2. The reason for this is that πpα does not identifythe same material of Bα at each instant of time. In other words, Pα corresponds to adifferent material point of Bα at each instant of time.

The problems in rigid body dynamics involving rolling or sliding rigid bodiesthat have received the most attention involve rigid bodies for which the vectors πpα

have relatively simple forms, specifically, rolling/sliding spheres, the sliding top, andthe rolling/sliding disks. The most famous example for which πpα is very difficultto express in a tractable form is the wobblestone or celt. As first reported by G. T.Walker in 1896 [223], this rigid body has the unique feature of reversing its directionof spin in a counterintuitive manner.∗

8.2.1 The Sliding Condition

For two bodies in contact, it is assumed that the contact persists and that the twobodies do not interpenetrate. These assumptions lead to the sliding condition:

vp1 · n = vp2 · n. (8.6)

We can express this condition in another form:

(v1 − v2) · n = (ω2 × πp2 − ω1 × πp1) · n.

∗ For further details on this interesting mechanical system, and more recent research on it, see [17]and references therein.

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8.3 A Canonical Function 241

The sliding condition implies that a certain relative velocity has components in onlythe tangential directions:

vs = vp1 − vp2 = vs1s1 + vs2s2. (8.7)

The velocity vs is often known as the sliding velocity and is important for specifyingthe friction forces at the point P.

8.2.2 The Rolling Condition

Rolling occurs when the velocity vectors of the point of contact for each body areidentical. In this case, we have the rolling condition:

vp1 = vp2.

Again, we can express this equation in another form:

v1 + ω1 × πp1 = v2 + ω2 × πp2. (8.8)

These three equations are equivalent to sliding condition (8.6) and the conditionthat the sliding velocity vs is zero.

Rolling condition (8.8) is equivalent to three scalar equations [cf. (8.3)]:

π1 = 0, π2 = 0, π3 = 0,

where, for i = 1, 2, 3,

πi = fi1 · v1 + fi2 · v2 + hi1 · ω1 + hi2 · ω2 + ei.

However, for two of these equations, say π2 = 0 and π3 = 0, it is not possible tofind functions �2 and �3 such that �2 = π2 and �3 = π3. In other words, two ofconstraints (8.8) are nonintegrable or nonholonomic. The one constraint of (8.8)that is integrable corresponds to the n component of (8.8). We shall shortly discusssome examples that illustrate this point.

8.3 A Canonical Function

For future purposes, it is convenient to consider two rigid bodies B1 and B2 andconstruct the general functional form of a possible integrable constraint or potentialenergy function that features the motions of both bodies. This function is presentedin (8.9) and, for future purposes, we also calculate its derivative. Our treatment herefollows [163].

As shown in Figure 8.1, the position vector of the center of mass Xα of the bodyBα is denoted by xα, where α = 1 or 2. Similarly, the rotation tensor of Bα is denotedby Qα. For each rigid body, we can define corotational bases:

Q1 =3∑

i=1

1ei ⊗ Ei, Q2 =3∑

i=1

2ei ⊗ Ei.

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It should be noted that we are using the same fixed basis {E1, E2, E3} to define thecorotational bases. The angular velocity vectors of the bodies are

ω1 = −12ε[Q1QT

1

], ω2 = −1

2ε[Q2QT

2

].

It should also be noted that the rotation tensor of B2 relative to B1 is

Q2QT1 =

3∑i=1

2ei ⊗ 1ei.

This tensor represents the rotation of B2 relative to an observer who is stationaryon B1.

A Scalar-Valued Function of the MotionsWe now consider a scalar function :

= (x1, x2, Q1, Q2, t) .

Clearly this function depends on the motions of both rigid bodies and time. Oneobserves functions of this type when representing integrable constraints on the mo-tions of rigid bodies and potential energies of rigid bodies.

As in the case of a particle, it is of interest to calculate . To calculate this timederivative, we invoke the chain rule:

= ∂

∂x1· v1 + ∂

∂x2· v2 + tr

(∂

∂Q1QT

1

)+ tr

(∂

∂Q2QT

2

)+ ∂

∂t.

Recalling our earlier discussion in Section 6.10 of derivatives of scalar functions ofrotation tensors, it is convenient to define the skew-symmetric tensor

Qα= 1

2

(∂

∂Qα

QTα − Qα

(∂

∂Qα

)T)

and an associated vector

ψQα= −ε [Qα

] .

Representations for these tensors and vectors, based on the parameterization of Qα,were discussed earlier. In particular, if the Euler angles νi are used to parameterizeQ1, say, then

ψQ1 =3∑

i=1

∂

∂νigi, ψQ1 · ω1 =

3∑i=1

∂

∂νiνi, (8.9)

where gi are the dual Euler basis vectors associated with the Euler angles.We can use the aforementioned skew-symmetric tensors to rewrite :

= ∂

∂x1· v1 + ∂

∂x2· v2 + tr

(Q1�

T1

)+ tr(Q2�

T2

)+ ∂

∂t

= ∂

∂x1· v1 + ∂

∂x2· v2 + ψQ1 · ω1 + ψQ2 · ω2 + ∂

∂t,

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8.4 Integrability Criteria 243

where

�α = QαQTα , ωα = −1

2ε[�α], (α = 1, 2) ,

are the angular velocity tensors and vectors of the rigid bodies.It is more efficient next to use the form of that involves vectors:

= ∂

∂x1· v1 + ∂

∂x2· v2 + ψQ1 · ω1 + ψQ2 · ω2 + ∂

∂t. (8.10)

With some minor rearrangements, we can eliminate v2 and ω2 in favor of the rel-ative velocity vectors v2 − v1 and ω2 − ω1 in the expression for . Equation (8.10),which first appeared in [163], will play a key role in examining potential forces andmoments.

Result (8.10) is rarely apparent in treatments of rigid body dynamics. This ispartially because specific parameterizations of xα and Qα are used. To elaborate onthis point, let

x1 =3∑

i=1

xiEi, x2 =3∑

i=1

yiEi,

and suppose that Q1 is parameterized by the Euler angles νi and the relative rotationtensor Q2QT

1 is parameterized by the Euler angles γi. That is,

ω1 =3∑

i=1

νigi, ω2 − ω1 =3∑

i=1

γini,

where gi and ni are the Euler basis vectors associated with νi and γi, respectively.Then,

= (x1, x2, Q1, Q2, t) = (xi, yj, ν

i, γk, t).

Furthermore,

= ˙

=3∑

i=1

(∂

∂xixi + ∂

∂yiyi + ∂

∂νiνi + ∂

∂γiγi)

+ ∂

∂t.

It is good exercise to compare this expression with (8.10) and identify the corre-sponding terms in both expressions.

8.4 Integrability Criteria

Earlier, in Section 1.8, we examined integrability criteria for constraints of the formf · v + e = 0 on the motion of a single particle. Here, we examine the correspondingcriteria for a constraint on the motions of two rigid bodies. In the dynamics of rigidbodies, situations involving systems of constraints are often inescapable: One onlyhas to think of the case of a rolling rigid body. It is also possible that, when a suffi-cient number of constraints are imposed on a rolling rigid body, the resulting system

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of constraints becomes integrable. Fortunately, there is a theorem, which is due toFrobenius, that provides a criterion for the integrability of a system of constraints.His criterion will be presented shortly. Before doing so, we discuss the situation ofa single constraint on a pair of rigid bodies.

A Single ConstraintFor rigid bodies, what is needed is a generalization of criterion (1.21) for a singleparticle to constraints of the form

π = 0,

where

π = f1 · v1 + f2 · v2 + h1 · ω1 + h2 · ω2 + e.

As mentioned earlier, this constraint is integrable if we can find an integrating fac-tor k and a function �(x1, x2, Q1, Q2, t) such that k� = π. Although the integrabilitycriterion is daunting in the number of algebraic calculations needed, for many prob-lems most of these calculations are trivial (but tedious).

The necessary and sufficient conditions for π = 0 to be integrable involve a setof up to 66 independent conditions. To establish these conditions, which are similarto those we discussed earlier for a single particle, we assume that x1 is parameterizedby use of a coordinate system {q1, q2, q3}, x2 is parameterized by use of a coordinatesystem {q4, q5, q6}, Q1 is parameterized by use of {ν1, ν2, ν3}, and Q2 is parameterizedby use of {ν4, ν5, ν6}. Thus the function π can be expressed as

π =3∑

i=1

(f1i qi + f2iqi+3 + h1iν

i + h2iνi+3)+ e.

We also define the intermediate functions

Wi = f1i, Wi+3 = f2i, Wi+6 = h1i, Wi+9 = h2i, W13 = e,

and variables

U i = qi, U i+3 = qi+3, U i+6 = νi, U i+9 = νi+3, U 13 = t,

where i = 1, 2, 3. It remains to define

IJKL = WJ SLK + WKSJL + WLSKJ, (8.11)

where SLK are the components of a skew-symmetric matrix:

SLK = ∂WL

∂UK− ∂WK

∂UL. (8.12)

In (8.11) and (8.12), the integer indices J , K, and L range from 1 to 13.With all the preliminaries taken care of, we are now in a position to state a

theorem that is due to Frobenius:

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8.4 Integrability Criteria 245

A necessary and sufficient condition for the constraint π = 0 to be integrable is that thefollowing 13

6 (13 − 1)(13 − 2) equations hold for all U1, . . . , U13:

IJKL = 0, for all J, K, L ∈ {1, . . . , 13}, J �= K �= L, J �= L. (8.13)

A proof of the theorem can be found in texts on differential equations (forexample, Section 161 of Forsyth [64]). We also note that only 66 of the 286 equa-tions IJKL = 0 are independent. Criterion (8.13) can be specialized to the case of aconstraint f · v + h · ω + e = 0 on a single rigid body. There, the indices J, K, L willrange from 1 to 7, and the number of independent conditions IJKL = 0 to verify willbe 10.

Systems of ConstraintsWhen rigid bodies are subject to several constraints, a criterion is available by whichthe possible integrability of the system of constraints can be evaluated. This criterionis known as Frobenius’ theorem. To simplify our exposition, we frame this criterionin terms of a single rigid body. Using the developments of the previous section, wecan easily extend this to the case of two rigid bodies.

Suppose we have a system of R ≤ 6 constraints on the motion of a single rigidbody:

π1 = 0, . . . , πR = 0, (8.14)

where

πA = fA · v + hA · ω + eA (A = 1, . . . , R).

Again, we assume that coordinates{q1, q2, q3

}have been chosen for the position

vector of the center of mass and parameters{ν1, ν2, ν3

}have been selected for the

rotation tensor of the rigid body. Thus each of the πA’s can be rewritten as

πA =3∑

i=1

(fAiqi + hAiν

i)+ eA.

Next, we define the following functions and variables:

WAi = fAi, WA(i+3) = hAi, WA7 = eA,

SALK = ∂WAL

∂UK− ∂WAK

∂UL, (8.15)

where we have used the following notation for the variables

Ui = qi, Ui+3 = νi, U7 = t. (8.16)

In (8.15) and (8.16), i = 1, 2, 3 and L, K = 1, . . . , 7. The functions WAL and SALK can

be used to construct two matrices, for example,

W =

⎡⎢⎢⎣

W11 · · · W17

.... . .

...

WR1 · · · WR7

⎤⎥⎥⎦ . (8.17)

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With the help of (8.15), it is easy to see that S1, . . . , SR are skew-symmetric matrices:SA

KL = −SALK. We also define the following seven-dimensional vectors:

a =

⎡⎢⎢⎣

a1

...

a7

⎤⎥⎥⎦ , b =

⎡⎢⎢⎣

b1

...

b7

⎤⎥⎥⎦ , x =

⎡⎢⎢⎣

x1

...

x7

⎤⎥⎥⎦ . (8.18)

We are now in a position to state the Frobenius integrability theorem:

For a single rigid body, the necessary and sufficient conditions for a system of R < 7constraints π1 = 0, . . . , πR = 0 to be integrable are for the following equations to hold

aT · (SAb) = 0 (A = 1, . . . , R),

for all values of the variables U1, . . . , U7, and for all distinct solutions a and b to theequation

Wx = 0,

that is, the seven-dimensional vectors a and b lie in the null space of W.

REMARKS ON FROBENIUS’ THEOREM. The statement of Frobenius’ remarkable theo-rem is adapted from Forsyth [63], and the reader is referred to this text for a clas-sical proof and to a paper by Hawkins [91] for additional historical perspectives.More modern proofs, featuring differential forms, are readily found; see, for exam-ple, [34, 63, 101, 189].

As SA are skew-symmetric, aT · (SAa) = 0 for all possible a. Often this result is

a key to using the theorem: For if W has a one-dimensional null space, then there isonly one vector, say b, satisfying Wb = 0, and then the theorem is trivially satisfied.

EXAMPLES. An example of the use of the theorem is discussed in the exercises atthe end of this chapter. There it is used to show that a rolling disk is generally sub-ject to nonintegrable constraints. However, when the disk is further constrained tobe vertical and its center of mass moves in a straight line, then the constraints onthe rolling disk become integrable. As discussed in Exercise 8.9, this is the familiarsituation from introductory dynamics classes.

A second example of the use of the theorem was mentioned earlier in conjunc-tion with two scleronomic constraints (1.23) on the motion of a single particle:

f11x + f12y + f13z = 0, f21x + f22y + f23z = 0,

where f1 =∑3k=1 f1kEk and f2 =∑3

k=1 f2kEk are functions of the position vector ofthe particle, and f2 × f1 �= 0. To apply Frobenius’ theorem to this case, some trivialmodifications to (8.15)–(8.18) are needed. First,

U1 = x, U2 = y, U3 = z, U 4 = t.

For this case, R = 2, and the indices L and K range from 1 to 4. It is left as an exerciseto establish that the 2 × 4 matrix W has a two-dimensional null space that is spanned

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8.5 Forces and Moments Acting on a Rigid Body 247

by the vectors

a = [0, 0, 0, 1]T, b = [g1, g2, g3, 0]T

,

where gk = (f1 × f2) · Ek. Now, as

SA =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0∂fA1

∂x2− ∂fA2

∂x1

∂fA1

∂x3− ∂fA3

∂x10

∂fA2

∂x1− ∂fA1

∂x20

∂fA2

∂x3− ∂fA3

∂x20

∂fA3

∂x1− ∂fA1

∂x3

∂fA3

∂x2− ∂fA2

∂x30 0

0 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

it follows that aT · (S1b) = 0 and aT · (S2b

) = 0. Thus the hypotheses of Frobenius’integrability theorem are satisfied, and we conclude that system of constraints (1.23)is integrable.

8.5 Forces and Moments Acting on a Rigid Body

Before discussing constraint forces and moments, we dispense with some prelimi-naries. The resultant force F acting on a rigid body is the sum of all the forces actingon a rigid body. Similarly, the resultant moment relative to a fixed point O, MO,is the resultant external moment relative to O of all of the moments acting on therigid body. We also denote the resultant moment relative to the center of mass X byM. These moments may be decomposed into two additive parts, the moment thatis due to the individual external forces acting on the rigid body and the applied ex-ternal moments that are not due to external forces. We refer to the latter as “pure”moments.

As an example, consider a system of forces and moments acting on a rigid body.Here, a set of L forces FK (K = 1, . . . , L) acts on the rigid body. The force FK acts atthe material point XK, which has a position vector xK. In addition, a pure momentMp, which is not due to the moment of an applied force, also acts on the rigid body(see Figure 8.4). For this system of applied forces and moments, the resultants are

F =L∑

K=1

FK,

MO = Mp +L∑

K=1

xK × FK,

M = Mp +L∑

K=1

(xK − x) × FK.

You should notice how the pure moment Mp features in these expressions.

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O

κt

XK

X

x

xK

πK Mp

FK

Figure 8.4. A force FK and a moment Mp acting on arigid body.

The mechanical power P of a force FK acting at a material point XK is definedto be

P = FK · xK.

Using the center of mass, we can obtain a different representation of this power.Specifically, we use the identity, xK = v + ω × πK, where πK = xK − x. It then fol-lows that

P = FK · xK = FK · v + (πK × FK) · ω.

In words, the power of a force is identical to the combined power of the same forceacting at the center of mass, and its moment, relative to the center of mass. Themechanical power of a pure moment Mp is defined to be

P = Mp · ω.

You should notice how this expression is consistent with the previous expression forthe mechanical power of a force FK.

Using the results for the mechanical power of a force FK and a pure momentMp, we find that the resultant mechanical power of the system of L forces and apure moment discussed previously is

P =L∑

K=1

FK · xK + Mp · ω = F · v + M · ω.

These results will play a key role in our future discussion of constraint forces andmoments and potential energies.

8.6 Constraint Forces and Constraint Moments

Given a system of constraints on the motion of one or more rigid bodies, a systemof forces and moments is required for ensuring that the constraints are enforced forall possible motions of the bodies that are compatible with the constraints. At issuehere is the prescription of these forces and moments.

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8.6 Constraint Forces and Constraint Moments 249

One of the major points of this chapter is the form of Lagrange’s prescriptionassociated with constraints of the form [cf. (8.3)]

π = 0,

where

π = f1 · v1 + f2 · v2 + h1 · ω1 + h2 · ω2 + e.

We showed earlier how most of the commonly used constraints in rigid body dy-namics could be written in this form.

Pedagogically, it is easiest to present some examples and then give Lagrange’sprescription for the constraint forces and constraint moments. However, this pre-scription is not universally applicable. For instance, as in particle dynamics, it doesnot give physically realistic constraint forces and constraint moments when dynamicfriction is present.

8.6.1 A Rigid Body Rotating About a Fixed Point

Consider the rigid body shown in Figure 8.5. The body is pin-jointed to the fixedpoint O. As a consequence of the pin joint, the body performs a fixed-axis rotationabout E3 = e3. Further, if the rotation of the body is known, then the motion of thecenter of mass is also known.

As a result of the pin joint, the angular velocity vector ω of the body is subjectto two constraints. These constraints can be expressed in a variety of manners. Forinstance,

ω · E1 = 0, ω · E2 = 0, (8.19)

or, equivalently,

ω · e1 = 0, ω · e2 = 0.

Now because O is fixed, there are three additional constraints on the motion of thebody:

v − ω × x = 0. (8.20)

O

X

x

E1

E2

e1

e2

Figure 8.5. A rigid body B that is performing a fixed-axis rotationabout the point O.

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In other words, the pin joint couples the rotational motion of the body with themotion of its center of mass.

To enforce constraints (8.19), we assume that two constraint moments are ex-erted by the pin joint on the body. These moments have components perpendicularto E3:

Mc1 = µ4E1, Mc2 = µ5E2,

where µ4 and µ5 are functions of time that are determined from the balance laws.These constraint moments are examples of pure moments. Our prescription of theconstraint moments assumes that the pin joint is frictionless. If friction were present,then they would have components in the E3 direction. Because the pin joint imposesthe constraints that O is fixed, it also supplies a reaction force R acting at O:

R =3∑

i=1

µiEi,

where µ1, µ2, and µ3 are functions of times that are determined from the balancelaws.

The reaction force and constraint moments are equipollent to a constraint forceFc acting at the center of mass of the rigid body and a constraint moment Mc relativeto the center of mass of the rigid body:

Fc =3∑

i=1

µiEi, Mc = µ4E1 + µ5E2 − x ×(

3∑i=1

µiEi

).

Notice that there are five constraints on the rigid body that were imposed by thepin joint and five unknown functions µ1, . . . , µ5 in the expressions for the constraintforces and moments.

8.6.2 A Sphere Rolling or Sliding on an Inclined Plane

The problem of a sphere rolling or sliding on a plane has a long history, in partbecause it is the basis for pool (billiards) and bowling. The main contributors to thisproblem in the 19th century were Gaspard G. de Coriolis (1792–1843) (see [41]) andEdward J. Routh (1831–1907) (see [184]). Studies on the dynamics of rolling sphereson surfaces of revolution are often known as Routh’s problem.∗

Consider the rigid body shown in Figure 8.6. The body in this case is movingon a fixed inclined plane. In the figure, the body is assumed to be a sphere. For thesphere, it is easy to see that πP = −RE3, where R is the radius of the sphere. Inaddition, you should notice that {s1, s2, n} = {E1, E2, E3}.

∗ The interested reader is referred to the informative account of Routh’s influence as a tutor andteacher in [226].

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P

g O

X

β

Sphere of mass m and radius R

Inclined plane

E1

E3

Figure 8.6. A rigid sphere moving on an inclined plane.

Let us first consider the case in which the sphere is assumed to be sliding. In thiscase, sliding condition (8.6), vP · n = 0, is simply

v · E3 = 0.

In addition, the slip velocity vs is

vs = vs1E1 + vs2E2 = v + ω × (−RE3) .

To enforce the sliding condition, a force Fc acts at the point P:

Fc = µ3E3 − µd||µ3E3|| vs

||vs|| ,

where µd is the coefficient of dynamic friction. As in the previous example, µ3 is afunction of time that is determined by the balance laws.

For the rolling sphere, there are three constraints:

vP · E1 = 0, vP · E2 = 0, vP · E3 = 0.

These constraints can be expressed in an alternative form:

v · Ei − R(ω × E3) · Ei = 0,

where i = 1, 2, or 3. To enforce the three constraints, we assume that a normalforce and a static friction force act at P. The sum of these two forces is a reactionforce R:

R =3∑

i=1

µiEi,

where µi are functions of time that are determined from the balance laws. Youshould notice that the reaction force R is equipollent to a force Fc = R acting atthe center of mass and a moment Mc = −RE3 × R relative to the center of mass. Arelated comment pertains to the sliding rigid body.

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g

P

O

X

ψ

φ

E1

E2

E3

e′′1

e′′1

e′′2

e1

e2

Figure 8.7. A thin circular disk moving with one point in contact with a rough horizontalsurface.

8.6.3 The Rolling Disk

As shown in Figure 8.7, a thin rigid circular disk of mass m and radius R rolls (with-out slipping) on a rough horizontal plane. The rotation tensor of the disk will bedescribed by use of a 3–1–3 set of Euler angles (see Subsection 6.8.2). This set hasthe advantage of having the singularities of the Euler angles coincide with the disklying flat on the horizontal plane (i.e., θ = 0, π).

For the disk, the position vector of the instantaneous point of contact P relativeto the center of mass X has the representations

πP = −Re′′2 = −R sin(φ)e1 − R cos(φ)e2

= −R (− cos(θ) sin(ψ)E1 + cos(θ) cos(ψ)E2 + sin(θ)E3) .

Taking the Cartesian components of the rolling condition, vP = v + ω × πP, we findthat the motion of the disk is subject to three constraints:

π1 = 0, π2 = 0, π3 = 0, (8.21)

where

π1 = x1 + Rφ cos(ψ) + Rψ cos(θ) cos(ψ) − Rθ sin(θ) sin(ψ),

π2 = x2 + Rφ sin(ψ) + Rψ cos(θ) sin(ψ) + Rθ sin(θ) cos(ψ),

π3 = x3 − Rθ cos(θ),

and x =∑3i=1 xiEi. You should notice that the last of these three constraints inte-

grates to x3 = R sin(θ) + constant. However, as is shown in one of the exercises atthe end of this chapter, system of constraints (8.21) is not integrable.

The rolling of the disk is possible because it is assumed that a static friction forceFf acts at P:

Ff = µ1E1 + µ2E2.

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Furthermore, this force and the normal force N = µ3E3 are equipollent to

Fc = Ff + N =3∑

i=1

µiEi, Mc = πP × Fc.

Here, µ1, µ2, and µ3 are determined from the balance laws.

8.6.4 Lagrange’s Prescription

For simplicity, we first discuss Lagrange’s prescription for the constraint forces andconstraint moments acting on a single rigid body.∗ Subsequently, the case of mul-tiple constraints in a system of two rigid bodies is discussed. We close this sectionwith an illustration of Lagrange’s prescription by using two pin-jointed rigid bodies.There we also find a generalization of Newton’s third law. Further illustration of theprescription applied to systems of rigid bodies can be found in Sections 11.3 and 11.4as well as in Exercise 11.4, which features two rolling rigid bodies and the Dynabee.

A SINGLE RIGID BODY. From the examples discussed in the previous sections, we as-sume that any constraint on a rigid body can be written as [cf. (8.3)]

π = 0,

where

π = f · v + h · ω + e.

Here, f, h, and e are functions of t, x, and Q. Lagrange’s prescription states that theconstraint force Fc and constraint moment Mc associated with this constraint is

Fc = µf, Mc = µh,

where µ is indeterminate. This form of the prescription is an obvious generalizationof the corresponding prescription for a single particle and system of particles.

For a system of R constraints on the motion of a rigid body, we have R con-straints of the form

πK = 0,

where K = 1, . . . , R, and

πK = fK · v + hK · ω + eK.

For this system of constraints, Lagrange’s prescription states that

Fc = µ1f1 + · · · + µRfR, Mc = µ1h1 + · · · + µRhR,

where µ1, . . . , µR are indeterminate. Notice that the prescription introduces R un-knowns for the R constraints.

∗ The developments presented in this section are based on the work of O’Reilly and Srinivasa [163].

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By examining the examples presented in this chapter, you should be able toverify that, apart from situations in which dynamic Coulomb friction is present, La-grange’s prescription gives physically realistic results.∗ You should also show that,unless e is zero for a single constraint, then the constraint force Fc and the constraintmoment Mc will collectively do work.

TWO RIGID BODIES. Suppose we have a system of R constraints acting on a system oftwo rigid bodies:

πK = 0,

where K = 1, . . . , R. We assume that the constraints are of the form [cf. (8.3)]:

πK = fK1 · v1 + fK2 · v2 + hK1 · ω1 + hK2 · ω2 + eK.

Here, fK1, fK2, hK1, hK2, and eK are functions of x1, x2, Q1, Q2, and t. Lagrange’s pre-scription states that the constraint forces and constraint moments that enforce theseR constraints are

Fc1 = µ1f11 + · · · + µRfR1,

Fc2 = µ1f12 + · · · + µRfR2,

Mc1 = µ1h11 + · · · + µRhR1,

Mc2 = µ1h12 + · · · + µRhR2. (8.22)

Here, Fcα is the resultant constraint force acting on Bα, Mcα is the resultant con-straint moment relative to the center of mass of Bα acting on Bα, and µ1, . . . , µR arefunctions of time that are determined from the balance laws for the system of tworigid bodies.

You should notice that, for each of individual R constraints, Lagrange’s pre-scription introduces a single unknown function of time: µK. Consequently, if thereare 10 constraints on the motions of the two rigid bodies, the prescription will intro-duce the unknowns µ1, . . . , µ10.

TWO PIN-JOINTED BODIES AND NEWTON’S THIRD LAW. For example, consider the caseof two rigid bodies connected by a pin joint shown in Figure 8.8. For this situation,there are five integrable constraints on the relative motion of the bodies:

1 = 0, . . . , 5 = 0,

∗ See, for example, the discussion of the constraint forces and moments on a rolling disk in Subsection8.6.3.

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O

Pin joint

x1

x2

X1

X2

πP1

πP2

Figure 8.8. The present configurations of two rigidbodies B1 and B2 that are connected by a pin joint.The axis of the pin joint is aligned with 2e3 = 1e3,and its position vector relative to the centers of massof the bodies are πP1 and πP2, respectively.

where

1 = v1 · E1 + (ω1 × πP1) · E1 − v2 · E1 − (ω2 × πP2) · E1,

2 = v1 · E2 + (ω1 × πP1) · E2 − v2 · E2 − (ω2 × πP2) · E2,

3 = v1 · E3 + (ω1 × πP1) · E3 − v2 · E3 − (ω2 × πP2) · E3,

4 = ω1 · 1e1 − ω2 · 1e1,

5 = ω1 · 1e2 − ω2 · 1e2.

We emphasize that the axis of the pin joint is aligned with 1e3 = 2e3, and its positionvectors relative to the centers of mass of the bodies are πP1 and πP2.

Using Lagrange’s prescription, (8.22), with the five constraints 1 =0, . . . , 5 = 0, we find that

Fc1 =3∑

i=1

µiEi,

Fc2 = −3∑

i=1

µiEi,

Mc1 = µ41e1 + µ51e2 + πP1 ×3∑

i=1

µiEi,

Mc2 = −µ41e1 − µ51e2 − πP2 ×3∑

i=1

µiEi. (8.23)

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Fc1

Fc2

Mr

−Mr

X1

X2

πP1

πP2

Figure 8.9. Free-body diagrams of the two rigid bodies B1 and B2 illustrating the constraintforces Fc1 and Fc2 and the reaction moment Mr. The reaction moment has the prescriptionMr = µ41e1 + µ51e2, and the constraint forces are equal and opposite [cf. (8.24)].

These expressions can be written in a more revealing fashion:

Fc1 = −Fc2,

Mc1 = µ41e1 + µ51e2 + πP1 × Fc1,

Mc2 = −µ41e1 − µ51e2 + πP2 × Fc2. (8.24)

As illustrated in Figure 8.9, Fc1 and Fc2 are the equal and opposite reaction forcesat the pin joint, and Mr = µ41e1 + µ51e2 and −µ41e1 − µ51e2 are the equal and op-posite reaction moments. Relative to the center of mass of each body, the reactionmoments Mc1 and Mc2 are not equal and opposite unless πP2 = πP1.∗

A NOTE OF CAUTION. For the cases of a frictionless joint between two bodies, fortwo bodies in rolling contact, and for two bodies in frictionless sliding contact,Lagrange’s prescription provides a very convenient method of specifying the con-straint forces and constraint moments. However, the prescription does not yieldphysically reasonable forces and moments for joints or contact where dynamic fric-tion is present.

8.7 Potential Energies and Conservative Forces and Moments

The presence of conservative forces and moments acting on rigid bodies is one of thekey features used to solve many problems. Although the definition of a conservativeforce originally arose in the dynamics of a single particle and is well understood, thesame cannot be said for conservative moments (see, for example, Antman [5] andO’Reilly and Srinivasa [163]). Indeed, as noted by Ziegler [234], for rigid bodies

∗ This was first noted in [163] and can be considered as a generalization of Newton’s third law to rigidbodies.

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8.7 Potential Energies and Conservative Forces and Moments 257

whose axis of rotation is not fixed, a constant moment is not necessarily conserva-tive. To this end, we start with three well-known examples of forces and their asso-ciated moments. These examples are followed by a discussion of Ziegler’s example.After the examples have been presented, a general treatment of conservative forcesand moments is given. Following this treatment, you should return to Ziegler’s ex-ample and convince yourself that a constant moment is not conservative.

Constant ForcesA constant force P acting on a rigid body is conservative. If this force acts at allmaterial points of the rigid body, then it is equipollent to a single force

∫R Pρdv

acting at the center of mass of the rigid body. The potential energy of this forceis − ∫R Pρdv · x. Notice that there is no moment (relative to the center of mass)associated with this force.

The most ubiquitous example of a constant force is a constant gravitationalforce acting at each material point. This force is equipollent to a force mgg acting atthe center of mass. Here g is the direction of the gravitational force. The potentialenergy of this force is −mgg · x

Spring ForcesIn many mechanical systems, a spring is used to couple the motions of two bodies.Consider the system of two rigid bodies shown in Figure 8.10. Here, a linear springof stiffness K and unstretched length L0 is connected to the material point Xs1 of thebody B1 and the material point Xs2 of the body B2.

The spring exerts a force Fs1 at the point Xs1 and an equal and opposite forceFs2 at the point Xs2:

Fs1 = −K(||xs1 − xs2|| − L0)xs1 − xs2

||xs1 − xs2|| ,

Fs2 = −K(||xs1 − xs2|| − L0)xs2 − xs1

||xs1 − xs2|| .

