This book discusses topics on D’Alembert’s principle, virtual work, Eulerian angles, Lagrange’s equation in generalized
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English Pages 982 [980] Year 2023
Table of contents :
Preface
Contents
About the Author
1 3D Kinematics of Rigid Bodies
1.1 Kinematics
1.1.1 Note
1.1.2 Note
1.1.3 Note
1.1.4 Note
1.1.5 Note
1.1.6 Note
1.1.7 Note
1.1.8 Note
1.1.9 Example
1.1.10 Example
1.1.11 Definition
1.1.12 Note
1.1.13 Note
1.1.14 Note
1.1.15 Note
1.1.16 Note
1.1.17 Note
1.1.18 Note
1.1.19 Note
1.1.20 Example
1.1.21 Note
1.1.22 Example 1
1.1.23 Note
1.1.24 Note
1.1.25 Note
1.1.26 Note
1.1.27 Note
1.1.28 Note
1.1.29 Note
1.1.30 Note
1.1.31 Note
1.1.32 Note
1.1.33 Example
1.1.34 Note
1.1.35 Note
1.1.36 Note
1.1.37 Example
1.1.38 Note
1.1.39 Note
1.1.40 Note
1.1.41 Example
1.1.42 Note
1.1.43 Note
1.1.44 Note
1.1.45 Note
1.2 Examples on Kinematics
1.2.1 Example
1.2.2 Note
1.2.3 Note
1.2.4 Note
1.2.5 Note
1.2.6 Note
1.2.7 Example
1.2.8 Example
1.2.9 Example
1.2.10 Example
1.2.11 Example
1.2.12 Example
1.2.13 Example
1.2.14 Example
1.2.15 Note
1.2.16 Note
1.2.17 Example
1.2.18 Example
1.2.19 Note
1.2.20 Example
1.2.21 Note
1.2.22 Note
1.2.23 Note
1.2.24 Example
1.2.25 Example
1.2.26 Example
1.2.27 Example
1.2.28 Example
1.2.29 Example
1.2.30 Example
1.2.31 Example
1.2.32 Example
1.2.33 Example
1.2.34 Example
1.2.35 Example
1.2.36 Example
1.2.37 Example
1.2.38 Example
1.2.39 Example
1.2.40 Note
1.2.41 Note
2 D’Alembert’s Principle
2.1 Principle of Virtual Work
2.1.1 Note
2.1.2 Note
2.1.3 Note
2.1.4 Note
2.1.5 Example
2.1.6 Example
2.1.7 Example
2.1.8 Example
2.1.9 Note
2.1.10 Example
2.1.11 Example
2.1.12 Example
2.1.13 Example
2.1.14 Example
2.1.15 Example
2.1.16 Example
2.1.17 Example
2.1.18 Example
2.1.19 Example
2.1.20 Example
2.1.21 Example
2.1.22 Example
2.2 D’Alembert’s Principle
2.2.1 Note
2.2.2 Note
2.2.3 Example
2.2.4 Note
2.2.5 Note
2.2.6 Note
2.2.7 Note
2.2.8 Example
2.2.9 Example
2.2.10 Example
2.2.11 Note
2.2.12 Example
2.2.13 Example
2.2.14 Example
2.2.15 Example
2.2.16 Example
2.3 Examples on Motion About a Fixed Axis
2.3.1 Example
2.3.2 Example
2.3.3 Example
2.3.4 Example
2.3.5 Example
2.3.6 Example
2.3.7 Example
2.3.8 Example
2.3.9 Example
2.3.10 Example
2.3.11 Example
2.3.12 Example
2.4 Examples on Compound Pendulum
2.4.1 Note
2.4.2 Example
2.4.3 Example
2.4.4 Note
2.4.5 Example
2.4.6 Example
2.4.7 Example
2.4.8 Example
2.4.9 Example
2.4.10 Example
2.4.11 Example
2.4.12 Example
2.4.13 Example
3 Momental Ellipsoid
3.1 General Equation of Second Degree in Three Variables
3.1.1 Note
3.1.2 Example
3.1.3 Note
3.1.4 Example
3.1.5 Note
3.1.6 Note
3.1.7 Note
3.1.8 Note
3.1.9 Note
3.1.10 Example
3.1.11 Example
3.1.12 Example
3.1.13 Example
3.1.14 Example
3.1.15 Example
3.1.16 Example
3.1.17 Example
3.1.18 Example
3.2 Momental Ellipsoid
3.2.1 Note
3.2.2 Note
3.2.3 Note
3.2.4 Example
3.2.5 Example
3.2.6 Example
3.2.7 Example
3.2.8 Note
3.2.9 Note
3.2.10 Note
3.2.11 Example
3.2.12 Example
3.2.13 Example
3.2.14 Example
3.2.15 Example
3.2.16 Note
3.2.17 Note
3.2.18 Note
3.2.19 Example
3.2.20 Note
3.2.21 Note
3.2.22 Note
3.3 EquiMomental Systems
3.3.1 Example
3.3.2 Example
3.3.3 Example
3.3.4 Example
3.3.5 Example
3.3.6 Example
3.3.7 Example
3.3.8 Example
3.3.9 Example
3.3.10 Example
3.3.11 Example
3.3.12 Example
3.3.13 Example
3.3.14 Example
3.3.15 Example
3.3.16 Note
3.3.17 Example
3.3.18 Example
3.3.19 Example
3.3.20 Example
3.3.21 Example
3.3.22 Example
3.3.23 Example
3.3.24 Example
3.3.25 Example
3.3.26 Example
3.3.27 Note
3.3.28 Example
3.3.29 Note
3.3.30 Example
3.3.31 Note
3.3.32 Note
3.3.33 Example
3.3.34 Example
3.3.35 Example
3.3.36 Example
3.3.37 Example
3.4 More Examples on Moment of Inertia
3.4.1 Note
3.4.2 Example
3.4.3 Example
3.4.4 Example
3.4.5 Example
3.4.6 Example
3.4.7 Example
3.4.8 Example
3.4.9 Example
3.4.10 Example
3.4.11 Example
3.4.12 Example
3.4.13 Note
3.4.14 Example
3.4.15 Example
3.4.16 Example
3.4.17 Example
4 Motion About a Fixed Axis
4.1 Motion About a Fixed Axis
4.1.1 Example
4.1.2 Note
4.1.3 Note
4.1.4 Example
4.1.5 Note
4.1.6 Note
4.1.7 Note
4.1.8 Note
4.1.9 Example
4.1.10 Example
4.1.11 Example
4.1.12 Example
4.1.13 Example
4.1.14 Note
4.1.15 Example
4.1.16 Example
4.1.17 Example
4.1.18 Example
4.1.19 Example
4.1.20 Example
4.1.21 Example
4.1.22 Example
4.1.23 Example
4.1.24 Example
4.1.25 Example
4.1.26 Example
4.1.27 Example
4.2 Examples on Motion About a Fixed Axis (Part I)
4.2.1 Example
4.2.2 Example
4.2.3 Example
4.2.4 Note
4.2.5 Example
4.2.6 Example
4.2.7 Note
4.2.8 Note
4.2.9 Note
4.2.10 Example
4.2.11 Note
4.2.12 Note
4.2.13 Example
4.2.14 Example
4.2.15 Note
4.2.16 Example
4.2.17 Example
4.2.18 Example
4.2.19 Example
4.2.20 Example
4.3 Examples on Motion About a Fixed Axis (Part II)
4.3.1 Example
4.3.2 Example
4.3.3 Example
4.3.4 Example
4.3.5 Example
4.3.6 Example
4.3.7 Example
4.3.8 Example
4.3.9 Example
4.3.10 Example
4.3.11 Example
4.3.12 Example
4.3.13 Note
4.3.14 Example
4.3.15 Example
4.3.16 Example
4.3.17 Example
4.3.18 Example
4.3.19 Example
4.3.20 Example
4.4 Examples on Motion About a Fixed Axis (Part III)
4.4.1 Note
4.4.2 Example
4.4.3 Example
4.4.4 Example
4.4.5 Example
4.4.6 Example
5 Motion in Two Dimensions (Finite Forces)
5.1 Miscellaneous Examples on 2D Motion (Part I)
5.1.1 Note
5.1.2 Example
5.1.3 Example
5.1.4 Example
5.1.5 Example
5.1.6 Example
5.1.7 Example
5.1.8 Example
5.1.9 Example
5.1.10 Example
5.1.11 Example
5.1.12 Example
5.1.13 Example
5.1.14 Example
5.1.15 Example
5.1.16 Example
5.1.17 Example
5.1.18 Example
5.1.19 Example
5.1.20 Example
5.2 Miscellaneous Examples on 2D Motion (Part II)
5.2.1 Example
5.2.2 Example
5.2.3 Example
5.2.4 Note
5.2.5 Note
5.2.6 Note
5.2.7 Note
5.2.8 Note
5.2.9 Example
5.2.10 Example
5.2.11 Example
5.2.12 Example
5.2.13 Example
5.2.14 Example
5.2.15 Example
5.2.16 Example
5.2.17 Example
5.2.18 Example
5.2.19 Example
5.3 Miscellaneous Examples on 2D Motion (Part III)
5.3.1 Note
5.3.2 Example
5.3.3 Note
5.3.4 Example
5.3.5 Example
5.3.6 Example
5.3.7 Example
5.3.8 Example
5.3.9 Example
5.3.10 Example
5.3.11 Example
5.3.12 Example
5.3.13 Example
5.3.14 Example
5.3.15 Example
5.3.16 Example
5.3.17 Example
5.3.18 Example
5.3.19 Example
5.3.20 Example
5.3.21 Example
5.3.22 Example
5.3.23 Example
5.3.24 Example
5.3.25 Example
6 Impulsive Forces
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
6.1.1 Example
6.1.2 Example
6.1.3 Example
6.1.4 Example
6.1.5 Example
6.1.6 Example
6.1.7 Example
6.1.8 Example
6.1.9 Example
6.1.10 Example
6.1.11 Example
6.1.12 Example
6.1.13 Example
6.1.14 Example
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
6.2.1 Note
6.2.2 Note
6.2.3 Note
6.2.4 Note
6.2.5 Example
6.2.6 Example
6.2.7 Example
6.2.8 Example
6.2.9 Example
6.2.10 Example
6.2.11 Example
6.2.12 Example
6.2.13 Example
6.2.14 Example
6.2.15 Example
6.2.16 Example
6.2.17 Example
6.2.18 Example
6.2.19 Example
6.2.20 Example
6.2.21 Example
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
6.3.1 Example
6.3.2 Example
6.3.3 Example
6.3.4 Example
6.3.5 Example
6.3.6 Example
6.3.7 Example
6.3.8 Example
6.3.9 Example
6.3.10 Example
6.3.11 Example
6.3.12 Example
6.3.13 Example
6.3.14 Example
7 Motion in Three Dimensions
7.1 3D Motion of a Rigid Body About a Fixed Point
7.1.1 Note
7.1.2 Note
7.1.3 Note
7.1.4 Note
7.1.5 Example
7.1.6 Example
7.1.7 Example
7.1.8 Example
7.1.9 Example
7.1.10 Example
7.1.11 Example
7.1.12 Note
7.1.13 Note
7.1.14 Note
7.1.15 Note
7.1.16 Note
7.1.17 Note
7.1.18 Note
7.1.19 Note
7.1.20 Note
7.1.21 Note
7.1.22 Note
7.1.23 Note
7.1.24 Note
7.1.25 Note
7.1.26 Note
7.1.27 Note
7.1.28 Example
7.1.29 Example
7.1.30 Note
7.1.31 Note
7.1.32 Note
7.1.33 Note
7.1.34 Note
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
7.2.1 Example
7.2.2 Example
7.2.3 Example
7.2.4 Example
7.2.5 Example
7.2.6 Example
7.2.7 Example
7.2.8 Note
7.2.9 Note
7.2.10 Example
7.2.11 Example
7.2.12 Example
7.2.13 Example
7.2.14 Example
7.2.15 Example
7.2.16 Example
7.2.17 Example
7.2.18 Example
8 Conservation Principles of A.M. and Energy
8.1 Conservation of Angular Momentum
8.1.1 Note
8.1.2 Example
8.1.3 Note
8.1.4 Example
8.1.5 Example
8.1.6 Example
8.2 Examples on Conservation of A.M.
8.2.1 Example
8.2.2 Example
8.2.3 Example
8.2.4 Example
8.2.5 Example
8.2.6 Example
8.2.7 Example
8.2.8 Example
8.2.9 Example
8.2.10 Example
8.2.11 Example
8.2.12 Example
8.2.13 Example
8.2.14 Example
8.2.15 Example
8.2.16 Example
8.2.17 Example
8.2.18 Example
8.2.19 Note
8.2.20 Note
8.2.21 Note
8.2.22 Example
8.3 Motion of a Top and Its Stability
8.3.1 Note
8.3.2 Note
8.3.3 Note
8.3.4 Note
8.3.5 Note
8.3.6 Note
8.3.7 Example
8.3.8 Note
8.3.9 Definition
8.3.10 Example
8.3.11 Note
8.3.12 Note
8.3.13 Note
8.3.14 Note
8.3.15 Note
8.3.16 Note
8.3.17 Note
8.3.18 Note
8.3.19 Note
8.3.20 Note
8.3.21 Note
References
Bibliography
Rajnikant Sinha
Advanced Newtonian Rigid Dynamics
Advanced Newtonian Rigid Dynamics
Rajnikant Sinha
Advanced Newtonian Rigid Dynamics
Rajnikant Sinha Hyderabad, India
ISBN 9789819920211 ISBN 9789819920228 (eBook) https://doi.org/10.1007/9789819920228 Mathematics Subject Classification: 70XX, 70Fxx, 70Hxx, 74XX © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #2101/04 Gateway East, Singapore 189721, Singapore
Preface
Newtonian Dynamics is a great subject humanity has discovered. Although Newton (1643–1727) had initiated the subject, it was nurtured and developed by so many other great mathematical physicists, such as Euler (1707–1783), D’Alembert (1717–1783), Lagrange (1736–1813), Hamilton (1805–1865), Jacobi (1804–1851) and so on. Many historians have opined that the development of Newtonian dynamics was the reason behind the great industrial revolution. How bodies move? Why a spinning top defy gravity? These riddles of nature can be understood only through the study of Newtonian physics. Just before the advent of quantum mechanics, there was a time when physicists began to convince that the science of dynamics is now complete. But new sciences, such as robotics and rocket science, are demanding results of dynamics in a big way. Motion of a top is a topic of dynamics, but is also applied in smartphones. Dynamics has numerous applications in modern appliances, such as gyroscopes. Great theoretical physicist Dirac was of the view that “no one can understand quantum mechanics well without solving tough problems in classical mechanics”. From experience, I have learnt that in order to make real progress in Newtonian dynamics, a student should tackle a few hundred problems in dynamics, which should be powerful as well as of varied nature. Although powerful problems can be had from old question papers of examinations conducted by prestigious universities of the world, there is a dearth of books of required nature. The first criteria of such a book should be that it should supply full solutions to all those problems, so that students could not get frustrated whenever (which is quite likely) they could not discover their answers themselves. Further, there is no right kind of book on rigid dynamics that can provide good theoretical foundation needed for engineering applications. Figure is a must for Newtonian dynamics, because force has direction. So, the second criteria of a book of required kind is that there should be abundant figures decorated with enough descriptions. The third but foremost quality of the book should be that the answers supplied should be natural and real time. This will enable students to learn how a successful attempt looks like. These voids prompted me to write this book.
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Preface
This textbook contains 357 problems, which are fully worked out, and 349 figures. Students should not be prompted at seeing those solutions until they have tried to their utmost ability. Only then their progress will be real. Mastering the art of drawing a good figure on a paper, related to some phenomenon occurring in three dimension, makes thing easier in a drastic way. This book also teaches along this line. It contains unprecedented number of figures with enough descriptions. This would make the “rate of page turning” greatly enhanced. Problems are well graded, varied and enormous. They will challenge the ingenuity of students. But, with this book, they will be assured of getting a friendly solution beneath a cover whenever their attempts don’t succeed. In this way, a student’s mastering rigid body dynamics would be ensured. Prerequisite for studying this book has been kept to a minimum. Wherever some uncommon knowledge in mathematics, such as gamma function, Dirichlet’s triple integral formula and geometry of conicoid, are needed we have derived them at the very place so that reader must not get bothered in searching them elsewhere. For pedagogical reasons and for the purpose of reinforcement, at few places, we have allowed repetition and deviation from the usual order of topics. In Chap. 1, we have developed necessary kinematics of rigid body. In the sequel, we have found Euler’s geometrical formula in terms of Eulerian angles. In Chap. 2, D’Alembert’s principle has been discussed in great detail. For later purpose in Chap. 8, the principle of virtual work has also been discussed here. In Chap. 3, the concept of momental ellipsoid of a point of a rigid body is discussed and 67 problems of various difficulty levels have been solved. Chapters 4–7 are largely devoted towards solving some powerful problems in rigid body dynamics in two and three dimensions. Chapter 6 is exclusively devoted to problems on impulsive forces—a topic of great importance in engineering. Results on angular velocity in three dimension are discussed in Chap. 7. Chapter 8 concerns conservation principles. Here we have tried to demonstrate how a judicious application of these principles makes calculation easy. In the final section of this chapter, first of all, we derive the celebrated “Lagrange’s equations in generalized coordinates”. As an application we have discussed motion of a spinning top in terms of Eulerian angles. Stability of its motion is also discussed here. This is considered as the culmination of Newtonian rigid body dynamics. Finally, on a personal note, I would like to thank my lovely wife, Bina, for her patient endurance and constant encouragement. Hyderabad, India
Rajnikant Sinha
Contents
1 3D Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Examples on Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1 42
2 D’Alembert’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 D’Alembert’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Examples on Motion About a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . 2.4 Examples on Compound Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . .
127 127 183 207 223
3 Momental Ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 General Equation of Second Degree in Three Variables . . . . . . . . . . 3.2 Momental Ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 EquiMomental Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 More Examples on Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . .
249 249 276 311 364
4 Motion About a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Motion About a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Examples on Motion About a Fixed Axis (Part I) . . . . . . . . . . . . . . . . 4.3 Examples on Motion About a Fixed Axis (Part II) . . . . . . . . . . . . . . . 4.4 Examples on Motion About a Fixed Axis (Part III) . . . . . . . . . . . . . .
387 387 442 484 538
5 Motion in Two Dimensions (Finite Forces) . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Miscellaneous Examples on 2D Motion (Part I) . . . . . . . . . . . . . . . . . 5.2 Miscellaneous Examples on 2D Motion (Part II) . . . . . . . . . . . . . . . . 5.3 Miscellaneous Examples on 2D Motion (Part III) . . . . . . . . . . . . . . .
563 563 612 654
6 Impulsive Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Examples on Impulsive Forces in 2D Motion (Part I) . . . . . . . . . . . . 6.2 Examples on Impulsive Forces in 2D Motion (Part II) . . . . . . . . . . . . 6.3 Examples on Impulsive Forces in 2D Motion (Part III) . . . . . . . . . . .
691 691 715 771
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Contents
7 Motion in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805 7.1 3D Motion of a Rigid Body About a Fixed Point . . . . . . . . . . . . . . . 805 7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point . . . 851 8 Conservation Principles of A.M. and Energy . . . . . . . . . . . . . . . . . . . . . . 8.1 Conservation of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Examples on Conservation of A.M. . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Motion of a Top and Its Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
891 891 900 926 972
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 973
About the Author
Rajnikant Sinha is a former Professor of Mathematics at Magadh University, Bodh Gaya, India. A passionate mathematician, Prof. Sinha has published numerous interesting research findings in international journals and four textbooks with Springer: Smooth Manifolds; Real Analysis and Complex Analysis, Vols. 1 and 2; and Galois Theory and Advanced Linear Algebra. His research focusses on topological vector spaces, differential geometry and manifolds.
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Chapter 1
3D Kinematics of Rigid Bodies
From earlier courses in elementary dynamics, we know that the dynamical equations of a phenomenon can only be written down, if we are conversant in kinematics. So this chapter first deals with threedimensional kinematics of rigid bodies. In the course we prove theorems due to Euler and Hamilton. Geometrical property of Eulerian angles has also been derived. For getting a grip over the subject, 34 problems of various difficulty levels have been solved.
1.1 Kinematics 1.1.1 Note Suppose that there is a rigid body. If we fix one point of the rigid body, the body is free to rotate about every line through this point. If we fix a second point of the body, the body limits its possible rotations. It can rotate about the line joining the two fixed points. If we fix a third point of the body, not collinear with the other two, it fixes the position of the body. Thus we see that the positions in space of any three noncollinear points of the body determine the position of the body. We know that nine coordinates are needed to determine the positions of the three points. But these three points belong to the rigid body, so the nine coordinates are connected by three relations, which express the constant distances between the chosen points. It follows that there are only six (= 9 − 3) independent variables. Conclusion A rigid body free from all constraints has six degree of freedom.
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 R. Sinha, Advanced Newtonian Rigid Dynamics, https://doi.org/10.1007/9789819920228_1
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2
1 3D Kinematics of Rigid Bodies
1.1.2 Note By a general displacement of a rigid body, we mean a displacement in which any three points O, P, Q are displaced to other positions O' , P ' , Q' . We shall try to show that the general displacement can be effected by a translation of the body as a whole (in which all points describe equal and parallel straight lines) together with a rotation about a definite axis. For this purpose, let us first translate point O to point O' . Suppose that after translation P occupies the position P1 and Q occupies the position Q1 . We have then to turn the body about some axis such that P1 occupies the position P ' and Q1 occupies the position Q' . It remains to find such an axis. Construction: Let us imagine a sphere fixed in the body, whose centre is O' . Suppose that the line O' P1 intersects the sphere at p1 , line O' Q1 intersects the sphere at q1 , line O' P ' intersects the sphere at p' , and line O' Q' intersects the sphere at q' . Thus p1 , q1 , p' , q' are points on the sphere. Let π be the plane passing through the centre O' of the sphere such that π bisects the great circle p1 p' . This bisector plane has an important property: (∗) For every axis O' n through O' which lies in the plane π, the effect of angle of rotation p1 np' is that p1 occupies the position p' . Let κ be the plane passing through the centre O' of the sphere such that κ bisects the great circle q1 q' . Let the planes π and κ intersect on the line O' a where a lies on the sphere. Let us compare the spherical triangles ap1 q1 and ap' q' : 1. Side p1 q1 is equal to side p' q' , because the body is rigid; 2. Side ap1 is equal to side ap' , because π bisects p1 p' ; 3. Side aq1 is equal to side aq' , because κ bisects q1 q' . It follows that spherical triangle ap1 q1 is congruent to spherical triangle ap' q' . By (∗), the effect of rotation angle p1 ap' about O' a is that p1 occupies the position ' p . Similarly, the effect of rotation angle q1 aq' about O' a is that q1 occupies the position q' . It suffices to show that angle p1 ap' is equal to angle q1 aq' . Since spherical triangle ap1 q1 is congruent to spherical triangle ap' q' ,( and they have one point common, it follows that angle ) p1 ap' = ∠p1 aq1 + ∠q1 ap' = ∠p' aq' + ∠q1 ap' = ∠q1 ap' + ∠p' aq' = ∠q1 aq' is equal to angle q1 aq' ∎ .
1.1.3 Note Suppose that a body is turning about a fixed point O. At any instant its axis of rotation will pass through O. The body in general does not continue to turn about the same axis. How to determine the position of the instantaneous axis of rotation at any instant?
1.1 Kinematics
3
Suppose that the directions of motion of two lines OA and OB in the body are known. Suppose that AA' and BB' are infinite displacements. Suppose that π is the plane through O and perpendicular to AA' . Clearly, the required axis must lie on π. Suppose that κ is the plane through O and perpendicular to BB' . Clearly, the required axis must lie on κ. Hence, the required axis is the line of intersection of π and κ.
1.1.4 Note Suppose that a rigid body undergoes two finite rotations about two concurrent lines OA and OB: 1. About an axis OA through a finite angle α. 2. About an axis OB through a finite angle β. By Sect. 1.1.2. the resultant displacement is a pure rotation, because O is fixed. It suffices to show that there exists an axis OC such that a single rotation about OC is equivalent to the net effect of the two rotations. Let us imagine a sphere S fixed in the rigid body. Let O be the centre of the sphere. Let A and B be the points in which the axes cut the sphere S. Let us draw great circle AB. Let us draw great circle AC1 and great circle AC2 on the opposite sides of AB such that angle C1 AB = angle BAC2 =
α . 2
It is clear that upon the first rotation, great circle AC1 will occupy the position of great circle AC2 . Let us draw great circle BC3 and great circle BC4 on opposite sides of BA such that angle C3 BA = angle ABC4 =
β . 2
Suppose that C is the intersection of AC1 and BC4 . Suppose that C ' is the intersection of AC2 and BC3 . Observe that upon the first rotation, C occupies the position of C ' . Next, upon the second rotation, C ' occupies the position of C. Thus the net effect of the two finite rotations is that C remains fixed (here, order of rotations is important). It follows that the required axis is OC. Now we want to find the angle of rotation about OC. The first rotation leaves A unchanged, and brings B to B1 such that angle BAB1 = α. The second rotation sends A to A' such that angle ABA' = β. Let us compare the spherical triangles A' BC and ABC. Observe that angle ABC = β2 = angle A' BC. Also AB = A' B, and BC is common. It follows that angle ACB = angle BCA' , and CA = CA' . Hence, the required rotation is
4
1 3D Kinematics of Rigid Bodies
( angle ACA
'
)
( ) = 2π − angle ACB + angle BCA' = 2π − (angle ACB + angle ACB) = 2π − 2(angle ACB)
,
and hence the required rotation is −2(angle ACB).
1.1.5 Note Successive rotations about three concurrent lines fixed in space, through twice the angles between the planes formed by them, restore a body to its original position. Proof: Let the lines OA, OB, OC cut a sphere of centre O in the corners of a spherical triangle ABC. By Sect. 1.1.4, rotation about OA, OB through angles 2A, 2B is equivalent to rotation about OC through angle −2(angleACB). This neutralizes the rotation about OC through angle 2C. ∎ This result is known as Hamilton’s theorem.
1.1.6 Note Assume that the two axes of rotations are parallel. We have to find the net effect of two rotations. Case I: When angles of rotations are not “equal but opposite”: We can think of as the geometrical construction given in Sect. 1.1.4, provided we keep in mind that 1. O is situated at infinity; 2. instead of spherical triangles, we have plane triangles perpendicular to the axes of rotation. Thus, by Sect. 1.1.4, the net effect of two rotations is a rotation about a parallel axis through a point C on the plane such that the angle of rotation about the new axis is α + β (Fig. 1.1). Fig. 1.1 2 2
2
2
1.1 Kinematics
5
Case II: When angles of rotations satisfy α = −β : Here, Sect. 1.1.4 does not apply. Here we shall argue differently. Since there is no motion parallel to the axes, we may confine our considerations to a plane perpendicular to the axes. Let us take a line L in such a plane. A rotation of the figure though an angle α about A turns L through an angle α. Again a rotation of the figure though an angle (−α) about B turns the same straight line to its original direction. Thus, the net effect of the two finite rotations leaves all straight lines in the plane unaltered in direction. It follows that the net effect of the two finite rotations is a translation in the direction at right angles to the axes of rotation. Conclusion of Case II It is expressed as (rot(α) about axisA) + (rot(−α) about parallel axisB) ≡ (transl. perp. to the axes) or (rot(α) about an axisB) + (transl. perp. to the axes) ≡ (rot(α) about parallel axisA).
In other words, a rotation about an axis together with a translation perpendicular to the axis is equivalent to a rotation about some parallel axis.
1.1.7 Note In Sect. 1.1.2, we have seen that a general displacement of a rigid body is equivalent to a translation together with a rotation about an axis. We shall try to show that the position of the axis of rotation can be so chosen that the translation is parallel to it. Such a combination of translation and rotation is called a screw. By pitch of the screw we mean the ratio of the distance of translation to the angle of rotation. Suppose that there is a rigid body which undergoes a general displacement. Let O and O' be the initial and final positions, respectively, of a point of the rigid body. Then −−→ by Sect. 1.1.2, the general displacement is equivalent to a translation OO' together with a rotation about an axis O' C ' through O' . Let us draw a line OC through O which is parallel to axis O' C ' . Thus OCO' C ' . Let N be the foot of perpendicular drawn from O' on OC. Thus NO' ⊥OC. It follows that NO' ⊥O' C ' . Since the general displacement −−→' ( −→ −→' ) is equivalent to a translation OO = ON + NO together with a rotation about an axis O' C ' through O' , the general displacement is equivalent to ) −→ ( −→ transl.NO' + transl.ON + rot.about O' C ' .
6
1 3D Kinematics of Rigid Bodies
−→ −→ Since transl.ON is parallel to O' C ' , it suffices to show that transl.NO' + rot.about O' C ' is equivalent to some rot.about an axis parallel to O' C ' . This is true by Sect. 1.1.6, because NO' ⊥OCandOCO' C ' . How to construct the position of the axis of the screw? It suffices to determine an axis L parallel to O' C ' such that the effect of rotation of given angle about L is that N occupies the position of O' . Let π be the plane perpendicular to the axis O' C ' and passing through O' . Clearly O' N lies on π. Let M be the midpoint of O' N . Let α be the angle of rotation. Let us draw a circle in plane π passing through N and O' such that arc NO' subtends an ' angle α at the centre C1 of the circle. Clearly, MC1 = O2N cot α2 . Thus the required axis of the screw is the line through C1 parallel to O' C ' .
1.1.8 Note When a rigid body is in motion with one point fixed, the instantaneous axis of rotation passes through this point. In the general case, the position of instantaneous axis of rotation varies both in the body and in space. Its locus in the body is a cone, and its locus in space is another cone. The two cones cut a unit sphere whose centre is the common vertex of the cones. Thus we get two curves. These curves correspond to the two pole curves in plane motion. We know that the motion of a plane lamina can be produced by the rolling of the pole curve fixed in the lamina on the pole curve fixed in the space. Similarly, the continuous motion of a body about a fixed point can be produced by the rolling of a cone fixed in the body on a cone fixed in space.
1.1.9 Example Consider a rigid rod sliding on a plane with its ends on two perpendicular straight lines CX and XY . − → Since the velocity of material point A is along AC, instantaneous axis of rotation must intersect the line through A perpendicular to CY . Similarly, instantaneous axis of rotation must intersect the line through B perpendicular to CX . It follows that instantaneous axis of rotation}passes through the point I (see Fig. 1.2). x = l cos φ Observe that , so x2 + y2 = l 2 , and hence the space centrode fixed y = l sin φ in space is a circle ⎡ whose centre is C and radius is l. Since the rod AB makes an angle π2 at I , the locus of I in the “extended body” of rod is a circle whose diameter is AB. Let us denote this circle by ⎡b . Observe that as the rod AB changes its position, ⎡b changes its position (in space) but its shape (determined by radius 2l ) remains unchanged. So if we imagine that,
1.1 Kinematics
7
Γ
Instantaneous centre of rotation Γ
Fig. 1.2
instead of body AB, our body is AB together with a circular wire ⎡b , then it is again a “rigid body”. Since I lies on ⎡b , and I lies on ⎡, ⎡b and ⎡ intersect at I . Since centre of ⎡b (that is, mid − point ofAB), centre of ⎡(that is, C), and I are collinear, circle ⎡b touches the circle ⎡ at I . Thus we have shown that at all positions of AB, ⎡b touches ⎡. We shall try to show that ⎡b rolls on ⎡. Observe that A approaches A' as φ → π2 . In the case of rolling without slipping, it suffices to show that arc A' I = arc AI . LHS = arc A' I = CI × ∠A' CI = AB × ∠A' CI = l × ∠A' CI , l RHS = arc AI = AO × ∠AOI = × ∠AOI 2 l = × (2∠ACI ) = l × ∠ACI = l × ∠A' CI . 2
1.1.10 Example The arms AC, CB of a wire bent at right angles slide upon two fixed circles in a plane. Show that the locus of the instantaneous centre in space is a circle, and that its locus in the body is a circle of double the radius of the space centrode (Fig. 1.3).
8
1 3D Kinematics of Rigid Bodies
Radius
Radius
Fig. 1.3
} x = (l sin θ ) cos θ Sol: Here, . So, y = (l sin θ ) sin θ that (
2x l
2x l
)2
= sin 2θ, and
( +
2y −1 l
2y l
= 1 − cos 2θ. It follows
)2 =1
or (
l x + y− 2 2
)2
( )2 l = (space centrode). 2
( ) This is a circle, whose centre is 0, 2l , and radius is 2l .
1.1 Kinematics
9
} x' = (l cos θ ) + b . So, Here, ' y = (l sin θ ) + a ( ' )2 ( )2 x − b + y' − a = l 2 (body centrode). This is a circle, whose centre is l( = double the radius of the space − centrode).
(b, a)
and
radius
is
1.1.11 Definition Suppose that there is a rigid body. Suppose that there is an axis. The rate of increase of the angle between two planes through the axis, one of which is fixed in the body while the other is fixed in space, is called the (magnitude of the) angular velocity of the body about the axis. By Sect. 1.1.6, (rot(α) about an axisB) + (transl. perp. to the axes) ≡ (rot(α) about parallel axisA),
so an angular velocity about an axis is equivalent to an equal angular velocity about a parallel axis together with a velocity perpendicular to the axis, and conversely.
1.1.12 Note Suppose that there is a rigid body. Suppose that there is an axis. Observe that the magnitude of the angular velocity together with sense of a rotation about an axis can be indicated by directed line segment along the axis. We can say that angular velocity is a vector localized in a line (the axis of rotation), provided we show that when they are so represented they obey the parallelogram law of composition. Recall that force is also an example of a vector localized in a line. Suppose that there is a body turning about an axis Oz with angular velocity ω. We can represent the angular velocity by a directed line segment localized in the axis of rotation. Let P be a point of the body. Let M be the foot of perpendicular drawn from P on Oz. Then the velocity of P is the vector − → ω × OP. − → −→ Suppose that the body has two angular velocities represented by AB and AC. Here AB and AC are intersecting axes. Let D be the point in space such that ABDC is a parallelogram. On using the distributive law of cross product, we find that the
10
1 3D Kinematics of Rigid Bodies
− → localized vector AD represents an angular velocity whose effect is same as the net − → − → effect of two angular velocities AB and AC.
1.1.13 Note Suppose that there is a rigid body. Suppose that the body has two angular velocities − → −→ AB and CD such that AB and CD are two parallel axes. Let the plane of the paper intersects the first axis at A and second axis at C. Further assume that AB and CD both are perpendicular to plane of the paper. Let E be a point on AC such that (AB)(AE) = (CD)(CE). It follows that E is at rest, and hence by Sect. 1.1.11, net effect is a pure rotation about a parallel axis through E. Let P be an arbitrary point on line AC. The net velocity of P is equal to (
) = (AB)(AE + EP) . (AB)(AP) + (CD)(CP) + (CD)(−CE + EP) = (AB + CD)EP Hence, for the net effect, we can draw a directed line segment localized in the parallel axis through E whose magnitude is AB + CD. Conclusion Analogous to force, angular velocity behaves like a localized vector.
1.1.14 Note In many cases, it is found convenient to refer the motion of a body to a “frame of reference” which is itself moving in a known manner. Then the motion of a body to a fixed “frame of reference” is its motion referred to the moving frame compounded with the motion of that frame. Let us consider a planar motion first (Fig. 1.4). Here x˙ , y˙ are the velocity components of point P referred to the moving frame. In other words, x˙ , y˙ are the “rate of departure” of the moving point P from the “point” which we regard as rigidly connected to the frame at P. Because of the motion of the frame, this “point” has a velocity ω(OP) perpendicular to OP. Hence the velocity of P referred to fixed frame is compounded of x˙ , y˙ , ω(OP). Observe that the components of velocity ω(OP) are −ωy, ωx. Suppose that u, v are the components of the velocity of P referred to axes fixed in space with which the moving axes are momentarily coincident. Then u = x˙ − ωy,
v = y˙ + ωx.
1.1 Kinematics
11 moving
Fig. 1.4
fixed axis
fixed axis
moving
1.1.15 Note We want to derive the formula in Sect. 1.1.14 by another method. Let M be the foot of perpendicular drawn from P on the xaxis (Fig. 1.5). In fixed frame: The velocity of P is equal to its velocity relative to M compounded −−→ with the velocity of M . Observe that velocity of M has components x˙ along OM and · −→ −→ ωx along MP. Observe that velocity of P relative to M has components y along MP − → and ωy along xO. So u = x˙ − ωy,
Fig. 1.5
v = y˙ + ωx.
moving
fixed axis
moving fixed axis
12
1 3D Kinematics of Rigid Bodies
Fig. 1.6
moving
fixed axis
moving fixed axis
1.1.16 Note − → Let OP represent any vector (for example, velocity, momentum, moment of − → momentum). Suppose that u, v are components OM and MP of OP (see Fig. 1.6) Here u˙ , v˙ are the components of rate of change of point P referred to the moving frame. In other words, u˙ , v˙ are the “rate of departure” of the moving point P from the “point” which we regard as rigidly connected to the frame at P. Because of the motion of the frame, this “point” has a velocity ω(OP) perpendicular to OP. Hence the velocity of P referred to fixed frame is compounded of u˙ , v˙ , ω(OP). Observe that the components of velocity ω(OP) are −ωv, ωu. Suppose that U , V are the components of the velocity of P referred to axes fixed in space with which the moving axes are momentarily coincident. Then U = u˙ − ωv, V = v˙ + ωu.
1.1.17 Note In Sect. 1.1.16, let us take the case when the vector is velocity. Then, the components of the acceleration of P referred to axes fixed in space with which the moving axes are momentarily coincident are d (˙x − ωy) − ω(˙y + ωx), dt
d (˙y + ωx) + ω(˙x − ωy), dt
that is, x¨ − (ωy ˙ + ω˙y) − ω(˙y + ωx), y¨ + (ωx ˙ + ω˙x) + ω(˙x − ωy),
1.1 Kinematics
13
that is, x¨ − ωy ˙ − 2ω˙y − ω2 x, y¨ + ωx ˙ + 2ω˙x − ω2 y.
1.1.18 Note Let the polar coordinates of point P be (r, θ ). Let rˆ be the unit vector in which only r increases, and let θˆ be the unit vector in which only θ increases. Our moving frame ˆ By θ is the only is made of line through O along rˆ , and line through O along θ. ˙ “angular coordinate”, ( ) so the moving axis is turning with angular velocity θ k which ˙ ˆ imparts velocity r θ θ. We want to apply Sect. 1.1.14. In order to bring the moving axes and “ momentarily” fixed axes coincident, we must put x = r, y = 0, ω = θ˙ in the formula of Sect. 1.1.14. The formulae u = x˙ − ωy, v = y˙ + ωx give ( ) u = r˙ , v = θ˙ r = r θ˙ .
1.1.19 Note By Sect. 1.1.18, let us take r for x, 0 for y, and θ˙ for ω. From Sect. 1.1.17. radial acceleration = r¨ − θ¨ 0 − 2θ˙ 0˙ − (θ˙ )2 r, transversal acceleration = 0¨ + θ¨ r + 2θ˙ r˙ − (θ˙ )2 0, that is, radial acceleration = r¨ − r(θ˙ )2 ,
transversal acceleration = r θ¨ + 2θ˙ r˙ =
1d( 2 ) r θ˙ . r dt
1.1.20 Example Show that if the axes of x and y rotate with angular velocities ω1 and ω2 , respectively, and ψ is the angle between them, the component accelerations of the point (x, y) parallel to the axes are (Fig. 1.7)
14
1 3D Kinematics of Rigid Bodies
Fig. 1.7
x¨ − x(ω1 )2 − (xω˙ 1 + 2˙xω1 ) cot ψ − (yω˙ 2 + 2˙yω2 ) csc ψ, and y¨ − y(ω2 )2 + (xω˙ 1 + 2˙xω1 ) csc ω + (yω˙ 2 + 2˙yω2 ) cot ψ. Sol: The acceleration of P(x, y) relative to origin O is acceleration of P(x, y) relative to origin N compounded with acceleration of N relative to origin O. In Fig. 1.7, R1 = x¨ − x(ω1 )2 , T1 = xω˙ 1 + 2˙xω1 ,
R2 = y¨ − y(ω2 )2 , T2 = yω˙ 2 + 2˙yω2 .
So required acceleration is (see Fig. 1.7) (R1 + T1 (− cot ψ)) + (−T2 csc ψ) along Ox, and R2 + T1 csc ψ along Oy, (that is, ) x¨ − x(ω1 )2 − (xω˙ 1 + 2˙xω1 ) cot ψ − (yω˙ 2 + 2˙yω2 ) csc ψ along Ox, and y¨ − y(ω2 )2 + (xω˙ 1 + 2˙xω1 ) csc ψ along Oy. Second Method Suppose that the axes make angles θ and φ with a fixed direction OX . It follows that θ˙ = ω1 , φ˙ = ω2 . Observe that ψ = φ − θ, and − → −→ − → r = OP ᷅ = ON ᷆ ᷇ + NP ᷄ = x(cos θ, sin θ ) + y(cos φ, sin φ), so dr = x˙ (cos θ, sin θ ) dt + xω1 (− sin θ, cos θ ) + y˙ (cos φ, sin φ) + yω2 (− sin φ, cos φ).
u(cos θ, sin θ ) + v(cos φ, sin φ) ≡
It follows that u cos θ + v cos φ = x˙ cos θ − xω1 sin θ + y˙ cos φ − yω2 sin φ, u sin θ + v sin φ = x˙ sin θ + xω1 cos θ + y˙ sin φ + yω2 cos φ, etc.
1.1 Kinematics
15
1.1.21 Note Let (r, θ, φ) be the polar coordinates of a point P. Let N be the foot of the perpendicular drawn from P on the plane Oxy, and let M be the foot of the perpendicular drawn from N on the Oxaxis. Here OP = r, ∠zOP = θ, ∠xON = φ. Let rˆ be the unit vector in the direction of θ, φ unchanged, θˆ be the unit vector in the direction of φ, r unchanged, φˆ be the unit vector in the direction of r, θ unchanged. Observe that rˆ = (sin θ cos φ, sin θ sin φ, cos θ ), ( (π ) (π ) (π )) θˆ = sin + θ cos φ, sin + θ sin φ, cos +θ 2 2 2 = (cos θ cos φ, cos θ sin φ, − sin θ ), rˆ ⊥θˆ , φˆ = rˆ × θˆ =
(sin θ cos φ, sin θ sin φ, cos θ ) ×(cos θ cos φ, cos θ sin φ, − sin θ )
= − sin2 θ sin φ − cos2 θ sin φ, cos2 θ cos φ + sin2 θ cos φ, sin θ cos θ sin φ cos φ − sin θ cos θ sin φ cos φ = (− sin φ, cos φ, 0). So ( ) ˙ cos θ θ˙ sin φ + sin θ cos φ φ, ˙ − sin θ θ˙ r˙ˆ = cos θ θ˙ cos φ − sin θ sin φ φ, = θ˙ (cos θ cos φ, cos θ sin φ, − sin θ ) + φ˙ sin θ (− sin φ, cos φ, 0) ˆ = θ˙ θˆ + φ˙ sin θ φ, ( ) ˙ − sin θ θ˙ sin φ + cos θ cos φ φ, ˙ − cos θ θ˙ θ˙ˆ = − sin θ θ˙ cos φ − cos θ sin φ φ, ˙ sin φ, cos φ, 0) = −θ˙ (sin θ cos φ, sin θ sin φ, cos θ ) + cos θ φ(− = −θ˙ rˆ + cos θ φ˙ φˆ ( ) ˙ − sin φ φ, ˙ 0 = −φ(cos ˙ φ˙ˆ = − cos φ φ, φ, sin φ, 0). It follows that ( ) ( ) ( ) ˆ i = i · rˆ rˆ + i · θˆ θˆ + i · φˆ φˆ = (sin θ cos φ)ˆr + (cos θ cos φ)θˆ + (− sin φ)φ, ( ) ( ) ( ) j = j · rˆ rˆ + j · θˆ θˆ + j · φˆ φˆ = (sin θ sin φ)ˆr + (cos θ sin φ)θˆ + cos φ φˆ ( ) ( ) ( ) k = k · rˆ rˆ + k · θˆ θˆ + k · φˆ φˆ = (cos θ )ˆr + (− sin θ )θˆ .
16
1 3D Kinematics of Rigid Bodies
Also φ˙ˆ = −φ˙ cos φi − φ˙ sin φj ⎛
⎞
= −φ˙ cos φ ⎝(sin θ cos φ)ˆr +(cos θ cos φ)θˆ + (− sin φ)φˆ ⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1 2 ⎛ ⎞ −φ˙ sin φ ⎝(sin θ sin φ)ˆr +(cos θ sin φ)θˆ + cos φ φˆ ⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
2
( ) ( ) = −φ˙ sin θ rˆ − φ˙ cos θ θˆ . So v=
⎛
( ) d( ) rˆr = r˙ rˆ + r r˙ˆ = r˙ rˆ + r θ˙ θˆ + φ˙ sin θ φˆ = r˙ rˆ + r θ˙ θˆ + r φ˙ sin θ φˆ dt ) d( dv = r˙ rˆ + r θ˙ θˆ + r φ˙ sin θ φˆ a= dt dt ( ) ( ) = r¨ rˆ + r˙ r˙ˆ + r˙ θ˙ θˆ + r θ˙ θˆ + r θ˙ θ˙ˆ ) ( + r˙ φ˙ sin θ φˆ + r φ¨ sin θ φˆ + r φ˙ cos θ θ˙ φˆ + r φ˙ sin θ φ˙ˆ ⎛
⎞⎞
⎛
⎛
⎞⎞
θ˙ θˆ +φ˙ sin θ φˆ ⎠⎠ + ⎝ ᷅ ᷆ ᷇ ᷄ r˙ θ˙ θˆ + ᷅ ᷆ ᷇ ᷄ r¨ rˆ +˙r ⎝ ᷅ ᷆ ᷇ ᷄ θ˙ rˆ + cos θ φ˙ φˆ ⎠⎠ = ⎝ ᷅ ᷆ ᷇ ᷄ r θ˙ θˆ +r θ˙ ⎝− ᷅ ᷆ ᷇ ᷄ ⎛
1
2
1
2
1
⎛
⎞⎞
) ( ) ⎟⎟ ⎜( ⎜ +⎝r˙ φ˙ sin θ φˆ + r φ¨ sin θ φˆ + r φ˙ cos θ θ˙ φˆ + r φ˙ sin θ ⎝ −φ˙ sin θ rˆ − φ˙ cos θ θˆ ⎠⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
2
( ) ( ) ( )2 ( )2 ( )2 = r¨ − r θ˙ − r φ˙ sin2 θ rˆ + 2˙r θ˙ + r θ¨ − r φ˙ sin θ cos θ θˆ ( ) + r˙ φ˙ sin θ + r θ˙ φ˙ cos θ + r˙ φ˙ sin θ + r φ¨ sin θ + r θ˙ φ˙ cos θ φˆ ( ) ( ) ( )2 ( )2 ( )2 = r¨ − r θ˙ − r φ˙ sin2 θ rˆ + 2˙r θ˙ + θ¨ − r φ˙ sin θ cos θ θˆ ( ( ) ) ˆ + 2r θ˙ φ˙ cos θ + 2˙r φ˙ + r φ¨ sin θ φ.
1.1 Kinematics
17
Conclusion ⎫ radial velocity = r˙ ⎬ , meridional velocity = r θ˙ ⎭ azimuthal velocity = r φ˙ sin θ
⎫ ˙ 2 sin2 θ ˙ 2 − r(φ) radial acceleration = r¨ − r(θ) ⎬ ˙ 2 sin θ cos θ meridional acceleration = 2˙r θ˙ + r θ¨ − r(φ) ¨ sin θ ⎭ azimuthal acceleration = 2r θ˙ φ˙ cos θ + (2˙r φ˙ + r φ) Another Method Observe the net effect of change in r and θ is a motion in meridian ( ) plane zON . Hence, velocity in meridian plane is r˙ rˆ + r θ˙ θˆ . Velocity due to change ) ( ˆ Hence the net in φ only is same as the velocity of N . This velocity is (r sin θ )φ˙ φ. velocity is ( ) ( ) ˆ r˙ rˆ + r θ˙ θˆ + r φ˙ sin θ φ. In cylindrical coordinates acceleration is ) ) ) ) ( d (ON ) ( )2 ( d2 ˆ ˙ ˙ ¨ sin θ rˆ + cos θ θ + 2 z¨ k + φ + (ON )φ φˆ (ON ) − (ON ) φ dt 2 dt ⎛( 2 ) )⎞ ( )2 ( d ˆ ˙ sin θ rˆ + cos θ θ ⎟ ( ) ⎜ dt 2 (r sin θ ) − (r sin θ ) φ ⎟ ⎜ = z¨ cos θ rˆ − sin θ θˆ + ⎜ ( ⎟. ) ⎠ ⎝ d (r sin θ ) + 2 φ˙ + (r sin θ )φ¨ φˆ dt ((
So radial acceleration is ) ( 2 ( )2 d 2 (r cos θ ) d ˙ sin θ cos θ + (r sin θ ) − (r sin θ ) φ dt 2 dt 2 ⎛ ( )⎞ ( ) d r˙ sin θ + r θ˙ cos θ d r˙ cos θ − r θ˙ sin θ ⎜ ⎟ = cos θ +⎝ dt ⎠ sin θ ( )2 dt ˙ −(r sin θ ) φ ⎛⎛ ⎞ ⎛
⎞⎞
˙ sin θ ⎠ − ⎝r˙ θ˙ sin θ + r θ¨ sin θ + r θ˙ θ˙ cos θ ⎠⎠ = cos θ ⎝⎝ ᷅r¨ cos ᷆ ᷇ θ ᷄ − r ᷅˙ θ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ⎛ ⎛⎛
1
2
⎞
⎛
2
3
4
⎞⎞⎞
⎜ ⎝⎝r¨ sin θ + r˙ θ˙ cos θ ⎠ + ⎝r˙ θ˙ cos θ + r θ¨ cos θ − r θ˙ θ˙ sin θ ⎠⎠⎟ ⎟ ⎜ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ +⎜ ⎟ sin θ 1 2 2 3 4 ⎠ ⎝ ( )2 −(r sin θ ) φ˙ ( )2 ( )2 = r¨ − r θ˙ − r φ˙ sin2 θ.
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1 3D Kinematics of Rigid Bodies
Next, meridional acceleration is ) ( 2 ( )2 d d 2 (r cos θ ) ˙ cos θ − sin θ + (r sin θ ) − (r sin θ ) φ dt 2 dt 2 ⎛ ( )⎞ ( ) d r˙ sin θ + r θ˙ cos θ ˙ d r˙ cos θ − r θ sin θ ⎟ ⎜ = − sin θ +⎝ dt ⎠ cos θ ( ) dt 2 −(r sin θ ) φ˙ ⎞ ⎛ ⎛⎛
⎞⎞
˙ sin θ ⎠ − ⎝r˙ θ˙ sin θ + r θ¨ sin θ + r θ˙ θ˙ cos θ ⎠⎠ = − sin θ ⎝⎝ ᷅r¨ cos ᷆ ᷇ θ ᷄ − r ᷅˙ θ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ⎛ ⎛⎛
1
2
⎞
⎛
2
3
4
⎞⎞⎞
⎜ ⎝⎝r¨ sin θ + r˙ θ˙ cos θ ⎠ + ⎝r˙ θ˙ cos θ + r θ¨ cos θ − r θ˙ θ˙ sin θ ⎠⎠⎟ ⎜ ⎟ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ +⎜ ⎟ cos θ 1 2 2 3 4 ⎝ ⎠ ( )2 ˙ −(r sin θ ) φ ( )2 = 2˙r θ˙ + r θ¨ − (r sin θ cos θ ) φ˙ . Finally, azimuthal acceleration is d (r sin θ ) φ˙ + (r sin θ )φ¨ ( ) ( dt ) = 2 r˙ sin θ + r θ˙ cos θ φ˙ + (r sin θ )φ¨ = 2r θ˙ φ˙ cos θ + 2˙r φ˙ + r φ¨ sin θ.
2
1.1.22 Example 1 For the motion of a particle on a right circular cone we put θ = α in Sect. 1.1.21: ⎫ velocity along generator = r˙ ⎬ , velocity along outward normal = 0 ⎭ ˙ velocity at right angle to axial plane = r φ sin α
⎫ ˙ 2 sin2 α acceleration along generator = r¨ − r(φ) ⎬ ˙ 2 sin α cos α . acceleration along inward normal = r(φ) ¨ sin α ⎭ acceleration at right angle to axial plane = (2˙r φ˙ + r φ) Example 2 For the motion on the surface of a sphere we put r = a in Sect. 1.1.21:
1.1 Kinematics
19
⎫ radial velocity = 0 ⎬ , meridional velocity = aθ˙ ⎭ ˙ azimuthal velocity = aφ sin θ )⎫ ( 2 ˙ 2 sin2 θ ⎬ ˙ + (φ) acceleration towards centre =( a (θ) ) ˙ 2 sin θ cos θ meridional acceleration = a θ¨ − (φ) ⎭ azimuthal acceleration = a(2θ˙ φ˙ cos θ + φ¨ sin θ )
1.1.23 Note Suppose that L is a straight line fixed in space and it passes through the fixed point O. Let rectangular axes Oxyz be turning about the line L. Let θ1 , θ2 , θ3 be the components of the instantaneous angular velocity of the “moving” frame Oxyz. Let P(x, y, z) be a point. Let N be the foot of the perpendicular drawn from P on the plane Oxy, and let M be the foot of the perpendicular drawn from N on the Oxaxis. Let u, v, w be the velocities of P referred to the fixed axes with which the moving axes are momentarily coincident. For velocities parallel to Ox The velocity of P relative to N is
z(θ2 ) ᷅ ᷆ ᷇ ᷄
+
due to rotation of frame Oxyz The velocity of N relative to M is −y(θ3 ) ᷅ ᷆ ᷇ ᷄
0 ᷅ ᷆ ᷇ ᷄ due to translation of Prelative to N + 0 ᷅ ᷆ ᷇ ᷄
due to rotation of frame Oxyz The velocity of M relative to O is 0 ᷅ ᷆ ᷇ ᷄ due to rotation of frame Oxyz So, the velocity of P relative to O is
(= −y(θ3 ))
due to translation of N relative to M + x˙ (= x˙ ) ᷅ ᷆ ᷇ ᷄ due to translation of M relative to O
z(θ2 ) + (−y(θ3 )) + x˙ . Thus, u = x˙ − y(θ3 ) + z(θ2 ). Similarly, v = y˙ − z(θ1 ) + x(θ3 ),
(= z(θ2 ))
w = z˙ − x(θ2 ) + y(θ1 ).
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1 3D Kinematics of Rigid Bodies
1.1.24 Note Suppose that L is a straight line fixed in space and it passes through the fixed point O. Let rectangular axes Oxyz be turning about the line L. Let θ1 , θ2 , θ3 be the components of the instantaneous angular velocity of the “moving” frame Oxyz. Suppose that there is a physical quantity (such as velocity, angular velocity, − → momentum, etc.) which is represented by vector OP. Let u, v, w be its components referred to moving frame Oxyz. Then the rate of displacement of the point represents the rates of change of the components of this vector. As in Sect. 1.1.23, the velocities of P relative to the fixed axes with which the moving axes are momentarily coincident are u˙ − v(θ3 ) + w(θ2 ), v˙ − w(θ1 ) + u(θ3 ), w˙ − u(θ2 ) + v(θ1 ). Thus, the components of the physical quantity referred to the fixed axes with which the moving axes are momentarily coincident are u˙ − v(θ3 ) + w(θ2 ), v˙ − w(θ1 ) + u(θ3 ), w˙ − u(θ2 ) + v(θ1 ).
1.1.25 Note In Sect. 1.1.24, let us take the case when the vector is velocity. Then, the components of the acceleration of P referred to the fixed axes with which the moving axes are momentarily coincident are d (˙x − y(θ3 ) + z(θ2 )) − (˙y − z(θ1 ) + x(θ3 ))(θ3 ) + (˙z − x(θ2 ) + y(θ1 ))(θ2 ), etc. dt
1.1.26 Note Let a body be turning about an instantaneous axis through a fixed point O. Let Oxyz be a turning rectangular axes with point O fixed. Let θ1 , θ2 , θ3 be the components of the angular velocity of moving frame in Oxyz. In place of physical quantity in Sect. 1.1.24, let us take instantaneous angular velocity Ω of the body. Let ω1 , ω2 , ω3 be its components about the moving set of axes, and ωx , ωy , ωz be its components about the “momentarily” fixed set of axes. Then, by Sect. 1.1.24, at the instant when the two sets of axes are coincident, by Sect. 1.1.24, ω˙ x = ω˙ 1 − ω2 (θ3 ) + ω3 (θ2 ), ω˙ y = ω˙ 2 − ω3 (θ1 ) + ω1 (θ3 ), ω˙ z = ω˙ 3 − ω1 (θ2 ) + ω2 (θ1 ).
1.1 Kinematics
21
If the moving axes are fixed in the body, then (θ1 , θ2 , θ3 ) = (ω1 , ω2 , ω3 ), and hence ωx = ω˙ 1 , ωy = ω˙ 2 , ωz = ω˙ 3 . Notation In what follows, when employing moving axes, as here, we use the suffixes x, y, z to denote components of a vector with regard to momentarily fixed set of axes, and suffixes 1, 2, 3 to denote components of a vector with regard to moving set of axes.
1.1.27 Note We may obtain the results of Sect. 1.1.20 by using the method of moving frame axes (see Sect. 1.1.24). This would be the third method of doing the same. Let (r, θ, φ) be the polar coordinates of point P. Let us take the moving axes such ˆ Observe that here we have that O1 lies along rˆ , O2 along θˆ , and O3 lies along φ. two “angular coordinates” namely θ and φ. Only change in θ contributes angular ˆ and only change in φ contributes angular velocity φk. ˙ Hence, angular velocity θ˙ φ, velocity of the rotating frame is ( ( (π ) ) ) ) ( ˙ + θ˙ φˆ = φ˙ cos θ rˆ + cos φk + θ θˆ + θ˙ φˆ = φ˙ cos θ rˆ − φ˙ sin θ θˆ + θ˙ φˆ . 2 We want to apply Sect. 1.1.23. Here θ1 = φ˙ cos θ, θ2 = −φ˙ sin θ, θ3 = θ˙ . In order to bring the moving axes and “ momentarily” fixed axes coincident, we must put x = r, y = 0, z = 0 in the formula of Sect. 1.1.23. The formulae u = x˙ − y(θ3 ) + z(θ2 ), v = y˙ − z(θ1 ) + x(θ3 ), w = z˙ − x(θ2 ) + y(θ1 ) give ( ) ( ) u = r˙ , v = r θ˙ , w = −r −φ˙ sin θ . Thus ⎫ radial velocity = r˙ ⎬ . meridional velocity = r θ˙ ⎭ azimuthal velocity = r φ˙ sin θ Again by Sect. 1.1.24,
22
1 3D Kinematics of Rigid Bodies
radial acceleration = ( )2 ( )( ) ( )( ) ( )2 d (˙r ) − r θ˙ θ˙ + r φ˙ sin θ −φ˙ sin θ = r¨ − r θ˙ − r φ˙ sin2 θ, dt )( ) ( ) d( ) ( r θ˙ − r φ˙ sin θ φ˙ cos θ + r˙ θ˙ meridional acceleration = dt ( )2 ˙ ˙ ¨ = r˙ θ + r θ − r φ sin θ cos θ + r˙ θ˙ ( )2 = 2˙r θ˙ + r θ¨ − r φ˙ sin θ cos θ, ) ( ) ( ) d( r φ˙ sin θ − r˙ −φ˙ sin θ + r θ˙ φ˙ cos θ dt = r˙ φ˙ sin θ + r φ¨ sin θ + r φ˙ θ˙ cos θ + r˙ φ˙ sin θ + r θ˙ φ˙ cos θ = 2˙r φ˙ sin θ + r φ¨ sin θ + 2r φ˙ θ˙ cos θ.
azimuthal acceleration =
1.1.28 Note Suppose that there are two sets of rectangular coordinates OXYZ and Oxyz with −→ −→ − → common origin O. Let eX , eY , eZ be unit vectors along OX , OY , OZ. Let ex , ey , ez − → − → − → be unit vectors along Ox, Oy, Oz. Let P be a point such that − → − → OP = U eX + V eY + W eZ , OP = uex + vey + wez . It follows that ) ( − → u = uex + vey + wez · ex = OP · ex = (U eX + V eY + W eZ ) · ex = (ex · eX )U + (ex · eY )V + (ex · eZ )W, etc. Hence ⎤⎡ ⎤ ⎡ ⎤ ⎡ U u ex · eX ex · eY ex · eZ ⎣ y ⎦ = ⎣ ey · eX ey · eY ey · eZ ⎦⎣ V ⎦ (∗). ez · eX ez · eY ez · eZ W z Let us denote ⎤ ex · eX ex · eY ex · eZ ⎣ ey · eX ey · eY ey · eZ ⎦ ez · eX ez · eY ez · eZ ⎡
by R. Thus
1.1 Kinematics
23
⎡ ⎤ ⎡ ⎤ u U ⎣ v ⎦ = R⎣ V ⎦, w W where ⎡
⎤ cos α11 cos α12 cos α13 R ≡ ⎣ cos α21 cos α22 cos α23 ⎦, cos α31 cos α32 cos α33 α11 denotes angle between ex , eX ; α12 denotes angle between ex , eY ; α13 denotes angle between ex , eZ ; etc. As above ⎡ ⎤ ⎡ ⎤ ⎡ u U cos α11 R−1 ⎣ v ⎦ = ⎣ V ⎦ = ⎣ cos α12 cos α13 w W ᷅
⎤⎡ ⎤ ⎡ ⎤ cos α21 cos α31 u u T⎣ ⎦ ⎦ ⎣ ⎦ = R cos α22 cos α32 v v . cos α23 cos α33 w w ᷆ ᷇ ᷄
It follows that R−1 = RT . Let A(X1 , Y1 , Z1 ), B(X2 , Y2 , Z2 ) be any two points. Observe that the distance ⎡ ⎤ ⎡ ⎤ X1 X2 between R⎣ Y1 ⎦ and R⎣ Y2 ⎦ is equal to Z1 Z2 ⎛ ⎡
⎤ ⎡ ⎤⎞T ⎛ ⎡ ⎤ ⎡ ⎤⎞ X1 X2 X1 X2 ⎝R⎣ Y1 ⎦ − R⎣ Y2 ⎦⎠ ⎝R⎣ Y1 ⎦ − ⎣ Y2 ⎦⎠ Z1 Z2 Z1 Z2 ⎛ ⎛⎡ ⎤ ⎡ ⎤⎞⎞T ⎛⎡ ⎤ ⎡ ⎤⎞ X1 X1 X2 X2 = ⎝R⎝⎣ Y1 ⎦ − ⎣ Y2 ⎦⎠⎠ R⎝⎣ Y1 ⎦ − ⎣ Y2 ⎦⎠ Z1 Z2 Z1 Z2 ⎛⎡ ⎤T ⎞ ⎛ ⎡ ⎤⎞ ⎡ ⎤T ⎤ ⎡ X1 − X2 X1 − X2 ( X1 − X2 X1 − X2 ) ⎟ ⎜⎣ = ⎝ Y1 − Y2 ⎦ RT ⎠R⎝⎣ Y1 − Y2 ⎦⎠ = ⎣ Y1 − Y2 ⎦ RT R ⎣ Y1 − Y2 ⎦ Z1 − Z2 Z1 − Z2 Z1 − Z2 Z1 − Z2 ⎡ ⎡ ⎤T ⎤ X1 − X2 ( ) X1 − X2 = ⎣ Y1 − Y2 ⎦ R−1 R ⎣ Y1 − Y2 ⎦ Z1 − Z2 ⎤ ⎡ ⎤T ⎡ ⎤ X1 − X2 X1 − X2 X1 − X2 X1 − X2 = ⎣ Y1 − Y2 ⎦ I ⎣ Y1 − Y2 ⎦ = ⎣ Y1 − Y2 ⎦ ⎣ Y1 − Y2 ⎦ Z1 − Z2 Z1 − Z2 Z1 − Z2 Z1 − Z2 ⎡
Z1 − Z2
⎤T ⎡
= the distance between (X1 , Y1 , Z1 ) and (X2 , Y2 , Z2 ).
24
1 3D Kinematics of Rigid Bodies
⎡
⎤ cos α11 cos α21 cos α31 Conclusion ⎣ cos α12 cos α22 cos α32 ⎦ represents a rigid body rotation about an cos α13 cos α23 cos α33 axis through O.
1.1.29 Note Let there be a rigid body. Let O be a fixed point in space. Suppose that the rigid body turns about O. Let us take a set of rectangular axes OX , OY , OZ fixed in the body. Imagine a unit sphere whose centre is O. Suppose that X , Y , Z all lie on the sphere. At a different orientation of the body, suppose that X occupies the position A, Y occupies the position B, and Z occupies the position C. Clearly, A, B, C also lie on the sphere. Let us join ZX by great circle. Since ∠ZOX = π2 , and OZ = OX = 1, arc ZX = π2 . Similarly arc XY = π2 , arc YZ = π2 . In short, OXYZ is an “octate” of the unit sphere with centre O. Similarly, OABC is an “octate” of the unit sphere. For convenience, let us think of octate OXYZ as our rigid body. We shall try to bring OXYZ to occupy the position OABC with the help of three “simple” rotations (see Fig. 1.8). For this purpose, let us join ZX , ZC, CA by great circles. Let us extend the arc ZC to the point K such that CK = π2 . Thus CK = CA = π2 . Let us join KA by great circle. First rotation: We rotate octate OXYZ aboutOZaxis by the angle XZC(say, ψ). This will result into bringing X on ZK. Fig. 1.8
2
2 2
1.1 Kinematics
25
Second rotation: We rotate in such a manner that Z occupies the position C. This will result into bringing X to K. Thus the new position of ZX is CK. Suppose that arc ZC = θ, that is, ∠ZOC = θ. Third rotation: We rotate in such a manner that K occupies the position A. Thus the new position of ZX is CA. Now since OXYZ and OABC are octates, Y occupies the position B. Thus octate OXYZ occupies the position OABC. Here, rotation is aboutOCaxis by the angle KCA(say, φ). Thus, the ordered triplet of three angles (ψ, θ, φ) determine the orientation of the rigid body OABC referred to the standard position OXYZ. Here, (ψ, θ, φ) are called Eulerian angles in zy–z convention.
1.1.30 Note Let there be a rigid body. Let O be a fixed point in space. Suppose that the rigid body turns about O. Let us take a set of rectangular axes OX , OY , OZ fixed in space. Let us take a set of rectangular axes OA, OB, OC fixed in the body. Let (ψ, θ, φ) be the Eulerian angles in zy–z convention of the orientation OABC referred to the standard position OXYZ (see Sect. 1.1.29). Suppose that OXYZ takes the position Ox' y' z ' after the first rotation, Ox' y' z ' takes the position Ox'' y'' z '' after the second rotation, and Ox'' y'' z '' takes the position Ox''' y''' z ''' (≡ OABC) after the third rotation. In the notation of Sect. 1.1.28, ⎡
⎤ ⎡ ⎤⎡ ⎤ u' ex' · eX ex' · eY ex' · eZ U ⎣ v' ⎦ = ⎣ ey' · eX ey' · eY ey' · eZ ⎦⎣ V ⎦ w' ez' · eX ez' · eY ez' · eZ W ᷅ ᷆ ᷇ ᷄ ⎡
⎤⎡ ⎤ cos ψ sin ψ 0 U = ⎣ − sin ψ cos ψ 0 ⎦⎣ V ⎦ 0 0 1 W ⎡ ⎤ U cos ψ + V sin ψ = ⎣ −U sin ψ + V cos ψ ⎦, W so ⎫ u' = U cos ψ + V sin ψ ⎬ v' = −U sin ψ + V cos ψ . ⎭ w' = W Here
26
1 3D Kinematics of Rigid Bodies
) ( ) ( U cos ψex' − sin ψey' + V sin ψex' + cos ψey' + W ez' = (U cos ψ + V sin ψ)ex' + (−U sin ψ + V cos ψ)ey' + W ez' = u' ex' + v' ey' + w' ez' = U eX + V eY + W eZ , ᷅ ᷆ ᷇ ᷄ so, for every U , V , W, ) ( ) ( U cos ψex' − sin ψey' + V sin ψex' + cos ψey' + W ez' = U eX + V eY + W eZ . This shows that ⎫ eX = cos ψex' − sin ψey' ⎬ eY = sin ψex' + cos ψey' . ⎭ eZ = ez' In the notation of Sect. 1.1.28, ⎤ ⎡ ⎤⎡ ' ⎤ cos θ 0 − sin θ u'' u ⎣ v'' ⎦ = ⎣ 0 1 0 ⎦⎣ v' ⎦ w'' w' sin θ 0 cos θ ⎡
so ⎫ u'' = u' cos θ − w' sin θ ⎬ . v'' = v' ⎭ '' ' ' w = u sin θ + w cos θ Here u' (cos θ ex'' + sin θ ez'' ) + v' ey'' + w' (− sin θ ex'' + cos θ ez'' ) + W ez' ( ) ( ) = u' cos θ − w' sin θ ex'' + v' ey'' + u' sin θ + w' cos θ ez'' = u'' ex'' + v'' ey'' + w'' ez'' = u' ex' + v' ey' + w' ez' , ᷅ ᷆ ᷇ ᷄ so, for every u' , v' , w' , u' (cos θ ex'' + sin θ ez'' ) + v' ey'' + w' (− sin θ ex'' + cos θ ez'' ) + W ez' = u' ex' + v' ey' + w' ez' . This shows that
1.1 Kinematics
27
⎫ ex' = cos θ ex'' + sin θ ez'' ⎬ . ey' = ey'' ⎭ ' '' '' ez = − sin θ ex + cos θ ez As above, ⎫ ex'' = cos φex''' − sin φey''' ⎬ ey'' = sin φex''' + cos φey''' . ⎭ ez'' = ez''' Observe that the angular velocity Ω is given by ˙ z' + θ˙ ey'' + φe ˙ z''' Ω = ψe ᷅ ᷆ ᷇ ᷄ ˙ ˙ z'' = ψ(− sin θ ex'' + cos θ ez'' ) + θ˙ ey' + φe ) ( ( ) ˙ z'' = ψ˙ − sin θ cos φex''' − sin φey''' + cos θ (ez''' ) + θ˙ ey' + φe ) ( ( ) ˙ z'' = ψ˙ − sin θ cos φex''' − sin φey''' + cos θ (ez''' ) + θ˙ ey'' + φe ) ( ( ) = ψ˙ − sin θ cos φex''' − sin φey''' + cos θ (ez''' ) ) ( ˙ z'' + θ˙ sin φex''' + cos φey''' + φe ⎞ ⎞ ⎛ ⎛ ⎟ ⎟ ⎜ ⎜ = ψ˙ ⎝− sin θ ⎝cos φex''' − sin φey''' ⎠ + cos θ (ez''' )⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
⎛
⎞
3
2
⎜ ⎟ ˙ ''' + θ ⎝sin φex''' + cos φey''' ⎠ + φe z ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
2
3
( ) ( ) = −ψ˙ sin θ cos φ + θ˙ sin φ ex''' + ψ˙ sin θ sin φ + θ˙ cos φ ey''' ( ) + ψ˙ cos θ + φ˙ ez''' , and hence ) ) ) ( ( ( Ω = θ˙ sin φ − ψ˙ sin θ cos φ ex''' + θ˙ cos φ + ψ˙ sin θ sin φ ey''' + φ˙ + ψ˙ cos θ ez''' .
− → − → −→ Thus, the angular velocity components ω1 , ω2 , ω3 along OA, OB, OC are given by ⎫ ω1 = θ˙ sin φ − ψ˙ sin θ cos φ ⎬ ω2 = θ˙ cos φ + ψ˙ sin θ sin φ . ⎭ ω3 = φ˙ + ψ˙ cos θ These are known as Euler’s Geometrical Equations.
28
1 3D Kinematics of Rigid Bodies
1.1.31 Note Let there be a rigid body. Let O be a fixed point in space. Suppose that the rigid body turns about O. Let us take a set of rectangular axes OX , OY , OZ fixed in space. Let us take a set of rectangular axes OA, OB, OC fixed in the body. Let (ψ, θ, φ) be the Eulerian angles in zy–z convention of the orientation OABC referred to the standard position OXYZ (see Sect. 1.1.29). Suppose that OXYZ takes the position Ox' y' z ' after the “first rotation”, Ox' y' z ' takes the position Ox'' y'' z '' after the “second rotation”, and Ox'' y'' z '' takes the position Ox''' y''' z ''' (≡ OABC) after the “third rotation”. In Sect. 1.1.30, we have seen that ⎫ eX = cos ψex' − sin ψey' ⎬ eY = sin ψex' + cos ψey' , ⎭ eZ = ez' so ⎫ ex' = cos(−ψ)eX − sin(−ψ)eY ⎬ ey' = sin(−ψ)eX + cos(−ψ)eY ⎭ ez' = eZ or ⎫ ex' = cos ψeX + sin ψeY ⎬ ey' = − sin ψeX + cos ψeY . ⎭ ez' = eZ In Sect. 1.1.30, we have seen that ⎫ ex' = cos θ ex'' + sin θ ez'' ⎬ , ey' = ey'' ⎭ ez' = − sin θ ex'' + cos θ ez'' so ⎫ ex'' = cos θ ex' − sin θ ez' ⎬ . ey'' = ey' ⎭ ez'' = sin θ ex' + cos θ ez' In Sect. 1.1.30, we have seen that ⎫ ex'' = cos φex''' − sin φey''' ⎬ ey'' = sin φex''' + cos φey''' , ⎭ ez'' = ez'''
1.1 Kinematics
29
so ⎫ ex''' = cos φex'' + sin φey'' ⎬ ey''' = − sin φex'' + cos φey'' . ⎭ ez''' = ez'' Observe that the angular velocity Ω is given by ˙ z' + θ˙ ey'' + φe ˙ z''' Ω = ψe ᷅ ᷆ ᷇ ᷄ ( ) ˙ Z ) + θ˙ ey' + φ(e ˙ z'' ) = ψ(e ˙ Z ) + θ˙ (− sin ψeX + cos ψeY ) = ψ(e ˙ + φ(sin θ ex' + cos θ ez' ) ˙ Z ) + θ˙ (− sin ψeX + cos ψeY ) = ψ(e ˙ + φ(sin θ (cos ψeX + sin ψeY ) + cos θ eZ ) ( ) ˙ = −θ sin ψ + φ˙ sin θ cos ψ eX ( ( ) ) + θ˙ cos ψ + φ˙ sin θ sin ψ eY + ψ˙ + φ˙ cos θ eZ , and hence ) ) ( ) ( ( Ω = −θ˙ sin ψ + φ˙ sin θ cos ψ eX + θ˙ cos ψ + φ˙ sin θ sin ψ eY + ψ˙ + φ˙ cos θ eZ .
−→ −→ − → Thus, the angular velocity components ωx , ωy , ωz along OX , OY , OZ are given by ⎫ ωx = −θ˙ sin ψ + φ˙ sin θ cos ψ ⎬ ωy = θ˙ cos ψ + φ˙ sin θ sin ψ . ⎭ ωz = ψ˙ + φ˙ cos θ
1.1.32 Note Let there be a rigid body. Let O be a fixed point in space. Suppose that the rigid body turns about O. Let us take a set of rectangular axes OX , OY , OZ fixed in space. Let us take a set of rectangular axes OA, OB, OC fixed in the body. Let (ψ, θ, φ) be the Eulerian angles in zy–z convention of the orientation OABC referred to the standard position OXYZ (see Sect. 1.1.29). Suppose that OXYZ takes the position Ox' y' z ' after the “first rotation”, Ox' y' z ' takes the position Ox'' y'' z '' after the “second rotation”, and Ox'' y'' z '' takes the position
30
1 3D Kinematics of Rigid Bodies
Ox''' y''' z ''' (≡ OABC) after the “third rotation”. In Sect. 1.1.30, we have seen that ⎫ eX = cos ψex' − sin ψey' ⎬ eY = sin ψex' + cos ψey' , ⎭ eZ = ez' ⎫ ex' = cos θ ex'' + sin θ ez'' ⎬ , ey' = ey'' ⎭ ez' = − sin θ ex'' + cos θ ez'' ⎫ ex'' = cos φex''' − sin φey''' ⎬ ey'' = sin φex''' + cos φey''' , ⎭ ez'' = ez'''
⎫ ex' = cos ψeX + sin ψeY ⎬ ey' = − sin ψeX + cos ψeY , ⎭ ez' = eZ ⎫ ex'' = cos θ ex' − sin θ ez' ⎬ , ey'' = ey' ⎭ ez'' = sin θ ex' + cos θ ez' ⎫ ex''' = cos φex'' + sin φey'' ⎬ ey''' = − sin φex'' + cos φey'' . ⎭ ez''' = ez''
It follows that ( ) ex''' = cos φex'' + sin φey'' = cos φ(cos θ ex' − sin θ ez' ) + sin φ ey' = cos φ(cos θ (cos ψeX + sin ψeY ) − sin θ (eZ )) + sin φ(− sin ψeX + cos ψeY ) = (cos φ cos θ cos ψ − sin φ sin ψ)eX + (cos φ cos θ sin ψ + sin φ cos ψ)eY + (− cos φ sin θ )eZ , hence cos XA = eX · ex''' = cos φ cos θ cos ψ − sin φ sin ψ . ᷅ ᷆ ᷇ ᷄ Thus cos XA = cos φ cos θ cos ψ − sin φ sin ψ. Next cos YA = eY · ex''' = cos φ cos θ sin ψ + sin φ cos ψ , ᷅ ᷆ ᷇ ᷄ and cos ZA = eZ · ex''' = − cos φ sin θ , ᷅ ᷆ ᷇ ᷄ etc.
1.1 Kinematics
31
1.1.33 Example Let the rectangular axes Oxyz be turning about some line through the fixed point O with an angular velocity whose components are θ1 , θ2 , θ3 . Find the angle between two positions of a moving line in a short interval of time δt, where l, m, n are the direction cosines of the line referred to moving axes. Sol: Let P be the position of the point at time t such that OP = 1, and coordinates of P referred to Oxyz be (l, m, n). As line moves point P moves on the unit sphere. Let P ' be the position of P at time t + δt. Suppose that coordinates of P ' referred to −→ Oxyz be (l + δl, m + δm, n + δn). Thus PP ' have components δl, δm, δn referred to −→ rotating frame Oxyz. It follows that PP ' have components ( ) ˙l − mθ3 + nθ2 (δt), (m ˙ − nθ1 + lθ3 )(δt), (˙n − lθ2 + mθ1 )(δt) referred to frame fixed in space and momentarily coincident with Oxyz. Since P, P ' are consecutive points on the unit sphere, −→  ' PP  arcPP ' ≈ (required angle) = ᷅radius ᷆ ᷇ 1 ᷄ / (( ) ) ˙l − mθ3 + nθ2 (δt) 2 + ((m = ˙ − nθ1 + lθ3 )(δt))2 + ((˙n − lθ2 + mθ1 )(δt))2 / ( )2 = δt ˙l − mθ3 + nθ2 + (m ˙ − nθ1 + lθ3 )2 + (˙n − lθ2 + mθ1 )2 .
1.1.34 Note Let us recall some facts about principal radii of curvature of a surface. The normals at consecutive points of a surface do not, as a rule, intersect. But at any point P there are two directions on the surface, at right angles to each other, such that the normal at a neighbouring point in either of these directions meets the normal at P. These directions are called the principal directions at P. A curve drawn on the surface, and possessing the property that the normal to the surface at consecutive points intersect, is called a line of curvature. It follows that through each point on the surface pass two lines of curvature cutting each other at right angles. The point of intersection of consecutive normals along a line of curvature at P is called a centre of curvature ( of the)surface, and its distance from P, measured in the direction of unit normal n = r1 H×r2 is called principal radius of curvature of the surface.
32
1 3D Kinematics of Rigid Bodies
1.1.35 Note A rigid body is rolling without sliding on a plane, and at any instant its angular velocity has components ω1 , ω2 along the tangents to the lines of curvature at the point of contact, and ω3 along the normal. Let Ox, Oy be tangents to the lines of curvature at the point of contact O of the body and the plane. In a short duration δt, the body undergoes a displacement compounded of three rotations ω1 δt, ω2 δt, ω3 δt. We may consider the effects of these rotations in any order. Since the normals at neighbouring points on a line of curvature intersect, the body may roll along Ox. Similarly, the body may roll along Oy. Case I: When the body rolls along Ox. Suppose that body has rolled along Ox through a small angle ω2 δt, and A is the point of contact at time t + δt. Suppose that the “original” point of contact of the body moves from O to P in duration δt (Fig. 1.9). Let M be the foot of perpendicular drawn from P in Ox, N be the foot of perpendicular drawn from P on the parallel to Oz through A. Since angular velocity has component ω1 along the Ox, it contributes to P a velocity ω1 (PM ) along negative yaxis. Since angular velocity has component ω2 parallel to Oy through A, it contributes to P a velocity ω2 j × (−(NP)i + (MP)k)(= ω2 (NP)k + ω2 (MP)i). Since angular velocity has component ω3 along the normal AC, it contributes to P a velocity ω3 (PN ) along negative yaxis.Thus the components of velocity of P parallel to axes are ω2 (MP), −ω1 (PM ) − ω3 (PN ), ω2 (NP). Let C be the centre of curvature corresponding to the line of curvature in the zxplane. It follows that CP ≈ CA(= R1 ). Also ∠PCN = ω2 δt. It follows that PN = CP sin(ω2 δt) ≈ R1 sin(ω2 δt) ≈ R1 (ω2 δt), and Fig. 1.9
Line of curvature
1.1 Kinematics
33
MP = CA − CN ≈ R1 − CN = R1 − CP cos(ω2 δt) ≈ R1 − R1 cos(ω2 δt) 1 = R1 (1 − cos(ω2 δt)) ≈ R1 (ω2 δt)2 . 2! Hence, the components of velocity of P parallel to axes are ( ) ( ) 1 1 2 2 ω2 R1 (ω2 δt) , −ω1 R1 (ω2 δt) − ω3 (R1 (ω2 δt)), ω2 (R1 (ω2 δt)), 2! 2! that is, −ω3 (R1 (ω2 δt)),
0,
ω2 (R1 (ω2 δt)),
that is, 0, −R1 ω2 ω3 δt, R1 (ω2 )2 δt. Case II: When the body rolls along Oy. As in case I, the components of velocity of P parallel to axes are −R2 ω1 ω3 δt,
R2 (ω1 )2 δt.
0,
It follows that the components of net velocity of P parallel to axes are 0 − R2 ω1 ω3 δt,
−R1 ω2 ω3 δt + 0,
R1 (ω1 )2 δt + R2 (ω1 )2 δt,
that is, −R2 ω1 ω3 δt,
−R1 ω2 ω3 δt,
R1 (ω1 )2 δt + R2 (ω1 )2 δt.
It follows that the point of the body which is at the point of contact has component accelerations R2 ω1 ω3 ,
−R1 ω2 ω3 ,
R1 (ω1 )2 δt + R2 (ω1 )2 .
1.1.36 Note If a body is rotated through right angles about xaxis, point (a, b, c) occupies the new position (a, −c, b). If a body is rotated through right angles about yaxis, point (p, q, r) occupies the new position (r, q, −p). If the first rotation is right angles about xaxis, and the second rotation is right angles about yaxis, then (a, b, c) occupies the new position (b, −c, −a). If the first rotation is right angles about yaxis, and the second rotation is right angles about xaxis, then (a, b, c) occupies the new position
34
1 3D Kinematics of Rigid Bodies
(c, a, b). This shows that if the rotations are finite angles, the order of the rotations about the axes is important.
1.1.37 Example ◦
◦
If a plane figure is rotated through 90 about fixed point A, and then through 90 ◦ about a fixed point B, the result is equivalent to a rotation of 180 about a certain fixed point C; find the position of C. Sol: Let O be the midpoint of AB. Let us take O as the origin, ( OA ) as the(positive ) direction of xaxis, and yaxis on the plane of the paper. Let A a2 , 0 , and B − a2 , 0 . (a a) ( a a) Here 2 , 2 −−−−−→ (0, 0) −−−−−→ − 2 , 2 , so the perpendicular bisector of the rot. (rot. about ) A ( ) about B ( ) ( ) line joining a2 , a2 and − a2 , a2 is x = 0. Here A a2 , 0 −−−−−→ A a2 , 0 −−−−−→ rot. about ( ( A) ( rot. about ) ) B A' − a2 , a , so the perpendicular bisector of the line joining a2 , 0 and − a2 , a is y−
( a − −a 0+a 2 =−2 x− 2 0−a
a 2
+ −a 2 2
)
or y−
a = x. 2
It follows that C is the point of intersection of lines x = 0 and y − a2 = x. Hence ( ) ( ( a) ) ( ) ◦ C 0, 2 . It remains to show that ∠ACA' = 180 , that is, a2 , 0 , 0, a2 , − a2 , a are collinear, that is,   a    2 0a 1   = 0.  0 1   2 −a a 1 2
 a →   ( ) )  2 0a 1  a2 a(a   −a +1 0+ = 0 = RHS. LHS =  0 2 1  = 2 2 4 −a a 1 2
1.1.38 Note Let the first finite rotation be about axis OA through angle 2α. Let the second finite rotation be about axis OB through angle 2β. Suppose that α, β are positive. Let us imagine a geometrical sphere whose centre is O. Suppose that A, B lie on the sphere. Let us choose the inward normal to the sphere as the positive direction.
1.1 Kinematics
35
Let us draw a great circle arc AC ' such that ∠BAC ' = α. Now since α is positive, and positive direction of normal to the sphere is along − → AO, AB, AC ' , AO are in righthanded orientation. Let us draw a great circle arc BC '' such that ∠ABC '' = β. Now since β is positive, and positive direction of normal to the sphere is along − → BO, BC '' , BA, BO are in righthanded orientation. It follows that AC ' and BC '' are on the same side of AB, and hence AC ' and BC '' will intersect at a point, say C. Thus ABC is a spherical triangle. Let us find a point C1 such that spherical triangles ABC and ABC1 are congruent but are on opposite sides of AB. Observe that due to the first finite rotation, the position of A remains unchanged. Let A' be the new position of A due to the second finite rotation. Thus, the net effect of two rotations is that A occupies the position A' . Observe that due to the first finite rotation, C1 occupies the position C. Due to the second finite rotation, C occupies the position C1 . Thus, the net effect of two rotations is that C1 remains unchanged. Hence, the net effect is a rotation about the OC1 axis. Since OC1 is the resultant axis of rotation, and the net effect of two rotations is that A occupies the position A' , the required angle of rotation for net effect is ∠AC1 A' . Observe that spherical triangles C1 BA and C1 BA' are congruent, and are on opposite sides of BC1 . Let us extend BC1 to C2 . We find that ∠AC1 A' = 2∠C2 C1 A = 2(π − ∠BC1 A) = 2π − 2∠BC1 A, and hence the required rotation is −2∠BC1 A.
1.1.39 Note Let ABC be a spherical trigonometry. By “cosine formula”, cos a = cos b cos c + sin b sin c cos A = cos b(cos a cos b + sin a sin b cos C) + sin b sin c cos A(bycosine formula) = cos b(cos a cos b + sin a sin b cos C) ( ) sin a + sin b sin C cos A(bysine formula) sin A ( ) = cos a cos2 b + sin a sin b cos b cos C + sin b(sin a sin C) cot A,
36
1 3D Kinematics of Rigid Bodies
so cos a = cos a cos2 b + sin a sin b cos b cos C + sin a sin b sin C cot A or cos a sin2 b = sin a sin b(cos b cos C + sin C cot A) or cot a sin b = cos b cos C + sin C cot A or cos b cos C = sin b cot a − sin C cot A. This is known as the cotangent formula.
1.1.40 Note Let the first finite rotation be about axis OA through angle 2α. Let the second finite rotation be about axis OB through angle 2β. Let us imagine a geometrical sphere whose centre is O. Suppose that A, B lie on the sphere. Let us choose the inward normal to the sphere as the positive direction. Let us draw a great circle arc AC ' such that ∠BAC ' = α. Let us draw a great circle arc BC '' such that ∠ABC '' = β. Suppose that AC ' and BC '' intersect at a point, say C. Thus ABC is a spherical triangle. Let us find a point C1 such that spherical triangles ABC and ABC1 are congruent but are on opposite sides of AB. By Sect. 1.1.38, the net effect of two rotations is a rotation about the OC1 axis (Fig. 1.10) ∠AC1 A' (≡ x). By triangle BAN , sin γ sin AN = , sin β sin π2 and hence sin AN = sin γ sin β. By triangle ANC1 , sin γ sin β sin AN sin AC1 = , x x = sin 2 sin 2 sin π2 ᷅ ᷆ ᷇ ᷄
1.1 Kinematics
37
Fig. 1.10
⁄
⁄
and hence sin AC1 =
sin γ sin β . sin 2x
On using cotangent formula (see Sect. 1.1.39) on triangle ABC1 , we get √ cos γ cos α = sin γ cot AC1 − sin α cot β = sin γ , ,, , √ )2 ( √ γ sin β 1 − sinsin x sin2 2 = sin γ − sin α cot β = sin γ sin γ sin β √ =
x 2
sin
sin2
x 2
− (sin γ sin β)2 sin β
− sin α cot β =
√ sin2
x 2
x 2
1 − sin2 AC1 − sin α cot β sin AC1 − (sin γ sin β)2
sin γ sin β
− sin α cot β
− (sin γ sin β)2 − sin α cos β sin β
and hence cos γ cos α =
√ sin2
x 2
− (sin γ sin β)2 − sin α cos β sin β
.
It follows that sin2
x − (sin γ sin β)2 = cos γ cos α sin β + sin α cos β 2
or sin2
x = (cos γ cos α sin β + sin α cos β)2 + (sin γ sin β)2 2
or sin
√ x = sin β (cos γ cos α + sin α cot β)2 + sin2 γ . 2
,
38
1 3D Kinematics of Rigid Bodies
1.1.41 Example Find the resultant rotation when a body revolves through a right angle in succession ◦ about two axes which are inclined to one another at an angle of 60 . Sol: Let us use the formula sin
/ x = sin β (cos γ cos α + sin α cot β)2 + sin2 γ 2
given in Sect. 1.1.40 with α = / 1 x sin = √ 2 2 ᷅
(
π ,β 4
=
π ,γ 4
π . 3
=
We get
/ / √ ( )2 ) 15 15 1 1 3 2 3 1 3 1 1 = , + = √ √ +√ 1 + = √ √ 2 2 4 4 4 2 2 2 2 2 8 ᷆ ᷇ ᷄
and hence, the required angle x is given by √ x = 2 sin
−1
15 . 4
1.1.42 Note During any motion, suppose that three points A, B, C fixed in the body move into the positions A' , B' , C ' , respectively. Let M be the midpoint of AA' , and N be the midpoint of BB' . Let us erect a perpendicular to AA' at M , and let us erect a perpendicular to BB' at N . We can rotate these perpendiculars till they intersect at a point, say O. Thus, OA = OA' , OB = OB' , OM ⊥AA' , ON ⊥BB' . Let us compare triangles AOB and A' OB' . Observe that ⎫ OA = OA' ⎬ . OB = OB' ⎭ ' ' AB = A B (since the body is rigid) Now by SSS, triangles AOB and A' OB' are congruent (Fig. 1.11). It follows that ' ' ' ' ∠AOA' + ∠A' OB = ᷅∠AOB = ᷆ ᷇∠A OB ᷄ = ∠A OB + ∠BOB ,
1.1 Kinematics
39
Fig. 1.11
and hence ∠AOA' + ∠A' OB = ∠A' OB + ∠BOB' . It follows that ∠AOA' = ∠BOB' . Since triangles AOB and A' OB' are congruent, we have ∠OBA = ∠OB' A' . Since A , B' , C ' are the new positions of A, B, C fixed in the body, and the body is rigid, triangles ABC and A' B' C ' are congruent, and hence ∠CBA = ∠C ' B' A' . Now since ∠OBA = ∠OB' A' , we have '
∠OBC = ᷅∠CBA − ∠OBA = ᷆ ᷇∠C ' B' A' − ∠OB' A ᷄' = ∠OB' C ' , and hence ∠OBC = ∠OB' C ' . Let us compare triangles OBC and OB' C ' . Observe that ⎫ OB = OB' ⎬ BC = B' C ' (since the body is rigid) , ⎭ ∠OBC = ∠OB' C ' so by SAS, triangles OBC and OB' C ' are congruent. It follows that OC = OC ' , and ' ' ' ' ∠COB' + ∠B' OB = ᷅∠COB = ᷆ ᷇∠C OB ᷄ = ∠C OC + ∠COB .
Hence ∠COB' + ∠B' OB = ∠C ' OC + ∠COB' . This shows that ' ∠A' OA = ᷅∠B' OB = ᷆ ᷇ ∠C OC ᷄ .
Now since the body is rigid, there exists a rotation about O, which brings A to A' , B to B' , and C to C ' . Thus O is the “centre of rotation”. Conclusion A body can be moved from one position into any other by a rotation about some point without translation. In short, at any instant, there exists an axis of
40
1 3D Kinematics of Rigid Bodies
pure rotation. For obtaining centre of rotation O, we erect perpendicular bisectors of AA' and BB' such that they intersect each other.
1.1.43 Note Since the conclusion of Sect. 1.1.42 is true for all finite displacements, it is valid for very small displacements. It follows that, in the case of uniplanar motion, a body may be moved into successive positions it occupies by successive instantaneous rotations about some centres. In order to obtain the position of the centre at any instant, we proceed as follows: Suppose that A and A' are successive positions of one point of the body. Suppose that B and B' are successive positions of the consecutive point of the body. Erect perpendiculars to AA' and BB' . In general, these perpendiculars will meet, because the motion is uniplanar. The point of intersection of these perpendiculars serves the purpose of centre of rotation. In the case of ordinary pendulum, the centre, or axis, of rotation is permanent. In the case of a wheel rolling in a straight line on the ground, the point of contact of the wheel with the ground is, for the moment, the centre of rotation. Thus, in this example, the centre of rotation is not permanent, but instantaneous. The instantaneous centre has two loci according to whether we consider its position with regard to the body, or its position with regard to the space. These two loci are called the body locus (or bodycentrode) and spacelocus (or spacecentrode) In the case of cartwheel, the successive points of contact are the points on the edge of the wheel, and hence the locus with regard to the body is the edge itself. This is a circle whose centre is that of the wheel. In space the points of contact are the successive points on the ground touched by the wheel, that is, a straight line on the ground. Thus, here, the body centrode is a circle and space centrode is a straight line. The motion of the body is given by the rolling of the body centrode, carrying the body with it, upon the space centrode.
1.1.44 Note A body has an angular velocity ω about an axis means every point of the body can be brought from its position at time t to its position at time t + δt by rotation around the axis through an angle ω(δt). A body has three angular velocities ω1 , ω2 , ω3 about three perpendicular axes Ox, Oy, Oz means that during interval of duration δt, the body is turned in succession through angles ω1 (δt), ω2 (δt), ω3 (δt) about theses axes. Convention ω1 is taken as positive when its effect is to turn the body in the direction from Oy to Oz.ω2 is taken as positive when its effect is to turn the body in
1.1 Kinematics
41
the direction from Oz to Ox.ω3 is taken as positive when its effect is to turn the body in the direction from Oz to Ox. For small δt, we shall try to show that it is immaterial in what order these rotations are performed, and hence they can be thought of taking place simultaneously. Let P(x, y, z) be any point of the body at time t. Let M be the foot of the perpendicular drawn from P on Ox. Let θ be the angle which PM makes with the xyplane. It follows that y = PM cos θ, z = PM sin θ. Suppose that a rotation ω1 (δt) is made about Ox, which results into sending P(x, PM cos θ, PM sin θ ) to P ' (x, PM cos(θ + ω1 (δt)), PM sin(θ + ω1 (δt))) ( ) ≈ P ' (x, PM (cos θ − (sin θ )ω1 (δt)), PM (sin θ + (cos θ )ω1 (δt))) Thus, a rotation ω1 (δt) about Ox moves the point (x, y, z) to the point (x, y − zω1 (δt), z + yω1 (δt)). Next a rotation ω2 (δt) about Oy moves the point (x, y − zω1 (δt), z + yω1 (δt)) to the point (x + (z + yω1 (δt))ω2 (δt), y − zω1 (δt), (z + yω1 (δt)) − xω2 (δt)) ≈ (x + zω2 (δt), y − zω1 (δt), (z + yω1 (δt)) − xω2 (δt)). Again, a rotation ω3 (δt) about Oz moves (x + zω2 (δt), y − zω1 (δt), (z + yω1 (δt)) − xω2 (δt)) to the point
the
point
(x + zω2 (δt)) − (y − zω1 (δt))ω3 (δt), (y − zω1 (δt)) + (x + zω2 (δt))ω3 (δt), z + yω1 (δt) − xω2 (δt) ≈ ((x + zω2 (δt)) − yω3 (δt), (y − zω1 (δt)) + xω3 (δt), z + yω1 (δt) − xω2 (δt)) = (x + zω2 (δt) − yω3 (δt), y − zω1 (δt) + xω3 (δt), z + yω1 (δt) − xω2 (δt)) ) ( (ω1 , ω2 , ω3 ) = (xy, z) + (δt) ×(x, y, z)
The symmetry of the final result shows that, for small rotations, the order of angular velocities is immaterial. By Sect. 1.1.36. for finite rotations, the order of angular velocities is material.
1.1.45 Note Let P(x, y, z) be any point of the rigid body. Let M be the foot of the perpendicular drawn fm P on Oxyplane, and let N be the foot of the perpendicular drawn from M on the xaxis. In the plane PMN , let us draw a line l through P in the plane PMN such that line l is perpendicular to PN . Suppose that line l intersects line MN at T (Fig. 1.12).
42
1 3D Kinematics of Rigid Bodies
,
Fig. 1.12
Suppose that the body has three angular velocities ω1 , ω2 , ω3 about three perpendicular axes Ox, Oy, Oz (see Sect. 1.1.44). Clearly, from Fig. 1.12, the ω1 rotation about Ox gives velocity ω1 (0, −z, y). Similarly, the ω2 rotation about Oy gives velocity ω2 (z, 0, −x), and the ω3 rotation about Oz gives velocity ω3 (−y, x, 0). If O is at rest, the net velocity is ω1 (0, −z, y) + ω2 (z, 0, −x) + ω3 (−y, x, 0) ) ( (ω1 , ω2 , ω3 ) = (ω2 z − ω3 y, ω3 x − ω1 z, ω1 y − ω2 x) = ×(x, y, z) If O is in motion, and u, v, w are the components of its velocity parallel to the fixed axes of coordinates, then the component velocities of P in space are
1.2 Examples on Kinematics 1.2.1 Example Prove that, if a spherical polygon turns about its angular points in succession, always remaining on the same spherical surface, and the angle turned through in each case is twice the angle of the polygon at that point, the polygon is restored to its original position. Sol: Let A0 A1 A2 A3 · · · An be a spherical polygon (a generalization of spherical triangle). Suppose that the first rotation is about A1 through 2∠A0 A1 A2 , the second rotation is about A2 through 2∠A1 A2 A3 , the third rotation is about A3 through 2∠A2 A3 A4 , etc.
1.2 Examples on Kinematics
43
1.2.2 Note Suppose that ABC is a spherical triangle. The great circle joining B, C divides the sphere into two hemispheres. Let us select the hemisphere which contains A, find the point A' such that A' B = π2 , and A' C = π2 . Obviously, AA' < π2 . In short, we say that A' is the pole of BC. Let B' be the pole of CA, and C ' be the pole of AB. The spherical triangle A' B' C ' is called the polar triangle of spherical triangle ABC.
