VECTOR DYNAMICS - epubs.surrey.ac.ukepubs.surrey.ac.uk/845649/1/VectorDynamics2ed_secure.pdfVECTOR DYNAMICS An introduction for engineering students Second Revised Edition ... It is

VECTOR DYNAMICS

An introduction for engineering students

Second Revised Edition

D. M. Birch

c©2009 University of Surrey, Guildford, Surrey, UK, GU2 7XH

Typeset with ProText LATEX using the American Mathematical Society amssymb packageftp://ftp.tex.ac.uk/tex-archive/systems/windows/protext/

No part of this publication may be reproduced, stored in a retrieval system, ortransmitted, in any form or by any means, electronic, mechanical, photocopying,

recording or otherwise, without the prior permission of the author.

This text is provided subject to the condition that it shall not, by way of trade orotherwise, be lent, re-sold, hired out or otherwise circulated without the author’s

prior consent in any form of binding or cover other than that in which it ispublished and without a similar condition including this condition being imposed

on the subsequent purchaser.

Help us continuously improve engineering education.The author welcomes any comments or feedback about this text.

Dr. D. M. Birch [email protected]

Cover photo: Kirsti Kivinen-Newman www.kirsti.caCover design: Andrea K. Lewis akl [email protected]

The author wishes to acknowledge Professor Emeritus Rom Knystautas,

who always said that the best textbook is the one you write yourself.

Law 1: Conservation of mass.The total mass of a closed system will remain constant.

Law 2: Conservation of momentum.The net external force acting upon a closed system will be

equal to the time rate of change of the total linearmomentum of the system.

Law 3: Conservation of energy.The net external work done upon a closed system will be

equal to the change in energy of the system.

i

Preface

It is a common and unfortunate misconception that the study of classical mechanicscan be reduced to the rote memorization of a handful of governing equations, andthe identification of the desired numerical data from problem statements. With thisapproach, the solution of simple, formulaic problems becomes trivial. However, wedo not live in a simple or formulaic world: the problems encountered by practic-ing engineers are rarely rectilinear or two-dimensional. A ‘top-down’ approach hastherefore been taken throughout this text, where every concept introduced has beendeveloped, by logical progression, from the very first principles. The generality ofthe result is then reduced (when necessary) by clearly stated assumptions and sim-plifications, and the need or justification for the simplifications is explained. Thesignificance of the results is then demonstrated and discussed, in order to extractphysical meaning from symbolic expressions. The emphasis of this text is on theuse of the concepts of classical mechanics to develop an understanding of physicalphenomena through mathematical modeling.

It is hoped that students will adopt the practice demonstrated here in approachingany engineering problem; that is, to begin from first principles, to make as manysimplifying assumptions as can be made without changing the nature of the problem,and to extract meaning (rather than just numerical results) from the equations. Itis also hoped that students will learn to question all things, and no longer merely‘accept’ and memorize governing equations presented to them. By adopting thispractice, the scope of engineering problems which the student can solve becomesessentially unlimited.

It is critical that students practice the application of the concepts introduced inthis text by solving problems. Though a very limited number of ‘practice’ problemsand worked examples are included in this text, nearly all textbooks on classical dy-namics will provide a large selection of problems with answers provided, and largevarieties of texts are usually available from university libraries. Since the problemspresented in each textbook are typically conceived specifically to be solved usingthe methods presented in that text, it is further recommended that students attempt

iii

iv Preface

problems from as many different textbooks as possible. In this way, students may beassured that they have adopted an analytical (rather than rote) approach to problem-solving.

Requirements:It is assumed that students have had some elementary introduction to classical rec-tilinear Newtonian mechanics, though this is not absolutely necessary. A rudimen-tary understanding of multivariable calculus is also assumed, though the multiple-independent-variable integration required for this text is reviewed in Appendix A.The Taylor series expansion is also used occasionally in this text, and an overview ofthis function is included as Appendix B. While mathematics is used extensively inthis text as a tool, the scope is limited primarily to simple integration, differentiationand symbolic algebra. All further specialized mathematical techniques or definitionsused in this text are introduced and developed within the text.

Contents

1 Review of fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Displacement, velocity and acceleration . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Displacement, velocity and acceleration as functions of time 21.2.2 Acceleration and velocity as a function of displacement . . . . 6

1.3 Other important considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4.1 Exponential acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4.2 Velocity as a function of displacement . . . . . . . . . . . . . . . . . . . 151.4.3 Sinusoidal acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.4 Acceleration as a function of displacement . . . . . . . . . . . . . . . 191.4.5 Linear acceleration with mixed initial conditions . . . . . . . . . . 21

1.5 Problem set 0: Velocity and acceleration . . . . . . . . . . . . . . . . . . . . . . . 24

2 Vector kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.1 Vector arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Vector addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.1.2 Scalar multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.1.3 Dot product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.1.4 Cross-product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2 Unit vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.1 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2.2 Cylindrical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.2.3 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.2.4 Converting between coordinate systems . . . . . . . . . . . . . . . . . 37

2.3 Time-derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.3.1 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.3.2 Cylindrical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.3 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.3.4 Normal-tangential coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.4 Relative motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

v

vi Contents

2.4.1 Universality of relative motion equations . . . . . . . . . . . . . . . . 532.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2.5.1 Computing velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.5.2 Generalized coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592.5.3 Relative motion: bug on a disc . . . . . . . . . . . . . . . . . . . . . . . . . 622.5.4 Relative motion: bug on a wheel . . . . . . . . . . . . . . . . . . . . . . . . 682.5.5 Angular velocity and axes of rotation . . . . . . . . . . . . . . . . . . . . 702.5.6 Relative motion in three dimensions . . . . . . . . . . . . . . . . . . . . 732.5.7 Constrained motion: bug on a ladder . . . . . . . . . . . . . . . . . . . . 822.5.8 Constrained motion: Slotted crank . . . . . . . . . . . . . . . . . . . . . . 872.5.9 Scotch yoke mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 922.5.10 Crank-slider mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 952.5.11 Crank-rocker mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

2.6 Problem-solving technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.7 Problem set 1: Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1242.8 Problem set 2: Relative motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

3 Linear momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1313.1 Conservation of linear momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

3.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1323.1.2 Implications of the law of conservation of linear momentum 1343.1.3 Newton’s three ‘Laws’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1353.1.4 Centre of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

3.2 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1423.2.1 Momentum and collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1433.2.2 Impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

3.3 Historical note: Origins of the conservation laws . . . . . . . . . . . . . . . . 1493.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

3.4.1 Elastic collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1513.4.2 Perfectly inelastic collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.4.3 The jet engine: an “explosive” collision . . . . . . . . . . . . . . . . . . 1563.4.4 Centre of mass of a mixed system . . . . . . . . . . . . . . . . . . . . . . 1583.4.5 Centre of mass of a solid object . . . . . . . . . . . . . . . . . . . . . . . . 1603.4.6 Centre of mass and linear momentum . . . . . . . . . . . . . . . . . . . 164

3.5 Problem set 3: Conservation of linear momentum . . . . . . . . . . . . . . . . 167

4 External forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1714.1 Free-body diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1714.2 Common applied forces, type 1: position-independent . . . . . . . . . . . . 173

4.2.1 Gravity (small-displacement assumption) . . . . . . . . . . . . . . . . 1734.2.2 Hydrostatic pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1774.2.3 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1814.2.4 Tension in an ideal string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1844.2.5 Dry friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

4.3 Common applied forces, type 2: position-dependent . . . . . . . . . . . . . . 192

Contents vii

4.3.1 Gravity: x = f (1/x2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1934.3.2 Springs: x = f (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2014.3.3 Viscous friction: x = f (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2044.3.4 Pressure drag: x = f (x2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2094.3.5 Pseudoforces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

4.4 Rockets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2144.4.1 Staged rockets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

4.5 Open systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2224.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

4.6.1 Projectile motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2274.6.2 Ideal strings and pulleys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2344.6.3 Kinetic friction with multiple objects . . . . . . . . . . . . . . . . . . . . 2384.6.4 Limits of static friction: rectilinear motion . . . . . . . . . . . . . . . 2414.6.5 Limits of static friction: constrained motion . . . . . . . . . . . . . . 2464.6.6 Simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2494.6.7 Normal force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2564.6.8 External forces and collisions . . . . . . . . . . . . . . . . . . . . . . . . . . 2594.6.9 Open systems: the falling chain . . . . . . . . . . . . . . . . . . . . . . . . 2644.6.10 Open systems: a jet engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

4.7 Problem-solving technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2714.8 Problem set 4: External forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2744.9 Problem set 5: Position-dependent forces . . . . . . . . . . . . . . . . . . . . . . . 279

5 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2835.1 Definition of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2835.2 Angular momentum of point masses . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

5.2.1 Demonstration of the law of conservation of angularmomentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

5.3 Angular momentum of solid bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . 2885.3.1 The inertia tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2935.3.2 Observations about angular momentum . . . . . . . . . . . . . . . . . . 295

5.4 Inertia tensors for common shapes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3005.4.1 Long slender rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3005.4.2 Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3025.4.3 Hollow cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3065.4.4 Cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3095.4.5 Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3145.4.6 Hollow sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3175.4.7 Rectangular block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319

5.5 Parallel axis theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3235.6 Torsional springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3255.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

5.7.1 Pulley with mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3275.7.2 Pure rolling motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3315.7.3 Pulley with multiple masses and additional forces . . . . . . . . . 340

viii Contents

5.7.4 Distributed forces and moments . . . . . . . . . . . . . . . . . . . . . . . . 3475.7.5 Rolling and slipping: a comparison . . . . . . . . . . . . . . . . . . . . . 3515.7.6 Combined rolling and slipping: backspin . . . . . . . . . . . . . . . . 3595.7.7 Simple clutch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3655.7.8 Rolling clutch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3745.7.9 Torsional pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3825.7.10 Reaction forces: swinging bar . . . . . . . . . . . . . . . . . . . . . . . . . . 3865.7.11 Compound problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3925.7.12 Relative motion and rolling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3995.7.13 Forces in a simple mechanism . . . . . . . . . . . . . . . . . . . . . . . . . 4075.7.14 Rotating imbalances: point masses . . . . . . . . . . . . . . . . . . . . . . 4185.7.15 Rotating imbalances: distributed masses . . . . . . . . . . . . . . . . . 4285.7.16 Gyroscopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4315.7.17 Constrained three-dimensional rotation . . . . . . . . . . . . . . . . . . 4365.7.18 Gimbal gyroscopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444

5.8 Problem-solving technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4515.9 Problem set 6: Angular momentum of particles . . . . . . . . . . . . . . . . . . 4545.10 Problem set 7: Angular momentum of bodies . . . . . . . . . . . . . . . . . . . 458

6 Energy methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4636.1 Conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

6.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4636.2 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465

6.2.1 Linear kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4656.2.2 Rotational kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468

6.3 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4736.3.1 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4736.3.2 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4756.3.3 Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476

6.4 Non-conservative forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4786.5 Common forces which do no work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4796.6 Application to mechanical systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4816.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483

6.7.1 Pulleys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4836.7.2 Pure rolling motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4866.7.3 Object dropped on a spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4886.7.4 Perfectly elastic collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4916.7.5 Pulley with mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4956.7.6 Kinetic friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4976.7.7 Spring-mass oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4996.7.8 Falling string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501

6.8 Problem-solving technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5056.9 Problem set 8: Conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . 507

Contents ix

A Review of multiple integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511A.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511A.2 Area integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512

A.2.1 Functional limits of integration . . . . . . . . . . . . . . . . . . . . . . . . . 513A.2.2 Areas bounded by cyclic functions . . . . . . . . . . . . . . . . . . . . . . 517A.2.3 Differential area elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519

A.3 Volume integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523A.3.1 Differential volume elements . . . . . . . . . . . . . . . . . . . . . . . . . . 524A.3.2 Functional limits of integration in 3 dimensions . . . . . . . . . . . 527

B Review of Taylor series expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531

C Table of inertia tensors for common shapes . . . . . . . . . . . . . . . . . . . . . . . 539

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543

List of Symbols

A Areaa Scalar magnitude of a linear accelerationaaa Linear acceleration vectoraa/baa/baa/b Linear acceleration vector of a relative to bc Centre of massC Viscous drag constante Orbital eccentricityeaeaea Unit vector pointing in direction ae′ae′ae′a Unit vector pointing in direction a attached to moving reference frameE EnergyEk Kinetic energyEkx Linear kinetic energyEkθ Rotational kinetic energyf An arbitrary functionF Scalar magnitude of a forceFFF Force vectorFBFBFB Buoyancy force vectorFextFextFext Force vector applied by the environment upon the systemFfFfFf Friction force vectorFgFgFg Gravitational force vectorFnFnFn Normal force vectorFTFTFT Tension force vectorg Small-displacement gravitational constant; g ≈ 9.81 m/s2 on EarthG Universal gravitational constant; G ≈ 6.67×10−11 N · m2/kg2

GGG Linear momentum vector; GGG = mvvvGsysGsysGsys Total linear momentum of a systemGbGbGb Momentum flux into an open systemh Immersed depthH HeightHaHaHa Angular momentum about the point aHsys/aHsys/aHsys/a Total angular momentum of a system about the point a

xi

xii List of Symbols

IaIaIa Nine-component inertia tensor, taken about the point ai Index of summationIii/a Moment of inertia about a point a along i-axis; i = x,y,zIi j/a Product of inertia about a point a; i = x,y,z, j = x,y,z, i �= jIsym/c Moment of inertia about the centre of mass of an axisymmetric body along

any transverse axisj Index of summationJ Scalar magnitude of an impulseJJJ Impulse vector; JJJ =

∫FFF dt

k Spring constantkθ Torsional spring constantK Pressure drag constantL Lengthm Massme Mass of the earth; me ≈ 5.97×1024 kgM Scalar magnitude of a momentMaMaMa Moment vector applied about a point aMext/aMext/aMext/a Moment vector applied about a point a by the environment upon the sys-

temn An integer numbero Fixed origin or reference frameo′ Moving origin or reference framep An arbitrary pointP Pressurer Scalar distance along ererer-axis; also radius of curvaturerrr Position vectorrarara Position vector of ara/b Scalar distance between a and bra/bra/bra/b Position vector of a relative to bre Radius of the Earth; re ≈ 6.37×106 mR RadiusS Trajectory, or path lengtht TimeUB Buoyancy potential energyUg Gravitational potential energyUk Spring potential energyV Volumev Scalar magnitude of a linear velocityvvv Linear velocity vectorva/bva/bva/b Linear velocity vector of a relative to bW WidthW Work; W =

∫FFF ·dSdSdS

Wext External workx Scalar distance along exexex-axisy Scalar distance along eyeyey-axis

List of Symbols xiii

z Scalar distance along ezezez-axisα Scalar magnitude of an angular accelerationααα Angular acceleration vectorαaxαaxαax Angular acceleration vector of a reference frameφ Angular position, used in spherical coordinatesθ Angular positionλ Linear or area density (in mass per length or mass per length2)μk Coefficient of kinetic friction (unitless)μs Coefficient of static friction (unitless)η Efficiency factorρ Volumetric density (in mass per length3)ξ A constant, or dummy variable used for integrationω Scalar magnitude of an angular velocityΩ Scalar magnitude of an angular velocity (used typically for 3-D rotation)ωωω Angular velocity vectorωaxωaxωax Angular velocity vector of a reference frame

Chapter 1Review of fundamentals

Summary A revision of the basic concepts of mathematics and kinematics; deriva-tives and integrals, definitions of displacement, velocity and acceleration, and theirderivation from first principles; solving simple initial-condition problems from firstprinciples using governing equations only.

1.1 Introduction

It is common practice at the secondary level for simple, one-dimensional mechanicsto be taught using what are commonly referred to as the ‘SUVAT’ equations; theseare simplified equations of motion valid only for the case of unconstrained, rectilin-ear motion with constant acceleration. These equations are typically memorized bythe student, and have the form

S =12

at2 +Ut

v =√

2aS+U2

where t is time, and S, v, a and U are displacement, velocity, acceleration andinitial velocity, respectively, and it is assumed that all motion begins at t = 0. Todemonstrate an understanding of one-dimensional rectilinear mechanics, then, stu-dents need only solve problems which require no more than substituting numericvalues for all of the variables into some variant of these SUVAT equations, yieldinga single numeric result.

Regrettably, this solution procedure cannot be applied successfully outside of thesecondary-level classroom, as real problems in engineering dynamics rarely lendthemselves to these idealized conditions. The SUVAT equations themselves are onlyused in secondary instruction, as students have not yet been introduced to integra-

1

2 1 Review of fundamentals

tion or differentiation and therefore lack the mathematical tools required to solveproblems without the aid of these ‘prefabricated’ equations.

In general, the rote memorization of governing equations is a very poor meansof developing the fundamental understanding required to solve complex problemsin any engineering discipline. It is strongly recommended that students cultivate thehighly important and versatile skill of solving problems from first principles, basedon the most general governing equations available. Not only does this drasticallyreduce the number of equations which must be memorized, it also dramatically re-duces the probability of error.

This chapter will provide a review of the most fundamental concepts of motionin order to demonstrate the proper approach to problem-solving.

1.2 Displacement, velocity and acceleration

1.2.1 Displacement, velocity and acceleration as functions of time

Consider an object moving along a straight path, as shown in Figure 1.1. The in-stantaneous position of the object may be completely described using the variablex, which represents the distance between the object and an arbitrary fixed reference(this distance is often referred to as the displacement). Here, x may be a function oftime or any other variable.

Fig. 1.1 Simple, one-dimensional rectilinear motion.

The velocity, (which is indicated using the symbol v), of the object is defined asthe time rate of change of position, so that

v(t) =dxdt

where t is time. The velocity will be positive if the value of x is increasing, and willbe negative if the value of x is decreasing (note that the parentheses here are used to

1.2 Displacement, velocity and acceleration 3

indicate a functional dependence). The acceleration (which shall be indicated usingthe symbol a), of the object is defined as the time rate of change of velocity, so that

a(t) =dvdt

=d2xdt2

The acceleration will be positive if the speed is increasing and negative if the speedis decreasing. It should also be noted here that both velocity and acceleration arefunctions of time as well. Since most dynamic analysis will yield accelerations, it isusually necessary to integrate the acceleration with respect to time. As an example,consider the special case of constant acceleration, in which a(t) = a0, where a0 is aconstant. Then,

a0 =dvdt

Since we may treat the quantity dt as a variable here, we can re-arrange this expres-sion as

dv = a0 dt

Integrating both sides,

∫dv =

∫a0 dt

v+A1 = a0t +A2

where A1 and A2 are constants of integration. For simplicity, we may combine theseconstants into a single constant of integration, as

v = a0t +(A2 −A1

)v = a0t +C0

where it is implicit that C0 = A2 −A1 (a more detailed introduction to the natureof integrals is presented in Appendix A). Since this equation must also be satisfiedin the initial condition (when t = 0 and v = v0), these instantaneous values may besubstituted to yield,

v0 = a0(0)+C0

We may then solve for the unknown constant C0 in terms of the known quantity v0

as


v0 = 0+C0

C0 = v0

so the constant of integration is just the initial velocity v0. Substituting back into thevelocity expression,

v = a0t +C0

v = a0t + v0

The displacement may then be obtained as a function of time by integrating oncemore with respect to time, so that

v =dxdt

dx = v dt

Integrating both sides, ∫dx =

∫v dt

Substituting our earlier result for v,

∫dx =

∫ (a0t + v0

)dt

x =12

a0t2 + v0t +C1

where C1 is a second combined constant of integration. Again, this equation mustalso be satisfied at the initial condition (when x = x0 and t = 0), so

x0 =12

a0(0)2 + v0(0)+C1

C1 = x0


The constant C1 is therefore simply the initial position x0. Substituting this resultback into the expression for displacement yields the familiar result,

x =12

a0t2 + v0t + x0 (1.1)

While Equation (1.1) is identical to the SUVAT displacement equation, by solvingfor the displacement by integration, any condition may be used to solve for theconstants of integration. Let us consider again the same example, but with the initialcondition unknown. Instead, let us say that the velocity v1 is known at some time t1,and the displacement x2 is known at some different time t2. Then,

v =∫

a0 dt = a0t +C0

as before. Now, to determine the constant C0 we need to use the condition at t1 �= 0:

v1 = a0t1 +C0

so, isolating and solving for C0,

C0 = v1 −a0t1

where both v1 and t1 are known quantities. Then, substituting this back into thevelocity equation,

v = a0t +(v1 −a0t1)

To obtain the displacement, this expression is integrated again with respect to time,so that

x =∫

v dt

=∫ (

a0t +(v1 −a0t1))

dt

x =12

a0t2 +(v1 −a0t1)t +C1

This time, we use the condition at t2 to solve for the constant of integration. At theinstant when x = x2, then, according to the information provided, t = t2. Since theearlier expression for x is always true, we may substitute in these values of x and tto obtain,

x2 =12

a0t22 +(v1 −a0t1)t2 +C1

Again, isolating and solving for C1 yields


C1 = x2 − 12

a0t22 − (v1 −a0t1)t2

Substituting this back into the earlier result,

x =12

a0t2 +(v1 −a0t1)t + x2 − 12

a0t22 − (v1 −a0t1)t2

x =12

a0(t2 − t2

2

)+(v1 −a0t1)(t − t2)+ x2

Only in the special case where t1 = t2 = 0, then v0 = v1 and x0 = x2, is it possible torecover Equation (1.1). The SUVAT equation would have failed in this case due toa violation of the assumption that the velocity and displacement ‘initial’ conditionsoccur simultaneously at time t = 0. Owing to the very large number of ways inwhich the initial conditions may be presented, it is not practical to simply committhe governing equations to memory.

1.2.2 Acceleration and velocity as a function of displacement

Often it is necessary to determine the velocity of an object when the object is ina particular position. Let us consider the case illustrated in Figure 1.1, where theobject again has a constant acceleration a0. To determine the velocity as a functionof position, we must eliminate time from the expression for velocity. Beginning withEquation (1.1),

x =12

a0t2 + v0t + x0

where the explicit dependence of x upon t has been omitted for simplicity. Takingthe first derivative with respect to time,

ddt

(x)=

ddt

(12

a0t2 + v0t + x0)

v = a0t + v0

Again, the dependence of v upon t has been implied. Isolating and solving for t,

t =v− v0

a0

This result may now be substituted into the displacement to yield


x =12

a0t2 + v0t + x0

=12

a0

(v− v0

a0

)2

+ v0

(v− v0

a0

)+ x0

=1

2a0

(v2 −2v0v+ v2

0

)+

1a0

(v0v− v2

0

)+ x0

=1

2a0

(v2 −2v0v+ v2

0 +2v0v−2v20

)+ x0

x =1

2a0

(v2 − v2

0

)+ x0

Now, the velocity may be expressed in terms of x independently from t, as

v =√

2a0(x− x0)+ v20 (1.2)

This result may also be familiar as one of the SUVAT equations. Again, Equation(1.2) will hold true if and only if (a) the acceleration is constant, and (b) the initialcondition of velocity and displacement are known at the same time t = 0.

On the other hand, it would be considerably less mathematically tedious if thetime dependence were to be eliminated from the expression before any integration,by substituting out the differential dt. This can be done by using the chain rule uponthe definition of acceleration to yield,

a(t) =dvdt

a(x) =dvdx

dxdt

Since dx/dt is the definition of velocity, this can be expressed as

a(x) = vdvdx

(1.3)

where no restrictions have been placed on a here at all. This relation is very use-ful, and will be used extensively in the chapters which follow. Both sides of thisexpression may now be multiplied by dx and integrated, so that

∫a dx =

∫v dv∫

a dx =12

v2 +C0


To solve for the constant C0, we must know what the velocity is at some initialcondition x0. Then,

∫ x0

0a dx =

12

v20 +C0

C0 =∫ x0

0a dx− 1

2v2

0

where v0 is the known value of velocity at the instant when x = x0. This result maythen be substituted back into the expression for acceleration to yield,

∫a dx =

12

v2 +C0∫a(x) dx =

12

v2 +∫ x0

0a dx− 1

2v2

0

Again, for the special case where a = a0 (a constant), then

∫a0 dx =

12

v2 +∫ x0

0a0 dx− 1

2v2

0

a0x =12

v2 +a0x0 − 12

v20

Isolating and solving for v, we find

v =√

2a0(x− x0)+ v20

which is the same result arrived at in Equation (1.2). By applying the chain rule asshown in Equation (1.3), it was possible to obtain the velocity as a function of theacceleration directly, without needing to solve for the time.

1.3 Other important considerations

There are a number of additional important considerations in the analysis of engi-neering systems which are not critical to the solution of secondary-level problems.These aspects of the analysis are therefore often neglected or ignored as unimportantby students. For the more complex problems encountered in real engineering appli-cations, these considerations become an important and integral part of the analysis.

1.3 Other important considerations 9

Working symbolically vs. working numerically

It is often tempting to substitute numbers into governing equations as early aspossible, in order to minimize algebraic complexity. However, doing so obscuresthe functional dependence of the result upon the various parameters. Evaluatingderivatives or integrals therefore becomes impossible, while the generality of theresult is entirely lost. Also, when working with either very large or very small num-bers, early substitution may result in errors due to numerical rounding. For example,consider the expression

C = (ξ 2 +η2)(ξ 2 −η2)

Given the values ξ = 3.54× 104 and η = 1.15, this expression may be evaluateddirectly, as

C =((3.54×104)2 +(1.15)2)((3.54×104)2 − (1.15)2)

= 1.57041×1018

Now let us say that we need to evaluate the expression,(Cξ 4 −1

)×1018

If we were to evaluate this numerically using the previous result,

(Cξ 4 −1

)×1018 =

(1.57041×1018(

3.54×104)4 −1

)×1018

=

(1.57041×1018

1.57041×1018 −1

)×1018

= (1−1)×1018

= 0

On the other hand, if we were to have solved the problem symbolically,

(Cξ 4 −1

)×1018 =

((ξ 2 +η2)(ξ 2 −η2)

ξ 4 −1

)×1018

=

(ξ 4 −η4

ξ 4 −1

)×1018


=

(1−(

η

ξ

)4

−1

)×1018

= −(

η

ξ

)4

×1018

= −(

1.153.54×104

)4

×1018

= −1.114

Solving numerically yielded a result of 0, while solving symbolically yielded a re-sult of −1.114. The numerical evaluation introduced a rounding error which wasdramatically magnified in the result.

Singular points

When manipulating algebraic expressions, another common oversight is the fail-ure to check for the existence of singular points. Singular points occur when theresult of an expression becomes infinite, usually as a result of carrying out an ‘ille-gal’ mathematical operation, such as dividing by zero or evaluating the logarithm ofa negative number. If one of these operations is carried out, the results of the anal-ysis may appear reasonable but will be entirely nonphysical. As a simple example,let us consider the variables A and B. If we let A = B and carry out some simplealgebraic manipulation,

A = B

A = B−A+AA

B−A=

B−A+AB−A

AB−A

= 1+A

B−AA

B−A− A

B−A= 1

0 = 1 ???

Using only simple algebra, we have managed to prove that 0= 1. The results are realnumbers- but are in this case very obviously nonsensical. The reason for this is thatwe had divided both sides of the expression by (B−A); since we began by definingB = A, this constituted a division by zero and rendered the expression singular. Anytime a division is carried out, it is good practice to consider whether or not it wouldbe possible for the denominator to equal zero. If it is possible, then a separate solu-tion should be derived for that special case. In the analysis of mechanical systems,this will result in additional insight into the behaviour of the system.

1.3 Other important considerations 11

Working with engineering units

Quantity Typical symbol SI UnitsDisplacement x mVelocity v m/sAcceleration a m/s2

Force F kg × m/s2

Torque or moment M kg × m2/s2

Pressure P kg / (m × s2)Density ρ kg / m3

Absolute viscosity μ kg / (m × s)Specific viscosity ν m2/sArea A m2

Volume V m3

Angle θ , φ (none)Angular velocity ω , Ω 1/sAngular acceleration α 1/s2

Linear momentum G kg × m /sAngular momentum H kg × m2/sMoment of inertia I kg × m2

Energy or work E, W kg × m2/s2

Power P kg × m2/s3

Linear spring constant k kg / s2

Torsional spring constant kθ kg × m/s2

Friction coefficient μ (none)Viscous drag factor C kg / sAerodynamic drag factor K kg / m

After finally reducing an expression to its simplest form and substituting in thenumeric values, many students do not consider the units attached. Instead, studentswill simply ensure that all of the numeric values used were in SI standard units(kg, m, s) and then assume that the result would be in the required units. In practice,however, the units provide the simplest and most effective check that the calculationsperformed were correct: when reduced algebraically, the units must combine to yieldthe required unit. For example, let us consider the expression,

F =

(21.3

kgm3

)(1.86

ms

)2

(0.028 m)2

where the quantity F is expected to be a force. Evaluating this expression yields,

F = (21.3)(3.46)(0.000784)kgm3

(ms

)2

m2

= 0.0578kg m4

m3 s2


= 0.0578kg m

s2

F = 0.0578 N

The result yielded units of mass × length / time2, which is equivalent to units offorce. Had any other combination of units emerged from the calculation, it wouldbe clear that a mistake had been made.

Also, it is only possible to add or subtract quantities with like units. Metres can-not be added to seconds, and kilograms cannot be subtracted from metres. Shouldany result require that quantities with dissimilar units be summed, this result isnecessarily incorrect. A table is included here which lists some common quanti-ties which may be encountered in problems of engineering dynamics, together withtheir associated units. It is also important to note that differentials (such as dx ordt) assume the same units as their arguments; for example, if x is a displacement inmetres, then the quantity dx would have units of metres as well.

1.4 Examples

1.4.1 Exponential acceleration

A ball is thrown toward the ground from a window at a height h above the pavement,with an initial velocity of v0. Because of the aerodynamic drag on the ball, theacceleration of the ball is given by the expression,

a(t) = a0e−κt

where a0 and κ are constant. Determine the velocity and height of the ball as a func-tion of time, and sketch the acceleration, velocity and height of the ball as functionsof time.

SOLUTION

First, let us define our origin as being on the ground, and define the positivedirection as upwards. Next, we begin with the definition of acceleration:

a(t) =dv(t)

dt

Then, multiplying both sides of this expression by dt and integrating yields,

1.4 Examples 13

Fig. 1.2 Exponential acceleration

v(t) =∫

a(t) dt

=∫

a0e−κt dt

v(t) = − 1κ

a0e−κt +C0

where C0 is a constant of integration. Note that the ball is accelerating downwards,so in our coordinate system, a0 < 0 and v0 < 0. To determine the value of the con-stant, we examine the initial condition; when t = 0, v(t) = v0. Since these valuesmust satisfy the equation,

v0 = − 1κ

a0e−κ(0) +C0

v0 = − 1κ

a0 +C0

C0 = v0 +1κ

a0

This result may then be substituted back into the expression for velocity to yield,


v(t) =− 1κ

a0e−κt +

(v0 +

1κ

a0

)

which is an expression of the velocity as a function of time. Next, to determine thedisplacement as a function of time, we look at the definition of velocity:

v(t) =dx(t)

dt

as before, we then multiply both sides by dt and integrate,

x(t) =∫

v(t) dt

=

∫− 1

κa0e−κt +

(v0 +

1κ

a0

)dt

x(t) =1

κ2 a0e−κt +

(v0 +

1κ

a0

)t +C1

where C1 is another constant of integration. To determine the constant of integration,we examine again the initial condition. The height of the ball is known to be h whent = 0; these two quantities must satisfy the displacement equation, so

h =1

κ2 a0e−κ(0) +

(v0 +

1κ

a0

)(0)+C1

h =1

κ2 a0 +C1

C1 = h− 1κ2 a0

Substituting this value for C1 back into the earlier expression for x(t) yields thedisplacement as a function of time,

x(t) =1

κ2 a0e−κt +

(v0 +

1κ

a0

)t +

(h− 1

κ2 a0

)

x(t) =1

κ2 a0

(e−κt −1

)+

(v0 +

1κ

a0

)t +h

as required.

1.4 Examples 15

Fig. 1.3 Nonlinear acceleration 1: Sketch of acceleration, velocity and altitude as functions oftime. Note that a0 and v0 were both assumed to be negative.

1.4.2 Velocity as a function of displacement

Demonstrate that, for the case of constant acceleration where

v =√

2a0(x− x0)+ v20,

the expression

a(x) = vdvdx

is satisfied.


SOLUTION

This is little more than an exercise in differentiation and algebraic manipulation.Begin by substituting together the two equations given:

vdvdx

=√

2a0(x− x0)+ v20

ddx

(√2a0(x− x0)+ v2

0

)

Now, the derivative may be evaluated, as

vdvdx

=12

√2a0(x− x0)+ v2

0

(2a0√

2a0(x− x0)+ v20

)

vdvdx

=2a0

2

√2a0(x− x0)+ v2

0√2a0(x− x0)+ v2

0

vdvdx

= a0 = a,

Q.E.D.

1.4.3 Sinusoidal acceleration

The scotch-yoke mechanism illustrated causes the piston to oscillate along a hori-zontal plane. The acceleration of the piston is given by the expression,

a = a0 sin(2π f t)

where f is the frequency of oscillation and is constant. When the acceleration is atits maximum value, the piston is instantaneously at rest, and when the accelerationis zero, the piston is at the position x0. Solve for the position of the piston as afunction of time, and sketch the acceleration, velocity and displacement of the pistonas functions of time. Which has a greater effect on the maximum displacement- thepeak acceleration a0 or frequency f ?

SOLUTION

We begin, as always, with the definition of acceleration:

a(t) =dv(t)

dt

1.4 Examples 17

Fig. 1.4 Sinusoidal acceleration

Multiplying both sides by dt and integrating yields,∫a(t)dt = v(t)

Substituting in the expression for acceleration provided,

v(t) =∫

a0 sin(2π f t) dt

v(t) = − a0

2π fcos(2π f t)+C0

In this problem, the ‘initial’ conditions given require a little thought before beingapplied. The velocity is known to vanish when the acceleration is maximum, andthe maximum possible value of acceleration is a0 and occurs when sin(2π f t) = 1.If sin(2π f t) = 1, then for this value of t, it must also be true that cos(2π f t) = 0.Therefore, at this instant in time,

v(t) = − a0

2π fcos(2π f t)+C0

(0) = − a0

2π f(0)+C0

C0 = 0

so the velocity of the piston is

v(t) =− a0

2π fcos(2π f t)


Next, we integrate the definition of velocity to obtain an expression for the positionas a function of time.

v(t) =dx(t)

dt∫v(t) dt = x(t)

Then, substituting in the expression for velocity already obtained,

x(t) =∫

− a0

2π fcos(2π f t) dt

x(t) = − a0

4π2 f 2 sin(2π f t)+C1

Now we need to solve for the constant C1. From the ‘initial’ condition given, x(t) =x0 when the acceleration is zero. If the acceleration is zero, then sin(2π f t) = 0.Substituting this into the previous result,

x0 = − a0

4π2 f 2 (0)+C1

C1 = x0

Substituting this value for C1 back into the result for the displacement,

x(t) =− a0

4π2 f 2 sin(2π f t)+ x0

From this expression, we see that the maximum displacement will occur whensin(2π f t) =−1. Then,

xmax =a0

4π2 f 2 + x0

so the maximum displacement is linearly proportional to a0 but inversely propor-tional to the square of the frequency- so the frequency has a greater effect on thepeak displacement than the amplitude of the acceleration.

1.4 Examples 19

Fig. 1.5 Nonlinear acceleration 2: Sketch of acceleration, velocity and displacement as functionsof time.

1.4.4 Acceleration as a function of displacement

An object of mass m is resting on a frictionless, horizontal plane. It is connected to afixed point with a spring of stiffness k, as illustrated. If the object is pulled a distanceL and released, the acceleration of the object will be given by the expression,

a =− km

x

where x is the distance between the origin and the object, measured positive to theright. Determine the velocity of the object at the instant that it strikes the origin.


Fig. 1.6 Acceleration as a function of displacement

SOLUTION

To solve this problem, we first recognize that the acceleration is provided as afunction of displacement, and the velocity is also required as a function of displace-ment. This problem is a good candidate, then, for the application of Equation (1.3).

a =dvdt

=dvdx

dxdt

= vdvdx∫

a dx =∫

v dv

Next, we can substitute in our expression for acceleration and integrate.

∫a dx =

∫v dv∫

− km

x dx =∫

v dv

−12

km

x2 +C0 =12

v2

where C0 is a constant of integration. To determine C0, we examine the initial con-dition: the object was released from rest (v = 0) when x = L. Therefore, these quan-tities must satisfy the expression for v, so

−12

km

x2 +C0 =12

v2

−12

km

L2 +C0 =12(0)2

1.4 Examples 21

C0 =12

km

L2

This result may then be substituted back into the expression for velocity, so

−12

km

x2 +C0 =12

v2

−12

km

x2 +12

km

L2 =12

v2

±v =

√km

(L2 − x2

)

Since the velocity is always to the left, and this direction was defined as negative,we select the negative solution. Finally, when the object reaches the origin, x = 0.At this instant, the above velocity equation is also satisfied, so

v = −√

km

(L2 − x2

)v = −

√km

(L2 −02

)v = −L

√km

which is the result required. It is interesting to note in this case that it is impossiblefor the velocity to be nonreal; therefore the absolute value of x must always be lessthan L. This condition is also required by the first law of thermodynamics, whichwill be covered later. It should also be noted that the result is somewhat nonphysical;in reality, the acceleration will change sign when x < 0, so the velocity too wouldchange sign and the solution would be different.

1.4.5 Linear acceleration with mixed initial conditions

An object is traveling in a straight line with an acceleration given by the expression,

a(t) = a0(t0 − t

)where a0 and t0 are constants. The velocity is known to have a value of v1 whent = t1, and the position x of the object is known to be x2 when t = t2. Solve for theposition x of the object as a function of time.


SOLUTION

Here, we have an expression for acceleration as a function of time. To solve thisproblem, we begin with the definition of acceleration,

a =dvdt∫

a dt =∫

dv

We may now substitute in the expression for acceleration which has been provided.

∫a0(t0 − t

)dt =

∫dv

a0

(t0t − 1

2t2)

= v+C0

where C0 is a constant of integration. Since we know v = v1 when t = t1, we can usethis initial condition to solve for C0, as

a0

(t0t1 − 1

2t21

)= v1 +C0

C0 = a0

(t0t1 − 1

2t21

)− v1

Substituting this into our previous result and solving for v,

a0

(t0t − 1

2t2)

= v+a0

(t0t1 − 1

2t21

)− v1

v− v1 = a0

(t0t − 1

2t2)−a0

(t0t1 − 1

2t21

)

v = a0t0t −a0t0t1 − 12

a0t2 +12

a0t21 + v1

To determine the displacement, we begin with the definition of velocity, as

v =dxdt

1.4 Examples 23∫v dt =

∫dx

Substituting in our result for v,

∫ (a0t0t −a0t0t1 − 1

2a0t2 +

12

a0t21 + v1

)dt =

∫dx

12

a0t0t2 −a0t0t1t − 16

a0t3 +12

a0t21 t + v1t = x+C1

where C1 is another constant of integration. However, we know that x = x2 whent = t2, so we can use this initial condition to solve for C1.

x2 +C1 =12

a0t0t22 −a0t0t1t2 − 1

6a0t3

2 +12

a0t21 t2 + v1t2

C1 =12

a0t0t22 −a0t0t1t2 − 1

6a0t3

2 +12

a0t21 t2 + v1t2 − x2

We may now substitute this back into the previous expression for x to obtain thedesired displacement as a function of time, as

x =12

a0t0t2 −a0t0t1t − 16

a0t3 +12

a0t21 t + v1t −C1

=12

a0t0t2 −a0t0t1t − 16

a0t3 +12

a0t21 t + v1t

−(12

a0t0t22 −a0t0t1t2 − 1

6a0t3

2 +12

a0t21 t2 + v1t2 − x2

)

Simplifying,

x =−16

a0(t3 − t3

2

)+

12

a0t0(t2 − t2

2

)+

(12

a0t21 −a0t0t1 + v1

)(t − t2

)+ x2

This is an expression for the displacement as a function of time, in terms of theknown quantities v1, t1, t2 and x2, as well as the given constants a0 and t0.


1.5 Problem set 0: Velocity and acceleration

Question 1

An object is traveling along a straight-line path in the x-direction with constantacceleration a0. The position of the object is known to be x1 at t = t1 and x2 att = t2. Solve for the position of the object as a function of time, starting from firstprinciples.

Answer:

x =12

a0(t2 − t2

1

)+

(x2 − x1 − 1

2a0(t22 − t2

1

)) t − t1t2 − t1

+ x1

Question 2

During a relay race, runner A is preparing to pass the baton to runner B. Becauseboth runners have never studied mechanics beyond the secondary level, they bothassume that they must only ever run with constant acceleration. At the instant whenrunner A passes the point x= 0, she has velocity v1 and acceleration aA. At that sameinstant, runner B begins to accelerate from rest with acceleration aB from her startingblock, which is located at x = x0. At the instant when the baton is passed (when bothrunners are at the same distance from x= 0), show that their instantaneous velocitiesare related by the expression,

v2A − v2

1

v2B

=aA

aB

(1+

2aBx0

v2B

)

where vA and vB are the velocities of runners A and B, respectively.

Question 3

A rocket is launched vertically, and is known to have an acceleration a given by theexpression

a =Ct

where C is a constant and t is the time elapsed since the instant of launch. The rockethas an initial velocity v = 0 at t = 0. The initial altitude of the rocket is unknown(the rocket was not necessarily launched at sea level), but a remote tracking stationshows that the altitude of the rocket (relative to mean sea level) is H1 at some time

1.5 Problem set 0: Velocity and acceleration 25

t1, and H2 at some later time t2. Solve for the constant C in terms of the availabledata.

Answer:

C = 6H2 −H1

t32 − t3

1

Question 4

A train applying its brakes will decelerate with an acceleration a which varies as afunction of the distance traveled x according to the expression,

a = a0(1− e−Kx)

where a0 and K are constants, and the train began applying its brakes at x = 0.Determine the velocity of the train as a function of position if it was traveling at aspeed of v0 when the brakes were first applied.

Answer:

v =

√2a0x+

2K

a0(e−Kx −1

)+ v2

0

Question 5

An object is vibrating along the x-axis such that its acceleration is given by theexpression

a = a0 sin(2π f0t)+a1 cos(2π f1t)

where the quantities a0, a1, f0 and f1 are all constant. Determine the position of theobject as a function of time, given that the object is at the position x = 0 when t = 0,and is instantaneously at rest at that moment.

Answer:

x =− 14π2

(a0

f 20

sin(2π f0t)+a1

f 21

cos(2π f1t)

)+

a0

2π f0t +

a1

4π2 f 21


Question 6

List 5 conditions in engineering practice when an object will have a constant,nonzero acceleration.

Chapter 2Vector kinematics

Summary An overview of vector addition and multiplication, introduction of unitvectors, definition of the Cartesian, cylindrical, spherical and generalized coordinatesystems, derivations of velocity and acceleration vectors in each coordinate system,and the derivation of the velocity and acceleration of a point relative to a movingand rotating reference frame.

2.1 Vector arithmetic

A vector is defined as a quantity which has both a magnitude and a direction associ-ated with it, such as velocity, acceleration or force. Quantities for which no directioncan be defined, such as mass, temperature or volume, are known as scalar quantities.

A vector can be thought of as an arrow floating in space. The arrow’s length isproportional to the magnitude of the vector quantity which it represents, and thearrow points in the same direction as that vector quantity which it represents. Thenotation used to indicate a vector quantity is typically a variable topped by a littlearrow or line, such as

→a or a, though vectors are more often indicated in printed texts

using bold, non-italic characters, such as aaa. Because a vector has both a magnitudeand a direction associated with it, the symbol aaa should be thought of as containingtwo separate pieces of information, rather than just one.

Since a vector contains only two pieces of information (the magnitude and direc-tion), then any vectors having the same magnitude and direction are said to be equal.In order for one vector to be equal to another vector, then, it is not necessary for thevectors to lie one on top of the other. Because vectors don’t change depending onwhere they lie, they can slide around or translate in space without changing.

The magnitude of a vector is typically indicated using sidebars, such as |aaa|, or bysimply not using the bold face, such as a. The magnitude |aaa| is a scalar quantity, andhas no direction associated with it. In other words, of the two pieces of informationcontained within the symbol aaa, the |aaa| operator extracts only the magnitude.

27

28 2 Vector kinematics

2.1.1 Vector addition

Vectors can be added one to the other, but the operation of ‘addition’ is not definedin the same way as it is for scalars, since the directions also need to be dealt with.Vector addition is a graphical process: to add the vector bbb to the vector aaa, we slidethe vector bbb until its butt-end touches the tip of aaa. The resultant vector, (aaa+bbb), whichwe shall call ccc, then begins at the butt-end of aaa and ends at the tip of bbb. Since both aaaand bbb can slide around without changing,

aaa+bbb = bbb+aaa = ccc

A vector can be made ‘negative’ by reversing its sense without changing its mag-nitude or direction; that is, swapping its butt-end for its tip without changing it inany other way. Vectors may then be subtracted from each other by recognizing

aaa−bbb = aaa+(−bbb)

The negation, addition and subtraction of vectors is illustrated in Figure 2.1.

Fig. 2.1 Graphical illustration of vector addition.

2.1.2 Scalar multiplication

Because there is more information contained in a vector than in an ordinary numberor scalar quantity, there are several ways to multiply vectors. The first and simplest

2.1 Vector arithmetic 29

form of vector multiplication is scalar multiplication, or the multiplication of a vec-tor by a scalar. Because the scalar cannot contribute to the direction of the vector(since it has no direction of its own), it simply increases the magnitude (or length) ofthe vector without affecting its direction. For example, if ccc = 3aaa, then the vector ccc isthree times longer than aaa but points in the same direction. The scalar multiplicationof a vector is illustrated in Figure 2.2.

Fig. 2.2 Graphical representation of the scalar product.

2.1.3 Dot product

The second form of vector multiplication is the dot product, which can only beperformed between two vectors and results in a scalar. The dot product is denotedas aaa ·bbb, and is defined as the magnitude of the projection of aaa upon bbb, scaled by themagnitude of bbb.

The vector aaa (the first argument of the dot product) can be expressed as the sumof any two vectors fff and ggg so that fff+ggg = aaa. While there are an infinite number ofpairs of vectors fff and ggg which satisfy the requirement that fff+ggg= aaa, there is only onepair of fff and ggg which can be chosen so that (i) fff lies parallel with bbb (which can beindicated fff ‖ bbb), and (ii) ggg is perpendicular to bbb (which can be indicated ggg ⊥ bbb). Forthis unique fff and ggg, then, aaa ·bbb = |bbb||fff|; in other words, the result of the dot product isthe scalar length of the vector bbb multiplied by the scalar length of the vector fff lyingparallel with bbb which simultaneously satisfies fff+ggg = aaa and fff ⊥ ggg. This property isillustrated in Figure 2.3.

From trigonometry, we recognize that the definition of the dot product may bemuch more simply expressed as

aaa ·bbb = |aaa| |bbb|cos(θ)

where θ is the angle subtended between aaa and bbb. Since |aaa|, |bbb| and θ do not de-pend on direction, it follows that (aaa ·bbb) = (bbb ·aaa). Also, if θ = 90◦ (i.e. aaa and bbb areperpendicular), then aaa ·bbb = 0.


Fig. 2.3 Graphical representation of the dot product; aaa ·bbb = |fff| |bbb|= |aaa| |bbb|cos(θ)

2.1.4 Cross-product

The third form of vector multiplication is the cross product. The cross product canalso only be performed between two vectors, but results in another vector. The crossproduct of the vectors aaa and bbb is denoted as aaa×bbb. The magnitude and direction ofthe cross product are defined separately.

For the magnitude of the cross-product, we express the vector aaa (the first ar-gument of the cross product) as the vector sum of any two vectors f1f1f1 and f2f2f2 sothat aaa = f1f1f1 + f2f2f2. Again, an infinite number of pairs of vectors f1f1f1 and f2f2f2 will satisfyaaa = f1f1f1 + f2f2f2, but only one pair exist such that (i) f1f1f1 ‖ bbb and (ii) f2f2f2 ⊥ bbb, as illustratedin Figure 2.4. The magnitude of ggg = aaa×bbb, then, is the scalar length of the vectorbbb multiplied by the scalar length of the vector f2f2f2 lying perpendicular to bbb whichsimultaneously satisfies aaa = f1f1f1 +f2f2f2 and f2f2f2 ⊥ f1f1f1.

Fig. 2.4 Graphical representation of the magnitude of the cross product; |aaa×bbb| = |f2f2f2| |bbb| =|aaa| |bbb|sin(θ)

From trigonometry, as before, we recognize that the magnitude of the cross prod-uct may be much more easily described as

|ggg|= |aaa×bbb|= |aaa| |bbb|sin(θ)

where θ is again the angle subtended between aaa and bbb.

2.1 Vector arithmetic 31

The direction of the cross product ggg = aaa×bbb is defined as being mutually perpen-dicular to aaa and bbb. This means that if a plane were defined such that both aaa and bbb layin that plane, ggg would be normal (or perpendicular) to that plane. The cross-productis therefore inherently a three-dimensional operator. By convention, the sense of gggis dictated by the right-hand rule: if you align the fingers of your right hand with aaaand then bend your fingers so that they are aligned with bbb (so that your fingers haveswept out the angle θ ), your thumb will point in the direction of aaa×bbb. It stands toreason, then, that

aaa×bbb =−bbb×aaa

The cross-product is illustrated graphically in Figure 2.5.

Fig. 2.5 Graphical representation of the cross product; ggg = aaa×bbb

While the cross-product initially seems to be an obscure construct with no ap-plication in the physical world, it is extremely useful in that it is a mathematicaltool which allows us to simply and efficiently deal with things which are rotating inthree-dimensional space.

The equations listed below are useful trigonometric vector identities, given thevectors aaa, bbb and ccc, and the scalar ξ .


ξ (aaa+bbb) = ξaaa+ξbbb

aaa ·bbb = |aaa| |bbb|cos(θ)

|aaa×bbb| = |aaa| |bbb|sin(θ)

aaa ·bbb = bbb ·aaaaaa×bbb = −bbb×aaa

aaa×aaa = 0

aaa ·aaa = |aaa|2ξaaa ·bbb = ξ (aaa ·bbb)

ξaaa×bbb = ξ (aaa×bbb)

(aaa+bbb) ·ccc = aaa ·ccc+bbb ·cccaaa× (bbb+ccc) = (aaa×bbb)+(aaa×ccc)

aaa×bbb×ccc = (aaa ·ccc)bbb− (aaa ·bbb)ccc

2.2 Unit vectors

A unit vector, usually denoted by the symbol eee with an appropriate subscript, is avector with a length of 1. Unit vectors are useful, since they allow a direction to begiven to a scalar without affecting the magnitude. For example, if a1 is a scalar, thena1 e1e1e1 is a vector quantity with magnitude a1, pointing in a direction parallel to e1e1e1.

Given three different unit vectors e1e1e1, e2e2e2 and e3e3e3, so long as these three unit vectorsdo not all lie in the same plane, any three-dimensional vector aaa may be expressed asa linear combination of these unit vectors, so that

aaa = a1 e1e1e1 +a2 e2e2e2 +a3 e3e3e3

where a1, a2 and a3 are scalar quantities. Typically, though, the unit vectors e1e1e1, e2e2e2

and e3e3e3 are defined in such a way that they are mutually orthogonal, which meansthat each unit vector is perpendicular to each other unit vector. Then, the magnitudeof the vector aaa is simply the square root of the sum of the squares of the scalarcoefficients of the unit vectors, or

|aaa|=√

a21 +a2

2 +a23

Because all practical engineering analysis takes place in three-dimensional space(time is usually considered as an independent variable), all engineering vectorquantities- from acceleration to magnetic field strength- can be expressed in thisform. The set of three unit vectors (e1e1e1,e2e2e2,e3e3e3) may then be said to define a coordi-nate system. It is extremely useful to have a single defined coordinate system andto express all vectors in terms of scalar multiples of the same unit vectors, since

2.2 Unit vectors 33

the magnitudes may then be treated independently of the directions. For example,consider the vectors aaa = a1 e1e1e1 +a2 e2e2e2 +a3 e3e3e3 and bbb = b1 e1e1e1 +b2 e2e2e2 +b3 e3e3e3; then

aaa+bbb = (a1 e1e1e1 +a2 e2e2e2 +a3 e3e3e3)+(b1 e1e1e1 +b2 e2e2e2 +b3 e3e3e3)

Since the vector sums can be performed in any order in the same way as scalararithmetic (see Figure 2.1), this can be re-arranged as

aaa+bbb = a1 e1e1e1 +b1 e1e1e1 +a2 e2e2e2 +b2 e2e2e2 +a3 e3e3e3 +b3 e3e3e3

and since scalar multiplication is distributive (see the list of trigonometric vectoridentities),

aaa+bbb = (a1 +b1) e1e1e1 +(a2 +b2) e2e2e2 +(a3 +b3) e3e3e3

The sum aaa+bbb may then be evaluated by dealing exclusively with the scalars a1,a2, a3, b1, b2 and b3, without having to work out any vector sums.

2.2.1 Cartesian coordinates

The most common and simplest coordinate system is the Cartesian system, namedfor the famed French mathematician Rene Descartes (1596-1650). In this system,the unit vectors exexex, eyeyey and ezezez lie along the edges of a cube, describing length, widthand depth (commonly known as the x, y and z axes). Two additional things must bedefined in order to describe the coordinate system: the location of the origin o (thepoint at which x = 0, y = 0 and z = 0), and the orientation of the axes (or which wayis defined as ‘up’ and which way is defined as ‘to the right’). The location of theorigin o and the orientation of the axes can be chosen at random, so long as they donot change. The origin must be fixed in space, and the axes cannot rotate.

The position of any point p lying at the location (x,y,z) may then be described interms of a vector that has its butt-end on the origin o and its tip on the point p. Thisposition vector can be represented by rp/orp/orp/o, indicating the vector rrr which describesthe location of the point ‘p’ with respect to the origin and coordinate system ‘o’.The three components of rp/orp/orp/o are then the distances along each of the unit vectorsexexex, eyeyey and ezezez (or the x, y and z axes) which you must travel to get from o to p (seeFigure 2.6). The position vector may then be written as,

rp/orp/orp/o = xexexex + yeyeyey + zezezez

It is useful to note that in the Cartesian coordinate system, all unit vectors arealways orthogonal to all other unit vectors. Consequently, the cross-product of anytwo unit vectors will always yield a third unit vector, since


Fig. 2.6 The Cartesian coordinate system and unit vectors.

e1e1e1 ×e2e2e2 =±|e1e1e1||e2e2e2|sin(90◦)e3e3e3 =±(1)(1)(1)e3e3e3

and since e3e3e3 is in a direction perpendicular to both e1e1e1 and e2e2e2. By convention, the x, yand z axes in a Cartesian coordinate system are oriented so that, by the right-hand-rule,

exexex ×eyeyey = ezezez

eyeyey ×ezezez = exexex

ezezez ×exexex = eyeyey

ezezez ×eyeyey = −exexex

eyeyey ×exexex = −ezezez

exexex ×ezezez = −eyeyey

2.2.2 Cylindrical coordinates

Another common and entirely equivalent coordinate system is the cylindrical sys-tem (which is also sometimes referred to as the polar system). This system is com-mon in engineering applications, as it is well suited for the analysis of rotating ma-chines. In this system, the unit vector ezezez is the same as in the Cartesian system, andnever moves. The unit vector ererer has its butt-end on the z-axis and is perpendicular tothat axis, but is free to rotate about the z-axis so that it can point in any direction. Theunit vector eθeθeθ is perpendicular to both ezezez and ererer, as shown in Figure 2.7. Unlike theCartesian system, these unit vectors are not absolute: while the z-axis remains fixedin space, the r axis is free to rotate around the z-axis. The angle θ , which representsthe angle through which the r-axis has rotated about the z-axis, does not have a unitvector associated directly with it, but the unit vectors ererer and eθeθeθ are both functions ofθ .

The position of any point p in this coordinate system may still be described bythe vector rp/orp/orp/o, however only the two unit vectors ererer and ezezez are required to describethe position of p, since the θ coordinate is included in ererer (ererer, by definition, alwayspoints in the ‘right’ direction).

2.2 Unit vectors 35

Fig. 2.7 The cylindrical coordinate system and unit vectors.

rp/orp/orp/o = rererer + zezezez

The z-component is the distance you have to go up the z-axis until you are lyingin the same z-normal plane as the point p. The r-component is the radius of the circlelying in that plane which passes through the point p. The angle θ is then the anglethrough which you must rotate along that circle to get from the reference θ = 0 tothe point p.

It is important to note here that the direction of the unit vectors ererer and eθeθeθ dependon the location of p, so unlike the Cartesian system, they are not absolute. The unitvectors can indeed change if the point p moves.

In a cylindrical system, the unit vectors are oriented so that,

ererer ×eθeθeθ = ezezez

eθeθeθ ×ezezez = ererer

ezezez ×ererer = eθeθeθ

ezezez ×eθeθeθ = −ererer

eθeθeθ ×ererer = −ezezez

ererer ×ezezez = −eθeθeθ

2.2.3 Spherical coordinates

A less common but still entirely equivalent coordinate system is the spherical sys-tem. In this system, the unit vector ererer has its butt-end on the origin, but is otherwisefree to point in any direction. The eφeφeφ unit vector is perpendicular to ererer and pointsdownward, while the eθeθeθ unit vector is perpendicular to both eφeφeφ and ererer. This is also arelative coordinate system, so the unit vectors are free to move (Figure 2.8).

The position of any point p can be described by the vector rp/orp/orp/o in this coordinatesystem, but only one unit vector is required to construct rp/orp/orp/o since ererer in a sphericalsystem is defined as always pointing directly from the origin to the point p. The unit


vector ererer is therefore a function of both θ and φ , so all of the positional informationrequired is contained within the one unit vector ererer.

rp/orp/orp/o = rererer

The scalar magnitude r describes the distance between the point p and the origin.Given only the magnitude r on its own, all we would know about the point p is thatit lies somewhere upon the surface of a sphere of radius r, centered upon the origino. The angle θ describes the angle subtended between ‘straight up’ and rp/orp/orp/o. Givenonly r and θ , the location of p could be narrowed to a horizontal circle lying on thesurface of the sphere (equivalent to a latitude), indicated in Figure 2.8 with a dottedline. The angle φ is the angle through which you must rotate along that circle to getfrom the reference φ = 0 to the point p, and is equivalent to a longitude.

Aside from its use in navigation, the spherical coordinate system is rarely usedoutside of pure mathematics, except for some highly specialized applications inspacecraft dynamics.

Fig. 2.8 Spherical coordinate system and unit vectors.

In a spherical system, the unit vectors are oriented so that,

eθeθeθ ×eφeφeφ = ererer

eφeφeφ ×ererer = eθeθeθ

ererer ×eθeθeθ = eφeφeφ

ererer ×eφeφeφ = −eθeθeθ

eφeφeφ ×eθeθeθ = −ererer

eθeθeθ ×ererer = −eφeφeφ

2.2 Unit vectors 37

2.2.4 Converting between coordinate systems

It is occasionally necessary to convert the components of a position vector rp/orp/orp/o frombeing in one coordinate system to being in another. To do this, we recognize firstthat a change in coordinate systems represents only a change in the way we describethe relative position of the origin o and the point p; the points themselves do notmove as a result of coordinate conversion.

Let us first consider the problem of converting a position vector rp/orp/orp/o from beingexpressed in cylindrical coordinates to being expressed in Cartesian coordinates.Recognizing that the vector rp/orp/orp/o cannot change as a result of the change in unitvectors,

rp/orp/orp/o = rererer + zezezez = xexexex + yeyeyey + zezezez

Coordinate conversion, then, reduces to the problem of expressing the Cartesianscalars x, y and z in terms of the cylindrical scalars r and z.

We may recognize immediately that since the ezezez unit vector is the same in boththe Cartesian and cylindrical systems, the z-component remains unchanged throughthe conversion. To express x and y in terms of r, we may use simple trigonometry(see Figure 2.9). By definition, x is the scalar distance between the origin and thepoint p in the direction parallel to exexex. However, this distance must also be equal tor cos(θ). Similarly, y is the scalar distance between the origin and the point p in thedirection parallel to eyeyey, and this distance must be equal to r sin(θ). Consequently,

x = r cos(θ)

y = r sin(θ)

Fig. 2.9 Converting from cylindrical coordinates to Cartesian coordinates.


Substituting this result into the expression for rp/orp/orp/o,

rp/orp/orp/o = rererer + zezezez = r cos(θ)exexex + r sin(θ)eyeyey + zezezez

Next, let us consider the conversion from the Cartesian system into the cylindricalsystem. To do this, we must find a way to express r in terms of x and y. This time,since r is simply the magnitude of the distance between the origin and the point pin the x− y plane,

r =√

x2 + y2

Substituting this back into our expression for rp/orp/orp/o,

rp/orp/orp/o = xexexex + yeyeyey + zezezez =√

x2 + y2 ererer + zezezez

Also, from the geometry, we see that

θ = tan−1(

yx

)

It should be noted that the tan−1 operator is the inverse tangent (sometimes indicatedas ‘atan’ or ‘atn’) and not the reciprocal of the tangent.

Similarly, to convert from the spherical coordinate system to the Cartesian coor-dinate system, we begin with the position vector,

rp/orp/orp/o = rererer = xexexex + yeyeyey + zezezez

(where it should be stressed again that the ererer unit vector in the spherical coordinatesystem is not the same as the ererer unit vector in the cylindrical system). Next, we needto find a way to express the scalar distances x, y and z in terms of r. Again, we mayturn to simple trigonometry (see Figure 2.10). To begin with, we may resolve rp/orp/orp/ointo two components: one component lying parallel to ezezez, and another componentlying somewhere in the x− y plane. The component lying parallel to ezezez has a lengthr cos(θ); by definition, this length must be equal to z. The component lying in thex− y plane has length r sin(θ), and may then be further resolved into componentslying parallel with exexex and eyeyey, so that

x =(r sin(θ)

)cos(φ)

y =(r sin(θ)

)sin(φ)

z = r cos(θ)

Substituting this into our expression for rp/orp/orp/o,

2.2 Unit vectors 39

Fig. 2.10 Converting from spherical coordinates to Cartesian coordinates.

rp/orp/orp/o = rererer = r sin(θ)cos(φ)exexex + r sin(θ)sin(φ)eyeyey + r cos(θ)ezezez

It is important to stress, again, that the unit vector ererer and the angle θ in the sphericalcoordinate system are not the same as they are in the cylindrical coordinate system.

To convert from Cartesian coordinates into spherical coordinates, it is only nec-essary to recognize that the length r is simply the magnitude of rp/orp/orp/o, so that

r =√

x2 + y2 + z2

And, substituting this back into the vector equation,

rp/orp/orp/o = xexexex + yeyeyey + zezezez =√

x2 + y2 + z2 ererer

The angles θ and φ may also be obtained from simple trigonometry, as

θ = tan−1(√

x2 + y2

z

)

φ = tan−1(

yx

)


2.3 Time-derivatives

The velocity of a point p in space is its time rate of change of position. Since theposition of point p is expressed in the form of the vector rp/orp/orp/o, the velocity vector ofthe point p is then simply given by

vp/ovp/ovp/o =ddt

rp/orp/orp/o = rp/orp/orp/o

where vp/ovp/ovp/o is the velocity of the point p relative to the origin and coordinate systemo. In dynamics, because of the heavy usage of time-derivatives, it is common to use adot over a variable as a shorthand for the time-derivative. Similarly, the accelerationvector of the point p is the time-derivative of the velocity vector, so

ap/oap/oap/o =ddt

( ddt

rp/orp/orp/o)= rp/orp/orp/o

where ap/oap/oap/o is the acceleration of the point p relative to the origin and coordinatesystem o, and the two dots over the position vector are shorthand for the secondderivative with respect to time.

2.3.1 Cartesian coordinates

In the Cartesian coordinate system, determining the velocity and acceleration of apoint is a trivial matter. First, we may express the position vector in terms of the unitvectors exexex, eyeyey, and ezezez, as

rp/orp/orp/o = xexexex + yeyeyey + zezezez (2.1)

where x, y and z are the scalar distances along each of the exexex, eyeyey and ezezez directionswhich you would have to travel to get from the butt-end of rp/orp/orp/o to its tip. Taking thetime derivative of this equation, then (remembering to use the product rule),

vp/ovp/ovp/o = rp/orp/orp/o = xexexex + x exexex + yeyeyey + y eyeyey + zezezez + z ezezez (2.2)

and since the unit vectors exexex, eyeyey and ezezez never change in the Cartesian system,

exexex = eyeyey = ezezez = 0 (2.3)

Substituting Equation (2.3) into Equation (2.2) yields the result

vp/ovp/ovp/o = xexexex + yeyeyey + zezezez

Repeating the derivative according to the product rule and substituting in Equation(2.3) again, the acceleration of the point p is given as

2.3 Time-derivatives 41

ap/oap/oap/o = xexexex + yeyeyey + zezezez

So, if the point p moves around in space so that the x, y and z coordinates inthe Cartesian system change as known functions of time (for example, x = 3+ 2t,y = 2+ 4t2, z = t) then the velocity is simply the time derivative of each of thecomponents (in our example, then, vp/ovp/ovp/o(t) = 2exexex +8teyeyey +1ezezez), and the accelerationis similarly the second time derivative of each of the components (in our example,ap/oap/oap/o(t) = 0exexex +8eyeyey +0ezezez).

2.3.2 Cylindrical coordinates

In the cylindrical coordinate system, determining the velocity and acceleration is alittle more complicated. Again, we begin by expressing the position vector in termsof the unit vectors- this time ererer and ezezez, as

rp/orp/orp/o = rererer + zezezez (2.4)

where r and z are the scalar distances along the ererer and ezezez axes, respectively (recallthat the unit vector eθeθeθ is not required in order to describe the position of p, sinceererer always points in the ‘right’ direction). Once again, we take the time-derivative ofthe position vector rp/orp/orp/o by applying the product rule, so that

vp/ovp/ovp/o = rp/orp/orp/o = rererer + r ererer + zezezez + z ezezez (2.5)

Now there is a problem- because of the way the coordinate system is defined, ifthe point p moves, then so does the unit vector ererer, so ererer �= 0. The magnitude is fixedand constant, equal to 1- but the direction changes. So what is the time-derivative ofa unit vector? Unit vectors do not change if they are just moved around, accordingto the definition of a vector (see Section 2.1)- so the only change in the unit vectormust come from a rotation of the axes.

Fig. 2.11 Determining the time-derivative of a unit vector.


Consider the unit vector ererer as the point p moves around, as illustrated in Figure2.11. The change in ererer between times t and t +Δ t can be determined by the simpleaddition of the vectors. From the geometry, assuming Δθ is small,

ererer(t +Δ t)−ererer(t) = |ererer|sin(Δθ)eθeθeθ ≈ Δθeθeθeθ (2.6)

since the magnitude of a unit vector is 1, and the sine of a small angle is equal tothe angle itself. The direction in which the result ererer(t +Δ t)−ererer(t) points was foundgraphically to be along eθeθeθ . Then, to approximate the derivative as a finite differencein time, we can divide both sides of Equation (2.6) by Δ t, so

ddt

ererer ≈ ererer(t +Δ t)−ererer(t)Δ t

=

(Δθ eθeθeθ

)Δ t

=Δθ

Δ teθeθeθ

ddt

ererer ≈ θeθeθeθ

Note that the approximations become equalities as Δ t → 0.Here, the quantity θ = dθ/dt (or the time rate of change of an angle) is known as

an angular velocity and is often referred to using the symbol ω . An angular velocityis, in essence, the speed with which a thing is rotating, having SI units of radiansper second. An angular velocity vector ωωω may be defined such that the magnitude isθ and the direction is parallel to the axis of rotation, in a sense given by the right-hand rule (if you curl your fingers of your right hand along the path described bythe rotating object, your thumb points in the direction of the angular velocity vector:see Figure 2.12).

Similarly, the quantity θ = d2θ/dt2 is known as an angular acceleration, whichhas units of radians per second2 and is typically indicated with the symbol α . Theangular acceleration vector ααα has a magnitude of θ and a sense determined by theright-hand rule.

In our analysis of a rotating unit vector above, we showed that dererer/dt = θeθeθeθ . Bycarrying out a similar analysis for the case of three-dimensional rotation, it may bedemonstrated that,

The time derivative of a moving unit vector e1e1e1 is equal to the cross-product of theangular velocity vector of the axes ωaxωaxωax and the unit vector itself, or

ddt

e1e1e1 =ωaxωaxωax ×e1e1e1 (2.7)


Fig. 2.12 Illustration of using the right-hand rule to determine the sense of an angular velocityvector.

Equation (2.7) represents one of the most important results in classical mechan-ics, and features prominently in kinematics, relative motion and mechanism analy-sis.

Going back to Equation (2.5), we can now solve for the time-derivatives of theunit vectors as

vp/ovp/ovp/o = rererer + r(ωaxωaxωax ×ererer)+ zezezez + z ezezez

Since the direction of ezezez never changes, ezezez = 0, as with Cartesian co-ordinates.Also, the only rotation is about the z-axis, and the speed of rotation is simply θezezez

(the unit vectors ererer and eθeθeθ are both rotating about the z axis with a speed of θ as thepoint p moves), so this can be re-expressed as

vp/ovp/ovp/o = rererer + r(θezezez ×ererer)+ zezezez

Pulling the scalar quantities out of the vector operations yields,

vp/ovp/ovp/o = rererer + rθ(ezezez ×ererer)+ zezezez

Recognizing from the right-hand rule that (ezezez ×ererer) = eθeθeθ (see Figure 2.7; recall thatsince both vectors have unit magnitude the resultant vector of the cross product willalso have unit magnitude), this can be expressed as

vp/ovp/ovp/o = rererer + rθeθeθeθ + zezezez (2.8)

The acceleration can be determined in the same manner, though the algebra doesbecome more complex. We begin by differentiating Equation (2.8) with respect totime, again applying the product rule.


ap/oap/oap/o = vp/ovp/ovp/o = rererer + rererer + rθeθeθeθ + rθeθeθeθ + rθ eθeθeθ + zezezez + zezezez

Applying Equation (2.7), and recognizing again that ezezez = 0,

ap/oap/oap/o = rererer + r(ωaxωaxωax ×ererer)+ rθ eθeθeθ + rθ eθeθeθ + rθ(ωaxωaxωax ×eθeθeθ )+ zezezez

and since ωaxωaxωax = θezezez,

ap/oap/oap/o = rererer + r(θezezez ×ererer)+ rθ eθeθeθ + rθ eθeθeθ + rθ(θezezez ×eθeθeθ )+ zezezez

Pulling the scalars out of the vector operations yields,

ap/oap/oap/o = rererer + rθ(ezezez ×ererer)+ rθ eθeθeθ + rθ eθeθeθ + rθ 2(ezezez ×eθeθeθ )+ zezezez

Since (ezezez ×ererer) = eθeθeθ and (ezezez ×eθeθeθ ) =−ererer, this can be reduced down to

ap/oap/oap/o = rererer + rθ eθeθeθ + rθ eθeθeθ + rθ eθeθeθ − rθ 2 ererer + zezezez

Then, gathering together the coefficients of the various unit vectors,

ap/oap/oap/o = (r− rθ 2)ererer +(2rθ + rθ)eθeθeθ + zezezez (2.9)

Equation (2.9) is a fairly useful result when dealing with problems which lendthemselves well to the cylindrical coordinate system. It is important to note that,unless further time-derivatives are to be taken, the fact that the vectors ererer and eθeθeθ

are moving is no longer relevant. The acceleration yielded by this equation will besimply resolved into components along the ererer, eθeθeθ and ezezez unit vectors the way theywere oriented when the position vector rp/orp/orp/o was defined.

2.3.3 Spherical coordinates

Since the unit vector ererer always points in the ‘right’ direction in a spherical coordinatesystem, the position vector rp/orp/orp/o was shown to be simply,


The difficulties, on the other hand, arise when taking the derivative of that unitvector. As with the cylindrical coordinate system, the velocity of the point in thespherical coordinate system can be expressed as the time-derivative of the positionvector. Applying the product rule yields,


vp/ovp/ovp/o =ddt

rp/orp/orp/o = rererer + rererer (2.10)

where we recall again that the unit vectors ererer and eθeθeθ in the spherical coordinatesystem are not the same as the unit vectors ererer and eθeθeθ in the cylindrical coordi-nate system. In spherical coordinates, evaluating the derivative ererer is somewhat morecomplicated.

Let us refer again to the illustration in Figure 2.8. The unit vectors in sphericalcoordinates are all functions of the two angles θ and φ . Since both these anglescan change independently with time, there will therefore be two components to theangular velocity vector of the axes, ωaxωaxωax. If θ changes with time, the axes will havean angular velocity of θ about the eφeφeφ axis. If φ changes as well, the axes will havean additional angular velocity of φ about the vertical axis. However, the direction‘vertical’ is not defined in the spherical coordinate system; all we have are ererer, eθeθeθ andeφeφeφ , and none of these point straight up. We must therefore resolve φ into componentsalong ererer and eθeθeθ (eφeφeφ cannot contribute, since it always lies in a horizontal plane).Then,

ωaxωaxωax = θeφeφeφ − φ sin(θ)eθeθeθ + φ cos(θ)ererer

We may then evaluate the time derivative of the unit vector ererer in spherical coordi-nates as,

ererer = ωaxωaxωax ×ererer

ererer =(θeφeφeφ − φ sin(θ)eθeθeθ + φ cos(θ)ererer

)×ererer

Applying the distributive property of the cross-product,

ererer = θ(eφeφeφ ×ererer)− φ sin(θ)(eθeθeθ ×ererer)+ φ cos(θ)(ererer ×ererer)

recognizing that (eφeφeφ ×ererer) = eθeθeθ , (eθeθeθ ×ererer) = −eφeφeφ and that the cross-product of anyvector with itself is zero so (ererer ×ererer) = 0, this can be finally re-written as

ererer = θeθeθeθ + φ sin(θ)eφeφeφ

Substituting this expression back into Equation (2.10),

vp/ovp/ovp/o = rererer + rθeθeθeθ + rφ sin(θ)eφeφeφ

The acceleration may now be calculated as the time-derivative of velocity,

ap/oap/oap/o = vp/ovp/ovp/o =ddt

(rererer + rθeθeθeθ + rφ sin(θ)eφeφeφ

)


For simplicity, let us consider each of the terms of the derivative individually. Forthe first term,

ddt

rererer = rererer + rererer

= rererer + r(ωaxωaxωax ×ererer

)= rererer + r

((θeφeφeφ − φ sin(θ)eθeθeθ + φ cos(θ)ererer

)×ererer

)= rererer + rθ(eφeφeφ ×ererer)− rφ sin(θ)(eθeθeθ ×ererer)+ rφ cos(θ)(ererer ×ererer)

= rererer + rθeθeθeθ − rφ sin(θ)(−eφeφeφ )+ rφ cos(θ)(0)

ddt

rererer = rererer + rθeθeθeθ + rφ sin(θ)eφeφeφ

For the second term of the derivative,

ddt

rθeθeθeθ = rθeθeθeθ + rθeθeθeθ + rθ eθeθeθ

= rθeθeθeθ + rθeθeθeθ + rθ(ωaxωaxωax ×eθeθeθ

)= rθeθeθeθ + rθeθeθeθ + rθ


)×eθeθeθ

)= rθeθeθeθ + rθeθeθeθ + rθ 2(eφeφeφ ×eθeθeθ )− rθ φ sin(θ)(eθeθeθ ×eθeθeθ )

+rθ φ cos(θ)(ererer ×eθeθeθ )

= rθeθeθeθ + rθeθeθeθ + rθ 2(−ererer)− rθ φ sin(θ)(0)+ rθ φ cos(θ)eφeφeφ

ddt

rθeθeθeθ = rθeθeθeθ + rθeθeθeθ − rθ 2ererer + rθ φ cos(θ)eφeφeφ

Finally, for the third term of the derivative,

ddt

rφ sin(θ)eφeφeφ = rφ sin(θ)eφeφeφ + rφ sin(θ)eφeφeφ + rφ θ cos(θ)eφeφeφ + rφ sin(θ)eφeφeφ

= rφ sin(θ)eφeφeφ + rφ sin(θ)eφeφeφ + rφ θ cos(θ)eφeφeφ + rφ sin(θ)(ωaxωaxωax ×eφeφeφ

)= rφ sin(θ)eφeφeφ + rφ sin(θ)eφeφeφ + rφ θ cos(θ)eφeφeφ

+rφ sin(θ)


)×eφeφeφ

)= rφ sin(θ)eφeφeφ + rφ sin(θ)eφeφeφ + rφ θ cos(θ)eφeφeφ

+rφ θ sin(θ)(eφeφeφ ×eφeφeφ )− rφ 2 sin2(θ)(eθeθeθ ×eφeφeφ )

+rφ 2 sin(θ)cos(θ)(ererer ×eφeφeφ )


= rφ sin(θ)eφeφeφ + rφ sin(θ)eφeφeφ + rφ θ cos(θ)eφeφeφ

+rφ θ sin(θ)(0)− rφ 2 sin2(θ)ererer + rφ 2 sin(θ)cos(θ)(−eθeθeθ )

ddt

rφ sin(θ)eφeφeφ = rφ sin(θ)eφeφeφ + rφ sin(θ)eφeφeφ + rφ θ cos(θ)eφeφeφ

−rφ 2 sin2(θ)ererer − rφ 2 sin(θ)cos(θ)eθeθeθ

The results of the derivatives of the three terms may now be combined to find theacceleration, as

ap/oap/oap/o =ddt

rererer +ddt

rθeθeθeθ +ddt

rφ sin(θ)eφeφeφ

ap/oap/oap/o = rererer + rθeθeθeθ + rφ sin(θ)eφeφeφ

+rθeθeθeθ + rθeθeθeθ − rθ 2ererer + rθ φ cos(θ)eφeφeφ

+rφ sin(θ)eφeφeφ + rφ sin(θ)eφeφeφ + rφ θ cos(θ)eφeφeφ

−rφ 2 sin2(θ)ererer − rφ 2 sin(θ)cos(θ)eθeθeθ

We may now collect together the scalar coefficients of the three unit vectors to yieldthe result,

ap/oap/oap/o =(r− rθ 2 − rφ 2 sin2(θ)

)ererer

+(2rθ + rθ − rφ 2 sin(θ)cos(θ)

)eθeθeθ

+(2rφ sin(θ)+2rθ φ cos(θ)+ rφ sin(θ)

)eφeφeφ

2.3.4 Normal-tangential coordinates

The normal-tangential or generalized coordinate system is another means of speci-fying the position, velocity and acceleration of a point p in three-dimensional space.This particular system is fairly exotic, and has few applications outside of pure math-ematics and certain branches of spacecraft dynamics and particle physics. The sys-tem is only defined when the point p has a given trajectory or path to follow. Inthis system, the position vector rp/orp/orp/o is specified in terms of the unit vector enenen, whichhas its butt-end on the point p and points directly away from the centre of the localradius of curvature of the trajectory, so it is locally perpendicular to the trajectory.The unit vector etetet, which is locally tangent to the trajectory, has its butt-end on p andpoints in the direction of motion (see Figure 2.13). Together, enenen and etetet define a planein space (called the osculating plane); since the trajectory is three-dimensional, this


plane can wobble around in space. The unit vector ebebeb is always normal to this plane.The sense of ebebeb is defined by the expression,

ebebeb = etetet ×enenen

Fig. 2.13 Normal-tangential or generalized coordinate system.

To understand how a local radius of curvature is defined, consider a very smallsegment of a path in three-dimensional space. If the segment is small enough, itcan be approximated as a small circular arc. This circular arc has a radius and acentre, though the centre could be floating anywhere in space. As the segment lengthdecreases, the approximation becomes more and more accurate until, as the segmentlength approaches zero, the path is perfectly defined by a circular arc.

Consider a point p following a path S in three-dimensional space. The locationof p may then be given as


in much the same way as was the case for spherical coordinates, but bearing in mindthat this is a different r altogether, and that the point o (the centre of the local radiusof curvature) is moving as well. Then, since etetet is defined as being in the direction ofmotion, the velocity of point p is trivially

vp/ovp/ovp/o = rp/orp/orp/o = Setetet (2.11)

where S is just the scalar speed of p. The acceleration can then be determined bydifferentiation, applying the product rule, as

ap/oap/oap/o = vp/ovp/ovp/o = Setetet + Setetet (2.12)

The derivative of the unit vector can be evaluated by again using Equation (2.7), so

etetet =ωaxωaxωax ×etetet = θebebeb ×etetet = θenenen (2.13)


where θ is the angular velocity of the vector rp/orp/orp/o. However, since the position of thepoint p has been determined by assuming that it was following a path defined by aninfinite series of infinitesimal circular arcs, the velocity along the arc can be foundfrom the radius of curvature and θ , as

θ =Sr

(2.14)

Substituting Equations (2.14) into (2.13) and then into Equation (2.12) yields theresult

ap/oap/oap/o = Setetet +S2

renenen (2.15)

Now, to determine the radius of curvature at a given point in the trajectory (givenonly the equation of the trajectory S(t), or some other equivalent expression), rec-ognize from Equations (2.13) and (2.14) that

etetet =Sr

enenen

or, equivalently

enenen

r=

ddt etetetddt S

=deeet

dS(2.16)

but, from Equation (2.11),

ddt

rp/orp/orp/o =dSdt

etetet

so, rearranging,

etetet =ddt rp/orp/orp/o

ddt S

=d

dSrp/orp/orp/o (2.17)

Substituting Equation (2.17) into (2.16),

enenen

r=

ddS

( ddS

rp/orp/orp/o)=

d2

dS2 rp/orp/orp/o

Looking at the magnitudes only, and recognizing that |enenen|= 1, we can solve for r as

r =1∣∣∣ d2

dS2 rp/orp/orp/o

∣∣∣Also, it can be shown that, for a trajectory S lying in the x− y plane such that S isdefined by the function y(x), the radius of curvature r is


r =

(1+( dy

dx

)2)3/2

d2ydx2

While it may seem arcane and unnecessarily complicated, the normal-tangentialcoordinate system is useful for describing motion when there is no convenient fixedobserver or fixed frame of reference. Consider, for example, a deep-space probe.Using the Earth as an ‘origin’ in its navigation calculations would not be practical,since the Earth also moves. The only reliable data available to the probe from its on-board sensors are the magnitudes of its own speed and acceleration S and S, whichit may determine from accelerometers and star-tracking cameras.

2.4 Relative motion

It is often necessary to determine the velocity or acceleration of one moving pointrelative to another moving point. Not only is relative motion analysis useful in aero-nautical and astronautical target-tracking applications, but it is also an importantstep in the dynamic analysis of any mechanism.

Fig. 2.14 Graphical representation of the fixed coordinate system o, the moving coordinate systemo′ and the position vectors rp/orp/orp/o, rp/o′rp/o′rp/o′ and ro′/oro′/oro′/o.

Consider a Cartesian coordinate system (defined by the unit vectors exexex, eyeyey andezezez) stuck to the origin o, which is fixed in space so that the unit vectors are alwayspointing in the same directions (See Figure 2.14). Now, consider a second Cartesiancoordinate system defined by the unit vectors e′xe′xe′x, e′ye′ye′y and e′ze′ze′z stuck to the origin o′ suchthat o′ is allowed to move around in space, and e′xe′xe′x, e′ye′ye′y and e′ze′ze′z are allowed to rotatearound so that they do not always point in the same direction. The location of thepoint p may still be defined in terms of the position vector rp/orp/orp/o, but it can equally bedefined in terms of the position vector rp/o′rp/o′rp/o′ , meaning the position of p relative to thecoordinate system o′. The position vector rp/o′rp/o′rp/o′ has its butt-end stuck to the movingorigin o′ and has its tip stuck to p. A third position vector, ro′/oro′/oro′/o, may similarly bedefined to describe the location of the moving origin o′ relative to the fixed origin o.

2.4 Relative motion 51

Since the orientation of the unit vectors stuck to both o and o′ is arbitrary, let usdefine the direction indicated by exexex as pointing from o to o′ at one particular instantin time, and let us define the direction indicated by e′xe′xe′x as pointing from o′ to p atthat same particular instant in time. We do this only to simplify the algebra that isto follow; the results would be identical regardless of which direction we chose fore′xe′xe′x, once all the sines and cosines were sorted out.

Considering the vector diagram in Figure 2.14 and comparing to the definition ofvector addition illustrated in Figure 2.1, it can be seen that

rp/orp/orp/o = ro′/oro′/oro′/o +rp/o′rp/o′rp/o′

Expressing this in terms of the unit vectors exexex and e′xe′xe′x (at our particular momentin time when everything just happens to line up),

rp/orp/orp/o = ro′/o exexex + rp/o′ e′xe′xe′x (2.18)

where ro′/o and rp/o′ are the magnitudes of the vectors ro′/oro′/oro′/o and rp/o′rp/o′rp/o′ , respectively(recall that non-boldface symbols indicate the magnitude of the vectors). The veloc-ity of the point p in space can then be expressed in terms of the position and velocityof the reference frame o′ by taking the time derivative of Equation (2.18), applyingthe product rule as before.

vp/ovp/ovp/o = ro′/o exexex + ro′/o exexex + rp/o′ e′xe′xe′x + rp/o′ e′xe′xe′x

Since o is fixed in space, exexex = 0. Substituting in Equation (2.7),

vp/ovp/ovp/o = ro′/o exexex + rp/o′ e′xe′xe′x + rp/o′(ωaxωaxωax ×e′xe′xe′x)

where ωaxωaxωax is the angular velocity of the moving unit vectors, and is therefore theangular velocity of the moving reference frame o′. Now, we can recognize thatro′/o exexex = vo′/ovo′/ovo′/o, and also rp/o′ e′xe′xe′x = vp/o′vp/o′vp/o′ . Then,

vp/ovp/ovp/o = vo′/ovo′/ovo′/o +vp/o′vp/o′vp/o′ +ωaxωaxωax × rp/o′ e′xe′xe′x

where the scalar rp/o′ has been moved inside the cross product. Finally, sincerp/o′ e′xe′xe′x = rp/o′rp/o′rp/o′ , the velocity can be expressed entirely in vector form as

vp/ovp/ovp/o = vo′/ovo′/ovo′/o +vp/o′vp/o′vp/o′ +ωaxωaxωax ×rp/o′rp/o′rp/o′

This important result allows us to compute the velocity vector vp/ovp/ovp/o of a point inspace knowing the translational velocity of a moving reference frame vo′/ovo′/ovo′/o, the an-gular velocity of that moving reference frame ωaxωaxωax, and the position rp/o′rp/o′rp/o′ and velocityvp/o′vp/o′vp/o′ of the point with respect to that moving reference frame.


The acceleration of the point in space may likewise be expressed in terms ofthe velocity and acceleration of the moving reference frame o′. The velocity wasdetermined in terms of the unit vectors exexex and e′xe′xe′x as,

vp/ovp/ovp/o = ro′/o exexex + rp/o′ e′xe′xe′x +ωaxωaxωax × rp/o′ e

′xe′xe′x

The acceleration can then be computed as the time-derivative of the velocity. Ap-plying the product rule,

ap/oap/oap/o = vp/ovp/ovp/o = ro′/o exexex + ro′/o exexex + rp/o′ e′xe′xe′x + rp/o′ e

′xe′xe′x

+ωaxωaxωax × rp/o′ e′xe′xe′x +ωaxωaxωax × (rp/o′ e

′xe′xe′x + rp/o′ e

′xe′xe′x)

Again, exexex = 0. Substituting in Equation (2.7) yields,

ap/oap/oap/o = ro′/o exexex + rp/o′ e′xe′xe′x + rp/o′(ωaxωaxωax ×e′xe′xe′x)+ωaxωaxωax × rp/o′ e

′xe′xe′x

+ωaxωaxωax ×(rp/o′ e

′xe′xe′x + rp/o′(ωaxωaxωax ×e′xe′xe′x)

)

Distributing the cross-products and moving the scalars into the products yields,

ap/oap/oap/o = ro′/o exexex + rp/o′ e′xe′xe′x +ωaxωaxωax × rp/o′ e

′xe′xe′x +ωaxωaxωax × rp/o′ e

′xe′xe′x

+ωaxωaxωax × rp/o′ e′xe′xe′x +ωaxωaxωax ×ωaxωaxωax × rp/o′ e

′xe′xe′x

Once again, by definition, ro′/o exexex = ao′/oao′/oao′/o and rp/o′ e′xe′xe′x = ap/o′ap/o′ap/o′ . If we define the an-gular acceleration of the axes as αaxαaxαax = ωaxωaxωax, the above expression can be reducedto

ap/oap/oap/o = ao′/oao′/oao′/o +ap/o′ap/o′ap/o′ +2ωaxωaxωax ×vp/o′vp/o′vp/o′

+αaxαaxαax ×rp/o′rp/o′rp/o′ +ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

This is another extremely important result, and each of the terms of this equationhas a special significance. The term (2ωaxωaxωax×vp/o′vp/o′vp/o′) is called the Coriolis acceleration,as it was first described by the French mathematician Gaspard-Gustave Coriolis(1792-1843). To understand the nature of the Coriolis acceleration, consider a pointon a rotating disc which is moving in a straight line away from the centre of thedisc. As the distance between the point and the centre of the disc increases, thelinear (tangential) speed of the point must also increase in order to keep up with thedisc. This increase in speed is the Coriolis acceleration. The Coriolis acceleration


also plays an important role in meteorology, as it affects how the air moves aroundthe Earth.

The term (ωaxωaxωax ×ωaxωaxωax × rp/o′rp/o′rp/o′) is the centripetal acceleration, and represents theinward acceleration felt by an object moving in a circular path at a constant speed;this is the same acceleration that causes a pilot to experience ‘g’s when executing abanked turn at constant airspeed. It should be noted that cross-products are alwayscarried out from right to left, so that

ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′ =ωaxωaxωax ×(ωaxωaxωax ×rp/o′rp/o′rp/o′

)(otherwise, ωaxωaxωax ×ωaxωaxωax = 0 and the centripetal acceleration would always be zero, bydefinition).

The quantity αaxαaxαax also requires some further discussion. We defined αaxαaxαax as thetime rate of change of ωωωax, while ωωωax may have both a magnitude and a directionwhich could change. Let us consider the case of an object spinning about any fixedaxis, so that

ωωωax = ω eieiei

where eieiei may be pointing in any direction, but does not change with time. Takingthe time-derivative of this angular velocity through the application of the productrule and Equation (2.7), then,

ωaxωaxωax = ω eieiei +ω eieiei

= ω eieiei +ω(ω eieiei ×eieiei

)= ω eieiei +ω2 (eieiei ×eieiei

)= ω eieiei +ω2 (0)

ωaxωaxωax = ω eieiei

Where we have used Equation (2.7). So, for the case of a rotation about any fixedaxis, the angular acceleration vector αaxαaxαax is in the same direction as the angularvelocity vector ωaxωaxωax and has a magnitude equal to the time rate of change of themagnitude of ωaxωaxωax. However, for cases where no single axis of rotation can be defined(or eieiei is moving), then αaxαaxαax will need to be evaluated as ω eieiei +ω eieiei.

2.4.1 Universality of relative motion equations

While the Cartesian, cylindrical and spherical coordinate systems provide a con-venient means to describe the motion of a point under specific circumstances, therelative motion equations may always be used, regardless of the manner in which the


point is moving. As a demonstration, let us consider a point p in three-dimensionalspace (a view of the x− y axis is illustrated in Figure 2.15). At the instant shown,the point p is some distance r from the origin o, at an angle θ from the x-axis.

Fig. 2.15 Demonstration of the universality of the relative motion equations.

Let us define a moving reference frame that has its origin o′ always lying onthe z-axis (and therefore always directly above or below the fixed origin o), and e′xe′xe′xalways points toward p. Then, the velocity of the point will be


Since o′ can only move vertically, vo′/ovo′/ovo′/o = ze′ze′ze′z. Then, since the point p will alwayslie on the e′xe′xe′x axis,

vp/ovp/ovp/o = ze′ze′ze′z +vp/o′vp/o′vp/o′ +ωaxωaxωax ×rp/o′rp/o′rp/o′

= ze′ze′ze′z + re′xe′xe′x +ω ezezez × re′xe′xe′x= ze′ze′ze′z + re′xe′xe′x +ωr(e′ze′ze′z ×e′xe′xe′x)

vp/ovp/ovp/o = ze′ze′ze′z + re′xe′xe′x +ωre′ye′ye′y

Here, we have assumed that the point is rotating about the origin (which is actuallythe ezezez axis) with an angular velocity of ω . Note that since the o′ coordinate systemis rotating in the x− y plane, e′ze′ze′z is always pointing in the same direction as ezezez.

Now, there are three important observations to be made about this result. First,since we defined e′xe′xe′x as always pointing from o to p, then, by definition, e′xe′xe′x = ererer in thecylindrical coordinate system with its origin on o. Second, the angular velocity ω isjust the time rate of change of θ , so ω = θ . Finally, e′ye′ye′y always points tangentiallyanti-clockwise (if the unit vector is translated over so that its butt-end lies on p).


Therefore, by definition, e′ye′ye′y = eθeθeθ . Substituting these observations into the relativemotion equation,

vp/ovp/ovp/o = ze′ze′ze′z + re′xe′xe′x +ωre′ye′ye′yvp/ovp/ovp/o = zezezez + rererer + rθeθeθeθ

which we can recognize immediately as the velocity in a cylindrical coordinate sys-tem. Similarly, we can obtain the acceleration from the relative motion equations.Again, we note that ao′/oao′/oao′/o must be along ezezez in the moving coordinate system whichwe defined, and that the point p is free only to move along the e′xe′xe′x axis.