It is easy to see that each of these forces is equipollent to a moment relative tothe center of mass and a force acting at the center of mass. The potential energyassociated with the spring is

Us = K2

(||xs1 − xs2|| − L0)2.

You should notice that this potential energy depends on the position vectors of bothmaterial points Xs1 and Xs2.

It is also convenient to note the identities

xs1 = Q1(Xs1 − X1

)+ x1, xs2 = Q2(Xs2 − X2

)+ x2,

where Xsα is the position vector of Xsα in the reference configuration of Bα. Usingthese identities, we can see that the forces, moments, and potential energy of thespring can be expressed as functions of the rotation tensors of both rigid bodies andthe position vectors of their centers of mass.

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O

Linear spring

x1

x2

X1

X2

Xs1

Xs2

πs1

πs2

Figure 8.10. The present configurationsof two rigid bodies B1 and B2 that areconnected by a spring.

Central Gravitational FieldsIn celestial and orbital mechanics, a standard problem is to consider the motion ofa body subject to a central force field. Such force fields date to Newton and arebased on his inverse-square force. You may recall that this conservative force isthe force exerted on a particle of mass m by a particle of mass M: F = −GMm

||r||2r

||r|| ,where r is the position vector of m relative to M and G is the universal gravitationalconstant.

For the force fields of interest, we consider two bodies B1 and B2 with massdensities per unit volume of ρ1 and ρ2, respectively. Every material point of B2 exertsan attractive force on each material point of B1 (see Figure 8.11). If we integratethese forces over all material points in B1 and B2, we will obtain the resultant forceexerted by B2 on B1. Similar integrations apply to the moment and potential energyof these forces. In short, the resultant gravitational force Fn and moment Mn on B1

that are due to B2 are

Fn = −∫R1

∫R2

G||x1 − x2||2

x1 − x2

||x1 − x2||ρ1dV1ρ2dV2,

Mn = −∫R1

∫R2

(x1 − x1) × G||x1 − x2||2

x1 − x2

||x1 − x2||ρ1dV1ρ2dV2.

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O

C

Rigid body B1

Rigid body B2

xx1

x2

X1

X2

Figure 8.11. Two rigid bodies B1 and B2 that exert mutual gravitational forces. The point C inthis figure denotes the center of mass of this system of rigid bodies.

The moment Mn is relative to the center of mass X1 of B1. The potential energyassociated with this force is

Un = −∫R1

∫R2

G||x1 − x2||ρ1dV1ρ2dV2.

Assuming that B2 is spherical and has a mass M, the expressions for the force, mo-ment, and potential energy simplify:

Fn = −∫R1

GM||x1 − x2||2

x1 − x2

||x1 − x2||ρ1dV1,

Mn = −∫R1

(x1 − x1) × GM||x1 − x2||2

x1 − x2

||x1 − x2||ρ1dV1,

Un = −∫R1

GM||x1 − x2||ρ1dV1.

It is important to notice that, even in this simplified case, the gravitational force canexert a moment on the rigid body B1.

To simplify the expressions as much as possible, we now use the fact that B1 is arigid body:

π = x1 − x1 = Q.

In addition, we assume that ||π|| is small relative to ||x1 − x2||. These assumptions al-low us to approximate the forces, moments, and potential energy associated with the

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central force field. After a substantial amount of manipulation, we find the results

Fn ≈ mg,

Mn ≈(

3GMR3

)c × (Jc) = −

(3GM

R3

)c × (Ec),

Un ≈ −GMmR

−(

GM2R3

)tr(J) +

(3GM2R3

)(c · (Jc)) , (8.25)

where J is the inertia tensor of B1 relative to its center of mass, E is the Euler tensorof B1 relative to its center of mass, m is the mass of B1,

mg = −GMmR2

c − 3GM2R4

(2J + (tr(J) − 5c · Jc) I) c,

and

R = ||x1 − x2|| , c = x1 − x2

||x1 − x2|| .

Notice that c points from the center of mass of the spherically symmetric body B2 tothe center of mass of B1. In addition, because of the presence of the inertia tensorJ, (8.25)1 does not correspond to Newton’s gravitational force on a particle of massm by another particle of mass M. Classical expressions (8.25) are used in the vastmajority of works on satellite dynamics (cf. Beletskii [16], Hughes [97], and Kane etal. [106]). The expression for the potential energy Un in the form shown in (8.25) iscredited to James Mac Cullagh (1809–1847).∗

In many of the works on the dynamics of a satellite about a fixed point O, itis common to approximate Fn with Fn = −GMm

R2 c. Such an approximation effectivelydecouples the motion of the center of mass X1 of B1 from its orientation, and one canthen conclude that X1 behaves like a particle in the one-body problem and solve forthe orientation of the satellite separately.† However, as pointed out by Barkin [12]and Wang et al. [224], this approximation violates angular momentum conservationand energy conservation. Studies on the dynamics of satellites for which Fn = mgcan be found in [12, 164, 224, 225]. These works show that steady motions of thesatellite are possible where the orbital plane of X1 does not contain the origin O.This is in contrast to the one-body problem where the orbital plane contains O.

Constant Moments That Are Not ConservativeTo see that a constant moment is not conservative, we recall Ziegler’s example [234].He considered a constant moment ME3. During a motion of a rigid body consistingof a rotation about E3 through −π rad, this moment does work equal to Mπ. How-ever, the same final orientation of the body can be achieved by a rotation about E1

through π rad, followed by a rotation about E2 through π rad.‡ Now, however, the

∗ Notes on his derivation of Un and Fn can be found in Propositions 4 and 5 of [1]. Comments byAllman on the derivation of Mn can also be found in [1].

† The one-body problem was discussed earlier in Section 2.8.‡ This is an example of the application of Rodrigues–Hamilton theorem (6.5) discussed in the exer-

cises at the end of Chapter 6.

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work done by the constant moment is zero! Consequently, the work done by themoment ME3 depends on the “path” taken by the body – and hence ME3 cannot beconservative.

In many courses on dynamics, the rotation of the rigid body is constrained to bea fixed-axis rotation and the rotations about E1 and E2 are not permitted. For thesecases, a constant moment ME3 is conservative. An alternative proof that a constantmoment is generally not conservative can be found in [160]. The proof presentedthere exploits the dual Euler basis.

General ConsiderationsIt is convenient at this point to give a general treatment of conservative forces andmoments in the dynamics of rigid bodies. Our discussion is in the context of two rigidbodies, but it is easily simplified to the case of one rigid body and easily generalizedto the case of N rigid bodies.

We assume that the most general form of the potential energy is

U = U (x1, x2, Q1, Q2) .

Clearly, this function depends on the motions of both rigid bodies and time. We cancalculate the time derivative of this function by using (8.10):

U = ∂U∂x1

· v1 + ∂U∂x2

· v2 + uQ1 · ω1 + uQ2 · ω2,

where uQαis a vector representing the derivative of U with respect to Qα. As dis-

cussed in Section 6.10, this vector has numerous representations, and the easiest touse arises when Q1 and Q2 are parameterized by sets of Euler angles. Denotingthese angles and their dual Euler basis vectors by

{γ1, γ2, γ3

}and

{g1, g2, g3

}, and{

ν1, ν2, ν3}

and{h1, h2, h3

}, respectively, the vectors uQ1 and uQ2 have the represen-

tations

uQ1 =3∑

k=1

∂U∂γk

gk, uQ2 =3∑

k=1

∂U∂νk

hk.

Consider the conservative forces Fconα and moments Mconα (relative to the re-spective centers of mass) associated with this potential. We assume that the workdone by these forces and moments is dependent on the initial and final configura-tions of the rigid bodies but is independent of the motions of the rigid bodies. Thisimplies that

−U = Fcon1 · v1 + Fcon2 · v2 + Mcon1 · ω1 + Mcon2 · ω2.

Substituting for U and collecting terms, we find that(Fcon1 + ∂U

∂x1

)· v1 +

(Fcon2 + ∂U

∂x2

)· v2 + (Mcon1 + uQ1 ) · ω1

+ (Mcon2 + uQ2 ) · ω2 = 0. (8.26)

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This can be interpreted as an equation for Fconα and Mconα that must hold for allmotions of the rigid bodies. Assuming that Fconα and Mconα are independent of thelinear and angular velocity vectors, we find that in order for (8.26) to hold for all vα

and ωα it is necessary and sufficient that

Fconα = − ∂U∂xα

, Mconα = −uQα. (8.27)

These are the expressions for the conservative forces and moments associated witha potential energy. It is left as an exercise for you to verify that these expressions areconsistent with the results presented earlier for the spring and central gravitationalforces.

An Example Featuring a Torsional SpringWe now present a very simple example featuring a torsional spring acting on a singlerigid body. This example illustrates representations (8.27) and makes use of the dualEuler basis. Even in this simple case, the conservative moment associated with thespring has a surprising feature.

Consider a single rigid body whose rotation tensor is parameterized by a set of3–1–3 Euler angles and suppose that a torsional spring acts on the body. The springis assumed to have a potential energy U = Kt

2 ψ2, where Kt is a torsional stiffness.Then, with the help of (8.27) and representation (8.9) for uQ in terms of Euler angleswe find that the conservative moment is

Mcon = −uQ = − ∂U∂ψ

g1

= −Ktψg1

= −Ktψ (cot(θ) (cos(ψ)E2 − sin(ψ)E1) + E3) .

The conservative force Fcon associated with this potential is 0 because the potentialis independent of x. It is interesting to notice that the components of Mcon are notentirely in the direction of E3.

8.8 Concluding Comments

The main results in this chapter involved prescriptions for constraint and conserva-tive forces and moments. For the former, we argued that Lagrange’s prescriptionprovided these expressions in cases in which dynamic Coulomb friction was absent.For instance, if a single rigid body is subject to a constraint that can be expressed as

f · v + h · ω + e = 0,

then Lagrange’s prescription states that

Fc = µf, Mc = µh.

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If we suppose that a single rigid body has a potential energy function

U = U(x1, x2, x3, γ

1, γ2, γ3) ,where xi are the Cartesian coordinates for x and γ i are Euler angles, then our de-velopments showed that the conservative force Fcon and conservative moment Mcon

associated with this potential are

Fcon = −3∑

i=1

∂U∂xi

Ei, Mcon = −3∑

i=1

∂U∂γ i

gi,

where gi are the dual Euler basis vectors associated with the Euler angles.

EXERCISES

8.1. Suppose the motion of a rigid body is subject to two constraints:

ω · g3 = 0, ω · g2 = 0.

Here, gi are the dual Euler basis for a set of Euler angles of your choice. Give a phys-ical interpretation of these constraints. What are the angular velocity vector, angularmomentum vector, and rotational kinetic energy of the resulting constrained rigidbody?

8.2. Consider a force P acting at a point P of a rigid body. In the present configura-tion, the point P has the position vector πP relative to the center of mass X of therigid body:

πP = xP − x.

In addition,

vP = xP = v + ω × πP.

(a) The mechanical power of P is P · vP. Show that this power has the equiva-lent representation

P · vP = P · v + (πP × P) · ω.

Using a free-body diagram, give a physical interpretation of this identity.

(b) A force P acting at the point P is said to be conservative if there exists apotential energy U = U(xP) such that

P = − ∂U∂xP

.

Show that this definition implies that

−U = P · v + (πP × P) · ω.

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(c) Show that the potential energy U = U(xP) can also be described as a func-tion of x, Q, X and P:

U = U(xP) = U(x, Q,P, X

).

Here, Q is the rotation tensor of the rigid body, and πP = QP. Using thisequivalence, show that

P = −∂U∂x

, πP × P = −uQ.

(d) Consider the case in which P represents a force that is due to a spring ofstiffness K and unstretched length L. One end of the spring is attached toa fixed point O. What are the functions U(xP) and U(x, Q,P, X) for thisforce P?

8.3. Consider two rigid bodies. The rotation tensor of the first rigid body is

Q1 =3∑

i=1

1ei ⊗ Ei.

The rotation tensor of the second rigid body is

Q2 =3∑

i=1

2ei ⊗ Ei.

Here, 1ei corotate with the first rigid body and 2ei corotate with the second rigidbody.

(a) Argue that the rotation tensor of the second body relative to the first bodyis

R = Q2QT1 =

3∑i=1

2ei ⊗ 1ei.

What is the rotation tensor of the first body relative to the secondbody?

(b) Show that the angular velocity vector of the second body relative to the firstbody is

ω = −12ε

[3∑

i=1

2oei ⊗2ei

],

where we have used the corotational derivative with respect to Q1:

oa= a − ω1 × a.

(c) Suppose a set of 3–2–1 Euler angles is used to parameterize Q1 and a setof 1–3–1 Euler angles is used to parameterize R. Show that the angular

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velocity vectors of both bodies and their relative angular velocity vectorhave the representations

ω1 = γ1E3 + γ21e′

2 + γ31e1,

ω2 = ν11e1 + ν2

2e′3 + ν3

2e1 + γ1E3 + γ21e′

2 + γ31e1,

ω = ν11e1 + ν2

2e′3 + ν3

2e1.

(d) Suppose the rotation of the second body relative to the first body is con-strained such that

ω = ν22e′

3.

Give a physical interpretation of the type of joint needed to enforce thisconstraint. In addition, give prescriptions for the constraint moments actingon both bodies.

8.4. As shown in Figure 8.12, a tippe top is a body with an axis of symmetry. Oneof its lateral surfaces can be approximated as a spherical surface of radius R andthe other is a cylinder of radius r. The top is designed so that the center of mass Xis located below the center of the sphere, and this feature leads to its ability to flipover (see Figure 8.13).∗

g

A

P

RX

E1

E3

e3

Figure 8.12. A tippe top moving ona rough horizontal plane.

The instantaneous point of contact of the spherical portion of the top with thehorizontal plane is denoted by P. The position vectors of P relative to X and thepoint A relative to X are

xP − x = −RE3 + le3, xA − x = he3,

respectively.

(a) Using a set of 3–1–3 Euler angles, establish expressions for ωi = ω · ei and�i = ω · Ei. For which orientations of the tippe top does this set of Eulerangles have singularities?

∗ The dynamics of the tippe top has been the subject of several investigations, and the interestedreader is referred to [20, 154, 156, 177, 202, 219] for discussions and references. Among other issues,these papers point out the important role played by friction forces acting at the point of contact.

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g

A

A

P

X

XE1

E3

e3

e3

ω0 ω1

Figure 8.13. The two steady motions of a tippe top: the upright position in which e3 = E3 andthe inverted state in which e3 = −E3.

(b) Show that the slip velocities of the point P of the tippe top have the repre-sentations

vs1 = x1 − �2 (R − l cos(θ)) + �3 (l sin(θ) cos(ψ)) ,

vs2 = x2 + �1 (R − l cos(θ)) + �3 (l sin(θ) sin(ψ)) ,

where vP = vs1E1 + vs2E2 and v =∑3k=1 xkEk.

(c) Suppose that the tippe top is sliding. Give a prescription for the constraintforce Fc and moment Mc that enforce the sliding constraint.

(d) Suppose that the tippe top is rolling. Give a prescription for the constraintforce Fc and moment Mc that enforce the rolling constraints.

(e) With the help of (8.13), show that two of the three constraints on a rollingtippe top are nonintegrable.

8.5. Consider the mechanical system shown in Figure 8.14. It consists of a rigid bodyof mass m that is free to rotate about a fixed point O. The joint at O does not permitthe body to have a spin. A vertical gravitational force mgE1 acts on the body. Theinertia tensor of the body relative to its center of mass C is

J0 = λ1E1 ⊗ E1 + (λ − mL20

)(E2 ⊗ E2 + E3 ⊗ E3) .

g

O

X

φ

θ

E1

E2

E3

e1

Figure 8.14. A pendulum problem.

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The position vector of the center of mass X of the body relative to O is L0e1.To parameterize the rotation tensor of the body, we use a set of 1–3–1 Euler

angles: g1 = E1, g2 = e′3 and g3 = e1. For these angles

ω = φE1 + θe′3 + ψe1

= (ψ + φ cos(θ))

e1 + (θ sin(ψ) − φ sin(θ) cos(ψ))

e2

+ (θ cos(ψ) + φ sin(θ) sin(ψ))

e3.

(a) Which orientations of the rigid body coincide with the singularities of theEuler angles?

(b) Derive expressions for the unconstrained potential U and kinetic T energiesof the rigid body.

(c) Show that the motion of the rigid body is subject to four constraints:

ψ = ω · g3 = 0, v − ω × (L0e1) = 0.

(d) Derive expressions for the constrained potential U and kinetic T energiesof the rigid body.

8.6. As shown in Figure 8.15, an axisymmetric rigid body is free to rotate about thefixed point O. The body, which has a mass m, has an inertia tensor

J = λae3 ⊗ e3 + λt (I − e3 ⊗ e3) .

The position vector of the center of mass of this body relative to O is

x = L1e3.

O

X

e3

E1

E2

E3

g

Figure 8.15. A rigid body that is free to rotate about the fixedpoint O. A set of 3–1–3 Euler angles is used to describe therotation tensor Q of this body.

(a) Using a set of 3–1–3 Euler angles, show that the angular velocity vector ω

also has the representation

ω = θe′′1 + ψ sin(θ)e′′

2 + (φ + ψ cos(θ))e′′3.

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(b) Show that the angular momentum H is

H = λtθe′′1 + λtψ sin(θ)e′′

2 + λa(φ + ψ cos(θ)

)e3.

(c) Show that the angular momentum of the center of mass relative to O is

mL21 (ω1e1 + ω2e2) ,

where ωi = ω · ei. What is the angular momentum of the rigid body relativeto O?

(d) Show that the kinetic energy T is

T = λt + mL21

2

(θ2 + ψ2 sin2(θ)

)+ λa

2

(φ + ψ cos(θ)

)2.

(e) A conservative moment that is due to a torsional spring is applied to therigid body. As a result, the total potential energy of the rigid body is

U = mgx · E3 + K2

ψ2.

Here, K is the torsional spring constant. What are the conservative forceFcon and moment Mcon acting on the rigid body?

8.7. Recall Ziegler’s example of a constant moment ME3 that was not conserva-tive. After choosing a set of Euler angles,

{γ1, γ2, γ3

}, show that you cannot find a

potential energy function U such that ME3 = −∑3i=1

∂U∂γi gi.∗

8.8. Suppose that a linear spring of stiffness K and unstretched length L0 is attachedto a point A on a rigid body and the other end of the spring is attached to a fixedpoint O. The position vector of A relative to the center of mass of the rigid body is

πA = Re3.

(a) With the assistance of a set of 3–1–3 Euler angles, show that

xA = (x1 + R sin(θ) sin(ψ)) E1 + (x2 − R sin(θ) cos(ψ)) E2

+ (x3 + R cos(θ)) E3.

In this equation, x =∑3k=1 xkEk.

(b) With the assistance of a set of 3–1–3 Euler angles, show that the potentialenergy of the spring is

U = U (x, Q) = K2

(√u − L0

)2,

where

u = (x1 + R sin(θ) sin(ψ))2 + (x2 − R sin(θ) cos(ψ))2

+ (x3 + R cos(θ))2.

∗ One solution to this problem can be found in [160].

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(c) Describe two equivalent methods to find the conservative force Fcon andconservative moment Mcon associated with the spring.

8.9. Consider the rolling disk discussed in Subsection 8.6.3 and consider constraints(8.21) on its motion. Here, we wish to show that this family of constraints is non-integrable by using Frobenius’ theorem.

To use the theorem, we first define the following seven variables:

U 1 = x1, U 2 = x2, U 3 = x3,

U 4 = φ, U 5 = ψ, U 6 = θ, U 7 = t.

(a) With the help of (8.15), compute the 3 × 7 matrix W. Determine a1, . . .,a4

that span the null space (kernel) of this matrix. You will find that one ofthese vectors is

a1 =[0 0 0 0 0 0 1

]T.

(b) With the help of (8.15), calculate the three 7 × 7 skew-symmetric matricesS1, S2, and S3 corresponding to π1, π2, and π3, respectively. You will findthat these matrices have at most two nonzero elements.

(c) Using the results of (a) and (b), show that Frobenius’ theorem implies thatthe rolling disk is subject to nonintegrable constraints. You should also indi-cate how your results can be used to conclude that the sliding disk is subjectto an integrable constraint.

(d) Suppose the rolling disk is subject to two additional integrable constraints:

x2 = 0, ψ = 0.

That is, the disk is constrained to roll vertically in a straight line. Computethe 5 × 7 matrix W and show that its null space is spanned by a1 and a2,where

a1 =[0 0 0 0 0 0 1

]T,

a2 =[−R 0 0 1 0 0 0

]T.

In addition, show that the skew-symmetric matrices S4 and S5 correspond-ing to the additional constraints are both zero. Finally, invoking Frobenius’theorem, show that the family of constraints π1 = 0, π2 = 0, π3 = 0, x2 = 0,and ψ = 0 is integrable.

8.10. A schematic of a Griffin grinding machine is shown in Figure 8.16.∗ The grainto be milled is placed in the bin, and a roller is designed to roll without slipping onthe inner wall of the grain bin. The normal force generated by the roller is substantialand crushes the grains. The motion of the roller is achieved by two drive shafts. The

∗ This figure is adapted from Arnold and Mauder [8] and Webster [227].

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270 Exercise 8.10

O

P

U

e3

E1

E2

E3

Universal joint

Grain bin Roller

Drive shaft I

Drive shaft II

ψ

φ

γ

Figure 8.16. Schematic of a Griffin grinding machine.

first shaft (drive shaft I) has an angular velocity vector ωI = ψE3. It is coupled bya universal joint at U to the second shaft. The roller is attached to the second shaft(drive shaft II) by a joint that allows it to have a rotation relative to the second shaftin the direction of e3. The basis {e1, e2, e3} corotates with the roller.

(a) Using a set of 3–1–3 Euler angles {ψ,π − γ, φ}, show that the angular veloc-ity vector of the roller has the representation

ω = ψ (sin(φ) sin(π − γ)e1 + cos(φ) sin(π − γ)e2 + cos(π − γ)e3)

−γ (cos(φ)e1 − sin(φ)e2) + φe3.

For which orientations of the roller does this set of Euler angles have singu-larities? What is the angular velocity vector ωII of drive shaft II, and what isthe angular velocity of the roller relative to drive shaft II?

(b) The center of mass X of the roller has a position vector relative to the fixedpoint U of

x = He3.

Establish an expression for the velocity vector v of the point X.

(c) The instantaneous point P of contact of the roller with the grain bin has thefollowing position vector relative to X:

πP = −r (cos(φ)e2 + sin(φ)e1) .

Using the identity vP = ω × πP + v, establish expressions for the compo-nents vP · ei.

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Exercise 8.10 271

(d) If vP = 0, then show that γ is a constant and the rotational speeds ψ and φ

are related:

φ =(

cos (γ0) − Hr

sin (γ0))

ψ.

In this equation, γ0 is the constant value of the angle γ.

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9 Kinetics of a Rigid Body

9.1 Introduction

In this chapter, the balance laws F = m ˙v and M = H for a rigid body are discussed.Several component forms of these laws are considered. For instance, the compo-nents of the balance of angular momentum with respect to the corotational basis ei,M · ei = H · ei lead to a set of equations [(9.9)] that are known as Euler’s equations.In a subsequent chapter, we shall show that M · gi = H · gi lead to Lagrange’sequations.

Once the balance laws have been discussed, a brief outline is given of the work–energy theorem, and it is shown how to establish energy conservation for a rigidbody. Our attention then turns to applications. In particular, we discuss the dynam-ics of a body rotating about a fixed point, moment-free motion of a rigid body,rolling and sliding spheres, and the dynamics of baseballs and footballs. Severalother applications are discussed in the exercises. Despite all these examples, ourconsideration of applications is far from exhaustive. At the conclusion of this chap-ter, some details are provided on some other interesting applications that have beenmodeled by use of a single rigid body.

9.2 Balance Laws for a Rigid Body

Euler’s laws for a rigid body can be viewed as extensions to Newton’s second law fora single particle. There are two laws, or postulates, the balance of linear momentumand the balance of angular momentum:

G = F, HO = MO. (9.1)

Here, HO is the angular momentum of the rigid body relative to a fixed point O andMO is the resultant external moment relative to O. As noted in Truesdell [216, 217],these laws are discussed in several of Euler’s works on rigid body dynamics and findtheir definitive form in Euler’s paper [56] that was published in 1776.∗

∗ The interested reader is referred to [135, 230], where additional commentaries and perspectives onthese balance laws can be found. For additional background on the problem of the precession ofthe equinoxes featured in [230], the text of Hestenes [93] is recommended.

272

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9.2 Balance Laws for a Rigid Body 273

In many cases, it is convenient to give an alternative description of the balanceof angular momentum. To do this, we start with the identity

HO = H + x × G.

Differentiating, and using the balance of linear momentum, we obtain

HO = H + v × G + x × G

= H + x × F.

Hence, invoking the balance of angular momentum, HO = MO, we find that

MO = HO = H + x × F.

However, the resultant moment relative to a fixed point O, MO, and the resultantmoment relative to the center of mass X, M, are related by∗

MO = M + x × F.

It follows that

M = H,

which is known as the balance of angular momentum relative to the center of massX. This form of the balance law is used in many problems for which the rigid bodyhas no fixed point O.

In summary, the balance laws for a rigid body are known as Euler’s laws. Twoequivalent sets of these laws are used. For the first set, the balance of angularmomentum relative to a fixed point O features

G = F, HO = MO. (9.2)

In the second set, the balance of angular momentum is taken relative to the centerof mass X:

G = F, H = M. (9.3)

It should be noted that

G = mv, H = Jω, HO = Jω + x × mv.

Here, O is the origin of the coordinate system used to define x.To determine the motion of the rigid body, it suffices to know x(t) and Q(t). To

obtain these results, Equations (9.2) or (9.3) must be supplemented by informationon the coordinate system used to parameterize x and the parameterization of Q.

∗ This may be seen from our previous discussion of a system of forces and moments acting on a rigidbody in Section 8.5.

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9.3 Work and Energy Conservation

The work–energy theorem for a rigid body equates the rate of change of kineticenergy to the mechanical power of the external forces and moments acting on therigid body. There are two equivalent forms of this theorem:

T = F · v + M · ω

=N∑

K=1

FK · vK + Mp · ω. (9.4)

The second form is the most useful for proving that energy is conserved in a specificproblem. After proving the theorem, we will close this section with a discussion ofenergy conservation.

Proving the Work–Energy TheoremThe difficulty in establishing the work–energy theorem lies in dealing with the an-gular momentum H = Jω. To overcome this, it is easiest to first show that

ω · (Jω) = 0. (9.5)

This result is established by use of an earlier result, J = �J − J�, and the identity�Tω = −ω × ω = 0: (

Jω) · ω = ((�J − J�) ω) · ω

= (�Jω) · ω − (J (ω × ω)) · ω

= Jω · (�Tω)+ (J (0)) · ω

= 0.

Now, as Jω · ω = 0,

H · ω =(

˙Jω)

· ω = (Jω) · ω = (Jω) · ω. (9.6)

In the last step, we invoked the symmetry of J. From (9.6), we can conclude that

H · ω = H · ω, (9.7)

a result that we shall presently invoke.To prove work–energy theorem (9.4), recall the Koenig decomposition for the

kinetic energy T of a rigid body:

T = 12

mv · v + 12

(Jω) · ω.

Differentiating T and using (9.5) and (9.7), we find that

T = G · v + 12

(H · ω + (Jω) · ω)

= G · v + H · ω.

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9.3 Work and Energy Conservation 275

Invoking the balances of linear and angular momentum, work–energy theorem(9.4)1 is established:

T = F · v + M · ω.

To establish the alternative form, (9.4)2, of the work–energy theorem, we con-sider a system of N forces FK acting at N material points XK of the rigid body and apure moment Mp acting on the rigid body. For such a system of forces and moments,we recall that

F =N∑

K=1

FK, M =(

N∑K=1

(xK − x) × FK

)+ Mp.

Noting that vK = v + ω × (xK − x), we find with some rearranging that

F · v + M · ω = Mp · ω +N∑

K=1

FK · vK.

This result establishes the alternative form, (9.4)2, of the work–energy theorem.

Energy ConservationIn most problems in rigid body dynamics in which there is no dynamic friction, thetotal energy E of the body is conserved. To show this, let us assume that the resultantconservative force Fcon and conservative moment (relative to the center of mass)Mcon acting on the body are associated with a potential energy U = U(x, Q):

Fcon = −∂U∂x

, Mcon = −uQ.

In addition, we assume that the body is subject to R constraints of the form

Lf · v + Lh · ω + Le = 0 (L = 1, . . . , R).

The constraint force and moment are prescribed by use of Lagrange’s prescription:

Fc =R∑

L=1

µL (Lf) , Mc =R∑

L=1

µL (Lh) .

Finally, we assume that the only forces and moments acting on the rigid body areconservative or constraint.

We now examine work–energy theorem (9.4)1 for the rigid body of interest:

T = F · v + M · ω

= Fc · v + Mc · ω + Fcon · v + Mcon · ω.

However,

Fc · v + Mc · ω =R∑

L=1

µL (Lf · v + Lh · ω) = −R∑

L=1

µL (Le) ,

Fcon · v + Mcon · ω = −∂U∂x

· v − uQ · ω = −U.

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Consequently,

T = −U −R∑

L=1

µL (Le) .

This implies that E = 0, where E = T + U, if∑R

L=1 µL (Le) = 0.In summary, one situation in which the total energy E is conserved arises when

1e = 0, . . . , Re = 0, and the constraint forces and constraint moments are prescribedby Lagrange’s prescription. This situation arises in most of the problems in rigidbody dynamics that are solvable analytically, and we shall shortly see severalexamples.

9.4 Additional Forms of the Balance of Angular Momentum

The balance of angular momentum H = M has several component forms and is oneof the most interesting equations in mechanics. In this section, several of these formsare discussed. First, we show that this equation is equivalent to

J3∑

i=1

ωiei + ω × Jω = M.

Next, we show that, if ei are the principal vectors of J, then we can find Euler’s cele-brated equations (9.9) As an intermediate result, we also indicate the correspondingcomponent form of these equations when ei are not principal vectors of J.

A Direct FormHere, we wish to show that H = M can be written as

Jα + ω × (Jω) = M.

To establish this result, we need to examine ω and J.First, we recall that

H = Jω.

Taking the derivative of this expression, we find

H = Jω + Jω.

To proceed, we need some identities. Specifically,

α = ω = ddt

3∑i=1

ωiei

=3∑

i=1

ωiei + ω ×(

3∑i=1

ωiei

)

=3∑

i=1

ωiei

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9.4 Additional Forms of the Balance of Angular Momentum 277

and

J = �J + J�T, Jω = �Jω − J�ω = ω × Jω.

Using the identities, we find that

H = Jω + Jω = ω × Jω + Jω

= J3∑

i=1

ωiei + ω × Jω.

In summary, H = M is equivalent to

J3∑

i=1

ωiei + ω × Jω = M.

This form of H = M is very useful when J is a constant tensor – for instance whendealing with rigid spheres and rigid cubes. It is also used to obtain conservationresults for H.

In passing, we note that the result α =∑3i=1 ωiei implies that for ω to be constant

it suffices that ωi are constant. This is in spite of the fact that ei may not be stationary.

A Component FormIf we choose an arbitrary basis {E1, E2, E3} for E

3, then the inertia tensors J0 and Jhave the representations

J0 =3∑

i=1

3∑k=1

JikEi ⊗ Ek, J =3∑

i=1

3∑k=1

Jikei ⊗ ek,

where ei = QEi. Consequently,

H = Jω =3∑

i=1

3∑k=1

Jikωkei.