1.2.3 Note Suppose that ABC is a spherical triangle. Let A' B' C ' be the polar triangle of ABC. It follows that A' is the pole of side BC, and hence A' B =
π π π , A' C = , A' A < . 2 2 2
B' C =
π π π , B' A = , B' B < , 2 2 2
C 'A =
π π π , C 'B = , C 'C < . 2 2 2
AB' =
π π π , AC ' = , AA' < , 2 2 2
Similarly,
and
It follows that
and hence A is the pole of side B' C ' . Similarly, B is the pole of side C ' A' , and C is the pole of side A' B' . Thus, ABC is the polar triangle of A' B' C ' . Conclusion If A' B' C ' is the polar triangle of ABC, then ABC is the polar triangle of A' B' C ' .
1.2.4 Note Suppose that ABC is a spherical triangle. Let A' B' C ' be the polar triangle of ABC. We denote B' C ' by a' , C ' A' by b' , A' B' by c' . Let us extend AB and AC such that B' C ' intersects AB at D, and B' C ' intersects AC at E. Since A' B' C ' is the polar triangle of ABC, by Sect. 1.2.3, ABC is the polar
44
1 3D Kinematics of Rigid Bodies '
triangle of A B' C ' , and hence A is the pole of B' C ' . Since B' C ' intersects AB at D, D lies on B' C ' . Now since A is the pole of B' C ' , we have AD = π2 . Similarly, AE = π2 . Since B' C ' intersects AB at D, B lies on AD. Similarly, C lies on AE. Since B lies on AD, C lies on AE, AD = π2 , and AE = π2 , we have ∠BAC = ∠DAE = DE. Thus DE = A. Since D lies on AB, and C ' is the pole of AB, we have C ' D = π2 . Since E lies on AC, and B' is the pole of AC, we have B' E = π2 . Since B' C ' intersects AB at D, D lies on B' C ' . Similarly E lies on B' C ' . Thus D, E lie on B' C ' . It follows that ( ( ) ) a' = B' C ' = B' E + DC ' − DE = B' E + DC ' − A ᷅ ᷆ ᷇ ᷄ ) (π (π π) + DC ' − A = + − A = π − A. = 2 2 2 Thus a' = π − A. Similarly, a = π − A' , etc. Conclusion If A' B' C ' is the polar triangle of ABC, then a' + A = π, a + A' = π, etc.
1.2.5 Note From Sect. 1.1.8, the continuous motion of a body about a fixed point can be produced by the rolling of a cone fixed in the body on a cone fixed in space. Suppose that the cones are right circular. Let OZ be the axis of the cone fixed in the space, and let OC be the axis of the cone fixed in the body. Suppose that α is the semiangle of the cone fixed in the space, and β is the semiangle of the cone fixed in the body. Here, cone fixed in the body rolls on the cone fixed in the space. Let OI be the “common generator” along which the two right circular cones touch each other. It follows that OI is the instantaneous axis of rotation of the rigid body. Also, the line OI rotates about the fixed axis OZ. Let Ω be the angular velocity of the line OI about OZ. (Here, Ω is called the angular velocity of precession.) Since OI is the instantaneous axis of rotation of the rigid body, and OC is a line fixed in the body, OC rotates about the axis OI . Let ω be the angular velocity of the line OC about OI . Let us take a point P on OC. Since ω is the angular velocity of the line OC about OI , and P lies on OC, ω is the angular velocity of the point P about OI , and hence the velocity of P is ⎧ ⎨ ω(OP) sin β inside paper in Fig (i) ω(OP) sin β outside paper in Fig (ii) ⎩ ω(OP) sin β outside paper in Fig (iii). Observe that OZ, OI , OC are coplanar, and Ω is the angular velocity of the line OI about OZ, Ω is the angular velocity of the line OC about OZ. Now since P lies on
1.2 Examples on Kinematics
45
Ω
i
ii
iii
Fig. 1.13
OC, Ω is the angular velocity of the point P about OZ, the velocity of P is Fig. 1.13 ⎧ ⎨ Ω(OP) sin(α + β) inside paper in Fig. (i) Ω(OP) sin(α − β) inside paper in Fig. (ii) ⎩ Ω(OP) sin(β − α) outside paper in Fig. (iii). It follows that ⎧ ⎨ Ω(OP) sin(α + β) = ω(OP) sin β in Fig. (i) Ω(OP) sin(α − β) = −ω(OP) sin β in Fig. (ii) ⎩ Ω(OP) sin(β − α) = ω(OP) sin β in Fig. (iii) or ⎧ ⎨ Ω sin(α + β) = ω sin β in Fig. (i) Ω sin(α − β) = −ω sin β in Fig. (ii) ⎩ Ω sin(β − α) = ω sin β in Fig. (iii). Suppose that ω' is the angular velocity of the rolling cone about its axis OC. Since OZ, OI , OC are always coplanar, the velocities of any point on OI due to angular velocities Ω and ω' are equal and opposite. It follows that Ω sin α = −ω' sin β. Similarly, Ω sin α = ω' sin β.
46
1 3D Kinematics of Rigid Bodies
1.2.6 Note Let O be the centre of the unit sphere on which the spherical triangle ABC is ' constructed. Let A' be the pole of side BC, B' be the pole of side CA, andC be the pole of side AB. Suppose that a rigid body undergoes two rotations: 1. About an axis OA' through a finite angle a. 2. About an axis OB' through a finite angle b.
) ( By Sect. 1.1.4, the net effect is a rotation through angle −2 angle A' DB' about D.
1.2.7 Example A right circular cone of semivertical angle α rolls without sliding on a plane. Show that, if ω is the angular velocity of the line of contact, the angular velocity of the cone is ω cot α. Sol: Here, we can apply the formula (see Sect. 1.2.5): Ω sin(α − β) = −ω sin βin Fig. (ii). We get ω sin
(π 2
) − α = −(angular velocity of the cone) sin α
or (angular velocity of the cone) = −ω cot α.
1.2.8 Example A circular disc of radius a rolls on a plane at a constant inclination α and its centre angular velocity Ω. Show that the angular velocity describes a circle of radius c with √ c+ c2 +4a2 of the disc about its axis is Ω (Fig. 1.14). 2a Sol: Here, we can apply the formula (see Sect. 1.2.5): Ω sin α = ω' sin β.
1.2 Examples on Kinematics
47
Fig. 1.14
We get ) (π π − α = ω' cos α, Ω sin = ω' sin 2 2 ᷅ ᷆ ᷇ ᷄ and hence ω' = Ω sec α. Here c cot α = a sin α, and hence c cos α = a − a cos2 α, or a sec2 α − c sec α − a = 0. It follows that √ c + c2 + 4a2 . sec α = 2a Thus, '
ω =Ω
c+
√ c2 + 4a2 . 2a
1.2.9 Example Two wheels, radii a and 2a, are freely mounted at the ends of an axle, at right angles to it, and their rims rest on a rough floor on which they roll, the distance apart of the points in contact being b. Relative to a frame of reference rotating about the vertical with the axle, the angular velocities of the wheels are ω in opposite sense. Prove that . the vertical plane containing the axle rotates with angular velocity 3aω b Sol: Let the required angular velocity be Ω about the “instantaneous centre of curvature”, say (0. − R, 0). It follows that (Fig. 1.15)
48
1 3D Kinematics of Rigid Bodies
ICC 2
2
Fig. 1.15
⎫ α, − sin α) ⎪ ⎪ (( b (0,) 0, Ω) ) = (( b ω(0, ) (− cos )) ⎬ × 0, 2 , 0 − 0, 2b − 2a sin α, 2a cos α × 0, 2 , 0 − (0, −R, 0) ⎪ (0, )0, Ω) (( ) = (( −b −ω(0, ) (− cos α, − sin α) )) ⎪ ⎭ × 0, 2 , 0 − 0, − 2b − a sin α, a cos α × 0, − 2b , 0 − (0, −R, 0) ⎫ ( ( ) ) ω(0, − cos α, − sin α) ⎪ b ⎪ − R + 2 Ω, 0, 0 = ⎬ ×(0, 2a sin α, −2a cos α) = ( ( ) ) −ω(0, − cos α, − sin α) ⎪ ⎪ ⎭ − R − 2b Ω, 0, 0 = ×(0, a sin α, −a cos α) ( ( ) ) } } −( R + 2b) Ω = ω2a R( + 2b Ω = −ω2a ) = = . R − 2b Ω = aω − R − 2b Ω = −aω It follows that Ω=
2b b b −ω3a , R= − = . b 3 2 6
1.2.10 Example Show that, if the instantaneous axis of rotation of a body free to move about a fixed point is fixed in the body, it is fixed in space. Sol: Here, we can apply the formula (see Sect. 1.2.5): Ω sin α = ω' sin β. ( From question, ω' = 0. We have to show that Ω =
ω' sin β sin α
)
= 0. This is clear.
1.2 Examples on Kinematics
49
1.2.11 Example Show that the acceleration of a particle moving in a plane is equivalent to components h2 r along the radius vector to the origin and ph2 dh along the tangent to the path, and p3 ρ ds h is the moment of the velocity about the origin. Find similar acceleration for the components in the direction of the radius vector and normal to the path. Sol: We have to show that ( ( )2 ) ( ) h dh h2 r ( ) r¨ − r θ˙ rˆ + 2˙r θ˙ + r θ¨ θˆ = 3 −ˆr + 2 t. p ρ p ds Since ( ) ( (π )) ( ) ˆ t = t · rˆ rˆ + t · θˆ θˆ = (1 · 1 cos φ)ˆr + 1 · 1 cos − φ θˆ = cos φ rˆ + sin φ θ, 2
it suffices to show that ( )2 2 1. r¨ − r θ˙ = − ph3 ρr + ph2 dh cos φ, ds h dh 2. 2˙r θ˙ + r θ¨ = p2 ds sin φ. Since   ( )         h = r × v = rˆr × v = rˆr × r˙ rˆ + r θ˙ θˆ  = r 2 θ˙ rˆ × θˆ  = r 2 θ˙ k = r 2 θ˙ , we have h = r 2 θ˙ . It follows that ( ) ( ) d (r 2 θ˙ ) d r 2 θ˙ r 2˙r θ˙ + r θ¨ dh 2r˙r θ˙ + r 2 θ¨ dt = = , = ds = s˙ s˙ ᷅ds ᷆ ᷇ ds ᷄ dt r 2˙r θ˙ +r θ¨ and hence dh = ( s˙ ) . ds For (2): Here
( ) r 2 θ˙ r 2˙r θ˙ + r θ¨ h dh sin φ = 2 RHS = 2 p ds p s˙ ( ) 2 ˙ r 2˙ ˙ ¨ r θ + r θ 2˙r θ˙ + r θ¨ θ˙ 1 rd θ r θ = r 4 θ˙ sin φ = 2 p s˙ ds s˙ s˙ p2 ( ( )2 ) ( )2 2˙r θ˙ + r θ¨ 1 1 dr = r 4 θ˙ + 4 2 2 r r dθ (˙s) ( ( ) ) ) ( ) r θ˙ + r θ¨ 1 1 r˙ 2 2˙r θ˙ + r θ¨ (( )2 4 ˙ 2 2˙ r θ˙ + (˙r )2 + 4 =r θ = 2 2 2 r r θ˙ (˙s) (˙s)
50
1 3D Kinematics of Rigid Bodies
=
2˙r θ˙ + r θ¨ 2 (˙s) = 2˙r θ˙ + r θ¨ = LHS. (˙s)2
This proves (2). For (1): Here RHS = −
h2 r h dh h dh h2 r r 2 + 2(r1 )2 − rr2 + 2 + cos φ = − cos φ p3 ρ p2 ds p3 (r 2 + (r )2 ) 23 p ds 1
( 2 ) 2 ( 2 )3 r θ˙ rr + 2(r1 )2 − rr2 r s˙ r 2 + 2(r1 )2 − rr2 h dh h dh =−( + cos φ = − ( )3 ( 3 )( ) ) 3 + p2 ds cos φ 2 ds p r sin φ 3 r 2 + (r1 )2 2 r 2 + (r1 )2 2 r rddsθ
( )2 θ˙ r 2 + 2(r1 )2 − rr2 h dh h dh (˙s)3 r 2 + 2(r1 )2 − rr2 cos φ = − cos φ + 2 + 2 = − ( )3 ( 3 ) ( )2 ) 23 p ds p ds ˙θ r θ˙ ( r 2 + (r1 )2 2 r s˙ r 2 + θr˙˙ =−
h dh h dh (˙s)3 r 2 + 2(r1 )2 − rr2 ( ˙ )3 (˙s)3 r 2 + 2(r1 )2 − rr2 ( ˙ )3 cos φ = − cos φ θ + 2 θ + 2 3 ( ) ( 2 ) 23 ˙ ˙ ( ) p ds p ds rθ r θ 2 2 2 s (˙ ) r θ˙ + (˙r )
)( )2 1( 2 h dh r + 2(r1 )2 − rr2 θ˙ + 2 cos φ r p ds ( ) ( )2 ( ) ( )2 r˙ d r˙ h dh 1 cos φ −r = − r2 + 2 θ˙ + 2 r d θ θ˙ p ds θ˙ ) ( ( )2 1 2 r˙ r¨ θ˙ − r˙ θ¨ 1 ( )2 h dh =− r +2 cos φ θ˙ + 2 − r ( )2 ˙θ ˙ r p ds θ θ˙ ) ( ( )2 h dh r˙ θ¨ (˙r )2 ˙ + 2 =− r θ +2 − r¨ + cos φ ˙ r p ds θ ( ) ) ( ( )2 r 2 θ˙ r 2˙r θ˙ + r θ¨ r˙ θ¨ r )2 (˙ + 2 − r¨ + cos φ = − r θ˙ + 2 r p s˙ θ˙ ( ) ) ( ( )2 r 2 θ˙ r 2˙r θ˙ + r θ¨ r˙ θ¨ (˙r )2 ˙ + =− r θ +2 − r¨ + cos φ r s˙ θ˙ (r sin φ)2 ( ) ) ( ( )2 r 2˙r θ˙ + r θ¨ dr r 2 θ˙ r˙ θ¨ r )2 (˙ +( − r¨ + = − r θ˙ + 2 )2 r s˙ ds θ˙ r · r ddsθ ( ) ) ( ( )2 r 2 θ˙ r 2˙r θ˙ + r θ¨ r˙ r˙ θ¨ (˙r )2 +( − r¨ + = − r θ˙ + 2 ) r ˙ 2 s˙ s˙ θ˙ r · r θs˙ ⎞ ⎛ ⎛ ⎞ 2 ¨ ⎟ ⎜ ( )2 ( ) r ˙ θ r (˙ ) ⎟ ⎝ 2˙r θ˙ + r θ¨ ⎠ r˙ = r¨ − r θ˙ 2 = LHS. ˙ = −⎜ ⎝r θ + 2 r −¨r + θ˙ ⎠ + ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ r θ˙ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1 2 =−
1
2
1.2 Examples on Kinematics
51
Second part: We have seen that acceleration =
h dh h2 r ( ) −ˆr + 2 t. 3 p ρ p ds
Assume that acceleration = aˆr + bn. We have to find a, b. Clearly, h2 r h dh − 3 + 2 (1 · 1 cos φ) = p ρ p ds
) ( ) ( ) h2 r ( ) h dh t · rˆ = aˆr + bn · rˆ = a + b n · rˆ −ˆ r + p3 ρ p2 ds ᷅ ᷆ ᷇ ᷄ (
)) ( (π + φ = a − b sin φ, = a + b 1 · 1 cos 2 so −
h dh h2 r + 2 cos φ = a − b sin φ. p3 ρ p ds
Again )) (π h2 r ( h2 r sin φ = − + φ 1 · 1 cos p3 ρ p3 ρ 2 ) ( 2 ( ) h dh h r( ) −ˆr + 2 t · n = aˆr + bn · n = 3 p ρ p ds ᷅ ᷆ ᷇ ᷄ )) ( ( ( ) π + φ + b = −a sin φ + b, = a rˆ · n + b = a 1 · 1 cos 2 so } 2 a sin φ − b + ph3 ρr sin φ = 0 ) ( 2 . cos φ = 0 a − b sin φ + ph3 ρr − ph2 dh ds Hence 2
a=
− ph3 ρr +
h dh p2 ds
cos φ +
h2 r p3 ρ
sin2 φ
=
h dh p2 ds
cos φ −
h2 r p3 ρ
cos2 φ
cos2 φ − sin2 φ + 1 h2 r h dh ds h2 r h dh h2 r h dh sec φ − 3 = 2 − 3 = 2 − 3 = 2 p ds p ρ p ds dr p ρ p dr p ρ
52
1 3D Kinematics of Rigid Bodies
or a=
h2 r h dh − 3 . 2 p dr p ρ
Also b=
h2 r p3 ρ
sin2 φ − sin φ
(
h2 r p3 ρ
−
h dh p2 ds
cos φ
)
− sin2 φ + 1 h dh sin φ cos φ h dh h h dh dh p2 ds = 2 tan φ = tan φ = sec φ = 2 cos φ p ds p(r sin φ) ds pr ds h dh h dh ds = . = pr ds dr pr dr
1.2.12 Example Point A of a simple pendulum (mass m, length l) in Fig. 1.16. moves with a constant acceleration a0 to the right. Determine the equation of motion of the bob. Sol: Let us introduce a “translating” ξ, ηcoordinate system at A (Fig. 1.17): Here the absolute acceleration abob of the bob is given by Fig. 1.16
Fig. 1.17
1.2 Examples on Kinematics
53
abob = (acceleration of bob relative to ξ, ηframe) + (acceleration of ξ, ηframe due to its translation) + (acceleration of ξ, ηframe due to its rotation) ( 2 ) d (l sin φ) d 2 (l cos φ) = eξ − eη dt 2 dt 2 + (acceleration of ξ, η frame due to its translation) + (acceleration of ξ, η frame due to its rotation) ( 2 ) d (l sin φ) d 2 (l cos φ) = eξ − eη + a0 eξ dt 2 dt 2 + (acceleration of ξ, η frame due to its rotation) ) ( 2 d (l sin φ) d 2 (l cos φ) + a0 eξ + 0 = e − e ξ η dt 2 dt 2 ( ( ) ( ) ) d φ˙ sin φ d φ˙ cos φ eξ + l eη + a0 eξ = l dt dt ⎞ ⎛ ( ) ( )2 l φ¨ cos φ − φ˙ sin φ eξ ⎟ ⎜ ) ⎠ + a0 eξ =⎝ ( ( )2 + l φ¨ sin φ + φ˙ cos φ eη ) (( ) ( )2 = l φ¨ cos φ − φ˙ sin φ + a0 eξ ( ) ( )2 + l φ¨ sin φ + φ˙ cos φ eη . Let T be the tension in the string. By Newton’s second law of motion, (( ) ) ( ) ( )2 ( )2 net force l φ¨ cos φ − φ˙ sin φ + a0 eξ + l φ¨ sin φ + φ˙ cos φ eη = mass of bob ) ( ) ( mg −eη + T − sin φeξ + cos φeη , = m and hence ⎫ ( ) ( )2 l φ¨ cos φ − φ˙ sin φ + a0 = −T msin φ ⎬ ( ) ( )2 φ ⎭ l φ¨ sin φ + φ˙ cos φ = −g + T cos m or ⎛⎛
⎞
⎞
⎞
⎛
( )2 ( )2 ⎜⎜ ⎟ ⎟ ⎟ ⎜ cos φ ⎝l ⎝φ¨ cos φ − φ˙ sin φ ⎠ + a0 ⎠ + sin φl ⎝φ¨ sin φ + φ˙ cos φ ⎠ = −g sin φ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
1
54
1 3D Kinematics of Rigid Bodies
or l φ¨ + a0 cos φ + g sin φ = 0.
1.2.13 Example The two discs in Fig. 1.18. rotate with constant angular velocities Ω and ω about their respective axes. Determine the absolute acceleration of point P at the instant shown. Sol: Suppose that x, y, z coordinate system is fixed in the big disc. We want to describe the acceleration of point P referred to x, y, z coordinate system. Here, velocity vr of P referred to axes fixed in space with which the moving axes of small disc are momentarily coincident is given by ( ( ( π )) π) ˙ − sin φ, cos φ) = rω(0, − sin φ, cos φ), , sin φ + vr = r φ˙ 0, cos φ + = r φ(0, 2 ᷄ ᷅ ᷆ ᷇ 2
and the acceleration ar of P referred to axes fixed in space with which the moving axes are momentarily coincident is given by ( ( ( ( )2 π )) π) ar = −r φ˙ (0, cos φ, sin φ) + r φ¨ 0, cos φ + , sin φ + 2 2 ᷄ ᷅ ᷆ ᷇ ( )2 ¨ − sin φ, cos φ) = −r φ˙ (0, cos φ, sin φ) + r φ(0, dω = −rω2 (0, cos φ, sin φ) + r (0, − sin φ, cos φ) dt = −rω2 (0, cos φ, sin φ) + 0 = −rω2 (0, cos φ, sin φ). Now, the Coriolis acceleration aC of P is given by Fig. 1.18
ℎ
Ω
1.2 Examples on Kinematics
55
(
Ω(0, 0, 1) aC = 2(Ω(0, 0, 1)) × vr = 2 ×rω(0, − sin φ, cos φ)
) = 2Ωrω(sin φ, 0, 0).
Acceleration af of the small disc is given by ) ( af = −(a + r cos φ)Ω2 (0, 1, 0). The absolute acceleration aP of P is given by aP = ar + aC + af = −rω2 (0, cos φ, sin φ) + 2Ωrω(sin φ, 0, 0) ᷆ ᷇ ᷄ ᷅ ) ( + −(a + r cos φ)Ω2 (0, 1, 0) ( ) = 2Ωrω sin φ, −rω2 cos φ − (a + r cos φ)Ω2 , −rω2 sin φ .
1.2.14 Example A horizontal circular platform (radius r) rotates with constant angular velocity Ω (Fig. 1.19.). A block (mass m) is locked in a frictionless slot at a distance a from the centre of the platform. At time t = 0 the block is released. Determine the velocity vr of the block relative to the platform when it reaches the rim of the platform. Sol: Suppose that x, y, z coordinate system is fixed in the disc. We want to describe the acceleration of point P referred to x, y, z coordinate system. Here, velocity vr of P(x, 0, 0) referred to axes fixed in space with which the moving axes of small disc are momentarily coincident is given by vr = x˙ (1, 0, 0) and the acceleration ar of P referred to axes fixed in space with which the moving axes are momentarily coincident is given by ar = x¨ (1, 0, 0). Fig. 1.19
Ω
56
1 3D Kinematics of Rigid Bodies
Now, the Coriolis acceleration aC of P is given by (
Ω(0, 0, 1) aC = 2(Ω(0, 0, 1)) × vr = 2 ×˙x(1, 0, 0)
) = 2Ω˙x(0, 1, 0).
Acceleration af of the slot is given by ) ( af = −xΩ2 (1, 0, 0). The absolute acceleration aP of P is given by ) ( aP = ar + aC + af = x¨ (1, 0, 0) + 2Ω˙x(0, 1, 0) + −xΩ2 (1, 0, 0) ᷅ ᷆ ᷇ ᷄ ( ) = x¨ − xΩ2 , 2Ω˙x, 0 . Let N be the reaction force on in the string. By Newton’s second law of motion, x¨ − xΩ2 = 0. It follows that x = A sinh Ωt + B cosh Ωt, where A, B are constants. Since x(t = 0) = a, we have A sinh Ω0 + B1 = a, and hence B = a. Thus x = A sinh Ωt + a cosh Ωt. It follows that x˙ (t) = AΩ cosh Ωt + aΩ sinh Ωt. Now since x˙ (t = 0) = 0, we have A = 0. Thus x = a cosh Ωt. It follows that vr = aΩ sinh Ωt(1, 0, 0). We have to find vr . Here a cosh Ωt = r, so the required velocity is /( ) √ r 2 aΩ − 1 = Ω r 2 − a2 . a
1.2.15 Note Let s be the arc length measured from a fixed point A on the curve, in the direction of motion, to the position of the particle P. We know that velocity v is given by
1.2 Examples on Kinematics
57 Tube
Fig. 1.20
ds ds d r dr = t, = v= dt ᷅ ᷆ ᷇ dt ᷄ dt dt where t denotes the unit tangent to the path of particle P(r). Thus v = s˙ t or v = vt. The acceleration a is given by dv d a= = (vt) = dt dt ᷅ ᷆ ᷇ ᷄
(
) ( ) ( ) ds d t ds d v dv t+v = v dt ds dt ds ds
) ( ) ) ( ( dv v2 dt dv = v t + v2 (κn) = v t + n, t+v v ds ds ds ρ where n denotes the unit inward normal to the path (Fig. 1.20). Let R be the reaction force on the particle acting inwardly. Let μ be the coefficient of friction. Let T , N be the components of the external force on the particle along the tangent and outward normal. It follows that } mv ddsv = T − μR 2 . m vρ = R − N Case: When a heavy particle slides upward on a curve in a vertical plane. The equations of motion are } mv ddsv = −mg sin ψ − μR 2 . m vρ = R − mg cos ψ It follows that mv or
v2 dv + μ · m = −mg sin ψ − μmg cos ψ ds ρ
58
1 3D Kinematics of Rigid Bodies
v
dv v2 + μ = −g sin ψ − μg cos ψ ds ρ
or v
v2 dv dψ + μ = −g sin ψ − μg cos ψ d ψ ds ρ
or ( ) v2 1 d v2 1 + μ = −g sin ψ − μg cos ψ 2 dψ ρ ρ or ( ) d v2 + (2μ)v2 = −2ρg(sin ψ + μ cos ψ). dψ ( ) Here, integrating factor is e∫ 2μd ψ = e2μψ . So its solution is v2 e2μψ = ∫ e2μψ · (−2ρg(sin ψ + μ cos ψ))d ψ.
1.2.16 Note Suppose that there is a smooth planar tube. Suppose that the tube rotates in its plane about fixed origin O. Suppose that ω is the angular velocity of the tube. Let A be a fixed point on the tube. Let a particle be constrained to move in the tube. Let P be the position of the particle at time t. Let χ be the angle which OA makes with a fixed direction Ox in space. Let ∠AOP = θ and OP = r (Fig. 1.21). Let s be the arc length measured from a fixed point A on the curve, in the direction of motion, to the position of the particle P. 1
̂
Fig. 1.21
1.2 Examples on Kinematics
59
Observe that the acceleration a of the particle is given by ( ) ( )) ( )2 ) 1 d ( 2( ˙ ˙ r χ˙ + θ θˆ a = r¨ − r χ˙ + θ rˆ + r dt ᷅ ᷆ ᷇ ᷄ ( ) ( )) ( )2 ) 1 d ( 2( r ω + θ˙ θˆ = r¨ − r ω + θ˙ rˆ + r dt ( ) ( ( )2 ) 1d( 2 ) 1d( 2 ) r ω + r θ˙ θˆ = r¨ − r ω + θ˙ rˆ + r dt r dt ) ) ( ( ) (( ( ) ( )2 ) 1d( 2 ) 1d( 2 ) r θ˙ θˆ + −2rωθ˙ rˆ − rω2 rˆ + r ω θˆ , = r¨ − r θ˙ rˆ + r dt r dt so ) ) ( (( ( ) ( ) ( )2 ) 1d( 2 ) 1d( 2 ) 2 ˙ ˆ ˙ ˆ ˙ r θ θ + −2rωθ rˆ − rω rˆ + r ω θ. a = r¨ − r θ rˆ + r dt r dt It follows that aω=0
) ( ( ( )2 ) 1d( 2 ) ˙ ˙ r θ θˆ = r¨ − r θ rˆ + r dt
and hence (
) a = aω=0 + −2rωθ˙ rˆ − rω2 rˆ + ᷅ ᷆ ᷇ ( dv = v t+ ds ( dv = v t+ ds ( dv = v t+ ds
(
) 1d( 2 ) r ω θˆ r dt ᷄
( ) ) 1d( 2 ) v2 r ω θˆ n − 2rωθ˙ rˆ − rω2 rˆ + ρ r dt ⎛ ⎞ ) v2 2 ⎝ 2˙r ω +r ω˙ ⎠θˆ ˙ n − ᷅2rω ᷆ ᷇θ ᷄rˆ −rω rˆ + ᷅ ᷆ ᷇ ᷄ ρ 1 1 ( ) 2 ) v ˆ n + 2ω r˙ θˆ − r θ˙ rˆ − rω2 rˆ + (r ω) ˙ θ. ρ
Thus ) ( ( ) v2 dv a = v t + n + 2ω r˙ θˆ − r θ˙ rˆ − rω2 rˆ + (r ω) ˙ θˆ (∗). ds ρ Since )) ( (π ( ) ( ) −φ t = t · rˆ rˆ + t · θˆ θˆ = (1 · 1 cos φ)ˆr + 1 · 1 cos 2
60
1 3D Kinematics of Rigid Bodies
( ) ( ) dr dθ θˆ = (cos φ)ˆr + (sin φ)θˆ = rˆ + r θˆ ds ds ( ( )) ( ) ) ( ( )) ( ) 1 1( 1 dt d θ dt dr r˙ rˆ + r θ˙ θˆ = θˆ = rˆ + r r˙ rˆ + r θ˙ θˆ , = ds dt ds dt v v v we have t=
) 1( r˙ rˆ + r θ˙ θˆ . v
Since )) ( ) ( (π ( ) + φ rˆ n = n · rˆ rˆ + n · θˆ θˆ = 1 · 1 cos 2 ( ) ( ) rd θ dr θˆ rˆ + + (1 · 1 cos φ)θˆ = (− sin φ)ˆr + (cos φ)θˆ = − ds ds ) ( ) ( ) ( ) ( ) dt dr −r 1 1( dt d θ rˆ + θˆ = θ˙ rˆ + r˙ θˆ = r˙ θˆ − r θ˙ rˆ , = −r ds dt ds dt v v v we have n=
) 1( r˙ θˆ − r θ˙ rˆ . v
Now, from (∗), ) ( v2 dv a = v t + n + 2ω(vn) − rω2 rˆ + (r ω) ˙ θˆ (∗∗). ds ρ Since )) ( (π ( ) ( ) +φ rˆ = t · rˆ t + n · rˆ n = (1 · 1 cos φ)t + 1 · 1 cos 2 ( ) ( ) dθ dr t− r n, n = (cos φ)t − (sin φ)n = ds ds we have ( rˆ =
) ( ) dθ dr t− r n. ds ds
Since )) ( ) ( ) ( (π −φ θˆ = t · θˆ t + n · θˆ n = 1 · 1 cos 2 ) ( ) ( dr rd θ t+ n, t + (1 · 1 cos φ)n = (sin φ)t + (cos φ)n = ds ds
1.2 Examples on Kinematics
61
we have θˆ =
(
) ( ) dr rd θ t+ n. ds ds
Now, from (∗∗), , (( ) ) ( ) ) dr v2 dθ dv 2 t− r n a = v t + n + 2ω(vn) − rω ds ρ ds ds (( ) ( ) ) rd θ dr + (r ω) ˙ t+ n ds ds (
or ) ( 2 ) ( v dr dv 2 dr 2 dθ 2 2 dθ − rω + r ω˙ t+ + 2ωv + r ω + r ω˙ n. a= v ds ds ds ρ ds ds Let R be the reaction force on the particle acting inwardly. Let T , N be the components of the external force on the particle along the tangent and inward normal. It follows that } ) ( dv 2 dr 2 dθ ( 2 m v ds − rω ds + r ω˙ ds )= T m vρ + 2ωv + r 2 ω2 ddsθ + r ω˙ dr =R+N ds or 2
m vρ
} 2 dθ mv ddsv = T + mrω2 dr − mr ω ˙ ds ds . = N − mr 2 ω2 ddsθ − mr ω˙ dr + (R − 2mωv) ds
Since ( ( ) ) dθ 2 dr 2 2 dθ 2 dr mrω t − mr ω n = mrω t − r n = mrω2 rˆ , ds ds ds ds ( ) ( ) dr dθ dr 2 dθ ˆ −mr ω˙ t − mr ω˙ n = −mr ω˙ r t + n = −mr ω˙ θ, ds ds ds ds we have ( ) ( 2) dv v mv t+ m n ds ρ ⎛
⎞
⎟ ⎜ ( ) ( ) ⎟ ⎜ = ⎜external force + mrω2 rˆ +mr ω˙ −θˆ + 2mωv(−n)⎟ + Rn. ᷅ ᷆ ᷇ ᷄⎠ ⎝ ᷅ ᷆ ᷇ ᷄ cenrifugal force
Coriolis force
62
1 3D Kinematics of Rigid Bodies
1.2.17 Example If the axes Ox, Oy revolve with constant angular velocity ω, and the component velocities of the point (x, y) parallel to the axes are ) ) ( ( ωy a2 − b2 ωx a2 − b2 and , a2 + b2 a2 + b2 prove that the point describes relatively to the axes an ellipse. Sol: Here ) ) ( ( ωy a2 − b2 ωx a2 − b2 = x˙ − ωy, = y˙ + ωx. a2 + b2 a2 + b2 So ) ( ) a2 − b2 a2 − b2 , y˙ = ωx −1 + 2 x˙ = ωy 1 + 2 a + b2 a + b2 (
or x˙ =
2a2 −2b2 ωy, y ˙ = ωx. a2 + b2 a2 + b2
It follows that −2b y˙ dy −b2 x 2 2 ωx = = a 2a+b2 = 2 dx x˙ a y ωy a2 +b2 ᷅ ᷆ ᷇ ᷄ 2
or b2 xdx + a2 ydy = 0. On integration, we get x2 y2 + = C, a2 b2 where C is a constant. This is an ellipse.
1.2 Examples on Kinematics
63
1.2.18 Example A particle moves with constant velocity V along a straight line fixed in a plane, while the plane turns around a fixed axis perpendicular to itself with constant angular velocity ω; prove that the path of the particle is the curve V θ = V cos−1
√ a + ω r 2 − a2 , r
where a is the least distance of the particle from the axis. Also, show that the accelerations along and at right angles to a perpendicular on this axis are, respectively, ( ) √ ω 2aV + r 2 ω 2V ω r 2 − a2 and . r r Sol: From Fig. 1.22, θ = cos−1
a + ωt, and(V t)2 + a2 = r 2 . r
Hence V θ = V cos−1
√ a + ω r 2 − a2 r
It suffices to show that ( )2 ω 2aV +r 2 ω) 1. r¨ − r θ˙ = − ( r , √ 2V ω r 2 −a2 ˙ ¨ 2. 2˙r θ + r θ = . r √ Since V t = r 2 − a2 , we have V =
d 1 r˙r d√ 2 r − a2 = √ , (2r˙r ) = √ (V t) = 2 2 2 dt r − a2 ᷅dt ᷆ ᷇ ᷄ 2 r − a
and hence Fig. 1.22 At time
At time
64
1 3D Kinematics of Rigid Bodies
V =√
r˙r r2
− a2
.
It follows that )√ ( 2 (˙r ) + r¨r r 2 − a2 − r˙r 2√r12 −a2 (2r˙r ) r˙r d d 0 = (V ) = √ = dt dt r 2 − a2 r 2 − a2 ᷅ ᷆ ᷇ ᷄ =
)(r 2 −a2 )−(r˙r )2 ((˙r )2 +r¨r√ r 2 −a2
r 2 − a2 ( ) ( ) ) ( 2 2 −(˙r )2 a2 + r¨r r 2 − a2 (˙r ) r − a2 + r¨r r 2 − a2 − (r˙r )2 = = ( )3 ( )3 r 2 − a2 2 r 2 − a2 2 or ) ( −(˙r )2 a2 + r¨r r 2 − a2 0= . ( )3 r 2 − a2 2 It follows that ) ( r¨r r 2 − a2 = (˙r )2 a2 or (˙r )2 a2 ). r¨ = ( 2 r r − a2 Since ωt = θ − cos−1
a r
we have ω=
d( a) −a d (ωt) −1 a˙r ˙ = θ − cos−1 = θ˙ − / ( a )2 r 2 r˙ = θ − r √r 2 − a2 dt dt r ᷅ ᷆ ᷇ ᷄ 1− r
and hence a˙r ω = θ˙ − √ . r r 2 − a2 It follows that
1.2 Examples on Kinematics
65
) ( d a˙r d (ω) = θ˙ − √ dt dt r r 2 − a2 ᷅ ᷆ ᷇ ᷄ ) ( √ √ r¨ r r 2 − a2 − r˙ r˙ r 2 − a2 + r 2√r12 −a2 (2r˙r ) ( ) = θ¨ − a r 2 r 2 − a2 ) ( √ √ 2 r˙ r¨ r r 2 − a2 − r˙ r˙ r 2 − a2 + √rr2 −a 2 ( ) = θ¨ − a r 2 r 2 − a2 √ (˙r )2 2r 2 −a2 r¨ r r 2 − a2 − √(r 2 −a2 ) ( ) = θ¨ − a r 2 r 2 − a2 ) ( ) ( r¨ r r 2 − a2 − (˙r )2 2r 2 − a2 ¨ =θ −a ( )3 r 2 r 2 − a2 2 ( ) (˙r )2 a2 − (˙r )2 2r 2 − a2 ¨ =θ −a ( )3 r 2 r 2 − a2 2 ( ) −2(˙r )2 r 2 − a2 2(˙r )2 ¨ = θ¨ − a ( ) 23 = θ + a 2 √ 2 r r − a2 r 2 r 2 − a2 0=
and hence 2(˙r )2 0 = θ¨ + a √ . r 2 r 2 − a2 It follows that (˙r )2 θ¨ = −2a √ . r 2 r 2 − a2 For 1: We have to show that ) ( 2aV ω a2 V 2 aV ω a2 V 2 (˙r )2 a2 2 2 ( ) − rω − − 3 = r¨ − r ω + 2 2 + 4 r r r r r r 2 − a2 )2 )2 ( ( aV a˙r = r¨ − r ω + 2 = r¨ − r ω + √ r r r 2 − a2 ) ( 2 ( )2 ω 2aV + r ω 2aV ω − rω2 , = r¨ − r θ˙ = − =− r r ᷅ ᷆ ᷇ ᷄ that is,
66
1 3D Kinematics of Rigid Bodies
2aV ω a2 V 2 2aV ω (˙r )2 a2 ) − rω2 − ( − 3 =− − rω2 , 2 2 r r r r r −a that is, a2 V 2 (˙r )2 a2 )= 3 , ( r r r 2 − a2 that is ( ) V 2 r 2 − a2 . (˙r ) = r2 2
This is clearly true, because V = For 2: We have to show that
√ r˙r . r 2 −a2
) ( 2a(˙r )2 (˙r )2 2˙r θ˙ − √ = 2˙r θ˙ + r −2a √ r r 2 − a2 r 2 r 2 − a2 √ √ r ω r 2 − a2 2 √r 2r˙−a 2V ω r 2 − a2 2 = 2˙r ω, = 2˙r θ˙ + r θ¨ = = r r ᷅ ᷆ ᷇ ᷄ that is, 2a(˙r )2 2˙r θ˙ − √ = 2˙r ω, r r 2 − a2 that is, ) ( ( ) 2a(˙r )2 a˙r ˙ . − √ = 2˙r ω − θ = 2˙r − √ r r 2 − a2 r r 2 − a2 ᷅ ᷆ ᷇ ᷄ This completes the proof.
1.2.19 Note Let OA be a line fixed in space. Suppose that Ox and Oy are rectangular axes, which are not fixed in space, but which revolve in any manner about O in their own plane. Let θ be the inclination of Ox to OA at time t. Let P be a moving point in the Oxyplane. Let M be the foot of perpendicular drawn from P on Ox. Let N be the foot of perpendicular drawn from P on Oy. Fig. 1.23
1.2 Examples on Kinematics
67
Acc.
Acc.
Acc.
Acc.
Fig. 1.23
Observe that the acceleration of P(x, y) along Ox is equal to (acc. of P relative to N ) + (acc. of N along Ox) = (acc. of M relative to O) + (acc. of N along Ox) ( 2 ) ( )2 d (OM ) ˙ = − OM · θ + (acc. of N along Ox) dt 2 ( 2 ) ( )2 d x ˙ = − x · + (acc. of N along Ox) θ dt 2 ( ( )2 ) + (acc. of N along Ox) = x¨ − x · θ˙ ) ( ( ) ( )2 d (ON ) ˙ ˙ ¨ = x¨ − x · θ θ + ON · θ − 2 dt ( ) ( ( )2 ) dy ˙ ¨ ˙ − 2 θ +y·θ = x¨ − x · θ dt ) ( ( ) ( )2 − 2˙yθ˙ + y · θ¨ , = x¨ − x · θ˙ so ( ) ( )2 ) ( − 2˙yθ˙ + y · θ¨ . acc. of P(x, y) along Ox = x¨ − x · θ˙ Similarly, ( ) ( )2 ) ( + 2˙xθ˙ + x · θ¨ . acc. ofP(x, y) alongOy = y¨ − y · θ˙
1.2.20 Example Show that the path of a point P which possesses two constant velocities u and v, the first of which is in a fixed direction and the second of which is perpendicular to the radius OP drawn from a fixed point O, is a conic whose focus is O and whose eccentricity is uv .
68
1 3D Kinematics of Rigid Bodies
Fig. 1.24
cos
2
, sin
2
1,0
Sol: From Fig. 1.24, net velocity of P(x, y) is equal to ( ( ( π) π )) , sin θ + u(1, 0) + v cos θ + (= (u − v sin θ, v cos θ )), 2 2 and hence dx dt
} = u − v sin θ , dy = v cos θ dt
or dr dt
} θ) cos θ − r ddtθ sin θ = d (r cos = u − v sin θ dt . dr sin θ + r ddtθ cos θ = d (r dtsin θ ) = v cos θ dt
It follows that dr dt dr dt
cos θ − r ddtθ sin θ
sin θ + r ddtθ cos θ
=
u − v sin θ v cos θ
or ⎛
⎞
v cos θ ⎝ ᷅cos ᷆ ᷇θdr ᷄ − ᷅r sin ᷆ ᷇θd θ ᷄⎠ = u(sin θdr + r cos θd θ ) ⎛
1
2
⎞
−v sin θ ⎝sin θdr ᷄ + ᷅r cos ᷆ ᷇θ d θ ᷄⎠ ᷅ ᷆ ᷇ 1
2
or vdr = u(sin θ dr + r cos θ d θ ) or (v − u sin θ )dr = ur cos θ d θ
1.2 Examples on Kinematics
69
or ln r =
u cos θ dr C = d θ = − ln(v − u sin θ ) + ln C = ln . v − ᷆ ᷇u sin θ ᷄ v − u sin θ ᷅r
It follows that ln r = ln
C v − u sin θ
or r=
C v − u sin θ
or C v
r
=1−
( u π) cos θ − . v 2
This is a conic whose focus is O, and eccentricity is uv .
1.2.21 Note Suppose that a particle moves in a plane with an acceleration which is always directed to a fixed point O in the plane. Let us choose O as the origin, and a fixed straight line OX as the initial line. Let the polar coordinates of moving point be (r, θ ). Since acceleration is always directed to O, 1d( 2 ) r θ˙ = 0. r dt It follows that r 2 θ˙ = h, where h is a constant. It follows that dθ 1 = hu2 , where u ≡ . dt r Let P be the acceleration of the particle directed towards O. So ( )2 d 2r dθ − r = −P(I ). 2 dt dt
70
1 3D Kinematics of Rigid Bodies
Since u = 1r , we have ( ( ) ) dr du 1 du 1 du d θ −1 du 2 =− 2 =− 2 = 2 hu = −h , dt u dt u d θ dt u dθ dθ and hence du dr = −h . dt dθ It follows that ( ) ( ) ( ) d du d 2r d dr d du = −h = = −h dt 2 dt dt dt dθ dt d θ ᷅ ᷆ ᷇ ᷄ ) ( ) ( )) ( ( d θ d 2u d 2u d 2u d θ d du · = −h · 2 = −h hu2 · 2 = −h2 u2 · 2 , = −h dt d θ d θ dt d θ dθ dθ and hence 2 d 2r 2 2d u = −h u . dt 2 dθ2
Now from (I), −h2 u2
d 2 u 1 ( 2 )2 − hu = −P dθ2 u
or d 2u P + u = 2 2 (II ). 2 dθ h u We know that ( ( ) )2 ( )2 )2 ( ) ( 1 du 1 1 1 dr 2 2 4 d u 2 4 −1 du 2 = 2 + 4 =u +u =u +u =u + , p2 r r dθ dθ u2 d θ dθ ᷅ ᷆ ᷇ ᷄
so 1 = u2 + p2 It follows that
(
du dθ
)2 .
1.2 Examples on Kinematics
d −2 dp = 3 p dθ dθ ᷅ =2
(
1 p2
71
)
( ( )2 ) d du d 2 u du du 2 = +2 = 2u u + dθ dθ dθ dθ dθ2 ᷆ ᷇ ᷄
) ( du d 2u du P u+ 2 =2 · dθ dθ d θ h2 u2
and hence du P −2 dp =2 · . p3 d θ d θ h2 u2 It follows that −1 dp P = 2 2 p3 du h u or P=
−1 h2 u2 3 p
( )2 ( )2 dp −1 dp −1 2 1 2 1 ) ( =h =h du r p3 d 1r r p3
dp dr −1 r2
=
h2 dp p3 dr
or h2 dp = P. p3 dr This is the pedal equation of the path.
1.2.22 Note Suppose that a particle P moves in a central orbit. Let P(r, θ ) be the position of the particle at time t and Q(r + Δr, θ + Δθ ) be the position of the particle at time t + Δt. So, rate of describing the sectorial area is equal to 1 OP · OQ sin P OQ area P OQ = lim 2 Δt→0 Δt→0 Δt Δt 1 1 r · (r + Δr) sin P OQ r · (r + Δr) sin Δθ = lim 2 = lim 2 Δt→0 Δt→0 Δt Δt sin Δθ Δθ 1 dθ 1 · = r · (r + 0)1 = lim r · (r + Δr) Δt→0 2 Δθ Δt 2 dt 1 2 1 = r θ˙ = h (see 1.2.21) 2 2
lim
72
1 3D Kinematics of Rigid Bodies
Recall that h is a constant. Conclusion In every central orbit, the sectorial area traced out by the radius vector to the centre of force is equal to the constant 21 h.
1.2.23 Note Suppose that a particle P moves in an ellipse under a force which is always directed towards its focus. Suppose that the equation to an ellipse referred to its focus is l = 1 − e cos θ. r It follows that u=
1 e − cos θ. l l
By Sect. 1.2.21 (II), ( ) ) ( ) ( 1 e 1 e e d2 1 e 1 = cos θ + − cos θ = − cos θ + − cos θ l l l l dθ2 l l l l d 2u P P 2 = +u = 2 2 = 2r , 2 ᷅d θ ᷆ ᷇ h u ᷄ h so P 1 = 2 r2. l h It follows that P=
μ h2 . , where μ ≡ r2 l
By Sect. 1.2.22, the sectorial area traced out by the radius vector to the centre of force is equal to the constant 21 h, and area of the ellipse is π ab, the time the particle takes to describe the whole arc of the ellipse is equal to ⎞ ⎛ π ab ⎝ πa 2π 3 ⎠ π ab π ab = / = √ a2 . = 1√ = / 1 μ 1 1 h μl 1 b2 2 2 μ μa 2 a 2
1.2 Examples on Kinematics
73
In short, we write 2π 3 T = √ a2. μ
1.2.24 Example If the motion of a point in a plane be referred to two axes one of which is fixed while the other revolves about the origin so that the line joining the origin to the point is equally inclined at an angle 21 θ to the axes, show that the component acceleration parallel to the fixed (ξ ) is ( ) ξ¨ − 2ξ˙ θ˙ + ξ θ¨ csc θ, and find the other component (Fig. 1.25) Sol: The component acceleration parallel to the fixed (ξ ) = (acc. of P relative to N ) + (acc. of N along ξ axis) = (acc. of M relative to O) + (acc. of N along ξ axis) = (acc. of M along ξ axis) + (acc. of N along ξ axis) ( ( ) ) d (ON ) ¨ ¨ ˙ ¨ = ξ + (acc. of N along ξ axis) = ξ + − 2 θ + ON · θ csc θ dt ( ) ( ) dη dξ ¨ ˙ ¨ ¨ ˙ ¨ = ξ − 2 θ + η · θ csc θ = ξ − 2 θ + ξ · θ csc θ, dt dt because ξ = η. The other component of acceleration
⁄ ⁄ fixed in space
Fig. 1.25
74
1 3D Kinematics of Rigid Bodies
= (acc. of P relative to M ) + (acc. of M along η axis) = (acc. of N relative to O) + (acc. of M along η axis) = (acc. of N along η axis) + (acc. of M along η axis) ( 2 ) ( )2 d (ON ) ˙ = (acc. of N along η axis) + 0 = − ON · θ dt 2 ( )2 ( )2 d 2η d 2ξ = 2 − η · θ˙ = 2 − ξ θ˙ , dt dt because ξ = η.
1.2.25 Example If the perpendiculars from a point P on the axes Ox, Oy of which Ox is fixed, and Oy revolves uniformly are ξ, η, prove that the accelerations of P parallel to the axes at the instant when they are perpendicular are η¨ − 2ξ˙ ω + ηω2 , and ξ¨ , the angular velocity of Oy being ω (Fig. 1.26). Sol: The acceleration parallel to the fixed axis
fixed in space
Fig. 1.26
1.2 Examples on Kinematics
75
= (acc. of P relative to N ) + (acc. of N along x axis) = (acc. of M relative to O) + (acc. of N along x axis) = (acc. of M along x axis) + (acc. of N along x axis) d 2 (OM ) d 2 (OM ) + of N along x axis) = (acc. 2 dt 2 ) ) ( dt ( d (ON ) ω + ON · ω˙ csc θ + − 2 dt ) ( d 2 (OM ) d (ON ) ω + ON · 0 csc θ = − 2 dt 2 dt d 2 (OM ) d (ON ) = − 2ω csc θ dt 2 dt d 2 (OM ) d (MP) = − 2ω csc θ dt 2 dt =
=
d 2 (NP) d 2 (OM ) d (ξ csc θ ) d (ξ csc θ ) = − 2ω csc θ − 2ω csc θ 2 2 dt dt dt dt
d 2 (η csc θ ) d (ξ csc θ ) − 2ω csc θ 2 dt dt ) ( ) ( ) d (ξ csc θ ) d( ˙ ˙ = η¨ csc θ + 2η˙ −θ csc θ cot θ + η −θ csc θ cot θ − 2ω csc θ dt dt ) ( d (ξ csc θ ) d = η¨ csc θ + 2η(−ω ˙ csc θ cot θ ) + η (−ω csc θ cot θ ) − 2ω csc θ dt dt ) ( d (ξ csc θ ) d cos θ − 2ω csc θ = η¨ csc θ − 2ηω ˙ csc θ cot θ − ηω 2 dt sin θ dt ) ( − sin3 θ − cos θ (2 sin θ cos θ ) ω = η¨ csc θ − 2ηω ˙ csc θ cot θ − ηω · sin4 θ d (ξ csc θ ) −2ω csc θ dt ) ( − sin3 θ − cos θ (2 sin θ cos θ ) ω = η¨ csc θ − 2ηω ˙ csc θ cot θ − ηω · sin4 θ ) ( −2ω csc θ ξ˙ csc θ + ξ (−ω csc θ cot θ ) ) ( − sin3 θ − cos θ (2 sin θ cos θ ) ω = η¨ csc θ − 2ηω ˙ csc θ cot θ − ηω · sin4 θ ) ( −2ω csc θ ξ˙ csc θ + ξ (−ω csc θ cot θ ) , =
and hence
76
1 3D Kinematics of Rigid Bodies
) ( −1 − 0 ω ¨ − 2ηω0 ˙ − ηω · (acc.parallel to the fixed axis)θ = π2 = η1 1 ( ) 2 ˙ ˙ −2ω1 ξ 1 + ξ (−ω0) = η¨ − 2ξ ω + ηω . The acceleration parallel to the moving axis = (acc. of P relative to M ) + (acc. of M along y axis) = (acc. of P relative to M ) + 0 = acc. of N relative to O ) ( 2 d (ON ) 2 = (acc.ofN along yaxis) = − ON · ω dt 2 ) ( d (ON ) ω + ON · ω˙ cot θ + 2 dt ( 2 ) ( ) d (ON ) d (ON ) 2 + 2 ω + ON · 0 cot θ − ON · ω dt 2 dt ) ( ) ( 2 d (ON ) d (MP) 2 ω cot θ = − MP · ω + 2 dt 2 dt ) ( ) ( 2 d (ON ) d (ξ csc θ ) 2 + 2 ω cot θ = − ξ csc θ · ω dt 2 dt ) ( d = ξ¨ csc θ + 2ξ˙ (−ω csc θ cot θ ) + ξ (−ω csc θ cot θ ) dt ) ( d (ON ) ω cot θ − ξ csc θ · ω2 + 2 dt ) ( d cos θ = ξ¨ csc θ + 2ξ˙ (−ω csc θ cot θ ) − ξ ω dt sin2 θ ( ) d (ON ) −ξ csc θ · ω2 + 2 ω cot θ dt ) ( − sin3 θ − cos θ (2 sin θ cos θ ) ¨ ˙ ω = ξ csc θ + 2ξ (−ω csc θ cot θ ) − ξ ω sin4 θ ) ( d (ON ) ω cot θ, −ξ csc θ · ω2 + 2 dt and hence (acc.parallel to the moving axis)θ = π2 ) ( d (ON ) 2 ω 0 = ξ¨ . −ξ 1 · ω + 2 dt
) ( 2 −1 − 0 ¨ = ξ1 + 0 − ξω 1
1.2 Examples on Kinematics
77
1.2.26 Example Figure 1.27 shows three bars AB, BC, CD, internally pin connected at B and C, externally hinges supported at A and D. Find the angular velocities of the bars BC, CD for the instant shown. Sol: Clearly, velocity vB of B is given by vB = (0.25)(6)
m m = 1.5 ↑ . s s
Suppose that ωCD is the angular velocity of bar CD. It follows that velocity vC of C perpendicular to CD is given by Fig. 1.28 / vC = ωCD (0.15)2 + (0.14)2 = ωCD 0.2. Suppose that ωBC is the angular velocity of bar BC. Here (BC)(ωBC )(←) is the velocity of C due to rotation of bar BC about B. It follows that
0.25 m, 6 rad/s
0.25 m
0.15 m
0.14 m
Fig. 1.27 0.14 m
(a)
Fig. 1.28
78
1 3D Kinematics of Rigid Bodies
) ( 0.15 = 1.6, 0.4ωBC = (0.25 + 0.15)ωBC = (BC)(ωBC ) = vB tan β = 1.5 ᷅ ᷆ ᷇ ᷄ 0.14
and hence ωBC = 4 rad/s Since 0.2 = 2.14, ωCD 0.2 = vC = vB sec β = 1.5 sec β = 1.5 ᷅ ᷆ ᷇ ᷄ 0.14 we have ωCD = 1.07 rad/s.
1.2.27 Example The centre C of an elliptic wire is moved with uniform velocity along a fixed line OY in its own plane while the wire is in contact at P with a fixed line OX perpendicular to OY . Show that the acceleration perpendicular to OX of the point of the wire in 1 contact at P varies as (CM1 )2 (PM1 )2 (OC) 5 where M is the foot of perpendicular drawn from P on the major axis of the ellipse. Figure 1.29 Sol: Let CO = p. Here equation of the ellipse is (X sin θ + (Y − p) cos θ )2 (X cos θ − (Y − p) sin θ )2 + = 1. a2 b2 For the point P, let us put Y = 0. We get (X cos θ − (0 − p) sin θ ) (X sin θ − (0 − p) cos θ ) + =1 a2 b2 or X 2 cos2 θ + 2X cos θ p sin θ + p2 sin2 θ a2 2 2 X sin θ − 2X sin θ p cos θ + p2 cos2 θ + = 1(∗). b2 This is a quadratic equation in X . Here disc. = 0, so
1.2 Examples on Kinematics
79
Ring at time
Fig. 1.29
(
) 2 cos θ p sin θ 2 cos θ p sin θ 2 − a2 b2 ( 2 ) 2 )( 2 p sin2 θ cos θ sin θ p2 cos2 θ −4 + + − 1 =0 a2 b2 a2 b2
or ⎞
⎛
⎛
⎞
2 2 ⎜ 2 2 2 ⎜ 1 ⎟ 1 1 1⎟ ⎟ − cos θ ⎜ p sin θ + p cos θ − 1⎟ p2 sin2 θ cos2 θ ⎜ + −2 ⎝ a4 ⎠ 4 2 b2 ⎠ 2 ⎝ 2 2 b a a a b ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1 2 1 ⎛ ⎞
−
2 2 2 2 ⎟ sin2 θ ⎜ ⎜ p sin θ + p cos θ −1⎟ = 0 ⎠ 2 b2 ⎝ a2 b ᷅ ᷆ ᷇ ᷄ 2
or
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1 3D Kinematics of Rigid Bodies
( ) ) ) ( ( 1 1 sin2 θ p2 sin2 θ cos2 θ p2 cos2 θ −p2 sin2 θ cos2 θ 2 2 2 − − 1 − − 1 =0 a b a2 b2 b2 a2
or ) ) ( ( 2p2 sin2 θ cos2 θ + cos2 θ p2 cos2 θ − b2 + sin2 θ p2 sin2 θ − a2 = 0 or )2 ( p2 cos2 θ + sin2 θ = b2 cos2 θ + a2 sin2 θ or p=
√
b2 cos2 θ + a2 sin2 θ .
Hence the equal root of (∗) is −(coefficient ofX ) ) ( 2 coefficient ofX 2 ) ( ) ( − 2 cos aθ 2p sin θ − 2 cos bθ 2p sin θ cos θ p sin θ a2 − b2 ) ( 2 = = 2 2 b cos2 θ + a2 sin2 θ 2 cosa2 θ + sinb2 θ ( ) ) cos θ sin θ ( cos θ p sin θ a2 − b2 = . = a2 − b2 p2 p Thus ) ) ( 2 2 cos θ sin θ ,Y = 0 . P X = a −b p (
On using {
x = X cos θ − (Y − p) sin θ y = X sin θ + (Y − p) cos θ
we have ⎜ P⎜ ⎝
or
) ⎞ ) ( 2 2 cos θ sin θ cos θ − (0 − p) sin θ, = a −b ⎟ p ⎟ ) ( ⎠ ( 2 ) 2 cos θ sin θ sin θ + (0 − p) cos θ = a −b p (
⎛ x y
1.2 Examples on Kinematics
⎛ ⎜ P⎜ ⎝
81
) ( ⎞ ) cos θ sin θ ( 2 cos θ + p sin θ, a − b2 ⎟ p ⎟ ) ( ⎠ ( 2 ) 2 cos θ sin θ sin θ − p cos θ = a −b p
x= y
or ) ) ) ( 2 ( 2 a − b2 sin2 θ − p2 a − b2 cos2 θ + p2 sin θ, y = cos θ P x= p p (
or ( 2 ) ( ) ⎞ a − b2 cos2 θ + b2 cos2 θ + a2 sin2 θ x = sin θ, ⎜ ⎟ p ⎜ ⎟ P⎜ ) ( 2 ( ) ⎟ 2 2 2 2 2 2 ⎝ ⎠ a − b sin θ − b cos θ + a sin θ y= cos θ p ⎛
or (Fig. 1.30) ) ( −b2 a2 sin θ, y = cos θ . P x= p p Thus CM =
√ b2 a2 sin θ, PM = cos θ, OC = p = b2 cos2 θ + a2 sin2 θ . p p
Hence
instantaneous centre of rotation
Fig. 1.30
82
1 3D Kinematics of Rigid Bodies
1 p2 1 1 1 p2 1 . = 4 2 4 = 4 4 2 2 5 a sin θ b cos2 θ p5 a b p sin2 θ cos2 θ (CM ) (PM ) (OC) Now, it suffices to show that ) ( ((acc. at P) · (unit vector alongOY axis)) p sin2 θ cos2 θ = const. From question, d√ 2 d (OC) dp = V = = b cos2 θ + a2 sin2 θ dt dt dt ᷅ ᷆ ᷇ ᷄ ( 2 ) 1 = √ b (−2 cos θ sin θ ) + a2 (2 cos θ sin θ ) θ˙ 2 2 2 2 2 b cos θ + a sin θ ) ) ( 2 ( 2 a − b2 cos θ sin θ a − b2 cos θ sin θ ˙ θ˙ θ= =√ p b2 cos2 θ + a2 sin2 θ so ) ( 2 a − b2 cos θ sin θ ˙ θ. V = p Here, angular velocity ω is given by V p V V V ) =( 2 = θ˙ , ω= = = 2 2 cos θ sin θ 2 cos θ sin θ CI OP a − b a − b ᷅ ᷆ ᷇ ᷄ p so ω = θ˙ . On using the formula VP = VC + ω × rP/C we get ) ) (( ) ( 2 2 cos θ sin θ , 0, 0 − (0, p, 0) a −b VP = V (0, 1, 0) + ω(0, 0, −1) × p ( ) (0, 0, −1) (( ) ) = V (0, 1, 0) + ω 2 × a − b2 cos θpsin θ , −p, 0 ( ) ( ) cos θ sin θ = V (0, 1, 0) + ω −p, − a2 − b2 ,0 p ) ( ( 2 ) cos θ sin θ , 0 = (−ωp, 0, 0). = −ωp, V − ω a − b2 p Since ) ( 2 a − b2 cos θ sin θ ω, V = p
1.2 Examples on Kinematics
83
we have d d 0 = (V ) = dt dt ᷅
(( ) ) a2 − b2 cos θ sin θ ω p ᷆ ᷇ ᷄
) p(− sin θ ω sin θ ω + cos θ cos θ ωω + cos θ sin θ ω) ( ˙ − cos θ sin θ ωp˙ = a2 − b2 , 2 p and hence ( ) p − sin θ ω2 sin θ + cos θ cos θ ω2 + cos θ sin θ ω˙ = cos θ sin θ ωp˙ or ( ) p2 cos 2θ ω2 + cos θ sin θ ω˙ = cos θ sin θ ωp˙p ᷅ ᷆ ᷇ ᷄ ) ωd( 2 ω d ( 2) p = cos θ sin θ b cos2 θ + a2 sin2 θ 2 dt 2 dt ) ω( 2 2 = cos θ sin θ a − b sin 2θ ω 2 ) )2 (( 2 1 1 ) a − b2 cos θ sin θ ω = ( ) (V p)2 . =( 2 2 2 a −b a − b2 = cos θ sin θ
It follows that ( ) p2 cos 2θ ω2 + cos θ sin θ ω˙ = (
1 ) (V p)2 a2 − b2
or V2 ) cos 2θ ω2 + cos θ sin θ ω˙ = ( 2 a − b2 or 2
V − cos 2θ ω2 (a2 −b2 ) . ω˙ = cos θ sin θ
On using the formula aP = aC + α × rP/C − ω2 rP/C we get
84
1 3D Kinematics of Rigid Bodies .
.
aP = 0 + ω ×rP/C − ω2 rP/C = ω × ( ) ( ) cos θ sin θ , −p, 0 −ω2 a2 − b2 p
) ( ( 2 ) cos θ sin θ , −p, 0 a − b2 p
V ( ) − cos 2θ ω2 ) ( 2 (a2 −b2 ) 2 cos θ sin θ = , −p, 0 (0, 0, −1) × a − b cos θ sin θ p , ) ( ( 2 ) 2 2 cos θ sin θ a −b , −p, 0 −ω p 2
and hence ) ( (acc. at P) · (unit vector alongOY axis) p sin2 θ cos2 θ ⎛
⎞ V2 − cos 2θ ω2 ( ) ( ) cos θ sin θ a2 −b2 ) ( a2 − b2 + ω2 p⎠ p sin2 θ cos2 θ =⎝ cos θ sin θ −p (( =
) ) ) ( 2 V2 2 2 2 2 ( ) − cos 2θ ω b − a + ω p sin2 θ cos2 θ a2 − b2
) ) ( ( = −V 2 − b2 − a2 cos 2θ ω2 + ω2 p2 sin2 θ cos2 θ ( ) = −V 2 sin2 θ cos2 θ − b2 − a2 cos 2θ (ω cos θ sin θ )2 + (ω cos θ sin θ )2 p2 ( )2 ( )2 ) ( 2 Vp Vp 2 ) + ( ) p2 = −V sin θ cos θ − b − a cos 2θ ( 2 a − b2 a2 − b2 ⎛ ( )2 ( )2 ⎞ ) ( p p ) − ( ) p2 ⎠ = −V 2 ⎝sin2 θ cos2 θ + b2 − a2 cos 2θ ( 2 a − b2 a2 − b2 2
2
2
) ) ) ( 2 −V 2 (( 2 2 2 2 2 2 2 4 =( )2 a − b sin θ cos θ + b − a cos 2θ p − p . a2 − b2 It suffices to show that (
a2 − b2
)2
) ( sin2 θ cos2 θ + b2 − a2 cos 2θ p2 − p4
)2
) ( sin2 θ cos2 θ + b2 − a2 cos 2θ p2 − p4
is a constant. Here (
a2 − b2
1.2 Examples on Kinematics
85
⎛
⎛
⎞
⎞
2 2 2 ⎠ = ⎝ ᷅ ᷆ ᷇ ᷄ a4 −2a2 b2 + ᷅ ᷆ ᷇ ᷄ b4 ⎠ sin2 θ cos2 θ + b2 cos 2θ ⎝b ᷅ 2 cos ᷆ ᷇ θ ᷄ +a sin θ 1
2
⎛
2
⎞
2 ⎠ −a2 cos 2θ ⎝b2 cos2 θ + ᷅a2 sin ᷆ ᷇ θ ᷄ 1
⎛
⎞
4 4 4 2 2 2 2 ⎠ −⎝ ᷅b4 cos ᷆ ᷇ θ ᷄ + a ᷅ sin ᷆ ᷇ θ ᷄ +2a b sin θ cos θ 2
1
) ( ( ) ( ) = −2a2 b2 sin2 θ cos2 θ + b2 cos 2θ a2 sin2 θ − a2 cos 2θ b2 cos2 θ ( ) − 2a2 b2 sin2 θ cos2 θ ⎞ ⎛ 2 ⎠ = a2 b2 ⎝−2 sin2 θ cos2 θ + ᷅cos 2θ sin2 θ − ᷆ ᷇ cos 2θ cos θ ᷄
( ) = a2 b2 −4 sin2 θ cos2 θ − cos 2θ cos 2θ ( ) = −a2 b2 sin2 2θ + cos2 2θ = −a2 b2 , which is a constant.
1.2.28 Example If the axes Ox, Oy revolve with constant angular velocity ω, and the component velocities of the point (x, y) parallel to the axes are ) ) ( ( ωy a2 − b2 ωx a2 − b2 and , a2 + b2 a2 + b2 prove that the point describes relatively to the axes an ellipse in the periodic time ) ( π a2 + b2 . ω ab Sol: This problem is a continuation of Sect. 1.2.17. Since / 2a2 dx 2a2 x2 2abω √ 2 = x˙ = 2 ωy = ω · b C − = Ca − x2 , 2 2 2 dt a2 a2 + b2 ᷅ a ᷆ ᷇+ b ᷄ a + b
86
1 3D Kinematics of Rigid Bodies
we have dx a2 + b2 . √ 2abω Ca2 − x2
dt = It follows that
√ x= { Ca
(required periodic time) = 4 x=0
1 a2 + b2 dx √ 2abω Ca2 − x2
)( )x=√Ca  a2 + b2 x −1  sin √ = 2 abω Ca x=0 ) ) ( 2 ( 2 a + b2 π a + b2 π = . =2 abω 2 abω (
1.2.29 Example The position of a point in a plane is determined by the coordinates x, r, where coordinate r is its distance from the origin and x is measured parallel to a fixed direction. If u, v denote the component velocities in the directions of x, r, express u, v in terms of x, r and their derivatives. Show that the components’ accelerations in the direction of x, r are u˙ +
xuv uv and v˙ − 2 . r r
Sol: Observe that Δ
r = rr, So (Fig. 1.31) Δ
·
Δ
Δ
·
v = r˙r + r r = r˙r + r θ(− sin θ, cos θ ). Suppose that Δ
(− sin θ, cos θ ) = Ai + Br. On taking dot product with ˆr, we get
1.2 Examples on Kinematics
87
Fig. 1.31 ̂ ̂
Δ
Δ
0 = (− sin θ, cos θ ) · r = Ai · r + B = A cos θ + B. ᷆ ᷇ ᷄ ᷅ On taking dot product with i, we get Δ
− sin θ = (− sin θ, cos θ ) · i = A + Bi · r = A + B cos θ. ᷆ ᷇ ᷄ ᷅ It follows that A cos θ + B = 0, A + B cos θ = − sin θ. Hence A = − csc θ, B = cot θ. Thus ( ( ) ·( · · ) · · ) v = ˙rr + r θ Ai + Br = r θ Ai + ˙r + r θ B r = −r θ csc θ i + ˙r + r θ cot θ ( ) · · x r r = −r θ √ i + ˙r + r θ √ r. r 2 − x2 r 2 − x2 Δ
Δ
Δ
Δ
Δ
Here ·
·
d x˙ r − x r x −1 x˙ r − x r −1 =√ , θ = cos−1 = / 2 ) ( 2 2 r r dt r r − x2 1 − xr ·
so
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1 3D Kinematics of Rigid Bodies
(· ) · · · r xr − xr r2 xr − xr u = −r θ √ =√ = , √ r r2 − x2 r 2 − x2 r2 − x2 r 2 − x2 (· ) · · · x xr − xr · x −1 x r − x r x · · · v = r +r θ √ = r +r √ = r− 2 √ r r − x2 r 2 − x2 r2 − x2 r 2 − x2 ( ( ) ) ) ( · · · · · · · r rr − xx r r2 − x2 − x x r − x r r r2 − x x r = = 2 = . r2 − x2 r − x2 r2 − x2 ·
1
r
Observe that Δ
v = ui + vr, so ) ( (· ) · · ) · d( dv · · = ui + vr = u i + v r + v r = u i + v r + v θ(− sin θ, cos θ ) acc. = dt ᷆ ᷇ ᷅dt ᷄ Δ
Δ
Δ
Δ
(· ·( )) · = u i + v r + v θ − csc θ i + cot θ r (· ) (· ) · · = u −v θ csc θ i + v +v θ cot θ r. Δ
Δ
Δ
Since u=
(· ) · r xr − xr r2 − x2
=
( ·√ ) r −r θ r2 − x2 r2 − x2
·
−r2 θ =√ , r2 − x2
we have θ˙ =
√ u r2 − x2 . −r2
Now ( ) ( ) ∂ 2Ω acc. = u˙ − vθ˙ csc θ i + v˙ + vθ˙ cot θ ∂u2 ( ) ( ) √ √ 2 2 u r −x u r 2 − x2 = u˙ − v csc θ i + v˙ + v cot θ rˆ −r 2 −r 2 ( ) ( ) √ √ u r 2 − x2 r x u r 2 − x2 = u˙ − v i + v˙ + v rˆ √ √ −r 2 −r 2 r 2 − x2 r 2 − x2 ( ( vux ) vu ) i + v˙ v − 2 rˆ . = u˙ + r r
1.2 Examples on Kinematics
89
1.2.30 Example The position of a point is given by the coordinates x, y, r, where x, y, z, r have their usual significance. Show that, if u, v, w denote the component velocities in the directions of x, y, r, the corresponding accelerations are u˙ +
uw vw xuw yvw , v˙ + and w˙ − 2 − 2 . r r r r
Sol: Observe that Δ
r = rr, so · d · · v = r r + r r = r r + r (sin θ cos φ, sin θ sin φ, cos θ ) dt ( ) · ˙ ˙ cos θ θ˙ sin φ + sin θ cos φ φ, ˙ − sin θ θ˙ = r r + r cos θ θ cos φ − sin θ sin φ φ, Δ
Δ
Δ
Δ
·
· = r r + r θ(cos θ cos φ, cos θ sin φ, − sin θ ) − rφ˙ sin θ (sin φ, − cos φ, 0). Δ
Observe that (cos θ cos φ, cos θ sin φ, − sin θ )    cos θ cos φ cos θ sin φ − sin θ     0 1 0    sin θ cos φ sin θ sin φ cos θ  =   i  1 0 0    0 1 0    sin θ cos φ sin θ sin φ cos θ     1 0 0    cos θ cos φ cos θ sin φ − sin θ     sin θ cos φ sin θ sin φ cos θ  +   j  1 0 0    0 1 0    sin θ cos φ sin θ sin φ cos θ     1 0 0    0 1 0    cos θ cos φ cos θ sin φ − sin θ  +   rˆ  1 0 0    0 1 0    sin θ cos φ sin θ sin φ cos θ 
90
1 3D Kinematics of Rigid Bodies
=
sin φ − sin θ cos φ i+ j+ rˆ = cos φ sec θ i + sin φ sec θ j − tan θ rˆ . cos θ cos θ cos θ
Now since v = r˙ rˆ + r θ˙ (cos θ cos φ, cos θ sin φ, − sin θ ) − r φ˙ sin θ (sin φ, − cos φ, 0), we have ( ) v = r˙ rˆ + r θ˙ cos φ sec θ i + sin φ sec θ j − tan θ rˆ − r φ˙ sin θ (sin φi − cos φj) ( ) ( ) = r θ˙ cos φ sec θ − r φ˙ sin θ sin φ i + r θ˙ sin φ sec θ + r φ˙ sin θ cos φ j ( ) + r˙ − r θ˙ tan θ rˆ , and hence Δ
v = ui + vj + wr, where u = r θ˙ cos φ sec θ − r φ˙ sin θ sin φ, v = r θ˙ sin φ sec θ + r φ˙ sin θ cos φ, w = r˙ − r θ˙ tan θ. It follows that ( ) ( ) v − r˙ rˆ acc. = u˙ i + v˙ j + wˆ ˙ r + wr˙ˆ = u˙ i + v˙ j + wˆ ˙r +w r ( ) ( ) = u˙ i + v˙ j + wˆ ˙ r + w θ˙ cos φ sec θ − φ˙ sin θ sin φ i . ( ) ( ) + θ˙ sin φ sec θ + φ˙ sin θ cos φ j + −θ˙ tan θ rˆ ( ( )) = u˙ + w θ˙ cos φ sec θ − φ˙ sin θ sin φ i ( ( )) ( ) . + v˙ + w θ˙ sin φ sec θ + φ˙ sin θ cos φ j + w˙ − wθ˙ tan θ rˆ ( ( ( )) ( ) u) = u˙ + w i + v˙ + w θ˙ sin φ sec θ + φ˙ sin θ cos φ j + w˙ − wθ˙ tan θ rˆ r ( ( ( ) v) u) = u˙ + w i + v˙ + w j + w˙ − wθ˙ tan θ rˆ . r r It suffices to show that xu yv + 2 = θ˙ tan θ. r2 r ) yv xu yv x( LHS = 2 + 2 = θ˙ cos φ sec θ − φ˙ sin θ sin φ + 2 r r r r
1.2 Examples on Kinematics
=
91
) y( ) x( θ˙ cos φ sec θ − φ˙ sin θ sin φ + θ˙ sin φ sec θ + φ˙ sin θ cos φ r r ⎛ ⎞ = sin θ cos φ ⎝θ˙ cos φ sec θ − φ˙ sin θ sin φ ⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1 2 ⎛ ⎞ + sin θ sin φ ⎝θ˙ sin φ sec θ + φ˙ sin θ cos φ ⎠ = θ˙ tan θ. ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
2
1.2.31 Example The component velocities u, v are measured along the families of orthogonal curves, prove that the corresponding accelerations are u˙ − κ1 uv + κ2 v2 and v˙ − κ2 uv + κ1 u2 , where κ1 , κ2 are the curvatures of the curves meeting at particles considered positive in the direction of the velocities. 2 Sol: Put p ≡ dy , q ≡ ddx2y . We know that dx d dx
( ) dy dx
q κ1 = ( =( )3 , ( )2 ) 23 1 + p2 2 dy 1 + dx d dx
( ) − dx dy
d2y 2
( dx )2
κ2 = ( =( ( )2 ) 23 dx 1 + − dy 1+
dy dx
(
) 23 = ( 1 dy dx
)2
dy d 2 y dx dx2
1+
( )2 ) 23
= pκ1 .
dy dx
Hence u˙ − κ1 uv + κ2 v2 = u˙ − κ1 uv + pκ1 v2 = u˙ − vκ1 (u − pv), v˙ − κ2 uv + κ1 u2 = v˙ − pκ1 uv + κ1 u2 = v˙ + uκ1 (u − pv). Here, the unit vector in the direction of first family of curves is
92
1 3D Kinematics of Rigid Bodies
⎛ ⎜ ⎜/ ⎝
⎞ ⎟ 1 ⎟ ( )2 , / ( )2 ⎠ = dy dy 1 + dx 1 + dx dy dx
( √
1 1 + p2
,√
p 1 + p2
) ) ( ≡ eˆ 1 .
It follows that the unit vector in the direction of second family of curves is (
−p
1
√ ,√ 1 + p2 1 + p2
) ) ( ≡ eˆ 2 .
Observe that ( ) (( ( ) ) ) cos θ + p sin θ −p cos θ + sin θ √ √ r = rˆr = r rˆ · eˆ 1 eˆ 1 + rˆ · eˆ 2 eˆ 2 = r eˆ 1 + eˆ 2 . 1 + p2 1 + p2
Next ) ) ( ( 1 dp d 1 p d d p d dx √ √ √ ,√ , e˙ˆ 1 = eˆ 1 = = dt dt dt dx dp 1 + p2 dp 1 + p2 1 + p2 1 + p2 ⎛
=
√
−2p dx dp ⎜ ⎝ ( )3 , dt dx 2 1 + p2 2
1 + p2 − p √2p 2 1 + p2
1+p2
⎞
⎟ ⎠ = x˙ (
dp dx
1 + p2
) 3 (−p, 1) = x˙ κ1 (−p, 1) 2
) ) ) (( ) ) ) (( ( ( = x˙ κ1 (−p)i + x˙ κ1 j = −˙xκ1 p i · eˆ 1 eˆ 1 + i · eˆ 2 eˆ 2 + x˙ κ1 j · eˆ 1 eˆ 1 + j · eˆ 2 eˆ 2 ⎛
⎞
⎛
⎞
⎜ ⎟ ⎜ ⎟ 1 p −p 1 ⎜ ⎟ ⎜ ⎟ = −˙xκ1 p⎜ √ eˆ 1 + √ eˆ 2 ⎟ + x˙ κ1 ⎜ √ eˆ 1 + √ eˆ 2 ⎟ ⎝ 1 + p2 ⎝ 1 + p2 1 + p2 ⎠ 1 + p2 ⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
1
=
√ x˙ κ1 p2 + x˙ κ1 √ eˆ 2 = x˙ κ1 1 + p2 eˆ 2 , 1 + p2
so √ e˙ˆ 1 = x˙ κ1 1 + p2 eˆ 2 . Next ) ( d d 1 −p √ ,√ = x˙ κ1 (−1, −p) = x˙ κ1 (−1)i − x˙ κ1 pj e˙ˆ 2 = eˆ 2 = dt dt 1 + p2 1 + p2
1.2 Examples on Kinematics
93
(( ) ) ) ) ) ) ( (( ( = −˙xκ1 i · eˆ 1 eˆ 1 + i · eˆ 2 eˆ 2 − x˙ κ1 p j · eˆ 1 eˆ 1 + j · eˆ 2 eˆ 2 ⎛
⎛
⎞
⎞
⎜ ⎜ ⎟ ⎟ p 1 −p 1 ⎜ ⎜ ⎟ ⎟ = −˙xκ1 ⎜ √ eˆ 1 + √ eˆ 2 ⎟ − x˙ κ1 p⎜ √ eˆ 1 + √ eˆ 2 ⎟ ⎝ 1 + p2 ⎝ 1 + p2 1 + p2 ⎠ 1 + p2 ⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
√ = −˙xκ1 1 + p2 eˆ 1 ,
1
so √ e˙ˆ 2 = −˙xκ1 1 + p2 eˆ 1 . Next / ( ( ( ))) 1 + p2 − sin θ θ˙ + (˙xq) sin θ + p cos θ θ˙ ( ) p − (cos θ + p sin θ ) √ · q˙x 1 + p2 d cos θ + p sin θ √ = 2 dt 1 + p2 1+p ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ θ θ˙ +˙xq sin θ + p cos θ θ˙ ⎠ + p2 ⎝− sin θ θ˙ + x˙ q sin θ + p cos θ θ˙ ⎠ ⎝ ᷅− sin ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ⎛
2
⎞
2
3
1
3
⎟ ⎜ − ⎝cos θ + p sin θ ⎠pq˙x ᷅ ᷆ ᷇ ᷄ 1
=
)3 ( 1 + p2 2
) ( ) ( − sin θ θ˙ 1 + p2 + x˙ q sin θ + p cos θ θ˙ 1 + p2 − (cos θ )pq˙x = ( )3 1 + p2 2 − sin θ + p cos θ q √ = θ˙ + ( ) 3 x˙ (sin θ − p cos θ ) 1 + p2 1 + p2 2 =−
sin θ − p cos θ √ θ˙ + κ1 x˙ (sin θ − p cos θ ), 1 + p2
Similarly ( ) ( ) ) ( ) ( sin θ − π2 − p cos θ − π2 d ( π) d cos θ − π2 + p sin θ − π2 √ √ θ− =− dt dt 2 1 + p2 1 + p2 ) ( )) ( ( π π − p cos θ − + κ1 x˙ sin θ − 2 2 or
94
1 3D Kinematics of Rigid Bodies
d sin θ − p cos θ − cos θ − p sin θ √ √ θ˙ + κ1 x˙ (− cos θ − p sin θ ). =− dt 1 + p2 1 + p2 Now since (
) cos θ + p sin θ −p cos θ + sin θ √ √ r=r eˆ 1 + eˆ 2 , 1 + p2 1 + p2 we have (
) cos θ + p sin θ −p cos θ + sin θ √ √ v = r˙ eˆ 1 + eˆ 2 1 + p2 1 + p2 ( ( ) ) d cos θ + p sin θ d −p cos θ + sin θ √ √ +r eˆ 1 + r eˆ 2 dt dt 1 + p2 1 + p2 cos θ + p sin θ ˙ −p cos θ + sin θ ˙ √ √ eˆ 1 + r eˆ 2 1 + p2 1 + p2 ( ) cos θ + p sin θ −p cos θ + sin θ √ √ = r˙ eˆ 1 + eˆ 2 1 + p2 1 + p2 ) ( sin θ − p cos θ +r − √ θ˙ + κ1 x˙ (sin θ − p cos θ ) eˆ 1 1 + p2 ( ) − cos θ − p sin θ √ θ˙ + κ1 x˙ (− cos θ − p sin θ ) eˆ 2 +r − 1 + p2 ) cos θ + p sin θ ( √ x˙ κ1 1 + p2 eˆ 2 +r √ 1 + p2 ) √ −p cos θ + sin θ ( √ −˙xκ1 1 + p2 eˆ 1 +r 1 + p2 +r
= uˆe1 + vˆe2 , where
1.2 Examples on Kinematics
95
) ( cos θ + p sin θ sin θ − p cos θ θ˙ + κ1 x˙ (sin θ − p cos θ ) u ≡ r˙ √ +r − √ 1 + p2 1 + p2 ) √ −p cos θ + sin θ ( √ +r −˙xκ1 1 + p2 , 1 + p2 ( ) −p cos θ + sin θ − cos θ + p sin θ ˙ √ √ v ≡ r˙ θ + κ1 x˙ (− cos θ − p sin θ ) +r − 1 + p2 1 + p2 ) cos θ + p sin θ ( √ +r √ x˙ κ1 1 + p2 . 1 + p2 Thus ⎛ u = r˙
⎞
cos θ + p sin θ sin θ − p cos θ √ θ˙ + κ1 x˙ (sin θ − p cos θ )⎠ + r ⎝− √ 2 ᷅ ᷆ ᷇ ᷄ 1+p 1 + p2 1
− r(−p cos θ + sin θ )˙xκ1 ᷅ ᷆ ᷇ ᷄ 1
cos θ + p sin θ sin θ − p cos θ = r˙ √ − r θ˙ √ , 1 + p2 1 + p2 ⎛ ⎞ −p cos θ + sin θ − cos θ − p sin θ √ √ θ˙ + κ1 x˙ (− cos θ − p sin θ )⎠ v = r˙ + r ⎝− ᷅ ᷆ ᷇ ᷄ 1 + p2 1 + p2 1
−p cos θ + sin θ − cos θ − p sin θ √ √ + r(cos θ + p sin θ )(˙xκ1 ) = r˙ − r θ˙ . 2 ᷅ ᷆ ᷇ ᷄ 1+p 1 + p2 1
Since v = uˆe1 + vˆe2 , we have ) ( ) ( acc. = u˙ eˆ 1 + ue˙ˆ 1 + v˙ eˆ 2 + ve˙ˆ 2 = u˙ eˆ 1 + v˙ eˆ 2 ) ( ) ( √ √ + u x˙ κ1 1 + p2 eˆ 2 + v −˙xκ1 1 + p2 eˆ 1 ( ) ) ( √ √ = u˙ − v˙xκ1 1 + p2 eˆ 1 + v˙ + u˙xκ1 1 + p2 eˆ 2 . It suffices to show that √ x˙ 1 + p2 = u − pv.
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1 3D Kinematics of Rigid Bodies
Here (
cos θ + p sin θ sin θ − p cos θ RHS = u − pv = r˙ √ − r θ˙ √ 2 1+p 1 + p2 ( ) − cos θ − p sin θ −p cos θ + sin θ √ √ − r θ˙ −p r˙ 1 + p2 1 + p2 ⎞⎞ ⎛ ⎛ 1 =√ sin θ ⎠⎠ r˙ ⎝cos θ + p sin θ −p⎝−p cos θ + ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1 + p2 1
⎛ ⎛
⎞
1
⎛
)
⎞⎞
⎠⎠ + r θ˙ ⎝−⎝sin θ − p cos θ ⎠ + p⎝− ᷅cos ᷆ ᷇ θ ᷄ −p sin θ ᷅ ᷆ ᷇ ᷄ 2
=√
1
2
( ( ) ( ( ) )) r˙ 1 + p2 cos θ + r θ˙ − 1 + p2 sin θ
1 + p2 √ ( ) √ d = 1 + p2 r˙ cos θ − r θ˙ sin θ = 1 + p2 (r cos θ ) dt √ √ dx = x˙ 1 + p2 = LHS. = 1 + p2 dt
1.2.32 Example A point moves in a plane with an angular velocity ω, and the plane is turning around the radius vector with an angular velocity ω' ; prove that the accelerations in the plane are r¨ − rω2 and r ω˙ + 2˙r ω, and that the acceleration perpendicular to the plane is rωω' . Sol: On using the formula a = a0 + α × r + ω × (ω × r), we get (Fig. 1.32) ( ) (( ( ) ( ) (( ) ( )) ( ) ) d ω' rˆ ( )2 ) × rˆr + ω' rˆ × ω' rˆ × rˆr a = r¨ − r θ˙ rˆ + 2˙r θ˙ + r θ¨ θˆ + dt ( ) (( ( ) ( ) ( ) ) d ω' rˆ ( )2 ) ˙ ¨ ˆ ˙ × rˆr + ω' rˆ × 0 = r¨ − r θ rˆ + 2˙r θ + r θ θ + dt ) d (ω' rˆ ) ( ) (( ) 2 ˆ × rˆr ˙ θ + = r¨ − rω rˆ + (2˙r ω + r ω) ) ( dt ' ( ) (( ) ) dω 2 ' d rˆ rˆ + ω × rˆr = r¨ − rω rˆ + (2˙r ω + r ω) ˙ + dt dt
1.2 Examples on Kinematics
97
) ) ( d ω' (( ( ) ) d rˆ ( ) rˆ × rˆr + ω' × rˆr r¨ − rω2 rˆ + (2˙r ω + r ω) ˙ θˆ + dt dt ) ) ( (( ( ) ) d rˆ × rˆr = r¨ − rω2 rˆ + (2˙r ω + r ω) ˙ θˆ + 0 + ω' dt ) ( ) ( ) (( ) ˙ θˆ + ω' θ˙ θˆ × rˆr = r¨ − rω2 rˆ + (2˙r ω + r ω) ) (( ( ) ( ) ) = r¨ − rω2 rˆ + (2˙r ω + r ω) ˙ θˆ + ω' ωθˆ × rˆr ) ( ) )( ( ˙ θˆ + rωω' θˆ × rˆ . = r¨ − rω2 rˆ + (2˙r ω + r ω) =
1.2.33 Example From Fig. 1.32, rP/O = Lˆn1 + H nˆ 3 + Rdˆ 1 − hˆn3 + r eˆ r . { } Let us denote the inertial frame nˆ 1 , nˆ 2{ , nˆ 3 by N } . This is orthogonal and rightˆ ˆ handed oriented. Let us denote the frame d1 , d2 , nˆ 3 by D. This is rotating in the frame N with angular velocity ωD/N given by ωD/N = θ˙ nˆ 3 . } orthogonal and righthanded oriented. Let us denote the frame { D is also ˆ eˆ r , eˆ φ , d1 by E. This is rotating in the frame D with angular velocity ωE/D given by
̂
̂
ℎ
Fig. 1.32
98
1 3D Kinematics of Rigid Bodies
ωE/D = φ˙ dˆ 1 . E is also orthogonal and righthanded oriented. Now since angular velocities are vectors, we have ωE/N = ωD/N + ωE/D = θ˙ nˆ 3 + φ˙ dˆ 1 . ᷅ ᷆ ᷇ ᷄ Since rP/O (t) = L(t)ˆn1 + H (t)ˆn3 + Rdˆ 1 − hˆn3 + r eˆ r , we have ) ) Nd ( d( rP/O (t) = L(t)ˆn1 + H (t)ˆn3 + Rdˆ 1 − hˆn3 + r eˆ r , dt dt
N
) N ( d where dt rP/O (t) denotes derivative of vector function rP/O (t) with respect to time t when looking from frame N , etc. Thus, ) ) Nd ( ) Nd ( ) Nd ( d( rP/O (t) = L(t)ˆn1 + H (t)ˆn3 + Rdˆ 1 dt dt dt dt N ( N ( ) ) d d hˆn3 r eˆ r − + dt ᷅dt ᷆ ᷇ ᷄ N
In frame N ≡{nˆ 1 ,ˆn2 ,ˆn3 }, h, nˆ 3 are constants
=
) d( L(t)ˆn1 ᷅dt ᷆ ᷇ ᷄ N
+
) N ( ) Nd ( ) d( d H (t)ˆn3 + r eˆ r Rdˆ 1 − 0 + dt dt dt
N
In frame N ≡{nˆ 1 ,ˆn2 ,ˆn3 }, nˆ 1 is constant
˙ n1 + = Lˆ
) d( H (t)ˆn3 ᷅dt ᷆ ᷇ ᷄ N
+
d ( ˆ ) Nd ( ) r eˆ r Rd1 + dt dt
N
In frame N ≡{nˆ 1 ,ˆn2 ,ˆn3 }, nˆ 3 is const.
˙ n1 + H˙ nˆ 3 + = Lˆ
N
d dt ᷅ ᷆ ᷇ ᷄
( ) Nd ( ) r eˆ r Rdˆ 1 + dt
R is a constant N (
) N ( ) ˙ n1 + H˙ nˆ 3 + R · d dˆ 1 + d r eˆ r = Lˆ dt ᷅dt ᷆ ᷇ ᷄ r is a constant
( ) N ( ) ˙ n1 + H˙ nˆ 3 + R · d dˆ 1 + r d eˆ r = Lˆ dt dt N
1.2 Examples on Kinematics
99
) (D ( ) N ( ) d ˆ d ˆ ˙ ˙ d1 + ω ND × d1 + r eˆ r = Lˆn1 + H nˆ 3 + R dt dt ⎛ ⎞ ⎜ ⎜ ˙ ˙ = Lˆn1 + H nˆ 3 + R⎜ ⎜ ⎝
d (ˆ ) d1 ᷅dt ᷆ ᷇ ᷄ D
⎟ N ( ) ⎟ d ˆ +ω × d1 ⎟ ⎟ + r dt eˆ r ⎠ D N
} { In frame D≡ dˆ1 ,dˆ 2 ,ˆn3 , dˆ1 is const.
) N ( ) ( ˙ n1 + H˙ nˆ 3 + R 0 + ωD/N × dˆ 1 + d r eˆ r = Lˆ dt ) (( ) N ( ) d ˙ n1 + H˙ nˆ 3 + R θ˙ nˆ 3 × dˆ 1 + r eˆ r = Lˆ dt ⎛ ⎞ ⎜ ˙ n1 + H˙ nˆ 3 + Rθ˙ ⎜ = Lˆ ⎜ ⎝
nˆ × dˆ ᷅ 3 ᷆ ᷇ ᷄1
} { Observe that D≡ dˆ1 ,dˆ 2 ,ˆn3 is righthanded oriented
( ) ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r Nd eˆ r = Lˆ dt ⎛ ⎜ ⎜ ˆ ˙ ˙ ˙ = Lˆn1 + H nˆ 3 + Rθ d2 + r ⎜ ⎜ ⎝
⎟ N ( ) d ⎟ eˆ r ⎟+r ⎠ dt
⎞ Ed ( ) eˆ r ᷅dt ᷆ ᷇ ᷄
⎟ ⎟ ⎟ + ωE/N × eˆ r ⎟ ⎠
} { In frame E ≡ eˆ r ,ˆeφ ,dˆ 1 , eˆ r is const.
) ( ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r 0 + ωE/N × eˆ r = Lˆ ) ) (( ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r θ˙ nˆ 3 + φ˙ dˆ 1 × eˆ r = Lˆ ) ( ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r θ˙ nˆ 3 × eˆ r = Lˆ ⎛
⎞
⎜ ⎜ + r φ˙ ⎜ ⎝
⎟ ⎟ ⎟ ⎠
dˆ 1 × eˆ r ᷅ ᷆ ᷇ ᷄
} { Observe that E≡ eˆ r ,ˆeφ ,dˆ 1 is right−handed oriented
100
1 3D Kinematics of Rigid Bodies
⎛
⎞
⎜ ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r θ˙ ⎜ = Lˆ ⎜ ⎝
nˆ × eˆ ᷅ 3 ᷆ ᷇ ᷄r
{ } Observe that D≡ dˆ 1 ,dˆ 2 ,ˆn3 is righthanded oriented
⎟ ⎟ ⎟ + r φ˙ eˆ φ ⎠
) ( ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r θ˙ dˆ 1 × dˆ 2 × eˆ r + r φ˙ eˆ φ = Lˆ ⎞ ⎛⎛ ⎜⎜ ⎜ ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r θ˙ ⎜ = Lˆ ⎜⎜ ⎝⎝
dˆ · eˆ ᷅ 1 ᷆ ᷇ ᷄r
} { E≡ eˆ r ,ˆeφ ,dˆ 1 is orthogonal
⎞
⎟ ) ⎟ ( ⎟ˆ ⎟ ⎟d2 − dˆ 2 · eˆ r dˆ 1 ⎟ + r φ˙ eˆ φ ⎠ ⎠
) ) ( ( ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 + r θ˙ 0dˆ 2 − dˆ 2 · eˆ r dˆ 1 + r φ˙ eˆ φ , = Lˆ so ) ( ) d( ˙ n1 + H˙ nˆ 3 + Rθ˙ dˆ 2 − r θ˙ dˆ 2 · eˆ r dˆ 1 + r φ˙ eˆ φ . rP/O (t) = Lˆ dt { } { } Since d 1 , d 2 , n3 and d 1 , er , eφ are righthanded orthogonal systems, and dˆ 1 N
Δ
Δ
Δ
Δ
Δ
Δ
Δ
Δ
Δ
Δ
is common to both the systems, d 2 , n3 , er , eφ are coplanar vectors:
Δ
Δ
It follows that d 2 · er = cos
(π 2
) + φ (= − sin φ), and hence
) d( ˙ 1 + H˙ n3 + Rθ˙ dˆ 2 + r θ˙ sin φ dˆ 1 + r φe ˙ φ. rP/O (t) = Ln dt
N
Δ
Δ
Δ
Again, it follows that N
d dt
(N
) ) d( rP/O (t) dt
) ) N ( N ( N ( ) ¨ n1 + H¨ nˆ 3 + R d θ˙ dˆ 2 + r d θ˙ sin φ dˆ 1 + r d φ˙ eˆ φ = Lˆ dt dt dt
1.2 Examples on Kinematics
101
) (D ( ) d ˆ ˆ ˆ ¨ ¨ ¨ ˙ D d2 + ω N × d2 = Lˆn1 + H nˆ 3 + Rθ d2 + Rθ dt )) ( (D ( ) ( ) d ˆ ˆ ˆ ¨ ˙ ˙ ˙ d1 + ω ND × d1 + r θ sin φ + θ φ cos φ d1 + θ sin φ dt )) ( (E d( ) eˆ φ + ω NE × eˆ φ + r φ¨ eˆ φ + φ˙ dt (D ( ) ) ¨ n1 + H¨ nˆ 3 + Rθ¨ dˆ 2 + Rθ˙ d dˆ 2 + θ˙ nˆ 3 × dˆ 2 = Lˆ dt )) ( (D ( ) ) ( d ˆ d1 + θ˙ nˆ 3 × dˆ 1 + r θ¨ sin φ + θ˙ φ˙ cos φ dˆ 1 + θ˙ sin φ dt )) ( (E ) d( ) ( eˆ φ + θ˙ nˆ 3 + φ˙ dˆ 1 × eˆ φ + r φ¨ eˆ φ + φ˙ dt ) ( ¨ n1 + H¨ nˆ 3 + Rθ¨ dˆ 2 + Rθ˙ 0 + θ˙ nˆ 3 × dˆ 2 = Lˆ (( )) ( ) + r θ¨ sin φ + θ˙ φ˙ cos φ dˆ 1 + θ˙ sin φ 0 + θ˙ nˆ 3 × dˆ 1 ( ) )) ( ( + r φ¨ eˆ φ + φ˙ 0 + θ˙ nˆ 3 + φ˙ dˆ 1 × eˆ φ ( ) ¨ n1 + H¨ nˆ 3 + Rθ¨ dˆ 2 + R θ˙ 2 nˆ 3 × dˆ 2 = Lˆ ) (( ) ( )2 + r θ¨ sin φ + θ˙ φ˙ cos φ dˆ 1 + θ˙ sin φ nˆ 3 × dˆ 1 ( )) ( + r φ¨ eˆ φ + φ˙ θ˙ nˆ 3 × eˆ φ + φ˙ dˆ 1 × eˆ φ ) (( ) ) ( ) ( ( ) ¨ n1 + H¨ nˆ 3 + Rθ¨ dˆ 2 + R θ˙ 2 −dˆ 1 + r θ¨ sin φ + θ˙ φ˙ cos φ dˆ 1 + θ˙ 2 sin φ dˆ 2 = Lˆ ))) ( ( ( + r φ¨ eˆ φ + φ˙ θ˙ nˆ 3 × eˆ φ + φ˙ −ˆer ( ( ( ) ( ) ) ( ) ) ¨ n1 + H¨ nˆ 3 + r θ¨ sin φ + θ˙ φ˙ cos φ − R θ˙ 2 dˆ 1 + Rθ¨ + r θ˙ 2 sin φ dˆ 2 = Lˆ )) ( ( + r φ¨ eˆ φ + r φ˙ θ˙ cos φ dˆ 1 − φ˙ eˆ r . Thus ( ( ) ) ( ) ¨ n1 + H¨ nˆ 3 + r θ¨ sin φ + θ˙ φ˙ cos φ − R θ˙ 2 + r φ˙ θ˙ cos φ dˆ 1 acc. = Lˆ ( ) ( )2 ( )2 + Rθ¨ + r θ˙ sin φ dˆ 2 − r φ˙ eˆ r + r φ¨ eˆ φ .