+αaxαaxαax ×rp/o′rp/o′rp/o′ +ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

= ze′ze′ze′z + re′xe′xe′x +2ω e′ze′ze′z × re′xe′xe′x+ω e′ze′ze′z × re′xe′xe′x +ω e′ze′ze′z ×ω e′ze′ze′z × re′xe′xe′x

= ze′ze′ze′z + re′xe′xe′x +2rω(e′ze′ze′z ×e′xe′xe′x)+rω(e′ze′ze′z ×e′xe′xe′x)+ω e′ze′ze′z × rω(e′ze′ze′z ×e′xe′xe′x)

= ze′ze′ze′z + re′xe′xe′x +2rω e′ye′ye′y+rω e′ye′ye′y +ω e′ze′ze′z × rω e′ye′ye′y

ap/oap/oap/o = ze′ze′ze′z + re′xe′xe′x +2rω e′ye′ye′y+rω e′ye′ye′y − rω2 e′xe′xe′x

Recalling again that, given the way in which we defined our moving coordinatesystem, e′xe′xe′x = ererer, e′ye′ye′y = eθeθeθ , e′ze′ze′z = ezezez, ω = θ and ω = θ ,

ap/oap/oap/o = ze′ze′ze′z + re′xe′xe′x +2rω e′ye′ye′y + rω e′ye′ye′y − rω2 e′xe′xe′x= zezezez + rererer +2rθ eθeθeθ + rθ eθeθeθ − rθ 2 ererer

ap/oap/oap/o = (r− rθ 2)ererer +(2rθ + rθ)eθeθeθ + zezezez

This is exactly the result obtained from the cylindrical coordinate system.


2.5 Examples

2.5.1 Computing velocities

A particle is following a trajectory in three-dimensional space along a path whichcan be described as

rp/orp/orp/o = β t cos(2πt)exexex +β t sin(2πt)eyeyey + γtezezez

where β and γ are scalars, and t is time. Compute the velocity of the particle as afunction of time in both Cartesian and cylindrical coordinates, and show that themagnitude of the velocity is the same in both cases.

Fig. 2.16 Computing velocities

SOLUTION

The velocity of a particle in Cartesian coordinates is

vp/ovp/ovp/o = xexexex + yeyeyey + zezezez

evaluating the derivatives,

x = −2πβ t sin(2πt)+β cos(2πt)

y = 2πβ t cos(2πt)+β sin(2πt)

z = γ

2.5 Examples 57

so the velocity is

vp/ovp/ovp/o =(−2πβ t sin(2πt)+β cos(2πt)

)exexex

+(2πβ t cos(2πt)+β sin(2πt)

)eyeyey

+ γezezez

and the magnitude of the velocity is

∣∣vp/ovp/ovp/o

∣∣ =√x2 + y2 + z2

=((−2πβ t sin(2πt)+β cos(2πt)

)2

+(2πβ t cos(2πt)+β sin(2πt)

)2+ γ2

)1/2

=(

4π2β 2t2 sin2(2πt)+β 2 cos2(2πt)−4πβ 2t sin(2πt)cos(2πt)

+4π2β 2t2 cos2(2πt)+β 2 sin2(2πt)+4πβ 2t sin(2πt)cos(2πt)+ γ2)1/2

=((

β 2 +(2πβ t)2)(sin2(2πt)+ cos2(2πt))+ γ2

)1/2

∣∣vp/ovp/ovp/o

∣∣ =√β 2 +(2πβ t)2 + γ2

Now, to convert to cylindrical coordinates, we know that

θ = tan−1(

yx

)

so we can solve for θ as

θ = tan−1(

yx

)

= tan−1(

β t sin(2πt)β t cos(2πt)

)= tan−1(tan(2πt)

)θ = 2πt

and we can solve for r from the expression

xexexex + yeyeyey + zezezez =(√

x2 + y2)ererer + zezezez


Solving for r,

r =√

x2 + y2

=

√(β t cos(2πt)

)2+(β t sin(2πt)

)2

=√

β 2t2(cos2(2πt)+ sin2(2πt)

)=√

β 2t2

r = β t

so the position vector can be expressed as

rp/orp/orp/o = β tererer + γtezezez

Then, the velocity can be determined according to the equation

vp/ovp/ovp/o = rererer + rθeθeθeθ + zezezez

Evaluating the derivatives,

r = β

θ = 2π

z = γ

Substituting,vp/ovp/ovp/o = βererer +(β t)(2π)eθeθeθ + γezezez

The magnitude of the velocity can then be computed as

∣∣vp/ovp/ovp/o

∣∣ =√r2 +(rθ)2 + z2

=

√β 2 +

((β t)(2π)

)2+ γ2

∣∣vp/ovp/ovp/o

∣∣ =√β 2 +(2πβ t)2 + γ2

which was the same result as obtained by using the Cartesian system, and was ar-rived at with considerable more ease.

2.5 Examples 59

2.5.2 Generalized coordinates

A rocket is fired from a stationary launch pad at (x,y) = (0,0). The rocket followsa path described by y = βx2, where y is the altitude, x is the horizontal distancefrom the launch pad, and β is a constant. The rocket telemetry shows that it isaccelerating constantly along its trajectory, so that the magnitude of its velocity isaS, where S is the distance it has traveled along its trajectory (which is not straight),and a is a constant. Compute the acceleration of the rocket as a function of x.

Fig. 2.17 Generalized coordinates

SOLUTION

Since we know things about the scalar speed of the rocket, this problem is a goodcandidate for the use of generalized coordinates. The acceleration in generalizedcoordinates is


renenen

The first thing we need is S. We already have S = aS, which is the magnitude of itsvelocity. For this, we can manipulate some derivatives:

S =dSdt

=dSdS

dSdt

=dSdS

S = aS

Substituting in the equation for S supplied,

S = aS = a(aS) = a2S

Since we need to find the acceleration as a function of x, we need to express this interms of x. To begin with, let’s look at some small segment of the path. The lengthof the segment dS can be expressed in terms of dx and dy as


dS =√

dx2 +dy2

=

√√√√dx2

(1+

(dydx

)2)

dS =

√1+

(dydx

)2

dx

but since we know that y = βx2,

dydx

= 2βx

and so

dS =√

1+(2βx)2 dx

S =

∫ √1+(2βx)2 dx+C

S2β

=∫ √(

12β

)2

+ x2 dx+C

where C is a constant of integration. We can look up the solution to this integralusing a table of integrals, and find that, for independent variable x and constant α ,∫ √

α2 + x2 dx =x2

√α2 + x2 +

α

2ln(x+√

α2 + x2)

recognizing for this problem that α = 1/(2β ),

S2β

=x2

√(1

2β

)2

+ x2 +12

(1

2β

)ln

(x+

√(1

2β

)2

+ x2

)+C

S = βx

√(1

2β

)2

+ x2 +12

ln

(x+

√(1

2β

)2

+ x2

)+C

Since we know that S = 0 at x = 0 (the rocket was launched from the pad, so S = 0there), by substituting x = 0 into this expression for S and requiring that S = 0, wecan solve for C as

2.5 Examples 61

0 = β (0)

√(1

2β

)2

+(0)2 +12

ln

((0)+

√(1

2β

)2

+(0)2

)+C

0 =12

ln

(√(1

2β

)2)+C

C = −12

ln

(1

2β

)

Substituting C back into the expression for S yields,

S = βx

√(1

2β

)2

+ x2 +12

ln

(x+

√(1

2β

)2

+ x2

)− 1

2ln

(1

2β

)

S = βx

√(1

2β

)2

+ x2 +12

ln

(2βx+

√1+4β 2x2

)

The last thing we need now is an expression for the radius of curvature. Since weknow that

r =

(1+( dy

dx

)2)3/2

d2ydx2

and, since the function y(x) is already known, we can solve for the derivatives easily:

dydx

= 2βx

d2ydx2 = 2β

so,

r =

(1+( dy

dx

)2)3/2

d2ydx2

=(1+4β 2x2)3/2

2β

And finally, assembling the terms in the acceleration equation,


renenen


= a2Setetet +a2S2

renenen

= a2

(βx

√(1

2β

)2

+ x2 +12

ln

(2βx+

√1+4β 2x2

))etetet

+2βa2

(1+4β 2x2)3/2

(βx

√(1

2β

)2

+ x2 +12

ln

(2βx+

√1+4β 2x2

))2

enenen

2.5.3 Relative motion: bug on a disc

A bug is walking on a disc rotating about its centre at a constant angular velocityof ω . The bug sets out from the disc’s hub, and walks straight outward at a constantspeed vbug toward the rim of the disc. Compute the velocity and acceleration of thebug as seen by a stationary observer, as a function of the distance r between the bugand the centre of the disc.

Fig. 2.18 Bug on a disc

SOLUTION

First, we have to define our origins and coordinate systems. Since we know thevelocity of the bug relative to the disc, it would make sense to have the movingreference frame o′ stuck to the disc. If o′ is stuck to the disc so that e′ze′ze′z passes throughthe centre of the disc, this will simplify the geometry of the problem. Similarly, thefixed reference frame o can be positioned so that the fixed axis ezezez also passes throughthe centre of the disc, so that e′ze′ze′z =ezezez. The other moving unit vectors, e′xe′xe′x and e′ye′ye′y shouldbe stuck to the disc so that e′xe′xe′x always points from the hub to the bug; this way, wewon’t need to use sines and cosines to describe the position of the bug. Rememberwhen orienting your axes that, by convention, exexex ×eyeyey = ezezez and ezezez ×eyeyey =−exexex.

2.5 Examples 63

Then, the velocity of the bug relative to a fixed observer vp/ovp/ovp/o is,


Now, we consider each of the terms in the above. The term vo′/ovo′/ovo′/o represents thevelocity of o′ relative to the fixed observer. However, though the o′ axes are rotating,the hub is not actually moving and o′ is always right on top of o. Therefore, vo′/ovo′/ovo′/o = 0.

Next, we consider vp/o′vp/o′vp/o′ . This represents the velocity of the bug relative to thedisc, which we know; the bug’s walking speed is provided in the problem descrip-tion, and the direction is always in a straight line along what we have defined as e′xe′xe′x.Therefore, vp/o′vp/o′vp/o′ = vbuge′xe′xe′x.

The angular velocity vector of the moving axes, ωaxωaxωax is also known. The movingaxes are stuck to the disc, and the angular velocity of the disc is provided in theproblem description. The magnitude of the angular velocity of the disc is givenas ω , and the direction is always along ezezez. The position vector of the bug relativeto the moving reference frame has a magnitude of r (as specified by the problemstatement), and, according to our definition, is always in the e′xe′xe′x direction, so rp/o′rp/o′rp/o′ =re′xe′xe′x. Then,


= 0+ vbug e′xe′xe′x +ωr (ezezez ×e′xe′xe′x)vp/ovp/ovp/o = vbug e′xe′xe′x +ωre′ye′ye′y

Since we can only describe the velocity of the bug at a single instant in time, wemight as well ensure that we define o′ such that the moving axes e′xe′xe′x and e′ye′ye′y lie righton top of the fixed axes exexex and eyeyey at that moment in time. Then, at that moment only,e′xe′xe′x = exexex and e′ye′ye′y = eyeyey, and

vp/ovp/ovp/o = vbug exexex +ωreyeyey

We were asked to determine the velocity relative to the fixed observer, so the resultscannot be presented in terms of some arbitrary moving unit vectors which we createdfor our own convenience. Because of how we defined o and o′, the conversion fromthe moving frame to the fixed one was very simple. Had we stuck o′ to the discsomewhere other than right on top of o, or had e′xe′xe′x pointed in a direction other thandirectly at the bug, there would have been a lot of sines and cosines involved inconverting from e′xe′xe′x and e′ye′ye′y to exexex and eyeyey, but the result (expressed in the universal,fixed coordinates) would have been exactly the same. This equivalence shall bedemonstrated later.

The acceleration of the bug relative to a fixed observer may be expressed as,



+ αaxαaxαax ×rp/o′rp/o′rp/o′ +ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

Again, we will look at each of these terms individually. First, ao′/oao′/oao′/o represents theacceleration of the moving origin o′ relative to the fixed reference frame o. Since oand o′ always lie one on top of the other, ao′/oao′/oao′/o = 0.

The term ap/o′ap/o′ap/o′ represents the acceleration of the bug relative to the moving frame(which is stuck to the disc). Since the bug is walking with constant velocity alongthe surface of the disc, ap/o′ap/o′ap/o′ = 0 as well.

In the term 2ωaxωaxωax ×vp/o′vp/o′vp/o′ , we have already determined that ωωωax = ω ezezez, and thatvp/o′vp/o′vp/o′ = vbug e′xe′xe′x.

The angular acceleration of the o′ reference frame, αaxαaxαax, is known to be zerobecause the rotational velocity of the reference frame is constant. Therefore, αaxαaxαax ×rp/o′rp/o′rp/o′ = 0.

Re-assembling these terms,

ap/oap/oap/o = 0+0+2ωvbug (ezezez ×e′xe′xe′x)+0+ω ezezez ×(ωr (ezezez ×e′xe′xe′x)

)= 2ωvbug e′ye′ye′y +ω ezezez ×ωre′ye′ye′y= 2ωvbug e′ye′ye′y +ω2r (ezezez ×e′ye′ye′y)

ap/oap/oap/o = 2ωvbug e′ye′ye′y −ω2re′xe′xe′x

If we again consider the instant when e′xe′xe′x = exexex and e′ye′ye′y = eyeyey, we can give the result,

ap/oap/oap/o = 2ωvbug eyeyey −ω2rexexex

The first term is the Coriolis acceleration, which disappears if vbug → 0. The secondterm is the centripetal acceleration, and is pointing in the −exexex direction (toward thehub).

Alternatively, we may also have determined the velocity and acceleration of thebug as a function of position (and indirectly, then, as a function of time). Ratherthan assuming that e′xe′xe′x is aligned with exexex at the ‘moment of interest’, we may equallysolve the problem when e′xe′xe′x is oriented at some arbitrary angle θ relative to the fixedcoordinate system, as illustrated in Figure 2.19.

Once again we will define a fixed coordinate system o with its origin at the centreof the disc. We shall define ezezez as being normal to the disc, pointing upwards; exexex andeyeyey will be fixed in space and oriented as illustrated. We will define a moving coordi-nate system o′ with its origin always coincident with o, but with its unit vectors e′xe′xe′xand e′ye′ye′y stuck to the surface of the disc and oriented so that e′xe′xe′x always points towardthe bug, as before. Once again ezezez is always parallel with e′ze′ze′z.

The velocity of the bug relative to a fixed observer is then,

2.5 Examples 65

Fig. 2.19 Example 4b


Once again, since the centre of the disc is not moving, vo′/ovo′/ovo′/o = 0. The e′xe′xe′x unit vectoralways points toward the bug, so rp/o′rp/o′rp/o′ = re′xe′xe′x. Similarly, since o′ is stuck to the disc,vp/o′vp/o′vp/o′ is just the velocity of the bug relative to the surface of the disc, so vp/o′vp/o′vp/o′ =vbug e′xe′xe′x. Substituting these expressions into the velocity equation, then,

vp/ovp/ovp/o = 0+ vbug e′xe′xe′x +ω ezezez × re′xe′xe′x

(Again, since o′ is stuck to the disc, ωaxωaxωax is simply the angular velocity ω of thedisc). We are now left with an equation with mixed ‘prime’ and ‘non-prime’ unitvectors; in order to carry out the addition and cross product, all of the unit vectorsmust be in the same coordinate system. We will therefore resolve everything into theo′ coordinate system (this choice was arbitrary; we could equally resolve everythinginto the o system). Since ezezez is always parallel with e′ze′ze′z, these two are interchangeable,and so

vp/ovp/ovp/o = vbug e′xe′xe′x +ω e′ze′ze′z × re′xe′xe′x= vbug e′xe′xe′x +ωr

(e′ze′ze′z ×e′xe′xe′x

)vp/ovp/ovp/o = vbug e′xe′xe′x +ωre′ye′ye′y

Now, to express this result from the point of view of a fixed observer, we mustresolve the moving unit vectors into components parallel to the fixed unit vectors.From the geometry, we see that

e′xe′xe′x = |e′xe′xe′x|cos(θ)exexex + |e′xe′xe′x|sin(θ)eyeyey

e′ye′ye′y = −|e′ye′ye′y|sin(θ)exexex + |e′ye′ye′y|cos(θ)eyeyey


e′ze′ze′z = |e′ze′ze′z|ezezez

Where it should be noted again that unit vectors have unit magnitude by definition,so this result reduces to

e′xe′xe′x = cos(θ)exexex + sin(θ)eyeyey

e′ye′ye′y = −sin(θ)exexex + cos(θ)eyeyey

e′ze′ze′z = ezezez

Substituting this into our earlier result,

vp/ovp/ovp/o = vbug e′xe′xe′x +ωre′ye′ye′y= vbug

(cos(θ)exexex + sin(θ)eyeyey

)+ωr

(−sin(θ)exexex + cos(θ)eyeyey)

vp/ovp/ovp/o =(vbug cos(θ)−ωr sin(θ)

)exexex +

(vbug sin(θ)+ωr cos(θ)

)eyeyey

Alternatively, we could have first resolved our expression for vp/ovp/ovp/o into compo-nents along exexex and eyeyey and then carried out the cross-product:

vp/ovp/ovp/o = vbug e′xe′xe′x +ω e′ze′ze′z × re′xe′xe′xvp/ovp/ovp/o = vbug


)+ω ezezez × r


)(2.19)

Distributing the cross-product,

vp/ovp/ovp/o = vbug(cos(θ)exexex + sin(θ)eyeyey

)+ωr cos(θ)

(ezezez ×exexex

)+ωr sin(θ)

(ezezez ×eyeyey

)vp/ovp/ovp/o = vbug


)+ωr cos(θ)

(eyeyey)+ωr sin(θ)

(−exexex)

Gathering terms,

vp/ovp/ovp/o =(vbug cos(θ)−ωr sin(θ)

)exexex +

(vbug sin(θ)+ωr cos(θ)

)eyeyey

which is exactly the same result as arrived at earlier.

Next, for the acceleration of the bug,

2.5 Examples 67



Since the centre of the disc is not moving, ao′/oao′/oao′/o = 0. Also, the bug is running withconstant speed relative to the disc, so ap/o′ap/o′ap/o′ = 0. The disc itself is rotating with con-stant speed, and since our moving axes are stuck to the disc, αaxαaxαax = 0 as well. Then,

ap/oap/oap/o = 0+0+2ωaxωaxωax ×vp/o′vp/o′vp/o′ +0+ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

ap/oap/oap/o = 2ω ezezez × vbug e′xe′xe′x +ω ezezez ×ω ezezez × re′xe′xe′x

Substituting ezezez = e′ze′ze′z, we can ensure that all of our unit vectors are in the same coor-dinate system so that the vector addition and multiplication may be carried out.

ap/oap/oap/o = 2ωvbug(e′ze′ze′z ×e′xe′xe′x

)+ω e′ze′ze′z ×ωr


)= 2ωvbug e′ye′ye′y +ω e′ze′ze′z ×ωre′ye′ye′y= 2ωvbug e′ye′ye′y +ω2r

(e′ze′ze′z ×e′ye′ye′y

)ap/oap/oap/o = 2ωvbug e′ye′ye′y −ω2re′xe′xe′x

Again, this must be expressed in components along the fixed unit vectors; substitut-ing, then,

ap/oap/oap/o = 2ωvbug e′ye′ye′y −ω2re′xe′xe′x= 2ωvbug

(−sin(θ)exexex + cos(θ)eyeyey)−ω2r


)ap/oap/oap/o =

(−2ωvbug sin(θ)−ω2r cos(θ))

exexex +(2ωvbug cos(θ)−ω2r sin(θ)

)eyeyey

Notice that for the special case of θ = 0, the earlier results (from when we assumedthat the moving unit vectors were parallel with the fixed unit vectors at the momentof interest) are recovered.


2.5.4 Relative motion: bug on a wheel

A cylindrical wheel of radius R is rolling along the ground without slipping, havinga constant angular velocity ω . Meanwhile, a bug walks along the surface of thewheel from the rim towards the hub with constant velocity vb relative to the wheel.Solve for the velocity and acceleration of the bug relative to a fixed observer whenthe bug is a distance b directly over the hub, as illustrated.

Fig. 2.20 Bug on a wheel

SOLUTION

To solve this problem, we will first define a moving coordinate system o′ whichis stuck to the hub of the wheel, rotating and translating with the wheel such that thee′ye′ye′y is always pointing toward the bug. We will then define a fixed coordinate systemo which is fixed in space but is instantaneously coincident with o′ at the instant ofinterest. If we define the point p as being the location of the bug, then


The first problem, then, is to determine the velocity vo′/ovo′/ovo′/o of the centre of thedisc relative to a fixed observer. To do this, let us define the point q as being theinstantaneous point of contact between the disc and the ground, as shown. We maythen apply the relative velocity equation to the point q, as,

vq/ovq/ovq/o = vo′/ovo′/ovo′/o +vq/o′vq/o′vq/o′ +ωaxωaxωax ×rq/o′rq/o′rq/o′

The point q may be considered equally as being either on the disc or on the ground.Two important observations may be made as a consequence: first, since the ground is

2.5 Examples 69

not moving, then the point q must be instantaneously at rest and vq/ovq/ovq/o = 0. Second,since the disc is not changing shape, and since both points q and o′ are stuck tothe disc, vq/o′vq/o′vq/o′ = 0. The velocity of the centre of the disc, vo′/ovo′/ovo′/o, remains unknown.Because the o′ coordinate system is stuck to the disc, the angular velocity of the axeswill be the same as the angular velocity of the disc, so ωaxωaxωax = −ω ezezez (the angularvelocity is negative because +ezezez points upward out of the page). Finally, since thepoint q lies directly underneath o′, rq/o′rq/o′rq/o′ =−Reyeyey. Substituting,


0 = vo′/ovo′/ovo′/o +0+(−ω ezezez

)× (−Reyeyey)

0 = vo′/ovo′/ovo′/o +ωR(ezezez ×eyeyey

)0 = vo′/ovo′/ovo′/o −ωRexexex

vo′/ovo′/ovo′/o = ωRexexex

Now that we have an expression for vo′/ovo′/ovo′/o, we may return to our expression forthe velocity of the point p.


Since vp/o′vp/o′vp/o′ represents the velocity of the bug relative to o′ and o′ is stuck to thesurface of the disc, vp/o′vp/o′vp/o′ =−vb e′ye′ye′y. Again, because o′ is stuck to the disc and rotatingwith the disc, the axes of the coordinate system will have the same angular velocityas the disc, and ωaxωaxωax =−ω ezezez. Finally, the position vector rp/o′rp/o′rp/o′ is simply the distancebetween the bug and the hub, as we defined e′ye′ye′y to always point toward the bug, sorp/o′rp/o′rp/o′ = b e′ye′ye′y.

Substituting these expressions into the relative velocity equation, recognizingthat at the moment of interest e′xe′xe′x = exexex and e′ye′ye′y = eyeyey (and that e′ze′ze′z = ezezez always),


= Rω exexex − vb e′ye′ye′y −ω ezezez ×b e′ye′ye′y= Rω exexex − vb eyeyey −ωb

(ezezez × eyeyey

)vp/ovp/ovp/o = Rω exexex − vb eyeyey +ωb exexex

Collecting terms, then,

vp/ovp/ovp/o = ω(R+b) exexex − vb eyeyey

Next, we consider the expression for relative acceleration,




Because the disc is moving at constant speed, we can recognize immediately thatao′/oao′/oao′/o = 0 and αaxαaxαax = 0. Furthermore, since the bug is running across the disc with aconstant velocity relative to the disk, we also recognize that ap/o′ap/o′ap/o′ = 0. Consequently,

ap/oap/oap/o = 0+0+2ωaxωaxωax ×vp/o′vp/o′vp/o′ +0+ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

= −2ω ezezez ×−vb e′ye′ye′y −ω ezezez ×−ω ezezez ×b e′ye′ye′y= 2ωvb

(ezezez ×eyeyey

)−ω ezezez ×−ωb(ezezez ×eyeyey

)= −2ωvb exexex −ω ezezez ×ωb exexex

= −2ωvb exexex −ω2b(ezezez ×exexex

)ap/oap/oap/o = −2ωvb exexex −ω2b eyeyey

Which is the result required.

2.5.5 Angular velocity and axes of rotation

Two points p and q are located a distance rp and rq from the centre c of a disc ro-tating with angular velocity ω , as illustrated. An observer is sitting over the point pand is translating along with the point p, but is not rotating with the disc. Demon-strate that the velocity vrelvrelvrel of the point q relative to the observer does not change,whether the disc is rotating about any of the points c, p or q.

SOLUTION

It should be pointed out here that the relative velocity vrelvrelvrel represents the velocityof the point q relative to an observer who is being dragged along with the disc overthe point p, but who is not rotating with the disc. If the observer had been rotatingwith the disc, the point q would always appear to be stationary.

First, let us consider the case where the disc is rotating about its centre. Let usdefine a moving coordinate system o′ which is stuck to the disc with the origin atits centre, and with the moving unit vectors e′xe′xe′x, e′ye′ye′y and e′ze′ze′z rotating with the disc. Atsome instant in time, the moving coordinates will be aligned momentarily with thefixed coordinates exexex, eyeyey and ezezez illustrated; at that instant, then, the velocity of thepoint p can be given by the expression,


2.5 Examples 71

Fig. 2.21 Angular velocity and axes of rotation

The fixed origin o can be anywhere. Since the centre of the disc is not moving,vo′/ovo′/ovo′/o = 0. Also, since both o′ and p are stuck to the disc, p cannot move within theo′ coordinate system and vp/o′vp/o′vp/o′ = 0. Then,

vp/ovp/ovp/o = 0+0+ω e′ze′ze′z × rp e′xe′xe′x= ωrp


)vp/ovp/ovp/o = ωrp e′ye′ye′y

This is the velocity of our observer. Similarly, the velocity of the point q is

vq/ovq/ovq/o = 0+0+ω e′ze′ze′z × rq e′xe′xe′x= ωrq


)vq/ovq/ovq/o = ωrq e′ye′ye′y

The velocity of the point q relative to our observer is then

vrelvrelvrel = vq/ovq/ovq/o −vp/ovp/ovp/o

=(ωrq e′ye′ye′y

)− (ωrp e′ye′ye′y)

vrelvrelvrel = ω(rq − rp

)e′ye′ye′y


Next, let us consider the case where the disc is rotating about the point p. We willnow stick our moving coordinate system o′ to the disc with the origin at p. Onceagain, we will consider the instant in time when e′xe′xe′x, e′ye′ye′y and e′ze′ze′z are aligned with exexex, eyeyey

and ezezez. Then, the velocity of the point q is given by,


Again, the origin o′ is fixed in space so vo′/ovo′/ovo′/o = 0. Again, since both o′ and q arestuck to the same disc, vq/o′vq/o′vq/o′ = 0. Then,

vq/ovq/ovq/o = 0+0+ωaxωaxωax ×rq/o′rq/o′rq/o′

= ω e′ze′ze′z ×(rq − rp

)e′xe′xe′x

= ω(rq − rp

)(e′ze′ze′z ×e′xe′xe′x

)vq/ovq/ovq/o = ω

(rq − rp

)e′ye′ye′y

We note that since the disc is rotating about p, that point is fixed in space by defini-tion; therefore, our observer is at rest and vp/ovp/ovp/o = 0. The velocity of q relative to theobserver is then,


=(ω(rq − rp

)e′ye′ye′y)− (0)


)e′ye′ye′y

as expected. Finally, let us consider the case where the disc is rotating about thepoint q. This time, we stick our moving coordinate system o′ to the disc with theorigin at q, and will consider the instant in time when the moving unit vectors arealigned with the fixed unit vectors. Then, the velocity of the point p will be givenby,


Since the point q is now fixed in space, vo′/ovo′/ovo′/o = 0. Also, since both p and q are againstuck to the same disc, vp/o′vp/o′vp/o′ = 0. Then,

vp/ovp/ovp/o = 0+0+ωaxωaxωax ×rp/o′rp/o′rp/o′

= ω e′ze′ze′z ×(rq − rp

)(−e′xe′xe′x)

= −ω(rq − rp

)(e′ze′ze′z ×e′xe′xe′x

)vp/ovp/ovp/o = −ω

(rq − rp

)e′ye′ye′y

2.5 Examples 73

This time, the point q is fixed in space and so vq/ovq/ovq/o = 0. The velocity of q relative tothe observer is then,


=(0)− (−ω

(rq − rp

)e′ye′ye′y)


)e′ye′ye′y

which is, again, the same relative velocity observed earlier.

This example has demonstrated an important concept: an angular velocity vectorcontains information about the magnitude and direction of rotation only; no infor-mation about the axis of rotation is provided by the vector. Since the velocity ofthe point q as seen by our observer depended only on the angular velocity vector,the location of the axis of rotation was irrelevant. As we noted earlier, vectors canalways be translated around in space without affecting their magnitude or direction.

2.5.6 Relative motion in three dimensions

Our poor bug has now found itself clinging for dear life to the tip of a wind turbineblade. The blades are rotating around the hub with an angular velocity of ω , andat the same time, the turbine shaft is being yawed into the wind with a constantangular velocity of Ω . The distance between the blade hub and the yaw axis is L,and the blade radius is R. Find the velocity and acceleration of the bug (relative toa stationary observer) as a function of the angle θ subtended between the turbineblade and the horizon.

SOLUTION

As always, the first thing to do is to define the coordinate systems. There are a fewdifferent systems which could be used to solve the problem with equal ease. Here,we will select the fixed reference frame o at the hub of the yawing axis, right at theintersection of the two axes of rotation. We will define eyeyey as pointing upwards, andezezez as pointing out toward the turbine hub along the turbine axle. The moving frameo′ we shall locate at the hub of the blades but with the axes stuck to the generator, sothat the o′ reference frame yaws with the generator, but does not rotate around theblade axis. We shall set e′ye′ye′y parallel to eyeyey, and let e′ze′ze′z always point along the turbineaxis toward the hub. As we have defined our coordinates, the angular velocity of our


Fig. 2.22 Relative motion in three dimensions

moving axes contains only one component of the rotation; the axes are not spinningwith the blades, so

ωaxωaxωax = Ω eyeyey

The velocity of the bug vp/ovp/ovp/o relative to a stationary observer can then be given as,


This time, none of the terms vanish. Let us address each one in turn, beginning withvo′/ovo′/ovo′/o. This term represents the velocity of the turbine blade hub relative to the fixedreference frame at the yaw hub, so

vo′/ovo′/ovo′/o = ωaxωaxωax ×ro′/oro′/oro′/o

= Ωeyeyey ×Lezezez

= ΩL(eyeyey ×ezezez)

vo′/ovo′/ovo′/o = ΩLexexex

Here, we have assumed that the turbine axis will be lying along the z-axis at themoment of interest; the problem did not require us to solve as a function of theyawing angle. However, it would have been just as easy to let ro′/oro′/oro′/o = Lsin(φ)exexex +Lcos(φ)ezezez. Next, the term vp/o′vp/o′vp/o′ represents the velocity of the bug with respect to ourreference frame which is rotating along with the generator (but not with the blades).To solve for this term, we recognize that, within the moving reference frame, the bug

2.5 Examples 75

is simply moving around a circular path of radius R at a constant angular velocityω; this situation is identical to the earlier example of the bug on the disc.

vp/o′vp/o′vp/o′ = ωe′ze′ze′z ×rp/o′rp/o′rp/o′

= ωe′ze′ze′z ×(Rcos(θ)e′xe′xe′x +Rsin(θ)e′ye′ye′y

)= ωR

(e′ze′ze′z × (cos(θ)e′xe′xe′x + sin(θ)e′ye′ye′y)

)vp/o′vp/o′vp/o′ = ωR

(cos(θ)e′ye′ye′y − sin(θ)e′xe′xe′x

)

Here, the position of the bug relative to the moving frame was resolved into compo-nents lying parallel with the e′xe′xe′x and e′ye′ye′y unit vectors. Finally, we can examine the lastterm in the relative velocity equation, recalling that, from the definition of our axes,eyeyey = e′ye′ye′y.

ωaxωaxωax ×rp/o′rp/o′rp/o′ = Ωeyeyey ×(Rcos(θ)e′xe′xe′x +Rsin(θ)e′ye′ye′y

)= ΩRe′ye′ye′y ×

(cos(θ)e′xe′xe′x + sin(θ)e′ye′ye′y

)ωaxωaxωax ×rp/o′rp/o′rp/o′ = −ΩRcos(θ)e′ze′ze′z

Pulling all the terms together, then,


= ΩLexexex +ωR(cos(θ)e′ye′ye′y − sin(θ)e′xe′xe′x

)−ΩRcos(θ)e′ze′ze′zvp/ovp/ovp/o =

(ΩL−ωRsin(θ)

)exexex +ωRcos(θ)e′ye′ye′y −ΩRcos(θ)e′ze′ze′z

Again, we can arbitrarily choose that the o axes and o′ axes are coincident at themoment of interest, so that e′xe′xe′x = exexex, e′ye′ye′y = eyeyey and e′ze′ze′z = ezezez. Then,

vp/ovp/ovp/o =(ΩL−ωRsin(θ)

)exexex +ωRcos(θ)eyeyey −ΩRcos(θ)ezezez

which gives us the velocity of the bug relative to a fixed observer in a stationaryreference frame, as required. Next, the acceleration of the bug is given by the ex-pression




Again, we’ll go through each of these terms one at a time. First, ao′/oao′/oao′/o is the acceler-ation of the moving origin o′ relative to the fixed frame. Since the point o′ is simplymoving around in a circle at constant angular velocity, it is experiencing centripetalacceleration only.

ao′/oao′/oao′/o = ωaxωaxωax ×ωaxωaxωax ×ro′/oro′/oro′/o

= Ωeyeyey ×Ωeyeyey ×Lezezez

= Ωeyeyey ×ΩL(eyeyey ×ezezez)

= Ωeyeyey ×ΩLexexex

= Ω 2L(eyeyey ×exexex)

ao′/oao′/oao′/o = −Ω 2Lezezez

Next, consider the term ap/o′ap/o′ap/o′ . This term represents the acceleration of the bug rela-tive to the moving reference frame. Since the bug is moving in a circle at constantspeed within the moving reference frame, it is experiencing a relative centripetalacceleration.

ap/o′ap/o′ap/o′ = ωe′ze′ze′z ×ωe′ze′ze′z ×(Rcos(θ)e′xe′xe′x +Rsin(θ)e′ye′ye′y

)= ωe′ze′ze′z ×ωRe′ze′ze′z ×


)= ωe′ze′ze′z ×ωR


)= ω2Re′ze′ze′z ×


)ap/o′ap/o′ap/o′ = ω2R

(−cos(θ)e′xe′xe′x − sin(θ)e′ye′ye′y)

which, reassuringly, indicates that the centripetal acceleration of the bug within themoving frame is in the direction of −rp/o′rp/o′rp/o′ . In this example, we were able to solveboth the ao′/oao′/oao′/o and ap/o′ap/o′ap/o′ terms essentially by inspection; however, the same resultmay have been obtained had we defined a point q coincident with o′, and solved forvq/o = vo′/o and aq/o = ao′/o as a separate, two-dimensional relative motion problem.

Next, we’ll consider the Coriolis term; thankfully, we’ve already worked out thevelocity of the bug relative to the moving frame. Recalling that eyeyey = e′ye′ye′y,

2ωaxωaxωax ×vp/o′vp/o′vp/o′ = 2Ωeyeyey ×ωR(cos(θ)e′ye′ye′y − sin(θ)e′xe′xe′x

)= 2ΩωReyeyey ×


)2ωaxωaxωax ×vp/o′vp/o′vp/o′ = 2ΩωRsin(θ)e′ze′ze′z

2.5 Examples 77

The angular acceleration term, αaxαaxαax × rp/o′rp/o′rp/o′ , is zero because the angular velocity ofthe moving axes does not change (it is of constant magnitude and always in thesame direction); that is, αaxαaxαax = 0. While the blades themselves experience an angularacceleration, recall that the axes are not stuck to the blades, but are only rotatingwith the hub. This leaves only the centripetal term,

ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′ = Ωeyeyey ×Ωeyeyey ×(Rcos(θ)e′xe′xe′x +Rsin(θ)e′ye′ye′y

)= Ωeyeyey ×ΩReyeyey ×


)= Ωeyeyey ×−ΩRcos(θ)e′ze′ze′z= −Ω 2Rcos(θ)(eyeyey ×e′ze′ze′z)

ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′ = −Ω 2Rcos(θ)exexex

Finally, we can gather together all of these terms, recognizing that at our moment ofinterest, the axes of o may be coincident with the axes of o′:

ap/oap/oap/o = ao′/oao′/oao′/o +ap/o′ap/o′ap/o′ +2ωaxωaxωax ×vp/o′vp/o′vp/o′ +αaxαaxαax ×rp/o′rp/o′rp/o′ +ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

= −Ω 2Lezezez +ω2R(−cos(θ)e′xe′xe′x − sin(θ)e′ye′ye′y

)+2ΩωRsin(θ)e′ze′ze′z

+0−Ω 2Rcos(θ)exexex

=(−ω2Rcos(θ)−Ω 2Rcos(θ)

)exexex −ω2Rsin(θ)eyeyey

+(−Ω 2L+2ΩωRsin(θ)

)ezezez

ap/oap/oap/o = −Rcos(θ)(Ω 2 +ω2)exexex −ω2Rsin(θ)eyeyey +(2ΩωRsin(θ)−Ω 2L

)ezezez

which is the acceleration of our intrepid bug relative to an observer in a fixed refer-ence frame as a function of the blade sweep angle, as required.

ALTERNATIVE SOLUTION

An alternative and equally valid approach would be to define a fixed origin oat the hub of the yawing axis as before, with eyeyey pointing upwards and ezezez alwayspointing along the turbine shaft towards the turbine hub. Then, a moving coordinateframe o′ can be defined with its origin on the blade hub such that e′ze′ze′z always pointsalong ezezez, and e′xe′xe′x always points at the bug. Defined in this way, the o′ coordinatesystem will rotate with the hub as before, but will also spin with the blades.

As always, the velocity of the bug vp/ovp/ovp/o relative to a stationary observer can begiven as,



Fig. 2.23 Relative motion in three dimensions: alternative solution

The term vo′/ovo′/ovo′/o represents the velocity of the turbine blade hub relative to a fixedobserver, which is the same whether or not the moving reference frame rotates withthe blades. Therefore, as before,

vo′/ovo′/ovo′/o = ΩLexexex

Since the bug is stuck to the turbine blade, and since the moving reference frame o′is also stuck to the turbine blade, then the bug is fixed within the moving o′ referenceframe (the position of the bug relative to the moving reference frame rp/o′rp/o′rp/o′ is alwaysR e′xe′xe′x). If the position of the bug in the moving reference frame does not change, thenvp/o′vp/o′vp/o′ = 0.

Since o′ is now stuck to the turbine blades, the angular velocity of the referenceframe now has two components: one component as a result of the rotation of theturbine hub about the vertical axis, and one component resulting from the rotationof the turbine blades about their axle. These angular velocities may be simply addedtogether, as

ωaxωaxωax = Ω eyeyey +ω e′ze′ze′z

It is important to note here that the ω component is about the moving e′ze′ze′z axis and notthe fixed ezezez axis, even if the two happen to be coincident at the moment of interest.Then, the velocity of the bug relative to a fixed observer can be expressed as


vp/ovp/ovp/o = ΩLexexex +0+(Ω eyeyey +ω e′ze′ze′z

)×R e′xe′xe′x

2.5 Examples 79

However, a cross-product cannot be easily evaluated between two different referenceframes; instead, let us express e′xe′xe′x and e′ze′ze′z in terms of exexex, eyeyey and ezezez. At the instantillustrated, from geometry,

e′xe′xe′x = cos(θ) exexex + sin(θ) eyeyey

e′ye′ye′y = −sin(θ) exexex + cos(θ) eyeyey


Where, as before, it was assumed that e′ze′ze′z and ezezez are coincident at the moment ofinterest, since we are not interested in resolving the components as a function of theyaw angle. Substituting this result into the above and carrying out the cross-product,

vp/ovp/ovp/o = ΩLexexex +(Ω eyeyey +ω e′ze′ze′z

)×R e′xe′xe′xvp/ovp/ovp/o = ΩLexexex +

(Ω eyeyey +ω ezezez

)×R(cos(θ) exexex + sin(θ) eyeyey

)

Distributing the cross-products and collecting scalar coefficients,

vp/ovp/ovp/o = ΩLexexex +ΩRcos(θ)(eyeyey ×exexex

)+ΩRsin(θ)

(eyeyey ×eyeyey

)+ ωRcos(θ)

(ezezez ×exexex

)+ωRsin(θ)

(ezezez ×eyeyey

)

Evaluating the cross-products,

vp/ovp/ovp/o = ΩLexexex −ΩRcos(θ) ezezez +ΩRsin(θ)(0)

+ ωRcos(θ) eyeyey −ωRsin(θ) exexex

Finally, by collecting terms, the velocity relative to a fixed observer may be ex-pressed as,

vp/ovp/ovp/o =(ΩL−ωRsin(θ)

)exexex +ωRcos(θ)eyeyey −ΩRcos(θ)ezezez

which is the same result obtained previously.To obtain the acceleration of the bug relative to a fixed observer, we begin again

with the general expression,



Once again, we will evaluate each of these terms one at a time. As before, o′ ismoving around in a circle, it is experiencing centripetal acceleration only.

ao′/oao′/oao′/o =−Ω 2Lezezez

Since the position of the bug relative to the moving reference frame rp/o′rp/o′rp/o′ is con-stant, the bug cannot be accelerating within the moving reference frame (simplyput, d/dt(vp/o′vp/o′vp/o′) = 0). Therefore, ap/o′ap/o′ap/o′ = 0.

Let us now consider the angular acceleration of the reference frame, αaxαaxαax. Fromthe definition,

αaxαaxαax = ωaxωaxωax

=ddt

(Ω eyeyey +ω e′ze′ze′z

)αaxαaxαax = Ω eyeyey +Ω eyeyey + ω e′ze′ze′z +ω e′ze′ze′z

Since eyeyey does not change direction, eyeyey = 0. Also, the magnitudes Ω and ω areconstant, so

αaxαaxαax = 0+0+0+ω e′ze′ze′z

Since e′ze′ze′z changes direction with time, the time derivative will be nonzero; usingEquation (2.7),

αaxαaxαax = ω e′ze′ze′z= ω

(ωaxωaxωax ×e′ze′ze′z

)= ω

((Ω eyeyey +ω e′ze′ze′z

)×e′ze′ze′z

)= ωΩ

(eyeyey ×ezezez

)+ω2(e′ze′ze′z ×e′ze′ze′z

)αaxαaxαax = ωΩ exexex +ω2(0)

recalling that e′ze′ze′z = ezezez at the instant of interest. Then,

αaxαaxαax ×rp/o′rp/o′rp/o′ = ωΩ exexex ×R e′xe′xe′x= ωΩ exexex ×R

(cos(θ) exexex + sin(θ) eyeyey

)= ωΩRcos(θ)

(exexex ×exexex

)+ωΩRsin(θ)

(exexex ×eyeyey

)= ωΩRcos(θ)(0)+ωΩRsin(θ) ezezez

αaxαaxαax ×rp/o′rp/o′rp/o′ = ωΩRsin(θ) ezezez

2.5 Examples 81

Finally, evaluating the centripetal term,

ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

=(Ω eyeyey +ω e′ze′ze′z

)×((Ω eyeyey +ω e′ze′ze′z)×R


))


)×(ΩRcos(θ)(eyeyey ×exexex

)+ΩRsin(θ)

(eyeyey ×eyeyey

)+ωRcos(θ)

(e′ze′ze′z ×exexex

)+ωRsin(θ)

(e′ze′ze′z ×eyeyey

))


)×(−ΩRcos(θ) ezezez +0+ωRcos(θ) eyeyey −ωRsin(θ) exexex

)= −Ω 2Rcos(θ)

(eyeyey ×ezezez

)+ΩωRcos(θ)

(eyeyey ×eyeyey

)−ΩωRsin(θ)(eyeyey ×exexex

)−ΩωRcos(θ)

(ezezez ×ezezez

)+ω2Rcos(θ)

(ezezez ×eyeyey

)−ω2Rsin(θ)(ezezez ×exexex

)= −Ω 2Rcos(θ) exexex +ωRcos(θ)(0)+ΩωRsin(θ) ezezez

−ΩωRcos(θ)(0)−ω2Rcos(θ) exexex −ω2Rsin(θ) eyeyey

Gathering terms,

ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′ =−Rcos(θ)(Ω 2 +ω2) exexex −ω2Rsin(θ) eyeyey +ΩωRsin(θ) ezezez

Now all of the above results may be substituted back into the acceleration equation:


= −Ω 2Lezezez +0+2ωaxωaxωax × (0)+ωΩRsin(θ) ezezez

+

(−Rcos(θ)

(Ω 2 +ω2) exexex −ω2Rsin(θ) eyeyey +ΩωRsin(θ) ezezez

)ap/oap/oap/o = −Rcos(θ)

(Ω 2 +ω2) exexex −ω2Rsin(θ) eyeyey +

(2ΩωRsin(θ)−Ω 2L

)ezezez

which is exactly the same result obtained using the previous method. As was demon-strated, the solution obtained for the velocity and acceleration relative to a fixedreference frame will not depend on the selection of the reference systems.


2.5.7 Constrained motion: bug on a ladder

Just as it thought it had escaped and settled for a quiet, retired life far from anyproblems in dynamics, our poor bug finds itself yet again clinging for dear life onto a moving object. This time, the bug is stuck on a ladder which is being pusheddown from a wall (as shown) so that the top end of the ladder slides along the wallwith a constant velocity v. The bug, frozen with terror, is holding stationary on theladder at a height ξ L up from the bottom. Find the velocity and acceleration of thebug relative to a fixed observer, as a function of the angle θ subtended between theladder and the floor.

Fig. 2.24 Bug on a ladder

SOLUTION

This time, let’s stick o at the junction between the floor and the wall so that exexex

points along the floor and eyeyey points up the wall. Then, we’ll stick o′ to the top of theladder (since the velocity of that point is known, we’ll already have vo′/ovo′/ovo′/o) so thatthe axes are fixed to the ladder and e′xe′xe′x points down the ladder, e′ye′ye′y points up and tothe right (as shown), and so ezezez must point out of the page. The velocity of the bugrelative to a fixed observer vp/ovp/ovp/o is


-but this time there is a problem. The angular velocity of the ladder (to which ourmoving axes are stuck) ωaxωaxωax is still unknown. We do know, however, that the bottomof the ladder is constrained in that it can only move sideways. So, since we havesome information about the velocity of the bottom of the ladder (the direction, if not

2.5 Examples 83

the magnitude), we shall begin by considering a point q stuck to the bottom of theladder. Then,


In this equation, we know the velocity vector of the origin o′ relative to o, vo′/ovo′/ovo′/o, is−v eyeyey. We also know that vq/o′vq/o′vq/o′ = 0 since the end of the ladder stays in the sameplace relative to the ladder (the ladder isn’t stretching or shrinking). We also knowthat the position vector pointing from o′ to q, rq/o′rq/o′rq/o′ , is Le′xe′xe′x. Finally, we know that theladder is rotating with some unknown angular velocity ω , but the direction is in theezezez direction. Putting this together,

vq/ovq/ovq/o =−veyeyey +0+ω ezezez ×Le′xe′xe′x

Now, in order to carry out the cross products more easily, it would help to convertall of the vectors to being in one reference frame. Resolving e′xe′xe′x, e′ye′ye′y and e′ze′ze′z into exexex, eyeyey

and ezezez,

e′xe′xe′x = cos(θ)exexex − sin(θ)eyeyey

e′ye′ye′y = sin(θ)exexex + cos(θ)eyeyey


Substituting this back into the previous result,

vq/ovq/ovq/o = −veyeyey +0+ω ezezez ×L(cos(θ)exexex − sin(θ)eyeyey

)= −veyeyey +ωLezezez ×

(cos(θ)exexex − sin(θ)eyeyey

)= −veyeyey +ωL

(cos(θ)eyeyey + sin(θ)exexex

)vq/ovq/ovq/o =

(ωLcos(θ)− v

)eyeyey +ωLsin(θ)exexex

This is an interesting result: since we know that the point q can only move along exexex,the scalar coefficient of eyeyey here must be equal to zero, and

vq/ovq/ovq/o = 0eyeyey + vq/o exexex

Substituting,

0eyeyey + vq/o exexex =(ωLcos(θ)− v

)eyeyey +ωLsin(θ)exexex

Equating the scalar coefficients of eyeyey,


ωLcos(θ)− v = 0

ω =v

Lcos(θ)

Since we’re not really interested in solving for the magnitude vq/o, there is no needto equate the scalar coefficients of exexex. Now that we have solved for ωaxωaxωax, we can goback to solving for the position of the bug.


= −veyeyey +0+v

Lcos(θ)ezezez × (L−ξ L)e′xe′xe′x

= −veyeyey +v

Lcos(θ)(L−ξ L)ezezez ×e′xe′xe′x

= −veyeyey +v

cos(θ)(1−ξ )ezezez ×


)vp/ovp/ovp/o = −veyeyey +

vcos(θ)

(1−ξ )(cos(θ)eyeyey + sin(θ)exexex

)

collecting together the terms and reducing, this yields our result,

vp/ovp/ovp/o = (1−ξ )v tan(θ)exexex −ξ veyeyey

which is the velocity of the bug as a function of the angle subtended between theladder and the ground, as required by the problem statement. Some quick checks tosee if this makes sense: if ξ = 1, the bug is sitting at the top of the ladder, and itsvelocity must then be −veyeyey; this is satisfied. If the bug is at the bottom of the ladder,ξ = 0 and the velocity of the bug must equal the velocity of q,

vq/ovq/ovq/o = 0eyeyey +ωLsin(θ)exexex

which, after some quick algebra, works out as well.

For the acceleration, we begin with the relative acceleration equation, startingonce again with an analysis of the point q in order to solve for the unknown angularacceleration of the ladder.

aq/oaq/oaq/o = ao′/oao′/oao′/o +aq/o′aq/o′aq/o′ +2ωaxωaxωax ×vq/o′vq/o′vq/o′ +αaxαaxαax ×rq/o′rq/o′rq/o′ +ωaxωaxωax ×ωaxωaxωax ×rq/o′rq/o′rq/o′

Since the velocity of o′ is constant down the wall, ao′/oao′/oao′/o = 0. Once again, becausethe ladder doesn’t stretch or shrink (the length remains the same), aq/o′aq/o′aq/o′ = vq/o′vq/o′vq/o′ = 0.Substituting,

aq/oaq/oaq/o = 0+0+2ωaxωaxωax ×0+α e′ze′ze′z ×L e′xe′xe′x +ω e′ze′ze′z ×ω e′ze′ze′z ×L e′xe′xe′x

2.5 Examples 85

= αL(e′ze′ze′z ×e′xe′xe′x

)+ω e′ze′ze′z ×ωL


)= αL e′ye′ye′y +ω e′ze′ze′z ×ωL e′ye′ye′y= αL e′ye′ye′y +ω2L


)aq/oaq/oaq/o = αL e′ye′ye′y −ω2L e′xe′xe′x

Resolving e′xe′xe′x and e′ye′ye′y into components along exexex and eyeyey (using our earlier result),

aq/oaq/oaq/o = αL(sin(θ)exexex + cos(θ)eyeyey

)−ω2L(cos(θ)exexex − sin(θ)eyeyey

)aq/oaq/oaq/o =

(αLsin(θ)−ω2Lcos(θ)

)exexex +

(αLcos(θ)+ω2Lsin(θ)

)eyeyey

Once again, we know that aq/oaq/oaq/o is constrained to move horizontally; the scalar coef-ficient of eyeyey must therefore vanish. Solving,

0 = αLcos(θ)+ω2Lsin(θ)

α = −ω2 sin(θ)cos(θ)

Now that α is known, we can solve for the acceleration of the point p, as


Since the bug is not moving on the ladder, ap/o′ap/o′ap/o′ = vp/o′vp/o′vp/o′ = 0. Substituting this intothe acceleration equation (along with some earlier results),

ap/oap/oap/o = 0+0+2ωaxωaxωax ×0+α e′ze′ze′z × (L−ξ L) e′xe′xe′x +ω e′ze′ze′z ×ω e′ze′ze′z × (L−ξ L) e′xe′xe′x= α(L−ξ L)


)+ω e′ze′ze′z ×ω(L−ξ L)


)= α(L−ξ L) e′ye′ye′y +ω e′ze′ze′z ×ω(L−ξ L) e′ye′ye′y= α(L−ξ L) e′ye′ye′y +ω2(L−ξ L)


)ap/oap/oap/o = α(L−ξ L) e′ye′ye′y −ω2(L−ξ L) e′xe′xe′x

Resolving back into fixed coordinates,

ap/oap/oap/o = α(L−ξ L)(sin(θ)exexex + cos(θ)eyeyey

)−ω2(L−ξ L)(cos(θ)exexex − sin(θ)eyeyey

)ap/oap/oap/o =

(α(L−ξ L)sin(θ)−ω2(L−ξ L)cos(θ)

)exexex


+(α(L−ξ L)cos(θ)+ω2(L−ξ L)sin(θ)

)eyeyey

Substituting in the known value of α ,

ap/oap/oap/o =

(−ω2 sin(θ)

cos(θ)(L−ξ L)sin(θ)−ω2(L−ξ L)cos(θ)

)exexex

+

(−ω2 sin(θ)

cos(θ)(L−ξ L)cos(θ)+ω2(L−ξ L)sin(θ)

)eyeyey

ap/oap/oap/o = ω2(L−ξ L)

(− sin(θ)

cos(θ)sin(θ)− cos(θ)

)exexex

+ω2(L−ξ L)

(− sin(θ)

cos(θ)cos(θ)+ sin(θ)

)eyeyey

ap/oap/oap/o = −ω2(L−ξ L)

(sin2(θ)

cos(θ)+

cos2(θ)

cos(θ)

)exexex

+ω2(L−ξ L)

(−sin(θ)+ sin(θ)

)eyeyey

ap/oap/oap/o = −ω2(L−ξ L)

(1

cos(θ)

)exexex +0 eyeyey

We may now substitute in our previous result for ω ,

ap/oap/oap/o = −(

vLcos(θ)

)2

(L−ξ L)

(1

cos(θ)

)exexex +0 eyeyey

ap/oap/oap/o = − v2

cos3(θ)

1−ξ

Lexexex

Alternatively, because all the terms in our earlier expression for the velocity vp/ovp/ovp/owere known, we could have arrived at the same result by differentiating. From thedefinition of acceleration,

ap/oap/oap/o =dvp/ovp/ovp/o

dt=

dvp/ovp/ovp/o

dθ

dθ

dt

= −ωdvp/ovp/ovp/o

dθ

= −(

vLcos(θ)

)d

dθ

((1−ξ )v tan(θ)exexex −ξ veyeyey

)

= −(

vLcos(θ)

)((1−ξ )v

1cos2(θ)

exexex −0eyeyey

)

2.5 Examples 87

ap/oap/oap/o = − v2

cos3(θ)

1−ξ

Lexexex

which was exactly the same result as obtained earlier. Note that, because of theway we have defined the axes in our problem, if θ increases with time, ω willbe in the −ezezez direction. Consequently, ω = −dθ/dt in this case. Still, the directdifferentiation was simple and straightforward. In many cases, however, this willnot be true.

2.5.8 Constrained motion: Slotted crank

A carriage carrying a slotted crank is made to slide across a horizontal surface witha constant velocity vc to the left. A horizontal rail is mounted a distance H above thecrank hinge, and a sleeve is caused to slide along the rail with a constant velocityvs to the right, as illustrated. The sleeve is also connected to the crank slot, so thatthe sleeve can also slide along the crank. At the instant illustrated, when the anglebetween the crank and vertical is θ , determine the angular velocity ω of the crank,the angular acceleration α of the crank, and the velocity of the pin relative to thecrank slot.

Fig. 2.25 Slotted crank


SOLUTION

First, we will define a fixed coordinate system o such that exexex is positive towardthe right and eyeyey is positive upwards. Then, we will define a moving origin o′ whichis stuck to the carriage at the hinge point, such that the unit vector e′ye′ye′y always pointstoward the pin and e′xe′xe′x always points upwards and to the right. The velocity of theslider pin P relative to a fixed observer will then be given by the expression,


However, some information is already known about some of these quantities. Sincethe pin is stuck to the slider, and the slider is constrained to move only to the right,the pin can only have a velocity in the exexex direction. Also, the moving origin o′ issliding across the ground, so it can also only have a velocity along exexex (though themagnitude will be negative). The crank is rotating in the plane of the page, so theangular velocity must be in the ezezez direction. Also, the relative position and veloc-ity between the pin and the crank, rp/o′rp/o′rp/o′ and vp/o′vp/o′vp/o′ , must be along the e′ye′ye′y direction.Because e′ye′ye′y always points along the crank slot, if P were to move in any other way,it would have to come out of the slot. Incorporating all of this information into thevelocity equation,


vs exexex +0 eyeyey = −vc exexex + vrel e′ye′ye′y +ω ezezez ×R e′ye′ye′yvs exexex +0 eyeyey = −vc exexex + vrel e′ye′ye′y +ωR

(ezezez ×e′ye′ye′y

)

where vrel is the scalar magnitude of the velocity of the pin relative to the crank, ω isthe scalar angular velocity of the crank (which will probably end up being negative),and R is the straight-line distance between the slider pin P and the crank hinge o′.The symbol R was defined for convenience only, since R can be expressed in termsof the given quantities H and θ from geometry, as

R =H

cos(θ)

The velocity equation above was in terms of both the moving coordinates and thefixed coordinates; in order to equate the terms, it is necessary to transform all ofthe unit vectors into a single reference frame. In this case, the ‘prime’ unit vectorscan be transformed into the fixed coordinate system by applying the simple rules oftrigonometry to the unit vector diagram shown in the illustration.



2.5 Examples 89


These values may now be substituted into the velocity equation. First, we will con-vert ezezez to e′ze′ze′z to carry out the cross-product, and then all of the moving unit vectorswill be transformed into the fixed frame.

vs exexex +0 eyeyey = −vc exexex + vrel e′ye′ye′y +ωR(ezezez ×e′ye′ye′y

)vs exexex +0 eyeyey = −vc exexex + vrel

(−sin(θ) exexex + cos(θ) eyeyey)+ωR


)= −vc exexex + vrel

(−sin(θ) exexex + cos(θ) eyeyey)−ωR e′xe′xe′x

vs exexex +0 eyeyey = −vc exexex + vrel(−sin(θ) exexex + cos(θ) eyeyey

)−ωR(cos(θ) exexex + sin(θ) eyeyey

)

Now we can consider separately the scalar coefficients of exexex and eyeyey on both sidesof this equation. To begin with, let us consider the coefficients of eyeyey; grouping thesetogether,

0 eyeyey = vrel cos(θ) eyeyey −ωRsin(θ) eyeyey

0 = vrel cos(θ)−ωRsin(θ)

vrel cos(θ) = ωRsin(θ)

vrel = ωRsin(θ)cos(θ)

which is the result required by the problem; however, the angular velocity is stillunknown. For this, we need to look at the scalar coefficients of exexex.

vs exexex = −vc exexex − vrel sin(θ) exexex −ωRcos(θ) exexex

vs = −vc − vrel sin(θ)−ωRcos(θ)

vs + vc = −vrel sin(θ)−ωRcos(θ)

However, we already have an expression for vrel ; substituting,

vs + vc = −(

ωRsin(θ)cos(θ)

)sin(θ)−ωRcos(θ)

vs + vc = −ωRsin2(θ)

cos(θ)−ωRcos(θ)


vs + vc = −ωRsin2(θ)

cos(θ)−ωR

cos2(θ)

cos(θ)

vs + vc = −ωR

(sin2(θ)+ cos2(θ)

cos(θ)

)

vs + vc = − ωRcos(θ)

ω = − 1R(vs + vc)cos(θ)

This is the answer to the first part of the question; however, the quantity R must beexpressed in terms of the given parameters.

ω = − 1R(vs + vc)cos(θ)

= −(

cos(θ)H

)(vs + vc)cos(θ)

ω = − 1H(vs + vc)cos2(θ)

The angular velocity was indeed negative, as expected. The relative velocity vrel

can be obtained by substituting this value for the angular velocity into the previousexpression,

vrel = ωRsin(θ)cos(θ)

=

(− 1

H(vs + vc)cos2(θ)

)(H

cos(θ)

)sin(θ)cos(θ)

vrel = −(vs + vc)sin(θ)

The result for vrel is also negative, indicating that the distance between the crankhinge and the slider pin is decreasing- which would also be expected. The simplic-ity of this final result also demonstrates again the advantage of solving problemssymbolically whenever possible.

Finally, we can solve for the angular acceleration of the slotted bar. From therelative acceleration equation,


However, since both the carriage and the slider have constant velocities, ap/oap/oap/o =ao′/oao′/oao′/o = 0. Also, we have already solved for ωaxωaxωax and vp/o′vp/o′vp/o′ above. Finally, while the

2.5 Examples 91

magnitude of the acceleration of the pin relative to the slider, ap/o′ap/o′ap/o′ , is unknown, thepin is constrained to move only within the slot and the slot is always parallel to e′ye′ye′y,the direction must be along e′ye′ye′y as well. Substituting,

0 = 0+arel e′ye′ye′y +2ω e′ze′ze′z × vrel e′ye′ye′y +α e′ze′ze′z ×R e′ye′ye′y +ω e′ze′ze′z ×ω e′ze′ze′z ×R e′ye′ye′y= arel e′ye′ye′y +2ωvrel


)+αR


)+ω e′ze′ze′z ×ωR


)= arel e′ye′ye′y −2ωvrel e′xe′xe′x −αR e′xe′xe′x −ω e′ze′ze′z ×ωR e′xe′xe′x= arel e′ye′ye′y −2ωvrel e′xe′xe′x −αR e′xe′xe′x −ω2R


)0 = arel e′ye′ye′y −2ωvrel e′xe′xe′x −αR e′xe′xe′x −ω2R e′ye′ye′y

Combining terms,

0 =(−2ωvrel −αR

)e′xe′xe′x +

(arel −ω2R

)e′ye′ye′y

Considering only the coefficients of e′xe′xe′x,

0 = −2ωvrel −αR

α = − 2R

ωvrel

We may now substitute the values for vrel and ω obtained earlier, and express R interms of H and θ .

α =−2cos(θ)H

(− 1

H(vs + vc)cos2(θ)

)(−(vs + vc)sin(θ)

)

Simplifying,

α =− 2H2 (vs + vc)

2 cos3 sin(θ)

Note here that we do not need to consider the coefficients of e′ye′ye′y. Also, we were ableto solve without resolving into components in the fixed coordinate system, since allthe terms were in the o′ coordinate system.

Alternatively, we may also have arrived at the same result by applying the chainrule to the earlier result for ω , as

α =dω

dt=

dω

dθ

dθ

dt=

dω

dθω

Substituting,


α =d

dθ

(− 1

H(vs + vc)cos2(θ)

)(− 1

H(vs + vc)cos2(θ)

)

=2H(vs + vc)cos(θ)sin(θ)

(− 1

H(vs + vc)cos2(θ)

)

α = − 2H2 (vs + vc)

2 cos3(θ)sin(θ)

Which is exactly the same result arrived at before. In general, however, the differen-tiation of terms will not work; it was only possible here because both vs and vc wereconstant.

2.5.9 Scotch yoke mechanism

A Scotch yoke mechanism consists of a crank-wheel A and a slider B. A pin on therim of the crank-wheel slides inside a vertical slot rigidly attached to the slider. Thecrank-wheel A has a radius R, and the pin joining the crank to the slider is at anangle of θ relative to the vertical as shown. Determine the velocity and accelerationof the slider B if the crank-wheel has an angular velocity of magnitude θ and anangular acceleration of magnitude θ .

Fig. 2.26 Scotch yoke mechanism

2.5 Examples 93

SOLUTION

The Scotch yoke is perhaps one of the simplest mechanisms, having only twomoving parts. In order to determine the velocity of the slider, it is sufficient to rec-ognize that the pin joining the crank-wheel to the slider is constrained to move withthe slider along the horizontal axis, though it is free to move relative to the slideralong the vertical axis. Consequently, the velocity of the slider must be equal to thehorizontal component of the velocity of the pin.

Let us define a fixed coordinate system o such that exexex points toward the right andeyeyey points upward. Then, ezezez points out of the plane of the figure, as illustrated. Then,we define a moving coordinate system o′ which is stuck to the crank, with its originat the centre. We shall also define the moving unit vector e′ye′ye′y such that it alwayspoints from o′ toward the point of interest p (which, in this case, is the pin joiningthe crank to the slider). Let e′xe′xe′x point downward and to the right, as illustrated; thene′ze′ze′z will always be parallel to ezezez, pointing out of the plane of the figure. From therelative velocity equation, then,

Fig. 2.27 Scotch yoke mechanism- the crank-wheel


Since the moving origin o′ is fixed to the centre of the crank, and since the hub of thecrank is not moving, then vo′/ovo′/ovo′/o = 0. Also, since the point p is also fixed to the crankand does not move within the o′ coordinate system, vp/o′vp/o′vp/o′ = 0 as well. Because themoving coordinate system is stuck to the crank, ωaxωaxωax will be the same as the angularvelocity of the crank θθθ . Substituting these values into the relative velocity equation,


vp/ovp/ovp/o = 0+0− θ ezezez ×R e′ye′ye′y= −θR


)vp/ovp/ovp/o = θR e′xe′xe′x

where we have noted here that e′ze′ze′z = ezezez. Also, note that because θ is anti-clockwise,the direction of ωaxωaxωax is −ezezez. We may now resolve the result into components alongthe fixed coordinate system, as




Substituting the above value of e′xe′xe′x into the relative velocity equation,

vp/ovp/ovp/o = θR e′xe′xe′x= θR


)vp/ovp/ovp/o = θRcos(θ)exexex − θRsin(θ)eyeyey

Finally, since the velocity of the slider is constrained to match the horizontal com-ponent of the velocity of the pin,

vbvbvb = θRcos(θ)exexex

This result demonstrates one of the most important characteristics of the Scotchyoke mechanism: it is capable of producing a purely sinusoidal linear motion, andis therefore used when a very smooth reciprocating action is required.

Next, we consider the acceleration. From the relative acceleration equation,


Again, because o′ is fixed in space, ao′/oao′/oao′/o = 0, and because the point p does not moverelative to the o′ coordinate system, ap/o′ap/o′ap/o′ = 0. Then,

ap/oap/oap/o = 0+0+2ωaxωaxωax ×0+(−θ ezezez

)×R e′ye′ye′y +(−θ ezezez

)× (−θ ezezez)×R e′ye′ye′y

= −θR(e′ze′ze′z ×e′ye′ye′y

)+(−θ e′ze′ze′z

)×−θR(e′ze′ze′z ×e′ye′ye′y

)= θR e′xe′xe′x +

(−θ e′ze′ze′z)× θR e′xe′xe′x

2.5 Examples 95

= θR e′xe′xe′x − θ 2R(e′ze′ze′z ×e′xe′xe′x

)ap/oap/oap/o = θR e′xe′xe′x − θ 2R e′ye′ye′y

Resolving e′xe′xe′x and e′ye′ye′y into components parallel to the fixed unit vectors exexex and eyeyey,

ap/oap/oap/o = θR(cos(θ) exexex − sin(θ) eyeyey

)− θ 2R(sin(θ) exexex + cos(θ) eyeyey

)ap/oap/oap/o =

(θRcos(θ)− θ 2Rsin(θ)

)exexex −

(θRsin(θ)+ θ 2Rcos(θ)

)eyeyey

Once again, since the motion of the slider can only track the horizontal componentof the motion of the pin,

ababab =(θRcos(θ)− θ 2Rsin(θ)

)exexex

Note, of course, that the same result may have been arrived at by directly differenti-ating the velocity, as

ababab = vbvbvb

=ddt

(θRcos(θ) exexex

)ababab = θRcos(θ) exexex + θ Rcos(θ) exexex − θ 2Rsin(θ) exexex + θRcos(θ) exexex

where we have applied the product rule. Since R is constant (the radius of the crankdoes not change with time), R = 0. Also, because the fixed unit vector exexex does notchange direction, exexex = 0 as well. Then,

ababab = θRcos(θ)exexex +0− θ 2Rsin(θ) exexex +0

ababab =(θRcos(θ)− θ 2Rsin(θ)

)exexex

as expected.

2.5.10 Crank-slider mechanism

A simple crank-slider assembly consists of a crank-wheel A, a slider B and a con-necting rod C. The slider B has a horizontal velocity vb and acceleration ab. Thecrank-wheel A has a radius R and the pin joining the wheel to the connecting rod is


at an angle of θ relative to the vertical as shown. The connecting rod has a lengthL. Determine the angular velocity and angular acceleration of the connecting rodas functions of θ .

Fig. 2.28 Crank-slider mechanism

SOLUTION

This is an example of a typical, simple mechanism, and is actually a fairly goodrepresentation of a reciprocating engine. This problem is very similar to the earlierbug-on-a-ladder example: there are constraints on both ends of the connecting rodC, though one end of the connecting rod is moving in circles rather than in a straightline. While the geometry may be fairly simple, the algebra involved in solving thisproblem symbolically can get extremely tedious.

To begin with, let us define the point p as the pin joining the connecting rodand the crank-wheel, and let us define the point q as the pin joining the connectingrod to the slider. In order to determine the angular velocity and angular accelerationof the connecting rod, we will need to apply the relative motion equations using acoordinate system stuck to the rod. Since all the points of interest are stuck to therod, vp/o′vp/o′vp/o′ = ap/o′ap/o′ap/o′ = 0 for any point of interest on the rod relative to any movingorigin o′ stuck to the rod. The only unknowns in the relative velocity or accelerationequation will therefore be

(1) the velocity (or acceleration) of o′;

(2) the velocity (or acceleration) of the point of interest

2.5 Examples 97

(3) ωaxωaxωax, and

(4) αaxαaxαax.

Since the problem is two-dimensional, each of the relative velocity and relative ac-celeration vector equations each represent a system of two equations (one for exexex andone for eyeyey). Therefore, we need to find two of these four unknowns before we cansolve the relative motion equations for the connecting rod.

Because the points p and q are constrained (p can only move in circles of radiusR about the centre of the crank-wheel, while q can only move along a horizontalline), it should be reasonably easy to find vp/ovp/ovp/o and vq/ovq/ovq/o (and, similarly, ap/oap/oap/o andaq/oaq/oaq/o). If we use one of these as our ‘point of interest’ on the connecting rod and theother as the moving origin, this will eliminate unknowns (1) and (2). Let us begin,then, by examining the crank-wheel in order to determine vp/ovp/ovp/o and ap/oap/oap/o.

Fig. 2.29 Crank-slider mechanism- the crank-wheel

Let us define a moving coordinate system which is stuck to the wheel, with theorigin o′ on the centre of the wheel. The moving unit vector e′ye′ye′y rotates along with thewheel and is always pointing from o′ toward p. The unit vector e′xe′xe′x is perpendicularto e′ye′ye′y and points downward and to the right at the instant of interest, while e′ze′ze′z alwayspoints out of the page. We will also define a fixed coordinate system o such that exexex

points to the right, eyeyey points up and ezezez points out of the page, as illustrated. Sincethey are not given, we will take magnitudes of the angular velocity and accelerationof the wheel as θ and θ , respectively, and treat these as unknowns for which we willhave to solve later. From the relative velocity equation,



Since the moving origin o′ is stuck to the centre of the wheel and the centre of thewheel is not moving, vo′/ovo′/ovo′/o = 0. Also, since both the moving coordinate system andthe point p are fixed to the same rigid wheel, vp/o′vp/o′vp/o′ = 0. Substituting,

vp/ovp/ovp/o = 0+0+(−θ)e′ze′ze′z ×Re′ye′ye′y= −θR


)vp/ovp/ovp/o = θRe′xe′xe′x

However, at the moment of interest,




Substituting,

vp/ovp/ovp/o = θRe′xe′xe′x= θR


)vp/ovp/ovp/o = θRcos(θ)exexex − θRsin(θ)eyeyey

Similarly, for the acceleration,


Again, the pin at the centre of the wheel is not moving and both the coordinatesystem o′ and the point p are stuck to the same wheel, so ao′/oao′/oao′/o = ap/o′ap/o′ap/o′ = 0.

ap/oap/oap/o = 0+0+2ωaxωaxωax ×0+αaxαaxαax ×rp/o′rp/o′rp/o′ +ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

= (−θ)e′ze′ze′z ×Re′ye′ye′y +(−θ)e′ze′ze′z × (−θ)e′ze′ze′z ×Re′ye′ye′y= −θR


)− θ e′ze′ze′z ×−θR(e′ze′ze′z ×e′ye′ye′y

)= θRe′xe′xe′x − θ e′ze′ze′z × θRe′xe′xe′x= θRe′xe′xe′x − θ 2R


)ap/oap/oap/o = θRe′xe′xe′x − θ 2Re′ye′ye′y

2.5 Examples 99

Converting back into the fixed coordinates,

ap/oap/oap/o = θR(cos(θ)exexex − sin(θ)eyeyey

)− θ 2R(sin(θ)exexex + cos(θ)eyeyey

)ap/oap/oap/o =


)exexex +

(−θRsin(θ)− θ 2Rcos(θ))

eyeyey

We now have expressions for the velocity and acceleration of the point p relativeto a fixed observer, in terms of the angular velocity and acceleration of the crankwheel (which remain unknown). Next, let us turn our attention to the point q. Sinceq is fixed to the slider B, and since the velocity and acceleration of the slider B isknown, we may simply write,

vq/ovq/ovq/o = vb exexex

aq/oaq/oaq/o = ab exexex

Finally, we direct our attention to the connecting rod. We will redefine the co-ordinate system o′ such that the moving origin is stuck to the connecting rod at thepoint q, and the moving unit vector e′ye′ye′y points along the connecting rod, always to-ward the point p. The unit vector e′xe′xe′x is perpendicular to e′ye′ye′y, pointing upwards and tothe right, while e′ze′ze′z again points out of the page. The fixed coordinate system remainsunchanged. For convenience, we shall define the angle θc as the angle subtendedbetween the connecting rod and the horizontal. Note that we could equally haveplaced o′ on top of point p and selected q as our point of interest; the result wouldbe identical.

Fig. 2.30 Crank-slider mechanism- the connecting rod


Applying the relative velocity equation,


Since the moving origin o′ is coincident with the point q, vo′/ovo′/ovo′/o = vq/ovq/ovq/o, which wehave already determined. Also, since the coordinate system o′ and the point q areboth stuck to the same rigid bar, vp/o′vp/o′vp/o′ = 0. Substituting,

vp/ovp/ovp/o = vb exexex +0+ω e′ze′ze′z ×Le′ye′ye′y= vb exexex +ωL


)vp/ovp/ovp/o = vb exexex −ωLe′xe′xe′x

where ω is the angular velocity of the connecting rod (which may be positive ornegative: the direction of rotation may not always be obvious). Once again, fromgeometry,

e′xe′xe′x = sin(θc)exexex + cos(θc)eyeyey

e′ye′ye′y = −cos(θc)exexex + sin(θc)eyeyey


Substituting,

vp/ovp/ovp/o = vb exexex −ωLe′xe′xe′x= vb exexex −ωL

(sin(θc)exexex + cos(θc)eyeyey

)vp/ovp/ovp/o =

(vb −ωLsin(θc)

)exexex −ωLcos(θc)eyeyey

However, we already have an expression for vp/ovp/ovp/o from our analysis of the crank-wheel:

vp/ovp/ovp/o = θRcos(θ)exexex − θRsin(θ)eyeyey

Because both of these expressions give the velocity of the point p relative to a fixedobserver in terms of unit vectors which cannot change, we can equate them together,so that

θRcos(θ)exexex − θRsin(θ)eyeyey =(vb −ωLsin(θc)

)exexex −ωLcos(θc)eyeyey

2.5 Examples 101

Note that the moving unit vectors were not the same during the analysis of the crank-wheel and the connecting rod, so the ‘prime’ coordinates from these two analysescould not be equated together. We may now use this expression to solve for theunknown θ . Since we know vb, let us first look at the coefficients of exexex.

θRcos(θ) = vb −ωLsin(θc)

θ =vb

Rcos(θ)−ω

LR

sin(θc)

cos(θ)

Is is important to remember here that θ is the angular velocity of the crank-wheel,whereas ω is the angular velocity of the connecting rod. Next, we can equate thecoefficients of eyeyey, substitute the above result for θ and solve for ω .

θRsin(θ) = ωLcos(θc) (2.20)(vb

Rcos(θ)−ω

LR

sin(θc)

cos(θ)

)Rsin(θ) = ωLcos(θc)

vbsin(θ)cos(θ)

−ωLsin(θc)sin(θ)cos(θ)

= ωLcos(θc)

Noting that we have constrained ourselves here to examining cases for whichcos(θ) �= 0; otherwise, the division would result in a singularity. Isolating ω ,

ω =vb

L1

cos(θc)cos(θ)sin(θ) + sin(θc)

(2.21)

However, we still have the unknown θc in this expression. Thankfully, this can beobtained from the geometry.

Fig. 2.31 Crank-slider mechanism- determining θc


Consider the triangle formed between the centre of the crank-wheel, the point pand the point q. The vertical line segment dividing the triangle under the point p hasa length Rcos(θ). However, this length must also be Lsin(θc). Consequently,

Rcos(θ) = Lsin(θc)

sin(θc) =RL

cos(θ) (2.22)

However, from the trigonometric identity,

sin2(θc)+ cos2(θc) = 1

cos(θc) =

√1− sin2(θc)

where we are restricting ourselves to 0 ≤ θc ≤ π/2. Substituting in the previousresult,

cos(θc) =

√1− R2

L2 cos2(θ)

These results may now be substituted into Equation (2.21) to eliminate θc, as

ω =vb

L1


=vb

L1(√

1− R2

L2 cos2(θ)

)cos(θ)sin(θ) +

(RL cos(θ)

)

ω =vb

Ltan(θ)√

1− R2

L2 cos2(θ)+ RL sin(θ)

where, for simplicity, we have multiplied through by tan(θ). Though cumbersome,this expression gives the value of ω as a function of vb, as required. As ω is positivefor positive vb, the connecting rod is rotating anti-clockwise when in the positionillustrated.

It may also be interesting to notice here that, given the expressions which wehave obtained above, it would equally be possible to solve for the velocity of theslider as a function of the angular velocity of the crank. From Equation (2.20),

2.5 Examples 103

θRsin(θ) = ωLcos(θc)

ω = θRL

sinθ

cosθc

Substituting this into Equation (2.21),

ω =vb

L1


θRL

sinθ

cosθc=

vb

L1


vb = θRsin(θ)

(cos(θ)sin(θ)

+sin(θc)

cos(θc)

)

Finally, from Equation (2.22),

sin(θc) =RL

cos(θ)

cos(θc) =

√1− R2

L2 cos2(θ)

where we have again applied the trigonometric identity. Substituting, then,

vb = θRsin(θ)

(cos(θ)sin(θ)

+sin(θc)

cos(θc)

)

vb = θRsin(θ)

(cos(θ)sin(θ)

+RL cos(θ)√

1− R2

L2 cos2(θ)

)

The relationship between the angular velocity of the crank and the linear velocityof the slider is not straightforward; a plot of this function is included below. How-ever, Two interesting observations may be made of this result. First, the velocityvb becomes nonreal (and therefore nonphysical) within the domain 0 ≤ θ ≤ 2π forthe case when R/L > 1. Furthermore, vb → ∞ within the domain when R/L → 1.Examining the illustration of the mechanism in Figure 2.28, the reason for this sin-gularity becomes clear. One end of the connecting rod is constrained to remain onthe horizontal centreline, and the other is constrained to follow the circumferenceof the crank. If the connecting rod C is shorter than the radius of the crank B (sothat R/L > 1), the connecting rod won’t be able to ‘reach’ far enough to follow thecircumference of the crank, and the mechanism will simply jam.


Fig. 2.32 Crank-slider mechanism- variation of vb with θ

It is also interesting to consider the case when R/L is very small. In this case,then,

vb ≈ θRsin(θ)

(cos(θ)sin(θ)

+(0)cos(θ)√

1− R2

L2 cos2(θ)

)

= θRsin(θ)

(cos(θ)sin(θ)

)vb ≈ θRcos(θ)

For R/L → 0, then, the motion becomes purely sinusoidal.

Continuing with our example, we next consider the acceleration of the connectingrod, repeating exactly the same procedure as for the velocity. We again begin withthe relative acceleration equation,


2.5 Examples 105

Because both the moving coordinate system o′ and the point p are stuck to the samebar, ap/o′ap/o′ap/o′ = vp/o′vp/o′vp/o′ = 0. Again, since o′ and q are coincident, ao′/oao′/oao′/o = aq/oaq/oaq/o, and wealready know that

aq/oaq/oaq/o = ab exexex

Substituting these values into the relative acceleration equation,

ap/oap/oap/o = ab exexex +0+2ωaxωaxωax ×0+α e′ze′ze′z ×Le′ye′ye′y +ω e′ze′ze′z ×ω e′ze′ze′z ×Le′ye′ye′y= ab exexex +αL


)+ω e′ze′ze′z ×ωL


)= ab exexex −αLe′xe′xe′x +ω e′ze′ze′z ×−ωLe′xe′xe′x= ab exexex −αLe′xe′xe′x −ω2L


)ap/oap/oap/o = ab exexex −αLe′xe′xe′x −ω2Le′ye′ye′y

where α is the magnitude of the angular acceleration of the connecting rod (which,like ω , may be negative). Once again, we express the moving unit vectors in termsof the fixed unit vectors, as

ap/oap/oap/o = ab exexex −αL(sin(θc)exexex + cos(θc)eyeyey

)−ω2L(−cos(θc)exexex + sin(θc)eyeyey

)= ab exexex −αLsin(θc)exexex −αLcos(θc)eyeyey +ω2Lcos(θc)exexex −ω2Lsin(θc)eyeyey

ap/oap/oap/o =(ab −αLsin(θc)+ω2Lcos(θc)

)exexex +

(−αLcos(θc)−ω2Lsin(θc))

eyeyey

This expression yielded the velocity of point p with respect to a fixed observer.However, we have already obtained an expression for ap/oap/oap/o from our analysis of thecrank-wheel:

ap/oap/oap/o =(θRcos(θ)− θ 2Rsin(θ)

)exexex +


eyeyey

Since these are two independent expressions for the same quantity using the samecoordinate system, we can equate them together, so that


)exexex +


eyeyey

=(ab −αLsin(θc)+ω2Lcos(θc)

)exexex +


eyeyey

(2.23)

Once again, we can equate together the coefficients of exexex to solve for the unknownθ , as


θRcos(θ)− θ 2Rsin(θ) = ab −αLsin(θc)+ω2Lcos(θc)

θ =ab

Rcos(θ)−α

LR

sin(θc)

cos(θ)+ω2 L

Rcos(θc)

cos(θ)+ θ 2 sin(θ)

cos(θ)

Next, we shall equate the coefficients of eyeyey.

−θRsin(θ)− θ 2Rcos(θ) =−αLcos(θc)−ω2Lsin(θc)

We may now substitute in the previous result for θ , as

−(

ab

Rcos(θ)−α

LR

sin(θc)

cos(θ)+ω2 L

Rcos(θc)

cos(θ)+ θ 2 sin(θ)

cos(θ)

)Rsin(θ)

−θ 2Rcos(θ)

= −αLcos(θc)−ω2Lsin(θc)

Distributing the multiplication,

− absin(θ)cos(θ)

+αLsin(θc)sin(θ)cos(θ)

−ω2Lcos(θc)sin(θ)cos(θ)

−θ 2Rsin2(θ)

cos(θ)− θ 2Rcos(θ)

= −αLcos(θc)−ω2Lsin(θc)

Simplifying and isolating α ,

−absin(θ)cos(θ)

+ω2L

(sin(θc)− cos(θc)

sin(θ)cos(θ)

)− θ 2R

sin2(θ)+ cos2(θ)

cos(θ)

= αL

(−cos(θc)− sin(θc)

sin(θ)cos(θ)

)

Simplifying,

−ab

Lsin(θ)cos(θ)

+ω2

(sin(θc)− cos(θc)

sin(θ)cos(θ)

)− θ 2 R

L1

cos(θ)

2.5 Examples 107

= α

(−cos(θc)− sin(θc)

sin(θ)cos(θ)

)

Next, we can substitute the expressions for sin(θc) and cos(θc) we obtained earlier,so that

−ab

Lsin(θ)cos(θ)

+ω2

(RL

cos(θ)−√

1− R2

L2 cos2(θ)sin(θ)cos(θ)

)− θ 2 R

Lcos(θ)

= α

(−√

1− R2

L2 cos2(θ)− RL

cos(θ)sin(θ)cos(θ)

)

Finally, we may express α as

α =

abL tan(θ)−ω2

(RL cos(θ)−

√1− R2

L2 cos2(θ) tan(θ)

)+ θ 2 R

Lcos(θ)√1− R2


where, as we had shown earlier,

ω =vb

Ltan(θ)√

1− R2


and

θ =vb

Rcos(θ)−ω

LR

sin(θc)

cos(θ)

=vb

Rcos(θ)−ω

LR

RL cos(θ)

cos(θ)

=vb

Rcos(θ)−ω

θ =vb

Rcos(θ)− vb

Ltan(θ)√

1− R2


Though incredibly cumbersome, we have derived algebraic expressions for ω andα (the angular velocity and angular acceleration of the connecting rod) which arefunctions of θ , R, L, vb and ab only, which are all known quantities.

As before, it would also be useful to determine the acceleration of the slider ab

as a function of the angular velocity of the crank. Let us consider the special case


when θ = 0, so that the crank is rotating with constant angular velocity θ . Then,recalling the combined relative acceleration equation (Equation 2.23),


)exexex +


eyeyey

=(ab −αLsin(θc)+ω2Lcos(θc)

)exexex +


eyeyey

and equating the scalar coefficients of eyeyey,

−θRsin(θ)− θ 2Rcos(θ) = −αLcos(θc)−ω2Lsin(θc)

−(0)Rsin(θ)− θ 2Rcos(θ) = −αLcos(θc)−ω2Lsin(θc)

α = θ 2 RL

cos(θ)cos(θc)

−ω2 sin(θc)

cos(θc)

Next, we can equate the scalar coefficient of exexex (again, considering only the specialcase of θ = 0) to solve for ab, as

θRcos(θ)− θ 2Rsin(θ) = ab −αLsin(θc)+ω2Lcos(θc)

(0)Rcos(θ)− θ 2Rsin(θ) = ab −αLsin(θc)+ω2Lcos(θc)

ab = −θ 2Rsin(θ)+αLsin(θc)−ω2Lcos(θc)

Substituting in our earlier result for α ,

ab = −θ 2Rsin(θ)+

(θ 2 R

Lcos(θ)cos(θc)

−ω2 sin(θc)

cos(θc)

)Lsin(θc)−ω2Lcos(θc)

= −θ 2Rsin(θ)+ θ 2Rcos(θ)sin(θc)

cos(θc)−ω2L

sin2(θc)

cos(θc)−ω2Lcos(θc)


cos(θc)− ω2L

cos(θc)

(sin2(θc)+ cos2(θc)

)ab = −θ 2Rsin(θ)+ θ 2Rcos(θ)

sin(θc)

cos(θc)− ω2L

cos(θc)

Next, we may use Equation (2.20) to eliminate ω , as

θRsin(θ) = ωLcos(θc)

2.5 Examples 109

ω = θRL

sinθ

cos(θc)

Substituting this into the previous result for ab,

ab = −θ 2Rsin(θ)+ θ 2Rcos(θ)sin(θc)

cos(θc)− ω2L

cos(θc)


cos(θc)− L

cos(θc)

(θ 2 R2

L2

sin2 θ

cos2(θc)

)

ab = θ 2Rsin(θ)

(−1+

cos(θ)sin(θ)

sin(θc)

cos(θc)− R

Lsinθ

cos3(θc)

)

From Equation (2.22), we already have shown

sin(θc) =RL

cos(θ)

cos(θc) =

√1− R2

L2 cos2(θ)

Substituting this into our equation for ab, then,

ab = θ 2Rsin(θ)

(−1+

cos(θ)sin(θ)

sin(θc)

cos(θc)− R

Lsinθ

cos3(θc)

)

= θ 2Rsin(θ)

(−1+

cos(θ)sin(θ)

RL cos(θ)√

1− R2

L2 cos2(θ)−

RL sinθ(

1− R2

L2 cos2(θ))3/2

)

ab = θ 2Rsin(θ)

(−1+

RL cos2(θ)

sin(θ)√

1− R2

L2 cos2(θ)−

RL sinθ(

1− R2

L2 cos2(θ))3/2

)

This result has yielded the acceleration of the slider in terms of the crank angle θfor the special case of θ = 0. As with the slider velocity, the relationship betweenthe slider acceleration and θ is fairly complicated; a plot of this function is includedbelow.

It is again interesting to note here that for the case of R/L → 1, the accelerationof the slider becomes infinite. This is to be expected, as we have already seen thatthe mechanism will no longer function if L < R. Also, if R/L is very small, then

ab ≈ θ 2Rsin(θ)(−1+0−0

)


Fig. 2.33 Crank-slider mechanism- variation of ab with θ

ab ≈ −θ 2Rsin(θ)

This is expected, as we saw that for small R/L,

vb ≈ θRcos(θ)

and, therefore,

ab =ddt

vb

≈ ddt

θRcos(θ)

ab ≈ θRcos(θ)+ θ Rcos(θ)− θ 2Rsin(θ)

where we have applied the product rule. Since the radius of the crank is constant,R = 0; also, since we have only considered above the case where θ = 0,

ab ≈ (0)Rcos(θ)+ θ(0)cos(θ)− θ 2Rsin(θ)

2.5 Examples 111

ab ≈ −θ 2Rsin(θ)

as expected.

2.5.11 Crank-rocker mechanism

A simple four-bar mechanism consists of a crank A of length La = 20 mm and arocker-arm C of length Lc = 40 mm, which are both pinned to the ground a distanceH = 70 mm apart. The crank and rocker-arm are joined by a connecting rod B oflength Lb = 80 mm. The crank is given a constant angular velocity of ω = 2 rad/s.Determine the angular velocity of the rocker-arm C as the angle θa of the crank Avaries from 0◦ to 360◦, in 45◦ increments.

Fig. 2.34 Crank-rocker mechanism

SOLUTION

This arrangement is known as a four-bar mechanism, and is the simplest possiblerigid-link mechanism (the ground is considered as one bar). This example is very


similar to the previous slider-crank mechanism, only here both sides of the connect-ing rod C are constrained to move in circles. Once again, we will begin with an anal-ysis of the crank. Also, as we saw in the previous example, the algebra involved indetermining the motion of linked components can become extremely complicated.Consequently, we will analyze this arrangement numerically.

Let us begin by considering the crank alone, and define the point p as the pinjoining the crank to the connecting rod. We shall define a moving coordinate systemo′ stuck to the crank with its origin on the pin joining the crank to the ground. Themoving unit vector e′ye′ye′y shall be always pointing from o′ to p, e′xe′xe′x shall be pointingdown and to the right as illustrated, and e′ze′ze′z shall be out of the page. The fixed unitvector eyeyey shall always be pointing upwards and exexex shall always be to the right.

Fig. 2.35 Crank-rocker mechanism- The crank

Then, from relative motion,


However, because the pin joining the crank to the ground at o′ never moves, vo′/ovo′/ovo′/o =0. Also, because both the point p and the moving origin o′ are stuck to the samerigid crank, vp/o′vp/o′vp/o′ = 0 as well. Finally, because the axes are stuck to the crank, ωaxωaxωax =−ω e′ze′ze′z, as specified by the problem statement. Substituting,

vp/ovp/ovp/o = 0+0−ω e′ze′ze′z ×La e′ye′ye′y= −ωLa


)vp/ovp/ovp/o = ωLa e′xe′xe′x

2.5 Examples 113

Next, since we will need this velocity when analyzing the connecting rod (which isnot well suited to the current ‘prime’ axes), we will resolve this result into the fixedcoordinates. From the problem geometry,

e′xe′xe′x = cos(θa) exexex − sin(θa) eyeyey

e′ye′ye′y = sin(θa) exexex + cos(θa) eyeyey


Substituting this into the previous result,

vp/ovp/ovp/o = ωLa(cos(θa) exexex − sin(θa) eyeyey

)vp/ovp/ovp/o = ωLa cos(θa) exexex −ωLa sin(θa) eyeyey

So, we can substitute in values at the required increments of θa to yield

θaθaθa vp/ovp/ovp/o (mm/s)0◦ 40.000 exexex +0.000 eyeyey45◦ 28.284 exexex −28.284 eyeyey90◦ 0.000 exexex −40.000 eyeyey135◦ −28.284 exexex −28.284 eyeyey180◦ −40.000 exexex +0.000 eyeyey225◦ −28.284 exexex +28.284 eyeyey270◦ 0.000 exexex +40.000 eyeyey315◦ 28.284 exexex +28.284 eyeyey

Next, we examine the connecting rod. Let us redefine the moving coordinatesystem so that the origin o′ is stuck to the connecting rod at p. We shall define thepoint q at the pin joining the connecting rod to the crank. The moving unit vector e′xe′xe′xshall always point parallel to the connecting rod, away from the point q. The movingunit vector e′ye′ye′y shall point upwards and to the right, as illustrated, and e′ze′ze′z points outof the plane of the page. For convenience, we shall define θb as the angle subtendedbetween the connecting rod and the horizontal.