Differentiating H, we find that

H = ˙Jω =3∑

i=1

3∑k=1

Jikωkei + ω ×(

3∑i=1

3∑k=1

Jikωkei

). (9.8)

If we equate this expression to M, we can find the component forms of H = M.However, except when ω has a simple form, it is not convenient to consider thiscomponent form of the equations. Examples of where (9.8) are used include mis-balanced rotors, where ω = θE3 and J13 �= 0 and/or J23 �= 0.∗ For cases not involvinga fixed axis of rotation, it is prudent to choose {E1, E2, E3} to be the principaldirections of J0.

∗ An example of such a system can be found in Section 7 of Chapter 9 in [159].

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The Principal Axis CaseIf we choose Ei to be the principal directions of J0, then this tensor has the familiarrepresentation

J0 =3∑

i=1

λiEi ⊗ Ei,

where λi are the principal moments of inertia. The vectors Ei are also known as theprincipal axes of the body in its reference configuration.

Defining ei = QEi, we find that the inertia tensor J = QJ0QT has the represen-tation J =∑3

i=1 λiei ⊗ ei. Consequently,

H = Jω =3∑

i=1

λiωiei.

Evaluating H, we find

H = Jω + ω × Jω

= J3∑

i=1

ωiei + ω ×(

3∑i=1

λiωiei

)

=3∑

i=1

λiωiei + (λ2 − λ1)ω1ω2e3 + (λ1 − λ3)ω1ω3e2 + (λ3 − λ2)ω3ω2e1.

In conclusion, H = M has the component form

λ1ω1 + (λ3 − λ2)ω3ω2 = M · e1,

λ2ω2 + (λ1 − λ3)ω3ω1 = M · e2,

λ3ω3 + (λ2 − λ1)ω1ω2 = M · e3. (9.9)

These equations, known as Euler’s equations, represent three first-order ordinarydifferential equations for ωi.

To determine the rotation tensor Q, it is necessary to supplement (9.9) by thethree first-order ordinary differential equations relating ω to Q,

ω = −12ε[QQT] .

For example, if a set of 3–2–1 Euler angles were used to parameterize Q, then thesedifferential equations would be⎡

⎢⎣ψ

θ

φ

⎤⎥⎦ =

⎡⎢⎣




⎤⎥⎦⎡⎢⎣

ω1

ω2

ω3

⎤⎥⎦ . (9.10)

You may wish to recall that the 3–2–1 Euler angles were discussed in Subsec-tion 6.8.1, and differential equations (9.10) can be inferred from the developmentsthere.

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9.5 Moment-Free Motion of a Rigid Body 279

9.5 Moment-Free Motion of a Rigid Body

Moment-free motion of a rigid body occurs when M = 0. Determining the motionis resolved by solutions of the balance laws,

G = F, H = 0,

for x and Q. It is common to focus exclusively on the balance of angular momen-tum and determine Q. In addition, although an analytical solution for Q was firstfound by Carl G. J. Jacobi (1804–1851) in 1849,∗ it is usual to focus on ω(t). Anotheringenious solution to this problem was presented by Poinsot in 1834 [172].†

The equations governing the components ωi = ω · ei of the angular velocity vec-tor are found from the three equations H · ei = 0 [cf. (9.9)]:

λ1ω1 + (λ3 − λ2)ω3ω2 = 0,

λ2ω2 + (λ1 − λ3)ω3ω1 = 0,

λ3ω3 + (λ2 − λ1)ω1ω2 = 0. (9.11)

It is easy to see that the solutions to these equations conserve the rotational kineticenergy Trot = 1

2 H · ω and the angular momentum vector H.Although there are several cases to consider, it suffices to consider three:

symmetric body: λ1 = λ2 = λ3;axisymmetric body: λ1 = λ2 �= λ3;asymmetric body: λ1 < λ2 < λ3.

For the axisymmetric body, when λ1 < λ3 the body is known as oblate. Whenλ1 = λ2 > λ3, the body is known as prolate. For the asymmetric body previouslydiscussed, e1 is known as the minor axis of inertia, e2 is known as the intermediateaxis of inertia, and e3 is known as the major axis of inertia. We now turn to discussingthe three cases and the solutions for ωi(t).

The Symmetric BodyFor the symmetric rigid body, (9.11) simplify to ωi = 0. In other words, the com-ponents of ω are constant. As ω =∑3

k=1 ωkek, this implies that ω is constant. If wechoose Q(t0) = I, then the axis of rotation q of the body is constant,‡ and so we find

Q(t) = cos(ν)(I − q ⊗ q) − sin(ν)εq + q ⊗ q,

∗ Jacobi’s solution is discussed at length in Section 69 of Whittaker [228] and Section 37 of Landauand Lifshitz [125].

† Discussions of Poinsot’s solution can be found in several texts, for instance, Marsden and Ratiu[138] and Routh [184].

‡ Other choices of Q(t0) are possible; however, these may not guarantee that the axis of rotation q ofQ is constant. For further details on this matter see [161].

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where the axis and angle of rotation are

q = ω (t0)∣∣∣∣ω (t0)∣∣∣∣ , ν = ∣∣∣∣ω (t0)

∣∣∣∣ (t − t0) ,

and ω(t) = ω (t0).For the symmetric body, any axis is a principal axis. Consequently, it is possible

to spin such a body at constant speed about any axis. We shall shortly see that thereare related results for axisymmetric and asymmetric rigid bodies.

The Axisymmetric BodyFor an axisymmetric body, it is convenient to define λt = λ1 = λ2 and λa = λ3. Theequations governing the components of angular velocity (9.11) simplify for thiscase to

ω1 = kt�ω2,

ω2 = −kt�ω1,

ω3 = �, (9.12)

where � = ω3(t0) is a constant and

kt = λt − λa

λt.

Differential equations (9.12) have a simple analytical solution:[ω1(t)

ω2(t)

]=[

cos(kt� (t − t0)) sin(kt� (t − t0))

− sin(kt� (t − t0)) cos(kt� (t − t0))

][ω1(t0)

ω2(t0)

]. (9.13)

In summary, ωi(t) have been calculated.There are some special cases to consider. First, notice that it is possible to rotate

the body at constant speed either about the e3 direction or about any axis in thee1 − e2 plane. All of these axes are principal axes of the body. Hence it is possibleto spin the body at constant speed about a principal axis.

An interesting feature about the axisymmetric body is that the component ofω in the direction of the axis of symmetry e3 is always constant. This occurs eventhough e3(t) may be quite complicated and is a consequence of the angular momen-tum H · e3 being conserved. The conservation of ω3(t) is one of the key results inrigid body dynamics and is extensively exploited in designing flywheels.

The Asymmetric BodyWhen the body is asymmetric, its principal moments of inertia are distinct. If wereexamine (9.11) for this case, then we find that if all but one ωi are zero, thenthe nonzero ωi will remain constant. For instance, if ω2 = 0 and ω3 = 0, then it ispossible for ω1 to have any value and for the equations of motion to preserve thisvalue. These results imply that it is possible to rotate the body about a principalaxis at constant speed under no applied moment M. Clearly, as with the other

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two types of rigid bodies, it is possible to spin the body at constant speed about aprincipal axis.

The Momentum SphereTo visualize the solutions of (9.11), a graphical technique is often used. This tech-nique dates to the mid-19th century. It is based on two facts: The solutions ωi(t)to (9.11) preserve the magnitude of H and the rotational kinetic energy Trot. As aresult, the solutions

hi(t) = H · ei = λiωi(t) (i = 1, 2, 3),

lie on the intersection of the constant surfaces h = h0 and Trot = TE. Here,

h2 = h21 + h2

2 + h23,

Trot = h21

2λ1+ h2

2

2λ2+ h2

3

2λ3,

and the values of h0 and TE are determined by the initial conditions ωi(t0).If we pick a value h0 of h, the surface h = h0 in the three-dimensional space

h1 − h2 − h3 is a sphere – the momentum sphere. Selecting a value of TE, we find thatthe surface Trot = TE in the three-dimensional space h1 − h2 − h3 is an ellipsoid –the energy ellipsoid. The intersection of the ellipsoid with the sphere is either a dis-crete set of points or a set of curves.∗ These intersections are the loci of hi(t). Forthe axisymmetric body, the intersections are shown in Figure 9.1(a). Correspondingrepresentative intersections for an asymmetric body are shown in Figures 9.1(b).†

These figures are among the most famous in dynamics.For the case presented in Figure 9.1(a), the energy ellipoid has an axis of revo-

lution (in this case the third axis). For a symmetric body, the energy ellipsoid degen-erates further into a sphere. This sphere coincides with the momentum sphere andso the graphical technique used to determine hi(t) [and ωi(t)] breaks down. How-ever, for the symmetric case, we found previously that ωi(t) = ωi(t0). Consequently,each point on the momentum sphere corresponds to a steady rotational motion ofthe rigid body.

Stability and Instability of the Steady RotationsWe can use the portrait of the trajectories of λiωi(t) on the momentum sphereto deduce some conclusions on the nature of the steady rotational motions of arigid body. Our discussion is very qualitative, and more rigorous presentationsof this topic can easily be found elsewhere. For example, analyses of the stabilityof the steady motions can be found in Hahn [85], Hughes [97], and Marsden andRatiu [138].

∗ For a more detailed discussion of these intersections, the text of Synge and Griffith [207] is highlyrecommended.

† These figures were kindly supplied by Patrick Kessler in the Spring of 2007.

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λ1ω1

λ1ω1

λ2ω2

λ2ω2

λ3ω3

λ3ω3

e2

e2

(a)

(b)

Figure 9.1. Trajectories of the components λiωi on the momentum sphere. The curves andpoints on the sphere are the intersection of the momentum sphere with the energy ellipsoid.For these figures two distinct rigid bodies are shown: (a) λ1 = λ2 = 4 and λ3 = 5, and (b) λ1 =2, λ2 = 4, and λ3 = 5. The figures on the right-hand side show a moment-free motion of thee2 vector that is corotating with a rectangular box motion that corresponds to one of thetrajectories on the sphere. For the trajectories and simulations shown in this figure, (9.10)and (9.11) were numerically integrated.

For the axisymmetric case, the trajectories shown in Figure 9.1(a) can be usedto infer that a steady rotation about the e3 axis is stable. By stability, we mean that,if we perturb the body’s rotation slightly from this steady state, then hi(t) [or equiv-alently ωi(t)] will remain close to the state (h1(t), h2(t), h3(t)) = (0, 0, h3s), whereh3s is the value of h3 corresponding to the steady rotation.∗ On the other hand, thetrajectories in Figure 9.1(a) show that steady rotations about any axis in the e1 − e2

plane do not satisfy this condition. Consequently, such steady rotations are unstable.For the asymmetric case, the trajectories shown in Figure 9.1(b) confirm the

previous statements that six steady rotations are possible. The trajectories in thisfigure also illustrate that the four steady rotations about e1 and e3 are stable, whereas

∗ Because hi = λiωi , it is trivial to ascribe results pertaining to hi to those for ωi .

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the pair of steady rotations about the e2 axis is unstable. That is, a rotation aboutthe intermediate axis of inertia is unstable, whereas those about the major (e3) andminor (e1) axes are stable.

Attitudes of the Rotational MotionsThe information we have thus far gleaned from the momentum sphere does not tellthe full story about the motion of the rigid body. What is missing is information onthe behavior of Q. To find this information, one can use Jacobi’s analytical solu-tions discussed earlier. Alternatively, one can choose a parameterization for Q andnumerically integrate the equations relating ωi to these parameterizations. For ex-ample, if a set of 3–2–1 Euler angles is used, then, in addition to integrating (9.11),(9.10) would also be integrated to determine φ(t), θ(t), and ψ(t). With the help ofthese results, ei(t) can be constructed and the motion of the body visualized.

Results from two distinct examples of the numerical integrations of (9.10) and(9.11) are shown in Figures 9.2, 9.3, and 9.4. One of these simulations correspondsto a perturbation of the steady rotation of a rigid body rotating about the principalaxes corresponding to its maximal moment of inertia. The resulting behaviorsof ωi(t) are displayed in the trajectory labeled (i) in Figure 9.2(b), whereas thebehaviors of the corotational basis vectors can be seen in Figure 9.3. It should beclear from the former figure that the perturbation to the steady motion does notappreciably alter ei(t) from their steady rotation behaviors. The easiest method ofvisualizing these results is to toss a book into the air with an initial rotation primarilyabout e3 and observe that, although the book will wobble, its instantaneous axis ofrotation does not wander far from its initial state.

In Figure 9.4, the behaviors of ek(t) corresponding to a trajectory of ωi(t) thatpasses close to the equilibrium (ω1, ω2, ω3) = (0, ω0, 0) that is labeled with a “star”

0

−5

−5

5

5

6(i)

(ii)

(a) (b)

X

ω0

ω1

ω2

ω3

E1

E2

E3

Figure 9.2. The moment-free motion of a rigid body: (a) a rigid body showing the principalaxes, (b) the components ωi(t) = ω · ei corresponding to two different sets of initial condi-tions: (i) ω(0) = 0.5E2 + 5.0E3 and (ii) ω(0) = 5.0E2 + 0.1E3. For these simulations, (9.10)and (9.11) were numerically integrated with the initial conditions Q(0) = I and the parame-ter values λ1 = 2, λ2 = 4, and λ3 = 5.

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−1

−1

−1

−1

−1

−1

1

1

1

1

1

1

(a) (b)

(c)

O O

O

e1

e2

e3

e1 · E1

e1 · E2

e1 · E3

e2 · E1

e2 · E2

e2 · E3

e3 · E1

e3 · E2

e3 · E3

Figure 9.3. Simulation results indicating the stability of the steady rotation of the rigid bodyabout the principal axis corresponding to the maximal axis of inertia: (a) the Ei componentsof e1(t), (b) the Ei components of e2(t), and (c) the Ei components of e3(t). These resultscorrespond to the trajectory labeled (i) in Figure 9.2(b).

are shown. It should be clear from the behavior of e2(t) shown in Figure 9.4(b) thate2(t), which is initially close to E2 at time t = 0, makes large excursions from itsinitial value. This is in contrast to the situation shown in Figure 9.3 and is indicativeof the instability of the steady moment-free motion of a rigid body about its inter-mediate axis of inertia. The easiest way to see this instability is to take a book andgive it an initial angular velocity about e2. One will see a wobbling motion whereω · e2 will periodically take positive and negative values. Interestingly, it is possibleto execute this motion and have the book perform a rotation about e1 or e3 by 180◦.This twisting motion was only recently noted and analyzed by Ashbaugh et al. [11].∗

∗ In [11], two distinct sets of Euler angles are used to avoid the singularities inherent in Euler angleparameterizations of rotation tensors. For the results presented in Figures 9.3 and 9.4, it was notnecessary to introduce a second set.

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9.6 The Baseball and the Football 285

−1

−1

−1

−1

−1

−1

1

11

1

1

1

(a)(b)

(c)

O

OO

e1

e2

e3

e1 · E1

e1 · E2

e1 · E3

e2 · E1

e2 · E2

e2 · E3

e3 · E1

e3 · E2

e3 · E3

Figure 9.4. Simulation results indicating the instability of the steady rotation of the rigid bodyabout the principal axis corresponding to the intermediate axis of inertia: (a) the Ei com-ponents of e1(t), (b) the Ei components of e2(t), and (c) the Ei components of e3(t). Theseresults correspond to the trajectory labeled (ii) in Figure 9.2(b).

9.6 The Baseball and the Football

Consider a sphere of mass m and radius R that is thrown into space with an initialvelocity v(t0), angular velocity ω(t0), and orientation Q(t0). We wish to determinethe motion x and Q of the sphere. As discussed by Tait [211], it was known toIsaac Newton that the rotation of the sphere as it moves through the ambient aircauses a curvature of the path of the center of the sphere. This feature resultsin interesting dynamics in a variety of sports ranging from golf to baseball andsoccer. Our interest here is to examine the curving of the path of the sphere. Wedo this by following several classical works on this problem: most notably Tait[211].

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XX

ωω

FMFM

vv

Figure 9.5. A rigid sphere whose center of mass is moving to the right with a velocity vectorv. When the ball is rotating clockwise the velocity of the air moving over the top of the ballis slower than the velocity of the air in contact with the bottom of the ball. From Bernoulli’sequation, the pressure on the top of the ball is greater than the pressure on the bottom ofthe ball and a net downward force FM results. The opposite occurs when the ball is rotatingcounterclockwise.

The Magnus ForceA key force experienced by the sphere is known as the lift or Magnus force (seeFigure 9.5),∗

FM = mBω × v,

where B is a positive constant. The sign of B is determined by use of Bernoulli’sequation.† Clearly, this force models the coupling between rotation and linear ve-locity. Recent research on free kicks in soccer has shown that there can be a transi-tion in the flow field from turbulent to laminar that causes dramatic changes in thetrajectory (see Carre et al. [25]). According to Ireson [100], for some free kicks insoccer, ||v|| = 25 m/s and mB ≈ 0.15716 kg.

Apart from gravity and the Magnus force, the other important force in this prob-lem is the drag force:

FD = −12ρf ACd (v · v)

v||v|| .

In this expression, Cd is the drag coefficient, ρf is the density of the fluid that thesphere is moving in, and A is the frontal area of the sphere in contact with the fluid:A = πR2.

Equations of MotionFor the system at hand, M = 0 and H = 2mR2

5 ω, so we find the important result thatω is constant:

ω(t) = ω (t0) .

∗ Credited to the German scientist Heinrich Gustav Magnus (1802–1870) in 1851 (see [133, 134]).† Bernoulli’s equation applies to inviscid fluid flow and states that the sum of the pressure p and

12 ρ f U2 is a constant. Here, U is the fluid flow velocity and ρ f is the fluid density.

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9.6 The Baseball and the Football 287

In other words, the angular velocity of the sphere does not change. As with thesymmetric body discussed earlier, we can easily solve for the rotation tensor of thesphere if we assume that Q (t0) = I:

Q(t) = cos(ν)(I − q ⊗ q) − sin(ν)εq + q ⊗ q,

where the axis and angle of rotation are

q = ω (t0)∣∣∣∣ω (t0)∣∣∣∣ , ν = ∣∣∣∣ω (t0)

∣∣∣∣ (t − t0) .

We shall see that solving for the motion of the center of mass in this problem is nottrivial.

The rotation tensor Q =∑3k=1 ek ⊗ Ek for the rigid body is parameterized by

a set of 3–1–3 Euler angles (see Subsection 6.8.2). The Euler basis vectors for thisparameterization have the representations⎡

⎢⎣g1

g2

g3

⎤⎥⎦ =

⎡⎢⎣

E3

e′1

e3

⎤⎥⎦ =

⎡⎢⎣



0 0 1

⎤⎥⎦⎡⎢⎣

e1

e2

e3

⎤⎥⎦

=

⎡⎢⎣

0 0 1

cos(ψ) sin(ψ) 0

sin(θ) sin(ψ) − sin(θ) cos(ψ) cos(θ)

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ . (9.14)

Further, the Euler angles are subject to the following restrictions: φ ∈ [0, 2π), θ ∈(0, π), and ψ ∈ [0, 2π). Using these Euler angles, we obtain

ω = �1E1 + �2E2 + �3E3

= (θ cos(ψ) + φ sin(θ) sin(ψ))

E1 + (θ sin(ψ) − φ sin(θ) cos(ψ))

E2

+ (ψ + φ cos(θ))

E3.

If we use a set of Cartesian coordinates for the position vector of the center of mass,x · Ei = xi, then we would find that

ω × v = (x3�2 − x2�3) E1 + (x1�3 − x3�1) E2 + (x2�1 − x1�2) E3. (9.15)

From the previous solution to the balance of angular momentum, we know that �i

are constant.The balance of linear momentum for the sphere provides the equation for the

motion of the center of mass. Evaluating F = G in Cartesian coordinates, we find

mx1 = (FD + FM) · E1,

mx2 = (FD + FM) · E2,

mx3 = −mg + (FD + FM) · E3.

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Ignoring the drag force and using (9.15), we find three differential equations forxi(t):

mx1 = mB (x3�2 − x2�3) ,

mx2 = mB (x1�3 − x3�1) ,

mx3 = mB (x2�1 − x1�2) − mg. (9.16)

For the general case, these equations can be integrated numerically to determinex(t).

The Path of the BallTurning our attention to a simple case, suppose ω (t0) = �10E1. From the previousanalysis, we know that ω is constant for this rigid body. Consequently, (9.16)simplifies to

mx1 = 0,

mx2 = −mBx3�10,

mx3 = mBx2�10 − mg.

The solution to these differential equations, assuming B�10 �= 0, is⎡⎢⎣

x1(t) − x1 (t0)

x2(t) − x2 (t0)

x3(t) − x3 (t0)

⎤⎥⎦ = A

⎡⎢⎣

x1 (t0)

x2 (t0) − gB�10

x3 (t0)

⎤⎥⎦+

⎡⎢⎢⎣

0g(t−t0)B�10

0

⎤⎥⎥⎦ , (9.17)

where

A =

⎡⎢⎢⎢⎣

t − t0 0 0

0 sin(B�10(t−t0))B�10

− 1−cos(B�10(t−t0))B�10

0 1−cos(B�10(t−t0))B�10

sin(B�10(t−t0))B�10

⎤⎥⎥⎥⎦ .

From (9.17), the trajectory of the sphere can be determined. Two important featuresare present. First, the spin �10 influences the forward speed x2 of the sphere. Second,it also affects the vertical position and speed. It is also of interest to compare (9.17)with the corresponding solution when the Magnus force is absent. In this case, thepath of the center of mass is the well-known parabolic trajectory:⎡

⎢⎣x1(t) − x1 (t0)

x2(t) − x2 (t0)

x3(t) − x3 (t0)

⎤⎥⎦ =

⎡⎢⎣

(t − t0) x1 (t0)

(t − t0) x2 (t0)

(t − t0) x3 (t0)

⎤⎥⎦−

⎡⎢⎢⎣

0

0g2 (t − t0)2

⎤⎥⎥⎦ .

Representative examples of the trajectories of a point launched from the origin areshown in Figure 9.6. For small values of B�10, we can see from this figure how thetrajectory differs from that in which the Magnus force is absent: A spin (�10) in

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9.7 Motion of a Rigid Body with a Fixed Point 289

10

15

30

−20

x3

x2

(i) (ii) (iii)

(iv)

(v)(vi)

Figure 9.6. The trajectories of a sphere that is launched from the origin with an initial velocityv (t0) = 10 (E2 + E3) and an initial angular velocity ω (t0) = �10E1. The trajectories showncorrespond to different values of B�10: (i) B�10 = −0.5, (ii) B�10 = −0.2, (iii) B�10 = 0.0,(iv) B�10 = 0.2, (v) B�10 = 0.5, and (vi) B�10 = 1.0. All of the trajectories are displayed fora period of 4 s and g = 9.81 m/s/s.

one direction will result in the ball “rising,” whereas a “dipping” effect can be ob-served by reversing the initial spin. However, we also observe that, for larger valuesof |B�10|, the behavior of the trajectories becomes unphysical: either through theappearance of a cusp or the reversal in the sign of x3.∗ Thus the prescription of theMagnus force may have a limited range of physical applicability.

Our formulation of the equations of motion for this problem are simplified bythe fact that the moment of inertia tensor for the sphere has a simple form. Indeed,it is interesting to compare the flight of a ball predicted by this model with that of aFrisbee. The interested reader is referred to [95, 97] where the equations of motionof a Frisbee are formulated and the lift and drag forces on the Frisbee computedfrom experiments.

9.7 Motion of a Rigid Body with a Fixed Point

The problem of a body that is free to rotate about a fixed point O, which is also amaterial point of the body, occupies a celebrated place in the history of mechanics.An example of such a body is shown in Figure 9.7, and related systems can be foundin the pendula in clocks and several types of spinning tops. In these systems, thereare three constraints on the motion of the rigid body and F = ma serves to determinethe constraint (reaction) forces at O that enforce these constraints. The remainingbalance law, MO = HO, has an interesting form and is used to determine the rotationtensor Q of the rigid body.

∗ As can be seen in Figure 4 of [211], motions of this type are also present in Tait’s analysis of thisproblem.

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gX

Xs

AO

Ball-and-socket joint

Linear spring

e1

e3

Figure 9.7. An example of a rigid body that is free to rotate about a fixed point O where thefixed point is a material point of the body. In this example, x − xO = he3 where h is a constant,and the body is also subject to conservative forces from the linear spring and gravity.

KinematicsFor the body of interest, the material point O is fixed at a point that we take to bethe origin: xO = 0. We assume that the position vector of the center of mass relativeto O has the representation

x − xO = L1e1 + L2e2 + L3e3,

where Li are constants. Differentiating this equation with respect to time, we seethat

v = ω × (L1e1 + L2e2 + L3e3) . (9.18)

This relation can be used to establish a convenient representation for the angularmomentum HO in terms of an inertia tensor JO for the body relative to O and thekinetic energy T of the rigid body.

To establish the representation for HO, we start with the relation for this quan-tity in terms of H and G:

HO = H + x × G

= Jω + m (L1e1 + L2e2 + L3e3) × (ω × (L1e1 + L2e2 + L3e3))

= JOω. (9.19)

For the final step in this result, we use the identity a × (b × c) = (a · c)b − (a · b)c.∗

The inertia tensor JO in (9.19) is

JO = J + m(L2

1 + L22 + L2

3

)I

− m (L1e1 + L2e2 + L3e3) ⊗ (L1e1 + L2e2 + L3e3) .

∗ As can be seen from (7.20), this identity was used earlier to establish the representation H = Jω.

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As expected, this expression for JO is in agreement with the one that we wouldobtain by using the parallel axis theorem [see (7.25)]. Starting with Koenig decom-position (7.21), a series of standard manipulations provides a convenient expressionfor the kinetic energy of the rigid body:

T = 12

(JOω

) · ω. (9.20)

The details are left as an exercise.We can show that JO is a positive-definite symmetric tensor, and consequently it

will have a set of principal axes. We now choose ei to be these axes, and, as a result,we can write

JO = λO1 e1 ⊗ e1 + λO

2 e2 ⊗ e2 + λO3 e3 ⊗ e3.

It is straightforward to show how λOi are related to the components of J and m, L1,

L2, and L3.

Constraint Forces and Constraint MomentsThe motion is subject to three (integrable) constraints:

1 = 0, 2 = 0, 3 = 0.

These constraints arise because the point O is fixed:

i = (x − L1e1 − L2e2 − L3e3) · Ei. (9.21)

Differentiating the constraints, we find Equation (9.18). Assuming that the joint atO is frictionless, we can easily use (9.18) with Lagrange’s prescription to show that

Fc = µ1E1 + µ2E2 + µ3E3, Mc = (−L1e1 − L2e2 − L3e3) × Fc.

Here, Fc and Mc are equipollent to the force Fc acting at the joint O. You may wishto recall that we considered the constraint forces and moments for the situation inwhich the joint at O was a pin joint in Subsection 8.6.1. For this case, Mc would havetwo additional components.

Equations of MotionFor this problem, it is convenient to follow a procedure of using F = ma to solve forthe three unknown components of Fc and to use MO = HO to solve for the motionof the body.

The balance law MO = HO can be written in components relative to the basis ei.Recalling that we are choosing this basis to be parallel to the principal axes of JO,we find the component form

λO1 ω1 + (λO

3 − λO2

)ω3ω2 = MO · e1,

λO2 ω2 + (λO

1 − λO3

)ω3ω1 = MO · e2,

λO3 ω3 + (λO

2 − λO1

)ω1ω2 = MO · e3. (9.22)

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These equations have obvious parallels to Euler’s equations (9.9). Indeed, if MO =0, then we can use the solutions for moment-free motion that we discussed previ-ously with a small number of modifications.

Equations (9.22) need to be supplemented by equations relating ω to Q in orderto solve for the rotation tensor of the body.∗ Once the rotation has been found, wecan then use F = ma, where a = α × x + ω × (ω × x), to determine R.

Euler–Poisson EquationsAn alternative formulation of the equations of motion for this case in which agravitational force −mgE3 is acting on the body is known as the Euler–Poissonequations.† For these equations, instead of using a set of 3–1–2 Euler angles toparameterize Q and then supplementing (9.22) with (7.28), we work with three ofthe nine components of Q.

By way of preliminaries, let us write Q = Q�0 in terms of components. Relativeto the basis Ei ⊗ Ek, we see that

⎡⎢⎣

Q11 Q12 Q13

Q21 Q22 Q23

Q31 Q32 Q33

⎤⎥⎦ =

⎡⎢⎣

Q11 Q12 Q13

Q21 Q22 Q23

Q31 Q32 Q33

⎤⎥⎦⎡⎢⎣

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

⎤⎥⎦ . (9.23)

Relations of this form for the time derivatives of the components of the rotationtensor were first found by Poisson in the early 19th century,‡ and are known as thePoisson kinematical relations. It is important to note that

E3 = Q31e1 + Q32e2 + Q33e3. (9.24)

Because of the moment that is due to gravity we shall subsequently need the com-ponents E3 · ei.

Note that, by differentiating (9.24) and using the identity ei = ω × ei, we woulddiscover that

Q31e1 + Q32e2 + Q33e3 = −3∑

k=1

Q3kω × ek.

These relations constitute three differential equations for Q3i – which are equivalentto those from (9.23). The coefficients Q3i = e3 · Ei are often known as the directioncosines of e3.

The equations of motion for the rigid body consist of three equations governingQ3i and the balance of angular momentum relative to O. Thus we combine the

∗ An explicit form of the differential equations can be found in an exercise at the end of this chapter:see (9.36).

† We shall discuss a third alternative, Lagrange’s equations of motion, in Chapter 10.‡ See Section 411 of his treatise [173].

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balance of angular momentum that is due to Euler and kinematical relations thatare due to Poisson. In component forms, the Euler–Poisson equations are

Q31 = Q32hO3

λO3

− Q33hO2

λO2

,

Q32 = Q33hO1

λO1

− Q31hO3

λO3

,

Q33 = −Q32hO1

λO1

+ Q31hO2

λO2

,

hO1 =(λO

2 − λO3

)λO

2 λO3

hO2hO3 + (mgE3 × x) · e1,

hO2 =(λO

3 − λO1

)λO

1 λO3

hO1hO3 + (mgE3 × x) · e2,

hO3 =(λO

1 − λO2

)λO

1 λO2

hO1hO2 + (mgE3 × x) · e3. (9.25)

Here, we have used the representations

HO =3∑

i=1

hOiei, hOk = λOk ωk (k = 1, 2, 3).

Additional developments and representations of Euler–Poisson equations (9.25)can be found, for example, in Beletskii [16] and Sudarshan and Mukunda [204].

ConservationsFor the problem of interest, we can use representation (9.20) for T to show thefollowing forms of the work–energy theorem:

T = MO · ω =N∑

K=1

FK · vK + Mp · ω.

If the applied forces and moments acting on the system are conservative, then, be-cause Fc acts at a point with zero velocity, it is easy to show that the total energy ofthe rigid body is conserved. This situation arises when a gravitational force acts onthe body.

If MO = 0, then HO is conserved. However, in the most common form of thisproblem a gravitational force −mgE3 acts on the rigid body. In this case, MO �= 0,but MO has no component in the E3 direction. It is easy to see for this case thatHO · E3 is conserved. If, in addition, the body has an axis of symmetry and λO

1 = λO2 ,

then you should be able to show with the help of (9.22) that HO · e3 is conserved.The problem in which the body has an axis of symmetry and MO = x × (−mgE3)

is often known as the (symmetric) Lagrange top.∗ The motion of this top conserves

∗ In the context of Lagrange’s equations of motion, we will discuss this problem in Section 10.8.