102
1 3D Kinematics of Rigid Bodies
1.2.34 Example The two discs in Fig. 1.33 rotate with constant angular velocities Ω and ω about their respective axes. Determine the absolute acceleration of point P at the instant shown. Sol: Actually, this problem was solved in Sect. 1.2.13. Here, we shall present another method. From Fig. 1.33, rP/O = adˆ 2 + hˆn3 + r eˆ r . } { . This is orthogonal and right, nˆ 3 by N } Let us denote the inertial frame nˆ 1 , nˆ 2{ handed oriented. Let us denote the frame dˆ 1 , dˆ 2 , nˆ 3 by D. This is rotating in the frame N with angular velocity ωD/N given by ωD/N = θ˙ nˆ 3 . { D is also } orthogonal and righthanded oriented. Let us denote the frame eˆ r , eˆ φ , dˆ 1 by E. This is rotating in the frame D with angular velocity ωE/D given by ωE/D = φ˙ dˆ 1 . E is also orthogonal and righthanded oriented. Now since angular velocities are vectors, we have ωE/N = ωD/N + ωE/D = θ˙ nˆ 3 + φ˙ dˆ 1 . ᷅ ᷆ ᷇ ᷄ Since Fig. 1.33 ̂ ̂ ( )
̂1
̂2 = ̇
ℎ ̂3 ̂1 Ω= ̇
( ) ̂1
̂2 ̂2
1.2 Examples on Kinematics
103
rP/O (t) = adˆ 2 + hˆn3 + r eˆ r , we have ) ) Nd ( d( rP/O (t) = adˆ 2 + hˆn3 + r eˆ r , dt dt
N
) N ( d where dt rP/O (t) denotes derivative of vector function rP/O (t) with respect to time t when looking from frame N , etc. Thus, ) Nd ( ) d( adˆ 2 + rP/O (t) = dt dt
d( ) hˆn3 dt ᷅ ᷆ ᷇ ᷄
N
N
+
d( ) r eˆ r dt
N
In frame N ≡{nˆ 1 ,ˆn2 ,ˆn3 }, h, nˆ 3 are constants ( ) N ( ) d ˆ d r eˆ r d2 − 0 + =a dt dt ) (D ( ) N ( ) d ˆ d eˆ r =a d2 + ω ND × dˆ 2 + r dt dt ⎞ ⎛ N
⎜ ⎜ ⎜ = a⎜ ⎜ ⎝
( ) dˆ 2 dt ᷅ ᷆ ᷇ ᷄ Dd
} { In frameD≡ dˆ 1 ,dˆ 2 ,ˆn3 ,dˆ 2 is const.
( ) = a 0 + ωD/N × dˆ 2 +
⎛ ⎜ ⎜ = aθ˙ ⎜ ⎝
Nd
dt
⎟ ⎟ Nd ( ) ⎟ eˆ r +ω D × dˆ 2 ⎟ + r N ⎟ dt ⎠
) (( ) Nd ( ) ( ) r eˆ r = a θ˙ nˆ 3 × dˆ 2 + r eˆ r dt
⎞
nˆ × dˆ ᷅ 3 ᷆ ᷇ ᷄2
{ } Observe that D≡ dˆ 1 ,dˆ 2 ,ˆn3 is right−handed oriented
⎟ N ( ) d ⎟ eˆ r ⎟+r ⎠ dt
⎛ ⎜ ) ( N ( ) ⎜ d ˆ ˆ ˙ ˙ eˆ r = −aθ d1 + r ⎜ = aθ −d1 + r ⎜ dt ⎝
⎞ d( ) eˆ r dt ᷅ ᷆ ᷇ ᷄ E
+ωE/N
} { In frame E≡ eˆ r ,ˆeφ ,dˆ 1 , eˆ r is const.
) ) (( ) ( = −aθ˙ dˆ 1 + r 0 + ωE/N × eˆ r = −aθ˙ dˆ 1 + r θ˙ nˆ 3 + φ˙ dˆ 1 × eˆ r ⎛
⎞
⎜ ) ( ⎜ = −aθ˙ dˆ 1 + r θ˙ nˆ 3 × eˆ r + r φ˙ ⎜ ⎝
⎟ ⎟ ⎟ ⎠
dˆ 1 × eˆ r ᷅ ᷆ ᷇ ᷄
} { Observe that E≡ eˆ r ,ˆeφ ,dˆ 1 is righthanded oriented
⎟ ⎟ × eˆ r ⎟ ⎟ ⎠
104
1 3D Kinematics of Rigid Bodies
⎞
⎛ ⎜ ⎜ = −aθ˙ dˆ 1 + r θ˙ ⎜ ⎝
nˆ 3 × eˆ r ᷅ ᷆ ᷇ ᷄
{ } Observe that D≡ dˆ 1 ,dˆ 2 ,ˆn3 is right−handed oriented
⎟ ⎟ ⎟ + r φ˙ eˆ φ ⎠
( ) = −aθ˙ dˆ 1 + r θ˙ dˆ 1 × dˆ2 × eˆ r + r φ˙ eˆ φ ⎞ ⎛⎛ ⎜⎜ ⎜⎜ = −aθ˙ dˆ 1 + r θ˙ ⎜⎜ ⎝⎝
dˆ 1 · eˆ r ᷅ ᷆ ᷇ ᷄
} { E≡ eˆ r ,ˆeφ ,dˆ 1 is orthogonal
⎞
⎟ ( ) ⎟ ⎟ˆ ⎟ ⎟d2 − dˆ 2 · eˆ r dˆ 1 ⎟ + r φ˙ eˆ φ ⎠ ⎠
( ( ) ) = −aθ˙ dˆ 1 + r θ˙ 0dˆ 2 − dˆ 2 · eˆ r dˆ 1 + r φ˙ eˆ φ , so ( ) ) d( rP/O (t) = −aθ˙ dˆ 1 − r θ˙ dˆ 2 · eˆ r dˆ 1 + r φ˙ eˆ φ . dt { } { } Since dˆ1 , dˆ 2 , nˆ 3 and dˆ 1 , eˆ r , eˆ φ are righthanded orthogonal systems, and dˆ 1 is common to both the systems, dˆ2 , nˆ 3 , eˆ r , eˆ φ are coplanar vectors: N
̂
3
̂ ̂
1
̂
2
It follows that dˆ 2 · eˆ r = cos φ, and hence ) d( rP/O (t) = −(a + r cos φ)θ˙ dˆ1 + r φ˙ eˆ φ . dt
N
Again, it follows that (N
) ) N ( N ( ) ) d( d d rP/O (t) = − φ˙ eˆ φ (a + r cos φ)θ˙ dˆ1 + r dt dt dt (D ( ) ) (( ) ) d ˆ ˆ ˆ ˙ ˙ ˙ ¨ = − −r φ sin φ θ + (a + r cos φ)θ d1 − (a + r cos φ)θ d1 + ω ND × d1 dt N
d dt
1.2 Examples on Kinematics
105
(
+ r φ¨ eˆ φ + φ˙
(E
d( ) eˆ φ + ω NE × eˆ φ dt
))
( ) ) (( ) = − −r φ˙ sin φ θ˙ + (a + r cos φ) θ¨ dˆ 1 − (a + r cos φ)θ˙ 0 + ω ND × dˆ 1 )) (E ( ) d( ) ( ˆ ˙ ˙ ˙ ¨ eˆ φ + θ nˆ 3 + φ d1 × eˆ φ + r φ eˆ φ + φ dt ) ( (( ) ) = − −r φ˙ sin φ θ˙ + (a + r cos φ)θ¨ dˆ 1 − (a + r cos φ)θ˙ θ˙ nˆ 3 × dˆ 1 ( ) )) ( ( + r φ¨ eˆ φ + φ˙ 0 + θ˙ nˆ 3 + φ˙ dˆ 1 × eˆ φ (( ) ( )2 ) = − −r φ˙ sin φ θ˙ + (a + r cos φ)θ¨ dˆ 1 − (a + r cos φ) θ˙ dˆ 2 )) ( ( + r φ¨ eˆ φ + φ˙ θ˙ nˆ 3 × eˆ φ + φ˙ dˆ 1 × eˆ φ ) (( ) ( )2 = − −r φ˙ sin φ θ˙ + (a + r cos φ)θ¨ dˆ 1 − (a + r cos φ) θ˙ dˆ 2 ))) ( ( ( + r φ¨ eˆ φ + φ˙ θ˙ nˆ 3 × eˆ φ + φ˙ −ˆer ) (( ) ( )2 = − −r φ˙ sin φ θ˙ + (a + r cos φ)θ¨ dˆ 1 − (a + r cos φ) θ˙ dˆ 2 )) ( ( + r φ¨ eˆ φ + r φ˙ θ˙ sin φ dˆ 1 − φ˙ eˆ r
( ) ( )2 ( ( ) ( )2 ) = 2 r φ˙ sin φ θ˙ − (a + r cos φ)θ¨ dˆ 1 − (a + r cos φ) θ˙ dˆ 2 + r φ¨ eˆ φ − r φ˙ eˆ r ) ( = (2(rω sin φ)Ω − (a + r cos φ)0)dˆ 1 − (a + r cos φ)Ω2 dˆ 2 + r0ˆeφ − rω2 eˆ r = 2rωΩ sin φ dˆ 1 − (a + r cos φ)Ω2 dˆ 2 − rω2 eˆ r (( ) ) ( ( ) ) = 2rωΩ sin φ dˆ 1 − (a + r cos φ)Ω2 dˆ 2 − rω2 eˆ r · dˆ 1 dˆ 1 + eˆ r · dˆ 2 dˆ 2 + eˆ r · nˆ 3 nˆ 3 ( ) = 2rωΩ sin φ dˆ 1 − (a + r cos φ)Ω2 dˆ 2 − rω2 0dˆ 1 + cos φ dˆ 2 + sin φ nˆ 3 ( ) = 2rωΩ sin φ dˆ 1 − (a + r cos φ)Ω2 + rω2 cos φ dˆ 2 − rω2 sin φ nˆ 3 .
1.2.35 Example A point moves on a straight line OP which is made to describe uniformly a right circular cone about axis OA, while OA sweeps out uniformly a right circular cylinder; find an expression for the acceleration of the point P in the direction OP Sol: From Fig. 1.34, rP/O' = Rdˆ 1 + hˆn3 + r eˆ r .
106
1 3D Kinematics of Rigid Bodies
′
= ̇ ̂
= ̇ ̂ ( ) ̂
̂
3
̂
( )
2
̂ ̂
1
1
̂
3
′
̂
̂
ℎ 2
1
Fig. 1.34
} { . This is orthogonal and rightLet us denote the inertial frame nˆ 1 , nˆ 2{ , nˆ 3 by N } ˆ ˆ handed oriented. Let us denote the frame d1 , d2 , nˆ 3 by D. This is rotating in the frame N with angular velocity ωD/N given by ωD/N = θ˙ nˆ 3 . } orthogonal and righthanded oriented. Let us denote the frame { D is also eˆ r , eˆ θ , eˆ φ by E. This is rotating in the frame D with angular velocity ωE/D given by ωE/D = φ˙ nˆ 3 . E is also orthogonal and righthanded oriented. Now since angular velocities are vectors, we have ( ) ωE/N = ωD/N + ωE/D = θ˙ nˆ 3 + φ˙ nˆ 3 = θ˙ + φ˙ nˆ 3 . ᷅ ᷆ ᷇ ᷄ Since rP/O' (t) = Rdˆ 1 (t) + hˆn3 + r(t)ˆer (t), we have ) ) Nd ( d( rP/O' (t) = Rdˆ 1 (t) + hˆn3 + r(t)ˆer , dt dt
N
1.2 Examples on Kinematics
107
) N ( d where dt rP/O' (t) denotes derivative of vector function rP/O' (t) with respect to time t when looking from frame N , etc. Thus, ) ) Nd ( d( rP/O' (t) = Rdˆ 1 + dt dt
d( ) hˆn3 ᷅dt ᷆ ᷇ ᷄
N
⎛ =R
Nd
dt
⎜ ⎜ ) ⎜ r eˆ r = R⎜ ⎜ dt ⎝
( ) dˆ 1 + 0 +
Nd
N
+
In frame N ≡{nˆ 1 ,ˆn2 ,ˆn3 }, h, nˆ 3 are constants
( ) dˆ 1 ᷅dt ᷆ ᷇ ᷄
(
Dd
{ } In frame D≡ dˆ 1 ,dˆ 2 ,ˆn3 , dˆ 1 is const.
d( ) r eˆ r dt
N
⎞
⎟ ⎟ Nd ( ) ⎟ r eˆ r +ω D × dˆ 1 ⎟ + N ⎟ dt ⎠
) Nd ( ) ( r eˆ r = R 0 + ωD/N × dˆ 1 + dt ) Nd ( ) ( Nd ( ) r eˆ r = Rθ˙ dˆ 2 + r eˆ r = Rθ˙ dˆ 2 + = R θ˙ nˆ 3 × dˆ 1 + dt dt ⎞
⎛
⎜ ⎜ = Rθ˙ dˆ 2 + ⎜ ⎝
(
Ed (
dt
)
r eˆ r + ω E N
( ) × r eˆ r
( ( ) ( )⎟ ( )) ⎟ r eˆ r +ω E × r eˆ r ⎟ = Rθ˙ dˆ 2 + r˙ eˆ r + ω E × r eˆ r N N ⎠ ᷅dt ᷆ ᷇ ᷄ In frame E≡{eˆ r ,ˆeθ ,ˆeφ }, eˆ r is const. Ed
) ( ( ( )) = Rθ˙ dˆ 2 + r˙ eˆ r + θ˙ + φ˙ nˆ 3 × r eˆ r )) ) ( ( )( ( )( = Rθ˙ dˆ 2 + r˙ eˆ r + r θ˙ + φ˙ nˆ 3 × eˆ r = Rθ˙ dˆ 2 + r˙ eˆ r + r θ˙ + φ˙ sin α eˆ φ . It follows that ) N ( N (N ( )) )( ( ) d d d ' rP/O (t) = Rθ˙ dˆ 2 + r˙ eˆ r + r θ˙ + φ˙ sin α eˆ φ dt dt dt =R ⎛ ⎜ ⎜ ⎜ = R⎜ ⎜ ⎝
N ( ( ) ) d ( ˆ ) Nd ( ) d r˙ eˆ r + sin α r θ˙ + φ˙ eˆ φ θ˙ d2 + dt dt dt ⎞
N
) θ˙ dˆ 2 ᷅dt ᷆ ᷇ ᷄ Dd
(
{ } In frame D≡ dˆ 1 ,dˆ 2 ,ˆn3 , dˆ 2 is const.
+ω D N
⎟ )⎟ N d ( ) Nd ( ( ) ) ⎟ ˆ ˙ r˙ eˆ r + sin α r θ˙ + φ˙ eˆ φ × θ d2 ⎟ + ⎟ dt dt ⎠ (
( ( )) N d ( ) N ( ( ) ) d r˙ eˆ r + sin α r θ˙ + φ˙ eˆ φ = R θ¨ dˆ 2 + θ˙ nˆ 3 × θ˙ dˆ 2 + dt dt )) N d ( ) ( N ( ( ) ) ( )2 ( d r˙ eˆ r + sin α r θ˙ + φ˙ eˆ φ = R θ¨ dˆ 2 + θ˙ −dˆ 1 + dt dt
)
108
1 3D Kinematics of Rigid Bodies ⎛
( ( )2 ) ⎜ ⎜ = R θ¨ dˆ 2 − θ˙ dˆ 1 + ⎜ ⎝
⎞
Nd ( ( ( ) ) ) ( )⎟ ⎟ r˙ eˆ r r θ˙ + φ˙ eˆ φ +ω E × r˙ eˆ r ⎟ + sin α N ⎠ dt dt ᷅ ᷆ ᷇ ᷄ In frame E≡{eˆ r ,ˆeθ ,ˆeφ }, eˆ r is const. Ed
( N ( ( ) ) ) ( )2 ) ( ( ( )) d r θ˙ + φ˙ eˆ φ = R θ¨ dˆ 2 − θ˙ dˆ 1 + r¨ eˆ r + θ˙ + φ˙ nˆ 3 × r˙ eˆ r + sin α dt ( N ( ( ) ) ) ( )2 ) ( ( ) d r θ˙ + φ˙ eˆ φ = R θ¨ dˆ 2 − θ˙ dˆ 1 + r¨ eˆ r + θ˙ + φ˙ r˙ sin α eˆ φ + sin α dt ) ( ( )2 ( ) = R θ¨ dˆ 2 − θ˙ dˆ 1 + r¨ eˆ r + θ˙ + φ˙ r˙ sin α eˆ φ ) (E ( ( ) ) ) ) d( ( ˙ ˙ ˙ ˙ r θ + φ eˆ φ + ω NE × r θ + φ eˆ φ + sin α dt ( ( )2 ) ( ) = R θ¨ dˆ 2 − θ˙ dˆ 1 + r¨ eˆ r + θ˙ + φ˙ r˙ sin α eˆ φ (( ( ( ( ( ) ( ) ) )) )) + sin α r˙ θ˙ + φ˙ + r θ¨ + φ¨ eˆ φ + θ˙ + φ˙ nˆ 3 × r θ˙ + φ˙ eˆ φ ( ( )2 ) ( ) = R θ¨ dˆ 2 − θ˙ dˆ 1 + r¨ eˆ r + θ˙ + φ˙ r˙ sin α eˆ φ ) (( ( ( )2 (( ( ( ) ( )) ) ) ) ) + sin α r˙ θ˙ + φ˙ + r θ¨ + φ¨ eˆ φ + r θ˙ + φ˙ nˆ 3 · eˆ r eˆ r + nˆ 3 · eˆ θ eˆ θ + nˆ 3 · eˆ φ eˆ φ × eˆ φ
( ( )2 ) ( ) = R θ¨ dˆ 2 − θ˙ dˆ 1 + r¨ eˆ r + θ˙ + φ˙ r˙ sin α eˆ φ
) (( ( )) ) ( )2 ( ( ) + sin α r˙ θ˙ + φ˙ + r θ¨ + φ¨ eˆ φ + r θ˙ + φ˙ cos α eˆ r + (− sin α)ˆeθ + 0ˆeφ × eˆ φ
( ( )2 ) ( ) = R θ¨ dˆ 2 − θ˙ dˆ 1 + r¨ eˆ r + θ˙ + φ˙ r˙ sin α eˆ φ (( ( )) )) ) ( )2 ( ( + sin α r˙ θ˙ + φ˙ + r θ¨ + φ¨ eˆ φ + r θ˙ + φ˙ − cos α eˆ θ + (− sin α)ˆer ( )2 ( )2 = −R θ˙ dˆ 1 + Rθ¨ dˆ 2 + r¨ eˆ r − r θ˙ + φ˙ sin2 α eˆ r )) )2 ) ( ( ( ( −r θ˙ + φ˙ sin α cos α eˆ θ + sin α 2˙r θ˙ + φ˙ + r θ¨ + φ¨ eˆ φ . Hence required radial acceleration ( =
) )2 )2 ( ( ( )2 −R θ˙ dˆ 1 + Rθ¨ dˆ 2 + r¨ eˆ r − r θ˙ + φ˙ sin2 α eˆ r − r θ˙ + φ˙ sin α cos α eˆ θ · eˆ r ( ( ) ( )) + sin α 2˙r θ˙ + φ˙ + r θ¨ + φ¨ eˆ φ
( )2 ( )2 = −R θ˙ dˆ 1 · eˆ r + Rθ¨ dˆ 2 · eˆ r + r¨ − r θ˙ + φ˙ sin2 α ( )2 ( )2 = −R θ˙ sin α cos φ + Rθ¨ sin α sin φ + r¨ − r θ˙ + φ˙ sin2 α
1.2 Examples on Kinematics
109
( )2 = r¨ − r ω + φ˙ sin2 α − Rω2 sin α cos φ + R0 sin α sin φ ( )2 = r¨ − r ω + ω' sin2 α − Rω2 sin α cos φ.
1.2.36 Example The velocity of a point moving in a plane is the resultant of two velocities v and v' along radii r, r ' measured from two fixed points at a distance a apart. Prove that the corresponding accelerations are (Fig. 1.35)
v˙ +
( ) ( )2 vv' r 2 − r ' + a2 2r 2 r '
'
and v˙ +
(( ) ) 2 vv' r ' − r 2 + a2 2(r ' )2 r
Sol: Here, ( ( )2 ( a )2 a )2 + y2 , r ' = x + + y2 . r2 = x − 2 2 Also (˙x, y˙ ) = v
1( a ) a ) 1( x − , y + v' ' x + , y , r 2 r 2
so ( ) v( a ) v' ( v v' a) x˙ = x− + ' x+ , y˙ = y + ' . r 2 r 2 r r We have to show that Fig. 1.35 ( , ) ′
(− , 0) 2 ( , 0) 2
.
110
1 3D Kinematics of Rigid Bodies
( )⎞ ( )2 ) ( vv' r 2 − r ' + a2 ⎠1 x − a, y (¨x, y¨ ) = ⎝v˙ + 2r 2 r ' r 2 (( ) ) 2 vv' r ' − r 2 + a2 1 ( a ) ,y , x + + v˙ ' + r' 2 2(r ' )2 r ⎛
that is,
( 1. x¨ = v˙ + ( 2. y¨ = v˙ +
( )) 2 vv' r 2 −(r ' ) +a2
( )) ( 2 ( ) ( ) vv' (r ' ) −r 2 +a2 ' a 1 x − 2 + v˙ + x + a2 . 2r 2 r ' r' 2(r ' )2 r ( )) ( )) ( 2 2 vv' r 2 −(r ' ) +a2 vv' (r ' ) −r 2 +a2 ' 1 1 y + v ˙ + y. 2r 2 r ' r r' 2(r ' )2 r 1 r
For 1: Since x˙ =
a ) v' ( a) v( x− + ' x+ , r 2 r 2
we have ( ( ) ( ) ) ( ( ) ( ( 1 1 a) a) a) a) d 1( ' 1 'd x¨ = v˙ x+ x+ x− x− +v + v˙ +v , r 2 dt r 2 r' 2 dt r ' 2
so it suffices to show that )⎞ ⎛ ( ( ' )2 ' 2 2 ) ) ( ( ( ( ) ) ( ) vv r − r + a a a d 1 d 1 ⎝ ⎠1 x − a + v' = x− x + v dt r 2 dt r ' 2 2r 2 r ' r 2 )⎞ ⎛ (( )2 ) ( vv' r ' − r 2 + a2 ⎠1 x+ a . +⎝ r' 2 2(r ' )2 r Observe that ) ( )( −1 ( −1 (( a) a) a) d 1( a) 1 = 2 r˙ · x − x− + · x˙ = 3 x − x˙ + y˙y x − dt r 2 r 2 r r 2 2 ( ( ) ) ) '( a v a 1 v . x− + ' x+ + r r 2 r 2 Similarly d dt
( ( ( ) ) )( 1 −1 (( a ) 1 v' ( a) a) a) a) v( = x ˙ + y˙ y x + + + x − . x + x + x + r' 2 2 2 r' r' 2 r 2 (r ' )3
Hence
1.2 Examples on Kinematics
111
) ( )( a) v v( a ) v' ( a) a) −v (( x ˙ + y˙ y x − + x − + x − x + r3 2 2 r r 2 r' 2 ( ) )( a ) v' v' ( a) a) a) v( −v' (( x˙ + y˙y x + + ' ' x+ + x− + ' 3 x+ 2 2 r r 2 r 2 (r ) ( ) ( ( ( )))( v v' a ) v' ( a) v( a) a) −v ( +y y + ' x− + ' x+ x− x− = 3 r 2 r 2 r 2 r r 2 ) ( a ) v' ( a) v v( x− + ' x+ + r r 2 r 2 )))( ( ( ) ( ( ) ' (( a v a) v v' a ) v' ( a) −v x+ +y y x− + ' x+ + ' 3 x+ + ' 2 r 2 r 2 r r 2 (r ) ( ) v' v' ( a) a) v( + ' ' x+ + x− r r 2 r 2 ( ( ) ( ))( v' a) −v v 2 a2 a2 x− x − ax + + y2 + ' x2 − + y2 = 3 r r 4 r 4 2 ) ( a ) v' ( a) v v( x− + ' x+ + r r 2 r 2 ( ( ) ( ))( 2 v' a) −v v 2 a a2 x− + y2 + ' x2 − + y2 x − ax + = 3 r r 4 r 4 2 ( ) a ) v' ( a) v v( x− + ' x+ + r r 2 r 2 ( ) ( ))( ( v' a) −v' v 2 a2 a2 x+ x − + y2 + ' x2 + ax + + y2 + ' 3 4 r 4 2 (r ) r ) ) ( ) '( '( v v v a a + ' ' x+ + x− r r 2 r 2 ⎞ ⎛ ( ( )) ( a) v⎜ v( a ) v' ( a )⎟ −v v ( 2 ) v' 2 a2 ⎟ x− r + ' x − + y2 x− = 3 + ⎜ + ' x+ ⎝ r r r 4 2 r ᷅r ᷆ ᷇ 2 ᷄ r 2 ⎠ ᷅ ᷆ ᷇ ᷄ LHS =
⎛
1
⎞ ( ) ) ( a) v( a )⎟ v ' ( )2 ( a ) v' ⎜ −v ' v 2 a2 v' ( ⎟ x+ + '⎜ + x− x+ x − + y2 + ' r ' + ⎝ ⎠ 3 r ' ' 4 r 2 r r 2 (r ) ᷅r ᷆ ᷇ 2 ᷄ 1
1
⎞
⎛
⎛ ⎞ ) '( 2 −v a)1⎜ v a vv' ⎟ v2 2 2 ⎠ ⎟ ⎜ ⎝ vr + x + y = x− − + + ' ⎠ 2 r ⎝ r 2 ᷅ ᷆ ᷇ ᷄ r ' 4 r r ᷅ ᷆ ᷇ ᷄ (
1
1
112
1 3D Kinematics of Rigid Bodies
⎛
⎞ ⎞ ( ) ) ( 2 ( ' ' 2 v' ⎟ a) 1 ⎜ ⎜ vv + −v ⎝ v x2 − a + y2 + v' r ' ⎠ + ⎟ + x+ 2 ⎝ ᷅ ᷆ ᷇ ᷄ ' 2 r r 4 r ' ᷄⎠ (r ' ) r ᷅ ᷆ ᷇ ⎛
2
2
( ) ) ( ( vv' a ) 1 −v v' 2 a2 2 x + + y = x− − 2 r r2 r' 4 r' ( ' )) ( ) ( ' 2 −v v 2 a a 1 vv 2 + ' 2 +y x − + x+ 2 r' r 4 (r ) r ( )) ( a2 a2 ' 2 2 2 2 ) ( −x + − y + x − ax + + y vv 4 4 a 1 = x− 2r' 2 r r (( ) ) a ) 1 vv + x+ 2 r' (
'
x2 + ax +
(
a2 4
+ y2 − x2 +
a2 4
− y2
(r ' )2 r
( ) a2 ' a2 ' ) ) ( ( ax + − ax vv vv 2 2 a 1 a 1 = x− + x+ 2 2 ' ' ' 2 r r r 2 r (r ) r ( ) )) ( ( ( a ) a2 − 2ax vv' ( a ) 2ax + a2 x− = + x+ 2rr ' 2 r2 2 (r ' )2 ( ( )2 (( ) ) ) ( )2 r ' − r 2 + a2 ( r 2 − r ' + a2 ( a) a) vv' x− x+ + = 2rr ' r2 2 2 (r ' )2 )
( )⎞ )⎞ ⎛ (( )2 ( )2 ( ) ) ( vv' r ' − r 2 + a2 vv' r 2 − r ' + a2 1 a ⎠ x− ⎠ 1 x + a = RHS. +⎝ =⎝ ( )2 ' 2 ' r 2 r 2 2r r 2 r' r ⎛
1.2.37 Example Prove that the meridian acceleration of a point moving on the surface of the anchor ring x = (c + a sin θ ) cos φ, y = (c + a sin θ ) sin φ, z = a cos θ ( )2 is aθ¨ − (c + a sin θ ) cos θ φ˙ , and determine the normal and transverse accelerations. Sol: Here ( ) x˙ = aθ˙ cos θ cos φ − (c + a sin θ )φ˙ sin φ,
1.2 Examples on Kinematics
113
(( ) ( ) ( )2 ) x¨ = a θ¨ cos θ cos φ − θ˙ sin θ cos φ − θ˙ cos θ φ˙ sin φ (( ) ( )2 ) − aθ˙ cos θ φ˙ sin φ + (c + a sin θ )φ¨ sin φ + (c + a sin θ ) φ˙ cos φ ( ) y˙ = aθ˙ cos θ sin φ + (c + a sin θ )φ˙ cos φ, ) (( ) ( )2 ) ( y¨ = a θ¨ cos θ sin φ − θ˙ sin θ sin φ + θ˙ cos θ φ˙ cos φ (( ) ) ( )2 + aθ˙ cos θ φ˙ cos φ + (c + a sin θ )φ¨ cos φ − (c + a sin θ ) φ˙ sin φ , ( ) z˙ = −a θ˙ sin θ , ( ) ( )2 z¨ = −a θ¨ sin θ + θ˙ cos θ . A vector in the direction of meridian to the anchor ring at point (x, y, z) is ) (π ) (π )) + θ cos φ, a sin + θ sin φ, a cos +θ 2 2 2 (= a(cos θ cos φ, cos θ sin φ, − sin θ )),
(
a sin
(π
so the unit vector in the direction of meridian to the anchor ring at point (x, y, z) is (cos θ cos φ, cos θ sin φ, − sin θ ). Hence, the meridian acceleration is equal to (¨x, y¨ , z¨ ) · (cos θ cos φ, cos θ sin φ, − sin θ ) ( (( )) ) ( )2 ( ) = cos θ cos φ a θ¨ cos θ cos φ − θ˙ sin θ cos φ − θ˙ cos θ φ˙ sin φ (( ) ( )2 ) − cos θ cos φ aθ˙ cos θ φ˙ sin φ + (c + a sin θ )φ¨ sin φ + (c + a sin θ ) φ˙ cos φ ( (( )) ) ( )2 ( ) + cos θ sin φ a θ¨ cos θ sin φ − θ˙ sin θ sin φ + θ˙ cos θ φ˙ cos φ (( ) ( )2 ) + cos θ sin φ aθ˙ cos θ φ˙ cos φ + (c + a sin θ )φ¨ cos φ − (c + a sin θ ) φ˙ sin φ ( ( )) ( )2 − sin θ −a θ¨ sin θ + θ˙ cos θ .
114
1 3D Kinematics of Rigid Bodies
Here, ( ) coefficient of θ¨ = a cos2 θ cos2 φ + a cos2 θ sin2 φ + a sin2 θ = a, ( ( )2 ) coefficient of θ˙ = −a sin θ cos θ cos2 φ −a sin θ cos θ sin2 φ + a sin θ cos θ = 0, ( ) coefficient of θ˙ φ˙ = −a cos2 θ sin φ cos φ − a cos2 θ sin φ cos φ + a cos2 θ sin φ cos φ + a cos2 θ sin φ cos φ = 0, ( ) coefficient of φ¨ = −(c + a sin θ ) sin φ cos θ cos φ + (c + a sin θ ) cos θ sin φ cos φ = 0, ( ( )2 ) coefficient of φ˙ = −(c + a sin θ ) cos θ cos2 φ −(c + a sin θ ) cos θ sin2 φ = −(c + a sin θ ) cos θ
.
Hence, the meridian acceleration is equal to ( )2 aθ¨ − (c + a sin θ ) cos θ φ˙ . A vector in the direction of normal to the anchor ring at point (x, y, z) is (a sin θ cos φ, a sin θ sin φ, a cos θ )(= a(sin θ cos φ, sin θ sin φ, cos θ )), so the unit vector in the direction of normal to the anchor ring at point (x, y, z) is (sin θ cos φ, sin θ sin φ, cos θ ). Hence, the normal acceleration is equal to (¨x, y¨ , z¨ ) · (sin θ cos φ, sin θ sin φ, cos θ ) ( (( )) ( ) ( )2 ) = sin θ cos φ a θ¨ cos θ cos φ − θ˙ sin θ cos φ − θ˙ cos θ φ˙ sin φ (( ) ( )2 ) − sin θ cos φ aθ˙ cos θ φ˙ sin φ + (c + a sin θ )φ¨ sin φ + (c + a sin θ ) φ˙ cos φ ( (( )) ) ( )2 ( ) + sin θ sin φ a θ¨ cos θ sin φ − θ˙ sin θ sin φ + θ˙ cos θ φ˙ cos φ (( ) ( )2 ) + sin θ sin φ aθ˙ cos θ φ˙ cos φ + (c + a sin θ )φ¨ cos φ − (c + a sin θ ) φ˙ sin φ
1.2 Examples on Kinematics
115
( ( )) ( )2 + cos θ −a θ¨ sin θ + θ˙ cos θ . Here, (
) coefficient of θ¨ = a sin θ cos θ cos2 φ
+ a sin θ cos θ sin2 φ = a sin θ cos θ, ( ( )2 ) coefficient of θ˙ = −a sin2 θ cos2 φ −a sin2 θ sin2 φ − a cos2 θ = −a, ( ) coefficient of θ˙ φ˙ = −a sin θ cos θ sin φ cos φ − a sin θ cos θ sin φ cos φ + a sin θ cos θ sin φ cos φ + a sin θ cos θ sin φ cos φ = 0, ( ( )2 ) coefficient of φ˙ = − sin θ cos2 φ(c + a sin θ ) − sin θ sin2 φ(c + a sin θ ) = − sin θ (c + a sin θ ), ( ) coefficient of φ¨ = − sin θ cos φ(c + a sin θ ) sin φ + sin θ sin φ cos φ(c + a sin θ ) = 0, so the normal acceleration is equal to ( )2 ( )2 a sin θ cos θ θ¨ − a θ˙ − sin θ (c + a sin θ ) φ˙ . A vector in the transverse direction to the anchor ring at point (x, y, z) is (
a sin θ cos
) (π ) ) + φ , a sin θ sin + φ , a cos θ (= a(− sin θ sin φ, sin θ cos φ, cos θ )), 2 2
(π
so the unit vector in the transverse direction to the anchor ring at point (x, y, z) is (− sin θ sin φ, sin θ cos φ, cos θ ). Hence, the transverse acceleration is equal to
116
1 3D Kinematics of Rigid Bodies
(¨x, y¨ , z¨ ) · (− sin θ sin φ, sin θ cos φ, cos θ ) ( (( )) ) ( )2 ) ( = − sin θ sin φ a θ¨ cos θ cos φ − θ˙ sin θ cos φ − θ˙ cos θ φ˙ sin φ (( ) ( )2 ) + sin θ sin φ aθ˙ cos θ φ˙ sin φ + (c + a sin θ )φ¨ sin φ + (c + a sin θ ) φ˙ cos φ ( (( )) ) ( )2 ( ) + sin θ cos φ a θ¨ cos θ sin φ − θ˙ sin θ sin φ + θ˙ cos θ φ˙ cos φ (( ) ( )2 ) + sin θ cos φ aθ˙ cos θ φ˙ cos φ + (c + a sin θ )φ¨ cos φ − (c + a sin θ ) φ˙ sin φ )) ( ( ( )2 + cos θ −a v sin θ + θ˙ cos θ . Here, ( ) coefficient of θ¨ = −a sin θ cos θ sin φ cos φ + a sin θ cos θ sin φ cos φ = 0, ( ( )2 ) coefficient of θ˙ = a sin2 θ sin φ cos φ − a sin2 θ sin φ cos φ = 0, ( ) coefficient of θ˙ φ˙ = a sin θ cos θ sin2 φ + a sin θ cos θ sin2 φ + a sin θ cos θ cos2 φ + a sin θ cos θ cos2 φ = 2a sin θ cos θ ( ( )2 ) coefficient of φ˙ = sin θ sin φ cos φ(c + a sin θ ) − sin θ sin φ cos φ(c + a sin θ ) = 0, ( ) coefficient of φ¨ = sin θ sin2 φ(c + a sin θ ) + sin θ cos2 φ(c + a sin θ ) = sin θ (c + a sin θ ), so the transverse acceleration is equal to ¨ 2a sin θ cos θ θ˙ φ˙ + sin θ (c + a sin θ )φ.
1.2.38 Example The position of a moving point is determined by the radii ξ1 , η1 , ζ1 of the three spheres which pass through it and touches three fixed rectangular coordinate planes at the origin. Find the component velocities u, v, w of the point in the directions of the outward normals to the spheres, and show that the component accelerations in the same directions are u˙ + v(ηu − ξ v) − w(ξ w − ζ u), and the two similar expressions.
1.2 Examples on Kinematics
117
Sol: Let (x, y, z) be the Cartesian coordinates of a moving point P at time t. Here, the sphere ( X−
1 ξ
)2 + (Y − 0)2 + (Z − 0)2 =
( )2 1 ξ
passes through the point (x, y, z), and hence x2 + y2 + z 2 −
2x = 0. ξ
It follows that x=
) ξ( 2 x + y2 + z 2 . 2
Similarly y=
) ) ζ( η( 2 x + y2 + z 2 , z = x2 + y2 + z 2 . 2 2
It follows that x2 + y2 + z 2 =
( ( ))2 ( η ( ))2 ( ζ ( ))2 ξ 2 x + y2 + z 2 x2 + y2 + z 2 x2 + y2 + z 2 + + 2 2 2
or 1=
) x2 + y2 + z 2 ( 2 ξ + η2 + ζ 2 4
or 2x 4 = x2 + y2 + z 2 = 2 . ξ ξ + η2 + ζ 2 ᷅ ᷆ ᷇ ᷄ Hence x=
2ξ . ξ 2 + η2 + ζ 2
Similarly, y=
ξ2
2η 2ζ ,z = 2 . 2 2 +η +ζ ξ + η2 + ζ 2
118
1 3D Kinematics of Rigid Bodies
It follows that x˙ = 2
( ) ( ) ξ˙ ξ 2 + η2 + ζ 2 − ξ 2ξ ξ˙ + 2ηη˙ + 2ζ ζ˙ . ( )2 ξ 2 + η2 + ζ 2
Similarly ) ( ) ( η˙ ξ 2 + η2 + ζ 2 − η 2ξ ξ˙ + 2ηη˙ + 2ζ ζ˙ , ( )2 ξ 2 + η2 + ζ 2 ( ) ( ) ζ˙ ξ 2 + η2 + ζ 2 − ζ 2ξ ξ˙ + 2ηη˙ + 2ζ ζ˙ . z˙ = 2 ( )2 ξ 2 + η2 + ζ 2
y˙ = 2
The unit outward normal vector aˆ 1 to the sphere (
1 X− ξ
)2
( )2 1 + (Y − 0) + (Z − 0) = ξ 2
2
at (x, y, z) is given by ( ) 1 aˆ 1 = 1 x − , y, z = (ξ x − 1, ξ y, ξ z). ξ ξ ᷅ ᷆ ᷇ ᷄ 1
Similarly, aˆ 2 = (ηx, ηy − 1, ηz), aˆ 3 = (ζ x, ζ y, ζ z − 1). Observe that aˆ 2 · aˆ 3 = (ηx, ηy − 1, ηz) · (ζ x, ζ y, ζ z − 1) ) ( = ηζ x2 + (ηy − 1)ζ y + ηz(ζ z − 1) = ηζ x2 + y2 + z 2 − ζ y − ηz ( ) ) η( 2y − ζ y − ηz = ζ y − ηz = ζ · x2 + y2 + z 2 = ηζ η 2 ) ζ( 2 −η · x + y2 + z 2 = 0, 2 etc. So, aˆ 1 , aˆ 2 , aˆ 3 are orthogonal unit vectors. It follows that (˙x, y˙ , z˙ ) = uaˆ 1 + vaˆ 2 + waˆ 3 , where
1.2 Examples on Kinematics
119
u ≡ (˙x, y˙ , z˙ ) · aˆ 1 , v ≡ (˙x, y˙ , z˙ ) · aˆ 2 , w ≡ (˙x, y˙ , z˙ ) · aˆ 3 . Observe that u ≡ (˙x, y˙ , z˙ ) · aˆ 1 = (˙x, y˙ , z˙ ) · (ξ x − 1, ξ y, ξ z) = x˙ (ξ x − 1) + y˙ ξ y + z˙ ξ z = ξ (x˙x + y˙y + z˙z ) − x˙ ) ξ d( 2 x + y2 + z 2 − x˙ 2 dt ( ) 4 ξ d 2ξ ξ˙ + 2ηη˙ + 2ζ ζ˙ = − x˙ = −2ξ ( )2 − x˙ 2 2 2 2 dt ξ + η + ζ ξ 2 + η2 + ζ 2 ( ) ( ) ξ˙ ξ 2 + η2 + ζ 2 − ξ 2ξ ξ˙ + 2ηη˙ + 2ζ ζ˙ 2ξ ξ˙ + 2ηη˙ + 2ζ ζ˙ = −2ξ ( )2 − 2 ( )2 ξ 2 + η2 + ζ 2 ξ 2 + η2 + ζ 2 2 = −ξ˙ · 2 , ξ + η2 + ζ 2 =
so ( ) x −2ξ˙ ˙ = −ξ · . u= 2 ξ + η2 + ζ 2 ξ Similarly ( ( ) ) y z −2ζ˙ −2η˙ ˙ = −η˙ · = −ζ · ,w = 2 . v= 2 ξ + η2 + ζ 2 η ξ + η2 + ζ 2 ζ Here, vel. = uaˆ 1 + vaˆ 2 + waˆ 3 , so ) ( ) ( ) ( d aˆ 1 d aˆ 2 d aˆ 3 + v˙ aˆ 2 + v + w˙ aˆ 3 + w , acc. = u˙ aˆ 1 + u dt dt dt and hence d aˆ 1 d aˆ 2 d aˆ 3 · aˆ 1 + v · aˆ 1 + w · aˆ 1 (acc.) · aˆ 1 = u˙ + u dt ᷆ ᷇ dt dt ᷅ ᷄ = u˙ + u
) 1d( d aˆ 3 d aˆ 2 aˆ 1 · aˆ 1 + v · aˆ 1 + w · aˆ 1 2 dt dt dt
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1 3D Kinematics of Rigid Bodies
= u˙ + u
1d d aˆ 3 d aˆ 2 d aˆ 3 d aˆ 2 · aˆ 1 + w · aˆ 1 = u˙ + v · aˆ 1 + w · aˆ 1 . (1) + v 2 dt dt dt dt dt
It suffices to show that 1. 2.
d aˆ 2 dt d aˆ 3 dt
· aˆ 1 = ηu − ξ v, · aˆ 1 = ζ u − ξ w.
For 1: Here d aˆ 2 d (ηx, ηy − 1, ηz) · aˆ 1 = · (ξ x − 1, ξ y, ξ z) dt dt = (ηx ˙ + η˙x, ηy ˙ + η˙y, ηz ˙ + η˙z ) · (ξ x − 1, ξ y, ξ z) = (ηx ˙ + η˙x)(ξ x − 1) + (ηy ˙ + η˙y)ξ y + (ηz ˙ + η˙z )ξ z ⎞ ⎛ ⎞ ⎛ ⎞ ⎛
⎛
⎞
˙ + ηy ˙ ⎠ξ y + ⎝ ηz ˙ + η˙z ⎠ξ z = ηx ˙ ⎝ ξ x −1⎠ + η˙x⎝ ξ x −1⎠ + ⎝ ηy ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
2
1
2
1
( ) = ηξ ˙ x2 + y2 + z 2 − ηx ˙ − η˙x + η · ξ (x˙x + y˙y + z˙z ) ( 2 ) 2 2 = η˙ · ξ x + y + z − ηx ˙ − η˙x + η · (u + x˙ ) ( 2 ) = ηu − ηx ˙ + η˙ · ξ x + y2 + z 2 = ηu − ηx ˙ + η˙ · 2x = ηu + ηx, ˙
2
so it suffices to show that ηx ˙ = −ξ v. ) ( 2x ξ y x2 +y2 +z 2 = yη˙ = yη˙ 2y RHS = −ξ v = −ξ −η˙ · = xη˙ = LHS. η η x2 +y2 +z 2 For 2: Similar to (1), we have ) ( d aˆ 3 · aˆ 1 d aˆ 3 d aˆ 3 d (0) d aˆ 3 d aˆ 1 · aˆ 1 = − · aˆ 1 = − · aˆ 1 = · aˆ 3 = ξ w − ζ u, − dt dt dt dt dt ᷅dt ᷆ ᷇ ᷄ and hence d aˆ 3 · aˆ 1 = ζ u − ξ w. dt
1.2.39 Example The centre of a smooth spherical bowl, of radius a, revolves with uniform angular velocity ω about a vertical axis at a distance c from the centre. Show that the meridian acceleration of a particle moving on the surface of the sphere is
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121
= ̇ ̂
3
̂
2
̂ ̂
Inertial frame ̇
≡ { ̂ 1 , ̂ 2 , ̂ 3 }. ̂
2
Frame
1
̂
1
̂
Frame
≡ { ̂ 1 , ̂ 2 , ̂ 3 }. ≡ { ̂ , ̂ , ̂ }.
All are righthanded oriented.
̂
Fig. 1.36
( )2 aθ¨ − a sin θ cos θ φ˙ + ω − cω2 cos φ cos θ, where θ is the angular distance of the particle from the lowest point of the sphere and φ is the angle between a vertical plane through the particle and the centre of the bowl and a plane through the centre and the axis. Determine also the normal and transverse accelerations (Fig. 1.36) Sol: By Sect. 1.1.22, velocity vD and acceleration aD of point P when looking from frame D are given by vD = a
( ) d (π − θ )ˆeθ + a sin(π − θ )φ˙ eˆ φ = −aθ˙ eˆ θ + a sin θ φ˙ eˆ φ , dt
and ( ) ( ) ( ) aD = −aθ˙ 2 − aφ˙ 2 sin2 θ eˆ r + −aθ¨ + aφ˙ 2 sin θ cos θ eˆ θ + aφ¨ sin θ + 2aφ˙ θ˙ cos θ eˆ φ
Since (
) ) ) ) ) ( ( ( d( ωnˆ 3 × rP/O + 2 ωnˆ 3 × vD + ωnˆ 3 · rP/O ωnˆ 3 − ωnˆ 3 · ωnˆ 3 rP/O dt = (−aθ˙ 2 − aφ˙ 2 sin2 θ)ˆer + (−aθ¨ + a2 φ˙ sin θ cos θ)ˆeθ aN = aD +
+ (aφ¨ sin θ + 2aφ¨ θ˙ cos θ)ˆeθ + 0 ) ( ) ( ) ) ( ( + 2 ωnˆ 3 × −aθ˙ eˆ θ + a sin θ φ˙ eˆ θ + ωnˆ 3 · rP/O ωnˆ 3 − ωnˆ 3 · ωnˆ 3 rP/O ,
we have
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1 3D Kinematics of Rigid Bodies
) ( ) ( ) ( aN · eˆ θ = −aθ¨ + aφ˙ 2 sin θ cos θ + 2 ωnˆ 3 × −aθ˙ eˆ θ + a sin θ φ˙ eˆ φ · eˆ θ ( ( )) ) ( + ωnˆ 3 · r OP ωnˆ 3 · eˆ θ − ωnˆ 3 · ωnˆ 3 rP/O · eˆ θ ) ( ( )) ( ) ( = −aθ¨ + aφ˙ 2 sin θ cos θ + 2 ωnˆ 3 · a sin θ φ˙ −ˆer ( ( )) + ωnˆ 3 · r OP ωnˆ 3 · eˆ θ − ω2 rP/O · eˆ θ ) ( ) ( = −aθ¨ + aφ˙ 2 sin θ cos θ − 2ωaφ˙ sin θ nˆ 3 · eˆ r ( ( )) + ωnˆ 3 · r OP ωnˆ 3 · eˆ θ − ω2 rP/O · eˆ θ ( ) = −aθ¨ + aφ˙ 2 sin θ cos θ − 2ωaφ˙ sin θ cos(π − θ ) ( ( )) + ωnˆ 3 · r OP ωnˆ 3 · eˆ θ − ω2 rP/O · eˆ θ ( ) = −aθ¨ + aφ˙ 2 sin θ cos θ + 2ωaφ˙ sin θ cos θ ( ( )) + ωnˆ 3 · r OP ωnˆ 3 · eˆ θ − ω2 rP/O · eˆ θ ( ) = −aθ¨ + aφ˙ 2 sin θ cos θ + 2ωaφ˙ sin θ cos θ ) ( )) ( π ( ) ( + ω2 nˆ 3 · cdˆ 1 + aˆer cos + θ − ω2 cdˆ 1 + aˆer · eˆ θ 2 ( ) 2 = −aθ¨ + aφ˙ sin θ cos θ + 2ωaφ˙ sin θ cos θ ( ) ( ) + ω2 nˆ 3 · cdˆ 1 + nˆ 3 · aˆer (− sin θ ) − ω2 cdˆ 1 + aˆer · eˆ θ ( ) = −aθ¨ + aφ˙ 2 sin θ cos θ + 2ωaφ˙ sin θ cos θ ( ) + ω2 (0 + a(− cos θ ))(− sin θ ) − ω2 cdˆ 1 · eˆ θ + aˆer · eˆ θ ⎛ ⎞ = ⎝−aθ¨ + aφ˙ 2 sin θ cos θ ⎠ + 2ωaφ˙ sin θ cos θ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1 1 ( ) + ᷅ω2 a sin ᷆ ᷇θ cos θ ᷄ −ω2 cdˆ 1 · eˆ θ + a0 1
)2 ( = −aθ¨ + a sin θ cos θ φ˙ + ω − cω2 dˆ 1 · eˆ θ . It suffices to show that dˆ 1 · eˆ θ = − cos φ cos θ. Observe that, from Fig. 1.36,
1.2 Examples on Kinematics
123
eˆ r = (sin θ ) cos φ dˆ 1 + (sin θ ) sin φ dˆ 2 − cos θ nˆ 3 , so ( ( ( ( ( π )) π )) π) cos φ dˆ 1 + sin θ − sin φ dˆ 2 − cos θ − nˆ 3 eˆ θ = sin θ − 2 2 2 = − cos θ cos φ dˆ 1 − cos θ sin φ dˆ 2 − sin θ nˆ 3 , and hence dˆ 1 · eˆ θ = − cos θ cos φ. Next ) ( ) ( ) ( aN · eˆ r = −aθ˙ 2 − aφ˙ 2 sin2 θ + 2 ωnˆ 3 × −aθ˙ eˆ θ + a sin θ φ˙ eˆ φ · eˆ r ( ( )) ( ) + ωnˆ 3 · r OP ωnˆ 3 · eˆ r − ωnˆ 3 · ωnˆ 3 rP/O · eˆ r ) ( ) ( = −aθ˙ 2 − aφ˙ 2 sin2 θ + 2ωnˆ 3 · aθ˙ eˆ φ + a sin θ φ˙ eˆ θ ( ( )) + ωnˆ 3 · r OP ωnˆ 3 · eˆ r − ω2 rP/O · eˆ r ) ( ) ( = −aθ˙ 2 − aφ˙ 2 sin2 θ + 2ω aθ˙ nˆ 3 · eˆ φ + a sin θ φ˙ nˆ 3 · eˆ θ ( )) ( ) ( + ω2 nˆ 3 · cdˆ 1 + aˆer (− cos θ ) − ω2 cdˆ 1 + aˆer · eˆ r ( ( ( ) π )) = −aθ˙ 2 − aφ˙ 2 sin2 θ + 2ω aθ˙ 0 + a sin θ φ˙ cos θ − 2 ( ) ( ) 2 2 −ω nˆ 3 · cdˆ 1 + nˆ 3 · aˆer cos θ − ω cdˆ 1 + aˆer · eˆ r ( ) = −aθ˙ 2 − aφ˙ 2 sin2 θ + 2ωa sin θ φ˙ sin θ ⎛ ⎞ ⎛
⎞
− ω2 ⎝0 + a(− cos θ )⎠ cos θ − ω2 ⎝cdˆ 1 · eˆ r + aˆer · eˆ r ⎠ ᷅ ᷆ ᷇ ᷄ ᷅ ᷆ ᷇ ᷄ 1
1
) ( ( ) = −aθ˙ 2 − aφ˙ 2 sin2 θ + 2ωaφ˙ sin2 θ − ω2 a sin2 θ − ω2 c dˆ 1 · eˆ r = −aθ˙ 2 − aφ˙ 2 sin2 θ + 2ωaφ˙ sin2 θ − ω2 a sin2 θ − cω2 (sin θ ) cos φ, and aN · eˆ φ ) ( ) ( ) ( = aφ¨ sin θ + 2aφ˙ θ˙ cos θ + 2 ωnˆ 3 × −aθ˙ eˆ θ + a sin θ φ˙ eˆ φ · eˆ φ ( ( )) ) ( + ωnˆ 3 · r OP ωnˆ 3 · eˆ φ − ωnˆ 3 · ωnˆ 3 rP/O · eˆ φ
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1 3D Kinematics of Rigid Bodies
) ( ) ( ) ( = aφ¨ sin θ + 2aφ˙ θ˙ cos θ + 2 ωnˆ 3 · −aθ˙ eˆ r ( )) ( + ωnˆ 3 · r OP ωnˆ 3 · eˆ φ − ω2 rP/O · eˆ φ
( ) = aφ¨ sin θ + 2aφ˙ θ˙ cos θ − 2ωaθ˙ (− cos θ ) ( ( )) ( ) + ω2 nˆ 3 · cdˆ1 + aˆer 0 − ω2 cdˆ1 + aˆer · eˆ φ ( ) ( ) = aφ¨ sin θ + 2aφ˙ θ˙ cos θ + 2ωaθ˙ cos θ − ω2 cdˆ1 · eˆ r + aˆer · eˆ r ( ) = aφ¨ sin θ + 2aφ˙ θ˙ cos θ + 2ωaθ˙ cos θ − ω2 (c(sin θ ) cos φ + a).
1.2.40 Note Suppose that there is a twisted curve. Suppose that the unit vector in the direction of tangent at a point on a curve be (l1 , m1 , n1 ). In other words, l1 , m1 , n1 are the direction cosines of tangent at a point on a curve. Similarly, suppose that l2 , m2 , n2 are the direction cosines of principal normal at a point on a curve, and l3 , m3 , n3 are the direction cosines of binormal at a point on a curve (Fig. 1.37) Sol: It follows that the unit vector along xaxis is (l1 , l2 , l3 ). This is constant in space. It follows that 0 = ˙l1 − l2 θ3 + l3 θ2 (∗), where θ1 , θ2 , θ3 are the components of the angular velocity of the triad. Observe that the corresponding triad of lines at a small distance ds along the arc is obtained by making a twist d τ (= θ1 dt) about the tangent and a twist d ∈(= θ3 dt) about the binormal. Hence, from (∗), Fig. 1.37 ( 3, 3, 3) Unit binormal
= twist about tangent
Unit principal normal ( 2, Unit tangent ( 1,
1,
1)
2,
2)
= twist about binormal
1.2 Examples on Kinematics
125
0 = dl1 − l2 (d ∈) + l3 0. Similarly 0 = dl2 − l3 d τ + l1 d ∈, 0 = dl3 − l1 0 + l2 d τ. Hence dl1 l2 l2 . = ds = ds (radius of curvature) d∈ Also l3 dl2 l1 l1 l3 = ds − ds = − , ds of torsion) of curvature) (radius (radius dτ d∈ and dl3 l2 l2 . = − ds = − ds (radius of torsion) dτ These are known as the Frenet formulae.
1.2.41 Note A point O moves along a twisted curve with velocity v, and the coordinates of P, referred to the principal axes of the curve at O, are ξ, η, ζ. Show that the velocities of P parallel to the moving axes are v + ξ˙ −
ζv ξv ηv ηv , η˙ − + , ζ˙ + , ρ σ ρ σ
where ρ and σ are the radii of curvature and torsion. Sol: Here we shall apply the formulae of Sect. 1.1.24 with (see Sect. 1.2.40) θ1 =
) ) ( ( dτ d∈ 1 ds 1 1 ds 1 d τ ds d ∈ ds = = v , θ2 = 0, θ3 = = = v . = = dt ds dt σ dt σ dt ds dt ρ dt ρ
The required velocities are ) ) ( ( ( ) ( ) ηv 1 v + ζ 0 = v + ξ˙ − , v + ξ˙ − ηθ3 + ζ θ2 = v + ξ˙ − η ρ ρ
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1 3D Kinematics of Rigid Bodies
( ( ) ( ) ) 1 1 ζv ξv v +ξ v = η˙ − + , η˙ − ζ θ1 + ξ θ3 = η˙ − ζ σ ρ σ ρ ( ( ) ) 1 ηv ˙ ˙ζ − ξ θ2 + ηθ1 = ζ˙ − ξ 0 + η v =ζ+ . σ σ
Chapter 2
D’Alembert’s Principle
The advancement of Newtonian dynamics is the result of D’Alembert’s principle. D’Alembert’s principle was enunciated by D’Alembert in his Trait e´ de Dvnamique in 1743. As we shall see that it is only a deduction from Third law of Newton’s laws of motion. Since the principle of virtual work is the counterpart of D’Alembert’s principle in statics (a precursor of dynamics), for completeness we have devoted one section on principle of virtual work. For getting a grip over the subject, 49 problems of various difficulty levels have been solved.
2.1 Principle of Virtual Work 2.1.1 Note Suppose that a particle moves in free space or in contact with a smooth curve or smooth surface. Let S be the component of the external force in the direction of the tangent to the path. Then, by Newton’s second law of motion, the equation of motion in the direction of tangent to the path is dv S = ((mass of the particle)(acc.)) · (unit tangent to the path) = m . .. . dt dv dr dv ·(unit tangent to the path) = m · =m · ˆt dt) ds) dt )) ( ( d vˆt ˆt + v ddsˆt ds · ˆt = m (dt ) · ˆt = m ddsv ds dt () (( ) )) dt ) ˆt + v ρ1 nˆ ds = m ddsv ds · ˆt dt (( dt ) ) ) d v ds ) = m ds dt ˆt · ˆt + mv ρ1 nˆ · ˆt ds dt (( ) ) ) d v ds ) 1 ds d v ds = m ds dt 1 + mv ρ 0 dt = m ds dt = m ddsv v. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 R. Sinha, Advanced Newtonian Rigid Dynamics, https://doi.org/10.1007/9789819920228_2
127
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2 D’Alembert’s Principle
Thus mv
dv = S. ds
It follows that { m
{ vd v =
Sds,
and hence ( 2 )final { final v  = Sds (Final kinetic energy) − (initial kinetic energy) = m 2 initial . .. final . = (work done by external force). Conclusion Work done by external force is equal to increase in kinetic energy. For conservative forces: Work done by force is equal to loss in potential energy, so increase in kinetic energy is equal to loss in potential energy. For reaction force on a smooth surface: Work done by force is equal to 0, because the direction of force is perpendicular to the displacement.
2.1.2 Note Suppose that there is a smooth surface. Let P be a particle moving on the surface. Let ┌ be the path of the particle. It follows that ┌ is a curve through P which lies on the surface. Let PT , PN , PB be the tangent, principal normal, binormal to the curve ┌. Let PL be the normal to the surface. Since PT , PN , PB are the tangent, principal normal, binormal to the curve ┌, PT is perpendicular to PN and PB. By definition, the normal plane, say σ, to the curve ┌ contains PN and PB. Since PT is tangent to the curve ┌, and ┌ lies on the surface, PT lies on the tangent plane. Now since normal PL is perpendicular to the tangent plane, PT is perpendicular to PL. Next since PT is perpendicular to PN and PB, PL, PN , PB are coplanar. They all lie in the plane σ whose normal is PT . Since PT lies on the tangent plane, we can draw a line PM on the tangent plane such that PM is perpendicular to PT . Now since PL, PN , PB are perpendicular to PT , the four lines PL, PN , PB, PM are coplanar. All the four lines lie in the plane σ whose normal is PT . By Meunier’s theorem of differential geometry, we have ρcosφ = ρ ' ,
2.1 Principle of Virtual Work
129
Fig. 2.1
where ρ ' denotes the radius of curvature of the curve ┌, ρ denotes the radius of curvature of the normal section touching ┌ at P, and φ is the angle between the normal section touching ┌ and the osculating plane of ┌. (Fig. 2.1). Observe that every normal section is perpendicular to tangent plane, PL lies on the normal section touching ┌ at P, PM lies on the tangent plane, the normal section and tangent plane intersects in PT , and PL, PM are perpendicular to PT , so PL ⊥ PM . Now since PL lies on the normal section touching ┌, PM is normal to the normal section touching ┌. Since, normal to the osculating plane of ┌ is parallel to PB, normal to the normal section touching ┌ is parallel to PM , and φ is the angle between the normal section touching ┌ and the osculating plane of ┌, we have ∠LPN = ∠MPB = φ . . .. . The acceleration of the particle has no component along binormal PB. The acceleration along PN to the path is ) ( v2 v2 = ' , ρ (radius of curvature of the path) and along PT is s¨. The component of acceleration along PL is equal to (
v2 ρ'
)
( ( 2) ( ) ) v v2 v2 cos ∠LPN = cos φ = cos φ = , ρ' ρ cos φ ρ
and component of acceleration along PM is equal to
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2 D’Alembert’s Principle
(
v2 ρ'
)
(
(
sin ∠LPN =
v2 ρ'
)
( sin φ =
v2 ρ cos φ
)
) v2 sin φ = tan φ . ρ
Hence, if R denotes the normal reaction measured inwards, and L, M are the components of the external force in the directions PL, PM , then ( 2 ) ( 2) v v = R + L, m tan φ = M . m ρ ρ In the determination of ρ, we sometimes use Euler’s theorem of differential geometry: 1 1 1 = cos2 θ + sin2 θ, ρ ρ1 ρ2 where ρ1 , ρ2 are the greatest and the least radii of curvatures of normal sections at the point P of the surface, θ is the inclination of the direction of motion to the direction of the least curvature of the surface. 1. In the case of cylinder, ρ1 becomes infinite, and ρ2 becomes the radius of the cylinder. So 1 1 = sin2 θ, ρ r where r denotes the radius of the cylinder and θ is the inclination of the direction of motion to the direction of the generator. Now the equation ( 2) v =R+L m ρ becomes ( 2 ) v m sin2 θ = R + L. r 2. When a particle moves on a smooth surface, under the action of no force except the reaction of the surface, the path is a geodesic on the surface. ( 2 ) ( 2 ) Reason: On using the equation m vρ tan φ = M , we get m vρ tan φ = 0, and hence tan φ = 0. This shows that the osculating plane at every point of the path contains the normal to the surface. 3. The result in (2) remains true, when the surface is rough. Reason: The friction acts along the path, and above result is obtained by resolving at right angles to the path, so it is unaffected by friction.
2.1 Principle of Virtual Work
131
2.1.3 Note Suppose that there is a smooth surface of revolution whose axis of revolution is vertical. Let P be a particle moving on the inner surface. Let ┌ be the path of the particle. It follows that ┌ is a curve through P which lies on the surface. We shall use cylindrical coordinates r, θ, z, where z increases in the vertically upwards direction. Let us take the zaxis as the axis of rotation (Fig. 2.2). In continuation with Sect. 2.1.2 Since velocity vector v is given by ) ) v = r˙ r + r θ˙ θ + zz , Δ
Δ
Δ
by Sect. 2.1.2, we have ) 1 ( 2 ) )2 1 m (˙r ) + r θ˙ + (˙z )2 + mgz = mv2 + mgz = const. 2 2 . .. . and hence ) 1 ( 2 ) )2 m (˙r ) + r θ˙ + (˙z )2 + mgz = const. 2
(∗).
Observe that the only force acting of the particle is the force of gravity, mg acting parallel to zaxis, and the normal reaction R whose line of action intersects the zaxis. It follows that moments of these forces about the zaxis vanish. Hence m
) d d) zˆ · (r × r˙ ) = mˆz · (r × r˙ ) = zˆ · (r × m¨r) = zˆ · (r × F) = 0 . . .. . dt dt
̂
Tangent to ̂ meridian curve ̂ ̂
Surface of revolution
Path of particle Meridian curve
Normal ̂
In one plane, = tan
Fig. 2.2
132
2 D’Alembert’s Principle
It follows that the moment of the velocity about the axis is a constant. Next since ) ( ) ( ) ) ) ) r 2 θ˙ = r θˆ · r˙ rˆ + r θ˙ θˆ + zˆz = z0 + r θˆ · v = zˆz × zˆ + rˆz × rˆ · v ) ) )) = zˆ × zˆz + rˆr · v ) ) = zˆ × r · v ) ) = zˆ × r · r˙ = zˆ · (r × r˙ ) = const. . .. . we have r 2 θ˙ = const. (say, h). Thus, the moment of the velocity about the axis is the constant h. Since the surface is a surface of revolution, we can suppose that equation of the surface is z = f (r). Now, by (∗), ) ( ( ) ) ' )2 ) 1 1 2 1 dr 2 ( 2g mh 1 + f (r) + 2 + 2 f (r) 2 r4 d θ r h ⎛ ⎞ ( ) ) ) (( 2 2⎟ 1 ⎜ h dr ' 1 ⎜ h dr ⎟ f = m⎜ 2 + 2 h2 + (r) ⎟ + mgf (r) ⎠ 2 ⎝ r dθ r r2 d θ . .. . . .. . 1 ()1 )2 ) )) )2 ) = 21 m θ˙ ddrθ + r12 h2 + θ˙ ddrθ f ' (r) + mgf (r) ) ( ) ) 1 1 2 = m (˙r )2 + 2 h2 + r˙ f ' (r) + mgf (r) 2 ( r ) )2 1 1 ) )2 ) = m (˙r )2 + 2 r 2 θ˙ + r˙ f ' (r) + mgf (r) = const. 2 r . .. . It follows that ( ) ) )2 ) 1 dr 2 ( 1 2g 1 + f ' (r) + 2 + 2 f (r) = const. 4 r dθ r h This is the differential equation of the projection of the particle’s path on a horizontal plane. Another Method Here, the components of acceleration are
2.1 Principle of Virtual Work
133
) )2 r¨ − r θ˙ ,
1d) 2 ) r θ˙ , z¨ . r dt
On resolving along the tangent to the meridian curve, we get ( ) (π ) ) )2 ) 1d) 2 ) π r θ˙ cos + z¨ cos r¨ − r θ˙ cos ψ + −ψ r dt 2 2 ) (π π − ψ + R cos = −g cos 2 2
(
or (
) )2 ) ) ) ) ) r¨ − r θ˙ cos ψ + r¨ f ' (r) + r˙ f '' (r) r˙ sin ψ ) ) ( ( ) )2 ) ) )2 ) d r˙ f ' (r) d 2 f (r) sin ψ = r¨ − r θ˙ = r¨ − r θ˙ cos ψ + cos ψ + sin ψ dt dt 2 ( ) ) )2 = r¨ − r θ˙ cos ψ + z¨ sin ψ = −g sin ψ . .. .
or ( ) )2 ) ) ) ) ) r¨ − r θ˙ cos ψ + r¨ f ' (r) + r˙ f '' (r) r˙ sin ψ = −g sin ψ or ( ) )2 ) 1 ) ) r¨ − r θ˙ + r¨ f ' (r) + (˙r )2 f '' (r) ' f (r) ( ) ) ) )2 ) dr ˙ + r¨ f ' (r) + (˙r )2 f '' (r) = r¨ − r θ dz ( ) ) ) )2 ) cot ψ + r¨ f ' (r) + (˙r )2 f '' (r) = −g = r¨ − r θ˙ . .. . or (
) )2 ) 1 r¨ − r θ˙ + r¨ f ' + (˙r )2 f '' = −g (∗). f'
Since there is no force at right angles to the meridian plane, we have 1d) 2 ) r θ˙ = 0, r dt and hence r 2 θ˙ = const (say, h).
134
2 D’Alembert’s Principle
It follows that r˙ =
dr h dr θ˙ = , 2 .d θ .. d θ r .
and hence ( ) ( )) (( ) (( )) 1 d 2r 1 d 2r h −2 dr −2 dr h dr ˙ + + r¨ = h r ˙ = h θ r3 dθ r2 d θ 2 r3 d θ r2 d θ r2 d θ 2 r2 . .. . ( ) ( ) −2h dr 2 h d 2r =h + 4 2 r5 d θ r dθ ( ) −2h2 dr 2 h2 d 2 r + 4 2. = r5 dθ r dθ It follows from (∗) that ( r¨
1+(f ' ) f'
2
)
) ( )2 ) ) 2 + ddrθ rh2 f '' = r¨ − hr 3 f1' + r¨ f ' + ddrθ ( ( )2 ) h 1 = r¨ − r 2 + r¨ f ' + (˙r )2 f '' = −g r f' . .. . −
h2 1 r3 f '
) h 2 '' f r2
or (
)( ) )2 ) ( ) ) ( 1+ f' −2h2 dr 2 h2 d 2 r dr h 2 '' h2 1 + 4 2 f = −g. − 3 '+ r5 dθ r dθ f' r f d θ r2
Multiplying by
f' , h2
⎛
we get ⎞
(
) ( ⎜ −2 dr ( ) ' )2 ) 1 1 d 2r ⎟ dr 1 2 ' '' f' ⎜ ⎟ − 3+ + 4 2⎟ 1+ f f f = −g ⎜ 5 ⎝r dθ r dθ ⎠ r d θ r2 h2 . .. . . .. . 1
)2
1
or ( ⎛ ) )2 ) ⎞( ) 2 ' '' 2 1+ f' ) ' )2 ) d 2 r f 1( f 1 gf ' ⎝ ⎠ dr 1 + f + − − = − (A). r4 dθ2 r4 r5 dθ r3 h2
2.1 Principle of Virtual Work
135
Let us multiply it by 2 ddrθ : )⎞ ( ⎛ ) ( ( )3 ( )3 4 1 + )f ' )2 ( ) 2 ' '' ) )2 f f 1 dr d r dr dr ⎠ +⎝ 4 2 2 1+ f' − 4 2 r dθ dθ r dθ dθ r5 −
1 dr gf ' dr = − 2 2 (∗). 2 3 r dθ h dθ
Since ( ) ( ) ) ( ) ' )2 ) dr 2 ) ' )2 ) dr d 2 r d 1( 1( 2 1+ f = 4 1+ f d θ r4 dθ r dθ dθ2 ( )( )2 dr dr 1 + 4 2f ' f '' r dθ dθ ) ( ) ( ) ' )2 ) dr 2 −4 dr ( 1 + f + r5 d θ dθ ) ( ( ) 2 ) ) dr d r 1 2 2 = 4 1+ f' r dθ dθ2 )⎞ ( ⎛ )3 ( )3 4 1 + )f ' )2 ' '' ( f f dr dr ⎠, − +⎝ 4 2 r dθ dθ r5 on integrating (∗) with respect to θ, we get ( ) ) ' )2 ) dr 2 1( 1 2g 1+ f + 2 = − 2 f + C. 4 r dθ r h This is the differential equation of the projection of the particle’s path on a horizontal plane. − → Let us resolve along the normal PL. We get ) (( ) ( 1 d ) )) (π ) )2 ) π +ψ + r 2 θ˙ cos + z¨ cos ψ cos m r¨ − r θ˙ 2 r dt 2 = −mg cos ψ + R cos 0 or ) ( ( ) )2 ) sin ψ + z¨ cos ψ + mg cos ψ. R = m − r¨ − r θ˙ By (A),
136
2 D’Alembert’s Principle
( ⎛ ) )2 ) ⎞( ) 2 ' '' 2 1+ f' ) ' )2 ) d 2 r f 1( f 1 gf ' ⎝ ⎠ dr 1 + f + − − 3 =− 2. 4 2 4 5 r dθ r r dθ r h Let us put r = a, where a is a constant. We get that ( ⎛ ) )2 ) ⎞ ' '' ' 2 1+ f' ) ' )2 ) f 1( f ⎠02 − 1 = − gf (a) 1+ f 0+⎝ 4 − 4 5 3 r r r a h2 or h2 = a3 gf ' (a). This is the condition that the path may be a circle of radius a. Another Method If the path is a circle of radius a, then m
) ) ) ) v2 = R sin ψr=a , R cos ψr=a = mg. a
Since )  )2 a θ˙ r=a g
)  )2 ) )2  (0)2 + a θ˙ r=a (˙r )2 + r θ˙  = =   ag ag r=a ) ) mv2 ) ) R sin ψ v2 ) r=a ) = tan ψr=a = a = = ag mg R cos ψr=a . .. .  dz  = = f ' (a) dr r=a
and  ) ) a2 θ˙ r=a = r 2 θ˙ r=a = h, . .. . we have a
) h )2 a2
g
= f ' (a)
or h2 = a3 gf ' (a).
2.1 Principle of Virtual Work
137
1. Let us denote the given surface of revolution, whose equation is z = f (r) by S. The particle moves on S. Assume that the particle is projected with velocity v and angle of projection α. By energy conservation principle, 1 2 mv + mgz = const, 2 or v2 = 2g(c − z), where c is a constant. Here, c depends on the initial conditions. Observe that the moment of) the velocity of particle about the   axis of the surface of revolution, that is, ) zˆ · (r × r˙ ) = · · · = r 2 θ˙ = h is less than zˆ r˙r (= 1rv = rv), so for the particle, √ 0 ≤ h ≤ rv = r 2g(c − z). . .. . ) ) When r 2 θ˙ = h = 0, we have θ˙ = 0, and hence θ = const. Thus, when h takes ) ) the value 0, the motion of the particle is in a vertical plane. When r 2 θ˙ = h = rv, we have / ) )2 r.θ˙.. = v. = (˙r )2 + r θ˙ + (˙z )2 , and hence (˙r )2 + (˙z )2 = 0. It follows that r˙ = 0 = z˙ , and hence there is no velocity component in rzplane. Thus, when h takes the value rv, and projectile velocity v is horizontal, it will remain in the horizontal plane. For the particle on surface S, we have h2 ≤ r 2 2g(c − z). Hence, for the particle on S, we have / rz=z0 ≥
h2 2g(c − z0 )
Since 0≤ we have
h2 r 2 2g
≤ c − z,
(B).
138
2 D’Alembert’s Principle
) h2 . 0≤z ≤ c− 2 r 2g (
2
Thus the particle cannot go above the circle z = c − r 2h2g . Let us denote the surface of revolution, whose equation is z = F(r), where F(r) ≡ c −
h2 1 2g r 2
by C. Since the equation of C can be written as ) ) (2g)r 2 z − (2gc)r 2 + h2 = 0, the surface of revolution C is “cubic” in r, z. Now, we have two surfaces of revolution, namely, S and C. The particle moves on S. At a given value of z = z0 , we shall try to compare the values of r on the two surfaces. For points on C, / rz=z0 =
h2 . 2g(c − z0 )
Thus, from (B), )
) ) ) rz=z0 for particle on S ≥ rz=z0 for points on C .
Since for the cubic surface C, h2 2 dz = (> 0), dr 2g r 3 for points on C, r increases as z increases. Also (see Fig. 2.3), )
) ) ) rz=z0 for points on surface C ≤ rz=z0 for particle on S .
Conclusion The cubic surface h2 = r 2 2g(c − z)
2.1 Principle of Virtual Work
139
Fig. 2.3
 
= 0
= 0
allowed zone for particle
0
Divides the given surface of revolution z = f (r) into several zones separated by horizontal circles, and the particle can only move with the given initial conditions on a zone of the given surface which is more remote from the axis than the corresponding part of the cubic surface. Also, the motion of the particle must clearly be confined to the zone on which it starts, so that it rises and fall between two horizontal circles. 2. We shall try to show that PL (see Fig. 2.2) is the radius of curvature of the normal section at right angles to the meridian plane through P. Here, suppose that equation to the surface of revolution is r ≡ (x, y, z) = (u cos φ, u sin φ, f (u))(u, φ are parameters). Its axis is the zaxis. Here r1 = (cos φ, sin φ, f1 ), r2 = (−u sin φ, u cos φ, 0), E ≡ r1 · r1 = 1 + (f1 )/2 , F ≡ r1 · r2 = 0, G ≡ r2 · r2 = u2 , √ H ≡ EG − F 2 = u 1 + (f1 )2 . Next, (
) (cos φ, sin φ, f1 ) r1 × r2 = = (−u cos φf1 , −u sin φf1 , u), ×(−u sin φ, u cos φ, 0) r11 = (0, 0, f11 ), r12 = (− sin φ, cos φ, 0), r22 = (−u cos φ, −u sin φ, 0), sin φf1 ,u) n = r1 H×r2 = (−u cos φf1 ,−u , L ≡ r11 · n = f11Hu , M ≡ r12 · n = 0, H 3 u2 f1 N ≡ r22 · n = H , T 2 ≡ LN − M 2 = f1 fH112u . On using the formula ρ=
E(du)2 + 2Fdud φ + G(d φ)2 L(du)2 + 2Mdud φ + N (d φ)2
140
2 D’Alembert’s Principle
of radius of curvature of normal section, radius of curvature of normal section along (u = const.) at P is equal to ⎛
/ 2
u G⎜ ⎝= 2 u N √ f1 u
1+(f1 )2
=u
⎞ 1 + (f1 ) ⎟ ⎠. f1 2
Observe that vector equation of normal line PL through P is r = (u cos φ, u sin φ, f (u)) + λn, and vector equation of axis of the surface of revolution is r = 0 + μˆz . Their shortest distance is equal to n×ˆz ((u cos φ, u sin φ, f (u)) − 0) · (n×ˆz) ( (−u cos φf ,−u sin φf ,u) ) 1 1 H n×ˆz  = (u cos φ, u sin φ, f (u)) · ×(0, 0, 1) 1 = H n×ˆz (u cos φ, u sin φ, f (u)) · (−u sin φf1 , u cos φf1 , 0) = 0,
and hence, the two lines intersect. For the point of intersection, we solve 0 + μˆz = (u cos φ, u sin φ, f (u)) + λn for μ, λ. It suffices to show that λ = (radius of curvature of normal section along (u = const.) at P). Here φf1 ⎫ 0 = u cos φ + λ −u cos ⎬ H φf1 . 0 = u sin φ + λ −u sin H ⎭ μ = f + λ Hu (
So λ =
H f1
=
u
√
1+(f1 )2 f1
) = radius of curvature of normal section along (u = const.) at P .
3. Let us assume that the particle is projected horizontally with velocity v at the initial position P. We shall try to obtain a condition so that the particle falls from horizon. We shall use cylindrical coordinates. We shall measure z vertically upwards. In Fig. 2.4, suppose that PN is the principal normal to the path in the initial position of the particle. Suppose that PB is the binormal to the path in the initial position of the particle. Since the path is horizontal at P, the tangent to the path is horizontal at P, and hence the plane, say π, containing PN and PB is vertical. Let ρ ' be the radius of curvature of the path at P. By (2), PL is the radius of curvature of the normal section at right angles to the meridian plane through P. It follows that tangent PT to the meridian curve, zaxis, lines PN , PL, PB all lie on the vertical plane π. Suppose that ∠NPL = φ. By Meunier’s theorem of differential geometry, we have
2.1 Principle of Virtual Work
141
̂
Surface of revolution
Path of particle Meridian curve
Fig. 2.4
(r csc ψ) cos φ = (KP csc ψ) cos φ = (PL) cos φ = (PL) cos(∠NPL) = ρ ' , . .. . so (r csc ψ) cos φ = ρ ' . Since the particle is projected horizontally, there is no acceleration along binormal PB, and hence the only acceleration in the plane is the osculating plane of the path at P. Resolving along tangent PT to the meridian curve, we get (
) v2 sin φ (r csc ψ) cos φ ( 2) π v R = sin φ = g sin ψ + cos = g sin ψ, ρ' m 2 .. . .
v 2 tan φ sin ψ = r
and hence v2 tan φ sin ψ = g sin ψ. r It follows that v2 tan φ = gr.
142
2 D’Alembert’s Principle
The condition that the particle falls below the horizontal circle is that the osculating plane (which contains PN ) is below the horizontal line PK, that is, ⎛ φ+ψ >
⎞
π⎜ ⎜that is, 2⎝
) (π ⎟ gr ⎟, − ψ = tan φ > tan = cot ψ ⎠ v2 . .. 2 .
that is, gr > v2 cot ψ. This is the required condition. 4. Definition An apse is a point on the central orbit at which the radius vector from the centre of force to the point has a maximum or minimum value. The length of the radius vector at an apse is called an apsidal distance. The angle between two consecutive apsidal distances is called an apsidal angle. At an apse, the radius vector is perpendicular to the tangent, that is, at an apse the particle moves at right angles to the radius vector. ) ) Proof By the definition of apse, r is maximum or minimum, so u ≡ 1r is maximum or minimum, and hence / 1 cot φ = r
1 − (r sin φ)2
It follows that φ =
( )2 / ( )2 / 1 1 1 1 du = 0. = − = − u2 = 2 2 r p r p .d θ.. .
π . 2
5. Suppose that the particle is describing a horizontal circle of radius a on the smooth surface of revolution. We can imagine that the motion is slightly disturbed by a small impulse in the meridian plane through the particle. It follows that h is unchanged. Put x ≡ r − a. Here x is small. Since the path is a circle of radius a, we have seen that h2 = a3 gf ' (a). On using Sect. 2.1.3(A), that is, ( ⎛ ) )2 ) ⎞( ) 2 ' '' 2 1+ f' ) ' )2 ) d 2 r f 1( f 1 gf ' ⎠ dr ⎝ 1 + f + − − = − , r4 dθ2 r4 r5 dθ r3 h2 we get
2.1 Principle of Virtual Work
( ) )2 ) d 2 (x + a) 1 1 + f ' (x + a) 4 dθ2 (x + a) ( ⎛ ) ' )2 ) ⎞( )2 ' '' 2 1 + f + a) (x f (x + a)f (x + a) ⎠ d (x + a) +⎝ − dθ (x + a)4 (x + a)5 −
1 gf ' (x + a) = − h2 (x + a)3
or (
) )2 ) d 2 (x + a) 1 + f ' (x + a) dθ2 ( ⎛ ) )2 ) ⎞( ) 2 1 + f ' (x + a) d (x + a) 2 ' '' ⎝ ⎠ + f (x + a)f (x + a) − x+a dθ −x − a = −
gf ' (x + a) (x + a)4 . h2
Since x is small, we can write ( ⎛ ) )2 ) ⎞ ( 2 1 + f ' (a) ) ' )2 ) d 2 x ⎠(0)2 − x − a 1 + f (a) + ⎝f ' (a)f '' (a) − dθ2 a ) ) ) g f ' (a) + xf '' (a) ) 4 =− a + 4a3 x 2 h or ) ) ) ' )2 ) d 2 x ) g f ' (a) + xf '' (a) ) 4 1 + f (a) a + 4a3 x −x−a =− 2 3 ' dθ a gf (a) . .. . ) ( f '' (a) f '' (a) x =− 1+x ' (a + 4x) ≈ −a − 4x − a ' f (a) f (a) (
or ( ) )2 ) d 2 x f '' (a) 1 + f ' (a) x − x − a = −a − 4x − a dθ2 f ' (a) or ) ( ( ) )2 ) d 2 x f '' (a) 1 + f ' (a) x = − 3 + a dθ2 f ' (a)
143
144
2 D’Alembert’s Principle
or d 2x = −μx, dθ2 where ''
μ≡
(a) 3 + a ff ' (a)
1 + (f ' (a))2
.
Here apsidal angle is equal to / ⎞ 1 + (f ' (a))2 π π ⎝ ⎠. = / √ '' (a) μ 3 + a ff ' (a) ⎛
The motion is stable, if μ > 0, that is, 3+a
f '' (a) > 0. f ' (a)
2.1.4 Note Suppose that there is a smooth sphere whose centre is the origin O, and radius a. We can treat the sphere as a surface of revolution whose axis of revolution is the vertical zaxis, and hence we can apply the results of Sect. 2.1.3. Let P be a particle moving on the inner surface of the sphere. The position of the particle can be defined by polar coordinates θ, φ. Here θ is measured from the downward vertical diameter. By Sect. 2.1.3, (a sin θ )2 φ˙ = h(a constant). Assume that the particle is projected with horizontal velocity u from the position θ = α. Now since the moment of the velocity about the axis is the constant h, we have ˙ u(a sin α) = h = (a sin θ )2 φ, . .. . and hence φ˙ =
u sin α . a sin2 θ
2.1 Principle of Virtual Work
145
By energy equation ( )) ( )) () ) () ) 2 2 α)(cos θ −cos α) 1 α ˙ 2 + u2 cos2 θ −cos = aθ˙ + u2 (cos θ +cossin a θ m 2 2 θ ⎛ 2 ⎞ sin θ ) ( ⎜) ) ( () ) )) u sin α 2 ⎟ 2 2 ⎟ 1 2 1 1 ⎜ 2 sin2 α ˙ ˙ = 2 m aθ + u sin2 θ − 1 = 2 m⎜ aθ + ⎟ − mu ⎝ ⎠ .2 .. . sin θ . .. . 1 1 )2 ) 1 2 1 () )2 ) − mu = −mg(a cos α − a cos θ ), = m aθ˙ + (a sin θ )φ˙ 2 .2 .. .
1 m 2
so ( ( )) ) )2 1 2 (cos θ + cos α)(cos θ − cos α) ˙ m aθ + u 2 sin2 θ = −mg(a cos α − a cos θ ) or ( ) ) )2 (cos θ + cos α)(cos θ − cos α) = 2ga(cos θ − cos α) aθ˙ + u2 sin2 θ For extreme values of θ (that is, for θ˙ = 0), we have ( u2
(cos θ + cos α)(cos θ − cos α) sin2 θ
) = 2ga(cos θ − cos α)
or )) ) ) (cos θ − cos α) u2 (cos θ + cos α) − 2ga 1 − cos2 θ = 0 or )) ) ) (cos θ − cos α) 2ga(cos θ )2 + u2 (cos θ ) + u2 cos α − 2ga = 0. It follows that cos θ = cos α )or ) f (θ ) ≡ 2ga(cos θ )2 + u2 (cos θ ) + u2 cos α − 2ga = 0. Thus, the extreme values of θ are
146
2 D’Alembert’s Principle
α, α1 ≡ cos−1 α2 ≡ cos−1
−2ga +
/ ) ) u4 − 8ga u2 cos α − 2ga 4ga
−2ga −
/ ) ) u4 − 8ga u2 cos α − 2ga 4ga
,
.
Clearly, α1 < α2 < π. Observe that f (0) = u2 (1 + cos α) (which is positive), f (π ) = u2 (−1 + cos α) (which ) 2 is negative), 2 ) 2 2 f (α) = 2u cos α − 2ga sin α = 2 u cos α − ga sin α . If u2 cos α < ga sin2 α, then f (α) is negative. Now since f (0) is positive, there exists θ lying between 0 and α such that f (θ ) = 0, and hence θ˙ = 0. This shows that if u2 cot α < ga sin α, the particle is confined to a zone between two horizontal circles (θ = α1 and θ = α) and it falls down the circle θ = α. If ga sin2 α < u2 cos α, then f (α) is positive. Now since f (π ) is positive, there exists θ lying between π and α such that f (θ ) = 0, and hence θ˙ = 0. Conclusion (i) If u2 cot α < ga sin α, the particle is confined to a zone between two horizontal circles and it falls down the circle θ = α. (ii) If ga sin α < u2 cot α, the particle is confined to a zone between two horizontal circles (θ = α and θ = α1 ) and it rises up the circle θ = α. (iii) If ga sin α = u2 cot α, the particle moves on the circle θ = α. Let R be the normal reaction experienced by the particle due to the inner surface of the particle. By Sect. 2.1.2, we have 2 u2 − 2g(cos α − cos θ ) = m a .
=
)1 2
) mu2 − mg(a cos α − a cos θ ) R − mg cos θ = a m .. .
R − g cos θ m
or R=
mu2 + mg(3 cos θ − 2 cos α). a
Another Method Here, cubic surface of revolution is (see Sect. 2.1.3(1)) (u(a sin α))2 = h2 = r 2 2g(c − z) . . .. .
2.1 Principle of Virtual Work
147
For the points of intersection of the cubic surface and the sphere, we have (u(a sin α))2 = (a sin θ )2 2g(c − (a − a cos θ )) or u2 sin2 α = 2g sin2 θ (c − a + a cos θ ). Next, since u2 = 2g(c − (a − a cos α)), we have ) ) u2 (cos θ − cos α)(cos θ + cos α) = u2 cos2 θ − cos2 α = u2 sin2 α − u2 · sin2 θ = 2g sin2 θ (a cos θ − a cos α) . .. . = 2ga sin2 θ (cos θ − cos α) ) ) = 2ga − 2ga cos2 θ (cos θ − cos α), and hence ) ) u2 (cos θ − cos α)(cos θ + cos α) = 2ga − 2ga cos2 θ (cos θ − cos α). It follows that the intersection of the cubic surface and the sphere is a collection of circles given by cos θ − cos α = 0oru2 (cos θ + cos α) = 2ga − 2ga cos2 θ, that is, ) ) θ = α, or 2ga(cos θ )2 + u2 (cos θ ) + u2 cos α − 2ga = 0, that is, θ = αorF(cos θ ) = 0. Conclusion By Sect. 2.1.3(1), the intersections of the cubic surface with the sphere are the circles θ = α, and the other circles determined by F(cos θ ) = 0. 1. We want to find the normal reaction R on the particle exerted by the sphere. Let us use cylindrical coordinates with z measured downwards from the centre. Here, the components of acceleration in “cylindrical coordinates” are
148
2 D’Alembert’s Principle
Fig. 2.5
( )2 dφ d 2 (a sin θ ) − sin θ , (a ) dt 2 dt
( ) dφ 1 d , (a sin θ )2 a sin θ dt dt
d 2 (a cos θ ) . dt 2
On resolving along OP direction (see Fig. 2.5), the equation of motion is (
+
( )2 ) ( ( ) ) dφ d 2 (a sin θ ) π 1 d 2 d φ cos π − sin θ cos − θ + sin θ (a ) (a ) dt 2 a sin θ dt dt 2 dt 2 d 2 (a cos θ ) −R + mg cos θ cos θ = m dt 2
or ( ) ) ) ) ) ) )2 d θ˙ cos θ d −θ˙ sin θ −R + mg cos θ − (sin θ ) φ˙ cos θ = a sin θ + a dt dt m or ⎞
⎛⎛
⎞
) )2 ⎟ ) )2 ⎟ ⎜⎜ ˙ ˙ a⎝⎝.θ¨ cos .. θ. − .θ ..sin θ.⎠ − (sin θ ) φ ⎠ sin θ ⎛
1
2
⎞
) )2 −R + mg cos θ ⎜ ˙ cos θ ⎟ −a⎝.θ¨ sin ⎠ cos θ = .. θ. + .θ .. m . 1
or
2
2.1 Principle of Virtual Work
149
) )2 ) )2 −R + mg cos θ −a θ˙ − a sin2 θ φ˙ = m or () ) ) )2 ) 2 R = mg cos θ + ma θ˙ + sin2 θ φ˙ . We have seen that aR R − mg cos θ − ag cos θ − u2 = a2 − u2 m ( ma ) )2 ) )2 ) − u2 = 2ga(cos θ − cos α), = a2 θ˙ + sin2 θ φ˙ . .. . and hence R = mg(3 cos θ − 2 cos α) +
mu2 . a
Another Method We shall apply the formula in Sect. 2.1.2. We get v2 R − mg cos θ = , a m where v2 = 2g(c − (a − a cos θ )), u2 = 2g(c − (a − a cos α)). It follows that a
R − mg cos θ − u2 = v2 − u2 = 2ga(cos θ − cos α), . .. . m
and hence a
R − mg cos θ = 2ga(cos θ − cos α) + u2 m
or R − mg cos θ = 2mg(cos θ − cos α) + or R = mg(3 cos θ − 2 cos α) +
mu2 . a
mu2 a
150
2 D’Alembert’s Principle
2. Observe that, for a given u, α, R increases as the particle falls down. Also, 2 R(θ = α) = mg + mua > 0. This show that when θ = α, the particle is in contact with the sphere. If it falls down, it remains in contact with the sphere. Now consider the situation, when the particle rises up from the circle θ = α. In this case, ag sin α < u2 cot α, and the particle is confined to a zone between two horizontal circles (θ = α and θ = α(1 ). It follows that. √ ) −1 1 −1 2 2 . Let us suppose that α = sin 3 = cos 3 We want to show that R(θ = α1 ) is positive, that is, 0 < (3 cos α1 − 2 cos α) +
u2 , ag
that is,
2 cos α −
2
u
⎞
) ( ⎟ π⎜ ⎜that is, gr = tan φ > tan π − ψ = cot ψ ⎟, ⎝ ⎠ 2 2 v . .. 2 .
that is, gr > v2 cot ψ. This is the required condition.
2.1.5 Example A spherical pendulum is started horizontally at the level of centre. Prove the equations of motion in azimuth (ψ) and depth (z), viz. ) ) 1. l 2 − z 2 {ψ˙ = ) V l, ) } 2. l 2 z˙ 2 = z 2g l 2 − z 2 − V 2 z , and verify that, if z = l cos θ, and ψt=0 = 0, then ) ( V √ Vt =√ sin θ sin ψ − cos θ . 2l 2gl
2.1 Principle of Virtual Work
155
Sol: Observe that the only forces acting of the particle are the force of gravity, mg acting parallel to zaxis, and the tension T of the string whose line of action intersects the zaxis. It follows that moments of these forces about the zaxis vanish. Hence, the moment of the velocity about the axis is a constant h, that is, (√ )2 l 2 − z 2 ψ˙ = h(a constant), where l is the length of the pendulum. Since, the spherical pendulum is started ( vt=0 ≡ V ) horizontally at the level of centre, we have )2 (√ )2 ) l − z 2 ψ˙ = l 2 − z 2 ψ˙ = .h =..V · .l ,
and hence )2 ) l − z 2 ψ˙ = V l. This proves (1). Here, energy equation is ( )2 ( )2 ) ( √ l − (˙z )2 + dtd l 2 − z 2 + √lV2 −z 2 ( ) )2 (√ ( √ )2 1 d 2 = m (˙z )2 + l − z2 + l 2 − z 2 ψ˙ − 2 dt . .. 1 m 2
1 mV 2 2
1 mV 2 = mgz . 2 .
So ( ) 2 2 l2 V 2 z2 z2 + lV2 −zl 2 − V 2 z )2 1 + l 2 −z (˙z )2 l 2 −z 2 + l 2 −z 2 = (˙ 2 ( ( )2 ( )2 ) 1 V l = (˙z )2 + − V 2 = 2gz √ (−2z˙z ) + √ 2 l 2 − z2 l 2 − z2 . .. . or ) ) (˙z )2 l 2 + V 2 z 2 = 2gz l 2 − z 2 or ) { ) } l 2 z˙ 2 = z 2g l 2 − z 2 − V 2 z . This proves (2). We have to show that
156
2 D’Alembert’s Principle
) ( V √ Vt =√ sin θ sin ψ − cos θ , 2l 2gl that is, √ / ( ) Vt V l 2 − z2 z sin ψ − , =√ l 2l 2gl l that is, √
l2
−
z2
) ( V √ Vt =√ sin ψ − z, 2l 2g
that is, √ ) ( V z Vt =√ √ sin ψ − . 2l 2g l 2 − z 2 Since ψt=0 = 0, it suffices to show that ) ( ) ( V V Vt · ψ˙ − =√ cos ψ − 2l 2l 2g
1 √ 2 z
√ √ √ l 2 − z 2 − z 21(−2z) l 2 −z 2 l 2 − z2
z˙ ,
that is, / ) ) ) 2g l 2 − z 2 − V 2 z ( V V ˙ =√ · ψ− √ √2 2 2l 2g 2g l − z
1 √ 2 z
√
l 2 − z2 +
3 2
√z l 2 −z 2
l 2 − z2
that is, /
)
2g l 2 − z 2
)
) ( V =V − V 2 z · ψ˙ − 2l
1 √ 2 z
√
3
2 l 2 − z 2 + √lz2 −z 2 z˙ , √ l 2 − z2
that is, ( ) ) ( 3 / ) 2 ) 1 V z =V √ + 2 2g l 2 − z 2 − V 2 z · ψ˙ − z˙ , 2l 2 z l − z2 that is / 2g
)
l2
−
z2
)
(
−
V 2z
V · ψ˙ − 2l
)
(
) l 2 − z 2 + 2z 2 ) z˙ , =V √ )2 2 z l − z2
z˙ ,
2.1 Principle of Virtual Work
157
that is, /
) Vl V − l 2 − z2 2l / ) } ) { )2 z 2g l − z 2 − V 2 z
) ) 2g l 2 − z 2 − V 2 z · (
=V
l 2 + z2 ) √ )2 2 z l − z2
(
l
,
that is, / 2g
)
l2
( =V
−
z2
)
−
) ) 2V l 2 − V l 2 − z 2 ) ) · l 2 − z 2 2l ) } )/ { ) 2 z 2g l − z 2 − V 2 z
V 2z
l 2 + z2 ) √ )2 2 z l − z2
l
,
that is, ( )√ ) ) V l 2 + z2 l 2 + z2 z ) ) =V √ ) ) . l l 2 − z 2 2l 2 z l 2 − z2 This is clearly true.
2.1.6 Example A particle moves under gravity on the surface of a smooth sphere of 1 m radius between two horizontal circles at depth 40 cm and 50 cm below the centre. Show that the velocity of the particle ranges between 404 cm/s and 428 cm/s. Sol: Observe that the only forces acting on the particle are the force of gravity, mg acting parallel to zaxis, and the normal reaction R whose line of action intersects the zaxis. It follows that moments of these forces about the zaxis vanish. Hence, the moment of the velocity about the axis is a constant h, that is, √
1002 − 402 vmin =
√
1002 − 502 vmax ,
or √ √ 14 × 6vmin = 15 × 5vmax . Here, energy equation is
158
2 D’Alembert’s Principle
1 1 m(vmax )2 − m(vmin )2 = mg(10cm) 2 2 or (√ )2 ) ( 15 × 5vmax 3 15 × 5 2 2 = (vmax ) 1 − = (vmax ) − √ (vmax ) 28 14 × 6 14 × 6 2
= (vmax )2 − (vmin )2 = 2g(10cm) . .. . = 2(980 cm)(10 cm) and hence / vmax =
2 × 980 × 10 × 28 cm/s = 428 cm/s 3
Next √ vmin = √
15 × 5
14 × 6
× 428 cm/s = 404 cm/s
See Fig. 2.8. Fig. 2.8
40 cm
100 cm 10 cm
2.1 Principle of Virtual Work
159
2.1.7 Example A particle moves under gravity on a smooth spherical surface whose centre is O. If A be a point at which the velocity is horizontal, prove that the path is symmetrical with respect to the vertical plane through OA, and hence infer that the whole path is included between two horizontal circles which it alternately touches. If those circles be nearly coincident, at an angular distance α from the lowest point of the sphere, find the apsidal angle (Fig. 2.9). Sol: Observe that the only forces acting on the particle are the force of gravity, mg acting parallel to zaxis, and the normal reaction R whose line of action intersects the zaxis. It follows that moments of these forces about the zaxis vanish. Hence, the moment of the velocity about the axis is a constant h, that is, (a sin θ )2 φ˙ = (a sin α)u. By the energy equation,
Fig. 2.9
160
2 D’Alembert’s Principle
( ( ) ) 1 2 d θ u sin α 2 1 2 sin2 α −1 ma + mu 2 d(φ a sin)2 θ 2( sin2 θ ) 2 dθ sin2 α 1 1 = ma2 φ˙ + mu2 −1 2 2 d(φ 2 sin ) θ ) ) ) ) 2 sin α 2 − 21 mu2 = 21 m a2 θ˙ + usin θ ( ) ) ( ) )2 1 1 (a sin α)u 2 = m aθ˙ + − mu2 2 a sin θ 2 )2 ) 1 2 1 () )2 ) − mu = mg(a cos θ − a cos α), = m aθ˙ + a sin θ φ˙ 2 .2 .. . and hence (
d θ u sin α d φ sin2 θ
(
)2 +u
2
) sin2 α − 1 = 2g(a cos θ − a cos α). sin2 θ
It suffices to show that this equation does not change, when φ is replaced by −φ. This is clearly true. By Sect. 2.1.3, apsidal angle is equal to / π 1 + (f ' (a sin α))2 / , '' (a sin α) 3 + a ff ' (a sin α) where f (r) ≡
√
a2 − r 2 .