To determine the velocity of the point q, we begin with the relative velocity equa-tion,


Since we have already solved for the velocity of the point p with respect to afixed observer, and since o′ is coincident with p, the velocity of our new moving


Fig. 2.36 Crank-rocker mechanism- The connecting rod

origin is simply vp/ovp/ovp/o. Also, since o′ and q are stuck to the same rigid bar, vq/o′vq/o′vq/o′ = 0.Note that the angular velocity of the connecting rod θb is unknown. Substituting,

vq/ovq/ovq/o = vp/ovp/ovp/o +0+ θb e′ze′ze′z ×−Lb e′xe′xe′x= vp/ovp/ovp/o − θbLb


)vq/ovq/ovq/o = vp/ovp/ovp/o − θbLb e′ye′ye′y

where we have already tabulated values of vp/ovp/ovp/o. Note that rp/o′rp/o′rp/o′ was in the negativee′xe′xe′x direction, as the e′xe′xe′x points away from q. Because the selection of the movingunit vectors is arbitrary, we are free to define them in any way we choose withoutaffecting the results when expressed in terms of the fixed unit vectors. Resolving themoving unit vectors into components along the fixed unit vectors,

e′xe′xe′x = cos(θb) exexex − sin(θb) eyeyey

e′ye′ye′y = sin(θb) exexex + cos(θb) eyeyey


Substituting into the previous result,

vq/ovq/ovq/o = vp/ovp/ovp/o − θbLb e′ye′ye′y

2.5 Examples 115

= vp/ovp/ovp/o − θbLb(sin(θb) exexex + cos(θb) eyeyey

)vq/ovq/ovq/o = vp/ovp/ovp/o − θbLb sin(θb) exexex − θbLb cos(θb) eyeyey

However, we have introduced the angle θb, which is unknown but may be deter-mined from the geometry of the problem. Consider the triangles formed by joiningthe ends of each bar with horizontal and vertical line segments, as illustrated. Thehorizontal distance described by the connecting rod B is clearly Lb cos(θb); how-ever, this length may also be determined from the horizontal distances described bythe other bars, as

Fig. 2.37 Crank-rocker mechanism- Problem geometry

Lb cos(θb) = H −Lc sin(θc)+La sin(θa)

where θc is the angle subtended between the rocker-arm C and the vertical. Sim-ilarly, the vertical distance described by the connecting rod may be expressed interms of the vertical distances described by the other bars, as

Lb sin(θb) = Lc cos(θc)−La cos(θa)

These two expressions then provide a system of two equations with two unknowns-θb and θc. Unfortunately, they cannot be solved explicitly in this form. Instead, wecan solve numerically by iteration. We begin by isolating θb from both expressions,as

Lb sin(θb)

Lb cos(θb)=

Lc cos(θc)−La cos(θa)

H −Lc sin(θc)+La sin(θb)


θb = tan−1

(Lc cos(θc)−La cos(θa)

H −Lc sin(θc)+La sin(θb)

)(2.24)

Then, we isolate θc from the first expression,

θc = sin−1

(HLc

+La

Lcsin(θa)− Lb

Lccos(θb)

)(2.25)

To obtain values for θb and θc, we begin by taking a ‘guess’ at θc. We then computeθb using Equation (2.24). Next, we substitute this value of θb to compute θc usingEquation (2.25). With this new value of θc, we can obtain a more accurate estimateof θb from Equation (2.24), and so forth. This process is repeated until the values ofθb and θc obtained stop changing.

Using this iterative process, we will obtain the results,

θaθaθa θbθbθb θcθcθc

0◦ 13.93◦ −11.02◦45◦ 18.22◦ 11.76◦90◦ 26.38◦ 27.28◦135◦ 37.30◦ 30.84◦180◦ 45.82◦ 20.87◦225◦ 42.47◦ −4.52◦270◦ 24.15◦ −35.09◦315◦ 14.06◦ −32.93◦

Note that there are many ways to solve the above system of equations for θb and θc,but this iterative technique is fairly simple and effective when numerical values areavailable. These results are also plotted in Figure 2.38.

Finally, we are able to return to our calculation of vq/ovq/ovq/o. Since we now have a valuefor θb corresponding to each value of θa, we can solve our earlier result numerically,as

vq/ovq/ovq/o = vp/ovp/ovp/o − θbLb sin(θb) exexex − θbLb cos(θb) eyeyey

vq/ovq/ovq/o =(vp/o)

x exexex +(vp/o)

y eyeyey − θbLb sin(θb) exexex − θbLb cos(θb) eyeyey

where (vp/o)x and (vp/o)y are the scalar magnitudes of the components of vp/ovp/ovp/o lyingalong exexex and eyeyey, respectively, which we have calculated and tabulated earlier. Forconvenience, we may combine our numerical results into the constants β1, β2, β3

and β4, so that

vq/ovq/ovq/o =(β1 +β2θb

)exexex +

(β3 +β4θb

)eyeyey (2.26)

2.5 Examples 117

Fig. 2.38 Crank-rocker mechanism- Calculation of θb and θc

where

β1 =(vp/o)

x

β2 = −Lb sin(θb)

β3 =(vp/o)

y

β4 = −Lb cos(θb)

Then, by substituting in values, we obtain the results tabulated below, where θb

remains unknown.

θaθaθa β1β1β1 β2β2β2 β3β3β3 β4β4β4

0◦ 40.000 -19.259 0.000 -77.64745◦ 28.284 -25.013 -28.284 -75.98990◦ 0.000 -35.546 -40.000 -71.669135◦ -28.284 -48.479 -28.284 -63.638180◦ -40.000 -57.372 0.000 -55.753225◦ -28.284 -54.016 28.284 -59.010270◦ 0.000 -32.730 40.000 -72.998315◦ 28.284 -19.435 28.284 -77.603


Finally, we examine the rocker arm C. We will redefine our moving coordinatesystem so that the origin o′ lies on the pin joining the rocker-arm to the ground.We shall define the moving unit vector e′ye′ye′y as being stuck to the rocker-arm, pointingparallel with the arm toward q; e′xe′xe′x shall be perpendicular to e′ye′ye′y, pointing downwardand to the right (as illustrated), and e′ze′ze′z points directly out of the page. Then, fromthe relative velocity equation,

Fig. 2.39 Crank-rocker mechanism- The rocker-arm

vq/ovq/ovq/o = vo′/ovo′/ovo′/o +vp/o′vp/o′vp/o′ +ωaxωaxωax ×rp/o′rp/o′rp/o′

However, the pin joining the rocker-arm to the ground is fixed to the ground, sovo′/ovo′/ovo′/o = 0. Also, because both o′ and q are joined to the same rigid rocker arm,vp/o′ = 0. Substituting,

vq/ovq/ovq/o = 0+0+ θc e′ze′ze′z ×Lc e′ye′ye′y= θcLc


)vq/ovq/ovq/o = −θcLc e′xe′xe′x (2.27)

where θc remains unknown. However, we have another expression for vq/ovq/ovq/o from ouranalysis of the connecting rod: Equations (2.26) and (2.27) therefore provide twoexpressions for vq/ovq/ovq/o which may be equated together in order to isolate and solve forθc. To do so, though, we will need to resolve Equation (2.27) into components alongthe common fixed coordinates, given that

2.5 Examples 119

e′xe′xe′x = cos(θc) exexex − sin(θc) eyeyey

e′ye′ye′y = sin(θc) exexex + cos(θc) eyeyey


Substituting,

vq/ovq/ovq/o = −θcLc e′xe′xe′x= −θcLc

(cos(θc) exexex − sin(θc) eyeyey

)vq/ovq/ovq/o = −θcLc cos(θc) exexex + θcLc sin(θc) eyeyey

Since we have already determined values of θc, we can evaluate this expressionnumerically. For convenience, let us define the constants γ1 and γ2, so that

vq/ovq/ovq/o = γ1θc exexex + γ2θc eyeyey (2.28)

where

γ1 = −Lc cos(θc)

γ2 = Lc sin(θc)

Evaluating the constants,

θaθaθa γ1γ1γ1 γ2γ2γ2

0◦ -39.262 -7.64745◦ -39.160 8.15390◦ -35.553 18.331135◦ -34.345 20.504180◦ -37.377 14.247225◦ -39.876 -3.153270◦ -32.727 -22.998315◦ -33.573 -21.745

We may now equate together Equations (2.26) and (2.28) for vq/ovq/ovq/o to solve for therequired θc, as

vq/ovq/ovq/o =(β1 +β2θb

)exexex +

(β3 +β4θb

)eyeyey

= γ1θc exexex + γ2θc eyeyey


Equating the coefficients of exexex,

β1 +β2θb = γ1θc

θb = −β1

β2+

γ1

β2θc

Next, equating the coefficients of eyeyey,

β3 +β4θb = γ2θc

We may now substitute in our previous result for θb, yielding

β3 +β4

(−β1

β2+

γ1

β2θc

)= γ2θc

β3 − β4β1

β2+

β4γ1

β2θc = γ2θc

θc

(γ2 − β4γ1

β2

)= β3 − β4β1

β2

Simplifying,

θc =

β3β4

− β1β2

γ2β4

− γ1β2

Finally, note that θc > 0 when the rocker-arm is rotating in a clockwise sense; there-fore, in our defined coordinate system, ωcωcωc = −θc ezezez. We may now easily evaluatethe angular velocity of the rocker-arm, as

θaθaθa ωcωcωc (rad/s)0◦ 1.07145◦ 0.89890◦ 0.444135◦ -0.135180◦ -0.769225◦ -1.465270◦ -0.800315◦ 0.754

2.6 Problem-solving technique 121

which is the result required. The angular velocity of the rocker-arm is also plottedas a function of θa in Figure 2.40 below.

Fig. 2.40 Crank-rocker mechanism- Angular velocity of the rocker arm as a function of θa.

The angular velocity of the rocker-arm C is clearly seen to be positive for 124◦ �θa � 288◦, and negative through the rest of the phase. The rotation of the rocker-armtherefore changes sense, resulting in a periodic (but clearly non-sinusoidal) motion.

It should be noted that we could similarly have solved for the angular accel-eration, by also equating together the relative accelerations of the points p and qobtained through the analyses of the three bars, though the algebra would be evenmore cumbersome.

2.6 Problem-solving technique

In solving problems in relative motion, the following basic technique may be gen-erally applied:

1: Identify a ‘point of interest’. The point of interest is typically the point beingtracked. In the case of constrained motion (such as in simple mechanisms), it isusually necessary to first track the motion of a point with known velocity magnitudeor direction. The ‘bug-on-a-ladder’ problem is an example of this sort of constrainedmotion.


2: Identify the coordinate systems. The problem may be considerably simplified bythe careful selection of coordinate systems. A good approach is usually to identifya rotating component (if one exists), and stick a moving reference frame o′ onto therotating component with the origin located on the axis of rotation. In this way, ωaxωaxωax

will be equal to the angular velocity vector of the rotating component. Also, it istypically easiest to select the orientations of the moving unit vectors exexex

′, eyeyey′ and ezezez

′such that one of them always points towards the ‘point of interest’, if this is possi-ble. The fixed coordinate system o may be located anywhere without affecting thevelocities or accelerations; however, if the results are not required as a function ofangular position, it will simplify the problem if the moving unit vectors are alignedwith their corresponding fixed unit vectors at the instant of interest. By including anarbitrary angle θ between unit vectors in the o and o′ coordinate systems, the resultsmay be expressed as a function of θ .

3: Write down the governing equations. The relative velocity and acceleration equa-tions are handy ones to remember.

4: Identify each of the terms in the governing equations. Write down all the infor-mation available about each of the terms in the relative velocity equations.

4.1: Magnitude and direction are both known. If both the magnitude and direc-tion of a vector is known, simply substitute in the values using the unit vectorswith which they are provided. For example,

vp/o′vp/o′vp/o′ = A exexex +B e′ze′ze′z

where A and B are the magnitudes given by the problem, and the directions havealso been specified by the problem.

4.2: Only direction is known. If the direction of a vector is known but the mag-nitude is not, then leave the magnitude as an unknown scalar and resolve intocomponents along your chosen unit vectors. For example, if the problem state-ment says that the relative velocity is at an angle β in the x− y plane, then youcan write

vp/o′vp/o′vp/o′ = ξ cos(β )e′xe′xe′x +ξ sin(β )e′ye′ye′y

where ξ is an unknown magnitude for which you will need to solve later.

4.3: Only magnitude is known. If the magnitude of a vector is known but the di-rection is not, then resolve the vector into components in your chosen coordinatesystem using an unknown angle. For example, if the problem statement says thatthe relative velocity has a magnitude V in the x− y plane, then you can write

vp/o′vp/o′vp/o′ =V cos(φ)e′xe′xe′x +V sin(φ)e′ye′ye′y


where φ is an unknown angle for which you will need to solve later.

4.4: Nothing is known. If neither the magnitude nor the direction of a vector isknown, then it is usually not helpful to resolve the vector into components alongyour chosen unit vectors. Instead, simply leave the vector in the equation as avector quantity.

5: Resolve all vectors into the same coordinate system. The vectors on both sides ofthe relative motion equations must be expressed in terms of the same unit vectors,though these may be either (e′xe′xe′x, e′ye′ye′y and e′ze′ze′z) or (exexex, eyeyey and ezezez).

6: Solve the system of equations for the unknowns. Each vector equation will thenyield three scalar equations with three unknowns (one each per unit vector exexex, eyeyey

and ezezez). The scalar coefficients of each of the unit vectors on the left- and right-hand sides of the equation must equal. If the problem yields more unknowns thanequations, then no solution can be found. This will occur if the problem was under-defined (so that insufficient information was provided to uniquely define the prob-lem) or if there exist more than one possible correct solution.


2.7 Problem set 1: Coordinate systems

Question 1

The trajectory of a particle in Cartesian coordinates is given by

rrr = 4t2 cos(2πt)exexex

+ 4t2 sin(2πt)eyeyey

+ 3t2ezezez

Find an expression for the trajectory in Cylindrical coordinates, and find θ as afunction of t. Find an expression for the trajectory in spherical coordinates, and findφ as a function of t. Show that θ is independent of t, and solve for it.

Answer:Cylindrical: rrr = 4t2 ererer +3t2 ezezez, θ = 2πtSpherical: rrr = 5t2 ererer, φ = 2πt, θ = 53◦

Question 2

A girl flying a kite observes that her taut kite string forms an angle of 78◦ from thehorizon at a heading of 17.5◦ east of true North. If the girl knows she has precisely24 m of kite string played out, determine (a) how far she would have to walk Eastand how far she would have to walk North in order to be standing immediatelybelow the kite; (b) the altitude of the kite, and (c) the straight-line distance that shehad to walk.

Answer:(a) 1.50 m East, 4.76 m North; (b) 23.48 m; (c) 4.99 m

Question 3

A ship is travelling along a straight-line path at a speed of 2 m/s. The 1-m diameterpropeller rotates at a rate of two revolutions per second. Solve for the position ofthe tip of the propeller as a function of time, using cylindrical coordinates withx = y = 0 at the hub of the propeller, and ezezez points in the direction that the shipis moving. You may select any initial position for the propeller. Convert this result

2.7 Problem set 1: Coordinate systems 125

into Cartesian coordinates, and solve for the magnitude of the position vector as afunction of time.

Answer:Cylindrical: rrr = 0.5 mererer +2t mezezez

Cartesian: rrr = 0.5cos(4πt)mexexex +0.5sin(4πt)meyeyey +2t mezezez

|rrr|=√

0.25+4t2 m

Question 4

Explain why cylindrical coordinate systems are so useful in engineering analysis.


2.8 Problem set 2: Relative motion

Question 1

Fig. 2.41 Question 1

The Royal Navy frigate HMS Vanquish is overflown by an unidentified aircraftin British territorial waters, and records the velocity of the UFO as

vvv = 210e′xe′xe′x +57e′ye′ye′y m/s

where the e′xe′xe′x direction is to starboard and e′ye′ye′y is forward, as illustrated. The officer ofthe watch reports the sighting, as well as the ship’s speed and bearing at the time ofthe encounter (which was 10 m/s at 50◦ East of true North). What was the UFO’svelocity relative to a fixed observer?

Answer:186.31exexex −118.01eyeyey m/s

2.8 Problem set 2: Relative motion 127


Question 2

An ice-hockey puck is shot forward by a player such that it has a constant linearvelocity v0, and a constant angular velocity ω about an axis parallel to the directionof motion rotating anti-clockwise, as illustrated. An unfortunate bug happened to besitting on the puck when it was struck, and began running in a straight line awayfrom the centre of the puck with constant linear velocity vb. Determine the velocityand acceleration of the bug relative to a fixed observer at the instant shown, whenthe bug is a distance b from the centre of the puck, at an angle θ relative to thedirection of motion of the puck.

Answer:

vvv =(v0 + vb cos(θ)

)exexex − vb sin(θ) eyeyey +ωbsin(θ) ezezez

aaa = ω2bsin(θ) eyeyey +2ωvb sin(θ) ezezez

where exexex is in the direction of motion of the puck, and ezezez is downwards.

Question 3

A slider is pushed up a rod by a slotted crank, as illustrated. As the crank rotateswith some constant angular velocity ω , a pin on the slider is pushed with some linearvelocity v by its contact with the sides of the slot in the crank. Solve for the linearvelocity of the slider and the acceleration of the pin relative to the crank. Expressthe results as functions of ω , the distance H between the crank centre of rotationand the rod, and the angle θ subtended between the crank and horizontal.

Answer:



v =ωH

cos2(θ)

arel =ω2H

cos(θ)

(1+2

sin2(θ)

cos2(θ)

)

Question 4

Our mysterious, 6-legged UFO is now flying over the midlands. The UFO is climb-ing at a constant rate so that its trajectory is at an angle φ from horizontal. A Ter-ritorial Army mobile tracking station is chasing the UFO, driving along the groundbeneath and behind the UFO with a constant horizontal velocity of vo. The mobiletracking system can measure the elevation angle θ of the tracking dish, the sweeprate ω of the dish, as well as the range R between the dish and the UFO. Find themagnitude of the velocity v of the UFO.

Answer:

v =ωR− vosin(θ)

cos(θ)sin(φ)− cos(φ)sin(θ)

2.8 Problem set 2: Relative motion 129



Question 5

An unfortunate bug finds himself standing on a millwheel of radius R when the millstarts up. The large, round stone rolls around on the ground as the horizontal shaftof length L is rotated around the central hub at some angular velocity Ω . The bugruns for its life straight toward the rim with a constant linear velocity (relative tothe stone) of v. For convenience, some coordinate systems have been defined in the


included figure. Show that the velocity and acceleration of the bug at the instant thatthe bug is lying on the eyeyey axis are

vp/ovp/ovp/o = −Ωrexexex +(ΩL+ v)eyeyey −ΩLrR

ezezez

ap/oap/oap/o = −Ω 2L

(2vΩL

+1

)exexex −Ω 2r

(L2

R2 +1

)eyeyey −2Ωv

LR

ezezez

where r is the distance between the bug and the stone hub. Hint: the instantaneouspoint of contact ‘q’ between the mill wheel and the floor is at rest, so vq/o = 0; youshould therefore be able to show that ωR = −ΩL, where ω is the angular velocityof the mill wheel about its own axis.

Question 6

Show that


given

aaa = ax exexex +ay eyeyey +az ezezez

bbb = bx exexex +by eyeyey +bz ezezez

ccc = cx exexex + cy eyeyey + cz ezezez

Question 7

Explain why it is possible for an object to move at a constant speed |vvv|, and yet stillhave nonzero acceleration.

Chapter 3Linear momentum

Summary Conservation of linear momentum within closed systems; internal andexternal forces; definition of the centre of mass for systems of particles and solidbodies; collisions, impulse and mean impact forces.

3.1 Conservation of linear momentum

There are three fundamental laws which govern all things, and nearly everything weknow about physics may be derived from these three simple laws.

Law 1: Conservation of mass. The total mass of a closed system will remainconstant.

Law 2: Conservation of momentum. The net external force acting upon aclosed system will be equal to the time rate of change of the total linearmomentum of the system.

Law 3: Conservation of energy. The net external work done upon a closedsystem will be equal to the change in energy of the system.

There is, of course, one important exception to these laws: mass and energy arenot conserved in a thermonuclear detonation. Since nuclear physics lies outside thescope of classical mechanics, it is assumed that the objects discussed in this sectionare not thermonuclear warheads at the moment of detonation.

The law of conservation of mass is fairly straight-forward, though it is importantnot to confuse mass with weight. The mass of an object is a measure of the resistanceof the object to being accelerated, while the weight of an object is the magnitude ofthe attractive force between the object and the Earth.

The law of conservation of momentum, on the other hand, will require someadditional explanation.

131

132 3 Linear momentum

3.1.1 Definitions

A system is defined in this context simply as ‘a collection of things’. There are norestrictions on what ‘things’ are being collected: particles, masses, machine parts,planets or takeaway pizzas. Everything which is not part of the system is togetherdefined as the environment. A closed system is defined as a system which does notgain anything from or lose anything to the environment. As an example, ‘all of thepeople on the university campus’ is not a closed system, since people can changefrom being part of the environment to being part of the system by simply enter-ing the campus grounds. On the other hand, ‘all the students of the university’ is aclosed system, since the people forming the system will remain part of the systemeven if they leave the campus, and visitors to the campus will not become part of thesystem (note, however, that this is only true over a relatively short period of time;over a longer period, the system would not be closed, as some students will grad-uate and new students will register, changing from being part of the environmentto being part of the system or vice-versa). Closed systems are often referred to asLagrangian systems, after the Italian mathematician Giuseppe Lodovico Lagrangia(1736-1813).

An external force is a force being applied between something which is part ofthe system and something which is part of the environment, and is indicated by thevector FextFextFext. By contrast, an internal force is a force acting between things whichare all part of the system. If our system again is defined as ‘all the students of theuniversity’, and a student is shoved out of the way by a visitor to the university,then that force (the shove) was an external one. If the student was shoved by anotherstudent, then the force was an internal one- regardless of where the shove took place.Gravity is an external force as well, as it is a force acting between the students(which are part of the system) and the Earth (which is part of the environment). Anet external force is the vector sum of all of the external forces acting between theenvironment and all the various things in the system.

The linear momentum of an object is defined as the product of its mass andits velocity, in SI units of kilogram-metres per second. The linear momentum isusually referred to by the symbol GGG. Then, the definition of linear momentum maybe expressed symbolically as

GGG = mvvv

where m is the mass of the object, vvv is its velocity vector. Since the linear momen-tum is the result of a scalar multiplied by a vector, the linear momentum is also avector quantity. The symbol used to represent the linear momentum may vary fromtextbook to textbook; here, we have adopted the notation of Meriam & Kraige.

The total linear momentum of a system is the vector sum of the linear momen-tums of all of the objects which are part of the system. The time rate of changeof the total linear momentum, then, is simply the time-derivative of the total linearmomentum. Finally, we may express the law of conservation of linear momentumsymbolically as

3.1 Conservation of linear momentum 133

ddt

(∑GGG

)= ∑FextFextFext

where the sigma operator (Σ ) here implicitly indicates a vector sum.

While this is a more general expression of the law of conservation of momentum,a well-known result may be obtained if we consider the case of a single object. If thesystem is defined so that it consists of a single object alone, then the summation isredundant. Furthermore, since the system only consists of the one object, any forceacting on that object will be an external force- so the subscript ‘ext’ may be droppedfrom the force vector.

ddt

GGG = FFF

If we further assume that the mass of object is constant, it may be pulled out of thederivative to yield,

ddt

(mvvv)= FFF

mddt

vvv = FFF

maaa = FFF

where aaa is the vector acceleration of the object. This far more familiar result is there-fore a constant-mass special case of the conservation of linear momentum. Further-more, we can note that, in the special case when there are no external forces actingupon the system (or when the net external force is exactly zero),

ddt

(∑GGG

)= ∑FextFextFext = 0

∑GGG = ∑(mvvv) =C

where C is a constant. This result requires that, in the absence of forces acting be-tween the system and the environment, the total momentum of the system cannotchange. As a consequence, if a system is initially at rest, then C = 0 and the centreof mass of the system must always remain at rest.


3.1.2 Implications of the law of conservation of linear momentum

Observation 1: Rockets can’t move. Absurd as it may sound, one of the implicationsof the law of conservation of momentum is that a rocket floating stationary in spacecannot actually move if it fires its engines. If we define the system as the rocketand everything which the rocket contains, then there can be no forces acting onthe rocket external to the system (we assume that there are no gravitational fields).If there can be no forces acting on the rocket externally, then the momentum ofthe rocket cannot change, and therefore the rocket cannot move no matter whatit does. Yet, if a rocket in space fires its engines, it will accelerate forward. Theapparent discrepancy between the result predicted by the law of conservation oflinear momentum and the result observed is because of how we defined ‘the rocket’.

Fig. 3.1 A rocket firing its engines. The exhaust plume remains part of the ‘rocket’.

The key is that we defined the system as the rocket and everything which therocket contains. The rocket is propelled forward by the expulsion of a high-speedjet of exhaust gases. Since the gases were inside the rocket when we defined thesystem, they remain part of the system throughout- much like students wanderingoff the campus in our earlier example. Once the rocket has fired its engines, thesystem includes the rocket plus its entire exhaust plume, as illustrated in Figure 3.1.If we were to compute the vector sum of the momentum of every particle of exhaustgas at a given point in time, the result would be exactly equal in magnitude andopposite in direction as the momentum of the rocket fuselage and everything left onboard. If, on the other hand, we defined the ‘rocket’ system as the rocket fuselageand payload only (that is, everything except the fuel), then the violent expulsion ofthe exhaust gases from the engine nozzle would constitute an external force actingbetween the system (the engine nozzle is part of the system) and the environment


(the exhaust gas, which is not part of the system), and the rocket can accelerate. So,as long as the ‘rocket’ includes all the fuel, rockets indeed don’t move.

Observation 2: The Earth moved when you accidentally dropped your keys. Imaginethat you have accidentally dropped your keys, thereby allowing them to acquiresome momentum before striking the ground. If we define the ‘system’ here as theEarth and everything which it contains, then both the keys and the Earth are partof the system. If we neglect the tidal pull of the moon and the gravitational pull ofthe sun (it is a reasonable approximation that these forces have little effect on theinteraction between the Earth and your keys), there are no external forces acting onthe system. The Earth, then, is just like the rocket: floating alone in space. Therefore,if there is no net external force acting on the system, the total momentum within thesystem must always be zero. So, if your keys were pulled toward the Earth withsome velocity, then the Earth has been likewise pulled upwards to meet your keysin order to satisfy the requirement that ΣGsysGsysGsys = 0. However, given the difference inmass between the keys and the rest of the Earth, the resulting change in velocity ofthe Earth would be unmeasurably small. In fact, any motion of any object with masson the surface of the Earth causes an unmeasurably small (but indeed real) changein the velocity of the Earth.

Observation 3: If you jump out of an airplane, the amount of momentum you impartupon the Earth is exactly the same whether or not you have a parachute. If youjump out of an airplane with a parachute, you will gently float down to the groundand impact at about 3 m/s. If you jump out of an airplane without a parachute,you will initially impact the ground at about 60 m/s and then bounce once or twice(depending on your body-mass index and the local ground conditions). If we definethe system again as the Earth and all that it contains, then you and the Earth areagain part of the same system. In the absence of external forces (we reasonablyneglect, as before, the tidal pull of the moon and the gravitational pull of the sun),the total momentum in the system must remain the same no matter what. Without theparachute, the bulk of your momentum is transferred to the ground in the collision.With the parachute, a small amount of momentum is still transferred to the Earthon touch-down, but the bulk of the momentum was transferred to the air by theparachute during your descent. Since we defined the system as being the Earth andall that it contains, the air is part of the system too.

3.1.3 Newton’s three ‘Laws’

In the study of classical mechanics, it is still common to refer to Sir Isaac Newton’s‘three laws of motion’, which first appeared in 1687 in his groundbreaking work onclassical mechanics, Philosophiæ Naturalis Principia Mathematica. While Newtonis indeed credited with the first demonstration of a law of conservation of linearmomentum, he is often misquoted.


The three ‘laws’, as they appear in the original text, are

Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uni-formiter in directum, nisi quatenus a viribus impressis cogitur statum illummutare.

Lex II: Mutationem motus proportionalem esse vi motrici impressæ, & fierisecundum lineam rectam qua vis illa imprimitur.

Lex III: Actioni contrariam semper & æqualem esse reactionem: sive cor-porum duorum actiones in se mutuo semper esse æquales & in partes con-trarias dirigi.

In translating these ‘laws’ from the Latin, many authors will skip directly to modernmathematical expressions. In order to preserve the original tone and intent, a morefaithful translation of the three ‘laws’ is included here:

Law I: Every body perseveres in a state of rest, or of uniform motion anddirection, unless it is caused to change that state by forces impressed uponit.

Law II: The change of motion is always proportional to the motive forceacting, and is in the direction of the line in which that force is acting.

Law III: Contrary to every action there is always an equal reaction: or theactions of two bodies upon each other are always equal, and oriented inopposing directions.

It may now be recognized that all three of these ‘laws’ are corollaries of the lawof conservation of linear momentum for closed systems. The second ‘law’ definesmomentum as a vector quantity, and indicates that the rate of change of momentumwill be a scalar multiple of the force. The third ‘law’ distinguishes between internaland external forces, and indicates that only external forces can change the momen-tum of a system. The first law is somewhat redundant, as it describes a special caseof the second ‘law’. It is also important to note that Newton’s ‘laws’ are incom-plete, in that they are only applicable to systems of two or less objects of constantmass, acted upon by a single force. For these reasons, it is inappropriate to think ofthese statements as laws; by definition, you cannot derive one ‘first principle’ fromanother.

Despite their historical significance and validity, Newton’s ‘laws’ are outdated,and should not be used in modern engineering analysis.

It is also interesting to note that, contrary to common belief, the expression‘F = ma’ does not appear anywhere in Newton’s original text.


3.1.4 Centre of mass

Let us consider again the case of the rocket firing its engines in space. According tothe law of conservation of linear momentum, the rocket system is not moving. How-ever, the rocket fuselage is moving in one direction and the rest of the system (theexhaust plume) is moving in the other direction. If the system as a whole is remain-ing stationary, then where, exactly, is it? To address this problem, let us examine thesimpler case of a blob of solid rocket fuel, as illustrated in Figure 3.2.

Fig. 3.2 A small blob of rocket fuel of mass mo is at rest before it explodes in space, releasing acloud of gas particles each with mass mi and velocity vector vivivi.

A small blob of rocket fuel of mass mo is initially at rest in space. It is thenignited without imparting any force upon it, causing it to explode into a cloud ofgas particles each with mass mi and velocity vector vivivi. If we define the system asthe blob of fuel, then since there were no external forces and the blob was initiallyat rest, the expanding cloud of exhaust gas must also be at rest. From the law ofconservation of momentum,

ddt

(∑GGG

)= ∑FextFextFext

Since there were no externally applied forces,

ddt

n

∑i=1

GiGiGi = 0

n

∑i=1

GiGiGi = CCC =GsysGsysGsys

where GiGiGi is the momentum of the ith particle, GsysGsysGsys is the momentum of the ‘system’which must remain constant, CCC is some constant vector quantity, and it is assumed


that there are a total of n gas particles released by the blast. Expanding the summa-tion and recognizing that GiGiGi = mivivivi,

n

∑i=1

(mivivivi) = m1v1v1v1 +m2v2v2v2 + ...+mnvnvnvn = msysvsysvsysvsys

where msys is the mass of the ‘system’ and GsysGsysGsys is the momentum vector of the‘system’. Now, if we wanted to, we could integrate both sides with respect to time,so that

n

∑i=1

∫(mivivivi)dt =

∫m1v1v1v1dt +

∫m2v2v2v2dt + ...+

∫mnvnvnvndt =

∫msysvsysvsysvsysdt

Since the masses of the particles do not change in time, the mass can be pulled outof the integral.

n

∑i=1

(mi

∫vivividt) = m1

∫v1v1v1dt +m2

∫v2v2v2dt + ...+mn

∫vnvnvndt = msys

∫vsysvsysvsysdt

However, we already know that the integral of the velocity vector with respect totime is equal to the position vector rrr, since vvv is defined as rrr. The constants of inte-gration are all zero, because in the case where ririri = 0 for all i, then rsysrsysrsys = 0 as well.Then, integrating,

n

∑i=1

(miririri) = m1r1r1r1 +m2r2r2r2 + ...+mnrnrnrn = msysrsysrsysrsys

Now, this result is interesting. The vector rsysrsysrsys points to the location in space wherethe ‘system’ is. So, as all of the gas particles fly apart after the blast and the systemremains stationary, this vector rsysrsysrsys tells us where the ‘system’ is while it remainsstationary. Isolating rsysrsysrsys,

rsysrsysrsys =∑(miririri)

msys=

m1r1r1r1 +m2r2r2r2 + ...+mnrnrnrn

msys

where the limits of summation have been omitted for clarity. Now, from the law ofconservation of mass, we know that the mass of the system cannot change so thatthe sum of all the little masses must always exactly equal msys, and that msys = mo.Then,

rsysrsysrsys =∑(miririri)

∑mi=

m1r1r1r1 +m2r2r2r2 + ...+mnrnrnrn

m1 +m2 + ...+mn

Since rsysrsysrsys and ririri are vector quantities and the summations cannot change the direc-tion of a unit vector, we can resolve rsysrsysrsys into components lying parallel to each ofthe exexex, eyeyey and ezezez axes. If we let ririri = xiexexex + yieyeyey + ziezezez, then


rsysrsysrsys =∑(mixi)

∑miexexex +

∑(miyi)

∑mieyeyey +

∑(mizi)

∑miezezez (3.1)

This is an extremely important result, and rsysrsysrsys is more commonly known as thecentre of mass of the system (and is commonly indicated as rcrcrc = xcexexex+yceyeyey+ zcezezez).The law of conservation of linear momentum then requires that, in the absence ofexternal forces, the centre of mass of a system must remain stationary.

Another important implication of this result is that the internal forces may beentirely ignored when treating systems upon which external forces are applied. Thecentre of mass of the system will accelerate in response to the applied externalforce, regardless of what the individual pieces of the system are doing. For example,consider a cloud of stationary particles floating in space, as illustrated in Figure 3.3.If a constant force FFF were applied to a single particle of mass ma, that particle wouldexperience an acceleration of aaaaaa, as

Fig. 3.3 A system of stationary particles with a force FFF applied to a single one.

FFF = GaGaGa

=ddt(mavavava)

= maddt

vavava

FFF = maaaaaaa

where we have implicitly defined our closed system as consisting of only the oneparticle, and va is the velocity of that particle. On the other hand, if we define oursystem as consisting of all of the particles, then the position vector of the centre ofmass of the system rcrcrc will be given by,


rcrcrc =∑miririri

∑mi

and the acceleration of the centre of mass of the system can be obtained by differ-entiating rcrcrc twice with respect to time, as

acacac =d2

dt2 rcrcrc

=d2

dt2

(∑miririri

∑mi

)

acacac =∑miririri

∑mi

where ririri is the acceleration of the ith particle, and the mass of each particles isassumed constant. However, since all of the particles but one remain at rest,

∑miririri = mararara

where rarara = aaaaaa by definition. Substituting this expression into our result for acacac,

acacac =∑miririri

∑mi

acacac =maaaaaaa

msysmaaaaaaa = msysacacac

where msys = Σmi is the total mass of the system. Substituting this back into ourresult from the analysis of the single particle,

FFF = maaaaaaa

FFF = msysacacac

which shows that the centre of mass of the system is accelerating as though thesystem were a single object of mass msys. A system of particles may therefore beeffectively reduced to a single particle with mass equal to the total mass of thesystem sitting on top of the centre of mass of the system, even if the individualparticles within the system are moving and exerting forces upon one another.

Let us now consider once again the blob of rocket fuel before the explosion, asillustrated in Figure 3.2. If we consider the blob as being a bunch of gas particles


‘glued’ together into one homogeneous mass so that the edges of the particles areno longer distinguishable, we can still find the centre of mass of the system. In thiscase, we still use Equation (3.1), but since the elements are no longer discrete, wehave to turn the sums into integrals. As the size of a particle approaches zero, theposition xi doesn’t change, but the mass gets very small. Then, m becomes a tinydifferential quantity dm, and Equation (3.1) becomes,

rcrcrc =

∫x dm∫dm

exexex +

∫y dm∫dm

eyeyey +

∫z dm∫dm

ezezez

If we can now assume that the rocket fuel had constant density ρ , then dm = ρdV ,where dV is a tiny differential volume.

rcrcrc =

∫x ρdV∫ρdV

exexex +

∫y ρdV∫ρdV

eyeyey +

∫z ρdV∫ρdV

ezezez

Since ρ is constant, it can be pulled out of the integrals. Also, in Cartesian coordi-nates, a small rectangular volume dV will be equal to (dx dy dz). Therefore,

rcrcrc =ρ

�x dx dy dz

ρ�

dx dy dzexexex +

ρ�

y dx dy dz

ρ�

dx dy dzeyeyey +

ρ�

z dx dy dz

ρ�

dx dy dzezezez

The densities may then be divided out, so that

rcrcrc =

�x dx dy dz�dx dy dz

exexex +

�y dx dy dz�dx dy dz

eyeyey +

�z dx dy dz�dx dy dz

ezezez

It is easily recognized that the denominator is the total volume of the blob.

Finally, it is important to notice that the acceleration of the centre of mass acacac ofour system of particles (whether or not they are glued together to form a solid object)does not depend on where the force FextFextFext is applied; it is only required that the forcebe applied to something which is part of the system. Consequently, if a force FFF isapplied anywhere on the surface of a solid object (or even within the volume of theobject itself), then the centre of mass of the object will have an acceleration vectoracacac, given by

FFF = msysacacac = mrcrcrc

(where m is the mass of the object) regardless of the point of application of the force.


3.2 Collisions

A collision is defined as an interaction between objects causing momentum to betransferred between the objects, whether or not they are a part of the same system.If we define a system such that it contains all of the objects which are interacting,then, since there are no interactions outside of the system, there are no externalforces and ΣGGG = 0. Under these circumstances,

∑mivivivi =C

where C is some constant. Alternatively, we could also write,

(∑mivivivi

)1 =(∑mivivivi

)2

where the subscripts ‘1’ and ‘2’ refer to two different instants in time t1 and t2.Expressing this in terms of the Cartesian unit vectors,

(∑mixiexexex +∑miyieyeyey +∑miziezezez

)1 =(∑mixiexexex +∑miyieyeyey +∑miziezezez

)2

where xi, yi and zi are the magnitudes of the velocity of the ith object along the exexex,eyeyey and ezezez axes, respectively. Gathering together the scalar coefficients of the unitvectors, this can be re-expressed as,

(∑mixiexexex

)1 =

(∑mixiexexex

)2(

∑miyieyeyey)

1 =(∑miyieyeyey

)2(

∑miziezezez)

1 =(∑miziezezez

)2

This indicates that the linear momentum is not only conserved, but it is also con-served independently along each of the three axes. There is no way to convert x-momentum into y-momentum without applying an external force. Note that the unitvectors may be divided out, so that

(∑mixi

)1 =

(∑mixi

)2(

∑miyi)

1 =(∑miyi

)2(

∑mizi)

1 =(∑mizi

)2

The masses should not be divided out of the equation, since there is no rea-son why the objects colliding cannot exchange mass with one another during thecollision. There may also be a different number of masses after the collision hasoccurred, as was the case in the example of the exploding blob of rocket fuel.

3.2 Collisions 143

There are three special types of collisions which will themselves be treated ingreater detail in Chapter 6:

A perfectly elastic collision is one in which the objects bounce off of eachother during the collision, and no kinetic energy is lost or gained by theobjects in the process.

A perfectly inelastic collision is one in which the objects collide and sticktogether thereafter.

An explosive collision is one in which the objects exert some internal forceagainst each other to fling themselves away from each other.

Our earlier case of a rocket spewing exhaust gas is an example of an explosivecollision (the exhaust gas is flinging itself away from the rocket nozzle), whereasyour keys striking the ground when you dropped them is an example of a perfectlyinelastic collision (the keys stuck to the ground after the impact).

It is also important to note that the objects do not necessarily have to come intophysical contact with each other in order for a collision to occur. Two magnets withlike polarities may bounce off each other without physically touching; likewise, astray asteroid whose trajectory in space has been altered by the gravity of a planethas ‘collided’ with the planet, even if the effect on the asteroid’s trajectory wassmall.

3.2.1 Momentum and collisions

The analysis of simple collisions between objects is probably best illustrated bydiscussing several typical examples. Consider two bodies A and B having massesma and mb and initial velocities (vavava)1 and (vbvbvb)1 such that

(vavava)1 = (xa)1 exexex +(ya)1 eyeyey +(za)1 ezezez

(vbvbvb)1 = (xb)1 exexex +(yb)1 eyeyey +(zb)1 ezezez

In the first instance, let us examine the case where the two bodies collide and carryon moving away from each other such that the post-collision velocity of A is knownto be (vavava)2, where

(vavava)2 = (xa)2 exexex +(ya)2 eyeyey +(za)2 ezezez

as illustrated in Figure 3.4. In this case, the post-collision velocity of B is unknown,and can be solved for.


Fig. 3.4 A typical collision.

To begin with, we define the system as consisting of bodies A and B. Since thereis no interaction with anything outside of the system, there are no external forces.Expressing the law of conservation of linear momentum in vector form,

∑GsysGsysGsys = 0

∑GsysGsysGsys = C0

where C0 is some constant. Since there are only two bodies in the system, the sumwill contain only two terms. This can be expressed explicitly as

GaGaGa +GbGbGb = C0

mavavava +mbvbvbvb = C0

ma(vavava)1 +mb(vbvbvb)1 = ma(vavava)2 +mb(vbvbvb)2

noting that (ma)1 = (ma)2 = ma as no mass is exchanged between the bodies. Sincewe want to determine the post-collision velocity of B, we can isolate that term as

(vbvbvb)2 =ma(vavava)1 +mb(vbvbvb)1 −ma(vavava)2

mb

(vbvbvb)2 =ma

mb

((vavava)1 − (vavava)2

)+(vbvbvb)1

This result can also be expressed in terms of the Cartesian unit vectors as

3.2 Collisions 145

(vbvbvb)2 =

(ma

mb

((xa)1 − (xa)2

)+(xb)1

)exexex

+

(ma

mb

((ya)1 − (ya)2

)+(yb)1

)eyeyey

+

(ma

mb

((za)1 − (za)2

)+(zb)1

)ezezez

This result will hold true for all collisions between two bodies, providing that nomass has been exchanged between them.

Let us next consider the case of a perfectly inelastic collision, in which body Ais stuck to body B in the final condition, as illustrated in Figure 3.5. Let us call theresultant body C; then, mc = ma+mb as required by the law of conservation of mass.The velocity of C after the collision is unknown, and can be solved for.

Fig. 3.5 A perfectly inelastic collision.

This time, from the law of conservation of linear momentum, we have

ma(vavava)1 +mb(vbvbvb)1 = mc(vcvcvc)2

ma(vavava)1 +mb(vbvbvb)1 = (ma +mb)(vcvcvc)2

Isolating (vcvcvc)2 yields,

(vcvcvc)2 =ma(vavava)1 +mb(vbvbvb)1

ma +mb

which, again, can be expressed in terms of the Cartesian unit vectors as


(vcvcvc)2 =ma(xa)1 +mb(xb)1

ma +mbexexex

+ma(ya)1 +mb(yb)1

ma +mbeyeyey

+ma(za)1 +mb(zb)1

ma +mbezezez

This result will be true of all perfectly inelastic collisions between two bodies. Itis very important to note here that the final velocity of the system (vcvcvc)2 does notdepend on the energy lost to material deformation during the collision. It does notmatter whether the objects A and B are made of putty or hardened steel; if they havethe same mass and initial velocity, and if they stick together after the collision, thefinal velocity must be the same.

Finally, let us consider the case where the two bodies collide with each otherand both come to a complete rest. In this case, only (vavava)1 is known; the initial ve-locity of body b is to be determined. Again, from the law of conservation of linearmomentum,


ma(vavava)1 +mb(vbvbvb)1 = ma(0)+mb(0) = 0

Isolating (vbvbvb)1 yields,

(vbvbvb)1 =−ma

mb(vavava)1

Since the ratio ma/mb is a scalar, this result shows that the vector (vbvbvb)1 must beparallel with the vector (vavava)1, but pointing in the opposite direction. The only wayfor the two bodies to come to rest after a collision is if the collision is head-on. Itis also interesting to note that this result is identical to the result for the perfectlyinelastic collision if the final velocity of the combined body C is zero. Consequently,it doesn’t matter what shape the masses become or whether or not they combine orexchange mass if the entire mass comes to rest following the collision.

3.2.2 Impulse

Let us consider more carefully the collision between two bodies. For the case ofbodies made of real materials, the bodies will have some finite elasticity and willtherefore be able to deform. All real materials are elastic, though some are muchmore elastic than others. Also, all real materials will deform when a force is applied

3.2 Collisions 147

to them, regardless of how stiff the material or how small the force. For example,if you deposited a grain of rice on the surface of a diamond, the diamond will sagand bend under the weight of the grain of rice. The amount of sag, however, will beunmeasurably small.

When deformable materials collide, the collisions last for some small but finiteamount of time Δ t. The magnitude of the internal force FFF exerted by one body onthe other will be a function of time during that small Δ t, as illustrated in Figure 3.6.

Fig. 3.6 Time-step illustration of the collision between two elastic bodies, showing a plot of themagnitude of the internal force vs. time.

The area under the function FFF(t) is defined as the impulse, which is denoted bythe symbol JJJ and has SI units of kg × m/s, which are also units of momentum.Mathematically, the impulse is given by the expression,

JJJ =∫ t0+Δ t

t0FFF(t) dt

where the collision begins at time t = t0. Since the force function FFF(t) is a vectorquantity, the impulse is a vector quantity as well. The average force FFF(t) exerted byone body upon the other may then be calculated as

FFF(t) =JJJ

Δ t

This average force represents the magnitude of the constant force one body wouldhave to apply upon the other during that same amount of time Δ t in order to generatethe same amount of impulse.

The significance of the impulse can be appreciated if we consider again the defi-nition of the linear momentum,

ddt

(∑GGG

)= ∑FextFextFext


If we now redefine our system so that it consists of one of the bodies only, then theforce applied to that body by the other ceases to be an internal force and becomesan external force. Considering then our new one-body system,

ddt

GGG = FFF(t)

Into this expression we can substitute the definition GGG = mvvv and integrate both sideswith respect to time from t = t0 to t = t0 +Δ t to yield,

∫ t0+Δ t

t0

(ddt

GGG

)dt =

∫ t0+Δ t

t0FFF(t) dt

GGG

∣∣∣∣t0+Δ t

t0

=∫ t0+Δ t

t0FFF(t) dt

(mvvv)2 − (mvvv)1 = JJJ

where the subscript 2 indicates the final condition at time to+Δ t, and the subscript 1indicates the initial condition at time to. So, if you know the time-history of the forceapplied to a body during a collision, you can calculate the change in the momentumof the body between t = t0 and t = t0 +Δ t. Much more simply, if you know theaverage force applied during the collision, then

(mvvv)2 − (mvvv)1 = FFF(t)Δ t

since we showed above that FFF(t) = JJJ/Δ t. This can also be expressed in terms of theCartesian unit vectors as

Δ(mx) exexex = Fx(t)Δ t exexex

Δ(my) eyeyey = Fy(t)Δ t eyeyey

Δ(mz) ezezez = Fz(t)Δ t ezezez

where the mean force vector is given by FFF(t) = Fx(t)exexex +Fy(t)eyeyey +Fz(t)ezezez. Then,considering only the scalar coefficients of each of the unit vectors,

Δ(mx) = Fx(t)Δ t

Δ(my) = Fy(t)Δ t

Δ(mz) = Fz(t)Δ t

3.3 Historical note: Origins of the conservation laws 149

This result indicates that if you can measure the change in momentum of an objectafter it has collided with something and you know how long the collision lasted,you can calculate the average force experienced by the object. Unfortunately, havingonly the change in momentum and the duration of the collision, it is impossible todetermine the shape of the function FFF(t) or the maximum force to which the objectwas subjected.

Fig. 3.7 Illustration of two force functions FFF(t) with the same impulse.

In fact, the shape of the function FFF(t) will depend heavily on the elasticity ofthe body. Given a collision causing the same change in momentum (and thereforethe same impulse and average external force), a very hard object will have a muchshorter interaction time and a much larger peak force, while a softer material willhave a longer interaction time and a much lower peak force. This occurs becausethe harder material will spring back more quickly. This phenomenon is illustratedgraphically in Figure 3.7.

3.3 Historical note: Origins of the conservation laws

Though they are used nearly universally in modern engineering analysis, the originsof the conservation laws are fairly obscure. The Greek philosopher Aristotle (384-322 BC) alluded to the law of conservation of mass (“one thing comes to be fromanother thing, and one sort of thing from another sort of thing,” Physica I:6) and thelaw of conservation of momentum (“everything that is in motion must be moved bysomething,” Physica VI:10). A formal law of conservation of mass was later explic-itly proposed by Epicurus (341-270 BC); in a letter to Herodotus, he wrote, “nothingcomes into existence from non-existence; [...] the universe has always been as it isnow, and always will be.” The law of conservation of mass was finally described,


explicitly and mathematically, by the French chemist Antoine-Laurent de Lavoisier(1743-1794), who was accidentally guillotined during the revolution.

The first proper mention of a law of conservation of momentum is sometimes at-tributed to the philosopher and physician Abu Ali Al-Hussein ibn-Sina (980-1037),who described a quantity which he called impetus, defined as mass × velocity. En-glish mathematician John Brehaut Wallis (1616-1703) analyzed the collision of in-elastic bodies in response to a call from the Royal Society in 1668, and also recog-nized the importance of this quantity in describing the interactions between objects.Though it was somewhat incomplete (as discussed earlier), the first formal and rig-orous mathematical definition of a law of conservation of momentum is typicallycredited to Sir Isaac Newton, as it appeared in his Philosophiæ Naturalis PrincipiaMathematica in 1687.

Though the early Greek philosophers had (in some sense) proposed a quantitywhich we would call ‘energy’, this tended to be very difficult for them to explicitlydefine. It is Galileo Galilei (1564-1642) who is usually credited with the first obser-vation of the conservation of energy during his experiments with pendulums, thoughhe did not explicitly propose a conservation law. Although he is better known forproviding us with the modern notation for differential calculus, the German math-ematician Gottfried Wilhelm Leibniz (1646-1716) was also the first to describe theprinciples of kinetic and potential energy, and proposed that the total energy in asimple mechanical system should be conserved. Many physicists had believed thatenergy was irrelevant in mechanical systems (or that the conservation of energycould be derived from the conservation of momentum); it was primarily the earlyengineers who recognized the importance of considering energy independently frommomentum.

Finally, in 1918, German mathematician Emmy Noether (1882-1935) proved thatany quantity which behaves in the same manner regardless of orientation will beconserved. Since the mass, momentum and energy of an object do not change de-pending on the orientation of an object, according to Noether’s so-called First The-orem, all three of these quantities must be conserved. Technically, Noether’s FirstTheorem supersedes the three conservation laws, as these can be derived directlyfrom it; however, the laws of conservation of mass, momentum and energy continueto be accepted and used as first principles in engineering analysis owing to theirsimplicity and familiarity. Described by Albert Einstein as one of the greatest math-ematicians ever to have lived, Noether made significant contributions to physicsand mathematics, and died tragically from complications following relatively minorsurgery.

3.4 Examples 151

3.4 Examples

3.4.1 Elastic collision

A 200-g ice hockey puck initially traveling at 15 m/s at an angle of 20◦ from hori-zontal is struck by a player’s stick. The puck leaves the player’s stick at a speed of20 m/s, 80◦ from horizontal. If the interaction between the puck and the stick lastedfor 0.02 seconds, compute (a) the average force exerted upon the puck by the player,and (b) the velocity imparted upon the 100-kg player by the impact. Assume that theice is a frictionless surface, and that the player was initially at rest.

Fig. 3.8 Elastic collision

SOLUTION

To solve this problem, we recognize that the average force can be determinedfrom the impulse, and the impulse can be determined by the change in momentumof the puck. Then, if the system is defined as containing the player and the puck,any change in momentum of the puck must equal the change in momentum of theplayer.

The first step, then, is to determine the change in the momentum of the puck. Todo this, we start by resolving the initial velocity of the puck into components lyingalong the exexex and eyeyey directions. This is easily done, since the magnitude and directionare given.

vp1vp1vp1 = x1exexex + y1eyeyey

vp1vp1vp1 = vi cos(θ1)exexex + vi sin(θ1)eyeyey

where vp1vp1vp1 is the initial velocity vector of the puck, vi is the initial scalar speed of thepuck (15 m/s), and θ1 is the angle subtended between the initial velocity vector andthe horizontal (20◦). So, the initial momentum vector Gp1Gp1Gp1 of the puck is


Gp1Gp1Gp1 = mpvp1vp1vp1

Gp1Gp1Gp1 = mpvi cos(θ1)exexex +mpvi sin(θ1)eyeyey

where mp is the mass of the puck (0.2 kg). Similarly, the final velocity of the puckcan be resolved as

vp2vp2vp2 = x2exexex + y2eyeyey

vp2vp2vp2 = v f cos(θ2)exexex − v f sin(θ2)eyeyey

where vp2vp2vp2 is the final velocity vector of the puck, v f is the final scalar speed of thepuck (20 m/s), and θ2 is the angle subtended between the final velocity vector andthe horizontal (80◦). Note that, since the direction of the vertical velocity componenthas reversed, so has the sign. Then, the final momentum vector of the puck Gp2Gp2Gp2 is

Gp2Gp2Gp2 = mpv f cos(θ2)exexex −mpv f sin(θ2)eyeyey

The change in the momentum vector can then be computed as

ΔGpGpGp = Gp2Gp2Gp2 −Gp1Gp1Gp1

=(mpv f cos(θ2)exexex −mpv f sin(θ2)eyeyey

)− (mpvi cos(θ1)exexex +mpvi sin(θ1)eyeyey)

ΔGpGpGp = mp(v f cos(θ2)− vi cos(θ1)

)exexex +mp

(−v f sin(θ2)− vi sin(θ1))

eyeyey

and all of the quantities in this equation are known. Next, we know from the impulse-momentum equation that

ΔGGG = FFFΔ t

We know that the duration of the collision Δ t = 0.02 seconds, and we have justcalculated the change in momentum during this time, so we can solve for the averageforce vector FFF = Fxexexex +Fyeyeyey as

FFF =ΔGGGΔ t

So,

Fx =mp

Δ t

(v f cos(θ2)− vi cos(θ1)

)

3.4 Examples 153

Fy =mp

Δ t

(−v f sin(θ2)− vi sin(θ1))

Finally, plugging in some numbers, we obtain for the x-component

Fx =(0.2 kg)(0.02 s)

((20 m/s)cos(80◦)− (15 m/s)cos(20◦)

)= −106.2 kg

ms2 =−106.2 N

and for the y-component,

Fy =(0.2 kg)(0.02 s)

(−(20 m/s)sin(80◦)− (15 m/s)sin(20◦))

= −248.3 kgms2 =−248.3 N

So the mean force applied to the puck is FFF = −106.2 N exexex − 248.3 N eyeyey, whichanswers part (a) of the question. For part (b), we now redefine the system to includethe hockey player as well. Then, the total change in momentum of the system mustbe zero since there are no external forces acting upon the system. Then, if we saythat the momentum of the player is GqGqGq,

ΔGsysGsysGsys = Δ(GpGpGp +GqGqGq) = 0

ΔGsysGsysGsys = (Gp2Gp2Gp2 +Gq2Gq2Gq2)− (Gp1Gp1Gp1 +Gq1Gq1Gq1) = 0

However, we know that the initial velocity (and therefore the initial momentum) ofthe player is zero.

(Gp2Gp2Gp2 +Gq2Gq2Gq2)− (Gp1Gp1Gp1 +0) = 0

Now we can isolate the final momentum vector of the player, so that

Gp2Gp2Gp2 +Gq2Gq2Gq2 −Gp1Gp1Gp1 = 0

Gq2Gq2Gq2 = Gp1Gp1Gp1 −Gp2Gp2Gp2

Expressing in terms of the Cartesian unit vectors, and recalling that GGG = mvvv


Gq2Gq2Gq2 = Gp1Gp1Gp1 −Gp2Gp2Gp2

mqvqxexexex +mqvqyeyeyey =(mpvi cos(θ1)exexex +mpvi sin(θ1)eyeyey

)− (mpv f cos(θ2)exexex −mpv f sin(θ2)eyeyey

)

where mq is the mass of the player, and vqx and vqy are the x- and y-components ofthe player’s velocity after the impact, respectively. Collecting the coefficients,

mqvqxexexex +mqvqyeyeyey =(mpvi cos(θ1)−mpv f cos(θ2)

)exexex

+(mpvi sin(θ1)+mpv f sin(θ2)

)eyeyey

Equating the scalar coefficients, for the x-component,

mqvqx = mpvi cos(θ1)−mpv f cos(θ2)

vqx =mp

mq

(vi cos(θ1)− v f cos(θ2)

)

Similarly, for the y-component,

mqvqy = mpvi sin(θ1)+mpv f sin(θ2)

vqy =mp

mq

(vi sin(θ1)+ v f sin(θ2)

)

Substituting in values, we obtain

vqx =0.2 kg100 kg

((15 m/s)cos(20◦)− (20 m/s)cos(80◦)

)= 0.021 m/s

vqy =0.2 kg100 kg

((15 m/s)sin(20◦)+(20 m/s)sin(80◦)

)= 0.050 m/s

So the final velocity vector of the player is vqvqvq = 0.021 m/s exexex +0.050 m/s eyeyey, whichis the answer to part (b) of the question.

3.4.2 Perfectly inelastic collision

A 100-g set of keys, initially at rest, is dropped on the surface of the Earth. Themutual gravitational pull between the two objects (the Earth and the keys) causes

3.4 Examples 155

the keys to strike the Earth with a velocity of 1.5 m/s. Compute the velocity of theEarth at the moment of impact if the mass of the Earth is 5.97×1024 kg.

Fig. 3.9 Perfectly inelastic collision

SOLUTION

First, we define the system as containing both the Earth and the keys. Then,according to the law of conservation of momentum, the total momentum of the Earthand the keys must remain constant throughout, since there are no forces acting oneither the Earth or the keys originating from outside the system. Remember herethat the force of gravity is acting between the parts of the system, so gravity is aninternal force and does not affect the total momentum within the system (gravity isdiscussed in detail in Chapter 4). Then,

ΔGsysGsysGsys = (Gk2Gk2Gk2 +Ge2Ge2Ge2)− (Gk1Gk1Gk1 +Ge1Ge1Ge1) = 0

where the subscripts e and k indicate the Earth and the keys, respectively, while thesubscripts 1 and 2 indicate the initial and final conditions, respectively. Since boththe keys and the Earth started from rest,

(Gk2Gk2Gk2 +Ge2Ge2Ge2)− (Gk1Gk1Gk1 +Ge1Ge1Ge1) = 0

(Gk2Gk2Gk2 +Ge2Ge2Ge2)− (0+0) = 0

Ge2Ge2Ge2 = −Gk2Gk2Gk2

Now we can substitute in the definition of momentum, and recognizing that all ofthe motion takes place along a single direction (which we shall call enenen, positivewhen pointing toward the Earth) we only have to look at coefficients of that one unitvector.


Ge2Ge2Ge2 = −Gk2Gk2Gk2

Ge2 enenen = −Gk2 enenen

meve2 enenen = −mkvk2 enenen

where mk is the mass of the keys and me is the mass of the Earth, and vk2 and ve2 arethe final velocities of the keys and the Earth, respectively. We may then divide bothsides by the unit vector enenen and isolate the final velocity of the Earth, as

ve2 =−mk

mevk2

Substituting in the numeric values,

ve2 =− 0.1 kg5.97×1024 kg

(1.5 m/s) =−2.5×10−26 m/s

The result is negative, so the Earth is moving upwards. Note that, at this speed, itwould take approximately 126 million years to travel a distance equivalent to themean diameter of a hydrogen atom.

3.4.3 The jet engine: an “explosive” collision

The Rolls-Royce Trent 1000 aero engine is a high-bypass turbofan with a flow rateof approximately 1300 kg/s. Compute the speed which two of these engines canimpart upon a 200,000 kg aircraft after two minutes, assuming a constant exhaustvelocity of 380 m/s and neglecting drag.

Fig. 3.10 The jet engine: an “explosive” collision

3.4 Examples 157

SOLUTION

This problem is essentially identical to the previous one. If we define the systemas the aircraft and the air around it, then since the only force acting on the aircraftis from the air, there are no external forces and the total momentum in the systemmust remain constant.

ΔGsysGsysGsys = (Ga2Ga2Ga2 +Ge2Ge2Ge2)− (Ga1Ga1Ga1 +Ge1Ge1Ge1) = 0

where the subscripts a and e indicate the air and the engine/aircraft, respectively,while the subscripts 1 and 2 indicate the initial and final conditions, respectively. Ifit is assumed that the system started from rest,

(Ga2Ga2Ga2 +Ge2Ge2Ge2)− (0+0) = 0

Ga2Ga2Ga2 +Ge2Ge2Ge2 = 0

Re-arranging and substituting in the definition of momentum,

Ge2Ge2Ge2 =−Ga2Ga2Ga2

meve2ve2ve2 =−mava2va2va2

ve2ve2ve2 =−ma

meva2va2va2

The mass of air squirted through two engines in two minutes is,

ma = 2mΔ t

where m is the engine mass flow rate (1300 kg/s) and Δ t is the running time (twominutes). Substituting this together,

ve2ve2ve2 =−2mΔ tme

va2va2va2

Again, since both the air and the aircraft are moving along the same axis, we onlyneed one unit vector, and again we shall call this unit vector enenen, which we shalldefine as being positive along the direction that the aircraft is moving. Then,

ve2enenen =−2mΔ tme

va2enenen

where the scalar quantities ve2 and va2 are the speeds of the aircraft and exhaust jet,respectively. Then, dividing out the unit vector and substituting in the values, we get

ve2 =−2(1300 kg/s)(120 s)

200,000 kg(−380 m/s) = 593 m/s


So the aircraft will be moving at 593 m/s. Given that this is Mach 1.8 at sea leveland most commercial aircraft have a maximum cruising speed of around Mach 0.83,the error introduced by neglecting the drag is large. Neglecting drag, after a 7-hourtransatlantic flight, the aircraft would reach a speed of 125,000 m/s, which is nearlyan order of magnitude larger than the Earth’s orbital escape velocity.

3.4.4 Centre of mass of a mixed system

Two bugs A and B (having masses ma and mb, respectively) sit on either end of auniform stick of length L and mass m0. Determine the location of the centre of massof the system.

Fig. 3.11 Centre of mass of a mixed system

SOLUTION

First, we shall define our system as consisting of both bugs and the stick. Weshall define a coordinate system o with the origin located on the left side of the stick(directly under the bug A), such that the exexex unit vector points to the right, and the eyeyey

unit vector points upwards. The location of the centre of mass rc = xc exexex + yc eyeyey +zc ezezez is then given by the expression,

rcrcrc =∑mixi

∑miexexex +

∑miyi

∑mieyeyey +

∑mizi

∑miezezez

Immediately we may recognize that because the stick is slender and because thebugs may be treated as discrete points, yi = zi = 0, so yc = zc = 0. Substituting thisinto our expression for rcrcrc,

3.4 Examples 159

rcrcrc =∑mixi

∑miexexex +0 eyeyey +0 ezezez

rcrcrc =∑mixi

∑miexexex

Therefore, we need only determine the location of each element of the system: thatis, the two bugs and the stick. Because the stick is an extended body, the stick maybe treated as a point mass located at its centre of mass. To determine the locationx0 of the centre of mass of the stick along the exexex-direction, we must integrate. Fromour earlier definition of the location of the centre of mass for extended bodies,

x0 =

∫x dm∫dm

however, because the stick is uniform and slender, we may say that dm = λ dx,where λ =m0/L is the linear density of the stick (in, say, kg/m), which is a constant.Then,

x0 =

∫ L0 λx dx∫ L0 λdx

=

∫ L0 x dx∫ L0 dx

=

(12 x2)∣∣L

0

x∣∣L0dx

=12

L2 −02

L−0

x0 =L2

So the centre of mass of the stick is at the geometric centre of the stick, as expected.Substituting this result into our earlier expression for the centre of mass of the entiresystem (and taking xa and xb as the locations of bugs A and B, respectively),

rcrcrc =maxa +mbxb +m0x0

ma +mb +m0exexex

=ma(0)+mb(L)+m0

(L2

)ma +mb +m0

exexex

rcrcrc =mbL+m0

(L2

)ma +mb +m0

exexex


Simplifying, the location of the centre of mass of the system can be given as,

rcrcrc =

mbm0

+ 12

1+ mam0

+ mbm0

L exexex

3.4.5 Centre of mass of a solid object

A solid body in the shape of a right triangle has uniform thickness t, and right-sidelengths of Lx and Ly. Compute the location of the centre of mass of this body if thedensity is uniform.

Fig. 3.12 Centre of mass of a solid object

Hint: the mathematical treatment of multiple integrals is reviewed in Appendix A.

SOLUTION

From the derivation of the location of the centre of mass (which we can callrcrcrc = xcexexex + yceyeyey + zcezezez),

rcrcrc =

∫x dm exexex +

∫y dm eyeyey +

∫z dm ezezez∫

dm

xc exexex + yc eyeyey + zc ezezez =

∫ρx dV exexex +

∫ρy dV eyeyey +

∫ρz dV ezezez∫

ρdV

xc exexex + yc eyeyey + zc ezezez =

�x dx dy dz exexex +

�y dx dy dz eyeyey +

�z dx dy dz ezezez�

dx dy dz

where ρ is the (constant) density of the object, m is mass and V is volume. Next, wecan equate the scalar coefficient of each of the unit vectors, so that

3.4 Examples 161

xc =


yc =


zc =


(3.2)

So, let us define a Cartesian coordinate system around our triangular object suchthat the origin lies at one corner, exexex points along the side of the triangle with lengthLx, eyeyey points along the side of the triangle with length Ly and ezezez points along thethickness.

To solve this expression, we will need to integrate along the x, y and z axes, sowe will need to first determine the limits of integration. This is accomplished simplyby finding the set of inequalities which completely describe the body. To begin, weknow that along the x-axis, the body is delimited by the vertical lines x = 0 andx = Lx, so the first inequality is

0 ≤ x ≤ Lx

In order to completely define the triangular shape, an additional two delimiting linesare required: a horizontal line at y = 0 and a diagonal line at y = (Ly/Lx)x, so thesecond inequality is

0 ≤ y ≤ Ly

Lxx

Because the body is of uniform thickness t, the volume may then be completelydescribed with the third inequality,

0 ≤ z ≤ t

These three inequalities therefore represent the limits of integration for this volume,so we may insert these limits into Equations (3.2) and solve. Let us begin by con-sidering the denominators, as these are common to all three expressions.

�dx dy dz →

∫ t

0

∫ Lx

0

∫ (Ly/Lx)x

0dy dx dz

=∫ t

0

∫ Lx

0y∣∣(Ly/Lx)x0 dx dz

=∫ t

0

∫ Lx

0

Ly

Lxx dx dz


=∫ t

0

12

Ly

Lxx2∣∣Lx

0 dz

=∫ t

0

12

LyLx dz

=12

LyLxz∣∣t0

=12

LyLxt

Which we recognize, thankfully, as the volume of a triangular prism. Note that theorder of integration needed to be changed, as there was an ‘x’ in the limits of they-integration. This term would itself need to be integrated with respect to dx, so they-integration was carried out first. The order of integration can be changed withoutaffecting the calculation since dV = dx dy dz = dy dx dz.

Now, we can address the numerators of Equations (3.2) one axis at a time. Alongthe x-axis,

�x dx dy dz →

∫ t

0

∫ Lx

0

∫ (Ly/Lx)x

0x dy dx dz

=∫ t

0

∫ Lx

0xy∣∣(Ly/Lx)x0 dx dz

=∫ t

0

∫ Lx

0

Ly

Lxx2 dx dz

=∫ t

0

13

Ly

Lxx3∣∣Lx

0 dz

=

∫ t

0

13

LyL2x dz

=13

LyL2xz∣∣t0

=13

LyL2xt

Now, to get the x-component of the vector pointing from the origin to the centre ofmass,

xc =


=13 LyL2

xt12 LyLxt

xc =23

Lx

3.4 Examples 163

Next, we evaluate the numerator for the y-component.

�y dx dy dz →

∫ t

0

∫ Lx

0

∫ (Ly/Lx)x

0y dy dx dz

=∫ t

0

∫ Lx

0

12

y2∣∣(Ly/Lx)x0 dx dz

=∫ t

0

∫ Lx

0

12

L2y

L2x

x2 dx dz

=∫ t

0

16

L2y

L2x

x3∣∣Lx

0 dz

=∫ t

0

16

L2yLx dz

=16

L2yLxz∣∣t0

=16

L2yLxt

Again, we divide this by the denominator to get the y-component of rcrcrc,

yc =


=16 L2

yLxt12 LyLxt

yc =13

Ly

Finally, we can evaluate the numerator of the z-component.

�z dx dy dz →

∫ t

0

∫ Lx

0

∫ (Ly/Lx)x

0z dy dx dz

=∫ t

0

∫ Lx

0yz∣∣(Ly/Lx)x0 dx dz

=∫ t

0

∫ Lx

0

Ly

Lxxz dx dz

=∫ t

0

12

Ly

Lxx2z∣∣Lx

0 dz


=∫ t

0

12

LyLxz dz

=14

LyLxz2∣∣t0

=14

LyLxt2

And, dividing by the denominator,

zc =


=14 LyLxt2

12 LyLxt

zc =12

t

which, reassuringly, is halfway through the body. Since the body is symmetric aboutthe z-axis, it was expected that the centre of mass would be halfway through thethickness; otherwise, we could have moved the location of mass within the bodyby simply switching around our axes. So, the location of the centre of mass of thetriangular body is,

rcrcrc =23

Lx exexex +13

Ly eyeyey +12

t ezezez

3.4.6 Centre of mass and linear momentum

Two bodies of mass ma and mb are joined together by a rigid massless pole oflength L and are set rotating with an angular velocity of ω . Demonstrate that linearmomentum is conserved.

SOLUTION

The first thing we need to recognize here is that while the two masses are rotating,if we define our system as the two masses and the pole, there are no external forcesapplied so the centre of mass of the system must be staying in exactly the sameplace. The only point in a rotating system which stays in exactly the same placethroughout is the centre of rotation, so therefore the bodies are rotating about theircommon centre of mass. So, all we need to do here is to establish where, exactly,the centre of mass is located; then, knowing where the centre of rotation is, we canwork out the velocities of the masses, and then show that GaGaGa +GbGbGb = 0.

3.4 Examples 165

Fig. 3.13 Centre of mass and linear momentum

To evaluate the location of the centre of mass of the system, let us define a co-ordinate system such that the origin lies on the body of mass ma, and the exexex unitvector lies parallel to the pole, pointing toward the body of mass mb. We notice thatthe masses all lie on the x-axis, so yc = zc = 0, and we only have to compute the x-coordinate of the location of the centre of mass of the combined system. Then, fora series of discrete masses,

xc =∑(ximi)

∑mi

=(0)ma +Lmb

ma +mb

xc = Lmb

ma +mb

We can now re-locate our origin at the centre of mass, so that the bodies are rotatingabout the origin. The locations of the masses will therefore be

rarara = −Lmb

ma +mbexexex

rbrbrb = L−Lmb

ma +mbexexex

The velocity of the bodies will then be,

vivivi = ωωω ×ri/ori/ori/o

= ωezezez × xiexexex

vivivi = ωxieyeyey


where xi is the x-coordinate of the ith body. Then,

vavava = ω(−L

mb

ma +mb

)eyeyey

vbvbvb = ω(L−L

mb

ma +mb

)eyeyey

The total linear momentum in the system is then,

GsysGsysGsys = mavavava +mbvbvbvb

= maω(−L

mb

ma +mb

)eyeyey +mbω

(L−L

mb

ma +mb

)eyeyey

= ωL(ma +mb)(−mamb +mb(ma +mb)−m2

b

)eyeyey

= 0

So, as required, we have demonstrated that the total linear momentum within therotating system is zero.

An additional interesting point has been raised here: any object which is rotatingin three-dimensional space with no external forces acting upon it will always rotateabout its centre of mass. By extension, then, as the moon orbits around the Earth,since the only force acting between them is their mutual gravitational attraction, themoon isn’t orbiting around the Earth at all, but rather around the centre of mass ofthe Earth-moon system.

3.5 Problem set 3: Conservation of linear momentum 167

3.5 Problem set 3: Conservation of linear momentum

Question 1


Determine the location of the centre of mass (xc,yc) of a thin half-circle of con-stant density. The radius of the half-circle is R.

Answer:

xc = 0 (on the centreline)

yc =4

3πR

Question 2



A 70 kg fisherman (who, incidentally, looks very much like a bug) paddles his50 kg, 2.5 m long canoe until the nose of the canoe just touches a dock. If thefisherman walks from his paddling position at the very back of the canoe to the noseof the canoe, over what distance will he have to jump in order to get out of the canoeand onto the dock? Assume that the canoe has a uniform density and is symmetric,and that the canoe can move in the water without friction.

Answer:

1.46 m

Question 3


Two figure-skaters A and B come together on the ice, execute a spin and thenskate apart. If the skaters have masses ma = 60 kg and mb = 68 kg, and pre-spinvelocities as illustrated, compute the final velocity of skater B if the final velocity ofskater A is as shown. The magnitudes of the velocities are:

|(vavava)1| = 1.70 m/s

|(vavava)2| = 1.80 m/s

|(vbvbvb)1| = 1.34 m/s

Assume that there is no friction on the ice.

3.5 Problem set 3: Conservation of linear momentum 169

Answer:

|(vbvbvb)2| = 1.47 m/s

θ = 7.4◦

Question 4

In the science-fiction film Eraser, Arnold Schwarzenegger is shown firing a futur-istic, rifle-like weapon which allegedly accelerates small aluminium projectiles tovelocities approaching the speed of light. Assuming the projectile is an aluminiumsphere of radius 1 mm and density 3000 kg/m3, and that it is fired at 0.1% of thespeed of light (which is 3 ×108 m/s), and assuming that Arnold has a mass of 130kg, what velocity will the rifle impart upon Arnold when it is fired?

Answer:

0.029 m/s

Question 5

The deep-space probe Voyager I (with a mass of 722 kg) was launched by NASAin 1977. Voyager I is currently the most distant man-made object and one of thefastest-moving man-made objects, with an approximate mean velocity of 32,000m/s relative to the Earth. On the other hand, the Boeing 747 is one of the largestand most successful commercial aircraft ever built, having a mass of approximately5×105 kg and a typical cruise speed of 240 m/s. What velocity must each of thesevehicles have imparted upon the Earth on takeoff? Why is this result likely to beinaccurate for the case of the Voyager I deep-space probe? The mass of the Earth isapproximately 5.97×1024 kg.

Answer:

Boeing 747 : 2.01×10−17 m/s

Voyager I : 3.87×10−18 m/s


Question 6

Two spacecraft approach each other and dock. Spacecraft A has mass 240 kg, andinitial velocity 270 m/s exexex − 130 m/s eyeyey + 41 m/s ezezez, while spacecraft B has mass680 kg, and initial velocity 120 m/s exexex + 100 m/s eyeyey − 29 m/s ezezez. After docking,spacecraft B transfers 50 kg of fuel to spacecraft A. The two spacecraft then undock,and spacecraft A is left with a final velocity of 240 m/s exexex − 60 m/s eyeyey − 8 m/s ezezez.Compute the final velocity of spacecraft B.

Answer:

122 m/sexexex +86 m/seyeyey −12 m/sezezez

Question 7

A child throws a tennis ball against a wall; the ball bounces off the wall and isthen caught again by the child. Explain how momentum is conserved through thisprocess, and how it is possible for the ball to bounce off the wall without apparentlyimparting any velocity upon the wall.

Chapter 4External forces

Summary Construction and use of free-body diagrams; discussion of appliedforces, including gravity (for both small and large deflections), buoyancy, tension,dry friction (static and kinetic), springs and restoring forces, wet friction, viscousfriction and pressure drag.

4.1 Free-body diagrams

The law of conservation of linear momentum requires that the time-rate of changeof the total momentum within a closed system be equal to the net external forceapplied by the environment upon the system. So far, we have limited our analysis tothe special case where GsysGsysGsys = 0. In cases where this is not true, in order to computethe time derivative of the total momentum within the system (or, equivalently, theacceleration of the centre of mass of a system with constant mass) at some point intime, it becomes necessary to evaluate the net external force acting upon the system.

One useful tool which can be used to help in calculating the net external forceis the free-body diagram. In a free-body diagram, the system is sketched, and theindividual applied external forces are drawn as vectors. These force vectors haveone end on the object within the system on which they act, and their direction is inthe direction of application of the force. A typical free-body diagram is illustratedin Figure 4.1.

This diagram not only helps visualize the forces acting upon a body in three-dimensional space, but also facilitates the decomposition of the force vectors intocomponents lying parallel to the appropriate unit vectors. The vector summation re-quired to compute the total or net external force can then be more easily carried out.For example, consider the object illustrated in Figure 4.1, as it is acted upon by theforces F1F1F1, F2F2F2, F3F3F3, F4F4F4, and F5F5F5. Resolving these forces into components acting alongthe unit vectors exexex and eyeyey and summing these components yields the net externalforce acting on the object in a convenient vector form, as

171

172 4 External forces

Fig. 4.1 A typical free-body diagram.

∑FextFextFext = F1F1F1 +F2F2F2 +F3F3F3 +F4F4F4 +F5F5F5

∑FextFextFext =(F2 +F3 cos(θ3)−F4 +F5 cos(θ5)

)exexex

+(F1 −F3 sin(θ3)−F5 sin(θ5)

)eyeyey

It may be noted that the precise point of application of the force within the system is,for the purposes of evaluating the linear momentum of the system, entirely irrelevant(see Chapter 3). However, the point of application of the force will be of significantimportance later when analyzing rotating systems (see Chapter 5).

From the law of conservation of linear momentum, if we define our system asthe mass m illustrated in Figure 4.1 and assume that the mass is constant, then

∑FextFextFext = GGG

=ddt(mvvv)

= mddt

vvv

∑FextFextFext = maaa

so, the acceleration vector aaa of the centre of mass of the system is simply

aaa =1m

(F2 +F3 cos(θ3)−F4 +F5 cos(θ5)

)exexex

+1m

(F1 −F3 sin(θ3)−F5 sin(θ5)

)eyeyey

With the acceleration defined, it is then possible to determine the velocity vvv andposition rrr of the centre of mass from the definitions of acceleration and displace-ment, as

4.2 Common applied forces, type 1: position-independent 173

vvv =∫

aaa dt

rrr =∫

vvv dt

In engineering practice, there are a number of common forces, each with uniquecharacteristics and properties, which may be encountered. The ability to identifythese forces when they occur, and apply them appropriately to a free-body diagram,is a critical skill and is rarely trivial.

4.2 Common applied forces, type 1: position-independent

4.2.1 Gravity (small-displacement assumption)

One of the most important forces in all engineering applications is gravity. Gravitycauses any object with mass to be attracted to all other objects with mass, though theforce is incredibly weak. Unless the mass of at least one of the objects is enormous,gravity may usually be neglected. The gravitational force has been of interest totheoretical physicists for decades, and is still not entirely understood; for engineers,though, it is sufficient to understand that it is proportional to the product of themasses affected and inversely proportional to the square of the distance betweenthem, and that it is always attractive. Mathematically, then, the gravitational forcevector FgFgFg pulling a body of mass m1 toward a body of mass m2 is

FgFgFg =−Gm1m2

r2 ererer

where G is a constant (not to be confused with linear momentum), the unit vector ererer

points from one mass to the other, and the negative convention here implies that theforce is always attractive.

This understanding of gravity was first put forth in Sir Isaac Newton’s work,PhilosophiæNaturalis Principia Mathematica, which was published in 1687. Inci-dentally, in order to derive the inverse-square law of gravity, Newton also had toinvent a new form of continuum mathematics, which we now commonly refer to ascalculus (though modern mathematics has embraced Leibniz’s far less tedious no-tation). Aside from a slight correction introduced by Albert Einstein 220 years laterto account for extreme velocities and masses, the governing equations of gravity inNewton’s Principia are still used to this day in calculating spacecraft trajectories.

The constant of proportionality G is called the universal gravitational constantand is equal to approximately 6.67× 10−11 N · m2/kg2. Owing to the difficulty ofmeasuring so small a quantity, the magnitude of G remained uncertain for more than


a century following the publication of the Principia. The determination of the valueof G is often erroneously attributed to the English scientist Henry Cavendish (1731-1810). Though his experiments are now recognized as the first to have provided dataaccurate enough to allow this value of G to be determined (to within about ±1%),the purpose of Cavendish’s now-famous experiments had actually been to determinethe mass of the Earth. Consequently, he cancelled the burdensome and irrelevantconstant G out of his equations. Cavendish undertook this effort not so much outof his own interest (he was primarily a chemist) but as a favour to ailing fellow-scientist and friend John Michell (1724-1793), a geologist, to whom he whimsicallywrote, “if your health does not allow you to go on [...], at least permit me the easierand less laborious employment of weighing the world.” Regrettably, Michell did notlive to see the results.

Consider an object with mass ma sitting at some distance h from surface of theEarth. We shall define a cylindrical coordinate system o with its origin at the centreat the earth, so that the vector rrr = r ererer describes the position of the object. Becausethe object is at an altitude h from the surface of the Earth, r = re +h, where re is theradius of the Earth (approximately 6.37×106 m). Then, The force felt by the objectwill be

FgFgFg =−Gmame

(re +h)2 ererer

where me is the mass of the Earth (approximately 5.97×1024 kg). Most engineeringapplications (with the exception of spacecraft navigation, which will be discussedseparately) will be within 20 km of the Earth’s mean radius, so (re + h) will bewithin ±0.3% of the mean radius, and (re +h)≈ re is a reasonable approximation.Then,

FgFgFg ≈−(Gme

r2e

)maererer

Since the mass and mean radius of the Earth are constants, all of the constants maybe combined, so that

FgFgFg ≈ −(Gme

r2e

)maererer

= −(6.67×10−11N ·m2/kg2)5.97×1024kg

(6.37×106)2m2 ma ererer

= −(9.81m/s2)maererer

which is the far more familiar result. For all engineering applications on the surfaceof the Earth, then, an object subjected only to the gravitational force will acceleratetoward the centre of the Earth (normal to the ground) at an approximately constant


rate of 9.81 m/s2; this acceleration is commonly symbolized as the lower-case g, sothat

FgFgFg =−mag eyeyey

where eyeyey points straight up from the ground.

Let us now consider a small cloud of objects (each of mass mi) sufficiently closeto the Earth that the gravitational force is independent of height. Let us define anorigin at the centre of the Earth, so that the position of each mass may be specifiedby the position vector ririri relative to the centre of the Earth (Figure 4.2). Gravitywill then cause a mutual attraction between the Earth and each of the objects ofmagnitude Fg/iFg/iFg/i, where

Fig. 4.2 An array of masses experiencing gravitational attraction toward the Earth.

Fg/iFg/iFg/i =−g mi er/ier/ier/i

where er/ier/ier/i is a position vector pointing along ririri. The total gravitational force Fg/totFg/totFg/totwill therefore be

Fg/totFg/totFg/tot = ∑(−g mi er/ier/ier/i

)However, since er/ier/ier/i has the same direction as ririri but with unit magnitude, it is equiv-alent to express er/ier/ier/i as

er/ier/ier/i =ririri

ri

where ri is simply the scalar magnitude of the vector ririri. Substituting this into theabove expression,


Fg/totFg/totFg/tot = ∑(−g mi

ririri

ri

)

We have assumed that the gravitational force is independent of position. Therefore,as before,

ri = re +h ≈ re

We may now substitute this approximation into our earlier result, yielding

Fg/totFg/totFg/tot = ∑(−g

miririri

ri

)

= −g∑(

miririri

re

)

= − gre

∑(

miririri

)

Fg/totFg/totFg/tot = −mtotgre

∑miririri

∑mi

where mtot = Σmi, the total mass of the objects (which is a constant). However, fromthe definition of the centre of mass,

rcrcrc =∑miririri

∑mi

Substituting,


rcrcrc


rc ececec

where rc is the scalar distance between the centre of the earth and the centre ofmass of the cloud of objects, and ececec is a unit vector pointing from the centre of theearth toward the centre of mass of the cloud of objects. Finally, since ri ≈ re for allobjects, then

rc ≈ re

Substituting this expression into the previous result,


re ececec


Fg/totFg/totFg/tot = −mtotgececec

This result indicates that, providing the altitudes are sufficiently small that g is con-stant, so far as gravity is concerned, any object or array of objects may be replacedby a point of equal mass located at its centre of mass. This is why the centre of massis sometimes referred to as the centre of gravity. It is very important to note, how-ever, that if g is not constant (or, equivalently, that the approximation ri ≈ rc ≈ re

fails), then the net force applied by gravity upon the system does not necessarily actfrom the centre of mass.

This analysis may equally be extended to solid bodies. In that case, the summa-tions may be replaced with integrals, and mi → dm, so that

∑Fg/totFg/totFg/tot =

∫−gererer dm

= −g∫

rrrr

dm

= −mtotgre

∫rrr dm∫dm

= −mtotgre

rcrcrc

= −mtotgre

re ececec

∑Fg/totFg/totFg/tot = −mtotg ececec

where again it has been assumed that r ≈ rc ≈ re.

4.2.2 Hydrostatic pressure

The hydrostatic pressure acting on an object is the pressure experienced by the sur-face of the object as a consequence of being immersed at a depth h in a fluid. Con-sider a plane object of arbitrary shape and surface area A, lying horizontally whileimmersed in a fluid, as illustrated in Figure 4.3. Because the fluid has no ‘stiffness’(as solids do), the object must support only the weight of the column of fluid imme-diately above itself.

We shall define a coordinate system such that the origin lies on the surface ofthe fluid, and the eyeyey unit vector points upwards. The object then lies at y =−h. Thecolumn of fluid above the object may be considered as a stack of an infinite numberof infinitesimally thin ‘slices’ of fluid, each of area A and thickness dy. The weightdFg of each slice is then,


Fig. 4.3 Illustration of the hydrostatic pressure acting on an immersed object.

dFg = ρ(A dy)g

where ρ is the density of the fluid, Ady is the volume of the slice, and g is the accel-eration due to gravity. To obtain the total weight Fg of the column of fluid, we onlyneed to integrate this expression. Because we are integrating over a physical domain,the limits of integration correspond to the bounds of the domain, or −h < y < 0, so

Fg =∫ 0

−hρgA dy

If the fluid is incompressible (which is a reasonable approximation for most liquidsin engineering applications) and the cross-sectional area A projected by the objectdoes not change with depth, then ρ and A are constant, and

Fg = ρgA∫ 0

−hdy

= ρgAy

∣∣∣∣0

−h

= ρgA(0+h)

Fg = ρgAh

The magnitude of the pressure P acting on the object is then given by,

P =Fg

A

=ρgAh

AP = ρgh


So, the pressure experienced by a surface immersed in an incompressible fluid issimply the product of the density of the fluid, the depth, and the acceleration due togravity.

Let us now consider the case of an object immersed in a compressible fluid, suchas a gas. In this case, the density cannot be removed from the integral, as it may varywith pressure. However, for the case of most gases used in engineering applications,at constant temperature the density is proportional to the pressure, so that

ρ = kP

where k is a constant which depends on the temperature and the type of gas (in airat standard sea-level conditions, k ≈ 1.28×10−5 m2/s2). Then,

Fg =∫ 0

−hρgA dy

P =Fg

A=

1A

∫ 0

−hρgA dy

=

∫ 0

−h(kP)g dy

P = kg∫ 0

−hP dy

Recall here that h represents the depth to which an object is immersed. Since gasescannot present a discrete ‘surface level’ which we can use as our origin, we willneed to adjust our limits of integration. The limits of integration must describe theextent of the column of gas above the object. If we define y0 to be the altitude ofthe object (or the distance between the object and, say, the bottom of the tank ofgas), then the lower bound of the column of gas will be y0. Since gases will expandto occupy whatever volume they are given, then the upper limit of the column ofgas will extend infinitely. The bounds of the column of gas above the object willtherefore be given by the inequality y0 < y < ∞. Replacing the limits of integration,then,

P = kg∫ ∞

y0

P dy

This result indicates that the function P is a multiple of its own integral: incidentally,we are already familiar with a function which has this property. Let us thereforeassume that P is of the form

P(y) =C0e−C1y


The negative exponent is assumed, as the pressure must decrease with increasingaltitude. Then, at our altitude of y = y0,

P = kg∫ ∞

y0

P dy

C0e−C1y0 = kg∫ ∞

y0

C0e−C1y dy

C0e−C1y0 = kg

(−C0

C1e−C1y

)∣∣∣∣∞

y0

e−C1y0 = − kgC1

(e−C1(∞)− e−C1(y0)

)

e−C1y0 = − kgC1

((0)− e−C1(y0)

)C1 = kg

Substituting this result into our expression for P(y),

P(y) =C0e−kgy

This expression yields the hydrostatic pressure experienced by an object immersedin a volume of gas. To solve for the unknown C0, we only need one point within thevolume of gas at which both the pressure and altitude are known.

As an example, let us consider the Earth’s atmosphere; we know that the sea-level pressure P0 = 101.3 kPa when the altitude y = 0. Using this condition to solvefor C0,

P0 = C0e−kg(0)

P0 = C0

So, the pressure in Earth’s atmosphere will vary with altitude according to the ex-pression,

P = P0e−kgy

Despite having assumed that both the acceleration due to gravity and the temper-ature of the air were constant, this expression actually provides a fairly good en-gineering approximation for the atmospheric pressure, as illustrated in Figure 4.4.The difference is due primarily to the change in temperature with altitude (which isdiscontinuous, and not easily represented mathematically).


Fig. 4.4 Comparison of results (with k assumed constant, taken at mean sea-level conditions) withthe standard atmospheric data published by the International Civil Aviation Organization.

4.2.3 Buoyancy

Buoyancy is the ‘floating force’ experienced by a body immersed in a fluid. Thediscovery of this principle of buoyancy is attributed to the Greek philosopher andmathematician Archimedes of Syracuse (287-212 BC), and pre-dates integral cal-culus by more than a thousand years. Consequently, buoyancy is often referred toas the Archimedes principle (Archimedes allegedly made the discovery while step-ping into his bath and became so excited that he ran through the streets shouting“Eureka!”- meaning, “I found it”. Regrettably, he neglected to put his clothes backon first.)

To understand the origin of the buoyant force FBFBFB, let us consider an object ofarbitrary shape submerged in a fluid. We can model the arbitrary shape of the objectas a collection of infinitely thin ‘straws’ each having a uniform cross-sectional areadA, as illustrated in Figure 4.5. Then, let us say that the fluid has density ρ and thatthe unit vector eyeyey points straight up out of the fluid.

As we have shown earlier, the hydrostatic pressure P at any point in an incom-pressible fluid is given by the expression,

P = ρgy

where y is the distance below the surface of the fluid. This pressure will exert a forcevector dF1dF1dF1 on the top surface of a given ‘straw’ of


Fig. 4.5 An arbitrary body immersed in a fluid.

dF1dF1dF1 = −P dA eyeyey

dF1dF1dF1 = −ρgy1 dA eyeyey

where y1 is the depth of the top of the straw. The pressure will always be actingto ‘crush’ the straw, so the force on the top of the straw will be acting downwards.Similarly, the bottom of the straw will be experiencing a force vector dF2dF2dF2, given by

dF2dF2dF2 = ρgy2 dA eyeyey

where y2 is the depth of the bottom of the straw. There will be no net horizontalpressure forces on the straw, since the pressure is everywhere the same at the samedepth (also, if there were an imbalance in horizontal forces, an immersed objectwould accelerate sideways). Therefore, we can obtain the net force on each straw,dFdFdF, as,

dFdFdF = dF1dF1dF1 +dF2dF2dF2

= −ρgy1 dA eyeyey +ρgy2 dA eyeyey

dFdFdF = ρg(y2 − y1

)dA eyeyey

We can then compute the total force on all the straws by integrating; however, weneed only integrate over the projected area in the x− z plane, Ay, since that areacontains all the straws.

dFdFdF = ρg(y2 − y1

)dA eyeyey

4.2 Common applied forces, type 1: position-independent 183∫dFdFdF =

∫ρg(y2 − y1

)dA eyeyey

FFF = ρg∫ (

y2 − y1)

dA eyeyey

where we have assumed that the fluid has constant density. Finally, we notice that ifwe wanted to compute the volume V of the body, we could do so by integrating thevolume of the straws, as

V =∫

dV =∫(y2 − y1)dA

This result can be substituted into the expression for the total force as

FFF = ρgV eyeyey

Now, we can notice that the volume of the submerged body is the same as the vol-ume of the fluid which it displaced; therefore, ρV is equivalent to the mass of thedisplaced fluid. Finally, we can conclude that the buoyant force exerted upon a sub-merged body is equal to the weight of the displaced fluid.