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P

g O

X

β

Sphere of mass m and radius R

Inclined plane

E1

E3

Figure 9.8. A rigid sphere moving on an inclined plane. The angle of inclination of the planeis β, and a gravitational force −mg cos(β)E3 + mg sin(β)E1 acts on the rigid body.

E, HO · e3, and HO · E3, and is one of the most famous mechanical systems. Indeed,as discovered by Lagrange,∗ analytical solutions for its equations of motion can befound. This discovery is remarkable, for if we relax the assumption that the bodyis symmetric (i.e., λO

1 = λO2 ), then analytical solutions are possible in only a handful

of special cases (see [128, 228]). Indeed, it is an interesting exercise to numericallyintegrate the equations of motion for the case in which MO = x × (−mgE3) and theλO

i ’s are distinct.

9.8 Motions of Rolling Spheres and Sliding Spheres

The problem of the sphere moving on a flat surface has several applications, bowlingand pool being the most famous. The most famous classical treatments of this prob-lem are due to Coriolis [41] and Routh [184], and generalizations of it occupy theliterature on nonholonomically constrained rigid bodies to date.† In our treatment,we assume that the surface is rough with a coefficient of static Coulomb friction ofµs and kinetic friction of µd. Of particular interest to us will be the transition be-tween rolling and sliding and our discussion is heavily influenced by Routh [184]and Synge and Griffith [207].

Consider the sphere moving on the surface shown in Figure 9.8. The radiusof the sphere is R, and the velocity of the point of contact of the sphere with theincline is

vP = v + ω × (−RE3) .

∗ See Section IX.34 of the Second Part of Lagrange’s Mecanique Analytique [121]. This analyticalsolution is also discussed by Whittaker [228].

† See, for example, Borisov and Mamaev [19], Frohlich [66], and Huston et al. [99]. The latter papersdiscuss the dynamics of bowling balls.

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9.8 Motions of Rolling Spheres and Sliding Spheres 295

Note the simple expression for πP here. Because the point P is the instantaneouspoint of contact,

vP · E3 = 0.

Consequently, this velocity field has the representations

vP = vs1E1 + vs2E2 = uc,

where u =√

v2s1 + v2

s2 is the slip velocity and c = vPu is the slip direction. When the

sphere is rolling, there are two additional constraints on vP and, as a result, vs1 = 0and vs2 = 0. For the rolling sphere, the slip direction is not defined.

The resultant force and moment on the sphere are

F = −mg cos(β)E3 + mg sin(β)E1 + NE3 + Ff , M = −RE3 × Ff .

When the sphere is rolling,

Ff = µ1E1 + µ2E2,

where µ1 and µ2 are unknowns. For the sliding sphere, on the other hand, we havethe classical prescription

Ff = −µd|N|c.

For convenience, we use the same notation as that of the friction forces for therolling and sliding spheres, but this should not cause confusion.

Rolling SphereWe determine the motion of the rolling sphere by using the balance laws and theconstraints vP = 0. Using Cartesian coordinates for x and setting ω =∑3

i=1 �iEi, wefind that these equations are

x1 = R�2,

x2 = −R�1,

x3 = 0,

mx1 = mg sin(β) + µ1,

mx2 = µ2,

0 = N − mg cos(β),

25

mR2�1 = Rµ2,

25

mR2�2 = −Rµ1,

25

mR2�3 = 0.

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To solve these equations, it is convenient to first determine the differential equationsgoverning �i. From the nine equations just listed, we can eliminate several variablesto find (

1 + 25

)mR2�1 = 0,

(1 + 2

5

)mR2�2 = mgR sin(β),

25

mR2�3 = 0.

These equations are easily solved:

ω (t) = ω (t0) + g sin(β) (t − t0)R

(1 + 2

5

)−1

E2. (9.26)

It is left as an exercise to determine x(t). When β = 0, you will find that the sphererolls in a straight line at constant speed.

Sliding SphereFor the sliding sphere, it is convenient to examine the differential equations for vP.Differentiating this velocity, we find that

vP = vs1E1 + vs2E2

= ˙v + α × (−RE3) .

Using the balances of linear and angular momentum, we substitute for ˙v and α tofind

mvs1 =(

1 + 52

)Ff · E1 + mg sin(β), mvs2 =

(1 + 5

2

)Ff · E2. (9.27)

After substituting for the friction force, these equations provide two differentialequations for the slip velocities.∗ It is convenient to express these equations as dif-ferential equations for u and the angle χ, where

c = cos(χ)E1 + sin(χ)E2, cos(χ) = vs1

u, sin(χ) = vs2

u.

When χ is constant, the slip direction c is constant. After some manipulations, wefind that (9.27) are equivalent to†

u = −µdg(

1 + 52

)+ g sin(β) cos(χ), uχ = −g sin(β) sin(χ). (9.28)

These differential equations have analytical solutions for χ(t) and u(t). It is also leftas an exercise to write the five differential equations governing x1, x2, and �i.

∗ It is left as an interesting exercise to nondimensionalize and numerically integrate (9.27) to deter-mine the behavior of the slip velocity components as the ratio of µd to sin(β) is varied.

† To obtain these equations we differentiated u2 = v2s1 + v2

s2 and sin(χ) = vs2u and then used (9.27).

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If we consider the simple case in which the incline is horizontal, then β = 0. Thedifferential equations for the slip velocity simplify considerably to

mvs1 = −µdmg(

1 + 52

)vs1

u, mvs2 = −µdmg

(1 + 5

2

)vs2

u.

From these equations, we will find that vs1 and vs2 always tend to zero. To see this,it is best to look at (9.28) and set β = 0:

u = −µdg(

1 + 52

), χ = 0. (9.29)

These equations have the solution

u (t) = u (t0) − µdg(

1 + 52

)(t − t0) , χ (t) = χ (t0) .

As a result, u will reach zero in a finite time T and the slip direction stays constant:

T = u (t0)µdg

(1 + 5

2

)−1

.

It can be shown that the path of the center of the sliding sphere is either fixed,a straight line, or a parabolic arc. Once u = 0, the sphere starts rolling. Now, asβ = 0, this implies that the sphere will roll at constant speed in a straight line (seeFigure 9.9). It is interesting to note that once the sphere starts rolling it will stayrolling. The transition between the parabolic path during sliding and the straightline path during rolling is a key to hook shots in bowling and massee shots in pool.∗

An example of this transition is shown in Figure 9.9(a).The factor of 2

5 in the equations of motion for rolling and sliding spheres is alsointeresting. It is related to the fact that the height of the “center of oscillation” Qof a sphere relative to the center of mass is 2R

5 . As discussed in Coriolis [41], Qis the point one aims for when hitting a cue ball so that it rolls without slippingimmediately after the impact of the tip of the cue with the ball.


We have touched on some problems in rigid body dynamics. There are several as-pects that we have not had the opportunity to address, and some of them are dis-cussed in the exercises and others in the references at the end of this book. Thetreatises of Appell [7], Papastavridis [169], Routh [184], and Whittaker [228], andthe splendid introductory text by Crabtree [42] are particularly recommended.

It is important to realize that, although rolling spheres and thrown baseballshave been analyzed for over a century, these problems are very rich. Indeed,a simple change in their kinematical features can lead to dramatically differentresults. One of the most celebrated instances of this change arises in a rolling sphere

∗ As discussed in Frohlich [66], bowling balls feature offset centers of mass and moment of inertiatensors that are not multiples of I. As a result, some of the intricacies of bowling are not explainedby our simple model of rolling spheres and sliding spheres.

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(a)

(b)

(c)

0

u(t)

u (t0)

t − t0O

O

O

E1

E1

E1

E2

E2

E2

x

x

x

Figure 9.9. Plot of slip speed u(t) as a function of time for a sphere that is initially sliding(and eventually rolls) on a rough horizontal plane. Three representative paths of the centerof mass of the sphere are also shown: (a) the path of the sliding center of mass is a parabolicarc; (b) the path of the sliding center of mass is a straight line; and (c) the center of mass isstationary while the sphere is sliding. For the rolling phases, the path of the center of mass isa straight line. The dashed part of these paths denotes the sliding phase of the motion.

where J �= 25 mR2I and πP = −RE3. Such a sphere is often known as a Chaplygin

sphere. Partially as a consequence of its asymmetry, the path of the point of contactof the Chaplygin sphere with the ground can be very intricate (see [19, 191] andreferences therein).

Two other celebrated examples of rigid bodies that exhibit interesting behaviorare Euler’s disk and the wobblestone (or celt) [42]. The former consists of a heavycircular cylinder that rolls and slides on a convex mirror. As the disk becomesincreasingly horizontal, a whirring sound is heard whose pitch increases. Eventually,the disk comes to a dramatic abrupt halt accompanied (some believe) by an impactof the disk with the convex mirror. The first analysis of this system was performedby Moffatt [143], and his controversial paper was followed by a series of workspromoting alternative mechanisms for the dramatic motion of Euler’s disk (see[110] and references therein). As mentioned earlier, the wobblestone is a rigidbody whose curved lateral surface rolls on a horizontal plane. At first glance, thecurved surface appears to be symmetric, but this is not the case, and, as a result, thewobblestone exhibits an unusual reversal of spin directions. The wobblestone has

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been the subject of several papers (see, for example, [129, 171]) and simulations; thepaper by Blackowiak et al. [17] is particularly recommended for a lucid explanationof the spin reversal mechanism.

Recent analyses of rigid body dynamics have focused on the stability and bi-furcation of their families of steady motions (see, for example, [128, 157]). Increas-ingly, some of these studies are evolving toward an examination of motions of thesesystems that (although not steady) are asymptotic to one or connect two steadymotions of the rigid body. For the interested reader, recent work on the tippe top[20, 177, 219], the possible jumping behavior of a spinning egg [22, 142, 194], and theattitude of a thrown tennis racket [11] are mentioned as examples of notable mod-ern analyses of rigid body dynamics. It is hoped that the exposition in this chapterwill enable you to explore and appreciate works of this type.

EXERCISES

9.1. Suppose a rigid body is rolling on a fixed surface under the influence of a grav-itational force −mgE3. Starting from the work–energy theorem for the rigid body,

T = F · v + M · ω,

prove that the total energy E of the rigid body is conserved. Prove that the totalenergy is also conserved if the rigid body is sliding on a smooth surface.

9.2. A rigid body of mass m is moving in space under the influence of an appliedforce Fa = Fae3 and an applied moment M = 0. Outline how you would determinethe attitude Q and motion of the center of mass of the rigid body.

9.3. The orientation of a rigid body relative to a fixed reference configuration isdefined by a rotation tensor Q. At time t0 this rotation tensor has the value Q(t0),and at time t1 this rotation tensor has the value Q(t1). Give a physical interpretationof the rotation tensor Q(t1)QT(t0). You should make use of the corotational basis inyour answer.

9.4. Consider a rigid body with a fixed point O. What are the three constraints onthe motion of this rigid body? Why is it sufficient to solve MO = HO to determinethe motion of this rigid body?

9.5. Solutions for ωi(t) have been determined for a rigid body dynamics problem.How would you determine Q(t) from this solution?

9.6. A rigid body has a potential energy U = U(x, γi

), where γ1, γ2, γ3 are the Euler

angles used to parameterize Q. If the conservative force F and conservative momentM are such that

−U = F · ˙x + M · ω, (9.30)

then verify that

F = −3∑

i=1

∂U∂xi

Ei, M = −3∑

i=1

∂U∂γi

gi,

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where xi = x · Ei and gi are the basis vectors for the dual Euler basis. Show thatthe gravitational potential energy Un for a rigid body orbiting a fixed sphericallysymmetric rigid body has the functional form U = U

(x, γi

).

9.7. As shown in Figure 9.8, a rigid sphere of mass m and radius R rolls (withoutslipping) on an inclined plane. The inertia tensors for the sphere are J = J0 = µI,where µ = 2mR2

5 .

(a) What are the three constraints on the motion of the sphere? Show that theseconstraints imply that

x1 − R�2 = 0, x2 + R�1 = 0, x3 = 0,

where xi = x · Ei, and �i = ω · Ei.

(b) With the help of the balance of linear momentum for the sphere, show that

Fc = (mR�2 − mg sin(β))

E1 − mR�1E2 + mg cos(β)E3.

(c) Show that the balance of angular momentum for the sphere and the resultsof (b) imply that

75

mR2�1 = 0,75

mR2�2 = mgR sin(β),25

mR2�3 = 0.

(d) Starting from the work–energy theorem for a rigid body, prove that thetotal energy E of the rolling sphere is constant, where

E = 7mR2

10�2

1 + 7mR2

10�2

2 + mR2

5�2

3

− mgx1 sin(β) + mgR cos(β).

(e) Why is the angular momentum H of the sphere in the E3 directionconserved?

(f) If, at time t = 0, the sphere is given an initial angular velocity ω(0) =∑3i=1 �i0Ei, then show that the angular velocity ω(t) is (9.26). What is the

angular acceleration vector α of the sphere?

(g) Suppose the sphere is placed on the inclined plane and released from restwith Q (0) = I. Verify that the sphere will start rolling and that the resultingattitude Q of the sphere corresponds to a fixed-axis rotation.

9.8. In a model for a rigid body flying through the air, the four primary forces onthe body are gravity, a lift force, a drag force, and a thruster force:

F = −mgE3 + mBω × v + f e1 − 12ρf ACd (v · v)

v||v|| ,

M = πt × f e1.

(a) Show that one of the four applied forces is conservative.

(b) Show that the lift force does no work.

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(c) If the thrust force acts at a point whose position vector relative to x is πt,then, with the assistance of the work–energy theorem, establish an expres-sion for E.

9.9. This famous problem is discussed in most books on satellite dynamics (see, forexample, Beletskii [16] or Hughes [97]). The 24 solutions subsequently discusseddate to Lagrange [116, 118] in the late 18th century. Among the remarkable featuresabout Lagrange’s extraordinary work on this topic in [118] is the (early) use of hiscelebrated equations of motion in the context of a rigid body and his clear discussionof a set of (what are now known as) 3–1–3 Euler angles. As shown in Figure 9.10,consider a rigid body B of mass m that is in motion in a central gravitational forcefield about a massive fixed body of mass M. The center of this force field is assumedto be located at a fixed point O. The force, moment, and potential energy of the fieldare given by approximations (8.25).

Rigid body B

Fixed body of mass M

O

X

E1

E2

E3

xFigure 9.10. Schematic of a rigid body of mass mthat is in orbit about a fixed symmetric body ofmass M.

(a) Verify that Mn = −x × Fn. What is the physical relevance of this result?

(b) Why are the angular momentum HO and the total energy E of the satelliteconserved?

(c) Using the balance of linear momentum, show that it is possible for the bodyto move in a circular orbit x = R0er about O with a constant orbital angularvelocity θ0, which is known as the modified Kepler frequency, ωKm:

θ0 = ωKm = ωK

√1 + 3

2R20m

(tr(J) − 3er · Jer), (9.31)

where the Kepler frequency was defined previously [see (2.12)]:

ω2K = GM

r30

.

In (9.31), er = cos(θ)E1 + sin(θ)E2, and this vector is an eigenvector of J.That is, er is parallel to one of the principal axes of the body.

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(d) Using the results of (c), show that a steady motion of the rigid body, that is,one in which ω = 0, is governed by the equation

ω × (Jω) = 3ω2Ker × (Jer). (9.32)

(e) Suppose that the body is asymmetric. That is, the principal values of J0 aredistinct. We seek solutions of (9.32) such that ω · er = 0. Show that thereare six possible solutions for ω that satisfy (9.32) and four possible solutionsfor er. Here, you should assume that J is known and as a result Q is known.As a result, there are 6 × 4 possible solutions of (9.32).

(f) Suppose that the body is such that J = µI, where µ is a constant. Show thatany constant ω satisfies (9.32) and consequently, any orientation of the rigidbody is possible in this case.

(g) Using the results of (e), explain why it is possible for an Earth-based ob-server to see the same side of a satellite in a circular orbit above theEarth.

9.10. Consider the problem of a sphere of mass m, radius R, and inertia tensor J0 =2mR2

5 I moving on a turntable, which is discussed by Gersten et al. [70], Lewis andMurray [127], and Pars [170], among others. The contact between the sphere andthe turntable is rough. In addition, the center O of the turntable is fixed and theturntable rotates about the vertical E3 with an angular speed �.

(a) Suppose that the sphere is rolling on the turntable. The position vector ofthe point of contact of the sphere with the turntable is πP = −RE3. Showthat the motion of the sphere is subject to three constraints:

v + ω × (−RE3) = �E3 × x. (9.33)

Using the representations ω =∑3i=1 �iEi and x =∑3

i=1 xiEi, show that thethree constraints imply that

x1 − R�2 + �x2 = 0, x2 + R�1 − �x1 = 0, x3 = 0.

(b) Assuming that a vertical gravitational force acts on the sphere, draw a free-body diagram of the sphere.

(c) Using a balance of linear momentum and with the assistance of the con-straints, show that the constraint force acting on the sphere is

Fc = mgE3 − m� (x2E1 − x1E2) + mR(�2E1 − �1E2

).

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Exercises 9.10–9.11 303

(d) Using a balance of angular momentum and with the assistance of the resultsof (a)–(c), show that the equations governing the motion of the sphere are

x1 = R�2 − �x2,

x2 = −R�1 + �x1,

x3 = 0,

�1 =(

m�Rµ + mR2

)x1,

�2 =(

m�Rµ + mR2

)x2,

�3 = 0. (9.34)

Here, µ = 2mR2

5 . Why are (9.34) sufficient to determine the motion(x(t), Q(t)) of the sphere?

(e) For the special case in which � = 0, show that the center of mass of thesphere will move in a straight line with a constant speed and that the angularvelocity vector ω of the sphere will be constant.

(f) Numerically integrate (9.34) for a variety of initial conditions. Is it possiblefor the sphere to fall off a turntable of radius R0? In choosing your ini-tial conditions (x(t0), v(t0),ω(t0)) make sure that they are compatible withrolling condition (9.33).

9.11. Recall the definition of the kinetic energy T of a rigid body:

T = 12

∫R

v · vρdv.

(a) Starting from the definition of T, prove the Koenig decomposition:

T = 12

mv · v + 12

H · ω.

(b) Establish the following intermediate results:

J = �J − J�,˙Jω·ω = H · ω,

˙Jω · ω = 2H · ω,

where the angular momentum H = Jω.

(c) Using the intermediate results and the balance laws, prove the work–energytheorem:

T = F · v + M · ω.

(d) If

F =K∑

i=1

Fi, M =K∑

i=1

(xi − x) × Fi + MP,

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304 Exercises 9.11–9.13

then show that

T =K∑

i=1

Fi · vi + MP · ω.

Give three examples of the use of this result for a single rigid body.

9.12. Establish the following theorem, which can be found in Section 44 of Euler[54]: “For any given (rigid) body, we can always find an axis, which passes throughits center of gravity, about which the body can rotate freely and uniformly.” Howis this result related to Euler’s later result that three distinct axes exist about whicha rigid body can rotate freely and at constant angular velocity? This result can befound in [53], where these axes were first termed principal axes of inertia.

9.13. Consider the tippe top shown in Figure 8.12 and discussed in Exercise 8.4.

(a) Continuing the earlier exercise, we follow Or [156] and suppose that thepoint P slides on the horizontal surface and experiences both Coulomb andviscous friction forces:

Ff = − (µk + µv ||vP||) ||N|| s,

where µk and µv are friction coefficients, N is the normal force, and s is theslip direction. Using a balance of linear momentum, show that the normalforce acting on the tippe top is

N = m(g + lθ sin(θ) + lθ2 cos(θ)

)E3.

(b) Starting from the work–energy theorem, prove that the total energy of asliding tippe top decreases with time whereas that for a rolling tippe top isconstant.

(c) If the inertia tensor of the tippe top is

J = λt (I − e3 ⊗ e3) + λae3 ⊗ e3,

then what are the differential equations governing the motion of the tippetop?

(d) For both rolling and sliding tippe tops, show that the angular momentumH · πP is conserved. This integral of motion, which is known as the Jellettintegral, was discovered in the 1870s by J. H. Jellett (1817–1888).∗

(e) When the tippe top is rolling on a horizontal surface, show that, in addi-tion to conservation of total energy, the following kinematical quantity isconserved:

I1 = ω23

(λaλt + mλa

((πP · e1)2 + (πP · e2)2

)+ mλt (πP · e3)2

). (9.35)

∗ A discussion of the history of this integral (and several others) can be found in Gray and Nickel[76].

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Exercises 9.13–9.14 305

Here, πP is the position vector of the instantaneous point of contact P (rel-ative to the center of mass X) of the tippe top with the horizontal surface.The integral of motion I1 was first established by Routh (see Section 243 of[184]) and can be described as the Routh integral. It is also known as theChaplygin integral [107, 115].

9.14. Consider a body that is free to move about one of its material points O thatis fixed (cf. Figure 9.7). The inertia tensor of the body relative to its center of massis J =∑3

i=1 λiei ⊗ ei, where {e1, e2, e3} is a corotational basis. The position vector ofthe center of mass X relative to O is

x − xO = he3,

where h is a constant. A linear spring of stiffness K and unstretched length L0 isattached to the body at the point XS and the other end is attached to a fixed pointA:

xs − xO = s1e1 + s3e3, xA − xO = LAE1.

In addition, a gravitational force −mgE3 acts on the body.

(a) Show that the velocity and acceleration vectors of the center of mass havethe representations

v = hω2e1 − hω1e2,

a = h (ω2 + ω1ω3) e1 − h (ω1 − ω2ω3) e2 − h(ω2

1 + ω22

)e3.

Here, ω =∑3k=1 ωkek.

(b) Show that the angular momentum of the rigid body relative to O is

HO = (λ1 + mh2)ω1e1 + (λ2 + mh2)ω2e2 + λ3ω3e3.

Show that the kinetic energy of the rigid body has the representation

T = 12

HO · ω.

(c) What are the three constraints on the motion of the rigid body? Give pre-scriptions for the constraint force Fc and constraint moment Mc that enforcethese constraints.

(d) Draw a free-body diagram of the rigid body.

(e) Using a balance of linear momentum, show that the reaction force at O is

Fc = mh (ω2 + ω1ω3) e1 − mh (ω1 − ω2ω3) e2

− mh(ω2

1 + ω22

)e3 + mgE3 − Fs,

where Fs is the spring force.

(f) Assuming that a set of 3–1–3 Euler angles is used to parameterize Q =∑3i=1 ei ⊗ Ei, show that the potential energy U = U (x, Q) of the rigid

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306 Exercise 9.14

body in this problem can be expressed as a function of these angles: U =U (ψ, θ, φ).

(g) Verify that the conservative force Fcon and conservative moment Mcon act-ing on the rigid body satisfy the identity

Fcon · v + Mcon · ω = MOcon · ω,

where MOcon is the resultant conservative moment acting on the body rela-tive to O. Using this result, explain why

MOcon = − ∂U∂ψ

g1 − ∂U∂θ

g2 − ∂U∂φ

g3.

(h) Show that the rotational motion of the rigid body is governed by the follow-ing equations:(

λ1 + mh2) ω1 = (λ2 + mh2 − λ3)ω2ω3 + MO · e1,(

λ2 + mh2) ω2 = (λ3 − λ1 − mh2)ω1ω3 + MO · e2,

λ3ω3 = (λ1 − λ2) ω1ω2 + MO · e3,⎡⎢⎣

ψ

θ

φ

⎤⎥⎦ =

⎡⎢⎣

sin(φ)cosec(θ) cos(φ)cosec(θ) 0


− sin(φ) cot(θ) − cos(φ) cot(θ) 1

⎤⎥⎦⎡⎢⎣

ω1

ω2

ω3

⎤⎥⎦ . (9.36)

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10 Lagrange’s Equations of Motion for a SingleRigid Body

10.1 Introduction

In a famous paper [118] on the dynamics of the Moon orbiting the Earth, which waspublished in 1782, Lagrange used his equations of motion in the form

ddt

(∂T∂qK

)− ∂T

∂qK= − ∂V

∂qK(K = 1, . . . , 6) , (10.1)

where V is the potential energy of the Moon and T is its kinetic energy. His interestlay in explaining oscillations (librations) in the attitude of the Moon as seen by anEarth-based observer. What is interesting is that he does not use Euler’s balancelaws, although these were available to him. One might ask what would have hap-pened had he used M = H and F = m ˙v instead of (10.1)? In this chapter, we willshow (among other matters) that his equations are equivalent to Euler’s balancelaws and so he would have arrived at the same conclusions.

We start this chapter by showing that the balance laws F = m ˙v and M = H fora rigid body are equivalent to Lagrange’s equations of motion:

ddt

(∂T∂qi

)− ∂T

∂qi= F · ai,

ddt

(∂T∂νi

)− ∂T

∂νi= M · gi. (10.2)

We illustrate this form of Lagrange’s equations by using a classical problem of asatellite orbiting a fixed body. Indeed, the problem we consider is equivalent to thatconsidered by Lagrange [118].

A second form of Lagrange’s equations of motion is also developed. This formallows broader classes of coordinate choices and, being the most general, is mostuseful in applications. The form of the celebrated equations is

ddt

(∂T∂uA

)− ∂T

∂uA= �A (A = 1, . . . , 6) , (10.3)

307

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308 Lagrange’s Equations of Motion for a Single Rigid Body

where

�A = F · ∂v∂uA

+ M · ∂ω

∂uA. (10.4)

Following the development of (10.3), we show how constraints may be incorporatedand illustrate this matter by using examples of rolling disks, sliding disks, and spin-ning tops.

When the system of constraints on the system is integrable, the constraint forcesand moments are prescribed by use of Lagrange’s prescription, and the coordinatesare chosen appropriately, we pleasantly find that we can decouple (10.3) into a setof equations involving the unconstrained motion and a set of equations for theconstraint forces and moments. The former equations are known as reactionless.We then discuss an approach to Lagrange’s equations that we have referred to asApproach II.

Much of the material concerning Lagrange’s equations in this chapter is basedon Casey [26, 28] supplemented with material on constraint forces and momentsthat is due to O’Reilly and Srinivasa [163].

10.2 Configuration Manifold of an Unconstrained Rigid Body

The motion of a rigid body can be decomposed into a translation x of the center ofmass X followed by a rotation Q. Here, the set of all vectors x is a three-dimensionalEuclidean space E

3, whereas the set of all rotation tensors Q is a three-dimensionalspace known as SO(3).∗ Thus the configuration manifold M of a rigid body is thespace of all vectors x and all rotation tensors Q. This space is the product of E

3 andSO(3):

M = E3 ⊕ SO(3).

The product ⊕ is a topological product. For example, E3 = E ⊕ E

2 = E ⊕ E ⊕ E. Itshould be clear that the dimension of M is 6. It should also be clear that the config-uration manifold can be considered as a submanifold of the configuration space S,which in this case is E

3 ⊕ E9. Here the three-dimensional space is the space contain-

ing the position vector x, and the nine-dimensional space is the space containing thesecond-order tensor Q.

To parameterize M, we can use any curvilinear coordinate system. In one ofthe forms to follow, we use q1, q2, and q3, to parameterize the position vector x ofthe center of mass and any set of Euler angles, ν1, ν2, and ν3, to parameterize therotation tensor Q. For the velocity vector of the center of mass, we then have

v =3∑

i=1

qiai,

∗ The space O(3) is the space of all orthogonal tensors. Hence, the S in SO(3) denotes “special”because rotation tensors have a determinant of +1.

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10.2 Configuration Manifold of an Unconstrained Rigid Body 309

where the covariant basis vectors

ai = ∂x∂qi

.

You should also recall that {ai} is a basis for E3 and that the Euler angles define the

Euler basis {gi}, which is also a basis for E3. In particular, we have

ω =3∑

i=1

νigi.

The Euler basis vectors are not linearly independent for certain values of the secondEuler angle. It is interesting to note that

gi = ∂ω

∂νi, ai = ∂v

∂qi.

These two identities are easily established.We can calculate the kinematical line-element ds for the configuration mani-

fold by using the given choice of Euler angles and curvilinear coordinates from thekinetic energy T:

ds =(√

2Tm

)dt

=(√

v · v + 1m

ω · (J · ω)

)dt.

Shortly an example will be presented of how to calculate the desired representationfor T.

As we are using curvilinear coordinates and Euler angles to parameterizethe motion of the rigid body, for certain points on the configuration manifoldsingularities in the parameters will arise. At these points, it is possible for thebody to be in motion, yet the value we will get for T will be zero. This situationviolates one of the chief attributes of T, namely that T is zero if and only if v = 0and ω = 0, i.e., the rigid body is instantaneously at rest. The singularities will alsoresult in errors when the equations of motion are being integrated numerically.For this reason, many computer codes use two or more sets of curvilinear coor-dinates and two sets of Euler angles for analyzing a given problem in rigid bodydynamics.

A Representation for the Kinetic EnergyWe now consider a specific example. First, we choose a set of spherical polarcoordinates R, �, and � to parameterize x and a set of 3–1–3 Euler angles to

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parameterize Q. For these Euler angles, we recall from Subsection 6.8.2 that theEuler basis {gi} has the representations

g1

g2

g3

=

E3

e′1

e3

=



0 0 1

e1

e2

e3

.

Using these results, we can readily compute the angular velocity vector:

ω = ψE3 + θe′1 + φe3

= (ψsin(φ) sin(θ) + θ cos(φ))

e1 + (ψcos(φ) sin(θ) − θ sin(φ))

e2

+ (ψcos(θ) + φ)

e3. (10.5)

For convenience, we choose Ei to be the principal axes of the body.∗ Hence,

J0 = λ1E1 ⊗ E1 + λ2E2 ⊗ E2 + λ3E3 ⊗ E3,

J = QJ0QT = λ1e1 ⊗ e1 + λ2e2 ⊗ e2 + λ3e3 ⊗ e3.

When these results are combined, the kinetic energy T of the rigid body has therepresentations

T = m2

v · v + 12ω · Jω

= m2

(R2 + R2�2 + R2 sin2(�)�2

)+ λ1

2

(ψsin(φ) sin(θ) + θ cos(φ)

)2+ λ2

2

(ψcos(φ) sin(θ) − θ sin(φ)

)2 + λ3

2

(ψcos(θ) + φ

)2.

It is left as an exercise to substitute this expression for T into the expression forthe kinematical line-element ds in order to find a measure of distance that the rigidbody travels along the configuration manifold M.

Singularities in the parameterization of the motion arise when θ = 0, π and � =0, π. To see the effects of these singularities on T, let us consider an instant in which� = 0 and θ = 0. At this instant, the preceding expression for T simplifies to

T = m2

(R2 + R2�2 + 0

)+ λ1

2

(0 + θ cos(φ)

)2 + λ2

2

(0 − θ sin(φ)

)2 + λ3

2

(ψ + φ

)2.

∗ If we do not make this choice, then we would find a more complicated representation for the ro-tational kinetic energy: ω · (Jω) = J011ω

21 + J022ω

22 + J033ω

23 + 2J012ω1ω2 + 2J023ω2ω3 + 2J013ω1ω3,

where ωi = ω · ei and J0ik = ei · (Jek) = Ei · (J0Ek).

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10.3 Lagrange’s Equations of Motion: A First Form 311

If, in addition, ψ = −φ, θ = R = � = 0, and � �= 0, then the preceding expressionfor T = 0, but the actual kinetic energy of the body is nonzero.