It follows that '
f (r) =
1(−2r) √ 2 a2 −r 2
=
√ −r , a2 −r 2
) ' ) f (a sin α) =
''
f (r) = −
−a sin α a cos α
√ 1 a2 −r 2 −r √ 1
2 a2 −r 2 a2 −r 2
(−2r)
=
−a2
( −1 = − tan α, f '' (a sin α) = a cos 3α.
Hence / √ π 1 + (f ' (a sin α))2 π 1 + (− tan α)2 / = / '' (a sin α) −1 3 + a ff ' (a 3α sin α) 3 + a −a cos tan α π π sec α =√ . =√ 2 2 3 + sec α csc α 3 cos α + csc α
a2 −r 2
3
)2
,
2.1 Principle of Virtual Work
161
2.1.8 Example A heavy particle P moves on the smooth inner surface of a fixed sphere of centre O, and the angle between OP and the downward vertical is denoted by θ. The particle is projected horizontally on the surface of a point at which θ = α (< π2 ). Prove that whatever be the velocity of projection, θ will not exceed π − α in the subsequent motion, and that, if 31 < sin α, the particle will not leave the surface (Fig. 2.10). Sol: By Sect. 2.1.4,
cos θmax =
−u2 −
/ ) ) u4 − 8ga u2 cos α − 2ga 4ga
(∗)
or θmax = cos−1
−u2 −
/
) ) u4 − 8ga u2 cos α − 2ga 4ga
.
Observe that d (θmax ) du ) ) 3 d (cos θmax ) 1 = 4ga = −2u − / 4u − 16ga cos αu , ) ) du 2 u4 − 8ga u2 cos α − 2ga . .. .
4ga(− sin θmax )
so when
Fig. 2.10
d (θmax ) du
= 0, we have
162
2 D’Alembert’s Principle
) 3 ) 1 −2u − / ) ) 4u − 16ga cos αu = 0 2 u4 − 8ga u2 cos α − 2ga or / ) ) − u4 − 8ga u2 cos α − 2ga = u2 − 4ga cos α. Now from (∗), we have cos θmax .
) ) −u2 + u2 − 4ga cos α = = − cos α = cos(π − α), 4ga .. .
so max(θmax ) = π − α. u
Let us suppose that sin−1
1 3
< α. We have to show that R > 0. By Sect. 2.1.4,
R = mg(3 cos θ − 2 cos α) + so it suffices to show that 0 < (3 cos θ − 2 cos α) + By energy equation
mu2 , a
u2 . ag
0 < v2 = u2 + 2ag(cos θ − cos α), . .. . so 2(cos α − cos θ )
= = h h, Mh + m(h − ε) (M + m)h , that is ⎛
⎞
⎞
⎛
Mh +m(h − ε)⎠ > h ⎝,,,, k 2 ⎝,,,, Mk 2 +m(h − ε)2 ⎠ that is, mk 2 (h − ε) > mh(h − ε)2 that is, h(OC) = k 2 > h(h − ε) = h(OG − ε) = h(OG − AG) = h(O A) , ,, , that is, OC > O A. ◼
This is known to be true.
4.2.4 Note For positive x, y, we define { Γ(x) ≡
u=∞
e−u u x−1 du,
u=0
{ B(x, y) ≡
t=1
t x−1 (1 − t) y−1 dt.
t=0
Observe that { Γ(x) · Γ(y) = = =
u=∞
e {u=0 v=∞
u
{ du ·
e−v v x−1 dv ·
v=0 ({ u=∞ u=0
−u x−1
u=∞
{ u=0 u=∞ ){
u=0
e−u u y−1 du
e−u u y−1 du e−u u y−1 du
v=∞
v=0
e−v v x−1 dv
448
4 Motion About a Fixed Axis v=∞ (
({ u=∞ )) e−v v x−1 e−u u y−1 du dv v=0 u=0 ) { v=∞ ({ u=∞ ( ( −v x−1 −u y−1 e v e u du dv = {
=
v=0
u=0
so { Γ(x) · Γ(y) =
v=∞ ({ u=∞ v=0
−v x−1
e v
) ( −u y−1 ( e u du dv.
u=0
For any positive u, v, let us introduce new variables z, t such that u = zt, and u v = z − zt. It follows that z = u + v, t = u+v . Clearly, 0 < t < 1, 0 < z. Thus   ∂u   ∂t dudv =  ∂v   ∂t
∂u ∂z ∂v ∂z
      z t   dtdz = z · dtdz. dtdz =   −z 1 − t  
Now Γ(x) · Γ(y) { v=∞ ({ = v=0
u=∞
e
−v x−1
v
(
e
−u y−1
u
(
) du dv =
u=0
t=0
,, {
= { =
z=∞ ({ t=1
) e−z (1 − t)x−1 z x+y−1 t y−1 dt dz
z=0 t=0 { t=1 z=∞ ( −z x+y−1
e
{
=
z=∞ ( { t=1 z=0
,
=
{
z=0 t=1
t=0 { t=1
z
) (1 − t)x−1 t y−1 dt dz
t=0
( ( (1 − t) (1 − t)x−1 t y−1 dt ·
{
z=∞
e−z z x+y−1 dz
z=0
t y−1 (1 − t)x−1 dt · Γ(x + y) = B(y, x) · Γ(x + y)
t=0
= B(x, y) · Γ(x + y).
Hence Γ(x) · Γ(y) = B(x, y) · Γ(x + y) or
e−(1−t)z ((1 − t)z)x−1 ) ( −zt ( y−1 e (zt) (zdt) dz ,
4.2 Examples on Motion About a Fixed Axis (Part I)
B(x, y) =
449
Γ(x) · Γ(y) . Γ(x + y)
Next, { t=1 Γ(x) · Γ(y) = B(x, y) = t x−1 (1 − t) y−1 dt Γ(x + y) t=0 { θ= π 2 ( (x−1 ( ( y−1 ( 2 ( sin2 θ 1 − sin2 θ d sin θ = =2
θ =0 { θ= π 2 θ =0
sin2x−1 θ cos2y−1 θ dθ,
so {
θ = π2
θ =0
sin2x−1 θ cos2y−1 θ dθ =
Γ(x) · Γ(y) . 2Γ(x + y)
The function Γ(x) is known as the gamma function, and B(x, y) is known as the beta function. Observe that )u=∞ { u=∞ ( { u=∞ e−u  e−u du Γ(x + 1) = u x e−u du = u x − xu x−1  −1 u=0 −1 u=0 u=0 { u=∞ = (0 − 0)x + x u x−1 e−u du = xΓ(x). u=0
So Γ(x + 1) = xΓ(x). Conclusion: For positive x, y, { Γ(x + 1) = xΓ(x),
θ = π2 θ =0
sin2x−1 θ cos2y−1 θ dθ =
Γ(x) · Γ(y) . 2Γ(x + y)
4.2.5 Example A rigid body is movable about a horizontal axis and starts from rest with its centre of mass in the horizontal plane through the axis. Show that the mean kinetic energy during one complete oscillation is to the greatest kinetic energy during the motion is ( ( ((2 ( ( 1 ((2 in the ratio of 2Γ 43 to Γ 4 , the mean being taken with regard to the time.
450
4 Motion About a Fixed Axis
Sol: Suppose that θ is the inclination of the plane in question with the vertical at time t. So, the equation of motion is (
( .. Mk 2 θ = −(Mg)(h sin θ )
or ..
θ =−
gh sin θ. k2
It follows that ( ) d (( (2 ) gh 2gh d θ˙ = 2θ˙ − 2 sin θ = 2 (cos θ ), dt k k dt , ,, , and hence ( (2 2gh θ˙ = 2 cos θ + C. k  Now since θ˙ θ= π = 0, we have C = 0, and hence 2
/ dθ 2gh √ ˙ =θ =− cos θ . dt k2 , ,, , It follows that the time T of one oscillation is given by T = 2√ ,
k 2gh
{
θ = π2 θ =− π2
,,
1 4k dθ = √ √ 2gh cos θ , 4k =√ 2gh
{
θ = π2 θ =0
{
θ = π2
1 dθ √ cos θ −1
(sin θ )0 (cos θ ) 2 dθ θ =0 ( 0+1 ( ( −12 +1 ) ·Γ 2 Γ 2 4k (1 1( =√ 2gh 2Γ + ( 1 ( 2( 1 ( 4 2k Γ 2 · Γ 4 ( ( =√ . 2gh Γ 34 Thus
4.2 Examples on Motion About a Fixed Axis (Part I)
451
( ( ( ( 2k Γ 21 · Γ 14 ( ( T =√ . 2gh Γ 43 We have to show that { t=T 1 ( t=0 2
(( (2  ) ( ( 3 ( )2 ( 2Γ 4 Mk 2 θ˙ dt 1 ( 2 (( (2  ( ( . = Mk θ˙  / T 2 Γ 41 θ =0
{ t=T 2gh { t=T ( (2 ˙ 2 cos θ dt t=0 θ dt  LHS = ( (2 = t=0 k   T θ˙  T 2gh cos θ  2 k θ =0 θ =0 { θ= π T { t=T { t= 4 −4 θ =02 (cos θ ) θ1˙ dθ 4 t=0 cos θ dt t=0 cos θ dt = = = T T T { θ = π2 1 −4 θ =0 (cos θ ) / 2gh √ dθ −
= = =
=
k 4 √2gh
{ θ = π2 θ =0
1
(sin θ )0 (cos θ ) 2 dθ T
Γ ( 21 )·Γ ( 41 ) √2k 2gh Γ ( 34 )
Γ ( 21 )·Γ ( 14 ) Γ ( 43 )
cos θ
T
1 3 k Γ ( 2 )·Γ ( 4 ) 2 √2gh Γ ( 45 )
Γ ( 21 )·Γ ( 34 ) 1 1 4 Γ( 4 )
k2
(
=
=
Γ ( 21 )·Γ ( 43 ) Γ ( 54 )
1 3 k Γ ( 2 )·Γ ( 4 ) 4 √2gh 5 2Γ ( 4 ) = T
Γ ( 21 )·Γ ( 41 ) Γ ( 34 )
( ( )2 2Γ 43 ( ( = RHS. Γ 14
4.2.6 Example A flywheel, turning with average angular velocity p, is acted on by a driving couple A sin2 pt, and a constant couple 21 A opposite the motion. Find the least moment of inertia required to make the difference between the greatest and the least angular p . velocities less than 100 Sol: Here equation of motion is I
( .. dω 1 A( A = I θ = A sin2 pt − A = − 1 − 2 sin2 pt = − cos 2 pt, dt 2 , 2 2 , ,,
452
4 Motion About a Fixed Axis
so ) ( A A sin 2 pt +C =− Iω = − sin 2 pt + C. 2 2p 4p Hence, the difference between the greatest and the least angular velocities ( ) A A A +C − − +C = . 4pI 4pI 2pI It is given that p A < , 2pI 100 that is,
50 A p2
< I. So, the required value is equal to
50 A . p2
4.2.7 Note Suppose that in place of a rigid body we have a lamina. Suppose that the lamina is compelled to move in the x yplane fixed in space. It is clear that the position of the lamina is determined if we know the position of a fixed point of the lamina together with the position of a fixed direction of the lamina. For the purpose of fixed point, we can take the centre of mass of the lamina. In order to determine the fixed direction of the lamina, we can take the inclination of the fixed direction of the lamina with a fixed direction in the x yplane. These quantities (say x, y, θ ) are called the coordinates of the lamina. If we determine them in terms of the time t, we have completely determined the motion of the body. By Sect. 2.2.4, the motion of centre of mass G(x, y) is given by M
Σ Σ d2x d2 y = X, 2 = Y. 2 dt dt
Suppose that x , (= x − x), y , (= y − y) are the coordinates of any point of the lamina relative to G. Then by Sect. 2.2.5, ) d (Σ m (moment about G of velocity relative to G of mass m) dt ( ( , , )) d Σ , dy , dx −y = m x dt dt dt ( , ,) Σ d , dy , dx x = m −y dt dt dt
4.2 Examples on Motion About a Fixed Axis (Part I)
453
) ( )) Σ (d( d dy , dx, x, · − y, · m dt dt dt dt 2 ,) Σ Σ ( d 2 y, ( ( d x x ,Y − y, X , = m x , 2 − y, 2 = dt dt , ,, ,
=
and hence ) d (Σ m (moment about G of velocity relative to G of mass m) dt Σ( ( = x , Y − y , X (∗). −→ −→ Suppose that G B has a fixed direction in the x yplane. Let G A be a fixed direction in the lamina. Suppose that ∠BG A = θ. Suppose that Gm is the line joining mass m and G. Clearly, the position of line Gm varies with time t. Suppose that ∠BGm = φ. Since A,.. G, m all lie in the rigid lamina, ∠AGm(= φ − θ ) is a constant, and hence .. φ˙ = θ˙ , φ = θ . Put Gm = r. Now since G, m lie in the rigid lamina, r is a constant, and hence the velocity relative to G of m is r
( ) dφ dθ d(∠BGm) =r =r dt dt dt
in the direction perpendicular to Gm. It follows that ( ) dθ (moment about G of velocity relative to G of m) = (Gm) r dt , ,, , ) ( dθ dθ = r2 , =r r dt dt and hence, from (∗), ( ) d dθ (M.I. of the lamina about Gz axis) dt dt ( ) d dθ = (M.I. of the lamina about Gz axis) dt dt ) ( d dθ Σ ( 2 ( = m r dt dt lamina ) ( d d(∠BG A) Σ ( 2 ( = m r dt dt lamina
454
4 Motion About a Fixed Axis
)) ( 2 d(∠BG A) m r dt lamina ) ( ) ( Σ( ( dθ d Σ x ,Y − y, X . m r2 = = dt lamina dt lamina , ,, ,
d = dt
(
Σ
Conclusion: The three dynamical equations for the motion of any body in one plane are Σ Σ d2x d2 y M 2 = X, 2 = Y, dt dt ( 2 ( .. Mk θ = (the moment about G of all external forces acting on the system).
4.2.8 Note Suppose that a uniform straight rod slides down in a vertical plane such that its ends is in contact with two smooth planes, one horizontal and the other vertical. Suppose that it started from rest at an angle α with the horizontal. The moving rod is in contact with a fixed surface, that is, a vertical wall. So a normal reaction, say R, is applied at the upper end of the rod by the smooth wall. Since the vertical wall is smooth, the direction of R is horizontal as shown in Fig. 4.27a. Similarly, a normal reaction, say S, is applied at the lower end of the rod as shown in Fig. 4.27a. Suppose that x, y are the coordinates of the centre of mass G. Since the rod moves in the x yplane, we can apply the results of Sect. 4.2.7. The three dynamical equations for the motion of the rod are M
d2x d2 y = R, M = S − Mg, dt 2 dt 2
and −M
a 2 .. θ= 3
( M
) ( ( d2 .. ) a2 ( 0 − θ = Mk 2 (π − θ ) = −R(a sin θ ) + S(a cos θ ) 3 dt 2 , ,, ,
( ( .. ) ) = a(S cos θ − R sin θ ) = a M y +g · cos θ − R sin θ ) ( ( .. ) ( .. ( = a M y +g cos θ − M x sin θ
4.2 Examples on Motion About a Fixed Axis (Part I)
455
(a)
2 direction fixed in body
direction fixed in space
(b) ≡ ( ) = 2
sin
1
direction fixed in body
direction fixed in space
Fig. 4.27
(( .. ) ) .. y +g cos θ − x sin θ (( 2 ) ( 2 ) ) d d = Ma sin θ + g cos θ − cos θ sin θ (a ) (a ) dt 2 dt 2 (( ) ( ) ) ( ( d( d( = Ma a θ˙ cos θ + g cos θ − −a θ˙ sin θ sin θ dt dt ⎞ ⎛ ⎛⎛⎛ ⎞ ⎞ ⎞ ( ( ( ( .. .. ⎜ ⎜⎜⎜ ⎟ g⎟ ⎟ ˙ 2 ˙ 2 cos θ ⎟ = Ma 2 ⎝⎝⎝θ, cos ⎠ sin θ ⎠ ,, θ, − ,θ ,,sin θ,⎠ + a ⎠ cos θ + ⎝θ, sin ,, θ, + ,θ ,, , = Ma
1
) ( .. g = Ma 2 θ + cos θ a
2
2
−M
) ( .. g a 2 .. θ = Ma 2 θ + cos θ 3 a
or
or ..
θ =− It follows that
3g cos θ. 4a
3
456
4 Motion About a Fixed Axis
) ( 3g dθ d (( (2 ) 3g d θ˙ − cos θ = − =2 (sin θ ), dt dt 4a 2a dt , ,, ,  ( (2 and hence θ˙ = − 3g sin θ + C. Now since θ˙ θ =α = 0, we have C = 2a hence
3g 2a
sin α, and
( (2 3g θ˙ = (sin α − sin θ )(∗). 2a Since ) ( ( d2x d2 d( −a θ˙ sin θ R = M 2 = M 2 (a cos θ ) = M dt dt , ,, dt , ( .. ) ( (2 = −Ma θ sin θ + θ˙ cos θ ) ) ) ( (( 3g 3g = −Ma − cos θ sin θ + (sin α − sin θ ) cos θ 4a 2a 3Mg = ((cos θ ) sin θ − 2(sin α − sin θ ) cos θ ) 4 3Mg = (3 sin θ − 2 sin α) cos θ, 4 we have 3Mg (3 sin θ − 2 sin α) cos θ. 4 ( ( It follows that R is zero, when θ = sin−1 23 sin α . Observe that for 0 < θ < ( ( sin−1 23 sin α , R is negative. Thus the upper end of the rod leaves the wall when ( ( θ = sin−1 23 sin α . Further, from (∗), we have R=
 θ˙ 
/
/ ) ( 2 3g g sin α. sin α − sin α = − =− θ =sin−1 ( 23 sin α ) 2a 3 2a /
Thus, as the rod leaves the wall, its angular velocity is instant horizontal velocity
g 2a
sin α. Also, at this
) ( ( (( d = −a θ˙ sin θ θ =sin−1 ( 2 sin α) = x˙θ =sin−1 ( 23 sin α) = (a cos θ ) 3 dt −1 2 θ =sin ( 3 sin α ) / / 2 1 g sin α · sin α = =a· 2ga sin3 α. 2a 3 3 (
4.2 Examples on Motion About a Fixed Axis (Part I)
457
Since ( ) ( d( d2 y d2 S = M 2 + Mg = M 2 (a sin θ ) + Mg = M a θ˙ cos θ + Mg dt dt dt , ,, , ) ( .. ( (2 = Ma θ cos θ − θ˙ sin θ + Mg ) ) ) ( (( 3g 3g = Ma − cos θ cos θ − (sin α − sin θ ) sin θ + Mg 4a 2a −3Mg = ((cos θ ) cos θ + 2(sin α − sin θ ) sin θ ) + Mg 4 −3Mg = (1 + (2 sin α − 3 sin θ ) sin θ ) + Mg 4 Mg = (1 + (−6 sin α + 9 sin θ ) sin θ ) 4 ) Mg ( = (3 sin θ − sin α)2 + cos2 θ 4
we have S=
( ( )) ( 2 Mg ( sin α . (3 sin θ − sin α)2 + cos2 θ 0 < θ < sin−1 4 3
The equations of motion now take a different form, because a discontinuity in motion has occurred. ( ( For 0 < θ < sin−1 23 sin α : M
d2x d2 y = 0, M 2 = S1 − Mg, 2 dt dt
and −M
a 2 .. θ= 3
( M
) ( ( d2 .. ) a2 ( 0 − θ = Mk 2 (π − θ ) = S1 (a cos θ ) 3 dt 2 , ,, ,
) ( 2 ( .. ) d (a sin θ ) = a M y +g · cos θ = a M + g · cos θ dt 2 ) ( ( d( a θ˙ cos θ + g cos θ = Ma dt (( .. ) g) ( (2 2 cos θ = Ma θ cos θ − θ˙ sin θ + a or −M
(( .. ) g) ( (2 a 2 .. cos θ θ = Ma 2 θ cos θ − θ˙ sin θ + 3 a
458
4 Motion About a Fixed Axis
or (( .. ) g) ( (2 1 .. cos θ θ = − θ cos θ − θ˙ sin θ + 3 a or (
) ( (2 g 1 .. cos θ + θ = sin θ cos θ θ˙ − cos θ. 3 a 2
It follows that ( ) ) ( ( (2 g 1 d (( (2 ) 2 cos θ + θ˙ = 2θ˙ sin θ cos θ θ˙ − cos θ 3 dt a or ) (( ) 1 ( (2 cos2 θ + θ˙ 3 ) ( (( (2 1 d (( (2 ) ( g + −2θ˙ sin θ cos θ θ˙ = −2θ˙ cos θ = cos2 θ + θ˙ 3 dt a , ,, ,
d dt
= −2
g d (sin θ ). a dt
Hence ( ) 1 ( (2 g cos2 θ + θ˙ = −2 sin θ + C. 3 a Now since /  g ˙θ  sin α, =− θ =sin−1 ( 23 sin α ) 2a we have ( ) 1 2g sin α 1 − sin2 α 3a 3 )) ( ( 1 g 4 sin α 1 − sin2 α = 3 3 2a (( )( / ( )2 ) )2 2 g g 2 1 = 1− − sin α sin α = −2 · sin α + C , + 3 3 2a a 3 , ,, ,
4.2 Examples on Motion About a Fixed Axis (Part I)
459
and hence C=
) ) ( ( 2g 2g 1 1 sin α 3 − sin2 α = sin α 1 − sin2 α . 3a 3 a 9
Thus ( ( ) ) 2g 1 1 ( (2 g cos2 θ + sin α 1 − sin2 α . θ˙ = −2 sin θ + 3 a a 9 It follows that ) ( ( ) 2g 1 4 (  (2 1 (  (2 g θ˙ θ =0 = −2 sin 0 + sin α 1 − sin2 α θ˙ θ =0 = cos2 0 + 3 3 a a 9 , ,, , ( ) 1 2g sin α 1 − sin2 α , = a 9 and hence  θ˙ 
/ θ =0
=
( ) 1 2 3g sin α 1 − sin α . 2a 9
Thus, the angular velocity of the rod when it reaches on the horizontal plane / =
( ) 1 3g sin α 1 − sin2 α . 2a 9
2
The equation M ddt x2 = 0 shows that in the second part of the motion, horizontal velocity is the constant ( ) / 1 x˙θ =sin−1 ( 23 sin α) = 2ga sin3 α . 3
4.2.9 Note In Sect. 4.2.8,for the first part of the motion, there are two normal reactions, namely, R and S. In Sect. 4.2.8, for the second part of the motion, there is one normal reaction, namely, S. In general, for each such contact there will be a normal reaction R, and corresponding to each R there will be a “geometric relation”expressing the fact that the
460
4 Motion About a Fixed Axis
velocity of point of contact on the moving body resolved along the normal to the fixed surface is 0. (For example see Sect. 4.2.16.)
4.2.10 Example Suppose that there are two smooth spheres. One of them is placed over the other in “unstable equilibrium” and the lower sphere resting on a smooth table. The system is slightly disturbed. Show that the spheres will separate when the line joining their centres makes an angle θ with the vertical given by the equation m cos3 θ − 3 cos θ + 2 = 0, m+M where M is the mass of the lower sphere, and m is the mass of upper sphere (Fig. 4.28). Sol: Here, the net force along horizontal direction is 0, so by Sect. 2.2.4, the acceleration of G is 0, and hence the horizontal velocity of G is constant. Now since the system is at rest initially, horizontal velocity of G is 0, and hence the vertical line through G is a line fixed in space. In our figure, we can take xaxis and yaxis as coordinate axes fixed in space. Now suppose that our system consists of only the lower sphere. In this case, the centre of mass is ( ( ( ) )) m C(−C G sin θ, a) = − (a + b) sin θ, a , m+M so ′
′
(fixed in space)
,
⏟ because
at time
=
Both spheres as one system
(fixed in space)
(
Fig. 4.28
+
)
=
+ ′
=
+ +
,
is the centre of mass of the system
,
(fixed in space)
′
4.2 Examples on Motion About a Fixed Axis (Part I)
( ( ) ) m d2 M 2 − (a + b) sin θ = −S sin θ, dt m+M
461
M
d2 (a) = R − Mg − S cos θ dt 2
or ) ( (2 Mm(a + b) ( .. θ cos θ − θ˙ sin θ = S sin θ, 0 = R − Mg − S cos θ. m+M Now suppose that our system consists of only the upper sphere. In this case, the centre of mass is ( (( ( ( , C C G sin θ, a + (a + b) cos θ = ,
) )) M (a + b) sin θ, a + (a + b) cos θ , m+M
so ) ) (( M d2 sin θ = S sin θ, + b) (a dt 2 m+M d2 m 2 (a + (a + b) cos θ ) = S cos θ − mg dt
m
or ( .. ) ) ( (2 ( (2 Mm(a + b) ( .. θ cos θ − θ˙ sin θ = S sin θ, m(a + b) − θ sin θ − θ˙ cos θ m+M = S cos θ − mg. Thus ) ( (2 Mm(a + b) ( .. θ cos θ − θ˙ sin θ = S sin θ (1) m+M ) ( .. ( (2 m(a + b) − θ sin θ − θ˙ cos θ = S cos θ − mg (2) 0 = R − Mg − S cos θ (3). From (1), (2), ) ( (2 Mm(a + b) ( .. θ cos θ − θ˙ sin θ m+M ) ) ( ( .. ( (2 = sin θ m(a + b) − θ sin θ − θ˙ cos θ + mg
cos θ
or ( .. ) ( (2 Mm(a + b) θ cos2 θ − θ˙ sin θ cos θ
462
4 Motion About a Fixed Axis
( .. ( ) ) ( (2 = (m + M) m(a + b) − θ sin2 θ − θ˙ sin θ cos θ + mg sin θ or (
( .. Mm(a + b) cos2 θ + (m + M)m(a + b) sin2 θ θ
( (2 + (−Mm(a + b) + (m + M)m(a + b)) sin θ cos θ θ˙ = (m + M)mg sin θ or ( ( .. ( (2 m(a + b) M + m sin2 θ θ +m 2 (a + b) sin θ cos θ θ˙ = (m + M)mg sin θ or (
( (2 ( .. (m + M)g sin θ. M + m sin2 θ θ +m sin θ cos θ θ˙ = a+b
˙ we get Multiplying by 2θ, (( (2 ) d (( M + m sin2 θ θ˙ dt ( ( ( (3 ( .. ( (m + M)g sin θ 2θ˙ = M + m sin2 θ 2θ˙ θ +m2 sin θ cos θ θ˙ = a+b , ,, , =
−2(m + M)g d (cos θ ) a+b dt
and hence (
(( (2 −2(m + M)g cos θ + C. M + m sin2 θ θ˙ = a+b
 Since θ˙ θ =0 = 0, we have C = (
2(m+M)g . a+b
Thus
(( (2 2(m + M)g M + m sin2 θ θ˙ = (1 − cos θ ). a+b
As the two spheres separate, we have S = 0, and hence from (1), ( (2 −m sin θ cos θ θ˙ +
(m+M)g a+b
sin θ
cos θ M + m sin θ 2(m+M)g (1 − cos θ ) ( (2 .. − a+b sin θ = θ cos θ − θ˙ sin θ = 0 2 , ,, , M + m sin θ 2
4.2 Examples on Motion About a Fixed Axis (Part I)
463
or ( (2 (m + M)g 2(m + M)g sin θ cos θ − −m sin θ cos2 θ θ˙ + (1 − cos θ ) sin θ = 0 a+b a+b or ( (2 −m(a + b) cos2 θ θ˙ + (m + M)g cos θ − 2(m + M)g(1 − cos θ ) = 0 or ( (2 −m(a + b) cos2 θ θ˙ − (m + M)g(2 − 3 cos θ ) = 0 or 2(m+M)g (1 − cos θ ) ( − (m + M)g(2 − 3 cos θ ) = 0 −m(a + b) cos2 θ (a+b M + m sin2 θ
or ( ( −m cos2 θ 2(m + M)g(1 − cos θ ) − M + m sin2 θ (m + M)g(2 − 3 cos θ ) = 0 or ( ( −m cos2 θ 2(1 − cos θ ) − M + m sin2 θ (2 − 3 cos θ ) = 0 or ( ( m cos2 θ 2(1 − cos θ ) + M + m − m cos2 θ (2 − 3 cos θ ) = 0 or ⎞
⎛
⎛
⎞
2 3 ⎠ 2 3 ⎠ ⎝ m ⎝2, cos ,, θ, − 2, cos ,, θ, + (M + m)(2 − 3 cos θ ) − m 2, cos ,, θ, − 3, cos ,, θ, = 0 1
2
1
or m cos3 θ + (M + m)(2 − 3 cos θ ) = 0 or m cos3 θ − 3 cos θ + 2 = 0. m+M
◼
2
464
4 Motion About a Fixed Axis
4.2.11 Note As in Sect. 4.2.10, if we have two moving bodies which are always in contact, there is a normal reaction R at the point of contact, and coresponding “geometric relation” expressing that the velocity of the point of contact of each body resolved along the “common normal” is the same. In general, for each reaction we get a “forced connection”, and a corresponding geometric equation. Thus the number of geometric equations is the same as the number of reaction forces.
4.2.12 Note In statics, there are three laws of friction: 1. the friction is a selfadjusting force, 2. friction tends to stop the relative motion of the point at which it acts, 3. friction cannot exceed a fixed multiple of the corresponding normal reaction, where μ is a quantity depending on the substances which are in contact. μ is called the coefficient of friction. In dynamics, the same laws hold. Although, it has been found that the value of μ slightly decreases as the relative velocity increases. The fundamental axiom of friction is that it will keep the point of contact at which it acts at relative rest if the amount of friction required is smaller than the “limiting friction”. It is important to note that ⎧ friction force required at the point of contact ⎪ ⎪ < coefficient of friction (for rolling) ⎨ normal force at the point of contact friction force required at the point of contact ⎪ ⎪ ⎩ coefficient of friction < (for sliding). normal force at the point of contact
In short, we can write {
< μ (for rolling) μ < FR (for sliding). F R
In any practical problem, we assume an unknown friction F in adirction opposite to what would be the direction of relative motion. Further we assume that the point of contact is at relative rest. This fact corresponds a geometric equation. Thus, to each friction there is a geometric equation expressing the fact that the point of contact is at relative rest. (For example, see Sect. 4.2.16).
4.2 Examples on Motion About a Fixed Axis (Part I)
465
4.2.13 Example A uniform solid cylinder is placed with its axis horizontal on a plane, whose inclination to the horizon is α. Show that the least coefficient of friction between it and the plane, so that it may roll and not slide, is 13 tan α. If the cylinder be hollow, and of small thickness, the least value is 21 tan α (Fig. 4.29). Sol: Let M be the mass of the cylinder, a be its radius. Let R be the normal reaction exerted by the plane on cylinder at the point of contact A. Let G be the centre of mass of the cylinder. Let μmin be the minimum coefficient of friction for rolling. We have to show that μmin =
1 tan α. 3
Here, equation of motions of centre of mass G are ..
x=
Mg sin α − F , M
R = Mg cos α.
For rolling, we have x = aθ. Hence ..
aθ =
Mg sin α − F . M
Further
Direction fixed in
plane At time
direction fixed in the body
Fig. 4.29
466
4 Motion About a Fixed Axis
( ) ) ( a 2 Mg sin α − F a 2 .. a(Mg sin α − F) a 2 + a 2 .. =M =M θ= M θ 2 2 Ma 2 4 ( 2 ( .. = Mk θ = net torque about horizontal axis through G , ,, , = 0 + 0 + Fa = Fa, so a(Mg sin α − F) = Fa 2 or Mg sin α = 3F or 1 μmin (Mg cos α) = μmin R = F = Mg sin α . 3 ,, , , Hence μmin (Mg cos α) =
1 Mg sin α 3
or μmin =
1 tan α. 3
◼
In the case of a hollow cylinder, we have Mk 2 = Ma 2 , and hence a(Mg sin α − F) = a F. It follows that 1 μmin (Mg cos α) = μmin R = F = Mg sin α 2 ,, , , and hence μmin =
1 tan α. 2
◼
4.2 Examples on Motion About a Fixed Axis (Part I)
467
4.2.14 Example A hollow cylinder rolls down a perfectly rough inclined plane in 1 min. Show that a solid cylinder will roll down the same distance in 52 s nearly, a hollow sphere in 55 s and a solid sphere in 50 s nearly. Sol: Suppose that α is the inclination of the plane to the horizon (Fig. 4.30). Let M be the mass of the cylinder, a be its radius. Let R be the normal reaction exerted by the plane on cylinder at the point of contact A. Let G be the centre of mass of the cylinder. Let μmin be the minimum coefficient of friction for rolling. Here, equation of motions of centre of mass G are ..
x=
Mg sin α − F , M
R = Mg cos α.
For rolling, we have x = aθ. Hence ..
aθ =
Mg sin α − F . M
Further ) ( Mg sin α − F k2 (Mg sin α − F) = Mk 2 a Ma ( 2 ( .. = Mk θ = net torque about horizontal axis through G , ,, , = 0 + 0 + Fa
Direction fixed in
plane At time
direction fixed in the body
Fig. 4.30
468
4 Motion About a Fixed Axis
so k2 (Mg sin α − F) = Fa a or ( Mg sin α = F
) a2 + 1 . k2
Next since ..
acc. = x = ,
Mg sin α − F ,, M , Mg sin α
a2 F 2 +1 = g sin α − k = g sin α − M M k2 a2 = g sin α − 2 g sin α = g sin α, k + a2 k2 + a2
which is a constant. Thus acc ⎧ a2 ⎧ ⎪ ⎪ g sin α for hollow cylinder ⎪ 1 ⎪ 2 ⎪ 2 ⎪ ⎪ a + a g sin α for hollow cylinder ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 ⎪ ⎪ a ⎪ ⎪ ⎪ ⎪ 2 g sin α for solid cylinder ⎪ ⎪ ⎪ ⎪ ⎨ 1 a2 + a2 ⎨ g sin α for solid cylinder 2 = = 3 3 a2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ g sin α for hollow sphere g sin α for hollow sphere ⎪ 2 2 ⎪ 2 ⎪ ⎪ 5 a +a ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 5 ⎪ ⎪ ⎪ ⎩ g sin α for solid sphere a2 ⎪ ⎪ g sin α for solid sphere 7 ⎩ 2 2 a + a2 5 Here (acc)(time taken)2 = const, we have (
) ) ( 1 2 g sin α (60 s)2 = g sin α (reqd time for solid cylinder)2 , 2 3
and hence / (reqd time for solid cylinder) =
1 3 × (60 s) ≈ 52 s. 2 2
4.2 Examples on Motion About a Fixed Axis (Part I)
469
Similarly / (reqd time for hollow sphere) = / (reqd time for solid sphere) =
1 5 × (60 s) ≈ 55 s, 2 3
1 7 × (60 s) ≈ 50 s. 2 5
4.2.15 Note Suppose that in place of a rigid body we have a lamina. Suppose that the lamina is compelled to move in the x yplane fixed in space. It is clear that the position of the lamina is determined, if we know the position of a fixed point of the lamina together with the position of a fixed direction of the lamina. For the purpose of fixed point, we can take the centre of mass of the lamina. In order to determine the fixed direction of the lamina, we can take the inclination of the fixed direction of the lamina with a fixed direction in the x yplane. These quantities (say x, y, θ ) are called the coordinates of the lamina. If we determine them in terms of the time t, we have completely determined the motion of the body. By Sect. 2.2.4, the motion of centre of mass G(x, y) is given by M
Σ Σ d2x d2 y = X, = Y. 2 2 dt dt
Suppose that x , (= x − x), y , (= y − y) are the coordinates of any point of the lamina relative to G. The kinetic energy of the lamina (( ) (( ) ( )2 ) ( )2 ) Σ1 dx 2 dx 2 dy dy 1Σ m = + m + = 2 dt dt 2 dt dt body body ⎞ ⎛( ( ( )2 ( ( , ( )2 d y + y¯ 1 Σ ⎝ d x , + x¯ ⎠ m + = 2 dt dt body (( ) ) ) ( , dx, 1Σ dy d x¯ 2 d y¯ 2 = m + + + 2 dt dt dt dt body (( ) ) ) ( , )2 ( )2 ( dx, 2 1Σ dy d x¯ 2 d x , d x¯ d y¯ dy , d y¯ = + m + +2 + +2 2 dt dt dt dt dt dt dt dt body
470
4 Motion About a Fixed Axis
(( (( ) ) ( , )2 ) ( )2 ) dx, 2 dx 2 dy dy 1Σ 1Σ = m + m + + 2 dt dt 2 dt dt body body ) ) ( ( , , Σ Σ dx dx dy d y + m m + dt dt dt dt body body ⎞ ⎛ (( (( ) ) ( , )2 ( )2 ) Σ ) dy dy dx, 2 dx 2 1Σ 1 ⎝ = m + + m⎠ + 2 dt dt 2 dt dt body body Σ ( d x , d x ) Σ ( dy , d y ) + m m + dt dt dt dt body body (( (( ) ) ( , )2 ( )2 ) ) dx, 2 dy dy dx 2 1Σ 1 = m + + + M 2 dt dt 2 dt dt body Σ ( d x , d x ) Σ ( dy , d y ) + m m + dt dt dt dt body body (( (( ) ) ( , )2 ( )2 ) ) dx, 2 dy dy dx 2 1Σ 1 = m + + + M 2 dt dt 2 dt dt body ⎞ ⎛ Σ ( dy , d y ) d x d ⎝Σ + mx , ⎠ + m dt dt dt dt body body (( ) (( ) ) ( , )2 ( )2 ) dx 2 dx, 2 dy dy 1Σ 1 m + + = + M 2 dt dt 2 dt dt body ) ( Σ dy , d y dx d m + (M(0)) + dt dt dt dt body (( ) (( ) ) ( , )2 ) ( )2 ) Σ ( , dx 2 dx, 2 dy dy dy d y 1Σ 1 m + + m = + M + 2 dt dt 2 dt dt dt dt body body (( (( ) ) ) ( ( ( ,) ) ) ) dy , 2 dy 2 dy dx, 2 dx 2 1Σ 1 dy Σ = m + + m + M + 2 dt dt 2 dt dt dt dt body body ⎞ ⎛ (( ) (( ) ( , )2 ) ( )2 ) dx 2 dx, 2 1Σ dy dy 1 d y d ⎝Σ = m + + my , ⎠ + M + 2 dt dt 2 dt dt dt dt body body (( ) (( ) ) ) ) ) ( ( 2 2 2 2 , , dx dx 1Σ dy dy 1 dy d = m + + + M + (M(0)) 2 dt dt 2 dt dt dt dt body (( ) (( ) ( , )2 ) ( )2 ) dx 2 dx, 2 dy dy 1Σ 1 m + + = + M 2 dt dt 2 dt dt body
4.2 Examples on Motion About a Fixed Axis (Part I)
471
(( ) ( , )2 ) dx, 2 dy 1Σ = m + 2 dt dt body
+ K.E. of a particle of mass M placed at G and moving with it 1Σ = m (velocity relative to G of m)2 2 body
+ K.E. of a particle of mass M placed at G and moving with it,
and hence, kinetic energy of the lamina =
1Σ m (velocity relative to G of m)2 2 body
+ K.E. of a particle of mass M placed at G and moving with it (∗). −→ −→ Suppose that G B has a fixed direction in the x yplane. Let G A be a fixed direction in the lamina. Suppose that ∠BG A = θ. Suppose that Gm is the line joining mass m and G. Clearly, the position of line Gm varies with time t. Suppose that ∠BGm = φ. Since A,.. G, m all lie in the rigid lamina, ∠AGm(= φ − θ ) is a constant, and hence .. φ˙ = θ˙ , φ = θ . Put Gm = r. Now since G, m lie in the rigid lamina, r is a constant, and hence the velocity relative to G of m is ( ) dφ dθ d(∠BGm) =r =r r (A) dt dt dt in the direction perpendicular to Gm. It follows from (∗) that kinetic energy of the lamina =
) ( 1Σ dθ 2 m r 2 body dt
+ K.E. of a particle of mass M placed at G and moving with it ) ( d(∠BG A) 2 1Σ m r = 2 body dt + K.E. of a particle of mass M placed at G and moving with it ( ( )2 ) 1Σ 2 d(∠BG A) = mr 2 body dt + K.E. of a particle of mass M placed at G and moving with it ( ) 1 d(∠BG A) 2 Σ 2 mr = 2 dt
472
4 Motion About a Fixed Axis
+ K.E. of a particle of mass M placed at G and moving with it ( ) 1 d(∠BG A) 2 = (M.I. of the lamina about Gz − axis) 2 dt + K.E. of a particle of mass M placed at G and moving with it. Thus, the kinetic energy of the lamina ) ( d(∠BG A) 2 1 (M.I. of the lamina about Gz − axis) 2 dt + K.E. of a particle of mass M placed at G and moving with it. =
Next, the moment of momentum about the origin O of the lamina ( () ( ( ) Σ ( Σ ( dy ( , ( d y, + y ( d x, + x ( , dx = −y − y +y = m x m x +x dt dt dt dt body body ) )) ( ( ( Σ ( ( dy , ( dx, ( dy dx − y, + y = m x, + x + + dt dt dt dt body ) ( ) (( Σ dy , dx, dy dx + x, − y, − y, m x, = dt dt dt dt body ) ( )) ( dy , dx, dy dx + x −y −y + x dt dt dt dt ) ,) Σ ( dy Σ ( dy , d x , , , , dx + −y −y = m x m x dt dt dt dt body body ) ) ( ( Σ Σ dy , dx, dy dx + −y −y + m x m x dt dt dt dt body body ⎛ ⎞ ,) Σ ( dy , Σ Σ d d x y x d − y, +⎝ = m x, mx , − my , ⎠ dt dt dt dt body body body ⎛ ⎞ ⎞⎞ ⎞ ⎛ ⎛ ⎛ Σ Σ Σ dy , Σ dx, y x d d ⎠ + ⎝x ⎝ −y + ⎝x m m m⎠ − y ⎝ m ⎠⎠ dt dt dt dt body body body body ⎛ ⎞ ) ) ( ( Σ dx dy , dx, ⎠ dy =⎝ − y, ·0− ·0 m x, + dt dt dt dt body ⎛ ⎞ ⎞⎞ ⎛ ⎛ ) ( Σ Σ d d dy dx ,⎠ , ⎠⎠ ⎝ ⎝ ⎝ M−y M + x my − y mx + x dt body dt body dt dt
4.2 Examples on Motion About a Fixed Axis (Part I)
⎛ ⎞ ( ) ( ) ,) Σ ( dy , d x d d dy dx , , ⎝ ⎠ = m x + x (0) − y (0) + M x −y −y dt dt dt dt dt dt body ⎛ ⎞ ) ( ,) Σ ( dy , dy dx , , dx ⎝ ⎠ −y −y = m x +M x dt dt dt dt body =
) Σ ( dy , dx, − y, m x, dt dt body
+ moment of momentum about O of a particle of mass M placed at G and moving with it Σ = (moment of momentum about centre of mass G of particle m) body
+ moment of momentum about O of a particle of mass M placed at G and moving with it ( ( ))) Σ( dθ = (dist. of m from G) m r dt body + moment of omentum about O of a particle of mass M placed at G and moving with it, (A) Σ ( dθ ) = + moment of momentum about O of a particle of r r dt body mass M placed at G and moving with it ( ) Σ dθ + moment of momentum about O of a particle of mr 2 = dt body mass M placed at G and moving with it ( ) Σ 2 d(∠BG A) + moment of momentum about O of a particle of mr = dt body mass M placed at G and moving with it d(∠BG A) Σ 2 = mr + moment of momentum about O of a particle of dt body mass M placed at G and moving with it d(∠BG A) = (M.I. of the lamina about Gzaxis) dt + moment of momentum about O of a particle of mass M placed at G and moving with it.
473
474
4 Motion About a Fixed Axis
Thus, the moment of momentum about the origin O of the lamina d(∠BG A) dt + moment of momentum about O of a particle of mass M placed at G(x, y) and moving with it. (∗∗)
= (M.I. of the lamina about Gzaxis)
Observe that, the moment of the impressed forces about the origin O of the lamina Σ (xY − y X ) ) Σ ( d2 y d2x m x 2 − y 2 , (see Sect. 2.2.2) = dt dt body ( ) ( )) Σ( ( ( ) ( ))) ( Σ d dy d dx dy dx d x − y = x m −y m = m dt dt dt dt dt dt dt body body ⎞ ⎛ ( ( ) ( )) d ⎝Σ dy dx ⎠ = x m −y m dt body dt dt
=
d (the moment of momentum about O of the lamina) dt ( ) d d(∠BG A) = (M. I. of the lamina about Gzaxis) dt dt d + (moment of momentum about O of a particle of mass M placed at G and dt moving with it) (by (∗∗)) ( ) d d(∠BG A) = (M.I. of the lamina about Gzaxis) dt dt ( ( ) )) ( d dy dx −y M + x M dt dt dt ( ) d d(∠BG A) = (M.I. of the lamina about Gzaxis) dt dt ( ( ))) ( ( ( )) d d dy dx − +M x y dt dt dt dt ( ) d d(∠BG A) = (M.I. of the lamina about Gzaxis) dt dt ( 2 2 ) d y d x +M x 2 −y 2 , dt dt
=
and hence, the moment of the impressed forces about the origin O of the lamina
4.2 Examples on Motion About a Fixed Axis (Part I)
475
) ( 2 ( .. d y d2x = Mk θ +M x 2 − y 2 . dt dt (
2
Again observe that, the moment of the impressed forces about the point (x0 , y0 ) of the lamina ( Σ Σ Σ Σ ) = Y − y0 X ((x − x0 )Y − (y − y0 )X ) = (xY − y X ) − x0 .. . (as above) ) ( Σ ( 2 Σ ) ( 2 ( .. d y d2x Y − y0 X = Mk θ +M x 2 − y 2 − x0 dt dt ) ( ( ) ( )) ( 2 ( 2 ( .. d2 y d2x d y d2x = Mk θ +M x 2 − y 2 − x0 M 2 − y0 M 2 dt dt dt dt ) ( ( 2 ( .. d2 y d2x = Mk θ +M (x − x0 ) 2 − (y − y0 ) 2 , dt dt hence, (the moment of the impressed forces about the point (x0 , y0 ) of the lamina) ) ( ( ( .. d2 y d2x = Mk 2 θ +M (x − x0 ) 2 − (y − y0 ) 2 . dt dt Conclusion For 2dimensional motion, the kinetic energy =
1 ( 2 (( (2 Mk θ˙ + K.E. of a particle of mass M placed at G and moving with it, 2
and (the moment of the impressed forces about the point (x0 , y0 ) ) ( ( ( .. d2 y d2x = Mk 2 θ +M (x − x0 ) 2 − (y − y0 ) 2 , dt dt where Mk 2 is the M.I. of the lamina about the Gzaxis, and θ denotes the inclination of a fixed direction in the lamina with a fixed direction in the x yplane.
4.2.16 Example Suppose that there is a uniform solid sphere, of mass M and radius a. Suppose that it starts from rest rolling down an inclined plane. Suppose that the plane is rough
476
4 Motion About a Fixed Axis
enough to prevent sliding. Suppose that α is the inclination of the plane with the horizon. Verify the “energy equation”, that is, K.E. of the sphere = work done by all forces. Sol: Let R be the normal reaction exerted by the plane on cylinder at the point of contact A. Let G be the centre of mass of the cylinder. Let F be the friction force necessary for rolling. In order to decide the direction of friction force on sphere at the point of contact A, we first imagine that F is absent. Now it is easy to see that the velocity of G has a component along positive x direction. Hence by the law Sect. 4.2.12(2), the direction of friction F on sphere is along the negative x direction. This is what is shown in Fig. 4.31. If we imagine A being attached to the sphere, the velocity of A is 0. This corresponds a “geometric condition”. Here, equation of motions of centre of mass G are ..
x=
Mg sin α − F , R = Mg cos α. M
For rolling, we have x = aθ. Hence ..
aθ =
Mg sin α − F . M
Further ) ( k2 2 Mg sin α − F (Mg sin α − F) = Mk a Ma
Normal reaction
At time Position of point of contact at time = 0.
direction fixed in the body path of
Fig. 4.31
attached to the sphere
Point of contact Common normal
4.2 Examples on Motion About a Fixed Axis (Part I)
477
( ( .. = Mk 2 θ = net torque about horizontal axis through G , ,, , = 0 + 0 + Fa so k2 (Mg sin α − F) = Fa a or (
) a2 Mg sin α = F 2 + 1 . k Next since d 2 (aθ ) Mg sin α − F F .. = g sin α − aθ = =x= = g sin α − 2 dt M , ,, M , ..
= g sin α −
Mg sin α a2 +1 k2
M
a2 k2 g sin α = g sin α, k2 + a2 k2 + a2
we have ..
aθ =
a2 g sin α = k2 + a2
a2 2a 2 5
+
a2
g sin α =
5 g sin α 7
or ..
θ=
5g sin α. 7a
It follows that dθ d (( (2 ) θ˙ =2 dt dt ,
(
) 10g dθ 5g sin α = sin α , 7a 7a dt ,, ,
and hence ( (2 10g θ˙ = (sin α)θ + C. 7a  Since the body starts from rest, we have θ˙ θ =0 = 0, and hence C = 0. Thus
478
4 Motion About a Fixed Axis
( (2 10g θ˙ = (sin α)θ. 7a By Sect. 4.2.14, (( (2 1( K.E. = Mk 2 θ˙ + K.E. of a particle of mass M placed at G and moving with it 2 , ,, , ( )( ) 10g 2a 2 1 M (sin α)θ 2 5 7a + K.E. of a particle of mass M placed at G and moving with it 2a Mg 1 2a Mg ˙ 2= = (sin α)θ + M(x) (sin α)θ 7 2 7 ( ) 10g 2a Mg 1 1 ( (2 + M a θ˙ = (sin α)θ + Ma 2 (sin α)θ 2 7 2 7a
=
and hence, change in K.E. = Mg(sin α)(aθ ). Observe that, the work done by gravity = (Mg)(aθ ) cos
(π 2
) −α .
By Sect. 4.2.9, velocity of point of contact on the moving body resolved along the normal to the fixed surface is 0. This is the “geometric relation” corresponding {to R. Thus, the work ( {done by R is 0. Since( friction F acts at A for zero duration, F · dr = zero duration F · vdt = 0 , and hence friction does no work in zero duration rolling. Thus, ) (π − α + 0 + 0 = Mg(sin α)(aθ ) the work done by all forces = (Mg)(aθ ) cos 2 , ,, , = change in K.E. Verified.
4.2.17 Example A cylinder rolls down a smooth plane whose inclination to the horizon is α, unwrapping, as it goes, a fine string fixed to the highest point of the plane. Find its acceleration and tension of the string (Fig. 4.32).
4.2 Examples on Motion About a Fixed Axis (Part I)
479 Direction fixed in
plane At time
direction fixed in the body
Fig. 4.32
Sol: Let M be the mass of the cylinder, a be its radius. Let R be the normal reaction exerted by the plane on cylinder at the point of contact A. Let G be the centre of mass of the cylinder. Let T be the tension in the string. Here, equation of motions of centre of mass G are ..
x=
Mg sin α − T , M
R = Mg cos α.
For rolling, we have x = aθ. Hence ..
aθ =
Mg sin α − T . M
Further ) ( Mg sin α − T k2 (Mg sin α − T ) = Mk 2 a Ma ( 2 ( .. = Mk θ = net torque about horizontal axis through G , ,, , = 0 + 0 + Ta so k2 (Mg sin α − T ) = T a a or
480
4 Motion About a Fixed Axis
(
) a2 Mg sin α = T 2 + 1 . k It follows that T =
k 2 Mg sin α . k2 + a2
Next since ..
acc. = x = ,
Mg sin α − T T = g sin α − M M ,, ,
= g sin α −
Mg sin α a2 +1 k2
M a2 g sin α, = 2 k + a2
= g sin α −
k2
k2 g sin α + a2
which is a constant. Thus acc. =
a2 g sin α. k2 + a2
4.2.18 Example One end of a thread, which is wound onto a reel, is fixed, and the reel falls in a vertical line. Its axis being horizontal and the unwound part being vertical. If the reel be a solid cylinder of radius a and weight W, show that acceleration of the reel is 2 g, and the tension of the thread is W3 . 3 Sol: This problem is a particular case of Sect. 4.2.17, with α = Ma 2 . 2
π , 2
Mg = W, Mk 2 =
4.2.19 Example A circular cylinder, whose centre of mass is at a distance c from its axis, rolls on a horizontal plane. If it be just started from a position of unstable equilibrium, show that the normal reaction of the plane when the centre of mass is in its lowest position 4c2 times its weight, where k is the radius of gyration about an axis is 1 + (a−c) 2 +k 2 through its centre of mass (Fig. 4.33).
4.2 Examples on Motion About a Fixed Axis (Part I)
481 Direction fixed in plane
Direction fixed in body
≡ ( )
Fig. 4.33
Sol: Here, G(aθ + c sin θ, a + c cos θ ), and since the motion is rolling, the instantaneous axis of rotation passes through the point of contact A(aθ, 0). Now the K.E. at time t =
( (2 1 (M.I. about the instantaneous axis of rotation) θ˙ 2
(( (2 ( ((( (2 1( 2 1( Mk + M(AG)2 θ˙ = Mk 2 + M a 2 + c2 − 2ac cos(π − θ ) θ˙ 2 2 (( (2 1 ( = M k 2 + a 2 + c2 + 2ac cos θ θ˙ . 2 It follows that 2Mgc = 0 + 0 + (Mg)(2c) = work done by normal reaction + work done by friction + work done by gravity = (work done by all forces) = K.E.θ=π − K.E.θ =0 , ,, , ((  (2 1 ( 2 M k + a 2 + c2 + 2ac cos π θ˙ θ =π 2 ((  (2 1 ( − M k 2 + a 2 + c2 + 2ac cos 0 θ˙ θ =0 2 ((  (2 1 ( ( 1 ( 2 = M k + a 2 + c2 − 2ac θ˙ θ =π − M k 2 + a 2 + c2 + 2ac (0)2 2 2 ((  (2 ((  (2 1 ( 1 ( = M k 2 + a 2 + c2 − 2ac θ˙ θ =π = M k 2 + (a − c)2 θ˙ θ =π , 2 2 =
and hence ( ((  (2 4gc = k 2 + (a − c)2 θ˙ θ =π .
482
4 Motion About a Fixed Axis
Here, equation of motions of centre of mass G(aθ + c sin θ, a + c cos θ ) are ( ( (2 ( (2 ) .. .. .. (a + c cos θ ) θ −c sin θ θ˙ = a θ +c cos θ θ − sin θ θ˙ =
−F d 2 (aθ + c sin θ ) = , 2 dt M, , ,,
( ( (2 ) d 2 (a + c cos θ ) .. R − Mg = −c sin θ θ + cos θ θ˙ = . (∗) 2 dt M , , ,, From (∗), we have  ( (  (2 ..  c θ˙ θ =π = −c sin π θ 
θ =π
(  (2 ) Rθ =π − Mg , = + cos π θ˙ θ =π M
and hence ) ( ( (  (2 ) 4gc Rθ =π = M g + c θ˙ θ=π = M g+c 2 k + (a − c)2 , ,, , ( = M g+
) 4gc2 . k 2 + (a − c)2
◼
4.2.20 Example Two equal cylinders, each of mass m, are bound together by an elastic string, whose tension is T , and roll with their axes horizontal down a rough plane of inclination α. Show that their acceleration is ( ) 2μT 2 g sin α 1 − , 3 mg sin α where μ is the coefficient of friction between the cylinders (Fig. 4.34). Sol: Here equations of motion of cylinder 2 are d2x 2T − S + (mg) sin α − F2 , = 2 dt ) m ( 1 2 .. ma θ = −(μS)a + a F2 , (mg) cos α = R2 + μS, 2 ..
aθ =
4.2 Examples on Motion About a Fixed Axis (Part I)
483
≡ ( )
2 Direction fixed in space 1
2
1
Direction fixed in body
2
1
=
Fig. 4.34
that is ..
ma θ = 2T − S + mg sin α − F2 , mg cos α = R2 + μS,
.. 1 ma θ = −μS + F2 . 2
Here equations of motion of cylinder 1 is −2T + S + (mg) sin α − F1 d2x = , 2 dt (m ) 1 2 .. ma θ = F1 a − (μS)a, (mg) cos α + μS = R1 , 2 ..
aθ =
that is ..
ma θ = −2T + S + mg sin α − F1 , mg cos α + μS = R1 ,
.. 1 ma θ = F1 − μS. 2
We have to show that ( ) 2μT 2 a θ = g sin α 1 − . 3 mg sin α ..
Since .. −2T + S + (mg) sin α − F1 2T − S + (mg) sin α − F2 =aθ = , m m
we have F1 − F2 = 2S − 4T . Since
484
4 Motion About a Fixed Axis
F1 − μS =
.. 1 ma θ = −μS + F2 , 2
we have F1 = F2 . Now since F1 − F2 = 2S − 4T, we have S = 2T . Next since 1 1 (mg sin α − F1 ) = (−2T + S + mg sin α − F1 ) 2 2 .. 1 = ma θ = F1 − μS = F1 − μ(2T ), ,2 ,, , we have 1 (mg sin α − F1 ) = F1 − μ(2T ) 2 or F1 =
mg sin α + 4μT . 3
Now since ..
ma θ = −2T + S + mg sin α − F1 = mg sin α − F1 , ,, , mg sin α + 4μT 3 2mg sin α − 4μT = 3
= mg sin α −
we have ..
aθ =
( ) 2μT 2 g sin α 1 − . 3 mg sin α
◼
4.3 Examples on Motion About a Fixed Axis (Part II) 4.3.1 Example A uniform rod is held in a vertical position with one end resting upon a horizontal table and, when released, rotates about the end in contact with the table. Show that, when it is inclined at an angle of 30◦ to the horizontal, the force of friction that must be exerted to prevent slipping is approximately 0.32 of the weight (Fig. 4.35). Sol:
4.3 Examples on Motion About a Fixed Axis (Part II)
485 Direction fixed in the body
≡ ( ) acc =
2 acc = ( ̇ )
̈
Fixed axis of rotation
Fig. 4.35
Here, the two equations of motion of centre of mass G are ( (2 (mg) cos θ − R cos θ − F sin θ a θ˙ = , m .. (mg) sin θ − R sin θ + F cos θ . aθ = m Case I when the horizontal plane is perfectly rough. In this case the axis of rotation is fixed, and it passes through the lower end O of rod. Hence (
) .. 1 2 ma + ma 2 θ = (mg)(a sin θ ) 3
or ..
θ=
3g sin θ. 4a
It follows that dθ d ( (2 θ˙ = 2 dt dt
(
3g sin θ 4a
)
and hence ( (2 3g cos θ. θ˙ = C − 2a  Now since θ˙ θ = π −α = 0, we have C = 2
3g 2a
sin α. This shows that
486
4 Motion About a Fixed Axis
( (2 3g θ˙ = (sin α − cos θ ). 2a Hence  θ˙ 
θ = π2
/ =
/
π) 3g ( sin α − cos = 2a 2
3g sin α. 2a
Case II when the horizontal plane is perfectly smooth. In this case the axis of rotation is never fixed. Hence the three equations of motion are ( (2 .. (mg) sin θ − R sin θ (mg) cos θ − R cos θ , aθ = , a θ˙ = m m
(
) 1 2 .. ma θ = R(a sin θ ) 3
From the last two equations, (mg) sin θ − aθ = , ,, m ..
(1 3
( .. ma θ
1 .. = g sin θ − a θ 3 ,
or ..
θ=
3g sin θ. 4a
It follows that d ( (2 dθ θ˙ = 2 dt dt
(
3g sin θ 4a
)
and hence ( (2 3g cos θ. θ˙ = C − 2a  Now since θ˙ θ= π −α = 0, we have C = 2
3g 2a
sin α. This shows that
( (2 3g θ˙ = (sin α − cos θ ). 2a Hence  θ˙ θ = π = 2
/
π) 3g ( sin α − cos = 2a 2
/
3g sin α. 2a
◼
4.3 Examples on Motion About a Fixed Axis (Part II)
487
4.3.2 Example A rough uniform rod, of length 2a, is placed on a rough table at right angles to its edge. If the centre of gravity be initially at a distance b beyond the edge, show that the rod will begin to slide when it has turned through an angle tan−1
μa 2 , a 2 + 9b2
where μ is the coefficient of friction (Fig. 4.36). Sol: Here, the equations of motion of G are ( ( ( ) ( ..) 2 m b θ˙ = −(mg) sin θ + F, m b θ = (mg) cos θ − R. Also ( 2 ) .. a a 2 + 3b2 .. 2 m θ = m + mb θ = (mg)(b cos θ ) 3 3 ,, , , or ..
θ=
3bg cos θ. a 2 + 3b2
It follows that d (( (2 ) dθ θ˙ =2 dt dt
(
) 3bg cos θ , a 2 + 3b2
≡ ( )
edge of table
Fig. 4.36
488
4 Motion About a Fixed Axis
and hence ( (2 θ˙ = C +
a2
6bg sin θ. + 3b2
 Now since θ˙ θ=0 = 0, we have C = 0. This shows that ( (2 θ˙ =
a2
6bg sin θ. + 3b2
Now since ( ) ( ( ( ) 6b2 mg 6bg 2 = −(mg) sin θ + F sin θ = mb sin θ = m b θ˙ 2 2 2 2 a + 3b a + 3b , ,, , we have 6b2 mg sin θ = −(mg) sin θ + F. a 2 + 3b2 Next since ( ) ( ..) 3b2 mg 3bg cos θ = mb cos θ = m b θ = (mg) cos θ − R a 2 + 3b2 a 2 + 3b2 , ,, , we have 3b2 mg cos θ = (mg) cos θ − R. a 2 + 3b2 At the verge of sliding (mg) sin θ +
( ) 6b2 mg 3b2 mg sin θ = F = μR = μ cos θ , cos θ − (mg) , ,, , a 2 + 3b2 a 2 + 3b2
and hence ( 1+
6b2 2 a + 3b2
)
( sin θ = μ 1 −
3b2 2 a + 3b2
or θ = tan−1
μa 2 . a 2 + 9b2
◼
) cos θ
4.3 Examples on Motion About a Fixed Axis (Part II)
489
4.3.3 Example A uniform rod is held at an inclination α to the horizon with one end in contact with a horizontal table whose coefficient of friction is μ. If it be then released show that sin α cos α (Fig. 4.37). it will commence to slide if μ < 31+3 sin2 α Sol: Here, equations of motion of centre of mass G are ( ( ( ) 2 = (mg) cos θ − R cos θ − F sin θ, m a θ˙ ( ..) m a θ = (mg) sin θ − R sin θ + F cos θ. Also 4 2 .. ma θ = 3
(
) .. 1 2 2 ma + ma θ = (mg)(a sin θ ) 3 , ,, ,
or ..
θ=
3g sin θ. 4a
It follows that dθ d (( (2 ) θ˙ =2 dt dt
(
) 3g sin θ , 4a
and hence Fig. 4.37
,2
≡ ( )
Fixed axis of rotation
490
4 Motion About a Fixed Axis
( (2 3g cos θ. θ˙ = C − 2a  Now since θ˙ θ= π −α = 0, we have C = 2
3g 2a
sin α, and hence
( (2 3g θ˙ = (sin α − cos θ ). 2a Next since ( ) ) ( 3mg 3g 3mg sin α − cos θ = ma (sin α − cos θ ) 2 2 2a ( ( ( ) 2 = (mg) cos θ − R cos θ − F sin θ , = m a θ˙ , ,, , we have 3mg 3mg sin α − cos θ = (mg) cos θ − R cos θ − F sin θ 2 2 or 5mg mg 3mg sin α − cos θ = −R cos θ − F sin θ (3 sin α − 5 cos θ ) = 2 2 ,2 ,, , or R cos θ + F sin θ =
mg (5 cos θ − 3 sin α). 2
Next since ( ) ( ..) 3g 3mg sin θ = ma sin θ = m a θ = (mg) sin θ − R sin θ + F cos θ , 4 4a , ,, , we have R sin θ − F cos θ =
mg sin θ. 4
It follows that mg sin θ R sin θ − F cos θ sin θ 4 , = mg = R cos θ + F sin θ 2(5 cos θ − 3 sin α) cos θ − 3 sin α) (5 , ,,2 ,
4.3 Examples on Motion About a Fixed Axis (Part II)
491
and hence sin θ R sin θ − F cos θ = . R cos θ + F sin θ 2(5 cos θ − 3 sin α) Now since Fθ = π2 −α = μmax Rθ = π2 −α , we have cos α − μmax sin α cos α = sin α + μmax cos α 2(5 sin α − 3 sin α) or (cos α − μmax sin α)4 sin α = cos α(sin α + μmax cos α) or ( ( ( ( 3 sin α cos α = μmax cos2 α + 4 sin2 α = μmax 1 + 3 sin2 α , ,, , or μmax =
3 sin α cos α 1 + 3 sin2 α
◼
4.3.4 Example The lower end of a uniform rod, inclined initially at an angle α to the horizon, is placed on a smooth horizontal table. A horizontal force is applied to its lower end of such a magnitude that the rod rotates in a vertical plane with constant angular velocity ω. Show that when the rod is inclined at an angle φ to the horizon, the magnitude of the force is mg cot φ − maω2 cos φ, where m is the mass of the rod (Fig. 4.38). Sol: Here, the equations of motion of centre of mass G are −maω2 sin θ = maω(−ω sin θ ) = m
and
d d 2 (a sin θ ) = F, (aω cos θ ) = m 2 dt , dt ,, ,
492
4 Motion About a Fixed Axis
Fig. 4.38 ,2
≡ ( )
−maω2 cos θ = −maω(ω cos θ ) = m
d d 2 (a cos θ ) = R − mg . (−aω sin θ ) = m 2 dt , dt ,, ,
Thus F = −maω2 sin θ, R − mg = −maω2 cos θ. Also ( 0=
) ( ) 1 2 1 2 .. ma 0 = ma θ = R(a sin θ ) − F(a cos θ ), 3 3 ,, , ,
so ( ( mg tan θ − maω2 sin θ = mg − maω2 cos θ tan θ = ,R tan,, θ = F,, and hence F = mg tan θ − maω2 sin θ. Thus Fθ = π2 −φ = mg cot φ − maω2 cos φ.
◼
4.3.5 Example A heavy rod, of length 2a, is placed in a vertical plane with its ends in contact with a rough vertical wall and an equally rough horizontal plane, the coefficient of friction
4.3 Examples on Motion About a Fixed Axis (Part II)
493
being tan ε. Show that it will begin to slip down if its initial inclination to the vertical is greater than 2ε (Fig. 4.39). Sol: Here equations of motion of G(a sin θ, a cos θ ) are ( ( (2 ) .. d 2 (a sin θ ) ma cos θ θ − sin θ θ˙ = −μR1 + R2 , =m dt 2 ,, , , ( ( (2 ) .. d 2 (a cos θ ) −ma sin θ θ + cos θ θ˙ = μR2 + R1 − mg =m dt 2 , ,, , Also ( ) 1 2 .. ma θ = −(μR1 )(a cos θ ) + R1 (a sin θ ) − R2 (a cos θ ) − (μR2 )(a sin θ ) 3 or (
) .. 1 ma θ = −(μR1 + R2 ) cos θ + (R1 − μR2 ) sin θ. 3
At( B(0, 2a cos θ ), )the velocity is −2a θ˙ sin θ and acceleration is ( (2 .. −2a θ sin θ + θ˙ cos θ , and at the verge of slipping, the velocity and accelera..
tion both are zero, so at the verge of slipping, θ˙ = 0 and θ = 0. Hence at the verge of slipping, ( ( ma cos θ0 − sin θ (0)2 = −μR1 + R2 , ( ( − ma sin θ0 + cos θ (0)2 = μR2 + R1 − mg, Fig. 4.39 2 2
1
≡ ( ) = tan 1
494
4 Motion About a Fixed Axis
(
) 1 ma 0 = −(μR1 + R2 ) cos θ + (R1 − μR2 ) sin θ. 3
Thus − μR1 + R2 = 0, μR2 + R1 − mg = 0, − (μR1 + R2 ) cos θ + (R1 − μR2 ) sin θ = 0. It follows that μ(μR1 ) + R1 − mg = 0, and hence R1 =
mg , 1 + μ2
R2 =
μmg . 1 + μ2
Thus ) ( ) ( mg mg μmg μmg cos θ + sin θ = 0 − μ + − μ 1 + μ2 1 + μ2 1 + μ2 1 + μ2 or ( ( −(μ + μ) cos θ + 1 − μ2 sin θ = 0 or 2μ 2 tan ε = tan 2ε. tan θ = = 1 − μ2 1 − tan2 ε , ,, , Hence θ = 2ε.
◼.
4.3.6 Example At the ends of a uniform beam are two small rings which slide on two equally rough rods, which are respectively horizontal and vertical. Obtain the value of the angular velocity of the beam, when it is inclined at any angle to the vertical, the initial inclination being α (Fig. 4.40). Sol: If we draw a circle with centre at G, and radius a, it will pass through O, and hence OG = a. Also, θ = ,∠AOG ,, = ∠O AG,. Here equation of motion of centre of mass G(x = OG cos(θ − ε) = a cos(θ − ε), y = OG sin(θ − ε) = a sin(θ − ε)),
4.3 Examples on Motion About a Fixed Axis (Part II)
495
Fig. 4.40
≡ ( ) ≡ tan
are ( ( (2 ) .. d2 −ma sin(θ − ε) θ + cos(θ − ε) θ˙ = m 2 (a cos(θ − ε)) = S − (mg) sin ε , , dt ,, , and ( ( (2 ) .. d2 ma cos(θ − ε) θ − sin(θ − ε) θ˙ = m 2 (a sin(θ − ε)) = R − (mg) cos ε . , dt ,, , Also, ( 2) 2 ( 2) 2 ( (π )) a d (∠O AG) a d θ = m = −R a sin − (θ + ε) + S(a sin(θ + ε)) m 2 2 3 dt 3 dt 2 , ,, , = −Ra cos(θ + ε) + Sa sin(θ + ε),
so a .. m θ = −R cos(θ + ε) + S sin(θ + ε) ,, , , 3 ( ( ) ( (2 ) .. = − cos(θ + ε) ma cos(θ − ε) θ − sin(θ − ε) θ˙ + (mg) cos ε ) ( ( ( (2 ) .. + (mg) sin ε + sin(θ + ε) −ma sin(θ − ε) θ + cos(θ − ε) θ˙ or m
.. a .. θ = −ma θ (cos(θ + ε) cos(θ − ε) + sin(θ + ε) sin(θ − ε)) 3
496
4 Motion About a Fixed Axis
( (2 + ma θ˙ (cos(θ + ε) sin(θ − ε) − sin(θ + ε) cos(θ − ε)) + mg sin ε sin(θ + ε) or m
( (2 .. a .. θ = −ma θ cos 2ε − ma θ˙ sin 2ε + mg sin ε sin(θ + ε) 3
or ( (2 .. .. g θ = −3 cos 2ε θ −3 sin 2ε θ˙ + 3 sin ε sin(θ + ε) a ( ( () ( (2 .. d θ˙ g = (1 + 3 cos 2ε) θ = −3 sin 2ε θ˙ + 3 sin ε sin(θ + ε) (1 + 3 cos 2ε) θ˙ dθ a , ,, , or ( ( ( ( 3g sin ε d θ˙ = −3 sin 2ε θ˙ + sin(θ + ε). (1 + 3 cos 2ε) dθ a θ˙ Put ξ ≡ θ˙ . We get 3 sin 2ε 3g sin ε dξ + ξ= sin(θ + ε). dθ 1 + 3 cos 2ε a(1 + 3 cos 2ε)ξ It follows that ξ
dξ 3g sin ε 3 sin 2ε sin(θ + ε) + ξ2 = dθ 1 + 3 cos 2ε a(1 + 3 cos 2ε)
Put ζ ≡ ξ 2 . We get 3 sin 2ε 3g sin ε 1 dζ + ζ = sin(θ + ε) 2 dθ 1 + 3 cos 2ε a(1 + 3 cos 2ε) or dζ + Aζ = B sin(θ + ε), dθ where A ≡
6 sin 2ε , 1+3 cos 2ε
B≡
6g sin ε . a(1+3 cos 2ε)
Put φ ≡ θ + ε. It follows that
dζ + Aζ = B sin φ. dφ
4.3 Examples on Motion About a Fixed Axis (Part II)
497
Hence ( (2 θ˙ = ζ = e−Aφ B , =
{
e Aφ e Aφ sin φdφ + C = Be−Aφ 2 ( A sin φ − cos φ) + C A +1 ,, ,
B (A sin(θ + ε) − cos(θ + ε)) + C. A2 + 1
Thus ( (2 θ˙ =
A2
B (A sin(θ + ε) − cos(θ + ε)) + C, +1
 where θ˙ θ = π −α = 0. It follows that C = − A2B+1 (A cos(ε − α) + sin(ε − α)). 2 Hence, the required angular velocity / =
B (A(sin(θ + ε) − cos(ε − α)) − cos(θ + ε) − sin(ε − α)). A2 + 1
4.3.7 Example Suppose that there are two spheres of radii a and b. Suppose that sphere of radius b is solid, and is placed at the top of the other one. Suppose that the later is fixed, and the sphere of radius b rolls down it. Show that it will slip when the common normal makes with the vertical an angle θ given by the equation 2 sin(θ − λ) = 5 sin λ(3 cos θ − 2), where λ is the angle of friction (that is, coefficient of friction is tan λ). Sol: Here path of the centre of mass G is a circle of radius a + b having centre C, so equation of motion of G are (Fig. 4.41) ( ( ( (2 ) .. ) = (mg) cos θ − R, m (a + b) θ = (mg) sin θ − F. m (a + b) θ˙ Also ( ) 2b(a + b) .. 2b 2b(a + b) (mg) sin θ − F =m (mg sin θ − F) = m θ 5 5 m(a + b) 5 ( ( ( ) ) ( 2b2 ( a ) .. 2b2 d 2 θ + ab θ = m 1+ θ= m 5 b 5 dt 2
498
4 Motion About a Fixed Axis
,
=
≡ ( ),
=
+
≡
( ),
≡
( )
Fig. 4.41
( ( ) ) 2b2 d 2 (θ + φ) 2b2 .. ψ = F ·b = m = m 5 dt 2 5 ,, , , so 2b (mg sin θ − F) = F · b 5 or F=
2mg sin θ. 7
Next since ( .. ) 2mg 5mg m (a + b) θ = (mg) sin θ − F = (mg) sin θ − sin θ = sin θ, 7 7 , ,, , we have ( .. ) 5mg sin θ m (a + b) θ = 7 and hence ..
θ=
5g sin θ. 7(a + b)
It follows that dθ d (( ( 2 ) θ˙ =2 dt dt
(
5g sin θ 7(a + b)
)
4.3 Examples on Motion About a Fixed Axis (Part II)
499
and hence ( (2 θ˙ = C −  Now since θ˙ θ=0 = 0, so C =
10g cos θ. 7(a + b)
10g , 7(a+b)
and hence
( (2 10g (mg) cos θ − R = θ˙ = (1 − cos θ ) . m(a + b) 7(a + b) , ,, , It follows that 10g (mg) cos θ − R = (1 − cos θ ) m(a + b) 7(a + b) or ) ( 10mg mg R = mg cos θ − (1 − cos θ ) = (17 cos θ − 10). 7 7 , ,, , At the verge of separation, ( mg ) 2mg sin θ = F = (tan λ)R = (tan λ) (17 cos θ − 10) , ,, , 7 7 or 2 sin θ = tan λ(17 cos θ − 10) or 2 sin(θ − λ) = ,2 sin θ cos λ − 2 cos θ sin λ,,= 15 cos θ sin λ − 10 sin λ, = 5 sin λ(3 cos θ − 2).
◼
4.3.8 Example A hollow cylinder, of radius a, is fixed with its axis horizontal. Inside it a solid cylinder, of radius b, whose velocity in its lowest position is given. If the friction between the cylinders be sufficient to prevent any sliding, find the motion (Fig. 4.42).
500
4 Motion About a Fixed Axis
=
≡ ( ),
=
+
≡
( ),
≡
( )
,
Direction fixed in moving cylinder
Direction fixed in space
Fig. 4.42
Sol: Here path of the centre of mass C is a circle of radius a − b having centre O, so equation of motion of C are ( ( .. ) ( (2 ) = −(mg) cos φ + R, m (a − b) φ = −(mg) sin φ + F. m (a − b) φ˙ Also, ( ) b(a − b) −(mg) sin φ + F b(a − b) .. b φ =m (−(mg) sin φ + F) = m 2 2 m(a − b) 2 ( ( ( 2 )( ) .. ( b2 ) d 2 a φ − φ a b b −1 φ = m = m 2 b 2 dt 2 ( 2) 2 ( 2) .. b d (ψ − φ) b = m = m θ = −F · b 2 2 dt 2 , ,, , so b (−(mg) sin φ + F) = −F · b 2 or F=
1 mg sin φ. 3
4.3 Examples on Motion About a Fixed Axis (Part II)
501
Next since ( .. ) −2 1 mg sin φ, m (a − b) φ = −(mg) sin φ + F = −(mg) sin φ + mg sin φ = 3 3 , ,, , we have ( .. ) −2 m (a − b) φ = mg sin φ 3 and hence ..
φ=
−2 g sin φ. 3(a − b)
It follows that ( ) −2 d (( (2 ) dφ ˙ g sin φ =2 φ dt dt 3(a − b) and hence ( (2 φ˙ = C +
4g cos φ. 3(a − b)
 Now since φ˙ φ=0 = Ω, so C = Ω2 −
4g , 3(a−b)
and hence
( (2 −(mg) cos φ + R 4g = φ˙ = Ω2 − (1 − cos φ) . m(a − b) 3(a − b) , ,, , It follows that −(mg) cos φ + R 4g = Ω2 − (1 − cos φ) m(a − b) 3(a − b) or ) ( 4 R = mg cos φ + m(a − b)Ω2 − mg(1 − cos φ) 3 , ,, , = and hence
mg (7 cos φ − 4) + m(a − b)Ω2 3
502
4 Motion About a Fixed Axis
g R = m (7 cos φ − 4) + m(a − b)Ω2 . 3 In order that the cylinder may just make complete revolutions, Rφ=π = 0, and / 11g , and hence the hence m g3 (7(−1) − 4) + m(a − b)Ω2 = 0, that is, Ω = 3(a−b) projectile velocity is equal to /
( / ) 11g 11g(a − b) = . (a − b) 3(a − b) 3 / If projectile velocity is less than
11g(a−b) , 3
then it will leave the cylinder where
R = 0, that is cos φ =
4 − g3 (a − b)Ω2 7
.
Equation of Energy: The formula ( (2 φ˙ = Ω2 −
4g (1 − cos φ) 3(a − b)
can be obtained by using equation of energy: (final K.E.) − (initial K.E.) = (work done by gravity) + (work done by normal reaction) + (work done by friction)
that is (( ( ) 3m ( (2 3m 3m 2 (a − b)2 φ˙ − Ω2 = (a − b)2 φ˙ − (a − b)2 (Ω)2 4 4 4 ( (2 3m = (a − b)2 φ˙ − (initial K.E.) 4 (2 ( (2 1 ( m = (a − b)2 φ˙ + m (b − a)φ˙ − (initial K.E.) 4 2 ( 2 )(( ) )2 1 ( (2 1 a b = m − 1 φ˙ + m (b − a)φ˙ − (initial K.E.) 2 2 b 2 ( 2 )(( ) )2 1 ( (2 1 a b = m − 1 φ˙ + m (b − a)φ˙ − (initial K.E.) 2 2 b 2
4.3 Examples on Motion About a Fixed Axis (Part II)
=
503
1 2
( 2 )( ( . ( (2 2 m b2 θ˙ + 21 m (b − a)φ˙ −(initial K.E.)
,
= −mg((a − b) − (a − b) cos φ) + 0 + 0 ,, ,
= −mg(a − b)(1 − cos φ). Thus (( ( ) 3m 2 (a − b)2 φ˙ − Ω2 = −mg(a − b)(1 − cos φ) 4 or ( (2 φ˙ = Ω2 −
4g (1 − cos φ). 3(a − b)
Small Oscillation: For small oscillation, equation ..
φ=
−2 g sin φ 3(a − b)
becomes ..
φ=−
2g φ. 3(a − b)
This is an S.H.M., whose time period / =/
2π 2g 3(a−b)
= 2π
3 2 (a
− b) . g
4.3.9 Example A disc rolls of inside a hollow circular cylinder whose axis is horizontal. The plane of the disc is vertical and perpendicular to the axis/of the cylinder. If, when in its lowest position, its centre is moving with a velocity 83 g(a − b). Show that the centre of the disc describe an angle φ about the centre of the cylinder in time /
( ) φ π 3(a − b) ln tan + . 2g 4 4
504
4 Motion About a Fixed Axis =
≡ ( ),
=
+
≡
( ),
≡
( )
,
Direction fixed in moving disc
Direction fixed in space
Fig. 4.43
Sol: Here path of the centre of mass C is a circle of radius a − b having centre O, so equations of motion of C are (Fig. 4.43) ( ( .. ) ( (2 ) = −(mg) cos φ + R, m (a − b) φ = −(mg) sin φ + F. m (a − b) φ˙ Also, ( ) b(a − b) −(mg) sin φ + F b(a − b) .. b φ =m (−(mg) sin φ + F) = m 2 2 m(a − b) 2 ( ( ( 2 )( ) .. ( b2 ) d 2 a φ − φ a b b −1 φ = m = m 2 b 2 dt 2 ( 2) ( 2) 2 .. b b d (ψ − φ) = m = m θ = −F · b 2 2 dt 2 , ,, , so b (−(mg) sin φ + F) = −F · b 2 or F= Next since
1 mg sin φ. 3
4.3 Examples on Motion About a Fixed Axis (Part II)
505
( .. ) −2 1 mg sin φ, m (a − b) φ = −(mg) sin φ + F = −(mg) sin φ + mg sin φ = 3 3 , ,, , we have ( .. ) −2 mg sin φ m (a − b) φ = 3 and hence ..
φ=
−2 g sin φ. 3(a − b)
It follows that ( ) −2 dφ d (( (2 ) ˙ g sin φ φ =2 dt dt 3(a − b) and hence ( (2 φ˙ = C +  Now since φ˙ φ=0 =
1 a−b
/
8 g(a 3
4g cos φ. 3(a − b)
4g 8g − − b), so C = = 3(a − b) 3(a − b) , ,, ,
and hence ( (2 φ 4g 8g cos2 . φ˙ = (1 + cos φ) = 3(a − b) 3(a − b) 2 , ,, , It follows that / φ˙ = −2
φ 2g cos 3(a − b) 2
or / 2g φ dt. sec dφ = −2 2 3(a − b) On integration,
4g , 3(a−b)
506
4 Motion About a Fixed Axis
(
φ
π + 2 2 ln tan 4 2
/
) + C1 = −2
2g t. 3(a − b)
Now since φt=0 = 0, we have C1 = 0, and hence (
φ π + ln tan 4 4
/
) =
2g t. 3(a − b)
It follows that / t=
) ( π φ 3(a − b) . ln tan + 2g 4 4
◼
Equation of Energy: Here the centre of mass C(x = (a − b) sin φ, y = (a − b) cos φ). So its velocity =
/ / ( (2 ( (2 ˙ ˙ 2 + ( y˙ )2 = (a − b)φ˙ cos φ + −(a − b)φ˙ sin φ = (a − b)φ, (x)
and hence its K.E. at time t (2 1 ( m (a − b)φ˙ + 2 (2 1 ( = m (a − b)φ˙ + 2
=
( 2) 1 b ( (2 θ˙ (by Sect. 4.2.15) m 2 2 ( 2 )(( ) )2 (2 a b 1 3 ( m − 1 φ˙ = m (a − b)φ˙ . 2 2 b 4
It follows that initial K.E. 3 = m 4
(/
8 g(a − b) 3
)2 = 2mg(a − b).
Further, sum of work done by all forces = (−mg)((a − b) − (a − b) cos φ) + 0 + 0 = −mg(a − b)(1 − cos φ), So, equation of energy is (2 3 ( m (a − b)φ˙ = 2mg(a − b) − mg((a − b)(1 − cos φ)) = mg((a − b)(1 + cos φ)) 4 , ,, ,
or
4.3 Examples on Motion About a Fixed Axis (Part II)
( (2 φ˙ =
507
4g (1 + cos φ). 3(a − b)
We have already encountered with this equation above.
4.3.10 Example A solid homogeneous sphere is rolling on the inside of a fixed hollow sphere, the two centres being always in the same vertical plane. Show that the smaller sphere will make complete revolutions if, when it is in its lowest position, the pressure on times its own weight (Fig. 4.44). it is greater than 34 7 Sol: Here path of the centre of mass C of smaller solid sphere is a circle of radius a − b having centre O, so equations of motion of C are ( ( .. ) ( (2 ) = −(mg) cos φ + R, m (a − b) φ = −(mg) sin φ + F. m (a − b) φ˙ Also, ( ) 2b(a − b) −(mg) sin φ + F 2b (−(mg) sin φ + F) = m 5 5 m(a − b)
=
≡ ( ),
=
+
≡
( ),
≡
( )
,
Direction fixed in moving disc
Direction fixed in space
Fig. 4.44
508
4 Motion About a Fixed Axis
) ( ) .. 2b(a − b) .. 2b2 ( a φ −1 φ =m = m 5 5 b ( ( ( ) ( 2 ) d2 a φ − φ 2b2 d 2 (ψ − φ) 2b b = m = m 5 dt 2 5 dt 2 ( 2 ) .. 2b = m θ = −F · b 5 , ,, , so 2b (−(mg) sin φ + F) = −F · b 5 or F=
2 mg sin φ. 7
Next since ( .. ) −5 2 mg sin φ, m (a − b) φ = −(mg) sin φ + F = −(mg) sin φ + mg sin φ = 7 7 , ,, , we have ( .. ) −5 mg sin φ m (a − b) φ = 7 and hence ..
φ=
−5 g sin φ. 7(a − b)
It follows that ( ) −5 dφ d (( (2 ) ˙ φ g sin φ =2 dt dt 7(a − b) and hence ( (2 φ˙ = C +
10g cos φ. 7(a − b)
 10g Now suppose that φ˙ φ=0 = Ω. So C = Ω2 − , and hence 7(a − b) , ,, ,
4.3 Examples on Motion About a Fixed Axis (Part II)
509
( ( (2 ) = −(mg) cos φ + R m (a − b) φ˙ ( (2 −(mg) cos φ + R 10g = φ˙ = Ω2 − (1 − cos φ) . m(a − b) 7(a − b) , ,, , Thus −(mg) cos φ + R 10g = Ω2 − (1 − cos φ) m(a − b) 7(a − b) or 10 R = (mg) cos φ + Ω2 m(a − b) − mg(1 − cos φ) 7 , ,, , = Ω2 m(a − b) +
mg (17 cos φ − 10) 7
or R = Ω2 m(a − b) +
mg (17 cos φ − 10). 7
When the smaller sphere just complete revolution, we have Rφ=π = 0, so 0 = Ω2 m(a − b) + and hence Ω2 =
27g . 7(a−b)
mg (−27), 7
Thus
mg 27g mg 17mg + R=m· (17 cos φ − 10) = (17 cos φ + 17) = (cos φ + 1). 7 7 7 7 , ,, , So required pressure = Rφ=0 =
17mg 34 mg. (1 + 1) = 7 7
◼
4.3.11 Example A circular plate rolls down the inner circumference of a rough circle under the action of gravity. The planes of both plate and circle are vertical. When the line joining their
510
4 Motion About a Fixed Axis
centres is inclined at an angle θ to the vertical, show that the friction between the bodies is sin3 θ (weight of the plate) (Fig. 4.45). Sol: Here path of the centre of mass C of circular plate is a circle of radius a − b having centre O, so equations of motion of C are ( ( ( (2 ) .. ) = −(mg) cos θ + R, m (a − b) θ = −(mg) sin θ + F. m (a − b) θ˙ Also, ( ) b(a − b) −(mg) sin θ + F b (−(mg) sin θ + F) = m 2 2 m(a − b) ( 2 )( ) .. b(a − b) .. a b =m −1 θ θ= m 2 2 b ( ( 2 ) 2( a b d bθ − θ = m = −F · b 2 dt 2 ,, , , so b (−(mg) sin θ + F) = −F · b 2 or
≡ ( )
,
Direction fixed in moving disc
Direction fixed in space
Fig. 4.45
4.3 Examples on Motion About a Fixed Axis (Part II)
F=
sin θ mg. 3
511
◼
4.3.12 Example A cylinder, of radius a, lies within a rough fixed cylinder of radius 2a. The centre of gravity of the cylinder is at a distance c from its axis, and the initial state is that of a stable equilibrium at the lowest point of the cavity. Show that the smallest angular velocity Ω with which the cylinder must be started that it may roll right round the cavity is given by ( Ω2 (a + c) = g 1 +
) 4(a + c)2 , (a − c)2 + k 2
where k is the radius of gyration about the centre of gravity. Find also the normal reaction between the cylinders in any position (Fig. 4.46). Sol: Here, G(x = a sin θ − c sin θ, a cos θ + c cos θ ). So the velocity of G / ( (2 ( (2 ˙ 2 + ( y˙ )2 = (a − c) cos θ θ˙ + (a + c) sin θ θ˙ (x) / = θ˙ a 2 + c2 − 2ac cos 2θ . =
/
≡ ( ) 2
,
Direction fixed in moving disc Direction fixed in space
Fig. 4.46
512
4 Motion About a Fixed Axis
Hence final K.E. (( (2 ( 1( 1 ( (2 ( 2 m θ˙ a + c2 − 2ac cos 2θ + mk 2 θ˙ 2 2 ( 1 ( (2 ( 2 2 2 = m θ˙ a + c − 2ac cos 2θ + k . 2 =
It follows that initial K.E. =
( ( 1 ( ( 1 m(Ω)2 a 2 + c2 − 2ac cos 2(0) + k 2 = m(Ω)2 (a − c)2 + k 2 . 2 2
Net work done by all forces = −mg((a + c) − (a cos θ + c cos θ )) + 0 + 0 = −mg(a + c)(1 − cos θ ). Thus equation of energy is ( 1 ( ( 1 ( (2 ( 2 m θ˙ a + c2 − 2ac cos 2θ + k 2 − m(Ω)2 (a − c)2 + k 2 2 2 = −mg(a + c)(1 − cos θ ) and hence ( 1 ( (2 ( 2 θ˙ a + c2 − 2ac cos 2θ + k 2 2 ( ( 1 = (Ω)2 (a − c)2 + k 2 − g(a + c)(1 − cos θ ). (∗) 2 On differentiation, we get ( 1 ( (2 .. ( θ˙ θ a 2 + c2 − 2ac cos 2θ + k 2 + θ˙ (2ac sin 2θ )2θ˙ = −g(a + c) sin θ θ˙ 2 or ( ( (2 .. ( θ a 2 + c2 − 2ac cos 2θ + k 2 + θ˙ (2ac sin 2θ ) = −g(a + c) sin θ. )2 ( ( 2 ) d (θ ) ) Observe that the accelerations of G relative to C are c d(θ and c along dt dt 2 −→ GC and perpendicular to it. Observe that the accelerations of C relative to O are )2 ( ( 2 ) −→ d (θ ) ) a d(θ and a along C O and perpendicular to it. dt dt 2 −→ It follows that the acceleration of G along C O
4.3 Examples on Motion About a Fixed Axis (Part II)
513
) (( ( ( ) ) ) ( 2 ) d(θ ) 2 d(θ ) 2 d (θ ) sin 2θ + a = c cos 2θ + c dt dt 2 dt ( ..) ) (( ( ( ) ( ( 2 2 cos 2θ + c θ sin 2θ + a θ˙ . = c θ˙ Hence m
( ..) ) ((( ( ( ) ( (2 ) 2 cos 2θ + c θ sin 2θ + a θ˙ = R − (mg) cos θ. c θ˙
If at θ = π, R = 0, we get m
( ..) ) ((( ( ( ) ( (2 ) 2 = 0 − (mg)(−1), c θ˙ 1 + c θ 0 + a θ˙
and hence ( (2 θ˙ =
g . c+a
Further, from (∗), we have ( 1 ( ( 1 g ( 2 a + c2 − 2ac1 + k 2 = (Ω)2 (a − c)2 + k 2 − g(a + c)2 2c+a 2 or g 4g(a + c) = Ω2 − c+a (a − c)2 + k 2 or ( Ω2 (a + c) = g 1 +
) 4(a + c)2 . (a − c)2 + k 2
◼
R can be found from ⎫ ( ( (2 ..( ⎪ θ a 2 + c2 − 2ac cos 2θ + k 2 + θ˙ (2ac sin 2θ ) = −g(a + c) sin θ ⎪ ( ( ( ( ( ( ⎪ 1 ˙ 2 2 2 − 2ac cos 2θ + k 2 = 1 (Ω)2 (a − c)2 + k 2 − g(a + c)(1 − cos θ ) ⎪ ⎬ θ a + c 2 2 .. ) (( ( 2 . ˙ m θ (c cos 2θ + a) + c θ sin 2θ = R − (mg) cos θ ⎪ ⎪ ( ) ⎪ ⎪ 2 ⎭ Ω2 (a + c) = g 1 + 4(a+c) 2 2 (a−c) +k
514
4 Motion About a Fixed Axis
4.3.13 Note Suppose that there is sphere of radius a. Suppose that there is an inclined plane whose inclination with the horizon is α. Suppose that the sphere moves on the inclined plane. Situation I: when friction F is equal to the maximum preventive force μR. Here the centre of mass is G(x, y = a), so its equations of motion are ..
..
M x = (Mg) sin α − μR, 0 = M y = −(Mg) cos α + R . , ,, , For the motion relative to centre of mass G, we have ( ,
2a 2 M 5
)
..
θ = (μR)a = (μ((Mg) cos α))a = Mgμa cos α, ,, ,
and hence (
2a 2 M 5
)
..
θ = Mgμa cos α.
It follows that ..
θ=
5gμ cos α (a constant). 2a
Hence θ˙ = ,
(
) ) (  5gμ 5gμ cos α t + θ˙ t=0 = cos α t + 0. 2a 2a ,, ,
It follows that θ˙ =
(
) 5gμ cos α t. 2a
Assume that θ t=0 = 0. It follows that ( ) 1 5gμ cos α t 2 . θ= 2 2a Since ..
M x = (Mg) sin α − μR,
R = (Mg) cos α,
4.3 Examples on Motion About a Fixed Axis (Part II)
515
we have ..
M x = (Mg) sin α − μ((Mg) cos α), and hence ..
x = (sin α − μ cos α)g (a constant). Hence x˙ = ((sin α − μ cos α)g)t + x ˙ t=0 = ((sin α − μ cos α)g)t + 0. , ,, , It follows that x˙ = ((sin α − μ cos α)g)t. Assume that xt=0 = 0. It follows that x=
1 ((sin α − μ cos α)g)t 2 . 2
that) velocity relative to G is (Observe ( ( )of) B(x, y = 0) 5gμ 5gμ ˙ a θ = a 2a cos α t = 2 cos α t along (−x)direction. Again, velocity of G(x, y = a) relative to O is x(= ˙ ((sin α − μ cos α)g)t) along (+x)direction. Hence, (velocity of B(x, y = 0) along xdirection(relative to O) )
= ((sin α − μ cos α)g)t + −5gμ cos α t 2 ,, , ) ( ) ( 7μ 5μ cos α gt = sin α − cos α gt. = sin α − μ cos α − 2 2
,
Thus (velocity of B(x, y = 0) along xdirection relative to O) ) ( 7μ cos α gt. (∗) = sin α − 2 Case I: when μ
0, = sin α − 2 , ,, , and hence, velocity of B(x, y = 0) is down the plane. Thus, the sliding occurs. In this case, equations of motion are, ( ) ⎫ 1 5gμ ⎪ θ= cos α t 2 ⎪ ⎬ 2 2a . ⎪ 1 ⎪ x = ((sin α − μ cos α)g)t 2 ⎭ 2 Here, 1 K.E. = M(((sin α − μ cos α)g)t)2 + 2 ( 1 2 2 = Mg t (sin α − μ cos α)2 + 2 Case II: when μ =
2 7
( )(( ) )2 5gμ 1 2a 2 M cos α t 2 5 2a ) 5μ2 cos2 α . 2
tan α. In this case, from (∗),
(velocity of (x, y = 0)along x − direction relative to O) = 0. Thus, point of contact is at rest. In this case, equations of motion are ( ) ⎫ 1 5g ⎪ sin α t 2 ⎪ θ= ⎬ 2 7a ( ) . ⎪ 1 5 ⎭ x= (sin α)g t 2 = aθ ⎪ 2 7 This motion is “pure rolling” because x = aθ and maximum friction μR is always being exerted. Here ) )2 ( )(( ) )2 5 5g 2a 2 1 sin α t M (sin α)g t + 7 2 5 7a ) (( )2 ( ) 5 1 10 2 1 2 2 2 2 5 2 sin α + sin α = Mg t sin α . = Mg t 2 7 49 2 7
1 K.E. = M 2
Case III: when
2 7
((
tan α < μ. In this case,
4.3 Examples on Motion About a Fixed Axis (Part II)
517
) ( 7μ cos α gt < 0, (velocity of B(x, y = 0)along x − direction) = sin α − 2 , ,, , and hence, velocity of B(x, y = 0)is up the plane. Now since friction is acting at B along up the plane,B will not come to rest. This contradicts the law of friction (see Sect. 4.2.12(2)). Hence this case is impossible. Situation II: when friction F is smaller than the maximum preventive force μR. In this case, we shall use Fig. 4.47b: Here the centre of mass is G(x, y = a), so its equations of motion are ..
..
M x = (Mg) sin α − F, 0 = M y = −(Mg) cos α + R . ,, , , For the motion relative to centre of mass G, we have
(a)
,
(fixed) ≡ ( ) ≡ ( )
(b)
,
(fixed) ≡ ( ) ≡ ( )
Fig. 4.47
518
4 Motion About a Fixed Axis
(
2a 2 M 5
)
..
θ = (F)a
or 2a .. .. M θ = F = (Mg) sin α − M x, , 5 ,, , and hence 2a .. .. θ = g sin α − x . 5 By Sect. 4.2.12, in this situation, the point of contact is at rest. So ( ( −a θ˙ + (x) ˙ = (velocity of B rel. to G in xdirection) + (velocity of G rel. to O in xdirection) = (velocity of B rel. to O in xdirection) = 0, , ,, , and hence dθ dx −a = 0. dt dt It follows that g sin α −
.. 2a .. .. θ = ,x =,,a θ,, 5
and hence ..