FBFBFB = ρgV eyeyey

It is important to remember that the density ρ here is the density of the fluid inwhich the object is immersed, and not the density of the object itself. Since thehydrostatic pressure increases in the same direction as gravity acts, the buoyant forcealways acts in the opposite direction as gravity. There are a couple of interestingimplications of this result:

There is no buoyancy in the absence of gravity. A balloon in the centre of a largemass of water within a zero-g environment will not float, nor would an anvil sink.They would both stay stationary in the mass of water.

The buoyant force is independent of depth. Since the value y does not appear in theexpression for the buoyant force, the force will be the same regardless of depth.However, as we have seen earlier, the density of gases increases with depth so thebuoyant force will also increase with depth. Liquids also exhibit a very small in-crease in density with depth (seawater, for example, increases in density by approx-imately 5% between the ocean surface and a depth of 10 km).

The buoyant force is independent of shape. So long as the volume of the ‘hole in thewater’ is the same, the force will be the same.


An object will only float if its density is smaller than that of the fluid in which itis immersed. Consider an object of volume V , mass ma and density ρa = ma/Vimmersed in a fluid of density ρ . Sketching a free-body diagram of this object wouldreveal,

∑FextFextFext = FBFBFB −FgFgFg

=(ρgV −mag

)eyeyey

=(ρgV − (ρaV )g

)eyeyey

∑FextFextFext = (ρ −ρa)gVeyeyey

If ρ > ρa, then ΣF > 0 and the body will accelerate upwards (which is how wedefine “floating”); if ρ < ρa, then ΣF < 0 and the body will accelerate downwards(which is how we define “sinking”). In the special case where ρ = ρa, then ΣF = 0and the object will be in static equilibrium while immersed. The object is then saidto be neutrally buoyant.

4.2.4 Tension in an ideal string

It is common in engineering analysis to approximate or ‘model’ flexible elementssuch as ropes, chains, strings, cords, cables and belts as ‘ideal strings’. An idealstring has two important characteristics: (i) it is perfectly inextensible, so that thelength will never increase no matter how large the forces applied, and (ii) it cannotsustain any bending at all, so it has no resistance to tangential loads. An ideal stringmay or may not be massless, depending on the nature of the problem at hand.

Tension is defined as the axial force exerted by an object resisting extension.Since an ideal string cannot support any tangential loads, the force applied by thestring will always be tensile (toward the middle of the string) and the force vectorwill always be parallel to the string, as illustrated in Figure 4.6. Also, since an idealstring cannot stretch, the tension is the same everywhere along the string, even if thestring passes over massless, frictionless pulleys or other ideal guides.

As an example, consider a block-and-tackle arrangement (also referred to as aparallel pulley arrangement), as illustrated in Figure 4.7. A force FL is supportedby a block, and the block is itself supported by an ideal string passing through npulleys; in the illustration, n = 2. The tension FT in the string is uniform along thestring, according to the definition of an ideal string. We may then define our systemas consisting of the block, the pulleys attached to the block and a little bit of thestring around the pulleys (as illustrated). The free-body diagram of the block maythen be sketched, and it becomes clear that


Fig. 4.6 Demonstration of the tension force vector applied by an ideal string.


∑FextFextFext = FTFTFT +FTFTFT +FTFTFT +FTFTFT −FLFLFL

= (2nFT −FL) eyeyey

where the unit vector eyeyey points straight upwards. If it is assumed that the system isnot accelerating, then

(2nFT −FL) eyeyey = 0

FT =1

2nFL

So, with a parallel pulley arrangement, it is possible to support a load 2n times largerthan the tension in the ideal string, where n is the number of pulleys on the block.

Next, let us say that the height of the block (as measured between the centres ofthe pulleys on the block and on the ceiling) is y, and the length of string on the end


being ‘pulled’ is l. According to the definition of the ideal string, the total length Lof the string cannot change, so we can say that

L = l +2n y+δ

where δ is the length of string wrapped around the pulleys and the length of stringpassing between stationary pulleys, which is always constant. Taking the time-derivative of this expression,

ddt

L =dldt

+2ndydt

+dδ

dt

0 =dldt

+2ndydt

+0

dydt

= − 12n

dldt

This result shows that the block will be raised at a speed 2n times slower than thespeed at which the free end of the string is being pulled (note that the signs areopposite because y is decreasing when l is increasing, and vice-versa). Similarly, ifthe string is pulled for some fixed time Δ t, the block will be raised by a distance 2ntimes smaller than the distance which the free end of the string has been pulled.

By taking the second derivative with respect to time, the acceleration of the blockmay be related to the acceleration of the free end of the string, as

d2ydt2 =− 1

2nd2ldt2

indicating, similarly, that the block will accelerate at a rate 2n times slower thanthe free end of the string. The relative rates of acceleration are required in order tocompute the tension in the string, as will be demonstrated in the examples.

This result is, of course, only valid for cases where the strings are all perpendic-ular to the block and to the ceiling. In the event that the strings meet the pulleys atsome angle (as illustrated in Figure 4.8), since the tension is always parallel withthe string, the forces will need to be resolved in order to calculate the vector sum. Ifwe define the unit vector eyeyey as pointing up and exexex pointing toward the right, then thesum of the external forces acting on the block-and-pulley system illustrated is

∑FextFextFext = (−nFT sin(θ1)+nFT sin(θ2)) exexex

+ (nFT cos(θ1)+nFT cos(θ2)−FL) eyeyey


where again, n indicates the number of pulleys, and θ1 and θ2 are the angles sub-tended between eyeyey and the two ropes by which each of the block’s pulleys is sus-pended, as illustrated. Note that in Figure 4.8, n = 1 for simplicity.


Looking at the scalar coefficients of exexex, it becomes immediately clear that theblock is only stable (and will not accelerate in the x-direction) if θ1 = θ2 = θ ; oth-erwise, the sum of the external forces in the x-direction will not be zero. So, if theblock is not accelerating sideways,

∑FextFextFext = (2nFT cos(θ)−FL) eyeyey

And, if it is further assumed that the acceleration in the vertical direction is alsozero,

0 = (2nFT cos(θ)−FL) eyeyey

2nFT cos(θ) = FL

FT =1

2ncos(θ)FL

This result has two important implications: first, to support a given load FL, therequired tension force FT is smallest when cos(θ) = 1, or when the strings are allaligned in the vertical direction. Second, for θ > 0, as the block approaches theceiling, cos(θ) approaches zero and the tension force required to support the loadbecomes infinite. It is therefore impossible to raise the load all the way to the ceilingin the configuration illustrated in Figure 4.8.


Consider now the pulley arrangement supporting a block of mass m, as illustratedin Figure 4.9. In this arrangement (referred to as a series pulley arrangement), thereare two separate pieces of string, so the two may have different tensions FT 1 andFT 2. Constructing a free-body diagram of the mass, the forces acting on the massmay be easily summed as

∑FextFextFext = (−Fg +2FT 2) eyeyey = mam/oam/oam/o

Fig. 4.9 A series pulley arrangement.

where am/oam/oam/o is the acceleration of the mass with respect to a fixed observer. From ourearlier examination of gravity, FgFgFg is known to be equal to −mg eyeyey. Now, constructinga free-body diagram of the pulley A,

∑FextFextFext = (−FT 2 +2FT 1) eyeyey

However, it may be assumed that the pulleys are massless unless otherwise specified.So, for the pulley, ΣFextFextFext = maaa = 0. Therefore,

(−FT 2 +2FT 1) eyeyey = 0

FT 2 = 2FT 1


mam/oam/oam/o = (−Fg +2FT 2) eyeyey

am/oam/oam/o =1m(−Fg +2(2FT 1)) eyeyey


am/oam/oam/o =1m(−Fg +4FT 1) eyeyey

Also, if the system is not accelerating,

0 =1m(−Fg +2(2FT 1)) eyeyey

0 = −Fg +2(2FT 1)

Fg = 2nFT 1

where n represents here the number of pulleys in the cascade. Therefore, in a se-ries arrangement, an applied tension can support a load 2n times greater. The seriesarrangement is consequently substantially more efficient than the parallel arrange-ment, though more mechanically complex.

4.2.5 Dry friction

Any real object sliding over a real surface will feel a force resisting its motion alongthe contact surface. This resisting force is called dry friction or Coulomb friction(after Charles-Augustin de Coulomb, 1736-1806). This resistance, surprisingly, de-pends only on the type of materials from which the object and surface are made andnot on the surface finish- so that polishing the interface between two surfaces willhave no effect on the friction between them. Centuries of experimental observation,however, have shown convincingly that

Ff = μFn

where Ff is the magnitude of the friction force (the direction will always be oppositeto the direction of motion), Fn is the magnitude of the force pressing the surfacestogether, and is often called the normal force (because it always acts in a directionnormal to the interface between the surfaces), and μ is called a coefficient of friction.

The reason for which dry friction occurs is still not entirely understood. It isknown, however, to be a phenomenon linked to intermolecular forces, and so fallsmore into the realm of atomic physics than classical mechanics. However, frictionis another very important force to be considered in engineering practice- in fact,according to the second law of thermodynamics, all machines necessarily must havesome amount of friction.

Consider a block of mass m sitting at rest on a frictional surface, as illustratedin Figure 4.10. An infinitesimally small force of magnitude F is applied at someangle θ relative to horizontal. The block will appear to ‘stick’ in place, with the


friction entirely balancing the applied force so that ΣFFF = 0. If |FFF| then increases, atsome critical force, the block will become ‘unstuck’ and will begin to accelerate inthe direction of the horizontal component of the applied force. The frictional forcecausing the block to initially stick is called static friction, while the friction whichacts against the motion while the block is moving is called kinetic friction, and thereis an important reason for the distinction.

Fig. 4.10 A block being acted upon by a constant applied force and dry friction.

Let us define a fixed Cartesian coordinate system o such that the unit vector exexex

points in the direction of motion of the block, and eyeyey points upward from the surface,as illustrated. Then, if we define our system as consisting of the block alone, drawinga free-body diagram of the mass and summing the forces acting on the block, weobtain

∑FextFextFext = FFF+FfFfFf +FnFnFn

∑FextFextFext =(F cos(θ)exexex −F sin(θ)eyeyey

)−Ff exexex +Fneyeyey

Collecting together the scalar coefficients of each of the unit vectors,

∑FextFextFext =(F cos(θ)−Ff

)exexex +

(Fn −F sin(θ)

)eyeyey

Applying the law of conservation of linear momentum,

GsysGsysGsys = ∑FextFextFext

ddt

mvvv =(F cos(θ)−Ff

)exexex +

(Fn −F sin(θ)

)eyeyey

mx exexex +my eyeyey =(F cos(θ)−Ff

)exexex +

(Fn −F sin(θ)

)eyeyey


We may now equate together the scalar coefficients of each unit vector indepen-dently. Starting with the eyeyey direction, and recognizing that the block can’t movevertically,

my = Fn −F sin(θ)

0 = Fn −F sin(θ)

Fn = F sin(θ)

so the normal force is equal to the component of FFF which is pressing the blockagainst the surface. Now, looking at the exexex direction and recalling that Ff = μFn,

mx = F cos(θ)−Ff

mx = F cos(θ)−μFn

Substituting in the value of Fn obtained from the balance of forces in the y-direction,

mx = F cos(θ)−μFn

mx = F cos(θ)−μF sin(θ)

x =Fm

(cos(θ)−μ sin(θ)

)

Let us assume that the force F is sufficiently small that the block remains ‘stuck’ tothe surface. Since the block is not moving, x = 0 and

0 =Fm

(cos(θ)−μ sin(θ)

)0 = cos(θ)−μ sin(θ)

μ =cos(θ)sin(θ)

Therefore, for the case of static friction, the coefficient of friction depends only onthe angle of application of the force F, and not on the nature of the materials.

If the applied force F is then slowly increased, at some critical instant the forcewill be just large enough that the block will become ‘unstuck’ and begin to slipacross the surface. At that critical instant, the coefficient of friction is referred to asthe coefficient of static friction, and is indicated as μs. Consequently, the block willmove if and only if


F cos(θ)> μsF sin(θ)

Or, in general, the applied forces along the direction of motion must exceed μsFn inorder for the motion to begin. The critical coefficient of static friction depends onlyon the types of materials being rubbed together.

Once the block begins to slip, then x> 0 and μ becomes a constant. This constantcoefficient of friction experienced once the block is moving is called the coefficientof kinetic friction, and is indicated as μk. The coefficient of kinetic friction alsodepends only upon the materials being rubbed together.

The most important distinction between static and kinetic friction is that staticfriction depends on the applied forces, while kinetic friction depends on the materi-als. In other words, the static friction force will assume whatever value is requiredin order to oppose all the applied forces and prevent motion from occurring, regard-less of the materials involved. The kinetic friction will always oppose motion, butits magnitude depends only on the materials- not on the forces it is opposing.

In addition, it is also important to remember that, for the case of static friction,the coefficient of friction μ is not necessarily constant. What we defined as the‘coefficient of static friction’, μs, is a measure of the critical force required to justget the object to begin to slip.

A table showing some typical values of μs and μk for common engineering ma-terials is included here. Interestingly, for most materials (with the notable exceptionof certain Aluminium alloys), μs > μk- which is why the materials feel as thoughthey become ‘unstuck’ so suddenly.

Material ...on material μsμsμs μkμkμk

Aluminium Aluminium 1.05−1.35 1.40Aluminium Mild steel 0.61 0.47Mild steel Brass 0.51 0.44Mild steel Bronze 0.35 0.34Mild steel Mild steel 0.74 0.57Hard steel Hard steel 0.78 0.42Titanium Aluminium 0.41 0.38Titanium Titanium 0.36 0.30

Teflon Teflon 0.04 0.04

4.3 Common applied forces, type 2: position-dependent

There are some forces which have magnitudes that are dependent upon the positionof the system. Because of this relationship, the momentum equation ΣFextFextFext = ΣGGGwill take on the form x = f (x), where f is some known function. As a result, isolat-ing and solving for the position as a function of time is no longer a matter of simpleintegration.

4.3 Common applied forces, type 2: position-dependent 193

4.3.1 Gravity: x = f (1/x2)

Consider again an object of mass ma a distance h above the surface of the Earth(which has mass and radius me and re, respectively). Let us now assume that h islarge relative to re, so that the approximation (re + h) ≈ re is no longer valid; thiswill be true of satellites and spacecraft. Under these circumstances, the accelerationdue to gravity will no longer be constant, so Fg �= mag.

To examine this case, let us locate the origin o of a cylindrical coordinate systemon the centre of the Earth such that the unit vector ererer is pointing at the mass (asillustrated in Figure 4.11). While the problem could be solved equally using rela-tive motion coordinates, cylindrical coordinates are well-suited to the formulationof this problem. The gravitational force FgFgFg acting upon the object is given by theexpression,

FgFgFg =−Gmame

r2 ererer

Fig. 4.11 Schematic of one mass orbiting another.

If we define our system as consisting only of the mass ma, then the gravitationalforce is an external one and the law of conservation of linear momentum requiresthat,

∑FextFextFext = GsysGsysGsys

FgFgFg =ddt(mava/ova/ova/o)

−Gmame

r2 ererer =ddt(mava/ova/ova/o)

where va/ova/ova/o is the velocity of the mass ma with respect to our fixed origin o. If wenow assume that the masses are constant, ma can be pulled out of the derivative toyield


−Gmame

r2 ererer = maddt(va/ova/ova/o)

−Gmame

r2 ererer +0 eθeθeθ = maaa/oaa/oaa/o (4.1)

where aa/oaa/oaa/o is the acceleration of the mass ma with respect to our fixed origin o. Forcylindrical coordinate systems, we have already shown in Chapter 2 that the accel-eration of a point within a cylindrical coordinate system is given by the expression,

aa/oaa/oaa/o = (r− rθ 2)ererer +(2rθ + rθ)eθeθeθ

Substituting this expression into the eθeθeθ -component of Equation (4.1),

−Gmame

r2 ererer +0 eθeθeθ = ma

((r− rθ 2)ererer +(2rθ + rθ)eθeθeθ

)

−Gmame

r2 ererer +0 eθeθeθ = ma(r− rθ 2)ererer +ma(2rθ + rθ)eθeθeθ

Equating the coefficients of the unit vectors,

−Gmame

r2 ererer = ma(r− rθ 2) ererer

0 eθeθeθ = ma(2rθ + rθ) eθeθeθ (4.2)

From the eθeθeθ scalar coefficient,

(2rθ + rθ) = 0

where we have divided both sides by ma. This expression is not entirely helpful inits current form; however, something interesting happens if we multiply both sidesby r.

2rrθ + r2θ = 0

We may recognize the left-hand side of this expression as the outcome of the ap-plication of the product rule with respect to time. To simplify the expression, let ustherefore integrate with respect to time.

∫2rrθ + r2θ dt =

∫0 dt

r2θ = C (4.3)


where C is some constant. This expression demonstrates that, for all orbital motions,this quantity C = r2θ is conserved. Next, we consider the scalar coefficients of ererer inEquation (4.2).

−Gmame

r2 = ma(r− rθ 2)

−Gme

r2 = r− rθ 2

However, Equation (4.3) demonstrated that for all orbital motions, r2θ =C. Substi-tuting,

−Gme

r2 = r− rθ 2 r3

r3

−Gme

r2 = r− C2

r3

So, finally, for all objects moving under the influence of gravity,

r =C2

r3 −Gme

r2 (4.4)

However, it would be more useful to have an expression for r as a function of θ , asthis would yield the trajectory of an orbiting object. To do this, we need to elimi-nate time from this expression (remember, r is a time-derivative). First of all, let usconsider the left-hand side of Equation (4.4). Using the chain rule, we can express ras

r =ddt

(drdt

)

=ddt

(drdθ

dθ

dt

)

r =ddt

(θ

drdθ

)

Since we are trying to eliminate all time-dependence, we may use Equation (4.3) toeliminate θ . Then,


r =ddt

(r2θ

r2

drdθ

)

r =ddt

(Cr2

drdθ

)(4.5)

Now, we may evaluate the time-derivative on the right-hand side using the productrule, as

r =Cr2

ddt

(drdθ

)−2

Cr3 r

drdθ

(4.6)

We can eliminate r in Equation (4.6) using Equation (4.5); all we need to do is tointegrate Equation (4.5) once with respect to time, as

r =ddt

(Cr2

drdθ

)∫

r dt =∫

ddt

(Cr2

drdθ

)dt

r =Cr2

drdθ

+C0

where C0 is a constant of integration. To solve for C0, consider the case of an ob-ject moving in circles: in this case, the expression for r must be satisfied as well.However, for circular motion, r = dr/dθ = 0 by definition. Then,

(0) =Cr2 (0)+C0

C0 = 0

and so

r =Cr2

drdθ

We can now substitute this expression into Equation (4.6) to eliminate r, as

r =Cr2

ddt

(drdθ

)−2

Cr3 r

drdθ

=Cr2

ddt

(drdθ

)−2

Cr3

(Cr2

drdθ

)drdθ


r =Cr2

ddt

(drdθ

)−2

C2

r5

(drdθ

)2

(4.7)

Next, we may recognize that

ddt

(drdθ

)=

d(dr/dθ)

dθ

dθ

dt

=d

dθ

(drdθ

)θ

ddt

(drdθ

)= θ

d2rdθ 2

We may again eliminate the time-dependent θ using Equation (4.3), as

ddt

(drdθ

)= θ

r2

r2

d2rdθ 2

ddt

(drdθ

)=

Cr2

d2rdθ 2

We may now express Equation (4.7) as,

r =Cr2

ddt

(drdθ

)−2

C2

r5

(drdθ

)2

=Cr2

(Cr2

d2rdθ 2

)−2

C2

r5

(drdθ

)2

r =C2

r4

d2rdθ 2 −2

C2

r5

(drdθ

)2

We now have an expression for r which is independent of time. We may now sub-stitute this into Equation (4.4) and solve for r as a function of θ , which will be anexpression for the trajectory of the orbiting object.

r =C2

r3 −Gme

r2

C2

r4

d2rdθ 2 −2

C2

r5

(drdθ

)2

=C2

r3 −Gme

r2


− 1r2

d2rdθ 2 +

2r3

(drdθ

)2

= −1r+G

me

C2 (4.8)

where we have divided both sides by −C2/r2. To solve this expression for r, let usmake the substitution,

ξ =1r

Then,

dξ

dθ=

ddθ

(1r

)=− 1

r2

drdθ

and

d2ξ

dθ 2 =d

dθ

(− 1

r2

drdθ

)

= − 1r2

d2rdθ 2 +

2r3

drdθ

drdθ

d2ξ

dθ 2 = − 1r2

d2rdθ 2 +

2r3

(drdθ

)2

However, we recognize that this is identically the left-hand side of Equation (4.8);we may therefore substitute,

− 1r2

d2rdθ 2 +

2r3

(drdθ

)2

= −1r+G

me

C2

d2ξ

dθ 2 = −ξ +Gme

C2 (4.9)

So, the second derivative of the function ξ is equal to the negative of the functionitself plus a constant. One function with which we are familiar exhibits this charac-teristic: let us say that ξ is of the form

ξ = A0 sin(A1θ +θ0)+A2

where A0, A1, A2 and θ0 are constants. In this expression for ξ , we may recognizethat the constants A1 and θ0 represent a frequency and phase shift. The frequencywill depend on the speed with which the object is orbiting, and the phase shift willdepend on the time at which the object began its orbit. Since we are only interestedin the trajectory of the object, we can arbitrarily select any values for A1 and θ0: the


trajectory will be the same for all values of A1 and θ0. For simplicity, then, let uschoose A1 = 1 and θ0 = 0. Then,

ξ = A0 sin(θ)+A2

And, differentiating,

dξ

dθ= A0 cos(θ)

d2ξ

dθ 2 = −A0 sin(θ)

Substituting into Equation (4.9),

d2ξ

dθ 2 = −ξ +Gme

C2(−A0 sin(θ))= −(A0 sin(θ)+A2

)+G

me

C2

0 = −A2 +Gme

C2

A2 = Gme

C2

So, our expression for ξ becomes

ξ = A0 sin(θ)+Gme

C2

Recalling that we defined ξ = 1/r,

1r= A0 sin(θ)+G

me

C2

r =C2

Gme

A0C2

Gmesin(θ)+1

Let us define the quantity e as

e = A0C2

Gme

Then,


r =e

A0

(1

esin(θ)+1

)

Though obtaining this result has been mathematically tedious, this expression de-scribes the trajectory of an object in orbit, where A0 and C are constants that dependon the initial position and velocity of the satellite. This result should be familiar asthe definition of a conical section in cylindrical coordinates; the quantity e is knownas the eccentricity of the orbit, and the trajectories may be classified according tothe value of e as shown in the table below. These different trajectories are also il-lustrated in Figure 4.12. For all cases where e < 1, the object will remain in orbit,while the cases where e ≥ 1 constitute collisions only (even though the object neveractually touches the Earth).

Fig. 4.12 Plot of some typical satellite trajectories as a function of orbital eccentricity. In all cases,C and A0 are selected so that the object is constrained to pass through the point (R, 0).

Condition Trajectorye > 1 Hyperbolice = 1 Parabolic

0 < e < 1 Elliptice = 0 Circular


4.3.2 Springs: x = f (x)

A linear spring will stretch or contract when an axial force is applied to it (an axialforce, by definition, is parallel to the spring). The force is linearly proportional tothe change in length of the spring, and will always act toward the initial positionof the spring (springs are therefore said to apply a restoring force). An ideal springcan stretch or contract infinitely, and the applied force will be linearly proportionalto the change in length throughout. Expressing this mathematically, for a spring ofinitial length L being stretched to some new length L+x under the action of an axialforce F (as illustrated in Figure 4.13), the force FkFkFk exerted by the spring is given bythe expression,

FkFkFk =−kx exexex

Fig. 4.13 A linear spring being stretched through the action of an axial force.

where the unit vector exexex is defined as being in the direction in which the springhas been extended. The minus sign here indicates that the force vector is always inthe direction opposing the displacement. This relation is often referred to as Hooke’sLaw, after Robert Hooke (1635-1703), the English physicist more commonly knownfor suggesting that Newton’s Principia was based upon his own work without duecredit (in fact, Newton’s famous statement that “if I have seen further it is by stand-ing on the shoulders of giants” is thought to have been less an expression of humilityand more a poke at Hooke, who was a very short man).

Real springs, on the other hand, tend to be either compressive or tensile; tryingto stretch a compressive spring or compress a tensile spring may result in nonlinearbehaviour. Real springs are also made of materials of finite strength. As a result, theywill typically be linear only over a short range of δ/L, and extension beyond thisrange will damage the spring and alter its characteristics. When approximating realsprings in real machines as ideal springs for the purposes of analysis, it is importantto ensure that the assumption of linearity is not violated.

Let us consider a block of mass m sitting on a frictionless surface, acted uponby an externally applied constant force F while being fixed in place by a spring, as


illustrated in Figure 4.14. If we define our system as consisting of only the block,then from the law of conservation of momentum,

∑GGG = maaa = ∑FextFextFext

where GGG is the momentum vector of the block, aaa is the acceleration vector of theblock, and ΣFextFextFext is the vector sum of the forces acting on the block (again, sincethere is only one object in the system, indeed all forces will be external to the sys-tem). Drawing the free-body diagram of the block, and defining the unit vector exexex

as being along the direction of travel of the block (as illustrated),

Fig. 4.14 A block being acted upon by a spring and an applied force.

maaa = FkFkFk +FFF

mx exexex = (−Fk +F) exexex

where x is the component of acceleration in the x-direction. By inspection, we seethat there are no forces and no acceleration in the y-direction. Now, knowing thatFk = kx, we can substitute and re-arrange, so that

mx exexex = (−Fk +F) exexex

mx = −kx+F

x+km

x− Fm

= 0

Next, we would like to determine the position of the block as a function of time,so we need to solve for x(t). Since we have both +x and +x terms on the left-handside of this expression, and since the entire expression must be equal to zero, thefunction x(t) (for which we need to solve) must satisfy the requirement x = −x towithin a few constants. Conveniently, the sine function does just that-


ddt

sin(x) = cos(x)

d2

dt2 sin(x) =ddt

cos(x) =−sin(x)

so we can surmise that a valid solution for our displacement function is x(t) =Asin(Bt +C)+D, where A, B, C and D are all constants to be determined. Then,

x = Asin(Bt +C)+D

x = ABcos(Bt +C)

x = −AB2 sin(Bt +C)

Now we can substitute these results into our earlier equation, so

x+km

x− Fm

= 0

(−AB2 sin(Bt +C))+

km

(Asin(Bt +C)+D

)− Fm

= 0

( km−B2)sin(Bt +C)+

kAm

(D− Fk) = 0

The only way for this equation to be satisfied at all times is if

B =

√km

D =Fk

However, with B and D specified as above, the equation will be satisfied for any val-ues of A and C; we therefore cannot yet determine unique values of these constants.Substituting the above values for B and D back into our equation for x(t),

x = Asin(Bt +C)+D

x = Asin

(√km

t +C

)+

Fk


What emerges now is easily recognized as a sinusoid with frequency (1/2π)√

k/mand amplitude A, phase C and offset F/k. To solve for A and C, then, all we need toknow is the location of the mass at some points in time. Unless a starting time for theproblem is explicitly specified, the point t = 0 can be set arbitrarily. We shall there-fore set t = 0 at whatever phase is required so that C = 0, in order to simplify oursolution. Because of the sinusoidal motion, this mass-spring arrangement is oftenreferred to as a simple oscillator, as it demonstrates the basic principle of vibration.

x = Asin

(√km

t

)+

Fk

There are some important observations which can be made about this result. Firstof all, the frequency of the motion will be the same regardless of the amplitude orapplied force. This is called the natural or resonant frequency of the system (some-times referred to as a mode or eigenmode). Secondly, if the amplitude is nonzeroat any one point in time, then the mass will keep moving indefinitely even if theapplied force F = 0. Third, the amplitude will remain constant for all t regardless ofk, m or F . For this reason, spring forces are said to be conservative.

Incidentally, the expression

x+km

x− Fm

= 0

is referred to as the equation of motion. An equation of motion is any expressionwhich relates the acceleration of an object to its position and/or velocity, and maynot always be integrable to yield a closed-form solution for the position as a functionof time. Any system which has an equation of motion of the form x+Ax+B = 0(where A and B are any constants) is said to be undergoing simple harmonic motion.

4.3.3 Viscous friction: x = f (x)

An object moving at very low speed through a fluid will experience a force opposingits motion as a result of the friction against the fluid, and this force is called wet orviscous friction. A marble slowly dropping through treacle and a block moving athigh speed over an oiled surface are both experiencing forms of viscous friction,since nowhere is there direct contact between two solid surfaces. In order to qualifyas viscous friction, either (i) the area of the body in contact with the fluid in a planefacing the direction of motion must be small, or (ii) the speed must be small. For amarble dropping through treacle, the speed is small; for the sliding block, the areain contact with the fluid (as seen from the direction of motion) is just a single line,and the area of a line is zero.

Viscous friction emerges from the equations of low-speed aerodynamics, whichare outside of the scope of this text. The viscous friction force is always proportional


to velocity and always opposes the direction of motion. Therefore, the friction forcecan be expressed as

FfFfFf =−Cx exexex

where exexex is defined as pointing in the direction of motion, and C is some constantwhich will depend on the fluid and the shape of the object.

Fig. 4.15 A body being pushed through a viscous medium by a constant force F.

Let us consider the case of a small sphere starting from rest and being forcedthrough a viscous fluid by a constant force F , as illustrated in Figure 4.15. We willdefine a fixed Cartesian coordinate system o such that exexex points in the direction ofmotion. Our system shall be defined as consisting of only the small sphere. Then,


maaa = FFF+FfFfFf

mxexexex = (F −Ff )exexex

Now we can divide out the unit vector and substitute in the earlier expression for theviscous friction,

mx = F −Cx

x+Cm

x =Fm

It would now be useful to be able to solve this equation of motion for the displace-ment as a function of time, x(t). To solve this, we first multiply through by eCt/m sothat

eCm t x+ e

Cm t(

Cm

)x = e

Cm t(

Fm

)


However, we may recognize on the left-hand side of this equation that,

eCm t x+ e

Cm t(

Cm

)x =

ddt

(xe

Cm t)

according to the product rule. Substituting this result into the previous expression,

eCm t x+ e

Cm t(

Cm

)x = e

Cm t(

Fm

)ddt

(xe

Cm t) = F

me

Cm t

Now, this expression may be directly integrated, as

xeCm t =

∫ (Fm

eCm t)dt

=Fm

∫ (e

Cm t)dt

xeCm t =

FC

eCm t +B

where B is some constant. We can actually determine the value of B, since we knowthat the sphere was initially at rest. Then, x = 0 when t = 0 and

(0)eCm (0) =

FC

eCm (0) +B

0 =FC+B

B = −FC

Substituting this value of B back into the previous result,

xeCm t =

FC

eCm t +B

xeCm t =

FC

eCm t − F

C

x =(e−

Cm t)F

Ce

Cm t − (e−C

m t)FC

x =FC−(

FC

)e−

Cm t


Now we can integrate again,

∫x dt =

∫ (FC−(

FC

)e−

Cm t

)dt

x =FC

t +

(mFC2

)e−

Cm t +B

where B is again some different constant. We can once again solve for this constantif we define the initial position of the sphere as being at x = 0. Then, x = 0 at t = 0and,

(0) =FC(0)+

(mFC2

)e−

Cm (0) +B

0 =mFC2 +B

B = −mFC2

Finally, substituting this result back into the previous expression for x, we obtain,

x =FC

t +

(mFC2

)e−

Cm t +B

x =FC

t +

(mFC2

)e−

Cm t − mF

C2

This is more conveniently re-arranged as

x =mFC2

(Cm

t + e−Cm t −1

)

Some interesting observations may be made from this result. First, if t is very large,e−Ct/m → 0 and Ct/m � 1 so

x ≈ mFC2

(Cm

t

)=

FC

t

So, for very large t, x is a linear function of t, and the derivative (x) is thereforeconstant. This means that after a long time the velocity will be constant, and thisconstant velocity x = F/C is called the terminal velocity of the body. The velocity


will also approach a constant for all t for the case of sufficiently large C, sinceCt/m � e−Ct/m and Ct/m � 1, these last two terms will nearly vanish from theexpression. So, if the treacle is thick enough, no matter how hard you push, thesphere won’t accelerate.

It is important to note here that shock absorbers (also known as dampers or dash-pots) work on the principle of viscous friction, and apply a force proportional to ve-locity which opposes motion. One of the most important mechanical arrangementsin engineering analysis is the spring-mass-damper assembly (as illustrated in Figure4.16). From the free-body diagram, we can easily substitute the forces into the linearmomentum equation, as

Fig. 4.16 A spring-mass-damper system.


FkFkFk +FfFfFf = maaa

−kxexexex −Cxexexex = mxexexex

We may now divide out the unit vector and re-arrange this expression as,

mx+Cx+ kx = 0

This expression is known as a second-order differential equation, and is one ofthe most important in all engineering analysis. This equation is used to describeanything which oscillates- from vibrating cables to signals in electronic circuits topressure in pipes. Solving for x as a function of time from the above equation wouldyield several possible solutions; one typical result is the under-damped oscillation,as illustrated in Figure 4.17.


Fig. 4.17 An under-damped oscillation.

4.3.4 Pressure drag: x = f (x2)

If an object is forced through a fluid with low viscosity, or if it is pushed at sufficientspeed, the object will begin to form a low-pressure wake of disturbed, turbulentfluid. Under these circumstances, the drag on the object is referred to as pressuredrag (since nearly all of the drag force experienced comes from the low-pressurewake). From the equations of fluid mechanics (which are, again, outside of the scopeof this text), the pressure drag can be shown to be proportional to the square of thespeed of the object so that

FDFDFD =−Kx2exexex

where exexex is defined as being in the direction of motion. The constant of proportion-ality K depends on the fluid and the geometry of the object.

Let us again consider an object being pushed through a fluid with constant forceF , as illustrated in Figure 4.15, except this time at a sufficiently high speed that thefriction force arises from pressure rather than viscous drag. We define the systemas consisting of the object only, and let exexex point in the direction of motion. Then,drawing a free-body diagram and summing the external forces yields,



maaa = FFF+FDFDFD

mxexexex = (F −FD)exexex

Dividing out exexex and substituting in the definition of pressure drag,

mx = F −Kx2

x =Fm− K

mx2

x =Fm

(1− K

Fx2)

Once again, it would be useful to solve this equation of motion for x. To do this,we need to convert this equation into a form which can be integrated. This equationseems like a good candidate for the method of separation of variables, so let us firstrecognize that

x =dxdt

=dxdx

dxdt

=dxdx

x

Substituting this into the previous result yields,

xdxdx

=Fm

(1− K

Fx2)

Now we can re-arrange this equation to separate the variables x and t,

x

1− KF x2

dx =Fm

dx

This equation suggests a substitution; let us say that

x =

√FK

sin(θ)

Then, the differential dx can be expressed in terms of dθ as

dx =

√FK

cos(θ)dθ

These values may now be substituted back into our equation with separated variablesto yield,

x

1− KF x2

dx =Fm

dx

4.3 Common applied forces, type 2: position-dependent 211√FK sin(θ)

1− KF

FK sin2(θ)

√FK

cos(θ)dθ =Fm

dx

sin(θ)cos(θ)

1− sin2(θ)dθ =

Km

dx

sin(θ)cos(θ)cos2(θ)

dθ =Km

dx

tan(θ)dθ =Km

dx

Finally, we have an equation which we can integrate.

∫tan(θ)dθ =

Km

∫dx

ln

(1

cos(θ)

)=

Km

x+C

where C is some constant dependent upon the initial conditions. Now, since θ is avariable we used just for convenience, we have to convert this expression back intobeing in terms of x only. To do this, we look back at our original substitution.

x =

√FK

sin(θ)

sin(θ) =

√KF

x

From trigonometry, then, we know that

cos(θ) =

√1− K

Fx2

Substituting this back into the earlier expression,

ln

(1

cos(θ)

)=

Km

x+C

ln

(1√

1− KF x2

)=

Km

x+C


If our object started from rest, then x = 0 when x = 0, and so our constant C = 0 aswell. Now, we can isolate x in the previous expression,

ln

(1√

1− KF x2

)=

Km

x+0

1√1− K

F x2= e

Km x

1− KF

x2 = e−2 Km x

x =

√FK

(1− e−2 K

m x)

This expression gives the velocity in terms of the position. The position as a functionof time may be calculated by performing a further integration, though, mathemati-cally, the result is horrendous and is beyond the scope of this discussion (it actuallyworks out to be the integral of a hyperbolic tangent).

Some interesting observations may still be made from this expression. First, forvery large x (which corresponds, of course, to very large t), the exponential termapproaches zero and x →√

F/K. Again, the body will achieve some constant ter-minal velocity after sufficient time has elapsed. Similarly, the exponential term willvanish at all times for very large K or very small m.

4.3.5 Pseudoforces

In some engineering disciplines, the linear momentum equation is expressed in aslightly different, but altogether equivalent, manner, as

The sum of the forces acting on a system in an inertial frame which is at restwith respect to the system must always be zero.

In other words, if we define a coordinate system o′ which is stuck to the object,the object is always at rest relative to o′. Consequently, the object cannot acceleratewithin o′. By definition, then, there can be no net external force acting upon theobject relative to the coordinate system o′. On the other hand, the coordinate systemitself may be accelerating; this would result in the object ‘feeling’ a force eventhough it it always at rest within o′. The ‘force’ the object feels is referred to as apseudoforce (because the force isn’t ‘real’). Pseudoforces are also known as inertialforces or d’Alembert forces after the French mathematician Jean le Rond d’Alembert(1717-1783).


To understand the concept of a pseudoforce, imagine sitting in an airplane withno windows. As far as you can tell, the airplane cabin never moves. As the airplanethrottles up for takeoff, you will feel a force pushing you into your seat. This forceyou feel isn’t ‘real’; in fact, an outside observer would be able to tell you that it’sactually the airplane seat pushing on you, and not some imaginary force pushing youinto the airplane seat. Locked in the cabin, though, you have no way of knowing this.If we were to draw a free-body diagram of you in your airplane seat, there wouldbe two forces acting: the ‘pseudoforce’ FPFPFP pushing you into your seat, and the sumof all the other external forces exerted upon you by the airplane, ΣFextFextFext, which areacting to keep you in equilibrium (relative to the airplane). Then,

∑FextFextFext +FPFPFP = maaa = 0

However, the pseudoforce will always be equal in magnitude to the time rate ofchange of the linear momentum but will act in the opposite direction, so that

FPFPFP =−GsysGsysGsys

so, the ‘pseudoforce’ statement of the linear momentum equation is simply

∑FextFextFext −GsysGsysGsys = 0

Clearly, the difference between the two statements of the law of conservation oflinear momentum is trivial. The use of pseudoforces tends to be more common inengineering disciplines involving static analysis, as it provides a means to incorpo-rate acceleration into a static analysis without requiring any reformulation of theequations. For the purposes of dynamic system analysis, the use of pseudoforces isdiscouraged as it can lead to confusion.

One of the most common pseudoforces is the so-called centrifugal force (from theLatin fuga, meaning ‘to flee from’), which arises as a result of centripetal accel-eration (from the Latin petere, meaning ‘to seek’). The centrifugal ‘force’, there-fore, should never be included in a free-body diagram when the standard (non-pseudoforce) expression of the law of conservation of linear momentum is used.

Interestingly, modern physics suggests that gravity itself is a pseudoforce. It isnow believed that any object with mass has the effect of distorting space, causingspace itself to accelerate away from the object. Since everything in the universe iscarried along with space, any reference system we define will also be acceleratingwith space. This acceleration results in a pseudoforce, and that pseudoforce is whatwe call gravity. However, since the theoretical warping of space-time lies far outsidethe scope of all engineering disciplines, gravity is always treated as a real force inclassical mechanics.


4.4 Rockets

In all of the above analyses, it was assumed that the mass of the body remainedconstant throughout. However, in engineering applications, it is often the case thatthe mass of an object changes with time (depending, of course, on how we defineour ‘object’). As a typical example of significant engineering importance, let usreconsider the case of a rocket.

The mass of the rocket changes as a function of time (as it blasts out high-speedexhaust gases), so GGG �= maaa. Then, if we define the system as consisting of the rocketchassis and payload only (and not the fuel), the exhaust plume exerts an externalforce upon the system in the form of the thrust FT .

Fig. 4.18 Example 6

Let us define a coordinate system o such that eyeyey is pointing in the direction of travelof the rocket. Drawing a free-body diagram of the system, we see that

∑FextFextFext = (FT −Fg)eyeyey

Since we defined the system as the rocket without the fuel, then the mass of thesystem is constant. Then, from the linear momentum equation,

∑FextFextFext = GsysGsysGsys = mveyeyey

FT −Fg = mv

FT −mg = mv

4.4 Rockets 215

Now, let us consider the exhaust gas particles. The force exerted by the exhaustplume upon the rocket must be equal in magnitude and opposite in direction as theforce exerted by the rocket upon the exhaust plume. Then, for a blob of exhaust gaswith mass mp starting with the same velocity as the rocket v and exiting the rocketnozzle with some other velocity u,

JJJ =∫

FFF dt = u mp eyeyey − v mp eyeyey∫F dt eyeyey = mp(u− v) eyeyey∫

F dt = mp(u− v)

according to the definition of impulse, which was discussed in Chapter 3. Differ-entiating both sides of this expression with respect to time, and assuming that thevelocity of the exhaust gas with respect to the rocket is constant (the relative velocityof the exhaust gas is u j = u− v),

F =ddt

(mp(u− v)

)= mpu j

FT = −mu j

Noting that FT = −F , since the force on the gas by the rocket is in the oppositedirection as the force on the rocket by the gas. Substituting this into the earlierresult from the linear momentum equation,

−mu j −mg = mv

− 1m

dmdt

u j −g =dvdt

−u jdmm

−g dt = dv

Both sides of this expression may then be integrated to yield,

−u j

∫dmm

−∫

g dt =∫

dv

−u j ln(m)−gt +C = v


where C is some constant of integration. If we assume that the rocket started at rest,then at t = 0, v = 0 and m = mo, and so

−u j ln(mo)−0+C = 0

C = u j ln(mo)

Substituting,

v = −u j ln(m)−gt +u j ln(mo)

v = u j ln(mo

m

)−gt

Here, it is important to note that the mass of the ‘system’ (that is, the rocket with-out the fuel) remained constant throughout. This expression is often referred to asthe Tsiolkovsky equation, after Konstantin Eduardovich Tsiolkovsky (1857-1935).A mathematics teacher by profession, Tsiolkovsky first published this solution in1897. At the time, Tsiolkovsky’s work was of little practical value and was largelyforgotten until three decades later, when technology was sufficiently advanced toput the theory into practice. A copy of Tsiolkovsky’s work was reputedly discoveredin the remains of the secret Luftwaffe rocket development facility at Peenemunde,Usedom, heavily annotated in the handwriting of Wernher von Braun (1912-1977).Widely regarded as a founding father of modern rocketry and acknowledged asthe inventor of the ballistic missile, von Braun’s job titles have included Sturm-bannfuhrer of the SS under the Nazi regime, Director of NASA in the US (1960-1970), and Editor of Popular Mechanics magazine.

4.4.1 Staged rockets

The advantage of staging in rocket design can also be seen from the above analysis.Staging is the process by which a rocket will jettison empty fuel tanks, redundantengines and sections of structure during ascent in order to reduce the required fuelload. Let us consider a rocket consisting of three stages, as illustrated in Figure 4.19.The three stages (numbered from the first to ignite to the last) contain fuel of massm1, m2 and m3, as well as a structural mass mr (for simplicity, we will assume thatall three stages have the same structural mass). At lift-off, then, the entire rocket willhave mass mo = m1 +m2 +m3 +n mr (where n is the number of stages; here n = 3).The initial velocity will again be zero. From earlier,

4.4 Rockets 217

−u j ln(m)−gt +C = v

Fig. 4.19 Example of a staged rocket ascent

where u j is the exhaust gas velocity (which we will assume to be constant always),m and v are the total mass and velocity of the rocket, respectively (which may bothchange with time) and C is a constant. Let us assume that the acceleration requiredto reach the required velocity is much greater than the acceleration due to grav-ity; under these circumstances, gravity may be neglected (which will significantlysimplify the analysis). Then,

−u j ln(m)+C = v

Substituting in the initial condition (m = mo, v = 0) for the staged rocket we cansolve for the constant C as

−u j ln(m)+C = v


−u j ln(m1 +m2 +m3 +n mr)+C = (0)

C = u j ln(m1 +m2 +m3 +n mr)

which is similar to the earlier result for a single-stage rocket. Then, at any timeduring the first-stage flight,

v = −u j ln(m)+u j ln(m1 +m2 +m3 +n mr)

v = u j ln

(m1 +m2 +m3 +n mr

m

)

When the fuel tanks in the first stage of the rocket have been depleted, the rocketwill have burned a mass m1 of fuel. Just before first stage separation, the mass ofthe rocket will be

m = mo −m1 = m2 +m3 +n mr

so, the final velocity of the rocket just before first stage separation v f/1 is

v f/1 = u j ln

(m1 +m2 +m3 +n mr

m

)

v f/1 = u j ln

(m1 +m2 +m3 +n mr

m2 +m3 +n mr

)

Next, the rocket jettisons the spent stage. Let us assume that the stage is simplyreleased, so that there is no momentum imparted upon the spent stage. The mass ofthe rocket at this point is

m = mo −m1 −mr = m2 +m3 +(n−1)mr

and the velocity of the rocket at the instant after the separation of stage 1 is the sameas the velocity at the instant before. The engines of stage 2 are then ignited, and therocket continues to accelerate. The velocity, as before, is given by the expression,

−u j ln(m)+C = v

We can now use as the ‘initial’ condition the instant after first stage separation, andsolve for the constant.

−u j ln(m)+C = v

−u j ln(m2 +m3 +(n−1)mr)+C = v f/1

4.4 Rockets 219

−u j ln(m2 +m3 +(n−1)mr)+C = u j ln

(m1 +m2 +m3 +n mr

m2 +m3 +n mr

)

Isolating the constant of integration,

C = u j ln(m2 +m3 +(n−1)mr)+u j ln

(m1 +m2 +m3 +n mr

m2 +m3 +n mr

)

C = +u j ln

((m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

m2 +m3 +n mr

)

Substituting into the earlier result, the velocity of the rocket at any time during thesecond stage burn is given by the expression,

v = −u j ln(m)+C

= −u j ln(m)+u j ln

((m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

m2 +m3 +n mr

)

v = u j ln

((m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

(m)(m2 +m3 +n mr)

)

When the fuel tanks in the second stage of the rocket have been depleted, the rocketwill have a total mass of

m = mo −m1 −m2 = m3 +(n−1)mr

so, the final velocity of the rocket just before first stage separation v f/2 is

v f/2 = u j ln

((m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

(m)(m2 +m3 +n mr)

)

v f/2 = u j ln

((m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

(m3 +(n−1)mr)(m2 +m3 +n mr)

)

The spent second stage will then separate, the third stage engines will ignite, and therocket will again continue to accelerate. The velocity of the third (and final) stagewill be

−u j ln(m)+C = v

though immediately after the second stage separation, the mass of the rocket hasdecreased again by an amount mr. Then, using the instant after separation as ourinitial condition for the third stage,


v f/2 = −u j ln(m)+C

u j ln

((m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

(m3 +(n−1)mr)(m2 +m3 +n mr)

)= −u j ln(m3 +(n−2)mr)

+C

Again, isolating the constant,

C = u j ln

((m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

(m3 +(n−1)mr)(m2 +m3 +n mr)

)+u j ln(m3 +(n−2)mr)

C = u j ln

((m3 +(n−2)mr)(m2 +m3 +(n−1)mr)(m1 +m2 +m3 +n mr)

(m3 +(n−1)mr)(m2 +m3 +n mr)

)

so the velocity at any time during the third stage flight is

v = −u j ln(m)+C

= −u j ln(m)

+u j ln


(m3 +(n−1)mr)(m2 +m3 +n mr)

)

v = u j ln


(m)(m3 +(n−1)mr)(m2 +m3 +n mr)

)

To extract some understanding from this burdensome algebraic expression, let usre-arrange the above result as follows:

v = u j ln

(mo

m

)+u j ln

(m3 +(n−2)mr

m3 +(n−1)mr

)+u j ln

(m2 +m3 +(n−1)mr

m2 +m3 +n mr

)

v = u j ln

(mo

m

)+u j ln

(m3 +(n−1)mr −mr

m3 +(n−1)mr

)+u j ln

(m2 +m3 +n mr −mr

m2 +m3 +n mr

)

v = u j ln

(mo

m

)+u j ln

(1− mr

m3 +(n−1)mr

)+u j ln

(1− mr

m2 +m3 +n mr

)

where we have recalled that m1 +m2 +m3 +n mr = mo and, in our example, n = 3.We may now recognize that, for any number of stages, the velocity during the laststage of flight is given by the expression,

4.4 Rockets 221

v = u j ln

(mo

m

)+u j

n−1

∑i=1

ln

(1− mr

m f/i

)(4.10)

where m f/i is the total mass of the rocket just before the separation of stage i (recallthat m is the instantaneous mass of the rocket at any point in time). This result isplotted in Figure 4.20, which also shows the velocity-mass curves for two rocketsof equal launch weight, where one consists of four stages and the other consists ofa single stage.

Fig. 4.20 Variation of velocity with mass for a single-stage and four-stage ideal rocket withmi/mr = 2.5.

Examining Equation (4.10), a number of interesting observations may be made.

If the structure of the rocket is massless, there’s no point to staging. If mr → 0, then

v → u j ln

(mo

m

)+u j

n−1

∑i=1

ln

(1− 0

m f/i

)

= u j ln

(mo

m

)+u j

n−1

∑i=1

ln

(1

)

= u j ln

(mo

m

)+u j

n−1

∑i=1

(0)

v → u j ln

(mo

m

)


which we recognize as the result for a simple, single-stage rocket with no structuralmass in the absence of gravity. Since 0 < mr/m f/i < 1 always, each term in thesummation will be negative. The summation will therefore reduce the maximumvelocity which can be achieved by the rocket. Consequently, the fastest possiblerocket is the one with no structural mass at all so that mr = 0.

The more stages a rocket has, the faster it can go. Real rockets clearly cannot bemassless. However, by keeping the individual terms of the summation as close tozero as possible, the velocity of the rocket will be maximized. To achieve this,mr/m f/i must be as small as possible. For rockets with structural mass, this canonly be done by keeping each individual stage as small as possible.

There is another important advantage to rocket staging which is unrelated to themass of the rocket structure. The efficient generation of thrust from escaping exhaustgas depends heavily upon the physical geometry of the rocket nozzle. However,the ‘ideal’ nozzle geometry changes with the ambient air pressure (and thereforealtitude). Consequently, the most efficient rocket nozzle would need to change itsgeometry with altitude (see Figure 4.21 for an example of how an optimally shapedrocket nozzle varies with altitude). While variable-geometry rocket nozzles do exist,the engine geometry may also be varied with altitude by staging: each stage is fittedwith a nozzle appropriate for the altitude at which it is ignited.

Fig. 4.21 Illustration of the variation of the optimal supersonic rocket nozzle geometry with stan-dard altitude, to scale (throat pressure is assumed constant).

4.5 Open systems

In the previous treatment, we needed to consider the rocket as a closed system withconstant mass in order to carry out our analysis. However, it is also possible to applythe law of conservation of linear momentum to an open system, which is a systemwhich may gain or lose mass. These systems are often referred to as being Eulerian,after the prolific Swiss mathematician Leonhard Paul Euler (1707-1783).

4.5 Open systems 223

Fig. 4.22 Demonstration of momentum flux.

Let us consider an empty box which is at rest and has some mass m, as illustratedin Figure 4.22. We will define our system as being the box and everything whichthe box contains. Now let us consider a solid, uniform rectangular bar (of cross-sectional area dA) passing through a hole and into the box. At any time, we shallsay that the bar pokes a distance x into the box. The bar then has a velocity vb = xwith respect to a fixed observer. Now, let us consider the expression of the law ofconservation of linear momentum as it applies to our system.

∑FextFextFext =ddt

GsysGsysGsys

Clearly, our box is not moving and must therefore have no momentum associatedwith it, even while the bar is entering it (to be pedantic, the centre of mass of thesystem may move because of the inclusion of the bar; however, there is no require-ment that the bar remains intact. Conservation of momentum must hold even if thebar is instantaneously vapourized upon entering the box, distributing its mass evenlyaround the interior). Therefore,

∑FextFextFext = GsysGsysGsys = 0

GsysGsysGsys = C

where C is some constant. Initially, the box was empty and at rest, and thereforecould have no momentum. Consequently, C = 0 and

GsysGsysGsys = 0


On the other hand, the bar does have some momentum associated with it, since it hasboth mass and nonzero velocity. As the bar enters the box, the bar and its momentumbecome part of our system. The system will therefore be acquiring momentum withtime, resulting in GsysGsysGsys > 0. However, this change in momentum is the result of achange in the contents of the system rather than the application of a force. It isimportant to recognize here that if there are no forces necessarily acting betweenthe box and the bar, so ΣFextFextFext = 0 and the box may remain at rest. For open systems,then, it is important to distinguish between the time rate of change of momentumof the system (which we know as GsysGsysGsys), and the momentum added to the system bythe environment, (which we shall call GbGbGb). This latter contribution is referred to asmomentum flux. For open systems, then

∑FextFextFext = GsysGsysGsys −GbGbGb

So, a more general expression of the law of conservation of linear momentum wouldbe,

The net external force acting upon a system will be equal to the differencebetween the time rate of change of the total linear momentum of the systemand the rate of momentum flux into the system.

For a closed system, the rate of momentum flux is zero by definition, and our originalstatement of the law of conservation of linear momentum is recovered.

Let us consider the term GbGbGb. We will assume that the bar in our example abovehas a uniform density ρ . Then, the mass mb of the portion of the bar inside the boxat any point in time is given by the expression,

mb = ρx dA

and the magnitude of the momentum of the portion of the bar inside the box is then

Gb = mbx = ρx x dA

The time-rate of change of the momentum inside the box may then be determined,as

Gb =ddt

Gb

=ddt

(ρx x dA

)Gb = ρ x2 dA+ρx x dA

Now, let us consider again the mass of the portion of the bar inside the box. The rateat which mass is being added to the system may be determined as,

4.5 Open systems 225

mb =ddt

mb

=ddt

(ρx dA

)mb = ρ x dA

We may now substitute this into the previous expression, so that

Gb = ρ x2 dA+ρx x dA

=(ρ x dA

)x+(ρx dA

)x

Gb = mbvb +mbvb

And so, for an open system, then, the law of conservation of linear momentum isexpressed as

∑Fext = Gsys −(mbvb +mbvb

)where vb is the velocity with which the bar is entering the system, and mb is themass of the bar inside the system. Now, if vb �= 0, there must necessarily be anexternal force being applied to the bar. This force is therefore also being appliedto the system, as the bar is assumed to be rigid. Consequently, this external forcemust necessarily be included in the ΣFext term on the left-hand side of the equation(effectively designating the quantity mbvb as a pseudoforce- in essence, saying thatthe box is accelerating into the bar, rather than vice-versa). Because of the way oursystem has been defined, then, vb = 0 always. Therefore, if the system is movingwith some velocity v,

∑Fext = Gsys −(mbvb +mb(0)

)∑Fext =

ddt(mv)− mbvb

∑Fext = mv+ vm− mbvb

where we have applied the product rule. Since mass is always conserved, the rate atwhich the mass of the system increases must be the same as the rate at which themass of the bar is entering the system, and so m = mb. The expression of the law ofconservation of linear momentum for our open system is then,


∑Fext = mv+mv− mvb (4.11)

The quantity m may be positive or negative, depending on whether mass is beingadded to or removed from the system. Finally, if we wish, we may express thevelocity of the bar relative to the velocity of the system vb/sys, as

vb = v+ vb/sys

Substituting this into Equation (4.11),

∑Fext = mv+mv− mvb

= mv+mv− m(v+ vb/sys

)= mv+mv− mv− mvb/sys

∑Fext = mv− mvb/sys

Significantly, this is exactly the same result obtained earlier for the case of a rocket,where we found that

−mu j −mg = mv

−mg = mv+ mu j

where u j was the velocity of the exhaust jet relative to the system, and the onlyexternal force acting was gravity. Recall that, for the case of the rocket, the systemwould be losing mass, so that m < 0. Gravity was also acting downwards, while ‘up’was defined as positive.

In our treatment of open systems above, we have considered only the case of lin-ear motion, and we have therefore developed Equation (4.11) as a scalar expression.If we were to develop it instead as a more general vector equation, the specific ge-ometry of the box would need to be considered (since the amount of mass enteringthe system will depend on the component of velocity normal to the surface of thesystem) and the bar must be allowed to deform (since conservation of momentumdoes not require that the bar only exit the system from a point directly opposite itspoint of entry). Under these assumptions, the bar becomes what we define as a fluid,and the vector expression of the law of conservation of linear momentum becomesthe Navier-Stokes equation- one of the fundamental governing equations of fluidmechanics (which is outside the scope of this text).

4.6 Examples 227

4.6 Examples

4.6.1 Projectile motion

An artillery shell having mass m is blasted from a cannon barrel with muzzle velocityvi and elevation angle θ . Determine the range of the artillery, the maximum altitudeof the shell and the time of flight, neglecting air resistance.

Fig. 4.23 Projectile motion

SOLUTION

Let us define our system as consisting of the artillery shell only, and let us like-wise define a coordinate system o with its origin located on the ground just at thepoint where the cannon fires the shell, with exexex pointing along the ground in thedirection of fire and eyeyey pointing upwards. Then, drawing a free-body diagram ofthe shell, we recognize that the weight of the shell is the only force acting upon it.Therefore,

∑FextFextFext = FgFgFg

∑FextFextFext = 0exexex −Fg eyeyey

From the momentum equation,

∑FextFextFext = GGG = mx exexex +my eyeyey


where x and y are the magnitudes of the accelerations along the exexex and eyeyey directions,respectively. Substituting together these results, we see that

0exexex −Fg eyeyey = mx exexex +my eyeyey

And, equating together the coefficients of the unit vectors,

0 = mx

−Fg = my

Let us first consider the exexex direction. We can integrate the acceleration to get thevelocity and position as a function of time, so that

0 = mx

x = 0∫x dt =

∫0 dt

x = C1

where C1 is a constant. The horizontal component of the velocity of the shell remainsconstant, and since we know the initial velocity vector v1v1v1 of the shell,

v1v1v1 = vi cos(θ) exexex + vi sin(θ) eyeyey

then we can say that

x =C1 = vi cos(θ)

To obtain the horizontal position as a function of time, we simply integrate thisexpression once more with respect to time.

∫x dt =

∫vi cos(θ) dt

x = vit cos(θ)+C2

where C2 is another constant of integration. Since we know that the shell is launchedfrom an initial position of x = 0 at time t = 0, it is therefore clear that C2 = 0 and

x = vit cos(θ)

4.6 Examples 229

Along the eyeyey direction, we can similarly equate the coefficients of the momentumequation to yield,

−Fg = my

−mg = my

y = −g

This expression can also be integrated twice to yield the vertical position as a func-tion of time, so that

∫y dt = −

∫g dt

y = −gt +C3

The constant of integration C3 can again be determined from the initial condition.Since we know the initial velocity along the eyeyey direction was vi sin(θ) at time t = 0,

vi sin(θ) = −g(0)+C3

C3 = vi sin(θ)

So, substituting this value of C3 into the earlier result,

y =−gt + vi sin(θ)

We can integrate this once again with respect to time in order to get the position yof the shell as a function of time, so that,

∫y dt =

∫ (−gt + vi sin(θ))

dt

y = −12

gt2 + vit sin(θ)+C4

where C4 is again a constant of integration. If we say that the shell was launchedfrom y = 0 at time t = 0,

0 = −12

g(0)2 + vi(0)sin(θ)+C4


C4 = 0

Substituting this value of C4 back into the previous result,

y = −12

gt2 + vit sin(θ)+C4

y = −12

gt2 + vit sin(θ)

So the position vector rrr of the shell as a function of time is given by

rrr =(vit cos(θ)

)exexex +

(−12

gt2 + vit sin(θ))

eyeyey

Now that we know the position of the projectile as a function of time, we cansolve for the range, maximum altitude and time-of-flight. To begin with, we recog-nize that the shell strikes the ground when y = 0 and t = t f > 0. so,

y = −12

gt2 + vit sin(θ)

0 = −12

gt2f + vit f sin(θ)

0 = −12

gt f + vi sin(θ)

t f =2vi sin(θ)

g

This result, then, is the required time-of-flight between firing and impact. To solvefor the range, we recognize that at t = t f , x = L; so, we simply substitute this time-of-flight into the component of the trajectory parallel to exexex, so that

x = vit cos(θ)

L = vi(2vi sin(θ)

g

)cos(θ)

L =2g

v2i sin(θ)cos(θ)

So now we have determined the time-of-flight and the range. Notice here that theproduct sin(θ)cos(θ) achieves a maximum for θ = 45◦, so the maximum possiblerange Lmax of the projectile is

4.6 Examples 231

Lmax =2g

v2i sin(45◦)cos(45◦)

Lmax =1g

v2i

All that remains is to determine the maximum altitude. There are two ways inwhich to do this: first, we can use the equation for the x-trajectory as a function oftime to isolate t as a function of x; substituting this into the y-trajectory will yield yas a function of x. Equating dy/dx = 0 will then yield the maximum altitude. Thiswould be, however, mathematically complex. More simply, we can recognize thatat the maximum altitude, y = 0. If we know the time th at which the y = 0, we cansubstitute this th into the equation of the y-trajectory to solve for the altitude at thattime.

y = −gt + vi sin(θ)

0 = −gth + vi sin(θ)

th =vi sin(θ)

g

Substituting this into the expression for the component of the trajectory parallel toeyeyey,

y = −12

gt2 + vit sin(θ)

H = −12

gt2h + vith sin(θ)

= −12

g(vi sin(θ)

g

)2+ vi(vi sin(θ)

g

)sin(θ)

= −v2i sin2(θ)

2g+

v2i sin2(θ)

g

H =v2

i sin2(θ)

2g

which is the maximum altitude of the shell, as required. It is interesting to notice thatthe maximum possible altitude Hmax is achieved for θ = 90◦ (the cannon is pointingstraight up), in which case


Hmax =v2

i sin2(90◦)2g

Hmax =1

2gv2

i

and also that Hmax = Lmax/2.

As an exercise, let us now see if we could get the same result by solving for thetrajectory of the shell as a function of the form y(x), as discussed earlier. We begin,then, by isolating t in the expression for x(t), as

x = vit cos(θ)

t =x

vi cos(θ)

Then, substituting back into the component of the trajectory lying along eyeyey,

y = −12

gt2 + vit sin(θ)

y = −12

g( x

vi cos(θ)

)2+ vi( x

vi cos(θ)

)sin(θ)

y = −( g

2v2i cos2(θ)

)x2 +

(tan(θ)

)x

so, clearly, the trajectory of the shell is parabolic. The maximum altitude is attainedwhen dy/dx = 0, so we can solve for the value of x (which we shall call xh) at thatpoint.

dydx

= 0 =ddx

(−( g

2v2i cos2(θ)

)x2

h +(tan(θ)

)xh

)

0 = −( g

v2i cos2(θ)

)xh +

sin(θ)cos(θ)

xh =v2

i cos(θ)sin(θ)g

Now, substituting this value back into the component of the trajectory lying parallelto exexex,

H = −( g

2v2i cos2(θ)

)x2

h +(tan(θ)

)xh

4.6 Examples 233

= −( g

2v2i cos2(θ)

)(v2i cos(θ)sin(θ)

g

)2+( sin(θ)

cos(θ)

)(v2i cos(θ)sin(θ)

g

)

= −v4i cos2(θ)sin2(θ)

2gv2i cos2(θ)

+v2

i sin2(θ)

g

= −v2i sin2(θ)

2g+

v2i sin2(θ)

g

H =v2

i sin2(θ)

2g

Which, thankfully, is the result we were expecting.

Finally, let us compute a value of H, then, for a typical artillery shell. The Mk.-IV Howitzer shell, for example, has a muzzle velocity vi = 517 m/s. Substituting,and taking θ = 45◦ (which is approximately the maximum possible elevation of aHowitzer barrel),

H =(517 m/s)2 sin2(45◦)

2(9.81 m/s2)

H = 6.8 km

which is reasonable; looking at the range, then, and again taking θ = 45◦ (for max-imum range),

Lmax =2g

v2i sin(θ)cos(θ)

=2

(9.81 m/s2)(517 m/s)2 sin(45◦)cos(45◦)

Lmax = 27.2 km

This result is suspicious; in practice, even the most experienced Mk-IV Howitzergun crews can achieve a maximum range of only about 15 km.

Here, an interesting point may be raised: our calculations have yielded reasonableresults for the altitude of an artillery shell, but the maximum range is overestimatedby a factor of nearly two. Therefore, the actual shape of the trajectory of an artilleryshell cannot be parabolic.

The apparent discrepancy between our calculations and the observed behaviourof an artillery shell comes about primarily because of aerodynamic drag. Because


the aerodynamic drag on a projectile acts in a direction opposing motion, the hori-zontal and vertical components of the drag will both vary as a function of position,altering the shape of the trajectory. As an example, Figure 4.24 shows the trajec-tories for a typical field artillery piece with varying θ , both with and without theeffects of aerodynamic drag.

Finally, because artillery shells are normally caused to spin, aerodynamic dragintroduces additional tumbling moments on the shell which can limit even furtherthe range (this effect is discussed in detail in Chapter 5).

Fig. 4.24 Comparison of the calculated trajectories for a 7.2-inch Mk.-IV Howitzer as a functionof barrel elevation angle, with and without aerodynamic drag (compressibility effects and spin havebeen neglected).

4.6.2 Ideal strings and pulleys

Two blocks A and B, with masses ma and mb, are connected together on an inclinedplane by a system of pulleys as illustrated below. If the plane is inclined by an angleθ , solve for the accelerations of the masses under the action of gravity.

SOLUTION

To solve this problem, we must first assume that the string is ideal and massless,and all the pulleys are massless and frictionless (or ideal as well). Then, notingthat there is only one string in the pulley arrangement, the tension will be the same

4.6 Examples 235

Fig. 4.25 Ideal strings and pulleys

everywhere along the string. First, let us look at mass B. We will define a fixedcoordinate system ob such that ey,bey,bey,b points straight up. Then, we define our systemas the block B and the pulley attached to the mass. From the free-body diagram, wesee that

Fig. 4.26 Ideal strings and pulleys- free-body diagram.

∑FextFextFext = FgbFgbFgb +FTFTFT +FTFTFT +FTFTFT

∑FextFextFext = (−Fgb +3FT )eybeybeyb


where Fgb is the gravitational force acting on block B. Then, since the mass of thesystem is constant, the momentum equation can be expressed as

GbGbGb = mbyb eybeybeyb = (−Fgb +3FT )eybeybeyb

mbyb = −mbg+3FT

yb = −g+3FT

mb

Next, we consider block A. Let us redefine our system to only contain block A andthe pulley to which it is attached. We will define the fixed coordinate system oa suchthat exaexaexa points upward along the plane while eyaeyaeya points normally upward from theplane. Then, from the free-body diagram,

∑FextFextFext = FgaFgaFga +FTFTFT +FTFTFT +FnFnFn

∑FextFextFext =(−Fga sin(θ)+2FT

)exaexaexa +

(−Fga cos(θ)+Fn)

eyaeyaeya

Since we know that the mass is constrained to remain on the surface of the plane,ya = 0 and so, from the momentum equation,

maya = 0 = −Fga cos(θ)+Fn

Fn = magcos(θ)

However, we are interested in the acceleration up the plane. The mass of the systemis constant, so looking at the exaexaexa coefficient of the momentum equation,

maxa = −Fga sin(θ)+2FT

maxa = −magsin(θ)+2FT

xa = −gsin(θ)+2FT

ma

So, we now know the acceleration of each block in terms of the tension in the string:

yb = −g+3FT

mb

4.6 Examples 237


ma

Since xa, yb and FT are all unknown, this constitutes a system of two equations inthree unknowns. To solve, we need one more equation relating these three variables.We can get this extra equation from the displacements; the motions of blocks a andb are restricted because they are connected to the system of pulleys.

If we consider the right-most string supporting the block B, it is clear that raisinga point on the string by some distance dL will cause the block to raise by a distancedL/3. Similarly, if we consider the upper string supporting the block A, lowering apoint on the string by the same distance dL down the incline will cause the blockA to lower down the incline by a distance dL/2. Then, since the right-most stringsupporting block B and the upper string supporting block A are the same bit ofstring, we can say that

dyb =13

dL

dxa =12

dL

Then, eliminating dL,

dyb =23

dxa

Both sides of this expression may then be divided by dt.

dyb

dt=

23

dxa

dt

yb =23

xa

This expression may then be differentiated once again with respect to time to yield

yb =23

xa

So, now we have a system of three equations and three unknowns:

yb = −g+3FT

mb


ma


yb =23

xa

Let us now isolate FT in terms of xa.

yb = −g+3FT

mb

23

xa = −g+3FT

mb

FT =2mb

9xa +

mbg3

Substituting this back into the earlier result and solving for yb,

xa = −gsin(θ)+22mb

9 xa +mbg

3

ma

xa = −gsin(θ)+49

mb

maxa +

23

mb

mag

xa =23

mbma

g−gsin(θ)

1− 49

mbma

And, finally,

yb =23

xa =23

mbma

g−gsin(θ)32 − 2

3mbma

4.6.3 Kinetic friction with multiple objects

Two frictional blocks A and B of masses ma and mb are linked together by means ofan ideal string and massless, frictionless pulley, as illustrated. A force is applied toblock B to cause it to accelerate to the right. Compute the acceleration of blocks.

SOLUTION

Let us begin with block A. We define a coordinate system o which is stuck tothe ground such that exexex is pointing to the right and eyeyey is pointing upwards. Then,we define our system to consist only of block A so that all forces acting on A areexternal to the system. Drawing a free-body diagram of the system, then, we see that

4.6 Examples 239

Fig. 4.27 Kinetic friction with multiple objects

∑FextFextFext = Ff1Ff1Ff1 +Fn1Fn1Fn1 +Fg1Fg1Fg1 +FTFTFT

∑FextFextFext = (Ff 1 −FT )exexex +(Fn1 −Fg1)eyeyey

We can now substitute this result into the momentum equation, remembering thatthe mass of the system is constant.


(Ff 1 −FT )exexex +(Fn1 −Fg1)eyeyey = maxa exexex +0eyeyey

(μk1Fn1 −FT )exexex +(Fn1 −mag)eyeyey = maxa exexex +0eyeyey

where we recall that for the case of kinetic friction, Ff = μkFn. Looking at the scalarcoefficients of eyeyey only, and recognizing that the block can’t move vertically,

Fn1 −mag = 0

Fn1 = mag

Now, we can examine the scalar coefficients of exexex and substitute in this value of Fn1,

μk1Fn1 −FT = maxa

μk1mag−FT = maxa


Here there are two unknowns: the acceleration of the block and the tension in thestring. Since the tension in the string is the same throughout, we’ll need that in ouranalysis of block B. Therefore, we will isolate the tension.

μk1mag−FT = maxa

FT = ma(μk1g− xa)

Now we may turn our attention to block B. Let us redefine our system so that itconsists only of block B. Then, drawing the free-body diagram of the system, wecan compute the net external force vector,

∑FextFextFext = FFF+Fn2Fn2Fn2 +Fg2Fg2Fg2 +Ff2Ff2Ff2 +FTFTFT +Ff1Ff1Ff1 +Fn1Fn1Fn1

∑FextFextFext = (F −Ff 1 −Ff 2 −FT )exexex +(Fn2 −Fn1 −Fg2)eyeyey

As with block A, the mass of the system is constant so the momentum equation maybe expressed as,


(F −Ff 1 −Ff 2 −FT )exexex +(Fn2 −Fn1 −Fg2)eyeyey = mbxb exexex +0eyeyey

Let us again consider first the scalar coefficients of the unit vector eyeyey.

Fn2 −Fn1 −Fg2 = 0

Fn2 −mag−mbg = 0

Fn2 = (ma +mb)g

So, the force pressing between block B and the ground is the combined weight ofthe two blocks, as we would expect. Now we can look at the scalar coefficients ofexexex.

F −Ff 1 −Ff 2 −FT = mbxb

F −μk1mag−μk2Fn2 −FT = mbxb

F −μk1mag−μk2(ma +mb)g−ma(μk1g− xa) = mbxb

4.6 Examples 241

Note here that we have substituted in previous results for Ff 1, as well as the valueof FT for which we had solved earlier. Finally, we may recognize that, because thetwo masses are connected together via a pulley, their motion is constrained. At alltimes, the speed and acceleration of the two blocks must be the same but in oppositedirections, so that xb =−xa. Then,

F −μk1mag−μk2(ma +mb)g−ma(μk1g− xa) = mbxb

F −μk1mag−μk2(ma +mb)g−ma(μk1g− xa) = mb(−xa)

F −μk1mag−μk2(ma +mb)g−μk1mag = −(ma +mb)xa

−F −2μk1mag−μk2(ma +mb)g(ma +mb)

= xa

and

F −2μk1mag−μk2(ma +mb)g(ma +mb)

= xb

Let us now consider the special case where μk1 = μk2 = μs, the coefficient of staticfriction between the surfaces. What would be the minimum tension required to causethe blocks to become ‘unstuck’? In this case, the acceleration of block B would be0, so

0 =F −2μsmag−μs(ma +mb)g

(ma +mb)

0 = F −2μsmag−μsmag−μsmbg

F = μsg(3ma +mb)

So, the mass of block A contributes three times more toward the static friction thanthe mass of block B.

4.6.4 Limits of static friction: rectilinear motion

A block B sits atop a wedge-shaped block A which is being accelerated horizontallyby a magnitude xa. If the critical coefficient of static friction between the blocks isμs, find the range of wedge angles θ over which the block B will remain ‘stuck’ tothe surface of Block A.


Fig. 4.28 Limits of static friction: rectilinear motion

SOLUTION

In this problem, there are two possible ways in which the block B may move: ifthe angle θ is too steep, it may slip downwards, and if the angle is too low, it mayslip upwards. The only difference in the analysis will be the direction in which thefriction acts, since the friction always opposes the direction of motion. Let us firstconsider the case where the angle is too high and the block slips downwards. In thiscase, the friction will act upwards along the slope.

Let us define a fixed coordinate system o such that the unit vector exexex pointsdownwards along the slope, and eyeyey points normally upwards from it. Then, we defineour system as consisting only of the block B. Drawing a free-body diagram of thesystem at the moment of slip, then,

∑FextFextFext = FgFgFg +FfFfFf +FnFnFn

∑FextFextFext =(Fg sin(θ)−Ff

)exexex +

(Fn −Fg cos(θ)

)eyeyey

However, we know that Fg = mbg and Ff = μsFn just at the moment of slip, so wemay substitute,

∑FextFextFext =(mbgsin(θ)−μsFn

)exexex +

(Fn −mbgcos(θ)

)eyeyey

Since the block B is ‘stuck’ to block A, the accelerations of block A and block B willbe the same for so long as they remain stuck. Since we are only interested in thecase when blocks A and B are stuck together, ababab = aaaaaa. However, we need to resolvethe acceleration vector into components parallel to our exexex and eyeyey unit vectors.

ababab = xcos(θ)exexex + xsin(θ)eyeyey

Now we may substitute the vector sum of the external forces and the accelerationvector into the momentum equation,

4.6 Examples 243

∑FextFextFext = GGG = mbababab(mbgsin(θ)−μsFn

)exexex +

(Fn −mbgcos(θ)

)eyeyey = mb

(xcos(θ)exexex + xsin(θ)eyeyey

)

The only unknown is the normal force. Since we are interested in the motion alongthe exexex direction, we can use the eyeyey coefficients to isolate and solve for the normalforce in terms of quantities which we do know.

Fn −mbgcos(θ) = mbxsin(θ)

Fn = mb(gcos(θ)+ xsin(θ)

)

This expression may now be substituted into the exexex coefficients of the momentumequation to yield,

mbgsin(θ)−μsFn = mbxcos(θ)

mbgsin(θ)−μs

(mb(gcos(θ)+ xsin(θ)

))= mbxcos(θ)

mbgsin(θ)−μsmbgcos(θ)−μsmbxsin(θ) = mbxcos(θ)

Because we defined Ff = μsFn, this expression will only be true just at the momentof slip. Therefore, if we isolate and solve for θ , we will have the value of θ at themoment of downward slipping.

mbgsin(θ)−μsmbgcos(θ)−μsmbxsin(θ) = mbxcos(θ)

(mbg−μsmbx)sin(θ) = (mbx+μsmbg)cos(θ)

sin(θ)cos(θ)

=mbx+μsmbgmbg−μsmbx

tan(θ) =x+μsgg−μsx

which is, again, only true at the moment of downward slip. So the critical angle fordownward slippage, which we shall call θd , is

θd = tan−1( x+μsgg−μsx

)


Let us now consider the case when the block just begins to slip upwards along theslope. In this case, the friction is acting down the slope, so the net external forceacting on the block at the moment of slip becomes,

∑FextFextFext =(Fg sin(θ)+Ff

)exexex +

(Fn −Fg cos(θ)

)eyeyey

∑FextFextFext =(mbgsin(θ)+μsFn

)exexex +

(Fn −mbgcos(θ)

)eyeyey

The acceleration vector of the block B remains unchanged:

ababab = xcos(θ)exexex + xsin(θ)eyeyey

So, the momentum equation now reads,

∑FextFextFext = GGG = mbababab(mbgsin(θ)+μsFn

)exexex +

(Fn −mbgcos(θ)

)eyeyey = mb

(xcos(θ)exexex + xsin(θ)eyeyey

)

Looking at the eyeyey direction, nothing has changed compared to the instant when theblock begins to slip downwards; consequently, the normal force will be exactly thesame as in the previous case.

Fn = mb(gcos(θ)+ xsin(θ)

)Equating the coefficients along the exexex direction and substituting this value of Fn,

mbgsin(θ)+μsFn = mbxcos(θ)

mbgsin(θ)+μs

(mb(gcos(θ)+ xsin(θ)

))= mbxcos(θ)

mbgsin(θ)+μsmbgcos(θ)+μsmbxsin(θ) = mbxcos(θ)

Again, this will only hold true at the moment of upwards slip. We can then isolateand solve for θ at that moment, as

mbgsin(θ)+μsmbgcos(θ)+μsmbxsin(θ) = mbxcos(θ)

(mbg+μsmbx)sin(θ) = (mbx−μsmbg)cos(θ)

sin(θ)cos(θ)

=mbx−μsmbgmbg+μsmbx

4.6 Examples 245

tan(θ) =x−μsgg+μsx

As before, we can define this critical angle at the moment of upwards slip as θu, sothat

θu = tan−1( x−μsgg+μsx

)And so, finally, the range of angles over which the block will remain ‘stuck’ is,

θu < θ < θd

tan−1( x−μsgg+μsx

)< θ < tan−1( x+μsg

g−μsx

)

There are a few interesting observations which may be made about this result. Firstof all, the mass divided out of the expression, so the critical slipping angle will be thesame regardless of the mass of the block B. Secondly, the critical slip-down and slip-up angles are identical apart from the sign of μs. Since the only thing that changedbetween the two critical cases was the direction of application of the frictional force,this makes sense; Ff went from −μsFnexexex to +μsFnexexex.

Finally, the results lead us to consider what happens if the acceleration x is verylarge. In that case,

θu = tan−1( x−μsgg+μsx

)≈ tan−1( xμsx

)= tan−1( 1

μs

)θd = tan−1( x+μsg

g−μsx

)≈ tan−1( x−μsx

)= tan−1(− 1

μs

)

This is interesting: according to the way we set up the problem, the wedge angle θcan range from 0◦ to 90◦; otherwise, the nature of the problem would change. Thecritical coefficient of static friction μs must be positive, so the result for θd at largeacceleration would seem impossible (there is no angle between 0 and 90◦ with anegative tangent). This suggests to us that as the acceleration gets increasingly large,the angle at which downward slip may occur also gets increasingly large until, forx → ∞, θd → 90◦ and our set-up of the problem was no longer valid. So, if youaccelerate fast enough, you can stick to a vertical surface. For very large x, then,

tan−1( 1μs

)< θ < 90◦

for no slippage to occur.


4.6.5 Limits of static friction: constrained motion

An object of mass m sits on the inside surface of a conical container at a radiusR from the axis, as illustrated. The sides of the conical container are at an angleθ relative to horizontal, and the conical container is spinning about its axis witha constant angular velocity ω . The object is held in place against the side of thecontainer by static friction. Determine the range of angular velocities over whichthe object will remain stationary relative to the container if the coefficient of staticfriction between the object and the container is μs.

Fig. 4.29 Limits of static friction: constrained motion

SOLUTION

To solve this problem, we need to solve for the frictional force as a function ofthe angular velocity, and then let Ff = μsFn; this is the condition for slipping to justbegin to occur. To do this, we start with the linear momentum equation.

∑FextFextFext = GsysGsysGsys = map/oap/oap/o

where ap/oap/oap/o is the acceleration vector of the object. From the free-body diagram, wesee that,

∑FextFextFext = FgFgFg +FnFnFn +FfFfFf

= −Fg eyeyey −Fn sin(θ) exexex +Fn cos(θ) eyeyey +Ff cos(θ) exexex +Ff sin(θ) eyeyey

∑FextFextFext =(Fn cos(θ)+μsFn sin(θ)−mg

)eyeyey +

(μsFn cos(θ)−Fn sin(θ)

)exexex

4.6 Examples 247

where it has been assumed for the moment that the object is just about to slip downthe incline, so friction is acting upwards. This expression, on its own, does not allowus to solve for ω; however, we know that ω is related to ap/oap/oap/o. We can thereforeobtain another expression for ap/oap/oap/o in terms of ω from relative motion. Let us definea coordinate system o′ which is stuck to the container with the origin on the axis ofthe container at the same height as the object; e′xe′xe′x always points at the object and e′ye′ye′yalways points straight up. At the instant of interest (illustrated), o′ is instantaneouslycoincident with our fixed reference frame, o. Then,


However, we can recognize immediately that o′ is rotating but not translating; there-fore ao′/oao′/oao′/o = 0. Furthermore, since the angular velocity of the container is constant,αaxαaxαax = 0. The object is stationary relative to the container up to and including theinstant that it first begins to slip, so ap/o′ap/o′ap/o′ = vp/o′vp/o′vp/o′ = 0. Consequently, the relativeacceleration equation reduces to

ap/oap/oap/o = 0+0+2ωaxωaxωax ×0+0×rp/o′rp/o′rp/o′ +ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

= ω eyeyey ×ω eyeyey ×R e′xe′xe′x= ω eyeyey ×−ωR ezezez

ap/oap/oap/o = −ω2R exexex

where we recognized that, at the instant of interest, exexex =e′xe′xe′x. Substituting these resultsback into the linear momentum equation,

∑FextFextFext = map/oap/oap/o(Fn cos(θ)+μsFn sin(θ)−mg

)eyeyey+(

μsFn cos(θ)−Fn sin(θ))

exexex = −mω2R exexex

Now, let us consider only the scalar coefficients of eyeyey.

Fn cos(θ)+μsFn sin(θ)−mg = 0

Fn(cos(θ)+μs sin(θ)

)= mg

Fn =mg

cos(θ)+μs sin(θ)

Next, we consider the scalar coefficients of exexex.


μsFn cos(θ)−Fn sin(θ) = −mω2R

Fn(μs cos(θ)− sin(θ)

)= −mω2R

However, we already have an expression for Fn; substituting,

mgcos(θ)+μs sin(θ)

(μs cos(θ)− sin(θ)

)= −mω2R

ω =

√gR

(sin(θ)−μs cos(θ)cos(θ)+μs sin(θ)

)

This is the critical angular velocity below which the object will slip down the incline.Next, for the case where the object slips up the incline, we reverse the direction ofthe friction and repeat the analysis. The expression for relative acceleration will notchange. From the free-body diagram, we see that,


= −Fg eyeyey −Fn sin(θ) exexex +Fn cos(θ) eyeyey −Ff cos(θ) exexex −Ff sin(θ) eyeyey

∑FextFextFext =(Fn cos(θ)−μsFn sin(θ)−mg

)eyeyey +

(−μsFn cos(θ)−Fn sin(θ))

exexex

Substituting into the linear momentum equation,

∑FextFextFext = map/oap/oap/o(Fn cos(θ)−μsFn sin(θ)−mg

)eyeyey+(−μsFn cos(θ)−Fn sin(θ))

exexex = −mω2R exexex

Equating the scalar coefficients of eyeyey, as before,

Fn cos(θ)−μsFn sin(θ)−mg = 0

Fn(cos(θ)−μs sin(θ)

)= mg

Fn =mg

cos(θ)−μs sin(θ)

Next, from the scalar coefficients of exexex,

4.6 Examples 249

−μsFn cos(θ)−Fn sin(θ) = −mω2R

Fn(−μs cos(θ)− sin(θ)

)= −mω2R

mgcos(θ)−μs sin(θ)

(−μs cos(θ)− sin(θ))= −mω2R

Isolating ω ,

ω =

√gR

(sin(θ)+μs cos(θ)cos(θ)−μs sin(θ)

)

This is the critical angular velocity above which the object will slip up the incline.Therefore, the range of angular velocities over which the object will remain on thesurface of the container is,√

gR

(sin(θ)−μs cos(θ)cos(θ)+μs sin(θ)

)< ω <

√gR

(sin(θ)+μs cos(θ)cos(θ)−μs sin(θ)

)

As expected, the only thing which has changed between the slip-down and slip-upcondition is the sign of μs.

4.6.6 Simple pendulum

An object of mass m is suspended from the ceiling by means of a thin massless stringof length L, and is allowed to swing in a plane. The object is raised initially so thatthe angle θ subtended between the string and vertical is θ0, and is then releasedfrom rest. Determine the angle θ of the string as a function of time.

SOLUTION

To begin with, we shall define our system as consisting of the object alone. Wewill define a fixed coordinate system o such that exexex is pointing to the right, and eyeyey ispointing upwards, as illustrated. From the law of conservation of linear momentum,


where ap/oap/oap/o is the acceleration of the object. Since the only external forces acting onthe object are gravity and the tension in the string, from the free-body diagram ofthe object,

FgFgFg +FTFTFT = map/oap/oap/o


Fig. 4.30 Simple pendulum

−mg eyeyey −FT sin(θ) exexex +FT cos(θ) eyeyey = map/oap/oap/o

−FT sin(θ) exexex +(FT cos(θ)−mg

)eyeyey = map/oap/oap/o

where both FT and ap/oap/oap/o are unknown. However, because the path of the object isconstrained (the string is forcing it to describe a circular arc), we can obtain anexpression for the acceleration vector of the object from relative motion. Let usdefine the moving coordinate system o′ which is stuck to the string with the originon the pin at the ceiling. We shall define e′xe′xe′x such that it always points toward theobject, and e′ye′ye′y such that it points up and to the right, as shown. Then, e′ze′ze′z points upout of the page. From the relative acceleration equation,


where the point p is stuck on the object. Since the object and the pin are both stuck tothe same reference frame and cannot move relative to each other, ap/o′ap/o′ap/o′ = vp/o′vp/o′vp/o′ = 0.Also, because the pin is fixed to the ceiling and does not move, ao′/oao′/oao′/o = 0 as well.Substituting,

ap/oap/oap/o = 0+0+2ωaxωaxωax ×0+ θ e′ze′ze′z ×L e′xe′xe′x + θ e′ze′ze′z × θ e′ze′ze′z ×L e′xe′xe′x= θL


)+ θ e′ze′ze′z × θL


)= θL e′ye′ye′y + θ e′ze′ze′z × θL e′ye′ye′y= θL e′ye′ye′y + θ 2L


)

4.6 Examples 251

ap/oap/oap/o = θL e′ye′ye′y − θ 2L e′xe′xe′x

However, in order to substitute the acceleration back into the linear momentumequation, we will need to resolve the unit vectors into components along exexex andeyeyey. From the geometry, we find that

e′xe′xe′x = sin(θ) exexex − cos(θ) eyeyey

e′ye′ye′y = cos(θ) exexex + sin(θ) eyeyey


Substituting this result into the relative acceleration equation,

ap/oap/oap/o = θL e′ye′ye′y − θ 2L e′xe′xe′x= θL


)− θ 2L(sin(θ) exexex − cos(θ) eyeyey

)ap/oap/oap/o =

(θLcos(θ)− θ 2Lsin(θ)

)exexex +

(θLsin(θ)+ θ 2Lcos(θ)

)eyeyey

We may now substitute this result for ap/oap/oap/o back into the linear momentum equation,so that

−FT sin(θ) exexex +(FT cos(θ)−mg

)eyeyey

= m

((θLcos(θ)− θ 2Lsin(θ)

)exexex +

(θLsin(θ)+ θ 2Lcos(θ)

)eyeyey

)

Equating the scalar coefficients of exexex,

−FT sin(θ) = mθLcos(θ)−mθ 2Lsin(θ)

FT = −mθLcos(θ)sin(θ)

+mθ 2L

And, equating the scalar coefficients of eyeyey,

FT cos(θ)−mg = mθLsin(θ)+mθ 2Lcos(θ)

However, we can eliminate the unknown tension force FT by substituting our earlierresult,


(−mθL

cos(θ)sin(θ)

+mθ 2L

)cos(θ)−mg = mθLsin(θ)+mθ 2Lcos(θ)

−mθLcos2(θ)

sin(θ)+mθ 2Lcos(θ)−mg = mθLsin(θ)+mθ 2Lcos(θ)

−θLcos2(θ)

sin(θ)−g = θLsin(θ)

θ

(sin(θ)+

cos2(θ)

sin(θ)

)= −g

L

θ

(sin2(θ)

sin(θ)+

cos2(θ)

sin(θ)

)= −g

L

θ = −gL

sin(θ)

Here, we have the magnitude of the angular acceleration as a function of the angle;however, since both of these quantities are functions of time, this equation cannotbe easily solved. Instead, we will make the approximation that the angle θ remainssmall. Then, we can make the substitution sin(θ)≈ θ , so that

θ ≈−gL

θ

So, the angle θ is some function of time such that its second derivative is equal to anegative multiple of the function itself. It is interesting to note that this expressionis of exactly the same form as seen earlier for the simple mass-spring system. Thesolution θ(t) will therefore be of the same form, so that

θ =C0 sin(C1t +C2)

where C0, C1 and C2 are all constants. We may also note that the frequency of oscil-lation will be C1/2π . We may now differentiate this expression to obtain the angularvelocity, as

θ =ddt

(C0 sin(C1t +C2)

)θ = C0C1 cos(C1t +C2)

However, we know that θ = 0 when the object is initially released at t = 0. We mayuse this initial condition to solve for C2, as

4.6 Examples 253

0 = C0C1 cos(C1(0)+C2)

0 = cos(C2)

C2 =π

2

Substituting this result into our earlier expression for θ ,

θ =C0 sin

(C1t +

π

2

)

However, from the initial condition, we know that θ = θ0 when t = 0. Substituting,

θ0 = C0 sin

(C1(0)+

π

2

)θ0 = C0(1)

C0 = θ0

Finally, we can differentiate our earlier expression for θ with respect to time onceagain, as

θ =dθ

dt

=ddt

(θ0C1 cos

(C1t +

π

2

))

θ = −θ0C21 sin

(C1t +

π

2

)

Substituting these results into our earlier expression for θ from the momentum equa-tion,

θ ≈ −gL

θ

−θ0C21 sin

(C1t +

π

2

)= −g

Lθ0 sin(C1t +

π

2)

C21 =

gL

C1 =

√gL


So, the frequency of oscillation of a pendulum is given by the expression

f =1

2πC1

f =1

2π

√gL

and is independent of the mass of the pendulum. Substituting our result for C1 backinto our expression for θ yields,

θ = θ0 sin

(√gL

t +π

2

)

where we must recall that we have assumed θ to be reasonably small. In practice, forthe case of a simple pendulum, the small-angle approximation produces reasonableresults for θ0 � 20◦, as illustrated in Figure 4.31.

Fig. 4.31 Comparison of the motion profile of the exact solution of a simple pendulum to thesmall-angle approximation (taking L = 1 m).

It should be noted that, when the small-angle approximation fails, the motion ofthe pendulum is no longer sinusoidal, even though it qualitatively appears to be. Todemonstrate the difference, let us assume that the motion is purely sinusoidal. Then,

θ = θ0 sin

(√gL

t +π

2

)√

gL

t +π

2= sin−1

(θ

θ0

)

4.6 Examples 255

Substituting this result into our earlier expression for the angular velocity,

θ = C0C1 cos(C1t +C2)

= θ0

√gL

cos

(√gL

t +π

2

)

θ = θ0

√gL

cos

(sin−1

(θ

θ0

))

However, from the trigonometric identity,

cos

(sin−1

(θ

θ0

))=

√θ 2

0 −θ 2

θ0

Substituting this identity into our previous result for θ ,

θ = θ0

√gL

(√θ 2

0 −θ 2

θ0

)

θ =

√gL

√θ 2

0 −θ 2

Squaring both sides of this expression and re-arranging,

(√gL

)2(θ 2

0 −θ 2) = θ 2

(θ

θ0√

g/L

)2

+

(θ

θ0

)2

= 1

This expression is clearly the equation of a circle with radius 1 in nondimensionalθ -θ space. These nondimensional variables are plotted in Figure 4.32 for the caseof θ0 = 50◦, showing that the motion of the pendulum does not describe a circle;consequently, the motion of the pendulum must be non-sinusoidal for larger θ0,despite the outward appearance of the motion profile in Figure 4.31.