10.3 Lagrange’s Equations of Motion: A First Form

In this section, we wish to establish Lagrange’s equations for an unconstrained rigidbody. Our proof is based on Casey [28], but our developments are not as generalas his. Resulting equations (10.6) were first shown by him to be equivalent to thebalances of linear and angular momentum for a rigid body.∗ There is also a strongsuggestion of the equivalence of Lagrange’s equations of motion and the balance ofangular momentum for a rigid body in Sections 8-2 and 8-6 of Greenwood [79].† Werefer to (10.6) as the first form of Lagrange’s equations of motion.

To start, we choose a set of curvilinear coordinates qi to parameterize x: x =x(q1, q2, q3). Next, we choose a set of Euler angles νi to parameterize the rotationtensor Q of the rigid body Q = Q(ν1, ν2, ν3). Then, as will be subsequently shown,Lagrange’s equations for the rigid body are

ddt

(∂T∂qi

)− ∂T

∂qi= F · ai,

ddt

(∂T∂νi

)− ∂T

∂νi= M · gi. (10.6)

Here, gi are the Euler basis vectors and ai are the basis vectors for E3 that are asso-

ciated with the curvilinear coordinate qi. For the preceding equations, you may wishto recall the results

v =3∑

i=1

qiai,∂v∂qi

= ai, ω =3∑

i=1

νigi,∂ω

∂νi= gi.

It should be evident from (10.6) that Lagrange’s equations for a rigid body havesimilarities to those we encountered earlier with particles. The main difference isthe balance of angular momentum.

If some of the forces and moments acting on the rigid body are conservative,then, for these conservative forces Fcon and moments Mcon, we have

Fcon = −3∑

i=1

∂U∂qi

ai, Mcon = −3∑

i=1

∂U∂νi

gi, (10.7)

where the potential energy function U has the representations

U = U (x, Q) = U(q1, q2, q3, ν1, ν2, ν3) .

Notice that Fcon · ai = − ∂U∂qi and Mcon · gi = − ∂U

∂ν i . Consequently, it is not necessaryto evaluate Fcon and Mcon in Lagrange’s equations; rather it suffices to evaluate the

∗ See, in particular, Theorems 4.2 and 4.4 of Casey [28].† Greenwood’s exposition is missing the intermediate result ∂T

∂γ i = H · gi .

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partial derivatives of U. It is remarkable that the situation with potential energies inrigid bodies is similar to that encountered in systems of particles.

It is also possible to write an alternative form of Lagrange’s equations of motionby using the Lagrangian L = T − U. Specifically,

ddt

(∂L∂qi

)− ∂L

∂qi= (F − Fcon) · ai,

ddt

(∂L∂νi

)− ∂L

∂νi= (M − Mcon) · gi. (10.8)

It is left as an (easy) exercise to show that (10.8) can be established from (10.6).

10.3.1 Proof of Lagrange’s Equations

To prove Lagrange’s equations, we need to exploit the Koenig decomposition anduse the angular velocity vector ω0 = QTω. The proof proceeds quickly after somepreliminary results have been addressed.

There are four steps in the proof. The first step involves parameterizing x byuse of a set of curvilinear coordinates and parameterizing Q by use of a set of Eulerangles νi. These parameterizations imply that the kinetic energy T is a function ofthese quantities and their time derivatives:

T = T(qi, qi, νk, νk) = 1

2mv · v + 1

2J0ω0 · ω0.

Here, the angular velocity vector ω0 = QTω and the inertia tensor J0 = QTJQ.We next consider the partial derivatives of T:

∂T∂qi

= ∂

∂qi

(12

mv · v)

= mv · ∂v∂qi

= mv · ai,

∂T∂qi

= ∂

∂qi

(12

mv · v)

= mv · ∂v∂qi

= mv · ai,

∂T∂νi

= ∂

∂νi

(12

J0ω0 · ω0

)= J0ω0 · ∂ω0

∂νi= J0ω0 · (QTgi

),

∂T∂νi

= ∂

∂νi

(12

J0ω0 · ω0

)= J0ω0 · ∂ω0

∂νi= J0ω0 · (QTgi

).

(10.9)

Here we have used the identities

ai = ∂v∂qi

, ai = ∂v∂qi

, gi = Q∂ω0

∂νi, gi = Q

∂ω0

∂νi. (10.10)

Shortly a derivation of these results will be given.

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10.3 Lagrange’s Equations of Motion: A First Form 313

For the third step, we examine (10.9)1,2 and note that G = mv. Hence,

ddt

(∂T∂qi

)− ∂T

∂qi= d

dt(G · ai) − G · ai

= G · ai

= F · ai.

Consequently, the first three Lagrange equations of motion have been established.The fourth step in the derivation is to note that, for any vector b,

H · b = Jω · b = QJ0QTQω0 · b = J0ω0 · QTb.

Using this result in conjunction with (10.9)3,4, we find that

ddt

(∂T∂νi

)− ∂T

∂νi= d

dt(H · gi) − H · gi

= H · gi

= M · gi.

Thus we have established the last three of Lagrange’s equations of motion.

10.3.2 The Four Identities

The proof of Lagrange’s equations is achieved by use of the four identities, (10.10).The first two of these results are similar to those we established for the single parti-cle. You should notice the presence of Q in (10.10)3,4. Unfortunately, ∂ω

∂νi �= gi.We first establish the easier results:

∂v∂qi

= ∂

∂qi

(3∑

k=1

qkak

)=

3∑k=1

δikak = ai

and

∂v∂qi

= ∂

∂qi

(3∑

k=1

qkak

)=

3∑k=1

qk ∂ak

∂qi

=3∑

k=1

qk ∂2x∂qi∂qk

=3∑

k=1

qk ∂

∂qk

(∂x∂qi

)

=3∑

k=1

qk ∂ai

∂qk

= ai.

Notice that we used the facts that ai are both independent of qk and the derivativesof x with respect to qi.

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We next consider the easier of the two remaining identities:

∂ω0

∂νi= ∂

∂νi

(QTω

) = ∂QT

∂νiω + QT

(∂ω

∂νi

)= 0ω + QT (gi)

= QTgi.

Some rearranging gives

gi = Q∂ω0

∂νi.

Notice that we used the fact that Q does not depend on νi to establish this result.For the final result, we again need to be cognizant of the fact that Q depends on

the Euler angles νi but not on νi. First, we note that

gi = ∂ω

∂νi= ∂

∂νi

(−1

2ε[QQT]) = ∂

∂νi

(−1

2ε

[3∑

k=1

νk ∂Q∂νk

QT

])

= −12ε

[∂Q∂νi

QT]

.

Next, we deduce, by differentiating QQT = I twice with respect to the Euler angles,that

∂Q∂νi

QT = −Q∂QT

∂νi,

∂Q∂νi

∂QT

∂νk= Q

∂QT

∂νi

∂Q∂νk

QT.

The previous results are now used to show the desired identity:

gi = ddt

(−1

2ε

[∂Q∂νi

QT])

=3∑

k=1

νk(

−12ε

[∂2Q

∂νi∂νkQT + ∂Q

∂νi

∂QT

∂νk

])

=3∑

k=1

νk(

−12ε

[Q(

QT ∂2Q∂νi∂νk

+ ∂QT

∂νi

∂Q∂νk

)QT])

=3∑

k=1

νk(

−12ε

[Q

∂

∂νi

(QT ∂Q

∂νk

)QT])

= −12ε

[Q

∂

∂νi

(QT

3∑k=1

νk ∂Q∂νk

)QT

]

= −12ε

[Q

∂�0

∂νiQT]

= Q(

−12ε

[∂�0

∂νi

])

= Q∂ω0

∂νi.

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10.4 A Satellite Problem 315

In the final stages of the proof we used the facts that ω0 is the axial vector of �0 =QTQ = QT�Q. We also used the identity

ε[QBQT] = det (Q) Q (ε [B]) .

This identity is one method relating ω0 to ω, and it was alluded to earlier whenangular velocity vectors were discussed [see (6.7)].

10.4 A Satellite Problem

As an example of a problem from rigid body dynamics for which there are no con-straints, we consider a satellite of mass m that is in orbit about a spherically sym-metric body of mass M (cf. Figure 10.1). We assume that the spherically symmetricbody is fixed and use its center of mass as the origin of the position vector of thecenter of mass of the satellite.

PreliminariesTo parameterize the motion of the center of mass of the rigid body we use a sphericalpolar coordinate system:

x = ReR

= R cos(φ)E3 + R sin(φ)er

= R cos(φ)E3 + R sin(φ) (cos(θ)E1 + sin(θ)E2)

and

v = ReR + R sin(φ)θeθ + Rφeφ.

You should be able to see from this equation what the covariant basis vectors ai are.We parameterize the rotation tensor Q by using a set of 1–2–3 Euler angles:

ω = ν1E1 + ν2e′2 + ν3e3.

You should notice that ei = QEi. The angular velocity vector of the body also hasthe representation

ω = ω1e1 + ω2e2 + ω3e3,

where

ω1 = ν2 sin(ν3) + ν1 cos(ν2) cos(ν3),

ω2 = ν2 cos(ν3) − ν1 cos(ν2) sin(ν3),

ω3 = ν1 sin(ν2) + ν3.

You should be able to infer the representations for gi from these results.We shall choose Ei to be the principal axes of the body in its reference config-

uration. Using this specification, it follows that the kinetic energy of the rigid body

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Rigid body B

Fixed symmetric body of mass M

O

θ

β

X

E1

E2

E3

e1

e2

e3

er

x

Figure 10.1. A rigid body orbiting a spherically symmetric body of mass M. The angle of lati-tude β = π

2 − φ.

has the representation

T = m2

(R2 + R2φ2 + R2 sin2(φ)θ2

)+ λ1

2(ν2 sin(ν3) + ν1 cos(ν2) cos(ν3))2

+ λ2

2(ν2 cos(ν3) − ν1 cos(ν2) sin(ν3))2

+ λ3

2(ν1 sin(ν2) + ν3)2

.

If the body has an axis of symmetry so that λ1 = λ2, then the expression for therotational kinetic energy will simplify considerably.

The sole force and moment acting on the rigid body is due to the central grav-itational force exerted on it by the body of mass M. These forces and moments areconservative and are associated with the potential energy

U = −GMmR

− GM2R3

tr(J) + 3GM2R3

(JeR) · eR. (10.11)

To express this potential energy in terms of the Euler angles νi, we need to use theresults

eR = cos(φ)E3 + sin(φ) cos(θ)E1 + sin(φ) sin(θ)E2,

E1 = cos(ν2) cos(ν3)e1 − cos(ν2) sin(ν3)e2 + sin(ν2)e3,

E2 = − sin(ν2) cos(ν3)e1 + sin(ν2) sin(ν3)e2 + cos(ν2)e3,

E3 = (sin(ν1) sin(ν3) − cos(ν1) sin(ν2) cos(ν3)) e1

+ (sin(ν1) cos(ν3) + cos(ν1) sin(ν2) sin(ν3)) e2 + cos(ν1) cos(ν2)e3.

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10.4 A Satellite Problem 317

Using these results, we can express eR in terms of the corotational basis, and then auseful expression for (JeR) · eR can be established. We can then derive the expres-sions for the force and moment associated with this potential energy by using repre-sentations (10.7). Alternatively, we can appeal to representations (8.25) describedin Chapter 8.

The Balance LawsFor the rigid body of interest, we have the balance laws

m ˙v = F

= −GMmR2

eR − 3GM2R4

(2J + ((λ1 + λ2 + λ3) − 5eR · JeR) I) eR,

H = M

=(

3GMR3

)eR × (JeR).

Here, we used our earlier representations, (8.25), for the conservative force andconservative moment associated with the gravitational potential. You should alsonotice that, if eR is an eigenvector of J, then JeR is parallel to eR. In this case, theso-called gravity-gradient torque M = 0. In addition, if R = R0er and θ = ω0, whereR0 and ω0 are constant, then the center of mass of the rigid body describes a circularorbit at a constant orbital speed ω0.

Lagrange’s Equations of MotionLagrange’s equations of motion for the satellite are the ai components of the balanceof linear momentum and the gi components of the balance of angular momentum:

ddt

(∂T

∂R

)− ∂T

∂R= F · eR

= −GMmR2

− 3GM2R4

((λ1 + λ2 + λ3) − 3eR · JeR) ,

ddt

(∂T

∂φ

)− ∂T

∂φ= F · Reφ

= −3GMR3

(eR · Jeφ) ,

ddt

(∂T

∂θ

)− ∂T

∂θ= F · R sin(φ)eθ

= −3GM sin(φ)R3

(eR · Jeθ) ,

ddt

(∂T∂νi

)− ∂T

∂νi= M · gi

=((

3GMR3

)eR × (JeR)

)· gi.

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In the last of these equations, i = 1, . . . , 3. It is left as an exercise to evaluate thepartial derivatives of T with respect to the coordinates and their velocities. Al-ternatively, one could also use (10.8) to write Lagrange’s equations by using theLagrangian L = T − U.

ConservationsBecause the only forces and moments acting on the rigid body are conservative,and there are no constraints on the motion of the rigid body, it is easy to see thatthe total energy E = T + U is conserved. In addition, the angular momentum HO isconserved. To see this notice that

HO = MO = ReR × F + M.

Substituting for F and M, one finds that MO = 0. Consequently HO is conserved.It is interesting to note that we cannot conclude that H is conserved for an ar-

bitrary motion of the rigid body – although it is conserved if the gravity-gradienttorque M is zero for a specific motion.

10.5 Lagrange’s Equations of Motion: A Second Form

Previously, we assumed that a set of coordinates q1, . . . , q6 had been chosen to pa-rameterize x and Q:

x = x(q1, . . . , q3) , Q = Q

(q4 = ν1, . . . , q6 = ν3) . (10.12)

We note that the covariant basis vectors associated with this choice of coordinatesare

ai = ∂v∂qi

, gi = ∂ω

∂q(i+3), i = 1, 2, 3.

With choice (10.12), we showed that Lagrange’s equations of motion have the fol-lowing form:

ddt

(∂T∂qi

)− ∂T

∂qi= F · ∂v

∂qi,

ddt

(∂T

∂q(i+3)

)− ∂T

∂q(i+3)= M · ∂ω

∂q(i+3). (10.13)

It is not too difficult to see that these equations can be written in a more compactform:

ddt

(∂T∂qA

)− ∂T

∂qA= F · ∂v

∂qA+ M · ∂ω

∂qA(A = 1, . . . , 6) . (10.14)

Form (10.14) of Lagrange’s equations is useful in several cases. Among them,

1. there are no constraints on the motion of the rigid body;2. the constraints (and the associated constraint forces and moments) on the rigid

body do not couple the rotational and translational degrees of freedom.

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10.5 Lagrange’s Equations of Motion: A Second Form 319

However, we shall need a more general form of Lagrange’s equations for otherapplications.

We now consider a new choice of coordinates:

x = x(u1, . . . , u6) , Q = Q

(u1, . . . , u6) . (10.15)

With the choice (10.15), we shall see that Lagrange’s equations of motion are

ddt

(∂T∂uA

)− ∂T

∂uA= �A, (10.16)

where the generalized forces are

�A = F · ∂v∂uA

+ M · ∂ω

∂uA(A = 1, . . . , 6) .

We shall also shortly discuss examples that use this form.To establish (10.16), we invoke the following identities∗:

ddt

(∂v∂uA

)= ∂v

∂uA,

ddt

(∂ω

∂uA

)= d

dt

(Q

∂ω0

∂uA

)= Q

∂ω0

∂uA,

where

ω = Qω0.

Using these identities and the decomposition of the kinetic energy,† we can use astraightforward set of manipulations to establish (10.16):

ddt

(∂T∂uA

)− ∂T

∂uA= d

dt

(mv · ∂v

∂uA+ Jω ·

(Q

∂ω0

∂uA

))

−(

∂T∂uA

= mv · ∂v∂uA

+ J0ω0 ·(

∂ω0

∂uA

))

= ddt

(G · ∂v

∂uA+ H ·

(Q

∂ω0

∂uA

))

−(

∂T∂uA

= G · ∂v∂uA

+ H ·(

Q∂ω0

∂uA

))

= G · ∂v∂uA

+ H ·(

Q∂ω0

∂uA

)

= F · ∂v∂uA

+ M · ∂ω

∂uA. (10.17)

This form of Lagrange’s equations is the starting point for most applications and adiscussion of Approach II.

We now turn to issues associated with the second form of Lagrange’s equationsin the presence of constraints.

∗ The proof of these results follows from (10.15) in a manner that is similar to the method by whichthe four identities were established in Subsection 10.3.2.

† Recall that Jω · ω = QJ0ω0 · ω = QJ0ω0 · Qω0 = J0ω0 · ω0.

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10.5.1 Changing Coordinates: An Identity

Recall that we are using two coordinate systems: u1, . . . , u6 and q1, . . . , q6. We havetacitly assumed that these coordinate systems are related by invertible functions:

qA = qA (u1, . . . , u6) .Then, as

qA =6∑

B=1

∂qA

∂uBuB,

we conclude that

∂qA

∂uB= ∂qA

∂uB(A = 1, . . . , 6, B = 1, . . . , 6) . (10.18)

These identities are often known as “cancellation of the dots.”It is a good exercise to show that (10.18) also hold when the coordinate trans-

formations are time dependent:

qA = qA (u1, . . . , u6, t).

The resulting identities are used in the literature with Approach II of Lagrange’sequations of motion.

10.5.2 Constraints and Constraint Forces and Moments

Suppose that an integrable constraint is imposed on the rigid body:

(q1, . . . , q6, t

) = 0.

We assume that we are at liberty to choose the coordinates u1, . . . , u6 to write thisconstraint in a simpler form, such as

u6 − f (t) = 0.

The question we wish to ask now is, what are the consequences for the right-handside of Lagrange’s equations?

The constraint = 0 implies that = 0. With some rearranging we find that = 0 can be expressed as

f · v + h · ω + e = 0,

where

f =3∑

i=1

∂

∂qiai, h =

3∑i=1

∂

∂q(i+3)gi, e = ∂

∂t.

For the present, it is most convenient to have representations for f and h in terms ofthe coordinates for x and Q, respectively, rather than in terms of u1, . . . , u6.

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To answer the question we just posed, we suppose that Lagrange’s prescriptionhas been used to prescribe the unknown constraint force and constraint moment:

Fc = µf = µ

3∑i=1

∂

∂qiai,

Mc = µh = µ

3∑i=1

∂

∂qi+3gi.

It is interesting to expand the constraint forces and moments for this case:

Fc · ∂v∂uA

+ Mc · ∂ω

∂uA=

3∑i=1

µ∂

∂qiai · ∂v

∂uA+

3∑i=1

µ∂

∂qi+3gi · ∂ω

∂uA

= µ

3∑i=1

3∑k=1

∂

∂qiai · ∂qk

∂uAak

+µ

3∑i=1

3∑k=1

∂

∂qi+3gi · ∂qk+3

∂uAgk

= µ

3∑i=1

3∑k=1

∂

∂qi

∂qk

∂uAδk

i + µ

3∑i=1

3∑k=1

∂

∂qi+3

∂qk+3

∂uAδk

i

= µ

3∑i=1

∂

∂qi

∂qi

∂uA+ µ

3∑i=1

∂

∂qi+3

∂qi+3

∂uA

= µ

3∑i=1

∂

∂qi

∂qi

∂uA+ µ

3∑i=1

∂

∂qi+3

∂qi+3

∂uA

= µ∂

∂uA. (10.19)

Notice that we used (10.18) in the penultimate step, and also expressed as a func-tion of u1, . . . , u6 and t:

= (q1, . . . , q6, t

) = (u1, . . . , u6, t

).

To avoid confusion where any may possibly arise, we ornament with a caret (hat)when it is expressed as a function of u1, . . . , u6, and t.

We are now in a position to make some important conclusions. For a constraint

= 0, where = (q1, . . . , q6, t

) = (u1, . . . , u6, t

),

Lagrange’s prescription yields

Fc · ∂v∂uA

+ Mc · ∂ω

∂uA= µ

∂

∂uA. (10.20)

Thus, in answer to our question, the presence of the constraint = 0 and the con-straint forces and moments it requires introduces a term µ ∂

∂uA on the right-hand side

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of Lagrange’s equations. This event is (reassuringly) similar to what arose previouslyin the case of a single particle and a system of particles.

10.5.3 Consequences of Lagrange’s Prescription and Integrable Constraints

Let us now explore some of the consequences of (10.20). Suppose we can expressthe integrable constraint as

u6 − f (t) = 0.

Then,

Fc · ∂v∂uA

+ Mc · ∂ω

∂uA= µδ6

A.

As a result, the constraint force and moment will contribute to only the Lagrange’sequation associated with u6. The familiar decoupling we found with a single particleand a system of particles thus holds for the rigid body!

10.5.4 Potential Energy and Conservative Forces and Moments

Suppose that a potential energy is associated with the rigid body:

U = U(q1, . . . , q6) = U

(u1, . . . , u6) .

The conservative forces and moments associated with this potential energy are

Fcon = −3∑

i=1

∂U∂qi

ai,

Mcon = −uQ = −3∑

i=1

∂U∂qi+3

gi.

Now suppose that we have chosen a new set of coordinates. Then, by setting =−U and µ = 1 in (10.19), we can show that

Fcon · ∂v∂uA

+ Mcon · ∂ω

∂uA= − ∂U

∂uA.

In summary, the conservative forces and moments appear in a familiar form on theright-hand side of Lagrange’s equations of motion.

10.5.5 Mechanical Power and Energy Conservation

If we have a single integrable constraint on the motion of the rigid body and apotential energy U, then the combined mechanical power P of these forces and

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moments is

P = Fc · v + Mc · ω − U

= µ

(−∂

∂t

)− U, (10.21)

where we have assumed that the constraint force and constraint moment areprescribed by using Lagrange’s prescription. Consequently, if the integrable con-straint is time independent, ∂

∂t = 0, and the only applied forces and moments areconservative, then, with the help of (10.21), we can surmise that

T = P = −U.

As a result, the total energy E = T + U is conserved when the integrable constraintis time independent and all the applied forces and moments are conservative.

To arrive at (10.21), a straightforward set of manipulations is needed:

P = (Fcon + Fc) · v + (Mcon + Mc) · ω

= µ

(3∑

i=1

∂

∂qiai · v +

3∑i=1

∂

∂qi+3gi · ω

)

−(

3∑i=1

∂U∂qi

ai · v +3∑

i=1

∂U∂qi+3

gi · ω

)

= µ

(3∑

i=1

∂

∂qiqi +

3∑i=1

∂

∂qi+3qi+3

)−(

3∑i=1

∂U∂qi

qi +3∑

i=1

∂U∂qi+3

qi+3

)

= µ

(6∑

A=1

∂

∂qAqA

)−(

6∑B=1

∂U∂qB

qB

)

= µ

( − ∂

∂t

)− U.

The final result, P = µ( − ∂

∂t

)− U, is what was used in writing (10.21).

10.5.6 Summary

Suppose we have a rigid body for which all of the applied forces and moments areconservative:

F = Fc − ∂U∂x

, M = Mc − uQ.

In addition, suppose that there are two constraints on the motion of the rigid body:

= u6 − f (t) = 0, f · v + h · ω + e = 0.

The first of these constraints is integrable, whereas the second is nonintegrable. Weassume that the constraint forces and moments associated with them are prescribed

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using Lagrange’s prescription:

Fc = µ1

3∑i=1

∂

∂qiai + µ2f, Mc = µ1

3∑i=1

∂

∂qi+3gi + µ2h. (10.22)

Turning to Lagrange’s equations of motion (10.16), we find that the equationswould read

ddt

(∂T∂uA

)− ∂T

∂uA= − ∂U

∂uA+ QA. (10.23)

As the only nonconservative forces acting on the body are due to the constraintforces and constraint moments, QA are simply

QA = Fc · ∂v∂uA

+ Mc · ∂ω

∂uA(A = 1, . . . , 6) .

Using a Lagrangian L = T − U and using prescriptions (10.22) for Fc and Mc, wefind that (10.23) can be expressed in the following form:

ddt

(∂L∂uB

)− ∂L

∂uB= µ2

(f · ∂v

∂uB+ h · ∂ω

∂uB

),

ddt

(∂L∂u6

)− ∂L

∂u6= µ1 + µ2

(f · ∂v

∂u6+ h · ∂ω

∂u6

).

In these equations, B = 1, . . . , 5. Notice that the right-hand sides of these forms ofLagrange’s equations are similar to those for a system of particles with the addedcomplication of the moment terms.

10.6 Lagrange’s Equations of Motion: Approach II

We examine Approach II applied to Lagrange’s equations for a rigid body whosemotion is constrained. Specifically, we parallel the discussion of Subsection 10.5.6and assume that the body is subject to one integrable constraint and one non-integrable constraint. Our discussion is easily generalized to multiple integrableand nonintegrable constraints. The results we discuss were first established byCasey [28]. The resulting form of Lagrange’s equations is the one most commonlyused in engineering and physics.

First, we assume that the six coordinates u1 . . . , u6 are chosen such that the in-tegrable constraint = 0 can be simply written as

= u6 − f (t).

Imposing this constraint, we can calculate the constrained kinetic T and potential Uenergies of the system. The former will be a function of u1, . . . , u5, u1, . . . , u5, and t,whereas the latter will be a function of u1, . . . , u5 and t. The nonintegrable constraintπ2 = 0 can be written in terms of the new coordinates and their velocities:

π2 =5∑

B=1

pBuB + e.

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10.7 Rolling Disks and Sliding Disks 325

Here p1, . . . , p5 and e are functions of t and u1, . . . , u5. We will use Lagrange’s pre-scription to specify the associated constraint forces and constraint moments [see(10.22)].

By following the same arguments used to establish Lagrange’s equations fora single particle by use of Approach II, we find that the equations governing themotion of the rigid body are

5∑B=1

pBuB + e = 0,

ddt

(∂T∂uB

)− ∂T

∂uB= F · ∂v

∂uB+ M · ∂ω

∂uB

= − ∂U∂uB

+ Fanc · ∂v∂uB

+ Manc · ∂ω

∂uB+ µ2 pB.

(10.24)

Here, B = 1, . . . , 5 and the constraint forces and moments associated with the inte-grable constraint are absent. In these equations, the forces and moments acting onthe body have been decomposed:

F = Fanc + Fc + Fcon,

M = Manc + Mc + Mcon.

For instance, Fanc are the applied nonconservative forces. The Magnus force FM =mBω × v, which is used in studies of the flight path of a baseball, is a good exampleof such a force.

10.7 Rolling Disks and Sliding Disks

As examples of constrained rigid bodies, we consider a rigid circular disk of massm and radius R that either rolls without slipping on a rough horizontal plane (cf.Figure 10.2) or slides on a smooth horizontal plane. It shall be assumed that theinertia tensor J0 has the representation

JO = λ (E1 ⊗ E1 + E2 ⊗ E2) + λ3E3 ⊗ E3. (10.25)

In other words, we are assuming that E3 is the axis of symmetry of the disk in itsreference configuration. The principal moments of inertia are λ3 = 2λ and λ = mR2

4 ,but we do not use these substitutions in order to enable an easier tracking of thealgebraic manipulations.

PreliminariesThe location of the center of mass of the disk can be parameterized by use of a setof Cartesian coordinates:

x =3∑

i=1

xiEi.

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g

P

O

X

ψ

φ

E1

E2

E3

e′′1

e′′1

e′′2

e1

e2

Figure 10.2. A circular disk moving with one point in contact with a horizontal plane.

The rotation tensor of the disk is described by a 3–1–3 set of Euler angles:

ω = ψE3 + θe′1 + φe3.

For this set of Euler angles, we recall, from (6.33) in Subsection 6.8.2, that the Eulerbasis vectors gi have the representations

g1 = E3

g2 = e′1

g3 = e3

=



0 0 1

e1

e2

e3

.

(10.26)

In addition, the Euler angles are subject to the restrictions ψ ∈ [0, 2π), θ ∈ (0, π),and φ ∈ [0, 2π).

Physically, the angle θ represents the inclination angle of the disk. When thisangle is π

2 , the disk is vertical. On the other hand, when θ = 0 or π, the disk is hor-izontal. In either of these situations, an entire surface of the disk is in contact withthe ground and the equations of motion that will be presented will not be valid.

ConstraintsAs discussed previously in Subsection 8.6.3, the motion of the rolling disk is subjectto three constraints because the velocity vector of the point of contact P is zero:

vP = v + ω × πP = 0. (10.27)

Earlier, we showed how these equations lead to the following three constraints [see(8.21)]:

π1 = 0, π2 = 0, π3 = 0.

We also found that the third of these constraints was integrable:

= x3 − R sin(θ) = 0, (10.28)

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whereas the remaining constraints were nonintegrable.∗ It should be clear that weexpress all three constraints on the rolling disk in the form f · v + h · ω = 0, and thisfacilitates the use of Lagrange’s prescription for Fc and Mc.

Coordinates and EnergiesMotivated by the presence of the integrable constraint x3 = R sin(θ), we now definea set of coordinates:

u1 = x1, u2 = x2, u3 = ψ,

u4 = θ, u5 = φ, u6 = x3 − R sin(θ). (10.29)

Given values of u1, . . . , u6, we can uniquely invert relations (10.29) to determine xi

and the Euler angles:

x1 = u1, x2 = u2, x3 = u6 + R sin(u4) ,

ψ = u3, θ = u4, φ = u5. (10.30)

It should be clear that u1, . . . , u5 are the generalized coordinates for both the rollingdisk and the sliding disk. For future reference, we compute that

v = u1E1 + u2E1 + (u6 + Ru4 cos(u4))E3,

ω = u3E3 + u4e′1 + u5e3. (10.31)

We also note that ω does not depend on u6 or its time derivative.We shall need the derivatives of v and ω with respect to the coordinates

u1, . . . , u5. With the help of (10.31), these can be obtained in a straightforwardmanner:

∂v∂u1

= E1,∂v∂u2

= E2,∂v∂u3

= 0,

∂v∂u4

= R cos (θ) E3,∂v∂u5

= 0,∂v∂u6

= E3, (10.32)

and

∂ω

∂u1= 0,

∂ω

∂u2= 0,

∂ω

∂u3= E3,

∂ω

∂u4= e′

1,∂ω

∂u5= e3,

∂ω

∂u6= 0. (10.33)

These vectors will soon feature in Lagrange’s equations of motion.Given the inertia tensor (10.25), representations (10.31), and the expressions

for ω · ek given in (10.5), an easy calculation shows that the (unconstrained) kinetic

∗ This was established by use of Frobenius’ theorem in one of the exercises at the end of Chapter 8.

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energy of the disk has the representation

T = m2

(x2

1 + x22 + (u6 + Rθ cos (θ)

)2)+ λ

2

(ψ2 sin2(θ) + θ2

)+ λ3

2

(φ + ψ cos(θ)

)2.

In addition, the potential energy of the disk is

U = mg(u6 + R sin(θ)

).

You should notice how the symmetry of the rigid body of interest simplifies therotational kinetic energy.

Constraint Forces and Constraint MomentsThe constraint forces and moments acting on the rolling disk can be prescribed byuse of Lagrange’s prescription. With the assistance of the expressions for πi, we find

Fc = µ3E3 + µ1E1 + µ2E2,

Mc = µ3(−R cos(θ)g2)

+µ1(R cos(θ) cos(ψ)g1 − R sin(θ) sin(ψ)g2 + R cos(ψ)g3)

+µ2(R cos(θ) sin(ψ)g1 + R sin(θ) cos(ψ)g2 + R sin(ψ)g3) .

The expressions for the dual Euler basis we use here can be inferred from (6.34) inSection 6.8.