θ=
5g sin α (a constant). 7a
Now θ˙ = ,
(
) ) (  5g 5g  ˙ sin α t + θ t=0 = sin α t + 0. 7a 7a ,, ,
It follows that θ˙ =
(
) 5g sin α t. 7a
4.3 Examples on Motion About a Fixed Axis (Part II)
519
Assume that θ t=0 = 0. Now, we have ) ( 1 5g sin α t 2 . θ= 2 7a Since ) 5g 5g x , =,,a θ, = a 7a sin α = 7 sin α ..
..
(
we have ..
x=
5g sin α (a constant). 7
Hence ( x˙ = ,
) ) ( 5g 5g sin α t + x ˙ t=0 = sin α t + 0. 7 7 ,, ,
It follows that ( x˙ =
) 5g sin α t. 7
Assume that xt=0 = 0. It follows that ( ) 1 5g sin α t 2 . x= 2 7 In this situation, equations of motion are ) ⎫ ( 1 5g ⎪ sin α t 2 ⎪ θ= ⎬ 2 7a ( ) . ⎪ 1 5g ⎭ sin α t 2 = aθ ⎪ x= 2 7 Here, point of contact is at rest, and x = aθ, but friction exerted is smaller than the maximum friction μR. So, the motion is rolling but not pure rolling. Here K.E. =
1 M 2
((
) )2 ( )(( ) )2 5g 5g 2a 2 1 sin α t + M sin α t 7 2 5 7a
520
4 Motion About a Fixed Axis
1 = Mg 2 t 2 2
((
5 sin α 7
)2
10 2 sin α + 49
) =
( ) 5 2 1 Mg 2 t 2 sin α . 2 7
Thus ) ( ⎧ 5μ2 2 1 ⎪ 2 2 2 2 ⎪ ⎪ Mg t (sin α − μ cos α) + cos α when F = μR and μ < tan α ⎪ ⎪ 2 2 7 ⎪ ⎪ ⎪ ⎨ ( ) 1 5 2 2 final K.E. = Mg 2 t 2 sin α when F = μR and μ = tan α ⎪ ⎪ 2 7 7 ⎪ ⎪ ( ) ⎪ ⎪ ⎪ 1 5 2 ⎪ 2 2 ⎩ Mg t sin α when F < μR. 2 7
Here, the work done by gravity when the centre has described a distance x ) (π − α = (Mg)(x) sin α = (Mg)(x) cos 2 ( ) ⎧ 1 2 ⎪ 2 ⎪ Mg ((sin α − μ cos α)g)t sin α when F = μR and μ < tan α ⎪ ⎪ 2 7 ⎪ ⎪ ⎪ ( ( ) ) ⎨ 2 1 5 = Mg (sin α)g t 2 sin α when F = μR and μ = tan α ⎪ 2 7 7 ⎪ ⎪ ( ( ) ) ⎪ ⎪ 1 5g ⎪ ⎪ ⎩ sin α t 2 sin α when F = μR (Mg) 2 7 ⎧1 2 ⎪ Mg 2 t 2 sin α(sin α − μ cos α) when F = μR and μ < tan α ⎪ ⎪ ⎪ 2 7 ⎪ ( ) ⎪ ⎨ 1 5 2 Mg 2 t 2 sin2 α when F = μR and μ = tan α = 2 7 7 ⎪ ⎪ ( ) ⎪ ⎪ ⎪ 1 5 ⎪ ⎩ Mg 2 t 2 sin2 α when F < μR. 2 7 In the case when F = μRand μ
5uμ, where u, ω are initial linear and angular velocities of the sphere, μ is the coefficient of friction, and α is the inclination of the plane. Sol: We assume the case when the friction (is μR (Fig. ( 4.50). ( ( Since velocity of point of contact B is x˙ − a θ˙ , we have x˙ − a θ˙ t=0 = u − a(−ω) = u + aω > 0. So the direction of friction μR is up the plane. Further (
) 2(Mg) sin α − 7μ((Mg) cos α) 2 sin α − 7μ cos α g= 2 2M 5μR 2(Mg) sin α − 7μR (Mg) sin α − μR = − = 2M M 2M ((μR) · a) (Mg) sin α − μR ( 2) −a = M M 2a 5
=
Since (acc. of B) = the required duration
.. .. .. (Mg) sin α − μR − a θ = x −a θ = (acc. of B) . , ,, , M
( 2 sin α−7μ cos α ( 2
Path of ≡ ( )↓ ̇
=0
( ,
g, and (vel. of B)t=0 = u − a(−ω) = u + aω,
direction fixed in body ,
=− = )
≡ ( )↓ ̇
=0
=
direction fixed in space
Fig. 4.50
526
4 Motion About a Fixed Axis
aΩ u + aω . = ( 2 sin α−7μ cos α ( = (3μ cos α − sin α)g g 2
◼
Here, equation of centre of mass G(x, y = a) are ..
..
M x = (Mg) sin α − μR, 0 = M y = R − (Mg) cos α , ,, , ..
and hence x = (sin α − μ cos α)g(a constant). Thus x˙ = u + (sin α − μ cos α)gt. Suppose that t0 is the time when sphere turns back in the course of its motion. So  u + (sin α − μ cos α)gt0 = 0 and 0 > θ˙ t=t0 . For the motion relative to centre of mass G, we have ( ,
2a 2 M 5
)
..
θ = (μR)a = (μ((Mg) cos α))a ,, ,
and hence, ..
θ=
5μg cos α .(a constant). 2a
 Now since θ˙ t=0 = −ω, we have ) ) ( (  −u 5μg cos α 5μg cos α  ˙ t0 = (−ω) + 0 > θ t=t0 = (−ω) + 2a 2a g(sin α − μ cos α) , ,, , ) ( −u 5μ cos α , = −ω + 2a (sin α − μ cos α) and hence ( 0 > −ω +
) −u 5μ cos α 2a (sin α − μ cos α)
or ( ω>
) u 5μ 2a (μ − tan α)
4.3 Examples on Motion About a Fixed Axis (Part II)
527
or 2aω(μ − tan α) > 5uμ.
◼
4.3.17 Example A sphere, of radius a, is projected up an inclined plane with velocity V and angular velocity Ω in the sense which would cause it to roll up. If V > aΩ and the coefficient of friction > 27 tan α. Show that the sphere will cease to ascend at the end of a time 5V +2aΩ , where α is the inclination of the plane. 5g sin α Sol: Here, it is clear that the friction is maximum, that is, μR(= μ((Mg) cos α)). We claim that the direction of friction is down the plane. If not, otherwise, suppose that the friction μMg cos α acts up the plane. We have to arrive at a contradiction. Here equations of motion are ..
M x = −(Mg) sin α + μMg cos α = −Mg(sin α − μ cos α), , ,, , (
2a 2 M 5
)
..
θ = −(μMg cos α)a.
It follows that ..
..
x = −g(sin α − μ cos α), θ = −
5μg cos α . 2a
Hence ) ( 5μg cos α t + Ω. x˙ = −g(sin α − μ cos α)t + V , θ˙ = − 2a It follows that (vel. of B) ) ) (( ( ( 5μg cos α ˙ t +Ω = x˙ + −a θ = (−g(sin α − μ cos α)t + V ) − a − 2a , ,, , ) ( 7 = −g sin α − μ cos α t + (V − aΩ) 2
528
4 Motion About a Fixed Axis
⎞
⎛
⎛ ⎞ ( )⎟ ⎜ 7 2 ⎟ ⎜ ⎠ = ⎜− g cos α tan α − μ ⎟t + ⎝,V − ,,aΩ, > 0, ⎠ ⎝ 2 7 , ,, , positive
and hence the direction of velocity of the point of contact is up the plane. It follows that friction acts down the plane. This contradicts the assumption. Hence, our claim is true (Fig. 4.51). The equations of motion are −Mg sin α − μ(Mg cos α) = −g(sin α + μ cos α), M
..
x= and hence
(a) Path of
direction fixed in body ,
≡ ( )↓ ̇
=0
( ,
=Ω = )
≡ ( )↑ ̇
=0
=
direction fixed in space
(b) Path of
direction fixed in body ,
direction fixed in space
Fig. 4.51
4.3 Examples on Motion About a Fixed Axis (Part II)
529
x˙ = V − g(sin α + μ cos α)t. Since ..
θ=
(μ(Mg cos α))a 2 M 2a5
=
5μg cos α , 2a
and hence θ˙ =
5μg cos α t + Ω. 2a
It follows that ) ( 7μ cos α ˙ t + (V − aΩ) . (rel. vel. of point of contact B) = x˙ − a θ = −g sin α + 2 , ,, , This shows that (rel. vel. of point of contact B) changes its sign from positive to negative at t = t1 , where ) ( 7μ cos α t1 + (V − aΩ) = 0, −g sin α + 2 and hence t1 =
2(V − aΩ) . g(2 sin α + 7μ cos α)
Also friction changes its direction from down the plane to up the plane at t = t1 . Here  5μg cos α t1 ≡ A. θ˙ t=t1 = Ω + 2a After rolling: Here ..
−Mg sin α + −Mg sin α + F a = M ( ) M ( ) 2 .. 2a .. x = −g sin α − θ = −g sin α − 5 5
x=
so
) .. ( 2 − M 2a5 θ
530
4 Motion About a Fixed Axis
( ) 7 .. x = −g sin α. 5 Hence, 5 x˙ = − g sin αt + a A. 7 So required time = t1 + t2 , where 0 = − 57 g sin αt2 + a A. Hence, ) ( 5μg cos α 7a A 7a Ω+ = t1 + t1 t1 + t2 = t1 + 5g sin α 5g sin α 2a ) ( 7 μ cos α 7aΩ t1 + 1+ = 5g sin α sin α 2 2(V − aΩ) 2 sin α + 7μ cos α 7aΩ + = 5g sin α 2 sin α g(2 sin α + 7μ cos α) V − aΩ 7aΩ + 5(V − aΩ) 7aΩ (5V + 2aΩ) + = = . = 5g sin α g sin α 5g sin α 5g sin α
◼
4.3.18 Example A sphere is projected up an inclined plane, for which μ = 17 tan α, with velocity V and an angular velocity Ω (in the direction in which it would roll up). If V > aΩ, show that friction acts downwards at first, and upward afterwards. Also prove that the whole time duration during which the sphere rises up is (Fig. 4.52) 17V + 4aΩ . 18g sin α Sol:
 ( ( ˙ t=0 > a θ˙ t=0 , velocity of B is along positive xSince (vel. ofB) = x˙ + −a θ˙ , and x direction, and hence friction acts downwards the plane. Also the friction is maximum. Hence equations of motion are ..
M x = −Mg sin α − μ(Mg cos α) = −Mg(sin α + μ cos α), , ,, , ( M
2a 2 5
Now since μ =
)
..
θ = a(μ((Mg) cos α)). 1 7
tan α, we have
4.3 Examples on Motion About a Fixed Axis (Part II)
531
direction fixed in space Path of
direction fixed in body ,
≡ ( )↑ ̇
=0
( ,
=Ω = )
≡ ( )↑ ̇
=0
=
> Ω=
̇
=0
Fig. 4.52 .. 5 g sin α 8 .. x = −g sin α, θ = . 7 14 a
It follows that ) ( ) 2 ( 8g 8g t , x˙ = − sin α t + V , x = V t + − sin α 7 7 2 ) ( ) 2 ( 5 g sin α t 5 g sin α t + Ω, θ = Ωt + . θ˙ = 14 a 14 a 2 Hence ) ( 3g ˙ (vel. of B) = x˙ − a θ = − sin α t + (V − aΩ) 2 Hence, vel. of B is up the plane till the time skidding for 0 < t
tan α). It has initially such a backward spin Ω that after a time t1 it starts moving uphill and continues to do so for a time t2 after which it once more descends. The motion being in a vertical plane at right angles to the given inclined plane, show that (Fig. 4.53) (t1 + t2 )g sin α = aΩ − V . Sol:
 ˙ t=0 , a θ˙ t=0 are positive, velocity of B is along positive Here (vel. of B) = x˙ + a θ˙ . x xdirection, and hence friction acts upwards the plane. Since velocity of B is nonzero, it is not rolling, and hence the friction is maximum. Thus, equations of motion during downhill are
(a)
direction fixed in space Path of
,
≡ ( )↑ ∴
̇ ( ,
=0
=Ω
= )
≡ ( )↑ ∴
(b)
Fig. 4.53
̇
=0
=
direction fixed in space Path of
,
534
4 Motion About a Fixed Axis ..
M x = Mg sin α − μ(Mg cos α) = −Mg cos α(μ − tan α), , ,, , (
( .. Ma 2 θ = −a(μ((Mg) cos α)).
Now we have ..
..
x = −g cos α(μ − tan α), θ = , ,, , negative
−μg cos α . a , ,, , negative
It follows that x˙ = (−g cos α(μ − tan α))t + V , θ˙ =
(
) −μg cos α t + Ω. a
α) If for some time t satisfying 0 < t < g cos α(μ−tan , (vel. of B) = 0, there V V commences rolling before time g cos α(μ−tan , and hence will never move uphill. So, α) V from question, there is no rolling before time g cos α(μ−tan α) (= tx=0 ˙ ). Hence
t1 =
V . g cos α(μ − tan α)
Observe that ) −μg cos α t1 + Ω a ) )( ( V −μg cos α +Ω = a g cos α(μ − tan α) −μV + aΩ(μ − tan α) = ≡ ω. a(μ − tan α)
 θ˙ t=t1 =
(
For uphill journey:
 ˙ t=0 = 0, θ˙ t=0 = ω > 0. This shows that Here (vel. of B) = x˙ − a θ˙ , x (vel. of B)t=0 is negative, and hence friction is uphill. Also friction is maximum. Now the equations are ( ( .. .. M x = −Mg sin α + μ(Mg cos α), Ma 2 θ = −(μMg cos α)a. Hence
4.3 Examples on Motion About a Fixed Axis (Part II)
535
⎞
⎛
−gμ cos α ⎟ ⎜ x˙ = ⎝g(− sin α + μ cos α)⎠t, θ˙ = t + ω. , ,, , a positive
Thus (vel. of B) = g((− sin α + μ cos α) + μ cos α)t − aω = g(− sin α + 2μ cos α)t − aω Observe that x ˙ (vel. of B)=0 = g(− sin α + μ cos α)
aω(tan α − μ) aω = > 0. g(− sin α + 2μ cos α) (tan α − 2μ)
Hence x ˙ (vel. of B)=0 > 0. It follows that before coming to rest, rolling occurs. aω Commencement of rolling after time g(− sin α+2μ : Now the equations of cos α) motion are ..
M x = −Mg sin α + F,
(
( .. Ma 2 θ = −Fa, x˙ = a θ˙ .
It follows that −g sin α +
.. −F F .. = x =aθ = M M
or F=
Mg sin α. 2
We must show that Mg sin α ≤ μR, that is, Mg sin α ≤ μ(Mg cos α), that is, 2 2 ≤ μ. This is clearly true, because tan α < μ. It follows that
tan α 2
..
..
x = −g sin α − (a) θ and hence 2 x˙ = x˙ + a θ˙ = −(g sin α)t + 2
aω(tan α − μ) . (tan α − 2μ)
So, tx=0 = ˙
α−μ) 2 aω(tan (tan α−2μ)
g sin α
=2
aω(tan α − μ) g sin α(tan α − 2μ)
536
4 Motion About a Fixed Axis
It follows that aω(tan α − μ) aω +2 t2 = g(− sin α + 2μ cos α) g sin α(tan α − 2μ) , ,, , = = = = =
aω(tan α − μ) aω +2 − cos αg(tan α − 2μ) g sin α(tan α − 2μ) ( ) aω −1 (tan α − μ) +2 g(tan α − 2μ) cos α sin α ( ) − sin α + 2(sin α − μ cos α) aω g(tan α − 2μ) cos α sin α aω(sin α − 2μ cos α) cos α sin αg(tan α − 2μ) −μV + aΩ(μ − tan α) a −μV + aΩ(μ − tan α) aω = . = g sin α g sin α a(μ − tan α) sin α(μ − tan α)
Hence −μV + aΩ(μ − tan α) V + g cos α(μ − tan α) sin α(μ − tan α) V sin α − μV cos α + aΩ(μ cos α − sin α) = g(μ cos α − sin α) sin α −V + aΩ V (sin α − μ cos α) + aΩ(μ cos α − sin α) = . = g(μ cos α − sin α) sin α g sin α
t1 + t2 =
◼
4.3.20 Example A uniform sphere of radius a is rotating about a horizontal diameter with angular velocity Ω is gently placed on a rough plane which is inclined at an angle α to the horizontal, the sense of rotation being such as to tend to cause the sphere move up the plane along the line of greatest slop. Show that, if the coefficient of friction is tan α, the centre of the sphere will remain at rest for a time 5g2aΩ , and will then sin α move downwards with acceleration 57 g sin α (Fig. 4.54). Sol:
 Here (vel. of B) = x˙ + a θ˙ . Now since x ˙ t=0 = 0, and θ˙ t=0 = Ω, we have (vel. of B)t=0 is positive, so initially, friction is μ(Mg cos α)(= tan α(Mg cos α) = Mg sin α) along uphill. Thus the equations of motion are
4.3 Examples on Motion About a Fixed Axis (Part II)
537
direction fixed in space Path of
,
≡ ( )↑ ̇
∴
( ,
=0
=Ω
= )
≡ ( ) ∴
̇
=0
=0
Fig. 4.54 ..
M x = (Mg) sin α − Mg sin α = 0, , ,, , ( ,
2a 2 M 5
)
..
θ = −(μ(Mg cos α))a = −Mga sin α ,, ,
that is x˙ = 0, θ˙ =
(
) −5g sin α t + Ω. 2a
Hence t(vel. of B)=0 = t0+a (( −5g sin α )t+Ω)=0 = 2a
This proves the first part. After time ..
M x = (Mg) sin α − F,
2aΩ , 5g sin α
( M
2a 2 5
)
2aΩ . 5g sin α
equations of motion are ..
θ = −Fa, x˙ + a θ˙ = 0.
It follows that g sin α −
and hence
( ) ( .. F −5F d( 7F .. = g sin α − +a = x +a θ = x˙ + a θ˙ = 0 . 2M M M2a ,dt ,, ,
538
4 Motion About a Fixed Axis
F=
2Mg sin α . 7
It follows that ..
M x = (Mg) sin α − , ,, ..
and hence x =
5g sin α . 7
5g sin α 2Mg sin α =M , 7 , 7
◼
4.4 Examples on Motion About a Fixed Axis (Part III) 4.4.1 Note Suppose that there is a nonuniform sphere of radius a having centre C. Suppose that G is the centre of mass of the sphere, where C G = c. Suppose that the sphere is placed on a rough horizontal plane such that C G is also horizontal. Suppose that k 2 is the radius of gyration of the sphere about the horizontal axis through G and perpendicular to C G. Suppose that the sphere rolls. Let us first try to find the formula for velocity of contact point B (Fig. 4.55). Here (vel. of B) = (vel. of B rel. to C) + (vel. of C(x, y = a)) ( ( = a θ˙ along negative x + (vel. of C(x, y = a)) ( ( = a θ˙ along negative x + (x˙ along positive x) (( ( ( = x˙ − a θ˙ along positive x .
, ≡ ( )↑ ( ,
= )
≡ ( )↑
Fig. 4.55
Direction fixed in space Direction fixed in body
4.4 Examples on Motion About a Fixed Axis (Part III)
539
In short, we write (vel. of B) = x˙ − a θ˙ . Since the sphere rolls, we have x˙ = a θ˙ . .. .. It follows that x = a θ . Now we try to write the “two differential equations” of motion—one for motion of centre of mass G, and another for motion relative to G. For the motion of G(x + c cos θ, a − c sin θ ), we have d 2 (x + c cos θ ) d 2 (a − c sin θ ) M = F, M = R − Mg, dt 2 dt 2 ( 2 ( .. Mk θ = R(c cos θ ) − F(a − c sin θ ), that is ( ( (2 ) .. ( ( (2 ) .. .. .. F = x +c − sin θ θ − cos θ θ˙ = a θ +c − sin θ θ − cos θ θ˙ , M , ,, , ( ( (2 ) .. − cM cos θ θ − sin θ θ˙ = R − Mg, (
( .. Mk 2 θ = R(c cos θ ) − F(a − c sin θ ).
On using the initial condition, we get  ..  a θ
 ( (  (2 ) ..  Fθ =0 = + c − sin 0 θ  − cos 0 θ˙ θ=0 , θ =0 θ =0 M  ) ( (  (2 ..  = Rθ =0 − Mg − cM cos 0 θ  − sin 0 θ˙ θ =0 (
( .. Mk 2 θ 
θ =0
θ =0
= Rθ =0 (c cos 0) − Fθ =0 (a − c sin 0)
or,   (  (2 ..  ..  Fθ =0 , −cM θ  a θ − c θ˙ θ =0 = = Rθ =0 − Mg, θ =0 θ =0 M  ( 2 ( .. Mk θ  = c Rθ =0 − a Fθ =0 . θ =0
or   ..  ..  Fθ =0 , −cM θ  a θ − c(0)2 = = Rθ =0 − Mg, θ =0 θ =0 M ( 2 ( .. Mk θ  = c Rθ =0 − a Fθ =0 . θ =0
Thus
540 .. a θ
4 Motion About a Fixed Axis
θ=0
=
) .. ( .. Fθ=0  = Rθ=0 − Mg, Mk 2 θ  = c Rθ=0 − a Fθ=0 . , −cM θ  θ=0 θ=0 M
It follows that ) ( ( Fθ =0 ( Fθ=0 = Rθ =0 − Mg, Mk 2 −c = c Rθ=0 − a Fθ =0 . a aM Hence k2
) ( F Fθ =0 = c Mg − c θ =0 − a Fθ =0 a a
or Fθ =0 =
cMg k2 a
+
c2 a
+a
=
acMg . k 2 + c2 + a 2
Next ) ( k2 + a2 acMg c = = Mg − Mg. a k 2 + c2 + a 2 k 2 + c2 + a 2
Rθ=0 Further
 θ .. 
Case I: when
ac k 2 +a 2
Fθ=0
θ =0
=
acMg k 2 +c2 +a 2
aM
=
cg . k 2 + c2 + a 2
( ( < μ. It follows that ac < μ k 2 + a 2 , and hence ( ( 2 k + a2 acMg = 2 0, ~ ~~ ~
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 R. Sinha, Advanced Newtonian Rigid Dynamics, https://doi.org/10.1007/9789819920228_5
563
564
5 Motion in Two Dimensions (Finite Forces) Direction fixed in space
Direction fixed in body
≡ ( )↑ ( ,
= )
≡ ( )↑ is point of contact. From question, ̇
=0
, ̇
=
Fig. 5.1.1(a) =0
= .
, F
Direction fixed in space
Direction fixed in body
≡ ( )↑ ( ,
= )
≡ ( )↑ is point of contact. From question, ̇
=0
=
, ̇
Fig. 5.1.1(b) =0
= .
, F
Direction fixed in space
Direction fixed in body
≡ ( )↓ ( ,
= )
≡ ( )↑ is point of contact. From question, ̇
=0
= − , ̇
Fig. 5.1.1(c) =0
= .
, F
Fig. 5.1
and hence the initial velocity of B is along the positive x. It follows that the friction F is equal to the maximum μR, and it acts along negative x. Now we try to write the “two differential equations” of motion—one for motion of centre of mass C, and another for motion relative to C. For the motion of C(x, y = a), we have () ) d 2 (a) a2 d 2 (x) = −μR, 0 = M = R − Mg , M θ¨ = F · a = (μMg)a M 2 dt 2 dt 2 ~ ~~ ~ ~ ~~ ~
5.1 Miscellaneous Examples on 2D Motion (Part I)
565
that is x¨ = −μg, θ¨ =
2μg . a
It follows that x˙ = v − μgt, θ˙ = ω +
2μg t. a
Since (vel. of B) = x˙ − a θ˙ ( acc. of B) = x¨ − a θ¨ = (−μg) − a ~ ~~ ~
()
2μg a
) = −3μg(< 0).
Since (vel.ofB)t=0 = v − aω, and (acc.ofB) = −3μg, after time v−aω 0), 3μg (> < t, pure rolling begins. Now the new (vel.ofB) will become 0, and hence for v−aω 3μg equations of motion are ) () d 2 (a) a2 d 2 (x) = −F, 0 = M = R − Mg , M M θ¨ = F · a , x¨ − a θ¨ = 0. 2 dt 2 dt 2 ~ ~~ ~ ~ ~~ ~ ( 2F ) = 0, and hence F = 0. Now no friction is required. It follows that −F − a Ma M Thus, the new equations of motion are x¨ = 0, θ¨ = 0
()
v − aω ≤t 3μg
)
It follows that ) () ) () ) () v − aω 2v + (aω) 2v + (v) v − aω =v− = < =v x˙ = v − μg 3μg 3 3 3 , the disc continues to roll with a constant velocity which is Thus, after time v−aω 3μg less than its initial velocity of the centre. Case II: when v is along positive x, ω is clockwise, and v < aω (see Fig. 5.1b): Here
566
5 Motion in Two Dimensions (Finite Forces)
(vel. of B) = (vel.ofBrel.toC) + (vel.ofC(x, y = a)) ( ) = a θ˙ along negativex + (vel.ofC(x, y = a)) ( ) = a θ˙ along negativex + (xalong ˙ positivex) (( ) ) = x˙ − a θ˙ along positivex . In short, we write (vel. of B) = x˙ − a θ˙ It follows that \ \ (vel. of B)t=0 = x ˙ t=0 − a θ˙ \t=0 = v − a θ˙ \t=0 = v − aω < 0 ~~ ~ ~
and hence the initial velocity of B is along the negative x. It follows that the friction F is equal to the maximum μR, and it acts along positive x. Now we try to write the “two differential equations” of motion—one for motion of centre of mass C, and another for motion relative to C. For the motion of C(x, y = a), we have ) () a2 d 2 (x) d 2 (a) M = μR, 0 = M = R − Mg , M θ¨ = −F · a = −(μMg)a, 2 dt 2 2 ~ dt ~~ ~ ~ ~~ ~
that is x¨ = μg, θ¨ = −
2μg a
It follows that x˙ = v + μgt, θ˙ = ω −
2μg t. a
Since (vel. of B) = x˙ − a θ˙ ( acc. of B) = x¨ − a θ¨ = (μg) − a ~~ ~ ~
()
−2µg a
) = 3μg(> 0).
Since (vel.ofB)t=0 = v − aω, and (acc.ofB) = 3μg, after time aω−v 0), 3μg (> aω−v (vel.ofB) will become 0, and hence for 3μg < t, pure rolling begins. Now the new equations of motion are
5.1 Miscellaneous Examples on 2D Motion (Part I)
567
) () d 2 (x) a2 d 2 (a) θ¨ = −F · a, x˙ − a θ¨ = 0., M = F, 0 = M = R − Mg , M 2 dt 2 2 ~ ~ ~ dt ~~ ~~ ~ ) ( F = 0, and hence F = 0. Now no friction is required. It follows that M − a −2F Ma Thus, the new equations of motion are x¨ = 0, θ¨ = 0
()
aω − v ≤t 3μg
)
It follows that ) () ) () ) () aω − v aω − v 2v + (aω) 2v + (v) =v x˙ = v + μg =v+ = > 3μg 3 3 3 Thus, after time aω−v the disc continues to roll with a constant velocity which is 3μg greater than its initial velocity of the centre. Case III: when v is along positive x, ω is anticlockwise (see Fig. 5.1c): Here (vel.ofB) = (vel.ofBrel.toC) + (vel.ofC(x, y = a)) ( ) = a θ˙ along negativex + (vel.ofC(x, y = a)) ( ) (( ) ) = a θ˙ along negativex + (xalong ˙ positivex) = x˙ − a θ˙ along positivex . In short, we write (vel. of B) = x˙ − a θ˙ It follows that \ \ (vel. of B)t=0 = x ˙ t=0 − a θ˙ \t=0 = v − a θ˙ \t=0 = v − a(−ω) = v + aω > 0, ~ ~~ ~
and hence the initial velocity of B is along the positive x. It follows that the friction F is equal to the maximum μR, and it acts along negative x. Now we try to write the “two differential equations” of motion—one for motion of centre of mass C, and another for motion relative to C. For the motion of C(x, y = a), we have ) () d 2 (x) d 2 (a) a2 M = −μR, 0 = M = R − Mg , M θ¨ = F · a = (μMg)a, 2 dt 2 2 ~ dt ~~ ~ ~ ~~ ~
that is
568
5 Motion in Two Dimensions (Finite Forces)
x¨ = −μg, θ¨ =
2μg . a
It follows that x˙ = v − μgt, θ˙ = −ω +
2μg t. a
Since (vel. of B) = x˙ − a θ˙ ( acc. of B) = x¨ − a θ¨ = (−μg) − a ~ ~~ ~
()
2μg a
) = −3μg(< 0).
Since (vel.ofB)t=0 = v + aω, and (acc.ofB) = −3μg, after time v+aω 0), 3μg (> v+aω (vel.ofB) will become 0, and hence for 3μg < t pure rolling begins. Now the new equations of motion are M
) () d 2 (a) d 2 (x) a2 θ¨ = F · a , x¨ − a θ¨ = 0 = −F, 0 = M = R − Mg , M 2 dt 2 2 ~ dt ~~ ~ ~ ~~ ~
( 2F ) It follows that −F − a Ma = 0, and hence F = 0. Now no friction is required. M Thus, the new equations of motion are x¨ = 0, θ¨ = 0
()
) v + aω ≤t . 3μg
It follows that ()
v + aω x˙ = v − μg 3μg
)
() =v−
v + aω 3
) =
2v − (aω) . 3
Subcase I: when aω < 2v. The disc continues to roll with a constant velocity along positive x.
2v−(aω) 3
Subcase II: when 2v < aω. The disc rolls back towards O with a constant velocity (aω)−2v along negative x. 3 We have seen that ) () v + aω . x˙ = v − μgt t < 3μg Observe that
v+aω 3μg
−
v μg
=
aω−2v 3μg
> 0, so
v μg
aω? (Fig. 5.2). Here ( ) ˙ negativex + (vel.ofC(x, y = a)) (vel.ofB) = (vel.ofBrel.toC) + (vel.ofC(x, y = a)) = a θalong ( ) (( ) ) = a θ˙ along negativex + (xalong ˙ positivex) = x˙ − a θ˙ along positivex .
In short, we write (vel. of B) = x˙ − a θ˙ It follows that \ \ ˙ t=0 − a θ˙ \t=0 = u − a θ˙ \t=0 = u − a(−ω) = u + aω > 0, (vel. of B)t=0 = x ~ ~~ ~
Direction fixed in space
Direction fixed in body
≡ ( )↓ ( ,
= )
≡ ( )↑ is point of contact. From question, ̇
=0
= − , ̇
Fig. 5.1.2 =0
= .
, F
Fig. 5.2
570
5 Motion in Two Dimensions (Finite Forces)
and hence the initial velocity of B is along the positive x. It follows that the friction F is equal to the maximum μR, and it acts along negative x. Now we try to write the “two differential equations” of motion—one for motion of centre of mass C, and another for motion relative to C. For the motion of C(x, y = a), we have M
) ( d 2 (x) d 2 (a) = −μR, 0 = M = R − Mg , Ma 2 θ¨ = F · a = (μMg)a, 2 2 dt ~~ ~ ~ ~ ~ dt ~~
that is x¨ = −μg, θ¨ =
μg a
It follows that x˙ = u − μgt, θ˙ = −ω +
μg t. a
Since (vel. of B) = x˙ − a θ˙ (acc. of B) = x¨ − a θ¨ = (−μg) − a ~ ~~ ~
()
μg g
) = −2μg(< 0)
Since (vel.ofB)t=0 = u + aω, and (acc.ofB) = −2μg, after time u+aω 0), 2μg (> < t, pure rolling begins. Now the new (vel.ofB) will become 0, and hence for u+aω 2μg equations of motion are M
) ( d 2 (a) d 2 (x) = −F, 0 = M = R − Mg , Ma 2 θ¨ = F · a , x˙ − a θ˙ = 0. 2 2 dt ~~ ~ ~ ~ ~ dt ~~
It follows that 0~ = x~~ ¨ − a θ~¨ =
−F M
−a
(
F Ma
)
= 0 and hence F = 0. Now no friction
is required. Thus, the new equations of motion are x¨ = 0, θ¨ = 0
()
) u + aω ≤t . 2μg
It follows that ) u + aω = u − μg ˙ t= u+aω (vel of centre when rolling commences) = x 2μg 2μg ) () u − (aω) u + aω = . =u− 2 2 ()
5.1 Miscellaneous Examples on 2D Motion (Part I)
571
Case I: When u < aω. So, u−(aω) (= x˙ = (vel.of the centre)) is negative, and hence 2 the ring rolls back towards O with a constant velocity u−(aω) of the centre. Observe 2 that u aω − u u + aω − = > 0, 2μg μg 2μg so
< u+aω . 2μg The required duration u 2g
( u
u+aω 2μg
)
+ 21 (−μg)
(
u+aω 2μg
)2
u + aω + aω−u 2μg 2 )⎞ ( ⎛ 1 u + 2 (−μg) u+aω 2μg uaω ⎝ ⎠ = 1+ aω−u 2μg 2 () ) 3u − aω u + aω 1+ = 2μg 2(aω − u)
=
=
u + aω u + aω (u + aω)2 = \ 2μg 2(aω − u) 4μg(aω − u)
Case II: When aω < u. Here, u−(aω) (= x˙ = (vel.of the centre)) is positive, and 2 hence the ring rolls along positive x with a constant velocity u−(aω) of the centre. 2
5.1.3 Example A uniform stick, of length 2a, hangs freely by one end, the other being closed to the ground. An angular velocity ω is then given to the stick, and when it has turned through a right angle the fixed end is let go. Show that on first touching the ground it will be in the upright position if. ω2 =
() ) p2 g 3+ , 2a p+1
where p is an odd multiple of π2 . Sol: (Fig. 5.3). Let ω' be the angular velocity of the stick when the fixed end is let go. By conservation principle of energy,
572
5 Motion in Two Dimensions (Finite Forces)
Vel =
′ ′
,2 Direction fixed in body Direction fixed in space
Fig. 5.3
() () ) ) ( ' )2 1 1 a2 a2 2 2 M + Ma M + Ma ω2 = −Mga. ω − 2 3 2 3 or 2a ( ' )2 2a 2 ω = ω −g 3 3 or / '
ω =
ω2 −
3g , 2a
which is a constant. If it touches the ground rotating =
π 2
ω'
If it touches the ground rotating =
3π 2
3π 2 ω'
=/
π 2
ω2 −
It takes the duration a under gravity. So
ω'
angle, then the time taken
.
angle, then the time taken =/
3π 2
ω2 −
etc. Suppose that it touches the ground rotating π 2
3g 2a
π 2
π 2
3g 2a
,
angle.
. During this duration, the centre of mass G falls a distance
5.1 Miscellaneous Examples on 2D Motion (Part I)
573
() π )2 () π )2 () ) ( ' ) π2 1 1 aπ 2 + (−g) ' + (−g) 2 ' −a = aω = ω' 2 ω 2 2 ω ~~ ~ ~ =a (π )
g
(π ) 2
( π )2
−
g
( π )2 2
2(ω' )2
2a 2 2 2aω2 − 3g ( π )2 (π ) a −g . =a 2 2 2 2aω − 3g
=a
−
2
Thus −a = a
(π ) 2
−g
( π )2 2
a 2aω2 − 3g
or −1 =
(π ) 2
−g
( π )2 2
1 2aω2 − 3g
or g
( π )2 2
(π ) 1 = +1 2aω2 − 3g 2
or ( )2 g π2 2aω − 3g = ( π ) +1 2 2
or ( ( π )2 ) g ω = .\ 3 + ( π )2 2a +1 2 2
574
5 Motion in Two Dimensions (Finite Forces) ≡ ( )↑ ( ,
Direction fixed in space
= )
impressed force
Direction fixed in body
̈= Fig. 5.1.4 ̇−
̇ = (vel of ) = 0
,
Fig. 5.4
5.1.4 Example A circular disc, of mass M, rolls in one plane upon a fixed horizontal plane, and its centre describes a straight line with uniform acceleration f. Find the magnitude and the line of action of the impressed force (Fig. 5.4). Sol: We have to find X, p. Here equations of motion are M f = ~M x¨~~= X~,
M
d2 (a) = R − Mg, dt 2
() M
) a2 θ¨ = X ( p), x˙ = a θ˙ . 2
It follows that () M f = X,
R = Mg,
) a2 M θ¨ = X ( p) 2
and ( ¨ f = ~x¨ = ~~a θ~ = a
and hence f =
2fp . a
X ( p) 2 M a2
) =
2(M f ) p 2fp 2X p = = , Ma Ma a
Thus, p = a2 .
5.1.5 Example A spindle of radius a carries a wheel of radius b, the mass of the combination being M and the moment of inertia I. The spindal rolls down a fixed track at an inclination α to the horizon, and, a string, wound round the wheel and leaving it at its under side, passes over a light pulley and has a mass m attached to the end which hangs vertically, the string between the wheel and pulley being parallel to the track. Show that the acceleration of the weight is Fig. 5.5.
5.1 Miscellaneous Examples on 2D Motion (Part I)
575
g(b − a)(Ma sin α + m(b − a)) . I + Ma 2 + m(b − a)2 Sol: Observe that (Fig.( 5.5). ) x˙ = a θ˙ , and b ax˙ = bθ˙ = x˙ + ξ˙ ~ ~~ ~ Hence () x˙
) b − 1 = ξ˙ . a
It follows that b−a x = ξ + C. a Since xt=0 = 0, ξ t=0 = 0 we have C = 0. Thus b−a x = ξ. a Since the spindle rolls, we can apply energy equation. We get (change inK. E.) =(Work done by gravity) + (Work done by tension) + (Work done by friction) +(Work done by normal reaction) ≡ ( )↑ Direction fixed in space Zero level Spindle of radius
( .
Direction fixed in body wheel of radius
fixed track
Fig. 5.5
≡ ( )↑
const

= ) =0
=0
= 0,
576
5 Motion in Two Dimensions (Finite Forces)
or ) (() () )() ) )2 ( )2 ( )2 I a 1 Ma 2 + I 1 ˙ ˙ ˙ M+ 2 +m ξ = ξ +m ξ 2 (b − a)2 2 a b−a ) ()() ) ) (( () )2 ) ( )2 1 I 1 1 1 ( )2 x˙ M + 2 (x) = = + m ξ˙ M(x) ˙ 2+ I ˙ 2 + m ξ˙ 2 a 2 2 a 2 ) ) ()() 1 ( )2 1 1 ( )2 2 ˙ ˙ + m ξ − 0 = (mgξ + Mg(x sin α)) + 0 + 0 + 0 = mgξ + Mg(x sin α) = M(x) ˙ + I θ 2 2 2 ~ ~~ ~ () ) () ) a Ma = mgξ + Mg ξ sin α = m + sin α gξ b−a b−a
Thus () ) ) () ( )2 1 Ma 2 + I Ma ˙ = m+ sin α gξ. + m ξ 2 (b − a)2 b−a On differentiation, we get () ) ) () 1 Ma 2 + I Ma ˙ ξ¨ = m + sin α g ξ˙ + m 2 ξ 2 (b − a)2 b−a Hence (reqd.acc.) = = ξ¨ = ~
( m+
Ma b−a Ma 2 +I (b−a)2
~~
) sin α g +m
~
=
g(b − a)(Ma sin α + m(b − a)) .\ I + Ma 2 + m(b − a)2
5.1.6 Example Three uniform spheres, each of radius a and of mass m, attract one another according to the law of inverse square of the distance. Initially they are placed on a rough horizontal plane with their centres forming a triangle whose sides are each of length 4a. Show that the velocity of their centres when they collide is / γ
5m , 14a
where γ is the constant of gravitation. Sol: In order to apply the principles of dynamics, we need a point fixed in space. Such a point is referred as the “origin”. Physically wise choice of origin makes calculation
5.1 Miscellaneous Examples on 2D Motion (Part I)
577
easier. Here we have three spheres. So we can take as a system in a number of ways – any one of them, any two of them, all the three. If we take any one of them as a system, then all the forces it experiences are “external forces”. If we take all the three as a system, then all the forces it experiences are “internal forces”. In this case, the net force experienced by their centre of mass, G, is zero, and hence the velocity of G is constant. Now since the three spheres are “placed” initially, the velocity of G is permanently zero. This shows that we can take G as the origin. From symmetry of the problem, we can say that at any time t, the centres of the spheres constitute an equilateral triangle.It follows that at each time t, the net force experienced by each sphere is directed towards the origin G. So, each sphere is moving under “central force” directed towards G. Suppose that at any time t, x is the distance between centres of the two spheres. Suppose that C1 is the centre of the sphere I, C2 is the centre of the sphere II, C3 is −−→ the centre of the sphere III. Then the net force experienced by sphere I is along C1 G and its magnitude ) ) () () m(m) m(m) ◦ cos 30 + γ cos 30◦ = γ x2 x2 ) () √ m2 1 m2 √ m2 γ . α = γ 2 3=γ 3 = √ x (2((C1 G) cos 30◦ ))2 3 (C1 G)2 (C1 G)2 Now since, this is a conservative force, the work done on sphere I x=2a
= ∫
x=4a
()
\ ) m2 γ γ m2 γ m 2 1 \\x=2a = √ √ (−d(GC1 )) = √ 2 3 (C1 G) 3 C1 G \x=4a 3
x 2
\ \x=2a \ 1 1 \x=2a 1 \ = γ m2 , = γ m 2 \\ \ 0 sec 30 x=4a x x=4a 4a
and hence work done on the whole system = γ m2
1 1 1 3 + γ m2 + γ m2 = γ m2 . 4a 4a 4a 4a
The change in K.E. of the whole system ⎞ ⎞ ⎞2 ⎛( ( ◦ ))\\ ( ) d x )2 () () ( ))\\ \ 2 ⎟ ⎟ ⎜⎜ 1 ◦ dt 2 sec 30 x 1 d 2a ⎟ ⎜ x=2a ⎟ + (· · · ) + (· · · )⎟ − 0 ⎜ \ =⎜ sec 30 ⎠ + m \ ⎠ ⎠ ⎝⎝ 2 m 5 ⎝ a 2 dt 2 x=2a ⎛⎛
=3
( ( () )() () ( () () ( () () ( )2 )2 ) )2 ) ))\\ ))\\ ))\\ 21 1 2 d x d x d x 1 1 ◦ ◦ ◦ \ \ \ = m sec 30 sec 30 m sec 30 + m \ \ \ 2 5 dt 2 2 dt 2 5 2 dt 2 x=2a x=2a x=2a )\ )2 () () ( ◦) \ 21 d x \ = . m sec 30 \ 10 dt 2 x=2a
Since there is rolling, we can apply the Energy equation. We get 21 m 10
() ()
)2 ))\\ d (x 3 ◦ \ sec 30 = γ m2 . \ dt 2 4a x=2a
578
5 Motion in Two Dimensions (Finite Forces)
Hence () required velocity =
/ ))\\ d (x 5γ m ◦ \ = sec 30 .\ \ dt 2 14a x=2a
5.1.7 Example A uniform sphere, of mass m and radius a, rolls on a horizontal plane. If the resistance of the air be represented by a horizontal force acting at the centre of the sphere equal v 2 and a couple about it equal to mμv 2 , where v is the velocity of the sphere to mλ a at any instant, and if V be the velocity at zero time, show that the distance described by the centre in time t is Fig. 5.6. 1 ln1 + AV t, A where A = 57 λ+μ . a Sol: (Fig. 5.6). Here equations of motion are m x¨ =
mλ 2 (x) ˙ + F, a
() R = mg,
m
) 2a 2 θ¨ = mμ(x) ˙ 2 − F(a), x˙ = a θ˙ . 5
It follows that ¨ = m(a θ)
mλ (a θ˙ )2 + F, a
()
) 2a 2 m θ¨ = mμ(a θ˙ )2 − Fa 5
or ˙ 2+ θ¨ = λ(θ)
F 5F 5 ˙ 2− , θ¨ = μ(θ) ma 2 2ma
It follows that Fig. 5.6 Direction fixed in space Direction fixed in body
( ̇ )2
( ̇ )2 ,
5.1 Miscellaneous Examples on 2D Motion (Part I)
θ¨ =
) 5 5( ˙ 2 μ(θ˙ )2 − θ¨ − λ(θ) 2 2
or 5 7 θ¨ = (λ + μ)(θ˙ )2 2 2 or () )2 x¨ x˙ = θ¨ = Aa(θ˙ )2 = Aa ~~ ~ ~ a a or dv = x¨ = A(x) ˙ 2 = Av 2 dt or dv = Adt. v2 On integration −1 = At + C. v It follows that C = −
−1 vt=0
=
−1 −V
. Hence.
AV t + 1 dt −1 −1 1 = = dx = = At + dx v V V ~ ~~ ~ dt
or Ad x = −
(AV )dt AV t + 1
On integration Ax = − ln(AV t + 1) + C1 . So C1 = Axt=0 . Hence
579
580
5 Motion in Two Dimensions (Finite Forces)
Ax = − ln(AV t + 1) + A xt=0 . It follows that  xt=0 − x(t) =
1 ln(AV t + 1).\ A
5.1.8 Example A uniform sphere rolls in a straight line on a rough horizontal plane and is acted upon by a horizontal force X at its centre. Show that the centre of the sphere moves as it would be if its mass were collected there and the force reduced to 57 X, and that the friction is equal to 27 X and is in a direction to that of X. (Fig. 5.7). Sol: Here equations of motion are (Fig. 5.7). M x¨ = F − X,
M
()
d2 (a) = R − Mg, dt 2
M
) 2a 2 θ¨ = −F(a), x˙ = a θ˙ . 5
It follows that ¨ = F − X, R = Mg, M(a θ) 2 (F − X ) = 5
()
2a M 5
)()
F−X Ma
)
() == ~
) 2a ¨ M θ = −F, 5 ~~ ~
and hence 2 (F − X ) = −F. 5
Fig. 5.7 Direction fixed in space Direction fixed in body
,
5.1 Miscellaneous Examples on 2D Motion (Part I)
Hence F = Now since
581
2X .\ 7
5X 2X ~M x¨ =~~F − X~ = 7 − X = − 7 , we have Mx =−
5X .\ 7
5.1.9 Example A man walks on a rough sphere so as to make it roll straight up a plane inclined at an angle α to the horizon, always keeping himself at an angle β from the highest point of the sphere. If the masses of the sphere and man, respectively, M and m, show that the acceleration of the sphere is (Fig. 5.8). 5g(m sin β − (M + m) sin α) . 7M + 5m(1 + cos(α + β)) Sol: Here equations of motion for sphere are Fig. 5.8 M x¨ = F − (Mg) sin α − S sin(α + β) − F1 cos(α + β), 0=M
d 2 (a) = R − (Mg) cos α − S cos(α + β) + F1 sin(α + β), dt 2 () ) 2a 2 M θ¨ = −F(a) − F1 (a). 5
Here equations of motion for man are m
d 2 (x + a sin(α + β)) = −(mg) sin α + S sin(α + β) + F1 cos(α + β), dt 2
0 = md Thus
2
(a+a cos(α+β)) dt 2
= −(mg) cos α + S cos(α + β) − F1 sin(α + β).
582
5 Motion in Two Dimensions (Finite Forces) direction fixed in space 1
≡ ( )↑ 1
( ,
= )
Path of direction fixed in body
≡ ( )↑ (vel of ) = ̇ −
Path of ( ,
̇=0
= ) , Fig. 5.1.9
Fig. 5.8 (
) x M 2a 5 a = −F − F1
)
M x = F − (Mg) sin α − S sin(α + β) − F1 cos(α + β)
,
) S sin(α + β) + F1 cos(α + β) − (mg) sin α − m x = 0 . S cos(α + β) − F1 sin(α + β) − (mg) cos α = 0
Hence S cos(α + β)(−(mg) cos α) − (−(mg) sin α − m x)(− ¨ sin(α + β)) =
F1 1 = (−(mg) sin α − mx) cos(α + β) − sin(α + β)(−(mg) cos α) −1
Thus S = cos(α + β)mg cos α + (mg sin α + m x) ¨ sin(α + β) = mg cos((α + β) − α) + m x¨ sin(α + β) = mg cos β + m x¨ sin(α + β), F1 = (mg sin α + m x) ¨ cos(α + β) − sin(α + β)mg cos α = mg sin(α − (α + β)) + m x¨ cos(α + β) = −mg sin β + m x¨ cos(α + β).
Further ) () 2a x¨ + M x¨ = −Mg sin α − S sin(α + β) − F1 (1 + cos(α + β)) M 5 a or () M
) 2a x¨ + M x¨ = −Mg sin α − (mg cos β + m x¨ sin(α + β)) sin(α + β) 5 α −(1 + cos(α + β))(−mg sin β + m x¨ cos(α + β))
5.1 Miscellaneous Examples on 2D Motion (Part I)
583
or 7M x¨ = −Mg sin α − mg cos β sin(α + β) − m x¨ sin2 (α + β) 5 +(1 + cos(α + β))mg sin β − (1 + cos(α + β))m x¨ cos(α + β) or x¨ =
−Mg sin α − mg cos β sin(α + β) + (1 + cos(α + β))mg sin β
7M 2 5 + m sin (α + β) + (1 + cos(α + β))m cos(α + β) −Mg sin α − mg sin((α + β) − β) + mg sin β 5g(m sin β − (M + m) sin α) = = .\ 7M 7M + 5m(1 + cos(α + β)) 5 + m + m cos(α + β)
5.1.10 Example A circular cylinder, of radius a and radius of gyration k, rolls inside a fixed horizontal cylinder of radius b. Show that the plane through the axes will move like a simple circular pendulum of length. ()
) k2 (b − a) 1 + 2 . a If the fixed cylinder be instead free to move about its axis, the corresponding pendulum will be of length (b − a)(1 + n), where n=
1+
k2 a2 b2 mk 2 a2 M K 2
,
m, M are, respectively, the masses of the inner and outer cylinders, and K is the radius of gyration of the outer cylinder about its axis (Fig. 5.9). Sol: Here equations of motion of point C of the inner cylinder are ) ( ¨ = −(mg) sin φ + F ˙ 2 = −(mg) cos φ + R, m((b − a)φ) m (b − a)(φ)
584
5 Motion in Two Dimensions (Finite Forces) =
≡ ( ),
=
−
≡
( ),
≡
Path of
( )
is the plane through the axes.
, , Direction fixed in moving cylinder
Direction fixed in space
Fixed point in space
Path of ,
, , Direction fixed in inner cylinder
Vertical Direction fixed in space
Direction fixed in outer cylinder
Fig. 5.9
Also, () ) b − a −(mg) sin φ + F k2 (−(mg) sin φ + F) =mk 2 a a m(b − a) () ) ) b ( b−a = mk 2 φ¨ = mk 2 − 1 φ¨ a a ) (b 2 ( ) d a φ − φ = mk 2 dt 2 ) d 2 (ψ − φ) ( 2 ) ( = mk 2 = mk θ¨ = −F · a dt 2 ~ ~~ ~ so
5.1 Miscellaneous Examples on 2D Motion (Part I)
585
k2 (−(mg) sin φ + F) = −F · a a or k 2 mg sin φ . k2 + a2
F= Next since
k 2 mg sin φ a 2 mg sin φ ¨ = −(mg) sin φ + F = −(mg) sin φ + m((b − a)φ) = − , ~ ~~ ~ k2 + a2 k2 + a2 we have ¨ =− m((b − a)φ)
k2
a2 mg sin φ + a2
and hence a2 g ) φ = −( 2 sin φ. k + a 2 (b − a) For small oscillations, equation a2 g ) φ¨ = − ( 2 sin φ k + a 2 (b − a) becomes φ¨ = − (
k2
a2 g ) φ. + a 2 (b − a)
So, OC moves like a simple circular pendulum of length =
g a2 g
(k 2 +a 2 )(b−a)
() ) k2 = (b − a) 1 + 2 .\ a
Second part: Here equations of motion of point C of the inner cylinder are ) ( ¨ = −(mg) sin φ + F. ˙ 2 = −(mg) cos φ + R, m((b − a)φ) m (b − a)(φ)
586
5 Motion in Two Dimensions (Finite Forces)
Also, equation for the relative motion of the inner cylinder about its centre of mass C is () () ) ) Fb2 −F(b) k2 k2 k2 (F − mg sin φ) + m (F − mg sin φ) − mb ((F − mg sin φ) − mbχ¨ ) = = 2 2 a MK a MK a () ) ) () ( ( ) b − a ) mk 2 −(mg) sin φ + F b − bχ¨ = mk 2 = φ¨ − χ¨ = mk 2 θ¨ = −F · a a m a a ~ ~~ ~
so ) () k2 Fb2 = −F · a (F − mg sin φ) + m a MK2 or F=
k 2 mg sin φ 2 2
k 2 + a 2 + m Mk Kb 2
.
Now, from ⎛ ⎞ 2 2 a 2 + m kMkb2 k 2 mg sin φ ⎝ ⎠mg sin φ, ¨ = −(mg) sin φ + F = −mg sin φ + = − m((b − a)φ) 2 2 2 2 ~~ ~ ~ k 2 + a 2 + m kMkb2 k 2 + a 2 + m kMkb2
we have ( φ¨ = −
2 2
a 2 + m kMkb2 1 b − a k 2 + a 2 + m Mk 2Kb22
)
(
2 2
a 2 + m Mk Kb 2 1 g sin φ ≈ − b − a k 2 + a 2 + m Mk 2Kb22
) gφ.
So, OC moves like a simple circular pendulum of length = (
(
g 2 2
a 2 +m k b 2 1 MK b−a k 2 +a 2 +m k 2 b2 2
) = (b − a) 1 + g
)
k2 2 2
k b a2 + m M K2
⎛ = (b − a)⎝1 +
1+
k2 a2 b2 mk 2 a2 M K 2
⎞ ⎠ = (b − a)(1 + n).[
MK
5.1.11 Example A uniform hoop has a fine string wound round it. The hoop is placed upright on a horizontal plane, and the string, leaving the hoop at its highest point, passes over a smooth pulley at a height above the plane equal to the diameter of the hoop and has a particle attached to the other end. Find the motion of the system, supposed to be all
5.1 Miscellaneous Examples on 2D Motion (Part I)
587
in one vertical plane; and show that whether the plane be smooth or rough the hoop will roll without slipping (Fig. 5.10). Sol: The equations motion of centre of mass G is (Fig. 5.10) d 2 (a) = R − Mg M x¨ = F + T , 0 = M 2 ~ dt ~~ ~ The equation motion relative to the centre of mass G is (
) Ma 2 θ¨ = −Fa + T a or Ma θ¨ = −F + T.
Also, since x˙ − a θ˙ = 0 we have F+T −F + T −F + T = x¨ = a θ¨ = a = , M Ma M and hence −F + T F+T = . M M It follows that F = 0. Now our equation is M x = T. ( From Fig. 5.11, the acceleration of mass m is a θ¨ = x¨ = the acceleration of mass m is =
Fig. 5.10
T M
mg − T . m
,
Direction fixed in the body
Direction fixed in space
)
. By Newton’s law,
588
5 Motion in Two Dimensions (Finite Forces)
=
((
2
((
)( ∙
))) )
,
Instantaneous axis of rotation
Fig. 5.11
It follows that mg m is M+m
T M
=
mg−T m
, and hence T =
Mmg . M+m
Hence the acceleration of mass
5.1.12 Example A disc rolls upon a straight line on a horizontal table, the flat surface of the disc being in contact with the plane. If v be the velocity of the centre of the disc at any instant, 27π v, where μ is the coefficient of friction show that it will be at rest after a time 64μg between the disc and table. Sol: Here, the plane of the paper is a horizontal table. Since the disc rolls upon a straight line on a horizontal table, the vertical line through the point of contact, B, of the disc with the given straight line, is the instantaneous axis of rotation for the disc. We shall apply 2.2.11(3). On taking torque about the instantaneous axis of rotation, we get Fig. 5.11 ()
) )) ) ( ) { θ =π { r=2a sin θ () ()() d v M a2 + Ma 2 = ((dr )(r · dθ)) g r μ 2 dt a πa 2 θ =0 r=0 ~ ~~ ~ ) ( ) () μMg 8a 3 θ =π 3 μMg θ =π r=2a sin θ 2 μMg θ =π (2a sin θ )3 ∫ ∫ ∫ ∫ sin θdθ dθ = = r dr dθ = 2 2 πa θ =0 π a θ =0 3 πa 2 3 θ =0 r=0 ( ) ) () θ = π2 μMg 16a 2 μMg 8a 3 μMg 32a 3 ∫ = 2 = , sin θ dθ = πa 2 3 π 3 3 π 9 θ =0
M
3a dv = 2 dt
M
and hence M
3a dv μMg 32a = 2 dt π 9
5.1 Miscellaneous Examples on 2D Motion (Part I)
589
It follows that dv μg 64 = . dt π 27 Hence, the required duration =
27π v. 64μg
5.1.13 Example A perfectly rough cylindrical grindstone, of radius a, is rotating with uniform acceleration about its axis, which is horizontal. If a sphere in contact with its edge can remain with its centre at rest. Show that the angular acceleration of the grindstone .. must not exceed 5g 2a Sol: Here, bψ = aθ. So −5 Mg sin φ −5 F −F · b −5 = bψ¨ = a θ¨ = a f, g sin φ = = =b 2 ~ ~~ ~ 2 2 M 2 M Mb2 5 and hence  f  =
5g 2a
sin φ ≤
5g 2a
\ (Fig. 5.12). sphere Centre of mass, rest. So
2,
of sphere is at
2
sin 90°
sin( − )
or
, Cylindrical grindstone 1
, Direction fixed in space
Direction fixed in space Angular acceleration ≡
Fig. 5.12
=
=
sin .
590
5 Motion in Two Dimensions (Finite Forces)
5.1.14 Example A perfectly rough ball is at rest within a hollow cylindrical garden roller, and the g(b − a), roller is then drawn along a level path with uniform velocity V . If V 2 > 27 7 show that the ball will roll completely round the inside of the roller, a and b being the radii of the ball and roller. Sol: Here, equations of motion are ) ( ¨ = −(mg) sin φ + F. ˙ 2 = −(mg) cos φ + R, m((b − a)φ) m (b − a)(φ) Also, equation for the relative motion of the ball about its centre of mass C is () () ) ) Fb2 −F(b) k2 k2 k2 (F − mg sin φ) + m (F − mg sin φ) − mb ((F − mg sin φ) − mbχ¨ ) = = 2 2 a MK a MK a () () ) ) 2 ) b − a ) ( ( −(mg) sin φ + F mk b = − bχ¨ = mk 2 φ¨ − χ¨ = mk 2 θ¨ = −F · a a m a a ~ ~~ ~
so () ) Fb2 k2 = −F · a. (F − mg sin φ) + m a MK2 Here equations of motion of centre of mass C of ball are Fig. 5.13 ( ) ˙ 2 = −(mg) cos θ + R, m((b − a)θ¨ ) = −(mg) sin θ + F m (b − a)(θ) Also, equation for the relative motion of the ball about its centre of mass C is Fig. 5.13
Cylindrical roller
,
vel = , ball Direction fixed in roller at zero time
Direction fixed in ball at zero time
5.1 Miscellaneous Examples on 2D Motion (Part I)
591
() () ) ) 2a d V 2a (F − mg sin θ ) = bm + F − mg sin θ 5 5 dt b 2a (bm φ¨ + F − mg sin θ ) = 5 () ) 2ma F − mg sin θ = bφ¨ + 5 m )() ) () b b − a F − mg sin θ 2a 2 φ¨ + = m 5 a a m(b − a) )() ) () ) () b b−a 2a 2 2a 2 φ¨ + θ¨ = m ψ¨ = −F · a. = m 5 a a 5 ~ ~~ ~ Hence 2a (F − mg sin θ ) = −F · a, 5 or F =
2mg sin θ . 7
Next since
m((b − a)θ¨ ) = −(mg) sin θ + F = −(mg) sin θ + ~~ ~ ~
5mg sin θ 2mg sin θ =− , 7 7
we have θ¨ = −
5g sin θ 7(b − a)
Hence 10g d d (( )2 ) = θ˙ (cos θ ). dt 7(b − a) dt It follows that ( )2 θ˙ = \ Now since θ˙ \θ =0 = Thus
a ψ˙ θ =0 −V b−a
10g cos θ + C. 7(b − a)
=
a0−V b−a
=
−V , b−a
we have C =
( −V )2 b−a
−
10g 7(b−a)
592
5 Motion in Two Dimensions (Finite Forces)
(() ) ) −V 2 −mg cos θ + R 10g 10g 2 = = (θ˙ ) = cos θ + − m(b − a) 7(b − a) b−a 7(b − a) ~~ ~ ~ Hence 10g −mg cos θ + R = cos θ + m(b − a) 7(b − a)
(()
−V b−a
)2
10g − 7(b − a)
)
or 10g −mg cos θ + R = cos θ + m 7
()
) V2 10g − . b−a 7
Since for minimum V , Rθ =π = 0, we have g=
−mg(−1) + 0 10g = (−1) + m 7 ~ ~~
()
10g (Vmin )2 − b−a 7
) = ~
20g (Vmin )2 − , b−a 7
and hence g=
20g (Vmin )2 − b−a 7
or / Vmin =
27g (b − a).\ 7
5.1.15 Example A solid uniform disc, of radius a, can turn freely about a horizontal axis through its centre, and an insect, of mass n1 that of the disc, starts from its lowest point and moves along the rim with constant velocity relative to the rim. Show that it will never get to the highest point of the disc if this constant velocity is less than 2/ 2ga(n + 2). n
5.1 Miscellaneous Examples on 2D Motion (Part I)
593
,
Insect Direction fixed in disc
Direction fixed in space Direction fixed in space
Fig. 5.14
Sol: Let m be the mass of the insect. Hence, mass of the disc is mn. Equations of motion of insect are Fig. 5.14 ) ( ¨ = F − (mg) sin θ m a(θ˙ )2 = R − (mg) cos θ, m(a θ) Equation of motion of disk is ) a2 ¨ φ = F(a). (mn) 2
()
From question, ) ( ( ( ))) ( = (vel of insect rel to rim ) = const , V a θ˙ + φ˙ = a θ˙ − − a φ˙ ~ ~~ ~
and hence F − mg sin θ F − (mg) sin θ 2F F − mg sin θ 2F(a) =a + = +a + a φ = a θ¨ + a φ¨ = 0 ~ ~~ ~ m mn m ma (mn)a 2
Thus F − mg sin θ 2F + =0 m mn or F= Next since
n mg sin θ. n+2
594
5 Motion in Two Dimensions (Finite Forces)
¨ = F − mg sin θ = m(a θ)
−2 n mg sin θ − mg sin θ = mg sin θ n+2 n+2
we have θ¨ =
−2 g sin θ n+2a
and hence d
(( ) ) 2 θ˙ dt
=
4g d (cos θ ). (n + 2)a dt
It follows that ( )2 4g cos θ + C. θ˙ = (n + 2)a ( \ )2 Suppose that θ t=0 = 0. It follows that C = θ˙ \t=0 −
4g (n+2)a
Hence
( )2 ( \ )2 4g 4g cos θ θ˙ = + θ˙ \t=0 − . (n + 2)a (n + 2)a For the initial angular momentum, we have ( (mn)
) ( ( ( \ ))) a 2 ( \\ ) φ˙ t=0 = (A. M. of disc )t=0 = ( A. M. of insect )t=0 = m a θ˙ \t=0 a, ~ ~~ ~ 2
we have ()
\ V − θ˙ \t=0 a
)
\ \ 2 \ = φ˙ \t=0 = θ˙ \\ n t=0 ~ ~~ ~
or \ V n . θ˙ \t=0 = a n+2 \ The insect will just get to the highest point when θ˙ \θ =π = 0, that is, the insect will just get to the highest point when
5.1 Miscellaneous Examples on 2D Motion (Part I)
595
( \ )2 4g cos π 4g + θ˙ \t=0 − = 0, (n + 2)a (n + 2)a that is, the insect will just get to the highest point when ( \ )2 θ˙ \t=0 =
8g , (n + 2)a
that is, the insect will just get to the highest point when ()
V n a n+2
)2 =
8g , (n + 2)a
that is, the insect will just get to the highest point when V2 =
8ga(n + 2) , n2
that is, the insect will just get to the highest point when V =
2/ 2ga(n + 2). \ n
5.1.16 Example Inside a rough hollow cylinder, of radius a and mass M, which is free to turn about its horizontal axis, is placed an insect of mass m. If the insect starts from the lowest generator walks in a plane perpendicular to the axis of the cylinder at a uniform rate v relatively to the cylinder, show that the plane containing it and the axis never makes with the upward drawn vertical an angle smaller than ⎞ 2 v Mk ⎠, 2 cos−1 ⎝ / 2a mga ( Mk 2 + ma 2 ) ⎛
where Mk 2 is the moment of inertia of the cylinder about its axis (Fig. 5.15). Sol: Equations of motion of insect are
596
5 Motion in Two Dimensions (Finite Forces)
, , Cylinder
Insect Direction fixed in cylinder
Direction fixed in space Direction fixed in space
Fig. 5.15
) ( ¨ = F − (mg) sin θ m a(θ˙ )2 = R − (mg) cos θ, m(a θ) Equation of motion of cylinder is (
) Mk 2 φ¨ = F(a)
From question, ( ) ( ( ( ))) a θ˙ + φ˙ = a θ˙ − − a φ˙ = (vel of insect rel to rim of cylinder) = const, v ~ ~~ ~
and hence F − mg sin θ F − mg sin θ F − (mg) sin θ Fa 2 F(a) = =a + +a + a φ¨ = a θ¨ + a φ¨ = 0 ~ ~~ ~ m m ma Mk 2 Mk 2
Thus Fa 2 F − mg sin θ =0 + m Mk 2 or F=(
Next since
g sin θ 1 m
+
a2 Mk 2
)=
m Mk 2 g sin θ . Mk 2 + ma 2
5.1 Miscellaneous Examples on 2D Motion (Part I)
597
m Mk 2 g sin θ −ma 2 m(a θ¨ ) = F − mg sin θ = − mg sin θ = mg sin θ, ~ ~~ ~ Mk 2 + ma 2 Mk 2 + ma 2 we have θ¨ =
−ma 2 g sin θ Mk 2 + ma 2 a
and hence d
(( ) ) 2 θ˙ dt
=
2ma 2 g d (cos θ ). 2 2 Mk + ma a dt
It follows that ( )2 θ˙ =
g 2ma 2 cos θ + C. 2 2 Mk + ma a
( \ )2 Suppose that θ t=0 = 0. It follows that C = θ˙ \t=0 − ( )2 θ˙ =
g 2ma 2 . Mk 2 +ma 2 a
Hence
( \ )2 g g 2ma 2 2ma 2 ˙\ cos θ + θ .(∗) − t=0 2 2 2 2 Mk + ma a Mk + ma a
For the initial angular momentum, we have (
Mk 2
)( \ ) ( ( ( \ ))) φ˙ \t=0 = (A. M. of cylinder )t=0 = (A. M. of insect )t=0 = m a θ˙ \t=0 a, ~ ~~ ~
we have \ \ ) \ ma 2 \\ \ \ θ˙ − θ˙ t=0 = φ˙ t=0 = a Mk 2 \t=0 ~ ~~ ~
(v
or \ Mk 2 v θ˙ \t=0 = . a Mk 2 + ma 2 Now, from (∗),,
598
5 Motion in Two Dimensions (Finite Forces)
)2 () v Mk 2 g g 2ma 2 2ma 2 2 ˙ cos θ + 0 < (θ ) = − 2 2 2 2 2 2 Mk + ma a a Mk + ma Mk + ma a ~~ ~ ~ or )2 () ) () v Mk 2 2ma 2 g 2 θ 2 sin < Mk 2 + ma 2 a 2 a Mk 2 + ma 2 ) ( 2 2 4 k or 4mag sin2 θ2 < av 2 MkM2 +ma 2. We have to show that ⎞ ⎛ 2 v Mk ⎠ < (π − θ ) 2 cos−1 ⎝ / 2a mga ( Mk 2 + ma 2 ) that is, ⎞ ) () 2 v Mk θ π −1 ⎝ ⎠ / − < cos 2a mga ( Mk 2 + ma 2 ) 2 2 ⎛
that is, ⎞ ⎛ 2 v π Mk θ ⎠ < − cos−1 ⎝ / 2 2 2a mga ( Mk 2 + ma 2 ) that is, ⎞
⎛
v Mk θ ⎠ < sin−1 ⎝ / 2 2a mga ( Mk 2 + ma 2 ) 2
that is, sin
that is,
v Mk 2 θ / < ( 2 2a mga Mk 2 + ma 2 )
5.1 Miscellaneous Examples on 2D Motion (Part I)
sin2
599
v2 θ M 2k4 ) ( < 2 2 4a mga Mk 2 + ma 2
that is, 4mag sin2
v2 θ M 2k4 < 2 . 2 a Mk 2 + ma 2
This has been shown to be true. \
5.1.17 Example A rough lamina, of mass M, can turn freely about a horizontal axis passing through its centre of gravity, the moment of inertia about this axis being Mk 2 . Initially the lamina was horizontal and a particle of mass m was placed on it at a distance c from the axis and then the motion was allowed to happen. Show that the particle will begin to slide on the lamina when the latter has turned through an angle μMk 2 , Mk 2 + 3mc2
tan−1 where μ is the coefficient of friction.
Sol: Observe that the path of particle m is a vertical circle, of radius a and centre G, where G is the centre of mass of the lamina. Let R, F be the normal reaction, and friction on the particle due to lamina. Let θ (≡ θ (t)) be the angle which the radius through m makes with horizontal (Fig. 5.16). Here, equations of motion of the particle are ( ( ) ) 2 = F − (mg) sin θ, m(c θ ) = (mg) cos θ − R. m c θ˙ The equation of motion of lamina is Fig. 5.16
Direction fixed in space
Direction fixed in lamina
600
5 Motion in Two Dimensions (Finite Forces)
(
Mk 2
( ) (mg) cos θ − R ) ¨ = R·c = Mk 2 (θ) mc ~ ~~ ~
Hence (
Mk 2
) (mg) cos θ − R = R·c mc
or R=
Mk 2 mg cos θ. mc2 + Mk 2
Now since (
) Mk 2 mg Mk 2 (θ¨ ) = R · c = cos θ · c, ~ ~~ ~ mc2 + Mk 2 we have (
) Mk 2 (θ¨ ) =
Mk 2 cmg cos θ mc2 + Mk 2
or θ¨ =
cmg cos θ. mc2 + Mk 2
Hence d d (( )2 ) 2cmg θ˙ = (sin θ ). 2 2 dt mc + Mk dt It follows that ( )2 θ˙ =
2cmg sin θ + C. mc2 + Mk 2
\ Now since θ˙ \θ=0 = 0, we have C = 0, and hence F − mg sin θ 2cmg = (θ˙ )2 = sin θ 2 + Mk 2 mc mc ~~ ~ ~ It follows that
5.1 Miscellaneous Examples on 2D Motion (Part I)
601
2cmg F − mg sin θ = sin θ mc mc2 + Mk 2 or () F = mg sin θ 1 +
) 2mc2 . mc2 + Mk 2
At the verge of slipping, mg sin θ
( ( ) ) Mk 2 mg 3mc2 + Mk 2 2mc2 = μ = mg sin θ 1 + cos θ = F = μR ~ ~~ ~ mc2 + Mk 2 mc2 + Mk 2 mc2 + Mk 2
or mg sin θ
() ) Mk 2 mg 3mc2 + Mk 2 = μ cos θ mc2 + Mk 2 mc2 + Mk 2
or tan θ =
μMk 2 .\ Mk 2 + 3mc2
5.1.18 Example A uniform beam, of mass M and length l, stands upright on perfectly rough ground; on the top of it, which is flat, rests a weight of mass m, the coefficient of friction between the beam and weight being μ. If the beam is allowed to fall on the ground, its inclination when the weight slips are given by Fig. 5.17 ()
) 4M M + 3m cos θ − sin θ = M + 2m. 3 6μ
Sol: ( ( ) ) 2 = (mg) cos θ − R, m(l θ ) = (mg) sin θ − F, m l θ˙
602
5 Motion in Two Dimensions (Finite Forces)
Fig. 5.17
Direction fixed in space
Direction fixed in body
≡ ( ) ,
and equation of motion of beam M is Fig. 5.17 M
l mg sin θ − F l 2 (mg) sin θ − F =M 3 m 3 ml ⎛ ( )2 ⎞ l () )2 ) () 2 l ⎟ l l 2 ⎜ = M θ = ⎝M +M sin θ + F(l), ⎠θ¨ = (Mg) 3 3 2 2 ~
~~
~
and hence M
() ) l l mg sin θ − F = (Mg) sin θ + F(l) 3 m 2
or () ) 1 mg sin θ − F = (Mg) sin θ + F M 3m 2 or F=
( ) −(Mg) 16 sin θ 1+
M 3m
=
−Mmg sin θ. 2(3m + M)
Now since M
) () ) () ) () Mg Mg l2 l −Mmg sin θ + F(l) = l sin θ + F = l sin θ + sin θ θ = (Mg) 3 2 2 2 2(3m + M) () ) Mg −m Mg 2m + M =l sin θ 1 + =l sin θ 2 3m + M 2 3m + M '
5.1 Miscellaneous Examples on 2D Motion (Part I)
603
we have M
Mg 2m + M l2 θ¨ = l sin θ 3 2 3m + M
or θ¨ =
2m + M 3g sin θ. 3m + M 2l
It follows that d
(( ) ) 2 θ˙ dt
=−
2m + M 3g d (cos θ ), 3m + M l dt
and hence ( )2 2m + M 3g cos θ + C. θ˙ = − 3m + M l \ Now since θ˙ \θ=0 = 0, we have C =
2m+M 3g . 3m+M l
Hence
mg cos θ − R 2m + M 3g ˙ 2= = (θ) (1 − cos θ ) . ml 3m + M ~~ l ~ ~ It follows that 2m + M 3g mg cos θ − R = (1 − cos θ ) ml 3m + M l or R = mg cos θ −
2m + M 3mg(1 − cos θ ). 3m + M
At the verge of slipping, ) () Mmg 2m + M sin θ = F = μR = μ mg cos θ − 3mg(1 − cos θ ) ~ ~~ ~ 2(3m + M) 3m + M or ) () 2m + M Mmg sin θ = μ mg cos θ − 3mg(1 − cos θ ) 2(3m + M) 3m + M
604
5 Motion in Two Dimensions (Finite Forces)
or M sin θ = 2μ((3m + M) cos θ − (2m + M)3(1 − cos θ )) or M sin θ = 2μ((9m + 4M) cos θ − 3(2m + M)) or 6μ(2m + M) = −M sin θ + 2μ(9m + 4M) cos θ or 2m + M =
9m + 4M −M sin θ + cos θ 6μ 3
or ()
) 4M M + 3m cos θ − sin θ = M + 2m.\ 3 6μ
5.1.19 Example A rough cylinder, of mass M, is capable of motion about its axis, which is horizontal; a particle of mass m is placed on it vertically above the axis and the system is slightly disturbed. Show that the particle will slip on the cylinder when it has moved through an angle θ given by μ(M + 6m) cos θ − M sin θ = 4mμ, where μ is the coefficient of friction (Fig. 5.18). Sol: Here, equations of motion of the particle are ) ( ¨ = (mg) sin θ − F m a(θ˙ )2 = (mg) cos θ − R, m(a θ) and the equation of motion of cylinder is
5.1 Miscellaneous Examples on 2D Motion (Part I) Fig. 5.18
605 Direction fixed in space
Direction fixed in cylinder
,
()
a (mg) sin θ − F M = 2 m
) () ) a 2 (mg) sin θ − F a2 M = M θ¨ = F · a. 2 ma 2 ~ ~~ ~
It follows that M
a mg sin θ − F = Fa 2 m
or (mg) sin θ − m(a θ ) = F =
Mm g sin θ M + 2m
or Mm 2m m(a θ¨ ) = (mg) sin θ − g sin θ = mg sin θ M + 2m ~ ~~ M + 2m ~ or θ¨ =
2m g sin θ M + 2m a
It follows that d
(( ) ) 2 θ˙ dt
=−
4m g d (cos θ ) M + 2m a dt
or ( )2 θ˙ = −
4m g cos θ + C. M + 2m a
606
5 Motion in Two Dimensions (Finite Forces)
\ Since θ˙ \θ =0 = 0, we have C =
4m g . M+2m a
Hence
4m g (mg) cos θ − R = (θ˙ )2 = (1 − cos θ ) . ma M + 2m ~~ a ~ ~ Thus 4m g (mg) cos θ − R = (1 − cos θ ). ma M + 2m a It follows that 4m (M + 2m) cos θ − 4m(1 − cos θ ) mg(1 − cos θ ) =mg R = (mg) cos θ − M + 2m M + 2m ~ ~~ ~ = mg
(M + 6m) cos θ − 4m , M + 2m
and hence R = mg
(M + 6m) cos θ − 4m . M + 2m
At the verge of slipping. ) () Mm (M + 6m) cos θ − 4m g sin θ = F = μR = μ mg ~ ~~ ~ M + 2m M + 2m or () ) Mm (M + 6m) cos θ − 4m g sin θ = μ mg M + 2m M + 2m or M sin θ = μ((M + 6m) cos θ − 4m) or μ(M + 6m) cos θ − M sin θ = 4mμ.\
5.1 Miscellaneous Examples on 2D Motion (Part I)
607
5.1.20 Example A hemisphere rests with its base on a smooth horizontal plane; a perfectly rough sphere is placed at rest on its highest point and is slightly displaced. Show that in the subsequent motion the angular velocity of the line joining the centres, when its inclination to the vertical θ, is θ 2 sin 2
/
5ng ) , c 7n − 5 cos2 θ (
and show also that the sphere will leave the hemisphere when θ satisfies the equation ) () 5 cos2 θ + 20 cos2 θ + 7(15 − 17n) cos θ + 70(n − 1) = 0, 5 3− n where c is the sum of the radii and n the ratio of the sum of the masses of the sphere and hemisphere to that of the sphere. [Use the principle of Linear Momentum and Energy.] (Fig. 5.19). Sol: Since there is no horizontal component of external forces on the system “m + M” by the conservation principle of linear momentum along horizontal direction, we have Direction fixed in space −( + )
Direction fixed in sphere , which was vertically downwards at time = 0.
≡ ( ), ≡ ( ), ,
= ∴
̇=
2
̇.
Vel = ( + ) ̇
,
1
Fig. 5.19
smooth
608
5 Motion in Two Dimensions (Finite Forces) ) )) ( (( M x˙ + m x˙ − m (a + b)θ˙ cos θ =
() M x˙ + m
) d (−((a + b) sin θ − x)) − 0 dt
= (x − comp. of linear momentum at time t) − (x − comp. of linear momentum at zero time ) = 0, ~ ~~ ~
and hence n x˙ − cθ˙ cos θ =
M +m x˙ − (a + b)θ˙ cos θ = 0 ~~ ~ ~ m
Thus n x˙ = cθ˙ cos θ. Here equation of energy for the system is (
1 M(x) ˙ 2+ 2
(
() )() ) )) ) 1 1 ( 2b2 d(θ + φ) 2 ˙ sin θ)2 + m −0 m (x˙ − (a + b)θ˙ cos θ)2 + (−((a + b)θ) 2 2 5 dt
(K.E.)final − ( K. E.)initial = work done by external forces = mg((a + b) − (a + b) cos θ ) ~ ~~ ~
or ) () ) 1 () 2b2 )( )2 )2 1 1 ( 2 ( M(x) ˙ 2+ m (x) m ˙ + (a + b)θ˙ − 2(a + b)x˙ θ˙ cos θ + θ˙ + φ˙ 2 2 2 5 = mg(a + b)(1 − cos θ ) or 1 M(x) ˙ 2+ 2
(
() )2 ) ) )2 1 ( 2 ( a θ˙ b2 m (x) ˙ + (a + b)θ˙ − 2(a + b)x˙ θ˙ cos θ + m θ˙ + 2 5 b
= mg(a + b)(1 − cos θ ) or () ) a )2 1 1 ( )2 2b2 ( 1+ ˙ 2 + m θ˙ (a + b)2 + (M + m)(x) 2 2 5 b ˙ − m(a + b)x˙ θ cos θ = mg(a + b)(1 − cos θ ) or
5.1 Miscellaneous Examples on 2D Motion (Part I)
609
7 ( )2 ˙ 2 + m θ˙ (a + b)2 − 2m(a + b)x˙ θ˙ cos θ = 2mg(a + b)(1 − cos θ ) (M + m)(x) 5 or M + m 2 7 ( )2 ˙ + θ˙ (a + b)2 − 2(a + b)x˙ θ˙ cos θ = 2g(a + b)(1 − cos θ ) (x) m 5 or ( )2 2 )2 ) 7 ( )2 θ˙ c ( 1( 7n − 5 cos2 θ = θ˙ c2 − cθ˙ cos θ 5n 5 n 7 ( )2 2 = θ˙ c − n(x) ˙ 2 5 7 ( )2 = n(x) ˙ 2 + θ˙ c2 − 2 x(n ˙ x) ˙ 5 7 = n(x) ˙ 2 + (θ˙ )2 c2 − 2 x(c ˙ θ˙ cos θ ) = 2gc(1 − cos θ ) = 4gc sin2 θ2 . 5 ~ ~~ ~ Thus ( ) ( )2 θ 7n − 5 cos2 θ c θ˙ = 20ng sin2 2 or θ θ˙ = 2 sin 2
/
5ng ) .\ c 7n − 5 cos2 θ (
Here, equations of motion of C2 (−((a + b) sin θ − x), (a + b) cos θ ) are ) (c( ) ) ) ( mc ( ˙ 2 sin θ sin θ + c(θ˙ )2 θ¨ cos θ sin θ + n − sin2 θ (θ˙ )2 = m θ¨ cos θ − (θ) n n ) ⎛ ( ˙ ⎞ θ ) () ( ) d cθ cos n d(x) ˙ 2 ⎝ ˙ ˙ 2 = m x¨ sin θ + (a + b)(θ) ˙ 2 sin θ + c(θ ) ⎠ = m sin θ + c(θ) =m dt dt ⎞ ⎞ ⎛⎛ ⎠ ˙ 2 ⎠ ˙ 2 ¨ ¨ = m ⎝⎝x¨ − (a + b)(cos ~ ~~θ θ~ + cos θ (θ) ) cos θ ~ ~~θ θ~ − sin θ (θ) ) sin θ + (a + b)(sin (( =m ~
and hence
) ( ) ) d2 d2 (−((a + b) sin θ − x)) sin θ − ((a + b) cos θ ) cos θ = (mg) cos θ − R, dt 2 dt 2 ~~ ~
610
5 Motion in Two Dimensions (Finite Forces)
( )( )2 ) mc ( = −R + (mg) cos θ.(∗) θ cos θ sin θ + n − sin2 θ θ˙ n Since (
) ( )2 θ 7n − 5 cos2 θ c θ˙ = 20ng sin2 , 2
we have ( ) ( )3 ( ) c 10 cos θ sin θ θ˙ + 7n − 5 cos2 θ 2θ˙ θ () ) ) ) θ d d (( 7n − 5 cos2 θ c(θ˙ )2 = 20ng sin2 = dt dt 2 ~ ~~ ~ = 20ng sin
θ θ cos θ˙ 2 2
and hence ) 10ng(1 − cos θ ) ( + 7n − 5 cos2 θ cθ 2 7n − 5 cos θ ( ) = 5 cos θ sin θ c(θ˙ )2 + 7n − 5 cos2 θ cθ¨ = 5ng sin θ . ~ ~~ ~
5 cos θ sin θ
Thus ) () ) ( 10ng(1 − cos θ ) 10(1 − cos θ ) 7n − 5 cos2 θ c θ = 5ng sin θ − 5 cos θ sin θ = 5ng sin θ 1 − cos θ 2 2 7n − 5 cos θ 7n − 5 cos θ = 5ng sin θ
7n − 10 cos θ + 5 cos2 θ 7n − 5 cos2 θ − 10 cos θ + 10 cos2 θ = 5ng sin θ . 7n − 5 cos2 θ 7n − 5 cos2 θ
Hence ( ) 7n − 10 cos θ + 5 cos2 θ . 7n − 5 cos2 θ cθ¨ = 5ng sin θ 7n − 5 cos2 θ From (∗), at the verge of leaving the hemisphere, R = 0, so at the verge of leaving the hemisphere, ) ) ( mc ( θ¨ cos θ sin θ + n − sin2 θ (θ˙ )2 = (mg) cos θ n or ( ) cθ¨ cos θ sin θ + n − sin2 θ c(θ˙ )2 = ng cos θ
5.1 Miscellaneous Examples on 2D Motion (Part I)
611
or 5ng sin θ
( ) 10ng(1 − cos θ ) 7n − 10 cos θ + 5 cos2 θ cos θ sin θ + n − sin2 θ = ng cos θ )2 ( 7n − 5 cos2 θ 7n − 5 cos2 θ
or ( ) ( ) ( ) 5 sin2 θ cos θ 7n − 10 cos θ + 5 cos2 θ + n − sin2 θ 10(1 − cos θ ) 7n − 5 cos2 θ ( ) = cos θ 49n 2 − 70n cos2 θ + 25 cos4 θ or ( ) 2 5 cos θ 7n − 10 cos θ + 5 cos2 θ − 5 cos3 θ (7n − 10 cos θ~ + ~5 cos ~~ θ~) ~ ~~ 2 1 ( ) ( ) +10n 7n − 5 cos2 θ − 7n cos θ + 5 cos3 θ − 10 7n − 5 cos2 θ − 7n cos θ + 5 cos3 θ ⎞ ⎞ ⎛ ⎛ 3 ⎠ 4 ⎠ 2 2 ⎝ +10 cos2 θ ⎝7n − 5~ cos cos θ~ + ~5 cos cos2 θ~ + ~25 cos ~~ θ~ − 7n ~~ θ~ − cos θ (49n − 70n ~~ θ~ ~ ~~ ~ ~~ 2
1
3
1
3
or ⎛
⎞
5 cos θ ⎝ 7n − 10 cos θ + 5 cos2 θ ) − 35n cos3 θ +10n(7n − 5 cos2 θ − 7n cos θ + 5 cos3 θ ⎠ ~~~~
~ ~~ ~
~ ~~ ~
3
2
1
~
~~
~
1
~ ~~ ~
~ ~~ ~
~ ~~ ~
2
3
1
3 2 cos θ~ + ~5 cos cos2 θ~ − ~49n 2~~cos θ~ = 0 −10(7n − 5~ cos ~~ θ~ − ~7n ~~ ~~ θ~) + 70n ~ ~~ 2
3
1
2
3
or ( ) (15n − 25) cos3 θ + 20n cos2 θ + 35n − 70n 2 + 70n − 49n 2 cos θ + 70n 2 − 70n = 0
or ) () 5 cos3 θ + 20 cos2 θ + 7(15 − 17n) cos θ + 70(n − 1) = 0.\ 5 3− n
612
5 Motion in Two Dimensions (Finite Forces)
5.2 Miscellaneous Examples on 2D Motion (Part II) 5.2.1 Example A thin hollow cylinder, of radius a and mass M, is free to turn about its axis, which is horizontal, and a similar cylinder, of radius b and mass m, rolls inside it without slipping, the axes of the two cylinders being parallel. Show that, when the plane of the two axes is inclined at an angle φ to the vertical, the angular velocity u of the larger is given by a 2 (M + m)(2M + m)u2 = 2gm 2 (a − b)(cos φ − cos α), provided both cylinders are at rest when φ = α. Fig. 5.20. Sol: We have to show that a 2 (M + m)(2M + m)(χ˙ )2 = 2gm 2 (a − b)(cos φ − cos α). Here, equations of motion of centre of mass C of the inner cylinder are ( ) ¨ = −(mg) sin φ + F ˙ 2 = −(mg) cos φ + R, m((a − b)φ) m (a − b)(φ) Also, equation for the relative motion of the inner cylinder about its centre of mass C is Fig. 5.20 Fixed point in space
Path of ,
, Direction fixed in inner cylinder
Vertical Direction fixed in space
Direction fixed in outer cylinder
5.2 Miscellaneous Examples on 2D Motion (Part II)
613
) ) () () −F · a F = b((F − mg sin φ) − ma χ ) = b (F − mg sin φ) − ma b (F − mg sin φ) + m M Ma 2 () ) ( ) ( )() a − b ) a mb2 −(mg) sin φ + F − aχ = mb2 φ − χ = mb2 θ¨ = −F · b = b m b b ~~ ~ ~
so ()
F b (F − mg sin φ) + m M
) = −F · b
or F=
m mM g sin φ m g sin φ = 2+ M 2M + m
Now from we have θ=
mM − 2M+m g sin φ
mb
=
−M g sin φ. 2M + m b
Next, from. () ) M +m mM ¨ = −(mg) sin φ + F = −mg sin φ + g sin φ = − m((a − b)φ) mg sin φ, ~ ~~ ~ 2M + m 2M + m
we have φ=−
1 M +m g sin φ. a − b 2M + m
It follows that 2 M +m d d (( )2 ) φ˙ g (cos φ) = dt a − b 2M + m dt or ( )2 φ˙ =
2 M +m g cos φ + C. a − b 2M + m
\ 2 M+m Now since φ˙ \φ=α = 0, we have C = − a−b g cos α, and hence 2M+m ( )2 φ˙ =
2 M +m g(cos φ − cos α). a − b 2M + m
614
5 Motion in Two Dimensions (Finite Forces)
Since mM 1 −F·a 1 χ= =− F =− g sin φ = 2 Ma Ma 2M + m ~ ~~Ma ~ () =
−m 1 2M + m a
()
) −m 1 (g sin φ) 2M + m a
)() ) 2M + m m a−b φ = φ, −(a − b) M +m M +m a
we have χ=
m a−b φ, M +m a
and hence χ˙ =
m a−b φ˙ + C1 . M +m a
\ Now since φ˙ \φ=α = 0, and χ˙ φ=α = 0, we have C1 = 0. It follows that χ˙ =
m a−b ˙ φ, M +m a
and hence (χ˙ )2 =
m2 (M + m)2
m2 (a − b)2 ( ˙ )2 (a − b)2 φ = 2 2 a a2 (M + m)
=
()
2 M +m g(cos φ − cos α) a − b 2M + m
1 2m 2 a − b g(cos φ − cos α). 2 M + m a 2M + m
Thus (χ˙ )2 =
1 2m 2 a − b g(cos φ − cos α) 2 M + m a 2M + m
or a 2 (M + m)(2M + m)(χ˙ )2 = 2gm 2 (a − b)(cos φ − cos α).\
)
5.2 Miscellaneous Examples on 2D Motion (Part II)
615
5.2.2 Example A perfectly rough solid cylinder, of mass m and radius r, rests symmetrically on another solid cylinder of mass M and radius R, which is free to turn about its axis which is horizontal. If m rolls down, show that at any time during the contact the angle φ which the line joining the centres make with the vertical is given by (R + r ) φ =
2(M + m) g sin φ. 3M + 2m
Find also the value of φ˙ when the cylinders separate (Fig. 5.21). Sol: Here, equation of motion for cylinder M is. ()
) 1 M R 2 ψ = F · R. 2
Also, equation of motion for cylinder m is Solid cylinder
Direction fixed in body
Direction fixed in body Direction fixed in space ≡ ( ),
≡ ( ),
≡
2
( )
( − ) = ( − ). ,
,
Solid cylinder 1
Direction fixed in space
Fixed point in space
Fig. 5.21
2 2
,
,
Couple Forward axle
Backward axle 1
Fig. 5.22
1
616
5 Motion in Two Dimensions (Finite Forces)
()
) 1 2 mr θ = F · r.(∗∗). 2
Equations for the centre of mass C2 of cylinder m are ( ( )2 ) = −S + (mg) cos φ, m((R + r ) φ) = −F + (mg) sin φ. m (R + r ) φ˙ From (∗), (∗∗) we have m+M m+M =F mM mM F r R 1 d2 F + = θ¨ + ψ¨ = == (r θ + Rψ) M ~~ 2 2 ~ 2 dt 2 ~ m
(mg sin φ − m(R + r )φ)
=
1 d2 R +r φ, (Rφ + r φ) = 2 dt 2 2
and hence (mg sin φ − m(R + r ) φ)
m+M R +r φ. = mM 2
Thus ()
1 m+M φ(R + r ) + 2 M
) =
m+M g sin φ M
or (R + r )φ¨ =
2(m + M) g sin φ. (∗) 2m + 3M
Hence d (( )2 ) −4(m + M) g d φ˙ = (cos φ) dt 2m + 3M R + r dt or ( )2 −4(m + M) g cos φ + C. φ˙ = 2m + 3M R + r \ Since φ˙ \φ=0 = 0, we have C =
4(m+M) g . 2m+3M R+r
Hence
( )2 4(m + M) g φ˙ = (1 − cos φ). 2m + 3M R + r
5.2 Miscellaneous Examples on 2D Motion (Part II)
617
Since () m(R + r )
) ( ) 4(m + M) g ˙ 2 = −S + (mg) cos φ (1 − cos φ) = m (R + r )(φ) 2m + 3M R + r ~~ ~ ~
we have ()
) 4(m + M) g m(R + r ) (1 − cos φ) = −S + (mg) cos φ. 2m + 3M R + r At the verge of separation, S = 0. So at the verge of separation ()
) 4(m + M) g (1 − cos φ) = mg cos φ 2m + 3M R + r
()
) 4(m + M) (1 − cos φ) = cos φ 2m + 3M
m(R + r ) or
or 4(m + M) − 4(m + M) cos φ = (2m + 3M) cos φ or 4(m + M) = (6m + 7M) cos φ or cos φ =
4(m + M) . 6m + 7M
Now since ( )2 4(m + M) g φ˙ = (1 − cos φ), 2m + 3M R + r we have \ φ˙ \
/ at the verge of separation
=
() ) 4(m + M) 4(m + M) g 1− 2m + 3M R + r 6m + 7M
618
5 Motion in Two Dimensions (Finite Forces)
/
4(m + M) g 2m + 3M 2m + 3M R + r 6m + 7M / 4(m + M) g . = 6m + 7M R + r
=
5.2.3 Example A locomotive engine of mass M has two pairs of wheels, of radius a, the moment of inertia of each pair about its axis being Mk 2 , and the engine exerts a couple L on the forward axle. If both pairs of wheels commence to roll without sliding when the engine starts, show that the friction between each of the front wheels and the line capable of being called into action must not be less than Fig. 5.22 L k2 + a2 . 2a 2k 2 + a 2 Sol: Here equation of motion of the forward axle is (
) Mk 2 θ = L − F · a − F · a.
Here equation of motion of backward axle is (
) Mk 2 θ = F1 · a + F1 · a.
Here, for the combined system, equation of motion of centre of mass is M(a θ ) = F + F − F1 − F1 =
) ( L − Mk 2 θ a
− 2F1 =
) ( L − Mk 2 θ a
and hence ) ( L − 2 Mk 2 θ M(a θ ) = a or θ= Since
L 1 . 2 M a + 2k 2
( −
) Mk 2 θ a
=
) ( L − 2 Mk 2 θ a
,
5.2 Miscellaneous Examples on 2D Motion (Part II)
619
() ) ( ) 1 L Lk 2 2 = Mk 2 θ¨ = L − 2Fa , = Mk a 2 + 2k 2 M a 2 + 2k 2 ~~ ~ ~
we have
a2
Lk 2 = L − 2Fa, + 2k 2
and hence ) () L a2 + k2 Lk 2 1 = L− 2 . F= 2a a + 2k 2 2a a 2 + 2k 2 Since ) () ( ) ( 2 ) 1 L Lk 2 = Mk 2 θ¨ = 2F1 a = Mk 2 2 2 2 a + 2k M a + 2k ~ ~~ ~
we have Lk 2 = 2F1 a a 2 + 2k 2 or F1 =
k2 L L a2 + k2 < .\ 2a a 2 + 2k 2 2a a 2 + 2k 2
5.2.4 Note Suppose that a particle, of mass m, is moving in an x yplane under the action of a force with components X, Y. Let (x, y) be the position of the particle at time t. We know that its equations of motion are m x = X, m y = Y. It follows that, for the time interval t1 ≤ t ≤ t2 ,
620
{ ~
t=t2
t=t1
5 Motion in Two Dimensions (Finite Forces)
{ t=t2 t=t2 d(m x) ) ( ) ( ˙ 2 dt = (m x) ˙ t=t ˙ t=t1 , = ∫ ˙ t=t2 − m x X dt = (m x)dt ¨ t=t1 = m x dt t=t1 t=t ~~ 1 ~
and hence ) ( ) ( ∫ X dt = m x ˙ t=t1 . ˙ t=t2 − m x
t=t2
t=t1
Similarly, ) ( ) ( ∫ Y dt = m y˙ t=t2 − m y˙ t=t1 .
t=t2
t=t1
The vector ()
t=t2
t=t2
t=t1
t=t1
)
∫ X dt, ∫ Y dt
) ( ) ( ) ( )) ˙ t=t1 , m y˙ t=t2 − m y˙ t=t1 m x ˙ t=t2 − m x ( ) ( ) = m x ˙ t=t2 , m y˙ t=t2 − m x ˙ t=t1 , m y˙ t=t1 = pt=t2 − pt=t1 ≡ /\p
=
((
is called the impulse on the particle during the time interval t1 ≤ t ≤ t2 . We have seen that impulse is equal to increment in the linear momentum.
5.2.5 Note Suppose that a particle, of mass m, is moving along xaxis. Suppose that at time t = 0, it is at rest, and at time t = 0, it is located at x = 0. Suppose that a force () () )) t X (t) ≡ A sin π τ acts on the particle during the time interval 0 ≤ t ≤ τ. Now the equation of motion of the particle is () () )) t . m x = A sin π τ It follows that
5.2 Miscellaneous Examples on 2D Motion (Part II)
() () () ))) 1 t m x˙ = A − cos π π + C. τ τ Now since x ˙ t=0 = 0, we have C = x˙ =
Aτ , π
and hence
() () () ))) t Aτ 1 − cos π . mπ τ
This shows that x=
() () () ))) τ t Aτ t − sin π + C1 . mπ π τ
Now since xt=0 = 0, we have C1 = 0, and hence () () () ))) t Aτ τ x= t − sin π . mπ π τ Since x˙ =
() () () ))) t Aτ 1 − cos π , mπ τ
change in velocity, /\u =
() () () ))) ( ( τ ))) Aτ 0 2 Aτ Aτ ( 1 − cos π − 1 − cos π = . mπ τ mπ τ mπ
Since x=
() () () ))) τ t Aτ t − sin π , mπ π τ
change in position, /\x () () () ))) ( ( τ ))) τ Aτ τ 0 Aτ 2 Aτ ( τ − sin π − 0 − sin π = . = mπ π τ mπ π τ mπ The average force, ( ( ))\t=τ ( ( ( t ))) A −τ cos π τt \t=0 ∫t=τ A −τ (−1 − 1) dt 2A t=0 A sin π τ π = = π = . = τ τ τ π Thus,
621
622
5 Motion in Two Dimensions (Finite Forces)
/\u =
∫t=τ ∫t=τ X dt t=0 X dt , /\x = t=0 τ. m 2m
t=τ
Conclusion If ∫ X dt(= τ (average vel.)) is a constant, and τ → 0, then the position t=0
of the particle remains unchanged, but there is a finite change in velocity. Further, the average force approaches infinity. Thus, we get an example of an infinite force acting for an infinitesimal duration such that there is no change in position, but there is an instantaneous change in velocity. Since no finite force can produce a discontinuous change in velocity, we have got a new concept of “force” which results into discontinuous change in velocity. Such a force is called impulsive force. Impulsive force does not cause displacement. It causes discontinuous change in velocity. In short, components of impulsive forces are given by ()
'
X = lim
t1 →t0
() ) ) t=t1 ' ∫ X dt , Y = lim ∫ Y dt .
t=t1
t=t0
t1 →t0
t=t0
Due to this simple connection between force X and impulsive force X ' , etc., it is unnecessary to repeat for impulsive forces results already known for ordinary forces. For example, impulsive reactions between the particles of rigid bodies obey the law of action and reaction. The idea of equivalent system of forces is valid for impulsive forces. The theory of moments of forces also applies to impulsive forces. Gravitational force remains finite as t1 → t0 , so gravity does not contribute to the impulsive force. Remark The dimension of impulsive force, MLT−1 , and the dimension of force, MLT−2 , are different, but both of them obey the same rules.