Fig. 4.32 Nondimensional velocity-displacement plot of a simple pendulum with θ0 = 50◦.

4.6.7 Normal force

A small marble of mass m sits atop a frictionless circular path of radius R. It is givenan initial push of velocity vo and slides down the circular surface under the actionof gravity. Compute the angle θ at which the marble leaves the surface.

Fig. 4.33 Normal force

4.6 Examples 257

SOLUTION

This question is essentially asking for the angle at which the normal force be-tween the path and the marble becomes zero. We begin, as always, by defining oursystem such that it consists only of the small marble. We stick a cylindrical coor-dinate system o onto the centre of the circular path so that ererer always points at themarble. We can then draw a free-body diagram of the marble, and find the sum ofthe forces acting on it:

∑FextFextFext = FgFgFg +FnFnFn

∑FextFextFext =(Fn −Fg cos(θ)

)ererer +Fg sin(θ)eθeθeθ

We also know that the acceleration of a point in a cylindrical coordinate system is,

ap/oap/oap/o = (r− rθ 2)ererer +(2rθ + rθ)eθeθeθ + zezezez

However, in our problem, r = r = z = 0 and r = R, so this reduces to

ap/oap/oap/o =−Rθ 2 ererer +Rθ eθeθeθ

Then, the momentum equation yields,

∑FextFextFext = GGG = map/oap/oap/o(Fn −Fg cos(θ)

)ererer +Fg sin(θ)eθeθeθ = m

(−Rθ 2 ererer +Rθ eθeθeθ

)(Fn −mgcos(θ)

)ererer +mgsin(θ)eθeθeθ = −mRθ 2 ererer +mRθ eθeθeθ

Now we can look at the scalar coefficients of the unit vectors. In the end, we wantto solve for Fn; we know everything in this equation except for θ and θ - and oneis just the integral of the other. Since there are fewer terms in the eθeθeθ direction, let’sstart there. We’ll solve for θ in terms of variable which we know.

mgsin(θ) = mRθ = mRdθ

dt= mR

dθ

dθ

dθ

dt= mRθ

dθ

dθ

gsin(θ) = Rθdθ

dθ∫gsin(θ)dθ =

∫Rθdθ

−gcos(θ) =12

Rθ 2 +C


θ 2 = −2gR

cos(θ)−C

where C is some constant of integration. Since we know that the initial linear veloc-ity of the marble is vo at θ = 0, the angular velocity at that point will be θo = vo/R.Substituting these values into the previous result and solving for C,

θ 2 = −2gR

cos(θ)−C

v2o

R2 = −2gR

cos(0)−C

C = − v2o

R2 − 2gR

Substituting this into our expression for θ 2,

θ 2 =−2gR

cos(θ)+v2

o

R2 +2gR

Now we can look at the scalar coefficients of ererer from our momentum equation:

Fn −mgcos(θ) = −mRθ 2

Fn −mgcos(θ) = −mR(−2g

Rcos(θ)+

v2o

R2 +2gR

)

At the moment when the marble leaves the surface, Fn = 0 so

0−mgcos(θ) = −mR(−2g

Rcos(θ)+

v2o

R2 +2gR

)gcos(θ) = −2gcos(θ)+

v2o

R+2g

cos(θ) =v2

o

3Rg+

23

So, the angle at which the normal force goes to zero and the marble leaves thesurface of the track is given by

θ = cos−1( v2o

3Rg+

23

)

4.6 Examples 259

Interestingly, the angle does not depend on the mass of the marble. Furthermore, ifthe initial velocity is very small,

θ = cos−1(23

)= 70.5◦

So, any object- regardless of shape or mass- would leave the surface of a frictionlesscylinder at an angle of 70.5◦ from vertical.

4.6.8 External forces and collisions

A ballistic projectile having mass ma and initial velocity v0 is fired horizontally intoa block of mass mb (initially at rest) and sticks to the block. The impact lasts for atime Δ t and then causes the block to slide some distance L along the ground afterthe impact, before coming to rest again. The coefficient of kinetic friction betweenthe block and the ground is μk, and the coefficient of static friction is negligiblysmall. Solve for (a) the velocity of the block at the instant after the impact, (b) themean internal force between the projectile and the block, and (c) the distance L.

Fig. 4.34 External forces and collisions

SOLUTION

To solve this problem, we must first determine the velocities using the principleof conservation of momentum; then, we can use the definition of impulse to solvefor the mean force of impact. Finally, we can draw a free-body diagram, determinethe acceleration and solve for the total displacement.

First, we consider the instant after impact; we assume that a time t0 has elapsedwhich is sufficiently large that t0 � Δ t but at the same time sufficiently small that


the displacement of the block is vanishingly small. Then, the velocity is zero andinstantaneously, the friction force FfFfFf = 0. Then, from the linear momentum equation,


0 = GsysGsysGsys

GsysGsysGsys = C0C0C0

where C0C0C0 is some constant which has both magnitude and direction. However, sincethe system is composed of a finite number of discrete masses,

GsysGsysGsys =GaGaGa +GbGbGb =C0C0C0

Let us define exexex as being positive to the right in the illustration given. Then, we canuse the initial condition (prior to impact) to determine the constant C0C0C0, as

(GaGaGa +GbGbGb

)1 =C0C0C0 =

(mava

)1 exexex +

(mbvb

)1 exexex

C0C0C0 =(mav0

)exexex +

(mb(0)

)C0C0C0 = mav0 exexex

where the subscript ‘1’ was used to indicate the initial condition. Substituting thisinto the previous result,

GsysGsysGsys = mava exexex +mbvb exexex = mav0 exexex

This equation must be satisfied at all times after the initial impact but before thebeginning of motion. Note that once the block begins to move, the friction forcewill act and the requirement that GsysGsysGsys = 0 will be violated. Since the projectile andthe block are stuck together after impact, then vavava = vbvbvb = v2 during this time. Then,

mava exexex +mbvb exexex = mav0 exexex

mav2 +mbv2 = mav0

v2 =ma

ma +mbv0

This velocity v2 is the velocity of the block the instant after impact, as required.Next, to determine the impact force, we look at the definition of impulse. Let usconsider the block; then, the impulse sustained by the block is given by

4.6 Examples 261

JaJaJa =∫ Δ t

0Fa(t)Fa(t)Fa(t)dt = FaFaFaΔ t

where FaFaFa is the force exerted upon the projectile by the block during the impact(which will be equal in magnitude but opposite in direction to the force exerted uponthe block by the projectile). However, this force is also related to the momentumthrough the definition,

FaFaFa =ddt

GaGaGa

Multiplying both sides by dt, integrating and substituting in the previous equationyields,

∫FaFaFa dt =

∫dGaGaGa

FaFaFaΔ t =(GaGaGa)

2 −(GaGaGa)

1

where the subscript ‘2’ here indicates the state after impact, and ‘1’ indicates thestate before impact. We can now solve for the average force FaFaFa.

FaFaFaΔ t =(GaGaGa)

2 −(GaGaGa)

1

FaFaFaΔ t =(mav2 exexex

)− (mav0 exexex)

FaFaFa =1

Δ tma(v2 − v0

)exexex

In order to express the result in terms of known quantities, the velocity v2 can besubstituted for the previous result,

FaFaFa =1

Δ tma

((ma

ma +mbv0

)− v0

)exexex

=1

Δ tmav0

(ma

ma +mb−1

)exexex

=1

Δ tmav0

(ma

ma +mb+

−ma −mb

ma +mb

)exexex

FaFaFa = − 1Δ t

v0

(mamb

ma +mb

)exexex


This result yields the force applied to the projectile by the block, and is in the −exexex

direction, as expected. Let us now determine the force imparted upon the block bythe projectile; as before,

∫FbFbFb dt =

∫dGbGbGb

FbFbFbΔ t =(GbGbGb)

2 −(GbGbGb)

1

FbFbFbΔ t =(mbv2 exexex

)− (mb(0))

FbFbFb =1

Δ tmbv2 exexex

Substituting for v2,

FbFbFb =1

Δ tmb

(ma

ma +mbv0

)exexex

FbFbFb =1

Δ t

(mamb

ma +mbv0

)exexex

FbFbFb = −FaFaFa

as expected.Once the block/projectile system begins to move, kinetic friction will begin to

slow the block. The block will decelerate until it comes to rest. A free-body diagramof the forces acting on the block once it begins to move is included as Figure 4.35.

Fig. 4.35 Free body diagram of the block/projectile system

Summing the forces acting on the block yields,

4.6 Examples 263

∑FextFextFext = GsysGsysGsys = (ma +mb)x exexex

−Ff exexex +Fn eyeyey −Fg eyeyey = (ma +mb)x exexex

−μkFn exexex +Fn eyeyey − (ma +mb)g eyeyey = (ma +mb)x exexex

Recalling that Ff = μkFn for dry friction. Equating the scalar coefficients of eyeyey,

Fn − (ma +mb)g = 0

Fn = (ma +mb)g

Next, we examine the scalar coefficients of exexex,

−Ff = (ma +mb)x

−μkFn = (ma +mb)x

Substituting in the previous result for Fn,

−μk(ma +mb)g = (ma +mb)x

x = −μkg

Since we are only interested in displacements and not in the time-of-flight, we canuse the relation,

x exexex =dxdx

dxdt

exexex = xdxdx

exexex∫x dx =

∫x dx∫

x dx =12

x2 +C1

where C1 is a constant of integration. Substituting our result for x and evaluating theintegral,

∫−μkg dx =

12

x2 +C1


−μkgx =12

x2 +C1

where the second constant of integration was just implicitly absorbed into C1. Tosolve for C1, we can use our ‘initial’ condition at the instant after the impact; theblock had not yet moved so x = 0, and we have already solved for the velocity atthis instant, so x = v2. Then,

−μkg(0) =12

v22 +C1

C1 = −12

v22

Substituting this back into the velocity equation,

−μkgx =12

x2 − 12

v22

When the block comes to rest, x = 0 and x = L; at that moment, then,

−μkgL =12(0)2 − 1

2v2

2

μkgL =12

v22

L =1

2μkgv2

2

Substituting the previous expression for v2,

L =1

2μkg

(mav0

ma +mb

)2

This is the maximum distance traveled by the block/projectile system after impact,as required.

4.6.9 Open systems: the falling chain

An ideal, uniform string of length L and mass m is draped over the edge of a surfacesuch that it is just about to begin slipping off the surface, as illustrated. The string isthen given an initial, infinitesimal nudge and it begins slipping off the surface. Thecoefficient of static friction and the coefficient of kinetic friction between the string

4.6 Examples 265

and the surface are both μ . Find the velocity of the string at the instant that theentire length has passed off of the surface.

Fig. 4.36 Open systems 1

SOLUTION

To begin, let us define the distance x as the length of string hanging over the edgeof the surface. Then, the length of string resting on the surface is L− x. Next, wewill define our system as containing only the string hanging over the edge. Sincethe mass of the string hanging over the edge will change, the contents of the systemwill change and the system is therefore open. Drawing a free-body diagram of oursystem,

∑FextFextFext = Fg1Fg1Fg1 +FTFTFT

∑FextFextFext = −m1g eyeyey +FT eyeyey

∑Fext = −m1g+FT

However, since the string is uniform and has constant density, we may define thelinear density of the string λ as

λ =mL


Then, the mass of any length x of string will be λx. If the string is falling over theedge of the surface with a velocity v1, the mass entering our system is also enteringwith a velocity vb = v1. Then, from the law of conservation of momentum for opensystems,

Fext = mv1 +mv1 − mvb

Fext = mv1 +m1v1 − mv1

Fext = m1v1

Substituting now the forces from the free-body diagram,

−m1g+FT = m1v1

FT = m1g+m1v1

FT = λxg+λxv1

Then, we recognize that the quantity x (which is always positive) is related to thequantity v1 (which is always negative in our coordinate system), as

v1 =−x

and so,

FT = λxg−λxx

This has given us an expression for the tension in the string just at the corner of thesurface. Let us now redefine our open system as consisting of the horizontal part ofthe string only, and let us say that the horizontal part of the string has a velocity v2.From the free-body diagram,

∑FextFextFext = Fg2Fg2Fg2 +FnFnFn +FTFTFT +FfFfFf

∑FextFextFext = −m2g eyeyey +Fn eyeyey +FT exexex −Ff exexex

Since the string is not moving in the vertical direction at all, there can be no netforce along eyeyey. Equating the scalar coefficients of eyeyey,

0 = −m2g eyeyey +Fn eyeyey

0 = −λ (L− x)g+Fn

4.6 Examples 267

Fn = λ (L− x)g

Now, eliminating the eyeyey components and examining the coefficients of exexex,

∑Fext exexex = 0+FT exexex −Ff exexex

∑Fext = FT −μFn

∑Fext =(λxg−λxx

)−μ(λ (L− x)g

)

We may now again apply the law of conservation of linear momentum for opensystems. However, it is important to notice that, as the string slides over the edge,the horizontal part of the string will be losing mass at a rate equal to that at whichthe vertical part is gaining mass. Therefore, m2 =−m1 =−m and

Fext = m2v2 +m2v2 − m2vb

= −mv2 +m2v2 + m2v2

= m2v2

Fext = λ (L− x)v2

where once again we have noted that the mass is being removed from the systemwith the same velocity of the system itself, so v2 = vb. Substituting our earlier resultfor Fext , and recognizing here that v2 = x,

(λxg−λxx

)−μ(λ (L− x)g

)= λ (L− x)v2

xg− xx−μ(L− x)g = (L− x)x

Significantly, the mass of the string has just been eliminated from the expressionand will therefore have no effect whatsoever on the motion of the string. We maynow arrange this expression in a form which may be integrated, as

xg− xx−μLg+μxg = Lx− xx

(1+μ)xg−μLg = Lx

(1+μ)xg−μLg = Ldxdx

dxdt

(1+μ)xg−μLg = Lxdxdx


We may now multiply both sides by dx and integrate.

((1+μ)xg−μLg

)dx = Lx dx∫ (

(1+μ)xg−μLg)

dx =∫

Lx dx

12(1+μ)g x2 −μgL x =

12

L x2 +C

Now, we need to evaluate the constant of integration C. To solve for this constant,we can use the initial condition, when the string was just about to slip. Let us saythat, at this instant, x= xi. Since the string is at rest at this instant, from the free-bodydiagram of the vertical part of the string,

∑FextFextFext = 0 = −m1g eyeyey +FT eyeyey

FT = m1g = (λxi)g

Also, from the horizontal part of the string, at the instant it is about to slip, Ff = μFn.Therefore,

∑FextFextFext = 0 = 0+FT exexex −Ff exexex

FT = Ff = μFn

(λxi)g = μ(λ (L− xi)g

)xi(1+μ) = μL

xi =μ

1+μL

So, substituting this back to solve for C (and recognizing that since the string startedfrom rest, x = 0 at this instant),

12(1+μ)g x2

i −μgL xi =12

L x2i +C

12

g(1+μ)

(μ

1+μL

)2

−μgL

(μ

1+μL

)=

12

L(0)2 +C

12

gL2 μ2

1+μ−gL2 μ2

1+μ= C

4.6 Examples 269

−12

gL2 μ2

1+μ= C

The equation of motion for the string is therefore,

12

g(1+μ) x2 −μgL x =12

L x2 −(

12

gL2 μ2

1+μ

)

This expression will be true of the motion of the string at all times. Then, at theinstant that the string has passed completely over the edge of the surface, x = L and

12

g(1+μ) x2 −μgL x =12

L x2 −(

12

gL2 μ2

1+μ

)12

g(1+μ)(L)2 −μgL(L) =12

L x2 −(

12

gL2 μ2

1+μ

)

x2 = g(1+μ)L−2μgL+gLμ2

1+μ

x2 = gL

((1+μ)−2μ +

μ2

1+μ

)

x2 = gL

((1+μ)(1+μ)

(1+μ)− 2μ(1+μ)

(1+μ)+

μ2

1+μ

)

x2 = gL

(1+2μ +μ2 −2μ −2μ2 +μ2

1+μ

)

x2 = gL

(1

1+μ

)

x =

√gL

(1

1+μ

)

So the velocity with which the string leaves the surface is dependent only upon thegravity, the length of the string and the coefficient of friction. Notice that, had thesurface been frictionless and μ = 0,

x =√

gL

On the other hand, if the string had been freely falling the same distance,

x =√

2gL


4.6.10 Open systems: a jet engine

The Rolls-Royce Trent 1000 aero engine is a high-bypass turbofan with a flow rateof approximately 1300 kg/s. Compute the speed which two of these engines canimpart upon a 200,000 kg aircraft after two minutes, assuming a constant exhaustvelocity of 380 m/s and neglecting drag.

Fig. 4.37 Open systems 2

SOLUTION

Let us define our open system as the aircraft and engines. Then, from the law ofconservation of linear momentum for open systems,

Fext = mv+mv− mvb

For convenience, let us express vb relative to v as

vb = v+ vb/sys

Then, if we define the positive direction as the direction in which the jet exhaust isblowing,

Fext = mv+mv− mvb

= mv+mv− m(v+ vb/sys)

Fext = mv− mvb/sys

If we neglect aerodynamic drag, there are no external forces acting upon thesystem and Fext = 0. Also, we assume that the air enters the engine with negligiblevelocity. Then,


0 = mv− mvb/sys

v =1m

mvb/sys

Substituting in values, and recalling that there are two engines on the aircraft,

v =−(2×1300 kg/s)(380 m/s)

(2×105 kg)

v = −4.94 m/s2

The mass flow rate m is negative because mass is squirting out of the system ratherthan into the system. All of the numerical values substituted into the above expres-sion are constant throughout the flight, so the result is constant as well. Therefore,

v =

∫v dt

v = vt +C

where C is a constant of integration. Since the aircraft is assumed to start from rest,v = 0 when t = 0, and therefore C = 0. After the two minute flight, then,

v = vt

= (−4.94 m/s2)(120 s)

v = −593 m/s

The result is negative because the aircraft is traveling in the opposite direction asthe engine exhaust. This is the same result as we obtained using the method ofcollisions, in Section (3.4.3).


In solving problems through the application of the law of conservation of linearmomentum, the following approach should be used:


1: Identify the system. It is important to be clear about how, exactly, your system isdefined. First, the definition of the system will determine which forces are external;forces acting between objects which are all part of the system may be entirely ig-nored in an application of the law of conservation of linear momentum. Also, thelaw of conservation of linear momentum relates the external forces to the motion ofthe system: if you have two blocks and define your system as the two blocks, thenapplying the law of conservation of linear momentum will yield the acceleration ofthe centre of mass of the two-block system- not the accelerations of the individualblocks. Consequently, when dealing with problems involving a series of intercon-nected objects, it is usually easiest to define the system as consisting of a singleblock, and repeat the analysis for each block (re-defining the system each time).

2: Define a coordinate system and axes. Next, you need to define a fixed coordinatesystem, as well as a system of unit vectors to define your axes. Your coordinatesystem may be located anywhere and oriented in any way- but these coordinatesmust be consistently used throughout.

3: Draw a free-body diagram of the system and compute the vector sum of the exter-nal forces acting on the system. Drawing a free-body diagram will always make iteasier to identify the forces. The equations can only be solved later on if the forcesare all resolved into components along the unit vectors selected.

4: Write down the governing equation. The linear momentum equation is a funda-mental first-principle, and is therefore a good equation to remember. Remember,this is a vector expression.

5: Identify each of the terms in the governing equation. Write down all the informa-tion available about each of the terms in the linear momentum equation. This mayinclude magnitudes and/or directions; it may help to resolve into components alongthe unit vectors defined.

5.1: Forces are known. If both the magnitudes and directions of all of the externalforces acting on the system are known, these may be substituted into the linearmomentum equation directly.

5.2: Acceleration is known. Similarly, if the magnitude and direction of the ac-celeration of the centre of mass of the system is known, these can be substitutedin as well.

5.3: Path is specified. If the path described by the motion of the centre of massof the system is defined, then some information about the acceleration (eithermagnitude or direction) is necessarily provided: you need to apply the relativeacceleration equation in order to most easily determine the known and unknowncomponents of the magnitude and direction of the acceleration vector.


6: Resolve all vectors into the same coordinate system. The vectors on both sides ofthe linear momentum equation must be expressed in terms of the same unit vectors,usually exexex, eyeyey and ezezez.

7: Solve. For the case of a single system, you should be able to solve for any un-knowns.

7.1: Single system. For cases of single systems acted upon by independent, exter-nal forces, the linear momentum vector equation will yield three scalar equations(one per direction along exexex, eyeyey and ezezez). Solve these simultaneously to eliminateyour unknowns.

7.2: Multiple linked systems. For cases of multiple linked systems (such asmasses attached with strings), you will have one vector momentum equation foreach mass. The equations will share common unknowns, and those unknownswill usually be associated with whatever is connecting the masses (such as thetension in the string). Since each system may have a different acceleration (as isthe case when masses are linked through pulleys), you will also need to equatetogether the accelerations of the various systems in order to solve for the un-knowns.


4.8 Problem set 4: External forces

Question 1


The blocks A and B, having masses ma = 10 kg and mb = 15 kg, respectively, aresuspended from a pulley with an ideal, massless string, as illustrated. The blocks arereleased from rest, and are acted upon by gravity. Determine (a) the acceleration ofthe blocks, and (b) the tension in the string.

Answer:

(a) a =mb −ma

mb +mag = 1.96 m/s2

(b) FT = mag

(1− ma −mb

ma +mb

)= 117.7 N

Question 2

The three blocks A, B and C (having masses ma, mb and mc, respectively) are con-nected by means of a pulley arrangement as illustrated. The block A is suspendedfreely, the block B slides on a frictional surface with coefficient of kinetic frictionμk and the block C rests on a frictionless inclined plane with inclination angle θ .Find (a) the acceleration of block A if it is moving upwards, and (b) the accelerationof block A if it is moving downwards.

4.8 Problem set 4: External forces 275

ma

mb

μ = μk

μ = 0

θ

mc


Answer:

(a) aa = g

(12

mcma

sin(θ)−μkmbma

−1

1+ mbma

+ 14

mcma

)

(b) aa = g

(12

mcma

sin(θ)+μkmbma

−1

1+ mbma

+ 14

mcma

)

Question 3

The two blocks A and B (having masses ma and mb, respectively) are connected bymeans of a pulley arrangement as illustrated. The block B is suspended freely whilethe block A slides on a frictional incline with inclination angle θ and coefficient ofkinetic friction μk. Find (a) the acceleration of block B if it is moving downwards,and (b) the acceleration of block B if it is moving upwards. Assume all pulleys aremassless and frictionless.

Answer:

(a) ab = g

(14

(sin(θ)−μk cos(θ)

)− mbma

116 +

mbma

)

(b) ab = g

(14

(sin(θ)+μk cos(θ)

)− mbma

116 +

mbma

)



Question 4


Block B (of mass mb) rests upon a horizontal surface, and block A (of massma) rests upon block B. The coefficient of kinetic friction between block B and theground is μkb, and the coefficient of kinetic friction between block A and block B isμka. A constant horizontal force of magnitude F is applied to block A. Both blocksbegin from rest. Find the minimum possible acceleration of block A required if blockA is to slip on the surface of block B.

Answer:

xa ≥ g

(μka

ma

mb−μkb

(ma

mb+1

))

4.8 Problem set 4: External forces 277

Question 5


A coyote (who, evidently, has never studied dynamics), in his ongoing effortsto capture a high-velocity desert bird, intends to ride a rocket-sled down a circulartrack of radius R to catch the bird. The rocket sled will provide a constant thrust T ,while the combined mass of the sled and the coyote is m. Help the luckless coyotewith his evil plan by determining (a) the velocity of the rocket sled as a function ofits angular position θ (as illustrated), and (b) the normal force exerted by the trackupon the sled. Assume the track is frictionless, and that the mass of fuel burned bythe rocket during the descent is negligible.

Answer:

(a) vvv =

√2R(T

mθ +gsin(θ)

)eθeθeθ

(b) FFF = −(2T θ +mgsin(θ))

ererer

Question 6

Our long-suffering bug (of mass m) now somehow has found himself clinging ontoan aero-engine rotor assembly while the system is undergoing its balancing tests.The engine component has a conical surface with a sectional angle of θ and isspinning with angular velocity ω . All is not lost for our poor bug, though: natureprovided him with sticky feet! The coefficient of static friction is μs. Solve for the



maximum speed of rotation at which our friend the bug will be able to keep his gripon the surface.

Answer:

ω =

√gR

(μs − tan(θ)

1+μs tan(θ)

)

Question 7

The coefficient of static friction between chalk and slate is much higher than thecoefficient of kinetic friction. Based on this observation, propose an explanation forwhy chalk often squeaks on a chalkboard.

4.9 Problem set 5: Position-dependent forces 279

4.9 Problem set 5: Position-dependent forces

Question 1


A small marble with mass m sits in the bottom of a hemispherical bowl of radiusR. After being given some small deflection, the marble begins to oscillate back andforth at the bottom of the bowl. Find the equation of motion for the marble anddetermine the frequency of oscillation, as well as the normal force between themarble and the bowl (as a function of velocity). What radius of bowl would berequired for a frequency of 1 per second (1 Hz)? Hint: for small angles, sin(θ)≈ θ .

Answer:

FT = m(g+Rθ 2)

θ +gR

θ = 0

f =1

2π

√gR

f = 1 → R = 0.25 m

Question 2

A small object of mass m and negligible volume is suspended from a cylindricalbuoy with radius R and negligible mass, and is immersed in a fluid with a density



of ρ . After some small initial tug on the mass, the buoy is observed to bob with aconstant frequency. Determine (a) the length of buoy which would be submerged ifthe acceleration of the buoy is zero (the buoy is in static equilibrium), and (b) thebobbing frequency.

Answer:

(a) y =m

ρπR2

(b) f =R2

√ρgπm

Question 3

A bowling ball with mass m = 6 kg and radius R = 10 cm is dropped from a greatheight. If the drag force on the bowling ball is given by the expression

FD =12

ρAxsCDx2

where ρ is the density of air (1.17 kg/m3), Axs is the largest cross-sectional area ofthe ball, CD is a constant (equal to 0.5) and x is the velocity. Determine the terminalvelocity of the ball.

4.9 Problem set 5: Position-dependent forces 281

Answer:

x = 80.1 m/s

Question 4

The bowling ball then lands in cold treacle and slowly sinks. The drag on the ball isgiven by the expression

FD = 3πμDx

where μ is the viscosity of treacle (≈ 10 kg/m·s), D is the diameter of the ball and xis the velocity. Determine the terminal velocity of the ball.

Answer:

x = 3.12 m/s

Question 5

A mass resting on a horizontal, frictionless surface is attached to a horizontal, idealspring. After an initial deflection, the mass is released and allowed to oscillate. Ex-plain how momentum is conserved through this process.

Chapter 5Angular momentum

Summary Derivation of the law of conservation of angular momentum as a specialcase of the law of conservation of linear momentum, and conservation of angularmomentum for systems of point masses; distributed masses and determining themoments and products of inertia; planar rotation, pure rolling motion, and combinedrolling and slipping; three-dimensional rotation and gyroscopic motion.

5.1 Definition of angular momentum

Angular momentum is a quantity derived from linear momentum, and is used tofacilitate the analysis of rotating systems. From the law of conservation of linearmomentum, we know that if the vector sum of the forces acting on a system is zero,then the centre of mass of the system remains stationary. However, consider the caseof an object subject to two equal but opposite forces applied at different positionson the object, as illustrated in Figure 5.1; in this case, the centre of mass will remainstationary but the object will begin to spin. If we wanted to determine the angularacceleration of the system as a result of the applied forces, the linear momentumequation,


would not be of very much help, yielding the trivial result 0 = 0. We need a way toapply this equation in such a way that it gives a useful (and nontrivial) representationof the motion of the system.

283

284 5 Angular momentum

Fig. 5.1 A solid body subjected to forces with a vector sum of zero.

5.2 Angular momentum of point masses

First of all, let us consider a system of objects of mass mi floating in space. If wedefine a fixed coordinate system o, we can describe the location of each object withthe position vector ririri. The law of conservation of linear momentum then requiresthat

∑FiFiFi = GsysGsysGsys =ddt

(∑(mivivivi)

)where vivivi is the velocity of the ith object, and FiFiFi is the external force acting on theith object. Not all objects need necessarily have forces acting upon them; in the casewhere there are no forces acting on a particular object, in that case, |FiFiFi| = 0. Now,one way to get a nontrivial result from the momentum equation when two equal butopposite forces are applied to the system is by taking the cross-product with rrr ofboth sides of the equation, so that

ririri ×∑FiFiFi = ririri × ddt

(∑(mivvvi)

)(5.1)

However, it can be seen that

ddt(rrr×mvvv) = rrr× d

dt(mvvv)+

drrrdt

× (mvvv)

= rrr× ddt(mvvv)+vvv× (mvvv)

= rrr× ddt(mvvv)+m(vvv×vvv)

ddt(rrr×mvvv) = rrr× d

dt(mvvv)+0

Recall that the cross-product of any vector with itself is zero, and vvv = rrr. Therefore,

5.2 Angular momentum of point masses 285

rrr× ddt(mvvv) =

ddt(rrr×mvvv)

Then, we can substitute this result into Equation (5.1). Note that, because of thedistributive property of the cross-product, it can be moved inside the summation.

∑(ririri ×FiFiFi

)=

ddt

(∑(ririri ×mivivivi)

)(5.2)

However, we recognize the quantity rrr×FFF as a moment (sometimes called a torque),which has units of force × length. A moment is represented by the vector MMM, andwill point in a direction perpendicular to both rrr and FFF, so that

MoMoMo = rf/orf/orf/o ×FFF

where rf/orf/orf/o is the position vector of the point of application of the force FFF with respectto some origin o, and MoMoMo is the moment about o produced by that force. The senseof the moment vector is determined by the right-hand rule. It is important to recallthat FFF is an external force, so only external forces may contribute to the moment MoMoMo.Consequently, we may refer to MoMoMo as an external moment.

Because of its importance, the vector quantity rrr×mvvv is defined as the angularmomentum of the system, and is indicated by the symbol HHH so that

HoHoHo = rp/orp/orp/o ×mvvv

where rp/orp/orp/o is the position vector of a mass with respect to some origin o, mvvv isthe linear momentum vector of that mass, and HoHoHo is the angular momentum aboutthe point o. Specifying the origin here is critical, since the angular momentum willbe different if it is computed about a different origin. The angular momentum ofa system may be thought of as a rotational equivalent of the linear momentum, inthe same way as a moment may be thought of as a rotational equivalent of a force.Substituting these two new definitions into Equation (5.2),

∑Mi/oMi/oMi/o =ddt

(∑Hi/oHi/oHi/o

)where Mi/oMi/oMi/o is the moment acting on the ith object about the origin o, and Hi/oHi/oHi/o isthe angular momentum of the ith object about the same origin o. The sum of theangular momentums of the individual objects in the system is just the total angularmomentum of the system, so this can be re-expressed more simply as

∑Mext/oMext/oMext/o = Hsys/oHsys/oHsys/o

where the subscript o is preserved, in order to indicate that both moments and angu-lar momentums must be evaluated about the same origin. This ‘alternative’ expres-


sion of the momentum equation is generally referred to as the law of conservationof angular momentum, and states:

The sum of the external moments about some point acting upon a closedsystem is equal to the time rate of change of the angular momentum of thesystem about that point.

It should be noted, however, that referring to the law of conservation of angularmomentum as a ‘law’ is something of a misnomer; to be pedantic, since the lawof conservation of angular momentum was derived from the law of conservation oflinear momentum, it does not technically qualify as a ‘first principle’. However, inpractical engineering analysis, the distinction is rarely (if ever) made, and the use ofconservation of angular momentum as a first principle is generally accepted.

5.2.1 Demonstration of the law of conservation of angularmomentum

Let us illustrate the law of conservation of angular momentum with a simple exam-ple. Consider a single small object with mass m attached to a string and swingingaround in a circular path of radius R1 at constant angular velocity ω1, as shown inFigure 5.2. The string is then shortened until the body is swinging around in a newcircular path of radius R2 with a new angular velocity ω2.

Fig. 5.2 A small mass swinging around on a string.

We define our system as consisting only of the object. Then, we define a cylin-drical coordinate system o with the origin fixed on the centre of rotation such thatthe unit vector ererer always points from the hub toward the mass and the unit vector eθeθeθ

5.2 Angular momentum of point masses 287

points in the direction of motion. Drawing a free-body diagram of the mass, we seethat the only force acting upon the mass is the tension in the string. The sum of themoments acting on the mass can then be computed, as

∑Mext/oMext/oMext/o = ∑(ririri ×FiFiFi

)= rererer ×−FTererer

= −rFT (ererer ×ererer)

= −rFT (0)

∑Mext/oMext/oMext/o = 0

Next, we can compute the angular momentum of the system about the same origin.

Hsys/oHsys/oHsys/o =ddt(rrr×mvvv)

=ddt(r ererer ×mv eθeθeθ )

=ddt

(mvr(ererer ×eθeθeθ )

)Hsys/oHsys/oHsys/o =

ddt(mvr ezezez)

where v is the scalar speed of the mass. We can now substitute these results into theangular momentum equation,

∑Mext/oMext/oMext/o = Hsys/oHsys/oHsys/o

0 =ddt(mvr ezezez)

C1C1C1 = mvr ezezez

where we have integrated both sides of this equation with respect to time, and C1C1C1 issome constant vector quantity. Since the mass m is also constant, it can be combinedwith C1C1C1 to form some new constant C2C2C2. Also, since we are interested only in themagnitude, we can take magnitudes of both sides of the equation to eliminate theunit vector ezezez.

C1C1C1

m=C2C2C2 = vr ezezez

|C2C2C2| = |vr ezezez|


vr = C2

This result indicates that the product of the linear velocity and the radius, vr, isalways constant. Consequently,

v1R1 = v2R2

We also know the linear speed v of a body translating in a circle,

vp/ovp/ovp/o = ωωω ×rp/orp/orp/o

= ωr (ezezez ×ererer)

= ωr eθeθeθ

|vp/ovp/ovp/o| = v = ωr

Substituting this result into the earlier result,

v1R1 = v2R2

(ω1R1)R1 = (ω2R2)R2

ω1R21 = ω2R2

2

So, the final angular velocity of the mass is

ω2 = ω1

(R1

R2

)2

This result will hold for any object of constant mass undergoing rotation about afixed point in the absence of external applied moments. Because the angular momen-tum must remain constant, this case may be considered as the rotational equivalentof a collision.

5.3 Angular momentum of solid bodies

A solid body may always be considered as a series of infinitesimally small ‘particles’or differential mass elements each having a tiny mass dm and position vector rp/orp/orp/orelative to a fixed coordinate system o. The tiny angular momentum dHodHodHo of each‘particle’ about the origin o is then,

5.3 Angular momentum of solid bodies 289

dHodHodHo = rp/orp/orp/o ×dmvp/ovp/ovp/o

The total angular momentum of the system about o is then

Hsys/oHsys/oHsys/o =∫

dHodHodHo =∫(rp/orp/orp/o ×vp/ovp/ovp/o)dm (5.3)

where the scalar mass dm was pulled out of the cross-product. This expression maybe evaluated fairly easily, providing that an origin o can be defined such that (a) thepoint o is fixed in space and does not move, (b) the object rotates about o, and (c)the geometry of the object can be conveniently integrated about o. Regrettably, thesethree conditions cannot always be met. Instead, let us consider an object undergoingsome arbitrary rotation in space. The position vector rp/orp/orp/o of any point on the objectmay then be expressed as the vector sum

rp/orp/orp/o = rc/orc/orc/o +rp/crp/crp/c

where rc/orc/orc/o is the position vector of the centre of mass of the body with respect to theorigin o (which is no longer necessarily a centre of rotation), and rp/crp/crp/c is the positionvector of the differential mass element dm with respect to the centre of mass of thebody, as illustrated in Figure 5.3. Then, the angular momentum of the system abouto may then be expressed as

Fig. 5.3 A solid mass subjected to a moment.

Hsys/oHsys/oHsys/o =∫ (

(rc/orc/orc/o +rp/crp/crp/c)×vp/ovp/ovp/o)

dm

=∫ (

(rc/orc/orc/o ×vp/ovp/ovp/o)+(rp/crp/crp/c ×vp/ovp/ovp/o))

dm

Hsys/oHsys/oHsys/o =∫(rc/orc/orc/o ×vp/ovp/ovp/o)dm+

∫(rp/crp/crp/c ×vp/ovp/ovp/o)dm


The position vector of the centre of mass of the system rc/orc/orc/o can be pulled out of thefirst integral, since it will be the same for every differential mass element.

Hsys/oHsys/oHsys/o =(rc/orc/orc/o ×

∫(vp/ovp/ovp/o)dm

)+∫(rp/crp/crp/c ×vp/ovp/ovp/o)dm

However, we already know that the linear momentum GsysGsysGsys of a system of objects isgiven by the expression,

GsysGsysGsys = ∑mivivivi

For a system of infinitesimal objects dm, this can be expressed in the form of anintegral, as

GsysGsysGsys =∫

vp/ovp/ovp/o dm


Hsys/oHsys/oHsys/o =(rc/orc/orc/o ×GsysGsysGsys

)+∫(rp/crp/crp/c ×vp/ovp/ovp/o)dm

The first term is referred to as the orbital angular momentum of the object, and isindependent of the shape of the object. We shall return to this term later. For now,we recognize that this expression will be much simplified if we evaluate our angularmomentum about the centre of mass of the object; then, our origin is placed on topof the centre of mass, rc/orc/orc/o = 0, rp/orp/orp/o = rp/crp/crp/c and vp/ovp/ovp/o = vp/cvp/cvp/c. Substituting,

Hsys/cHsys/cHsys/c =(rc/orc/orc/o ×GsysGsysGsys

)+

∫(rp/crp/crp/c ×vp/ovp/ovp/o)dm

=(0×GsysGsysGsys

)+∫(rp/crp/crp/c ×vp/cvp/cvp/c)dm

Hsys/cHsys/cHsys/c =∫(rp/crp/crp/c ×vp/cvp/cvp/c)dm

Interestingly, this is identical to the result obtained for an object rotating about afixed origin o (instead of the centre of mass c), as shown in Equation (5.3). This ex-pression will hold true even if the centre of mass is moving, since we never requiredit to be fixed in space. For simplicity, then, we may observe that

Hsys/oHsys/oHsys/o =∫(rp/orp/orp/o ×vp/ovp/ovp/o)dm

providing that the point o is either a stationary point or a centre of mass.

Let us now say that the object is rotating about the point o with some angularvelocity ωωω . Then,


vp/ovp/ovp/o =ωωω ×rp/orp/orp/o

Substituting this into the angular momentum equation yields,

Hsys/oHsys/oHsys/o =∫(rp/orp/orp/o ×vp/ovp/ovp/o)dm

=∫(rp/orp/orp/o ×ωωω ×rp/orp/orp/o)dm

where we have assumed that our centre of rotation o is either a stationary point ora centre of mass. However, one of the vector trigonometric identities of the crossproduct is that, for three vectors aaa, bbb and ccc,


So the angular momentum can be expanded to


(rp/orp/orp/o ·rp/orp/orp/o)ωωω − (rp/orp/orp/o ·ωωω)rp/orp/orp/o)

dm

Now, if we say that rp/orp/orp/o = (xexexex + yeyeyey + zezezez) and ωωω = (ωxexexex +ωyeyeyey +ωzezezez), thisprevious result can be written as,


(x2 + y2 + z2)(ωxexexex +ωyeyeyey +ωzezezez)

− (xωx + yωy + zωz)(xexexex + yeyeyey + zezezez))

dm

This expression can now be expanded further to yield an 18-term expression,


x2ωxexexex + x2ωyeyeyey + x2ωzezezez + y2ωxexexex + y2ωyeyeyey + y2ωzezezez

+ z2ωxexexex + z2ωyeyeyey + z2ωzezezez − x2ωxexexex − xyωxeyeyey − xzωxezezez

− yxωyexexex − y2ωyeyeyey − yzωyezezez − zxωzexexex − zyωzeyeyey − z2ωzezezez)

dm

Gathering together the scalar coefficients of the unit vectors and simplifying,


(y2 + z2)ωx − yxωy − zxωz)

exexex dm

+∫ (

(x2 + z2)ωy − xyωx − zyωz)

eyeyey dm


+∫ (

(x2 + y2)ωz − xzωx − yzωy)

ezezez dm

If the body is assumed to be rigid, then all of the differential mass elements dm willbe stuck together. Consequently, they must all have the same angular velocity vectorωωω = (ωxexexex +ωyeyeyey +ωzezezez). Since ω is constant for all of the infinitesimal massesdm, it may be pulled out of the integrals.

Hsys/oHsys/oHsys/o =(ωx

∫(y2 + z2)dm+ωy

∫(−yx)dm+ωz

∫(−zx)dm

)exexex

+(ωy

∫(x2 + z2)dm+ωx

∫(−xy)dm+ωz

∫(−zy)dm

)eyeyey

+(ωz

∫(x2 + y2)dm+ωx

∫(−xz)dm+ωy

∫(−yz)dm

)ezezez

These resulting integrals are of significant practical importance. The first integralin each coefficient is called the moment of inertia, and is indicated by the symbol Iwith a repeated subscript.

Ixx =∫(y2 + z2)dm

Iyy =∫(x2 + z2)dm

Izz =∫(x2 + y2)dm

The other integrals are each referred to as a product of inertia, and are indicated bythe same symbol I but with two subscripts,

Ixy = Iyx =∫(−xy)dm

Iyz = Izy =∫(−yz)dm

Ixz = Izx =∫(−xz)dm

It is also important to note that the moments and products of inertia are functions ofthe geometry alone, and are independent of the rate of rotation. Then, given thesedefinitions, the angular momentum may be re-expressed as

Hsys/oHsys/oHsys/o = (Ixx/oωx + Ixy/oωy + Ixz/oωz)exexex


+ (Ixy/oωx + Iyy/oωy + Izy/oωz)eyeyey

+ (Ixz/oωx + Iyz/oωy + Izz/oωz)ezezez

Since the moments and products of inertia will also depend on the location of theorigin, an additional subscript ‘o’ was added to indicate the origin used.

5.3.1 The inertia tensor

Together, the three moments of inertia and three products of inertia can be arrangedin the form of a matrix. A matrix is simply an arrangement of numbers or variables,much like a vector, except that it can have many rows and columns (while vectorsin three-dimensional space have three rows and only one column). A matrix is in-dicated by a bold character, just like a vector. Then, we may define the matrix IIIas,

III =

⎡⎣ Ixx Ixy Ixz

Iyx Iyy Iyz

Izx Izy Izz

⎤⎦

There is a special class of matrix, called a tensor, which has some very specificmathematical properties. The matrix III is one of these; consequently, III is often re-ferred to as the inertia tensor. Because of the way in which matrices are multipliedtogether, there is an easier way for us to express the earlier result for Hsys/oHsys/oHsys/o. Con-sider a 3×3 matrix AAA and a vector BBB = (Bxexexex +Byeyeyey +Bzezezez). The multiplication isdefined as follows:

ABABAB =

⎡⎣A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤⎦⎛⎝Bx

By

Bz

⎞⎠=

⎛⎝A11Bx +A12By +A13Bz

A21Bx +A22By +A23Bz

A31Bx +A32By +A33Bz

⎞⎠

Note that, though the vector BBB and the resulting vector ABABAB are written vertically,each element is still assumed to be multiplied by an appropriate unit vector, so that

BBB =

⎛⎝Bx

By

Bz

⎞⎠= Bx exexex +By eyeyey +Bz ezezez

Also,

ABABAB =

⎛⎝A11Bx +A12By +A13Bz

A21Bx +A22By +A23Bz

A31Bx +A32By +A33Bz

⎞⎠


= (A11Bx +A12By +A13Bz)exexex

+ (A21Bx +A22By +A23Bz)eyeyey

+ (A31Bx +A32By +A33Bz)ezezez

Furthermore, in standard notation, matrices are indicated using square brackets [ ],while vectors are indicated using round brackets ( ) or brace brackets { }. From theabove, it is clear that the angular momentum may be expressed as the product of theinertia tensor and the angular velocity vector, or

Hsys/oHsys/oHsys/o = (Ixx/oωx + Ixy/oωy + Ixz/oωz)exexex

+ (Ixy/oωx + Iyy/oωy + Izy/oωz)eyeyey

+ (Ixz/oωx + Iyz/oωy + Izz/oωz)ezezez

Hsys/oHsys/oHsys/o =

⎡⎣ Ixx/o Ixy/o Ixz/o

Iyx/o Iyy/o Iyz/o

Izx/o Izy/o Izz/o

⎤⎦⎛⎝ωx

ωy

ωz

⎞⎠= IoωIoωIoω

where the point o is either a centre of mass or a stationary point. So, by expressingthe moments of inertia and products of inertia as a matrix, using the rules of ma-trix multiplication, we are provided with a convenient short-hand way of writing atedious, nine-term equation using only two symbols:

Hsys/oHsys/oHsys/o = IoωIoωIoω

This expression may be even further simplified if the axes are selected with care. Ifwe chose to work in a moving coordinate system o′ such that the unit vectors exexex

′, eyeyey′

and ezezez′ are stuck to the object, the terms in the inertia tensor will not change as the

object moves. Then, if we’re lucky enough to have an object with Ixz = Iyz = Ixy = 0,the angular momentum may be further simplified to

Hsys/oHsys/oHsys/o = Ixx/oωx exexex′+ Iyy/oωy eyeyey

′+ Izz/oωz ezezez′

In fact, it will be shown that the products of inertia of many common shapes areindeed zero if the origin o is stuck to the centre of mass and the axes are selectedappropriately. Since all of the moments of inertia are constant in time (because theaxes are stuck to the body, the orientation of the object within the coordinate systemdoes not change),

Hsys/oHsys/oHsys/o =ddt

(Ixx/oωx exexex

′+ Iyy/oωy eyeyey′+ Izz/oωz ezezez

′)


Hsys/oHsys/oHsys/o = Ixx/oddt(ωx exexex

′)+ Iyy/oddt(ωy eyeyey

′)+ Izz/oddt(ωz ezezez

′)

Distributing the time-derivative using the product rule,

Hsys/oHsys/oHsys/o = Ixx/oωx exexex′+ Ixx/oωx exexex

′

+Iyy/oωy eyeyey′+ Iyy/oωy eyeyey

′

+Izz/oωz ezezez′+ Izz/oωz ezezez

′

This reduced expression only holds true if all of the simplifying assumptions wemade are true; to summarize, (a) the point o about which the moments and angularmomentum are being calculated is either a centre of mass or a stationary point; (b)the axes are stuck to the object and are rotating along with it, and (c) the productsof inertia are all zero.

For the even further simplified case of planar rotation (that is, the case of an x−ylamina or body of extrusion rotating in its own plane, as illustrated below in Figs.5.4 and 5.5), ωx = ωy = 0 and ezezez

′ is always pointing in the same direction (and istherefore constant, and so ezezez

′ = 0). Then, in this further simplified case, the time-derivative of our horrible eighteen-term angular momentum equation reduces to thesingle term,

Hsys/oHsys/oHsys/o = Izz/oωz ezezez′

5.3.2 Observations about angular momentum

If we express the law of conservation of angular momentum in terms of the inertiatensor, we see that, for a system consisting of one body,

∑Mext/oMext/oMext/o = Hsys/oHsys/oHsys/o =ddt(IoIoIoωωω)

Comparing this to the expression of the law of conservation of linear momentum asapplied to a system of one body,

∑FextFextFext = GsysGsysGsys =ddt(mvvv)

the parallel between the two becomes evident.


The inertia tensor is a rotational equivalent of mass. Remembering that the mass ofan object and the weight of an object are two entirely separate concepts, the massis best defined as a measure of the resistance of an object to being accelerated. Thelarger the ‘mass’, the harder you have to push on the object to get it to accelerate ata given rate. The inertia tensor, in the same way, is a measure of the resistance of anobject to being angularly accelerated.

Mass is a scalar quantity, since the resistance of the object to being accelerateddoes not change if you push in a different direction. The inertia tensor, on the otherhand, changes both with orientation (it can be harder to spin an object in one direc-tion than in another) and origin (it can be harder to spin the object about one pointthan about another). The values of the moments and products of inertia depend onhow we defined our coordinate system. If we were to change the orientation of ourcoordinates so that exexex becomes eyeyey, for example, then Ixx would become Iyy- and thesetwo values are not necessarily the same. We defined our origin as lying on either thecentre of mass or a stationary point- if it were elsewhere, then not only would wehave violated the assumption that rc/orc/orc/o ×GsysGsysGsys = 0 (so we would be missing terms-remember the ‘orbital angular momentum’?), but the values of all of the terms in IIIwould also be different.

Therefore, in practice, it is important to ensure that the moments and products ofinertia were computed about a point through which was either a centre of mass ora stationary point, and that the axes used to evaluate the integrals in III are the sameaxes in which the angular velocity vector ωωω is expressed.

It is also important to note that if the object rotates within the coordinate system(in other words, if the axes are not stuck to the object), the inertia tensor will changewith time. In order for the inertia tensor to remain constant, then, the tensor mustbe defined in terms of unit vectors which are stuck to the body. Since the angularvelocity must be defined in the same coordinate system as the inertia tensor, theangular velocity must also be defined in terms of unit vectors which are stuck to thebody.

Planar rotation problems are a special case of the law of conservation of angularmomentum. Consider an infinitely thin object lying in the x− y plane. Let us de-fine a Cartesian coordinate system o′ as being stuck to the object, with the originat the centre of mass c. For all differential mass elements dm, then, z = 0. Conse-quently, Ixx/c = Iyy/c = Ixz/c = Iyz/c = 0. If the thin object is then rotated in the sameplane about its centre of mass, ωωω = ωezezez, as illustrated in Figure 5.4. The angularmomentum about the centre of mass c, then, reduces to

Hsys/cHsys/cHsys/c = IcωIcωIcω =

⎡⎣ 0 Ixy/c 0

Iyx/c 0 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz

⎞⎠= Izz/cωz ezezez

as we had seen earlier. Because we defined our coordinate system such that theorigin o is stuck to the centre of mass c and the unit vectors e′xe′xe′x, e′ye′ye′y, and e′ze′ze′z rotate alongwith the mass, the law of conservation of angular momentum may be expressed as


Fig. 5.4 A planar lamina rotating in the x− y plane.

∑Mext/cMext/cMext/c =ddt(Izz/cωz e′ze′ze′z)

Since Izz/c will not change as the object rotates (the coordinate system is stuck tothe mass), and since the unit vector e′ze′ze′z will not change either,

∑Mext/cMext/cMext/c = Izz/cωz ezezez

where ωz is the planar angular acceleration.

A prism and a lamina with the same total mass and cross-sectional shape will havethe same angular momentum when rotated about their long axis. Consider a prismwith a uniform cross-section lying with its long axis along the ezezez direction, as il-lustrated in Figure 5.5. In this case, because the prism has a uniform cross-section,the centre of mass will lie halfway along its length. The x- and y- coordinates of thecentre of mass would remain the same as for a lamina with the same cross-sectionalshape. If we place our origin at the centre of mass of the prism c, with the axes stuckto the prism, then the product of inertia Ixz/c can be expressed as,

Ixz/c =∫(−xz)dm

If we assume the prism has a constant density ρ , then dm = ρ dz dy dx. Then,

Ixz/c = ρ∫ z2

z1

∫ y2

y1

∫ x2

x1(−xz)dx dy dz

where arbitrary limits of integration have been assumed. We can express this integralin a slightly different fashion, as

Ixz/c = ρ∫ z2

z1

(∫ y2

y1

∫ x2

x1(−xz)dx dy

)dz


Fig. 5.5 A prism rotating about its long axis.

Since the cross-sectional shape will be the same for all z, the limits of the x and yintegrations will not depend on z. Therefore, the x and y integrals can be pulled outof the z integration.

Ixz/c = ρ

(∫ y2

y1

∫ x2

x1(−x)dx dy

)∫ z2

z1z dz

Since the integration will be carried out across the z = 0 plane, the z integration mayequivalently be split into two parts.

Ixz/o = ρ

(∫ y2

y1

∫ x2

x1(−x)dx dy

)(∫ z2

0z dz+

∫ 0

z1z dz

)

and finally, because the centre of mass lies halfway along the length of the prism,z1 =−z2, so

Ixz/c = ρ

(∫ y2

y1

∫ x2

x1(−x)dx dy

)(∫ −z1

0z dz+

∫ 0

z1z dz

)

= ρ

(∫ y2

y1

∫ x2

x1(−x)dx dy

)(12(z2)

∣∣∣∣−z1

0+

12(z2)

∣∣∣∣0

z1

)

= ρ

(∫ y2

y1

∫ x2

x1(−x)dx dy

)(12(z2

1 −0)+12(0− z2

1)

)

= ρ

(∫ y2

y1

∫ x2

x1(−x)dx dy

)(12(z2

1 − z21)

)Ixz/c = 0


A similar treatment would show that Iyz/c = 0 as well. Then, computing the angularmomentum of the cylinder,

Hsys/cHsys/cHsys/c = IcωIcωIcω =

⎡⎣ Ixx/c Ixy/c 0

Iyx/c Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz

⎞⎠= Izz/cωz ezezez

Which is exactly the same result as obtained for a lamina. Now, looking at Izz/c,

Izz/c =∫(x2 + y2)dm

= ρ∫ z2

z1

(∫ y2

y1

∫ x2

x1(x2 + y2)dx dy

)dz

Izz/c = ρH

(∫ y2

y1

∫ x2

x1(x2 + y2)dx dy

)

where H = z2 − z1 is simply the length of the prism. So, for two objects with thesame cross-sectional shape to have the same moment of inertia Izz/c, then the quan-tity ρH must be the same. Then, for two such objects A and B with the same cross-sectional shape,

ρaHa = ρbHb

since both objects have the same cross-sectional area, Aa = Ab = A, then we canmultiply both sides of this expression by A

ρaHaA = ρbHbA

Notice that the quantity HA is the volume of the object V . Then,

ρaVa = ρbVb

ma = mb

Since by definition, the density multiplied by the volume is the mass. So, we havejust demonstrated that two cylindrical objects with the same cross-sectional shapewill have the same angular momentum about their long axes if both objects have thesame mass. This result holds true even for the case of the lamina, where H → 0.

The moments and products of inertia are additive. Since the moments and productsof inertia are integrals, the III tensor of a block with a hole in it will be simply theIII tensor for the block less the III tensor for the hole (if the hole had been filled withmass, of course).


5.4 Inertia tensors for common shapes

As we have already observed, the moments and products of inertia of an objectdepend only on the geometry of the object. We have also found that the mathematicaltreatment of rotating systems is much less complicated if the origin is defined atthe centre of mass of the rotating object. Furthermore, if the axes are stuck to theobject, then the moments and products of inertia of that object will always be thesame- regardless of how the object is oriented or how the object is rotating. It wouldtherefore be valuable to evaluate the inertia tensors for shapes which are commonin engineering applications, with the origin at the centre of mass and the axes stuckto the objects.

Again, it is important to remember that because the axes are stuck to the object,the angular velocity vector will always need to be resolved into components in themoving reference frame o′ (so that ωωω = ωxe′xe′xe′x+ωye′ye′ye′y+ωze′ze′ze′z). When taking the time-derivative of the angular momentum, then, we will have to take the time-derivativesof the moving unit vectors e′xe′xe′x, e′ye′ye′y and e′ze′ze′z as well.

Evaluating the inertia tensors of solid objects will almost universally require thesolution of multiple integrals. For a review of the construction and mathematicaltreatment of multiple integrals, see Appendix A.

The moments of inertia for common shapes computed in this section are summa-rized, for easy reference, in Appendix C.

5.4.1 Long slender rod

Consider a slender rod of length L, having a constant density ρ and very small cross-sectional area A, as illustrated in Figure 5.6. We shall define our origin o as beingat the centre of mass of the rod (which will be at the middle of the rod), and theunit vector ezezez lies parallel to the rod. The coordinate system o may be considered asbeing ‘stuck’ to the rod, since neither the coordinate system nor the rod are actuallymoving.

Fig. 5.6 A long, thin rod.

5.4 Inertia tensors for common shapes 301

The total mass m of the rod, then, is simply

m = ρA L

Since both the cross-sectional area and the density are constant, the product ρA willalso be constant. For simplicity, then, let us define λ as the linear density, or densityper unit length, so that

λ = ρA =mL

A tiny element of the rod will therefore have a length dz and a mass dm = λ dz.Finally, the volume occupied by the rod is defined by the inequality,

−L2≤ z ≤ L

2

This inequality will determine the limits of integration.Since x = y = 0 at all points along the mass, it is clear that Izz/c = Ixy/c = Ixz/c =

Iyz/c = 0. So, let us begin with Ixx/c. The subscripts ‘c’ here indicate that the originused is the centre of mass of the object.

Ixx/c =

∫(y2 + z2)dm

=∫ L/2

−L/2(0+ z2)λdz

=13(z3)∣∣L/2−L/2λ

=13

λ(L3

8− −L3

8

)Ixx/c =

13

λL3

4

However, the total mass of the bar m = λL, so

Ixx/c =13(λL)

L2

4

Ixx/c =1

12mL2

Similarly, for Iyy/c,


Iyy/c =∫(x2 + z2)dm

Iyy/c =∫ L/2

−L/2(0+ z2)λdz

So the result must necessarily be the same, and Iyy/c = Ixx/c. The inertia tensor IcIcIc

for the long slender rod lying on the z-axis and rotating about its centre of mass istherefore,

IcIcIc =

⎡⎣ Ixx/c Ixy/c Ixz/c

Iyx/c Iyy/c Iyz/c

Izx/c Izy/c Izz/c

⎤⎦=

⎡⎣ 1

12 mL2 0 00 1

12 mL2 00 0 0

⎤⎦

5.4.2 Cylinder

Consider now a circular cylinder of mass m, radius R and length H. Let us placeour origin o at the centre of mass of the cylinder, oriented such that the ezezez unitvector is parallel to its length, as illustrated in Figure 5.7. If we work in cylindricalcoordinates, then a small differential element of mass will have depth dz, width dr,and side length r dθ . The small element will therefore have mass

dm = ρr dθ dr dz

Fig. 5.7 A circular cylinder.

where ρ is the density of the cylinder. The limits of integration will again be definedby the inequalities describing the volume bounded by the cylinder, which are


0 ≤ r ≤ R

0 ≤ θ ≤ 2π

−H2≤ z ≤ H

2

Again, a more detailed discussion of the cylindrical volume element and volumeintegrals is included as Appendix A.

Let us first consider the moments of inertia. Beginning with Izz/c,


=

∫ (r2 cos2(θ)+ r2 sin2(θ)

)dm

=∫ H/2

−H/2

∫ 2π

0

∫ R

0r2 ρr dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

∫ R

0r3 dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

14

R4 dθ dz

= ρ∫ H/2

−H/2(2π)

14

R4 dz

= ρ(H2− −H

2)

π

2R4

= ρHπ

2R4

= ρ(πR2H)12

R2

Izz/c =12

mR2

Next, we consider Iyy/c,

Iyy/c =∫(x2 + z2)dm

=∫ H/2

−H/2

∫ 2π

0

∫ R

0

(r2 cos2(θ)+ z2)ρr dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

∫ R

0

(r3 cos2(θ)+ rz2)dr dθ dz


= ρ∫ H/2

−H/2

∫ 2π

0

(14

R4 cos2(θ)+12

R2z2)dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

(14

R4(12+

12

cos(2θ))+

12

R2z2)

dθ dz

= ρ∫ H/2

−H/2

(14

R4(12

θ +14

sin(2θ))+

12

R2z2θ

)∣∣2π

0 dz

= ρ∫ H/2

−H/2

(18

R4(2π)+12

R2z2(2π))

dz

= ρ(π

4R4z+

π

3R2z3)∣∣H/2

−H/2

= ρ

(π

4R4(H

2− −H

2

)+

π

3R2(H3

8− −H3

8

))

= ρ(π

4R4H +

π

3R2 H3

4

)= ρ(πR2H)

(14

R2 +1

12H2)

Iyy/c =14

mR2 +1

12mH2

And now for Ixx/c,

Ixx/c =∫(y2 + z2)dm

=∫ H/2

−H/2

∫ 2π

0

∫ R

0

(r2 sin2(θ)+ z2)ρr dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

∫ R

0

(r3 sin2(θ)+ rz2)dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

(14

R4 sin2(θ)+12

R2z2)dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

(14

R4(12− 1

2cos(2θ)

)+

12

R2z2)

dθ dz

= ρ∫ H/2

−H/2

(14

R4(12

θ − 14

sin(2θ))+

12

R2z2θ

)∣∣2π

0 dz

Ixx/c = ρ∫ H/2

−H/2

(18

R4(2π)+12

R2z2(2π))

dz

which leads to exactly the same result as before, so,

Ixx/c =14

mR2 +1

12mH2


Now for the products of inertia. Let us start with Ixy/c.

Ixy/c =∫(−xy)dm

=∫ H/2

−H/2

∫ 2π

0

∫ R

0

(−r2 sin(θ)cos(θ))

ρr dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

∫ R

0

(−r3 sin(θ)cos(θ))

dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

(−14

R4 sin(θ)cos(θ))

dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

(−18

R4 sin(2θ))

dθ dz

= ρ∫ H/2

−H/2

( 116

R4 cos(2θ))∣∣2π

0 dz

= ρ∫ H/2

−H/2

( 116

R4(1−1))

dz

Ixy/c = 0

Next, we’ll consider Ixz/c

Ixz/c =∫(−xz)dm

=∫ H/2

−H/2

∫ 2π

0

∫ R

0

(−rzcos(θ))

ρr dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

∫ R

0

(−r2zcos(θ))

dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

(−13

R3zcos(θ))

dθ dz

= ρ∫ H/2

−H/2

(−13

R3zsin(θ))∣∣2π

0 dz

= ρ∫ H/2

−H/2

(−13

R3z(0−0))

dz

Ixz/c = 0

Finally, we’ll consider Iyz/c; owing to the symmetry of the bar, there is good reasonto suspect that Iyz/c will be zero as well.


Iyz/c =∫(−yz)dm

=∫ H/2

−H/2

∫ 2π

0

∫ R

0

(−rzsin(θ))

ρr dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

∫ R

0

(−r2zsin(θ))

dr dθ dz

= ρ∫ H/2

−H/2

∫ 2π

0

(−13

R3zsin(θ))

dθ dz

= ρ∫ H/2

−H/2

(13

R3zcos(θ))∣∣2π

0 dz

= ρ∫ H/2

−H/2

(13

R3z(1−1))

dz

Iyz/c = 0

So, for a solid cylinder of mass m, height H and radius R , the inertia tensor aboutthe centre of mass is:

IcIcIc =


Iyx/c Iyy/c Iyz/c

Izx/c Izy/c Izz/c

⎤⎦=

⎡⎣ 1

4 mR2 + 112 mH2 0 0

0 14 mR2 + 1

12 mH2 00 0 1

2 mR2

⎤⎦

Of course, as the height H of the cylinder becomes much larger than the radius R,the cylinder will begin to approximate a slender rod (for which R is negligible andIzz/c → 0). The accuracy of applying the slender-rod approximation is plotted inFigure 5.8, which shows the ratio of Iyy/c of a slender rod to Iyy/c of a solid cylinderof similar mass as a function of the cylinder aspect ratio H/R. This plot suggeststhat a cylinder may be reasonably approximated as a slender rod for H/R � 10.

5.4.3 Hollow cylinder

Since the moments and products of inertia are additive, we can simply subtract onecylindrical ‘hole’ of radius Ri from a solid cylinder of radius Ro. Considering Izz/cfirst, we saw from the case of the solid cylinder that,

Izz/c = ρHπ

2R4

So, for the hollow cylinder of inner radius Ri, outer radius Ro and length H,


Fig. 5.8 Accuracy of using the slender-rod approximation for the moment of inertia of a cylinder.

Izz/c = (Izz/c)outside − (Izz/c)hole

= ρHπ

2R4

o −ρHπ

2R4

i

= ρHπ

2(R4

o −R4i )

= ρHπ

2(R2

o −R2i )(R

2o +R2

i )

Izz/c =12

m(R2o +R2

i )

let us next consider Iyy/c. From symmetry, we also know that Ixx/c = Iyy/c. From thecase of the solid cylinder, we know that,

Iyy/c = ρ(π

4R4H +

π

3R2 H3

4

)

For the hollow cylinder, then,

Iyy/c = ρ(π

4R4

oH +π

3R2

oH3

4

)−ρ(π

4R4

i H +π

3R2

iH3

4

)= ρπH

(14

R4o +

112

R2oH2 − 1

4R4

i −112

R2i H2)

= ρπH(R2o −R2

i )(1

4(R2

o +R2i )+

112

H2)


Iyy/c =14

m(R2o +R2

i )+1

12mH2

Since the products of inertia for both the outside of the cylinder and for the hole arezero, the sum will be zero as well. Then, the inertia tensor about the centre of massfor a hollow cylinder of mass m, inner radius Ri and outer radius Ro is

IcIcIc =

⎡⎣ 1

4 m(R2o +R2

i )+1

12 mH2 0 00 1

4 m(R2o +R2

i )+1

12 mH2 00 0 1

2 m(R2o +R2

i )

⎤⎦

Notice that for the special case of a very thin cylindrical shell, Ro ≈ Ri = R and

IcIcIc =


Iyx/c Iyy/c Iyz/c

Izx/c Izy/c Izz/c

⎤⎦=

⎡⎣ 1

2 mR2 + 112 mH2 0 0

0 12 mR2 + 1

12 mH2 00 0 mR2

⎤⎦

Of course, a ‘real’ cylindrical shell must always have an outer radius larger than itsinner radius, no matter how thin the wall. Therefore, the inertia tensor for a hollowcylinder may always be used for real-world cylindrical shells. However, the accuracyof the thin-shell approximation (taking the shell radius R as the average of Ri andRo) is fairly good for Ri/Ro � 0.8. This is illustrated in Figure 5.9, in which theratio of Izz/c of a thin cylindrical shell to Izz/c of a hollow cylinder of similar mass isplotted as a function of Ri/Ro.

Fig. 5.9 Accuracy of using the cylindrical-shell approximation for the moment of inertia of ahollow cylinder.


5.4.4 Cone

The inertia tensor of a cone of constant density ρ may also be fairly easily deter-mined using cylindrical coordinates. Let us consider a cone of mass m, height Hand maximum radius R lying along the z-axis, as illustrated in Figure 5.10. We shalldetermine the moments and products of inertia about the apex of the cone o ratherthan about the centre of mass, since o is easier to define geometrically.

Fig. 5.10 A cone.

To begin with, we must recognize that for a cone, the limits of integration willnot be constant. Instead, the volume occupied by the cone is defined in terms of theinequalities,

0 < z < H

0 < θ < 2π

0 < r <RH

z

(for more details on determining functional limits of integration, see Appendix A).We will begin with Ixx/o, recalling that, in cylindrical coordinates, dm= ρr dr dθ dz.

Ixx/o =∫(y2 + z2)dm

=∫ H

0

∫ 2π

0

∫ zR/H

0

(r2 sin2(θ)+ z2)ρr dr dθ dz

= ρ∫ H

0

∫ 2π

0

∫ zR/H

0

(r3 sin2(θ)+ rz2) dr dθ dz


= ρ∫ H

0

∫ 2π

0

(14

r4 sin2(θ)+12

r2z2)∣∣∣∣zR/H

0dθ dz

= ρ∫ H

0

∫ 2π

0

(14

z4 R4

H4 sin2(θ)+12

z4 R2

H2

)dθ dz

= ρ∫ H

0

(14

z4 R4

H4

(θ

2− 1

4sin(2θ)

)+

12

z4 R2

H2 θ

)∣∣∣∣2π

0dz

= ρ∫ H

0

(14

z4 R4

H4

(2π

2− 1

4sin(4π)

)+

12

z4 R2

H2 (2π)

)dz

= ρ∫ H

0

(14

πz4 R4

H4 +πz4 R2

H2

)dz

= ρ( 1

20πz5 R4

H4 +15

πz5 R2

H2

)∣∣∣∣H

0

=1

20ρπH5 R4

H4 +15

ρπH5 R2

H2

Ixx/o =1

20ρπHR4 +

15

ρπH3R2

However, we know that the mass of a cone of constant density ρ is given by theexpression,

m =13

ρπR2H

Substituting,

Ixx/o =1

20ρπHR4 +

15

ρπH3R2

=

(13

ρπR2H

)3

20R2 +

(13

ρπR2H

)35

H2

Ixx/o =3

20mR2 +

35

mH2

Next, we will consider Iyy/o.

Iyy/o =∫(x2 + z2)dm

=∫ H

0

∫ 2π

0

∫ zR/H

0

(r2 cos2(θ)+ z2)ρr dr dθ dz

= ρ∫ H

0

∫ 2π

0

∫ zR/H

0

(r3 cos2(θ)+ rz2) dr dθ dz


= ρ∫ H

0

∫ 2π

0

(14

r4 cos2(θ)+12

r2z2)∣∣∣∣zR/H

0dθ dz

= ρ∫ H

0

∫ 2π

0

(14

z4 R4

H4 cos2(θ)+12

z4 R2

H2

)dθ dz

= ρ∫ H

0

(14

z4 R4

H4

(θ

2+

14

sin(2θ))+

12

z4 R2

H2 θ

)∣∣∣∣2π

0dz

= ρ∫ H

0

(14

z4 R4

H4

(2π

2+

14

sin(4π))+

12

z4 R2

H2 (2π)

)dz

= ρ∫ H

0

(14

πz4 R4

H4 +πz4 R2

H2

)dz

= ρ( 1

20πz5 R4

H4 +15

πz5 R2

H2

)∣∣∣∣H

0

=1

20ρπH5 R4

H4 +15

ρπH5 R2

H2

Iyy/o =1

20ρπHR4 +

15

ρπH3R2

This is exactly the same result obtained for Ixx/o, which we could have predictedowing to the symmetry of the cone. Therefore,

Iyy/o =3

20mR2 +

35

mH2

Now, we will consider Izz/o.

Izz/o =

∫(x2 + y2)dm

=∫ H

0

∫ 2π

0

∫ zR/H

0

(r2)ρr dr dθ dz

= ρ∫ H

0

∫ 2π

0

∫ zR/H

0r3 dr dθ dz

= ρ∫ H

0

∫ 2π

0

14

r4

∣∣∣∣zR/H

0dθ dz

= ρ∫ H

0

∫ 2π

0

14

z4 R4

H4 dθ dz

= ρ∫ H

0

14

z4 R4

H4 θ

∣∣∣∣2π

0dz

= ρ∫ H

0

14

z4 R4

H4 (2π) dz


= ρ∫ H

0

12

πz4 R4

H4 dz

=1

10ρπz5 R4

H4

∣∣∣∣H

0

=1

10ρπH5 R4

H4

Izz/o =1

10ρπHR4

Again, we can re-organize this result to substitute in the mass of the cylinder, as

Izz/o =110

ρπHR4

=

(13

ρπR2H

)3

10R2

Izz/o =310

mR2

Now for the products of inertia. Since the cone is symmetric about the z-axis, weexpect all of the products of inertia to vanish. Let us start with Ixy/o.

Ixy/o =∫(−xy)dm

= −∫ H

0

∫ 2π

0

∫ zR/H

0

(r2 sin(θ)cos(θ)

)ρr dr dθ dz

= −ρ∫ H

0

∫ 2π

0

∫ zR/H

0r3 sin(θ)cos(θ) dr dθ dz

= −ρ∫ H

0

∫ 2π

0

14

r4 sin(θ)cos(θ)

∣∣∣∣zR/H

0dθ dz

= −ρ∫ H

0

∫ 2π

0

14

z4 R4

H4 sin(θ)cos(θ) dθ dz

= −ρ∫ H

0

14

z4 R4

H4

(−1

2cos2(θ)

)∣∣∣∣2π

0dz

= −ρ∫ H

0

14

z4 R4

H4

(−1

2+

12

)dz

Ixy/o = 0

Next, we will consider Iyz/o.


Iyz/o =∫(−yz)dm

= −∫ H

0

∫ 2π

0

∫ zR/H

0

(r sin(θ)z

)ρr dr dθ dz

= −ρ∫ H

0

∫ 2π

0

∫ zR/H

0r2 sin(θ)z dr dθ dz

= −ρ∫ H

0

∫ 2π

0

13

r3 sin(θ)z

∣∣∣∣zR/H

0dθ dz

= −ρ∫ H

0

∫ 2π

0

13

z3 R3

H3 sin(θ)z dθ dz

= ρ∫ H

0

13

z4 R3

H3 cos(θ)

∣∣∣∣2π

0dz

= ρ∫ H

0

13

z4 R3

H3 (1−1) dz

Iyz/o = 0

Finally, we will consider Ixz/o.

Ixz/o =∫(−xz)dm

= −∫ H

0

∫ 2π

0

∫ zR/H

0

(r cos(θ)z

)ρr dr dθ dz

= −ρ∫ H

0

∫ 2π

0

∫ zR/H

0r2 cos(θ)z dr dθ dz

= −ρ∫ H

0

∫ 2π

0

13

r3 cos(θ)z

∣∣∣∣zR/H

0dθ dz

= −ρ∫ H

0

∫ 2π

0

13

z3 R3

H3 cos(θ)z dθ dz

= −ρ∫ H

0

13

z4 R3

H3 sin(θ)

∣∣∣∣2π

0dz

= −ρ∫ H

0

13

z4 R3

H3 (0) dz

Ixz/o = 0

So, for a solid cone of mass m, maximum radius R and height H, the inertia tensorabout the apex is:


IcIcIc =

⎡⎣ Ixx/o Ixy/o Ixz/o

Iyx/o Iyy/o Iyz/o

Izx/o Izy/o Izz/o

⎤⎦=

⎡⎣ 3

20 mR2 + 35 mH2 0 0

0 320 mR2 + 3

5 mH2 00 0 3

10 mR2

⎤⎦

5.4.5 Sphere

Let us now consider a solid sphere of radius R. We will put the origin on the centreof mass of the sphere (which is at the centre of the sphere), and since the sphere issymmetric about all possible orientations of the axis, we know that Ixx/c = Iyy/c =Izz/c. Also, because the sphere is symmetric about all possible axes, we also knowthat Ixy = Ixz = Iyz = 0.

Now, let us determine the mass of the infinitesimal differential mass elementdm. If we use a radius line to slice a wedge-shaped segment from the sphere, theradius line would first cut vertically along a north-south latitude line through someangle dθ , describing a small arc length r dθ along the surface of the sphere. Then,the radius line would cut horizontally along an east-west longitude line, sweepingthrough some angle dφ . The radius of the circle along which the longitude cut wasmade is r sin(θ), so the length of the cut is r sin(θ) dφ . So, at some radius r fromthe centre of the sphere, our thin wedge would have an infinitesimal cross-sectionalarea dA of

dA =(r sin(θ)dφ

)(r dθ)

= r2 sin(θ)dφ dθ

Consequently, the volume of a small mass element would be

dm = ρ dA dr = ρr2 sin(θ)dφ dθ dr

where ρ is the density of the sphere, and is assumed constant. This is illustratedgraphically in Figure 5.11. In spherical coordinates, the volume bounded by a sphereis described by the inequalities,

0 ≤ r ≤ R

0 ≤ θ ≤ π

0 ≤ φ ≤ 2π


Fig. 5.11 A spherical mass element.

which, as before, will determine our limits of integration.From the coordinate system conversions, we know that

x = r sin(θ)cos(φ)

y = r sin(θ)sin(φ)

So,

x2 + y2 = r2 sin2(θ)cos2(φ)+ r2 sin2(θ)sin2(φ)

= r2 sin2(θ)(cos2(φ)+ sin2(φ)

)x2 + y2 = r2 sin2(θ)

So, we can then compute the moments of inertia. Let’s work in the x− y plane.


=∫ π

0

∫ R

0

∫ 2π

0

(r2 sin2(θ)

)(ρr2 sin(θ)dφ dr dθ

)= ρ

∫ π

0

∫ R

0

∫ 2π

0

(r4 sin3(θ)

)dφ dr dθ

= ρ∫ π

0

∫ R

02π(r4 sin3(θ)

)dr dθ


= ρ∫ π

02π(1

5R5 sin3(θ)

)dθ

Izz/c =25

ρπR5∫ π

0sin3(θ)dθ

From our trusty table of integrals, we see that

∫sin3(x)dx =

112

(cos(3x)−9cos(x)

)This poses a slight problem: the cosine goes negative when the mass element isbelow the z = 0 plane. Since we’re dealing with a real, physical quantity whichmust always be positive, just to be sure there isn’t any ‘phantom’ or nonphysicalnegatives hanging about, let’s break the integral up so that

∫ π

0sin3(θ)dθ =

∫ π/2

0sin3(θ)dθ +

∫ π

π/2sin3(θ)dθ

Finally, since the function sin3(x) is symmetric,

∫ π/2

0sin3(θ)dθ +

∫ π

π/2sin3(θ)dθ = 2

∫ π

π/2sin3(θ)dθ

So, substituting this into the previous result,

Izz/c =25

ρπR5∫ π

0sin3(θ)dθ

=25

ρπR5(

2∫ π

π/2sin3(θ)dθ

)

=45

ρπR5(

112

(cos(3θ)−9cos(θ)

))∣∣∣∣π

π/2

=1

15ρπR5(cos(3θ)−9cos(θ)

)∣∣ππ/2

=1

15ρπR5(−1+9−0+0

)=

815

ρπR5

=(4

3πR3ρ

)25

R2

Izz/c =25

mR2


Recall that the volume of a sphere is 4πR3/3. Finally, the inertia tensor about thecentre of mass of a solid sphere of mass m and radius R is

IcIcIc =


Iyx/c Iyy/c Iyz/c

Izx/c Izy/c Izz/c

⎤⎦=

⎡⎣ 2

5 mR2 0 00 2

5 mR2 00 0 2

5 mR2

⎤⎦

5.4.6 Hollow sphere

We can now easily find the moments of inertia for a hollow sphere, in the sameway as we solved for the moments of inertia for a hollow cylinder. Again, from thesymmetry, we know that Ixx/c = Iyy/c = Izz/c, and Ixy/c = Ixz/c = Iyz/c = 0. From thesolid sphere, we know

Izz/c =8

15ρπR5

So, for a hollow sphere of mass m, inner radius Ri and outer radius Ro, we againsubtract the ‘hole’.

Izz/c = (Izz/c)outside − (Izz/c)hole

=8

15ρπR5

o −815

ρπR5i

=8

15ρπ(R5

o −R5i )

=8

15ρπ(R3

o −R3i )

R5o −R5

i

R3o −R3

i

Izz/c =25

mR5

o −R5i

R3o −R3

i

So the inertia tensor about the centre of mass for the hollow sphere of mass m, innerradius Ri and outer radius Ro is

IcIcIc =


Iyx/c Iyy/c Iyz/c

Izx/c Izy/c Izz/c

⎤⎦=

⎡⎢⎢⎢⎢⎣

25 m

R5o−R5

iR3

o−R3i

0 0

0 25 m

R5o−R5

iR3

o−R3i

0

0 0 25 m

R5o−R5

iR3

o−R3i

⎤⎥⎥⎥⎥⎦


For the special case of a spherical shell, let us define the functions f (r) and g(r)such that

f (r) = r5 −R5i

g(r) = r3 −R3i

Then, we can expand f (r) and g(r) each as a Taylor series about Ri, yielding

f (r) = (R5i −R5

i )+5R4i (Ro −Ri)+ ...

g(r) = (R3i −R3

i )+3R2i (Ro −Ri)+ ...

(A review of the Taylor series and its development is included as Appendix B).Since, for a thin spherical shell, Ri → Ro, the expansion about Ri will hold for Ri <r < Ro, and the higher order terms will become negligibly small. Then,

f (r) ≈ (0)+5R4i (Ro −Ri)+(0)

g(r) ≈ (0)+3R2i (Ro −Ri)+(0)

Therefore,

Izz/c =25

mR5

o −R5i

R3o −R3

i

=25

f (r)g(r)

=25

(5R4

i (Ro −Ri)

3R2i (Ro −Ri)

)

Izz/c =25

5R4i

3R2i

However, for the case of an infinitely thin spherical shell, Ro = Ri = R. Finally, then,

Izz/c =23

mR2

The inertia tensor about the centre of mass for a thin spherical shell of mass m andradius R is then,


IcIcIc =


Iyx/c Iyy/c Iyz/c

Izx/c Izy/c Izz/c

⎤⎦=

⎡⎣ 2

3 mR2 0 00 2

3 mR2 00 0 2

3 mR2

⎤⎦

As with the thin cylindrical shell, a thin spherical shell is also just a special caseof the hollow sphere. Once again, by plotting the ratio between Izz/c for a sphericalshell and Izz/c for a hollow sphere of similar mass as a function of Ri/Ro (Figure5.12), we see that the thin-shell approximation will apply reasonably well for real-world hollow spheres (with the shell radius R taken as the average of Ri and Ro) forRi/Ro � 0.8.

Fig. 5.12 Accuracy of using the spherical-shell approximation for the moment of inertia of ahollow sphere.

5.4.7 Rectangular block

Let us finally consider the case of a rectangular block of mass m, having height H,length L and width W , as illustrated in Figure 5.13. A small mass element dm wouldthen have a mass of ρ dx dy dz, where ρ is the density of the block. The volumebounded by the block is defined by the inequalities,

−W2

≤ x ≤ W2

−L2≤ y ≤ L

2

−H2≤ z ≤ H

2


Fig. 5.13 A rectangular block.

Starting with the moments of inertia again, let us consider Ixx first.

Ixx =∫(y2 + z2)dm

=∫ H/2

−H/2

∫ L/2

−L/2

∫ W/2

−W/2(y2 + z2)ρ dx dy dz

= ρ∫ H/2

−H/2

∫ L/2

−L/2(y2W + z2W )dy dz

= ρ∫ H/2

−H/2

(13

y3W + yz2W)∣∣L/2

−L/2 dz

= ρ∫ H/2

−H/2

( 112

L3W + z2LW)

dz

= ρ( 1

12zL3W +

13

z3LW)∣∣H/2

−H/2

= ρ( 1

12HL3W +

13

H3

4LW)

= (ρHLW )( 1

12L2 +

112

H2)Ixx =

112

m(L2 +H2)

Next, we consider Iyy.


Iyy =∫(x2 + z2)dm

=∫ H/2

−H/2

∫ L/2

−L/2

∫ W/2

−W/2(x2 + z2)ρ dx dy dz

= ρ∫ H/2

−H/2

∫ L/2

−L/2

(13

x3 + xz2)∣∣W/2−W/2 dy dz

= ρ∫ H/2

−H/2

∫ L/2

−L/2

(13

W 3

4+ z2W

)dy dz

= ρ∫ H/2

−H/2

( 112

yW 3 + yz2W)∣∣L/2

−L/2 dz

= ρ∫ H/2

−H/2

( 112

LW 3 + z2LW)

dz

= ρ( 1

12zLW 3 +

13

z3LW)∣∣H/2

−H/2

= ρ( 1

12HLW 3 +

13

H3

4LW)

= (ρHLW )( 1

12W 2 +

112

H2)Iyy =

112

m(W 2 +H2)

Now, we consider Izz.

Izz =∫(x2 + y2)dm

=∫ H/2

−H/2

∫ L/2

−L/2

∫ W/2

−W/2(x2 + y2)ρ dx dy dz

= ρ∫ H/2

−H/2

∫ L/2

−L/2

(13

x3 + xy2)∣∣W/2−W/2 dy dz

= ρ∫ H/2

−H/2

∫ L/2

−L/2

(13

W 3

4+ y2W

)dy dz

= ρ∫ H/2

−H/2

( 112

yW 3 +13

y3W)∣∣L/2

−L/2 dz

= ρ∫ H/2

−H/2

( 112

LW 3 +13

L3

4W)

dz

= ρ( 1

12zLW 3 +

112

zL3W)∣∣H/2

−H/2

= ρ( 1

12HLW 3 +

112

HL3W)


= (ρHLW )( 1

12W 2 +

112

L2)Izz =

112

m(W 2 +L2)

Finally, we can consider the products of inertia; let us start with Ixy.

Ixy =∫(−xy)dm

=∫ H/2

−H/2

∫ L/2

−L/2

∫ W/2

−W/2(−xy)ρ dx dy dz

= −ρ∫ H/2

−H/2

∫ L/2

−L/2

(12

x2y)∣∣W/2

−W/2 dy dz

= −ρ∫ H/2

−H/2

∫ L/2

−L/2

(18(W 2 −W 2)y

)dy dz

Ixy = 0

So the product of inertia Ixy vanishes. It should be fairly clear that, for reasons ofsymmetry, if Ixy = 0 then Iyz = Ixz = 0 as well. So, the inertia tensor of a solid blockabout its centre of mass is

IcIcIc =


Iyx/c Iyy/c Iyz/c

Izx/c Izy/c Izz/c

⎤⎦=

⎡⎣ 1

12 m(L2 +H2) 0 00 1

12 m(W 2 +H2) 00 0 1

12 m(W 2 +L2)

⎤⎦

It is interesting to note that, for all of the shapes considered, the products ofinertia were all zero. This happened because of the way in which the axes wereselected; one of the ‘special’ mathematical properties which defines a tensor is thatall of the off-diagonal components will go to zero if the axes are selected in such away as to ensure symmetry. Since we defined our axes in just this way, the productsof inertia all vanished. Had we oriented the axes differently (say, had we said thatezezez points diagonally across the opposite corners of the block), then the products ofinertia would have been nonzero.

Having no products of inertia in the inertia tensor will reduce the number ofterms in the angular momentum equation, and make the calculations much easier. Ifpossible, it is always best to try and stick the axes to the body in such a way as toensure the maximum amount of symmetry about the axes.

5.5 Parallel axis theorem 323

5.5 Parallel axis theorem

In the preceding section, we determined the inertia tensors for common shapes abouttheir centres of mass (with the exception of the cone, where III was determined aboutits apex for mathematical convenience). These results are therefore only applicablefor cases where the objects are rotating about their centres of mass; clearly, in en-gineering practice, objects are very seldom constrained to only rotate about theircentre of mass. Still, it would be convenient if we could find a way to translate mo-ments and products of inertia from being about a centre of mass to being about someother point.

Let us consider the moments and products of inertia of a body about some pointo which is not the centre of mass of the body, as illustrated in Figure 5.14. In thiscase, we may express the position vector rp/orp/orp/o of each differential mass element dmas a sum of two vectors, so that

Fig. 5.14 Decomposition of the rp/orp/orp/o vector into vectors relative to the centre of mass.

rp/orp/orp/o = rc/orc/orc/o +rp/crp/crp/c

Recall that rc/orc/orc/o is the position vector of the centre of mass relative to the point o, andrp/crp/crp/c is the position vector of the mass element relative to the centre of mass. Then,if rp/orp/orp/o = xp/oexexex + yp/oeyeyey + zp/oezezez,

xp/o = xc/o + xp/c

yp/o = yc/o + yp/c

zp/o = zc/o + zp/c

Let us now consider the moment of inertia Izz/o about the point o.


Izz/o =∫(x2

p/o + y2p/o)dm

=∫ (

(xc/o + xp/c)2 +(yc/o + yp/c)

2)dm

Izz/o =∫ (

x2c/o + x2

p/c +2xc/oxp/c + y2c/o + y2

p/c +2yc/oyp/c)

dm

This integral can be divided into four parts, so that

Izz/o =∫ (

x2c/o + y2

c/o

)dm+

∫ (x2

p/c + y2p/c

)dm

+∫

2xc/oxp/c dm+∫

2yc/oyp/c dm

Also, since the location of the centre of mass of the object relative to o will be thesame for all the differential mass elements, xc/o and yc/o can be pulled out of theintegrals.

Izz/o = (x2c/o + y2

c/o)∫

dm+∫ (

x2p/c + y2

p/c

)dm

+ 2xc/o

∫xp/c dm+2yc/o

∫yp/c dm

However, because the point c is the centre of mass, the location of the centre of massrelative to the point c, xc/c, is zero. Therefore, from the definition of the centre ofmass,

xc/c = 0 =

∫xp/c dm

m

0 =∫

xp/c dm

Similarly, ∫yp/c dm = 0

Substituting this result into the equation for Izz/o,

Izz/o = (x2c/o + y2

c/o)∫

dm+∫ (

x2p/c + y2

p/c

)dm

+ 2xc/o(0)+2yc/o(0)

5.6 Torsional springs 325

Izz/o = m(x2c/o + y2

c/o)+∫ (

x2p/c + y2

p/c

)dm

Izz/o = m(x2c/o + y2

c/o)+ Izz/c

where Izz/c is the moment of inertia about the centre of mass. Therefore, if you knowthe moment of inertia about the centre of mass, you can easily calculate the momentof inertia about some other point o knowing only the distance between the centre ofmass and o, so long as the orientation of the coordinate system does not change. Asimilar treatment may be applied to the other moments of inertia to yield,

Ixx/o = Ixx/c +m(y2c/o + z2

c/o)

Iyy/o = Iyy/c +m(x2c/o + z2

c/o)

Izz/o = Izz/c +m(x2c/o + y2

c/o)

And for the products of inertia,

Ixy/o = Ixy/c −m(xc/oyc/o)

Ixz/o = Ixz/c −m(xc/ozc/o)

Iyz/o = Iyz/c −m(yc/ozc/o)

where c is the centre of mass of the object, m is the total mass of the object, ando is any other point in space (which does not necessarily have to be within thebody). These transformations are known as the parallel axis theorem, because thetransformation may only be applied when the axes of the coordinate system areunaltered.

5.6 Torsional springs

Let us consider now an ideal spring of stiffness k constrained to lie along a circularpath of radius L. One end of the spring is fixed, while the other end is attached toa thin bar. The thin bar is pinned at the centre o of the circular arc, as illustrated inFigure 5.15. Let us define a cylindrical coordinate system with its origin at the pino, such that ererer always points along the bar and eθ points in the direction in whichthe spring extends, as illustrated. As the spring stretches and contracts, the springis always described by a circular arc of radius L. If the bar is deflected by somesmall angle θ , the spring will extend a small amount Δs along its circular path inthe direction of eθeθeθ , so that


Fig. 5.15 Demonstration and schematic representation of a torsional spring.

Δsss = Lθ eθeθeθ

The force vector exerted by the spring upon the bar as a result of the deflection, then,is

FkFkFk = −k Δsss

FkFkFk = −k(Lθ) eθeθeθ

Next, we recall that the definition of the moment vector produced by the force FkFkFk

about the point o is given by the expression,

Mk/oMk/oMk/o = L ererer ×FkFkFk

where L ererer is the position vector of the point of application of the force FkFkFk. Substi-tuting in our earlier result,

Mk/oMk/oMk/o = L ererer ×−k(Lθ) eθeθeθ

= −kL2θ(ererer ×eθeθeθ )

Mk/oMk/oMk/o = −kL2θ ezezez

This expression indicates that the spring moment is proportional to the angular de-flection. By convention, any spring with this property is said to be a linear torsional

5.7 Examples 327

spring. The constant of proportionality, kL2, is therefore referred to as the torsionalspring constant or torsional stiffness and is often represented by the symbol kθ . Fortorsional springs, then,

Mk/oMk/oMk/o =−kθ θ ezezez

where ezezez is in the direction of angular displacement (according to the right-handrule). The moment is negative because the spring force will always be in a directionopposite to the deflection. Note that if the spring is not constrained to lie on a circularpath, this expression will not hold true. In practice, it is fairly difficult to get realsprings to extend and contract along circular arcs; instead, real torsional springstend to be spiral or helical in shape.

5.7 Examples

The best way to illustrate the practical application of the law of conservation ofangular momentum to rotating systems in engineering mechanics is through theanalysis of several classical examples.

5.7.1 Pulley with mass

Let us first consider the example of a block of mass mb suspended by an ideal stringwhich is, in turn, wound around a disc with radius R and mass ma, as illustrated inFigure 5.16. As the block accelerates downwards under the action of gravity, it mustalso start the pulley spinning.

We will define our system first as consisting only of the block. We will also definea Cartesian coordinate system o which is stationary, with eyeyey pointing downwards.Then, drawing a free-body diagram of the block,

∑FextFextFext = (Fg −FT ) eyeyey

From the law of conservation of linear momentum,

∑FextFextFext = GsysGsysGsys = mbaaa

(Fg −FT ) eyeyey = mbxexexex +mbyeyeyey

From the scalar coefficients of exexex, we recognize immediately that x = 0. Looking,then, at the scalar coefficients of eyeyey,


Fig. 5.16 A block suspended from a pulley with mass.

mbg−FT = mby

y = g− 1mb

FT

We have no further information about the magnitude of the tension force FT . Forthis, we need to examine the pulley. Let us redefine our system to consist only ofthe pulley, with a coordinate system o′ stuck to the centre of the pulley and rotatingwith the pulley. At the instant of interest, we can say that the moving unit vectorse′xe′xe′x, e′ye′ye′y and e′ze′ze′z align exactly with the stationary unit vectors exexex, eyeyey and ezezez. Then, fromthe free-body diagram,

∑Mext/cMext/cMext/c = ∑(rrr×FFF)

= rTrTrT ×FTFTFT

= Re′xe′xe′x ×FT e′ye′ye′y= RFT (e′xe′xe′x ×e′ye′ye′y)

∑Mext/cMext/cMext/c = RFT e′ze′ze′z

Now, we can compute the angular momentum of the pulley. We know the momentof inertia of a disc about its centre (in a coordinate frame stuck to the disc), andωωω = ωz e′ze′ze′z, so

5.7 Examples 329

HcHcHc = IcωIcωIcω =

⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz

⎞⎠= Izz/cωz e′ze′ze′z

This result can then be substituted into the angular momentum equation,

∑Mext/cMext/cMext/c = Hsys/cHsys/cHsys/c =ddt(IcωIcωIcω)

RFT e′ze′ze′z =ddt(Izz/cωz e′ze′ze′z)

It is again important to stress that ΣMMM = HHH only when evaluated about a point whichis either a centre of mass or a point which is not moving, because we have neglectedthe orbital angular momentum. We can now evaluate the time-derivative by applyingthe product rule,

RFT e′ze′ze′z =ddt(Izz/c)ωz e′ze′ze′z + Izz/cωz e′ze′ze′z + Izz/cωz e′ze′ze′z

Since we evaluated III along axes which were stuck to the disc, III will not change intime. Also, e′ze′ze′z always points in the same direction even as the axes rotate, so it willnot change in time either.