If we express the dual Euler basis vectors gi in terms of e′′i and expand the ex-

pressions for Mc, we will find that Mc = πP × Fc. This implies, not surprisingly, thatFc and Mc are equipollent to a force R = Fc acting at the instantaneous point ofcontact P. The E3 component of this force is the normal force, whereas the remain-ing components constitute the friction force. Thus Lagrange’s prescription yields aphysically reasonable set of constraint forces and constraint moments.

Imposing the Integrable ConstraintIf we impose the integrable constraint, then u6 = 0 and the resulting constrainedpotential and kinetic energies are

U = mgR sin(θ),

T = m2

(x2

1 + x22

)+ λ

2

(ψ2 sin2(θ)

)+ 1

2

(λ + mR2 cos2(θ)

)θ2

+ λ3

2

(φ + ψ cos(θ)

)2.

It is important to notice that T = T(x1, x2, x1, x2, θ, φ, θ, ψ). In other words, u6 hasbeen eliminated.

The Rolling Disk’s Equations of MotionLagrange’s equations of motion for the rolling disk can now be easily obtained. Wefirst note that the resultant force F acting on the disk is composed of a constraint

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force and a conservative force: F = Fc − mgE3. The resultant moment M consistsentirely of the constraint moment: M = Mc

Using (10.6), and with the help of (10.32) and (10.33), we find that

ddt

(∂T∂x1

)−(

∂T∂x1

)= �1,

ddt

(∂T∂x2

)−(

∂T∂x2

)= �2,

ddt

(∂T

∂ψ

)− ∂T

∂ψ= �3,

ddt

(∂T

∂θ

)− ∂T

∂θ= �4,

ddt

(∂T

∂φ

)− ∂T

∂φ= �5,

ddt

(∂T∂u6

)−(

∂T∂u6

)= �6. (10.34)

In these equations,

�1 = F · E1 + M · 0 = µ1,

�2 = F · E2 + M · 0 = µ2,

�3 = F · 0 + M · E3 = R cos(θ) (µ1 cos(ψ) + µ2 sin(ψ)) ,

�4 = F · R cos(θ)E3 + M · e′1

= −mgR cos(θ) + R sin(θ) (µ2 cos(ψ) − µ1 sin(ψ)) ,

�5 = F · 0 + M · e3 = µ1R cos(ψ) + µ2R sin(ψ),

�6 = F · E3 + M · 0 = µ3 − mg.

Evaluating the partial derivatives of T in (10.34) and imposing the integrable con-straint u6 = 0, we find, with some minor rearranging, that

mx1 = µ1,

mx2 = µ2,

ddt

(λ sin2(θ)ψ + λ3

(φ + ψ cos(θ)

)cos(θ)

)= �3,

λθ + λ3(φ + ψ cos(θ)

)ψ sin(θ) − λψ2 sin(θ) cos(θ) = �4,

ddt

(λ3(φ + ψ cos(θ)

)) = �5,

md2

dt2(R sin(θ)) = µ3 − mg. (10.35)

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These equations are supplemented by the constraints,

x3 − R sin(θ) = 0, π1 = 0, π2 = 0, (10.36)

to form a closed system of equations for the nine unknowns xi, φ, θ, ψ , and µi.It is important to note that systems of equations (10.35) and (10.36) are not

readily integrated numerically. In particular, they cannot immediately be written inthe form y = f(y), which is required for most numerical integrators – such as thosebased on a Runge–Kutta scheme.

Conservations for the Rolling DiskThe easiest method to see that the rolling disk’s total energy E is conserved isto use the alternative form of the work–energy theorem. Two forces act on therolling disk: the gravitational force −mgE3 and the constraint force Fc that acts at P.Consequently,

T = Fc · vP − mgE3 · v.

However, vP = 0, and mgE3 · v = ddt (mgx3); consequently,

ddt

(T + mgx3) = 0.

Because E = T + mgx3, this implies that the total energy of the disk is conserved.The proof of energy conservation presented here applies to any rolling rigid bodyunder a gravitational force.

Surprisingly, there are two other conserved quantities associated with therolling disk. These were discovered by Appell [6], Chaplygin [36], and Korteweg[113] in the late 19th century (see [19, 43, 107, 157]). Unfortunately, as withRouth integral (9.35) for the tippe top, their physical interpretation is still an openquestion.

The Sliding Disk’s Equations of MotionThe equations governing the motion of the sliding disk on a smooth horizontal planecan be obtained from (10.35) and (10.36). Specifically, one sets µ1 = µ2 = 0 andignores the constraints π1 = 0 and π2 = 0.

It is instructive, however, to use Approach II. For the sliding disk there is nononintegrable constraint, and the constraint force and constraint moment are pre-scribed by use of Lagrange’s prescription, so (10.24) simplifies to

ddt

(∂T∂uA

)− ∂T

∂uA= − ∂U

∂uA+ Fc · ∂v

∂uA+ Mc · ∂ω

∂uA. (10.37)

Here, A = 1, . . . , 5, Fc = µ3E3, and Mc = −Re′1 × Fc.

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10.8 Lagrange and Poisson Tops 331

Using (10.29) in conjunction with (10.37), we find that the resulting governingequations are

mx1 = 0,

mx2 = 0,

ddt

(λ sin2(θ)ψ + λ3

(φ + ψ cos(θ)

)cos(θ)

)= 0,

λθ + λ3φψ sin(θ) + (λ3 − λ) ψ2 sin(2θ)2

= −mgR cos(θ),

ddt

(λ3(φ + ψ cos(θ)

)) = 0. (10.38)

Notice that we could also use these equations to arrive at the equations of motionfor the rolling disk. First, we need to supplement the constraints and, second, weneed to append the constraint forces and moments associated with the integrableconstraints to the right-hand side of (10.38).

The equations governing the motion of the sliding disk are clearly far simplerthan those for the rolling disk. Indeed, these equations can be written in theform y = f(y), where y is a column vector with 10 rows. These equations canthen be integrated numerically by use of a standard numerical integrator. Anynumerical simulations of these equations should ensure that the linear momentumG, angular momenta H · e3 and H · E3, and total energy E of the sliding disk areconserved.

Configuration ManifoldThe configuration manifold M for the rolling disk and the sliding disk are identical:

M = E2 ⊕ SO(3).

Here, x1 and x2 are coordinates for E2 and the Euler angles are coordinates for

SO(3). That is, both disks have five degrees-of-freedom and five generalized cordi-nates. The kinematical line-element ds is

ds =√

2Tm

dt,

where T is the constrained kinetic energy defined earlier.You should notice that the nonintegrable constraints on the rolling disk do not

affect the configuration manifold. This is identical to a situation we encounteredearlier with the single particle subject to nonintegrable constraints.

10.8 Lagrange and Poisson Tops

There are two classical problems in rigid body dynamics involving spinning tops. Inthe first, known as the Lagrange top, the axisymmetric rigid body is free to rotate

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g

O

X

E1

E2

E3

e1

e2

e3

Figure 10.3. The Lagrange top: an axisymmetric rigid body that is free to rotate about a fixedpoint O. The image on the left-hand side is a portrait of Joseph-Louis Lagrange.

about a fixed point O under the action of a gravitational force (see Figure 10.3). Thisproblem was discussed by Lagrange as an illustration of his equations of motion in1789.∗ A few years later, Poisson [173] considered the same problem except now theapex of the top was free to move on a horizontal surface (see Figure 10.4). Poisson’sinterest in this problem arose in part because a spinning top can in principle be usedto determine the vertical on a ship at sea, and hence was considered to be potentiallyuseful in navigation.

In this section, we do not pursue the complete exposition of the equations ofmotions for the two tops; rather, we focus on choices of coordinates for them andbriefly discuss how their equations of motion can be found.

The Lagrange Top: Coordinates, Constraints, and EnergiesFor the Lagrange top, we assume that the position vector of its center of mass rela-tive to the point O is

x = x0 + L1e1 + L2e2 + L3e3.

Here, Lk are constants. Further, we assume that the rotation tensor of the top isparameterized by a set of 3–1–3 Euler angles.† That is, the angular velocity vectorof the top is ω = ψE3 + θe′

1 + φe3. You should notice that the singularities ofthese Euler angles occur when E3 = ±e3, that is, when θ = 0 or π and the top isvertical.

∗ See Section IX.34 of the Second Part of Lagrange’s Mecanique Analytique [121].† This set of Euler angles has been used extensively in the present chapter and is discussed in Subsec-

tion 6.8.2.

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g

O

X

P

E1

E2

E3

e1

e2

e3

Figure 10.4. The Poisson top: an axisymmetric rigid body that moves with one point P incontact with a horizontal surface.

Motivated by the presence of three integrable constraints xO = 0, we now definea set of coordinates:

u1 = ψ, u2 = θ, u3 = φ, u4 = x1 −(

3∑k=1

Lkek

)· E1,

u5 = x2 −(

3∑k=1

Lkek

)· E2, u6 = x3 −

(3∑

k=1

Lkek

)· E3. (10.39)

In these equations, ek are functions of the Euler angles, but the lengthy expressionsare not recorded here. Given values of u1, . . . , u6, we can uniquely invert relations(10.39) to determine xi and the Euler angles:

x1 = u4 +(

3∑k=1

Lkek

)· E1, x2 = u5 +

(3∑

k=1

Lkek

)· E2,

x3 = u6 +(

3∑k=1

Lkek

)· E3, ψ = u1, θ = u2, φ = u3. (10.40)

Using (10.40), we can easily compute that

v = u4E1 + u5E2 + u6E3 + ω ×3∑

k=1

Lkek.

ω = u1E3 + u2e′1 + u3e3. (10.41)

You should notice that v will depend on the Euler angles and their time derivatives.This is entirely due to our choice of coordinates. In place of x and the three Eulerangles, we are using the Euler angles and the three components of xO as ourcoordinates.

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For future reference, we shall need several vectors. To calculate these vectors itis convenient to recall that ∂ω

∂ψ= E3, ∂ω

∂θ= e′

1, and ∂ω∂φ

= e3. First, we calculate

∂v∂ψ

= E3 ×3∑

k=1

Lkek,∂v∂θ

= e′1 ×

3∑k=1

Lkek,∂v∂φ

= e3 ×3∑

k=1

Lkek,

∂v∂u4

= E1,∂v∂u5

= E2,∂v∂u6

= E3, (10.42)

and

∂ω

∂u1= E3,

∂ω

∂u2= e′

1,∂ω

∂u3= e3,

∂ω

∂u4= 0,

∂ω

∂u5= 0,

∂ω

∂u6= 0. (10.43)

These vectors can be used to compute the right-hand sides of Lagrange’s equationsof motion.

We assume that the top has an axis of symmetry, and hence its inertia tensor J0

has the representation

J0 = λaE3 ⊗ E3 + λt (I − E3 ⊗ E3) .

With the help of (10.41), we find that the kinetic energy of the top has the represen-tation

T = m2

((u4)2 + (u5)2 + (u6)2)

+ m(u4E1 + u5E2 + u6E3

) ·(

ω ×3∑

k=1

Lkek

)

+(

λt + mL22 + mL2

3

2

)ω2

1 +(

λt + mL21 + mL2

3

2

)ω2

2

+(

λa + mL21 + mL2

2

2

)ω2

3.

For convenience, we have not substituted for ωi = ω · ei in terms of the Euler an-gles and their derivatives. The needed expressions for ωi are given in (10.5) in Sec-tion 10.2. The potential energy of the top is

U = mg

(u3 +

(3∑

k=1

Lkek

)· E3

).

This energy is due entirely to the gravitational force.There are three constraints on the motion of Lagrange’s top: u4 = 0, u5 = 0, and

u6 = 0. That is, xO = 0. These constraints are integrable and imply that the general-ized coordinates for the Lagrange top are the three Euler angles. Thus the configura-tion manifold M for the top is SO(3). A straightforward calculation with the three

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constraints vO · Ek = 0, where O is the fixed point of the top, shows that the con-

straint force Fc =∑3k=1 µkEk and constraint moment Mc = −

(∑3k=1 Lkek

)× Fc. As

expected, this system of forces and moments is equipollent to a force Fc acting at O.The constrained kinetic and potential energies are

T =(

λt + mL22 + mL2

3

2

)ω2

1 +(

λt + mL21 + mL2

3

2

)ω2

2

+(

λa + mL21 + mL2

2

2

)ω2

3,

U = mg

(3∑

k=1

Lkek

)· E3.

It is not too difficult to see that 2T = HO · ω and the angular momentum HO canbe easily found by use of the parallel axis theorem. We have kept Lk distinctand nonzero to help elucidate some of our kinematical developments. If we nowimpose the standard assumptions that L1 = 0 = L2, then our results for T andU = mgL cos(θ) are in accord with those found in many dynamics textbooks.

The Poisson Top: Coordinates, Constraints, and EnergiesFor the Poisson top, it is convenient to choose a different coordinate system:

u1 = ψ, u2 = θ, u3 = φ,

u4 = x1, u5 = x2, u6 = x3 −(

3∑k=1

Lkek

)· E3. (10.44)

It is left to the reader to compare (10.44) with (10.39) and to compute ∂v∂uA and ∂ω

∂uA .In contrast to the Lagrange top, the Poisson top is subject to a single integrable

constraint u6 = 0. A straightforward calculation with the constraint vP · E3 = 0,where P is the point of contact of the top with the horizontal surface, shows that the

constraint force Fc = µ1E3 and constraint moment Mc = −(∑3

k=1 Lkek

)× µ1E3.

As expected, this force and this moment are equipollent to a force Fc acting at P.The development of expressions for T and U follow in a similar manner to those

discussed previously for the Lagrange top, and so their constrained expressions aremerely summarized:

T =(

λt + mL22 + mL2

3

2

)ω2

1 +(

λt + mL21 + mL2

3

2

)ω2

2

+(

λa + mL21 + mL2

2

2

)ω2

3 + m2

(x2

1 + x22

),

U = U = mg

(3∑

k=1

Lkek

)· E3.

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336 Exercise 10.1

You should observe from these expressions that the generalized coordinates for thePoisson top are ψ , θ, φ, x1, and x2. Thus this system has a five-dimensional config-uration manifold M. Indeed, the configuration manifold of this top is identical tothat for the sliding disk.

Comments on the Equations of MotionWe can find the equations of motion for the Lagrange and Poisson tops by usingApproach II and without calculating ∂v

∂uA and ∂ω∂uA . There are three reasons for this.

First, the constraint forces and moments are compatible with Lagrange’s prescrip-tion. Second, the coordinates have been chosen so that the integrable constraintsare each described in terms of one coordinate, and finally the applied forces actingon these tops are conservative. Thus the form of Lagrange’s equations of motionfeaturing the Lagrangian L = T − U can be used. In the interest of brevity, explicitequations of motion are not given here: They will be similar to the earlier results forthe disks.

For both tops, it is easy to see that the total energy E is conserved. The Pois-son top also features linear momenta conservations: G · E1 and G · E2. If we nowimpose the conditions, L1 = L2 = 0 for both tops, then we find that the Lagrangetop features two angular momenta conservations: HO · E3 and HO · e3, whereas thePoisson top features the angular momenta conservations H · E3 and H · e3.


Lagrange’s equations of motion have been illustrated for a variety of systems featur-ing a single rigid body. When there are no constraints on the rigid body or when theconstraints are “ideal,” this formulation of the equations of motion provides a set ofdifferential equations that can be integrated to determine the motion. However, fornonintegrably constrained rigid bodies or for rigid bodies when dynamic friction ispresent, Lagrange’s equations of motion are not very attractive, and it is often bestto simply examine the balances of linear and angular momentum. As demonstratedin the examples discussed in Chapter 9, with some insight and patience, F = m ˙v andM = H can lead to a tractable set of equations from which the motion of the rigidbody can be inferred.

EXERCISES

10.1. Consider the mechanical system shown in Figure 10.5. It consists of a rigidbody of mass m that is free to rotate about a fixed point O. A vertical gravitationalforce mgE1 acts on the body. The inertia tensor of the body relative to its center ofmass C is

J0 = λ1E1 ⊗ E1 + (λ − mL20)(E2 ⊗ E2 + E3 ⊗ E3).

The position vector of the center of mass X of the body relative to O is L0e1.

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Exercise 10.1 337

g

O

X

ψ

θ

E1

E2

E3

e1

Figure 10.5. A whirling axisymmetric rigid body.

(a) To parameterize the rotation tensor of the body, we use a set of 1–3–1 Eulerangles: g1 = E1, g2 = e′

3, and g3 = e1. Show that

ω = ψE1 + θe′3 + φe1

= (φ + ψ cos(θ))

e1 + (θ sin(φ) − ψ sin(θ) cos(φ))

e2

+ (θ cos(φ) + ψ sin(θ) sin(φ))

e3.

(b) Because the point O is fixed, we define the coordinates u4, . . . , u6 to be thecoordinates of O:

x = u4E1 + u5E2 + u6E3 + L0e1,

and we choose u1 = ψ , u2 = θ, and u3 = φ. With this choice of the coordi-nates, show that

∂v∂u1

= E1 × L0e1,∂v∂u2

= e′3 × L0e1,

∂v∂u3

= 0,

∂v∂u4

= E1,∂v∂u5

= E2,∂v∂u6

= E3, (10.45)

and∂ω

∂u1= E1,

∂ω

∂u2= e′

3,∂ω

∂u3= e1,

∂ω

∂u4= 0,

∂ω

∂u5= 0,

∂ω

∂u6= 0. (10.46)

These results won’t be needed in the remainder of this problem.

(c) Derive expressions for the unconstrained potential U and kinetic T energiesof the rigid body. These expressions should be functions of the coordinatesu1, . . . , u6 and, where appropriate, their time derivatives.

(d) In what follows, the motion of the rigid body is subject to four constraints:

φ = ω · g3 = 0, v − ω × (L0e1) = 0.

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338 Exercises 10.1–10.2

Assuming the joint at O is frictionless, give prescriptions for the constraintforce and moment acting on the rigid body.

(e) What are the generalized coordinates for this system, and why is the config-uration manifold a sphere?

(f) Prove that the total energy of the rigid body is conserved.

(g) Show that the (constrained) kinetic and potential energies of the body are

T = 12

(λ1ψ

2 cos2(θ) + λ(θ2 + ψ2 sin2(θ)))

, U = mgL0 cos(θ). (10.47)

In addition, using Approach II, write Lagrange’s equations of motion forthe rigid body:

ddt

(∂L

∂θ

)− ∂L

∂θ= 0,

ddt

(∂L

∂ψ

)− ∂L

∂ψ= 0, (10.48)

where L is the Lagrangian.

(h) Argue that the solutions ψ(t) and θ(t) to (10.48) determine the motion ofthe rigid body. You should also discuss why it was not necessary to calculateLagrange’s equations of motion for all six coordinates.

10.2. Consider the simple model for an automobile shown in Figure 10.6. It consistsof a single rigid body of mass m. The moment of inertia tensor of the rigid bodyis J =∑3

i=1 λiei ⊗ ei. Here, the inertia tensor J and mass m include the masses andinertias of the wheels, suspension, engine, and occupants. Interest is restricted to thecase in which the front wheels are sliding while the rear wheels are rolling. In otherwords, the front wheels’ brakes are locked. To model the rolling of the rear wheelsin this simple model, it is assumed that

vQ · e2 = 0, (10.49)

where πQ = −L1e1 − L2e2 is the position vector of Q relative to the center of massX of the rigid body. Constraint (10.49) is often known as Chaplygin’s constraint (see[151, 165, 186]).

Q

Rigid body of mass m

O

X

φ

E1

E1E2

e1

e1

e2

Figure 10.6. A rigid body model for an automobile moving on a horizontal plane.

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Exercises 10.2–10.3 339

(a) Assume that the rigid body is performing a fixed-axis rotation through anangle φ about E3, that the motion of its center of mass is planar, and that(10.49) holds. Using parameterizations of x and Q of your choice, establishexpressions for the four constraints on the motion of the rigid body.

(b) Verify that one of the constraints in (a) is nonintegrable.

(c) Using Lagrange’s prescription, what are the constraint forces Fc and mo-ments Mc acting on the rigid body?

(d) Show that the motion of the rigid body is governed by the equations

mx1 = −µ4 sin(φ), mx2 = µ4 cos(φ), λ3φ = −µ4L1,

x1 sin(φ) − x2 cos(φ) = −L1φ, (10.50)

where xi = x · Ei.

(e) Show that (10.50) allow the center of mass of the rigid body to move in astraight line without the body rotating, and prove that the total energy ofthe body is conserved.

10.3. As shown in Figure 10.7, a circular rod of mass m, length L, and radius Rslides on a smooth horizontal plane. We assume that the inertia tensor J0 has therepresentation

JO = λ (E1 ⊗ E1 + E3 ⊗ E3) + λ2E2 ⊗ E2.

The sole external applied force acting on the rod is gravitational: −mgE3.

g

O

X

Horizontal plane

Circular cylinder

E1

E2

E3

e1 e2

e3

Figure 10.7. A circular rod moving on a horizontal plane.

(a) Assuming that the rotation tensor of the rod is described by a 3–2–3 set ofEuler angles, which orientations of the rod coincide with the singularities ofthis set of Euler angles?

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340 Exercises 10.3–10.4

(b) For this problem, we use Cartesian coordinates to describe the position ofthe center of mass. That is, u1 = x1, u2 = x2, u3 = x3, u4 = ψ , u5 = θ, andu6 = φ. For this set of coordinates, calculate ∂v

∂uA and ∂ω∂uA . To write the equa-

tions of motion of this rigid body, under which circumstances can you avoidusing these 12 vectors?

(c) Show that the unconstrained kinetic energy of the rod is

T = m2

(x2

1 + x22 + x2

3

)+ λ

2

(θ sin(φ) − ψ sin(θ) cos(φ)

)2+ λ2

2

(θ cos(φ) + ψ sin(θ) sin(φ)

)2+ λ

2

(φ + ψ cos(θ)

)2,

where x = x1E1 + x2E2 + x3E3.

(d) Show that the two constraints on the motion of the rigid body can be writtenin the form

v · E3 = 0, ω · g3 = 0.

(e) Using Lagrange’s prescription, what are the constraint force Fc and con-straint moment Mc that enforce the two constraints? With the assistance ofa free-body diagram of the sliding rod, give physical interpretations of Fc

and Mc.

(f) Show that the six Lagrange’s equations of motion for the rod yield the fol-lowing differential equations for the motion of the rod:

mx1 = 0, mx2 = 0, λ2θ = 0, λψ = 0. (10.51)

(g) Show that the six Lagrange’s equations of motion for the rod also yieldsolutions for the constraint force and constraint moment:

Fc = mgE3, Mc = −λ2ψ θ sin(θ)g3.

By expressing g3 in terms of e′1, you should also be able to establish a simple

expression for Mc.

(h) Give physical interpretations of the solutions to (10.51) and show that theyallow the cylinder to have motions where ω is not constant.

10.4. As shown in Figure 9.8, a sphere of mass m and radius R is free to move on arough inclined plane. Here, we reexamine this problem, which was discussed earlierin Section 9.8, by using Lagrange’s equations of motion. The instantaneous point ofcontact of the sphere with the plane is denoted by P. You will need to recall that

vP = v + ω × πP, (10.52)

where v is the velocity vector of the center of mass X of the sphere, and πP is theposition vector of P relative to X.

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Exercises 10.4–10.5 341

(a) Using a set of 3–1–3 Euler angles, show that the slip velocities of the sphereare

vs1 = x1 − Rθ sin(ψ) + Rφ sin(θ) cos(ψ),

vs2 = x2 + Rθ cos(ψ) + Rφ sin(θ) sin(ψ),

where vP = vs1E1 + vs2E2 and v =∑3k=1 xkEk.

(b) What is the constraint on a sliding sphere? Give prescriptions for the con-straint force Fc and moment Mc that enforce this constraint.

(c) Show that at least one of the three constraints on a rolling sphere isnonintegrable. Give prescriptions for the constraint forces Fc and momentsMc that enforce these constraints.

(d) If the unconstrained kinetic energy of the sphere is

T = mR2

5

(ψ2 + θ2 + φ2 + 2ψφ cos(θ)

)+ m2

(x2

1 + x22 + x2

3

),

then what are Lagrange’s equations of motion for the sliding sphere?

(e) What alterations need to be made to the Lagrange’s equations of motionfor (d) so that they now apply to the rolling sphere?

(f) Starting from (10.52) and using balances of linear and angular momentum,show that

mvs1 =(

1 + 52

)Ff · E1 + mg sin(β),

mvs2 =(

1 + 52

)Ff · E2,

where Ff is the friction force acting on the sphere. If the sphere is rolling,then what is Ff ?

10.5. Consider a rigid body of mass m that is free to rotate about a fixed point O.The position vector of the center of mass X of the rigid body relative to O is x − xO =L0e3, where L0 is a constant. The inertia tensor of the body relative to its center ofmass is J =∑3

k=1 λkek ⊗ ek. A vertical gravitational force −mgE3 acts on the rigidbody.

(a) Using a set of 3–1–3 Euler angles, establish an expression for the rotationalkinetic energy Trot of the rigid body.

(b) What are the three constraints on the motion of the rigid body?

(c) Using Lagrange’s prescription, give prescriptions for the constraint force Fc

and moment Mc acting on the rigid body.

(d) In terms of the Euler angles and a set of Cartesian coordinates xk = x · Ek,prescribe a set of six coordinates u1, . . . , u6 such that the three integrableconstraints on the motion of the rigid body can be expressed as

i = 0 (i = 1, 2, 3) ,

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342 Exercises 10.5–10.6

where

1 = u4, 2 = u5, 3 = u6.

Notice that the generalized coordinates for this rigid body are the Eulerangles.

(e) Calculate the following 12 vectors:

∂v∂uA

,∂ω

∂uA(A = 1, . . . , 6) .

What is the potential energy U = U(u1, . . . , u6

)of the rigid body?

(f) What are Lagrange’s equations of motion for the generalized coordinates?In your solution, show that the contributions to the right-hand sides of theseequations that are due to Fc and Mc sum to zero.

(g) With the assistance of the balance laws, show that

Fc = mgE3 + mL0e3.

What is the corresponding result for Mc?

(h) Prove that the total energy E of the rigid body is conserved.

10.6. In this exercise, we reconsider the satellite problem discussed earlier in Sec-tion 10.4. Here, we employ a coordinate system that is popular in the satellite dy-namics community (see, for example, [164, 187, 203]).

Referring to Figure 10.1, to parameterize x, we use a spherical polar coordinatesystem, R, θ, and β, where β is the latitude and θ is the longitude. For this coordinatesystem, we define the following unit vectors:

er = cos(θ)E1 + sin(θ)E2, eR = cos(β)er + sin(β)E3,

eβ = cos(β)E3 − sin(β)er, eθ = cos(θ)E2 − sin(θ)E1.

The rotation tensor Q of the satellite is a transformation from the fixed basis {Ei}to the corotational basis {ei}. To parameterize Q, it is standard in satellite dynamicsstudies to exploit the rotation θ about E3 inherent in the definition of the sphericalpolar coordinates. To do this, we use the following decomposition:

Q = Q2Q1,

where the rotation tensor Q1 is

Q1 = cos(θ)(I − E3 ⊗ E3) − sin(θ)εE3 + E3 ⊗ E3.

The rotation tensor Q2 will be parameterized by a set of 1–2–3 Euler angles. The firstof the Euler angles corresponds to a counterclockwise rotation through an angle ν1

about the vector er. Similarly, the second rotation corresponds to a counterclockwiserotation through an angle ν2 about cos

(ν1)

eθ + sin(ν1)

E3 and the last rotation isabout e3 through a counterclockwise angle ν3.

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Exercises 10.6–10.7 343

(a) Starting from the representation x = ReR, show that the velocity vector v ofthe center of mass of the satellite has the representation

v = ReR + R cos(β)θeθ + Rβeβ.

(b) Starting from the representation Q =∑3i=1 ei ⊗ Ei, show that e1 = Q2er,

e2 = Q2eθ, and e3 = Q2E3.

(c) With the help of three relative angular velocity vectors, show that the angu-lar velocity vector ω of the satellite has the representation

ω = θE3 + ν1er + ν2 (cos(ν1) eθ + sin

(ν1)E3

)+ ν3e3. (10.53)

(d) Show that the angular velocity vector of the satellite has the representation

ω = ω1e1 + ω2e2 + ω3e3,

where

ω1 = θ(sin(ν1) sin(ν3) − cos(ν1) sin(ν2) cos(ν3)

)+ ν2 sin(ν3) + ν1 cos(ν2) cos(ν3),

ω2 = θ(sin(ν1) cos(ν3) + cos(ν1) sin(ν2) sin(ν3)

)+ ν2 cos(ν3) − ν1 cos(ν2) sin(ν3),

ω3 = θ cos(ν1) cos(ν2) + ν1 sin(ν2) + ν3.

(e) Suppose the coordinates chosen for the satellite are u1 = R, u2 = θ, u3 = β,u4 = ν1, u5 = ν2, and u6 = ν3. With this choice, give expressions for the 12vectors

∂v∂uA

,∂ω

∂uA(A = 1, . . . , 6) .

(f) The potential energy associated with the gravitational force and momentcan be found in Equation (10.11). With the choice of u1, . . . , u6, it can beshown that ∂U

∂θ= 0. Using this result, show that ∂T

∂θis conserved. Prove that

this conservation corresponds to conservation of HO · E3.

(g) Using Lagrange’s equations, establish the equations governing the motionof the satellite.

10.7. Consider a body that is free to move about one of its material points O, whichis fixed (cf. Figure 9.7). This system was discussed previously in Exercises 9.15 and10.5.

(a) Calculate the constrained Lagrangian L and the equations of motion [see(10.8)]:

ddt

(∂L∂γi

)− ∂L

∂γi= 0, (10.54)

where γ1 = ψ , γ2 = θ, and γ3 = φ are the 3–1–3 set of Euler angles used toparameterize Q.

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344 Exercises 10.7–10.8

(b) Show that equations of motion (10.54) are equivalent to HO · gi = MO · gi,where

{g1, g2, g3

}are the Euler basis vectors.

(c) Compare equations of motion (10.54) with (9.36), which were discussed inExercise 9.14.

10.8. Recall the definition of the force �A given by (10.4):

�A = F · ∂v∂uA

+ M · ∂ω

∂uA.

Now suppose we have a force P that acts at a point XP on a rigid body. If the positionvector of XP is xP, then show that∗

P · ∂vP

∂uA= P · ∂v

∂uA+ ((xP − x) × P) · ∂ω

∂uA.

How can this identity be used to simplify the calculation of �A?

∗ This identity is used extensively in many treatments of analytical mechanics (see, for example, Kaneand Levinson [105] and Kane et al. [106]).

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PART FOUR

SYSTEMS OF RIGID BODIES

345

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346

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11 Introduction to Multibody Systems

11.1 Introduction

In this chapter, we examine systems of rigid bodies. Our goal is simply to discoverhow our previous developments can be used to obtain the equations of motion forthese systems. As you might imagine, the equations of motion can be very complex,and judicious component selections from the balance laws are often needed to ex-tract the equations of motion. This is illustrated with the example of a rategyroscope.

The presentations here are limited in scope and we do not have the opportunityto discuss many interesting systems featuring several rigid bodies such as the dual-spin spacecraft, bicycles, gyrocompasses, and the Dynabee in detail. As discussedin [86], a dual-spin satellite has the ability to reorient itself in an environmentwhere the resultant moment on the satellite is negligible. This ability has beenused in communications satellites and was employed in the Galileo spacecraft. Thisspacecraft was launched in 1989 and some 6 years later began its orbits of the planetJupiter. These orbits were designed so that the spacecraft could capture data onsome of the largest moons of Jupiter; Galileo’s mission was a remarkable success.∗

The Dynabee (or Rollerball) was invented in the early 1970s by Archie Mishler[141] and features spinning a rotor to speeds in excess of 5000 rpm by carefullyrotating an outer casing (housing). This novel gyroscopic device is discussed in anexercise at the end of this chapter.