5.2.6 Note Suppose that there is a system. Let X, Y be the total components of external forces in the direction of O x, O y axes. Let Mx , M y be the components of linear momentum in the direction of O x, O y axes. We know that d My d Mx = X, = Y. dt dt It follows that () () () ) ) ) t=t1 t=t1 d M x dt = lim (Mx (t1 ) − Mx (t0 )) ≡ /\Mx , X ' = lim ∫ X dt = lim ∫ t1 →t0 t=t0 t1 →t0 t=t0 t1 →t0 dt
5.2 Miscellaneous Examples on 2D Motion (Part II)
623
and hence X ' = /\Mx . Similarly, Y ' = /\M y . Conclusion The sudden change in the linear momentum of a system is equal to the total external impulsive force. In short, we write /\M = F' .
5.2.7 Note By 4.2.7, for the motion of any body in one plane M
E d2x = X. 2 dt
Suppose that impulsive forces act for the time interval 0 ≤ t ≤ T . It follows that t=T
∫
t=0
()
) ) E { t=T (E ) E (){ t=T d2x M 2 dt = X dt = X ', X dt = dt t=0 ~~ ~ ~ t=0
where X ' is the impulse of the force acting at any point (x, y). Let u, v be the velocities of the centre of mass parallel to axes just before the impulsive forces act, and u ' , v ' be the velocities of the centre of mass parallel to axes just after the impulsive forces act. It follows that \ () ) () ) t=T t=T d ( ) dx d x \t=T d2x dt = M \\ = M u' − u X ' = ∫ M 2 dt = M ∫ dt dt dt t=0 t=0 dt t=0 and hence ( ) E ' M u' − u = X. Similarly, ( ) E ' M v' − v = Y. Conclusion Change in the linear momentum of the mass M, supposed collected at the centre of mass, in any direction is equal to the sum of impulses in that direction.
624
5 Motion in Two Dimensions (Finite Forces)
5.2.8 Note By 4.2.7, for the motion of any body in one plane, E(
(M.I. of the lamina about Gz axis) θ =
) x 'Y − y' X .
lamina
Suppose that impulsive forces acts for the time interval 0 ≤ t ≤ T . It follows that (
{ ( ) \ ) t=T t=T Mk θ˙ \t=0 = Mk 2 ∫ θ dt = 2
t=0
~
t=T
( Mk
2
)
¨ = θdt
t=0
{
t=T
t=0
E (
) ) x Y − y X dt , '
'
lamina
~ () () ) ) E t=T ( E E( t=T t=T ) ) ∫ x ' Y − y ' X dt = x ' ∫ Y dt − y ' ∫ X dt = = x 'Y ' − y' X ' lamina
t=0
lamina
~~
(
t=0
t=0
lamina
where X ' , Y ' are the components of impulse of the force acting at any point (x, y). Let ω, ω' be the angular velocities of the body just before the impulsive forces act, and just after the impulsive forces act. It follows that (
E ( )( ) \t=T ) ) ( Mk 2 ω' − ω = Mk 2 θ˙ \t=0 = x 'Y ' − y' X ' ~
lamina
~~
~
and hence E ( )( ) ) ( x 'Y ' − y' X ' (change produced in the angular momentum about the centre of mass) = Mk 2 ω' − ω = ~
lamina
~~
= (moment about the centre of mass of impulses of the force)
Conclusion The change produced in the A.M. about the centre of mass is equal to the moment about the centre of mass of the impulses of forces.
5.2.9 Example A uniform rod AB, of length 2a, is lying on a smooth horizontal plane and is struck by a horizontal blow, of impulse P, in a direction perpendicular to the rod at a point distant b from its centre. Find the motion.
~
5.2 Miscellaneous Examples on 2D Motion (Part II)
625
Sol: Let u ' be the velocity of the centre of mass G perpendicular to the rod after the blow. Let ω' be the angular velocity of the rod about G after the blow. Let M be the mass of the rod. Now by 5.2.7, By 5.2.8, M
) a2 ' a2 ( ' ω − 0 = P(b). ω =M 3 ~ 3 ~~ ~
Thus just after blow, G will move with velocity 3Pb . Ma 2
P , M
and rod’s angular velocity is
5.2.10 Example A uniform rod at rest is struck by a blow at right angles to its length at a distance x from its centre. Find the point about which it will begin to turn. Sol: Let u ' be the velocity of the centre of mass G perpendicular to the rod after the blow. Let ω' be the angular velocity of the rod after the blow. Let M be the mass of the rod, 2a be the length of the rod. Now, by 5.2.7, ( ) Mu ' = = M u ' − 0 = P ~~ ~ ~ By 5.2.8, M
) a2 ( ' a2 ' ω =M ω − 0 = P(x). 3 ~~ ~ ~ 3
Let K be the required point, where G K = ξ. It follows that velocity of K is 0, that is, 3P xξ P P(x) ' ' . = 2 ξ = ω ξ = u = 2 a ~ ~~ ~ M Ma M3 2
xξ a Hence, 3P = MP or ξ = 3x . This determines the position of the required point Ma 2 K . By 4.2.15, the K.E. acquired
() =
) () ) a 2 ( ' )2 1 ( ' )2 1 −0 M u + M ω 2 2 3
626
5 Motion in Two Dimensions (Finite Forces)
)2 () )2 () )( a2 P(x) P 1 1 M + = M 2 2 M 2 3 M a3 ) () 3x 2 1 P2 P2 3P 2 x 2 .(∗) 1 + = = + 2 M 2Ma 2 2M a2 When end A is fixed: Then the resulting angular velocity ω1 =
P(a + x) 2 M a3
+
Ma 2
=
3P(a + x) . 4Ma 2
By 4.2.15, the K.E. acquired (
) ) ) () () )() 1 3P(a + x) 2 1 a2 3P(a + x) 2 M a M = + −0 2 4Ma 2 2 3 4Ma 2 () )2 2 2 3P(a + x) = Ma 3 4Ma 2 2( x )2 3P 1+ .(∗∗) = 8 M a The ratio of the energies given by (∗) and (∗∗) =
( ) 2 ) ( 1 + 3x 4 a 2 + 3x 2 a2 . ( )2 = P2 3(a + x)2 1 + ax M
P2 2M 3 8
Let us find the least value of ) ( ) () 4 a 2 + 3x 2 6ax + 2a 2 . f (x) ≡ = 4 1 − 3(a + x)2 3(a + x)2 Here ( ) a f ' (x) = 06x(a + x)2 = a2 + 3x2 2(a + x)3x(a + x) = a2 + 3x2 x = . 3 So the least value of the ratio of the energies given by (∗) and (∗∗) is obtained when x = a3 . ( ) a2 2 a) 4 a + 3 9 = ( f x= )2 = 3 3 a + a3 (
16 3 16 3
= 1.
5.2 Miscellaneous Examples on 2D Motion (Part II)
627
This shows that K.E. generated, when rod is free, is always greater than when the end A is fixed, and x /= a3 .
5.2.11 Example A rod, of mass m, is moving in the direction of its length on a smooth horizontal plane with velocity u. A second perfectly rough rod, of same mass and length 2a, which is in the same vertical plane as the first rod, is gently placed with one end on the first rod. If the initial inclination of the second rod to the horizontal be α, show that Fig. 5.23. ( ) 3u 2 sin2 α = 4ga(1 − sin α) 5 + 3 cos2 α . Sol: Let R be the impulse reaction at O to the inclined rod. Here, the impulse experienced by plank is m(u − v) along negative xdirection. It follows, by Newton’s third law, that the horizontal impulse experienced by the inclined rod at the lower end is m(u − v) along positive xdirection. Suppose that the vertical impulse experienced by the inclined rod at the lower end is R along positive ydirection. Thus we have three unknowns, velocity v, angular velocity ω, and impulse reaction R. (Fig. 5.23). By 5.2.7 and 5.2.8, the three equations of motion of rod are m((v − aω sin α) − 0) = m(u − v), m(aω cos α − 0) = R, () 2 ) a m (ω − 0) = (m(u − v))(a sin α) − R(a cos α). 3 Here, we have three unknowns, velocity v, angular velocity ω, and impulse reaction R. Let us determine them. Here maω = (m(u − v)) sin α − R cos α = (m(v − aω sin α)) sin α − (maω cos α) cos α = m(v sin α − aω). 3 ~~ ~
~
Hence Fig. 5.23
628
5 Motion in Two Dimensions (Finite Forces)
aω = 3v sin α − 3aω or ω=
3 v sin α. 4a
Since m((v − aω sin α) − 0) = m(u − v), we have u+a u + aω sin α = v= ~~ 2 ~ ~
(
3 v sin α 4a
)
sin α
2
=
u + 43 v sin2 α . 2
It follows that v=
u + 43 v sin2 α 2
or v=
4u u 4u . = = 5 + 3 cos2 α 8 − 3 sin2 α 2 − 43 sin2 α
Thus. () ) 3 4u 3u sin α 4u , ω= sin α = . v= 2 2 5 + 3 cos α 4a 5 + 3 cos α a 5 + 3 cos2 α Also () R = ma
sin α 3u a 5 + 3 cos2 α
) cos α =
3mu sin α cos α . 5 + 3 cos2 α
Next () m(u − v) = m u −
4u 5 + 3 cos2 α
) = mu
1 + 3 cos2 α . 5 + 3 cos2 α
Suppose that θ is the inclination of the inclined rod at time t. By 4.2.15, the equation of energy of the system “inclined rod + plank” is () 2 ) ) () )2 ( )2 ) 1 a ( )2 1 1 (( m(x) ˙ 2+ m x˙ − a θ˙ sin θ + a θ˙ cos θ m θ˙ + 2 2 2 3 () () () 2 ) )) ) 1 1 2 1 ( a 2 2 2 − = −mg(a sin θ − a sin α) mv + m (v − aω sin α) + (aω cos α) + m ω 2 2 2 3
5.2 Miscellaneous Examples on 2D Motion (Part II)
629
or
( − v2 +
( (
( )2 ( )2 a 2 ( )2 θ˙ ˙ 2 + x˙ − a θ˙ sin θ + a θ˙ cos θ + (x) 3 )
(v − aω sin α)2 + (aω cos α)2 +
(
) )) a2 = −2g(a sin θ − a sin α) ω2 3
or 2(x) ˙ 2 − 2 xa ˙ θ˙ sin θ +
() ) 4a 2 ( )2 4a 2 2 θ˙ − 2v 2 − 2vaω sin α + ω = −2g(a sin θ − a sin α) 3 3
or 2(x) ˙ 2 − 2 xa ˙ θ˙ sin θ +
4a 2 ( )2 θ˙ 3
( () − 2
)2 ) () ) () 3u sin α 4u 4u a sin α −2 5 + 3 cos2 α 5 + 3 cos2 α a 5 + 3 cos2 α () )2 4a 2 3u sin α + = −2g(a sin θ − a sin α) 3 a 5 + 3 cos2 α
or 2(x) ˙ 2 − 2 xa ˙ θ˙ sin θ +
) ( 1 4a 2 ( )2 2 2 2 2 2 θ˙ − ( )2 32u − 24u sin α + 12u sin α = −2ga(sin θ − sin α) 3 2 5 + 3 cos α
or 2(x) ˙ 2 − 2 xa ˙ θ˙ sin θ +
( ) 4a 2 ( )2 4u 2 2 θ˙ − ( )2 8 − 3 sin α = −2ga(sin θ − sin α) 2 3 5 + 3 cos α
2(x) ˙ 2 − 2 xa ˙ θ˙ sin θ +
) ( 4u 2 4a 2 ( )2 2 θ˙ − ( )2 5 + 3 cos α = −2ga(sin θ − sin α) 3 5 + 3 cos2 α
or
or 2(x) ˙ 2 − 2 xa ˙ θ˙ sin θ +
4u 2 4a 2 ( )2 θ˙ − = −2ga(sin θ − sin α).(∗) 3 5 + 3 cos2 α
630
5 Motion in Two Dimensions (Finite Forces)
For the system “inclined rod + plank”, there is no external horizontal force, so horizontal momentum of the system is conserved. Thus ( ) m 2 x˙ − a θ˙ sin θ = m x˙ + m(x˙ − a θ˙ sin θ ) = mu + 0 = mu . ~ ~~ ~ Hence 2 x˙ − a θ˙ sin θ = u. Now from (*), () 2
u + a θ˙ sin θ 2
)2
( ) 4a 2 ( )2 4u 2 θ˙ − = −2ga(sin θ − sin α) − u + a θ˙ sin θ a θ˙ sin θ + 3 5 + 3 cos2 α
or 1 2 1 2 ( )2 2 4a 2 ( )2 4u 2 θ˙ − u − a θ˙ sin θ + = −2ga(sin θ − sin α) 2 2 3 5 + 3 cos2 α or () ) 1 2 1 2 4 2 ( )2 4u 2 u + − sin θ + a θ˙ − = −2ga(sin θ − sin α). 2 2 3 5 + 3 cos2 α \ Suppose that θ˙ \θ = π = 0. We get 2
) () ( π ) 4 2 2 1 2 π 1 4u 2 u + − sin2 + a (0) − = −2ga sin − sin α 2 2 2 3 5 + 3 cos2 α 2 or ( ) u 2 3 cos2 α − 3 −3u 2 sin2 α 4u 2 1 ( ) = ( ) = u 2 − = −2ga(1 − sin α) 5 + 3 cos2 α~~ 2 5 + 3 cos2 α 2 5 + 3 cos2 α ~2 ~ or ( ) 3u 2 sin2 α = 4ga(1 − sin α) 5 + 3 cos2 α .\
5.2 Miscellaneous Examples on 2D Motion (Part II)
631
5.2.12 Example A rough wedge, of mass M and inclination α, is free to move on a smooth horizontal plane; on the inclined face is placed a uniform cylinder, of mass m. Show that the acceleration of the centre of the cylinder down the face, and relative to it, is Fig. 5.24 2g sin α
M +m . 3M + m + 2m sin2 θ
Sol: We have to show that Fig. 5.24 x¨ = 2g sin α
M +m . 3M + m + 2m sin2 θ
In the system “M + m” there is no component of horizontal external force, and hence the centre of mass of the combined system describes the path OY. It follows that d 2ξ d2 M ξ¨ = M 2 = m 2 (x cos α − ξ ) = m(x¨ cos α − ξ¨ ). dt~~ ~ dt ~ Thus M ξ¨ = m(x¨ cos α − ξ¨ ) or (M + m)ξ¨ = m x¨ cos α Here, equations of motion of cylinder are m(x¨ − (ξ¨ ) cos α) = (mg) sin α − F, It follows that Fig. 5.24
,2
()
) a2 m θ¨ = F · a, x = aθ 2
632
5 Motion in Two Dimensions (Finite Forces)
( ) ma d 2 ax m ma x= θ¨ = F == 2 2 2 dt 2 ~~ ~ ~ = (mg) sin α − m( x − (ξ ) cos α) () ) m x cos α = mg sin α − m x − cos α M +m and hence () ) m x cos2 α m x = mg sin α − m x − 2 M +m or m x cos2 α 3 x = g sin α + 2 M +m or x=
g sin α 3 2
−
m cos2 α M+m
=
M +m 2(M + m)g sin α = 2g sin α .\ 2 3(M + m) − 2m cos α 3M + m + 2m sin2 θ
5.2.13 Example A uniform circular ring moves on a rough curve under the action of no force, the curvature of the curve being everywhere less than that of the ring. If the ring be projected from a point A of the curve without rotation and begin to roll at B, then the angle between the normals at A and B is ln 2 . μ Sol: Let μ be the coefficient of friction, R be the reaction, and ρ be the radius of curvature of the rough curve. Let m be the mass of the ring. Here, equations of motion of the point of contact are ) () ) () 2 ) () (˙s ) dψ dψ = m (˙s )2 =m = R. m(s) = −(μR), m s˙ dt ds ρ ~ ~~ ~
5.2 Miscellaneous Examples on 2D Motion (Part II)
633
) ( Also, for the motion of ring, ma 2 θ¨ = (μR)a. From the first two equations, Fig. 5.25 )) () () dψ m(¨s ) = −μ m s˙ dt or d s¨ dψ (ln(˙s )) = = −μ dt s ˙ ~ ~~ dt~ Hence ln(˙s ) = −μψ + ln C or s˙ = Ce−μψ . \ It is given that θ˙ \t=0 = 0. Suppose that s˙ ψ=0 = V . It follows that C = V . Hence s˙ = V e−μψ . Since Here, lower end of the inclined rod rests ,2 on the horizontal plank, so axis of rotation of the inclined rod passes Let sudden change through its lower end. in angular velocity of rod be from 0 to ∴ Sudden change in velocity of is . cos ) + from (0,0) to (− sin , ( , 0). Direction fixed in space
( − )
,2
≡ ( )
Direction fixed in space
Fig. 5.25
Direction fixed in body Direction fixed in space
Let sudden change in velocity of plank be from to . ( − )
Direction fixed in body Direction fixed in space
634
5 Motion in Two Dimensions (Finite Forces)
( 2 ) ma θ¨ = (μR)a = (−m(¨s ))a ~ ~~ ~ \ we have a θ¨ = −¨s and hence a θ˙ = C − s˙ . Since s˙ t=0 = V , and θ˙ \t=0 = 0, we have C = V , and hence a θ˙ = V − s˙ . We have to show that ψs˙−a θ˙ =0 =
ln 2 , μ
that is, ψs˙−(V −˙s )=0 =
ln 2 , μ
that is, ψ2(V e−μψ )−V =0 =
ln 2 , μ
that is, ψeμψ =2 =
ln 2 . μ
This is clearly true.
5.2.14 Example A uniform rod has one end fastened by a pivot to the centre of a wheel which rolls on a rough horizontal plane, the other extremity resting against a smooth vertical wall at right angles to the plane containing the wheel and rod. Show that the inclination θ of the rod to the vertical, when it leaves the wall, is given by the equation 9M cos3 θ + 6m cos θ − 4m cos α = 0, where M and m are the masses of the wheel and rod, and α is the initial inclination to the vertical when the system was at rest (Fig. 5.26). Sol: Here, the energy equation of the system “M + m” is
5.2 Miscellaneous Examples on 2D Motion (Part II)
635
direction fixed in space
axis fixed in space ,
≡ ( )
direction fixed in cylinder
≡ ( )
Free to move axis fixed in wedge
≡ ( )
Acc. of moving frame = ̈ axis fixed in wedge Free to move axis fixed in space
cos
−
Fig. 5.26
() 3M cos2 θ + m
) 2 2 ( )2 a θ˙ 3
() 2 ) ( )2 1 ( )2 a 2 ˙ = 3Ma cos θ θ + m + m(a) θ˙ 2 3 ()() () 2 ) ) ) ( )2 ( )2 1 a = 3Ma 2 cos2 θ θ˙ + m + m(I G)2 θ˙ −0 2 3 ) () 2
2
)2 ( )2 1 ( + change in K.E.of rod Ma 2 cos2 θ θ˙ + M 2a cos θ θ˙ 2 ( () )2 () )2 ) ( )2 1 d d + change in K.E.of rod = Ma 2 cos2 θ θ˙ + M (2a sin θ ) + (b) 2 dt dt ) () ( )2 1 = Ma 2 cos2 θ θ˙ + M(vel.of G 1 )2 + change in K.E.of rod 2 )() ) ( ( )2 b2 2a 1 1 M = cos θ θ˙ + M(vel.of G 1 )2 + change in K.E.of rod 2 2 b 2 )() () ) ( ( ))2 b2 d 2a sin θ − 2a sin α 1 1 M + M(vel.of G 1 )2 + change in K.E.of rod = 2 2 dt b 2 ) ) ) (( ( b2 ( ˙ )2 1 1 M φ + M(vel.of G 1 )2 − 0 + change in K.E.of rod = 2 2 2 =
= = change in K. E. of wheel + change in K. E. of rod = work done by external forces = work done by gravity on rod ~ ~~ ~ = (mg)(downwards displacement of G) = (mg)((b + a cos α) − (b + a cos θ )) = mga(cos α − cos θ ).
Hence, ) () 2 2 ( )2 2 3M cos θ + m a θ˙ = mga(cos α − cos θ ) 3 or
636
5 Motion in Two Dimensions (Finite Forces)
( )( )2 3mg 9M cos2 θ + 2m θ˙ = (cos α − cos θ ) a On differentiation, we get ( ) 3mg ˙ 2 + 9M cos2 θ + 2m 2θ˙ θ¨ = sin θ θ˙ ((9M(−2 cos θ sin θ ))θ˙ )(θ) a or ) ( 9M cos2 θ + 2m θ − 9M cos θ sin θ
3mg α − cos θ ) a (cos 9M cos2 θ + 2m
) ( 3mg ˙ 2= sin θ .. = 9M cos2 θ + 2m θ¨ − 9M cos θ sin θ (θ) 2a ~ ~~ ~ Thus ( ( )2 ) 2a 9M cos2 θ + 2m θ¨ = 54Mmg(cos α − cos θ) cos θ sin θ + 3mg sin θ 9M cos2 θ + 2m
or ( )2 2a 9M cos2 θ + 2m θ = 27Mmg(2 cos α − cos θ ) cos θ sin θ + 6m 2 g sin θ Here, the three equations of motion of rod are ( ( )2 ) d2 = m 2 (a sin θ ) = R − X ma cos θ θ − sin θ θ˙ ~ dt ~~ ~ ( ( )2 ) d2 = m 2 (b + a cos θ ) = Y − mg −ma sin θ θ + cos θ θ˙ ~ dt ~~ ~ ()
a2 m 3
) θ = −X (a cos θ ) + Y (a sin θ ) − R(a cos θ ).
When the rod leaves the wall: We have R = 0 and hence ( ( ( )2 ) ( )2 ) + mg. X = −ma cos θ θ − sin θ θ˙ , Y = −ma sin θ θ + cos θ θ˙ Also, () 2 ) a m θ = −X (a cos θ ) + Y (a sin θ ) 3
5.2 Miscellaneous Examples on 2D Motion (Part II)
637
or ( ( ( ) )) ¨ = −3X cos θ + 3Y sin θ = − 3 −ma cos θ θ − sin θ θ˙ 2 cos θ ma θ ~ ~~ ~ ( ( ) ( )2 ) + 3 −ma sin θ θ + cos θ θ˙ + mg sin θ. Hence ( ( ( ) ) g ) sin θ θ¨ = 3 cos θ θ¨ − sin θ (θ˙ )2 cos θ + 3 − sin θ θ¨ + cos θ (θ˙ )2 + a ~~ ~ ~ ˙ 2+ = 3 cos 2θ θ¨ − 6 sin θ cos θ (θ)
3g sin θ a
It follows that (1 − 3 cos 2θ )
3mg (cos α − cos θ ) 27Mmg(2 cos α − cos θ ) cos θ sin θ + 6m 2 g sin θ + 6 sin θ cos θ a ( )2 2 9M cos2 θ + 2m 2a 9M cos θ + 2m 3g ˙ 2= = (1 − 3 cos 2θ)θ¨ + 6 sin θ cos θ(θ) sin θ a ~ ~~ ~
or ( )( ) 4 − 6 cos2 θ 9Mm(2 cos α − cos θ ) cos θ sin θ + 2m 2 sin θ ( ) + sin θ cos θ 12m(cos α − cos θ ) 9M cos2 θ + 2m ( )2 = 2 sin θ 9M cos2 θ + 2m or (
) 2 − 3 cos2 θ (9Mm(2 cos α − cos θ) cos θ + 2m 2 ) + cos θ6m(cos α − cos θ)(9M cos2 θ + 2m ) ~ ~~ ~ ~~~~ ~ ~~ ~ ~~~~ 1
2
1
2
= 81M 2 cos4 θ + 4m 2 + 36Mm cos2 θ
or (( ) ) 9Mm 2 − 3 cos2 θ (2 cos α − cos θ ) cos θ + 6 cos3 θ (cos α − cos θ ) (( ) ) +2m 2 2 − 3 cos2 θ + 6 cos θ (cos α − cos θ ) = 81M 2 cos4 θ + 4m 2 + 36Mm cos2 θ
or
638
5 Motion in Two Dimensions (Finite Forces)
⎛
⎞
2 2 2 ⎠ 9Mm cos θ ⎝2 cos α(2 − ~3 cos ~ ~~α~ − cos ~ ~~ θ~) ~~ θ~) − (2 − ~3 cos ~~ θ~) cos θ + 6 cos θ (cos 1
1
2
2
( ) 2 +2m 2 −9 cos2 θ + 6 cos θ cos α = 81M 2 cos4 θ + ~~~~ 4m 2 + 36Mm ~~cos θ~ ~ 3
1
or ( ) 9Mm cos θ −3 cos3 θ + 2(2 cos α − cos θ ) − 4 cos θ + 6m 2 cos θ (2 cos α − 3 cos θ ) = 81M 2 cos4 θ
or ( ) 9Mm cos θ −3 cos3 θ + 2(2 cos α − 3 cos θ ) + 6m 2 cos θ (2 cos α − 3 cos θ ) = 81M 2 cos4 θ
or 6Mm(2 cos α − 3 cos θ ) + 2m 2 (2 cos α − 3 cos θ ) = 27M 2 cos3 θ + 9Mm cos3 θ or 2m((2 cos α − 3 cos θ ))(3M + m) = 9M cos3 θ (3M + m) or 2m((2 cos α − 3 cos θ )) = 9M cos3 θ or 9M cos3 θ + 6m cos θ − 4m cos α = 0.\
5.2.15 Example Rope is coiled around a drum of radius a. Two wheels each of radius b a fitted to the ends of the drum, and the wheels and drum form a rigid body having a common axis. The system stands on level ground and a free end of the rope, after passing under ◦ the drum, is inclined at an angle of 60 to the horizon. If a force P be applied to the rope, show that the drum starts to roll in the opposite direction, its centre having acceleration
5.2 Miscellaneous Examples on 2D Motion (Part II)
639
P(2a − b)b ) , ( 2M b2 + k 2 where M is the mass of the system and k its radius of gyration about the axis (Fig. 5.27). Sol: Here equation of motion of centre of mass G is P(a) − F(b) − F(b) b (Pa − 2Fb) =Mb 2 k Mk 2 = Mb(θ ) =M
d2 P ◦ F − P cos 60~ = 2F − . (bθ ) = ~M x¨ = F + ~~ dt 2 2
Hence b P (Pa − 2Fb) = 2F − 2 k 2 or 1 + ab2 M x + P2 P k 2 + 2ab . = F = P 2 2bk 2 = 2 4 k 2 + b2 2 + k2 ~ ~~ ~
Direction fixed in space
Projectile velocity ≡
Position of the ring at zero time Direction fixed in ring Position of the ring at time , arc
= ≡
≡ ( ) ( )
≡ ( )
Fig. 5.27
640
5 Motion in Two Dimensions (Finite Forces)
Hence M x + P2 P k 2 + 2ab = 2 4 k 2 + b2 or Mx =
P 2ab − b2 P k 2 + 2ab P P b(2a − b) = − = .\ 2 2 2 2 2 k +b 2 2 k +b 2 k 2 + b2
5.2.16 Example A thin circular cylinder, of mass M and radius b, rests on a perfectly rough horizontal plane and inside it is placed a perfectly rough sphere of mass m and radius a. The system is disturbed in a plane perpendicular to the generators of the cylinder. Obtain the equations of finite motion and two first integrals of them. If the motion be small, show that the length of the simple equivalent pendulum is 14M(b − a) . 10M + 7m Sol: Here, the two equations of motion of centre C1 of sphere are Fig. 5.28 ( ) ( )2 ˙ 2 m (b − a) θ˙ − b ψ sin θ = = m( (b − a)(θ) ~ ~~ ~
~ m((b − a) θ +b ψ cos θ ) = m( ~
acc of C1 −−−−−→ along rel to C
()()
(b)θ¨ ~~~~
+
acc of C1 rel to C transverse to C1 C
() m
2a 5
) ) () () 2 d (bψ) sin θ ) = R − (mg) cos θ, + − 2 dt ~~ ~ ~
2)
~
−−→ acc of C along C1 C
~~
~
) ) d2 (bψ) cos θ ) = −(mg) sin θ − F. . 2 dt ~~ ~
acc of C transverse to C1 C
~~
~
φ = F(a)
or 2m d 2 2m 2m d 2 (aφ) = F − a)θ + bψ) = ((b ((b − a) θ +b ψ) = 2 5 5 dt 2 ~ 5 dt ~~ ~ = −(mg) sin θ − m((b − a) θ +b ψ cos θ )
5.2 Miscellaneous Examples on 2D Motion (Part II)
641 Direction fixed in space
Direction fixed in space
Direction fixed in rod ≡ ( ),
≡ ( )
= 2 sin
− 2 sin
,2 , Direction fixed in wheel 1
Direction fixed in wheel at zero time
Instantaneous axis of rotation of rod passes through .
rolling
Direction fixed in space
( sin , + cos )
Direction fixed in space
Direction fixed in space
Direction fixed in rod ≡ ( ),
≡ ( )
= 2 sin
− 2 sin
,2 , Direction fixed in wheel 1
Direction fixed in wheel at zero time
rolling
Direction fixed in space Forces experienced by the rod
Instantaneous axis of rotation of rod passes through . ( sin , + cos )
Fig. 5.28
or 2 ((b − a) θ +b ψ) = −g sin θ − (b − a) θ −b ψ cos θ 5 or 7(b − a) θ +b(2 + 5 cos θ ) ψ = −5g sin θ.(∗) Here, the equation of motion of centre C of cylinder is ()
) d2 Mb ψ = M (bψ) = −F1 + R sin θ + F cos θ dt 2 ~ ~~ ~ Also (
or
) Mb2 ψ = −F(b) + F1 (b)
642
5 Motion in Two Dimensions (Finite Forces) Mbψ¨ = F1 − F = (R sin θ + F cos θ − Mb ψ) − F = R sin θ + F(cos θ − 1) − Mb ψ ~ ~~ ~ ( ( ) ) ( )2 = m (b − a) θ˙ − b ψ sin θ + (mg) cos θ sin θ + F(cos θ − 1) − Mb ψ
= (m((b − a)(θ˙ )2 − bψ¨ sin θ ) + (mg) cos θ ) sin θ + (− (mg) sin θ −m((b − a)θ¨ + bψ¨ cos θ ))(cos θ − 1) ~ ~~ ~ ~ ~~ ~ ~ ~~ ~ ~ ~~ ~ 1
2
2
1
( )2 ( ) − Mbψ¨ = m(b − a) sin θ θ˙ − m(b − a)(cos θ − 1) θ + −mb sin2 θ − mb cos θ (cos θ − 1) − Mb ψ ~ ~~ ~ 1
+mg(cos θ sin θ − sin θ (cos θ − 1)) ( )2 = m(b − a) sin θ θ˙ − m(b − a)(cos θ − 1) θ +b(−m + m cos θ − M) ψ +mg sin θ
and hence ( )2 Mb ψ = m(b − a) sin θ θ˙ − m(b − a)(cos θ − 1) θ +b(−m + m cos θ − M) ψ +mg sin θ
or ( )2 b(m − m cos θ + 2M) ψ = m(b − a) sin θ θ˙ − m(b − a)(cos θ − 1) θ +mg sin θ or ( )2 b ψ(2M + m(1 − cos θ )) − m(b − a) sin θ θ˙ + m(b − a)(cos θ − 1) θ = mg sin θ (∗∗)
or ) () ) ( ( )2 7(b − a) θ +b(2 + 5 cos θ ) ψ b ψ(2M + m(1 − cos θ )) − m(b − a) sin θ θ˙ + (1 − cos θ ) θ = m −5
or () () ) ) ( )2 7θ m(2 + 5 cos θ ) − m(b − a) sin θ θ˙ + (1 − cos θ ) θ − =0 b ψ 2M + m(1 − cos θ ) + 5 5
or () () ) ) ) () ( )2 7 2 + cos θ θ = 0 b ψ 2M + m − m(b − a) sin θ θ˙ − 5 5 or () ) ( ) ( )2 7 2 b ψ 2M + m + m(b − a) θ +m(b − a) − sin θ θ˙ + cos θ θ = 0 5 5
5.2 Miscellaneous Examples on 2D Motion (Part II)
or () ) ) 7 2 d( cos θ θ˙ = 0. b ψ 2M + m + m(b − a) θ +m(b − a) 5 5 dt On integration, we get ) () 2 7 b 2M + m ψ˙ + m(b − a)θ˙ + m(b − a) cos θ θ˙ = C 5 5 or ( ( ) ) b 2M + 75 m ψ˙ + m(b − a)θ˙ 25 + cos θ = C. This is a first integral. From (∗), for small θ, we have 7(b − a) θ +b(2 + 5 cos 0) ψ = −5gθ and hence 5 (b − a) θ +b ψ = − gθ. 7 From (∗∗), for small θ, we have ( )2 b ψ(2M + m0) − m(b − a)0 θ˙ + m(b − a)(0) θ = mgθ or () ) 5 2M −(b − a) θ − gθ = 2Mbψ¨ = mgθ ~~ ~ ~ 7
and hence θ =−
mg + 10 Mg 7 θ 2M(b − a)
or θ =−
7m + 10M gθ. 14M(b − a)
So, the length of the simple equivalent pendulum is
643
644
5 Motion in Two Dimensions (Finite Forces)
14M(b − a) .\ 7m + 10M Here, change in kinetic energy of the cylinder is (
1 M 2
(()
d (bψ) dt
)2 )
) )( )2 ( )2 1( 2 + Mb ψ˙ − 0 = Mb2 ψ˙ . 2
Here, change in kinetic energy of the sphere is (
=m
(()
) )2 () )2 ) () ) d 2a 2 ( )2 d 1 ˙ m φ + −0 ((b − a) sin θ + bψ) + (b − (b − a) cos θ ) dt dt 2 5 ) ( 2 )2 )2 ( a ( )2 1 ( φ˙ +m = m (b − a) cos θ θ˙ + bψ˙ + (b − a) sin θ θ˙ 2 5 () )2 ) ( )2 ( )2 1 ( 1 d = m (b − a)2 θ˙ + b2 ψ˙ + 2b(b − a) cos θ θ˙ ψ˙ + m (aφ) 2 5 dt () )2 ) ( ( ) ) d 1 1 ( 2 2 = m (b − a)2 θ˙ + b2 ψ˙ + 2b(b − a) cos θ θ˙ ψ˙ + m ((b − a)θ + bψ) 2 5 dt ) )2 ( ) ( ) 1( 1 ( 2 ˙ 2 2 ˙ 2 ˙ ˙ = m (b − a) θ + b ψ + 2b(b − a) cos θ θ ψ + m (b − a)θ˙ + bψ˙ 2 5
1 m 2
(
7 10 (b
( )2 − a)2 θ˙ +
7 2 b 10
) ( )2 ( ) ψ˙ + b(b − a) 25 + cos θ θ˙ ψ˙ + 25 b(b − a)θ˙ ψ˙ .
Here, work done by external forces is = work done by gravity on cylinder + work done by gravity on sphere = 0 − (mg)((b − a) − (b − a) cos θ ) = −mg(b − a)(1 − cos θ ). Thus, the equation of energy is ⎛
⎞ () ) ⎜ 7 ⎟ ( )2 2 7 2 ˙ 2 2 ⎟ ˙ 2 +m ⎜ Mb2 (ψ) b (ψ) +b(b − a) + cos θ θ˙ ψ˙ + b(b − a)θ˙ ψ˙ ⎟ = −mg(b − a)(1 − cos θ ) ⎜ (b − a)2 θ˙ + ⎝ 10 ⎠ 10 5 5 ~ ~~ ~ ~ ~~ ~ 1
()
1
) ( )2 ( )2 7m 7 ˙ M+ + m(b − a)2 θ˙ b ψ 10 10 () ) 2 ˙ ˙ = −mg(b − a)(1 − cos θ ). + mb(b − a)ψ θ cos θ + 5 2
This is also a first integral.
5.2 Miscellaneous Examples on 2D Motion (Part II)
645
5.2.17 Example A uniform sphere, of mass M, rests on a rough plank of mass m which rests on a rough horizontal plane, and the plank is suddenly set in motion with velocity u in the direction of its length. Show that the sphere will first slid, and then roll, on the plank, and that the whole system will come to rest in time mu , μg(m + M) where μ is the coefficient of friction at each of the points of contact. Sol: (Fig. 5.29). Since the sphere has no motion in the vertical direction, the reaction exerted by plank on sphere is equal to Mg (upward). Again since the plank has no motion in the vertical direction, and the net downward force experienced by plank is (Mg + mg), the reaction exerted by horizontal plane on plank is equal to (Mg + mg) (upward). Since the reaction exerted by plane on plank is (Mg + mg) (upward), and the direction of motion of plank is +x, the friction experienced by plank due to plane is μ(Mg + mg)(= (M + m)μg) along −x direction. This shows that the acceleration of the centre of mass of “m + M ” is −μg, and hence the direction of motion of centre of mass of “plank + sphere” is along −x. Now since the direction of motion of centre of mass of plank is along +x, the direction of motion of centre of mass of “sphere” is −x. Let us assume that the friction μ(Mg) on plank due to sphere acts along −x direction. See Fig. 5.16b. Here, equation of motion of plank is m x¨ = −μ(Mg) − μ(Mg + mg) = −μ(2M + m)g. ~ ~~ ~ Here, equation of motion of centre of mass G(−(ξ − x), · · ·) of sphere is Direction fixed in body Direction fixed in space ( , 0,0) rolling
≡ ( ) ≡ ( ) = .
Fig. 5.29
646
5 Motion in Two Dimensions (Finite Forces)
()
−μ(2M + m)g −ξ M m
)
() 2 ) d ξ = M(x − ) = M (−(ξ − x)) = μ(Mg), dt 2 ~~ ~ ~
Thus −μ(2M + m)g − ξ = μg m or ξ=
−2μ(M + m)g . m
Also ()
2a 2 M 5
) θ = (μ(Mg))a.
It follows that θ=
5 μg. 2a
θ˙ =
5 μgt. 2a
\ Next since θ˙ \t=0 = 0 we have
Since m x¨ = −μ(2M + m)g and x ˙ t=0 = u, we have x˙ = u −
μ(2M + m)g t. m
It follows that the plank will stop at time Since
um . μ(2M+m)g
M(x¨ − ξ¨ ) = μ(Mg), we have
5.2 Miscellaneous Examples on 2D Motion (Part II)
647
)\ () \ d (ξ − x) \\ + μgt = 0 + μgt = μgt, x˙ − ξ˙ = − dt t=0 ~~ ~ ~
and hence x˙ − ξ˙ = μgt. Since () ) ) () 5 μ2(M + m)g 5 μ(2M + m)g μgt = a μgt = a θ˙ < ξ˙ = x˙ − μgt = u − t − μgt = u − t ~ ~~ ~ 2 2a m m
we have 5 μ2(M + m)g μgt < u − t 2 m or t
0)
Direction fixed in sphere ( − ,⋯)
1(
fr = (
+
,⋯)
(plank)
Direction of motion of plank
)
Direction fixed in space
Horizontal plane point fixed in plank
Direction fixed in space
≡ ( ) > 0, ≡ ( ) > 0,
Direction fixed in space
≡ ( ) > 0. ̇
,
Direction fixed in sphere
=0
̇+
=
(> 0).
(−( − )) < ̇
(−( − ), ⋯ ) that is ̇ < ̇. (sliding) ( fr = ( fr = (
+
)
1(
,⋯)
(plank) Direction of motion of plank
)
) Horizontal plane Direction fixed in space
point fixed in plank
Direction fixed in space
−
Fig. 5.31
ξ˙ =
(m) M
μgt.
Here one equation of motion of sphere is ()
μ(mg) m x¨ + M
)
d2 = m(x¨ + ξ¨ ) = m 2 (x + ξ ) = −μ(mg) . ~ dt ~~ ~
It follows that x¨ +
μ(mg) = −μg M
652
5 Motion in Two Dimensions (Finite Forces)
or x¨ = −
M +m μg. M
Hence ) ) () () M +m M +m μgt = u − μgt. x˙ = x ˙ t=0 − M M ~~ ~ ~ Thus () x˙ = u −
) M +m μgt. M
Here another equation of motion of sphere is ()
) 2a 2 m θ¨ = (μ(mg))a 5
It follows that θ¨ =
5 μg. 2a
Hence \ 5 5 μgt = ω + μgt. θ˙ = θ˙ \t=0 + 2a ~ ~~ 2a ~ Thus θ˙ = ω +
5 μgt. 2a
Here, for sliding () ) ) () 5 M +m 5 μgt = a~ θ˙~~ < x~˙ = u − μgt, aω + μgt = a ω + 2 2a M and hence 5 aω + μgt < u − 2
()
) M +m μgt M
5.2 Miscellaneous Examples on 2D Motion (Part II)
653
or t
2M 7M+2m
) () d2 2a 2 M ξ = F, m(x + ξ ) = m 2 (x + ξ ) = −F , m θ = Fa, x˙ = a θ˙ . 5 ~~ ~ ~ dt Since ) ) () () d ( ) d ˙ ¨ ξ ξ ξ = −m(a θ + ξ ), = −m aθ + M ξ = F = −m(x + ) = −m ˙ + (x) ~ ~~ ~ dt dt
we have M ξ¨ = −m(a θ¨ + ξ¨ ) and hence ξ¨ =
−ma θ¨ M +m
Since () M
)
−ma θ¨ = M ξ¨ = F = ~ ~~ ~ M +m
( ) 2 m 2a5 θ¨ a
() ) 2a = m θ¨ , 5
we have ) () ) 2a −ma ¨ θ = m θ¨ , M M +m 5 ()
−ma ¨ and hence θ¨ = 0 Next since ξ¨ = M+m θ , we have ξ¨ = 0. Since x˙ = a θ˙ and θ¨ = 0 we have x¨ = 0 Thus, after rolling, x, ˙ ξ˙ , θ˙ are constants. \
654
5 Motion in Two Dimensions (Finite Forces)
5.3 Miscellaneous Examples on 2D Motion (Part III) 5.3.1 Note Let O A be a fixed line in space. Let O x y be a rectangular coordinate system. See Fig. 5.32. Let P be a moving point. Suppose that rectangular axes O x and O y are rotating about O. Thus O x and O y are not fixed in space. They revolve in any manner about the origin O in its own plane. Here Fig. 5.32 (velocity of P relative to O along O x direction) = (velocity ofPrelative toMalongO xdirection) + (velocity ofMrelative toOalongO xdirection) d (O M) dt d = (velocity ofN relative toOalongO xdirection) + (O M) dt ) d d (π d d ˙ + θ + (x) = x˙ − y θ, = −(O N ) (∠ AO N ) + (O M) = −y dt dt dt 2 dt = (velocity ofPrelative toMalongO xdirection) +
and (velocity ofPrelative toOalongO ydirection) = (velocity ofPrelative toMalongO ydirection) + (velocity ofMrelative toOalongO ydirection) d (∠ AO M) dt d = (velocity ofN relative toOalongO ydirection) + (O M) (∠ AO M) dt d d d d ˙ = (O N ) + (O M) (∠ AO M) = (y) + (x) (θ ) = y˙ + x θ. dt dt dt dt
= (velocity ofPrelative toMalongO ydirection) + (O M)
5.3.2 Example Show that the path of a point P which possesses two constant velocities u and v, the first of which is in a fixed direction, and the second of which is perpendicular to the radius O P drawn from a fixed point O, is a conic whose focus is O and whose eccentricity is uv . Fig. 5.33. Sol: By 5.3.1, Fig. 5.33
5.3 Miscellaneous Examples on 2D Motion (Part III)
655
Direction fixed in space
≡ ( ), ≡ ( ) ̇
,
=0
=
(> 0)
Direction fixed in sphere ( − ,⋯)
1(
,⋯)
Direction of motion of board Direction fixed in space
Horizontal surface
point fixed in plank
Direction fixed in space
(board)
≡ ( ) > 0, ≡ ( ) > 0,
Direction fixed in space
≡ ( ) > 0. ̇
, Direction fixed in sphere
̇+
=0
=
(> 0).
(−( − )) < ̇
(−( − ), ⋯ ) that is ̇ < ̇. (sliding) ( fr = (
)
Direction fixed in space
)
1(
(board) Direction of motion of board
,⋯)
point fixed in board
Horizontal surface
Direction fixed in space
−
Fig. 5.32
r˙ =
d(O P) d − (0)θ˙ = u cos θ , r θ˙ = (0) + (O P)θ˙ = v − u sin θ dt dt ~ ~~ ~ ~ ~~ ~
and hence dr u cos θ r˙ = = ˙ r dθ v~~ − u sin θ~ ~r θ It follows that
656
5 Motion in Two Dimensions (Finite Forces) Direction fixed in space Direction fixed in sphere , ( + ,⋯)
(board)
(
)
1( , ⋯ ) fr = ( )
table point fixed in Direction fixed in space board
fr = (
)
Direction fixed in space
Fig. 5.33
dr d(v − u sin θ ) u cos θ = dθ = − . v− v − u sin θ ~r ~~u sin θ ~ Hence ln r = ln c − ln(v − u sin θ ) It follows that r=
c v − u sin θ
or c = v − u sin θ r or c v
r
=1+
(u )
( π) cos θ + . v 2
This represents a conic, whose focus is O and whose eccentricity is uv .\
5.3 Miscellaneous Examples on 2D Motion (Part III)
657
5.3.3 Note Let O A be a fixed line in space. Let O x y be a rectangular coordinate system. See Fig. 5.34. Let P be a moving point. Suppose that rectangular axes O x and O y are rotating about O. Thus O x and O y are not fixed in space. They revolve in any manner about the origin O in its own plane. Here Fig. 5.34 (acceleration ofPrelative toOalongO xdirection) = (acceleration ofPrelative toMalongO xdirection) + (acceleration ofMrelative toOalongO xdirection) ( )2 ) () d d2 = (acceleration ofPrelative toMalongO xdirection) + (∠ AO M) (O M) − (O M) dt dt 2 ( () )2 ) d2 d M) − M) AO M) = (acceleration ofN relative toOalongO xdirection) + (O (O (∠ dt dt 2 () () ) ( 2 )2 ) d d d 1 d M) − M) AO M) − (O (O (O N )2 (∠ AO N ) + (∠ O N dt dt dt dt 2 ) 1 d ( 2 ) ( 2 ˙ ˙ y θ + x¨ − x(θ ) =− y dt
and (acceleration ofPrelative toOalongO ydirection) = (acceleration ofPrelative toMalongO ydirection) + (acceleration ofMrelative toOalongO ydirection) () ) d 1 d = (acceleration ofPrelative toMalongO ydirection) + (O M)2 (∠AO M) O M dt dt () ) 1 d 2 d = (acceleration ofN relative toOalongO ydirection) + (O M) (∠AO M) O M dt dt ( () )2 ) ) () ( ) 1 d ( 2 ) d2 d d 1 d ˙ 2 + (O M)2 (∠AO M) = y¨ − y(θ) (∠AO N ) x θ˙ . (O N ) − (O N ) = + dt 2 dt O M dt dt x dt
Fig. 5.34
Direction fixed in space
658
5 Motion in Two Dimensions (Finite Forces)
5.3.4 Example A smooth straight thin tube revolves with uniform angular velocity ω in a vertical plane about one extremity which is fixed. If at zero time the tube be horizontal, and a particle inside it be at a distance a from the fixed end, and be moving with velocity V along the tube, show that its distance at time t is Fig. 5.35 () a cosh(ωt) +
g V − ω 2ω2
) sinh(ωt) +
g sin ωt. 2ω2
Sol: Fig. 5.35 Here one equation of motion is () ) ( ) d(0) −(mg) sin θ 2 ˙ = −g sin(ωt) r¨ − r ω − 2 θ + 0(0) = dt m or r¨ − r ω2 = −g sin(ωt) It follows that r = A cosh(ωt) + B sinh(ωt) +
g sin ωt. 2ω2
Hence r˙ = Aω sinh(ωt) + Bω cosh(ωt) +
g cos ωt. 2ω
It is given that r˙ t=0 = V So V = Bω + Hence B =
2V ω−g . 2ω2
g . 2ω
It is given that r t=0 = a. So a = A. Thus
Fig. 5.35
Direction fixed in space
5.3 Miscellaneous Examples on 2D Motion (Part III)
r = a cosh(ωt) +
659
2V ω − g g sinh(ωt) + sin ωt.\ 2 2ω 2ω2
5.3.5 Example A vessel steams at a constant speed v along a straight line while another vessel, steaming at a constant speed V , keeps the first always exactly abeam. Show that the path of either vessel relative to the other is a conic section of eccentricity Vv . Sol: At time t, let the position of first vessel be A, and the position of second vessel be B. Let iˆ be the unit vector in the direction of the given straight line. Thus the direction of iˆ is fixed in space. Let AB = r, and θ be the angle which AB makes with iˆ. Here Fig. 5.36 V θˆ = (velocity of B relative to origin O) = ( velocity of B relative to A) + ( velocity of A relative to origin O) ~ ~~ ~ ( ) ( ) = r˙rˆ + r θ˙ θˆ + (velocity of Arelative to origin O) = r˙rˆ + r θ˙ θˆ + v iˆ ( ) ( ) ( ) ˆ = r˙rˆ + r θ˙ θˆ + v cos θ rˆ − sin θ θˆ = (˙r + v cos θ )ˆr + r θ˙ − v sin θ θ,
so ( ) ˆ V θˆ = (˙r + v cos θ )ˆr + r θ˙ − v sin θ θ, and hence r˙ + v cos θ = 0, r θ˙ − v sin θ = V. It follows that Fig. 5.36
Direction fixed in space
660
5 Motion in Two Dimensions (Finite Forces)
r˙ dr d −v cos θ = = − (ln(V + v sin θ )) = ˙ r dθ V~~+ v sin θ~ dθ ~r θ
and hence r=
C . V + v sin θ
It follows that C V
r
=1−
( v π) cos θ + . V 2
This shows that the path of B relartive to A is a conic whose eccentricity is
v V
.
5.3.6 Example A boat, which is rowed with constant velocity u, starts from a point A on the bank of a river which flows with a constant velocity nu; it points towards a point B on the other bank exactly opposite to A. Find the equation to the path of the boat (Fig. 5.37). Sol: Let d be the breadth of the river. So r˙ = (nu) sin θ − u, r θ˙ = (nu) cos θ. It follows that Fig. 5.37 r˙ dr n sin θ − 1 (nu) sin θ − u = . = = ˙ r dθ (nu) cos θ n cos θ rθ ~~ ~ ~ Hence { () ln r = ~
) 1 1 1 n sin θ − 1 dθ = ∫ tan θ dθ − ∫ sec θ dθ = ln sec θ − ln(sec θ + tan θ ) + ln C n cos θ n n n ~~ ~
or rn =
C (sec θ )n . sec θ + tan θ
5.3 Miscellaneous Examples on 2D Motion (Part III)
661 ≡ ( ), ≡ ( ), ̇=
= 0.
, ̈=0
Fixed direction in space
Fig. 5.37 Fig. 5.38
̂ ̂ ̂
Now since r θ=0 = d, we have d n = C. Hence r n (cos θ )n =
dn . sec θ + tan θ
This is the required equation of path. If n = 1, then the path is r = represents a parabolic path whose focus is B.
d . 1+sin θ
This
5.3.7 Example An insect crawls at a constant rate u along the spoke of a cartwheel, of radius a, the cart moving with velocity v. Find the acceleration along and perpendicular to the spoke Fig. 5.38. Sol: Here
662
5 Motion in Two Dimensions (Finite Forces)
−→ −→ −→ ( acc. of P rel. to origin O along C P) = ( acc. of P rel. to centre C along C P) + ( acc. of C rel. to origin O along C P) ~~ ~ ~ ( ( v )2 ) ) ( −→ ˙ 2 = −r (θ˙ )2 = −r ˙ 2 + 0 = 0 − r (θ) = r¨ − r (θ˙ )2 + ( acc. of C rel. to origin O along C P) = r¨ − r (θ) a
Next −→ −→ −→ (acc. of P rel. to O perp. to C P) = ( acc. of P rel. to C perp. to C P) + ( acc. of C rel. to O perp. to C P) ~ ~~ ~ v v −→ ¨ + ( acc. of C rel. to origin O along C = (2˙r θ˙ + r θ) P) = (2˙r θ˙ + r θ¨ ) + 0 = 2˙r θ˙ + r 0 = 2˙r = 2u . a a
5.3.8 Example The velocity of the particle along and perpendicular to the radius from a fixed origin are λr and μθ. Find the path and show that the accelerations, along and perpendicular to the radius vector, are λ2 r −
( μ) μ2 θ 2 , andμθ λ + r r
Sol: Here we have r˙ = λr, and hence r = Aeλt . Since r θ˙ = μθ, so
μ dθ = = ~θ dt ~~ r~
μ Aeλt
.
It follows that μ μ 1 r μ 1 λt e == = ln(Bθ ) = r. A −λ A −λ A −λ A2 ~ ~~ ~ Hence μ
r
Bθ = e −λA2 . This is the required equation of path. The radial acceleration = r¨ − r (θ˙ )2 =
d 1 1 ˙ 2 = λ2 r − (μθ )2 , (λr ) − r (θ˙ )2 = λ˙r − r (θ˙ )2 = λ(λr ) − r (θ˙ )2 = λ2 r − (r θ) dt r r
and the transversal acceleration =
() ) ( μθ μ) 1 d ( 2 ) 1 d 1 d μ r θ˙ = λr ·θ + r· = μθ λ + .\ (r (μθ )) = μ (r θ ) = r dt r dt r dt r r r
5.3 Miscellaneous Examples on 2D Motion (Part III)
663
5.3.9 Example A point starts from the origin in the direction of the initial line with velocity ωf and moves with constant angular velocity ω about the origin and with constant negative radial acceleration − f. Show that the rate of growth of the radial velocity is never positive, but tends to the limit zero, and prove that the equation of path is ) ( ω2 r = f 1 − e−θ . Sol: Here, − f = r¨ − r ω2 , so r = Aeωt + Be−ωt +
f . ω2
It follows that r˙ = Aωeωt − Bωe−ωt . Since r t=0 = 0, we have A + B = − ωf2 . Since r˙ t=0 = and hence A = 0, B = − ωf2 . Thus
f , we have ω
A− B =
f , ω2
) () ) ) f ( f f f ( r = 0eωt + − 2 e−ωt + 2 = 2 1 − e−ωt = 2 1 − e−θ .\ ω ω ω ω It follows that the rate of growth of radial velocity =r =−
f −θ e < 0, ω2
and lim r = 0.\ t→∞
5.3.10 Example A point P describes a curve with constant velocity and its angular velocity about a given point O varies inversely as the distance from O. Show that the curve is an equiangular spiral whose pole is O, and that acceleration of the point is along the normal at P and varies inversely as O P.
664
5 Motion in Two Dimensions (Finite Forces)
Sol: Here, / ( )2 β (˙r )2 + r θ˙ = α, and θ˙ = (α, β are constants). r It follows that dr r˙ r˙ = = = r dθ β r θ˙ √
/ α2 − β 2 , β
α 2 −β 2
and hence r = Ae β θ . This is an equiangular spiral whose pole is O. It suffices to show that / α 2 − β 2 r˙ dr r˙ = = = β β r dθ r θ˙ \ \ \ (π )\ \ 1 d (r 2 θ˙ ) \ \ \ \ r dt \ −φ \=\ = cot φ = \tan ( )2 \\ \ r − r θ˙ ~~ 2 ~ ~ \ \ \ \ β r˙ \ 1 d (βr ) \ \ \ \ r dt \ \ \ r =\ \=\ \ ( ( ) ) 2 2 \ r − r θ˙ \ \ r − r θ˙ \ \ \ \ \ \ β r˙ \ \ β r˙ \ \\ r˙ \\ /α 2 − β 2 \ r \ \ r \ \ .\ =\ \= − \= \=\ \ r − β2 \ \ 0 − β2 \ \ β \ β r
r
5.3.11 Example A point P describes an equiangular spiral with constant angular velocity about the pole. Show that its acceleration varies as O P and is in a direction making with the tangent at P the same constant angle that O P makes. Sol: Let the equation to the path of P be r = Ae(cot α)θ . Put θ˙ = ω, where ω is a constant. We have to show /( )2 ( )2 )2 ( r −r θ˙ + 2˙r θ˙ + r θ /( LHS =
r ) ( )2 ω2 cot2 α r − r ω2 + (2(ω cot αr )ω + r 0)2 =ω
/ 2 (
r cot2 α + 1
)2
= const. = ω2
/ (
cot2 α − 1
= ω2 csc2 α, which is a constant.
)2
+ (2(cot α))2
5.3 Miscellaneous Examples on 2D Motion (Part III) Fig. 5.39
665 (pole)
Initial line
Observe that tan(angle between acc.andO P) =
2˙r θ˙ + r θ 2˙r ω + r 0 ( )2 = r − r ω2 ˙ r −r θ 2(ω cot αr )ω + r 0 2 cot α ( ) = cot 2 α − 1 ω2 cot 2 α r − r ω2 2 sin α cos α = tan 2α, = cos 2α
=
so 2α is the angle between acc. and O P. Since tan α =
rω r θ˙ r dθ = = = tan( angle between tangent and O P), ω cot αr r˙ ~dr ~~ ~ 2
α is the angle between tangent and O P. Figure 5.39
5.3.12 Example A point moves in a given straight line on a plane with constant velocity V , and the plane moves with constant angular velocity ω about an axis perpendicular to itself through a given point O of the plane. If the distance of O from the given straight line be a, show that the path of the point in space is given by the equation / a V Vθ = r 2 − a 2 + cos−1 , ω ω r referred to O as pole (Fig. 5.40). Sol: Here
666
5 Motion in Two Dimensions (Finite Forces)
≡ ( ),
≡ ( )
, are constants vel. =
Wheel is rolling. =
̇,
∴ ̈ = 0, vel. =
Acc. of
̇= ̈ = 0,
is 0.
Fig. 5.40
/ a r = a 2 + (V t)2 , θ = ωt + cos−1 . r So RHS =
/ a a V V V Vθ r 2 − a 2 + cos−1 = V t + cos−1 = V t + (θ − ωt) = = LHS.\ ω r ω r ω ω
5.3.13 Example A straight smooth tube revolves with angular velocity ω in a horizontal plane about one extremity which is fixed. If at zero time a particle inside it be at a distance a from the fixed end and moving with velocity V along the tube, show that its distance at time t is Fig. 5.41. a cosh ωt +
V sinh ωt. ω
Sol: Fig. 5.41 Here one equation of motion is () ) ) ( d(0) r −r ω2 − 2 θ˙ + 0(0) = 0 dt or r −r ω2 = 0.
5.3 Miscellaneous Examples on 2D Motion (Part III)
667
Fig. 5.41
1
(
2
̇)
2 ̈ − ( ̇)
Fig. 5.42
1
(
2
Direction of acceleration
̇)
Direction of tangent 2 ̈−
̇
2
It follows that r = A cosh(ωt) + B sinh(ωt). Hence r˙ = Aω sinh(ωt) + Bω cosh(ωt). It is given that r˙ t=0 = V So V = Bω. Hence B =
V . ω
It is given that r t=0 = a. So a = A. Thus r = a cosh(ωt) +
V sinh(ωt).\ ω
5.3.14 Example A thin straight smooth tube is made to revolve upwards with a constant angular velocity ω in a vertical plane about one extremity O. When it is in a horizontal
668
5 Motion in Two Dimensions (Finite Forces)
position a particle is at rest in it at a distance a from the fixed end O. If ω be very ( ) 13 6a nearly (Fig. 5.42). small, show that it will reach O in time gω Sol: Here one equation of motion is () ) ( ) d(0) −(mg) sin θ 2 ˙ r¨ − r ω − 2 = −g sin(ωt) θ + 0(0) = dt m ~~ ~ ~ or r −r ω2 = −g sin(ωt). It follows that r = A cosh(ωt) + B sinh(ωt) +
g sin ωt. 2ω2
Hence r˙ = Aω sinh(ωt) + Bω cosh(ωt) +
g cos ωt. 2ω
It is given that r˙ t=0 = 0 So 0 = Bω + Hence B =
−g . 2ω2
g . 2ω
It is given that r t=0 = a. So a = A. Thus
r = a cosh(ωt) +
g −g sinh(ωt) + sin ωt. 2 2ω 2ω2
It follows that 0 = a cosh(ω( tr =0 )) + = a cosh(ω( tr =0 )) +
−g g sinh(ω( tr =0 )) + sin ω( tr =0 ) 2 2ω 2ω2
g −g sinh(ω( tr =0 )) + sin ω( tr =0 ) 2 2ω 2ω2
= a cosh(ω( tr =0 )) +
g (sin ω( tr =0 ) − sinh(ω( tr =0 ))) 2ω2
5.3 Miscellaneous Examples on 2D Motion (Part III)
g ≈ a cosh(ω( tr =0 )) + 2ω2 = a cosh(ω( tr =0 )) −
669
()() ) () )) (ω( tr =0 ))3 (ω( tr =0 ))3 ω( tr =0 ) − − ωt + 3! 3!
g (ω( tr =0 ))3 ω2 3!
≈ a1 −
gω t 6 ( r =0 ).
Hence () tr =0 =
6a gω
) 13
.\
5.3.15 Example A particle is at rest on a smooth horizontal plane which commences to turn about a straight line lying in itself with constant angular velocity ω downwards. If a be the distance of the particle from the axis of rotation at zero time, show that the body will leave the plane at time t given by the equation Fig. 5.43. a sinh(ωt) +
g g cosh(ωt) = 2 cos ωt. 2ω2 ω
Sol: Here equations of motion are Fig. 5.43 ) () ) ( d(ω) (mg) sin θ d(0) 2 r −r ω = r¨ − r ω − 2 ω+0 = = g sin(ωt), dt dt m ~~ ~ ~ 2
) () ) dr d(ω) −R + mg cos(ωt) d 2 (0) 2 = − 0ω + 2 ω + r 2˙r ω = 2 dt dt dt m ~ ~~ ~ ()
Thus r −r ω2 = g sin(ωt), 2ωr˙ =
−R + mg cos(ωt) . m
Since r −r ω2 = g sin(ωt).
670
5 Motion in Two Dimensions (Finite Forces)
Fig. 5.43
≡ ( ),
Fixed in space
It follows that r = A cosh(ωt) + B sinh(ωt) −
g sin ωt. 2ω2
Hence r˙ = Aω sinh(ωt) + Bω cosh(ωt) −
g cos ωt. 2ω
It is given that r˙ t=0 = 0 So 0 = Bω − Hence B =
g . 2ω2
g . 2ω
It is given that r t=0 = a. So a = A. Thus
r = a cosh(ωt) +
g g sinh(ωt) − sin ωt. 2 2ω 2ω2
Also −R+mg cos(ωt) m
2ω
g g = r˙ = aω sinh(ωt) + cosh(ωt) − cos ωt. 2ω~~ 2ω ~ ~
Thus −R + mg cos(ωt) = a2ω2 sinh(ωt) + g cosh(ωt) − g cos ωt. m
=
5.3 Miscellaneous Examples on 2D Motion (Part III)
671
Hence g cos(ω( t R=0 )) =
−0 + mg cos(ω( t R=0 )) = a2ω2 sinh(ω( t R=0 )) + g cosh(ω( t R=0 )) − g cos ω( t R=0 ) m ~ ~~ ~
Thus 2 cos(ω( t R=0 )) =
a2ω2 sinh(ω( t R=0 )) + cosh(ω( t R=0 )) g
or g g cos(ω( t R=0 )) = a sinh(ω( t R=0 )) + cosh(ω( t R=0 )).\ ω2 2ω2
5.3.16 Example A particle falls from rest within a straight smooth tube which is revolving with uniform angular velocity ω about a point O in its length, being acted on by a force equal to mμ(distance) towards O. Show that the equation of its path in space is (/ r = a cosh
) (/ ) ω2 − μ μ − ω2 θ or r = a cos θ , ω2 ω2
< > according as μ ω2 or μ ω2 . If μ = ω2 , show that the path is a circle (Fig. 5.44). Sol: Here one equation of motion is Fig. 5.44
≡ ( ), ≡ ( ), ̇=
= 0.
, ̈=0
Fixed direction in space
672
5 Motion in Two Dimensions (Finite Forces)
) () ( ) −mμr d(0) (−ω) + 0(0) = r¨ − r (−ω)2 − 2 dt m ~~ ~ ~ or ) ( r − ω2 − μ r = 0. It follows that ) (/ ) ⎫ ω2 − μt + B sinh ω2 − μt i f ω2 > μ ⎪ ⎪ ⎬ (/ ) (/ ) 2 2 2 r = A cos μ − ω t + B sin μ − ω t i f ω < μ ⎪. ⎪ ⎭ r = At + Bi f μ = ω2 .
r = A cosh
(/
Hence (/ ) (/ ) ⎫ / / r˙ = A ω2 − μ sinh ω2 − μt + B ω2 − μ cosh ω2 − μt i f ω2 > μ ⎪ ⎪ ⎬ (/ ) (/ ) / / 2 2 2 2 2 r˙ = −A ω − μ sin μ − ω t + B ω − μ cos μ − ω t i f ω < μ ⎪. ⎪ ⎭ r˙ = Ai f ω2 = μ It is given that r˙ t=0 = 0. So / ⎫ 0 = B /ω2 − μi f ω2 > μ ⎬ 0 = B μ − ω2 i f ω2 < μ . ⎭ A = 0i f ω2 = μ Thus ) (/ ⎫ r = A cosh ω2 − μt i f ω2 > μ ⎪ ⎪ ⎬ (/ ) 2 2 r = A cos μ − ω t i f ω < μ ⎪ ⎪ ⎭ r = Bi f ω2 = μ or ) (/ ⎫ r = A cosh ω2 − μ ωθ i f ω2 > μ ⎪ ⎪ ⎬ (/ ) θ 2 2 r = A cos μ − ω ω i f ω < μ ⎪.\ ⎪ ⎭ pathisacir clei f ω2 < μ
5.3 Miscellaneous Examples on 2D Motion (Part III)
673
5.3.17 Example A particle is placed at rest in a rough tube at a distance a from one end, and the tube starts rotating with a uniform angular velocity ω about this end. Show that the distance of the particle at time t is ae−ωt tan ε(cosh(ωt · sec ε) + sin ε sinh(ωt · sec ε)), where tan ε is the coefficient of friction (Fig. 5.45). Sol: Here equations of motion are ) () ( ) ) ( ) () d(0) −(tan ε)R R dr d 2 (0) r −r ω2 = = r¨ − r ω2 − 2 ω + 0(0) = , 2˙r ω = − 0ω2 + 2 ω + r (0) = ' , 2 dt m dt m dt ~~ ~ ~ ~ ~~ ~
and hence r −r ω2 =
−(tan ε)(2m r˙ ω) = −2ω tan εr˙ . m
Thus r +2ω tan εr˙ − ω2 r = 0. For C.F., we solve
≡ ( ), ≡ ( ), ̇=
= 0.
, ̈=0
Fixed direction in space
Fig. 5.45
674
5 Motion in Two Dimensions (Finite Forces)
m 2 + (2ω tan ε)m − ω2 = 0. Hence m=
−2ω tan ε ±
√ 4ω2 tan2 ε + 4ω2 = (− tan ε ± sec ε)ω. 2
Thus r = e−(ω tan ε)t (A cosh(ω sec εt) + B sinh(ω sec εt)) It follows that ) ( r˙ = −(ω tan ε)e−(ω tan ε)t (A cosh(ω sec εt) + B sinh(ω sec εt)) +e−(ω tan ε)t (ω sec ε)(A sinh(ω sec εt) + B cosh(ω sec εt)). Now since r˙ t=0 = 0, we have 0 = −ω tan ε A + ω sec ε B, and hence A sin ε = BThus r = e−(ω tan ε)t A(cosh(ω sec εt) + sin ε sinh(ω sec εt)). Now since r t=0 = a, we have a = A. Thus r = ae−(ω tan ε)t (cosh(ω sec εt) + sin ε sinh(ω sec εt)).\
5.3.18 Example One end A of a rod is made to revolve with uniform angular velocity ω in the circumference of a circle of radius a, while the rod itself revolves in the opposite direction about that end with the same angular velocity. Initially the rod coincides with a diameter and a smooth ring capable of sliding freely along the rod is placed at the centre of the circle. Show that the distance of the ring from A at time t is Fig. 5.46. a (4 cosh(ωt) + cos(2ωt)). 5 Sol: Let m be the mass of the ring (Fig. 5.46). Since the ring is smooth, we have
5.3 Miscellaneous Examples on 2D Motion (Part III)
675 ≡ ( ), ≡ ( ), ̇=
= 0.
, ̈=0 =
Fixed direction in space
Fig. 5.46 ()() 2 )) ) () ) ( d (r ) dω d(0) r −r ω2 + aω2 cos 2ωt = − r (ω)2 − 2 ω+0 + aω2 cos 2ωt 2 dt dt dt ) ( ( ) −→ = acceleration ofPrelative toAalong A Pin inertial frame + aω2 cos 2ωt ) ( −→ = acceleration ofPrelative toAalong A Pin inertial frame ) ( −→ + acceleration ofArelative toOalong A Pin inertial frame ) ( friction on ring at P 0 −→ = = 0, = acceleration ofPrelative toOalong A Pin inertial frame = = m m ~~ ~ ~
and hence r −r ω2 + aω2 cos 2ωt = 0. It follows that r −r ω2 = −aω2 cos 2ωt. Hence r = A cosh(ωt) + B sinh(ωt) + (P.I.). Observe that 1 1 1 cos(2ωt) = cos(2ωt), (cos(2ωt)) = D 2 − ω2 −5ω2 −(2ω)2 − ω2
676
5 Motion in Two Dimensions (Finite Forces)
hence ) ( r = A cosh(ωt) + B sinh(ωt) + −aω2
1 cos(2ωt) −5ω2
or r = A cosh(ωt) + B sinh(ωt) +
a cos(2ωt). 5
Now since r t=0 = a, we have a = A + a5 . Thus r˙ =
a(2ω) 4a ω sinh(ωt) + Bω cosh(ωt) − sin(2ωt). 5 5
Since r˙ t=0 = 0, we have 0 = Bω. Thus r=
a 4a cosh(ωt) + cos(2ωt). 5 5
5.3.19 Example P Q is a tangent at Q to a circle of radius a, P Q is equal to ρ and makes an angle θ with a fixed tangent to the circle. Show that the acceleration of P along and perpendicular to Q P are, respectively Fig. 5.47. ) ( ( )2 ( )2 1 d ρ 2 θ˙ ρ −ρ θ˙ + a θ and + a θ˙ . ρ dt Sol: Fig. 5.47 Here, ( ) −→ acceleration ofPrelative toOalong Q Pin inertial frame ⎛⎛ ( ( ( ( ) )2 ⎞ ( ) ) )⎞ ( ) d θ − π2 d 2 θ − π2 d2ρ d(a) d θ − π2 ⎝ ⎝ ⎠ = ρ −ρ θ˙ 2 + a θ , ⎠ = −ρ + 2 +a dt dt dt dt 2 dt 2
and
5.3 Miscellaneous Examples on 2D Motion (Part III)
677 ≡ ( ), ≡ ( ),
= 0.
̇=− , ̈=0
Fig. 5.47
(
) −→ acceleration ofPrelative toOperpendicular to Q Pin inertial frame
⎛⎛
2
d a = ⎝⎝ 2 − a dt
( ) ( ( ( ) )2 ⎞ ( ) ( ) )⎞ 2 ( )2 1 d ρ θ˙ d θ − π2 d θ − π2 d 2 θ − π2 ⎠ − 2 d(ρ) ⎠ ˙ = −a θ − +ρ .\ dt dt dt ρ dt dt 2
5.3.20 Example Two particles of masses m and m ' , connected by an elastic string of natural length a, are placed in a smooth tube of small bore which is made to rotate about a fixed point in its length with angular velocity ω. The coefficient of elasticity of the string ' 2 ω . Show that, if the particles are initially just at rest relative to the tube and is 2mm m+m ' the string is just taut, their distance apart at time t is 2a − a cos ωt. Sol: Here, equations of motion are Fig. 5.48 ξ −ξ ω2 = Hence
−T T 2mm ' ω2 . , η −ηω2 = ' , T = λ((ξ − η) − a), λ = m m m + m'
678
5 Motion in Two Dimensions (Finite Forces) ≡ ( ), ≡ ( ), ̇=
, ̈ = 0,
Fixed direction in space
Fig. 5.48
d 2 (ξ − η) 0 friction on ring at P − ω2 (ξ − η) = = = 0, 2 dt m ~~ m ~ ~ = −T =−
m + m' m + m' = −(λ((ξ − η) − a)) mm ' mm '
2mm ' ω2 m + m' = −2ω2 ((ξ − η) − a). − η) − a) ((ξ m + m' mm '
Thus d 2 (ξ − η) − ω2 (ξ − η) = −2ω2 ((ξ − η) − a) dt 2 or d 2 ((ξ − η) − 2a) + ω2 ((ξ − η) − 2a) = 0. dt 2 It follows that (ξ − η) − 2a = A cos ωt + B sin ωt. Hence ξ˙ − η˙ = −ω A sin ωt + Bω cos ωt.
= 0. =
5.3 Miscellaneous Examples on 2D Motion (Part III)
679
\ Since ξ˙ \t=0 = 0, η ˙ t=0 = 0, we have B = 0. Thus (ξ − η) − 2a = A cos ωt. Since (ξ − η)t=0 = a, we have A = −a. Thus (ξ − η) = 2a − a cos ωt.
5.3.21 Example A weight can slide along the spoke of a wheel, whose mass may be neglected. The weight is attached to the centre of the wheel by means of a light spring. When the . If the wheel is started wheel is fixed, the period of oscillation of the weight is 2π n √
to rotate freely with angular velocity 6n5511 , prove that the greatest extension of the spring is onefifth of its original length. Sol: Let m be the mass of the weight. Let λ be the spring constant of the spring. Let l be the natural length of the spring. From question, 2π 2π =/ , n λ m
and hence n 2 =
λ . m
Here one equation of motion is ( )2 λ(r − l) r −r θ˙ = − . m
Next, since wheel has negligible mass, we have ( ) 0 1 d r 2 θ˙ = = 0, r dt m ~ ~~ ~ ( √ ) ( \ ) ( \ ) and hence r~2 θ˙ =~~const.~ = ( r t=0 )2 θ˙ \t=0 = l 2 θ˙ \t=0 = l 2 6n5511 . Thus Fig. 5.49 ( r 2 θ˙ = l 2
√ ) 6n 11 . 55
680
5 Motion in Two Dimensions (Finite Forces)
Non Inertial frame
2
Inertial frame
Fig. 5.49
Next since
r −l 4
2
36n −3 r 55 × 5
⎛ ( √ ) ⎞2 l 2 6n5511 ⎠ = r¨ − r (θ) ˙ 2 = − λ(r − l) = −n 2 (r − l). = r −r ⎝ r2 m ~ ~ ~~
we have r −l 4
36n 2 −3 r = −n 2 (r − l) 55 × 5
It follows that ) ( () ) d (˙r )2 dr 4 36n 2 −3 =2 l r − n 2 (r − l) . dt dt 55 × 5 Thus () )) () 2 2 r −2 4 36n 2 r −n − lr + C. (˙r ) = 2 l 55 × 5 −2 2 2
Now since r˙ r =l = 0, we have ( C = −2 l 4 ~
( )) ) () ( ) 36n 2 l −2 36n 2 1 l2 n2l 2 36 239 2 2 − n2 − l(l) + = − 1 n2l 2 = − n l , = −2 l 2 55 × 5 −2 2 55 × 5 −2 2 55 × 5 275 ~~ ~
5.3 Miscellaneous Examples on 2D Motion (Part III)
681
and hence )) () () 2 r 239 2 2 36n 2 r −2 − n2 − lr − n l . (˙r )2 = 2 l 4 55 × 5 −2 2 275 It suffices to show that r˙ r =l+ 5l = 0. Here ⎞⎞ ⎛ ( )2 ( )−2 6l 6l () ) 2 6l 36n 5 5 ⎟⎟ 239 2 2 ⎜ ⎜ − n2 ⎝ −l n l r˙ r =l+ l = 2⎝l 4 ⎠⎠ − 55 × 5 −2 2 5 275 5 ( (() ) )) ) () 6l 2 12l 2 −25 − 396 + 660 n2 239 2 2 239 2 2 = −l 2 − − − n2 − n l = n2l 2 n l = 0.\ 11 5 5 275 275 275 ()
)2
⎛
5.3.22 Example A uniform chain AB is placed in a straight tube O AB which revolves in a horizontal plane, about the fixed point O, with uniform angular velocity ω. Show that the motion of the middle point of the chain is the same as would be the motion of a particle placed at this middle point. Also, show that the tension of the chain at any point P is 21 mω2 (A P)(P B), where m is the mass of a unit length of the chain Fig. 5.50. Sol: Let C be the midpoint of the chain AB. Hence OC = x + l. Here equation of motion of chainelement at P is Fig. 5.50 x − (x + ξ )ω2 =( x + 0) − (x + ξ )ω2 = ( x + ξ ) − (x + ξ )ω2 =
(T + dT ) − T d 2 (x + ξ ) − (x + ξ )ω2 = 2 dt m(dξ ) ~ ~~ ~
=
dT . m(dξ )
Thus, equation of motion of chainelement at P is x −(x + ξ )ω2 =
dT . m(dξ )
Hence, equation of motion of midelement of chain is
682
5 Motion in Two Dimensions (Finite Forces)
Non Inertial frame
Inertial frame Fixed in space
Fixed tangent
Fig. 5.50
\ 1 dT \\ x −(x + l)ω = .(∗) m dξ \mid element of chain 2
Further 0 =0 − 0 = TB − T A { ξ =2l { ξ =2l ( ) x¨ − (x + ξ )ω2 mdξ = dT = ξ =0 ξ =0 ~ ~~ ~ ξ =2l ) ξ =2l ( = m x −xω2 ∫ dξ − mω2 ∫ ξ dξ ξ =0
ξ =0
) ( ) = m x −xω2 (2l) − mω2 2l 2 . (
Thus. ) ( ) ( 0 = m x −xω2 (2l) − mω2 2l 2 . It follows that (
) x −xω2 − ω2 l = 0.
Now since equation of motion of chainelement at P is
5.3 Miscellaneous Examples on 2D Motion (Part III)
683
dT ω2 l − ξ ω2 = x¨ − (x + ξ )ω2 = m(dξ ) ~~ ~ ~
we have ( T = mω2 lξ − ~~
~
ξ2 2
) = ~
1 1 1 mω2 ξ (2l − ξ ) = mω2 (A P)(AB − A P) = mω2 (A P)(P B). 2 2 2
This proves the second part. Since () ) ξ2 , T = mω2 lξ − 2 we have dT = mω2 (l − ξ ), dξ and hence \ dT \\ = mω2 (l − l) = 0. dξ \mid element of chain Thus, from (∗), equation of motion of midelement of chain is x −(x + l)ω2 = 0. Observe that the equation of motion of a particle placed at C is x −(x + l)ω2 =
d 2 (x + l) d 2 (OC) 2 − + l)ω = − (OC)ω2 = 0 (x 2 dt 2 dt ~ ~~ ~
This proves the first part.
5.3.23 Example A perfectly rough plane turns with uniform velocity ω about a horizontal axis lying in its plane. Initially when the plane was horizontal a homogeneous sphere was in contact with it, and at rest relative to it at a distance a from the axis of rotation. Show
684
5 Motion in Two Dimensions (Finite Forces)
Fig. 5.51
′
Direction fixed in space
Fixed point
Fig. 5.52
that at time t the distance of the point of contact from the axis of rotation was (/ a cosh
5 ωt 7
)
√
35g sinh + 12ω2
(/
5 ωt 7
) −
5g sin ωt. 12ω2
Find also when the sphere leaves the plane (Fig. 5.51). Sol: Here equations of motions are ( ) m b φ −(a + bφ)ω2 = m
((
~
) () )) d 2 (a + bφ) 2 − 2 d(b) (−ω) + (b) d(−ω) = (mg) sin(ωt) − F , − (a + bφ)(−ω) dt dt dt 2 ~~ ~
and ( ) ˙ m −bω2 − 2bφω =m
Hence
((
) () )) d(−ω) d(a + bφ) d2b = −(mg) cos(ωt) + R. − b(−ω)2 + 2 (−ω) + (a + bφ) 2 dt dt dt
5.3 Miscellaneous Examples on 2D Motion (Part III)
b φ −(a + bφ)ω2 = g sin(ωt) −
685
) R F ( , ω + 2φ˙ bω = g cos(ωt) − . m m
Also () () )) 2 2 φ = F(b). m b 5 Thus 2 F = g sin(ωt) − b φ +(a + bφ)ω2 bφ¨ = 5 m ~ ~~ ~ or 2 b φ = g sin(ωt) − b φ +(a + bφ)ω2 5 or 7 b φ −bω2 φ = g sin(ωt) + aω2 5 or 5 5 5 φ − ω2 φ = g sin(ωt) + aω2 . 7 7b 7b Hence ) (/ ) (/ 5 5 ωt + B sinh ωt + (P.I.). φ = A cosh 7 7 Observe that ) ( ) 1 5 1 5 5g 5 g sin(ωt) + aω2 = aω2 e0t (sin(ωt)) + 5 5 7b 7b 7b D 2 − ω2 7b D 2 − 7 ω2 7 ( ) 5g 1 5 1 a 5g e0t = aω2 sin(ωt) − . = ) 5 (sin(ωt)) + ( 5 2 2 2 2 7b −ω − ω 7b b −12bω2 ω 0 − 7 7 ()
1
D 2 − 57 ω2
Thus (/ φ = A cosh It follows that
5 ωt 7
)
(/ + B sinh
5 ωt 7
) −
5g a sin(ωt) − . 2 12bω b
686
5 Motion in Two Dimensions (Finite Forces)
/
5 φ˙ = A ω sinh 7
(/
5 ωt 7
)
/ +B
5 ω cosh 7
(/
/ \ Now since φ˙ \t=0 = 0, we have 0 = B 57 ω − φt=0 = 0, we have A = ab . Thus a φ = cosh b
(/
5 ωt 7
)
√
35g + sinh 12bω2
(/
5 ωt 7
5g , 12bω
5 ωt 7
) −
5g cos(ωt). 12bω
and hence B =
) −
√ 35g . 12bω2
Since
5g a sin(ωt) − . 12bω2 b
Hence (/ a + bφ = a cosh
5 ωt 7
)
√
35g + sinh 12ω2
(/
5 ωt 7
) −
5g sin ωt.\ 12ω2
5.3.24 Example If a uniform heavy right circular cylinder, of radius a, be rotated about its axis and laid gently on two rough horizontal rails, at the same level 2a sin α apart, so that the axis of cylinder is parallel to the rail, show that the cylinder will remain in contact with both rails if μ < tan α, but will initially rise on the one rail if μ > tan α. Fig. 5.52. Sol: (Fig. 5.52). Here one equation of motion is (
) Ma 2 ω˙ = −(μR)(a) − (μS)(a) = −μa(R + S),
so Ma ω˙ = −μ(R + S). Also (S cos α + R cos α) + ((μS) sin α + (−μR) sin α) = Mg, (S sin α + (−μR) cos α) = (R sin α + (μS) cos α). Thus
5.3 Miscellaneous Examples on 2D Motion (Part III)
687
} (cos α − μ sin α)R + (cos α + μ sin α)S − Mg = 0 . (−μ cos α − sin α)R + (sin α − μ cos α)S = 0 It follows that R S = Mg(sin α − μ cos α) Mg(μ cos α + sin α) =
1 . (cos α − μ sin α)(sin α − μ cos α) − (cos α + μ sin α)(μ cos α + sin α)
Hence R=
Mg(sin α − μ cos α) Mg(μ cos α + sin α) ,S = . −2μ −2μ
If μ < tan α, then R and S have the same sign, and hence the cylinder will remain in contact with both rails. If μ > tan α, then R and S have opposite signs, and hence the cylinder will initially rise on the one rail \
5.3.25 Example A billiard ball, struck low, slides and rotates until uniform motion presently ensues. Investigate te motion, and show that the point of the ball, which is threetenth of the diameter below the top, has a velocity which remains of the same magnitude all the time Fig. 5.53. Sol: Fig. 5.53. Suppose that \ x ˙ t=0 = u, θ˙ \t=0 = ω.
Fig. 5.53
≡ ( ), ≡ ( ), = const. 2 ̇=
, ̈ = 0,
+ Fixed direction in space
=
688
5 Motion in Two Dimensions (Finite Forces)
Here equations of motion are (Figs. 5.54, 5.55 and 5.56). () ) 2a 2 m x = −μ(mg), m θ = (μ(mg))a. 5 So \ 5μg 5μg t =ω+ t. x˙ = x ˙ t=0 − μgt = u−−μgt, θ˙ = θ˙ \t=0 + 2a 2a
Point fixed in space
Direction fixed in the body
( = ,
Direction fixed in space
Fig. 5.54
2 sin ,
−
Fig. 5.55
+
,
= )
5.3 Miscellaneous Examples on 2D Motion (Part III)
689
, Direction fixed in body Direction fixed in body =
= (
)
Direction fixed in space
Direction fixed in space ≡ ( )
Fig. 5.56
It follows that (velocity at the point three − tenth of the diameter below the top) = x˙ +
which is a constant. \
2a 2aω θ˙ = u + , 5 5
Chapter 6
Impulsive Forces
In daily life, impulsive forces are more prevalent than force. In colloquial language, impulsive forces are referred as “blow”. In contrast to finite force, their reason is infinite force, and it is measured in terms of change in momentum resulted. For getting a grip over the subject, 44 problems of various difficultylevels have been solved.
6.1 Examples on Impulsive Forces in 2D Motion (Part I) 6.1.1 Example Two uniform rods AB, BC are freely jointed at B and laid on a horizontal table. AB is struck by a horizontal blow of impulse P in a direction perpendicular to AB at a distance c from its centre. The length of AB, BC being 2a and 2b and their masses ' M and M , find the motion immediately after the blow. Sol: Let u 1 be the velocity of the centre of mass of the rod AB just after the blow, and ω1 be the angular velocity of the rod AB just after the blow. Let u 2 be the velocity of the centre of mass of the rod BC just after the blow, and ω2 be the angular velocity of the rod BC just after the blow. There will be an impulsive action between the rods AB, BC at B when the blow is struck. Let its impulse be Q, in opposite directions on the rods. We shall try to determine the five quantities u 1 , ω1 , u 2 , ω2 , Q. (Fig. 6.1). For AB: ( 2) M(u 1 ) − M(0) = P − Q, M a3 ω1 = P(c) − Q(a) Here, upwards velocity of B of rod AB is u 1 + aω1 . For BC:
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 R. Sinha, Advanced Newtonian Rigid Dynamics, https://doi.org/10.1007/9789819920228_6
691
692
6 Impulsive Forces 1
1
2
1
2
2
Fig. 6.1
) ( 2 M ' (u 2 ) − M ' (0) = Q, M ' b3 ω2 = Q(b) . Here, upwards velocity of B of rod BC is u 2 + bω2 . Since the two rods are freely jointed at B, P−Q 3(Pc − Qa) −4Qa + 3Pc + Pa = + = u 1 + aω1 Ma M Ma = (upwards velocity of B of rod AB) = (upwards velocity of B of rod BC) ~~ ~ ~ Q 3Q 4Q = u 2 + bω2 = ' + ' = ' M M M and hence −4Qa + 3Pc + Pa 4Q = ' Ma M or (−4Qa + 3Pc + Pa)M ' = 4Q Ma or ( ) P(3c + a)M ' − Q4a M ' + M = 0. or ) ( P(3c + a)M ' P M' 3c Q= = . 1+ 4a(M ' + M) 4(M ' + M) a ( ' ) . Since M b3 ω2 = Q, we have ( 2) M a3 ω1 = . Since ω2 = M3' b Q. Since M(u 1 ) = P − Q, we have u 1 = P−Q M '
Now since M (u 2 ) = Q, we have u 2 =
P(c) − Q(a), we have ω1 =
3(Pc−Qa) . Ma 2
Q M'
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
693
6.1.2 Example Three equal uniform rods AB, BC, C D are hinged freely at their ends, B and C, so as to form three sides of a square and are laid on a smooth table. The end A is struck by a horizontal blow P at right angles to AB. Show that the initial velocity of A is 19 times that of D, and that the impulsive actions at B and C are respectively 5P 12 P and 12 . Sol: (Fig. 6.2). Let m be the mass of each rod. Since impulse P acts perpendicular to AB, the instantaneous axis of rotation passes through the line AB. It follows that velocity at −→ −→ B of AB is along C B in our figure. This shows that BC is translated along C B. For AB: ( 2) m(u 1 ) − m(0) = P − Q, m a3 ω1 = P(a) + Q(a) For BC: m(u 2 ) − m(0) = R − Q. For C D: 2 2
3
1 1
Fig. 6.2
1
3
3
694
6 Impulsive Forces
( 2) m(u 3 ) − m(0) = −R, m a3 ω3 = R(a) For joint B: P−Q 3(P + Q) −2P − 4Q Q−R = − = u 1 − aω1 = −u 2 = , ~~ ~ ~ m m m m so R − 5Q = 2P. For joint C: −R 3R −4R R−Q = − = u 3 − aω3 = u 2 = , ~ ~~ ~ m m m m so Q = 5R = 5(2P + 5Q). ~ ~~ ~ Thus 24Q = −10P. Hence Q =
−5 P. 12
Now since
P −25 P + 2P = − . R = 5Q + 2P = ~ ~~ ~ 12 12 Since u1 =
P − −5 P 17 P P−Q 12 = = . m m 12 m
Since ( ) −5 7P 3 3 P+ P = . ω1 = (P + Q) = ma ma 12 4ma Since u3 =
P P −R = 12 = . m m 12m
Since ( ) 3 P P 3 R= − =− . ω3 = am am 12 4am Now
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
velocity of A u 1 + aω1 u 1 + aω1 u 1 + aω1 = P = = = −P −P velocity of D u 3 + aω3 + 4m 12m 6m
695 7P 17 P + 4m 12 m −P 6m
=
38P 12m −P 6m
= −19.∎
6.1.3 Example AB, BC are two equal similar rods freely hinged at B and lie in a straight line on a smooth table. The end A is struck by a blow perpendicular to AB. Show that the resulting velocity of A is 27 times that of B. Sol: Let u 1 be the velocity of the centre of mass of the rod AB just after the blow, and ω1 be the angular velocity of the rod AB just after the blow. Let u 2 be the velocity of the centre of mass of the rod BC just after the blow, and ω2 be the angular velocity of the rod BC just after the blow. There will be an impulsive action between the rods AB, BC at B when the blow is struck. Let its impulse be Q, in opposite directions on the rods. We shall try to determine the five quantities u 1 , ω1 , u 2 , ω2 , Q (Fig. 6.3). For AB: ( 2) M(u 1 ) − M(0) = P − Q, M a3 ω1 = −P(a) − Q(a). Here, upwards velocity of B of rod AB is u 1 + aω1 . For BC: ( 2) M(u 2 ) − M(0) = Q, M a3 ω2 = Q(a). Here, upwards velocity of B of rod BC is u 2 + aω2 . Since the two rods are freely hinged at B, 1
1
2
1
2
2
Fig. 6.3
696
6 Impulsive Forces
−4Q − 2P P−Q 3(−P − Q) = + = u 1 + aω1 M M M = (upwards velocity of B of rod AB) = (upwards velocity of B of rod BC) ~~ ~ ~ 3Q 4Q Q + = = u 2 + aω2 = M M M and hence 4Q −4Q − 2P = M M or Q=
−P . 4
( ) Since M a3 ω2 = Q, we have ( 2) 3 M a3 ω1 = Q. Since M(u 1 ) = P − Q, we have u 1 = P−Q . Since ω2 = Ma M Now since M(u 2 ) = Q, we have u 2 =
−P(a) − Q(a), we have ω1 =
−3(P+Q) . Ma
u 1 − aω1 velocity of A = = velocity of B u 1 + aω1 =
Q . M
Now
P−Q M P−Q M
− +
−3(P+Q) M −3(P+Q) M
2P + −P 2P + Q 4P + 2Q 7 (4 ) = = = .∎ −2P − 4Q −P − 2Q −2 −P − 2 −P 4
6.1.4 Example Two uniform rods, AB of length 2a and BC of length 2b, are smoothly jointed at B and placed in a horizontal line. The rod BC is struck at G by a blow at right angles to it. Find the position of G so that the angular velocities of AB and BC may be equal in magnitude. Sol: (Fig. 6.4). Let λ be the mass/length of rod AB and BC both. For AB: ( 2) M(u 1 ) − M(0) = −Q, M a3 ω1 = −Q(a). Here, upwards velocity of B of rod AB is u 1 + aω1 . For BC:
6.1 Examples on Impulsive Forces in 2D Motion (Part I) 1
697
′ 2
1
2 1 2
Fig. 6.4
) ( 2 M ' (u 2 ) − M ' (0) = Q + P, M ' b3 ω2 = Q(b) − P(x). Here, upwards velocity of B of rod BC is u 2 + bω2 . Since the two rods are freely hinged at B, −Q 3(−Q) −4Q = + = u 1 + aω1 M M M = (upwards velocity of B of rod AB) = (upwards velocity of B of rod BC) ~ ~~ ~ Q+P 3(Qb − P x) 4bQ + b P − 3x P = u 2 + bω2 = + = , M' bM ' bM ' and hence 4bQ + b P − 3x P −4Q = M bM ' or −4bQ M ' = 4bQ M + b P M − 3x P M or Q=
3x − b 3x − b M λ(2a) 3x − b a P= P= P ' 4b M + M 4b λ(2a) + λ(2b) 4b a + b
When ω1 = ω2 : We have −3Q 3(Qb − P x) = Ma M ' b2 or −Q (Qb − P x) = (λa)a (λb)b2 or
698
6 Impulsive Forces
( ( ) 3x − b a ) ( ) 3x − b a P a 2 + b2 = P a 2 b + b3 = Q a 2 b + b3 = a 2 x P 4 a+b 4b a + b or ) 3x − b 1 ( 2 a + b2 = ax 4 a+b or ) ( (3x − b) a 2 + b2 = 4(a + b)ax or ( ) ( ) x 3a 2 + 3b2 − 4(a + b)a = b a 2 + b2 or ) ( b a 2 + b2 x= 2 . 3b − 4ab − a 2 When ω1 = −ω2 : We have 3(Qb − P x) −3Q =− Ma M ' b2 or −Q (Qb − P x) =− (λa)a (λb)b2 or ( ) ( ) 3x − b a 3x − b Pa(a − b) = P a 2 b − b3 = Q a 2 b − b3 = a 2 x P 4 4b a + b or (3x − b)(a − b) = 4ax or x(−a − 3b) = b(a − b) or x=
b(b − a) . a + 3b
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
699
6.1.5 Example Two equal uniform rods, AB and AC, are freely hinged at A and rest in a straight line on a smooth table. A blow is struck at B perpendicular to the rods. Show that the kinetic energy generated is 47 times what it would be if the rods were rigidly fastened together at A. Sol: Let u 1 be the velocity of the centre of mass of the rod AB just after the blow, and ω1 be the angular velocity of the rod AB just after the blow. Let u 2 be the velocity of the centre of mass of the rod AC just after the blow, and ω2 be the angular velocity of the rod AC just after the blow. There will be an impulsive action between the rods AB, AC at A when the blow is struck. Let its impulse be Q, in opposite directions on the rods. We shall try to determine the five quantities u 1 , ω1 , u 2 , ω2 , Q (Fig. 6.5). For AB: ( 2) M(u 1 ) − M(0) = P − Q, M a3 ω1 = −P(a) − Q(a). Here, upwards velocity of A of rod AB is u 1 + aω1 . For AC: ) ( a2 ω2 = Q(a). M(u 2 ) − M(0) = Q, M 3 Here, upwards velocity of A of rod AC is u 2 + aω2 . Since the two rods are freely hinged at A, P−Q 3(−P − Q) −4Q − 2P = + = u 1 + aω1 M M M = (upwards velocity of A of rod AB) = (upwards velocity of A of rod AC) ~~ ~ ~ Q 3Q 4Q = u 2 + aω2 = + = M M M and hence 1
1
2
1
2
2
Fig. 6.5
700
6 Impulsive Forces
−4Q − 2P 4Q = M M or Q=
−P . 4
) ( Since M a3 ω2 = Q, we have ( 2) 3 ω2 = Ma Q. Since M(u 1 ) = P − Q, we have u 1 = P−Q . Since M a3 ω1 = M Now since M(u 2 ) = Q, we have u 2 =
−P(a) − Q(a), we have ω1 =
−3(P+Q) . Ma
Q . M
Here, (Fig. 6.6)
( ) ) ( 1 a2 1 M(u 1 )2 + M K.E. of AB + K.E. of AC = (ω1 )2 2 2 3 ( ( ) 2) 1 1 a + M(u 2 )2 + M (ω2 )2 2 2 3 ) ) 1 ( a2 ( = M (u 1 )2 + (u 2 )2 + M (ω1 )2 + (ω2 )2 2 ( 6 (( ( ( )2 ( )2 ) ) )2 ) 1 Q 3 P−Q −3(P + Q) 2 a2 = M + + Q +M 2 M M 6 Ma Ma ) ) 1 ( 1 ( = (P − Q)2 + Q 2 + (−3(P + Q))2 + (3Q)2 2M ( 6M ) (( ( ( ( ) ) ) ) ) 1 −P 2 −P 2 −P 2 −P 2 3 = + + P− P+ + 2M 4 4 2M 4 4 ) 2 2 2 ( 13 3P 5 7P P + = . = 2M 8 2M 8 4M For BC: ) (2a)2 ω = P(2a). (2M)(u) − (2M)(0) = P, (2M) 3 (
2
Fig. 6.6
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
Hence. K.E. =
( P )2 1 2 (2M) 2M =
+
1 2
(
2 (2M) (2a) 3
)
(ω) = 2
701
P2 4M
+
1 2
(
2 (2M) (2a) 3
)(
)2 P(2a) 2
(2M) (2a) 3
.