RFT e′ze′ze′z = 0+ Izz/cωz e′ze′ze′z +0

RFT = Izz/cωz

We can isolate and solve for FT from this expression, so that

FT = ωzIzz/c

R

We may now substitute this result back into the linear momentum equation.

y = g− 1mb

FT

= g− 1mb

(ωz

Izz/c

R

)= g− ωz

Izz/c

mbR


Now, if we wanted to solve for the acceleration of the block y, we would need toexpress ωz in terms of y as well. For this, we recognize that the length of string Sreleased from the pulley is equal to the distance y traveled by the block. Since thestring was wrapped around the pulley,

S = y = Rθ

where θ is the angle through which the pulley rotated. Taking the time-derivativesof both sides of this expression and recognizing that θ = ωz,

y = Rθ = Rωz

y = Rθ = Rωz

We may now express ωz in terms of y, as

ωz =1R

y


y = g− ωzIzz/c

mbR

y = g− ( 1R

y) Izz/c

mbR

y(1+

Izz/c

mbR2

)= g

y =g(

1+Izz/c

mbR2

)

For a cylinder, we know that Izz/c = mR2/2, so

y =g(

1+ 12

maR2

mbR2

)y =

g(1+ 1

2mamb

)

Notice that if the mass of the pulley ma → 0 then y → g, and the block simply goesinto free-fall. It is also interesting to notice that the radius of the pulley has no effecton the acceleration of the block.

5.7 Examples 331

5.7.2 Pure rolling motion

Consider a round object sitting atop an inclined plane at angle θ , as illustrated inFigure 5.17. The round object has radius R and mass m, and inertia tensor about itscentre of mass IcIcIc (with the axes oriented as illustrated in the figure). If the objectrolls down the incline without slipping, then it is said to be undergoing pure rollingmotion. A block sliding down the incline would undergo pure slipping, while arolling object which is also skidding would be undergoing some combination ofrolling and slipping.

Fig. 5.17 Illustration of a round object rolling down an inclined plane.

If the rolling object is not slipping at all, the instantaneous point of contact pbetween the rim of the object and the surface must be at rest relative to the surface.The instantaneous point of contact will move around the rim as the object rolls, butat the instant when the point p on the rim touches the surface, that point must havevelocity vp/ovp/ovp/o = 0. Drawing a free-body diagram of the object, we see that there arethree forces acting upon it- gravity, the normal force, and friction. There must befriction acting on the body, since without friction, it would simply slide down thesurface without rolling. Since the friction acts at the instantaneous point of contactand the point of contact is at rest, the friction must be static. The static friction willhave just the right magnitude to ensure that the point of contact does not slip, soFf �= μFn.

So, if we define our system as consisting of the object only,


∑FextFextFext =(Fg sin(θ) exexex −Fg cos(θ) eyeyey

)+(−Ff exexex

)+(Fn eyeyey

)


where we have defined exexex as pointing down the surface and eyeyey as pointing normallyup from the surface, as illustrated. The unit vector ezezez, then, points into the page.Collecting together the scalar coefficients of each of the unit vectors, then,

∑FextFextFext = (Fg sin(θ)−Ff ) exexex +(Fn −Fg cos(θ)) eyeyey

From the law of conservation of linear momentum,

∑FextFextFext = GsysGsysGsys = macacac

where acacac = xcexexex + yceyeyey is the acceleration of the centre of mass. As we showedearlier, this result holds true regardless of whether or not the object begins to rotate.Then,

(Fg sin(θ)−Ff ) exexex +(Fn −Fg cos(θ)) eyeyey = m(xcexexex + yceyeyey)

However, the object is constrained to remain on the surface, so yc = 0. Looking atthe scalar coefficients of eyeyey,

Fn −Fg cos(θ) = myc

Fn −mgcos(θ) = 0

Fn = mgcos(θ)

Now, looking at the scalar coefficients of exexex,

Fg sin(θ)−Ff = mxc

mgsin(θ)−Ff = mxc

xc = gsin(θ)− 1m

Ff

Since the friction is static, we have no further information about the magnitude ofFfFfFf. In order to solve for Ff , then, we require another equation: for this, we can turnto the law of conservation of angular momentum.

Let us therefore consider the rotation about the object’s centre of mass c. Lookingagain at our free-body diagram,

∑Mext/cMext/cMext/c = ∑(rrr×FFF) = (rg/crg/crg/c ×FgFgFg)+(rf/crf/crf/c ×FfFfFf)+(rn/crn/crn/c ×FnFnFn)

where rg/crg/crg/c, rf/crf/crf/c, and rn/crn/crn/c are position vectors pointing from an origin at c to the pointof application of FgFgFg, FfFfFf and FnFnFn, respectively. Since gravity acts on the centre of mass,

5.7 Examples 333

rg/crg/crg/c = 0; also, since the normal force is parallel to rn/crn/crn/c, the cross product betweenthem will be zero.

∑Mext/cMext/cMext/c = 0+rf/crf/crf/c ×FfFfFf +0

=(−R eyeyey

)× (−Ff exexex)

= RFf(eyeyey ×exexex

)∑Mext/cMext/cMext/c = −RFf ezezez

The result is negative, because ezezez was defined as positive into the page, and thefriction exerts an anticlockwise (out-of-the-page) moment on the object. Then, fromthe law of conservation of angular momentum,

∑Mext/cMext/cMext/c = Hsys/cHsys/cHsys/c =ddt(IcωcIcωcIcωc)

where ωcωcωc is the angular velocity vector of the system, and the subscript c has beenadded just to remind us that the rotation must be about the centre of mass of theobject. Evaluating the angular momentum, then,

HHH = IcωcIcωcIcωc =

⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz


where ωz is the magnitude of the z-component of the angular velocity vector of theobject (which is still unknown), and e′ze′ze′z is stuck to the object and pointing in thesame direction as ezezez at the instant of interest.

It is very important to note here that we assumed that the angular velocity vectorof the object is pointing in the +ezezez direction, when clearly the object will be rollingdownhill (the −ezezez direction). All this means is that the magnitude of ωz will benegative. The reason for this is only to avoid confusion- had the circumstances beenmore complex, or if there had been more applied forces, the object may have rolledeither uphill or downhill, and we wouldn’t know which until we solved the problem.By choosing the unknown ωz in the +ezezez direction, then the sign of ωz will tellus immediately the direction of rotation as well. As a general rule, then, unknownvectors should always be defined as pointing in a positive direction; the sign of theunknown magnitude, then, will indicate the direction.

Furthermore, we may note that the radial symmetry of the object was assumed(thereby causing the products of inertia Ixy, Ixz and Iyz to all go to zero); nonzeroproducts of inertia would cause there to be resultant moments about axes other thanezezez, which would mean that the object would ‘wobble’ on the way down.

Taking the time-derivative of HHH and applying the product rule,


Hsys/cHsys/cHsys/c =ddt(Izz/cωz e′ze′ze′z)

Hsys/cHsys/cHsys/c = ωzddt(Izz/c) e′ze′ze′z + Izz/cωz e′ze′ze′z + Izz/cωz e′ze′ze′z

However, since the axes are stuck to the body, the moment of inertia will not changeand

ddt

Izz/c = 0

Also, the unit vector e′ze′ze′z always points into the page, so e′ze′ze′z = 0 as well. Therefore,

Hsys/cHsys/cHsys/c = Izz/cωz e′ze′ze′z

So, substituting this result back into the angular momentum equation,

∑Mext/cMext/cMext/c = Hsys/cHsys/cHsys/c

−RFf ezezez = Izz/cωz ezezez

ωz = −RFf

Izz/c

The sign is negative, again, because the object will be accelerating rolling down theplane (or, as we had expected, ωz < 0). However, we still don’t know anything aboutthe magnitude of the friction. We do have two equations, though, with the frictionin it: the linear momentum equation, which yielded xc as a function of Ff , and theangular momentum equation, which yielded ωz as a function of Ff . Since xc, ωz andFf are all unknowns, we have three unknowns and two equations. We could solvefor everything if we only had one more equation with these three variables. For this,we look at the definition of pure rolling motion:

vp/ovp/ovp/o = 0

where we recall that p is the instantaneous point of contact between the object andthe surface. However, from relative motion, if we stick our moving axes onto thebody with the origin o′ at the centre of mass c,


We already have shown that, by virtue of the pure rolling motion, vp/ovp/ovp/o = 0. Further-more, we notice that vo′/ovo′/ovo′/o is the velocity of the centre of mass of the object, whichwe already know to be vcvcvc = xc exexex. Also, since the point p and the moving origin o′are both stuck to the same object and the object is rigid, they cannot move relativeto each other and so vp/o′vp/o′vp/o′ = 0. Substituting, then,

5.7 Examples 335

0 = vcvcvc +0+ωz e′ze′ze′z ×−R e′ye′ye′yvcvcvc = Rωz(e

′ze′ze′z ×e′ye′ye′y)

xc exexex = −Rωze′xe′xe′x

xc = −Rωz

where we have again remembered that, at our instant of interest, e′xe′xe′x = exexex. This resultis encouraging; since we know that the magnitude ωz will be negative, the quan-tity xc will therefore be positive, and the object isn’t rolling uphill. To get anotherexpression relating xc to ωz, we can take the time derivative of this expression.

ddt

xc = − ddt

(Rωz)

xc = −Rωz

It is important to note here that we have applied the relative velocity equation anddifferentiated in order to obtain the acceleration. While the relative accelerationequation may have been applied to yield xc directly, the only information providedby the condition of pure rolling is that the point of contact is instantaneously at rest.No information about ap/oap/oap/o is implied by the condition of pure rolling, so the relativeacceleration equation could not be solved.

We now have three equations and three unknowns in xc, ωz and Ff :

xc = gsin(θ)− 1m

Ff

ωz = −RFf

Izz/c

xc = −Rωz

Let us first eliminate friction from the linear momentum equation:

Ff = −ωzIzz/c

R

xc = gsin(θ)− 1m

(−ωz

Izz/c

R

)

xc = gsin(θ)+ ωzIzz/c

mR


Now we can eliminate the angular acceleration:

ωz = − xc

R

xc = gsin(θ)+

(− xc

R

)Izz/c

mR

xc

(1+

Izz/c

mR2

)= gsin(θ)

xc =gsin(θ)(1+

Izz/c

mR2

)

The angular acceleration was then −xc/R, or

ωz =−gsin(θ)

R(1+

Izz/c

mR2

)As expected, the angular velocity turned out to be negative, and the object is indeedrolling downhill. We could also determine the velocity of the block as a function oftime by integrating this expression with respect to time, as

∫ωz dt = ωz =

∫ −gsin(θ)

R(1+

Izz/c

mR2

) dt

ωz =−gsin(θ)

R(1+

Izz/c

mR2

) t +C

where we recognize that none of the terms in the expression for ωz were functionsof time, and C is the constant of integration. If we say that the object was releasedfrom rest so that ωz = 0 at t = 0, then

0 =−gsin(θ)

R(1+

Izz/c

mR2

) (0)+C

C = 0

and so the angular velocity is given by the expression

ωz =−gsin(θ)

R(1+

Izz/c

mR2

) t

5.7 Examples 337

and the direction is, of course, along ezezez. Alternatively, we could solve for the linearvelocity of the centre of mass vc as a function of the distance L traveled in the exexex

direction (down the slope). The acceleration of the centre of mass is given by,

xc = −Rωz

= −R

(−gsin(θ)

R(1+

Izz/c

mR2

))

xc =gsin(θ)(1+

Izz/c

mR2

)

Next, we recognize that

xc =dvc

dt

=dvc

dxc

dxc

dt

xc =dvc

dxcvc

Substituting this result into the earlier expression,

xc =gsin(θ)(1+

Izz/c

mR2

)dvc

dxcvc =

gsin(θ)(1+

Izz/c

mR2

)vc dvc =

gsin(θ)(1+

Izz/c

mR2

) dxc

where we have separated our variables. Finally, we can integrate both sides, recog-nizing that no terms on the right-hand-side of the expression are functions of xc.

∫vc dvc =

∫gsin(θ)(1+

Izz/c

mR2

) dxc

12

v2c =

gsin(θ)(1+

Izz/c

mR2

) xc +C


vc =

√√√√ 2gsin(θ)(1+

Izz/c

mR2

) xc +2C

where C again is our constant of integration. If the object started off at rest, thenvc = 0 at x = 0 and

0 =

√√√√ 2gsin(θ)(1+

Izz/c

mR2

) (0)+2C

C = 0

So the velocity of the object as a function of the displacement of the centre of massxc is

vc =

√√√√ 2gsin(θ)(1+

Izz/c

mR2

) xc

and when the object has rolled a distance L down the incline from its starting posi-tion, the velocity of the centre of mass will be,

vc =

√√√√2gLsin(θ)(1+

Izz/c

mR2

)in the +exexex direction (downhill).

In analyzing this problem, we elected to stick our moving origin o′ to the centreof mass of the object; this was an entirely arbitrary decision. In our earlier derivationof angular momentum, we found that the definition of angular momentum was thesame if the origin was either a centre of mass or a stationary point. With equal ease,then, we could have solved the problem with the moving origin stuck to the objecton a stationary point, if one could be found. Because the object is undergoing purerolling motion, we know that the instantaneous point of contact p must be stationary.As an exercise, then, let us solve the problem again, this time with our moving origino′ stuck to the object at the point p. Again, we will assume that the stationary andmoving unit vectors all align with each other at the instant of interest.

Since the point p is a stationary point,

∑Mext/pMext/pMext/p = Hsys/pHsys/pHsys/p =ddt(IpωpIpωpIpωp)

taking moments about p, since this is the point of application of both the normalforce and the friction force, rfrfrf = rnrnrn = 0.

5.7 Examples 339

∑Mext/oMext/oMext/o = rg/prg/prg/p ×FgFgFg

=(R e′ye′ye′y)× (Fg sin(θ) exexex −Fg cos(θ) eyeyey

)= RFg sin(θ)

(e′ye′ye′y ×e′xe′xe′x

)−RFg cos(θ)(e′ye′ye′y ×e′ye′ye′y

)= −RFg sin(θ) e′ze′ze′z −RFg cos(θ)

(0)

∑Mext/oMext/oMext/o = −RFg sin(θ) e′ze′ze′z

where again we have recalled that e′xe′xe′x = exexex and e′ye′ye′y = eyeyey at our instant of interest. Fromthe parallel axis theorem,

Izz/o = Izz/c +mR2

The angular velocity vector can be moved around without changing it, so the angularvelocity about the point p will be the same as the angular velocity about this pointo. This could also be demonstrated from relative motion, since

vc/ovc/ovc/o = vo′/ovo′/ovo′/o +vc/o′vc/o′vc/o′ +ωaxωaxωax ×rc/o′rc/o′rc/o′

where vc/ovc/ovc/o = vcvcvc, the velocity of the centre of mass of the object; vo′/ovo′/ovo′/o = 0 since themoving origin is on a stationary point, and again vc/o′vc/o′vc/o′ = 0 since both c and o′ arestuck to the same rigid object. Substituting,

vcvcvc = 0+0+ωaxωaxωax ×R e′ye′ye′y−Rωz exexex = Rωax(ezezez ×eyeyey)

−Rωz exexex = −Rωax exexex

ωax = ωz

as we had anticipated. Finally, substituting these results into the angular momentumequation,

∑Mext/pMext/pMext/p =ddt(IpωpIpωpIpωp)

−RFg sin(θ)ezezez =ddt

((Izz/c +mR2)(ωz)ezezez

)−mgRsin(θ) = (Izz/c +mR2)ωz

ωz =−mgRsin(θ)Izz/c +mR2


ωz =−gsin(θ)

R+Izz/cmR

ωz =−gsin(θ)

R(1+

Izz/c

mR2

)

Which, thankfully, is the same result obtained earlier. If the object had been a cylin-der, then Izz/c = mR2/2, and this result would reduce to

ωz =−gsin(θ)

R(1+ 1

2mR2

mR2

)= −2

3gR

sin(θ)

So, for the case of a cylindrical object rolling down a hill under the action of gravity,

x =−ωzR =23

gsin(θ)

This indicates that the acceleration depends only on the angle of the ramp- and noton either the mass or the radius of the disc.

In general, for the analysis of systems in which a stationary point may be identi-fied, there will always be these two approaches available. If you select the centre ofmass as your origin, you will need to solve the linear momentum and angular mo-mentum equations simultaneously, with an additional constraint linking the linearand rotational motion (in our example, this constraint was that xc = −Rωz). If youselect the stationary point as your origin, you will not need to use the linear momen-tum equation, but you will need to translate the inertia tensor from being about thecentre of mass to being about the location of the stationary point, using the parallelaxis theorem.

If no stationary point can be identified, then there is no choice but to put theorigin on the centre of mass. If you select any point other than a stationary point ora centre of mass as your origin, then HpHpHp �= IpωIpωIpω and you will need to consider theorbital angular momentum as well. This, however, is seldom necessary (since allobjects must have a centre of mass).

5.7.3 Pulley with multiple masses and additional forces

An object of mass m1 is suspended from an ideal, inextensible string. The stringwinds without slipping over a solid cylindrical pulley of mass m2 and radius R, andthen attaches to an object of mass m3 which rests on a surface inclined at an angle θ

5.7 Examples 341

Fig. 5.18 Multiple blocks attached to a pulley with mass.

relative to horizontal, as illustrated in Figure 5.18. The coefficient of kinetic frictionbetween the third object and the incline is μk. Let us analyze this system and solvefor the accelerations.

First, we will assume that the objects are moving in such a way as to lower thesuspended mass; if it turns out that this could not happen (if, for example, m3 wastoo large), then we will end up with an impossible result.

Next, we can recognize that the object of mass m1 can only move vertically andthe object of mass m3 can only move up and down the incline. Therefore, it wouldbe convenient to define unit vectors ex1ex1ex1 and ex3ex3ex3 which are aligned with the directionof travel of the objects of mass m1 and m3, respectively. Since the two objects aredirectly coupled, then, the scalar magnitude of the acceleration of the object of massm1 must be equal to the scalar magnitude of the acceleration of the object of massm3, or x1 = x3. We will also define the unit vectors ez3ez3ez3 as pointing into the page.

We begin our analysis by isolating the first object, defining our system as the firstobject (of mass m1) only. From the law of conservation of linear momentum,

∑FextFextFext = GsysGsysGsys = m1x1 ex1ex1ex1

Drawing a free-body diagram of this system (as illustrated in Figure 5.19),

∑FextFextFext = m1x1 ex1ex1ex1

Fg1Fg1Fg1 +FT1FT1FT1 = m1x1 ex1ex1ex1

m1g ex1ex1ex1 −FT 1 ex1ex1ex1 = m1x1 ex1ex1ex1


Fig. 5.19 Multiple blocks attached to a pulley with mass- free body diagrams.

We can now divide out the unit vector and work exclusively with scalar quantities.Also, since the object m1 ‘communicates’ with the pulley via FT 1, it will simplifythe maths if we isolate this quantity for future use.

m1g−FT 1 = m1x1

FT 1 = m1g−m1x1

Next, we examine the pulley. The pulley is being acted upon by the tension in thestring, gravity, and a reaction force FRFRFR (of unknown magnitude and direction) beingapplied to the centre of the pulley by the pulley support. A free-body diagram of thepulley has been sketched in Figure 5.19.

The most important thing to note here is that the tension in the two halves of thestring is not the same; that is, FT 1 �= FT 2. Since the string does not slip on the surfaceof the pulley, the string may be considered as ‘glued’ to the pulley- and is thereforeeffectively behaving as two separate strings. Alternatively, if FT 1 were equal to FT 2,since both are acting tangentially but in opposite directions upon the pulley, therewould be no net imbalance in moments and the pulley would never turn.

So, we will re-define our system to contain only the pulley. We will also definea new coordinate system having unit vectors ex2ex2ex2 pointing upward, ey2ey2ey2 pointing tothe right, and ez2ez2ez2 pointing into the page, as illustrated. Note that ez2ez2ez2 = −ez3ez3ez3. Then,applying the law of conservation of linear momentum to the new system,


FT1FT1FT1 +FT2FT2FT2 +Fg2Fg2Fg2 +FRFRFR = m2a2a2a2

5.7 Examples 343

However, the centre of mass of the pulley is constrained to remain in the same placeby the frame to which it is attached; therefore, a2a2a2 = 0 and

FT1FT1FT1 +FT2FT2FT2 +Fg2Fg2Fg2 +FRFRFR = 0

This expression will enable us to determine the reaction force at the pin (which weshall do later). However, the linear momentum equation does not yield any usefulinformation about the motion. Instead, we must consider the law of conservation ofangular momentum for the pulley.

∑Mext/oMext/oMext/o = HoHoHo

The pulley is described as a solid cylinder, and it rotates about its centre of mass cexclusively in the ez2ez2ez2 direction. If we define the unit vectors e′x2e′x2e′x2, e′y2e′y2e′y2 and e′z2e′z2e′z2 as beingstuck to the disc with the origin at c (and parallel to ex2ex2ex2, ey2ey2ey2 and ez2ez2ez2, respectively, atthe instant of interest),


⎡⎣ 1

4 m2R2 + 112 m2H2 0 0

0 14 m2R2 + 1

12 m2H2 00 0 1

2 m2R2

⎤⎦⎛⎝ 0

0ωz2

⎞⎠

HcHcHc =12

m2R2ωz2 e′z2e′z2e′z2

where ωz2 e′z2e′z2e′z2 is the instantaneous angular velocity vector of the pulley. Then, sincee′z2e′z2e′z2 never changes its direction (it always points into the page),

∑Mext/cMext/cMext/c = HcHcHc

FT 1R e′z2e′z2e′z2 −FT 2R e′z2e′z2e′z2 =ddt

(12


)

FT 1R e′z2e′z2e′z2 −FT 2R e′z2e′z2e′z2 =12


Next, we notice that since the pulley is ‘rolling’ on the string,

x1 = Rωz2

Then, substituting this into the previous result and dividing out the unit vector,

FT 1R−FT 2R =12

m2R2(

x1

R

)


FT 1 −FT 2 =12

m2x1

However, we can substitute into this expression our earlier result for FT 1, so

(m1g−m1x1

)−FT 2 =12

m2x1

Again, the pulley can only ‘communicate’ with the object on the incline by meansof T2; it therefore seems reasonable to isolate this quantity for future use.

m1g−m1x1 −FT 2 =12

m2x1

FT 2 = m1g−m1x1 − 12

m2x1

Finally, we consider the object on the incline. Let us again redefine our systemas consisting of the object m3 only. Then, from the law of conservation of linearmomentum,

∑FextFextFext = GsysGsysGsys = m3x3 ex3ex3ex3

and, from the free-body diagram,

∑FextFextFext = FT2FT2FT2 +FnFnFn +Fg3Fg3Fg3 +FfFfFf

∑FextFextFext = FT 2 ex3ex3ex3 +Fn ey3ey3ey3 −Fg3 sin(θ) ex3ex3ex3 −Fg3 cos(θ) ey3ey3ey3 −μkFn ex3ex3ex3

Substituting,

FT 2 ex3ex3ex3 +Fn ey3ey3ey3 −Fg3 sin(θ) ex3ex3ex3 −Fg3 cos(θ) ey3ey3ey3 −μkFn ex3ex3ex3 = m3x3 ex3ex3ex3

Equating the scalar coefficients of ey3ey3ey3,

Fn ey3ey3ey3 −Fg3 cos(θ) ey3ey3ey3 = 0 ey3ey3ey3

Fn = m3gcos(θ)

Then, equating the scalar coefficients of ex3ex3ex3,

FT 2 ex3ex3ex3 −Fg3 sin(θ) ex3ex3ex3 −μkFn ex3ex3ex3 = m3x3 ex3ex3ex3

5.7 Examples 345

However, we already have expressions for FT 2 (from the analysis of the pulley) andFn (from the coefficients of ey3ey3ey3); also, we have selected out coordinate systems suchthat x1 = x3. Substituting,

FT 2 −Fg3 sin(θ)−μkFn = m3x3

m1g−m1x1 − 12

m2x1 −m3gsin(θ)−μkm3gcos(θ) = m3x1

We may now isolate and solve for the acceleration x1, as

(m3 +m1 +

12

m2)x1 = m1g−m3gsin(θ)−μkm3gcos(θ)

x1 =

m1m3

− (sin(θ)+μk cos(θ))

1+ m1m3

+ 12

m2m3

g

According to our definition, x1 must be positive. Otherwise, the object m1 wouldbe raised rather than lowered, and the direction in which we applied the frictionalforce upon object m3 would be incorrect. It is clear from the above that x1 would bepositive if and only if

m1

m3> sin(θ)+μk cos(θ)

If this inequality is not satisfied, then the analysis would have to be repeated with thefrictional force applied in the opposite direction (in this case, it would be sufficientto simply reverse the sign of μk). So, if the object m1 is being raised,

m1

m3< sin(θ)+μk cos(θ)

and

x1 =

m1m3

− (sin(θ)−μk cos(θ))

1+ m1m3

+ 12

m2m3

g

Finally, in the special case where

m1

m3= sin(θ)+μk cos(θ)

the object is not accelerating at all, as x1 = 0. It should be noted, however, that thisresult is only valid if x1 �= 0; otherwise, the friction would become static, Ff �= μkFn

and the problem would be fundamentally different.


Let us now return to the reaction force acting on the pulley, FR. From the linearmomentum equation,

0 = FT1FT1FT1 +FT2FT2FT2 +Fg2Fg2Fg2 +FRFRFR

0 = −FT 1 ex2ex2ex2 −FT 2 ex3ex3ex3 −m2g ex2ex2ex2 +FRFRFR

FRFRFR = FT 1 ex2ex2ex2 +FT 2 ex3ex3ex3 +m2g ex2ex2ex2

where it is important to remember that our system was defined as containing onlythe pulley, and all of the forces are taken as they are acting upon the pulley. However,coefficients of the unit vector ex2ex2ex2 cannot be added to coefficients of the unit vectorex3ex3ex3, as they are not parallel. The unit vector ex3ex3ex3 must therefore be decomposed intocomponents parallel to ex2ex2ex2 and ey2ey2ey2. From geometry,

ex3ex3ex3 = cos(θ) ey2ey2ey2 + sin(θ) ex2ex2ex2

Substituting,

FRFRFR = FT 1 ex2ex2ex2 +FT 2(cos(θ) ey2ey2ey2 + sin(θ) ex2ex2ex2

)+m2g ex2ex2ex2

FRx ex2ex2ex2 +FRy ey2ey2ey2 =(FT 1 +FT 2 sin(θ)+m2g

)ex2ex2ex2 +FT 2 cos(θ) ey2ey2ey2

where we have decomposed FRFRFR into components along ex2ex2ex2 and ey2ey2ey2. We will nowexamine the scalar coefficients of ex2ex2ex2; recall that we have already determined valuesfor FT 1 and FT 2 in terms of x1.

FRx ex2ex2ex2 =

((m1g−m1x1)+

(m1g−m1x1 − 1

2m2x1

)sin(θ)+m2g

)ex2ex2ex2

FRx = m1g−m1x1 +(m1g−m1x1 − 1

2m2x1

)sin(θ)+m2g

FRx = m1(1+ sin(θ)

)(g− x1)+m2

(g− 1

2x1 sin(θ)

)

where the expression for x1 may be substituted into the above if desired. Similarly,examining the scalar coefficients of ey2ey2ey2,

FRy ey2ey2ey2 = FT 2 cos(θ) ey2ey2ey2

FRy =(m1g−m1x1 − 1

2m2x1

)cos(θ)

5.7 Examples 347

where again the scalar acceleration x1 could be substituted into this expression.

5.7.4 Distributed forces and moments

It is occasionally necessary to analyze systems having distributed loads or pres-sures, rather than the discrete point loads considered so far. In these cases, the lawof conservation of linear momentum and the law of conservation of angular momen-tum may still be applied, but the net external forces and moments will need to beintegrated rather than summed.

Consider the case of a sluice gate of height H, width W and mass m. The gate ispinned at the centre of its sides, and is used to hold a reservoir of water also of depthH, as illustrated in Figure 5.20. At some instant in time, the gate is released andallowed to rotate freely about the pins. In order to design the pins, it is necessary tobe able to predict the load in the pins and the angular acceleration of the sluice gateat the instant it is released.

Fig. 5.20 Distributed forces and moments: a sluice gate.

We shall define our system as consisting of the sluice gate and pins only. Let usdefine the fixed coordinate system o which has its origin on the centre of mass of thegate. The unit vector eyeyey shall point upwards, exexex shall point to the right, and ezezez shallpoint out of the plane of the figure. To determine the load in the pins at the instantshown, we begin with the law of conservation of linear momentum.



Drawing the free-body diagram of the sluice gate, we find

∑FextFextFext = FxFxFx +FyFyFy +FPFPFP +FgFgFg

where FxFxFx and FyFyFy are the horizontal and vertical components of the reaction forceat the pins, and FPFPFP is the net resultant force acting upon the gate as a result of theapplied pressure. From Section 4.2.2, we know that the pressure P acting on animmersed object is given by the expression,

P = ρgh

where ρ is the density of the fluid and h is the depth to which the object is immersed.In order to solve the problem, it would be useful to express the pressure as a functionof the vertical distance y from the centre of mass of the gate. To do this, we recognizethat an object at a depth h in the liquid will be at a position y = H/2−h relative too. The pressure may then be expressed as

P = ρg

(H2− y

)

If we consider an infinitesimally thin strip of the gate at some location y relative tothe origin, this strip will have an exposed area dA = W dy. The differential forceelement acting on that strip as a result of the applied pressure will then have adifferential magnitude dFP, given by the expression

dFP = P dA

dFP = ρg

(H2− y

)(W dy)

To obtain the magnitude of the total force acting on the gate, then, we need onlyintegrate our expression for dFP over the ranges of y bounded by the gate, as

FP =∫ H/2

−H/2dFP

=∫ H/2

−H/2ρg

(H2− y

)(W dy)

= ρgW∫ H/2

−H/2

(H2− y

)dy

= ρgW

(H2

y− 12

y2)∣∣∣∣

H/2

−H/2

5.7 Examples 349

= ρgW

(H2

4− 1

2H2

4+

H2

4+

12

H2

4

)

FP =12

ρgWH2

Substituting this result into our earlier expression for the net external force,

∑FextFextFext = FxFxFx +FyFyFy +FPFPFP +FgFgFg

∑FextFextFext = FxFxFx +FyFyFy +12

ρgWH2 exexex −mg eyeyey

Then, substituting this expression into the linear momentum equation,

∑FextFextFext = macacac

FxFxFx +FyFyFy +12

ρgWH2 exexex −mgeyeyey = 0

where we have recognized that acacac = 0 because the centre of mass of the systemremains stationary on the axis of the pins. Equating the scalar coefficients of eyeyey, itbecomes immediately clear that Fy = mg; from the scalar coefficients of exexex, we findthat the only load on the pins at the moment the gate is released is then

FxFxFx =−12

ρgWH2 exexex

Next, to determine the angular acceleration of the gate at the instant illustrated,we begin with the angular momentum equation,

∑Mext/cMext/cMext/c = HcHcHc =ddt

IcIcIcωcωcωc

where ωcωcωc is the angular velocity vector of the gate as it rotates about its pin. Ex-amining the free-body diagram (Figure 5.20), we notice that the only force whichis not acting directly through the centre of mass is the distributed force applied as aresult of the hydrostatic pressure. Therefore, the pressure force is the only externalapplied force which may be able to produce a moment about the centre of mass.Considering again an infinitesimally thin strip of the gate at a location y relative tothe origin, the small moment dMcdMcdMc produced about o will be given by

dMcdMcdMc = rPrPrP ×dFPdFPdFP

dMcdMcdMc = y eyeyey ×P dA exexex


where we have recognized that the force applied as a consequence of the hydrostaticpressure would be acting in the positive exexex direction. Substituting our earlier resultsfor P and dA and evaluating the cross-product,

dMcdMcdMc = y eyeyey ×ρg

(H2− y

)(W dy) exexex

= ρgWy

(H2− y

)(eyeyey ×exexex

)dy

dMcdMcdMc = −ρgW

(H2

y− y2)

ezezez dy

(5.4)

The net external moment may then be obtained by integrating over the domain in yoccupied by the gate, as

∑Mext/cMext/cMext/c =∫ H/2

−H/2dMcdMcdMc

= −∫ H/2

−H/2ρgW

(H2

y− y2)

ezezez dy

= −ρgW

(H4

y2 − 13

y3)∣∣∣∣

H/2

−H/2ezezez

= −ρgW

(H3

16− 1

3H3

8− H3

16− 1

3H3

8

)ezezez

∑Mext/cMext/cMext/c =1

12ρgWH3 ezezez

Next, we need to evaluate the angular momentum of the gate at the instant of interest.If we define a moving coordinate system o′ such that the moving unit vectors e′xe′xe′x, e′ye′ye′yand e′ze′ze′z are stuck to the gate but are also parallel, respectively, to exexex, eyeyey and ezezez at theinstant of interest, then

IcωcIcωcIcωc =

⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz


where ωz is the scalar magnitude of the angular velocity of the gate at the instant ofinterest. Substituting this result and our expression for the net external moment intothe angular momentum equation,

5.7 Examples 351


112

ρgWH3 ezezez =ddt

Izz/cωz e′ze′ze′z1

12ρgWH3 ezezez = Izz/cα e′ze′ze′z

where α = ωz, and we have recognized that the direction in which e′ze′ze′z points neverchanges and is therefore constant. The moment of inertia Izz/c is also constant, asthe axes are stuck to the object, and so the rotation of the object does not change itsorientation or position relative to the o′ coordinate system. If we consider the gateas a rectangular plate, then

Izz/c =112

mH2

(the moments of inertia for a rectangular prism are tabulated in Appendix C; notethat for a very thin plate, L → 0). Substituting,

112

ρgWH3 ezezez = Izz/cα e′ze′ze′z

112

ρgWH3 ezezez =

(1

12mH2

)α e′ze′ze′z

α = ρgWH

m

where we note that, at the instant of interest, e′ze′ze′z = ezezez. We may further recognizem/WH as being the area density λ of the gate; the angular acceleration may there-fore be expressed simply as,

α =ρ

λg

which, interestingly, is entirely independent of the dimensions of the gate.

5.7.5 Rolling and slipping: a comparison

Having already considered kinetic friction in Chapter 4 and pure rolling motion inthis chapter, it would be useful to compare the two. Let us examine the case of around object of radius R, mass m and uniform density, on a surface inclined at anangle θ as illustrated in Figure 5.21. The object may either be placed on its edgeand allowed to roll down the surface, or it can be placed on its side and allowed toslip.


Fig. 5.21 Comparison between rolling and slipping.

To begin, let us consider the case where the object rolls down the incline. We shalldefine our system as the round object only, and orient our coordinate system suchthat exexex points down the slope and eyeyey points normally upward from the slope, asillustrated. Then, ezezez points out of the page. From the free-body diagram (see Figure5.22), then

Fig. 5.22 Free-body diagrams for gravity-driven rolling and slipping.



)−Ff exexex +Fn eyeyey

Collecting together the scalar coefficients, this may equally be expressed as,

∑FextFextFext =(Fg sin(θ)−Ff

)exexex +

(Fn −Fg cos(θ)

)eyeyey

5.7 Examples 353

This result may now be substituted into the law of conservation of linear momentum,so that

∑FextFextFext = GsysGsysGsys = macacac(Fg sin(θ)−Ff

)exexex +

(Fn −Fg cos(θ)

)eyeyey = m

(xc exexex + yc eyeyey

)

where acacac = xc exexex + yc eyeyey is the acceleration of the centre of mass c of the object. Be-cause the object is constrained to always be in contact with the surface, the verticalcomponent of the acceleration must vanish. Examining, then, the scalar coefficientsof eyeyey,


Fn −mgcos(θ) = 0

Fn = mgcos(θ)

Considering next the scalar coefficients of exexex,


mgsin(θ)−Ff = mxc

Ff = mgsin(θ)−mxc

Again, because pure rolling motion is driven by static friction, we have no furtherinformation about the magnitude of Ff . In order to eliminate this unknown, there-fore, it is necessary to apply the law of conservation of angular momentum. To beginwith, we can evaluate the sum of the external forces acting on the object about thecentre of mass; from the free-body diagram,

∑Mext/cMext/cMext/c =(rf/crf/crf/c ×FfFfFf

)+(rg/crg/crg/c ×FgFgFg

)+(rn/crn/crn/c ×FnFnFn

)where rf/crf/crf/c, rg/crg/crg/c and rn/crn/crn/c are position vectors pointing from the centre of mass of theobject to the points of application of FfFfFf, FgFgFg and FnFnFn, respectively. Then,

∑Mext/cMext/cMext/c =(−R eyeyey ×−Ff exexex

)+(0×FgFgFg

)+(−R eyeyey ×Fn eyeyey

)= RFf

(eyeyey ×exexex

)+(0)−RFn

(eyeyey ×eyeyey

)= −RFf ezezez +

(0)−RFn

(0)

∑Mext/cMext/cMext/c = −RFf ezezez


Alternatively, we may have immediately recognized that friction is the only forcenot acting through the centre of mass and therefore capable of exerting an externalmoment about the centre of mass. Substituting in our earlier result for Ff ,

∑Mext/cMext/cMext/c =−R(mgsin(θ)−mxc

)ezezez

Next, from the law of conservation of angular momentum,

∑Mext/cMext/cMext/c = Hsys/cHsys/cHsys/c =ddt

(IcIcIcωcωcωc

)where IcIcIc is the inertia tensor of the object evaluated about c in our coordinate system,and ωcωcωc is the angular velocity of the system. Let us now consider the argument ofthe derivative, as

IcωcIcωcIcωc =

⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz


where we have recognized here that, as the object rolls downhill (in the +exexex direc-tion), the angular velocity vector ωcωcωc = ωz e′ze′ze′z, where ωz is the scalar magnitude ofthe angular velocity and e′ze′ze′z is a unit vector stuck to the object and pointing in thesame direction as ezezez. The magnitude of the angular velocity here must necessarilyend up being negative, since the object will be rotating in a clockwise direction asit rolls downhill, and, according to the right-hand-rule, clockwise rotation would beindicated by a vector pointing into the page, or in the −e′ze′ze′z direction. It is also sig-nificant to note here that we have assumed the object to be symmetric about the x, yand z axes, so that the products of inertia Ixy, Izx and Iyz all go to zero.

Next, we can evaluate the time-derivative of Hsys/cHsys/cHsys/c, as

ddt

Hsys/cHsys/cHsys/c =ddt

(Izz/cωz e′ze′ze′z

)ddt

Hsys/cHsys/cHsys/c = Izz/cωz e′ze′ze′z + Izz/cωz e′ze′ze′z + Izz/cωz e′ze′ze′z

where we have applied the product rule. However, because the axes are stuck to theobject, then, by definition, Izz/c must be constant and Izz/c = 0. Also, because e′ze′ze′z isalways pointing out of the page, e′ze′ze′z = 0 as well. Then,

ddt

Hsys/cHsys/cHsys/c = Izz/cωz e′ze′ze′z

Substituting this expression and our earlier result for ΣMext/cMext/cMext/c into the angular mo-mentum equation,

5.7 Examples 355

∑Mext/cMext/cMext/c =ddt

(IcIcIcωcωcωc

)−R(mgsin(θ)−mxc

)ezezez = Izz/cωz e′ze′ze′z

−R(mgsin(θ)−mxc

)= Izz/cωz

−mRgsin(θ)+mRxc = Izz/cωz

where we have divided through by the unit vector, recognizing again that e′ze′ze′z = ezezez.Here we have a single expression with two unknowns: ωz and xc. However, onceagain we may express ωz in terms of xc using the relative motion equation. If wedefine the point p as the instantaneous point of contact between the object and theground (which must be at rest), and if we define a moving coordinate system o′ withits origin stuck to the centre of the object so that the moving unit vectors e′xe′xe′x, e′ye′ye′y ande′ze′ze′z are aligned with the fixed unit vectors exexex, eyeyey and ezezez, respectively, at the momentof interest, then


0 = xc exexex +0+ωz e′ze′ze′z ×−R e′ye′ye′y0 = xc exexex −ωzR


)0 = xc exexex +ωzR e′xe′xe′x

xc = −ωzR

where xc exexex is the velocity of the centre of mass of the object as it rolls down theincline, and we have divided through by the unit vector exexex (which is identical to theunit vector e′xe′xe′x at the instant of interest). Recall here that vp/o′vp/o′vp/o′ = 0 because the pointp and the point o′ are stuck to the same object, so that their relative positions cannotchange. Taking the time-derivative of both sides of this expression,

ddt

xc = − ddt

ωzR

xc = −ωzR

ωz = − xc

R

This expression may now be substituted into the angular momentum equation toyield,


−mRgsin(θ)+mRxc = Izz/cωz

−mRgsin(θ)+mRxc = Izz/c

(− xc

R

)−mR2gsin(θ)+mR2xc = −Izz/cxc

We may now isolate and solve for the acceleration of the object, as

xc(mR2 + Izz/c

)= mR2gsin(θ)

xc =gsin(θ)

1+Izz/c

mR2

Next, let us consider the case where the object is placed on its side and allowedto slip down the incline without rolling. Examining the free-body diagram in thiscase (see Figure 5.22),



)−Ff exexex +Fn eyeyey

Immediately, we may recognize that this is identical to the result obtained for thecase when the object was allowed to roll. Once again, combining the scalar coeffi-cients and substituting this result into the law of conservation of linear momentum,

∑FextFextFext = GsysGsysGsys = macacac(Fg sin(θ)−Ff

)exexex +

(Fn −Fg cos(θ)

)eyeyey = m

(xc exexex + yc eyeyey

)

Equating together the scalar coefficients of eyeyey,


Fn −mgcos(θ) = 0

Fn = mgcos(θ)

5.7 Examples 357

which is, once again, identical to the earlier result for the case when the object isallowed to roll. However, when the object is allowed to slip, the friction will bekinetic. Considering, then, the scalar coefficient of exexex,


mgsin(θ)−μkFn = mxc

Substituting in our earlier result for Fn,

mgsin(θ)−μk(mgcos(θ)

)= mxc

xc = g(sin(θ)−μk cos(θ)

)

It is important to note here that this expression permits the case of xc < 0, whichwould correspond physically to the object being pulled up the hill by the action offriction. Since friction may only oppose motion, the minimum possible value of theacceleration is xc = 0. In this case,

0 = g(sin(θ)−μk cos(θ)

)sin(θ)cos(θ)

= μk

θ = tan−1(μk)

So, for cases when θ ≤ tan−1(μk), the object will be held in place by static friction,and will not accelerate at all. We are, of course, assuming here that the coefficient ofstatic friction is equal to the coefficient of kinetic friction, though in general μs > μk,as discussed in Chapter 4.

We may now compare the accelerations obtained between the cases when theobject is allowed to roll and when the object is made to slip, as

xc =

⎧⎪⎪⎨⎪⎪⎩

gsin(θ)

1+Izz/cmR2

for rolling

g(sin(θ)−μk cos(θ)

)for slipping

Let us now consider the special case of θ = 0, so that the object is sitting on ahorizontal surface. In this case, then,


xc =

⎧⎨⎩

0 for rolling

−μkg for slipping

For the case of pure rolling, the acceleration goes to zero, as expected. For the caseof slipping, the acceleration has a negative value, which is clearly nonphysical; thisis, however, expected since for the case of θ = 0, clearly θ ≤ tan−1(μk) and theobject would remain stationary.

Next, let us consider the special case of θ = 90◦, so that the object is releasedadjacent to a vertical wall and essentially allowed to fall. Because of the way weformulated the solution (that is, requiring that xc =−Rωz), the object is still requiredto ‘roll’ down the wall despite the absence of any normal force. In this case, then,

xc =

⎧⎪⎨⎪⎩

g

1+Izz/cmR2

for rolling

g for slipping

Here, we recognize that an object released adjacent to a wall and allowed to ‘slip’down the surface will be subjected to no normal force, and therefore the friction ex-perienced by the object will be zero. Consequently, the object will simply fall freelywhile adjacent to the surface. When the object is constrained to roll, the accelerationof the object will be less than g, and the larger the moment of inertia of the object,the smaller its acceleration will be.

From these results, we see that for small θ , the rolling object will accelerate morequickly, while for large θ , the slipping object will accelerate more quickly. It standsto reason, then, at some critical angle θ0, the acceleration will be the same for bothrolling and slipping. Then,

xc =gsin(θ0)

1+Izz/c

mR2

= g(sin(θ0)−μk cos(θ0)

)1

1+Izz/c

mR2

= 1−μkcos(θ0)

sin(θ0)

cos(θ0)

sin(θ0)=

1μk

Izz/c

mR2

1+Izz/c

mR2

θ0 = tan−1

(μk

(mR2

Izz/c+1

))

These results are perhaps more clearly illustrated in Figure 5.23, in which xc isplotted against θ for both the cases of rolling and slipping, and the critical pointsdiscussed above are labeled.

5.7 Examples 359

Fig. 5.23 A plot of the acceleration of an object against surface inclination for the cases of rollingand slipping.

5.7.6 Combined rolling and slipping: backspin

Let us now consider the case of a billiard ball with mass m, radius R and inertiatensor III shot across a felt surface with coefficient of kinetic friction μk, as illustratedin Figure 5.24. The ball is given an initial linear velocity vo and ‘backspin’ angularvelocity ωo. This is not a case of pure rolling motion, the instantaneous point ofcontact is not at rest, and the frictional force is due to kinetic friction.

Fig. 5.24 Illustration of a round object simultaneously rolling and slipping.

So, we will define our system as consisting of the ball alone, and stick a coordi-nate system to the centre of the ball so that, at our moment of interest, exexex points in


the direction of motion and eyeyey points straight up. The unit vector ezezez, then, points outof the page in the illustration. In this case, no stationary point can be identified, sowe have no choice but to use the centre of mass as our origin. Drawing a free-bodydiagram of the ball, then, we see that


∑FextFextFext = −Ff exexex +(Fn −Fg)eyeyey

Notice that the friction is acting in the direction opposite to the motion of the instan-taneous point of contact: the ball is moving to the right and spinning anti-clockwise,and friction will always act to oppose the relative motion of two slipping surfaces.If the ball had been spinning clockwise (over-spin), friction would have been inthe +exexex direction. We can now substitute the vector sum of the forces into the lawconservation of linear momentum,


−Ff exexex +(Fn −Fg)eyeyey = mxc exexex +myc eyeyey

where x and y are the components of acceleration of the centre of mass along exexex

and eyeyey, respectively. We will first look at the scalar coefficients of eyeyey. We know thatyc = 0 because the centre of mass of the ball can’t accelerate in a vertical direction.

Fn −Fg = myc

Fn −mg = 0

Fn = mg

Now, we can look at the scalar coefficients of exexex.

−Ff = mxc

−μkFn = mxc

−μk(mg) = mxc

xc = −μkg

So now we know the acceleration of the centre of mass of the ball. However, sincethe friction is slowing the slipping speed of the instantaneous point of contact, it will

5.7 Examples 361

eventually stop slipping; when that happens, the friction will no longer be kinetic,the motion will become pure rolling, and everything changes. We need to know,then, when this is going to occur. Let us now consider the instant when pure rollingmotion just begins to happen. If we call the instantaneous point of contact betweenthe ball and the surface p, then at the moment that pure rolling begins,


We recognize that vp/ovp/ovp/o = 0, since the instantaneous point of contact is at rest inpure rolling motion. Also, vo′/ovo′/ovo′/o = xc exexex, the linear velocity of the centre of mass ofthe ball. Since both p and the origin o′ are stuck to the same rigid ball, vp/o′vp/o′vp/o′ = 0.Substituting,

0 = xc exexex +0+(ωz e′ze′ze′z

)× (−R e′ye′ye′y)

−xc exexex = −Rωz(e′ze′ze′z ×e′ye′ye′y

)xc exexex = Rωz

(−e′xe′xe′x)

xc = −Rωz

The negative sign is not a worry, since ωz will be negative and xc will be positive, asexpected. So, we need to solve for xc as a function of time, solve for ω as a functionof time, set xc =−Rω and then solve for the time at which this happens. We begin,then, by solving for xc.

xc = −μkg∫xc dt = −

∫μkg dt

xc = −μkgt +C1

where C1 is some constant of integration. However, we know that when t = 0, xc =vo, so

vo = −μkg(0)+C1

C1 = vo

so the velocity of the centre of mass of the ball is given by the expression,

xc =−μkgt + vo


For the angular velocity, we need to start with the external moments. With the systemand axes defined as before,

∑Mext/cMext/cMext/c =−Ff Rezezez

Since our origin is stuck to the centre of mass of the object,

HHH = IcωcIcωcIcωc =

⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz


Substituting these results into the law of conservation of angular momentum,

∑Mext/cMext/cMext/c = HcHcHc =ddt(IcωIcωIcω)

−Ff R ezezez =ddt


)−μkmgR ezezez = Izz/cωz ezezez

ωz = −μkgmRIzz/c

where it has been recognized that e′ze′ze′z = ezezez, and does not change with time. Now tosolve for the angular velocity as a function of time,

∫ωz dt = −

∫μkg

mRIzz/c

dt

ωz = −μkgmRIzz/c

t +C2

Where again, C2 is some constant of integration. Since we know that the angularvelocity is ωo (which is positive since it is in the anticlockwise direction) at timet = 0, we can solve for C2.

ωo = −μkgmRIzz/c

(0)+C2

C2 = ωo

So the angular velocity of the ball will be given by the expression,

ωz =−μkgmRIzz/c

t +ωo

5.7 Examples 363

Now we can solve for the instant in time tp when x =−Rωz and pure rolling motionjust begins to occur.

x = −Rωz

−μkgtp + vo = −R(−μkg

mRIzz/c

tp +ωo)

Isolating and solving for tp,

μkgtp +μkgmR2

Izz/ctp = Rωo + vo

tp =Rωo + vo

μkg(

mR2

Izz/c+1)

So the velocity at that instant is

xc = −μkgt + vo

= −μkg

(Rωo + vo

μkg(

mR2

Izz/c+1))+ vo

= −Rωo + vo

mR2

Izz/c+1

+ vo

=−Rωo − vo + vo

mR2

Izz/c+ vo

mR2

Izz/c+1

xc =vo

mR2

Izz/c−Rωo

mR2

Izz/c+1

It is interesting to note that this result does not depend on the coefficient of kineticfriction; the speed would be the same on ice as it would be on sandpaper.

Once the ball has achieved pure rolling motion, the instantaneous point of contactbetween the ball and the felt will be at rest, and the friction will instantaneouslyswitch from being kinetic to being static. Static friction will provide whatever forceis required to ensure that the instantaneous point of contact o remains at rest, orto ensure that x = −Rωz. However, at the instant when pure rolling motion begins,the condition x =−Rωz has already been met; therefore, the static friction does notneed to apply any force at all in order for the object to purely roll and Ff = 0.


Since the static friction force is the only one acting in the direction of motionand this force is zero once pure rolling motion is achieved, there are no net forcesor moments acting to change the speed of the ball. Consequently, the velocity willremain constant after that time. The velocities and frictional force are sketched as afunction of time in Figure 5.25.

Fig. 5.25 Sketch of the velocity and spin speed for a ball which was given an initial forwardvelocity and backspin.

Let us now see under what conditions the ball would come to a dead stop, so thatx = 0 just as pure rolling motion is about to begin. Then,

xc = 0 =vo

mR2

Izz/c−Rωo

mR2

Izz/c+1

0 = vomR2

Izz/c−Rωo

Rωo

vo=

mR2

Izz/c

5.7 Examples 365

so the critical backspin velocity ratio Rωo/vo for which the ball will come to restis mR2/Izz/c. For a sphere, Izz/c = 2mR2/5, so the critical velocity ratio for a realbilliard ball would be

Rωo

vo=

52

Surprisingly, this result is independent of the mass of the ball, the radius of the balland even the frictional coefficient of the felt.

5.7.7 Simple clutch

A simple clutch mechanism consists of two flat discs which engage through theaction of friction. Consider the two solid, uniform discs A and B. Disc B has massmb and radius Rb, and rests on a frictionless surface. Disc A has mass ma and radiusRa (with Ra ≤ Rb) and is suspended directly above disc B. Discs A and B have initialangular velocities ωai and ωbi, respectively. At some instant, disc A is allowed todrop onto disc B, and if ωai �= ωbi, kinetic friction will act between the discs. Thecoefficient of kinetic friction between the discs is μk.

Fig. 5.26 A simple clutch

We recognize that when the two discs first come into contact, they will slip; onedisc will accelerate while the other decelerates, until the angular velocity of the twodiscs is the same and they move as a single body. At that moment, friction ceases toact as the two surfaces will no longer be slipping. Let us define a fixed coordinatesystem such that exexex is to the right and eyeyey is up, as illustrated. To begin with, weshall define our system as consisting of the disc A alone. Then, applying the linearmomentum equation to our system,


Fig. 5.27 Free-body diagrams of the discs.


FnaFnaFna +FgaFgaFga = maaaaaaa

Fna eyeyey −Fga eyeyey = ma(xa exexex + za ezezez

)

where aaaaaa is the acceleration vector of the centre of mass of the disc A. We have notincluded the frictional force, because, for reasons of symmetry, it can exert no netforce. It is clear that the disc cannot move vertically, so it can only accelerate in theexexex and/or ezezez directions. By equating the scalar coefficients of the unit vectors, wecan see that

Fna = Fga = mag

xa = 0

za = 0

since there are no exexex or eyeyey terms on the left-hand side of the expression, and thereare no exexex terms on the right-hand side. From this result, we observe that the centreof mass of the disc will not move; the disc will only spin in place.

Next, we consider the angular momentum equation. Taking moments about thecentre of mass of the disc,


Mf eyeyey = HcHcHc

Evaluating the angular momentum of the disc,

5.7 Examples 367


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

θa

0

⎞⎠

HcHcHc = Iyy/cθa e′ye′ye′y

where θa is the angular velocity of disc A. Also, because the disc is rotating in aplane, e′ye′ye′y = eyeyey. Substituting this result into the angular momentum equation,


Mf eyeyey =ddt

(Iyy/cθa

)eyeyey

Mf = Iyy/cθa

where we note that the only term in the derivative which changes with time is theangular velocity. We may now substitute in the moment of inertia of a disc aboutits centre of mass, which is available from Appendix C. Note here that we havearbitrarily selected our axes in this problem so that the unit vector eyeyey is perpendicularto the surface of the disc; in Appendix C, the axes are arbitrarily selected so that ezezez

is perpendicular to the surface of the disc. Therefore, our y-axis corresponds to thetable’s z-axis, and we must use Izz/c from the table for our Iyy/c.

Mf =12

maR2aθa

The magnitude of the friction moment Mf can be determined by considering a smallarea element dA on the surface of disc A. Because the disc is level and rigid, theweight of the disc will be equally distributed over its surface, resulting in a similarlydistributed normal force (or pressure) given by

Pn =Fn

Aa=

Fn

πR2a

where Aa is the surface area of disc A. Then, for a small element of area dA, thesmall component of the normal force dFn ‘felt’ by the element will be

dFn = Pn dA =Fn

πR2a

dA

The small component of the friction force dFf acting on the element is then

dFf = μk dFn


= μk( Fn

πR2a

dA)

= μk(mag

πR2a

dA)

dFf = μkmag1

πR2a

dA

where we have substituted in our earlier result for Fn. The small moment dMf ex-erted by dF about the centre of the circle is simply,

dMf = r dFf = μkmagr1

πR2a

dA

where r is the radius at which the element is located. Because dA is a smallarea element on a circle, we can substitute dA = r dr dθ and then integrate over thebounds of a circle (that is, 0 < r < Ra and 0 < θ < 2π; for a detailed derivation, seeAppendix A).

dMf = μkmagr1

πR2a

(r dr dθ

)∫

dMf =∫ 2π

0

∫ Ra

0μkmagr2 1

πR2a

dr dθ

Mf =∫ Ra

0

(μkmagr2 1

πR2a

θ

)∣∣∣∣2π

0dr

=∫ Ra

02πμkmagr2 1

πR2a

dr

=

(23

πμkmagr3 1πR2

a

)∣∣∣∣Ra

0

=23

πμkmagR3a

1πR2

a

Mf =23

Raμkmag

This result indicates that the frictional moment behaves exactly as a concentratedfriction force applied at a radius of 2Ra/3. While interesting, the magnitude of thefriction moment is not required in order to solve this problem.

Continuing with our analysis, we can obtain an expression for the angular accel-eration of the disc from the angular momentum equation, and then integrate it withrespect to time in order to determine the angular velocity.

5.7 Examples 369

Mf =12

maR2aθa

θa = 2Mf

maR2a

However,

θa =dθa

dt∫θa dt =

∫dθa

Substituting our earlier result for θ and integrating,

∫2

Mf

maR2a

dt =∫

dθa

2Mf

maR2a

t = θa +C0

where C0 is some constant of integration. Since we know that this equation must besatisfied when disc A is first released, we shall define that instant as t = 0, and θa =ωai at that instant. We may then solve for C0 by substituting this initial condition, as

2Mf

maR2a(0) = ωai +C0

C0 = −ωai

The angular velocity of the disc A is then given by the expression,

2Mf

maR2a

t = θa −ωai

θa = 2Mf

maR2a

t +ωai

Next, we consider the disc B. We shall redefine our system to consist only of thedisc B and retain the same coordinate system. Because the centre of mass of thedisc is not moving, and because the disc is resting on a frictionless surface, there isno useful information which can be obtained from the linear momentum equation.


Since there are no external forces acting on disc B aside from the friction againstdisc A, the angular momentum equation yields,


−Mf eyeyey = HcHcHc

where we recognize that the frictional moment applied by disc A upon disc B willbe equal and opposite to the moment applied by disc B upon disc A. Evaluating theangular momentum of disc B,


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

θb

0

⎞⎠

HcHcHc = Iyy/cθb e′ye′ye′y

where θb is the angular velocity of disc B. Once again, we note that e′ye′ye′y = eyeyey forplanar rotation. Substituting HcHcHc back into the angular momentum equation,


−Mf eyeyey =ddt

(Iyy/cθb

)eyeyey

−Mf = Iyy/cθb

Once again, we may substitute the moment of inertia of a solid disc about its centreof mass, so

−Mf =12

mbR2bθb

θb = −2Mf

mbR2b

We may now integrate this expression with respect to time to obtain the angularvelocity of the disc B, as

θb =dθb

dt

5.7 Examples 371∫θb dt =

∫dθb∫

−2Mf

mbR2b

dt =∫

dθb

−2Mf

mbR2b

t = θb +C1

where C1 is another constant of integration. Since θb = ωbi when t = 0, we can usethis initial condition to solve for C1, as

−2Mf

mbR2b

(0) = ωbi +C1

C1 = −ωbi

So the angular velocity of the disc B is given by

−2Mf

mbR2b

t = θb −ωbi

θb = −2Mf

mbR2b

t +ωbi

Finally, at the instant that the discs stop slipping, the final velocity ω f of the discsmust be the same, so that θa = θb = ω f . We can therefore solve for the critical timetc when the discs just stop slipping by equating θa = θb, and then use that value oftc to solve for ω f . To begin,

θa = θb

2Mf

maR2a

tc +ωai = −2Mf

mbR2b

tc +ωbi

tc =ωbi −ωai

2Mf(

1maR2

a+ 1

mbR2b

)

So, in the final condition, θa = ω f and t = tc. We may therefore substitute,

θa = 2Mf

maR2a

t +ωai


ω f = 2Mf

maR2a

tc +ωai

ω f = 2Mf

maR2a

(ωbi −ωai

2Mf(

1maR2

a+ 1

mbR2b

))+ωai

ω f =ωbi −ωai

1+ maR2a

mbR2b

+ωai

This expression may be simplified further, as

ω f = ωai

ωbiωai

+ maR2a

mbR2b

1+ maR2a

mbR2b

This result provides the final angular velocity of both discs once their surfaces haveceased to slip.

ALTERNATIVE SOLUTION

On the other hand, we may equally have defined our system as consisting of bothdiscs together. In that case, the frictional force between the discs is an internal force.Since friction is the only force which exerts a moment about the centre of mass ofthe two-disc system, there are no external moments acting on the system. Then, forour combined system,

∑Mext/cMext/cMext/c = ∑HcHcHc = 0(HcHcHc)

a +(HcHcHc)

b = C

where C is some constant. From our earlier analysis, we know that

(HcHcHc)

a =(Iyy/c)

aθa eyeyey(HcHcHc)

b =(Iyy/c)

bθb eyeyey

Substituting,

(Iyy/c)

aθa eyeyey +(Iyy/c)

bθb eyeyey = C(Iyy/c

)aθa +

(Iyy/c

)bθb = C

5.7 Examples 373

This expression must hold true both during the initial condition at during the finalcondition, so

(Iyy/c)

aωai +(Iyy/c)

bωbi = C(Iyy/c)

aω f +(Iyy/c)

bω f = C

Equating these two expressions together to eliminate C,

(Iyy/c)

aωai +(Iyy/c)

bωbi =(Iyy/c)

aω f +(Iyy/c)

bω f

ω f =

(Iyy/c

)aωai +

(Iyy/c

)bωbi(

Iyy/c

)a +(Iyy/c

)b

Substituting in our expressions for the moments of inertia,

ω f =12 mbR2

bωbi +12 maR2

aωai12 mbR2

b +12 maR2

a

And, simplifying,

ω f = ωai

ωbiωai

+ maR2a

mbR2b

1+ maR2a

mbR2b

which is exactly the same result as obtained earlier. While it may have been farless mathematically taxing to consider the system as consisting of both discs, it isimpossible to determine the magnitude of the internal friction force when solving inthis way.

Let us consider the case of two identical discs, so that ma = mb and Ra = Rb.Then,

ω f =ωai +ωbi

2

so the final angular velocity of the system becomes the average of the two initialvelocities. Additionally, we observe that for the case of two identical discs, if ωai =−ωbi, so that the initial angular velocities of the two discs are equal in magnitudebut opposite in sense, then ω f = 0 and the system comes to a dead stop.


5.7.8 Rolling clutch

A rolling clutch may be modeled as two discs A and B suspended one above theother, as illustrated. Disc A has mass ma, radius Ra and initial angular velocity ωai,while disc B has mass mb, radius Rb and initial angular velocity ωbi. At some instant,disc A is released so that it drops down onto disc B, and it is assumed that thesurfaces will initially slip.

Fig. 5.28 A rolling clutch.

Let us apply the linear momentum equation to disc A. We shall define our systemas consisting of the disc only, and our coordinate system shall be oriented so that exexex

points to the right and eyeyey points upwards, as illustrated. Then,

∑FextFextFext = GsysGsysGsys = maaaaaaa

FfFfFf +FnFnFn +FgaFgaFga +FaxFaxFax = ma(0)

−Ff exexex +Fn eyeyey −Fg eyeyey +Fax exexex = 0

where Ff is the friction force acting between the two discs, and we have assumedthat the friction is acting to the left. The force Fax is the horizontal force applied bythe pin to the disc which prevents the disc from moving sideways; note that therecan be no vertical force applied by the pin to the disc, as the pin is sitting in a verticalslot. Equating the coefficients of eyeyey,

Fn −Fg = 0

Fn = Fg = mag

5.7 Examples 375

Next, we can equate the coefficients of exexex,

−Ff +Fax = 0

Fax = Ff = μkFn

Fax = μkmag

where μk is the coefficient of kinetic friction between the discs. Here, we have ob-tained two important results: first, we determined the magnitude of the horizontalload in the pin supporting disc A, and second, we have determined the magnitude ofthe normal force, from which we were able to calculate the magnitude of the frictionforce.

Next, we can apply the law of conservation of angular momentum to the disc A,so that


−Ff Ra ezezez = HcHcHc

The angular momentum of the disc is then given by,


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0θa

⎞⎠

HcHcHc = Izz/cθa e′ze′ze′z

where θa is the angular velocity of the disc A. Substituting into the angular momen-tum equation,

−Ff Ra ezezez =ddt

HcHcHc

−μkmagRa ezezez =ddt

(Izz/cθa ezezez

)−μkmagRa ezezez = Izz/cθa ezezez

where we have recognized that the only component of the angular momentum whichvaries in time is θa; the unit vector e′ze′ze′z = ezezez, and is constant (this will always be the


case for an object rotating in a plane). However, since the disc is solid and assumedto be of constant density,

−μkmagRa =

(12

maR2a

)θa

θa = −2μkg1

Ra

where we have divided through by the unit vector ezezez. We may now integrate theacceleration with respect to time to determine the angular velocity of the disc A, as

θa =∫

θa dt

=∫

−2μkg1

Radt

θa = −2μkg1

Rat +C0

where C0 is some constant of integration. If we assume that the disc was releasedwhen time t = 0, then at that instant θa = ωai, the initial angular velocity of the disc.We can use this initial condition to solve for C0, as

ωai = −2μkg1

Ra(0)+C0

C0 = ωai

The angular velocity of disc A is therefore given by the expression,

θa =−2μkg1

Rat +ωai

Next, we shall consider the disc B. We will redefine our system as consisting ofthe disc B only, and retain the same coordinates as we used previously. The linearmomentum equation then yields,

∑FextFextFext = GsysGsysGsys = mbababab

FfFfFf +FgFgFg +FbxFbxFbx +FbyFbyFby +FnFnFn = mb(0)

Ff exexex −Fg eyeyey +Fbx exexex +Fby eyeyey −Fn eyeyey = 0

5.7 Examples 377

where Fbx and Fby are the horizontal and vertical components of the force applied tothe disc B by the pin holding it in place. Once again, the acceleration of the centre ofmass of the disc B is zero, because it never moves. Also, note that the forces appliedto disc B by disc A (Ff and Fn) are equal and opposite to the forces applied to discA by disc B, so

Fn = mag

Ff = μkmag

Equating the scalar coefficients of exexex,

Ff +Fbx = 0

Fbx =−μkFn

Fbx =−μkmag

This yields an expression for the horizontal component of the reaction force on thepin supporting disc B. Equating the scalar coefficients of eyeyey,

−Fg +Fby −Fn = 0

−mbg+Fby −mag = 0

Fby = (ma +mb)g

We have now solved for the reaction forces in the pin supporting disc B. Finally, weapply the angular momentum equation.


−Ff Rb ezezez = HcHcHc

The angular momentum of the disc is then given by,


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0θb

⎞⎠

HcHcHc = Izz/cθb e′ze′ze′z


where θb is the angular velocity of the disc B. Substituting into the angular momen-tum equation, and recalling that e′ze′ze′z = ezezez,

−Ff Rb ezezez =ddt

HcHcHc

−μkmagRb ezezez =ddt

(Izz/cθb ezezez

)−μkmagRb = Izz/cθb

where we have divided through by the unit vector ezezez. We may now substitute themoment of inertia of the disc, as

−μkmagRb =

(12

mbR2b

)θb

θb = −2μkma

mb

1Rb

g

We may now integrate this previous expression with respect to time in order to solvefor the angular velocity of the disc B.

θb =∫

θb dt

=∫ (

−2μkma

mb

1Rb

g

)dt

θb = −2μkma

mb

1Rb

gt +C1

where C1 is another constant of integration. Once again, since we assumed that t = 0at the instant the disc A was released, then at this instant we know that θb = ωbi.Using this initial condition to solve for C1,

ωbi = −2μkma

mb

1Rb

g(0)+C1

C1 = ωbi

The angular velocity of the disc B is then given by the expression,

θb =−2μkma

mb

1Rb

gt +ωbi

5.7 Examples 379

As with the previous example, kinetic friction will act until the instant that purerolling motion begins; then, from that instant onward, Ff = 0 and θa = θb = 0. Todetermine the final velocity of the discs, then, we must identify that instant whenthe surfaces stop slipping. At that instant, the tangential velocity of the rims of thediscs must be equal and the senses of rotation must be opposite, so

Raωa f =−Rbωb f

where ωa f and ωb f are the constant, final velocities of discs a and b once purerolling motion begins. Since we already have expressions for θa and θb, and sincewe know that the above expression will only hold true at the critical time tc whenpure rolling motion begins,

Raωa f = −Rbωb f

Ra

(−2μkg

1Ra

tc +ωai

)= −Rb

(−2μk

ma

mb

1Rb

gtc +ωbi

)

2μkma

mbgtc +2μkgtc = ωaiRa +ωbiRb

tc =1

2μkgωaiRa +ωbiRb

1+ mamb

We have now determined the critical time at which pure rolling motion begins; wemay now substitute this time into our result for θa to determine the final angularvelocity of the disc A, as

ωa f = −2μkg1

Ratc +ωai

= −2μkg1

Ra

(1

2μkgωaiRa +ωbiRb

1+ mamb

)+ωai

ωa f = ωai

(1−

1+ ωbiωai

RbRa

1+ mamb

)

Determining the final angular velocity of the disc B is then trivial, as

ωb f =Ra

Rbωa f

ωb f = ωaiRa

Rb

(1−

1+ ωbiωai

RbRa

1+ mamb

)


Once pure rolling motion begins, the angular velocities of the discs will remainconstant at these values.

It is interesting to note here that the results are independent of the coefficient ofstatic friction; this indicates that the above expressions for the final angular veloc-ities will hold true regardless of which direction friction is acting. If we were tochange the direction in which friction was acting, the only effect upon the results isto change the sign of μk. Since μk does not appear in our expressions for the angularvelocities, then the change in the sign of μk will have no effect.

It is also interesting to note that if disc A is identical to disc B, then ma = mb = mand Ra = Rb = R, so the final angular velocity of disc A would be

ωa f = ωai

(1−

1+ ωbiωai

RR

1+ mm

)

ωa f =12

ωai

(1− ωbi

ωai

)

Two interesting possibilities emerge from this expression. First, if ωai =−ωbi, then

ωa f =12

ωai

(1+

ωai

ωai

)ωa f = ωai

This indicates that, for the case of two identical discs with initial angular velocitieswhich are equal in magnitude but opposite in sense, suddenly allowing contact be-tween the two discs will not change the velocities of the discs. Indeed, pure rollingmotion will begin immediately and there will be no friction at all (a quick checkwill show that tc = 0 under these circumstances).

Second, if ωai = ωbi, then

ωa f =12

ωai

(1− ωai

ωai

)ωa f = 0

so if the identical discs are given equal initial angular velocities, they will both cometo a dead stop.

We may also use this example to demonstrate a very important point. The onlymoment acting on each disc is the friction, so if we define our system as consisting of

5.7 Examples 381

both discs, then the friction becomes an internal force, and for each disc, ΣMext/cMext/cMext/c =0. Then, for the system consisting of both discs, let us say that

∑Mext/cMext/cMext/c = ∑HcHcHc = 0

∑HcHcHc = C(HcHcHc)

a ezezez +(HcHcHc)

b ezezez = C

where C is some constant. Substituting our earlier results for HcHcHc,

(Izz/c)

aθa ezezez +(Izz/c)

bθb ezezez = C(Izz/c)

aθa +(Izz/c)

bθb = C

Since we know that, initially, θa = ωai and θb = ωbi, we can solve for C as

(Izz/c)

aωai +(Izz/c)

bωbi =C

Substituting this into our previous result,

(Izz/c)

aθa +(Izz/c)

bθb = C(Izz/c)

aθa +(Izz/c)

bθb =(Izz/c)

aωai +(Izz/c)

bωbi

Then, in our final condition, θa = ωa f and θb = ωb f . However, since pure rollingmotion is taking place, as we had shown before,

Raωa f =−Rbωb f

Then, in our final condition,

(Izz/c)

aωa f +(Izz/c)

bωb f =(Izz/c)

aωai +(Izz/c)

bωbi(Izz/c)

aωa f −(Izz/c)

b

Ra

Rbωa f =

(Izz/c)

aωai +(Izz/c)

bωbi

ωa f

((Izz/c)

a −Ra

Rb

(Izz/c)

b

)=(Izz/c)

aωai +(Izz/c)

bωbi

Then, for the case of two identical discs, (Izz/c)a =(Izz/c)b =(Izz/c) and Ra =Rb =R,so


ωa f

(Izz/c −

RR

Izz/c

)= Izz/cωai + Izz/cωbi

ωa f

(0

)= Izz/c

(ωai +ωbi

)0 = ωai +ωbi

ωai = −ωbi ???

This is clearly impossible, as we are free to set any initial conditions ωai and ωbi wechoose. There is no constraint which requires that the initial velocities be equal andopposite (in fact, from our earlier analysis, we know that if the discs are identicaland ωai = −ωbi, pure rolling begins immediately- while this is an interesting case,it is not an absolute requirement).

The reason for these nonsensical results was that we confused our systems in ourdefinition of the problem. From the law of conservation of angular momentum,


where the left-hand side of this expression represents the sum of the external mo-ments acting on the system about the centre of mass of the system. Since we definedour system as consisting of the two discs, the centre of mass of the system is some-where between the centres of the discs. If the two discs are identical, the centre ofmass of the combined system will be located at the point of contact. The horizontalcomponents of the reactions at the pins, Fax and Fbx, therefore contribute externalmoments, and ΣMext/cMext/cMext/c �= 0.

5.7.9 Torsional pendulum

Let us consider a disc of mass m and radius R which has been mounted on a tor-sional spring of stiffness kθ , as illustrated in Figure 5.29. The disc is given an initialangular deflection of θ0 and is then released from rest at that angle. We shall definea coordinate system o with its origin at the centre of the disc, oriented so that theunit vector ezezez is pointing straight up out of the surface of the disc, with exexex and eyeyey inthe plane of the disc, as shown. We shall further define a moving coordinate systemo′ which is stuck to the surface of the disc, and oriented so that e′xe′xe′x, e′ye′ye′y and e′ze′ze′z arecoincident with exexex, eyeyey and ezezez at our moment of interest.Then, from the angular momentum equation,


5.7 Examples 383

Fig. 5.29 A torsional pendulum.

The only moment acting on the disc is the torsional spring moment, so if the disc isangularly deflected by an angle θ ,

∑Mext/oMext/oMext/o =−kθ θ ezezez

where we have assumed that the moment exerted by the spring vanishes at θ = 0, sothat there was no tension in the spring initially. The angular momentum of the discmay then be evaluated, as

HoHoHo = IoωIoωIoω =

⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0θ

⎞⎠

HoHoHo = Izz/cθ e′ze′ze′z

where we have noted here that ezezez = e′ze′ze′z, since the two are always parallel. Also, sincethe origin o (coincident with o′) is located at the centre of the disc, we recognizedthat Izz/o = Izz/c. Substituting these results into the angular momentum equation,


−kθ θ ezezez =ddt

(Izz/cθ e′ze′ze′z

)


Since the o′ coordinate system is stuck to the disc, Izz/c is constant within this refer-ence frame. Also, because e′ze′ze′z does not change its orientation as the disc rotates, it isconstant as well. Therefore,

−kθ θ ezezez =ddt

(Izz/cθ e′ze′ze′z

)−kθ θ ezezez = Izz/c

ddt

(θ)

e′ze′ze′z

θ = − Izz/c

kθθ

Since the governing equation for the motion of the disc is of the form x+Ax = 0(where A is any constant), we can recognize that the disc will be undergoing simpleharmonic motion. Consequently, the angular displacement θ(t) must be of the form,

θ(t) = Asin(Bt +C)+D

where A, B, C and D are constants. Then,

θ = ABcos(Bt +C)

θ = −AB2 sin(Bt +C)

Substituting this into our previous result,

θ = − Izz/c

kθθ

(Asin(Bt +C)+D

)= − Izz/c

kθ

(−AB2 sin(Bt +C))

sin(Bt +C)

(1−B2 Izz/c

kθ

)+

DA

= 0

The only way for this equation to hold true at all times is if

B =

√kθ

Izz/c

D = 0

5.7 Examples 385

Substituting this into our expression for the angular displacement,

θ(t) = Asin(Bt +C)+D

θ(t) = Asin

(√kθ

Izz/ct +C

)+0

For the constants A and C, we must look at the initial conditions. First, we know thatthe disc was released from rest, so θ = 0 at t = 0. The angular velocity of the disccan be determined, as

θ =ddt

Asin

(√kθ

Izz/ct +C

)

θ = A

√kθ

Izz/ccos

(√kθ

Izz/ct +C

)

Substituting in our initial condition for the angular velocity,

(0) = A

√kθ

Izz/ccos

(√kθ

Izz/c(0)+C

)

0 = cos(C)

C =π

2

So our angular displacement becomes,

θ(t) = Asin

(√kθ

Izz/ct +

π

2

)

Finally, the position of the disc is θ = θ0 when t = 0, so

θ0 = Asin

(√kθ

Izz/c(0)+

π

2

)

θ0 = Asin

(π

2

)θ0 = A


So, the angular displacement of the disc is given by the expression,

θ(t) = θ0 sin

(√kθ

Izz/ct +

π

2

)

Equivalently, since we know that the moment of inertia of a disc about its centre ofmass along the normal axis is Izz/c = mR2/2, we may substitute,

θ(t) = θ0 sin

(1R

√2

kθ

mt +

π

2

)

A symmetric object mounted on a torsional spring, as examined here, is knownas a torsional pendulum. From the above results, we can see that this simple devicewill produce sinusoidal motion with a frequency f given by the expression

f =1

2π

√kθ

Izz/c

The ‘restoring force’ which is driving the motion is provided by the spring, whereasgravity drives suspended pendulums. Consequently, torsional pendulums will oscil-late with the same frequency regardless of their orientation. Because of this property,the torsional pendulum was widely used to regulate the mechanisms of early pocket-watches; in fact, the first serious examination of torsional pendulums for timekeep-ing is attributed to the Dutch astronomer Christiaan Huygens (1629-1695), thoughRobert Hooke famously accused Huygens of stealing his work on the subject duringone particularly animated meeting of the Royal Society in 1675. Though the recordsfrom the meetings of the Royal Society during this period have curiously vanished,an unofficial copy- in Hooke’s own distinctive handwriting- was unearthed in 2006which supported Hooke’s version of events.

5.7.10 Reaction forces: swinging bar

Another common requirement is to calculate the reaction forces on a pin aroundwhich a solid object is being caused to rotate. As an example, let us consider a longslender bar of mass m and length L suspended from a pin at one end. Released fromrest in a horizontal position, it swings downward about the pin through some angleθ , as illustrated in Figure 5.30.

Let us define a cylindrical coordinate system o which is stuck on to the bar suchthat ererer always points along the bar and eθeθeθ points in the direction of motion. The unitvector ezezez, then, points into the page in the illustration. If we define our system asconsisting of the bar only, from the free-body diagram we see that,

5.7 Examples 387

Fig. 5.30 Reaction forces on a pin supporting a swinging object.

∑FextFextFext = FgFgFg +FRFRFR

=(Fg sin(θ) ererer +Fg cos(θ) eθeθeθ

)+(FRr ererer +FRθ eθeθeθ

)∑FextFextFext =

(FRr +Fg sin(θ)

)ererer +

(FRθ +Fg cos(θ)

)eθeθeθ

where we have resolved the reaction force vector FRFRFR into the components FRrFRrFRr andFRθFRθFRθ lying along the ererer and eθeθeθ directions, respectively (as illustrated). We have donethis because both the magnitude and the direction of FRFRFR are unknown. Substitutingthis result into the law of conservation of linear momentum,

∑FextFextFext = GsysGsysGsys = macacac(FRr +Fg sin(θ)

)ererer +

(FRθ +Fg cos(θ)

)eθeθeθ = macacac

where acacac is the acceleration vector of the centre of mass of the bar. But, in a cylin-drical system, we know that

aaa = (r− rθ 2)ererer +(2rθ + rθ)eθeθeθ

so this result may also be substituted into the linear momentum equation to yield,

(FRr +Fg sin(θ)

)ererer +

(FRθ +Fg cos(θ)

)eθeθeθ

= m((rc/o − rc/oθ 2)ererer + (2rc/oθ + rc/oθ)eθeθeθ

)


Since the centre of mass c of the bar cannot move along the bar as the bar swings,rc/o = (L/2) and rc/orc/orc/o = rc/orc/orc/o = 0. Then,

(FRr +Fg sin(θ)

)ererer +

(FRθ +Fg cos(θ)

)eθeθeθ

= m

((0− L

2θ 2)ererer +

(2(0)θ +

L2

θ)

eθeθeθ

)

Simplifying,

(FRr +Fg sin(θ)

)ererer +

(FRθ +Fg cos(θ)

)eθeθeθ =−m

L2

θ 2 ererer +mL2

θ eθeθeθ

Now we can solve for the reaction forces; starting with the scalar coefficients of ererer,

FRr +Fg sin(θ) = −mL2

θ 2

FRr = −mgsin(θ)−mL2

θ 2

And from the scalar coefficients of eθeθeθ ,

FRθ +Fg cos(θ) = mL2

θ

FRθ = mL2

θ −mgcos(θ)

We still don’t know anything about the angular velocity and acceleration of the bar,θ and θ , though. For this, we need to consider the angular momentum. Because ourorigin is a stationary point, the law of conservation of angular momentum gives,

∑Mext/oMext/oMext/o = Hsys/oHsys/oHsys/o =ddt(IoωoIoωoIoωo)

From our free-body diagram,

∑Mext/oMext/oMext/o = FgL2

cos(θ) ezezez

Noting, of course, that the reaction forces produce no moments about o. Just becauseit’s easier to work in Cartesian coordinates, for the time being let us stick a Cartesiancoordinate system onto the bar such that exexex always points along ererer, and ezezez remains

5.7 Examples 389

unchanged. In this reference frame, θ =(0exexex+0eyeyey+ωzezezez). The angular momentumof the bar can then be evaluated as


⎡⎣ 0 0 0

0 Iyy/o 00 0 Izz/o

⎤⎦⎛⎝ 0

0ωz

⎞⎠= Izz/oωz ezezez

We immediately recognized that Ixx/o and the cross-products would be zero, becausey = z = 0 everywhere along the bar. Since we know that the z-moment of inertia ofa bar about its centre of mass is given by

Izz/c =1

12mL2

We may apply the parallel axis theorem to get the moment of inertia about one end:

Izz/o = mr2c/o + Izz/c

= m(L

2

)2+

112

mL2

=14

mL2 +1

12mL2

Izz/o =13

mL2

So the angular momentum of the rod about the point o is

HoHoHo =13

mL2θ ezezez

Alternatively, we may have retrieved this result directly from Appendix C. Substi-tuting these results into the angular momentum equation,

∑Mext/oMext/oMext/o =ddt(IωIωIω)

FgL2

cos(θ) ezezez =ddt

(13

mL2θ ezezez)

mgL2

cos(θ) =13

mL2θ

θ =32

gL

cos(θ)

We can now do some integrating to solve for θ too; however, we want it as a functionof θ and not of time, so


θ =dθ

dt=

dθ

dθ

dθ

dt= θ

dθ

dθ=

32

gL

cos(θ)∫θ dθ =

∫32

gL

cos(θ)dθ

12

θ 2 =32

gL

sin(θ)+C

θ 2 = 3gL

sin(θ)+C

where C is some constant of integration. Since the bar was released from rest, θ = 0when θ = 0, so C = 0.

θ 2 = 3gL

sin(θ)

We can now substitute these equations for θ and θ into the earlier expressionsfor the reaction forces. Starting with the radial reaction force,

FRr = −mgsin(θ)−mL2

θ 2

= −mgsin(θ)−mL2

(3

gL

sin(θ))

= −mgsin(θ)− 32

mgsin(θ)

FRr = −52

mgsin(θ)

And now for the tangential reaction force,

FRθ = mL2

θ −mgcos(θ)

= mL2

(32

gL

cos(θ))−mgcos(θ)

=34

mgcos(θ)−mgcos(θ)

FRθ = −14

mgcos(θ)

The vector reaction force acting on the pin is therefore

FRFRFR =−52

mgsin(θ) ererer − 14

mgcos(θ) eθeθeθ

5.7 Examples 391

Again, it is interesting to note that the length of the bar is irrelevant; the reactionforce in the pin only depends on the mass and the instantaneous angle.

If we wanted to find the angle at which the maximum and minimum forces occur,we could take the derivative of |FRFRFR| with respect to θ and set the result equal to zero.To make the derivative easier, we can also recognize that |FRFRFR|2 will be at a localextreme at the same time as |FRFRFR|.

|FRFRFR|2 =(−5

2mgsin(θ)

)2+(−1

4mgcos(θ)

)2

=254

m2g2 sin2(θ)+1

16m2g2 cos2(θ)

The extremes will occur when the derivative of |FRFRFR|2 is zero, so

0 =d

dθ

(254

m2g2 sin2(θ)+116

m2g2 cos2(θ))

=d

dθ

(254

sin2(θ)+1

16cos2(θ)

)=

254

(2sin(θ)cos(θ)

)+

116

(−2cos(θ)sin(θ))

=998

sin(θ)cos(θ)

0 = sin(θ)cos(θ)

so the maxima and minima will occur when θ = 0 and π/2. We still don’t knowwhich is which, though. So, we can go back to the magnitude of the force andsubstitute in these values of θ . First, at θ = 0,

|FRFRFR|2 =(−5

2mgsin(0)

)2+(−1

4mgcos(0)

)2

|FRFRFR| = 14

mg

Then, at θ = π/2,

|FRFRFR|2 =(−5

2mgsin(π/2)

)2+(−1

4mgcos(π/2)

)2

|FRFRFR| = 52

mg


So the range of the magnitude of the reaction force to which the pin will be subjectedis

14≤ |FRFRFR|

mg≤ 5

2

This information would be of great importance in the design of that pin joint.

5.7.11 Compound problem

So far, the examples illustrated the application of the law of conservation of angularmomentum to the exclusion of the principles covered in the previous chapters. Inengineering practice, this rarely happens. Therefore, let us now examine a problemwhere we will need to apply angular momentum, relative motion and rectilinearmechanics. Consider a solid, homogeneous disc of mass m connected to the ceilingby means of an ideal string wound around its perimeter, as illustrated in Figure 5.31.A bug of negligible mass is sitting on the bottom of the disc. At some instant in time,the disc is released from rest, and at the same time, the bug is startled and begins torun straight towards the centre of the disc with a constant velocity v0 relative to thesurface of the disc.

Fig. 5.31 A compound problem.

Let us say we wanted to determine the velocity of the bug relative to a fixedobserver after the disc had undergone one complete revolution. We will need to usethe relative motion equation,

5.7 Examples 393


However, none of the terms (with the exception of vp/o′vp/o′vp/o′) are immediately available.Let us define the moving coordinate system o′ such that it is stuck to the hub ofthe disc with −e′ye′ye′y always pointing towards the bug. The fixed coordinate system wedefine to coincide with the moving coordinate system at the instant when the bugis immediately below the disc hub, as illustrated. We will define ezezez = e′ze′ze′z as alwaysfacing out of the plane of the disc. Then, vo′/ovo′/ovo′/o is the velocity of the centre of the discafter one revolution, ωaxωaxωax is the angular velocity of the disc after one revolution, andrp/o′rp/o′rp/o′ is the position of the bug relative to the centre of the disc after one revolution.Thankfully, vp/o′vp/o′vp/o′ is simply v0e′ye′ye′y.

To begin with, let us consider vo′/ovo′/ovo′/o. To determine the velocity of the centre ofthe disc after a single revolution, we will need the angular acceleration of the discfrom the conservation of angular momentum. However, this will require first that wesolve for the tension FT in the string. Therefore, we begin with the linear momentumequation; from the free-body diagram shown in the figure,


FT eyeyey −Fg eyeyey = my eyeyey

FT = my+Fg

FT = my+mg

where Fg is the gravitational force acting on the disc. Now, to determine the angularacceleration we use the angular momentum equation. The angular momentum of thedisc about its centre of mass, Hc, is given by the expression


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz


where the axes are stuck to the object. Then, from conservation of angular momen-tum,

∑McMcMc = HcHcHc =ddt


)since the axes are stuck to the disc, Izz/c is constant; also, because e′ze′ze′z always pointsout of the plane of the figure, it is also constant. Therefore,

∑McMcMc = Izz/cωz e′ze′ze′z

From the free-body diagram, we can obtain an expression for the sum of the externalmoments acting on the disc about the centre of mass. Noting that, since gravity actsthrough the centre of mass, rrr×FgFgFg = 0 and


∑McMcMc = R e′xe′xe′x ×FTFTFT

= R e′xe′xe′x ×FT e′ye′ye′y= FT R

(e′xe′xe′x ×e′ye′ye′y

)∑McMcMc = FT R e′ze′ze′z

Substituting this result into the angular momentum equation,

FT R e′ze′ze′z = Izz/cωz e′ze′ze′z(my+mg)R = Izz/cωz

where we have substituted the previous result for FT . However, since the disc is‘rolling’ down the string, y = −Rωz. Also, from Appendix C, we can find that themoment of inertia of a disc about its centre is Izz/c = mR2/2. Substituting this intothe previous result,

(my+mg)R =12

mR2ωz

(−mRωz +mg)R =12

mR2ωz

−mR2ωz +mgR =12

mR2ωz

32

mR2ωz = mgR

ωz =23

gR

Now, to solve for the angular velocity after one complete revolution, when θ = 2π ,

θ =dθ

dt=

dθ

dθ

dθ

dt= θ

dθ

dθ∫θ dθ =

∫θ dθ

Substituting in the earlier result (recalling that, by definition, ω = θ ) and evaluatingthe integral,

5.7 Examples 395∫23

gR

dθ =∫

θ dθ

23

gR

θ =12

θ 2 +C0

To solve for the constant of integration C0, we use the initial condition that the diskwas released from rest at θ = 0. Then,

23

gR

0 =12

02 +C0

C0 = 0

So the angular velocity of the disc is given by the expression,

23

gR

θ =12

θ 2

And after one full revolution, when θ = 2π ,

23

gR(2π) =

12

ω2z

ωz =

√83

πgR

Again, since the disc is ‘rolling’ down the string,

vo′/ovo′/ovo′/o = −vc eyeyey =−ωzR eyeyey

=

√83

πgR

R

vo′/ovo′/ovo′/o =

√83

πgR

The next thing we need to determine is rp/o′rp/o′rp/o′ ; for this, we need to know how far thebug has run after one complete revolution. Since he is running at constant speed,it is sufficient to know the time required for the disc to complete one revolution.Therefore, we need to integrate the acceleration twice with respect to time.

θ =∫

θ dt


=∫

23

gR

dt

θ =23

gR

t +C1

where C1 is a constant of integration. Again, since the disc was released from rest,C1 = 0. Then,

θ =dθ

dt=

23

gR

t

θ =∫

23

gR

t dt

θ =13

gR

t2 +C2

where C2 is yet another constant of integration. Because we define θ = 0 at themoment the disc begins to move, C2 = 0. Finally,

θ =13

gR

t2

To solve for the time tc required to complete one revolution, we isolate t and setθ = 2π .

θ =13

gR

t2

t =

√3

Rg

θ

tc =

√6π

Rg

So the bug has moved a total distance Sbug inward from the disc rim of

Sbug =Votc = v0

√6π

Rg

So, after one complete revolution of the disc, the bug will again be immediatelybelow the hub (though closer to the centre). The bug’s position vector is therefore,

rp/o′rp/o′rp/o′ =−(

R− v0

√6π

Rg

)e′ye′ye′y

5.7 Examples 397

Finally, all of the above results may be substituted back into the relative motionequation,


= −√

83

πgR eyeyey + v0e′ye′ye′y +√

83

πgR

ezezez ×−(

R− v0

√6π

Rg

)e′ye′ye′y

=

(v0 −

√83

πgR

)eyeyey +

(−√

83

πgR

(R− v0

√6π

Rg

))(ezezez ×e′ye′ye′y

)

=

(v0 −

√83

πgR

)eyeyey +

(√83

πgR

R− v0

√6π

Rg

√83

πgR

)exexex

vp/ovp/ovp/o =

(√83

πgR−4πv0

)exexex +

(v0 −

√83

πgR

)eyeyey

The acceleration of the bug relative to a fixed observer may also be obtained after asingle revolution relatively easily. As with any problem of relative motion, we beginwith the relative acceleration equation.


Each of these terms may then be evaluated one at a time. To begin with, ao′/oao′/oao′/o isthe acceleration of the centre of the disc relative to a fixed observer. As with thevelocity, since the disc is ‘rolling’ down the string,

ao′/oao′/oao′/o = −ωzR eyeyey

= −(

23

gR

)R eyeyey

ao′/oao′/oao′/o = −23

g eyeyey

where we have substituted our earlier result for ωz evaluated when θ = 2π . Next, weconsider ap/o′ap/o′ap/o′ . Since the bug is moving at a constant velocity v0 along the surface ofthe disc, it is not accelerating relative to the disc and so ap/o′ap/o′ap/o′ = 0. Then, the Coriolisterm 2ωaxωaxωax ×vp/o′vp/o′vp/o′ may be worked out fairly easily, since both the angular velocityand the relative velocity of the bug are already known at the moment of interest.