11.2 Balance Laws and Lagrange’s Equations of Motion

We are interested in examining the dynamics of a system of N rigid bodies,B1, . . . ,BN. For the body BK, we denote its mass by mK, its center of mass by XK,the position of its center of mass by xK, its inertia tensor relative to XK by JK, andits angular velocity by ωK. Using these quantities, we can define the kinetic energyTK, angular momentum HK, and linear momentum GK in the usual manner:

GK = mK ˙xK, HK = JKωK, TK = 12

GK · ˙xK + 12

HK · ωK.

∗ A history of the Galileo spacecraft’s mission was recently written by Meltzer [140].

347

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Setting K = 1, . . . , N we can obtain expressions for the kinematic quantities associ-ated with each of the N rigid bodies. Constraints on, and potential energies for, thesystem of rigid bodies can be inferred from our developments in Chapter 8 and weshall adopt them here.

For the system of rigid bodies, we can define the center of mass X. The positionvector of this point is defined by

x = 1∑NK=1 mK

(N∑

K=1

mKxK

).

The linear momentum G of the system of rigid bodies is the sum of the individuallinear momenta, and G can be related to the momentum of the center of mass:

G =(

N∑K=1

mK

)˙x =

N∑K=1

mK ˙xK.

In many problems featuring systems of rigid bodies, it is necessary to calculate theangular momentum of the system relative to a point, say A: HA. This point is oftenthe center of mass of the system or a fixed point. To compute HA, we use a standardidentity [(7.15)] applied to each individual rigid body:

HA =N∑

K=1

(HK + (xK − xA) × mK ˙xK) .

Finally, the kinetic energy of the system of N rigid bodies is simply the sum of the ki-netic energies of the individual bodies: T = T1 + · · · + TN. The resulting expressionfor T can be tremendously complex, and often symbolic manipulation packages,such as Mathematica or Maple, are used.

We find the governing equations for the system of N rigid bodies by using thebalances of linear and angular momenta for each of the rigid bodies:

m1 ¨x1 = F1, . . . , mN ¨xN = FN,

H1 = M1, . . . , HN = MN. (11.1)

Here, FK is the resultant force acting on the Kth rigid body and MK is the resultantmoment relative to the center of mass XK of the rigid body. Equations (11.1) yielddifferential equations for the motion of the bodies and expressions for the constraintforces and moments.

As shown in [30], balance laws (11.1) are equivalent to Lagrange’s equations ofmotion for the system of rigid bodies. For these equations, we denote by u1, . . . , u6N

the coordinates chosen to parameterize the position vectors xK and rotation tensorsQK. Then Lagrange’s equations of motion are

ddt

(∂T∂uA

)− ∂T

∂uA= �A (A = 1, . . . , 6N) , (11.2)

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11.3 Two Pin-Jointed Rigid Bodies 349

where v = ˙x and

�A =N∑

K=1

FK · ∂vK

∂uA+ MK · ∂ωK

∂uA. (11.3)

It is emphasized that using either (11.1) or (11.2) will lead to equivalent equationsof motion for the system of rigid bodies.

As in our discussion of Lagrange’s equations of other systems, it can be shownthat, if the constraint forces and moments are prescribed by Lagrange’s prescription,the system of constraints on the system is integrable, and the coordinates u1, . . . , u6N

are chosen appropriately, then Lagrange’s equations of motion decouple into twosets of equations: one from which the constraint forces and moments are easily de-termined, and the other from which motion of the system can be computed. Theproof of this can be found in [30], and the interested reader is referred to this paperfor details. Here, we are primarily content to use the result, but before doing so itis important to comment on the expressions for the generalized forces that followfrom (11.3). Suppose that the system of rigid bodies is subject to C integrable con-straints and that the coordinates are chosen so that these constraints can be simplyrepresented as

u6N−C+1 = f 1(t), . . . , u6N = f C(t).

We also suppose that the constraint forces FcK and constraint moments McK asso-ciated with these constraints are prescribed by Lagrange’s prescription. It can beshown that

N∑K=1

FcK · ∂vK

∂uA+

N∑K=1

McK · ∂ωK

∂uA= 0 (A = 1, . . . , 6N − C) . (11.4)

That is, the constraint forces and constraint moments do not contribute to the gen-eralized forces �1, . . . , �6N−C. As a result, our expressions (11.3) for the forces �A

are identical to those found in other treatments of Lagrange’s equations of motionfor systems of rigid bodies that are available.∗

11.3 Two Pin-Jointed Rigid Bodies

As our first example, we return to the case of two bodies, B1 and B2, that are pinjointed. This system was discussed earlier in Subsection 8.6.4 and is illustrated inFigure 8.8. Our goal is to outline how the equations of motion for this system can beestablished by using Lagrange’s equations of motion.

We start with the kinematics. As discussed earlier, the pin joint introducesfive constraints on the motions of B1 and B2. Mindful of these constraints, weuse Cartesian coordinates to parameterize x1, a set of Euler angles γ1, γ2, γ3 toparameterize Q1, and the angle θ to parameterize the relative rotation tensor Q2QT

1 .The angle θ represents the rotation of B2 relative to B1 and the axis of rotation

∗ See, for example, Baruh [14], Greenwood [80], and Kane [104].

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associated with this rotation is e3: Q2QT1 = L (θ, e3). It is prudent to define the

following seven coordinates:

uk = x1 · Ek, uk+3 = γk, u7 = θ,

where k = 1, 2, 3. We also assume that

πP1 = L1e1, πP1 = (L2) 2e1,

where L1 and L2 are constants. We also note that the corotational basis vectorsfor B1 are {e1 = 1e1, e2 = 1e2, e3 = 1e3}, and the basis vectors {2e1, 2e2, 2e3 = e3}corotate with B2.

From the coordinates chosen, we have the following representations:

v1 =3∑

i=1

xiEi,

v2 = v1 + ω1 × L1e1 + (ω1 + θe3)× (L2) 2e1,

ω1 =3∑

i=1

γigi,

ω2 =3∑

i=1

γigi + θe3.

These results can then be used to compute the constrained kinetic energy T of thesystem. This energy is the sum of the kinetic energies of B1 and B2, and in the interestof brevity, we do not record its full expression here.

Expressions for the constraint forces and moments acting on the system thatenforce the constraints associated with the pin joint were presented in Equation(8.23). Because these quantities are prescribed by Lagrange’s prescription, and be-cause of our choice of coordinates, we can use Approach II and immediately writethe equations of motion for this system∗:

ddt

(∂T∂uA

)− ∂T

∂uA= �A (A = 1, . . . , 7) , (11.5)

where

�A =2∑

K=1

FK · ∂vK

∂uA+ MK · ∂ωK

∂uA.

It is emphasized that Fc1, Fc2, Mc1, and Mc2 do not contribute to �1, . . . , �7. Indeed,it is a good exercise to verify that

Fc1 · ∂v1

∂uA+ Fc2 · ∂v2

∂uA+ Mc1 · ∂ω1

∂uA+ Mc2 · ∂ω2

∂uA= 0 (A = 1, . . . , 7) .

∗ Technically, we have shown that Approach II works only for a single rigid body, a single particle,and a system of particles. As mentioned previously, the corresponding result for a system of rigidbodies was established in [30], and we make use of it here.

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11.4 A Single-Axis Rate Gyroscope 351

You may have noticed that these identities are special cases of (11.4). Once the ap-plied forces and moments are prescribed and T calculated, the seven second-orderdifferential equations of (11.5) suffice to determine the motion of the system of tworigid bodies. The resulting system of differential equations can be formidable, andoften additional constraints and geometric simplifications are employed.

An alternative formulation of the equations of motion would be to consider thebalance of linear and angular momenta for each of the rigid bodies:

m1x1 = F1,

m2x2 = F2,

H1 = M1,

H2 = M2. (11.6)

This would yield a set of coupled equations for the constraint reactions µ1, . . . , µ5

[cf. (8.23)] and u1, . . . , u7. The advantages of Lagrange’s equations of motion in itsautomatic selection of linear combinations of Equations (11.6) should be obvious.

11.4 A Single-Axis Rate Gyroscope

As shown in Figure 11.1, a single-axis rate gyroscope is mounted on a platform P .The gyroscope consists of a gimbal G that is free to rotate relative to P through anangle α about the axis e1 and a rotor R that is free to rotate relative to G through anangle β about the axis e2. In addition to the pair of bearings connecting the gimbaland platform, a spring of stiffness K and a dashpot with a damping coefficient c aresuspended between G and P .

α

β

K c

P

RG

e1e2

r1

r2

r3

t1t2

t3

Figure 11.1. Schematic of a single-axis rate gyroscope. This devices consists of a rotor R thatis suspended by a gimbal G over a platform P . When the platform rotates about t3 the rotorwill want to dip, and this tendency is resisted by the spring and dashpot shown in the figure.

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The principle of operation of the gyroscope is as follows: Suppose that the plat-form is given an angular velocity � about the vertical t3 direction; then the rotorwill seek to rotate about a horizontal direction e1. This tendency to rotate will becounterbalanced by the spring, but will result in an angle of deflection α = α0. Aswe shall subsequently see, ∗

α0 ∝ �.

Consequently, if the gyroscope is calibrated, then a measurement of α0 can be usedto determine �. In practice, however, α0 is sensitive to many other effects that canoften serve to corrupt the measurement of �. A wonderful discussion of these andother matters on rate gyroscopes can be found in Arnold and Maunder [8]. Here,we follow some of their developments and outline how (11.7) can be established.Before doing so, we note that three single-rate gyroscopes mounted at right anglesto each other on a rigid body B can be used to measure ωi = ω · ei. Integratingωi(t) with the help of equations of the form (7.28) can then be used to determinethe attitude Q of B. Such a system is known as a “strap-down inertial navigationsystem.”

Kinematics

By way of notation, the bases vectors {t1, t2, t3}, {e1, e2, e3}, and {r1, r2, r3} corotatewith P , G, and R, respectively. All of these bases are right-handed, and, with thehelp of a fixed Cartesian basis {E1, E2, E3}, can be used to define the rotation tensorof each of the three rigid bodies. In the analysis that follows, it is assumed that thereis negligible friction between the rotor and the gimbal. The position vectors of thecenter of mass of the gimbal and rotor are assumed to be coincident and denotedby x.

The rotation tensor QP =∑3i=1 ti ⊗ Ei of the platform is parameterized by a set

of 3–1–3 Euler angles. Denoting the angular velocity vector of the platform by ωp,the gimbal by ωg, and the rotor by ωr, with the help of (6.32), we can easily establishrepresentations for these three angular velocity vectors:

ωp = φt3 + θt′1 + ψE3,

ωg = ωp + αe1,

ωr = ωg + βe2.

We denote the inertia tensors of the rotor and gimbal relative to their centers ofmass by

Jr = λ1 (I − r2 ⊗ r2) + λ2r2 ⊗ r2,

Jg = ρ1e1 ⊗ e1 + ρ2e2 ⊗ e2 + ρ3e3 ⊗ e3,

∗ See, in particular, (11.9).

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11.4 A Single-Axis Rate Gyroscope 353

respectively. Notice that the symmetry of the rotor allows its inertia tensor to beeasily written in either the corotational basis for the gimbal or rotor:

Jr = λ1 (I − r2 ⊗ r2) + λ2r2 ⊗ r2

= λ1 (I − e2 ⊗ e2) + λ2e2 ⊗ e2.

The angular momentum of the rotor relative to its center of mass and the angularmomentum of the rotor and gimbal relative to their coincident centers of massare

Hr = Jrωr,

H = Hr + Hg

= Jgωg + Jrωr

= ((ρ1 + λ1) e1 ⊗ e1 + (ρ2 + λ2) e2 ⊗ e2 + (ρ3 + λ3) e3 ⊗ e3) ωg

+ λ2βe2.

The symmetry of the rotor is partially responsible for the simplicity of theexpression for H.

Constraint Forces and Constraint Moments

There are five constraints on the motion of the gimbal relative to the rotor:

vr = vg,

(ωr − ωg) · r1 = 0,

(ωr − ωg) · r3 = 0.

We assume that the bearings on the gimbal supporting the rotor are frictionless, andthis allows us to use Lagrange’s prescription (8.22) to prescribe the constraint forcesand constraint moments:

Fcr = −Fcg =3∑

k=1

µkek,

Mcr = −Mcg = µ4e1 + µ5e3.

It is a good exercise to show that the force Fcr and moment Mcr are equipollent totwo reaction forces R1 and R2 that act on opposite ends of the rotor. This impliesthat Lagrange’s prescription yields physically reasonable results.

Governing Equations for the Gyroscope

We can use a balance of angular momentum of the rotor relative to its center ofmass to establish an interesting conservation. First, we note that Mr = Mcr and this

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moment has no component in the e2 = r2 direction. It is also easy to see that Hr ·e2 = 0. Consequently, h = Hr · e2 is conserved, where

h = Jrωr · r2 = λ2β + λ2ωp · e2. (11.7)

You should notice from this conservation that we can change the relative speed β ofthe rotor by changing the angular velocity of the platform in the e2 = r2 direction.Generally, it is assumed that

β � ωp · e2.

Consequently, β is assumed to be constant during the operation of the gyroscope.Examining the resultant force and moment on the gimbal–rotor system, we

easily see that the constraint moments at the bearings where the gimbal is attachedto the platform have no moment in the e1 direction. Consequently, it is prudentto consider the balance of angular momentum H = M for the gimbal–rotor systemand examine the e1 component of this equation. After approximating the momentin the e1 direction that is due to the spring and dashpot as −(Kα + cα)Ae1, where Ais a constant whose dimensions are length squared, we find a differential equationgoverning the angle α:

(λ1 + ρ1) α + KAα + cAα = λ2βω3 − (λ1 + ρ1) ω1

+ (λ2 + ρ2 − (λ3 + ρ3)) ω2ω3, (11.8)

where ωp =∑3i=1 ωiti.

Operation of the Gyroscope

Differential equation (11.8) informs us how α changes when the platform is moved.It should be transparent that the angle α is sensitive to all of the components of ωp.For an operational gyroscope, the rotor is usually spun to a sufficiently high speed β,that λ2βω3 dominates the right-hand side of (11.8). Consequently, we approximate(11.8) by

(λ1 + ρ1) α + KAα + cAα = λ2βω3. (11.9)

Now suppose initially that α = 0 (the rotor is horizontal), β is large, and ω3 = 0.It should be easy to see from (11.9) that the rotor will remain horizontal. Now ifω3 is given a constant nonzero value ω30, then α(t) changes from its rest value andwill vary with time. It is left as an exercise to integrate (11.9) and to observe that,if λ2βω3 is constant and cA > 0, then the solution α(t) will eventually settle to aconstant value α0. Equation (11.9) can be used to relate this value to the constantangular speed of the platform:

ω30 =(

KA

2λ2β

)α0. (11.10)

This equation is the explanation for the governing principle behind the operation ofthe gyroscope.

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Exercise 11.1 355

Many devices that are similar to this gyroscope can be found: in particular, thegyrocompass, the gyrohorizon, and the dual-axis rate gyroscope. Space prevents usfrom discussing them here, but excellent treatments can be found in the texts byArnold and Maunder [8], Crabtree [42], and Ginsberg [71]. Although these mechan-ical systems can seem dated when compared with modern microelectromechanicalgyroscopes, the mechanical design featured in many of their practical implementa-tions is ingenious.


We have touched here on two examples that illustrate how the equations of motionfor a system of rigid bodies can be determined. Fortunately, for many physical exam-ples, Lagrange’s prescription is applicable and Lagrange’s equations can be used todetermine the equations of motion: a theme that has been a central thread through-out this book and is extensively exploited in formulations of multibody systems.

Alternative forms of Lagrange’s equations are available for nonholonomic sys-tems. These equations, of which the most well known are the Gibbs–Appell equa-tions and Boltzmann–Hamel equations, enable a further elimination of the con-straint forces and moments associated with the nonintegrable constraints. Suchequations of motion are beyond the scope of this book, and there is a variety oftexts and surveys that cover them, for example, Baruh [14], Greenwood [80], Hamel[87], Karapetyan and Rumyantsev [108], Neımark and Fufaev [151], Papastavridis[167–169], and Udwadia and Kalaba [218], to name but a few. Most other forms ofthe equations of motion are derivable from Lagrange’s equations of motion (11.2)and so, by use of the developments in this book [and the help of (11.3)], they toocan be related to the balances of linear and angular momenta.

EXERCISES

11.1. As shown in Figure 11.2, a rigid plate of mass m is attached at a point A by apin joint to a slender rod. The rod has a length L, one of its ends, O, is fixed, and itrotates about the E3 axis with an angular speed ψ = �.

(a) With the help of a set of 3–1–3 Euler angles, show that the five constraintson the motion of the plate can be written as

−Lψe′1 = ˙x + ω × πA, ω · g1 = �, ω · g3 = 0, (11.11)

where x is the position vector of the center of mass of the plate, ω is theangular velocity of the plate, and πA is the position of A relative to the X.

(b) Which orientations of the plate coincide with the singularities of the Eulerangles?

(c) Give prescriptions for the forces Fc and moments Mc that ensure that thefive constraints are enforced. Illustrate your prescriptions with a free-bodydiagram.

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356 Exercises 11.1–11.2

ψg

O

A

X

Rectangular plate

Slender rod of length L

E1

E2

E3

e1

e2

Figure 11.2. A rigid plate that is pin jointed at A to a slender rod. The rod is supported bybearings at O that enable it to freely rotate about E3. Note that a gravitational force −mgE3

acts on the plate.

(d) Show that the mechanical power of the constraint forces and moments isnonzero if � �= 0.

(e) If the vector πA and the moment of inertia tensor J of the plate relative toits center of mass are

πA = −ae2, J =3∑

i=1

λiei ⊗ ei, (11.12)

then establish an expression for the angular momentum HA and show thatHA �= JAω unless � = 0.∗

(f) Outline how you would establish the equations of motion of the plate.

11.2. Following (2.1), the impulse momentum form of the balances of linear andangular momentum for a rigid body are

G (t1) − G (t0) =∫ t1

t0F(τ)dτ,

H (t1) − H (t0) =∫ t1

t0M(τ)dτ. (11.13)

Here, we wish to apply these equations to an impact scenario. Thus t1 − t0 → 0, butG(t) and H(t) can be discontinuous in this limit. We assume that the impact occursat time t, and so we consider limits where t1 = t + σ ↘ t and t0 = t − σ ↗ t.

∗ Hint: You may find the identity a × (b × c) = (a · c) · b − (a · b) · c helpful.

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Exercises 11.2–11.3 357

As in treatments of discontinuities in continuum mechanics, we introduce thefollowing notation to denote the change (or jump) in a function f (u(t), u(t), t) attime t:

[[f ]] = limσ→0

f (u(t + σ), u(t + σ), t + σ) − f (u(t − σ), u(t − σ), t − σ) ,

where σ > 0.

(a) Suppose a set of coordinates u1, . . . , u6 is used to parameterize the motionof the rigid body. Then, what are the conditions under which[[

∂v∂uK

]]= 0,

[[∂ω

∂uK

]]= 0, (11.14)

where K = 1, . . . , 6? Give a physical interpretation of the conditions youhave found.

(b) Assuming that (11.14) holds, by taking appropriate limits show that (11.13)can be written as [[

∂T∂qK

]]= �K, (11.15)

where K = 1, . . . , 6 and the generalized impulsive forces are

�K = limσ→0

(∫ t+σ

t−σ

F(τ)dτ

)· ∂v∂uK

(t) + limσ→0

(∫ t+σ

t−σ

M(τ)dτ

)· ∂ω

∂uK(t) .

For assistance in establishing (11.15), calculations (10.17) might be helpful.An alternative derivation of (11.15) for a system of particles can be foundin Section 15.2 of Synge and Griffith [207].

(c) Show how the six equations of (11.15) can be used to determine the dynam-ics of a rod impacting a smooth horizontal surface.

11.3. Consider, once again, the robotic arm shown in Figure 7.9. As discussed in anearlier exercise, the robotic arm has a mass m and moment of inertia tensor relativeto its center of mass,

J0 = λ1E1 ⊗ E1 + λ2E2 ⊗ E2 + λ3E3 ⊗ E3.

A system of motors, which is not shown in Figure 7.9, is used to actuate the rotationof the robotic arm. The rotation consists of

1. a rotation about the E3 axis through an angle ψ ,

2. a rotation about g2 = e′1 = cos(ψ)E1 + sin(ψ)E2 through an angle θ.

Another system of actuators prescribes the motion of the point P of the robotic arm.The position vector of the center of mass of the arm relative to P is

x − xP = L2

e3.

(a) Interpreting the angles ψ and θ as members of a 3–1–3 set of Euler angleswhere φ = 0, show that the dual Euler basis vectors, g1, g2, and g3, are not

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358 Exercises 11.3–11.4

orthonormal. For which positions of the robotic arm are these vectors notdefined?

(b) Noting that the motion of the point P is prescribed, xP = u(t), and that therotation of the arm is prescribed, what are the six constraints on the motionof the arm?

(c) Show that the angular momentum H (relative to its center of mass) of thearm is

H = λ1θe1 + λ2ψ sin(θ)e2 + λ3ψ cos(θ)e3.

(d) Draw a free-body diagram of the robot arm. In your free-body diagram,include the gravitational force −mgE3.

(e) Show that the force needed to actuate the motion of the center of mass is

Fact = mgE3 + mu + mL2

(2ψ θ cos(θ) + ψ sin(θ)

)e1

+ mL2

(ψ2 sin(θ) cos(θ) − θ

)e2 − mL

2(θ2 + ψ2 sin2(θ))e3.

(f) Show the moment Mc needed to actuate a slewing maneuver of the roboticarm is

Mact = mL2

e3 × (gE3 + u) + mL2

2ψ0θ0 cos(θ)e2

− mL2

4ψ2

0 sin(θ) cos(θ)e1 + ω × H.

During the slewing maneuver, θ = θ0, ψ = ψ0, and consequently ω is con-stant.

(g) Show that the time rate of change in the total energy E of the robotic armis equal to the work done by Fact and Mact:

E = Fact · u + Mact · ψE3 + Mact · θe1.

11.4. As mentioned earlier, a recently invented wrist exerciser, which is known asthe Dynabee, Powerball, or Rollerball, features a heavy rotor that can roll withoutslipping on a circular track [82, 94]. The track is part of a housing that can be heldand rotated. As analyzed in [82], effectively rotating the track allows the rotor tospin-up. The resulting spin-up requires that the holder provide an ever-increasingmoment and, as a result, the holder exercises several muscles in the arm and wrist.In the problem discussed here, we consider the formulation of the constraints on therotor and the housing.

As illustrated in Figure 11.3, we define a set of vectors {et1, et2, et3} to corotatewith the track and the set of vectors {er1, er2, er3} to corotate with the rotor. We alsodefine an intermediate set of vectors

a1 = er1, a2 = et3 × er1, a3 = et3.

We emphasize that these vectors are not corotational.

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Exercise 11.4 359

α

γ

TR

et1

et2

et3 = a3

er1 = a1

er3

Figure 11.3. Schematic of a Dynabee showing the circular track T and the rotor R. The anglesα and γ shown in this figure are used to parameterize the rotation of the rotor relative to thetrack. This figure is adapted from [82].

(a) The rotation tensor Q =∑3k=1 etk ⊗ Ei of the housing is assumed to be pa-

rameterized by a set of 3–1–3 Euler angles (ψ, θ, φ). Referring to Figure11.3, what are the consequences of constraining ψ = −φ and θ to a constantvalue θ0?

(b) The rotation tensor R of the rotor relative to the track (housing) is assumedto be parameterized by a set of 3–1–2 Euler angles (α, γ, µ ≈ 0). We empha-size that

R = er1 ⊗ et1 + er2 ⊗ et2 + er3 ⊗ et3. (11.16)

Using the aforementioned sets of Euler angles, establish representations forthe angular velocity vectors of the rotor ωr and the track ωr.

(c) Referring to Figure 11.4, we assume that the rotor has two points in con-tact with the track. For convenience, we define a fourth set of unit vectors{eπ1, eπ2, eπ3} such that eπ1 is parallel to the line segment joining P and Q,and eπ2 = a2. If β is the angle between er1 and eπ1, then show that

β = tan−1(

Ra

Rt

),

where Ra is the radius of the rotor’s axle and Rt is the radius of the track.

(d) Assuming rolling at the points P and Q, show that these constraints implythat

vr = vt, (ωr − ωt) × eπ1 = 0,

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360 Exercises 11.4–11.5

T

T

R

β

P

Q

X

et3 = a3

er1 = a1

eπ1

eπ3

Figure 11.4. Schematic showing the two points of contact P and Q of the rotor and the track.For our analysis, the centers of mass of the rotor R and housing T are assumed to coincideat the point X. In this figure, the radius Ra of the axle of the rotor and the angle µ have beenexaggerated.

where vr and vt are the velocity vectors of the centers of mass of the rotorand housing, respectively. In addition, show that these constraints implythat

γ = −(

Rt

Ra

)α. (11.17)

As in [82], we assume that µ ≈ 0.

(e) Assuming that the motion of the housing is completely prescribed, showthat the system of rigid bodies consisting of the housing and the rotor issubject to 11 independent constraints. Using Lagrange’s prescription, giveprescriptions for the constraint forces and moments acting on the two rigidbodies R and T .

(f) Argue that the eπ1 component of the balance of angular momentum forthe rotor relative to center of mass X gives a differential equation for themotion of the rotor.

(g) When the rotor is steadily rotating, ψ = α. Explain why (11.17) then impliesthat the rotor will be spinning faster than the housing.

11.5. Consider a rigid body and assume that its rotation tensor Q is parameterizedby a set of 3–1–3 Euler angles (ψ, θ, φ). Suppose that, after an interval of time t1 − t0,

e3 (t1) = e3 (t0) . (11.18)

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Exercise 11.5 361

(a) Show that the rotation of the rigid body during the time interval t1 − t0 hasthe representation

Q (t1) QT (t0) = L (φ (t1) − φ (t0) , e3 (t0)) .

(b) Show that the measurement of a single rate gyro ω · e3 from a rate gyro-scope is insufficient to determine φ (t1) − φ (t0).∗

(c) Give an example of a motion of a rigid body where (11.18) holds.†

(d) For your example in (c), imagine the base point of the vector e3 is fixed, andplot the coordinates of the tip of this vector. The tip lies on the unit sphere.Using a result that can be found in Section 123 of Kelvin and Tait [109] andthat was independently rediscovered by Goodman and Robinson [73] andLevi [126], show that the area enclosed by the closed curve traced by the tipof e3(t) on the unit sphere is related to the angle φ (t1) − φ (t0).

(e) Show that the results of (d) are related to Codman’s paradox discussed inExercise 6.8.

∗ You may wish to recall (7.22) from Exercise 7.4.† For assistance with this matter, the papers of Montgomery [144] and O’Reilly [158] might be helpful.

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Appendix: Background on Tensors

A.1 Introduction

A tensor is a linear mapping that transforms a vector into another vector. It is com-mon practice to use matrices to represent these transformations. In this appendix,another quantity, known as a tensor, is introduced to achieve the same purpose.We generally denote tensors by uppercase boldfaced symbols: for example, A, andsymbolize the transformation of a vector a by A to a vector b as

b = Aa.

The advantages of using tensors are that they are often far more compact than ma-trices, they are easier to differentiate, and their components transform transparentlyunder changes of bases. The latter feature is very important in rigid body kinemat-ics. To this end, the introductory section, 6.2, of Chapter 6, contains an example inwhich tensor and matrix notations are contrasted and, in the interests of brevity, thereader is referred to this section for further motivation on the advantages of tensors.

We also note that the material presented in this Appendix is standard back-ground for courses in continuum mechanics, and, in compiling them, the primarysources were Casey [26, 28], Chadwick [35], and Gurtin [84].

A.2 Preliminaries: Bases, Alternators, and Kronecker Deltas

Here, Euclidean three-space is denoted by E3. For this space, we define a right-

handed fixed orthonormal basis {E1, E2, E3}. We also use another right-handed or-thonormal basis {p1, p2, p3}. This basis is not necessarily fixed.∗ Lowercase italicLatin indices, such as i, j, and k, will range from 1 to 3.

You may also wish to recall that a set of vectors {b1, b2, b3} is orthonormal ifbi · bk = 0 when i �= k and bi · bk = 1 when i = k. Further, a set of vectors {b1, b2, b3}

∗ This basis serves to represent corotational bases, other time-varying bases such as {er, eθ, E3}, andfixed bases such as {E1, E2, E3}.

362

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A.3 The Tensor Product of Two Vectors 363

is right-handed if the following scalar triple product is positive:

[b1, b2, b3] = b3 · (b1 × b2).

Because we will be using copious amounts of dot products, it is convenient todefine the Kronecker delta δik:

δi j ={

1, i �= j

0, i = j.

Clearly,

pi · pk = δik.

We also define the alternating (or Levi–Civita) symbol εi jk:

ε123 = ε312 = ε231 = 1,

ε213 = ε132 = ε321 = −1,

εi jk = 0 otherwise.

In words, εi jk = 1 if i jk is an even permutation of 1, 2, 3, εi jk = −1 if i jk is an oddpermutation of 1, 2, 3, and εi jk = 0 if either i = j, j = k, or k = i. We also note that

[pi, pj, pk] = εi jk.

It is easy to verify this result by using the definition of the scalar triple product.

A.3 The Tensor Product of Two Vectors

The tensor (or cross-bun) product of any two vectors a and b in E3 is defined by

(a ⊗ b) c = (b · c) a, (A.1)

where c is any vector in E3. That is, a ⊗ b projects c onto b and multiplies the re-

sulting scalar by a. Clearly, a ⊗ b transforms c into a vector that is parallel to a.Equivalently, we can define a related tensor product:

c (a ⊗ b) = (a · c) b.

Both tensor products are used in this book. You should notice that a ⊗ b provides alinear transformation of any vector c that it acts on. For example, if a = E1, b = E2,and c = E2, then

(E1 ⊗ E2) E2 = E1, E2 (E1 ⊗ E2) = 0.

It is a useful exercise to consider other choices of c here.The tensor product of a and b has some useful properties. First, if α and β are

any two scalars, and a, b, and c are any three vectors, then

(αa + βb) ⊗ c = α (a ⊗ c) + β (b ⊗ c) ,

c ⊗ (αa + βb) = α (c ⊗ a) + β (c ⊗ b) .

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364 Background on Tensors

These properties follow from the definition of the tensor product. To prove them,one merely shows that the left- and right-hand sides of the identities provide thesame transformation of any vector d.

A.4 Second-Order Tensors

A second-order tensor A is a linear transformation of E3 into itself. That is, for any

two vectors a and b and any two scalars α and β,

A (αa + βb) = αAa + βAb,

where Aa and Ab are both vectors in E3. To check if two second-order tensors A

and B are equal, it suffices to show that Aa and Ba are identical for all vectors a.The tensor a ⊗ b is a simple example of a second-order tensor.

It is standard to define the following composition rules for second-order tensors:

(A + B)a = Aa + Ba, (αA)a = α(Aa), (AB)a = A(Ba),

where A and B are any second-order tensors, a is any vector, and α is any scalar.We also define the identity tensor I and the zero tensor O:

Ia = a, Oa = 0,

where a is any vector.