P2 P2 2a 2 P 2 3P 2 P2 = + + = . 2 4M 4M 4M M (2M) (2a) 3
Hence, the required ratio =
7P 2 4M P2 M
=
7 .∎ 4
6.1.6 Example Two equal uniform rods, AB and BC, are freely jointed at B and turn about a smooth joint at A. When the rods are in a straight line, ω being the angular velocity of AB and u the velocity of the centre of mass of BC, BC impinges on a fixed inelastic obstacle at a point D. Show that the rods are instantaneously brought to rest if B D = 2a
2u − aω , 3u + 2aω
where 2a is the length of either rod. Sol: (Fig. 6.7). It follows that (2a)ω = u + aω1 Suppose that both the rods are brought to rest after impinge. For AB:
2
2 1
Fig. 6.7
702
6 Impulsive Forces
(
) ( ) a2 a2 2 2 M + Ma 0 − M + Ma ω = −Q(2a). 3 3
For BC: ) ( ) a2 a2 0− M ω1 = Q(a) − P(x). M(0) − M(u) = Q + P, M 3 3 (
Thus 2a 2a ω = Q, −Mu = Q + P = M ω + P, ~ ~~ ~ 3 3 ) ) ( ( a 2 (2a)ω − u a2 a ω1 = Q(a) − P(x) =− M −M (2aω − u) = − M 3 3 a 3 ~ ~~ ~ ) ( ) ( 2a 2a = M ω a + M ω + Mu x. 3 3 M
Hence −a(2aω − u) = 2a 2 ω + (2aω + 3u)x or x=
−a(4aω − u) 2aω + 3u
Hence BD = a + x = a +
a(4u − 2aω) 2a(2u − aω) −a(4aω − u) = = .∎ 2aω + 3u 2aω + 3u 2aω + 3u
6.1.7 Example Two rods AB and BC, of lengths 2a and 2b and of masses proportional to their lengths, are freely jointed at B and are lying in a straight line. A blow is communicated to the end A. Show that the resulting kinetic energy when the system is free is to the energy when C is fixed as (4a + 3b)(3a + 4b) : 12(a + b)2 . Sol: Let λ be the mass/length of each rod. Let u 1 be the velocity of the centre of mass of the rod AB just after the blow, and ω1 be the angular velocity of the rod AB just after the blow. Let u 2 be the velocity of the centre of mass of the rod BC just after the blow, and ω2 be the angular velocity of the rod BC just after the blow. There will be an impulsive action between the rods AB, BC at B when the blow is struck.
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
703
1
1
2
1
2
2
Fig. 6.8
Let its impulse be Q, in opposite directions on the rods. We shall try to determine the five quantities u 1 , ω1 , u 2 , ω2 , Q (Fig. 6.8). For AB: ) ( a2 ω1 = −P(a) − Q(a). (λ(2a))(u 1 ) − (λ(2a))(0) = P − Q, (λ(2a)) 3 Here, upwards velocity of B of rod AB is u 1 + aω1 . For BC: ) ( b2 ω2 = Q(b). (λ(2b))(u 2 ) − (λ(2b))(0) = Q, (λ(2b)) 3 Here, upwards velocity of B of rod BC is u 2 + bω2 . Since the two rods are freely jointed at B, P−Q 3(−P − Q) −4Q − 2P = + = u 1 + aω1 λ(2a) λ(2a) λ(2a) = (upwards velocity of B of rod AB) = (upwards velocity of B of rod BC) ~ ~~ ~ Q 3Q 4Q = u 2 + bω2 = + = λ(2b) λ(2b) λ(2b) and hence 4Q −4Q − 2P = λ(2a) λ(2b) or b(−2Q − P) = 2a Q or
704
6 Impulsive Forces
Q=
−b P . 2(a + b)
) ( Q Now since (λ(2b))(u 2 ) = Q, we have u 2 = λ(2b) . Since λ(2b) a3 ω2 = Q, P−Q 3 we have ω2 = λ(2b)a Q. Since λ(2a)(u 1 ) = P − Q, we have u 1 = λ(2a) . Since ( ) −3(P+Q) a2 λ(2a) 3 ω1 = −P(a) − Q(a), we have ω1 = λ(2a)a . Here, (Fig. 6.9) ) ( ) a2 1 1 2 2 λ(2a) K.E.of AB + K.E.ofBC = λ(2a)(u 1 ) + (ω1 ) 2 2 3 ) ( ) ( b2 1 1 λ(2b) + λ(2b)(u 2 )2 + (ω2 )2 2 2 3 ) )) ( 1 (( a(aω1 )2 + b(bω2 )2 = λ a(u 1 )2 + b(u 2 )2 + λ 3 ( ( ( ( )2 )2 ) ) ) ) ( ( P−Q Q 3Q 2 1 3(−P − Q) 2 =λ a +b +b +λ a λ(2a) λ(2b) 3 λ(2a) λ(2b) ( ) ( ) Q2 1 (−3(P + Q))2 1 (P − Q)2 (3Q)2 + + + = 4λ a b 12λ a b ⎛( ⎞ ⎛ ( ( ( )2 )2 )2 )2 ⎞ −b P −b P −b P −b P P − P + 2(a+b) 2(a+b) 2(a+b) 2(a+b) 3⎜ 1⎜ ⎟ ⎟ + + = ⎝ ⎠+ ⎝ ⎠ 4λ a b 4λ a b (
⎛( =
2
P ⎜ ⎝ 4λ
2a+3b 2(a+b)
a
(
)2 +
b 2(a+b)
b
)2 ⎞
⎛( 2
⎟ 3P ⎜ ⎠+ ⎝ 4λ
2a+b 2(a+b)
a
(
)2 +
b 2(a+b)
b
)2 ⎞ ⎟ ⎠
( ( ) ) 3P 2 P2 (2a + 3b)2 (2a + b)2 + b + + b a a 16λ(a + b)2 16λ(a + b)2 2 ) ( P 16a 2 + 28ab + 12b2 = 16λ(a + b)2 a
=
=
) ( 2 P 2 (4a + 3b)(a + b) P2 P 2 (4a + 3b) 2 . = + 7ab + 3b 4a = 4λ(a + b)a 4λ(a + b)2 a 4λ(a + b)2 a
For AB: ) ( a2 ω3 = −P(a) − Q ' (a). (λ(2a))(u 3 ) − (λ(2a))(0) = P − Q ' , (λ(2a)) 3 Here, upwards velocity of B of rod AB is u 3 + aω3 . For BC:
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
705
3 ′
3
4
2
1
′
Fig. 6.9
( ) b2 4b2 2 ω4 = (λ(2b)) + (λ(2b))b ω4 = Q ' (2b) (λ(2b)) 3 3 ~~ ~ ~ or ω4 =
3Q ' . 4b2 λ
Here, upwards velocity of B of rod BC is (2b)ω4 . Since the two rods are freely jointed at B, ) ( 3 −P − Q ' −4Q ' − 2P P − Q' = + = u 3 + aω3 . λ(2a) λ(2a) λ(2a) = (upwards velocity of B of rod AB) = (upwards velocity of B of rod BC) ~~ ~ ~ ' 3Q = (2b)ω4 = 2bλ and hence 3Q ' −4Q ' − 2P = λ(2a) λ(2b) or ( ) b −4Q ' − 2P = 3a Q ' or Q' =
−2b P . 3a + 4b
Now since (λ(2a))(u 3 ) = P − Q ' , we have u 3 = 3 −P−Q ' ) . −P(a) − Q ' (a), we have ω = ( 3
Here,
(λ(2a))a
P−Q ' . λ(2a)
( ) 2 Since (λ(2a)) a3 ω3 =
706
6 Impulsive Forces
) ( ) a2 1 1 2 2 λ(2a)(u 3 ) + λ(2a) K.E.of AB + K.E.of BC = (ω3 ) 2 2 3 ) ( b2 1 λ(2b) + λ(2b)b2 (ω4 )2 + 2 3 ⎛ ( ( ) )2 ⎞ ( ( ' )2 2 ) 3 −P − Q ' P − Q 1 1 a ⎠ = ⎝ λ(2a) + λ(2a) 2 λ(2a) 2 3 (λ(2a))a )( ( ) 3Q ' 2 b2 1 λ(2b) + λ(2b)b2 + 2 3 4b2 λ )2 ( )2 1 3 ( 3 ( ' )2 1 P − Q' + P + Q' + Q = 2 λ(2a) 2 λ(2a) 4bλ ) ( ( ) ( ) −2b P 2 1 3 −2b P 2 −2b P 2 3 1 P − 3a+4b + P+ + = 2 λ(2a) 2 λ(2a) 3a + 4b 4bλ 3a + 4b ( 3a+6b )2 )2 ( 1 3 1 3b 2 1 2 3a+4b 2 3a + 2b + P + = P P 2 λ(2a) 2 λ(2a) 3a + 4b λ (3a + 4b)2 ( ) 3(3a + 2b)2 P2 (3a + 6b)2 + + 3b = 4a 4a λ(3a + 4b)2 ( ) 2 2 3(a + 2b) 3P (3a + 2b)2 + +b = 4a 4a λ(3a + 4b)2 (
=
( ) 3P 2 12a 2 + 28ab + 16b4
=
4aλ(3a + 4b)2 3P 2 (a + b) aλ(3a + 4b)
=
( ) 3P 2 3a 2 + 7ab + 4b4 aλ(3a + 4b)2
=
3P 2 (3a + 4b)(a + b) aλ(3a + 4b)2
.
So the required ratio =
P 2 (4a+3b) 4λ(a+b)a 3P 2 (a+b) aλ(3a+4b)
=
(4a + 3b)(3a + 4b) .∎ 12(a + b)2
6.1.8 Example Three equal rods, AB, BC, C D, are freely jointed and placed in a straight line on a smooth table. The rod AB is, struck at its end A by a blow which is perpendicular
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
707
to its length. Find the resulting motion. Further show that the velocity of the centre of mass of AB is 19 times that of C D, and its angular velocity 11 times that of C D. Sol: (Fig. 6.10). For AB: ) ( a2 ω1 = −P(a) − Q(a). M1 (u 1 ) − M1 (0) = P − Q, M1 3 Here, upwards velocity of B of rod AB is u 1 + aω1 . For BC: ) ( b2 ω2 = Q(b) − R(b). M2 (u 2 ) − M2 (0) = Q + R, M2 3 Here, upwards velocity of B of rod BC is u 2 + bω2 . Here, upwards velocity of C of rod BC is u 2 − bω2 . For C D: ) ( c2 ω3 = R(c). M3 (u 3 ) − M3 (0) = R, M3 3 Here, upwards velocity of C of rod C D is −(u 3 + cω3 ). Since B is a hinge, Q+R 3(Q − R) 4Q − 2R = + = u 2 + bω2 = u 1 + aω1 ~ ~~ ~ M2 M2 M2 −3(P + Q) −2P − 4Q P−Q + = , = M1 M1 M1 we have 2Q − R −P − 2Q = M2 M1 or (2M1 + 2M2 )Q − M1 R + M2 P = 0. 1 2 3 1 2
2
1
3
3
1 2 3
Fig. 6.10
708
6 Impulsive Forces
Since C is a hinge, −2Q + 4R Q+R 3(Q − R) = − = u 2 − bω2 = −(u 3 + cω3 ) ~~ ~ ~ M2 M2 M2 =−
R 3R −4R − = , M3 M3 M3
we have Q − 2R 2R = M2 M3 or M Q = (2M2 + 2M3 )R = (2M2 + 2M3 ) ~ 3 ~~ ~
(2M1 + 2M2 )Q + M2 P M1
or Q=
2M2 (M2 + M3 ) P. M3 M1 − 4(M2 + M3 )(M1 + M2 )
It follows that R=
M3 Q. 2(M2 + M3 )
Also u3 =
R 3R Q+R 3(Q − R) P−Q −3(P + Q) , ω3 = , ω2 = , ω1 = , u2 = , u1 = . M3 M3 c M2 M2 b M1 M1 a
From question M1 = M2 = M3 ≡ M, and a = b = c. Hence 1 1 4 P, R = Q = P, −15 4 −15 1 1 1 1 19 −11 P P P P P P u 3 = −15 , ω3 = −5 , u 2 = −3 , ω2 = −5 , u 1 = 15 , ω1 = 5 . M Ma M Ma M Ma Q=
These determine the resulting motion. Next u1 vel. of the centre of mass of AB = = vel. of the centre of mass of C D −u 3 Also
19 15
P M 1 15 P M
= 19. ∎
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
709
1 1 1
3
2 1
2
⁄2
⁄2
Fixed in space
⁄2
1
Fig. 6.11
angular vel. of AB ω1 = = angular vel. of C D ω3
−11 5
P Ma 1 −5 P Ma
= 11. ∎
6.1.9 Example Three equal uniform rods are placed in a straight line are freely jointed and move with a velocity v perpendicular to their lengths. If the middle point of the middle rod , be suddenly fixed, show that the ends of the other two rods will meet in time 4πa 9v where a is the length of each rod (Fig. 6.11). Sol: The symmetry of the problem indicates that ω2 is neither positive nor negative, and hence ω2 = 0. From question, u 2 = 0. Thus, just after G 2 is suddenly fixed, rod BC comes to rest. Hence the perpendicular at B to rod AB is line fixed in space. Hence the angular velocity of rod AB =
∫x=a x=0
(( M a
) ) dx v x
2 M a3
=
( M ) a2 v 2 a 2 M a3
=
3v . 2a
Hence the required duration 2π 3 3v 2a
=
4πa .∎ 9v
6.1.10 Example Two equal uniform rods, AB and AC, are freely jointed at A, and are placed on a smooth table so as to be at right angles. The rod AC is struck by a blow at C in
710
6 Impulsive Forces
a direction perpendicular to itself. Show that the resulting velocities of the middle points of AB and AC are in the ratio 2 : 7 (Fig. 6.12). Sol: For AC: ) ( a2 ω = X (a) − I (a). M(u) − M(0) = I + X, M 3 −→ Here, velocity of A of rod AC is u + aω along AB. For B A: M(v) − M(0) = −X. −→ Here, velocity of A of rod B A is v along AB. Since A is a hinge, 4X − 2I −X I+X 3(X − I ) = + = ~u + aω ~~ = v~ = M , M M M we have X =
2I 5
. We have to show that
v u
= ± 27 . Here
( ) − 2I5 −2 M(v) −X v = = = = .∎ 2I u M(u) I+X 7 I+ 5
Fig. 6.12
1
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
711
6.1.11 Example Two uniform rods, AB and AC, are freely jointed at A, laid on a smooth horizontal table so that the angle B AC is a right angle. The rod AB is struck by a blow P at , B in a direction perpendicular to AB. Show that the initial velocity of A is 4m2P ' +m ' where m, m are the masses of AB, AC respectively. Sol: (Fig. 6.13). For AB: ( 2) a ω = X (a) − P(a). m(u) − m(0) = P + X, m 3 −→ Here, velocity of A of rod AB is u + aω along AC. For C A: m ' (v) − m ' (0) = −X. −→ Here, velocity of A of rod C A is v along AC. Since A is a hinge, 4X − 2P −X P+X 3(X − P) = + = ~u + aω = v~ = ' , ~~ m m m m we have X =
2m ' P . 4m ' +m
So velocity of A
Fig. 6.13
1 ′
712
6 Impulsive Forces
( ' ) 2m P − 4m ' +m −X 2P =v= ' = =− ' .∎ ' m m 4m + m
6.1.12 Example AB, C D Are two equal and similar rods connected by a string BC. AB, BC, C D form three sides of a square. The point A of the rod AB is struck a blow in a direction perpendicular to the rod. Show that the initial velocity of A is 7 times that of D. Sol: Let T be the tension in the string (Fig. 6.14). Let m be the mass of each rod. Since impulse P acts perpendicular to AB, the instantaneous centre of rotation lies on the line AB. It follows that velocity at B of −→ AB is along BC in our figure. For AB: ( 2) a ω1 = P(a) − T (a). m(u 1 ) − m(0) = P + T , m 3 −→ Here, velocity of B of rod AB is u 1 − aω1 along BC. For C D:
2
3
1 1
Fig. 6.14
1
3
3
6.1 Examples on Impulsive Forces in 2D Motion (Part I)
713
) a2 m(u 3 ) − m(0) = T , m ω3 = −T (a). 3 (
−→ Here, velocity of C of rod C D is −(u 3 − aω3 ) along BC. Now since the string BC is tight, we have P+T 3(P − T ) −2P + 4T = − = u 1 − aω1 = −(u 3 − aω3 ) ~~ ~ ~ m m m =− and hence LHS =
T −3T −4T + = , m m m
−2P+4T m
=
−4T m
u 1 + aω1 = u 3 + aω3
or P = 4T . To prove that
P+T m T m
+ +
3(P−T ) m −3T m
=
u 1 +aω1 u 3 +aω3
= ±7.
4P − 2T 2P 2(4T ) =− +1=− +1 −2T T T
= −7 = RHS.∎
6.1.13 Example Three particles of equal mass are attached to the ends, A and C and the middle point B of a light rigid rod ABC, and the system is at rest on a smooth table. The particle C is struck a blow at right angles to the rod. Show that the energy communicated to the system when A is fixed, is to the energy communicated when the system is free as 24 to 25 (Fig. 6.15). Sol: Here the energy communicated to the system (Fig. 6.16) )2 ( ) 1 ( 2) 2 1 ( 5 ( 2) 2 5 ( 2) 2P 2 P(2a) 2 2 = ma ω + m(2a) ω = ma ω = ma = . 2 2 2 2 5m ma 2 + m(2a)2
Here equations are
Fig. 6.15
714
6 Impulsive Forces 0
Fig. 6.16
(m + m + m)v = P, ω0 =
P(a) . ma 2 + ma 2
The energy communicated to the system )2 ) )( 1( 2 1( 2 1 P2 1 P(a) 2 2 2 2 ma + ma (ω0 ) = + ma + ma = (m + m + m)v + 2 2 2 3m 2 ma 2 + ma 2 =
1 P2 5P 2 1 P2 P2 P 2a2 + + = . = 2 2 2 3m 2 ma + ma 6m 4m 12m
Hence the required ratio =
2P 2 5m 5P 2 12m
=
24 .∎ 25
6.1.14 Example A uniform straight rod, of length 2 m and mass 2 kg has at each end a mass 1 kg and at its middle point a mass of 4 kg. One of the 1 kg masses is struck a blow at right angles to the rod and this end starts off with velocity of 5 m/s. Show that the other end of the rod begins to move in the opposite direction with a velocity of 2.5 m/s. Sol: Here a = 1 m, M = 2 kg, and v + aω0 = 5m/s (upwards). We have to show that 5
m m − 2(1 m)ω0 = (v + aω0 ) − 2aω0 = v − aω0 = −2.5 s s~ ~ ~~
rad (Fig. 6.17). that is, ω0 = 15 4 s Here equations are
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
715
0
= 1 kg
= 1 kg
1
= 4 kg
Fig. 6.17
) ( m (8kg) 5 − (1 m)ω0 = (8kg)v = ((m + m 1 + m) + M)v = P , ~~ ~ ~ s P(a) ω0 = ( ) 2 ma 2 + ma 2 + M a3 ~ ~~ ~ = (
P
) = 8 kg a 3
( ) ) ) (8kg) 5 ms − (1 m)ω0 3 ( m 3( m ( ) 5 − (1 m)ω0 = 5 − (1 m)ω0 = 8 a s 1m s 3 kg a
1 = 15 − 3ω0 , s
so ω0 = 15 1s − 3ω0 . Hence ω0 =
15 rad .∎ 4 s
6.2 Examples on Impulsive Forces in 2D Motion (Part II) 6.2.1 Note The problem of the motion of smooth spheres after collision is determined by two facts, namely. 1. The principle of conservation of linear momentum, 2. The relative velocity of spheres along C1 C2 immediately after the impact is (−e)(the relative velocity of spheres along C1 C2 before the impact). Here, e is a positive constant depending on the substance of the spheres, and C1 , C2 are the centres of the spheres. The constant e is called the coefficient of restitution. For hard substances, for example steel, e ≈ 1. For soft substances, e ≈ 0. When a substance is described as “perfectly elastic”, it is understood that e = 1. When a substance is described as inelastic, it is understood that e = 0.
716
6 Impulsive Forces
6.2.2 Note '
'
Suppose that two spheres, of masses m, m , and initial velocities u, u , undergo into ' direct impact. Let v, v be their velocities after impact. Here we suppose that their motion is along the line of centres. By 6.2.1(1), mv + m ' v ' = mu + m ' u ' . Next by 6.2.1(2), ) ( v − v ' = (−e) u − u ' . It follows that v' v = m ' · e(u − u ' ) − (−(mu + m ' u ' ))(−1) (−(mu + m ' u ' ))1 − m · e(u − u ' ) 1 = m(−1) − m ' 1 or ) ( ) ) ( ) ( ( −m ' · e u − u ' + mu + m ' u ' mu + m ' u ' + m · e u − u ' ' v= ,v = . m + m' m + m' Observe that the impulse between the spheres that reduces the velocity of the first sphere from u to vv (
) ( )) ( −m ' · e u − u ' + mu + m ' u ' = mv − mu = m − mu m + m' ) ( ) ( ) ) ( m ( −m ' · e u − u ' + mu + m ' u ' − m + m ' u ' m+m ) ( ( )) m ( −m ' · e u − u ' + m ' u ' − u = ' m+m ( ) mm ' u ' − u = (1 + e). m + m'
=
Thus, the impulse between the spheres that reduces the velocity of the first sphere from u to v ( ) the impulse between the spheres that reduces the velocity of the = (1 + e). first sphere from u to v when e = 0
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
717
6.2.3 Note Let us assume that the direction of motion of the two spheres are not along the line ' of centres. Let us resolve the velocities of each sphere along CC and perpendicular to it. Let u, U be the components of velocity of the “first sphere” before impact. Let ' ' u , U be the components of velocity of the “second sphere” before impact. Let v, V ' ' be the components of velocity of the “first sphere” after impact. Let v , V be the components of velocity of the “second sphere” after impact. Pictorially, in oblique collision, these data are represented as follows: (Fig. 6.18). ' ' If the spheres are smooth, then U = V, U = V : (Fig. 6.19). Let us try to calculate the loss of kinetic energy due to “oblique impact” in the case of smooth spheres. Observe that, in oblique impact, the square of the vel. of the first sphere before impact is u 2 + U 2 , the square of the vel. of the second sphere before impact is ( ' )2 ( ' )2 u + U , the square of the vel. of the first sphere after impact is v 2 + U 2 , the ( ' )2 ( ' )2 square of the vel. of the second sphere after impact is v + U . So the loss of K.E. of the system
Vel. of second sphere after impact
Vel. of first sphere after impact ′
′
′ ′
Vel. of first sphere before impact
Fig. 6.18
Vel. of second sphere before impact
′
′
718
6 Impulsive Forces
′ ′
′
′ ′
′
Fig. 6.19
) ( ) ) 1 (( ) ) 1 (( ) ( )2 ) ( )2 ) 1 ( 2 1 ( 2 2 2 m v + U 2 + m ' v' + U ' m u + U 2 + m' u' + U ' − 2 2 2 2 ) ( ) ( 1 2 1 ' (( ' )2 ) 1 2 1 ' (( ' )2 ) mv + m v mu + m u − = 2 2 2 2 (
=
= (the loss of K.E. of the system in the case of direct impact).
6.2.3.1
Problem
Show that the K.E. of a system of particles is equal to the K.E. of the whole mass moving with the velocity of their centre of mass plus the K.E. of the particles in their motion relative to the centre of mass. Proof: Let M be the total mass of the system of particles. Let x, y, z be the coordinates of the centre of mass G. Here K.E. of the system = 1∑ m = 2 1∑ = m 2
(( ((
) 1∑ ( 2 m (x) ˙ + ( y˙ )2 + (˙z )2 2 )2
) d( ' x +x dt d ( ') ˙ x +x dt
( +
)2
( +
)2
) d( ' y +y dt d ( ') ˙ y +y dt
( +
)2
( +
)2 )
) d( ' z +z dt d ( ') ˙ z +z dt
)2 )
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
1∑ m = 2
+x˙
∑
m
(((
719
) ) ) )) ( ( d ( ') 2 d ( ') 2 d ( ') 2 x y z + + dt dt dt
d ( ' ) ˙ ∑ d ( ' ) ˙ ∑ d ( ' ) 1 ∑ (( ˙ )2 ( ˙ )2 ( ˙ )2 ) x +y y +z z + m x + y + z m m dt dt dt 2
( ( ) ) ) )) d ( ') 2 d ( ') 2 d ( ') 2 x y z + + dt dt dt 1 (∑ )(( ˙ )2 ( ˙ )2 ( ˙ )2 ) ˙ ˙ ˙ +x(0) m + y(0) + z(0) + x + y + z 2
1∑ = m 2
(((
= (K.E. of the particles in their motion relaive to the centre of mass) +(K.E. of the whole mass moving with the velocity of their centre of mass).∎ On using 6.2.3.1, (the K.E. of the system before “direct impact”) ( =
( ) ( ) ) ( ' ' )2 1 mu + m ' u ' 2 1 ' ' mu + m ' u ' 2 1( ' ) mu + m u m u− m m + m u + − + 2 m + m' 2 m + m' 2 m + m'
(
( )2 ( ) ( ) ) 1 ( ' )2 u − u ' 2 1 ' 2 u ' − u 2 1 mu + m ' u ' + mm = + m m 2 m + m' 2 m + m' 2 m + m' ( ) )2 ( ' 2 1 1 mu + m ' u ' ' u−u = mm + , 2 m + m' 2 m + m' hence 2 2 u−u ' mu+m ' u ' (K.E. of the system before “direct impact”) = 21 mm ' (m+m)' + 21 ( m+m ' ) . Similarly, 2 2 v−v ' mv+m ' v ' ) (K.E. of the system after “direct impact”) = 1 mm ' ( ) + 1 ( 2
(
−m ' ·e(u−u ' )+(mu+m ' u ' ) m+m '
m+m '
2
mu+m ' u ' )+m·e(u−u ' ) −( m+m '
m+m '
)2
1 mm ' 2 m + m' ) ( −m ' ·e(u−u ' )+(mu+m ' u ' ) mu+m ' u ' )+m·e(u−u ' ) 2 + m' ( m+m ' m+m ' 1 m + 2 m + m' ( ( )) )2 ( ' 2 1 mu + m ' u ' 1 ' −e u − u + . = mm 2 m + m' 2 m + m'
=
Hence, the loss of K.E. of the system just after “direct impact”
720
( =
6 Impulsive Forces
( ( ( )2 )2 ) ( ( ( ))2 )2 ) u − u' −e u − u ' 1 1 1 mu + m ' u ' 1 mu + m ' u ' ' ' mm mm + + − 2 m + m' 2 m + m' 2 m + m' 2 m + m'
=
)2 ( ) 1 mm ' ( u − u ' 1 − e2 .∎ ' 2m+m
Hence, K.E. of the system relative to c.m. just before “direct impact” ) ( ) ( mu + m ' u ' 2 1 ' ' mu + m ' u ' 2 1 + m u − = m u− 2 m + m' 2 m + m' ( ) ( ) )2 1 mm ' ( 1 ( ' )2 u − u ' 2 1 ' 2 u ' − u 2 m u − u' + m = = m m 2 m + m' 2 m + m' 2 m + m' 1 (loss of K.E. of the system just after direct impact) 1 − e2 1 = (the loss of K.E. of spheres in oblique impact). 1 − e2 =
Conclusion The loss of K.E. of spheres in oblique impact ) ( = 1 − e2 (K.E. of the system relative to c.m. just before “direct impact”.
6.2.4 Note Let us propose the following experimental result: For impacts of bodies, even other than spheres, the relative velocities, before and after impact, of points of the two bodies that come into contact resolved along the common normal at the point of contact are in the ratio 1 : (−e). This is known as the generalized Newton’s rule.
6.2.5 Example A sphere moving with velocity u strikes a smooth plane in a direction making an angle θ with the normal to the plane. It rebounds with velocity v in the direction making an angle φ with the normal to the plane. Show that cotφ = ecotθ (Fig. 6.20). Sol: Since the plane is smooth, we have
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
721
Common Normal +
Fig. 6.20
u sin θ = v sin φ. Now by 6.2.4, −
u sin θ cos φ = (−v cos φ) sin φ
= (vel.of A rel. to B after impact along normal) = (−e)(vel.of A rel. to B before impact along normal) ~ ~~ ~ = (−e)(u cos θ ),
and hence −
u sin θ cos φ = (−e)(u cos θ ) sin φ
or cot φ = e cot θ.∎
6.2.6 Example '
Two equal spheres of mass m are suspended by vertical strings so that they are in contact with their centres at the same level. A third equal sphere of mass m falls vertically and stricks the other two simultaneously so that their centres at the instant of impact form an equilateral triangle in a vertical plane. If u is the velocity of m just before impact, find the velocities just after impact and impulsive tension in the ' strings. Let e be the coefficient of restitution between m , m.
722
6 Impulsive Forces
Sol: Let I be the impulsive reaction between the shaded sphere and the upper sphere. Let T be the impulsive tension in the string. (Figs. 6.21, 6.22, 6.23, 6.24, 6.25, 6.26, 6.27, 6.28, 6.29 and 6.30). ' Here m is suspended by a vertical string. Because of this “constraint”, just after ' ' impact, m will move horizontally, say with velocity v . Suppose that v is the velocity of m after impact. From symmetry of the figure, the direction of v is vertical. Thus, the shaded sphere and sphere of mass m will undergo oblique impact. Hence, by 6.2.4, ) ( (v cos 30◦ ) − v ' cos 60◦ = (−e)(u cos 30◦ − 0) It follows that √
√ 3 3 1 − v ' = −eu 2 2 2
v or
v' −
√
3v =
√ 3eu.(1)
(6.1)
′
′
Fig. 6.21
′
′
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
723
1
2
2
1
Fig. 6.22
0
+( + ) 2 + = 3 3
Fig. 6.23
0
Fig. 6.24
For the upper sphere: √ ◦ ◦ m(v − u) = ~mv − mu = −I cos ~~ 30 − I cos 30~ = − 3I, and hence √ m(v − u) = − 3I.
(6.2)
724
6 Impulsive Forces
Vel. =
cos 30° − cos 60°
′
′
60°
Fig. 6.25
′ ′ ′
Direction of common normal
Point of contact
Fig. 6.26
For the shaded sphere: I m ' v ' = ~m ' v ' − 0 ~~ = I sin 30~◦ = , T = I cos 30◦ 2 and hence 2m ' v ' = I.
(6.3)
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
725
′ ′ ′
Direction of common normal
Point of contact
Fig. 6.27
√ T =I
3 . 2
(6.4)
Here we have four unknowns, namely I, T , v.v ' , and four equations. √ √ 3v = 3eu.
(6.5)
√ m(v − u) = − 3I.
(6.6)
v' −
From (3) and (4), 2m '
(√ √ ) m(v − u) 3eu + 3v = 2m ' v ' = √ , − 3 ~~ ~ ~
so 6m ' (eu + v) = m(u − v). It follows that v= Now from (1),
m − 6em ' u. 6m ' + m
726
6 Impulsive Forces
vel. =
Common normal
≡ ( ) ( = 0,
=
cos )
Line fixed in space Direction fixed in space Fig. 6.28
'
√
√ ) ) ) 3 ( ( ' m − 6em ' 3eu + 3 e 6m + m + m − 6em ' u u = 6m ' + m 6m ' + m ~~ ~ √
(
v = ~ √ 3(e + 1)m = u. 6m ' + m
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
727
(vel. rel. to ) = (2 )
2
,2
,2
2 2
vel. = (2 )
,2
1
,2
1
1
Common normal Fig. 6.29
(
Next from (2), m(v − u) = I = √ − 3
m
m−6em ' u 6m ' +m
√ − 3
−u
)
( ) √ mu −6m ' (1 + e) 2 3m ' m(1 + e) = = u. √ 6m ' + m − 3(6m ' + m)
Now, from (4), √ √ 3 2 3m ' m(1 + e) 3m ' m(1 + e) T = · u = u. 2 6m ' + m (6m ' + m)
728
6 Impulsive Forces
,2
,2
1 2
1
1 1
,2
1
,2
1
1
2
2
Common normal Fig. 6.30
6.2.7 Example A body of mass m 1 + m 2 is split into two parts of masses m 1 and m 2 by an internal explosion which generates kinetic energy E. Show that if after explosion the parts move in the same line as before, their relative speed is
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
/
729
2E(m 1 + m 2 ) . m1m2
Sol: Here equations are ( (m 1 + m 2 )u = m 1 v1 + m 2 v2 , E =
) 1 1 1 2 2 m 1 (v1 ) + m 2 (v2 ) − (m 1 + m 2 )u 2 . 2 2 2
We have to show that (v1 − v2 )2 =
2E(m 1 + m 2 ) m1m2
that is, ) ) (( m 1 m 2 (v1 − v2 )2 = 2E(m 1 + m 2 ) = (m 1 + m 2 ) m 1 (v1 )2 + m 2 (v2 )2 − (m 1 + m 2 )u 2 ~~ ~ ~
( ) = (m 1 + m 2 )m 1 (v1 )2 + (m 1 + m 2 )m 2 (v2 )2 − ((m 1 + m 2 )u)2 ) ( = (m 1 + m 2 )m 1 (v1 )2 + (m 1 + m 2 )m 2 (v2 )2 − (m 1 v1 + m 2 v2 )2 ) ) ( ( = (m 1 + m 2 )m 1 − (m 1 )2 (v1 )2 + (m 1 + m 2 )m 2 − (m 2 )2 (v2 )2 − 2m 1 m 2 v1 v2 = m 1 m 2 (v1 )2 + m 1 m 2 (v2 )2 − 2m 1 m 2 v1 v2 = m 1 m 2 (v1 − v2 )2 .∎
6.2.8 Example A light rigid rod ABC has three particles each of mass m attached to it at A, B, C. The rod is struck by a blow P at right angles to it at a point distant from A equal to BC. Prove that the kinetic energy of the set up is 1 P 2 a 2 − ab + b2 , 2 m a 2 + ab + b2 where AB = a, BC = b. Sol: Here equations are
730
6 Impulsive Forces
(m + m + m)v = P, ω0 =
m
( 2a+b )2 3
P
( +m a−
( 2a+b 3 ) 2a+b 2 3
−b
)
( + m (a + b) −
) 2a+b 2 3
.
The kinetic energy of set up ⎛ ( ) ) ⎞ ( 2a + b 2 2a + b 2 +m a− m ⎟ 1 1⎜ 3 3 ⎟ ⎜ = (m + m + m)v 2 + ⎜ ⎟(ω0 )2 )2 ( ⎠ 2 2⎝ 2a + b + m (a + b) − 3 ( )2 P 2 2(a−b) 3 1 1 P2 + = ) ) ) ( ( ( 2 3m 2 m 2a+b 2 + m a − 2a+b 2 + m (a + b) − 2a+b 2 3 3 3 1 P2 1 P2 2P 2 (a − b)2 2P 2 (a − b)2 ) ( ) = + ( + 2 3m 2 3m 6m a 2 + ab + b2 m (2a + b)2 + (a − b)2 + (a + 2b)2 ) ( 2(a − b)2 P2 1+ 2 = 6m a + ab + b2
=
=
1 P 2 a 2 − ab + b2 .∎ 2 m a 2 + ab + b2
Another Method: The motion of the rod is completely determined by its angular velocity ω0 and the linear velocity u of any point. For simplicity, let us choose that point C. The velocity of B is u + bω0 (upwards) and the velocity of A is u + (a + b)ω0 (upwards). Here the two unknowns are u, ω0 . Let us resolve at right angles to rod, and equate the total momentum to the impulse: m(3u + (a + 2b)ω0 ) = m(u + (a + b)ω0 ) + m(u + bω0 ) + mu = P ~ ~~ ~ For obtaining another equation, let us take moments of momentums about a point. For simplicity, let us choose that point O. We get ( ) ( ( ) ) 2m (b − a)u + b2 ω0 = m 2u(b − a) + 2b2 ω0 = m(u(b − a + b − a) + ((a + b)b − b(a − b))ω0 ) = (m(u + (a + b)ω0 ))(b) + (−(m(u + bω0 ))(a − b)) + (−(mu)(a)) = (moment of impulsePaboutO) = 0. ~ ~~ ~
Thus } 3u + (a + 2b)ω0 − mP = 0 . (b − a)u + b2 ω0 = 0
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
731
It follows that u=
P 2 b m
3b2 − (a + 2b)(b − a) P (a − b) . = 2m 2 a + b + ab
=
P 2 b m a 2 + b2 +
ab
, ω0 =
− mP (b − a) 3b2 − (a + 2b)(b − a)
Hence the K.E. created ) ( 1 1 1 2 2 2 m(u + (a + b)ω0 ) + m(u + bω0 ) + m(u) = 2 2 2 ( )2 P 2 P b − b) 1 m m (a + (a + b) 2 = m 2 2 a + b2 + ab a + b2 + ab ( ( )2 )2 P 2 P P 2 b b − b) 1 1 m m (a m +b 2 + m 2 + m 2 2 a + b2 + ab a + b2 + ab 2 a + b2 + ab =
P 2a4 P 2 a 2 b2 P 2 b4 1 1 1 ( )2 + ( )2 + ( ) 2 m a 2 + b2 + ab 2 m a 2 + b2 + ab 2 m a 2 + b2 + ab 2 ( ) ( ) 1 P 2 a 2 + b2 − ab 1 P 2 a 4 + a 2 b2 + b4 = ( ) = ( ) 2 m a 2 + b2 + ab 2 2 m a 2 + b2 + ab 2 ( )( ) 1 P 2 a 2 + b2 + ab a 2 + b2 − ab ( ) = .∎ 2 m a 2 + b2 + ab
6.2.9 Example Three equal particles A, B, C of mass m are placed on a smooth horizontal plane. A ◦ is joined to B and C by light threads AB, AC and the angle B AC is 60 . An impulse −→ I is applied at A in the direction B A. Find the initial velocities of√ the particles and −→ show that A begins to move in a direction making an angle tan−1 73 with B A. Sol: For A: Equations of impulsive motion are mu − m(0) = I − T − T ' cos 60◦ , mv − m(0) = T ' sin 60◦ For B: Equation of impulsive motion is mu − m(0) = T .
732
6 Impulsive Forces
For C: Equation of impulsive motion is ( √
1 3 −u m v 2 2
)
2 = m(v cos 30◦ − u cos 60◦ ) − m(0) = −T ' = − √ mv, ~~ ~ ~ 3
so ( √ ) 1 3 −2 −u m v = √ mv 2 2 3 or (√ v
2 3 +√ 2 3
) =
u 2
or √ 3u . v= 7 Since (
'1
'1
) 1 2 √ mv 2 3
mu = I − T − T = I − mu − T = I − mu − 2~ 2 ~~ ~ (√ ) 3u 1 1 8mu , = I − mu − √ mv = I − mu − √ m =I− 7 7 3 3 we have mu = I −
8mu , 7
√ 3u Now since v = and hence u = = 7 ~ ~~ ~ Here the required angle 7 I . 15 m
= tan
−1
v = tan−1 u
√
7 3( 15 7
I m
) =
√
√
3 I 15 m 7 I 15 m
√ 3 I , 15 m
= tan
−1
3 .∎ 7
we have v =
√
3 I . 15 m
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
733
6.2.10 Example A uniform sphere, rotating with an angular velocity ω clockwise about an axis perpendicular to the plane of motion of its centre, impinges on a horizontal plane. Find the resulting change in its motion. Sol: Here equations of impulsive motion are ) ) ( 2a 2 2a 2 ' ω − M ω = F(a) Mv − Mv = −R, Mu − Mu = −F, M 5 5 '
(
'
that is, ) ) ( ) ( ) ( 2a ( ' ω − ω = F = M u − u' , M v − v ' = R, M u − u ' = F, M ~~ ~ ~ 5 and hence ) ( ) ( 2a ω' − ω = 5 u − u ' . Here A is the point of contact, and the plane is rough enough to prevent sliding, just after impact, the velocity of A is zero, and hence u ' − aω' = 0. Let e be the coefficient of restitution between the sphere and the plane. By Newton’s law, ( ' ) v − 0 = (−e)(v − 0), or '
v = −ev. '
'
'
'
We assume that u, v, ω are known. We want to determine u , v , ω . Since v = ' −ev, and v is known, v becomes known. Since ) ( ) 7 ' 5u 5( ' ' ' = ~u ' − aω u − aω − = u − aω + u − u ~~ = 0~, 2 2 2 we have 7 ' 5u u − aω − = 0, 2 2 and hence
734
6 Impulsive Forces
u' =
2aω + 5u . 7
Next since ( ) ) ( 2u − 2aω 2aω + 5u M =M u− = M u − u' = F , 7 7 ~ ~~ ~ we have F=
2M (u − aω). 7
Observe that vertical velocity after impact is ev (upward). Next since ) ( Mv(1 + e) = M(v − (−ev)) = M v − v ' = R , ~ ~~ ~ the normal impulsive reaction is Mv(1 + e). Case I when u = aω. Now since ) ( 2M M u − u' = F = (u − aω) = 0 7 ~~ ~ ~ '
There is no friction, and u = u. Since )) ( ( μ M v − v ' = μR = F = 0 ~ ~~ ~ '
'
we have v = v. Since u = u, we have ) ( ) ( 2a ω' − ω = 5 u − u ' = 0, ~~ ~ ~ '
and hence ω = ω. In short, u, v, ω are unaltered after impact. Case II when aω < u. Now since ) ( 2M M u − u' = F = (u − aω) > 0 7 ~~ ~ ~ '
friction F acts along ←, and u < u. Since friction F acts along ←, the point of contact A moves along → on the plane surface. Since )) ( ( > 0~, μ M v − v ' = μR = ~F ~~ '
'
we have v < v. Since u < u, we have
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
735
) ( ) ( 2a ω' − ω = 5 u − u ' > 0, ~ ~~ ~ '
and hence ω > ω. In short, u, v, are decreased, and ω is increased after impact. Case III when u < aω. Now since ) ( 2M M u − u' = F = (u − aω) < 0 7 ~~ ~ ~ '
friction F acts along →, and u < u . Since friction F acts along →, the point of contact A moves along ← on the plane surface. Since )) ( ( < 0~, μ M v − v ' = μR = ~F ~~ '
'
we have v < v . Since u < u , we have ) ( ) ( 2a ω' − ω = 5 u − u ' < 0, ~~ ~ ~ '
and hence ω < ω. In short, u, v, are increased, and ω is decreased after impact.
6.2.11 Example A uniform sphere, rotating with an angular velocity ω anticlockwise about an axis perpendicular to the plane of motion of its centre, impinges on a horizontal plane. Find the resulting change in its motion. Sol: Here equations of impulsive motion are ) ) ( 2a 2 2a 2 ' ω − M ω = −F(a) Mv − Mv = −R, Mu − Mu = −F, M 5 5 '
'
(
that is, ) ( ) ) ) ( ( 2a ( ' ω − ω = −F = −M u − u ' , M v − v ' = R, M u − u ' = F, M ~ 5 ~~ ~ and hence ( ) ( ) 2a ω' − ω = −5 u − u ' .
736
6 Impulsive Forces
Here A is the point of contact, and plane is rough enough to prevent sliding, just after impact, velocity of A is zero, and hence u ' + aω' = 0. Let e be the coefficient of restitution between the sphere and the plane. By Newton’s law, ( ' ) v − 0 = (−e)(v − 0), or v ' = −ev. We assume that u, v, ω are known. We want to determine u ' , v ' , ω' . Since v ' = −ev, and v is known, v ' becomes known. Since ) ( ) 7 ' 5u 5( ' u + aω − = u ' + aω − u − u ' = ~u ' + aω ~~ = 0~, 2 2 2 we have 7 ' 5u u + aω − = 0, 2 2 and hence u' =
−2aω + 5u . 7
Next since M
( ) ( ) −2aω + 5u 2u + 2aω =M u− = M u − u' = F , 7 7 ~ ~~ ~
we have F=
2M (u + aω). 7
Observe that vertical velocity after impact is ev (upward). Next since ) ( Mv(1 + e) = M(v − (−ev)) = M v − v ' = R , ~ ~~ ~ the normal impulsive reaction is Mv(1 + e).
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
737
6.2.12 Example A rod, of length 2a, is held in a position inclined at an angle α to the vertical, and is then let fall on to a smooth inelastic horizontal plane. Show that the end which hits the plane will leave it immediately after the impact if the height through which the rod falls is greater than ( )2 1 a sec α csc2 α 1 + 3 sin2 α . 18 Sol: Suppose that u, ω are the vertical velocity and angular velocity just after the impact. Suppose that V is the vertical velocity just before impact. Let R be the impulse of the reaction of the plane just after the impact. Hence ) ( 2) ( 2) a a a2 ω= m ω− m 0 = R(a sin α) . mu − mV = −R, m 3 3 3 ~ ~~ ~ (
Here A is the point of contact, and impact is on a smooth inelastic horizontal plane, so vertical velocity of A after impact is zero, and hence ( V−
) ( a ) 1 + sin α aω = V − ω − (aω) sin α 3 sin α 3 sin α (
=V−
) 2 m a3 ω
a sin α
− (aω) sin α = V −
R − (aω) sin α m
m mV − R − (aω) sin α = u − (aω) sin α = 0 . = ~ ~~ ~ m
Thus 3V sin α V ). )= ( ω= ( 1 a 1 + 3 sin2 α a 3 sin α + sin α Since the impact is inelastic, the end of the rod remains in contact with the plane. Here equation of motion of G is ( ( )2 ) d 2 (a cos θ ) =m −ma sin θ θ¨ + cos θ θ˙ = −mg + S , dt 2 ~~ ~ ~ and hence ( ( )2 ) = mg − S. ma sin θ θ¨ + cos θ θ˙
738
6 Impulsive Forces
Also ( 2) ( ( ( )2 )) a θ¨ = S(a sin θ ) = mg − ma sin θ θ¨ + cos θ θ˙ m (a sin θ ). 3 ~~ ~ ~ Hence ( 2) ( ( ( )2 )) a m θ¨ = mg − ma sin θ θ¨ + cos θ θ˙ (a sin θ ) 3 or ( ( )2 ) a θ¨ = 3g sin θ − 3a sin2 θ θ¨ + cos θ sin θ θ˙ or (
1 + 3 sin2 θ
( ) ) S(a sin θ ) 2 m a3
( )2 + 3 cos θ sin θ θ˙
( ) ( )2 3g sin θ = 1 + 3 sin2 θ θ¨ + 3 cos θ sin θ θ˙ = . a ~ ~~ ~ It follows that ( ) ( ) S(a sin θ ) ( )2 3g sin θ 2 .(∗) 1 + 3 sin θ + 3 cos θ sin θ θ˙ = a2 a m3 Suppose that the end which hits the plane will leave it immediately after the impact, when height is H. It follows that. Sθ =α = 0. Put θ = α in (∗). We get ( 2 ) cos α sin3 α cos α sin3 α = 27 0 + 2g H ( ( ) )2 2 a 2 1 + 3 sin2 α a 2 1 + 3 sin2 α ( )2 3V sin α cos α sin3 α 2 ) = 3 cos α sin α(ω)2 > 27V ( )2 = 3 cos α sin α ( a 1 + 3 sin2 α a 2 1 + 3 sin2 α 54g H
(  )2 = 3 cos α sin α θ˙ θ =α ( ) ( ) (0)(a sin α) (  )2 3g sin α 2 = 1 + 3 sin α , + 3 cos α sin α θ˙ θ =α = a2 a m3 ~~ ~ ~ and hence
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
54g H
739
cos α sin3 α 3g sin α . ( )2 > 2 2 a a 1 + 3 sin α
It follows that H>
( )2 1 a sec α csc2 α 1 + 3 sin2 α .∎ 18
6.2.13 Example Four equal rods, each of mass m and length 2a, are freely jointed at their ends so as to form a rhombus. The rhombus falls with a diagonal vertical, and is moving with velocity V when it hits a fixed horizontal inelastic plane. Find the motion of the rods immediately after the impact, and show that their angular velocities are each equal to V sin α 3 ( ), 2 a 1 + 3 sin2 α where α is the angle each rod makes with the vertical. 1 It is also shown that the impact destroys a fraction 1+3sin of the kinetic energy 2 α just before the impact. Sol: Here, from symmetry, 0 = (horizontal velocity of C) ~ ~~ ~ = (horizontal velocity of C relative to B) + (horizontal velocity of B) = (horizontal velocity of C relative to B) + ((2a)ω1 ) cos α = ((2a)ω2 ) cos α + ((2a)ω1 ) cos α, Hence 0 = ((2a)ω2 ) cos α + ((2a)ω1 ) cos α. It follows that ω2 = −ω1 .
740
6 Impulsive Forces
Next (horizontal velocity of G 2 ) = (horizontal velocity of G 2 relative toB) + (horizontal velocity ofB) = (horizontal velocity of G 2 relative to B) + ((2a)ω1 ) cos α = (aω2 ) cos α + ((2a)ω1 ) cos α
= (a(−ω1 )) cos α + ((2a)ω1 ) cos α = aω1 cos α(→), and (vertical velocity ofG 2 ) = (vertical velocity ofG 2 relative toB) + (vertical velocity ofB) = (vertical velocity ofG 2 relative toB) + ((2a)ω1 ) sin α = −(aω2 ) sin α + ((2a)ω1 ) sin α
= −(a(−ω1 )) sin α + ((2a)ω1 ) sin α = 3aω1 sin α(↓). From symmetry, Y = 0. (Figs. 6.31, 6.32, 6.33, 6.34, 6.35, 6.36, 6.37, 6.38, 6.39 and 6.40). On taking moments about B for the rod BC, we get Fig. 6.31
,2
1
vel. =
2
vel. = 3
1
sin 1
1
1
cos
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
741
,2
1
vel. =
2
vel. = 3
1
1
cos
sin 1
1
1 1
,2
1
1
2
2
vel. = Common normal Fig. 6.32
( ma 2 ω1
( ) ) ) 2 1 ( − 4 sin2 α + mV a sin α = ma 2 ω1 − + −3 sin2 α + cos2 α 3 3
+ mV a sin α = −m
( ) a2 ω1 + ma 2 ω1 −3 sin2 α + cos2 α + mV a sin α 3
) ( ( 2) a = − m ω1 + (−a sin α, −a cos α, 0) × (m(aω1 cos α, 3aω1 sin α, 0)) + mV a sin α 3
1
cos
742
6 Impulsive Forces
′
′
′
′ ′
Fig. 6.33
2 3
′
′
Fig. 6.34 ) ( ( 2) a −−→ ω1 + BG 2 × (m(velocity ofG 2 after impact)) − (−(mV )(a sin α)) = ((2a) cos α)X , = − m 3 ~ ~~ ~
and hence ( ma 2 ω1
) 2 − 4 sin2 α + mV a sin α = (2a) cos α X. 3
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
743
, ,
′ ′
Fig. 6.35 Fig. 6.36
,2
√2 ′ ′
or ( ω1 Also,
) 2X V 2 − 4 sin2 α = cos α − sin α. 3 ma a
744
6 Impulsive Forces
,
′ ′ ′
Direction of common normal
Point of contact
Fig. 6.37
Vel.=
Vel.= (1 + ) Horizontal range Fig. 6.38
,
′
Fixed rough point
Fig. 6.39
′
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
745
,
ℎ
Fig. 6.40
X 1 + X = m(aω1 cos α), −Y1 = m(3aω1 sin α) − mV . Observe that (change in A.M. of rod AB about A) + (change in A.M. of rod BC about A) = X ((2a) cos α + (2a) cos α) = 4a X cos α, ~~ ~ ~ ) ( ) ) ξ =2a (( m a2 2 + ma ω1 − ∫ dξ V (ξ sin α) (change in A.M. of rod AB about A) = m 3 2a ξ =0 ~ ~~ ~ 4a 2 mV sin α ( 2 ) 4a 2 =m ω − 2a = m ω − mV a sin α, 3 1 2a 3 1
and (change in A.M. of rodBCabout A) = (A.M. of rodBCabout A) ) ) ξ =2a (( m dξ V (ξ sin α) − ∫ 2a ξ =0 = (A.M. of rodBCabout A) − mV a sin α = (A.M. of rodBCaboutB) −→ + AB × m(vel.ofG 2 ) − mV a sin α = (A.M. of rodBCaboutB) + (2a)(sin α, − cos α) × m(aω1 cos α, 3aω1 sin α, 0) − mV a sin α
746
6 Impulsive Forces
( ) = (A.M. of rodBCaboutB) + 2a 2 mω1 3 sin2 α + cos2 α − mV a sin α ) ( ( 2) a −−→ ω1 + BG 2 × (m(velocity of G 2 after impact)) = − m 3 ( ) 2 + 2a mω1 3 sin2 α + cos2 α − mV a sin α ) ( ( 2) a ω1 + (−a sin α, −a cos α, 0) × (m(aω1 cos α, 3aω1 sin α, 0)) = − m 3 ( ) +2a 2 mω1 3 sin2 α + cos2 α − mV a sin α = −m
( ( ) ) a2 ω1 + ma 2 ω1 −3 sin2 α + cos2 α + 2a 2 mω1 3 sin2 α + cos2 α − mV a sin α 3
= −m
( ) a2 8 ω1 + ma 2 ω1 3 sin2 α + 3 cos2 α − mV a sin α = ma 2 ω1 − mV a sin α, 3 3
so ( ) ( ) 4a 2 8 2 ω1 − mV a sin α + ma ω1 − mV a sin α 4ma ω1 − 2mV a sin α = m 3 3 ~ ~~ ~ 2
= 4a X cos α. Hence 2maω1 − mV sin α = 2X cos α or 2X ma
cos α − 2 3
V a 2
sin α
− 4 sin α
X V sin α + cos α . = ω1 = 2a ma ~ ~~ ~
It follows that V 2X cos α − sin α = ma a
(
( ) ) 2 2 V X 2 2 − 4 sin α sin α + − 4 sin α cos α 3 2a 3 ma
or X
( ( ( ) )) ) 1 4 cos α 1 2 cos α 4 cos α ( 1 + 3 sin2 α = X + sin2 α = X 1− − 2 sin2 α 3ma ma 3 ma 3
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
747
( ) ( ( ) ) 2 V 2 2 cos α V cos α − − 4 sin2 α = − 4 sin2 α sin α + sin α =X ma 3 ma 3 2a a ~~ ~ ~ ) ( ) ( 4 V 2V = sin α − 2 sin2 α = sin α 2 − 3 sin2 α . a 3 3a Thus (
X=
2V sin α 2 − 3 sin2 α 3a ) ( 4 cos α 1 + 3 sin2 α 3ma
) =
2 − 3 sin2 α 3 cos2 α − 1 mV mV tan α tan α = . 2 2 1 + 3 sin2 α 1 + 3 sin2 α
Now since X V V sin α + cos α = sin α + ω1 = 2a ma 2a ~ ~~ ~ =
mV 2
cos α−1 tan α 31+3 sin2 α 2
ma
cos α
V 3 cos2 α − 1 sin α 3V V sin α + sin α = .∎ 2 2a 2a 2a 1 + 3 sin α 1 + 3 sin2 α
where α is the angle each rod makes with the vertical. Next since 3 cos2 α − 1 mV tan α = X 1 + X = m(aω1 cos α) X1 + ~ ~~ ~ 2 1 + 3 sin2 α ) ) ( ( sin α 3mV sin α cos α 3V cos α = =m a 2a 1 + 3 sin2 α 2 1 + 3 sin2 α we have 3 cos2 α − 1 mV 3mV sin α cos α tan α − 2 2 1 + 3 sin α 2 1 + 3 sin2 α ) ( mV tan α 3 cos2 α − 1 mV sin α ) ). = ( 3 cos α − = ( 2 cos α 2 1 + 3 sin α 2 1 + 3 sin2 α
X1 =
Again since ) ) ( ( sin α 3V sin α Y1 = −m(3aω1 sin α) + mV = −m 3a ~~ ~ ~ 2a 1 + 3 sin2 α + mV =
mV −9 sin2 α + mV 2 1 + 3 sin2 α =
mV 3 cos2 α − 1 mV 2 − 3 sin2 α = . 2 1 + 3 sin2 α 2 1 + 3 sin2 α
748
6 Impulsive Forces
Thus, the final K.E. ( 2 ( 2 ) ) a a 1 1 2 2 2 m + ma (ω1 ) + m + ma (ω1 )2 = 2 3 2 3 ~~ ~ ~ ~~ ~ ~ ⎛
K.E.of AB
K.E.of AD
⎞
⎜ 1 ( a2 ) )⎟ 1 ( ⎜ ⎟ +⎜ m (ω1 )2 + m (aω1 cos α)2 + (3aω1 sin α)2 ⎟ ⎝2 ⎠ 3 2 ~ ~~ ~ K.E.ofBC
⎛
⎞
⎜ 1 ( a2 ) )⎟ 1 ( ⎜ ⎟ +⎜ m (ω1 )2 + m (aω1 cos α)2 + (3aω1 sin α)2 ⎟ ⎝2 ⎠ 3 2 ~ ~~ ~ K.E.ofC D
) 4 2 a2 ma (ω1 )2 + m (ω1 )2 + (aω1 cos α)2 + (3aω1 sin α)2 3 3 ( ( ) ) 2 2 2 2 8 2 2 2 5 + cos α + 9 sin α = ma (ω1 ) + 8 sin α = ma (ω1 ) 3 3 ( ( )2 )2 ( ) 8 2 3V 3V sin α 1 8 2 ma sin α 1 + 3 sin α = = ma 2 3 2a 1 + 3 sin2 α 3 2a 1 + 3 sin2 α sin2 α = 6mV 2 1 + 3 sin2 α )( ) ( 3 sin2 α 1 1 1 1 mV 2 + mV 2 + mV 2 + mV 2 = 2 2 2 2 1 + 3 sin2 α 3 sin2 α .∎ = (the initial K.E.) 1 + 3 sin2 α (
=
Remark: Since we are considering only the change in the motion produced by the blow, the finite external forces (in this case, weights of the rods) do not come into the equations. The reason is that blow occurs for a very short duration, and finite forces produce negligible effect during the very short duration.
6.2.14 Example A body, whose mass is m, is acted upon at a given point P by a blow of impulse X. ' If V and V be the velocities of P in the direction of X just before and just after the
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
749
action of X, show( that the)change in K.E. of the body, that is, the work done on it by ' the impulse, is 21 V + V X. Sol: Let k be the radius of gyration of the body about an axis through the centre of mass G and perpendicular to the x yplane. Here equations are ) ) ( ( mu ' − mu = −X, mv ' − mv = 0, mk 2 ω' − mk 2 ω = −y ' X. It follows that u' − u =
)( ( ) −X ' , v = v, mk 2 ω' − ω = −y ' X. m
Also from question, V = (velocity ofPin the direction of X just before the action of X ) ~ ~~ ~ = (velocity of P relative to G in the direction of X just before the action of X ) +(velocity ofGin the direction of X just before the action of X ) = (velocity ofPrelative toGin the direction of X just before the action of X ) + (−u)
( ( −→)) = componentin the direction of X of ω(0, 0, −1) × G P − u ( ( ( ))) = componentin the direction of X of ω(0, 0, −1) × x ' , y ' , 0 − u ) ( ( ( ))) ( = componentin the direction of X of ω y ' , · · · , · · · − u = −ωy ' − u, and V ' = (velocity ofPin the direction of X just after the action of X ) ~ ~~ ~ = (velocity ofPrelative toGin the direction of X just after the action of X ) +(velocity ofGin the direction of X just after the action of X ) ) ( = (velocity ofPrelative toGin the direction of X just after the action of X ) + −u ' ( ( −→)) = componentin the direction of X of ω' (0, 0, −1) × G P − u '
750
6 Impulsive Forces
( ( ))) ( = componentin the direction of X of ω' (0, 0, −1) × x ' , y ' , 0 − u ' ( ( ( ))) ( ) = componentin the direction of X of ω' y ' , · · · , · · · − u ' = −ω' y ' − u ' . Thus u ' + y ' ω' = −V ' , u + y ' ω = −V . The change in K.E. of the body ) (( ) ) ) ) 1 ( 2 1 ( 2 )( ' )2 1 (( ' )2 ( ' )2 ) 1 ( 2) 2 2 m u + v m u +v − + mk ω = + mk ω 2 2 2 2 ) (( ) ) ) (( ) )( )2 ) ) 1 ( 2 1( 1( 1 (( ' )2 − m u + (v)2 + mk 2 ω' m u + v 2 + mk 2 ω2 = 2 2 2 2 ) ( )( ' ) 1 ( 2 )( ' )( ' ) 1 ( ' ) −X 1 ( ' = m u + u u − u + mk ω + ω ω − ω = m u + u 2 2 2 m (( )( ))( ) 1 mk 2 ω' − ω ω' + ω + 2 ((
) 1( )( ) −X (( ' )) ) ( −1 ( ' X u + u + −y ' X ω' + ω = u + y ' ω' + u + y ' ω 2 2 2 ) ( )) −X (( ' u + y ' ω' + u + y ' ω = 2
=
=
) ( ) ) 1( ) )) −X (( −X (( −V ' + u + y ' ω = −V ' + (−V ) = V + V ' X.∎ 2 2 2
6.2.15 Example A horizontal inelastic rod falls without rotation, being inclined at any angle to the horizon, and hits a smooth peg at a distance from its upper end equal to onethird of its length. Show that the lower end begins to descend vertically. Here, equations of impulsive motion are ) ) ( a2 2a , mu − m(u cos α) = 0, mv − m(u sin α) = −X, m (ω − 0) = X a − 3 3 '
that is,
'
(
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
u ' = u cos α, v ' = u sin α −
751
X X ,ω = . m ma
The velocity of lower end ( ) = v ' (cos α, − sin α, 0) + u ' (− sin α, − cos α, 0) + ω(0, 0, 1) × a(− sin α, − cos α, 0)
( ) = v ' cos α − u ' sin α + ωa cos α, −v ' sin α − u ' cos α − ωa sin α, 0 ) ( ) ( ) (( X X X cos α − (u cos α) sin α + a cos α, − u sin α − sin α u sin α − m ma m ( ) X − (u cos α) cos α − a sin α, 0 ma =
= (0, −u, 0) = u(0, −1, 0). ∎
6.2.16 Example A light string is wound round the circumference of a uniform reel, of mass m, radius a, and radius of gyration k about its axis. The free end of the string being tied to a fixed point, the reel is lifted up and let fall so that, at the moment when the string becomes tight, the velocity of the centre of the reel is vertical. Find the change in the motion and show that the impulsive tension is mu ·
k2 . k2 + a2
Sol: Here equations of impulsive motion are ( ) ( ) mu ' − mu = −T , mv ' − m0 = 0, mk 2 ω − mk 2 0 = T (a), u ' = aω, and hence T T a a aω = u ' = u − , v ' = 0, ω = = (u − aω) 2 . 2 m m k k ~ ~ ~~ ~ ~~ ~ Since ω = (u − aω) ka2 , we have au u' =ω= 2 , 2 a a ~~ + k ~ ~
752
6 Impulsive Forces
and hence u ' = Next since
a2 u . a 2 +k 2
a2 we have T = mu ·
au T a =ω= 2 2 +k ~ ~~m k ~
k2 .∎ k 2 +a 2
6.2.17 Example A square plate, of side 2a, is falling with velocity u, a diagonal being vertical, when an inelastic string attached to the middle point of an upper edge becomes tight in a vertical position. Show that the impulsive tension of the string is 47 Mu, where M is the mass of the plate. Verify the result of 6.2.14. Sol: Here equations of impulsive motion are ( Mu ' − Mu = −X, Mv ' − M0 = 0, M
) ( ) ) ( a2 + a2 a2 + a2 a ω− M 0=X √ . 3 3 2
Thus Xa
√ 3X X 2 u = u − , v ' = 0, ω = . √ 2 = 2a M 2 2Ma M 3 ~ ~~ ~ '
Here, velocity ofP ( ) ) (1, −1, 0) −→ ( = v ' , u ' , 0 + ω(0, 0, −1) × G P = v ' , u ' , 0 + ω(0, 0, −1) × a √ 2 ) (
) ωa ( ) ωa ( ωa ωa = v ' , u ' , 0 + √ (−1, −1, 0) = 0, u ' , 0 + √ (−1, −1, 0) = − √ , u ' − √ , 0 . 2 2 2 2
From question, ωa u ' − √ = (vertical component of velocity ofP) = 0, ~~ ~ 2 ~ so
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
( ωa X = u' = √ = u− M 2 ~ ~~ ~
√3X 2 2Ma
√
2
753
) a
=
3X , 4M
3X X and hence u − M = 4M . Thus X = 4Mu .∎ ( ) 3u 7 3 4Mu ' = It follows that u = 4M . Next 7 7
( ) √ 3 4Mu 3X 3 2u 7 . ω= √ = √ = 7a 2 2Ma 2 2Ma ~~ ~ ~ By 6.2.14, the change in K.E. ))) ) ( ( ( ( 1 ωa 1 ωa ' ' X= = √ −u−u X (−u) + − u − √ 2 2 2 2 It suffices to show that ( ) 1 ( 2a 2 ) ) 1 1 (( ' )2 2 M u +0 + M ω2 − Mu 2 2 2 3 2 ) ) ( ) ( ( 1 2a 2 1 (( ' )2 ( ' )2 ) 1 1 ωa ' 2 2 M u + v M ω − Mu = + = √ −u−u X 2 2 3 2 2 2 ~ ~~ ~ ( ) 4Mu 1 ωa = √ − u − u' 2 7 2 that is ⎛ ( √ )2 ⎞ ( )2 ) ( 2 3u −28 2 12 3 2u ⎠ 9 −4u 2 2a = u = u2 + −1 =⎝ + − u2 7 49 49 49 7 3 7a ) ) ( ( ( ' )2 2a 2 2 ωa 2 ' 4u ω −u = √ −u−u = u + 3 7 2 ~ ~~ ~ ⎛( √ ) ⎞ 3 2u ( ) ( ) a 7a 3 3 4u 2 −4u 2 3u ⎠ 4u = −1− = .Verified =⎝ √ −u− 7 7 7 7 7 7 2
754
6 Impulsive Forces
6.2.18 Example If a hollow lawn tennis ball of elasticity e has on striking the ground, supposed perfectly rough, a vertical velocity u and an angular velocity ω about a horizontal axis, find its angular velocity after impact and prove that the range of the rebound will be 4 aω eu. 5 g Sol: Here equations of impulsive motion are ( ( ) ) 2a 2 2a 2 Mv ' − M0 = −F = −(1R), Mu ' − Mu = −R, M ω' − M ω = −F(a) ~ ~~ ~ 3 3
that is, ) ) ( ) ( 2a ( ' ω − ω = −F = −R = −M u − u ' , Mv ' = −R, M u − u ' = R, M ~~ ~ ~ 3 and hence ) ( ) ( −v ' = u − u ' , 2a ω' − ω = −3 u − u ' . Here A is the point of contact, and the plane is rough enough to prevent sliding, just after impact, velocity of A is zero, and hence v ' + aω' = 0. Here e is the coefficient of restitution between the ball and the ground. So, by Newton’s rule, ( ' ) u − 0 = (−e)(u − 0), or u ' = −eu. Since aω +
) ( ) ) 5 5( 3( −aω' = aω + v ' = v ' + aω − −v ' 2 2 2 ) ( ( ) 3 ' = v ' + aω − u − u ' = ~v ' + aω ~~ = 0~, 2
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
755
we have 2 (angular velocity after impact) = ω' = ω . ~ ~~ 5 ~ Since ( a
) 2 ω = aω' = ~−v ' =~~u − u~' = u − eu = (1 + e)u, 5
we have ) 2 a ω = (1 + e)u. 5 (
The required horizontal range = ((1 + e)u)(time of flight) = ((1 + e)u) We have to show that known to be true.
2u 2 e(e g
+ 1) =
4 aω eu, 5 g
2(eu) 2u 2 = e(e + 1). g g
that is, u(e + 1) = 25 aω. This is
6.2.19 Example An imperfectly elastic sphere descending vertically comes in contact with a fixed rough point, the impact taking place at a point distant α from the lowest point, and the coefficient of elasticity being e. Find the motion, and show that the sphere will start moving horizontally after the impact if / α = tan
−1
7e 5
Sol: Here equations of impulsive motion are ( Mu ' − M(u cos α) = −R, Mv ' − M(u sin α) = −F , ~
Thus
~~
~
) ( ) 2a 2 2a 2 M ω− M 0 = −F(a). 5 5
756
u' =
6 Impulsive Forces
( ( ) ) 5 Mv ' − M(u sin α) 5 v ' − u sin α −R −F + u cos α, v ' = + u sin α, ω = = M M 2Ma 2a
Also ( ' ) ( ) v + ωa, −u ' , 0 = v ' , −u ' , 0 + ω(0, 0, 1) × (0, −a, 0) ( ) −→ = v ' , −u ' , 0 + ω(0, 0, 1) × G A = vel.ofAafter impact ~ ~~ ~ Now since A is a fixed rough point, ) ( 5 v ' − u sin α 7v ' − 5u sin α ' =v + = ~v ' + ωa ~~ = 0~ . 2 2 α . Hence ω = −5u7asin α . So v ' = 5u sin 7 By Newton’s rule, (−e)(u cos α) = u ' . Suppose that the sphere starts moving horizontally after the impact. So
( ((−e)(u cos α)) cos α +
5u sin α 7
)
sin α = u~ ' cos α +~~ v ' sin α = 0~,
and hence −e cos2 α +
5 sin2 α = 0. 7
or tan2 α =
7e .∎ 5
6.2.20 Example A billiard ball at rest on a horizontal table and is struck by a horizontal blow in a vertical plane passing through the centre of the ball. If the initial motion is one of pure rolling, find the height of the point of struck above the table. (There is no impulsive friction.) Sol: Here, equations of impulsive motion are ) ( ) ( 2a 2 2a 2 ω− M 0 = −X (a − h), M(aω) − M(0) = X, M 5 5
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
757
that is, ) 2a 2 ω = −X (a − h) = −(M(aω))(a − h). M(aω) = X, M 5 ~~ ~ ~ (
Thus (
) 2a 2 M ω = −(M(aω))(a − h) 5
or 2a = −(a − h) 5 Hence h =
7a . 5
6.2.21 Example A rough imperfectly elastic ball is dropped vertically, and when its velocity is V, a man suddenly moves his racket forward in its own plane with velocity U, and thus subjects the ball to pure cut in a downward direction making an angle α with the horizon. Show that, on striking the rough ground, the ball will not proceed the point of impact, provided (Figs. 6.41, 6.42, 6.43, 6.44, 6.45, 6.46, 6.47, 6.48, 6.49, 6.50, 6.51, 6.52, 6.53, 6.54, 6.55, 6.56, 6.57, 6.58 and 6.59). (
) a2 (U − V sin α)(1 − cos α) > (1 + e) 1 + 2 V sin α cos α. k Sol: Here, the moment of momentum about A is conserved. So ) ( 7a 2 ω(0, 0, −1) Ma(0, 0, −V1 sin α + U1 cos α) + M 5 ) ( 2a 2 2 + Ma ω(0, 0, −1) = a(sin α, − cos α, 0) × M(U1 , −V1 , 0) + M 5 ) ( 2a 2 −→ + Ma 2 ω(0, 0, −1) = AG × M(U1 , −V1 , 0) + M 5 = (A.M.about Aafter jerk) = (A.M.about Abefore jerk) ~ ~~ ~ −→ = AG × M(O, V , 0) = a(sin α, − cos α, 0) × M(O, V , 0) = Ma(0, 0, V sin α).
758
6 Impulsive Forces
,
1
−
1
racket
Point of contact with racket
Fig. 6.41
,
−
1
1
2 2 2
2
Rough ground
Fig. 6.42
2
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
759
Direction fixed in space Direction fixed in space
Direction fixed in body ≡ ( ), ̇ (0) = .
, Path of centre of mass
Fig. 6.43
60°
1
1
1
1
Fig. 6.44
This shows that −V1 sin α + U1 cos α − or
k2 ω = V sin α a
760
6 Impulsive Forces
/2
Fig. 6.45
( , )
Fig. 6.46
U1 cos α − V1 sin α = where k 2 ≡
7a 2 . 5
k2 ω + V sin α, a
From question,
k2 + a2 ω + V sin α = a
(
) k2 ω + V sin α + ωa = (U1 cos α − V1 sin α) + ωa a
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
761
4
3
3 4
3
vel. =
−
vel. =
+ 2
4
2
1
1 2 1
Fig. 6.47 3 2
3 3
3
3
2
2
4 4
=
3
+
3
1
1
1 1
Fig. 6.48
1
2
=
1
−
1,
2
=
3
−
3
762
6 Impulsive Forces
(vel. rel. to ) = (2 )
,2
2
,2
2 2
vel. = (2 )
,2
1
,2
1
1
Common normal Fig. 6.49
= (U1 , −V1 , 0) · (cos α, sin α, 0) + ωa = (velocity of point of contact A) = U , ~~ ~ ~ so ω= Also by Newton’s rule,
a(U − V sin α) . k2 + a2
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
763
,2
,2
1 2
1
1 1
,2
1
,2
1
1
2
2
Common normal Fig. 6.50
−U1 sin α − V1 cos α = (U1 , −V1 , 0) · (− sin α, cos α, 0) = (−e)(V cos α), ~ ~~ ~ so U1 sin α + V1 cos α − eV cos α = 0.
764
6 Impulsive Forces
Fig. 6.51
,2
1
vel. =
2
vel. = 3
1
1
cos
sin 1
1
Next since U1 cos α − V1 sin α + (ωa − U ) = 0, we have U1 =
cos α(ωa − U ) − eV cos α sin α = cos α(eV sin α − ωa + U ). −1
Also V1 =
−eV cos2 α − sin α(ωa − U ) = eV cos2 α + sin α(ωa − U ). −1
Here, the moment of momentum about B is conserved. So ( ) ) ( k2 Ma 0, 0, U2 − ω2 = Ma(0, 0, U2 ) + Mk 2 ω2 (0, 0, −1) a ) ( = a(0, −1, 0) × M(U2 , V2 , 0) + Mk 2 ω2 (0, 0, −1) ( ) 2a 2 −→ = BG × M(U2 , V2 , 0) + M + Ma 2 ω2 (0, 0, −1) 5 = (A.M.aboutBafter jerk) = (A.M.aboutBbefore jerk) ~ ~~ ~
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
765
,2
1
vel. =
2
vel. = 3
1
1
cos
sin 1
1
1 1
,2
1
1
2
2
vel. =
1
cos
Common normal Fig. 6.52
) ( −→ = BG × M(U1 , −V1 , 0) = a(0, −1, 0) × M(U1 , −V1 , 0) + Mk 2 ω(0, 0, −1) ( ) k2 = Ma(0, 0, U1 ) + Mk 2 ω(0, 0, −1) = Ma 0, 0, U1 − ω . a Thus U2 −
k2 k2 ω2 = U1 − ω. a a
766
6 Impulsive Forces
2 2
2
2
2 1 1 2
2 1, 2
3
1
2 1
1
1 1 2 2, 2
4
Common normal Fig. 6.53
=
2
,
,
1
= =
=
=
=
= = First disc second disc Fig. 6.54
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
767
1
1 1
2
2 2
,
Fig. 6.55
Common normal Fig. 6.56
768
6 Impulsive Forces
Instanteneous centre of rotation
( ̇
≡ ( )  ̇
=0
=0
=
− (
̇ ) cos
̇ (
̇ ) sin
− cos
̇ ) sin
Circular path of with centre and radius .
=0 sin
− Fixed direction in space Fixed point in space
Fig. 6.57 Direction fixed in space ,
Direction fixed in space
1
Fig. 6.58
6.2 Examples on Impulsive Forces in 2D Motion (Part II)
769 Solid sphere
1
=
2
(condition for rolling)
1
+ 2
2
′
,
′
Direction fixed in space
1
1
,
Solid sphere
sin + rolling
cos
1
Fig. 6.59
If, on striking the rough ground, the ball does not proceed the point of impact, we have ( 1+
( ) U2 − U1 + a2 a2 − U + aω = U + a U 2 1 2 k2 k2 k2 a
k2 a ω
) = U2 + aω2 = 0, and U2 < 0, ~ ~~ ~
and hence ) k2 k2 k2 = cos α(eV sin α − ωa + U − + 1 U = U − ω ω 0> ) 2 1 a2 a a ~ ~~ ~ ( ) k2 = eV sin α cos α − ω a cos α + + U cos α. a (
Thus ) ( ) )( k2 k2 a(U − V sin α) a cos α + = eV sin α cos α + U cos α < ω a cos α + a a k2 + a2 ~~ ~ ~ (
=
a 2 cos α + k 2 a 2 cos α + k 2 U − sin αV k2 + a2 k2 + a2
770
6 Impulsive Forces
or ( ) a 2 cos α + k 2 (1 + e)a 2 cos α + k 2 (1 + e cos α) V sin α = e cos α + V sin α k2 + a2 k2 + a2 a 2 cos α + k 2 = eV sin α cos α + sin αV k2 + a2
ga , and hence, as above, the sphere leaves the second edge immediately, etc. ∎
6.3.2 Example An equilateral triangle, formed of uniform rods freely hinged at their ends, is falling freely with one side horizontal and uppermost. If the middle point of this side be suddenly√ stopped, show that the impulsive actions at the upper and lower high are in the ratio 13 : 1. Sol: Let m be the mass of each rod. Let 2a be the length of each rod. We have to show that √ √ X2 + Y 2 13 √ . = 2 2 1 (X 1 ) + (Y1 ) From symmetry, Y1 = −Y1 , and hence Y1 = 0. Thus it sufficed to show that X 2 + Y 2 = 13(X 1 )2 . For the rod BC: Equations of impulsive motion are mV = Y − Y1 , X + X 1 = 0. √ Hence, it suffices to show that Y 2 = 12X 2 , that is, Y = ±2 3X. On taking moment about G, we get X (a sin 60◦ ) + Y (a cos 60◦ ) = X 1 (a sin 60◦ ) = (−X )(a sin 60◦ ), ~ ~~ ~ √ and hence Y = −2X 3.∎
6.3.3 Example A lamina in the form of an equilateral triangle ABC lies on a smooth plane. Suddenly it receives a blow at A in a direction parallel to BC, which causes A to move with velocity V . Determine the instantaneous velocities of B and C and describe the subsequent motion of the lamina. Sol: Let m be the mass of the lamina. Let a be the length of each side of the triangle.
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
773
Here equations are u−0= m ( a tan 30◦ )2 + m ( a 3 2 3 2
~~
√
=
) ( 12 I √ I a2 sec 30◦ 3 )2 = )2 m ( a )2 = ( a ma m 2 tan 30◦ tan 30◦ + 3 2 tan 30◦ ~
I (AG)
ω−0= ~
0 I ,v − 0 = , m m
4 3I ma
so √ 4 3I I . u = , v = 0, ω = m ma Also (
) I −→ , 0, 0 + ω(0, 0, −1) × G A m ) ) ( ( (a ) I I ◦ , 0, 0 + ω(0, 0, −1) × sec 30 (0, 1, 0) = , 0, 0 + ω(0, 0, −1) = m 2 m ( ) a × √ (0, 1, 0) 3 √ ) ) ( ( 4 3I I a , 0, 0 + = (0, 0, −1) × √ (0, 1, 0) m ma ) ) 3 ) ( ( ( (V , 0, O) = (velocity ofA) =
=
4I I , 0, 0 + (0, 0, −1) × (0, 1, 0) = m m
4I I , 0, 0 + (1, 0, 0) = m m
5I , 0, 0 . m
It follows that V =
5I . m
Next ( (velocity ofB) =
) I −→ , 0, 0 + ω(0, 0, −1) × G B m
) ( ( √ ) ) (a ) 1 I 3 I ◦ , 0, 0 + ω(0, 0, −1) × sec 30 ,− ,0 = , 0, 0 + ω(0, 0, −1) = − m 2 2 2 m ) ( )( √ a 1 3 × √ ,− ,0 − 2 2 3 (
774
6 Impulsive Forces
) ) ( √ ) ) (I ) √ √ 4 3I a ( a ( I , 0, 0 + ω √ −1, 3, 0 = , 0, 0 + = √ −1, 3, 0 m m ma 2 3 2 3 ) ( ( ) √ 2I I , 0, 0 + −1, 3, 0 = m m ) V( ) √ √ I( −1, 2 3, 0 = −1, 2 3, 0 . = m 5 (
Next ( (velocity of C) =
) I −→ , 0, 0 + ω(0, 0, −1) × GC m
) ( ( ) ) (a ) √3 1 I I , 0, 0 + ω(0, 0, −1) × sec 30◦ ,− ,0 = , 0, 0 + ω(0, 0, −1) m 2 2 2 m ) ( )( √ a 1 3 × √ ,− ,0 2 2 3 ) ) ( √ ) ( ) (I ) √ √ 4 3I I a ( a ( , 0, 0 + ω √ −1, − 3, 0 = , 0, 0 + = √ −1, − 3, 0 m m ma 2 3 2 3 ) ( ( ) √ 2I I , 0, 0 + −1, − 3, 0 = m m (
=
=
) V( ) √ √ I( −1, −2 3, 0 = −1, −2 3, 0 . m 5
6.3.4 Example A rectangular lamina, whose sides are of length 2a and 2b, is at rest when one corner is caught and suddenly made to move with prescribed speed V in the plane of lamina. Show that the greatest angular velocity which can thus be imparted to the lamina is 3V √ 4 a 2 + b2 Sol: Here equations of impulsive motion are (
) ) (√ a 2 + b2 Mu − M0 = I cos θ, Mv − M0 = −I sin θ, M a 2 + b2 sin(θ + α) . ω=I 3
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
775
Thus u=
−I sin θ I cos θ ,v = , M M
( ) a 3I b sin θ √ ω= √ + cos θ √ a 2 + b2 a 2 + b2 M a 2 + b2 ~~ ~ ~ ( ) I sin θ I cos θ 3 3 a +b = 2 = 2 (a(−v) + bu). a + b2 M M a + b2 It is given that −→ V (cos θ, sin θ, 0) = (velocity ofA) = (u, −v, 0) + ω(0, 0, −1) × G A ~~ ~ ~ = (u, −v, 0) + ω(0, 0, −1) × (−a, b, 0) = (u, −v, 0) + ω(b, a, 0) = (u + bω, −v + aω, 0). So V cos θ = u + bω, V sin θ = −v + aω. Since 3 3 ω= 2 (a(−v) + bu) = 2 (a(V sin θ − aω) + b(V cos θ − bω)) 2 a + b a + b2 ~ ~~ ~ 3V (a sin θ + b cos θ ) = − 3ω, a 2 + b2 we have 3V ) (a sin θ + b cos θ ). ω(θ ) ≡ ω = ( 2 4 a + b2 We have to show that 3V ω dω = √ , dθ =0 4 a 2 + b2 that is,  (a sin θ + b cos θ )  = 1, √  a 2 + b2 (a cos θ−b sin θ )=0 This is clearly true.∎
776
6 Impulsive Forces
6.3.5 Example Four freelyjointed rods, of the same material and thickness, form a rectangle of sides ' 2a and 2b and of mass M . When lying in this form on a horizontal plane an inelastic particle of mass M moving with velocity V in a direction perpendicular to the rod of length 2a impinges on it at a distance c from its centre. Show that the kinetic energy lost in the impact is
1 M
1 2 V (2 ). c2 + M1 ' 1 + 3a+3b a+3b a 2
Sol: Mass of rod AB =
M' M 'a (2a) = (≡ m 1 ). 2a + 2b + 2a + 2b 2(a + b) '
M b Similarly, mass of rod BC is 2(a+b) (≡ m 2 ). Since the direction of velocity of the centre of mass of the system just before impact is along yaxis, the direction of velocity of the centre of mass of the system just after the impact is along yaxis, This shows that velocities and angular velocities of G 1 , G 2 , G 3 , G 4 are as shown in the figure. For the system “four rods + particle M”: Since in collision, linear momentum along yaxis is conserved, we have M V = (l.m.along y − axis before collision of the system) = (l.m. along y − axis after collision of the system) ~ ~~ ~ = m 1 u + m 2 (u + aω) + m 1 u + m 2 (u − aω) + M(u + cω) = (2m 1 + 2m 2 )u + M(u + cω) = M ' u + M(u + cω).
and hence M V = M ' u + M(u + cω). Since in collision, angular momentum about G is conserved, we have (M V )c = (a.m. aboutG before collision of the system) = (a.m. aboutGafter collision of the system) ~ ~~ ~ ) ) ) ) ( ( ( ( a2 a2 ω + (a(m 2 (u + aω)) + 0) + 0 + m 1 ω = 0 + m1 3 3 +(−a(m 2 (u − aω)) + 0) + (M(u + cω))c ( ) M 'a a2 2 ω + 2a 2 m 2 ω + (M(u + cω))c = ω a2 3 3 2(a + b) ) ( M'b ω + Mc(u + cω) + 2a 2 2(a + b) = 2m 1
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
777
M ' a 2 bω M ' ωa 2 (a + 3b) M ' a3 ω + + Mc(u + cω) + Mc(u + cω) = 3(a + b) 3(a + b) (a + b) ' 2 ( ) M ωa (a + 3b) = + c M V − M'u 3(a + b)
=
and hence M V c = M 'ω
( ) a 2 (a + 3b) + c M V − M 'u 3(a + b)
or ) ( M V − Mcω a 2 (a + 3b) , = cu = c ω 3(a + b) M + M' ~~ ~ ~ Hence ω=
MV c a 2 (a+3b) 3(a+b) (M
+ M ' ) + Mc2
.
Next since (
'
'
)
M V = M u + M(u + cω) = u M + M + Mc ~~ ~ ~
(
MV c a 2 (a+3b) 3(a+b) (M
)
+ M ' ) + Mc2
we have ( ) a 2 (a+3b) M + M' MV 3(a+b) = ' u= M' + M M + M a 2 (a+3b) (M + M ' ) + Mc2 3(a+b) ) ( a 2 (a + 3b) M + M ' MV = ' M + M a 2 (a + 3b)(M + M ' ) + Mc2 3(a + b) MV −
M 2 V c2
a 2 (a+3b) ' 2 3(a+b) (M+M )+Mc
= MV
a 2 (a + 3b) . a 2 (a + 3b)(M + M ' ) + Mc2 3(a + b)
Now change in kinetic energy ( =
) ( ( ) ) ( ( ) ) 1 1 a2 a2 1 1 1 2 2 2 2 m1u + m1 ω + m 2 (u + aω) + m1u + m1 ω2 2 2 3 2 2 2 3 ) ( 1 1 1 2 + m 2 (u − aω) + M(u + cω)2 − M V 2 2 2 2
778
6 Impulsive Forces
) ( ( ) 1 a2 1 ω2 + m 2 u 2 + a 2 ω2 + M(u + cω)2 − M V 2 = m1u + m1 3 2 2 ) ( 1 1 1 2 = (m 1 + m 2 )u + m 1 + m 2 a 2 ω2 + M(u + cω)2 − M V 2 3 2 2 2
( = −
) ) ( M 'b M 'a M 'a 1 1 M 'b + u2 + a 2 ω2 + M(u + cω)2 + 2(a + b) 2(a + b) 2(a + b) 3 2(a + b) 2
1 MV 2 2
( ) )2 1 1 ( M ' 2 M ' a+3b 3 u + a 2 ω2 + M V − M 'u − M V 2 = 2 2(a + b) 2M 2 ( )2 a 2 (a + 3b) M' MV 2 2 a (a + 3b)(M + M ' ) + Mc2 3(a + b) )2 ) ( ( M ' a+3b MV c 2 3 a a 2 (a+3b) + 2(a + b) (M + M ' ) + Mc2 =
3(a+b)
(
+
1 a 2 (a + 3b) M V − M' M V 2 2M a (a + 3b)(M + M ' ) + Mc2 3(a + b)
)2 −
1 MV 2 2
3 1 ' 2 2 4 2 ' 2 2 2 2 2 M M V a (a + 3b) 2 (a + b)M (a + 3b)a M V c = ( )2 + ( )2 ) ) ( ( a 2 (a + 3b) M + M ' + Mc2 3(a + b) a 2 (a + 3b) M + M ' + Mc2 3(a + b)
+
( )2 a 2 (a + 3b) MV 2 1 1 − M' 2 − MV 2 2 a (a + 3b)(M + M ' ) + Mc2 3(a + b) 2 ) ( 2 M ' M 2 V 2 a 2 (a + 3b) a2 (a + 3b) + 23 (a + b)c2 = ( )2 a 2 (a + 3b)(M + M ' ) + Mc2 3(a + b) ( )2 MV 2 Ma 2 (a + 3b) + Mc2 3(a + b) 1 2 2 +( )2 − M V 2 ' 2 2 a (a + 3b)(M + M ) + Mc 3(a + b)
( 2 )2 ) ( 2 M ' M 2 V 2 a 2 (a + 3b) a2 (a + 3b) + 23 (a + b)c2 + M2V M 2 a 2 (a + 3b) + c2 3(a + b) = ( ) ( )2 a 2 (a + 3b) M + M ' + Mc2 3(a + b) −
1 MV 2 2
( )( )) ( M 2 21 V 2 a 2 (a + 3b) + c2 3(a + b) M ' a 2 (a + 3b) + M a 2 (a + 3b) + c2 3(a + b) = ) ( )2 ( a 2 (a + 3b) M + M ' + Mc2 3(a + b) 1 − MV 2 2
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
779
( ) ) ( 2 M 2 21 V 2 1 + c 2 3(a+b) M 2 21 V 2 a 2 (a + 3b) + c2 3(a + b) a (a+3b) 1 1 ( ) = 2 − MV 2 − MV 2 = ( ) c2 3(a+b) 2 2 a (a + 3b) M + M ' + Mc2 3(a + b) ' M+M +M 2
⎛
( M 1+
2
c 3(a+b)
)
a (a+3b)
⎞
=
a 2 (a+3b) 1 M V 2⎝ − 1⎠ 2 2 (M + M ' ) + M ac 23(a+b) (a+3b)
=
−M ' 1 MV 2 = 2 2 (M + M ' ) + M ac 23(a+b) (a+3b)
1 M
−1V2 ( 2 ) .∎ c2 + M1 ' 1 + 3a+3b 2 a+3b a
6.3.6 Example Four equal uniform rods, AB, BC, C D, D E, are freely jointed at B, C, D and lie on a smooth table in the form of a square. The rod AB is struck by a blow at a right angle to AB from the inside of the square. Show that the initial velocity of A is 79 times that of E. Sol: Let m be the mass of each rod. Let 2a be the length of each rod. For rod AB: ( 2) a mu 1 = I + X 1 , m ω1 = I a − X 1 a. 3 For rod BC: m(u 1 − aω1 ) = mu 2 = X 2 − X 1 . ~~ ~ ~ For rod C D: (
) a2 m((u 1 − aω1 ) + aω3 ) = m(u 2 + aω3 ) = mu 3 = −X 2 − X 3 , m ω3 = X 2 a − X 3 a. ~ ~~ ~ 3
For rod D E: m(((u 1 − aω1 ) + aω3 ) + aω3 ) = m((u 2 + aω3 ) + aω3 ) = m(u 3 + aω3 ) = mu 4 = X 3 . ~ ~~ ~
Thus ( a) mu 1 = I + X 1 , m ω1 = I − X 1 , 3
780
6 Impulsive Forces
m(u 1 − aω1 ) = X 2 − X 1 , ( a) m(u 1 + a(ω3 − ω1 )) = −X 2 − X 3 , m ω3 = X 2 − X 3 , 3 m(u 1 + a(−ω1 + 2ω3 )) = X 3 . Now on adding the four equations we get m(4u 1 + a(−3ω1 + 3ω3 )) = mu 1 + m(u 1 − aω1 ) + m(u 1 + a(ω3 − ω1 )) + m(u 1 + a(−ω1 + 2ω3 )) = (I + X 1 ) + (X 2 − X 1 ) + (−X 2 − X 3 ) + X 3 = I. Hence m(4u 1 + 3a(−ω1 + ω3 )) = I. Since ( a) mu 1 = I + X 1 , m ω1 = I − X 1 , 3 we get ( a) mu 1 + m ω1 = 2I. 3 On adding ( a) m(−u 1 + aω1 ) = X 1 − X 2 , m ω3 = X 2 − X 3 , m(u 1 + a(−ω1 + 2ω3 )) = X 3 , 3
we get ) ( a) 7 m aω3 = X 1 = I − m ω1 . 3 3 ~~ ~ ~ (
Thus (
7 1 m aω1 + aω3 3 3 Thus
) = I.