2ωaxωaxωax ×vp/o′vp/o′vp/o′ = 2ωz ezezez × v0 eyeyey


= 2

(√83

πgR

)ezezez × v0 e′ye′ye′y

= 2v0

√83

πgR

(ezezez ×e′ye′ye′y

)2ωaxωaxωax ×vp/o′vp/o′vp/o′ = −2v0

√83

πgR

exexex

since e′ye′ye′y = eyeyey at the instant of interest. The angular acceleration term may also bedetermined, as both αaxαaxαax and rp/o′rp/o′rp/o′ are already known as well.

αaxαaxαax ×rp/o′rp/o′rp/o′ = ωz ezezez ×−(

R− v0

√6π

Rg

)e′ye′ye′y

=23

gR

ezezez ×−(

R− v0

√6π

Rg

)e′ye′ye′y

=

(−2

3g+

23

v0gR

√6π

Rg

)(ezezez ×e′ye′ye′y

)

αaxαaxαax ×rp/o′rp/o′rp/o′ =

(23

g− v0

√83

πgR

)exexex

Finally, the centripetal term may be evaluated,

ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′ = ωz ezezez ×ωz ezezez ×−(

R− v0

√6π

Rg

)e′ye′ye′y

=

√83

πgR

ezezez ×√

83

πgR

ezezez ×−(

R− v0

√6π

Rg

)

=

√83

πgR

ezezez ×−(√

83

πgR

R− v0

√83

πgR

√6π

Rg

)(ezezez ×e′ye′ye′y

)

=

√83

πgR

ezezez ×(√

83

πgR−4πv0

)exexex

=

(√83

πgR

√83

πgR−4πv0

√83

πgR

) (ezezez ×exexex

)

ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′ =

(83

πg−4πv0

√83

πgR

)eyeyey

5.7 Examples 399

Now we can substitute all of the above terms into the relative acceleration equation,as


=

(−2

3g eyeyey

)+0+

(−2v0

√83

πgR

exexex

)

+

((23

g− v0

√83

πgR

)exexex

)+

((83

πg−4πv0

√83

πgR

)eyeyey

)

ap/oap/oap/o =

(−2v0

√83

πgR+

23

g− v0

√83

πgR

)exexex +

(−2

3g+

83

πg−4πv0

√83

πgR

)eyeyey

This result may be further reduced to,

ap/oap/oap/o =

(23

g−3v0

√83

πgR

)exexex +

(8π −2

3g−4πv0

√83

πgR

)eyeyey

which is the acceleration of the bug relative to a fixed observer after a single revo-lution of the disc, as required by the problem statement. Once the angular velocitiesand accelerations of the disc and the position of the bug relative to the disc had beendetermined, the process of evaluating the acceleration of the bug was tedious butfairly straightforward.

5.7.12 Relative motion and rolling

In section 4.6.4, we considered an object which was caused to slide either up ordown a wedge as a result of the acceleration of the wedge. Let us now reconsider thisproblem, though for the case where the object rolls rather than slides. A roller B ofmass mb rests on the surface of a wedge-shaped block A of mass ma and inclinationθ . The wedge rests on a frictionless, horizontal surface, and a force of magnitude Fis applied to the wedge, as illustrated in Figure 5.32.

In order to analyze this system, we must recognize that the angular accelerationof the roller B will depend on the acceleration of its centre of mass relative to theacceleration of the surface on which it is rolling. Because the wedge itself is ac-celerating, it shall therefore be necessary at some point to apply the relative motionequations in order to determine the accelerations of the objects. Also, we shouldnote that because the wedge is moving, the instantaneous point of contact betweenthe roller and the surface is not a stationary point.

We shall begin by defining our system as the roller alone. We shall define thefixed coordinate system o such that exexex points to the right and eyeyey points upwards;


Fig. 5.32 Relative motion and rolling.

then, ezezez will point out of the plane of the figure, as illustrated in Figure 5.33. Fromthe free-body diagram,

Fig. 5.33 Relative motion and rolling- free-body diagrams.

∑FextFextFext = FfFfFf +FnbFnbFnb +FgbFgbFgb

= Ff cos(θ) exexex −Ff sin(θ) eyeyey +Fnb sin(θ) exexex +Fnb cos(θ) eyeyey −mbg eyeyey

∑FextFextFext =(Ff cos(θ)+Fnb sin(θ)

)exexex +

(−Ff sin(θ)+Fnb cos(θ)−mbg)

eyeyey

where Ff is the scalar magnitude of the static friction force acting between thewedge and the roller, Fnb is the scalar magnitude of the normal force acting betweenthe wedge and the roller, and we have assumed here that the roller is rolling up the

5.7 Examples 401

wedge. Because the friction between the wedge and the roller is static, no furtherinformation about this force is yet available. Then, applying the law of conservationof linear momentum to our system,

∑FextFextFext = GsysGsysGsys = mbababab

where it is important to note here that ababab is the vector acceleration of the centre ofmass of the roller. Substituting in our earlier result for ΣFextFextFext,

mbxb exexex +mbyb eyeyey =(Ff cos(θ)+Fnb sin(θ)

)exexex

+(−Ff sin(θ)+Fnb cos(θ)−mbg

)eyeyey

In this expression, we must retain both the horizontal and vertical components of theacceleration, since the direction of the acceleration of the roller is not yet known.Equating the scalar coefficients of exexex,

mbxb = Ff cos(θ)+Fnb sin(θ)

Fnb =mbxb −Ff cos(θ)

sin(θ)(5.5)

Then, equating the scalar coefficients of eyeyey,

mbyb =−Ff sin(θ)+Fnb cos(θ)−mbg

Substituting in our result for Fnb,

mbyb = −Ff sin(θ)+

(mbxb −Ff cos(θ)

sin(θ)

)cos(θ)−mbg

mbyb = −Ff sin(θ)+mbxbcos(θ)sin(θ)

−Ffcos2(θ)

sin(θ)−mbg

mbyb = −Ffsin2(θ)+ cos2(θ)

sin(θ)+mbxb

cos(θ)sin(θ)

−mbg

Ff = mbxb cos(θ)−mbyb sin(θ)−mbgsin(θ)

Recall that, in the above result, the magnitude of the friction force Ff is still un-known; in order to eliminate this unknown, we need another expression relating theacceleration of the roller to the forces applied to it. The law of conservation of an-gular momentum must therefore be applied. Again, from the free-body diagram, we


can evaluate the sum of the external moments acting on the roller about its centre ofmass as

∑Mext/cMext/cMext/c = Ff R ezezez

where we have immediately recognized that the only external force applied to theroller which can exert a moment about the centre is the static friction force, and thatthe resulting moment is in the +ezezez direction according to the right-hand rule. Next,we may evaluate the angular momentum of the system about its centre of mass, as


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ω

⎞⎠= Izz/cω e′ze′ze′z

where ω is the angular velocity of the roller as it rolls up the incline, and the unitvector e′ze′ze′z is stuck to the roller, pointing out of the plane of the illustration in Figure5.32. Since the roller only rolls in a plane, e′ze′ze′z = ezezez. From the law of conservation ofangular momentum, then,

∑Mext/cMext/cMext/c = HcHcHc =ddt

IcωIcωIcω

Ff R ezezez =ddt

(Izz/cω e′ze′ze′z

)Ff R = Izz/cω

If we assume that the roller is a solid cylinder, then Izz/c = mbR2/2, and so

Ff R =12

mbR2ω

Ff =12

mbRω

We may now substitute in our earlier result for Ff , as

mbxb cos(θ)−mbyb sin(θ)−mbgsin(θ) =12

mbRω

xb cos(θ)− yb sin(θ)−gsin(θ) =12

Rω

However, we are still not in a position to solve this problem, as there are three un-knowns in this expression: xb, yb and ω . In order to continue, then, we need another

5.7 Examples 403

expression linking the angular acceleration of the roller to its linear acceleration.For this, we turn to the relative acceleration equation.

Let us define the moving coordinate system o′ with its origin stuck to the wedgesuch that e′xe′xe′x points down the surface of the wedge and e′ye′ye′y points normally up fromthe surface, as illustrated in Figure 5.34. Since we are interested in the accelerationof the centre of mass of the roller, we will define our point of interest p as being atthe centre of the roller. From the relative acceleration equation,

Fig. 5.34 Relative motion and rolling- moving coordinates.


where ap/oap/oap/o is the acceleration of the centre of mass of the roller relative to a fixedobserver; however, we have already defined the components of this acceleration, as

ap/oap/oap/o = ababab = xb exexex + yb eyeyey

Similarly, then, because o′ is stuck to the wedge, ao′/o is simply the horizontal accel-eration of the wedge relative to a fixed observer. Because the wedge is not rotatingat all, ωaxωaxωax = αaxαaxαax = 0 (remember, we stuck the o′ coordinate system to the wedge,not to the roller). Substituting,


xb exexex + yb eyeyey = xa exexex +arel e′xe′xe′x +2(0)×vp/o′vp/o′vp/o′ +0×rp/o′rp/o′rp/o′ +0×0×rp/o′rp/o′rp/o′

xb exexex + yb eyeyey = xa exexex +arel e′xe′xe′x

where xa is the magnitude of the acceleration of the wedge, and we have defined thequantity arel as the acceleration of the centre of the roller relative to the surface ofthe wedge. Because the roller is undergoing pure rolling motion, we know that


arel =−Rω

Note that if the magnitude arel is positive, then, in the coordinate system we havedefined, ω will be negative. Furthermore, we can resolve e′xe′xe′x into components alongexexex and eyeyey, as

e′xe′xe′x = cos(θ) exexex − sin(θ) eyeyey

Substituting these expressions into the relative acceleration equation,

xb exexex + yb eyeyey = xa exexex −Rω(cos(θ) exexex − sin(θ) eyeyey

)xb exexex + yb eyeyey =

(xa −Rω cos(θ)

)exexex +Rω sin(θ) eyeyey

Equating the scalar coefficients of the unit vectors,

xb = xa −Rω cos(θ)

yb = Rω sin(θ) (5.6)

Substituting this result back into the angular momentum equation,

xb cos(θ)− yb sin(θ)−gsin(θ) =12

Rω

(xa −Rω cos(θ)

)cos(θ)− (Rω sin(θ)

)sin(θ)−gsin(θ) =

12

Rω

xa cos(θ)−Rω(cos2(θ)+ sin2(θ)

)−gsin(θ) =12

Rω

xa cos(θ)−gsin(θ) =32

Rω (5.7)

We have now reduced this problem to a single equation with two unknowns: xa andω . In order to solve, we will need another expression relating these two variables.Since we have not yet considered the motion of the wedge, let us turn our attentionthere. Redefining the system as consisting of the wedge only, then, from the earlierfree-body diagram (Figure 5.33),

∑FextFextFext = FFF+FgaFgaFga +FnaFnaFna −FfFfFf −FnbFnbFnb

= F exexex −mag eyeyey +Fna eyeyey −Ff cos(θ) exexex +Ff sin(θ) eyeyey

−Fnb sin(θ) exexex −Fnb cos(θ) eyeyey

5.7 Examples 405

∑FextFextFext =(F −Ff cos(θ)−Fnb sin(θ)

)exexex

+(−mag+Fna +Ff sin(θ)−Fnb cos(θ)

)eyeyey

Note that the forces applied to the roller by the wedge are equal in magnitude butopposite in sense to the forces applied to the wedge by the roller. Applying the lawof conservation of linear momentum to our system, then,


as our system consists of a single object of constant mass. Substituting in our ex-pression for ΣFextFextFext,(F−Ff cos(θ)−Fnb sin(θ)

)exexex+

(−mag+Fna+Ff sin(θ)−Fnb cos(θ))

eyeyey =maxa exexex

Equating the scalar coefficients of eyeyey,

0 = −mag+Fna +Ff sin(θ)−Fnb cos(θ)

Fna = mag−Ff sin(θ)+Fnb cos(θ)

And, equating the scalar coefficients of exexex,

maxa = F −Ff cos(θ)−Fnb sin(θ)

Ff =F −maxa −Fnb sin(θ)

cos(θ)

However, we already had an expression for Fnb; from Equation (5.5),

Fnb =mbxb −Ff cos(θ)

sin(θ)

Substituting,

Ff =F −maxa −Fnb sin(θ)

cos(θ)

Ff =

F −maxa −(

mbxb−Ff cos(θ)sin(θ)

)sin(θ)

cos(θ)

Ff cos(θ) = F −maxa −mbxb +Ff cos(θ)

0 = F −maxa −mbxb


However, from Equation (5.6),

xb = xa −Rω cos(θ)

Substituting,

0 = F −maxa −mbxb

0 = F −maxa −mb(xa −Rω cos(θ)

)ω =

−F +maxa +mbxa

mbRcos(θ)

At last, we may substitute this into our earlier expression for xa from the angularmomentum equation (Equation 5.7),


Rω


R

(−F +maxa +mbxa

mbRcos(θ)

)23

xa cos2(θ)− 23

gsin(θ)cos(θ) = − Fmb

+ma

mbxa + xa

Finally, isolating and solving for the acceleration xa,

xa =

Fmb

− 23 gsin(θ)cos(θ)

1+ mamb

− 23 cos2(θ)

This result shows that the linear acceleration of the wedge does not depend onthe radius of the roller, but only on its mass. Also, notice that if F = 0, the result isnecessarily negative: if the wedge and the roller were released from rest, as the rollerrolled downwards, it would push the wedge toward the left (the −exexex direction).

If we were to consider the special case when the roller does not roll up or downthe surface but remains stationary relative to the wedge, then ω = 0. Substitutingthis into Equation (5.7),


Rω

x0 cos(θ)−gsin(θ) =32

R(0)

5.7 Examples 407

x0 = gsin(θ)cos(θ)

This expression yields the critical acceleration x0 for which the roller will remainstationary on the surface of the wedge. Substituting this into our general expres-sion for xa, we may obtain the critical applied force F0 required to hold the rollerstationary on the surface of the wedge, as

xa =

Fmb


1+ mamb

− 23 cos2(θ)

x0 =

F0mb


1+ mamb

− 23 cos2(θ)

gsin(θ)cos(θ)

=

F0mb


1+ mamb

− 23 cos2(θ)

gma

mb

sin(θ)cos(θ)

+gsin(θ)cos(θ)

− 23

gsin(θ)cos(θ) =F0

mb− 2

3gsin(θ)cos(θ)

The critical force F0 is therefore given by the expression,

F0 = mbgsin(θ)cos(θ)

(1+

ma

mb

)

5.7.13 Forces in a simple mechanism

Let us now consider a similar but more complex system. We shall compute the re-action forces acting on the hub of solid disc of mass ma and radius R, which is sup-ported by a pin through its centre. One end of a slender bar of mass mb and length Lis pinned to the rim of the disc, while the other end of the bar is pinned to a massless,frictionless slider, as illustrated in Figure 5.35. Gravity is acting downwards.

This example is actually a good demonstration of a force analysis for a sim-ple mechanism; notice that the system is identical to the crank-slider mechanismanalyzed in Chapter 2. Because this is a constrained mechanical system, the accel-eration of the disc (and therefore the forces subjected to the disc) will depend on theacceleration of the bar. Any force analysis must therefore begin with determiningthe linear acceleration of the centres of mass of all of the objects involved, as wellas their angular accelerations.

For simplicity, let us consider the instant when the bar is horizontal. To begin, letus define the point p as being at the pin joining the disc to the bar, and the point qas being at the pin joining the bar to the slider. We shall define a fixed coordinate


Fig. 5.35 Constrained motion with forces.

system o such that the unit vector exexex points toward the right and eyeyey points upwards.The unit vector ezezez therefore points out of the page. Considering the disc, we candefine a rotating coordinate system o′ which is stuck to the disc at its centre sothat the unit vector e′xe′xe′x always points from the centre of the disc to the point p.The moving unit vector e′ye′ye′y shall be perpendicular to e′xe′xe′x, pointing upwards. Then, atthe instant when the bar is horizontal (as illustrated in Figure 5.36), the fixed androtating unit vectors will all be aligned, and may be used interchangeably.

The velocity of the point p with respect to a fixed observer may then be deter-mined, as


The centre of the disc never moves, so vo′/ovo′/ovo′/o = 0. Also, since the points p and o′ areboth stuck to the same rigid disc, vp/o′vp/o′vp/o′ = 0. Substituting,

vp/ovp/ovp/o = 0+0+ θ e′ze′ze′z ×R e′xe′xe′x= θR


)vp/ovp/ovp/o = θR eyeyey

where we were able to make the substitution eyeyey = e′ye′ye′y, because the moving and fixedunit vectors are instantaneously aligned at the instant of interest. The angle θ wehave defined as the angle swept by the rotation of the disc, as illustrated. Next, wecan determine the acceleration of the point p when the bar is horizontal, as

5.7 Examples 409

Fig. 5.36 Constrained motion with forces- defining coordinates.


Again, because the centre of the disc never moves and both the points p and o′ arestuck to the disc, ao′/oao′/oao′/o = ap/o′ap/o′ap/o′ = vp/o′vp/o′vp/o′ = 0. Substituting,

ap/oap/oap/o = 0+0+2θ e′ze′ze′z ×0+ θ e′ze′ze′z ×R e′xe′xe′x + θ e′ze′ze′z × θ e′ze′ze′z ×R e′xe′xe′x= θR


)+ θ e′ze′ze′z × θR


)= θR e′ye′ye′y + θ e′ze′ze′z × θR e′ye′ye′y= θR e′ye′ye′y + θ 2R


)ap/oap/oap/o = θR eyeyey − θ 2R exexex

where again we have recognized that the moving and fixed coordinates are inter-changeable at the instant of interest. Next, we consider the bar. We shall redefineour moving coordinate system so that the moving origin o′ is stuck to the bar at thepoint q, and the moving coordinates once again coincide with the fixed coordinatesat the instant of interest. Once again, the velocity of the point p with respect to afixed observer is



Because the points p and o′ are both stuck to the same rigid disc, vp/o′vp/o′vp/o′ = 0. However,since o′ is free to slide horizontally, the magnitude of vo′/ovo′/ovo′/o remains unknown. Forconvenience, we shall define θb as the angle subtended between the bar and thehorizontal. Substituting,

vp/ovp/ovp/o = xb e′xe′xe′x +0+ θb e′ze′ze′z ×(−L e′xe′xe′x

)= xb e′xe′xe′x − θbL


)vp/ovp/ovp/o = xb exexex − θbL eyeyey

where xb is the horizontal, linear velocity of the point o′ (which is unknown). Sincewe now have two expressions for vp/ovp/ovp/o, we may equate them together, as

θR eyeyey = xb exexex − θbL eyeyey

We may now equate together the coefficients of exexex, as

xb = 0

So, at the instant when the bar is horizontal, the slider is momentarily at rest. Equat-ing the coefficients of eyeyey,

θR = −θbL

θb = −RL

θ

We have now obtained an expression for the unknown angular velocity of the bar.Next, we can consider the acceleration of the point p, as


Once again, since the points p and o′ are both stuck to the same rigid bar, ap/o′ap/o′ap/o′ =vp/o′vp/o′vp/o′ = 0. However, the point o′ is free to accelerate horizontally with the slider.Substituting,

ap/oap/oap/o = xb e′xe′xe′x +0+2ωaxωaxωax ×0+ θb e′ze′ze′z ×(−L e′xe′xe′x

)+ θb e′ze′ze′z × θb e′ze′ze′z ×

(−L e′xe′xe′x)

= xb e′xe′xe′x − θbL(e′ze′ze′z ×e′xe′xe′x

)− θb e′ze′ze′z × θbL(e′ze′ze′z ×e′xe′xe′x

)= xb e′xe′xe′x − θbL e′ye′ye′y − θb e′ze′ze′z × θbL e′ye′ye′y

5.7 Examples 411

= xb e′xe′xe′x − θbL e′ye′ye′y − θ 2b L(e′ze′ze′z ×e′ye′ye′y

)= xb exexex − θbL eyeyey + θ 2

b L exexex

ap/oap/oap/o =(xb + θ 2

b L)

exexex − θbL eyeyey

We can now equate our two expressions for the acceleration of the point p when thebar is horizontal, as

θR eyeyey − θ 2R exexex =(xb + θ 2

b L)

exexex − θbL eyeyey

We may now equate the scalar coefficients of exexex, as

−θ 2R = xb + θ 2b L

xb = −θ 2R− θ 2b L

Substituting our earlier result for θb,

xb = −θ 2R−(−R

Lθ

)2

L

= −θ 2R− R2

Lθ 2

xb = −θ 2R

(1+

RL

)

And, finally, equating the coefficients of eyeyey in the earlier expression,

θR eyeyey = −θbL

θb = −RL

θ

Note that this result may have also been obtained by simple differentiation, since

θb = −RL

θ

ddt

(θb

)= − d

dt

(RL

θ

)

θb = −RL

θ


Notice that the angular velocity and acceleration of the bar will always be oppositein sense to the angular velocity and acceleration of the disc, as expected.

In order to do any force analysis of the bar, we will need to know both the angularacceleration of the bar θb as well as the acceleration of the centre of mass c of thebar. We must therefore apply the relative acceleration equation one last time.

ac/oac/oac/o = ao′/oao′/oao′/o +ac/o′ac/o′ac/o′ +2ωaxωaxωax ×vc/o′vc/o′vc/o′ +αaxαaxαax ×rc/o′rc/o′rc/o′ +ωaxωaxωax ×ωaxωaxωax ×rp/o′rp/o′rp/o′

Since the points c and o′ are both stuck to the same rigid rod, ac/o′ac/o′ac/o′ =vc/o′vc/o′vc/o′ = 0. Then,

ac/oac/oac/o = xb e′xe′xe′x +0+2ωaxωaxωax ×0+ θb e′ze′ze′z ×(−L

2e′xe′xe′x

)+ θb e′ze′ze′z × θb e′ze′ze′z ×

(−L

2e′xe′xe′x

)

= xb e′xe′xe′x − θbL2


)− θb e′ze′ze′z × θbL2


)= xb e′xe′xe′x − θb

L2

e′ye′ye′y − θb e′ze′ze′z × θbL2

e′ye′ye′y

= xb e′xe′xe′x − θbL2

e′ye′ye′y − θ 2b

L2


)= xb exexex − θb

L2

eyeyey + θ 2b

L2

exexex

ac/oac/oac/o =(xb + θ 2

bL2

)exexex − θb

L2

eyeyey

However, we may substitute into this expression our earlier results for xb, θb and θb,as

ac/oac/oac/o =

(−θ 2R

(1+

RL

)+

(−R

Lθ

)2 L2

)exexex −

(−R

Lθ

)L2

eyeyey

ac/oac/oac/o = −Rθ 2(

1+12

RL

)exexex +

12

Rθ eyeyey

Knowing the accelerations of the centres of mass of the disc and the bar, as wellas the angular accelerations of the disc and the bar, we may now carry out a properforce analysis of each component. Let us begin with the disc. Applying the law ofconservation of linear momentum to the disc, and defining our system as the discalone and using the same coordinates as earlier,


5.7 Examples 413

where aaaaaa is the acceleration of the centre of mass of the disc. However, since thecentre of the disc remains fixed, aaaaaa = 0. So, examining the free-body diagram of thedisc when θ = 90◦ (Figure 5.37),

Fig. 5.37 Constrained motion with forces- free-body diagrams.

FextFextFext = ma(0)

FRxFRxFRx +FRyFRyFRy +FAxFAxFAx +FAyFAyFAy −FgFgFg = 0

FRx exexex +FRy eyeyey +FAx exexex +FAy eyeyey −mag eyeyey = 0(FRx +FAx

)exexex +

(FRy +FAy −mag

)eyeyey = 0

where FRx and FRy are the horizontal and vertical components of the reaction forceat the hub, while FAx and FAy are the horizontal and vertical components of the forceapplied to the disc by the bar, and all four of these forces are unknown. Equating thecoefficients of exexex,

FRx +FAx = 0

FAx = −FRx (5.8)

Similarly, equating the coefficients of eyeyey,

FRy +FAy −mag = 0


FAy = mag−FRy (5.9)

Note that the forces FAx, FRx, FAy and FRy remain unknown. Next, we may apply thelaw of conservation of angular momentum to the disc.


The angular momentum HcHcHc of the disc about its centre of mass is easily evaluatedas


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz


However, in this case, ωz = θ . Also, the moment of inertia of a disc about its centreof mass is already known (and can be found in Appendix C). Substituting,

HcHcHc =12

maR2θ ezezez

Also, from the free-body diagram, we may evaluate the moments acting on the discabout the centre of mass when θ = 90◦, as

∑Mext/cMext/cMext/c = RFAy ezezez

Substituting these results into the angular momentum equation,

Mext/cMext/cMext/c = HcHcHc =ddt

(HcHcHc)

RFAy ezezez =ddt

(12

maR2θ ezezez

)

RFAy ezezez =12

maR2θ ezezez

FAy =12

maRθ

The unit vector remains unchanged by the differentiation as the disc is rotating ina plane (the direction of ωωω is constant), so the only time-dependent variable in theexpression is θ . We may now substitute in our earlier result for FAyFAyFAy from Equation(5.9), as

(mag−FRy

)=

12

maRθ

5.7 Examples 415

Isolating the unknown θ ,

θ =2gR

−2FRy

maR(5.10)

However, we still have the unknown θ . To eliminate this unknown, we apply thelaw of conservation of angular momentum to the bar. Let us redefine the systemas consisting of the bar only. Using the same coordinate system as in our earlierrelative motion analysis for the bar,


FAyR ezezez +FByR ezezez = HcHcHc

where we have taken the moments about the centre of mass when θ = 90◦. Theforce FBy is the vertical component of the force applied to the bar by the slider,which is unknown. We also note that the force applied to the bar by the disc is equalin magnitude but opposite in direction to the force applied to the disc by the bar.Since there is no obvious stationary point, our expression of the law of conservationof angular momentum can only be applied about the centre of mass. Evaluating theangular momentum, then,


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ 0

0ωz


which is similar to the result obtained for the disc. However, ωz here is the angularvelocity of the bar θb, which is not the same as the angular velocity of the disc. Aswith the disc, the moment of inertia of a bar about its centre of mass has alreadybeen solved, and is available in Appendix C. Substituting,

HcHcHc =112

mbL2θb ezezez

We may substitute this result into the angular momentum equation.

FAyR ezezez +FByR ezezez = HcHcHc

FAyR ezezez +FByR ezezez =ddt

(1

12mbL2θb ezezez

)

FAyR ezezez +FByR ezezez =1

12mbL2θb ezezez

FAyR+FByR =1

12mbL2θb


However, from our relative motion analysis we have already found an expression forθb in terms of θ , and we can substitute FAy once again with our result from Equation(5.9), yielding

(mag−FRy

)R+FByR =

112

mbL2(−R

Lθ

)

mag−FRy +FBy = − 112

mbLθ

Since we are not particularly interested in the force FBy, we can isolate this for latersubstitution.

FBy = FRy −mag− 112

mbLθ (5.11)

Finally, we will apply the law of conservation of linear momentum to the bar.

∑FextFextFext = GsysGsysGsys = mbacacac

where acacac = ac/oac/oac/o is the acceleration of the centre of mass of the bar. Examining thefree-body diagram when θ = 90◦, then,

FAxFAxFAx +FAyFAyFAy +FgFgFg +FBxFBxFBx +FByFByFBy = mbacacac

−FAx exexex −FAy eyeyey −mbgeyeyey +FBx exexex +FBy eyeyey = mbacacac(−FAx +FBx)

exexex +(−FAy −mbg+FBy

)eyeyey = mbacacac

where FBx is the horizontal component of the force applied to the bar by the slider.However, we already have an expression for acacac = ac/oac/oac/o for the bar, from our relativemotion analysis. Substituting,

(−FAx +FBx)

exexex +(−FAy −mbg+FBy

)eyeyey = mb

(−Rθ 2

(1+

12

RL

)exexex +

12

Rθ eyeyey

)(5.12)

Equating the coefficients of eyeyey,

−FAy −mbg+FBy =12

mbRθ

We may now substitute in our earlier results for FAy (Equation 5.9) and FBy (Equation5.11), yielding

5.7 Examples 417

−(mag−FRy)−mbg+

(FRy −mag− 1

12mbLθ

)=

12

mbRθ

−(2ma +mb)g+2FRy = mb

(1

12L+

12

R

)θ

Next, we eliminate the final unknown θ by substituting our result from the analysisof the disc (Equation 5.10),


(1

12L+

12

R

)(2gR

−2FRy

maR

)


(16

LR+1

)(g− FRy

ma

)

Having now eliminated all of the unknowns, we can finally solve for FRy, as


(16

LR+1

)g− mb

ma

(16

LR+1

)FRy

FRy

(2+

mb

ma

(16

LR+1

))= mb

(16

LR+1

)g+(2ma +mb)g

FRy = mbg16

LR +2 ma

mb+2

2+ mbma

(16

LR +1

)

This expression yields our result for the vertical component of the force applied tothe disc by the pin at its hub. Note that FRy > 0, indicating that the force is actingupwards and therefore that the disc support is applying an upwards external forceupon the disc. The disc is therefore applying an equal but downwards load upon thesupport, so that the pulley support is in compression when the mechanism is in theposition illustrated.

For the horizontal reaction force on the disc FRx, we equate the coefficients of exexex

in Equation (5.12).

−FAx +FBx =−mbRθ 2(

1+12

RL

)

We can equate FAx to FRx using Equation (5.8), but this still leaves the unknown FBx.To solve for this force, we need to briefly consider the slider. Since the slider has nomass, then the sum of external forces acting on the slider must be zero. Because theslider is free to move horizontally without friction, the only force acting horizontallyon the slider is FBx. Therefore, if we take our system as being only the slider, thehorizontal component of the linear momentum equation will yield


−FBx = (0)x = 0

Substituting this into our earlier expression,

−FAx +FBx = −mbR

(1+

12

RL

)θ 2

−(−FRx)+0 = −mbR

(1+

12

RL

)θ 2

FRx = −mbR

(1+

12

RL

)θ 2

which is the horizontal component of the reaction force at the hub of the disc. Themagnitude FRx is negative, indicating that the horizontal component of the forceapplied to the disc by the support is acting to the left at the instant shown, so thedisc is pulling the support toward the right. There remains an unknown θ , but thisquantity cannot be determined from the information given; the instantaneous angu-lar velocity of the disc when θ = 0 will depend on both (a) the value of θ at whichthe system was released, and (b) any initial angular velocity imparted upon the discwhen it was released. The sign of θ , however, will not affect the sign of FRx. Noticethat if the disc was released from rest at θ = 0, then θ = 0 and FRx = 0.

To solve this problem, we first recognized that we would need to know the an-gular acceleration of the bar and the acceleration of the centre of mass of the bar.Since we were looking for the reaction forces acting on the disc, it was necessary toexpress the acceleration of the bar in terms of the acceleration of the disc; to do this,we applied the equations of relative motion. Next, we applied the linear momentumequation and the angular momentum equation to both the disc and the bar, whichallowed us to eliminate all of the unknowns.

5.7.14 Rotating imbalances: point masses

So far, all of the examples considered have involved the analysis of objects un-dergoing planar (two-dimensional) rotation. Next, let us consider a case which isfundamentally three-dimensional.

Consider a thin, massless shaft of length L having two point objects of mass moffset by a distance R from the shaft centre, as illustrated in Figure 5.38. The endsof the shaft are mounted in the fixed bearings A on the left and B on the right. Themasses are on opposite sides of the shaft, a distance L/4 from each end, and theshaft rotates with a constant angular velocity ω . It would be useful to be able todetermine the reaction forces at the bearings for this simple rotating system.

Because the shaft is not translating in space and because the shaft has no angularacceleration along any axis, the analysis of the shaft reduces to a simple problem in

5.7 Examples 419

Fig. 5.38 Imbalanced rotating point masses.

statics. However, the difficulty arises in determining the loads applied to the shaft bythe objects; when the shaft is rotating, these loads may not be simply due to gravity.To solve this problem, we will first need to examine the masses, apply the law ofconservation of linear momentum to the masses in order to determine the forcesapplied to the masses by the shaft, and then realize that the force applied by theshaft to the mass is equal in magnitude but opposite in direction to the force appliedby the mass to the shaft.

Let us begin by examining the static case, where ω = 0. If we define our systemas consisting of the shaft and both the point masses only, then the only externalforces acting on the system are gravity and the reactions at the bearings. Becausethe shaft is free to spin, let us consider the moments about the shaft axis caused bygravity.

∑Mext/oMext/oMext/o =−magRsin(θ) ezezez +−mbgRsin(θ) ezezez

where ma is the mass of the point object closest to the bearing A, and mb is the massof the point object closest to B. However, since ma = mb = m,

Mext/oMext/oMext/o = −mgRsin(θ) ezezez +−mgRsin(θ) ezezez

Mext/oMext/oMext/o = 0


so the shaft will not rotate. Finally, since the bearings hold the shaft in place, it can-not accelerate in any direction and the acceleration of the system is zero. Therefore,

∑FextFextFext = GcGcGc

−mag eyeyey −mbg eyeyey +FAy eyeyey +FBy eyeyey +FAx exexex +FBx exexex = 0

−2mg eyeyey +FAy eyeyey +FBy eyeyey +FAx exexex +FBx exexex = 0

where FAy and FBy are the vertical components of the reaction forces at A and B,respectively, and FAx and FBx are the horizontal components of the reaction forcesat A and B, respectively. From symmetry, we also may deduce that FAy = FBy andFAx = FBx. Substituting,

−2mg eyeyey +FAy eyeyey +FAy eyeyey +FAx exexex +FAx exexex = 0

−2mg eyeyey +2FAy eyeyey +2FAx exexex = 0


2FAx = 0

FAx = FBx = 0

and equating the coefficients of eyeyey,

−2mg+2FAy = 0

FAy = FBy = mg

So each bearing is supporting an entirely vertical load of magnitude mg, as wouldbe expected. The shaft is in static equilibrium.

Let us now consider the case when the shaft is rotating with a constant angularvelocity ω . To begin with, let us define our system as the object closest to A only.Then, from the law of conservation of linear momentum,


where the point p is on the object closest to A, so ap/oap/oap/o is the acceleration vectorof that object with respect to a fixed observer. There are two external forces acting

5.7 Examples 421

Fig. 5.39 Free-body diagram of the object.

on the object: gravity and the tension in the bar connecting the object to the shaft(which can be divided into a horizontal component FT x and a vertical componentFTy). From the free-body diagram (Figure 5.39), then,

∑FextFextFext = FgFgFg +FTxFTxFTx +FTyFTyFTy

map/oap/oap/o = −mg eyeyey +FT x exexex +FTy eyeyey

map/oap/oap/o = FT x exexex +(−mg+FTy

)eyeyey

Because the path of the point object is constrained (it must move in circles), we canfind another expression for ap/oap/oap/o using relative motion. We shall define a fixed coor-dinate system o such that ezezez lies parallel to the shaft, pointing to the right in Figure5.38; eyeyey points upward and exexex points into the page, as illustrated. Next, we define amoving origin o′ which is stuck to the shaft, located on the shaft axis adjacent to theobject. The moving unit vector e′ye′ye′y shall always point from o′ to the mass; e′xe′xe′x shall beperpendicular to e′ye′ye′y, pointing down and to the right (when looking along the shaftaxis toward A), and e′ze′ze′z shall be parallel to ezezez. Then,


Because the mass is stuck to the shaft and cannot move relative to the shaft, ap/o′ap/o′ap/o′ =vp/o′vp/o′vp/o′ = 0. While the shaft rotates, the point o′ does not change its location in space;therefore ao′/oao′/oao′/o = 0. Also, because the moving coordinate system is stuck to the shaftand rotating with constant speed ω , ωaxωaxωax = ω e′ze′ze′z and αaxαaxαax = 0. Substituting,


= ω e′ze′ze′z ×ω e′ze′ze′z ×R e′ye′ye′y= ω e′ze′ze′z ×ωR


)= ω e′ze′ze′z ×−ωR e′xe′xe′x


= −ω2R(e′ze′ze′z ×e′xe′xe′x

)ap/oap/oap/o = −ω2R e′ye′ye′y

We now have an expression for the acceleration of the mass closest to A. We maynow substitute this back into the linear momentum equation, as


)eyeyey

m(−ω2R e′ye′ye′y

)= FT x exexex +

(−mg+FTy)

eyeyey

where the only unknowns are the components of the tension force, FT x and FTy.However, in order to equate the two sides of this expression, we must first resolve themoving unit vectors into components along the fixed unit vectors. From geometry,

e′xe′xe′x = cos(θ) exexex − sin(θ) eyeyey

e′ye′ye′y = sin(θ) exexex + cos(θ) eyeyey


Substituting,

m(−ω2R e′ye′ye′y

)= FT x exexex +

(−mg+FTy)

eyeyey

−mω2R(sin(θ) exexex + cos(θ) eyeyey

)= FT x exexex +

(−mg+FTy)

eyeyey

−mω2Rsin(θ) exexex −mω2Rcos(θ) eyeyey = FT x exexex +(−mg+FTy

)eyeyey


FT x =−mω2Rsin(θ)

and equating the coefficients of eyeyey,

−mω2Rcos(θ) = −mg+FTy

FTy = −mω2Rcos(θ)+mg

5.7 Examples 423

Finally, because the shaft is applying a force FT x exexex+FTy eyeyey upon the mass, the massmust simultaneously be applying a force −FT x exexex −FTy eyeyey upon the shaft. The totalforce FTAFTAFTA applied by the mass closest to A upon the shaft is therefore

FTAFTAFTA = mω2Rsin(θ) exexex +(mω2Rcos(θ)−mg

)eyeyey

Next, let us consider the object closest to B. We can immediately recognize thatthe free-body diagram of the object closest to B will be identical to the free-bodydiagram of the object closest to A, so the linear momentum equation will yield thesame result,



)eyeyey

though it is very important to note here that the horizontal and vertical componentsof the tension force in the bar connecting the object to the shaft, FTxFTxFTx and FTyFTyFTy, arenot necessarily the same as they were for the object closest to A.

Once again, we can obtain an additional expression for ap/oap/oap/o from relative motion.Let the moving unit vectors e′xe′xe′x, e′ye′ye′y and e′ze′ze′z point in exactly the same directions asbefore, but let us re-define the moving origin o′ as being stuck to the shaft adjacentto the object closest to B. The point p is now located on the object closest to B aswell. Then,


Once again, the mass is stuck to the shaft and cannot move relative to the shaft, andthe point o′ is fixed in space, so ap/o′ap/o′ap/o′ =vp/o′vp/o′vp/o′ = ao′/oao′/oao′/o = 0. Also, as before, ωaxωaxωax = ω e′ze′ze′zand αaxαaxαax = 0. Substituting,


= ω e′ze′ze′z ×ω e′ze′ze′z ×−R e′ye′ye′y= ω e′ze′ze′z ×−ωR


)= ω e′ze′ze′z ×+ωR e′xe′xe′x= ω2R


)ap/oap/oap/o = ω2R e′ye′ye′y

Substituting this result into the linear momentum equation,



)eyeyey

m(ω2R e′ye′ye′y

)= FT x exexex +

(−mg+FTy)

eyeyey

Resolving the moving coordinates into components along the fixed axes,

mω2R(sin(θ) exexex + cos(θ) eyeyey

)= FT x exexex +

(−mg+FTy)

eyeyey

mω2Rsin(θ) exexex +mω2Rcos(θ) eyeyey = FT x exexex +(−mg+FTy

)eyeyey


FT x = mω2Rsin(θ)


mω2Rcos(θ) = −mg+FTy

FTy = mω2Rcos(θ)+mg

As before, we may therefore say that the object closest to B is exerting a force onthe shaft equal to −FT x exexex −FTy eyeyey, so that the total resultant force FTBFTBFTB on the shaftis

FTBFTBFTB =−mω2Rsin(θ) exexex +(−mω2Rcos(θ)−mg

)eyeyey

Finally, let us define our system as the shaft itself. The external forces acting onthe shaft are therefore the forces applied by the objects to the shaft FTAFTAFTA and FTBFTBFTB,as well as the reaction forces at the bearings, as illustrated in Figure 5.40. Since theshaft is fixed in space, and because it is not undergoing any angular accelerationalong any axis,

∑FextFextFext = 0

∑Mext/qMext/qMext/q = 0

where the sum of the external moments may be taken about any point q. Beginningwith the forces, from the free-body diagram,

∑FextFextFext = FTAFTAFTA +FTBFTBFTB +FAxFAxFAx +FAyFAyFAy +FBxFBxFBx +FByFByFBy

5.7 Examples 425

Fig. 5.40 Free-body diagram of the shaft.

0 = FTAFTAFTA +FTBFTBFTB +FAx exexex +FAy eyeyey +FBx exexex +FByeyeyey

Substituting in the previous values for FTAFTAFTA and FTBFTBFTB,

∑FextFextFext = FTAFTAFTA +FTBFTBFTB +FAxFAxFAx +FAyFAyFAy +FBxFBxFBx +FByFByFBy

0 = mω2Rsin(θ) exexex +(mω2Rcos(θ)−mg

)eyeyey −mω2Rsin(θ) exexex

+(−mω2Rcos(θ)−mg

)eyeyey +FAx exexex +FAy eyeyey +FBx exexex +FBy eyeyey

Collecting terms,

0 =(FAx +FBx

)exexex +

(−2mg+FAy +FBy)

eyeyey


0 = FAx +FBx

FAx = −FBx


0 = −2mg+FAy +FBy

FAy = 2mg−FBy


We have obtained two equations with four unknowns (FAx, FAy, FBx and FBy); tosolve, we will need another two equations. For this, we may use the moments. Tak-ing moments about the point A,

∑Mext/AMext/AMext/A = ∑(rFrFrF ×FextFextFext

)where rFrFrF is the position vector of the point of application of each force. Then,

∑Mext/AMext/AMext/A = 0×FAx exexex +0×FAy eyeyey +L4

ezezez ×FTAFTAFTA

+3L4

ezezez ×FTBFTBFTB +L ezezez ×FBx exexex +L ezezez ×FBy eyeyey

Once again, there is no net moment, and we may eliminate FTAFTAFTA and FTBFTBFTB by substi-tuting the earlier results,

0 = 0+0+L4

ezezez ×(mω2Rsin(θ) exexex +

(mω2Rcos(θ)−mg

)eyeyey)

+3L4

ezezez ×(−mω2Rsin(θ) exexex +

(−mω2Rcos(θ)−mg)

eyeyey)

+L ezezez ×FBx exexex +L ezezez ×FBy eyeyey

Distributing the cross-products,

0 =L4

mω2Rsin(θ)(ezezez ×exexex

)+

L4

(mω2Rcos(θ)−mg

)(ezezez ×eyeyey

)−3L

4mω2Rsin(θ)

(ezezez ×exexex

)+

3L4

(−mω2Rcos(θ)−mg)(

ezezez ×eyeyey)

+FBxL(ezezez ×exexex

)+FByL

(ezezez ×eyeyey

)

And, carrying out the cross-products,

0 =L4

mω2Rsin(θ) eyeyey − L4

(mω2Rcos(θ)−mg

)exexex

−3L4

mω2Rsin(θ) eyeyey − 3L4

(−mω2Rcos(θ)−mg)

exexex

+FBxL eyeyey −FByL exexex

Collecting terms and simplifying,

5.7 Examples 427

0 =(−L

4mω2Rcos(θ)+

L4

mg+3L4

mω2Rcos(θ)+3L4

mg−FByL)

exexex

+(L

4mω2Rsin(θ)− 3L

4mω2Rsin(θ)+FBxL

)eyeyey

0 =(L

2mω2Rcos(θ)+Lmg−FByL

)exexex

+(−L

2mω2Rsin(θ)+FBxL

)eyeyey

Finally, we may equate the coefficients of each of the unit vectors to zero to solvefor the unknowns; beginning with exexex,

0 =L2

mω2Rcos(θ)+Lmg−FByL

FBy =12

mω2Rcos(θ)+mg


0 = −L2

mω2Rsin(θ)+FBxL

FBx =12

mω2Rsin(θ)

We now have a result for the reaction forces at B. To obtain the reaction forces at A,we need only substitute these values into the earlier result,

FAx = −FBx

FAx = −12

mω2Rsin(θ)

and

FAy = 2mg−FBy

= 2mg− (12

mω2Rcos(θ)+mg)

FAy = mg− 12

mω2Rcos(θ)


Therefore, the reaction forces at A and B may be expressed as

FAFAFA = −12

mω2Rsin(θ) exexex +(mg− 1

2mω2Rcos(θ)

)eyeyey

FBFBFB =12

mω2Rsin(θ) exexex +(1

2mω2Rcos(θ)+mg

)eyeyey

A few very interesting observations may be made from this result.

The reactions at the bearings are independent of the shaft length. This seems bizarre,but because we assumed the shaft to be perfectly rigid (it wasn’t deflecting orstretching as it was rotating), and because the shaft was supported on both ends,there was no way for the bearings to feel how long the shaft is.

If the shaft is not rotating, the trivial static result is recovered. If we set ω = 0,then FAFAFA = FBFBFB = mg eyeyey, which is what we had found in our earlier analysis of thestationary system.

If the shaft is rotating, then the reactions at the bearings depend on the angularvelocity and angular position of the masses. Because the shaft is rotating and theangle θ is constantly changing, the reactions at the bearings will be changing as afunction of time. In fact, this is what gives rise to the sensation of vibration when animbalanced load is caused to spin.

The fluctuating component of the reaction force is proportional to R. As R → 0,FAFAFA → mg eyeyey and FBFBFB → mg eyeyey, again recovering the trivial static result. The time-dependence of the reaction forces is due entirely to the distance between the massesand the axis of rotation.

5.7.15 Rotating imbalances: distributed masses

Let us now consider the case of a solid disc of near-zero thickness, mass m andradius R mounted through its centre on a shaft at some angle β from perpendicular,as illustrated in Figure 5.41. The shaft is rotating with a constant angular velocityω . The shaft is mounted through the centre of mass of the disc, so clearly the centreof mass cannot accelerate and ΣFextFextFext = 0. However, the disc may transmit momentsto the shaft if it is rigidly welded in place.

We can compute the sum of the moments transmitted from the disc to the shaftfrom the law of conservation of angular momentum. Let us define our system asconsisting of the disc only; then, any moments transmitted by the shaft onto the disc

5.7 Examples 429

Fig. 5.41 An imbalanced rotating mass.

are external ones. We will define a coordinate system o fixed in space such that ezezez

points along the shaft, and we will define an additional coordinate system o′ stuckto the disc such that e′ze′ze′z is normal to the disc at all times, as shown in the illustration.Then, since c is a centre of mass,

∑Mext/cMext/cMext/c = Hsys/cHsys/cHsys/c =ddt(IcωIcωIcω)

Remember that one of the rules of tensor multiplication is that both the origin andthe axes have to be the same for all the vectors and tensors. Since IcIcIc will only beconstant if the axes are stuck to the disc, it would be more convenient to resolve ωωωinto the o′ coordinate system. We can choose the direction of e′xe′xe′x so that it points inthe direction of tilt; then the component of ezezez in the x′ − y′ plane (which is just thesurface of the disc) will always be in the −e′xe′xe′x direction. Then,

ωωω = ω ezezez =−ω sin(β )e′xe′xe′x +ω cos(β )e′ze′ze′z

Now we can compute the angular momentum about the point c, in the rotating o′coordinate system.


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝−ω sin(β )

0ω cos(β )

⎞⎠

IcωIcωIcω =−Ixx/cω sin(β )e′xe′xe′x + Izz/cω cos(β )e′ze′ze′zSo, substituting this into the angular momentum equation,

∑Mext/cMext/cMext/c =ddt(IcωIcωIcω)


∑Mext/cMext/cMext/c =ddt

(−Ixx/cω sin(β )e′xe′xe′x + Izz/cω cos(β )e′ze′ze′z)

but ω is constant, β is constant, and III was evaluated in axes stuck to the body so itwill be constant too.

∑Mext/cMext/cMext/c = −Ixx/cω sin(β )e′xe′xe′x + Izz/cω cos(β )e′ze′ze′z= −Ixx/cω sin(β )(ωaxωaxωax ×e′xe′xe′x)+ Izz/cω cos(β )(ωaxωaxωax ×e′ze′ze′z)

where the time-derivative of a moving unit vector is given by Equation (2.7). Tominimize the length of these expressions, we will evaluate the cross products indi-vidually.

ωaxωaxωax ×e′xe′xe′x =(−ω sin(β )e′xe′xe′x +ω cos(β )e′ze′ze′z

)×e′xe′xe′x= −ω sin(β )(e′xe′xe′x ×e′xe′xe′x)+ω cos(β )(e′ze′ze′z ×e′xe′xe′x)= ω cos(β )e′ye′ye′y

ωaxωaxωax ×e′ze′ze′z =(−ω sin(β )e′xe′xe′x +ω cos(β )e′ze′ze′z

)×e′ze′ze′z= −ω sin(β )(e′xe′xe′x ×e′ze′ze′z)+ω cos(β )(e′ze′ze′z ×e′ze′ze′z)= ω sin(β )e′ye′ye′y

So, substituting these derivatives back into the angular momentum equation,

∑Mext/cMext/cMext/c = −Ixx/cω sin(β )(ω cos(β )e′ye′ye′y

)+ Izz/cω cos(β )

(ω sin(β )e′ye′ye′y

)∑Mext/cMext/cMext/c = ω2 sin(β )cos(β )(Izz/c − Ixx/c)e′ye′ye′y

For a thin disc, Izz/c = mR2/2 and Ixx/c = mR2/4;

∑Mext/cMext/cMext/c = ω2 sin(β )cos(β )(1

2mR2 − 1

4mR2)e′ye′ye′y

=14

mR2ω2 sin(β )cos(β )e′ye′ye′y

5.7 Examples 431

So there is a resultant moment on the disc by the shaft about an axis pointing alongthe rotating unit vector e′ye′ye′y. To determine the reactions at the bearings (if desired),it would only be necessary to again consider the shaft as a static system. Whereaswith the point masses there were two external applied forces upon the shaft by theobjects, in this case, there would be a single external moment applied to the centreof the shaft by the disc (which would be −ΣMext/cMext/cMext/c, and would be time-dependent).Note also that if β → 0, then ΣMext/cMext/cMext/c → 0 as well.

5.7.16 Gyroscopes

A simple gyroscope or top may be approximated as a disc of mass m and radiusR mounted centrally on a thin shaft of length L and set spinning with a constantangular velocity ω , as illustrated in Figure 5.42. The axis of the gyroscope formsa small angle φ relative to the vertical, and the contact between the shaft and theground (which we shall define as point o) is assumed to be able to transmit nomoments (like a ball-and-socket joint).

Fig. 5.42 A simple gyroscope.

Let us define our system as consisting of the disc and the shaft. Since the point ofcontact between the shaft and the ground is at rest, we will be able to neglect orbitalangular momentum about that point and so the angular momentum equation is

∑Mext/oMext/oMext/o = Hsys/oHsys/oHsys/o =ddt(IωIωIω)

Let us define a coordinate system o′ which is stuck to the disc. The origin of o′will be coincident with o, and the axes will be oriented so that e′ze′ze′z always points


upwards along the shaft and, at some instant, e′xe′xe′x points along the surface of the disctowards the ground and e′ye′ye′y points straight into the page in the illustration. Taking themoments about o at this instant, then,

∑Mext/oMext/oMext/o = FgLsin(φ)e′ye′ye′y

Because the normal force is acting through the point o, it can exert no momentsabout that point. Next, we can compute the angular momentum about o in the rotat-ing coordinate system assuming that ωωω = ωze′ze′ze′z.


⎡⎣ Ixx/o 0 0

0 Iyy/o 00 0 Izz/o

⎤⎦⎛⎝ 0

0ωz

⎞⎠= Izz/oωz e′ze′ze′z

Substituting these results into the angular momentum equation,

∑Mext/oMext/oMext/o =ddt(HoHoHo)

FgLsin(φ)e′ye′ye′y =ddt(Izz/oωz e′ze′ze′z)

= Izz/oωz e′ze′ze′z= Izz/oωz(ωaxωaxωax ×e′ze′ze′z)

= Izz/oωz(ωz e′ze′ze′z ×e′ze′ze′z)

FgLsin(φ)e′ye′ye′y = 0 ???

This is an impossibility! Neither Fg or L are zero, and φ is not necessarily zero either.Since the only assumption we made was that ωωω = ωze′ze′ze′z, this assumption thereforecannot be valid, which tells us that it is impossible for the gyroscope to have anangular velocity only along its shaft.

Let us instead now say that the gyroscope also has an angular velocity Ω causingit to rotate about the vertical ezezez axis; this is the only other axis about which it mightrotate without immediately crashing into the ground. However, this does give usanother problem- to resolve Ω ezezez into our rotating coordinate system, we wouldneed to know the spin angle θ of the gyroscope (where ωz = θ ), since there will becomponents of Ω along both e′xe′xe′x and e′ye′ye′y that will change with θ . This would lead tosome really horrifying three-dimensional geometry and algebra.

To avoid this and make the calculations much easier, we can assume that thegyroscope is axisymmetric about its shaft. Then, IoIoIo will not depend on θ . If we thenredefine o′ so that e′ze′ze′z always points upwards along the shaft but the axes are notspinning around with the body, in the angular momentum equation, IoIoIo will still beconstant and may still be pulled out of the derivative.

Resolving the combined angular velocity vector ω e′ze′ze′z +Ω ezezez into this newly re-defined o′ system,

5.7 Examples 433

ωωω +ΩΩΩ = ω e′ze′ze′z +Ω(cos(φ)e′ze′ze′z − sin(φ)e′xe′xe′x

)=(ω +Ω cos(φ)

)e′ze′ze′z −Ω sin(φ)e′xe′xe′x

So, computing the new angular momentum vector,


⎡⎣ Ixx/o 0 0

0 Iyy/o 00 0 Izz/o

⎤⎦⎛⎝ −Ω sin(φ)

0ωz +Ω cos(φ)

⎞⎠

=−Ixx/oΩ sin(φ)e′xe′xe′x + Izz/o(ωz +Ω cos(φ)

)e′ze′ze′z

And the new angular momentum equation becomes

∑Mext/oMext/oMext/o =ddt

III(ωωω +ΩΩΩ)

FgLsin(φ)e′ye′ye′y =ddt

(−Ixx/oΩ sin(φ)e′xe′xe′x + Izz/o(ωz +Ω cos(φ)

)e′ze′ze′z)

FgLsin(φ)e′ye′ye′y = −Ixx/oΩ sin(φ) e′xe′xe′x + Izz/o

(ωz +Ω cos(φ)

)e′ze′ze′z

Remember now that our coordinate system is fixed to the axis of the gyroscopebut not spinning along with it, so, ωaxωaxωax = Ω cos(φ)e′ze′ze′z −Ω sin(φ)e′xe′xe′x. Evaluating thederivatives of the unit vectors (using Equation 2.7), then,

e′xe′xe′x = ωaxωaxωax ×e′xe′xe′x=(Ω cos(φ)e′ze′ze′z −Ω sin(φ)e′xe′xe′x

)×e′xe′xe′x= Ω cos(φ)(e′ze′ze′z ×e′xe′xe′x)−Ω sin(φ)(e′xe′xe′x ×e′xe′xe′x)= Ω cos(φ)e′ye′ye′y

e′ze′ze′z = ωaxωaxωax ×e′ze′ze′z=(Ω cos(φ)e′ze′ze′z −Ω sin(φ)e′xe′xe′x

)×e′ze′ze′z= Ω cos(φ)(e′ze′ze′z ×e′ze′ze′z)−Ω sin(φ)(e′xe′xe′x ×e′ze′ze′z)= Ω sin(φ)e′ye′ye′y

Substituting this back into the angular momentum equation,

FgLsin(φ)e′ye′ye′y = −Ixx/oΩ sin(φ) e′xe′xe′x + Izz/o

(ωz +Ω cos(φ)

)e′ze′ze′z


= −Ixx/oΩ sin(φ)(Ω cos(φ)e′ye′ye′y

)+ Izz/o

(ωz +Ω cos(φ)

)(Ω sin(φ)e′ye′ye′y)

Combining the terms and dividing out the unit vector,

FgLsin(φ) = −Ixx/oΩ 2 sin(φ)cos(φ)+ Izz/oωzΩ sin(φ)+ Izz/oΩ 2 sin(φ)cos(φ)

= (Izz/o − Ixx/o)Ω2 sin(φ)cos(φ)+ Izz/oωzΩ sin(φ)

If we assume the gyroscope to be a uniform solid disc on a slender, massless shaft,Izz/o = mR2/2, and, from the parallel axis theorem, Ixx/o = mL2 + mR2/4. Also,Fg = mg. Substituting,

mgLsin(φ) =

(12

mR2 −mL2 − 14

mR2)

Ω 2 sin(φ)cos(φ)+

(12

mR2)

ωzΩ sin(φ)

gsin(φ) =

(14

R2

L2 −1

)LΩ 2 sin(φ)cos(φ)+

(12

R2

L2

)LωzΩ sin(φ)

Some interesting observations may be made from this equation. First of all, if theangle φ is zero and the gyroscope is perfectly vertical, the angular momentum equa-tion yields a trivial result (0 = 0). However, as soon as any tiny angle developsbetween the axis and vertical, the axis of the gyroscope will begin to rotate aroundconically with some speed Ω . This secondary rotation is called precession, and willalways happen since the φ = 0 condition is unstable. Secondly, all of the terms in thisequation may remain constant without violating the law of conservation of angularmomentum- so a gyroscope which is inclined at some angle φ while both spinningand precessing is stable in that condition (which is called steady-state precession).

If we assume that φ > 0, then we can divide both sides of the above expressionby Lsin(φ) to yield,

gL=

(14

R2

L2 −1

)Ω 2 cos(φ)+

(12

R2

L2

)ωzΩ (5.13)

From this expression we can observe another important property of spinning gyro-scopes. If the gyroscope is unpowered, then the angular velocity ωz of the rotatingdisc will gradually decrease as a result of friction and/or drag. While we have notconsidered the more general (and mathematically complex) case of ωz �= 0, we maystill make a few observations by assuming that the angular accelerations are verysmall. As ωz decreases, so does the angular momentum of the gyroscope; conse-quently, it will be ‘less able to resist’ the moment applied by gravity and the gyro-scope will begin to tip over. This motion will increase the angle φ , decreasing the

5.7 Examples 435

quantity cos(φ). Therefore, to satisfy Equation (5.13) above, the precession rate Ωmust increase. This is why spinning tops always eventually tip over, and wobblearound faster as they do.

Another good example of precessional motion is an artillery shell; these projec-tiles are pointed on the front in order to reduce the drag on them. In order to keepthem pointed forward and prevent them from tumbling (which would increase dragsignificantly), they are caused to spin. This makes it more difficult for the wind load-ing to cause the shell to tumble, but as a consequence of the spinning and the windloading, the shell starts to precess. As the spin speed decays as a result of frictiondrag, the precession angle increases- and this is effectively what limits the range-not the size of the explosive charge used to fire the shell. So-called ‘training shells’are used for experimental testing and training of gun crews, and are fitted with littlefins to deliberately slow the rotation rate and thereby severely limit the range (seeFigure 5.43).

Fig. 5.43 A sketch comparing live artillery rounds to training artillery rounds.

It can also be demonstrated from Equation (5.13) that a spinning top only stopsspinning because the side strikes the ground; the expression can equally be satisfiedwith φ = 90◦, as

gL=

(14

R2

L2 −1

)Ω 2 cos(90◦)+

(12

R2

L2

)ωzΩ

gL=

(14

R2

L2 −1

)Ω 2(0)+

12

R2

L2 ωzΩ

g =12

R2

LωzΩ

Ω =2gLR2ωz


5.7.17 Constrained three-dimensional rotation

Let us now examine a case where the angular velocity of an object is forced tochange direction as a function of time. Consider a disc of mass m and radius R,pinned at its centre to a massless shaft of length L. The other end of the shaft ispinned to a swivel support, so that the disc is free to roll. The disc rests on a conicalsurface of inclination θ , as illustrated in Figure 5.44. The central (vertical-axis) shaftis caused to rotate with a constant angular velocity ΩΩΩ, and the disc rolls withoutslipping (with an angular velocity ωωω) This is actually a good representation of aform of grain mill.

Fig. 5.44 Constrained three-dimensional rotation: the conical grain mill.

To begin, because the disc is rolling without slipping, the instantaneous point ofcontact between the disc and the surface must be at rest. Using the relative velocityequation, then, it is possible to express ω in terms of Ω ; we can thereby entirelyeliminate the variable ω from the problem.

Let us define a moving coordinate system o′ with its origin stuck to the centre ofthe swivel pin so that it rotates with the shaft but not with the disc; this will, again,avoid some horrifying three-dimensional geometry. The e′xe′xe′x unit vector shall pointalong the shaft, away from the disc; e′ye′ye′y shall point upward and parallel to the planeof the disc, and e′ze′ze′z must then point in the direction of motion of the centre of the disc,as shown. The fixed unit vector exexex shall point to the right, eyeyey shall point upwards,

5.7 Examples 437

and ezezez will therefore point out of the plane of the page. If the point p is defined asbeing at the centre of the disc, then,


Because the point o′ is not translating, vo′/ovo′/ovo′/o = 0. Both p and o′ are stuck to the sameshaft, and the shaft does not deform, so vp/o′vp/o′vp/o′ = 0 as well. The moving coordinatesystem does not spin about the shaft with the disc, so ωaxωaxωax = Ωeyeyey. Substituting,

vp/ovp/ovp/o = 0+0+Ω eyeyey ×−L e′xe′xe′xvp/ovp/ovp/o = −ΩL

(eyeyey ×e′xe′xe′x

)

In order to carry out the cross-product, the moving unit vectors must be expressedin terms of the stationary ones. From geometry, then, at the instant of interest,




Also, because the angular momentum is easiest to determine in components parallelto unit vectors stuck to the rotating object, it is convenient to express the fixed unitvectors in terms of the moving ones. At the moment of interest, then,

exexex = cos(θ) e′xe′xe′x − sin(θ) e′ye′ye′yeyeyey = sin(θ) e′xe′xe′x + cos(θ) e′ye′ye′yezezez = e′ze′ze′z

Arbitrarily, we will select the fixed coordinate system to solve our equation for vp/ovp/ovp/o;substituting our expressions for e′xe′xe′x and e′ye′ye′y into our earlier result,

vp/ovp/ovp/o = −ΩL(eyeyey ×e′xe′xe′x

)= −ΩL

(eyeyey ×


))= −ΩLcos(θ)

(eyeyey ×exexex

)−ΩLsin(θ)(eyeyey ×eyeyey

)= ΩLcos(θ)

(ezezez)−ΩLsin(θ)

(0)

vp/ovp/ovp/o = ΩLcos(θ) ezezez


So far, we have entirely ignored ω . Since the only angular velocity in the relativevelocity equation is ωaxωaxωax, in order to relate ω to Ω using the relative velocity equa-tion, we must temporarily redefine ωaxωaxωax so that it includes ω . To this end, we willredefine the moving origin o′ so that it is coincident with the point p, and stuck tothe disc. The moving unit vectors e′xe′xe′x, e′ye′ye′y and e′ze′ze′z will have exactly the same orienta-tion as before (at this particular instant of interest), but will now also be spinningaround with the disc. Then,

ωaxωaxωax = ωωω +ΩΩΩ

ωaxωaxωax = ω e′xe′xe′x +Ω eyeyey

Let us now define q as the instantaneous point of contact between the disc andthe surface. Applying the relative velocity equation with our temporarily redefinedcoordinate system,

vq/ovq/ovq/o = vp/ovp/ovp/o +vq/pvq/pvq/p +ωaxωaxωax ×rq/prq/prq/p

recalling, of course, that o′ and p are coincident. Since the disc is undergoing purerolling motion, q will be instantaneously at rest and vq/ovq/ovq/o = 0. Also, both p and q arefixed to the same rigid disc, so vq/pvq/pvq/p = 0. Substituting, then,

0 = ΩLcos(θ) ezezez +0+(ω e′xe′xe′x +Ω eyeyey

)×−R e′ye′ye′y= ΩLcos(θ) ezezez −ωR


)−ΩR(eyeyey ×e′ye′ye′y

)= ΩLcos(θ) ezezez −ωR e′ze′ze′z −ΩR

(eyeyey ×

(−sin(θ) exexex + cos(θ) eyeyey))

= ΩLcos(θ) ezezez −ωR e′ze′ze′z +ΩRsin(θ)(eyeyey ×exexex

)−ΩRcos(θ)(eyeyey ×eyeyey

)= ΩLcos(θ) ezezez −ωR e′ze′ze′z −ΩRsin(θ) ezezez −ΩRcos(θ)

(0)

0 = ΩLcos(θ)−ωR−ΩRsin(θ)

where we have recognized that, at the moment of interest, e′ze′ze′z = ezezez. Isolating ω , then,

ω = Ω

(LR

cos(θ)− sin(θ)

)

The angular velocity of the object may therefore be expressed as

ωωω +ΩΩΩ = ω e′xe′xe′x +Ω eyeyey

5.7 Examples 439

= Ω

(LR

cos(θ)− sin(θ)

)e′xe′xe′x +Ω

(sin(θ) e′xe′xe′x + cos(θ) e′ye′ye′y

)ωωω +ΩΩΩ = Ω

LR

cos(θ) e′xe′xe′x +Ω cos(θ) e′ye′ye′y

Because the law of conservation of angular momentum is expressed in terms of themoving unit vectors, we have decomposed eyeyey into components lying parallel withe′xe′xe′x and e′ye′ye′y for simplicity.

With ωωω now eliminated, we can apply the law of conservation of angular momen-tum. Let us define the system as consisting of the disc and the shaft only. We willthen stick a moving coordinate system o′ to the upper end of the shaft, as before;e′xe′xe′x will point along the shaft away from the disc, e′ye′ye′y will be oriented upwards andperpendicular to the shaft, and e′ze′ze′z will point in the direction of motion of the centreof the disc. Again, the moving unit vectors will rotate around with the shaft with anangular velocity of ΩΩΩ, but will not spin with the disc about the shaft. Because o′ is astationary point, we can use this as the origin of the system and thereby ensure thatthe orbital angular momentum vanishes.

From the law of conservation of angular momentum, then,

∑Mext/o′Mext/o′Mext/o′ = Ho′Ho′Ho′ =ddt

(Io′Io′Io′ωωω

)We may now consider the free-body diagram of the wheel and the shaft (illustratedin Figure 5.45). Because the wheel is undergoing pure rolling motion at constantspeed, the magnitude of the static friction force between the disc and the surfacewill be zero; then, because the disc is not accelerating along the z-axis, there can beno reaction force at the pivot along that direction either. The reaction forces at thepivot along the x- and y-axes, FxFxFx and FyFyFy, cannot produce any moments about o′ asthey are acting at this point. The pivot joint and shaft can only transmit moments tothe wheel along the y-axis, and since Ω is constant, then there can be no momentabout this axis either. The only moments acting on the system about the point o′,then, are resulting from the weight of the wheel and the normal force between thewheel and the surface.

∑Mext/o′Mext/o′Mext/o′ = ∑rrr×FextFextFext

= −L e′xe′xe′x ×−Fg eyeyey +(−L e′xe′xe′x −R e′ye′ye′y

)×Fn e′ye′ye′y= −L e′xe′xe′x ×−mg

(sin(θ) e′xe′xe′x + cos(θ) e′ye′ye′y

)− (L e′xe′xe′x +R e′ye′ye′y)×Fn e′ye′ye′y

= mgLsin(θ)(e′xe′xe′x ×e′xe′xe′x

)+mgLcos(θ)


)−FnL


)−FnR(e′ye′ye′y ×e′ye′ye′y

)= mgLsin(θ)

(0)+mgLcos(θ) e′ze′ze′z −FnL e′ze′ze′z −FnR

(0)

∑Mext/o′Mext/o′Mext/o′ =(mgLcos(θ)−FnL

)e′ze′ze′z


Fig. 5.45 Constrained three-dimensional rotation: free-body diagram.

The angular momentum about the origin, Ho′Ho′Ho′ , may equally be evaluated, as

Ho′Ho′Ho′ = Io′ωIo′ωIo′ω =

⎡⎣ Ixx/o′ 0 0

0 Iyy/o′ 00 0 Izz/o′

⎤⎦⎛⎝Ω L

R cos(θ)Ω cos(θ)

0

⎞⎠

Ho′Ho′Ho′ = Ixx/o′ΩLR

cos(θ) e′xe′xe′x + Iyy/o′Ω cos(θ) e′ye′ye′y

Here, we have expressed everything in components along the moving unit vectorse′xe′xe′x, e′ye′ye′y and e′ze′ze′z. Differentiating the angular momentum vector with respect to time,

Ho′Ho′Ho′ =ddt

(Io′ωIo′ωIo′ω

)Ho′Ho′Ho′ =

ddt

(Ixx/o′Ω

LR

cos(θ) e′xe′xe′x + Iyy/o′Ω cos(θ) e′ye′ye′y)

Because the axes were stuck to the shaft and not to the disc itself, it could be possiblethat dIo′Io′Io′/dt �= 0; however, if it is assumed that the disc is axisymmetric, then theposition of the object will not change relative to the moving coordinate system asthe disc spins. The terms of the inertia tensor will therefore be constant. Then, sinceΩ is constant as well,


cos(θ) e′xe′xe′x + Iyy/o′Ω cos(θ) e′ye′ye′y

5.7 Examples 441

Because e′xe′xe′x and e′ye′ye′y change direction as the disc moves, the time-derivatives of theseunit vectors do not vanish. From Equation (2.7), then,

e′xe′xe′x = ωaxωaxωax ×e′xe′xe′x= Ω eyeyey ×e′xe′xe′x= Ω eyeyey ×


)= Ω cos(θ)

(eyeyey ×exexex

)+Ω sin(θ)

(eyeyey ×eyeyey

)= −Ω cos(θ) ezezez +Ω sin(θ)

(0)

e′xe′xe′x = −Ω cos(θ) ezezez

and, similarly,

e′ye′ye′y = ωaxωaxωax ×e′ye′ye′y= Ω eyeyey ×e′ye′ye′y= Ω eyeyey ×

(−sin(θ) exexex + cos(θ) eyeyey)

= −Ω sin(θ)(eyeyey ×exexex

)+Ω cos(θ)

(eyeyey ×eyeyey

)= Ω sin(θ) ezezez +Ω cos(θ)

(0)

e′ye′ye′y = Ω sin(θ) ezezez

Substituting this into our earlier result for Ho′Ho′Ho′ ,


cos(θ)(−Ω cos(θ) ezezez

)+ Iyy/o′Ω cos(θ)

(Ω sin(θ) ezezez

)= −Ixx/o′Ω

2 LR

cos2(θ) ezezez + Iyy/o′Ω2 cos(θ)sin(θ) ezezez

Ho′Ho′Ho′ = Ω 2 cos2(θ)

(Iyy/o′

sin(θ)cos(θ)

− Ixx/o′LR

)ezezez

Finally, substituting this result for Ho′Ho′Ho′ and our earlier expression for ΣMext/o′Mext/o′Mext/o′ intothe law of conservation of angular momentum,

∑Mext/o′Mext/o′Mext/o′ = Ho′Ho′Ho′(mgLcos(θ)−FnL

)e′ze′ze′z = Ω 2 cos2(θ)

(Iyy/o′

sin(θ)cos(θ)

− Ixx/o′LR

)ezezez

Fn = mgcos(θ)−Ω 2 cos2(θ)

(Iyy/o′

Lsin(θ)cos(θ)

− Ixx/o′

R

)


The moments of inertia for a disc have been solved earlier, and are listed in Ap-pendix C (though it is important to notice that the axes are defined differently inthis problem). Note that we will need to apply the parallel axis theorem in order toobtain Iyy/o′ , as the centre of mass is offset by a distance L from the origin used.

Fn = mgcos(θ)−Ω 2 cos2(θ)

((14

mR2

L+

112

mH2

L+m

L2

L

)sin(θ)cos(θ)

− 12

mR2

R

)

Fn = mgcos(θ)− 12

mRΩ 2 cos2(θ)

((12

RL+

16

H2

LR+2

LR

)sin(θ)cos(θ)

−1

)

where H is the thickness of the disc. Let us first consider the special case whenθ = 0 and the shaft is horizontal. Then,

Fn = mg+12

mRΩ 2

In the special case when the shaft is horizontal, then, moment incurred as a conse-quence of the changing direction of ωωω always acts to increase the normal force, asFn > 0 regardless of the magnitude of Ω . Also, if the disc stops spinning (so thatΩ = 0), then Fn = mg, as expected.

Let us next consider the special case where the disc is very small. If the dimen-sions of the disc are negligible, then the disc begins to approximate a point massat the end of the shaft, and the angular momentum of the disc will vanish from ourexpression for Fn. Let us assume, therefore, R L and H L. Then, a number ofthe terms in our expression for the normal force will become negligibly small, as

Fn = mgcos(θ)− 12

mRΩ 2 cos2(θ)

((12

RL+

16

H2

LR+2

LR

)sin(θ)cos(θ)

−1

)

= mgcos(θ)− 12

mΩ 2 cos2(θ)

((12

R2

L+

16

H2

L+2L

)sin(θ)cos(θ)

−R

)

Fn ≈ mgcos(θ)−mΩ 2Lcos(θ)sin(θ)

The first term on the right-hand side of this expression is clearly the moment abouto′ resulting from the weight of the disc; the second term is the moment about o′produced by the product of the mass m of the disc, the square of the angular velocityΩ and the distance Lcos(θ) between the disc and the axis of rotation. We mayrecognize this as the surface-normal component of the centripetal term (mΩΩΩ×ΩΩΩ×rrr).

5.7 Examples 443

When θ = 0, the centripetal acceleration is entirely parallel to the surface, and so Fn

remains unaffected.Next, we may observe that there are both positive and negative terms in our

general expression for Fn; it is therefore possible for conditions to exist under whichFn = 0 and the disc will lose contact with the surface. Under these circumstances,then,

Fn = 0 = mgcos(θ)− 12

mRΩ 20 cos2(θ)

((12

RL+

16

H2

LR+2

LR

)sin(θ)cos(θ)

−1

)

mgcos(θ) =12

mRΩ 20 cos2(θ)

((12

RL+

16

H2

LR+2

LR

)sin(θ)cos(θ)

−1

)

Ω0 =

√2g

Rcos(θ)

((12

RL+

16

H2

LR+2

LR

)sin(θ)cos(θ)

−1

)− 12

where Ω0 is the critical angular velocity of the central pivot at which the disc justbegins to lose contact with the surface. As we have found earlier, it is impossible forthe disc to lift off the surface when θ = 0; substituting this value into our expressionfor Ω0, we find

Ω0 =

√−2g

R

This imposed condition of θ = 0 yields an imaginary critical angular velocity, whichis clearly nonphysical and is therefore impossible, as expected. In order for the discto be able to lose contact with the surface, then, Ω0 must remain real. Therefore, thecondition, (

12

RL+

16

H2

LR+2

LR

)sin(θ)cos(θ)

−1 > 0

must be met; otherwise, Fn > 0 for all θ . Re-arranging,

(12

RL+

16

H2

LR+2

LR

)tan(θ) > 1

tan−1

(2

RL + 1

3H2

LR +4 LR

)< θ

Finally, if it is further assumed that H is small (or that the disc is very thin), then forit to be possible for some Ω0 to exist at which the disc could lose contact with thesurface,


θ > tan−1

(2

RL +4 L

R

)

This expression is also plotted in Figure 5.46. Interestingly, the plot has a localmaximum at R/L = 2, meaning that for a conical mill of the type illustrated inFigure 5.44, a radius-to-length ratio of 2 will allow the largest possible cone angle(at 26◦) while ensuring that the stone can never leave the surface, even for very largeΩ .

Fig. 5.46 Constrained three-dimensional rotation: Range of angles at which Fn = 0 can occur, asa function of R/L (for the special case of small H).

This analysis has revealed that the normal force acting between the disc andthe ground is affected by three separate mechanisms: the moment resulting fromgravitational acceleration of the disc (in the +ezezez direction), the moment resultingfrom the centripetal acceleration of the disc (in the −ezezez direction), and the momentresulting from the angular acceleration caused by the change in direction of ωωω (inthe +ezezez direction). Consequently, the centripetal acceleration is the only influencewhich may cause the disc to leave the surface, and this could only happen within acertain range of R/L and θ .

5.7.18 Gimbal gyroscopes

A gimbal is a set of concentric rings pinned together at 90◦, as illustrated in Fig-ure 5.47. If a gyroscope is mounted inside a gimbal, two important things are true

5.7 Examples 445

of the gyroscope: first, no external moments can be transmitted to the gyroscopethrough the gimbal mechanism, and second, the centre of mass of the gyroscope liessimultaneously on all the axes of rotation, so gravity cannot exert a moment either.

Fig. 5.47 A gyroscope mounted in a gimbal. The axis of the gyroscope is free to rotate (so φ canchange), while the centre of mass of the gyroscope is always coincident with the pivot point. Sincethe centre of mass and the pivot point are the same, gravity can never exert a moment about thepivot point.

Let us now define the coordinate system o′ which is stuck to the gyroscope axis(but not to the gyroscope itself), as before, such that e′ze′ze′z always points upward alongthe gyroscope axis; e′xe′xe′x always points downwards along the surface of the gyroscope(as illustrated) and e′ye′ye′y always points in the direction of precession. Let φ again bethe inclination angle of the gyroscope axis, while ωz and Ω are the spin and preces-sion rates, respectively. If we now consider the more general case where φ �= 0, theangular velocity vector of the gyroscope can be given as

ωωω = ωz e′ze′ze′z + φ e′ye′ye′y +Ω ezezez

However, we have mixed ‘primed’ and ‘unprimed’ unit vectors. For simplicity, wewill resolve ezezez into components along the moving unit vectors, as

ezezez =−sin(φ)e′xe′xe′x + cos(φ)e′ze′ze′z

Substituting this into the expression for the angular velocity of the gyroscope,

ωωω = ωz e′ze′ze′z + φ e′ye′ye′y +Ω(−sin(φ)e′xe′xe′x + cos(φ)e′ze′ze′z

)ωωω = −Ω sin(φ)e′xe′xe′x + φ e′ye′ye′y +

(ωz +Ω cos(φ)

)e′ze′ze′z


If we assume that the gyroscope is symmetric about the z-axis, then IcIcIc is constantwithin our moving reference frame. The angular momentum of the gyroscope aboutits centre of mass is then, in our moving reference frame,


⎡⎣ Ixx/c 0 0

0 Iyy/c 00 0 Izz/c

⎤⎦⎛⎝ −Ω sin(φ)

φωz +Ω cos(φ)

⎞⎠

=−Ixx/cΩ sin(φ)e′xe′xe′x + Iyy/cφ e′ye′ye′y + Izz/c

(ωz +Ω cos(φ)

)e′ze′ze′z

Since we have used a centre of mass as our origin,


=ddt

(−Ixx/cΩ sin(φ)e′xe′xe′x + Iyy/cφ e′ye′ye′y + Izz/c

(ωz +Ω cos(φ)

)e′ze′ze′z

)

∑Mext/cMext/cMext/c = −Ixx/cΩ sin(φ)e′xe′xe′x − Ixx/cΩφ cos(φ)e′xe′xe′x − Ixx/cΩ sin(φ) e′xe′xe′x+Iyy/cφ e′ye′ye′y + Iyy/cφ e′ye′ye′y+Izz/c

(ωz + Ω cos(φ)−Ωφ sin(φ)

)e′ze′ze′z + Izz/c

(ωz +Ω cos(φ)

)e′ze′ze′z

where we have applied the product rule. For simplicity, let us evaluate each of theunit vector derivatives separately. Again, it is important to note that, since the axesare not spinning along with the gyroscope,

ωaxωaxωax = φ e′ye′ye′y +Ω ezezez

= φ e′ye′ye′y +Ω(−sin(φ)e′xe′xe′x + cos(φ)e′ze′ze′z

)ωaxωaxωax = −Ω sin(φ)e′xe′xe′x + φ e′ye′ye′y +Ω cos(φ)e′ze′ze′z

Then, using Equation (2.7), we can evaluate e′xe′xe′x as,

e′xe′xe′x = ωaxωaxωax ×e′xe′xe′x=(−Ω sin(φ)e′xe′xe′x + φ e′ye′ye′y +Ω cos(φ)e′ze′ze′z

)×e′xe′xe′x= −Ω sin(φ)

(e′xe′xe′x ×e′xe′xe′x

)+ φ(e′ye′ye′y ×e′xe′xe′x

)+Ω cos(φ)


)= −Ω sin(φ)

(0)+ φ(−e′ze′ze′z

)+Ω cos(φ)

(e′ye′ye′y)

e′xe′xe′x = Ω cos(φ)e′ye′ye′y − φ e′ze′ze′z

5.7 Examples 447

Next, for e′ye′ye′y,

e′ye′ye′y = ωaxωaxωax ×e′ye′ye′y=(−Ω sin(φ)e′xe′xe′x + φ e′ye′ye′y +Ω cos(φ)e′ze′ze′z

)×e′ye′ye′y= −Ω sin(φ)


)+ φ(e′ye′ye′y ×e′ye′ye′y

)+Ω cos(φ)


)= −Ω sin(φ)

(e′ze′ze′z)+ φ(0)+Ω cos(φ)

(−e′xe′xe′x)

e′ye′ye′y = −Ω cos(φ)e′xe′xe′x −Ω sin(φ)e′ze′ze′z

Finally, for e′ze′ze′z,

e′ze′ze′z = ωaxωaxωax ×e′ze′ze′z=(−Ω sin(φ)e′xe′xe′x + φ e′ye′ye′y +Ω cos(φ)e′ze′ze′z

)×e′ze′ze′z= −Ω sin(φ)

(e′xe′xe′x ×e′ze′ze′z

)+ φ(e′ye′ye′y ×e′ze′ze′z

)+Ω cos(φ)

(e′ze′ze′z ×e′ze′ze′z

)= −Ω sin(φ)

(−e′ye′ye′y)+ φ(e′xe′xe′x)+Ω cos(φ)

(0)

e′ze′ze′z = φ e′xe′xe′x +Ω sin(φ)e′ye′ye′y

These results may now be substituted into the expression for Mext/cMext/cMext/c above, as

∑Mext/cMext/cMext/c = −Ixx/cΩ sin(φ)e′xe′xe′x − Ixx/cΩφ cos(φ)e′xe′xe′x−Ixx/cΩ sin(φ)

(Ω cos(φ)e′ye′ye′y − φ e′ze′ze′z

)+Iyy/cφ e′ye′ye′y + Iyy/cφ

(−Ω cos(φ)e′xe′xe′x −Ω sin(φ)e′ze′ze′z)

+Izz/c


)e′ze′ze′z

+Izz/c(ωz +Ω cos(φ)

)(φ e′xe′xe′x +Ω sin(φ)e′ye′ye′y

)

Expanding this expression,

∑Mext/cMext/cMext/c = −Ixx/cΩ sin(φ)e′xe′xe′x − Ixx/cΩφ cos(φ)e′xe′xe′x−Ixx/cΩ 2 sin(φ)cos(φ)e′ye′ye′y + Ixx/cΩφ sin(φ)e′ze′ze′z+Iyy/cφ e′ye′ye′y − Iyy/cΩφ cos(φ)e′xe′xe′x − Iyy/cΩφ sin(φ)e′ze′ze′z+Izz/cωz e′ze′ze′z + Izz/cΩ cos(φ)e′ze′ze′z − Izz/cΩφ sin(φ)e′ze′ze′z+Izz/cωzφ e′xe′xe′x + Izz/cωzΩ sin(φ)e′ye′ye′y + Izz/cΩφ cos(φ)e′xe′xe′x+Izz/cΩ 2 sin(φ)cos(φ)e′ye′ye′y


Collecting the coefficients of the unit vectors,

∑Mext/cMext/cMext/c =(−Ixx/cΩ sin(φ)− Ixx/cΩφ cos(φ)− Iyy/cΩφ cos(φ)+ Izz/cωzφ

+Izz/cΩφ cos(φ))e′xe′xe′x

+(−Ixx/cΩ 2 sin(φ)cos(φ)+ Iyy/cφ + Izz/cωzΩ sin(φ)+ Izz/cΩ 2 sin(φ)cos(φ)

)e′ye′ye′y

+(Ixx/cΩφ sin(φ)− Iyy/cΩφ sin(φ)+ Izz/cωz + Izz/cΩ cos(φ)− Izz/cΩφ sin(φ)

)e′ze′ze′z

However, we assumed that the gyroscope was symmetric about the z-axis. Let ussay, then, that the moment of inertia about the centre of mass along any axis in thex−y plane is Isym; then, Ixx/c = Iyy/c = Isym/c. Making this substitution into the aboveand simplifying,

∑Mext/cMext/cMext/c =(−Isym/cΩ sin(φ)−2Isym/cΩφ cos(φ)+ Izz/cωzφ + Izz/cΩφ cos(φ))e′xe′xe′x

+(−Isym/cΩ 2 sin(φ)cos(φ)+ Isym/cφ + Izz/cωzΩ sin(φ)+ Izz/cΩ 2 sin(φ)cos(φ)

)e′ye′ye′y

+(Izz/cωz + Izz/cΩ cos(φ)− Izz/cΩφ sin(φ)

)e′ze′ze′z

This result may be further simplified, as

∑MMM =

(−Isym/c

(Ω sin(φ)+2Ωφ cos(φ)

)+ Izz/cφ

(ωz +Ω cos(φ)

))e′xe′xe′x

+

(Isym/c

(φ −Ω 2 sin(φ)cos(φ)

)+ Izz/cΩ sin(φ)

(ωz +Ω cos(φ)

))e′ye′ye′y

+ Izz/c


)e′ze′ze′z (5.14)

where all moments are about the centre of mass. This general expression holds truefor all axisymmetric objects undergoing rotation about their centre of mass, andis used extensively in spacecraft dynamics. Equation (5.14) is often referred to asEuler’s equation, after Leonhard Paul Euler (1707-1783), though an alternative con-vention is typically used describing the angles. The notation used here is comparedto the standard Euler notation in the table included. The inclination angle is alsosometimes referred to as the angle of nutation as well.

5.7 Examples 449

Notation: Ours Euler’sSpin angle − ψSpin rate ω ψSpin acceleration ω ψInclination angle φ θInclination rate φ θInclination acceleration φ θPrecession angle − φPrecession rate Ω φPrecession acceleration Ω φ

For the special case of our gyroscope mounted in a gimbal, recall that ΣMext/cMext/cMext/c =0. Under these conditions,

−Isym/c(Ω sin(φ)+2Ωφ cos(φ)

)+ Izz/cφ

(ωz +Ω cos(φ)

)= 0

Isym/c(φ −Ω 2 sin(φ)cos(φ)

)+ Izz/cΩ sin(φ)

(ωz +Ω cos(φ)

)= 0

ωz + Ω cos(φ)−Ωφ sin(φ) = 0

(5.15)

These three expressions must be satisfied at all times if there are no external mo-ments applied to the gyroscope. Re-arranging the first equation, we can isolate Izz/cas

Izz/c = Isym/cΩ sin(φ)+2Ωφ cos(φ)

φ(ωz +Ω cos(φ)

)where we have assumed that φ �= 0 and ωz +Ω cos(φ) �= 0. This result may then besubstituted into the second equation to eliminate Izz/c as

0 = Isym/c


)+ Izz/cΩ sin(φ)

(ωz +Ω cos(φ)

)= Isym/c


)+Isym/c

Ω sin(φ)+2Ωφ cos(φ)

φ(ωz +Ω cos(φ)

) Ω sin(φ)(ωz +Ω cos(φ)

)

= Isym/c(φ −Ω 2 sin(φ)cos(φ)+

Ω

φΩ sin2(φ)+2Ω 2 sin(φ)cos(φ)

)0 = φ +Ω 2 sin(φ)cos(φ)+

Ω

φΩ sin2(φ)


where we were able to divide through by Isym/c, since Isym/c �= 0 by definition. Sum-marizing, then, the three conditions satisfied by gyroscopes subjected to no externalmoments now reduce to two:

φ +Ω 2 sin(φ)cos(φ)+Ω

φΩ sin2(φ) = 0


Curiously, the moments of inertia of the object do not matter at all- only the rates ofrotation.

Let us consider finally the two assumptions which we had made. First, if φ = 0,the gyroscope is undergoing steady-state precession; this is a condition which wehave already examined. Second, in the special case where ωz +Ω cos(φ) = 0, thenEquations (5.15) reduce to

−Isym/c(Ω sin(φ)+2Ωφ cos(φ)

)+ Izz/cφ

(0)= 0

Isym/c(φ −Ω 2 sin(φ)cos(φ)

)+ Izz/cΩ sin(φ)

(0)= 0


Recalling that Isym/c �= 0, this can be re-arranged to read,

Ω sin(φ)+2Ωφ cos(φ) = 0

φ −Ω 2 sin(φ)cos(φ) = 0


(5.16)

The third equation here is actually redundant, as

ωz + Ω cos(φ)−Ωφ sin(φ)

=ddt

(ωz +Ω cos(φ)

)

However, in this special case it is assumed that ωz +Ω cos(φ) = 0, so this expres-sion will be identically zero. Similarly, the first line in Equation (5.16) may be re-arranged to read,


Ω

Ω=−2φ

cos(φ)sin(φ)

However, it may be recognized here that both sides of this expression are derivativesof common functions, so that this may be re-expressed as

ddt

(ln(ΩC)

)= −2

ddt

ln(sin(φ)

)ddt

(ln(ΩC)

)=

ddt

ln

(1

sin2(φ)

)

where C is an arbitrary constant. Both sides of this expression may then be integratedto yield,

ln(ΩC) = ln

(1

sin2(φ)

)

Ω =1

C sin2(φ)

This expression for Ω may then be substituted into the second line of Equation(5.16) to yield,

φ −Ω 2 sin(φ)cos(φ) = 0

φ −(

1

C sin2(φ)

)2

sin(φ)cos(φ) = 0

φ =cos(φ)

C2 sin3(φ)

thereby reducing the three lines of Equation (5.16) to a single expression, for thespecial case where ωz +Ω cos(φ) = 0.


Though the types of problems encountered are varied, the same basic and methodi-cal approach may be taken.

1: Identify the rotating system. Each rotating object must be treated separately. Theaxis of rotation and centre of rotation must be identified as well.


2: Define appropriate coordinate system(s). Define a fixed origin o wherever con-venient, and fixed unit vectors exexex, eyeyey and ezezez which do not move. Define a rotatingreference frame o′ with the origin on a stationary point or centre of mass, and withthe moving unit vectors e′xe′xe′x, e′ye′ye′y and e′ze′ze′z stuck to the rotating object. If the object isaxisymmetric, the moving axes may not need to spin with the object along its ownaxis.

3: Resolve the angular velocity vector into the rotating o′ coordinate system. Becareful: if the angular velocity vector moves around, make sure that the resolvedcomponents are expressed in terms of the moving quantities or angles. Do not sim-ply resolve the angular velocity vector at some useful instant in time!

4: Determine the angular momentum of the object. Matrix-multiply an inertia tensorby the angular velocity vector (which has been resolved into components in themoving reference frame). It is best to do this symbolically and solve for the actualterms of the inertia tensor later on; most of the terms will usually disappear, so youwill probably not have to work them all out in the end. For axisymmetric objects, theproducts of inertia are usually zero. Remember that the resulting angular momentumwill be in terms of the same unit vectors as the angular velocity- that is, the rotatinge′xe′xe′x, e′ye′ye′y and e′ze′ze′z. If (a) the object is symmetric about each of the axes, (b) the unit vectorsare stuck to the rotating object and the origin is on the axis of rotation, and (c) theorigin is either a stationary point or centre of mass, then the angular momentumreduces to

HoHoHo = Ixx/oωxexexex′+ Iyy/oωyeyeyey

′+ Izz/oωzezezez′

5: Take the time-derivative of the angular momentum. Use the product rule wher-ever necessary, and take the time-derivatives of the moving unit vectors as well.Remember that the time derivative of a unit vector is the cross-product of the angu-lar velocity of the reference frame with the unit vector itself. The angular velocityof the reference frame is not the same as the angular velocity of the rotating objectif the reference frame does not spin along with the object.

6: Work out the required terms of the inertia tensor of the object. It is vitally im-portant to remember that the origin must be the same point o′, and that o′ must beeither the centre of mass or the stationary point about which the object is rotating.The axes in the inertia tensor calculations must be oriented along the moving e′xe′xe′x, e′ye′ye′yand e′ze′ze′z unit vectors, or the moments of inertia will become functions of time. Often,you can simply look up the terms of the inertia tensor in a reference text. Apply theparallel axis theorem if required, to ensure that the origin is at the correct point.

7: Identify and compute any external moments. The external moments must be cal-culated about the same point o′ as the inertia tensor.

8: Assemble the angular momentum equation. The net external moment acting onthe object about the point o′ is given by the expression,



The moments may change direction in time, if the object is rotating about an axiswhich is not an axis of symmetry. It may be sufficient to just solve the angularmomentum equation at one moment of interest- but only after the derivative HoHoHo hasbeen evaluated.

9: Solve for any other unknowns as required. The problem statement may placeadditional constraints or conditions on the rotation, such as in the case of pure rollingmotion. Express the additional equations in terms of the unknowns in the angularmomentum equation, and substitute them in to solve.


5.9 Problem set 6: Angular momentum of particles

Question 1


A very small object of mass m swings around along a circular path on a horizontalplane with a constant angular velocity ω . The object is constrained to move along acircular path (with radius r) by a string with some tension T . If T is very graduallydecreased, the radius of the path of the object will increase, and ω will change. Findthe rate of change of ω with respect to r.