A.5 A Representation Theorem for Second-Order Tensors

It is convenient at this stage to establish a representation for any second-order ten-sor A. The main result we wish to establish is that

A =3∑

j=1

3∑k=1

Akjpk ⊗ pj, (A.2)

where

Aik = (Apk) · pi, aj =3∑

k=1

Akjpk, A =3∑

j=1

aj ⊗ pj.

Here, Aik are known as the components of A relative to the basis {p1, p2, p3}.Initially, it is convenient to interpret tensors by use of the representation A =∑3

j=1 aj ⊗ pj. In this light, A transforms pk into ak. Hence, if we know what A doesto three orthonormal vectors, we can write its representation immediately.

To establish representation (A.2), we note that Api is a vector. Consequently,it can be written as a linear combination of the basis vectors p1, p2, and p3:

Ap1 = A11p1 + A21p2 + A31p3,

Ap2 = A12p1 + A22p2 + A32p3,

Ap3 = A13p1 + A23p2 + A33p3.

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A.5 A Representation Theorem for Second-Order Tensors 365

The order of the indices i and k for the scalars Aik is important. Given any vectorb =∑3

i=1 bipi, we find that

Ab =3∑

j=1

A(bjpj) =3∑

j=1

bj(Apj)

=3∑

j=1

bj

(3∑

k=1

Akjpk

)=

3∑j=1

3∑k=1

bj (Akjpk)

=⎛⎝ 3∑

j=1

3∑k=1

Akjpk ⊗ pj

⎞⎠ 3∑

i=1

bipi

=⎛⎝ 3∑

j=1

3∑k=1

Akjpk ⊗ pj

⎞⎠ b.

In the next-to-last step, we used the definition of the tensor product of two vectors.In summary, we have shown that

Ab =3∑

j=1

3∑k=1

(Akjpk ⊗ pj) b.

As this result holds for all vectors b, and A is assumed to be a linear transformation,we conclude (A.2) is a valid representation for A.

A Representation for a Linear Transformation

We can use representation theorem (A.2) to establish expressions for the transfor-mation induced by a second-order tensor A. The result will be similar to a familiarmatrix multiplication. To proceed, recall that

A =3∑

j=1

3∑k=1

Akjpk ⊗ pj.

The transformation induced by A on a vector b is readily computed:

Ab =⎛⎝ 3∑

j=1

3∑k=1

Akjpk ⊗ pj

⎞⎠ b =

3∑j=1

3∑k=1

Akjbjpk.

If we define c = Ab, then it is easy to see that the components ci = c · pi of c are

ci =3∑

j=1

Aijbj. (A.3)

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Expressed in matrix notation, (A.3) should be familiar:⎡⎢⎣

c1

c2

c3

⎤⎥⎦ =

⎡⎢⎣

A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤⎥⎦⎡⎢⎣

b1

b2

b3

⎤⎥⎦ . (A.4)

It should be clear from (A.4) that the identity tensor has the representation I =∑3i=1 pi ⊗ pi. For familiar choices of the basis

{p1, p2, p3

}, we can use this result to

show that

I = E1 ⊗ E1 + E2 ⊗ E2 + E3 ⊗ E3

= er ⊗ er + eθ ⊗ eθ + E3 ⊗ E3

= eR ⊗ eR + eθ ⊗ eθ + eφ ⊗ eφ.

Several other examples of this representation I =∑3i=1 pi ⊗ pi appear in this book.

A Representation for a Product of Two Second-Order Tensors

We now turn to the important result of the product of two second-order tensors Aand B. The product AB is defined here to be a second-order tensor C. First, let

A =3∑

i=1

3∑k=1

Aikpi ⊗ pk, B =3∑

i=1

3∑k=1

Bikpi ⊗ pk, C =3∑

i=1

3∑k=1

Cikpi ⊗ pk.

We now solve the equations

Ca = (AB)a,

where a is any vector for the nine components of C. Using the arbitrariness of a, weconclude that

Cik =3∑

j=1

AijBjk.

This result is identical to that used in matrix multiplication. Indeed, if we definethree matrices whose components are Cik, Aik, and Bik, then we find the representa-tion ⎡

⎢⎣C11 C12 C13

C21 C22 C23

C31 C32 C33

⎤⎥⎦ =

⎡⎢⎣

A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤⎥⎦⎡⎢⎣

B11 B12 B13

B21 B22 B23

B31 B32 B33

⎤⎥⎦ .

In this expression, the components of the three tensors are all expressed in the samebasis. It is straightforward to establish a representation for the product BA.

Finally, we consider the product of two second-order tensors a ⊗ b and c ⊗ d.Using the representation theorem for both tensors, and then taking their product,

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A.6 Functions of Second-Order Tensors 367

we find that

(a ⊗ b)(c ⊗ d) = a ⊗ d(b · c).

This result is the easiest way to remember how to multiply two second-order tensors.

A.6 Functions of Second-Order Tensors

Symmetric and Skew-Symmetric Parts of a Second-Order Tensor

The transpose AT of a second-order tensor A is defined, for all vectors a and b, as

b · (Aa) = (ATb) · a. (A.5)

If we consider the second-order tensor c ⊗ d, we can use the definition of the trans-pose to show that

(c ⊗ d)T = d ⊗ c, (a ⊗ b + c ⊗ d)T = b ⊗ a + d ⊗ c.

These results will prove to be very useful.Given any two second-order tensors A and B, it can be shown that (AB)T =

BTAT. If A = AT, then A is said to be symmetric. On the other hand, A is skew-symmetric if A = −AT. Any second-order tensor B can be decomposed into the sumof a symmetric second-order tensor and a skew-symmetric second-order tensor:

B = 12

(B + BT)+ 1

2

(B − BT) .

Skew-symmetric tensors that feature prominently in this book include the angularvelocity tensors. These tensors are intimately related to angular velocity vectors.The primary examples of symmetric tensors we encounter are the Euler tensors E0

and E and the inertia tensors J0 and J.We next examine representations for the symmetric and skew-symmetric parts

of a second-order tensor A. Using the definition (Aa) · b = (ATb) · a, and the arbi-trariness of a and b, it can be shown that

AT =3∑

i=1

3∑k=1

Akipi ⊗ pk =3∑

i=1

3∑k=1

Aikpk ⊗ pi

where

A =3∑

i=1

3∑k=1

Aikpi ⊗ pk.

As a second-order tensor A is symmetric if A = AT, we find that

A = AT if Aik = Aki.

This implies that a symmetric second-order tensor has six independent components.Similarly, A is skew-symmetric if AT = −A:

A = −AT if Aik = −Aki.

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Notice that this result implies that a skew-symmetric second-order tensor has threeindependent components.

Invariants, Determinants, and Traces

There are three scalar quantities associated with a second-order tensor that are in-dependent of the right-handed orthonormal basis used for E

3. Because these quan-tities are independent of the basis, they are known as the (principal) invariants ofa second-order tensor. Given a second-order tensor A, the invariants, IA, IIA, andIIIA, of A are defined as∗

[Aa, b, c] + [a, Ab, c] + [a, b, Ac] = IA[a, b, c],

[a, Ab, Ac] + [Aa, b, Ac] + [Aa, Ab, c] = IIA[a, b, c],

[Aa, Ab, Ac] = IIIA[a, b, c],

where a, b, and c are any three vectors.The first invariant is known as the trace of a tensor, and the third invariant is

known as the determinant of a tensor:

tr(A) = IA, det(A) = IIIA. (A.6)

We shall now see why this terminology is used. First, recall that

[Aa, b, c] + [a, Ab, c] + [a, b, Ac] = tr(A)[a, b, c],

[Aa, Ab, Ac] = det(A)[a, b, c].

If we choose a = p1, b = p2, and c = p3, where {p1, p2, p3} is a right-handed or-thonormal basis for E

3, then we have the intermediate results

[p1, p2, p3] = 1,

[Ap1, p2, p3] + [p1, Ap2, p3] + [p1, p2, Ap3] =3∑

i=1

(Api) · pi,

[Ap1, Ap2, Ap3] = det(A).

Using these results, we find that the trace of A is

tr(A) =3∑

i=1

(Api) · pi = A11 + A22 + A33.

∗ Our discussion here closely follows Chadwick [35].

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A.6 Functions of Second-Order Tensors 369

A similar result holds for the trace of a matrix. Further, we find that

det(A) = [Ap1, Ap2, Ap3]

=3∑

i=1

3∑j=1

3∑k=1

Ai1Aj2Ak3[pi, pj, pk]

=3∑

i=1

3∑j=1

3∑k=1

εi jkAi1Aj2Ak3. (A.7)

Recall that the determinant of a 3×3 matrix whose components are Bik is

det

⎛⎜⎝⎡⎢⎣

B11 B12 B13

B21 B22 B23

B31 B32 B33

⎤⎥⎦⎞⎟⎠ =

3∑i=1

3∑j=1

3∑k=1

εi jkBi1Bj2Bk3. (A.8)

Comparing (A.7) with (A.8) we see that we can calculate the determinant of a tensorA by representing the tensor using a right-handed orthonormal basis and then usinga standard result from matrices:

det(A) = det

⎛⎜⎝⎡⎢⎣

A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤⎥⎦⎞⎟⎠ .

We also note in passing the useful result that tr(a ⊗ b) = a · b.

Inverses and Adjugates

The inverse A−1 of a second-order tensor A is the second-order tensor that satisfies

A−1A = AA−1 = I.

For the inverse to exist, det(A) �= 0. Taking the transpose of this equation, we findthat the inverse of the transpose of A is the transpose of the inverse.

The adjugate A∗ of a second-order tensor A is the second-order tensor thatsatisfies

A∗(a × b) = Aa × Ab,

where a and b are arbitrary vectors. If A is invertible, then this definition yields

A∗ = det(A)(A−1)T.

Notice how this result simplifies if A has a unit determinant.

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Eigenvalues and Eigenvectors

The characteristic values (or eigenvalues or principal values) of a second-order ten-sor A are defined as the roots λ of the characteristic equation of A:

det(λI − A) = 0.

The three roots of this equation are denoted λ1, λ2, and λ3. On expanding the equa-tion, we find that

λ3 − IAλ2 + IIAλ − IIIA = 0.

That is,

λ1λ2λ3 = det(A) = IIIA,

λ1λ2 + λ2λ3 + λ1λ3 = 12

(tr(A)2 − tr(A2)) = IIA,

λ1 + λ2 + λ3 = tr(A) = IA.

It should be noted that if a tensor is positive-definite, then all of its eigenvalues arestrictly positive.

The eigenvector (or characteristic direction or principal direction) of a tensorA is the vector u that satisfies

Au = λu, (A.9)

where λ is a root of the characteristic equation. A second-order tensor has threeeigenvectors. To determine these eigenvectors, we express (A.9) with the help of(A.4) and then use standard techniques from linear algebra.

Let us now consider a simple example:

J0 = ma2

12E1 ⊗ E1 + mb2

12E2 ⊗ E2 + mc2

12E3 ⊗ E3.

By inspection, this tensor has the eigenvalues ma2

12 , mb2

12 , and mc2

12 and the correspond-ing eigenvectors are E1, E2, and E3. We also note that, if a, b, and c are nonzero,then J0 is positive-definite.

A.7 Third-Order Tensors

We shall find it necessary to use one example of a third-order tensor. A third-ordertensor transforms vectors into second-order tensors and may transform second-order tensors into vectors. We can parallel all the developments for a second-ordertensor that we previously performed. However, here it suffices to note that, withrespect to a basis {p1, p2, p3}, any third-order tensor A has the representation

A =3∑

i=1

3∑j=1

3∑k=1

Ai jkpi ⊗ pj ⊗ pk.

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A.7 Third-Order Tensors 371

We also define the tensor products:

(a ⊗ b ⊗ c) [d ⊗ e] = a(b · d)(c · e), (a ⊗ b ⊗ c)d = a ⊗ b(c · d). (A.10)

Other tensor products can also be defined. The presence of the brackets [·] in (A.10)should be noted.

The main example of a third-order tensor we use is known as the alternator ε:

ε =3∑

i=1

3∑j=1

3∑k=1

εi jkpi ⊗ pj ⊗ pk

= p1 ⊗ p2 ⊗ p3 + p3 ⊗ p1 ⊗ p2 + p2 ⊗ p3 ⊗ p1

− p2 ⊗ p1 ⊗ p3 − p1 ⊗ p3 ⊗ p2 − p3 ⊗ p2 ⊗ p1.

This tensor has some useful features. First, if A is a symmetric tensor, then it is aproductive exercise to show that ε [A] = 0. Second, suppose that c =∑3

k=1 ckpk is avector; then

εc =3∑

i=1

3∑j=1

3∑k=1

εi jkpi ⊗ pjck

= c3(p1 ⊗ p2 − p2 ⊗ p1)+c2(p3 ⊗ p1 − p1 ⊗ p3) + c1(p2 ⊗ p3 − p3 ⊗ p2), (A.11)

which is a skew-symmetric tensor.The fact that ε acts on a vector to produce a skew-symmetric tensor enables us

to define a skew-symmetric tensor C for every vector c and vice versa:

C = −εc, c = −12ε [C] .

The vector c is known as the axial vector of C. It is a good exercise to verify that, ifC has the representation

C = c21(p2 ⊗ p1 − p1 ⊗ p2) + c32(p3 ⊗ p2 − p2 ⊗ p3) + c13(p1 ⊗ p3 − p3 ⊗ p1),

then, with the help of (A.10)1,

c = −12ε [C] = c21p3 + c13p2 + c32p1.

We also note the important result that

Ca = (−εc)a = c × a.

This result allows us to replace cross products with tensor products, and vice versa.

EXAMPLES. It is prudent to provide some examples here. First, suppose we are givena skew-symmetric tensor

� = � (E2 ⊗ E1 − E1 ⊗ E2) .

Then,

ε [�] = −2�E3,

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and we conclude that the axial vector of � is �E3. It is useful to verify that

(� (E2 ⊗ E1 − E1 ⊗ E2)) a = �E3 × a

for all vectors a.Similarly, given a vector γE1, we can compute that

εγE1 = γE2 ⊗ E3 − γE3 ⊗ E2.

It follows from this result that

γE1 × b = (γE3 ⊗ E2 − γE2 ⊗ E3) b

for all vectors b.

A.8 Special Types of Second-Order Tensors

There are three types of second-order tensors that play an important role in rigidbody dynamics: proper-orthogonal tensors, symmetric positive-definite tensors, andskew-symmetric tensors.

Orthogonal Tensors

A second-order tensor L is said to be orthogonal if LLT = LTL = I. That is, thetranspose of an orthogonal tensor is its inverse. It also follows that det(L) = ±1. Anorthogonal tensor has the unique property that La · La = a · a, and so it preservesthe length of the vector that it transforms.

Some examples of orthogonal tensors include

I, −I, E1 ⊗ E1 + E2 ⊗ E2 − E3 ⊗ E3.

The last of these examples constitutes a reflection in the plane x3 = 0.

Proper-Orthogonal Tensors

A second-order tensor Q is said to be proper-orthogonal if QQT = QTQ = I anddet(Q) = 1. Euler’s theorem states that this type of tensor is equivalent to a rotationtensor (see Section 7.2). Proper-orthogonal second-order tensors are a subclass ofthe second-order orthogonal tensors. Indeed, it can be shown that any second-orderorthogonal tensor is either a rotation tensor or can be obtained by multiplying arotation tensor by −I.

Using a result that is known as Euler’s formula, any rotation tensor can be writ-ten as

Q = cos(θ)(I − p ⊗ p) − sin(θ)εp + p ⊗ p,

where θ is a real number and p is a unit vector. The variable θ is known as the(counterclockwise) angle of rotation, and p is known as the axis of rotation. Weexamine several examples of rotation tensors in Chapter 6.

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A.9 Derivatives of Tensors 373

Symmetric, Positive-Definite, Second-Order Tensors

A tensor A is said to be positive-definite if Aa · a > 0 for all a �= 0 and Aa · a = 0 if,and only if, a = 0. A consequence of the definition is that a skew-symmetric second-order tensor can never be positive-definite. Examples of positive-definite tensors inmechanics include the Euler tensors and the inertia tensors.

If A is positive-definite, then it may be shown that all three of its eigenvaluesare positive, and, furthermore, the tensor has the representation

A = λ1u1 ⊗ u1 + λ2u2 ⊗ u2 + λ3u3 ⊗ u3,

where Aui = λiui. That is, ui is an eigenvector of A with the eigenvalue λi. Thisrepresentation is often known as the spectral decomposition.

A.9 Derivatives of Tensors

We shall often encounter derivatives of tensors. Suppose a tensor A has the repre-sentation

A =3∑

i=1

3∑k=1

Aikpi ⊗ pk

and that the components of A and the vectors pi are functions of time. The timederivative of A is defined as

A =3∑

i=1

3∑k=1

Aikpi ⊗ pk +3∑

i=1

3∑k=1

Aikpi ⊗ pk +3∑

i=1

3∑k=1

Aikpi ⊗ pk.

Notice that we differentiate both the components and the basis vectors.For example, consider the tensor

A = er ⊗ E1 + eθ ⊗ E2 + E3 ⊗ E3.

Then, after expressing er and eθ in terms of E1, E2, and E3 and differentiating, wefind that

A = θeθ ⊗ E1 − θer ⊗ E2.

It is left as an exercise to show that A is a rotation tensor and that AAT is a skew-symmetric tensor.

We can also define a chain rule and product rules. Suppose A = A(q(t)), B =B(t), and c = c(t). Then,

A = ∂A∂q

q,

ddt

(AB) = AB + AB,

ddt

(Ac) = Ac + Ac.

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374 Exercises A.1–A.7

If we have a function ψ = ψ(A), then the derivative of this function with respect toA is defined to be the second-order tensor

∂ψ

∂A=

3∑i=1

3∑k=1

∂ψ

∂Aikpi ⊗ pk.

In addition, if the vectors pi are constant, then

ψ =3∑

i=1

3∑k=1

∂ψ

∂AikAik = tr

(∂ψ

∂AAT)

.

This result is used when we consider constraints on the motions of rigid bodies.

EXERCISES

A.1. Consider a = E1 and b = E2. For any vector c =∑3i=1 ciEi, show that (a ⊗

b)c = c2E1 and c(a ⊗ b) = c1E2.

A.2. Consider a = E1 and b = E2. For any vector c =∑3i=1 ciEi, show that (a ⊗ b −

b ⊗ a)c = c2E1 − c1E2.

A.3. Using the definition of the transpose (A.5), verify that (a ⊗ b)T = b ⊗ a.

A.4. If a = 10E1 and b = 5E2, verify that (a ⊗ b)u = 50E1(E2 · u) and (b ⊗ a)u =50E2(E1 · u).

A.5. If a = er and b = eθ, then show that (a ⊗ b)u = er(eθ · u) and (b ⊗ a)u = eθ(er · u).

A.6. Using the definition of the alternator, show that

−εE3 = E2 ⊗ E1 − E1 ⊗ E2.

Where is this result used in rigid body kinematics? Verify that E3 × a =(E2 ⊗ E1 − E1 ⊗ E2) a for any vector a.

A.7. Show that the following matrix multiplication,⎡⎢⎣

er

eθ

ez

⎤⎥⎦ =

⎡⎢⎣

cos(θ) sin(θ) 0


0 0 1

⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦ ,

can be written as

er = PE1, eθ = PE2, ez = PE3,

where

P = cos(θ) (E1 ⊗ E1 + E2 ⊗ E2) + sin(θ) (E2 ⊗ E1 − E1 ⊗ E2) + E3 ⊗ E3

= cos(θ)(I − E3 ⊗ E3) − sin(θ)εE3 + E3 ⊗ E3.

To show the second representation of P, it is helpful to first show that

P = er ⊗ E1 + eθ ⊗ E2 + ez ⊗ E3.

Finally, verify that P is a proper-orthogonal tensor.

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Exercises A.8–A.14 375

A.8. Show that the matrix multiplication⎡⎢⎣

eR

eφ

eθ

⎤⎥⎦ =

⎡⎢⎣

cos(θ) sin(φ) sin(θ) sin(φ) cos(φ)

cos(θ) cos(φ) sin(θ) cos(φ) − sin(φ)


⎤⎥⎦⎡⎢⎣

E1

E2

E3

⎤⎥⎦

can be written as

eR = RE3 , eφ = RE1 , eθ = RE2,

where

R = cos(θ) sin(φ)E1 ⊗ E3 + sin(θ) sin(φ)E2 ⊗ E3 + cos(φ)E3 ⊗ E3

+ cos(θ) cos(φ)E1 ⊗ E1 + sin(θ) cos(φ)E2 ⊗ E1

− sin(φ)E3 ⊗ E1 − sin(θ)E1 ⊗ E2 + cos(θ)E2 ⊗ E2.

To see this, it is helpful to first show that

R = eR ⊗ E3 + eφ ⊗ E1 + eθ ⊗ E2.

A.9. Verify that the tensor R in the previous exercise is proper-orthogonal.

A.10. Give an example that illustrates that tensor multiplication is not commuta-tive, i.e., AB �= BA.

A.11. Verify that, if B is a second-order tensor, then

ε[B] = 12ε[B − BT] .

A.12. Verify that

−ε[c ⊗ d] = −12ε[c ⊗ d − d ⊗ c] = d × c,

where c and d are any two vectors.

A.13. Suppose that A and B are skew-symmetric second-order tensors. Verify that

a · b = 12

tr(ABT) , (A.12)

where a and b are the axial vectors of A and B, respectively.

A.14. Suppose that A is a skew-symmetric second-order tensor, a is its axial vector,and C is a symmetric second-order tensor. Verify that

−12ε [AC + CA] = (tr (C) I − C) a.

This identity can be used to establish an expression for the angular momentum Hof a rigid body in terms of its Euler tensor E and angular velocity vector ω: H =(tr(E)I − E) ω.

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376 Exercise A.15

A.15. Let A be a tensor and b and c be any two vectors. Show that(tr(A)I − AT) (b × c) = Ab × c + b × Ac.

Hint: Perhaps the quickest way to prove this result is to first argue that it suffices topick b = E3 and c = c1E1 + c2E2. Substituting into the identity then yields an easyway to verify the result.

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[2] S. L. Altmann. Rotations, Quaternions, and Double Groups. Oxford SciencePublications, The Clarendon Press, Oxford University Press, New York, 1986.

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388

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Index

Aircraft stability, 190Alternating symbol, 363Alternator, 363, 370Angular acceleration vector, 211Angular momentum

particle, 5rigid body

relative to a fixed point O, 218relative to a point A, 218relative to center of mass, 218, 223

system of particles, 104Angular velocity

tensor, 168vector, 168, 193, 195, 201, 211

Appell, P., 330Areal velocity, 5, 27, 47, 66Atlas, 80Axial vector, 371

Balance of angular momentumrigid body, 272

Balance of linear momentumparticle, 33, 84rigid body, 272

Barycentric coordinate system, 133Bernoulli’s equation, 286Bernoulli, J., 128Boltzmann, L., 20Boltzmann–Hamel equations, 355Bryan angles, 190, 191Bryan, G. H., 184

Cardan angles, 190, 191Cardano, G., 184Cayley, A., 201Center of mass

rigid body, 215system of particles, 105

Center of oscillation, 297Changing coordinates, 320Chaos, 151

Chaplygin integral, 305Chaplygin sphere, 298Chaplygin, S. A., 298, 330, 338Chart, 80Chasles, M., 209Christoffel symbols, 94, 123Components

contravariant, 11covariant, 11

Configuration manifold, 82, 142, 151, 308, 331Configuration space, 114Configurations of a rigid body, 206Conservation

angular momentum, 46, 49, 151, 293, 318, 335,343

energy, 45, 46, 151, 275, 293, 318, 322, 330, 335,343

linear momentum, 45Constraints, 74

ball-and-socket joints, 238Chaplygin, 338holonomic, 20ideal, 92impulsive, 59independence, 26integrability, 243integrability criteria, 23, 32integrable, 15, 39, 75, 107, 238, 290, 328, 333

rheonomic, 20scleronomic, 20

multiple, 26, 109, 245, 269nonholonomic, 21nonintegrable, 21, 39, 76, 107, 109, 243, 252piecewise integrable, 23pin joints, 238rigid bodies, 237rolling rigid bodies, 239, 358sliding rigid bodies, 239static Coulomb friction, 43, 57, 63, 95system of particles, 116systems of, 26, 109, 245, 269

389

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390 Index

Contravariant basis vectors, 11, 115Coordinate curve, 9Coordinate surface, 9Coordinates

Cartesian, 6, 105curvilinear, 9, 115cylindrical polar, 6, 29ellipsoidal, 99generalized, 83, 142–144, 308, 331helix, 29, 87, 96parabolic, 28, 92, 99spherical polar, 7, 28, 95, 105, 315, 342

Coriolis, G. G., 250, 294Corotational

basis, 212derivative, 169, 178, 226tensor, 226vector, 213, 217, 226

Covariant basis vectors, 10, 115Curvilinear coordinates, 80, 115

D’Alembert, J. Le Rond, 71, 128D’Alembert’s principle, 103, 128Degrees-of-freedom, 84, 331Dual Euler basis, 183, 186, 189, 199, 261, 263, 328Dual-spin spacecraft, 347Dumbbell satellite, 140Dynabee, 347, 358

Euler angles1–3–1 set, 204, 264, 2673–1–2 set, 2333–1–3 set, 188, 251, 265, 267–269, 325, 332, 340,

341, 357, 3583–2–1 set, 184, 204, 235, 2643–2–3 set, 198, 230, 339angular velocity vector, 181asymmetric set, 191body-fixed, 204singularities, 183, 187, 189, 190, 228, 267space-fixed, 204switching sets of, 283symmetric set, 191twelve sets, 190

Euler basis, 181Euler tensors, 219, 372Euler’s disk, 298Euler’s equations, 278Euler, L., 34, 163, 171, 204, 207, 304Euler–Lagrange equations, 130Euler–Poisson equations, 292

Forcescentral, 66conservative, 36, 110, 117, 256, 264, 322constraint, 39, 108, 116, 248, 291, 320, 328drag, 285generalized, 85, 111, 125, 319, 344, 349generalized impulsive, 357

gravitational, 37, 133, 140, 257, 316Magnus, 286, 325spring, 37

Foucault, J. B. L., 4Frenet triad, 60, 87Frisbee, 289Frobenius

integrability theorem of, 26, 246, 269Frobenius, F. G., 24, 26, 245

Galileo spacecraft, 347Gauss’ principle of least constraint, 103, 129Gauss, C. F., 128, 193Gibbs, J. W., 171Gibbs–Appell equations, 355Griffin grinding machine, 269Gyroscope, 351

Hamilton, Sir W. R., 71, 130, 152, 193Harmonic oscillator, 134Helicoid, 96Hertz, H., 20, 82Hoberman sphere, 193

Impact of a rigid body, 356Impulse momentum, 33, 356Inertia tensors, 219, 372

change of bases, 231parallel axis theorem, 231, 290principal axes of inertia, 220, 304

Jacobi integral, 159Jacobi’s criterion, 25Jacobi, C. G. J., 24, 82, 130, 279Jellett integral, 304Jellett, J. H., 304Jumping, 298

Kelvin, Lord, 361Kepler frequency, 51, 301Kepler’s laws, 47Kepler, J., 47Kinematical line-element, 121, 142, 144, 151, 308,

331Kinetic energy, 135

Koenig decomposition, 224, 312particle, 5rigid body, 224system of particles, 104

Koenig, J. S., 224Korteweg, D. J., 330Kronecker delta, 362

Lagrangeprinciple of, 40

Lagrange multipliers, 128Lagrange top, 331Lagrange’s equations of motion

rigid body, 312, 318, 324

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Index 391

single particle, 71contravariant form, 94covariant form, 72, 93

system of particles, 113, 119contravariant form, 125covariant form, 125

Lagrange’s prescription, 40, 107, 108, 116, 128,253, 320

Lagrange, J.-L., 39, 47, 71, 103, 152, 188, 301, 307,332

Lagrange–D’Alembert principle, 40Lagrangian, 73, 118Lagrangian reduction, 48Levi–Civita, T., 82, 93Line of nodes, 188Linear momentum

particle, 5rigid body, 217system of particles, 105

Mac Cullagh, J., 260Magnus, H. G., 286Manifold, 80

Riemannian, 82Moment potential, 197, 256, 262, 268Moment-free motion of a rigid body, 279Moments

conservative, 256, 322constant, 260constraint, 248, 291, 320, 328gravitational, 257, 316torsional spring, 268

Momentum sphere, 281Motion of a rigid body with a fixed point, 289, 305,

343

N-body problem, 159Navigation, 194Newton’s third law, 109, 110, 254Newton, Sir I., 34, 47, 285Nondimensionalization, 51, 56

Pendulum and cart, 143Phase portrait, 51Planar double pendulum, 157Poincare, H., 152Poinsot, L., 163, 213, 279Poisson kinematical relations, 292Poisson top, 235, 331Poisson, S. D., 213, 331Principal axes, 220, 277, 304Principal moments of inertia, 220Principle of virtual work, 128Procrustes problem, 194

Quaternions, 193

Reference frame, 3corotational, 217

inertial, 3Relative angular velocity

tensor, 178vector, 178, 182

Representative particle, 113, 123Ricci, M. M. G., 82, 93Riemann, G. F. B., 82Rigid body motion, 206Robotic arm, 357Rollerball (Powerball), 347, 358Roller coaster, 60Rolling disk, 228, 251, 269, 325Rolling rod, 339Rolling sphere, 250, 294, 300, 340

turntable, 302Rotation

angle of, 173axis of, 173, 213Chasles’ theorem, 209Codman’s paradox, 204, 361composition of, 172, 211

Rodrigues formula, 201constant angular velocity motions, 227direction cosines, 168Euler parameters, 193Euler’s formula, 172Euler’s representation, 171

angular velocity vector, 175Euler’s theorem, 176, 207Euler–Rodrigues symmetric parameters, 193,

201fixed-axis case, 164Gibbs vector, 192instantaneous axis of, 213matrix representation, 173, 185, 193quaternions, 193relative, 241, 264Rodrigues vector, 192Rodrigues–Hamilton theorem, 204, 260screw axis of, 213

Routh integral, 305Routh, E. J., 71, 250, 294Routh–Voss equations of motion, 71, 78Routhian reduction, 48

Satellite dynamics, 156, 301, 315, 342Screw motion, 209Sliding disk, 228, 325Sliding sphere, 250, 294, 340Slip velocity, 297SO(3), 308Spectral decomposition theorem, 372Spinning eggs, 298Stability of motion, 281Static friction criterion, 89Steady orbital motions, 302, 317Stick–slip, 63, 138Strap-down inertial navigation system, 351

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392 Index

Tait angles, 190, 191Tait, P. G., 184, 285, 361Tensor product

second-order and vector, 366third-order and second-order,

370third-order and vector, 370two second-order tensors, 364, 366two vectors, 363

Tensorsadjugates, 369angular velocity, 168derivatives, 373determinant, 368eigenvalues, 370eigenvectors, 370invariants, 368inverses, 369orthogonal, 372proper-orthogonal, 166, 176, 372representation theorem for, 364rotation, 163, 171second-order, 364

skew-symmetric, 169, 367symmetric, 367symmetric positive-definite, 372third-order, 370trace, 368

Three-body problem, 151Tippe top, 265, 304Two-body problem, 47, 133

Universal joint, 269

Vehicle dynamics, 107, 338Voss, A., 71

Wahba problem, 194Walker, G. T., 240Wobblestone, 240, 298Work–energy theorem

rigid body, 274single particle, 46system of particles, 104

Ziegler, H., 256, 260

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