Answer:dω

dr=−2

ω

r

Question 2

Our long-suffering friend the bug has now been vindictively blasted into Earth orbitby a disgruntled engineering student. The bug’s orbital trajectory is elliptical, with aminor axis R1 of 11,720 km and a major axis R2 of 13,520 km. At perigee (or closestEarth approach), the bug has an altitude H of 390 km and a speed of 33,800 km/h.(a) Show that gravity cannot exert an external moment upon the bug about the centreof the Earth; (b) determine the speed of the bug at the point A and at apogee (shownas the point B), and (c) starting from the law of conservation of angular momentum(and using your result from part a), demonstrate that the quantity r2θ is constant if

5.9 Problem set 6: Angular momentum of particles 455

the bug is described in a cylindrical system with its origin on the centre of the Earth.The mean radius of the Earth Re is approximately 6371 km.

Answer:

vA = 11,300 km/h

vB = 19,550 km/h


Question 3

Two small objects of equal mass are sitting in a slotted disc at equal distance r fromthe centre, as illustrated. The disc is rotating about its centre with an angular velocityω = 8 rad/s. The objects are latched in a position such that r = R1. At some instantin time, the latch is released and the objects roll out to assume a new position atr = R2 = 2R1. As a result of the motion of the objects, the angular velocity of thedisc will change. Solve for the new angular velocity. There are no torques externallyapplied to the disc.

Answer:



ω = 2 rad/s

Question 4


A tetherball of mass m is tethered to a cylindrical pole of radius b by means of amassless string of length L. The ball is struck so that it has an initial angular velocityθ of ωo, as illustrated. (a) Show that the quantity rθ is constant; (b) show that thetension in the string FT is given by the expression

FT =mL2ω2

o

L−bθ

5.9 Problem set 6: Angular momentum of particles 457

and (c) show that the radial component of velocity r cannot be determined using theprinciples of angular momentum.

Question 5

Demonstrate that angular momentum is conserved (about an arbitrary origin) forthe case of a small object travelling along a straight line at constant speed, in theabsence of externally applied forces. (Hint: examine the geometry of the problem).


5.10 Problem set 7: Angular momentum of bodies

Question 1


A yo-yo may be considered as a solid, homogenous cylinder of constant thicknessand total mass m and outer radius Ro, with string-winding drum of negligible massand inner radius Ri. An ideal string is wound around the drum, the yo-yo is placedon a level surface, and the string is pulled with a constant tension FT so that theyo-yo rolls without slipping. The string may pass over the top of the winding drum(configuration 1) or under the bottom of the winding drum (configuration 2), asillustrated. Solve for the acceleration of the centre of mass of the yo-yo in both ofthese configurations. Under which configuration will the tension need to be greaterin order to achieve the same acceleration?

Answer:

a1 =23

FT

m

(1+

Ri

Ro

)

a2 =23

FT

m

(1− Ri

Ro

)

5.10 Problem set 7: Angular momentum of bodies 459


Question 2

An object A (with mass ma) hangs freely, and is connected to another object C(with mass mc) by means of an ideal string. The object C is resting on a horizontal,frictional surface with a coefficient of kinetic friction μk. Between objects A and C,the string passes over a pulley B, which may be modeled as a solid, homogenouscylinder with mass mb and radius R. The string does not slip on the pulley. Computethe acceleration of the object A. Hint: even though the string is ideal, because itdoes not slip on the pulley, the pulley is effectively ‘clamped’ onto the string, so thetension in the vertical bit may be different from the tension in the horizontal bit!

Answer:

aa = gμk

mcma

−1mcma

+ 12

mbma

+1

Question 3

A mining ore rail car (with mass m1) carries a uniform, homogenous load of moltenmetal in a rectangular bucket measuring L×H (with total mass m2). The car travelsalong a straight, level track as illustrated. The ore bucket is hinged on the rearwardcorner, and rests on a stop on the forward corner. Find (a) the force in the stop Fs asa function of the thrust P provided by the wheels, and (b) the maximum thrust thewheels can provide before the load of molten metal begins to tip over. Hint: the orebucket will be on the verge of tipping over when Fs → 0.

Answer:



Fs =12

m2g− 12

m2

m1 +m2

HL

P

Pmax = (m1 +m2)LH

g

Question 4


A uniform, slender bar of mass m and length L is allowed to slide down a fric-tionless right-angle corner from a vertical position, while a different bar with thesame length and mass is pinned at one end and is allowed to tip over. If the two bars

5.10 Problem set 7: Angular momentum of bodies 461

are simultaneously released from rest with θ = 90◦, demonstrate that the two barswill strike the ground at exactly the same time. (This may be achieved by showingthat the angular accelerations and initial conditions are the same).

Answer:

α =32

gL

cos(θ)

Question 5


A thin half-circle of uniform density and thickness is suspended from the ceiling,pinned at its centre as illustrated. The half-circle is released from rest with θ = 0. (a)Show that the moment of inertia about o of the half-disc (in a direction perpendicularto the disc) is the same as for a whole disc with the same mass; (b) solve for the radialand tangential components of the reaction force on the pin as a function of θ .

Answer:

Izz/o =12

mR2

Fr = mgsin(θ)(1+

649π2

)Fθ = mgcos(θ)

(1− 32

9π2

)


Question 6

Explain why it is impossible to balance on a bicycle unless it is moving, and why itis possible to steer a bicycle without touching the handlebars.

Chapter 6Energy methods

Summary Introduction to the law of conservation of energy; external work, linearkinetic energy, rotational kinetic energy, gravitational potential energy and springpotential energy; non-conservative forces.

6.1 Conservation of energy

Though it is the last of the three conservation laws discussed here, the law of con-servation of energy can be exceedingly useful in solving problems in engineeringmechanics. Though somewhat more limited in applicability than the law of conser-vation of momentum, it provides a much simpler way of solving certain types ofproblems.

The law of conservation of energy states that the net external work done upon aclosed system will be equal to the change in energy of the system. Let us considernow the statement of this law.

6.1.1 Definitions

A system, as before, is a collection of objects. A closed system, again, is a collectionof objects which always remain part of the system (a more detailed discussion ofsystems and closed systems is included in Chapter 3). The concepts of work andenergy are a little more difficult to define.

Work, in classical mechanics, is essentially the act of pushing something. If youapply a force to an object and that object moves in response to the force which youhave applied, you have done work on it. Consider an object being acted upon by aforce vector FFF. That object then responds by moving along some small displacementvector ΔSSS following some path S (which is not necessarily a straight line). The smallamount of work ΔW done by the force FFF, then, is

463

464 6 Energy methods

ΔW = FFF ·ΔSSS

which is just the magnitude of the force multiplied by the magnitude of the com-ponent of the displacement vector lying parallel to the force vector. The result ofthe dot product is a scalar, so work is a scalar quantity and is independent of thesense of the force. The units will be force × distance, which are also units of energy(Newton-metres or Joules, in the SI system). Two important observations may bemade at this point: first, a force applied to a stationary object does no work, sinceΔSSS = 0, and second, a force applied perpendicular to the direction of travel doesno work, since the dot product would be zero (though you are pushing the object,it is not moving in response to the push). When varying forces are applied alongarbitrary trajectories, the definition of work may be expressed as a path integral,

W =∫

SFFF ·dSdSdS

where the integration is carried out along the path S. External work is simply workdone by the environment on the system, or by the system on the environment. Workdone by an external force, therefore, is external work.

Energy is even more difficult to define, but, for the purposes of classical mechan-ics, it may be thought of as the ability to do work. An object is said to have a certainamount of energy E if, through any means, it can be made to do that same amountE of work on something else. Exactly how the object does the work is irrelevant-even if it is impossible in a practical sense. For example, a piece of wood has en-ergy associated with it, since you can burn the wood to drive a steam engine anduse the engine to push something; a hot lump of metal has energy associated with it,since you can attach it to a thermocouple and use the voltage to power a small motorwhich, in turn, can be used to push something, and a loud noise has energy associ-ated with it, since the pressure waves can be used to vibrate a membrane, which canthen be used to push something else.

Since energy can take many forms, we have to be careful in how we define a‘closed system’. So long as all energy is either stored inside the system or usedto push around objects within the system, then no external work has been done toor done by the system. However, because of the many forms in which energy mayoccur, external work may be done without there being any external forces. If a hotobject floating in space is allowed to cool, then it is radiating heat energy to theenvironment. If the system had been defined as consisting of the object alone, thenthe object would be doing external work on the environment even though there areno forces acting on the object. No external work is done only if no energy can leavethe system.

Energy has the same units as work, and can occur in many forms- such as chem-ical (the burning wood), thermal (the hot metal) or sound (the loud noise). Whilethe law of conservation of energy requires that the total amount of energy within aclosed system remain the same if no external work is done, it does allow the energywithin the system to change forms. Burning the wood changes chemical energy into

6.2 Kinetic energy 465

thermal energy; attaching a thermocouple to a heat source changes thermal energyinto electrical energy, and so forth. If we were to place a stick of dynamite insidean armoured, insulated box and define our system as the box and all that it contains,then the total amount of energy within the box will remain the same before and afterthe dynamite is detonated; the detonation merely converts chemical potential energyto thermal, electromagnetic and sound pressure energy.

The laws of thermodynamics place certain restrictions on the law of conservationof energy- namely that when energy is converted from one form to another, it isimpossible to convert all of it to only one other form. You therefore cannot convertenergy from one form to another and then back again without ‘losing’ energy alongthe way to some third form (this is referred to as the concept of irreversibility).However, these losses are typically neglected in classical mechanics as they can belumped all together at the end of an engineering analysis into some sort of efficiencyfactor, usually denoted by the symbol η . Determining efficiency factors is outsideof the scope of this text. Here, it will be assumed that any form of energy maybe converted completely into any other form of energy, without restriction (so youcould turn a pile of charcoal and ash back into a tree by shouting at it loudly enough).

6.2 Kinetic energy

6.2.1 Linear kinetic energy

Let us consider an object A of mass ma with velocity vavava. This object could be usedto push a second object B (thereby doing work) if it collides with it, as illustrated infigure 6.1. If we define our system as consisting only of the original object A, thenthe force applied to it by the object B during the collision is an external force. Fromthe definition of impulse (see Section 3.2.2), during the collision,

Fig. 6.1 A simple collision.


JJJ =∫

FextFextFext dt = mavavava

So, if we consider only a small time dt during the collision, the small change in thevelocity of object A, dvadvadva, will be given by

FextFextFext dt = madvadvadva

FextFextFext = madvadvadva1dt

Let us now take the dot product of both sides with the small displacement vectordSbdSbdSb, which is the displacement of object B caused by the collision during that smalltime dt.

FextFextFext ·dSbdSbdSb = madvadvadva1dt

·dSbdSbdSb

Now we must make a number of observations. First, we notice that the force appliedto object B by object A, FbFbFb, must be equal in magnitude but opposite in direction asthe force applied to object A by object B, so FbFbFb =−FextFextFext. Next, during the collision,the two objects are in contact so dSb = dSa. So,

−FbFbFb ·dSbdSbdSb = madvadvadva1dt

·dSadSadSa

FbFbFb ·dSbdSbdSb = −madvadvadva · dSadSadSa

dtFbFbFb ·dSbdSbdSb = −madvadvadva ·vavava

Since vavava and dvadvadva must be in the same direction (that is, along the line of action ofFextFextFext),

FbFbFb ·dSbdSbdSb = −mava dva∫SFbFbFb ·dSbdSbdSb = −

∫S

mava dva∫SFbFbFb ·dSbdSbdSb = −1

2mav2

a

∣∣S


We recognize the left-hand side of this result as being the external work done bythe system upon the object B by means of the contact force. According to the lawof conservation of energy, then, our system must have lost an amount of energyequal in magnitude to the external work done. The right-hand side must thereforerepresent the change in energy of the system. If the velocity of block A decreasesas a result of the collision, the right side of this equation will be positive, and sothe work done on object B by the contact force (that is, the left-hand side of theequation) will be positive as well.

The scalar quantity Ek, defined as

Ek =12

mv2

therefore represents the maximum possible amount of work which an object withmass m and scalar speed v can perform on something else; this quantity is thereforereferred to as the kinetic energy of the object. An object with mass m which changesfrom some speed vi to some other speed v f has experienced a change in kineticenergy ΔEk of

ΔEk =12

mv2f −

12

mv2i

If v f < vi, ΔEk will be negative, as the object has lost kinetic energy. It is importantto note that the quantity v is a scalar magnitude, and does not depend on direction;v = |vvv|. Therefore, v cannot be negative. If an object has an initial velocity vector vivivi

and changes to a velocity vector vfvfvf such that vfvfvf =−vivivi, then ΔEk = 0.Alternatively, we may have arrived at the same result by using the law of conser-

vation of linear momentum and the definition of work done by a force; We know thatthe work done (and therefore the kinetic energy transferred) in pushing an object isgiven by the expression,

W = Ek =∫

FFF ·dsdsds

However, we also know that, for a single force acting on a single object,

FFF = GGG

Substituting these expressions together,

Ek =∫

FFF ·dsdsds

Ek =∫

dGdGdGdt

·dsdsds


However, since 1/dt is essentially a scalar quantity, this can be moved to the otherside of the dot product.

Ek =∫

dGdGdG · dsdsdsdt

=∫

d(GGG) ·vvv

=∫

mdvdvdv ·vvv

= m∫

vvv ·dvdvdv

Ek =12

mvvv ·vvv

It should be noted here that the differential operator, d, distributes in the same wayas a derivative (so that d(xy) = x dy+ y dx). Here we have obtained the same resultfor the kinetic energy, since

vvv ·vvv = |v|2

as indicated in the list of vector identities included in Chapter 2.

6.2.2 Rotational kinetic energy

Let us say that a force F is being used to push a small object of mass m tangen-tially around a circular path, as illustrated in Figure 6.2. If we define our system asconsisting of the object only, the external work W done by the force is given by

Fig. 6.2 An object translating along a circular path under the action of an applied force.


W =∫

SFFF ·dSdSdS

W =∫

FFF · r dθ eθeθeθ

where eθeθeθ is defined as being in the direction in which the object is traveling, tan-gential to the circular path. If FFF is perpendicular to rrr (where rrr is the position vectorof the object relative to the centre of the circular path),

W =∫

Fr dθ

W =∫

M dθ

where M = |MMM|= |rrr×FFF| is the magnitude of the moment introduced by the force FFFabout the centre of the circle. If the object were then to collide with something, itcould transfer a maximum energy Ek of

Ek =12

mv2

Ek =12

mr2ω2

where ω is the angular velocity of the object and r is the radius of the circular path.This is referred to as the rotational kinetic energy of the object.

Let us now consider a solid lamina rotating in a plane about some point o. Thelamina may be considered as being made up of an infinite number of infinitesimallysmall mass elements dm, each at radius r from the axis of rotation and all havingangular velocity ω . The total kinetic energy contained in the rotating lamina maythen be determined by integration, as

Ek =∫

12

ω2r2 dm

Ek =12

ω2∫

r2 dm

However, the integral should be immediately recognizable as the moment of inertiaabout o along an axis parallel to ω . So, for an object rotating in a plane,


Ek =12

Izz/oω2z

where it has been assumed that ωωω = ω ezezez.Let us consider now an object rotating about its centre of mass while the centre of

mass translates in space; recall that this motion is typical of an object upon which noexternal forces are acting. The velocity v of each differential mass element may thenbe expressed as a vector sum of two components: the translational velocity of thecentre of mass vcvcvc and rotational velocity of the mass element relative to the centreof mass vp/cvp/cvp/c. Then, for each differential mass element, the scalar speed is

v =√

v2c + v2

p/c

where the quantities in the argument are the magnitudes of the velocity vectors. So,the total kinetic energy of the lamina, Ek is given by

Ek =∫

12

v2 dm

=∫

12

(√v2

c + v2p/c

)2dm

=∫

12

(v2

c + v2p/c

)dm

=∫

12

(v2

c +ω2z r2)dm

Ek =12

v2c

∫dm+

12

ω2z

∫r2 dm

where vc and ωz could be pulled out of the integrals because they are the same forevery dm, and r is the scalar distance between each differential mass element dmand the centre of mass of the lamina. The integral in the first term is simply the totalmass of the lamina, while the integral in the second term is the moment of inertia ofthe lamina about its centre of mass along an axis normal to the lamina.

Ek =12

mv2c +

12

Izz/cω2z

So, the total kinetic energy of an object translating and rotating about its centre ofmass is equal to the sum of its translational kinetic energy and rotational kineticenergy.

Finally, let us examine the more general case of rotation with ωx �= 0 and ωy �= 0.This will occur if the object is rotating simultaneously about more than one axis,such as the gyroscope illustrated in Figure 5.42. To begin the analysis, let us considerthe work done by a single external moment upon a solid object; if the centre of mass


of the object remains stationary, all of this work will be converted into rotationalkinetic energy. Therefore,

W = Ek =∫

MMM ·dθdθdθ

However, for a solid object acted upon by a single external moment, the law ofconservation of angular momentum requires that

MoMoMo = HoHoHo

where the point o is a stationary point or a centre of mass. Since there is no net forceacting on the object, the centre of mass will remain stationary; therefore the centre ofmass must be the centre of rotation as well, so o is coincident (and interchangeable)with c. Substituting these results together,

Ek =∫

McMcMc ·dθdθdθ

=∫

dHcdHcdHc

dt·dθdθdθ

=

∫dHcdHcdHc · dθdθdθ

dt

Ek =∫

d(IcIcIcωωω) ·ωωω

Here, it is helpful to assume that the axes of a moving coordinate system o′ are stuckto the object with the origin at c. The angular velocity vector must then be expressedin terms of the unit vectors of o′, as

ωωω = ωx e′xe′xe′x +ωy e′ye′ye′y +ωz e′ze′ze′z

IcIcIc is evaluated in the rotating reference frame; since the object is fixed within therotating reference frame, IcIcIc is constant. Then,

Ek =∫

d(IcIcIcωωω) ·ωωω

=∫

IcIcIcdω ·ωωω

=∫

IcIcIcωωω ·dω

Ek =12

ωωω ·IcIcIcωωω


We notice that this can be re-expressed in terms of the angular momentum vector,as

Ek =12

ωωω ·HcHcHc

Recalling that

HcHcHc = (Ixx/cωx + Ixy/cωy + Ixz/cωz)e′xe′xe′x+(Ixy/cωx + Iyy/cωy + Izy/cωz)e′ye′ye′y+(Ixz/cωx + Iyz/cωy + Izz/cωz)e′ze′ze′z

Applying the rules of tensor multiplication (described in Chapter 5), we can evaluateEk, then, as

Ek =12

(Ixx/cω2

x + Ixy/cωxωy + Ixz/cωxωz)

+12

(Iyx/cωyωx + Iyy/cω2

y + Iyz/cωyωz)

+12

(Izx/cωzωx + Izy/cωzωy + Izz/cω2

z

)

Or, combining terms and recalling that Ixy = Iyx, Ixz = Izx and Iyz = Izy,

Ek =12

(Ixx/cω2

x + Iyy/cω2y + Izz/cω2

z

)+Ixy/cωxωy + Ixz/cωxωz + Iyz/cωyωz

Or, if we assume that the object is symmetric about the axes, then Ixy = Iyz = Ixz = 0.Under these circumstances, then,

Ek =12

Ixx/cω2x +

12

Iyy/cω2y +

12

Izz/cω2z

This result indicates that, providing that our assumptions are valid (that is, that (i)the object to be rotating about its centre of mass, (ii) the axes are stuck to the objectand rotating along with it, (iii) the origin is at the centre of mass of the object, and(iv) the object is symmetric about the axes), the total kinetic energy of an objectundergoing a three-dimensional rotation is equal to the sum of the kinetic energiesassociated with each component of the rotation.

6.3 Potential energy 473

6.3 Potential energy

Energy which has been stored within the system is referred to as potential energy,as it can be released at any time to cause something to be pushed. Potential energy istypically given the symbol U with an appropriate subscript. There are several waysin which energy may be stored within a system in classical mechanics; two of themost common will be discussed in greater detail below.

6.3.1 Gravity

Let us consider some object with mass ma moving about in the vicinity of the Earth,as illustrated in Figure 6.3. Let us define a system consisting of the object only.Then, if the distance between the object and the centre of the Earth is r, then theattractive force between the object and the Earth is given by the law of universalgravitation,

Fig. 6.3 An object translating along an arbitrary path within a gravitational field.

FgFgFg =−Gmame

r2 ererer

where me is the mass of the Earth and G is the universal gravitational constant, whileererer always points from the centre of the Earth to the object. The work done, then, indisplacing the system from any point 1 to any other point 2 along some path S is

W =∫

SFgFgFg ·dSdSdS

= −∫

S

Gmame

r2 ererer ·dSdSdS


W = −Gmame

∫S

1r2

(ererer ·dSdSdS

)

It can be recognized here that (ererer ·dSdSdS) is just the component of the small displace-ment vector dS which is parallel to ererer. In other words, (ererer ·dSdSdS) = dr. Then,

W = −Gmame

∫S

1r2 dr

W = Gmame( 1

r2− 1

r1

)

A number of interesting observations may be made from this result. First of all, thework done is independent of the actual path taken- it doesn’t matter how the objectgot from the distance r1 to the distance r2. This property of path-independence alsomeans that, regardless of the path taken, if the end of the path is at the same pointas the beginning of the path, r2 = r1 and W = 0, so no work is done if you end upwhere you started. Incidentally, any force which is path-independent is said to beconservative, since if you do work getting from point 1 to point 2, you get the sameamount of energy back when you return from point 2 to point 1. Mathematically,this conservative property can be written as∮

FFF ·dSdSdS = 0

or, The total work done moving around a closed path in a conservative field is zero.Magnetism and electrostatic attraction are other examples of conservative forces towhich this rule applies.

Let us now consider the case of an object which is near enough to the Earth thatthe changes in r are small compared to the radius of the Earth. Then, we can saythat r = re +h where re is the radius of the Earth, and h is the altitude of the object.Then,

W = Gmame

(1r2

− 1r1

)

= Gmame

(1

re +h2− 1

re +h1

)

W = Gmame

(h1 −h2

(re +h2)(re +h1)

)

However, re +h ≈ re, so


W = Gmame

(h1 −h2

r2e

)

= ma

(Gme

r2e

)(h1 −h2)

W = −mag(h2 −h1)

where we have recalled from our discussion of gravity in Chapter 4 that the accel-eration due to gravity near the Earth g = Gme/r2

e . So, the work you need to do inorder to move an object by some altitude Δh is −mgΔh. If the object is movingdownwards (so that Δh is positive), the work done is negative, meaning that you’regetting energy out of the system rather than putting energy into the system. Conse-quently, the gravitational potential energy Ug of a system is defined as

Ug = mgh

where h is the height of the object relative to some arbitrary datum. The gravita-tional potential is only really defined as a difference, since you may define a datumas being anywhere at all. The quantity h can be negative if it is below the datum (thisis not really a problem, since re +h is still positive).

6.3.2 Buoyancy

Energy may also be stored in buoyancy in much the same way as with gravity. Whilework is done to raise something in a gravitational field, work is similarly done tolower something in a fluid. The buoyant force is opposite of gravity, so the buoyantpotential energy UB for an object which is completely immersed is also the oppositeof gravity:

UB =−FBh

where FB is the buoyant force (the weight of the displaced fluid) and h is the distancefrom the surface of the fluid. The quantity is negative, since all values of h will benegative (as with gravity, h was defined as positive when above the datum).

It is important to note here that conservative forces, in energy analysis, do not doexternal work. Since no energy is ‘lost’ in moving around, the work is consideredas being simply ‘stored’ in the system as potential energy. Potential energy, then, isjust a simple technique for avoiding complex integrals.


6.3.3 Springs

Let us now consider the case of a spring. If a spring with spring constant k lyingalong the exexex direction is compressed by some small amount x, then the work donein compressing the spring is

W =∫

SFkFkFk ·dSdSdS

=∫

S−kxexexex ·dSdSdS

Since the displacement will also be along the exexex direction, dS = dx. Therefore,

W =

∫−kxexexex ·dx exexex

=∫

−kx dx(exexex ·exexex

)=∫

−kx dx

W = −12

kx2

where it was assumed that the spring started from its rest position. From this ex-pression, it is apparent that springs are conservative as well; the amount of workrequired to compress the spring by the amount x does not depend on the path takento get there; you could have compressed by an amount of 50x and then returned tox- and all of the extra work you did would have been delivered back by the spring.Also, if you compress a spring back and forth a few times, returning afterward tothe point at which you started, you would have done no net work. Because springforces are conservative, the work done by a spring may be considered as a storedpotential energy. The potential energy within the spring Uk is thus the negative ofthe work done to the spring, and is given by

Uk =12

kx2

where x here is the amount by which the spring has been expanded or compressed.This quantity represents the maximum possible work which could be done by aspring which has been compressed by an amount x.

In order for a spring to contribute potential energy to a system, it must be part ofthe system- otherwise, the spring will do external work upon the system. However,


in most mechanical energy analysis, you may simply include all possible springs inyour system by definition.

For the case of torsional springs, consider again the idealized torsional springillustrated in Figure 6.4. The torsional spring is idealized as a linear spring con-strained to follow a circular path of radius L; we shall define the unit vector ererer

as pointing radially outward, and eθeθeθ pointing in the direction of the deflection, asshown. The work done in deflecting the spring through some angle dθ , then, is

Fig. 6.4 Schematic representation of a torsional spring.

W =

∫SFkFkFk ·dSdSdS

=∫

S−kΔseθeθeθ ·dSdSdS

W =∫

S−k(Lθ)eθeθeθ ·dSdSdS

However, because the path is a circular segment, dSdSdS = L dθ eθeθeθ . Substituting,

W =∫

−kLθ eθeθeθ ·L dθ eθeθeθ

=∫

−kL2θ dθ(eθeθeθ ·eθeθeθ

)=

∫−kL2θ dθ


W = −12

kL2θ 2

From our analysis of torsional springs in Chapter 5 we know that, by convention,the quantity kL2 is defined as kθ , the torsional spring constant. Therefore, we maysay that the potential energy stored within a torsional spring is

Uk =12

kθ θ 2

where again we recognize that the energy stored in the spring is the negative of thework done to the spring.

6.4 Non-conservative forces

Let us now examine a force which we know to be non-conservative. From experi-ence, we know that if we do work in pushing an object across a frictional surface,we cannot get this work back from the object at some later time; furthermore, if wedo work in pushing an object around in a circle so that the object finishes exactlywhere it started, we would have had a net expenditure in energy.

Consider an object of mass m allowed to slide along a surface with coefficient ofkinetic friction μk, as illustrated in Figure 6.5. From our discussion in Chapter 4, weknow that the magnitude of the frictional force Ff is given by the expression

Ff = μkFn

Fig. 6.5 Kinetic friction acting on a sliding block.

where Fn is the normal force. The direction in which friction acts is always oppositeto the velocity. If we define our system as consisting of the block only, drawing aquick free-body diagram, we see that there are only three forces acting upon theblock: gravity, the normal force and friction. The system is assumed to be moving

6.5 Common forces which do no work 479

around in a horizontal plane, so there is no gravitational potential being stored withinthe system. Also, the normal force is always perpendicular to the direction of travel,so Fn · dS = 0, and no external work is done by the normal force (this is examinedin detail later in this chapter).

Friction, on the other hand, does do external work upon the system.

W =∫

SFf ·dS

W =∫

S−μFnevevev ·dSdSdS

where we define the unit vector evevev as always being parallel to the velocity vector.However,

evevev ·dSdSdS =vvv|vvv| ·dSdSdS

=1|vvv|

dSdSdSdt

·dSdSdS

=1dSdt

1dt

dS2

evevev ·dSdSdS = dS

where dS is a small, scalar segment of the path followed by the object. Substitutinginto the previous result,

W =∫

S−μFndS

W = −μFnS

So, the work done by friction upon the system depends on the path length, not justthe change in the object’s position. Therefore, moving the object around in a circleof radius R from one point back to the same point will diminish the energy in thesystem by an amount −μFn(2πR).

6.5 Common forces which do no work

There are a number of forces which, by their nature, can do no work upon a system.The first which we shall examine is the normal force.


Fig. 6.6 Illustration of the normal force acting on an object.

Let us consider an object of mass m moving with some constant velocity v alongthe x-axis, as illustrated in Figure 6.6. From the free-body diagram, we see thatthere are only two forces acting on the object: gravity, which acts downwards fromthe centre of mass of the object, and the normal force, which acts upwards upon thelower surface of the object. Since the object is moving along a straight line alongthe x axis, then dSdSdS = dxexexex. The work done by the normal force, then, is given by

W =

∫FnFnFn ·dsdsds

=∫

Fn eyeyey ·dxexexex

=∫

Fn(eyeyey ·exexex

)dx

=∫

Fn(0)dx

W = 0

Because the normal force is always perpendicular to the direction of motion, it cannever do any work.

The second force which we shall examine is static friction. Let us again considera block sitting on a horizontal surface. A force FFF is applied to the block in an attemptto move it, but this applied force is entirely counteracted by a static friction forceFfFfFf, as illustrated in Figure 6.7. Since the block may not necessarily be on the vergeof slipping, Ff �= μFn.

In this case, the object is at rest by definition. Consequently, v = 0. However,

vvv =dSdSdSdt

Therefore,

6.6 Application to mechanical systems 481

Fig. 6.7 Illustration of static friction acting on an object.

dSdSdSdt

= 0

dSdSdS = 0

(since dt �= 0). Because the object is not moving, the time-rate of change of the po-sition along the trajectory must be zero. The work done by static friction, therefore,is

W =∫

FfFfFf ·dSdSdS

=∫

FfFfFf · (0)W = 0

This result indicates that a static friction force can never do any work, regardless ofits magnitude. Similarly, we can determine the work done by the applied force FFF, as

W =∫

FFF ·dSdSdS =∫(0)FFF = 0

In other words, so long as an object is stationary, no work can be done to it by anyforce, regardless of magnitude or direction.

6.6 Application to mechanical systems

Let us assume that we have a closed system consisting of a number of objects, andthat no energy is radiated by the system to the environment (or by the environmentto the system) other than by means of applied, external forces. All of the energy inthe system, then, will be mechanical.


In the previous section, we have derived expressions for a number of differenttypes of mechanical energy associated with the system. The total energy containedwithin the system, (Esys)1, at some initial point in time t1, may be given by theexpression,

(Esys)1 =(∑Ek

)1 +(∑U

)1

(Esys)1 =(∑Ekx

)1 +(∑Ekθ

)1 +(∑Ug

)1 +(∑Ub

)1 +(∑Uk

)1

where we have distinguished between the linear kinetic energy Ekx and the rotationalkinetic energy Ekθ (so that the total kinetic energy Ek = Ekx +Ekθ ). At some laterpoint in time t2, the system will similarly contain an amount of energy (Esys)2, suchthat

(Esys)2 =(∑Ekx

)2 +(∑Ekθ

)2 +(∑Ug

)2 +(∑Ub

)2 +(∑Uk

)2

The change in energy of the system between the times t1 and t2 will then be,

ΔEsys = (Esys)2 − (Esys)1

ΔEsys =

((∑Ekx

)2 +(∑Ekθ

)2 +(∑Ug

)2 +(∑Ub

)2 +(∑Uk

)2

)

−((

∑Ekx)

1 +(∑Ekθ

)1 +(∑Ug

)1 +(∑Ub

)1 +(∑Uk

)1

)

ΔEsys =

((∑Ekx

)2 −(∑Ekx

)1

)+

((∑Ekθ

)2 −(∑Ekθ

)1

)

+

((∑Ug

)2 −(∑Ug

)1

)+

((∑Ub

)2 −(∑Ub

)1

)

+

((∑Uk

)2 −(∑Uk

)1

)

Or, alternatively, this result may be rearranged to yield the entirely equivalent (butmore compact) expression,

ΔEsys = Δ ∑Ekx +Δ ∑Ekθ +Δ ∑Ug +Δ ∑Ub +Δ ∑Uk

However, the law of conservation of energy requires that the change in energy ofa closed system be equal to the net external work done upon the system. Since wehave assumed that all of the external work is being done by external forces, then

Wext =∫ S2

S1FextFextFext ·dsdsds

6.7 Examples 483

where the integration is carried out over a path S, where S1 is the position at time t1and S2 is the position at time t2. From the law of conservation of energy, then,

Wext = ΔEsys

And, substituting our earlier results,

∫ S2

S1FextFextFext ·dsdsds = Δ ∑Ekx +Δ ∑Ekθ +Δ ∑Ug +Δ ∑Ub +Δ ∑Uk

This final expression, then, yields an equation relating the positions and velocitiesof the objects at two specific, given points in time to the external forces applied tothe system.

6.7 Examples

Energy methods are useful for solving problems in which velocities and/or positionsare required at some point in time, when the velocities and positions are known atsome other point in time. These techniques are not usually recommended whenvelocities or displacements are required as functions of time, or when internal forcesneed to be determined.

6.7.1 Pulleys

An object of mass m is being raised by means of the pulley arrangement shown. If theforce applied to the string is FT , and the object is released from rest in the positionshown, determine the velocity of the block after it has been raised by a distance h.

SOLUTION

We shall define our system here as consisting of the block, the pulleys and thestring. If we assume the string to be ideal and massless, it can have no potentialenergy associated with it. Similarly, if we assume all the pulleys are massless, therecan be no kinetic energy associated with them. If we set our gravitational datum atthe initial position of the object, then the initial energy of our system is

(Esys)1 = Ek +Ug = 0

since the object begins at rest at y = 0. In the final condition, the object will havebeen raised by a distance h above the datum, and will have some velocity v, so


Fig. 6.8 Pulleys.

(Esys)2 = Ek +Ug

(Esys)2 =12

mv2 +mgh

Also, while the object is being raised, an external force FT is being applied to thesystem. This force is doing external work to the system, so that

Wext =∫ L

0FTFTFT ·dSdSdS

where dSdSdS is a small vector distance along the path followed by the free end of thestring, and L is the total distance over which the free end of the string has beenpulled. Since strings can only be in tension, the angle subtended between FTFTFT and dSdSdSmust always be zero, so

Wext =∫ L

0FT dS

= FT S∣∣L0

= FT (L−0)

Wext = FT L

However, because of the pulleys, if the free end of the string is pulled a distance L,the object will be raised a distance h = L/3. Substituting,

Wext = 3FT h

6.7 Examples 485

So, from the law of conservation of energy, then,

Wext = (Esys)2 − (Esys)1

3FT h =12

mv2 +mgh−0

v =

√6FT

hm−2gh

which is the result required.

On the other hand, had we approached this same problem by applying the lawof conservation of linear momentum, we would begin with the linear momentumequation,

∑FextFextFext = GsysGsysGsys = maaa

where aaa is the acceleration of the centre of mass of the object. From the free-bodydiagram, then,

3FTFTFT +FgFgFg = maaa

3FT eyeyey −mg eyeyey = maaa

aaa =(3FT

1m−g)

eyeyey

a = 3FT1m−g

We may now solve for the velocity, as

a =dvdt

=dvdh

dhdt

a = vdvdh∫

a dh =

∫v dv

Substituting our expression for a,

∫3FT

1m−g dh =

∫v dv


3FThm−gh =

12

v2 +C

where C is some constant of integration. Since we know that the object began fromrest when h = 0,

3FT0m−g(0) =

12(0)2 +C

C = 0

Substituting this into our earlier expression,

3FThm−gh =

12

v2

v =

√6FT

hm−2gh

which, thankfully, is the same result obtained from the application of the law ofconservation of energy.

6.7.2 Pure rolling motion

An object of mass m and radius R is released from rest at the top of a surface ofinclination θ , upon which it rolls without slipping. Determine the velocity of thecentre of the disc after it has rolled a distance L down the surface of the inclineif the moment of inertia of the object about its centre of mass (in the direction ofrotation) is Izz/c.

Fig. 6.9 Pure rolling motion.

6.7 Examples 487

SOLUTION

Pure rolling motion is well-suited to the application of energy methods, as thestatic friction does no work. Because the instantaneous point of contact is always atrest, dS = 0. Equally, the normal force acting on the rolling object does no work, asFnFnFn ·dSdSdS = 0.

Let us define our system as consisting of the disc alone, and let us set our gravi-tational datum at the initial position of the object. Since the object is released fromrest, Ek = 0, and since it is on the gravitational datum, Ug = 0 as well. Then, thetotal energy at that initial point is

(Esys)1 = Ek +Ug = 0

At some later state, the object will have descended some distance L along the plane,and the corresponding change in altitude would be −Lsin(θ). Additionally, the ob-ject’s centre of mass will have some linear velocity vc while the object will be ro-tating about its centre of mass with some angular velocity ωz. Then, at that laterstate,

(Esys)2 = Ek +Ug

=12

mv2c +

12

Izz/cω2z −mgLsin(θ)

But, since the object is rolling without slipping, vc = Rωz and

(Esys)2 =12

mv2c +

12

Izz/c

(vc

R

)2

−mgLsin(θ)

=12

mv2c +

12

Izz/c

R2 v2c −mgLsin(θ)

=12

m

(1+

Izz/c

mR2

)v2

c −mgLsin(θ)

The law of conservation of energy requires that


And since the only forces acting on the rolling object are gravity (which is storedinside the system, so it is not considered as doing external work), the normal force(which is perpendicular to the direction of travel, so it does no work) and friction(which acts on the stationary point of contact, so it does no work), there is no exter-nal work being done to or by the system. Then,


(Esys)1 = (Esys)2

Substituting,

0 =12

m

(1+

Izz/c

mR2

)v2

c −mgLsin(θ)

vc =

√√√√2gLsin(θ)

1+Izz/c

mR2

Which, again, is the same as the result obtained from the application of the law ofconservation of linear momentum and the law of conservation of angular momen-tum.

6.7.3 Object dropped on a spring

An object of mass m is released from rest to slide down a frictionless inclined plane.Some distance L down the plane, the object strikes a spring-loaded platform andis brought to rest. What is the maximum speed achieved by the object, and what isthe total distance traveled by the object from the time it is released to the time itcomes to rest, if the plane is at an inclination angle of θ and the spring has a springconstant of k?

Fig. 6.10 Object dropped on a spring.

SOLUTION

To begin with, we identify that energy is being exchanged between gravitationalpotential, spring potential, and kinetic. If we define our system as consisting of theblock and the spring, then, the law of conservation of energy requires that

6.7 Examples 489


where the subscripts 1 and 2 refer to the initial and final conditions, respectively.Since there is no external work being done on the system,

Esys = Ek +Ug +Uk =C

where Esys is the total energy in the system and C is some constant. Note that, sincegravity is considered as a potential energy and can be stored within the system, itis not considered an external force from an energy point of view. You could disre-gard the gravitational potential energy and treat gravity as an external work, but theresults would be identical.

For the first part of the problem, we need to recognize that the object will ac-celerate down the incline until it strikes the spring; then, it will begin to decelerate.So, the maximum speed is achieved at the instant that it strikes the spring. If we setour gravitational datum at the spring surface, then, initially, all of the energy in thesystem is in gravitational potential.

Esys = Ek +Ug +Uk = (0)+mgLsin(θ)+(0) = mgLsin(θ)

Next, we consider the instant the object strikes the spring. Since h = 0, there isno gravitational potential left in the system. Also, since the spring is not yet com-pressed, there is no spring potential in the system either. Then,

Esys = Ek +Ug +Uk =12

mv2 +(0)+(0) =12

mv2

Then, since the total energy within the system must remain constant,

mgLsin(θ) =12

mv2

v =√

2gLsin(θ)

We could have arrived at the same result using the conservation of linear momentum;if we define the exexex direction as being down the incline, then

∑FextFextFext = mxexexex = mgsin(θ)exexex

x =dxdt

=dxdx

dxdt

= xdxdx

= gsin(θ)


∫ L

0x dx =

∫ L

0gsin(θ)dx

12

x2 = gLsin(θ)

x = v =√

2gLsin(θ)

However, the application of the law of conservation of energy was considerablymore straightforward. For the second part of the problem, we need to consider thepoint at which the object comes to rest, when the spring is fully compressed. If wesay that the distance down the incline which the spring is compressed is x, then thetotal energy in the system at that point is

Esys = Ek +Ug +Uk = (0)+mg(−xsin(θ)

)+

12

kx2 =C

The object has no velocity at the point when it comes to rest, so the kinetic energyin the system then is zero. The object has, however, sunk below our original datum(the uncompressed level of the spring), so the system has acquired some negativegravitational potential. Finally, energy has been stored in the spring. Since the to-tal energy in the system cannot have changed from our initial condition (still, noexternal work has been done),

mgLsin(θ) = mg(−xsin(θ)

)+

12

kx2

12sin(θ)

kx2 −mgx−mgL = 0

Solving this quadratic,

x =mg±

√m2g2 + 2mgLk

sin(θ)

ksin(θ)

x =mgsin(θ)

k

(1+

√1+

2Lkmgsin(θ)

)

The positive solution was selected because, in our formulation of the problem, werequired x to be positive (otherwise we would have had positive gravitational poten-tial when the object dipped below the datum). The total distance D traveled by theobject down the incline, then, is

6.7 Examples 491

D = L+ x = L+mgsin(θ)

k

(1+

√1+

2Lkmgsin(θ)

)

6.7.4 Perfectly elastic collisions

An object A of mass ma and initial speed vo collides with an object B of mass mb, ini-tially at rest. What are the final velocities of both masses, if the collision is perfectlyelastic?

Fig. 6.11 Perfectly elastic collisions.

SOLUTION

If we define our system as consisting of both masses, then there are no externalforces acting upon the system. Then, from the law of conservation of linear momen-tum, we know that


mavo exexex +(0) = ma(vavava)2 +mb(vbvbvb)2

where the subscripts 1 and 2 have been used to indicate the initial and final condi-tions, respectively, and exexex points in the initial direction of travel of object A. Fromthis equation alone, it is not possible to solve for the final velocity. However, per-fectly elastic collisions are collisions in which no energy is lost in the permanentdeformation of the objects colliding. Therefore, in classical mechanics, energy isconserved in one of these collisions. Then, recognizing that all of the energy in thesystem is kinetic and no external work is done to the system,



Esys = Eka +Ekb =C

where Eka and Ekb are the kinetic energies associated with objects A and B, respec-tively, and C is some constant. Then, in the initial condition,

(Esys)1 = (Eka)1 +(Ekb)1 =12

mav2o +(0)

While in the final condition,

(Esys)2 = (Eka)2 +(Ekb)2 =12

ma(v2a)2 +

12

mb(v2b)2

Noting, of course, that no information is provided about the directions of thevelocity- only the magnitudes. Then,

(Eka)1 +(Ekb)1 = (Eka)2 +(Ekb)2

12

mav2o =

12

ma(v2a)2 +

12

mb(v2b)2

The linear momentum equation and the energy equation have now provided twoequations for the two unknowns: (va)2 and (vb)2. We can now solve these equations.Let us start by isolating (vb)2 from the momentum equation.

mavo exexex = ma(vavava)2 +mb(vbvbvb)2

mavo = ma(va)2 +mb(vb)2

(vb)2 =ma

mb

(vo − (va)2

)

Since we have eliminated the unit vector without knowing the direction of (va)2, wemust therefore allow this quantity to assume either positive or negative values. Onthe other hand, we know that (vb)2 must be a positive magnitude, since there is noway that the object B can get sucked back toward the object A after the collision; thiswould violate the law of conservation of linear momentum. Now, we can substitutethis result into the energy equation,

12

mav2o =

12

ma(v2a)2 +

12

mb(v2b)2

12

mav2o =

12

ma(v2a)2 +

12

mb(ma

mb

)2(vo − (va)2

)2

6.7 Examples 493

v2o = (v2

a)2 +mb

ma

(ma

mb

)2(vo − (va)2

)2

v2o = (v2

a)2 +ma

mbv2

o +ma

mb(v2

a)2 −2ma

mbvo(va)2

0 =(1+

ma

mb

)(v2

a)2 −(2

ma

mbvo)(va)2 + v2

o

(ma

mb−1)

We may now solve this last quadratic for (va)2. However, we will not know whetherthe result should be the positive or the negative intercept.

(va)2 =1

2(1+ ma

mb

)(2ma

mbvo ±

√4

m2a

m2b

v2o −4v2

o

(ma

mb−1)(

1+ma

mb

) )

=1

2(1+ ma

mb

)(2ma

mbvo ±

√4

m2a

m2b

v2o −4v2

o

(m2a

m2b

−1) )

=1

2(1+ ma

mb

)(2ma

mbvo ±

√4v2

o

(m2a

m2b

− m2a

m2b

+1) )

=1

2(1+ ma

mb

)(2ma

mbvo ±2vo

)

(va)2 = vo

( mamb

±1mamb

+1

)

If the ± uncertainty were to be positive, then (va)2 > vo; this would be a violationof the law of conservation of linear momentum. Consequently,

(va)2 = vo

( mamb

−1mamb

+1

)

Substituting this expression for (va)2 back into the linear momentum equation, wecan solve for the velocity of the object B after the collision,

(vb)2 =ma

mb

(vo − (va)2

)

=ma

mb

(vo − vo

( mamb

−1mamb

+1

))

= voma

mb

(1−

mamb

−1mamb

+1

)


= voma

mb

( mamb

+1mamb

+1−

mamb

−1mamb

+1

)

= voma

mb

( mamb

+1− mamb

+1mamb

+1

)

= voma

mb

(1+1

mamb

+1

)

(vb)2 = vo

(2 ma

mbmamb

+1

)

So, we have found that,

(va)2 = vo

( mamb

−1mamb

+1

)

(vb)2 = vo

(2 ma

mbmamb

+1

)

Examining these results, some very interesting observations may be made.

No matter the masses, (va)2 < vo. This is encouraging, as the object A is not gainingenergy from the impact with object B; this would violate both the laws of conser-vation of energy and conservation of momentum. We did, however, force this resultthrough our resolution of the ± uncertainty.

If ma < mb, then (va)2 < 0. If the mass of object A is less than the mass of objectB, object A bounces off of object B, reversing its direction of motion. The velocityimparted upon object B will be lower than the original velocity of object A. In thelimiting case where ma → 0 or mb → ∞, (va)2 =−vo and (vb)2 = 0), as expected.

If ma > mb, then (va)2 > 0. If the mass of object A is greater than the mass of objectB, then object A will essentially knock object B out of its way, and continue on inthe same direction. Object B, on the other hand, will be propelled forward with avelocity greater than that of object A. In the limit as mb → 0, then ma/mb � 1 and(vb)2 → 2vo.

If ma = mb, object B and object A reverse roles. If the two objects are of equal mass,then ma/mb = 1 and so (va)2 = 0. This indicates that the object A will come to adead stop as a result of the impact, acquiring the same velocity as object B prior tothe impact. Object B, on the other hand, acquires a velocity of (vb)2 = vo, which isidentical to the initial velocity of object A. The system consisting of the two objectstherefore ends in a state which is essentially the reverse of the initial state.

6.7 Examples 495

It is important to note again that there will always be some directional ambiguitywhen the law of conservation of energy is used to solve for velocities, since the ki-netic energy is independent of the direction of motion. Some assumptions (or logicalarguments) will always need to be made in order to eliminate the ± uncertainties.

6.7.5 Pulley with mass

An object B of mass mb hangs from an ideal massless string which is wound arounda pulley A of mass ma and radius R. Determine the speed of the object B when it hasfallen down a distance L after being released from rest.

Fig. 6.12 Pulley with mass.

SOLUTION

First, we define our system as consisting of the block and the pulley only. Then,we recognize that the total energy in the system will consist of linear kinetic, rota-tional kinetic and gravitational potential only. The total energy in the system Esys istherefore given by

Esys = Eka +Ekb +Ug

Recognizing that there is no external work being done on the system, from the lawof conservation of energy,


Wext = 0 = (Esys)2 − (Esys)1

Esys =C

where C is some constant. At the initial condition, nothing is moving. If we set ourgravitational datum at the initial location of the object B, then the total energy justat the start of motion is

(Esys)1 = (Eka)1 +(Ekb)1 +(Ug)1 = 0+0+0 = 0

At the final condition, the block is moving, the pulley is spinning and the heightof the block has changed. We may dispense with the subscripts 1 and 2, since allmotion will refer to the final condition only.

(Esys)2 = (Eka)2 +(Ekb)2 +(Ug)2

=12

Izz/cω2a +

12

mbv2b +mbg(−L)

If we assume that the pulley is a solid cylinder, then Izz/c = mR2/2. Also, since thestring is rolling off the pulley, ωa = vb/R.

(Esys)2 =14

maR2(vb

R

)2+

12

mbv2b −mbgL

And, from the law of conservation of energy,

(Esys)1 = (Esys)2

0 =14

maR2(vb

R

)2+

12

mbv2b −mbgL

0 =14

ma

mbv2

b +12

v2b −gL

(14

ma

mb+

12

)v2

b = gL

vb =

√2gL

12

mamb

+1

This is also the result which we obtained using the law of conservation of angularmomentum in Chapter 5. We saw then that,

6.7 Examples 497

vb =g

1+ 12

mamb

So, we can integrate once to get

vb =dvb

dt= vb

dvb

dh=

g

1+ 12

mamb∫ (vb)2

0vb dvb =

∫ L

0

g

1+ 12

mamb

dh

12

v2b =

gL

1+ 12

mamb

vb =

√2gL

12

mamb

+1

As expected.

6.7.6 Kinetic friction

A block of mass m is released from rest some distance L up a frictionless planeinclined at an angle of θ . At the bottom of the incline, it slides along a horizontalsurface with coefficient of kinetic friction μk. Determine the distance S which theblock will slide before coming to rest.

Fig. 6.13 Kinetic friction.


SOLUTION

Here, we define our system as the block only. However, this time friction will doexternal work to the system. The amount of work done by friction Wf will be

Wf =∫ S

0FfFfFf ·dSdSdS

=∫ S

0−Ff dS

= −Ff S

since friction will be acting opposite to the displacement. So now, recognizing thatour system will have both gravitational potential and kinetic energy, we can writethe law of conservation of energy as


In our initial condition, since the block is released from rest, all of the energy is thesystem is in the gravitational potential. If we set our gravitational datum at the levelof the horizontal part of the path,

(Esys)1 =Ug = mgLsin(θ)

In our final condition, the block is at rest at the gravitational datum, so

(Esys)2 = 0

Substituting this into the energy equation,


−Ff S = 0−mgLsin(θ)

μkmgS = mgLsin(θ)

μkS = Lsin(θ)

S =Lμk

sin(θ)

where it has been immediately recognized that Ff = μkFn = μkmg. While it wouldhave been possible to solve for S using only the linear momentum equation, thiswould have required (a) Integrating the momentum equation and solving for the

6.7 Examples 499

velocity as a function of position as the block slid down the frictionless slope; (b)integrating the momentum equation with friction and solving for the velocity as afunction of position as the block slid across the horizontal plane, using the resultsfrom (a) as an initial condition, and (c) setting the result from (b) equal to zero andsolving for S. Clearly, the energy technique was much simpler.

6.7.7 Spring-mass oscillator

A block of mass m lying on a horizontal plane is attached to a spring with stiffnessk. The block is pulled back some initial distance xo and is released. Solve for thevelocity of the block as a function of position.

Fig. 6.14 Spring-mass oscillator.

SOLUTION

Here we have a classical spring-mass oscillator, as was examined previously.Energy methods provide a much simpler means of analyzing this situation. First,we define our system as consisting of the block and the spring. Since there is noexternal work done to the system and the block remains in a horizontal plane, thesystem will contain only kinetic energy and spring potential. In the initial condition,when the block is just released, it has no initial velocity- but there is an initial springpotential, so

(Esys)1 =(Ek)

1 +(Uk)

1

(Esys)1 = 0+12

kx2o

In any final condition, there will be both kinetic energy and spring potential, so


(Esys)2 =(Ek)

2 +(Uk)

2

(Esys)2 =12

mv2 +12

kx2

Now, from the law of conservation of energy,


0 =12

mv2 +12

kx2 − 12

kx2o

v =

√km

√x2

o − x2

Recognizing that v = x, a solution of the form x = AsinBt+C (where A, B and C areconstants) will satisfy this equation. Since we already know the result from previouswork, we know the frequency should be

√k/m, and we know that the maximum

amplitude can’t be more than xo. Since there’s no way for energy to leave the system,the maximum displacement also shouldn’t be less than xo, and the system shouldoscillate about x = 0. We also know that x = xo at t = 0, so let us guess that

x(t) = xo sin

(√km

t +π

2

)

and see if this works out. Starting with this guess,

x =

√km

xo cos

(√km

t +π

2

)

x2 =km

x2o cos2

(√km

t +π

2

)

x2 =km

x2o

(1− sin2

(√km

t +π

2

))

x2 =km

(x2

o − x2o sin2

(√km

t +π

2

))

x2 =km(x2

o − x2)

x = v =

√km

√x2

o − x2

Exactly as we had expected.

6.7 Examples 501

6.7.8 Falling string

An ideal, uniform string of length L and mass m is draped over the edge of a surfacesuch that it is just about to begin slipping off the surface, as illustrated. The string isthen given an initial, infinitesimal nudge and it begins slipping off the surface. Thecoefficient of static friction and the coefficient of kinetic friction between the stringand the surface are both μ . Find the velocity of the string at the instant that theentire length as passed off of the surface.

Fig. 6.15 Falling string.

SOLUTION

First of all, we shall define our system as consisting of the entire string. At anypoint in time, the string will have both kinetic energy and potential energy. In addi-tion, the friction force will do work upon the environment, removing energy fromour system. Then, from the law of conservation of energy,


Let us first consider the initial energy of the system. In the initial state, the stringis at rest, so the only energy will be gravitational potential. For convenience, let usdefine the surface as our gravitational datum. Then, if we say that the vertical partof the string has mass ma and the horizontal part of the string has mass mb, the totalgravitational potential of the system will be


Ug = magha +mbghb

where ha and hb are the vertical positions of the centres of mass of the vertical andhorizontal parts of the string, respectively. Since the string is uniform, it will have aconstant linear density λ = m/L such that any length of string x will have mass λx.If we define x as the length of string hanging over the edge of the surface, then,

ha = − x2

hb = 0

where it should be noted that the centre of mass of the vertical part of the stringis below the datum. Then, substituting this into the expression for the gravitationalpotential,

Ug = magha +mbghb

= −(λx)g

(x2

)+(λ (L− x)

)g(0)

Ug = −12

λgx2

If we define xi as the initial length of string hanging over the edge of the surface, theinitial gravitational potential of the string will be,

(Ug)1 = (Esys)1 =−12

λgx2i

Next, we will consider the final energy of the system. In the final state, the stringwill have both kinetic energy and gravitational potential. The gravitational potentialmay be obtained by substituting x = L into our expression for Ug, as

(Ug)2 =−12

λgL2

since x= L at the instant that the end of the string passes over the edge of the surface.For the kinetic energy of the system in the final state, we recognize that the entirestring will be moving with the same velocity v. The kinetic energy will therefore begiven by

(Ek)2 =12

mv2

(Ek)2 =12(λL)x2

6.7 Examples 503

where we have recognized that v= x. Finally, the external work done by friction maybe determined. Let us define the unit vector exexex as being in the direction of motion ofthe horizontal part of the string, and let us say that S is the distance traveled by theend of the string along the surface. Then,

Wext =∫ L−xi

0FFF ·dSdSdS

=∫ L−xi

0−Ff exexex ·dS exexex

Wext =∫ L−xi

0−μFndS

where Fn is the normal force. However, from a free-body diagram of the horizontalpart of the string, we would see that |Fn| = |mbg|. Since the string is ideal and in-extensible, the distance covered by the end of the string along the surface must beequal to the distance covered by the beginning of the string; therefore, S = (x− xi).The limits of integration will therefore need some adjustment, since

S = 0 → x = xi

S = (L− xi) → x = L

Substituting all this into the above,

Wext =∫ L−xi

0−μFndS

=∫ L

xi

−μmbgdx

=∫ L

xi

−μλ (L− x)gdx

=

∫ L

xi

−μλ (L− x)gdx

= −μλ

(Lx− 1

2x2)

g

∣∣∣∣L

xi

= −μλ

(L2 − 1

2L2 −Lxi +

12

x2i

)g

Wext = −μλ

(12

L2 −Lxi +12

x2i

)g


Now we can substitute our results for (Ug)1, (Ug)2 and (Ek)2 into the energy equa-tion, as


Wext = (Ug)2 +(Ek)2 − (Ug)1

−μλ

(12

L2 −Lxi +12

x2i

)g = −1

2λgL2 +

12(λL)x2 +

12

λgx2i

12

Lx2 = −μ

(12

L2 −Lxi +12

x2i

)g+

12

gL2 − 12

gx2i

x2 = (1−μ)gL+2μgxi − (1+μ)gL

x2i

Finally, we need an expression for xi. Initially, the system is at rest and in static equi-librium. Just before the string is about to slip, the tension in the string is equal to theweight of the hanging part of the string. Therefore, recognizing that the horizontalpart of the string will be in static equilibrium,

∑Fext = 0 = FT −Ff

FT = Ff

mag = μFn = μmbg

(λxi)g = μ(λ (L− xi)g

)xi(1+μ) = μL

xi =μ

1+μL

Substituting this into our earlier result,

x2 = (1−μ)gL+2μgxi − (1+μ)gL

x2i

= (1−μ)gL+2μg

(μ

1+μL

)− (1+μ)

gL

(μ

1+μL

)2

= (1−μ)gL+2gLμ2

1+μ−gL

μ2

1+μ

= gL

((1−μ)(1+μ)

1+μ+

μ2

1+μ

)

= gL

(1−μ2 +μ2

1+μ

)


x2 = gL

(1

1+μ

)

x =

√gL

(1

1+μ

)

As expected, this is the same result as was obtained using the conservation of linearmomentum for open systems, as demonstrated in Section (4.6.9). Since the energyapproach deals directly with displacements and velocities, the integration was some-what more simple.


In solving the above problems, the same basic technique has been applied:

1: Identify the system. Similar to the application of the law of conservation of mo-mentum, the law of conservation of energy may only be applied to a closed system.

2: Ensure that no energy is unaccounted for. Energy methods may only be usedwhen all energy sources and sinks are known; collisions, for example, may onlybe solved with energy methods if they are known to be perfectly elastic, or if the‘efficiency’ of the collision is known.

3: Solve for any external work done. The external work done is simply

Wext =∫

SFextFextFext ·dSdSdS

4: Solve for the initial energy in the system. At some initial point in time, add to-gether all of the linear kinetic, rotational kinetic, gravitational potential and springpotential energies.

(Esys)1 = ∑Ekx +∑Ekθ +∑Ug +∑Uk

5: Solve for the final energy in the system. At some final time, add together again allof the energies, as was done at the initial condition.

6: Substitute into the law of conservation of energy and solve. All of the previousresults may then be substituted into the energy equation,


The unknown quantity required may then be isolated.


7: Introduce the law of conservation of momentum or the law of conservation ofmass if required. If there are too many unknowns, the other equations may be sub-stituted in as well, as needed.

6.9 Problem set 8: Conservation of energy 507

6.9 Problem set 8: Conservation of energy

Use energy methods to solve the following problems.

Question 1


A small object with density ρ = 120 kg/m3 is immersed at a depth h = 1 m ina fluid of density ρw = 600 kg/m3 and negligible viscosity. The object is releasedfrom rest, floats up towards the surface and then pops up out of the fluid. Solve for(a) the velocity of the object as the object just reaches the surface of the fluid, and(b) the maximum altitude H to which the object will rise above the surface of thefluid. Hint: since the viscosity of the fluid is negligible, there will be no drag.

Answer:

(a) v =

√2gh

(ρw −ρ

ρ

)= 8.9 m/s

(b) H = h

(ρw −ρ

ρ

)= 4 m

Question 2

Find the equation of motion of a simple pendulum with length L and mass m usingenergy methods. Compare your result to that obtained from the momentum analysis,



in Section 4.6.6. Explain why it is impossible to use energy methods to determinethe magnitude of the tension in the string.

Answer:θ +

gL

θ = 0

Question 3


6.9 Problem set 8: Conservation of energy 509

An object of mass m is launched with an initial velocity of vo up an inclined planewith inclination angle θ . The coefficient of kinetic friction between the object andthe plane is μk. Solve for the maximum altitude H attained by the object.

Answer:

H =v2

o

2g(1+ μk

tan(θ)

)

Question 4


The coyote has repaired his rocket sled, and will attempt yet again to capturethe high-velocity desert bird. The sled travels on a circular track of radius R, andwill provide a constant thrust T . The combined mass of the sled and the coyote ism. Help the luckless coyote with his evil plan by determining the velocity of therocket sled as a function of its angular position θ (as illustrated). Explain why it isimpossible to determine the normal force exerted by the track upon the sled usingenergy methods. Assume the track is frictionless, and that the mass of fuel burnedby the rocket during the descent is negligible.

Answer:

vvv =

√2R(T

mθ +gsin(θ)

)eθeθeθ



Question 5

A yo-yo may be considered as a solid, homogenous cylinder of constant thicknessand total mass m and outer radius Ro, with string-winding drum of negligible massand inner radius Ri. The yo-yo is released from rest and allowed to fall a distance htoward the ground. Show that the velocity of the centre of the yo-yo is related to thedistance it has fallen according to the expression,

v =

√gh

12 +

14

(RoRi

)2

Question 6

Imagine that, magically, our universe is suddenly altered so that (a) static frictiondoes work, and (b) the normal force does work. Discuss some of the bizarre ways inwhich things would be different in day-to-day life.

Appendix AReview of multiple integrals

Summary A brief review of multiple integrals, as they are commonly applied in theanalysis of dynamic systems.

A.1 Definitions

It is common, especially when analyzing the dynamics of solid objects, to carry outintegrals over a three-dimensional mass. In its purest sense, an ‘integral’ is an ab-stract mathematical operation; however, it may be used in engineering applicationsto represent a very real, physical quantity.

Consider a small cube of material of length Δx, as illustrated in Figure A.1. Ifwe were to line up a row of these small cubes along the x-axis, the length of therow will simply be L = nΔx, where n is the number of cubes. To express this resultmathematically,

L = ∑ Δx

Fig. A.1 A one-dimensional row of differential elements.

511

512 A Review of multiple integrals

This result will hold true regardless of how small the cubes are, or how many ofthem are lined up. However, in the extreme case when n → ∞, then Δx → 0 and thesum cannot be evaluated directly. Instead, we remember that

limΔx → 0 ∑Δx =

∫ L

0dx = x

∣∣L0 = L

or, an integral is defined as the sum of an infinite number of infinitely small quan-tities. The limits of the integral are the positions on the x-axis of the left and rightsides of the row of cubes; that is, the first cube is sitting on x = 0, and the last cubeis sitting on x = L. To be pedantic, this result indicates that the length of the row ofcubes is equal to the distance between the origin and the point on the x-axis uponwhich the last cube is sitting.

The infinitesimal quantity dx is referred to as a differential. It has a vanishinglysmall magnitude, but otherwise may be treated as any other symbolic quantity; itmay be added, multiplied or distributed as usual. If A and B are scalar constants,

(A+B)dx = A dx+B dx

A = B → A dx = B dx

A.2 Area integrals

Let us now consider a two-dimensional array of m× n cubes, arranged in the x− yplane as illustrated in Figure A.2. The area A covered by the cubes can be simplywritten as,

A = ∑(

∑Δx

)Δy

= ∑(

n Δx

)Δy

= m n Δx Δy

As the size of the little cubes approaches zero, again, the expression becomes inde-terminate. Expressing the area as an integral, then, yields

A =∫ H

0

(∫ L

0dx

)dy

A.2 Area integrals 513

Fig. A.2 A two-dimensional array of differential elements.

A =∫ H

0

∫ L

0dx dy

=∫ H

0x∣∣L0 dy

=∫ H

0L dy

= L y∣∣H0

A = LH

as expected.

A.2.1 Functional limits of integration

In the previous example, the integration could be carried out in any order; however,if the boundary of the area is not along one of the axes but rather is a specifiedfunction, the order of integration must be carefully selected.

let us consider the case of the right-triangular area shown in Figure A.3. The areaof the triangle is therefore described by the inequalities,

0 < x < L

0 < y <HL

x

Since the area is completely bounded by these inequalities, they also describe thelimits of integration along each of the axes. The area dA covered by each of the littlecubes is simply dx dy, so


Fig. A.3 A triangular area.

A =∫

dA =∫ H

L x

0

∫ L

0dx dy

Therefore, the area of the triangle may be determined as

A =

∫ HL x

0

∫ L

0dx dy

=∫ H

L x

0x∣∣L0 dy

=∫ H

L x

0L dy

= L y∣∣H

L x0

A = LHL

x ???

This result clearly does not make any sense; the area of the triangle is a constant, andwill not change depending on where you are standing when you look at it- therefore,it cannot be a function of x. As a simple rule of thumb, if you have a multiple integralwhere one of the quantities being integrated also appears in one of the limits of oneof the integrals, the integral with the variable in the limits must be carried out first.Then,

A =∫ L

0

∫ HL x

0dy dx


=∫ L

0y∣∣H

L x0 dx

=∫ L

0

HL

xdx

=12

HL

x2∣∣L0

A =12

HL

which we recognize as the correct area of a right triangle.As a second example, consider the parabolic half-segment illustrated in Figure

A.4. The parabolic segment has a defined maximum height and length, but no fur-ther information is given. In order to determine the area, we must first express thearea included by the shape as a pair of inequalities such that only one of them isfunctionally dependent upon a variable. Then,

Fig. A.4 A parabolic segment.

0 < x < L

0 < y <C0x2 +C1x+C2

where C0, C1 and C2 are constants. Since the point (x,y) = (0,H) lies on theparabola,


y = C0x2 +C1x+C2

H = C0(0)2 +C1(0)+C2

C2 = H

Next, the peak of the parabola occurs at (x,y) = (0,H) as well; therefore, the deriva-tive must be zero at this point. Evaluating the derivative,

ddx

(C0x2 +C1x+H

)= 2C0x+C1

Then, at (x,y) = (0,H),

2C0x+C1 = 0

2C0(0)+C1 = 0

C1 = 0

Finally, we know that the point (x,y) = (L,0) lies on the parabola, so at that point,

y = C0x2 +H

0 = C0(L)2 +H

C0 = − HL2

so the two inequalities may be expressed explicitly as,

0 < x < L

0 < y <− HL2 x2 +H

As before, then, using these inequalities as the limits of integration, we can solvefor the area A of the parabola segment as

A =∫

dA

=∫ L

0

∫ − HL2 x2+H

0dy dx

=∫ L

0y∣∣− H

L2 x2+H

0 dx


=∫ L

0

(− H

L2 x2 +H

)dx

=

(−1

3HL2 x3 +Hx

)∣∣∣∣L

0

= −13

HL2 L3 +HL

A =23

HL

A.2.2 Areas bounded by cyclic functions

Let us consider now the area bounded by the curve y= sin(x), as illustrated in FigureA.5. We will describe the area bounded by the sine wave using the inequalities,

0 < x < 2π

0 < y < |sin(x)|

Fig. A.5 A single cycle of a sine wave .

Recall that the wavelength of the function y = sin(x) is 2π . So, as before, we candetermine the area A bounded by the sine wave as,


A =∫ 2π

0

∫ sin(x)

0dy dx

=∫ 2π

0y∣∣sin(x)0 dx

=∫ 2π

0sin(x)dx

= −cos(x)∣∣2π

0

= −cos(2π)+ cos(0)

= −1+1

A = 0 ???

This makes no sense, since the area bounded by the curve is very clearly nonzero.The reason for this apparent discrepancy is that the area integration will sum to-gether all of the area above y= 0, but subtract all the area below y= 0; it is thereforepossible to have negative area. In order to compute the area shaded in Figure A.5,then, we must break the area into two parts. Then,

A =∫ π

0

∫ sin(x)

0dy dx+

∫ 2π

π

∫ 0

sin(x)dy dx

This integral can now be evaluated, as

A =

∫ π

0

∫ sin(x)

0dy dx+

∫ 2π

π

∫ 0

sin(x)dy dx

=∫ π

0y∣∣sin(x)0 dx+

∫ 2π

πy∣∣0sin(x) dx

=∫ π

0sin(x) dx+

∫ 2π

π−sin(x) dx

= −cos(x)∣∣π0 + cos(x)

∣∣2π

π

= 1+1+1+1

A = 4

Alternatively, we could have recognized that the area is symmetric about x = π , sothat

A = 2∫ π

0

∫ sin(x)

0dy dx

= 2∫ π

0y∣∣sin(x)0 dx


= 2∫ π

0sin(x) dx

= −2cos(x)∣∣π0

= −2(−1−1)

A = 4

Similarly, we could also have recognized that the area inside 0 < x < π/2 is thesame as the area inside π/2 < x < π , so we could equally have determined the areaas

A = 4∫ π/2

0

∫ sin(x)

0dy dx

= 4∫ π/2

0y∣∣sin(x)0 dx

= 4∫ π/2

0sin(x) dx

= −4cos(x)∣∣π/20

= −4(0−1)

A = 4

as expected.

A.2.3 Differential area elements

There is more than one possible way to describe the small differential area elementdA. While it is convenient to describe areas bounded by straight lines and rectilinearfunctions using rectangular area elements, it would be considerably more difficultto describe, for example, a circle. Since circles are very common in engineeringapplications, it would be convenient to define an area element which lends itselfreadily to circular areas.

Consider the circular area bounded by the inequality x2 + y2 < R2. We can breakthis area up into a number of quasi-rectangular elements by dividing the area usingradii and concentric circles, as illustrated in Figure A.6. Each area element dA wouldthen be described by a small angle dθ , and would have a radial length of dR. Thearea of the element is therefore

dA =

(π(R2

o −R2i

))(dθ

2π

)


Fig. A.6 A circular segment area element.

where Ri and Ro are the inner and outer radii of the area element, respectively. Thefirst term represents the area of a round annulus, while the second term reduces thisto only that part of the annulus which is within a wedge described by the angle dθ .This expression may be simplified, as

dA =12

(R2

o −R2i

)dθ

dA =12

(Ro +Ri

)(Ro −Ri

)dθ

For the case of a very small element, Ro ≈ Ri = r. Also, by definition, Ro −Ri = dr.Substituting,

dA =12

(2r)(

dr)

dθ

dA = r dr dθ

It is interesting here to notice that as the area element gets infinitely small, the shapeof the element approaches that of a rectangle, and the area of the element is simplythe product of the lengths of the adjacent sides.

The area inside the circle can be described by the inequalities,


0 < r < R

0 < θ < 2π

so the area of the circle is given by the expression,

A =∫

dA =∫ 2π

0

∫ R

0r dr dθ

=∫ 2π

0

12

r2∣∣R0 dθ

=∫ 2π

0

12

R2 dθ

=12

R2θ∣∣2π

0

=12

R2(2π)

A = πR2

which is, as we recognize, the area of a circle. Of course, it would have been equallypossible to compute the area using rectangular elements; let us consider a quartercircle lying along the positive x and y axes (Figure A.7). To approach this problemin the same way as we approached the triangle, we would need an equation for thecircle in the form of y = f (x); for this, let us examine the definition of the circle.

x2 + y2 = R2

y = ±√

R2 − x2

Since we are only interested in the positive area, we will consider only the positivesolution. Then, the quarter circle is defined by the inequalities,

0 < x < R

0 < y <√

R2 − x2

and therefore the area of the full circle may be obtained as

A = 4∫ R

0

∫ √R2−x2

0dy dx


Fig. A.7 A circular segment area element.

A = 4∫ R

0y∣∣√R2−x2

0 dx

A = 4∫ R

0

√R2 − x2 dx

We now turn to our trusty table of integrals, finding that

∫ √R2 − x2 dx =

12

(√R2 − x2 x+R2 tan−1

(x√

R2 − x2

))

Substituting this into the previous result,

A =42

(√R2 − x2 x+R2 tan−1

(x√

R2 − x2

))∣∣∣∣∣R

0

= 2

(√R2 −R2 R+R2 tan−1

(R√

R2 −R2

))

−2

(√R2 −02 0+R2 tan−1

(0√

R2 −02

))

= 2

(R2 π

2

)−2

(0

)

A = πR2

A.3 Volume integrals 523

Recalling that tan(π/2)→ ∞. Clearly, determining the area of a circle using rectan-gular elements is possible, but the integration is considerably more complex.

A.3 Volume integrals

A volume integral is just a three-dimensional extension of an area integral. Con-sider the stacked array of m× n× p cubes illustrated in Figure A.8. The volume Voccupied by the stack of cubes may be expressed as

Fig. A.8 A three-dimensional array of differential elements.

V = ∑

(∑(

∑Δx

)Δy

)Δz

= ∑

(∑ pΔx Δy

)Δz

= ∑npΔx Δy Δz

V = mnpΔx Δy Δz

Again, as the size of the little cubes approaches zero, the expression becomes inde-terminate. Instead, we may express the sums instead as integrals; however, for the


limits of integration, the volume occupied by the stack of cubes must be expressedin the form of three inequalities (as it is a three-dimensional stack).

0 < x < L

0 < y < H

0 < z <W

Then, expressing the volume summations as integrals,

V =∫ W

0

(∫ H

0

(∫ L

0dx

)dy

)dz

Note that the parentheses are redundant. The integrals may then be evaluated todetermine the volume of the stack.

V =∫ W

0

∫ H

0

∫ L

0dx dy dz

=∫ W

0

∫ H

0x∣∣L0dy dz

=

∫ W

0

∫ H

0L dy dz

=∫ W

0Ly∣∣H0 dz

=∫ W

0LH dz

= LHz∣∣W0

V = LHW

as expected.

A.3.1 Differential volume elements

As with the area elements, volume elements dV may take on different geometriesin order to simplify the integration. In addition to the rectangular volume element,cylindrical volume elements are very commonly used in engineering. A cylindricalvolume element is identical to a circular area element (as described earlier) buthaving an out-of-plane depth of dz, as illustrated in Figure A.9. As with the circular


area element, then, as the volume of the cylindrical volume element approacheszero, the shape begins to approximate a cube, so

Fig. A.9 A cylindrical volume element.

dV = r dr dθ dz

Consider the solid cylinder of radius R and height H lying about the z-axis, as illus-trated. The volume described by the cylinder may be expressed as inequalities in thevariables r, θ and z as

0 < r < R

0 < θ < 2π

0 < z < H

so the volume of the cylinder may be determined by integration as

V =

∫dV =

∫ H

0

∫ 2π

0

∫ R

0r dr dθ dz

=

∫ H

0

∫ 2π

0

12

r2∣∣R0 dθ dz

=∫ H

0

∫ 2π

0

12

R2 dθ dz


=∫ H

0

12

R2θ∣∣2π

0 dz

=∫ H

0πR2 dz

= πR2z∣∣H0

V = πR2H

which we recognize as the volume of a cylinder. Of course, it would also be possibleto determine the volume of a cylinder by using the rectangular area elements dV =dx dy dz, but this would again lead to very complicated integration.

A third differential volume element is occasionally used in engineering practice,though it is not encountered very frequently. Spherical volume elements are gen-erating by slicing up a sphere using lines of latitude and longitude. If the latitude(distance from the south pole) is given by the angle θ , then the vertical side of thevolume element will have length r dθ . If the longitude (distance east of Greenwich)is given by the angle φ , the distance between the vertical axis and the volume el-ement is r sin(θ), so the horizontal side of the volume element will have lengthr sin(θ) dφ (the longitude circles have a radius which is smaller than r, and the hor-izontal side of the element is just a small arc length of a longitude circle). Then, thevolume of the spherical element is

Fig. A.10 A spherical volume element.

dV =(r dθ

)(r sin(θ) dφ

)(dr)


dV = r2 sin(θ) dr dθ dφ

As was shown for the case of the cylindrical volume element, the shape of a spher-ical volume element approaches that of a cube as the element becomes infinitesi-mally small. The volume occupied by the sphere of radius R is described by theinequalities,

0 < r < R

0 < θ < π

0 < φ < 2π

Note that we define the latitude so that θ = 0 at the south pole, and θ = π at thenorth pole. Then, the volume of the sphere may be integrated.

V =∫

dV =∫ 2π

0

∫ π

0

∫ R

0r2 sin(θ) dr dθ dφ

=∫ 2π

0

∫ π

0

13

r3 sin(θ)∣∣R0 dθ dφ

=∫ 2π

0

∫ π

0

13

R3 sin(θ) dθ dφ

=∫ 2π

0−1

3R3 cos(θ)

∣∣π0 dφ

=

∫ 2π

0

23

R3 dφ

=23

R3φ∣∣2π

0

V =43

πR3

which we recognize as the volume of a sphere.

A.3.2 Functional limits of integration in 3 dimensions

Let us now consider the cone with a base of radius R and height H, lying along thez-axis with its apex at the origin, as illustrated in Figure A.11. Since this object isaxisymmetric about the z-axis, this suggests that cylindrical volume elements would


be the easiest to integrate. The volume occupied by the cone is described by theinequalities,

0 < r <C0z+C1

0 < θ < 2π

0 < z < H

Fig. A.11 A cone.

where C0 and C1 are constants to be determined. We recognized here that the outerradius of the cone increases linearly with z. To determine the constants, we may useany two points known to be on the surface of the cone. In particular, two good pointsto use would be the apex (r,z) = (0,0) and the bottom rim (r,z) = (R,H). Startingwith the apex,

r = C0z+C1

0 = C0(0)+C1

C1 = 0

Then, substituting this result and looking at the point on the topmost rim,


r = C0z

R = C0(H)

C0 =RH

Substituting this into our system of inequalities, the cone may be completely de-scribed by

0 < r <RH

z

0 < θ < 2π

0 < z < H

We may now integrate the volume element dV in order to obtain the volume of thecone, as

V =∫

dV =∫ H

0

∫ 2π

0

∫ RH z

0r dr dθ dz

=

∫ H

0

∫ 2π

0

12

r2∣∣ R

H z0 dθ dz

=∫ H

0

∫ 2π

0

12

R2

H2 z2 dθ dz

=∫ H

0

12

R2

H2 z2θ∣∣2π

0 dz

=∫ H

0π

R2

H2 z2 dz

=13

πR2

H2 z3∣∣H0

=13

πR2

H2 H3

V =13

πR2H

As expected. Note that the dz integration was carried out last, as the limits of inte-gration of one of the other variables were functions of z.

Appendix BReview of Taylor series expansions

Summary A brief review of the polynomial expansion of arbitrary functions

The Taylor polynomial expansion, first proposed by English mathematician BrookTaylor (1685-1731), allows any function of a single variable to be approximatedby a polynomial. This approximation can significantly simplify many mathematicaloperations, since polynomials are themselves one of the simplest functions, and areeasily integrated and differentiated.

Let us say that we have an arbitrary function f (x) (which may be horribly com-plicated), and that we would like a simpler expression for the value of f (x) nearsome point x = a. To begin with, we look at the definition of the integral:

f (ξ ) =∫

d f (ξ )dξ

dξ (B.1)

where ξ is just a dummy variable. Alternatively, this may be expressed as a definiteintegral over the range a < ξ < x, as

f (x)− f (a) =∫ x

a

d f (ξ )dξ

dξ

From this result, we may isolate the function f (x), as

f (x) = f (a)+∫ x

a

d f (ξ )dξ

dξ

f (x) = f (a)+ I0 (B.2)

531

532 B Review of Taylor series expansions

where, for simplicity’s sake, we have made the convenient substitution,

I0 =∫ x

a

d f (ξ )dξ

dξ

Now, let us look more closely at the integral I0. Our ultimate goal is to represent thefunction f (x) as a series (with no integrated terms), so we want to somehow extractterms from the integral. To do this, we can use the definition of the integral presentedin Equation (B.1) by taking advantage of the product rule. Without changing I0, wemay add and subtract a term, as

I0 =∫ x

a

(d f (ξ )

dξ− (x−ξ )

d2 f (ξ )dξ 2 +(x−ξ )

d2 f (ξ )dξ 2

)dξ

From the product rule, though, we know that,

ddξ

((x−ξ )

d f (ξ )dξ

)=−d f (ξ )

dξ+(x−ξ )

d2 f (ξ )dξ 2

Substituting this into the previous expression,

I0 =

∫ x

a

(d f (ξ )

dξ− (x−ξ )

d2 f (ξ )dξ 2 +(x−ξ )

d2 f (ξ )dξ 2

)dξ

=∫ x

a

(− d

dξ

((x−ξ )

d f (ξ )dξ

)+(x−ξ )

d2 f (ξ )dξ 2

)dξ

I0 = −∫ x

a

ddξ

((x−ξ )

d f (ξ )dξ

)dξ +

∫ x

a

((x−ξ )

d2 f (ξ )dξ 2

)dξ

However, we immediately notice that the first integral is in exactly the same form asEquation (B.1): it is the integral of a derivative. Consequently, we can express thisas,

I0 =−((x−ξ )

d f (ξ )dξ

)∣∣∣∣x

a+∫ x

a

((x−ξ )

d2 f (ξ )dξ 2

)dξ

To evaluate this expression, we recognize that a and x are related, as they lie on thesame axis. The location of a may be expressed in terms of the location of x, as

a = x+C

where C is some constant, as illustrated in Figure B.1. Then,

da = d(x+C)

B Review of Taylor series expansions 533

Fig. B.1 An arbitrary function f (ξ ).

da = dx

Furthermore, for simplicity, let us also make the substitution,

I1 =∫ x

a

((x−ξ )

d2 f (ξ )dξ 2

)dξ

Then, evaluating our earlier derivative,

I0 = −((x−ξ )

d f (ξ )dξ

)∣∣∣∣x

a+∫ x

a

((x−ξ )

d2 f (ξ )dξ 2

)dξ

= −(x− x)d f (x)

dx+(x−a)

d f (a)da

+ I1

I0 = (x−a)d f (a)

dx+ I1

We may now substitute this result back into Equation (B.2) to yield,

f (x) = f (a)+ I0

f (x) = f (a)+(x−a)d f (a)

dx+ I1 (B.3)


Next, let us consider the integral I1. Once again, we wish to extract terms from theintegral. Having observed how the product rule may be exploited, we may again addand subtract a term from I1 without changing its value, as

I1 =∫ x

a

((x−ξ )

d2 f (ξ )dξ 2

)dξ

I1 =∫ x

a

((x−ξ )

d2 f (ξ )dξ 2 − 1

2(x−ξ )2 d3 f (ξ )

dξ 3 +12(x−ξ )2 d3 f (ξ )

dξ 3

)dξ

From the product rule, we observe that,

ddξ

(12(x−ξ )2 d2 f (ξ )

dξ 2

)=−(x−ξ )

d2 f (ξ )dξ 2 +

12(x−ξ )2 d3 f (ξ )

dξ 3

Substituting this result into our expression for I1,

I1 =∫ x

a

((x−ξ )

d2 f (ξ )dξ 2 − 1

2(x−ξ )2 d3 f (ξ )

dξ 3 +12(x−ξ )2 d3 f (ξ )

dξ 3

)dξ

=∫ x

a

(− d

dξ

(12(x−ξ )2 d2 f (ξ )

dξ 2

)+

12(x−ξ )2 d3 f (ξ )

dξ 3

)dξ

I1 = −∫ x

a

ddξ

(12(x−ξ )2 d2 f (ξ )

dξ 2

)dξ +

∫ x

a

(12(x−ξ )2 d3 f (ξ )

dξ 3

)dξ

once again, we notice that the first integral is in the form of Equation (B.1) and iseasily evaluated, as

I1 =−(

12(x−ξ )2 d2 f (ξ )

dξ 2

)∣∣∣∣x

a+∫ x

a

(12(x−ξ )2 d3 f (ξ )

dξ 3

)dξ

For simplicity again, we can make the substitution,

I2 =∫ x

a

(12(x−ξ )2 d3 f (ξ )

dξ 3

)dξ

Evaluating I1, then,

I1 = −(

12(x−ξ )2 d2 f (ξ )

dξ 2

)∣∣∣∣x

a+ I2

= −12(x− x)2 d2 f (x)

dx2 +12(x−a)2 d2 f (a)

da2 + I2


I1 =12(x−a)2 d2 f (a)

dx2 + I2

Substituting this result back into Equation (B.3) yields,

f (x) = f (a)+(x−a)d f (a)

dx+ I1

f (x) = f (a)+(x−a)d f (a)

dx+

12(x−a)2 d2 f (a)

dx2 + I2 (B.4)

Repeating the process yet again on the integral I2,

I2 =∫ x

a

(12(x−ξ )2 d3 f (ξ )

dξ 3

)dξ

=∫ x

a

(12(x−ξ )2 d3 f (ξ )

dξ 3 − 12

13(x−ξ )3 d4 f (ξ )

dξ 4 +12

13(x−ξ )3 d4 f (ξ )

dξ 4

)dξ

=∫ x

a

(− d

dξ

(12

13(x−ξ )3 d3 f (ξ )

dξ 3

)+

12

13(x−ξ )3 d4 f (ξ )

dξ 4

)dξ

= −(

12

13(x−ξ )3 d3 f (ξ )

dξ 3

)∣∣∣∣x

a+∫ x

a

(12

13(x−ξ )3 d4 f (ξ )

dξ 4

)dξ

I2 =12

13(x−a)3 d3 f (a)

dx3 +∫ x

a

(12

13(x−ξ )3 d4 f (ξ )

dξ 4

)dξ

Substituting this result into Equation (B.4),

f (x) = f (a)+(x−a)d f (a)

dx+

12(x−a)2 d2 f (a)

dx2 + I2

f (x) = f (a)+(x−a)d f (a)

dx+

12(x−a)2 d2 f (a)

dx2 +12

13(x−a)3 d3 f (a)

dx3 +(...)

The process of extracting terms from the integral may be carried on indefinitely; itis clear from this that the integral will breed terms which will be of the form,

1k!(x−a)k dk f (a)

dxk

where k is an integer index. The function f (x) may therefore be expressed explicitlyas a summation, in the form


f (x) =n

∑k=1


dxk + In

where In is a remainder of the form,

In =∫ x

a

(1n!(x−ξ )n dn+1 f (ξ )

dξ n+1

)dξ

This integral will tend to vanish for sufficiently large n (owing primarily to the factor1/n!). Therefore, so long as n is reasonably large,

f (x)≈n

∑k=1


dxk

This is the familiar form of Taylor’s series expansion. Since this approximation forthe function f (x) was developed for the sole purpose of determining the value off (x) near the point x = a, the error in the approximation will increase as x movesfurther away from a.

Examples are given in Figures B.2 and B.3 of how the summation approachesf (x) near x = a for increasingly large values of n. Note that the number of termsrequired to obtain a certain range of agreement between the function and the Taylorseries depends on the function itself (some functions may require hundreds of termsfor good agreement over a fairly small range).

Fig. B.2 Graphical representation of the Taylor expansion of y = ex about a = 1, showing n = 2and n = 4.


Fig. B.3 Graphical representation of the Taylor expansion of y = sin(x) about a = π/2, showingn = 6 and n = 12.

Appendix CTable of inertia tensors for common shapes

Slender rod

Ixx/c = 0

Iyy/c =1

12mL2

Izz/c =1

12mL2

Ixy/c = Iyz/c = Ixz/c = 0

Slender rod, about one end

Ixx/o = 0

Iyy/o =13

mL2

Izz/o =13

mL2

Ixy/o = Iyz/o = Ixz/o = 0

539

540 C Table of inertia tensors for common shapes

Solid cylinder

Ixx/c =14

mR2 +1

12mH2

Iyy/c =14

mR2 +1

12mH2

Izz/c =12

mR2


Hollow cylinder

Ixx/c =14

m(R2o +R2

i )+112

mH2

Iyy/c =14

m(R2o +R2

i )+112

mH2

Izz/c =12

m(R2o +R2

i )


Cylindrical shell

Ixx/c =12

mR2 +1

12mH2

Iyy/c =12

mR2 +1

12mH2

Izz/c = mR2


C Table of inertia tensors for common shapes 541

Sphere

Ixx/c =25

mR2

Iyy/c =25

mR2

Izz/c =25

mR2


Hollow sphere

Ixx/c =25

mR5

o −R5i

R3o −R3

i

Iyy/c =25

mR5

o −R5i

R3o −R3

i

Izz/c =25

mR5

o −R5i

R3o −R3

i


Spherical shell

Ixx/c =23

mR2

Iyy/c =23

mR2

Izz/c =23

mR2


542 C Table of inertia tensors for common shapes

Block

Ixx/c =1

12m(L2 +H2)

Iyy/c =1

12m(W 2 +H2)

Izz/c =1

12m(W 2 +L2)


Cone, about its apex

Ixx/o =320

mR2 +35

mH2

Iyy/o =320

mR2 +35

mH2

Izz/o =310

mR2

Ixy/o = Iyz/o = Ixz/o = 0

Index

AccelerationAngular, 42As function of displacement, 6Centripetal, 53, 213Coriolis, 52Definition, 3Nonlinear, 12, 16Of a system, 140Relative, 52

Archimedes of Syracuse, 181Aristotle, 149Artillery, 227, 435

Block-and-tackle, 184Buoyancy, 181

Cavendish, Henry, 174Centre of mass, 139, 158, 160, 177Clutch

Rolling, 374Simple, 365

Collisions, 142, 465Conservation of momentum and, 143Examples, 151, 155, 156Perfectly elastic, 491Types of collisions, 143

Conservation laws, 131Constrained motion, 87Coordinates

Cartesian, 33Converting between, 37Cylindrical, 34Moving, 50Normal-tangential, 48, 59Spherical, 35

Coriolis, Gaspard-Gustave, 52Cross-product, 30

d’Alembert, Jean le Rond, 212Dampers, 208de Coulomb, Charles-Augustin, 189Descartes, Rene, 33Differentials

Cylindrical, 519Spherical, 526

Dot-product, 29Double integrals, 512Drag, 12

Effect of, 234Pressure, 209Viscous, 204

Efficiency, 465Einstein, Albert, 150, 173Energy, 464

Buoyant potential, 475Frictional losses, 479Gravitational potential, 475Linear kinetic, 467Rotational kinetic, 469Spring potential, 476Total kinetic, 470

Epicurus, 149Equation of motion, 204Euler angles, 448Euler, Leonhard Paul, 222, 448

First principle, 136Force

Conservative, 204, 475Distributed, 347External vs. internal, 132Normal, 189, 256, 479Reaction, 386Resolving, 171

543

544 Index

Restoring, 201Four-bar mechanism, 111Free-body diagram, 171Frequency, 204Friction

Coefficients of, 189, 192Dry, 189, 241, 259Static, 246Wet, 204Work done by, 479, 497

Galileo, 150Gravity, 173

Centre of, 177Inverse-square law, 173Orbital trajectory, 200Universal constant, 174

Gyroscope, 431

Hooke, Robert, 201, 386Huygens, Christiaan, 386

Impulse, 147Impulse-momentum relation, 149

Inertia tensor, 293, 296of a cone, 313of a cylindrical shell, 308of a hollow cylinder, 308of a hollow sphere, 317of a rod, 302of a solid block, 322of a solid cylinder, 306of a solid sphere, 317of a spherical shell, 318Table of, 539

Initial conditions, 5Irreversibility, 465

Lagrangia, Giuseppe Lodovico, 132Leibniz, Gottfried Wilhelm, 150, 173Limits of integration, 528Losses, 465

Mass, 131, 296Matrix, 293Mechanisms, 16, 92, 96, 407Michell, John, 174Moment, 285Moment of inertia, 292Momentum

Angular, 283Linear, 132of a lamina, 295, 299of a prism, 295, 299

Orbital angular, 290, 296Momentum flux, 224

Newton’s ‘laws’ of motion, 136Newton, Isaac, 135, 150, 173, 201Noether, Emmy, 150Numerical resolution, 9Nutation, 448

Orthogonality, 32Oscillator, 19, 204, 382, 499

Parallel axis theorem, 325, 389, 434Pendulum, 249

Torsional, 382Precession, 434Pressure

Atmospheric, 180Drag, 209Hydrostatic, 177, 181, 347

Principia Mathematica, 135, 150, 173, 201Product of inertia, 292Projectile motion, 227, 435Pulleys, 184, 234

With mass, 327, 392, 495

Radius of curvature, 48Rectilinear motion, 2Relative motion, 50

Crank-slider, 96Example of, 62, 68, 73, 82, 87Four-bar mechanism, 111Scotch yoke, 92Universality of, 54

Right-hand rule, 31, 42, 285Rockets, 134, 214

Staging, 216Rolling, 331, 395, 487

Backspin, 359Rotating imbalances, 418, 428

Scalars, 27Scotch yoke mechanism, 16, 92Series expansion, 318, 531Sina, Abu Ali, 150Singular points, 10Spring, ideal, 201Spring, torsional, 325String, ideal, 184SUVAT equations, 1

Derivation of, 3System

Definition, 132Open, 222, 265, 270

Index 545

Taylor, Brooke, 531Tension, 184Tensor, 293Torque, 285Trajectory, 47Tsiolkovsky, Konstantin, 216

Unit vectors, 32Resolving, 65

Units, 12Table of, 11

Vector, magnitude of, 27Vectors, 27

Addition, 28Cross-product, 30Dot-product, 29

Position, 33Scalar multiplication, 29Time-derivatives of, 40Translation, 27Unit vectors, 32Vector identities, 31

VelocityAngular, 42Definition, 2Relative, 51Terminal, 207, 212

Vibration, 203Von Braun, Wernher, 216

Wallis, John Brehaut, 150Weight, 131Work, 463

Documents

VECTOR DYNAMICS - epubs.surrey.ac.ukepubs.surrey.ac.uk/845649/1/VectorDynamics2ed_secure.pdfVECTOR DYNAMICS An introduction for engineering students Second Revised Edition ... It is