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
781
⎫ m(4u 1 + 3a(−ω1 + ω3 )) = I ⎬ . 3mu 1 + (ma)ω1 = 6I ⎭ m(aω1 + 7aω3 ) = 3I Since ) ( ) ) ( 4 24 9 6I − (ma)ω1 I+ − − maω1 = 4 8+ 7 3 7 3 ) ( 3I − maω1 = m(4u 1 + 3a(−ω1 + ω3 )) = I , +3a −mω1 + ~ ~~ ~ 7a
(
we have 29I =
50 maω1 . 3
Since (
87I 50
) + 7maω3 = m(aω1 + 7aω3 ) = 3I , ~ ~~ ~
we have maω3 =
9I . 50
Since 3mu 1 +
87I = 3mu 1 + (ma)ω1 = 6I , ~ ~~ ~ 50
we have mu 1 =
71I 50
We have to show that u 1 + aω1 = 79. u 1 + a(−ω1 + 2ω3 ) LHS =
mu 1 + maω1 = mu 1 − maω1 + 2maω3
71I 50
71I + 87I 50 50 ( ) 87I − 50 + 2 9I 50
=
158 = 79.∎ −16 + 18
782
6 Impulsive Forces
6.3.7 Example Four equal rods, each of mass m and length 2a, are freely jointed at their ends so as to form a rhombus. The rhombus falls with a diagonal vertical, and is moving with velocity V when it hits a fixed horizontal inelastic plane. Find the motion of the rods immediately after the impact, and show that their angular velocities are each equal to V sin α 3 ( ), 2 a 1 + 3 sin2 α where α is the angle each rod makes with the vertical. 1 of the kinetic energy just Show also that the impact destroys a fraction 1+3sin 2 α before the impact. Sol: Although we have worked out this problem in 6.2.13, we shall supply here another solution: From symmetry, 0 = (horizontal velocity ofC) = (horizontal velocity ofCrelative toB) ~ ~~ ~ + (horizontal velocity ofB) = (horizontal velocity ofCrelative toB) + ((2a)ω1 ) cos α = ((2a)ω2 ) cos α + ((2a)ω1 ) cos α. Hence 0 = ((2a)ω2 ) cos α + ((2a)ω1 ) cos α. It follows that ω2 = −ω1 . Next (horizontal velocity ofG 2 ) = (horizontal velocity ofG 2 relative toB) + (horizontal velocity ofB) = (horizontal velocity ofG 2 relative toB) + ((2a)ω1 ) cos α = (aω2 ) cos α + ((2a)ω1 ) cos α
= (a(−ω1 )) cos α + ((2a)ω1 ) cos α = aω1 cos α(→), and
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
783
(vertical velocity ofG 2 ) = (vertical velocity ofG 2 relative toB) + (vertical velocity ofB) = (vertical velocity ofG 2 relative toB) + ((2a)ω1 ) sin α = −(aω2 ) sin α + ((2a)ω1 ) sin α
= −(a(−ω1 )) sin α + ((2a)ω1 ) sin α = 3aω1 sin α(↓). From symmetry, Y = 0. On taking moments about B for the rod BC, we get ( ma 2 ω1
( ) )) 1 ( 2 − 4 sin2 α + mV a sin α = ma 2 ω1 − + −3 sin2 α + cos2 α 3 3
+ mV a sin α = −m ( ( = − m ~
) ( a2 ω + ma 2 ω1 −3 sin2 α + cos2 α + mV a sin α 3 1 )
) a2 ω1 + (−m(3aω1 sin α)(a sin α)) + m(aω1 cos α)(a cos α) − (−(mV )(a sin α)) = ((2a) cos α)X , 3 ~~ ~
and hence ( ma ω1 2
) 2 2 − 4 sin α + mV a sin α = (2a) cos α X. 3
or ( ω1
) 2 V 2X 2 − 4 sin α = cos α − sin α. 3 ma a
Also, X 1 + X = m(aω1 cos α), −Y1 = m(3aω1 sin α) − mV . On taking moments about A for the two rods AB, AC, we have ) ( ) 8 4a 2 ω1 − mV a sin α + ma 2 ω1 − mV a sin α 3 3 ) ( 4a 2 = m ω − (mV )(a sin α) 3 1 ) (
4ma 2 ω1 − 2mV a sin α = m
((( +
m
) a2 (−ω1 ) + (m(3aω1 sin α))(a sin α) + (m(aω1 cos α))(3a cos α) 3
−(mV )(a sin α))) = (change in A.M. of rod AB about A) + (change in A.M. of rod BC about A) = X ((2a) cos α + (2a) cos α) ~~ ~ ~
= 4Xa cos α,
784
6 Impulsive Forces
so 2maω1 − mV sin α = 2X cos α. Thus 2X ma
cos α − 2 3
V a 2
sin α
− 4 sin α
V X = ω1 = sin α + cos α . 2a ma ~ ~~ ~
It follows that V 2X cos α − sin α = ma a
(
( ) ) 2 2 V X 2 2 − 4 sin α sin α + − 4 sin α cos α 3 2a 3 ma
or X
( ( ( ) )) ) 1 4 cos α 1 2 cos α 4 cos α ( 1 + 3 sin2 α = X + sin2 α = X 1− − 2 sin2 α 3ma ma 3 ma 3
) ( ) ) 2 2 V cos α V 2 2 − 4 sin α = − 4 sin α sin α + sin α 3 ma 3 2a a ~~ ~ ( ) ( ) 4 V 2V − 2 sin2 α = sin α 2 − 3 sin2 α . = sin α a 3 3a
(
2 cos α − =X ma ~
(
Thus (
X=
2V sin α 2 − 3 sin2 α 3a ( ) 4 cos α 1 + 3 sin2 α 3ma
) =
2 − 3 sin2 α 3 cos2 α − 1 mV mV tan α tan α = . 2 2 1 + 3 sin2 α 1 + 3 sin2 α
Now since X V V sin α + cos α = sin α + ω1 = 2a ma 2a ~ ~~ ~ =
mV 2
cos α−1 tan α 31+3 sin2 α 2
ma
cos α
V 3 cos2 α − 1 sin α 3V V sin α + sin α = .∎ 2 2a 2a 2a 1 + 3 sin2 α 1 + 3 sin α
where α is the angle each rod makes with the vertical. Next since 3 cos2 α − 1 mV tan α = X 1 + X = m(aω1 cos α) X1 + ~ ~~ ~ 2 1 + 3 sin2 α ) ) ( ( sin α 3mV sin α cos α 3V cos α = =m a 2a 1 + 3 sin2 α 2 1 + 3 sin2 α
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
785
we have X1 =
( ) 3 cos2 α − 1 3mV sin α cos α mV mV sin α 3 cos2 α − 1 ) 3 cos α − − = ( tan α 2 2 2 2 1 + 3 sin α 2 cos α 1 + 3 sin α 2 1 + 3 sin α
mV tan α ). = ( 2 1 + 3 sin2 α
Again since ) ) ( ( sin α 3V sin α + mV Y1 = −m(3aω1 sin α) + mV = −m 3a ~ ~~ ~ 2a 1 + 3 sin2 α mV −9 sin2 α + mV = 2 1 + 3 sin2 α =
mV 3 cos2 α − 1 mV 2 − 3 sin2 α = . 2 1 + 3 sin2 α 2 1 + 3 sin2 α
Thus, the final K.E. ( 2 ( 2 ) ) a a 1 1 2 2 2 m + ma (ω1 ) + m + ma (ω1 )2 = 2 3 2 3 ⎛
K.E.of AB
K.E.of AD
⎞ ( ) ⎜1 )⎟ a2 1 ( ⎜ ⎟ +⎜ m (ω1 )2 + m (aω1 cos α)2 + (3aω1 sin α)2 ⎟ ⎝2 ⎠ 3 2 K.E.of BC
) 4 2 a 2 2 2 2 = ma (ω1 ) + m (ω1 ) + (aω1 cos α) + (3aω1 sin α) 3 3 ( ( ) ) 5 8 + cos2 α + 9 sin2 α = ma 2 (ω1 )2 + 8 sin2 α = ma 2 (ω1 )2 3 3 ( ( )2 )2 ( ) 8 2 3V sin α 1 8 2 3V 2 sin α 1 + 3 sin α = ma = ma 2 3 2a 1 + 3 sin α 3 2a 1 + 3 sin2 α sin2 α = 6mV 2 1 + 3 sin2 α )( ) ( 3 sin2 α 1 1 1 1 2 2 2 2 mV + mV + mV + mV = 2 2 2 2 1 + 3 sin2 α 2 3 sin α .∎ = (the initial K.E.) 1 + 3 sin2 α (
2
786
6 Impulsive Forces
6.3.8 Example A rectangle formed of four uniform rods freely jointed at their ends is moving on a smooth horizontal plane with velocity V in a direction along one of its diagonals which is perpendicular to a smooth inelastic vertical wall on which it impinges. Show that the loss of energy due to the impact is V2 1 m 1 +m 2
+
3 cos2 α 3m 1 +m 2
+
3 sin2 α m 1 +3m 2
,
where m 1 and m 2 are the masses of the rods and α is the angle the above diagonal makes with side of mass m 1 . Sol: Let m 1 be the mass of rod AB. Let m 2 be the mass of rod BC. Observe that (· · · , −V1 ) + ((bω2 ) sin α, (bω2 ) cos α) + ((aω1 ) cos α, −(aω1 ) sin α) = (vel.ofOrel.toA) + (vel.ofG 1 rel.toO) + (vel.of Arel.toG 1 ) = (0, 0), ~ ~~ ~ so a b V1 = (bω2 ) cos α − (aω1 ) sin α = (bω2 ) √ − (aω1 ) √ . 2 2 2 a +b a + b2 Hence ab(ω2 − ω1 ) =
√ a 2 + b2 V1 .(1)
On taking moments about O of all the four rods, we get ) ) ( ( a2 ω1 + (m 1 V1 )(a cos α) − b(m 1 (bω2 )) − m1 3 ) ( ( ) b2 + − m2 ω2 + (m 2 V1 )(a cos α) − a(m 2 (aω1 )) 3 ) ) ( ( a2 ω1 − (m 1 V1 )(a cos α) − b(m 1 (bω2 )) + − m1 3 ) ) ( ( b2 ω2 − (m 2 V1 )(a cos α) − a(m 2 (aω1 )) = 0 + − m2 3 or
(6.7)
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
787
) ( ( ) ) ) ( ( a2 b2 ω1 − m 1 b(bω2 ) + 2 − m 2 ω2 − m 2 a(aω1 ) = 0 2 − m1 3 3 or (m
) ) (m 2 + m 2 a 2 ω1 + + m 1 b2 ω2 = 0 3 3 1
or ) (√ a 2 + b2 V1 + ω1 (m 1 + 3m 2 )a ω1 + (3m 1 + m 2 )b ab 2
2
= (m 1 + 3m 2 )a 2 ω1 + (3m 1 + m 2 )b2 ω2 = 0 ~ ~~ ~
(6.8)
or ( ) b√ 2 a + b2 V1 . (3) (m 1 + 3m 2 )a 2 + (3m 1 + m 2 )b2 ω1 = −(3m 1 + m 2 ) a On taking moments about A of two rods AB and BC, we get ) (( (( ) ) a2 b2 −m 2 −m 1 ω1 + (m 1 V1 )(a sin α) + ω2 + (m 2 V1 )(a sin α) 3 3 − (m 2 (aω1 ))(2a)) = ((m 1 V )(a sin α) + (m 2 V )(a sin α)) + X 2 (AC) or ⎞
⎛ 2
⎛
⎞
⎞
⎛
2
⎟ ⎜ b ⎟ ⎜ ⎟ ⎜ a m 1 ⎝− ω1 + V1 a sin α ⎠ + m 2 ⎝− ω2 + V1 a sin α −2a 2 ω1 ⎠ = ⎝ m 1 + m 2 ⎠V a sin α ~~~~ ~~~~ ~ ~~ ~ ~ ~~ ~ 3 3 1
2
1
2
+ 2X 2 (a cos α + b sin α)
or a2 b2 ω1 − m 2 ω2 − m 1 a sin α(V − V1 ) − m 2 a sin α(V − V1 ) − 2m 2 a 2 ω1 3 3 = 2X 2 (a cos α + b sin α)
− m1
or −a 2 ω1
(m
) b2 + 2m 2 − m 2 ω2 − (m 1 + m 2 )a sin α(V − V1 ) = 2X 2 (a cos α + b sin α) 3 3 1
788
6 Impulsive Forces
or ) m ( −(m + 3m )a 2 ω ) 2 1 2 1 + 2m 2 − − (m 1 + m 2 )a sin α(V − V1 ) 3 3 3m 1 + m 2 = 2X 2 (a cos α + b sin α)
− a 2 ω1
(m
1
or ( ) m1 m 2 (m 1 + 3m 2 ) 2 b − − 2m 2 + a ω1 − (m 1 + m 2 )a √ (V − V1 ) 2 3 3(3m 1 + m 2 ) a + b2 √ = 2X 2 a 2 + b2 or −m 1 (3m 1 + m 2 ) − 6m 2 (3m 1 + m 2 ) + m 2 (m 1 + 3m 2 ) 2 a ω1 3(3m 1 + m 2 ) √ b − (m 1 + m 2 )a √ (V − V1 ) = 2X 2 a 2 + b2 a 2 + b2 or √ −3(m 1 )2 − 18m 1 m 2 − 3(m 2 )2 2 b a ω1 − (m 1 + m 2 )a √ (V − V1 ) = 2X 2 a 2 + b2 3(3m 1 + m 2 ) a 2 + b2
or −
√ ab (m 1 )2 + 6m 1 m 2 + (m 2 )2 2 a ω1 − (m 1 + m 2 ) √ (V − V1 ) = 2X 2 a 2 + b2 . (3m 1 + m 2 ) a 2 + b2
(6.9)
On taking moments about B of rod BC, we get (
) b2 −m 2 ω2 − (m 2 V1 )(b cos α) = −(m 2 V )(b cos α) + (X 2 (2b) sin α + Y2 (2b) cos α) 3
or m2 − 3
(
−(m 1 + 3m 2 )a 2 ω1 b(3m 1 + m 2 )
) − m 2 V1 cos α = −m 2 V cos α + 2(X 2 sin α + Y2 cos α)
or ) ( a b a m 2 (m 1 + 3m 2 ) 2 . = 2 X2 √ + Y2 √ a ω1 + m 2 (V − V1 ) √ 3b(3m 1 + m 2 ) a 2 + b2 a 2 + b2 a 2 + b2
(6.10)
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
789
On taking moments about D of rod C D, we get (
) a2 −m 1 ω1 + (m 1 V1 )(a sin α) = (m 1 V )(a sin α) + (Y2 ((2a) sin α) − X 2 ((2a) cos α)) 3
or ( ) a b a b = 2 Y2 √ − X2 √ −m 1 ω1 − m 1 (V − V1 ) √ . (6.11) 3 a 2 + b2 a 2 + b2 a 2 + b2 Thus, by b(5) − a(6), we have ) m 2 (m 1 + 3m 2 ) 2 a a ω1 + m 2 (V − V1 ) √ 3b(3m 1 + m 2 ) a 2 + b2 ( ) √ a b − a −m 1 ω1 − m 1 (V − V1 ) √ = 2X 2 a 2 + b2 3 a 2 + b2 (
b
or (
) √ m 2 (m 1 + 3m 2 ) m 1 2 ab a ω1 + (m 1 + m 2 )(V − V1 ) √ + = 2X 2 a 2 + b2 . 3(3m 1 + m 2 ) 3 a 2 + b2
or √ 3(m 1 )2 + 2m 1 m 2 + 3(m 2 )2 2 ab a ω1 + (m 1 + m 2 )(V − V1 ) √ = 2X 2 a 2 + b2 . 3(3m 1 + m 2 ) a 2 + b2
(6.12)
Hence, by (7) and (4), 3(m 1 )2 + 2m 1 m 2 + 3(m 2 )2 2 ab a ω1 + (m 1 + m 2 )(V − V1 ) √ 2 3(3m 1 + m 2 ) a + b2 =−
ab (m 1 )2 + 6m 1 m 2 + (m 2 )2 2 a ω1 − (m 1 + m 2 ) √ (V − V1 ) 2 (3m 1 + m 2 ) a + b2
or 6(m 1 )2 + 20m 1 m 2 + 6(m 2 )2 2 ab a ω1 + 2(m 1 + m 2 )(V − V1 ) √ =0 2 3(3m 1 + m 2 ) a + b2 or 2(3m 1 + m 2 )(m 1 + 3m 2 ) 2 ab a ω1 + 2(m 1 + m 2 )(V − V1 ) √ =0 2 3(3m 1 + m 2 ) a + b2 or
790
6 Impulsive Forces
b (m 1 + 3m 2 ) =0 aω1 + (m 1 + m 2 )(V − V1 ) √ 3 a 2 + b2 or (m 1 + 3m 2 )aω1 = −3(m 1 + m 2 )(V − V1 ) sin α.
(6.13)
Now since (m 1 + 3m 2 )a 2 ω1 + (3m 1 + m 2 )b2 ω2 = 0 we have −(3m 1 + m 2 )b2 ω2 b = −3(m 1 + m 2 )(V − V1 ) sin α = −3(m 1 + m 2 )(V − V1 ) √ 2 a a + b2 ~~ ~ ~
or a (3m 1 + m 2 )bω2 = 3(m 1 + m 2 )(V − V1 ) √ 2 a + b2 or (3m 1 + m 2 )bω2 = 3(m 1 + m 2 )(V − V1 ) cos α.
(6.14)
Next since (
) a cos α b sin α + 3m 1 + m 2 m 1 + 3m 2 ) ( ) ( −3(m 1 + m 2 )(V − V1 ) sin α 3(m 1 + m 2 )(V − V1 ) cos α −b =a 3m 1 + m 2 m 1 + 3m 2 √ = a(bω2 ) − b(aω1 ) = ab(ω2 − ω1 ) = a 2 + b2 V1 , ~ ~~ ~ 3(m 1 + m 2 )(V − V1 )
we have ( V1 = 3(m 1 + m 2 )(V − V1 ) ~ = 3(m 1 + m 2 )(V − V1 ) and hence
(
√ a a 2 +b2
cos α
3m 1 + m 2 ~~
+
√ b a 2 +b2
m 1 + 3m 2
cos α sin α + 3m 1 + m 2 m 1 + 3m 2 2
sin α
2
)
) ~
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
791
) ( ) ( cos2 α cos2 α V1 sin2 α sin2 α − V1 . =V + + 3(m 1 + m 2 ) 3m 1 + m 2 m 1 + 3m 2 3m 1 + m 2 m 1 + 3m 2 Thus ) ( ) ( 1 cos2 α cos2 α sin2 α sin2 α = V1 + V + + 3m 1 + m 2 m 1 + 3m 2 3(m 1 + m 2 ) 3m 1 + m 2 m 1 + 3m 2 or ( V1 =
V
cos2 α 3m 1 +m 2
1 3(m 1 +m 2 )
+
+
sin2 α m 1 +3m 2
cos2 α 3m 1 +m 2
+
)
sin2 α m 1 +3m 2
.
By 6.2.14, the change in K.E. of the body =
1 1 ((−V ) + V1 )(impulse) = (−V + V1 )((2m 1 + 2m 2 )(−V )) = (V − V1 )(m 1 + m 2 )V 2 2
(
⎛ = ⎝V − ( =
1 3(m 1 +m 2 ) 2 1 + 3mcos1 +mα 2 3(m 1 +m 2 )
V
cos2 α 3m 1 +m 2
1 3(m 1 +m 2 )
+
sin α m 1 +3m 2 2
+
sin2 α m 1 +3m 2
+ 3mcos1 +mα 2 + ) 2
)
sin2 α m 1 +3m 2
(m 1 + m 2 )V 2 =
⎞ ⎠(m 1 + m 2 )V V2 1 m 1 +m 2
+
3 cos2 α 3m 1 +m 2
+
3 sin2 α m 1 +3m 2
.∎
6.3.9 Example Of two inelastic discs with milled edges, each of mass m and radius a, one is rotating with angular velocity ω round its centre O which is fixed on a smooth plane, and the other is moving without spin in the plane with velocity v directed towards O. Find the motion immediately afterwards, and show that the energy lost by the impact is ( ) 1 a 2 ω2 2 m v + . 2 5 Sol: From the second disc, F = mu − 0, andX = 0 − (−mv). Also ( 2) a m ω2 − 0 = Fa = (mu)a 2 ~~ ~ ~
792
6 Impulsive Forces
Thus a F = mu, X = mv, ω2 = u. 2 From the first disc, ( ( 2) ) ( ( 2) ) a a − m ω1 − − m ω = Fa = (mu)a 2 2 ~ ~~ ~ Thus a (ω − ω1 ) = u. 2 Also, 3u = u + 2u = u + aω2 = aω1 = aω − 2u. ~ ~~ ~ Thus u=
aω . 5
2u 2 aω = ω − ( a5 ) = Next since a2 (ω − ω1 ) = u, we have ω1 = ω − ~ ~~ a~ aω 2u 2 a = ( a5 ) = 2ω ω = u, we have ω2 = . 2 2 5 a ~ ~~ ~ Thus, K.E. lost
3ω . 5
Since
( ( ) (( ( ) ( ( )) ) ) ) 1 1 1 1 2 1 2 a2 a2 a2 2 2 2 = m ω + mv − m m (ω1 ) + (ω2 ) + mu 2 2 2 2 2 2 2 2 (( ( )) ( ) )( ) ( )( ) 2 2 ( 2 2 2 1 3ω 2ω aω )2 1 1 1 1 a a a = + + m m ω2 + mv 2 − m m 2 2 2 2 2 5 2 2 5 2 5
=
( 2) a 3ma 2 ω2 1 ma 2 ω2 ma 2 ω2 1 1 m ω2 + mv 2 − − (9 + 4 + 2) = mv 2 + 2 2 2 100 2 4 20 ( ) 1 ma 2 ω2 a 2 ω2 1 = m v2 + .∎ = mv 2 + 2 10 2 5
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
793
6.3.10 Example A uniform circular disc, of mass M and radius a, is rotating with uniform angular velocity ω on a smooth plane and impinges normally with any velocity u upon a rough rod, of mass m, resting on the plane. Show that the angular velocity of the disc is immediately reduced to M +m ω. M + 3m Sol: For the impulsive motion of rod, mv1 − m0 = −F. For the impulsive motion of disc, ) ) ( a2 + a2 a2 + a2 ω2 − M ω = −F(a). Mv2 − M0 = F, M 4 4 (
Thus ( a) mv1 = −F, Mv2 = F, M (ω − ω2 ) = F. 2 Also F −F − aω2 = v2 − aω2 = v1 = , ~~ ~ ~ M m and hence ( M
a) Mmaω2 aω2 . = (ω − ω2 ) = F = 1 1 2 M +m +m M ~ ~~ ~
It follows that ω − ω2 =
2mω2 M +m
or ω2 =
M +m ω.∎ M + 3m
794
6 Impulsive Forces
6.3.11 Example An elliptic disc, of mass m, is dropped in a vertical plane with velocity V on a perfectly rough horizontal plane. Show that the loss of kinetic energy by the impact is ) k 2 + p2 1( 1 − e2 mV 2 2 , 2 k + r2 where r is the distance of the centre of the disc from the point of contact, p is the central perpendicular on the tangent, and e is the coefficient of elasticity. Sol: Here equations of impulsive motion of disc are √ ) ( ) ( mu − m0 = F, mv − mV = −R, mk 2 ω − mk 2 0 = R r 2 − p 2 − F( p). Thus √ ) ( mu = F, m(V − v) = R, mk 2 ω = R r 2 − p 2 − F( p). Also √ r 2 − p2 = ((v − (r ω) sin φ) − 0) = (−e)(V − 0) . v − (r ω) ~ ~~ ~ r Thus √ ω r 2 − p 2 − v = eV. Further u − pω = u − (r ω)
p = u − (r ω) cos φ = (horizontal vel.ofpoint of contactA) = 0 . ~ ~~ ~ r
Thus u = pω. Since √ u√ 2 r − p 2 − v = ω r 2 − p 2 − v = eV , ~~ ~ ~ p we have
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
v=
795
u√ 2 r − p 2 − eV . p
Since ( ) √ √ ( ( 2) u ) = mk 2 ω = R r 2 − p 2 − F( p) = R r 2 − p 2 − (mu)( p) mk p ~~ ~ ~ √ = (m(V − v)) r 2 − p 2 − mup, we have ( ( √ ))√ √ k2u u 2 2 2 2 = (V − v) r − p − up = V − r − p − eV r 2 − p 2 − up. p p ~ ~~ ~ Thus √ √ ) k2u u( ur 2 = V (1 + e) r 2 − p 2 − r 2 − p 2 − up = V (1 + e) r 2 − p 2 − p p p ~~ ~ ~ or u=
√ pV (1 + e) r 2 − p 2 . k2 + r 2
Here, by 6.2.14, the change in K.E. of the body =
1 1 (V + (−eV ))(impulse) = ((1 − e)V )R. 2 2
It suffices to show that R = (1 + e)mV
k 2 + p2 . k2 + r 2
Since / / / ( ) mk 2 ω = R r 2 − p2 − F( p) = R r 2 − p2 − (mu)( p) = R r 2 − p2 − (m( pω))( p) ~ ~~ ~
we have √ ) ( R r 2 − p2 = m k 2 + p2 ω Thus, it suffices to show that
796
6 Impulsive Forces
) ( m k 2 + p2 ω k 2 + p2 √ = (1 + e)mV 2 k + r2 r 2 − p2 that is, √ u r 2 − p2 , = ω = (1 + e)V 2 2 p ~ ~~ k + r ~ that is, √ p r 2 − p2 . u = (1 + e)V k2 + r 2 This is known to be true.
6.3.12 Example Two similar ladders, of mass m and length 2a, smoothly hinged together at the top, are placed on a smooth floor and released from rest when their inclination to the horizon is α. When their inclination is θ they are brought to rest by tightening the string, of length l which joins similarly situated rings. Show that the jerk in the string is / 2 a ga(sin α − sin θ ). 2m cot θ l 3 Sol: Let P be the point of AB to which the string is attached. Clearly, l = (A P) cos θ. 2 Let us observe that the velocity of G is −a θ˙ in the downward direction perpendicular to OG. Equation of Energy for the motion of rod AB is ( )2 2 ma 2 θ˙ 3 ( ) ( ) ) ( ) ( ))2 1 )2 1 ( (  1 ( )2 1 a 2 ( )2 a 2 (  θ˙ θ˙ t=0 + m a θ˙ + m a θ˙ t=0 m m − 2 3 2 2 3 = 2 = (mg)(a sin(π − α) − a sin(π − θ )) + 0 + 0 ~~ ~ = mga(sin α − sin θ ).
~
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
797
and hence / 3g(sin α − sin θ ) ˙θ = − . 2a Equations of impulsive motion of rod AB are ) ) (( (( ) ) d m −a θ˙ sin θ = m a (π − θ ) sin(π − θ ) = T − X , dt ~ ~~ ~ ) ) (( d ma θ˙ cos θ = m a (π − θ ) cos(π − θ ) = Y , dt ~~ ~ ~ ( ( ) ( ) ) 2 2 2 d a a a − m θ˙ = m 0 = −X (a sin θ ) + Y (−a cos θ ) − T (G P sin θ ) (π − θ ) − m 3 3 dt 3 ~ ~~ ~ ) ( ( (( ) )) (( ) ) ˙ ˙ = − m a θ sin θ + ~~~~ T (a sin θ ) + m a θ cos θ (−a cos θ ) (( ) ) −T A P − ~~~~ a sin θ = −ma 2 θ˙ − T ( A P) sin θ ) ( l Tl sec θ sin θ = −ma 2 θ˙ − tan θ = −ma 2 θ˙ − T 2 2
Thus (
) a2 Tl m θ˙ = ma 2 θ˙ + tan θ 3 2
or ) ( )( / 2a 2 3g(sin α − sin θ ) 2ga(sin α − sin θ ) = −m ma − 3 3 2a ( 2) 2a Tl tan θ . = −m θ˙ = 3 2 ~ ~~ ~ /
Hence ( / ) / 2 2ga(sin α − sin θ ) 2 a ga(sin α − sin θ ). ∎ T = cot θ ma = 2m cot θ l 3 l 3
798
6 Impulsive Forces
6.3.13 Example A sphere of mass m falls with velocity V on a perfectly rough inclined plane of mass M and angle α which rests on a smooth horizontal plane. Show that the vertical velocity of the centre of the sphere immediately after the impact is 5(M + m)V sin 2α 7M + 2m + 5m sin2 α the bodies being all supposed perfectly inelastic. Sol: Here equations are m(v − V ) = m(−v) − (−mV ) = F sin α + R cos α , mu = mu − 0 = −F cos α + R sin α , ~~ ~ ~ ~~ ~ ~
( ) ( ) ( ) 2a 2 2a 2 2a 2 m ω= m ω− m 0 = −F(a) . 5 5 5 ~~ ~ ~ Thus m(v − V ) = F sin α + R cos α, mu = −F cos α + R sin α, ( ) 2a m ω = −F = cos α(−F cos α + R sin α) − sin α(F sin α + R cos α) 5 ~ ~~ ~ = cos α(mu) − sin α(m(v − V )). Thus 2a ω = u cos α − sin α(v − V ). 5 Since external force acts along vertical direction, conservation of linear momentum holds along horizontal direction. It follows that 0 = mu + M(−u 1 ). Since the impact is perfectly inelastic, after the impact, the point of contact A is at rest on the inclined plane, and hence (−u 1 , 0) = (u + aω cos α, v + aω sin α), (geometrical condition). Thus
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
( v + aω sin α = 0, u − v cot α = u +
−v sin α
)
799
mu cos α = u + aω cos α = −u 1 = − . ~~ ~ ~ M
Thus m+M u = v cot α. M Next since ( ) 2a −v 2a −2v = = ω = u cos α − sin α(v − V ) 5 sin α 5 a sin α ~~ ~ ~5 ) ( Mv cot α cos α − sin α(v − V ) = m+M we have α Mv cos −2v sin α = + sin α(v − V ) 5 sin α m+M 2
or 2(m + M) + 5m sin2 α + 5M 7M + 2m + 5m sin2 α =v 5(m + M) sin α 5(m + M) sin α 2 2(m + M) + 5(m + M) sin α + 5M cos2 α =v 5(m + M) sin α ) ( M cos2 α 2 + sin α + = V sin α . =v 5 sin α m + M sin α ~ ~~ ~
v
Hence v=
5(M + m)V sin2 α .∎ 7M + 2m + 5m sin2 α
6.3.14 Example A sphere, of mass m, is resting on a perfectly rough horizontal plane. A second ' sphere, of mass m , falling vertically with velocity V stricks the first. Both spheres are inelastic and perfectly rough and the common normal at the point of contact makes an angle γ with the horizon. Show that the vertical velocity of the falling sphere will be instantaneously redued to
800
6 Impulsive Forces
) ( V m + m' 7 m 5
sec2 γ + m ' + 27 m ' tan2
(π 4
+
γ ). 2
Show also that the lower sphere will be set spinning in any case. Sol: Here, equations of impulsive motion of lower sphere is 7 7 7 ( u1 ) mau 1 = m a 2 = m a 2 ω1 5 ( 5 )a 5 ) ( 2 2 = m a 2 + ma 2 ω1 − m a 2 + ma 2 0 = F(a sin γ + a) − R(a cos γ ) . 5 5 ~~ ~ ~ Thus 7 mu 1 = F(sin γ + 1) − R cos γ . 5 Here, equations of impulsive motion of upper sphere are ) ( ) ( ' ' '2 2 − F = m u − m (V cos γ ), R = −m v − m (−V sin γ ), m b ω 5 ) ( 2 − m ' b2 0 = F(b). 5 '
'
Thus 2 F = m ' (V cos γ − u), R = m ' (V sin γ − v), m ' bω = F. 5 Since the impact is inelastic, after the impact, the point of contact A is at rest and hence (u, v) + bω(−1, 0) = (u 1 sin γ , −u 1 cos γ ) + aω1 (1, 0). Thus u − bω = u 1 sin γ + aω1 = u 1 sin γ + u 1 = u 1 (1 + sin γ ), v = −u 1 cos γ . ~~ ~ ~ It follows that ( u − bω = or
) −v (1 + sin γ ) cos γ
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
801
R(1 + sin γ ) + 27 F cos γ R(1 + sin γ ) + F cos γ V (1 + sin γ ) − = V (1 + sin γ ) − m' m' ) ( 5F cos γ −b 2bm '
= V (1 + sin γ ) − ⎛
R(1 + sin γ ) + F cos γ ' ⎞m
⎛⎛ ⎞ ⎞ R F −bω cos γ = ⎝V sin γ − ' ⎠(1 + sin γ ) + ⎝⎝V cos γ − ' ⎠ − bω⎠ cos γ ~ ~~ ~ m ~ ~~ ~ m 1
1
= v(1 + sin γ ) + (u − bω) cos γ = 0 . ~~ ~ ~ Thus R(1 + sin γ ) +
7 F cos γ = m ' V (1 + sin γ ) 2
or 2(1 + sin γ )R + 7 cos γ F − 2m ' V (1 + sin γ ) = 0. Since V sin γ −
) ( 5(F(sin γ + 1) − R cos γ ) R cos γ , = − = v = −u cos γ 1 ~~ ~ ~ m' 7m
we have 7mm ' V sin γ − 7m R = −5m ' F cos γ (sin γ + 1) + 5m ' R cos2 γ or ( ) 7m + 5m ' cos2 γ R − 5m ' cos γ (1 + sin γ )F − 7mm ' V sin γ = 0. Also 2(1 + sin γ )R + 7 cos γ F − 2m ' V (1 + sin γ ) = 0. Hence 5m ' cos γ (1 + sin γ )2m ' V (1 + sin γ ) + 7mm ' V sin γ · 7 cos γ ) R= ( 7m + 5m ' cos2 γ 7 cos γ + 5m ' cos γ (1 + sin γ )2(1 + sin γ ) 10(1 + sin γ )m ' (1 + sin γ ) + 49m sin γ ) = m' V ( 7m + 5m ' cos2 γ 7 + 5m ' (1 + sin γ )2(1 + sin γ )
802
6 Impulsive Forces
10(1 + sin γ )m ' (1 + sin γ ) + 49m sin γ ) , = m' V ( 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 and F=
( ) −7mm ' V sin γ 2(1 + sin γ ) + 7m + 5m ' cos2 γ 2m ' V (1 + sin γ ) ( ) 7m + 5m ' cos2 γ 7 cos γ + 5m ' cos γ (1 + sin γ )2(1 + sin γ ) ) ( −7m sin γ + 7m + 5m ' cos2 γ 2m ' V (1 + sin γ ) ) ( . = cos γ 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2
Here, required vertical velocity ( ) ( ) F R = u cos γ + v sin γ = V cos γ − ' cos γ + V sin γ − ' sin γ m m R F = V − ' cos γ − ' sin γ m m ( ) ⎛ ⎞ −7m sin γ + 7m + 5m ' cos2 γ 2V + sin γ (1 ) ⎝ ⎠ cos γ =V− ( ) cos γ 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 ( ) 10(1 + sin γ )m ' (1 + sin γ ) + 49m sin γ − V( sin γ ) 2 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ ) ( ) ⎛ ⎞ ' −7m sin γ + 7m + 5m cos2 γ ⎠ = V − ⎝2V (1 + sin γ ) ( ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 ( ) 10(1 + sin γ )m ' (1 + sin γ ) + 49m sin γ − V( sin γ ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 ( ) −14m sin γ (1 + sin γ ) + 2(1 + sin γ ) 7m + 5m ' cos2 γ + 10(1 + sin γ )2 sin γ m ' + 49m sin2 γ =V −V ) ( 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 =V −V
−14m sin γ (1 + sin γ ) + 14m(1 + sin γ ) + 10(1 + sin γ )2 (1 − sin γ )m ' + 10(1 + sin γ )2 sin γ m ' + 49m sin2 γ ( ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 =V −V
− 14m sin γ (1 + sin γ ) + 14m(1 + sin γ ) +10(1 + sin γ )2 m ' + 49m sin2 γ ( ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 14m cos2 γ + 10(1 + sin γ )2 m ' + 49m sin2 γ ( ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 ) ( 7m + 5m ' cos2 γ 7 − 14m cos2 γ − 49m sin2 γ =V ( ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2
=V −V
35m ' cos2 γ − 14m cos2 γ + 49m cos2 γ =V( ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 35m ' cos2 γ + 35m cos2 γ =V( ) 7m + 5m ' cos2 γ 7 + 10m ' (1 + sin γ )2 ( ( ) ) ' 2 V m + m' V m + m cos γ = = ' 2 2 ' 7m+5m cos γ 7m sec γ +5m + 27 m ' (1 + sin γ )2 + 27 m ' sec2 γ (1 + sin γ )2 5 5 ( ( ) ) V m + m' V m + m' = )2 = ( ⎛ γ ⎞2 7 m sec2 γ + m ' + 2 m ' 1+sin γ 2 tan 2 cos γ 7 5 ⎜ 1+ 1+tan2 γ ⎟ 7 m sec2 γ + m ' + 2 m ' ⎜ 2 ⎟ ⎟ 7 ⎜ 5 ⎝ 1−tan2 γ2 ⎠ γ 1+tan2 2
6.3 Examples on Impulsive Forces in 2D Motion (Part III)
=
803
( ) V m + m' = .∎ ) 7 m sec2 γ + m ' + 2 m ' tan2 ( π + γ ) γ 2 7 4 2 5 7 m sec2 γ + m ' + 2 m ' 1+tan 2 γ 7 5 1−tan ( ) V m + m' (
2
Suppose that the lower sphere is not spinning. Then 5(F(sin γ +1)−R cos γ ) 7m
a
=
u1 = ω1 = 0, ~ ~~ ~ a
and hence '
γ )+49m sin γ ) ) m ' V (10(1+sin' γ )m2 (1+sin 10(1 + sin γ )m ' (1 + sin γ ) + 49m sin γ cos γ 7m+5m cos γ 7+10m ' (1+sin γ )2 ( ( )) = ) 2m ' V (1+sin γ ) −7m sin γ +(7m+5m ' cos 2 2(1 + sin γ ) −7m sin γ + 7m + 5m ' cos2 γ γ
(
cos γ
1 + sin γ R = = F cos γ ~~ ~ ~
Thus 2(1 + sin γ )2 10(1 + sin γ )2 m ' + 49m sin γ = 7m(1 − sin γ ) + 5m ' cos2 γ cos2 γ or (
10(1 + sin γ )2 m ' +49m sin γ ~~ ~ ~
)
( ) cos2 γ = 2(1 + sin γ )2 7m(1 − sin γ ) + 5m ' cos2 γ ~ ~~ ~
Or 49m sin γ cos2 γ = 14m(1 + sin γ )2 (1 − sin γ ) or 7 sin γ = 2(1 + sin γ ) or sin γ =
2 . 5
Thus, lower sphere will be set spinning in all cases, except when γ = sin−1 25 ∎.
Chapter 7
Motion in Three Dimensions
Angular velocity is a concept very crucial to the theory of motion of a rigid body in three dimensions. Angular velocity is associated not only with a direction but with a fixed straight line in space. Further it can be compounded into three angular velocities. These features make angular velocity a predominant concept in this chapter. For getting a grip over the subject, 24 problems of various difficultylevels have been solved.
7.1 3D Motion of a Rigid Body About a Fixed Point 7.1.1 Note To fix the position of a point in space we need its three coordinates. This fact is expressed as saying: It has three “degrees of freedom”. If one condition is given (for example, a relation between its coordinates so that it lies on a fixed surface), then we say that it has two degrees of freedom and one of “constraint”. If two conditions are given (for example, two relations between its coordinates so that it must be on a curve), then we say that it has one degree of freedom and two of constraint. A rigid body, free to move, has six degrees of freedom. Reason: Its position is completely determined when the position of its three points are given. The nine coordinates of these three points are connected by three relations expressing the invariable lengths of the three points joining them. Hence, in all, the body has 6 degrees of freedom. A rigid body with one point fixed has 6 − 3(= 3) degrees of freedom, and hence 3 of constraint. A rigid body with two of its points fixed, is free to move about an axis. Hence it has one degree of freedom. Reason: The six coordinates of these two points are © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 R. Sinha, Advanced Newtonian Rigid Dynamics, https://doi.org/10.1007/9789819920228_7
805
806
7 Motion in Three Dimensions
equivalent to five constraining conditions, since the distance between the two points is a constant. A rigid body has its position determined when we know the three coordinates of a given point G of it, together with the angles which any two lines, say G A and G B, fixed in the body make with the axes of coordinates. Caution: If G and G A only were known, then the body might revolve ( about )G A. Suppose that (l, m, n) are the direction cosines of line G A, and l , , m , , n , are the direction cosines of line G B. We have. 2 2 1. l(2 + ) m (+ ,n)2 = (1,, )2 , 2 2. l + m + n = 1, 3. ll , + mm , + nn , = cos(∠AG B).
It follows that, as before, six quantities, namely three coordinates and three angles must be known to fix the position of the body.
7.1.2 Note Suppose that a rigid body undergoes uniplanar motion. During any motion, suppose that three points A, B, C fixed in the body move into the positions A, , B , , C , respectively. Let M be the mid point of line segment A A, . Let N be the mid point of line segment B B , . Let us draw perpendicular on A A, at M. Again let us draw perpendicular on B B , at N . Let A A, be not parallel to B B , . It follows that these perpendiculars will intersect. Let O be the point of intersection of these perpendiculars. It follows that O A = O A, ,
O B = O B , , ∠O A A, = ∠O A, A, ∠O B B , = ∠O B , B.
Since A, , B , , C , are the new positions of A, B, C, and A, B, C are points of the rigid body, we have AB = A, B , , ∠C B A = ∠C , B , A, . It follows that △O AB is congruent to △O A, B , , and hence ∠O AB = ∠O A, B , , ∠AB O = ∠A, B , O, ∠B O A = ∠B , O A, . Observe that ∠AO A, = ∠AO B + ∠B O A, = ∠B O A + ∠B O A, = ∠B , O A, + ∠B O A, = ∠B , O A, + ∠A, O B = ∠B , O B = ∠B O B , , so
7.1 3D Motion of a Rigid Body About a Fixed Point
807
∠AO A, = ∠B O B , . Observe that ∠O BC = ∠O B A − ∠ABC = ∠AB O − ∠ABC = ∠A, B , O − ∠ABC = ∠A, B , O − ∠C B A = ∠A, B , O − ∠C , B , A, = ∠A, B , O − ∠A, B , C , = ∠C , B , O = ∠O B , C , . Thus ∠O BC = ∠O B , C , . Next since O B = O B , , and BC = B , C , , we find that △O BC is congruent to △O B , C , , and therefore OC = OC , . Also ∠C O B = ∠C , O B , . Next ∠C OC , = ∠C O B + ∠B OC , = ∠C , O B , + ∠B OC , = ∠B OC , + ∠C , O B , = ∠B O B , = ∠AO A, . Thus ∠AO A, = ∠B O B , = ∠C OC , . It follows that the same rotation about O which brings A to A, and B to B , brings any point C to its new position. Thus O is the required position (Fig. 7.1). Conclusion In uniplanar motion, at any instant there exists an axis of pure rotation, that is, a body can be moved from one position to any other position by a rotation about some point without any translation, provided A A, ∦BB, . Fig. 7.1
′
′
′
808
7 Motion in Three Dimensions
If A A, B B , , then the motion is one of the simple translation. In this case, the corresponding point O is situated at infinity. Since the conclusion is true for all finite displacements, it is true for very small displacements. Hence, a body in uniplanar motion may be moved into the successive positions it occupies, by successive instantaneous rotations about some centres. In order to determine the position of the point O at any instant, let A and A, be successive positions of one point, and B and B , be successive positions of another point of the body. Now erect perpendiculars to A A, and B B , . These perpendiculars meet at O.
7.1.3 Note In the case of a simple pendulum, the centre of rotation and the axis of rotation is “permanent”. In the case of a wheel rolling in a straight line on the ground, the point of contact of the wheel on the ground is, for the instant, the centre of rotation. Hence the centre of rotation is “instantaneous”. The instantaneous centre has two loci according to whether we consider its position with regard to the body, or in space. Thus, in the case of the cartwheel, the successive points of contact are the points on the edge of the wheel. Their locus with regard to the body is the edge itself, that is, a circle whose centre is the centre of the wheel. In space, the points of contact are the successive points on the ground touched by the wheel, that is a straight line on the ground. These two loci are called the bodylocus and spacelocus. Bodylocus is also called bodycentrode, and spacelocus is also called spacecentrode. In the case of a wheel rolling in a straight line on the ground, edge of the wheel is the bodycentrode, and the straight line on the ground is the spacecentrode.
7.1.4 Note Suppose that there is a rigid body in uniplanar motion. Let C1 , , C2 , , C3 , , C4 , , · · · be the successive points of its bodycentrode, and corresponding successive points of its spacecentrode be C1 , C2 , C3 , C4 , · · · . At any instant, C1 , coincides with C1 such that the body is for that instant moving about C1 as centre. When the body has turned through the small angle δθ, the point C2 , coincides with C2 and becomes the new centre of rotation. A rotation about C2 through a small angle brings C3 , to C3 , and then a small rotation about C3 brings C4 , to C4 , etc.
7.1 3D Motion of a Rigid Body About a Fixed Point
809
Bodycentrode
′ 1
′ 4
′ 3
′ 2
′ 5
′ 5 ′ 4
Bodycentrode
′ 3
1
′ 2
2
Spacecentrode 3
4
5
5
4
3
2
′ 1 1
Spacecentrode
Fig. 7.2
In the case of a wheel, the points C1 , , C2 , , C3 , , C4 , , · · · lie on the edge of the wheel, and the points C1 , C2 , C3 , C4 , · · · lie on the ground (Fig. 7.2). Clearly, the motion of the body can be given by the rolling of the bodycentrode, carrying the body with it, upon the spacecentrode.
7.1.5 Example For the sake of reinforcement, we shall reproduce Sect. 7.1.1.9. Consider a rigid rod sliding on a plane with its ends on two perpendicular straight lines C X and X Y. −→ Since the velocity of the material point A is along AC, the instantaneous axis of rotation must intersect the line through A perpendicular to CY. Similarly, the instantaneous axis of rotation must intersect the line through B perpendicular to C X. It follows that instantaneous axis of rotation that passes through the point I (see Fig. 7.3). { x = l cos φ , so x 2 + y 2 = l 2 , and hence the spacecentrode fixed Observe that y = l sin φ in space is a circle Γ whose centre is C and radius is l. Since the rod AB makes an angle π2 at I, the locus of I in the “extended body” of the rod is a circle whose diameter is AB. Let us denote this circle by Γb . Observe that as the rod AB changes its position, Γb changes its position (in space) but its shape (determined by radius 2l ) remains unchanged. So if we imagine that, instead of body AB, our body is AB together with a circular wire Γb , then it is again a “rigid body”. Since I lies on Γb , and I lies on Γ, Γb and Γ intersect at I. Since centre of Γb (that is, mid − point of AB), centre of Γ(that is, C), and I are collinear, circle Γb touches the circle Γ at I. Thus we have shown that at all positions of AB, Γb touches Γ. We shall try to show that Γb rolls on Γ. Observe that A approaches A, as φ → π2 . In the case of rolling without slipping, it suffices to show that
810
7 Motion in Three Dimensions ′
( = cos Γ
,
= sin )
Instantaneous centre of rotation Γ
′
Fig. 7.3
arcA, I = arcAI. LHS = arc A, I = C I × ∠A, C I = AB × ∠A, C I = l × ∠A, C I, l l RHS = arcAI = AO × ∠AO I = × ∠AO I = × (2∠AC I ) 2 2 = l × ∠AC I = l × ∠A, C I. ◼
7.1.6 Example The end A of a rod A P is compelled to move on the yaxis, and the rod itself always passes through the point (b, 0). Clearly, the locus of P passes through the point (b, 0). Show that when the other end P of the rod coincides with (b, 0), the tangent at (b, 0) to the locus of P coincides with the direction of rod (Fig. 7.4). Sol: Here / / b2 + c2 = (length of the rod) = A P = (ξ − 0)2 + (η − (c − h))2 (h is a parameter). , , ,,
7.1 3D Motion of a Rigid Body About a Fixed Point
811
ℎ
Locus of ( , 0)
( , )
Fig. 7.4
Since P(ξ, η) lies on the line joining A(0, c − h) and (b, 0), we have ξ η + = 1. b c−h It follows that η
b2 + c2 = ξ 2 + (η − (c − h))2 , c − h = ,
,,
1−
and hence (
bη b +c =ξ + η− b−ξ 2
2
)2
2
or ( b +c =ξ + 2
2
2
ξη b−ξ
)2 .
So the equation of locus of P(ξ, η) is y=−
x − b/ 2 b + c2 − x 2 . x
ξ b
,
=
bη , b−ξ
812
7 Motion in Three Dimensions
Hence √ dy x − b d b2 + c2 − x 2 b/ 2 2 2 =− 2 b +c −x − . dx x x dx Thus  b/ −c dy  = − 2 b2 + c2 − b2 − 0 = . ◼  d x (b,0) b b
7.1.7 Example The end A of a rod A P is compelled to move on a given straight line CY , while the rod itself always passes through a fixed point B. Obtain the position of the centre of instantaneous motion, and the body and space—centrodes. Sol: Let C be the foot of perpendicular drawn from B on CY. The instantaneous motion of A is along CY, so that the instantaneous centre O lies on the perpendicular AO. By Sect. 7.1.6, the point B of the rod is, for the moment, moving in the direction of AB, so O lies on the perpendicular O B to AB. From Fig. 7.5, the polar equation of bodycentrode is r = (a sec φ) sec φ, that is, r=
2a . 1 + cos 2φ
(7.1)
axis for spacecentrode ( tan ) tan
( sec ) sec Pole for bodycentrode tan
sec
tan axis for spacecentrode Initial line for bodycentrode
Fig. 7.5
7.1 3D Motion of a Rigid Body About a Fixed Point
813
From Fig. 7.5, the equations of spacecentrode is x = a + (a tan φ) tan φ, y = a tan φ. (φ is a parameter) Hence (y) x = a + y tan φ = a + y . , ,, , a Thus, the Cartesian equation of spacecentrode is y 2 = a(x − a).
(7.2)
This is a parabola whose vertex is B(b, 0). Thus, the motion is given by the rolling of the curve (7.1), carrying the rod with it, upon the parabola (7.2). This motion is known as the conchoidal motion.
7.1.8 Example Obtain the position of the centre of instantaneous motion, and the body and space– centrodes in the following cases: 1. A rod AB moves with its ends upon two fixed straight lines not at right angles, 2. A rod AB moves so that its end A describes a circle, of centre O and radius a, while B is compelled to move on a fixed straight line passing through O. (Connecting rod motion.) Compare the velocities of A and B. 3. Two rods AB, B D are hinged at B. AB is hinged to a fixed point A and revolves round A. B D always passes through a small fixed ring at C which is free to rotate about C. (Oscillating cylinder motion.) 4. The middle point G of a rod AB is forced to move on a given circle while at the same time the rod passes through a small ring at a fixed point C of the circle, the ring being free to rotate. Sol: (1) At A and B draw perpendiculars to O X and OY, and let them meet in P. The motions of A and B are instantaneously along AX and B O, so P is the instantaneous centre of rotation (Fig. 7.6). Let us draw a circle Γ whose diameter is O P. Now since ∠O A P = π2 , Γ passes through A. Similarly Γ passes through B. Thus Γ circumscribes quadrilateral O A P B.
814
7 Motion in Three Dimensions
Instantaneous centre
bodycentrode
spacecentrode
Fig. 7.6
From △AB P, a a AB AB AB AP = = = = = sin X OY sin AO B sin AO B sin(π − AO B) ,sin A P B ,, sin AB P, AP AP = = = A P csc ω = O P, sin AO P sin ω so OP =
a . sin X OY
Now since sin Xa OY is a constant, the O P is a constant. Thus the locus of P with regard to space is a circle whose radius is sin Xa OY and centre is O. Thus, the spacecentrode is a circle whose radius is sin Xa OY and centre is O. Observe that ∠A P B = π − ∠AO B = π − ∠X OY Now since ∠X OY is a constant, ∠A P B is independent of P. This shows that ( ( )) the locus of P with regard to the body is a circle whose radius is O2P = 21 sin Xa OY . Thus, the bodycentrode is a circle whose radius is 2 sin aX OY . (2) See Fig. 7.7.
7.1 3D Motion of a Rigid Body About a Fixed Point
Direction fixed in space
815
( + ℎ) cot
(0, + ℎ)
( = ( + ℎ) cot
,
=
+ ℎ)
Instantaneous centre of rotation
ℎ ( cos
, sin )
Direction fixed in space
Fig. 7.7
Here equations of the spacecentrode are x = (a + h) cot φ,
y = a + h, l = ,
/
/ = a 2 + (a + h)2 − 2a sin φ(a + h).
(a cos φ)2 + (a sin φ − (a + h))2 ,, ,
Thus l 2 = a 2 + (a + h)2 − 2a sin φ(a + h) = a 2 + y 2 − 2a sin φy , ,, , ( ) y 2 2 = a + y − 2a / y x 2 + y2 Thus, the Cartesian equation of the spacecentrode is (2 )2 ( ) l − a 2 − y 2 x 2 + y 2 = 4a 2 y 4 .
816
7 Motion in Three Dimensions
Bodycentrode:
Direction fixed in space
Instantaneous centre of rotation
Initial line Direction fixed in body Direction fixed in space
From △B AO, ( ) / (π ) sin π2 − θ l 1 − (sin B I A)2 = cos B I A = sin − B I A = sin B O A = l = cos θ, 2 a , ,, , a
so ( (sin B I A) = 1 − 2
l cos θ a
)2 .
From △B AO, sin(θ + B I A) l ·l (sin θ cos B I A + cos θ sin B I A) = r r sin(π − (θ + B I A)) sin B AI = ·l = · l = sin B I A, r ,, , , r so l sin θ cos B I A = (r − l cos θ ) sin B I A. It follows that ( ) l 2 sin2 θ 1 − sin2 B I A = l 2 sin2 θ cos2 B I A = (r − l cos θ )2 sin2 B I A, ,, , , and hence ( 1−
l cos θ a
)2
l 2 sin2 θ . = sin2 B I A = 2 2 l sin θ + (r − l cos θ )2 ,, , ,
7.1 3D Motion of a Rigid Body About a Fixed Point
817
Initial line Direction fixed in space Direction fixed in body Initial line
(Pole for body − centrode )
(Pole for space − centrode)
Direction fixed in space
Fig. 7.8
Thus the polar equation of the bodycentrode is ( 1−
l cos θ a
)2 =
l 2 sin2 θ l 2 sin2 θ + (r − l cos θ )2
or l2 (r − l cos θ )2 2 cos θ = . a2 l 2 sin2 θ + (r − l cos θ )2 Ratio of velocities of A and B : (Fig. 7.8). From △B AI, l l (π ) · sin AB I r − a = AI = · sin AB I = sin B I A sin − BOI , ,, , 2 (π ) l l l ) · sin AB I = (π = · sin AB I = · sin − OBA cos θ cos θ 2 sin 2 − θ / l l 1 − (sin O B A)2 · cos O B A = = cos θ cos θ / ( ) sin B O A 2 l 1− OA · = cos θ AB
818
7 Motion in Three Dimensions
/ l = cos θ
( ) sin θ 2 1 /2 1− a· = l − a 2 sin2 θ , l cos θ
so the polar equation of the spacecentrode is (r − a) cos θ =
/ l 2 − a 2 sin2 θ .
From △B AI, ) ( sin π2 − φ cos φ R AI sin AB I (π ) = = = = cos θ l AB sin AI B sin − θ , ,, , 2 = = =
(O B cos φ+O A cos O AB)−O A cos O AB OB l+a cos ψ OB
cos θ
cos θ
=
=
AB−O A cos O AB OB
cos θ
l+a cos ψ OB (O B cos θ +AB cos O AB)−AB cos O AB OB
=
=
l−a cos(π −ψ) OB
cos θ
l+a cos ψ OB O A−AB cos O AB OB
l + a cos ψ l + a cos ψ = . a − l cos(π − ψ) a + l cos ψ
Hence, the polar equation of the bodycentrode is R l + a cos ψ = . l a + l cos ψ Let ω be the instantaneous angular velocity of the rod. Hence, (vel.of A) = ω(I A). Similarly, (vel.of B) = ω(I B). Hence, ) ( sin π2 − φ vel.of A sin I B A IA cos φ cos φ = = = = = vel.of B I B sin I AB sin ψ sin ψ sin(θ + φ) ,, , , cos φ 1 = = . sin θ cos φ + cos θ sin φ sin θ + cos θ tan φ
7.1 3D Motion of a Rigid Body About a Fixed Point
819 Instantaneous centre of rotation Direction fixed in space Initial line for spacecentrode
Pole for spacecentrode
Fig. 7.9
Thus vel.of B = sin θ + cos θ tan φ = sin θ + cos θ √ sin φ 2 1−(sin φ) vel.of A ,, , , ( ) sin θ ·a = sin θ + cos θ / l sin θ 2 = sin θ + cos θ √ 2a sin2 θ 2 = sin θ 1 + √ 2a cos2 θ 2 . l −a sin θ l −a sin θ 1−( l ·a ) (3) Fig. 7.9. From triangle ABC, / c c a 2 + c2 − 2ac cos θ = BC = · sin θ = · sin θ sin ,, ABC , , sin(π − I BC) c c c · sin θ = I C · sin θ = I C · sin θ = sin I BC IB r −a =
c(r − a) sin θ c(r − a) sin θ . =√ 2 IC r + c2 − 2r c cos θ
Thus / c(r − a) sin θ a 2 + c2 − 2ac cos θ = √ 2 r + c2 − 2r c cos θ This gives the spacecentrode.
820
7 Motion in Three Dimensions Instantaneous centre of rotation Direction fixed in body Initial line for bodycentrode
Pole for bodycentrode
Fig. 7.10
Bodycentrode: (Fig. 7.10). From right triangle ABC, c2 = a 2 + (BC)2 − 2a(BC) cos(π − φ) = a 2 + (BC)2 + 2a(BC) cos φ , ,, , = a 2 + (R cos φ)2 + 2a(R cos φ) cos φ or /
c2 − a 2 sin2 φ = (R + a) cos φ.
This gives bodycentrode. (4) Fig. 7.11.
7.1.9 Example The arms AC, C B of a wire bent at right angles slide upon two fixed circles in a plane. Show that the locus of the instantaneous centre in space is a circle, and that its locus in the body is a circle of double the radius of the spacecontrode. Sol: Here, O I = C1 C2 , which is a constant, and O is fixed relative to the body, so the bodycentrode is a circle of radius C1 C2 . It is clear that the bodycentrode is a circle of diameter C1 C2 . ◼ (Fig. 7.12).
7.1 3D Motion of a Rigid Body About a Fixed Point
821
Instantaneous centre of rotation
Spacecentrode bodycentrode
Fig. 7.11
Point fixed in body Pole for bodycentrode Initial line for bodycentrode Pole for spacecentrode
Instantaneous centre of rotation
Fig. 7.12
2
1
Spacecentrode
Initial line for spacecentrode
822
7 Motion in Three Dimensions
Fig. 7.13 2
=4
(0, ) (0, ) ( , 0)
7.1.10 Example A straight thin rod moves in any manner in a plane. Show that, at any instant, the direction of motions of all its particles are tangent to a parabola. (Fig. 7.13). Sol: Here, the slope of the tangent to the parabola from R(0, ν) =−
a−0 a = , 0−ν ν
and hence the equation of the tangent to the parabola from R(0, ν) is y−ν =
a (x − 0) ν
or y=
(a ) ν
a x + ( a ) . (∗) ν
touches the parabola y 2 = 4ax. Thus the required parabola is y 2 = 4ax, where (a, 0) is the instantaneous centre of rotation. ◼
7.1.11 Example AB, BC, C D Are three bars connected by joints at B and C, and with the ends A and D fixed, and the bars are capable of motion in one plane. Show that the angular velocities of the rods AB and C D are as B O · DC is to AB · C O, where O is the point of intersection of AB and C D (Fig. 7.14).
7.1 3D Motion of a Rigid Body About a Fixed Point
823
Instantaneous centre of rotation
′
′
Fig. 7.14
Sol: Here, angular velocity of B angular velocity of AB = angular velocity of C D angular velocity of C D =
ω·0B angular velocity of B AB = angular velocity of C angular velocity of C D
=
ω·O B AB ω·OC CD
=
CD · OB . ◼ AB · OC
7.1.12 Note Suppose that there is a planar body moving in its plane. Let G be its centre of mass. Let u, v be the component velocities along x and yaxes of the centre of mass. Let ω be the angular velocity of the body. Let P be any point of the body whose coordinates referred to G be (x, y). Let P G be inclined at θ angle to the axis of x. Then the component velocities of P along x and yaxes are u − yω(= u − (P G sin θ )ω), v + xω(= v − (P G cos θ )ω).
824
7 Motion in Three Dimensions
Fixed in space
Planar shaped moving body
Fixed in space
Fixed in space
Fig. 7.15
If P is the instantaneous centre of rotation, then P is a point of no velocity, and hence u − yω = 0, v + xω = 0. ( ) Thus the coordinates of instantaneous centre of rotation are −v , u relative to G ω ω (Fig. 7.15). Coordinates of centre of no acceleration: Components of accelerations of any point P relative to G {
{ ( )2 −→ −→ (P G)ω2 along P G (P G) θ˙ along P G = −→ −→ (P G)θ¨ perpendicular to P G (P G)ω˙ perpendicular to P G.
Hence components of accelerations of point P relative to O { =
( ) ( ) u˙ − ((P G)ω2 ) cos θ + ((P G)ω) ˙ cos( π2 + θ) along x − axis ˙ sin π2 + θ along y − axis v˙ − (P G)ω2 sin θ + ((P G)ω) { 2 u˙ − (P G cos θ )ω − (P G sin θ )ω˙ along x − axis = v˙ − (P G sin θ )ω2 + (P G cos θ )ω˙ along y − axis { u˙ − (x)ω2 − (y)ω˙ along x − axis = 2 + (x)ω˙ along y − axis v˙ − { (y)ω 2 ˙ along x − axis u˙ − ω x − ωy = v˙ + ωx ˙ − ω2 y along y − axis.
For the coordinates of centre of no acceleration, we have
7.1 3D Motion of a Rigid Body About a Fixed Point
825
{ ω2 x + ωy ˙ − u˙ = 0 . ωx ˙ − ω2 y + v˙ = 0 Thus x=
ω˙ ˙ 2 ˙ v − uω , −ω4 − (ω) ˙ 2
y=
−ω˙ u˙ − v˙ ω2 . −ω4 − (ω) ˙ 2
Hence the coordinates of centre of no acceleration are ) ( 2 ˙ v ω˙ u˙ + v˙ ω2 uω ˙ − ω˙ , . ω4 + (ω) ˙ 2 ω4 + (ω) ˙ 2
7.1.13 Note Suppose that in place of a rigid body we have a lamina. Suppose that the lamina is compelled to move in the x yplane. By Sect. 7.2.2.7, (M.I. of the lamina about Gz axis)θ¨ = , =
lamina
=
lamina
,, Σ Σ lamina
(xY − y X ) − x
Σ (
(xY − y X ) − x M
Y+y d2 y dt 2
Σ ( Σ ) x ,Y − y, X = ((x − x)Y − (y − y)X )
)
Σ
X= (
+y M
, Σ
(
(xY − y X ) − x M
lamina
d2x dt 2
lamina
d2 y dt 2
) +y
Σ
X
)
) d2 y d2x = (moment of the forces about O) − M x 2 − y 2 , dt dt (
so ) ( 2 d2x d y M x 2 − y 2 + Mk 2 θ¨ = (moment of the forces about O), dt dt where k is the radius of gyration about centre of mass. Since (moment of the forces about P(x0 , y0 )) = (moment of the forces about O) ) −→ (Σ Σ +P O × X, Y, 0 , ,, , (Σ Σ ) = (moment of the forces about O) − (x0 , y0 , 0) × X, Y, 0 ( Σ Σ ) = (moment of the forces about O) − x0 Y − y0 X
826
7 Motion in Three Dimensions
( ( ) ( )) d2 y d2x = (moment of the forces about O) − x0 M 2 − y0 M 2 dt dt ( 2 ( ( ( 2 ) 2 ) 2 )) d y d x d y d x = M x 2 − y 2 + Mk 2 θ¨ − x0 M 2 − y0 M 2 dt dt dt dt ( 2 2 ) d y d x ¨ = M (x − x0 ) 2 − (y − y0 ) 2 + Mk 2 θ, dt dt we have (
d2 y d2x (moment of the forces about (x0 , y0 )) = M (x − x0 ) 2 − (y − y0 ) 2 dt dt
) ¨ + Mk 2 θ.
7.1.14 Note Suppose that in place of a rigid body we have a lamina. Suppose that the lamina is compelled to move in the x yplane. Let G be its centre of mass. Let u, v be the component velocities along x and yaxes of G. Let ω be the angular velocity of the body. Let P be any point of the body whose coordinates referred to G be (x, y). Suppose that P is the instantaneous centre of rotation, and L is the moment of the forces about P. By Sect. 7.1.13, ) ( L = M((−x)˙v − (−y)u) ˙ + Mk 2 ω˙ = M y u˙ − x v˙ + k 2 ω˙ , ,, , (( ) ) ( ) u −v 2 =M u˙ − v˙ + k ω˙ , by 7.1.12 ω ω ) ( ) u u˙ + vv˙ M( 2 = k (2ωω) ˙ + (2u u˙ + 2v˙v ) = M k 2 ω˙ + ω 2ω ( (/ )2 ) ) M d( 2 2 M d 2 2 2 2 2 2 = u +v k ω +u +v = k ω + 2ω dt 2ω dt ) M d( 2 2 k ω + ((P G)ω)2 (because P is the instantaneous centre of rotation) = 2ω dt ) ) ) M d (( 2 M d( = k + (P G)2 ω2 = (k1 )2 ω2 , 2ω dt 2ω dt ( / ) where k1 ≡ k 2 + (P G)2 is the radius of gyration about the instantaneous centre P. Thus L=
) M d( (k1 )2 ω2 . (Equation of moment of momentum) 2ω dt
7.1 3D Motion of a Rigid Body About a Fixed Point
827
Case I: when / the instantaneous centre P is fixed in the body. Here P G is a constant, and hence k 2 + (P G)2 (= k1 ) is a constant. Hence L=
d ( 2) M ˙ ω = M(k1 )2 ω. (k1 )2 2ω dt
Case II: when the instantaneous centre P is not fixed in the body. Here k1 is not a constant. Put P G ≡ r So (( 2 ) ) ( ) ) ( M d M k + r 2 ω2 = 2ω 2r r˙ ω2 + k 2 + r 2 2ωω˙ L = 2ω dt) ) ( ( 2 ( ( )) = M r r˙ ω + k + r 2 ω˙ = M(r r˙ ω) + M k 2 + r 2 ω˙ = M(k1 )2 ω˙ + M(r r˙ ω). Thus, the moment of the forces about instantaneous centre rotation in coplanar motion = M(k1 )2 ω˙ + M(r r˙ ω).
7.1.15 Note Suppose that there is a rigid body. We know that its position in space is determined when the position in space of three points of it are known. Suppose that O is a fixed point of the body. Let α, β be the given points of the body. Let us draw a spherical surface whose centre is O. Suppose that the radius from O to α meet the sphere at A, and the radius from O to β meet the sphere at B. Suppose that, when the body has been moved into a second position, A and B go to A, and B , respectively. Let us join the points A, B, A, , B , by great circles. Now since the body is rigid, arc AB = arcA, B , . Let D be the midpoint of arcA A, . Let E be the midpoint of arcB B , . Suppose that the great circles through D and E perpendicular to arcA A, and arcB B , meet in C (Fig. 7.16). It follows that arcC A = arcC A, , arcC B = arcC B , . Fig. 7.16
′ ′
828
7 Motion in Three Dimensions
Next since arc AB = arcA, B , , the spherical triangle ABC is congruent to the spherical triangle A, B , C, and hence , , , , ∠AC A, − ∠BC A, = ,∠AC B = ,,∠A C B, = ∠BC B − ∠BC A
Hence ∠AC A, = ∠BC B , . This shows that the rotation about AC which brings A to A, will bring B to B . Thus, the three points O, A, B have been brought into their second positions O, A, , B , by the same rotation about OC. Hence any other point P will be brought into its second position by the same rotation. Conclusion One point of a rigid body being fixed, the body can be transferred from one position into other position by a rotation about a suitable axis. This result is due to Chasles. Chasles (15.11.1793–18.12.1880) was a great French mathematician. ,
7.1.16 Note Suppose that there is a rigid body. We know that its position in space is determined when the position in space of three points of it are known. Suppose that the three points O, A, B of the body have been brought into their second positions O , , A, , B , . Give the whole body the translation, without any rotation, which brings O to O , . Now O being kept fixed. The same rotation about some axis O , C, which brings A and B into their final positions, will bring any other point of the body into its final position. Conclusion Generally, every displacement of a rigid body is compounded of 1. some motion of translation, wherby every particle has the same translation as any assumed point O, 2. some motion of rotation about some axis passing through O. Clearly, these motions are independent, and hence can take place in any order or simultaneously.
7.1.17 Note Suppose that there is a rigid body. Let O x, O y, Oz be three mutually perpendicular lines. Suppose that during three successive intervals of δt time the body is turned in succession through angles ω1 (δt), ω2 (δt), ω3 (δt) about the axes O x, O y, Oz respectively.
7.1 3D Motion of a Rigid Body About a Fixed Point
829
(Convention: ω1 is taken as positive when its effect is to turn the body in the direction from O y to Oz. ω2 is taken as positive when its effect is to turn the body in the direction from Oz to O x. ω3 is taken as positive when its effect is to turn the body in the direction from O x to O y.) We take δt to be small, so that (δt)2 can be neglected, etc. Let P(x, y, z) be any point of the body. Let M be the foot of the perpendicular drawn from P on the line O x. Let θ be the angle between P M and the plane x O y. It follows that x = O M, y = M P cos θ, z = M P sin θ. Suppose that a rotation ω1 (δt) is made about O x so that P goes to P , . Hence P , (x, y + δy, z + δz), where y + δy = M P cos(θ + ω1 (δt)) = M P(cos θ cos(ω1 (δt)) − sin θ sin(ω1 (δt))) , ,, , ≈ M P(cos θ 1 − sin θ sin(ω1 (δt))) ≈ M P(cos θ − sin θ (ω1 (δt))) = (M P cos θ ) − (M P sin θ )(ω1 (δt)) = y − (M P sin θ )(ω1 (δt)) = y − z(ω1 (δt)), and z + δz = M P sin(θ + ω1 (δt)) = M P(sin θ cos(ω1 (δt)) + cos θ sin(ω1 (δt))) , ,, , ≈ M P(sin θ 1 + cos θ sin(ω1 (δt))) ≈ M P(sin θ + cos θ (ω1 (δt))) = (M P sin θ ) + (M P cos θ )(ω1 (δt)) = z + (M P cos θ )(ω1 (δt)) = z + y(ω1 (δt)). Thus a rotation ω1 (δt) about O x moves the point P(x, y, z) to the point P , (x, y − z(ω1 (δt)), z + y(ω1 (δt))). Similarly a rotation ω2 (δt) about O y moves the point (x, y, z) to the point (x + z(ω2 (δt)), y, z − x(ω2 (δt))). Hence a rotation ω2 (δt) about P , (x, y − z(ω1 (δt)), z + y(ω1 (δt))) to the point
Oy
moves
the
point
P ,, (x + (z + y(ω1 (δt)))(ω2 (δt)), y − z(ω1 (δt)), (z + y(ω1 (δt))) − (x)(ω2 (δt))) ≈ P ,, (x + z(ω2 (δt)), y − z(ω1 (δt)), z + y(ω1 (δt)) − (x)(ω2 (δt))). Similarly a rotation ω3 (δt) about Oz moves the point (x, y, z) to the point (x − y(ω3 (δt)), y + x(ω3 (δt)), z).
830
7 Motion in Three Dimensions
Hence a rotation ω3 (δt) about Oz moves the point P ,, (x + z(ω2 (δt)), y − z(ω1 (δt)), z + y(ω1 (δt)) − (x)(ω2 (δt))) to the point P ,,, ((x + z(ω2 (δt))) − (y − z(ω1 (δt)))(ω3 (δt)), (y − z(ω1 (δt))) +(x + z(ω2 (δt)))(ω3 (δt)), z + y(ω1 (δt)) − (x)(ω2 (δt))) ≈ P ,,, ((x + z(ω2 (δt))) − (y)(ω3 (δt)), (y − z(ω1 (δt))) +(x)(ω3 (δt)), z + y(ω1 (δt)) − (x)(ω2 (δt))) = P ,,, (x + (ω2 z − ω3 y)δt, y + (ω3 x − ω1 z)δt, z + (ω1 y − ω2 x)δt). The symmetry of the final result shows that, if the square of δt be neglected, the rotations about the axes might have been made in any order. This result justifies the following. Definition A body is said to have three angular velocities ω1 , ω2 , ω3 about three perpendicular axes O x, O y, Oz if during three successive intervals of time δt the body is turned in succession through angles ω1 (δt), ω2 (δt), ω3 (δt) about these axes. Conclusion When a body has three instantaneous angular velocities, for indefinitely small rotations, the rotations may be treated as taking place in any order and, therefore as taking place simultaneously. We shall see later that this result does not hold for finite rotations.
7.1.18 Note Suppose that there is a rigid body. Suppose that O X and O X 1 be two lines fixed in space. Suppose that the body undergoes two angular velocities ω1 and ω2 about the axes O X, O X 1 respectively. Let the angular velocity ω1 about the axes O X be −→ represented by vector O A, where A is a point on its axis of rotation O X. Let the −→ angular velocity ω2 about the axes O X 1 be represented by vector O B, where B is a point on its axis of rotation O X 1 . Let us complete the parallelogram O AC B. Let us take a point P on OC. Let M be the foot of perpendicular drawn from P on O X. Let N be the foot of the perpendicular drawn from P on O X 1 (Fig. 7.17). By Sect. 7.1.17, the rotations ω1 (δt) and ω2 (δt) about O A and O B respectively would move P through a small distance perpendicular to the plane of the paper which = (P M(ω1 (δt))) + (−P N (ω2 (δt))) = (P M · ω1 − P N · ω2 )δt = (P M · O A − P N · ω2 )δt ( )) ( 1 1 P M · O A + − P N · O B δt = (P M · O A − P N · O B)δt = 2 2 2 )) ( ( 1 = 2 ar(△O P A) + − P N · O B δt = 2(ar(△O P A) − ar(△O P B))δt 2
7.1 3D Motion of a Rigid Body About a Fixed Point Fig. 7.17
831 1
2
1
= (2 · ar(△O P A) + (−2 · ar(△O P B)))δt (−→ −→ −→ −→) ((−→ −→) −→) = (δt) O A × O P + O B × O P = (δt) O A + O B × O P (−→ −→) = (δt) OC × O P = 0. Thus, the rotations ω1 (δt) and ω2 (δt) about O A and O B respectively would not move P. Now since P in any point of OC, rotations ω1 (δt) and ω2 (δt) about O A and O B respectively would not move any point of OC. Thus, we have shown that OC is the new instantaneous axis of rotation. Suppose that ω is the resultant angular velocity about OC. Hence (ω(δt))(O A sin AOC) = (ω(δt))(per. dist. of A from OC) (rot. ω(δt) about OC would move A a dist.) = (rot. ω1 (δt) and ω2 (δt) = about O A and O B resp. would move A a dist.) , ,, , = (ω1 (δt))(per. dist. of A from O X ) + (ω2 (δt))(per. dist. of A from O B) = (ω1 (δt))(0) + (ω2 (δt))(per. dist. of A from O B) = (ω2 (δt))(O A sin AO B) = ω2 (δt)O A sin(π − AO B) = ω2 (δt)O A sin O AC. Thus (ω(δt))(O A sin AOC) = ω2 (δt)O A sin O AC or resultant angular velocity about OC ω sin O AC OC OC OC . = = = = = AC OB ω2 ω2 ω2 sin AOC ,, , , Thus (resultant angular velocity about OC) = OC.
832
7 Motion in Three Dimensions
Now since OC is the new axis of rotation, the resultant angular velocity is −→ represented by OC. Conclusion Likewise forces or linear velocities, two angular velocities are compounded by parallelogram law, provided their axes of rotation are intersecting lines. Thus, an angular velocity ω about a line O P is equivalent to an angular velocity ω cos x O P about the axis O x, and an angular velocity ω sin x O P about the axis perpendicular to O x and in the plane P O x. Also angular velocities ω1 , ω2 , ω3 about three rectangular axes O x, O y, Oz are equivalent to the angular velocity / (ω1 )2 + (ω2 )2 + (ω3 )2 about a line O P, where P(ω1 , ω2 , ω3 ). In short, angular velocity about a line is a “localized vector” in that line.
7.1.19 Note Suppose that there is a rigid body. Suppose that the body has two angular velocities, ω1 and ω2 , about two parallel axes. We want to find the motion. Let us take any point P of the body. Suppose the plane of the paper is perpendicular to both the axes of rotation, and it passes through P. Let O1 , O2 be the point of intersection of the paper with the axes of rotation. Suppose that the plane of the paper is the x yplane and the positive direction of zaxis is along the outward direction of the paper (Fig. 7.18). The velocities of P are (O1 P)ω1 along P K 1 , and (O2 P)ω2 along P K 2 . Here P K 1 ⊥O1 P, P K 2 ⊥O2 P. Let N be a point on O1 O2 which divides O1 O2 in the ratio ω2 : ω1 . Thus O1 N ω2 = . N O2 ω1 Observe that the net velocity of P Fig. 7.18
2
1 1
2 2
1
7.1 3D Motion of a Rigid Body About a Fixed Point
833
( −−→) ( −−→) = ω1 (0, 0, −1) × O1 P + ω2 (0, 0, −1) × O2 P ( −−→ −−→) = (0, 0, −1) × ω1 O1 P + ω2 O2 P ( (−−→ −→) −−→) = (0, 0, −1) × ω1 O1 N + N P + ω2 O2 P (−−→ −→)) ( (−−→ −→) = (0, 0, −1) × ω1 O1 N + N P + ω2 O2 N + N P (( −−→ −→) ( −−→ −→)) = (0, 0, −1) × ω1 O1 N + ω1 N P + ω2 O2 N + ω2 N P (( −−→ −→) ( −−→ −→)) = (0, 0, −1) × ω2 N O2 + ω1 N P + ω2 O2 N + ω2 N P (( −−→ −→) ( −−→ −→)) = (0, 0, −1) × −ω2 O2 N + ω1 N P + ω2 O2 N + ω2 N P ( −→ −→) = (0, 0, −1) × ω1 N P + ω2 N P −→ = ((ω1 + ω2 )(0, 0, −1)) × N P. This shows that P moves as it would be if it had an angular velocity ω1 + ω2 about N . Conclusion Two angular velocities ω1 and ω2 about two parallel axes through O1 and O2 are equivalent to an angular velocity ω1 + ω2 about an axis which divides the distance O1 O2 in the ratio ω2 : ω1 .
7.1.20 Note In Sect. 7.1.19, suppose that angular velocities are unlike, and ω1 > ω2 . Then N −−→ −−→ divides O1 O2 externally such that ω1 O1 N = ω2 O2 N , and the resultant angular velocity is ω1 − ω2 . (see Fig. 7.19). Exceptional Case: when the angular velocities are unlike, but ω1 = ω2 (≡ ω, say). In this case, N is situated at infinity, and the resultant angular velocity is zero. Then, the resultant motion of P is a linear velocity. Proof In this case the velocities of P are perpendicular to O1 P, O2 P, and proportional to O1 P, O2 P. Here, the resultant velocity of P
2
2 1
Fig. 7.19
1
1
2
834
7 Motion in Three Dimensions
{ = { = {
( ) ( ) ((O1 P)ω) cos( π2 − α ) − ((O2 P)ω) cos( π2 − β) along x − axis ((O1 P)ω) sin π2 − α + ((O2 P)ω) sin π2 − β along y − axis ((O1 P)ω) sin α − ((O2 P)ω) sin β along x − axis ((O1 P)ω) cos α + ((O2 P)ω) cos β along y − axis
ω((O1 P) sin α − (O2 P) sin β) along x − axis ω((O1 P) cos α + (O2 P) cos β) along y − axis ) ) { (( O1 P ω sin sin β sin α − P) sin β along x − axis (O 2 β
= =
ω((O1 P) cos α + (O2 P) cos β) along y − axis ) ) { (( O2 P ω sin α sin β sin α − (O2 P) sin β along x − axis = ω((O1 P) cos α + (O2 P) cos β) along y − axis { 0 along x − axis = ω((O1 P) cos α + (O2 P) cos β) along y − axis { 0 along x − axis = ω(O1 O2 ) along y − axis.
Thus, the resultant velocity of P is equal to ω(O1 O2 ), its direction is perpendicular to O1 O2 . In short, we write: (the resultant velocity of P) = ω · O1 O2 ↓ .
7.1.21 Note Suppose that there is a rigid body. Suppose that the body has an angular velocity ω about an axis. Suppose that L is a line parallel to the axis of rotation. Suppose that a is the distance between axis of rotation and the line L . Let us take any point P of the body. Suppose the plane of the paper is perpendicular to the line and axis of rotation, and it passes through P. Let O1 , O2 be the point of intersection of the paper with the axis of rotation and the line L . Here O1 O2 = a. Observe that, as we change P, all planes through P and perpendicular to the axis of rotation and the line L are parallel. Suppose that the plane of the paper is the x O1 yplane and the positive direction of zaxis is along the outward direction of the paper. Clearly, the directions of O x, O y, Oz are independent of positions of P. The velocity of P is (O1 P)ω along P K . Here P K ⊥ O1 P. Observe that the velocity of P (−−−→ −−→) −−→ = ω(0, 0, −1) × O1 P = ω(0, 0, −1) × O1 O2 + O2 P
7.1 3D Motion of a Rigid Body About a Fixed Point
835
Fig. 7.20
1
2
−−−→ −−→ = ω(0, 0, −1) × O1 O2 + ω(0, 0, −1) × O2 P −−→ = ωa(0, −1, 0) + ω(0, 0, −1) × O2 P. (Fig. 7.20). So, for every point P of the body, (the velocity of P) =
−−→ + ω(0, 0, −1) × O2 P . ωa(0, −1, 0) , ,, , ,, , , constant velocity given by velocity given by linear velocity angular velocity ω · O1 O2 perpendicular to O1 O2 ω about O2
Conclusion An angular velocity ω about an axis is equivalent to an angular velocity ω about a parallel axis distant a from the former together with a linear velocity ωα perpendicular to O1 O2 .
7.1.22 Note In practice, the content of Sect. 7.1.19 can be remembered most easily by taking the point P on O1 O2 (Fig. 7.21). Here ω2 ω1 + ω2 ω1 + ω2 ω1 , = = = N O2 + O1 N O1 O2 NO O N , 2 ,, 1 , so ω2 ω1 + ω2 ω1 = . = N O2 O1 N O1 O2
Fig. 7.21 1 1
2 2
836
7 Motion in Three Dimensions
Observe that the net velocity of P = ω1 · O1 P + ω2 · O2 P = ω1 (O1 N + N P) + ω2 · O2 P = (ω1 · O1 N + ω1 · N P) + ω2 · O2 P = (ω2 · O2 N + ω1 · N P) + ω2 · O2 P = ω2 (O2 N + O2 P) + ω1 · N P = ω2 · N P + ω1 · N P = (ω1 + ω2 )N P.
7.1.23 Note In practice, the content of Sect. 7.1.20 can be remembered most easily by taking the point P on O1 O2 (Fig. 7.22). Here ω2 ω1 − ω2 ω1 − ω2 ω1 = , = = N O2 O1 N N O2 − O1 N O1 O2 ,, , , so ω2 ω1 − ω2 ω1 = . = N O2 O1 N O1 O2 Observe that the net velocity of P = ω1 · O1 P − ω2 · O2 P = ω1 (−O1 N + N P) − ω2 · O2 P = (−ω1 · O1 N + ω1 · N P) − ω2 · O2 P = (−ω2 · O2 N + ω1 · N P) − ω2 · O2 P = −ω2 (O2 N + O2 P) + ω1 · N P = −ω2 · N P + ω1 · N P = (ω1 − ω2 )N P. Exceptional Case: when the angular velocities are unlike, but ω1 = ω2 (≡ ω, say) (Fig. 7.23). Observe that the net velocity of P
1 1
2 2
Fig. 7.22
7.1 3D Motion of a Rigid Body About a Fixed Point
837
(a) 1 2
1
2
(b) 1 2
Fig. 7.23 Fig. 7.24 1
2
= ω · O1 P − ω · O2 P = ω(O1 P − O2 P) = ω · O1 O2 = a constant velocity perpendicular to O1 O2 ↓ .
7.1.24 Note In practice, the content of Sect. 7.1.21 can be remembered most easily by taking the point P on O1 O2 (Fig. 7.24). Observe that the velocity of P = ω · O1 P = ω(O1 O2 + O2 P) = ω · O1 O2 + ω · O2 P. So, it is equivalent to a linear velocity ω · O1 O2 perpendicular to O1 O2 together with an angular velocity ω about O2 .
7.1.25 Note By Sect. 7.1.16, every instantaneous motion of a rigid body is equivalent to a translational velocity of any point O, together with an angular velocity about a straight line passing through O. −→ Suppose that the linear velocity v of the rigid body is along O A, the axis of angular velocity ω is Oz (Fig. 7.25). From the above chain of figures, we have tried to show that the motion is equivalent to a linear velocity vcosθ along Ozaxis, and an angular velocity ω about an axis parallel to Ozaxis and passing through O , .
838
7 Motion in Three Dimensions sin
′
Here, . Put ∠
is equivalent to, by 7.1.21
Vel.=
∙
′
=
sin ′
sin
cos
sin
is equivalent to
′
cos
sin
sin is equivalent to
′
cos
Fig. 7.25
lies in the plane
≡ ,
′
≡
sin
.
7.1 3D Motion of a Rigid Body About a Fixed Point
839
Conclusion The instantaneous motion of a body may be a screw motion, that is, to a linear velocity along a certain line together with an angular velocity about the line. This construction is known as the Poinsot’s central axis theorem. Poinsot (03.01.1777–05.12.1859) was a great French mathematician and physicist. Remark Angular velocity in “rigid dynamics” corresponds to a force in “statics”. Also linear velocity in “rigid dynamics” corresponds to a couple in “statics”.
7.1.26 Note Suppose that O is a point of a rigid body. Let O x, O y, Oz be three mutually perpendicular axes fixed in space. 1. Suppose that the body is rotated through a right angle about O x axis. This would bring any point P on Oz to a position on the negative axis of y. Now a second rotation about O y would not alter its position. 2. Suppose that the body is rotated first through a right angle about O y axis. This would bring P to a position on the axis of x. Now a second rotation about O x would not alter its position. Conclusion If the rotations are through “finite angles”, then the order of rotations about the axes is important. Remark According to Sect. 7.1.17, if the rotations are through “indefinitely small angles”, then the order of rotations about the axes is immaterial.
7.1.27 Note Suppose that there is a rigid body. Suppose that the rotations be through angles 2α and 2β about axes O A and O B are in the directions shown in Fig. 7.26. Fig. 7.26
′
840
7 Motion in Three Dimensions
On the “geometrical sphere with O as centre” let us draw great circle AC such that ∠B AC = α. Here, the direction of AC is taken the same direction as the rotation about O A. Again, let us draw great circle BC such that ∠ABC = β. Here the direction of BC has taken the opposite direction as the rotation about O B. Let us take a point C , on the other side of AB, symmetrical with C, such that ∠C , AC = 2α, and ∠C , BC = 2β. Observe that a rotation of the body through an angle 2α about O A would bring OC , into the position OC, and a second rotation 2β about O B would bring OC back again into the position OC , . Hence, the net effect of the two component rotations would be that the position of OC , is unaltered. This shows that OC , is the resultant axis of rotation. (Similarly, if the rotations had been first about O B and secondly about O A, then OC is the resultant axis of rotation.) Observe that A is unaltered by the first rotation 2α about O A. Suppose that the second rotation 2β about O B brings A to P. It follows that ∠AB P = 2β, arcA P = 2β, arcB A = arcB P, ∠B A P = ∠B P A. Since OC , is the resultant axis of rotation, and the net rotation brings A to P, the magnitude of the net rotation is ∠AC , P(≡ x), and , p ≡ arcC A ,, = arcC , P, . ,
Thus, the rotation x about OC , brings A to P. Put arcB A ≡ γ . Thus, arcB P = γ . We want an expression of ∠AC , P in terms of α, β, γ (Fig. 7.27). Since rotation x about OC , brings A to P, and rotation 2β about O B brings A to P, great circle C , B, when extended will perpendicularly bisect arc A P at N . Also 1 x ∠AC , N = ∠AC , P = . , ,,2 , 2 On using sine formula on the spherical triangle AN B, we get Fig. 7.27
2 ′
2
7.1 3D Motion of a Rigid Body About a Fixed Point
841
sin γ sin AN = sin β = sin β sin γ . sin π2 ,, , , On using sine formula on the spherical triangle AN C , , we get x sin p x sin AN = sin sin p. π = sin 2 2 sin 2 ,, , , Thus sin β sin γ = sin
x sin p, 2
or sin p =
sin β sin γ sin 2x
On using cotangent formula (see Sect. 1.1.39) on the spherical triangle ABC , , we get / cos γ cos α = sin γ cot p − sin α cot β = sin γ ,, , ,
sin2
x 2
− sin2 β sin2 γ
sin β sin γ
and hence /
sin2
x 2
− sin2 β sin2 γ sin β
= cos γ cos α + sin α cot β
or sin2
x − sin2 β sin2 γ = (cos α sin β cos γ + sin α cos β)2 2
or sin
/ x = sin β (cos γ cos α + sin α cot β)2 + sin2 γ . 2
This is the required expression.
− sin α cot β
842
7 Motion in Three Dimensions New position of
New position of
After 1st rotation
(0,0)
(0,2)
After 2nd rotation
(2, −2)
(0, −2)
Fig. 7.28
7.1.28 Example ◦
◦
If a plane figure is rotated through 90 about a fixed point A, and then through 90 ◦ (in the same sense) about a fixed point B, the result is equivalent to a rotation of 180 about a certain fixed point C. Find the position of C. Sol: Suppose that initially the positions of A, B are A(0, 0), B(2, 0) (Fig. 7.28). Since the net transformation of (0, 0) is (2, −2), we have C(1, −1). Verification: The middle point of (0, 0), (2, −2) and the middle point of (2, 0), (0, −2) are the same point C(1, −1).
7.1.29 Example When the rotations are each through two right angles, show that resultant axis of rotation is perpendicular to the plane through the two component axes, and the resultant angle of rotation is equal to twice the angle between them. Sol: By Sect. 7.1.27, the resultant axis of rotation is OC , . Also C , is a pole of arcAB. It follows that resultant axis of rotation OC , is perpendicular to the plane through the two component axes O A and O B (Fig. 7.29). By Sect. 7.1.27, the net rotation x is given by sin
/ x = sin β (cos γ cos α + sin α cot β)2 + sin2 γ , 2
Fig. 7.29
′
−
⁄2 ⁄2
2
⁄2
⁄2
7.1 3D Motion of a Rigid Body About a Fixed Point π , 2
where α =
β=
π , 2
π x sin = sin 2 2 ,
843
γ = ∠AO B. Thus
/(
cos γ cos
π π )2 π + sin2 γ = sin γ , + sin cot 2 2 2 , ,,
and hence x = 2γ = 2 · ∠AO B. ◼
7.1.30 Note Suppose that there is a rigid body that undergoes three instantaneous angular velocities: ω1 about axis of rotation O x, ω2 about axis of rotation O y, ω3 about axis of rotation Oz. Let P(x, y, z) be a point of the body. From Fig. 7.30, we see that the angular velocity ω1 about the axis of rotation O x contributes to P the velocity {
−→ −ω1 z along O y − → ω1 y along Oz.
Similarly, the angular velocity ω2 about the axis of rotation O y contributes to P the velocity {
− → −ω2 x along Oz −→ ω2 z along O x
1
vel. = =( =(
1
1
) cos ( − 2
∙
=(
1
=(
) sin
∙ 1
=
1
∙
=(
( , , )
) cos
∙
=(
) =(
1
1
1
Fig. 7.30
∙
∙
) sin ( − 2 =(
) cos =
)
∙
1
1
) sin
∙
1
1
)
1
∙
)
844
7 Motion in Three Dimensions
and, the angular velocity ω3 about the axis of rotation Oz contributes to P the velocity {
−→ −ω3 y along O x −→ ω3 x along O y.
Thus, if O is in motion, and u, v, w are the components of the velocity parallel to the fixed axes of coordinates, then the velocity of P is ⎧ −→ ⎪ ⎨ u + ω2 z − ω3 y along O x −→ v + ω3 x − ω1 z along O y ⎪ − → ⎩ w + ω1 y − ω2 x along Oz. In short, we write: (v1 , v2 , v3 ) = (u 1 , u 2 , u 3 ) + (ω1 , ω2 , ω3 ) × (x, y, z) or v = u + ω × r.
7.1.31 Note Suppose that there is a rigid body. Let O be a fixed point in space. Let O x, O y, Oz be the rectangular coordinate axes fixed in space. Let the body be moving about O. Suppose that the body undergoes three instantaneous angular velocities: ωx about axis of rotation O x, ω y about axis of rotationO y, ωz about axis of rotation Oz. The moment of momentum of the body about the O xaxis Σ( ( dz ) ( ( dy ))) y m + −z m = dt dt body ) ) ( Σ ( ( Σ ) dy dy dz −z = , by 7.1.30 = m y 0 + ωx y−−ω y x − z m y dt dt dt body body Σ ( ( ) ) = m y ωx y − ω y x − z(0 + ωz x − ωx z) , by 7.1.30 body
=
Σ( ( ( )) ) ωx m y 2 + z 2 −−ω y (mx y) − ωz (mzx) body
= ωx
Σ ( Σ Σ ) m y2 + z2 − ωy mx y − ωz mzx body
body
body
7.1 3D Motion of a Rigid Body About a Fixed Point
= ωx A − ω y
Σ
mx y − ωz
body
Σ
845
mzx, by 3.1.5
body
= ωx A − ω y F − ωz E by 3.1.5 = Aωx − Fω y − Eωz . (C aut i on : Since O x, O y, Oz are fixed in space, and the body moves with respect to them, A, B, C, D, E, F are in general variables.) Thus, (moment of momentum of the body about the O x − axis) = Aωx − Fω y − Eωz . Similarly, (moment of momentum of the body about the O y − axis) = Bω y − Dωz − Fωx , and (moment of momentum of the body about the Oz − axis) = Cωz − Eωx − Dω y . Next, the kinetic energy of the body (( ) ( )2 ( )2 ) Σ 1 dy dz dx 2 + + m 2 dt dt dt body ( ) ( ( )2 ) ( )2 1 Σ dy 2 dz = m 0 + ω y z − ωz y + + 2 dt dt body ( ( )2 ) ( )2 dz 1 Σ 2 = m ω y z − ωz y + (0 + ωz x − ωx z) + 2 dt
=
body
)2 ( )2 ) 1 Σ (( = m ω y z − ωz y + (ωz x − ωx z)2 + 0 + ωx y − ω y x 2 body
( ( )) ( ) ( ( )) ( ( )) 1 Σ( 2 m z 2 + x 2 + (ωz )2 m x 2 + y 2 = (ωx )2 m y 2 + z 2 + ω y 2 body ) −2ω y ωz (myz) − 2ωz ωx (mzx) − 2ωx ω y (mx y) ) ) ) ( ( Σ ( )2 1 Σ 1 Σ 1 Σ ( 2 = (ωx )2 m y2 + z2 + ωy m z 2 + x 2 + (ωz )2 m z + x 2 − ω y ωz myz 2 2 2 body body body body Σ Σ mzx − ωx ω y mx y − ωz ω x body
body
1 = (ωx )2 A 2 ) 1 ) Σ ( 1 ( )2 Σ ( 2 + ωy m z + x 2 + (ωz )2 m z2 + x 2 2 2 body
body
846
7 Motion in Three Dimensions
− ω y ωz
Σ
myz − ωz ωx
body
Σ
mzx − ωx ω y
body
Σ
mx y
body
1 (ωx )2 A 2 Σ Σ Σ 1 1 ( )2 myz − ωz ωx mzx − ωx ω y mx y + ω y B + (ωz )2 C − ω y ωz 2 2 =
1 ( )2 1 = (ωx )2 A + ω y B + 2 2
(
body
body
body
)2 Σ Σ 1 ωz C − ω y ωz D − ωz ω x mzx − ωx ω y mx y 2 body
body
1 ( )2 1 1 = (ωx )2 A + ω y B + (ωz )2 C − ω y ωz D − ωz ωx E − ωx ω y F. 2 2 2
Conclusion (moment of momentum the O x − axis) = Aωx − Fω y − Eωz , etc. ) ( of the body(about )2 (K.E. of the body) = 21 A(ωx )2 + B ω y + C(ωz )2 − 2Dω y ωz − 2Eωz ωx − 2Fωx ω y .
7.1.32 Note Suppose that there is a rigid body. Let O be a point fixed in space. Let the body be moving about O. It follows that we can also treat O as a point in the rigid body. Let O x , , O y , , Oz , be rectangular coordinate axes fixed in body. Now since the body is moving, O x , , O y , , Oz , are in general not fixed in space. Suppose that the body undergoes three instantaneous angular velocities: ω1 about axis of rotation O x , , ω2 about axis of rotationO y , , ω3 about axis of rotation Oz , . Let O x, O y, Oz be the rectangular coordinate axes fixed in space. Suppose that the body undergoes three instantaneous angular velocities: ωx about axis of rotation O x, ω y about axis of rotationO y, ωz about axis of rotation Oz. We choose O x, O y, Oz such that, at the instant under consideration, O x, O y, Oz are coincident with O x , , O y , , Oz , . It follows that, at the instant under consideration, ωx = ω1 , ω y = ω2 , ωz = ω3 . Hence, at the instant under consideration, the expressions for the moments of momentum in Sect. 7.1.31 are (moment of momentum of the body about the O x , − axis) = Aω1 − Fω2 − Eω3 , etc. Also,
7.1 3D Motion of a Rigid Body About a Fixed Point
(K.E. of the body) =
847
) 1( A(ω1 )2 + B(ω2 )2 + C(ω3 )2 − 2Dω2 ω3 − 2Eω3 ω1 − 2Fω1 ω2 , 2
where A, B, C are now the moments of inertia, and D, E, F are products of inertia about the axes fixed in the body and moving with it. If the later axes are the principal axes at O, then D, E, F vanish, and ) ( moment of momentum of the body about the O x , − axis = Aω1 − 0ω2 − 0ω3 = Aω1 , etc. ,, , ,
Also, ) 1( A(ω1 )2 + B(ω2 )2 + C(ω3 )2 − 2(0)ω2 ω3 − 2(0)ω3 ω1 − 2(0)ω1 ω2 (K.E. of the body) = 2 , ,, , ) ( 1 2 2 2 = A(ω1 ) + B(ω2 ) + C(ω3 ) . 2
7.1.33 Note Let ax 2 + by 2 + cz 2 = 1 be any conicoid. We know that its centre is the origin. Let Ax + By + C z = 0 be any plane through the centre. Suppose that its “conjugate diameter” is x y z = = . l m n
(7.3)
Hence Ax + By + C z = 0 is the locus of all chords parallel to (7.3). Suppose that (α, β, γ ) is a point of the locus. Hence, the sum of roots of the quadratic equation a(α + rl)2 + b(β + r m)2 + c(γ + r n)2 = 1 is 0. Thus
848
7 Motion in Three Dimensions
a(2αl) + b(2βm) + c(2γ n) = (the coefficient of r ) = 0 . ,, , , or alα + bmβ + cnγ = 0. Hence, the locus of (α, β, γ ) is alx + bmy + cnz = 0. But Ax + By + C z = 0 is the locus of (α, β, γ ), so B C A = = . al bm cn Now, the “conjugate diameter” y z x = = l m n becomes x y z = = . A/a B/b C/c Conclusion For the central conicoid ax 2 + by 2 + cz 2 = 1, the conjugate diameter to the central plane Ax + By + C z = 0 is x y z = = A/a B/b C/c
7.1 3D Motion of a Rigid Body About a Fixed Point
849
7.1.34 Note Suppose that there is a rigid body. Let O be a point fixed in space. Let O x, O y, Oz be the rectangular coordinate axes fixed in space. Let the body be moving about O. It follows that we can also treat O as a point of the rigid body. Suppose that the body is acted upon by the given blows. Suppose that, just before the action of the blow, ωx is the angular velocity of the body about O x, ω y is the angular velocity of the body about O y, and ωz is the angular velocity of the body about Oz. Suppose that, just after the action of the blow, ωx , is the angular velocity of the body about O x, ω y , is the angular velocity of the body about O y, and ωz , is the angular velocity of the body about Oz. It follows that the net angular velocity of the body, just after ( ) the action of the blow, is in the direction whose direction ratios are ωx , , ω y , , ωz , . By Sect. 7.1.31, just before the action of the blows, (moment of momentum of the body about the O x − axis) = Aωx − Fω y − Eωz , (moment of momentum of the body about the O y − axis) = Bω y − Dωz − Fωx , (moment of momentum of the body about the Oz − axis) = Cωz − Eωx − Dω y . Again, by Sect. 7.1.31, just after the action of the blows, (moment of momentum of the body about the O x − axis) = Aωx, − Fω,y − Eωz, , (moment of momentum of the body about the O y − axis) = Bω,y − Dωz, − Fωx, , (moment of momentum of the body about the Oz − axis) = Cωz, − Eωx, − Dω,y . Hence, (the change in moment of momentum of the body about the O x − axis) ( ) ( ) ( ) = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz , (the change in moment of momentum of the body about the O y − axis) ) ( ) ( ) ( = B ω,y − ω y − D ωz, − ωz − F ωx, − ωx , (the change in moment of momentum of the body about the Oz − axis) ) ( ) ( ) ( = C ωz, − ωz − E ωx, − ωx − D ω,y − ω y .
850
7 Motion in Three Dimensions
If L denotes the moments of the blows about the O xaxis, then ) ( ) ( ) ( L = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz . Similarly, ) ( ) ( ) ( M = B ω,y − ω y − D ωz, − ωz − F ωx, − ωx , and ( ) ( ) ( ) N = C ωz, − ωz − E ωx, − ωx − D ω,y − ω y . If axes are the principal axes at O, then D, E, F vanish, and hence ) ( L = A ωx, − ωx ,
( ) M = B ω,y − ω y ,
( ) N = C ωz, − ωz .
Since, L , M, N denote the moments of the blows about O xaxis, O yaxis, Ozaxis, the “impulsive” couple acts along the direction ratios (L , M, N ). Thus the plane of “impulsive couple” is L x + M y + N z = 0. Since axes are the principal axes at O, D, E, F vanish, and hence the momental ellipsoid at O is Ax 2 + By 2 + C z 2 = k. (a central conicoid) By Sect. 7.1.33, conjugate diameter to the central plane L x + M y + N z = 0 is y z x = = L/(A/k) M/(B/k) N /(C/k) or x y z ( L ) = ( M ) = ( N ). A
B
C
Special Case: When the body starts from rest: In this case, ωx = 0, ω y = 0, ωz = 0. It follows that ωx, =
M N L , ω,y = , ωz, = . A B C
By Sect. 7.1.18, the net angular velocity, the blows, acts in ( just after )(the(action of)) the direction whose direction ratios are ωx, , ω,y , ωz, = LA , MB , CN .
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
851
Conclusion Suppose that a rigid body is fixed at a point O and initially at rest. Then, due to the action of blows, the body begins to turn about the diameter of the momental ellipsoid at O which is conjugate to the plane of the impulsive couple.
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point 7.2.1 Example A lamina is in the form of a quadrant of a circle O H O , whose centre is H. (see Fig. 7.31) Calculate A, B, C, D, E, F. Sol: From symmetry, we can suppose that G(x, a − x) is the centre of mass of the quadrant O H O , . From symmetry, x = y. Let ρ be mass/area of the lamina. Let M be mass of the quadrant lamina. Clearly, )) { x=a ( (√ {x=a(/ 2 − x2 · dx ) a x ρ x=0 4 2 − x2 · dx x ( 2) x= a = π a2 ρ π4a x=0 , ,, , {x=a ) ( 2 )1 ( −2 = a − x 2 2 d a2 − x 2 2 πa x=0 ( ) 3 x=a ) −2 a 2 − x 2 2  4a −4 ( = 0 − a3 = = .  3 2 2 πa 3πa 3π  2 x=0
Next Fig. 7.31 − ̅
̅ ′
̅
852
7 Motion in Three Dimensions
( ) 2 2 B = (M.I. about O y − axis) = 41 (4M)a , = Ma 4 4 ( ( ) ) ) ( 2 ) ( 2 2 2 Ma 2 1 (4M)a 2 C= 4 + M (x) + (a − x) = 2 − M(2ax) + Ma 2 − M (x) + (x) 2 ( ( )) ( ) 2 2 2 4a 8 = 3Ma − M 2a 3π − 8Ma = Ma 2 23 − 3π = 3Ma , 2 2 3π ( ) Ma 2 ( ) 8 8 2 5 2 3 A = C − B = Ma 2 − 3π − 4 = Ma 4 − 3π . Since the zcoordinate of each point of the lamina is 0, we have D = 0, E = 0. Next ) ( ˜ F= (r sin θ )(a − r cos θ ) πMa2 (dr · r dθ ) 4 quadrant ˜ ( 2 ) 4M 3 ar sin θ − r sin θ cos θ dr dθ = π a2 ( 3) ( ) {quadrant π 2 2 2 2 2 a 4M a 4 1 θ = 2 sin 2θdθ = 4Ma + Ma − Ma = 5Ma . 1 − = 4Ma = 4M πa 3 π a 2 4 2 θ =0 3π 4π (−2) 3π 2π 6π
7.2.2 Example A lamina is in the form of a quadrant of a circle O H O , whose centre is H (see Fig. 7.32), has one extremity O of its arc and is struck by a blow P at the other extremity O , perpendicular to its plane. Find its motion. Sol: By Sect. 7.1.34, ( ) ( ) ( ) L = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz , ( ) ( ) ( ) M = B ω,y − ω y − D ωz, − ωz − F ωx, − ωx , ( , ) ( , ) ( , ) N = C ωz − ωz − E ω x − ω x − D ω y − ω y . Here, they take the form: Fig. 7.32
′
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
853
( ) ( ) ( ) ⎫ P.a = A ωx, − 0 − F ω,y − 0 − E ωz, − 0 ⎪ ) ( ) ( )⎬ ( −P.a = B ω,y − 0 − D ωz, − 0 − F ωx, − 0 ) ( ) ( ) ⎪ ( 0 = C ωz, − 0 − E ωx, − 0 − D ω,y − 0 ⎭ or ⎧ , , , ⎪ ⎨ Aωx − Fω y − Eωz = Pa Bω,y − Dωz, − Fωx, = −Pa ⎪ ⎩ Cω, − Eω, − Dω, = 0, z x y where, by Sect. 7.2.1, ( A = Ma
2
D = 0,
) 8 5 − , 4 3π
) ( Ma 2 8 2 3 B= , C = Ma − , 4 2 3π 5Ma 2 F= . 6π
E = 0,
Thus, equations are ⎧ , , ⎪ ⎨ Aωx − Fω y − Pa = 0 , Fωx − Bω,y − Pa = 0 ⎪ ⎩ Cωz, = 0, Thus, ωz, = 0. Also ωx, =
F · Pa − B · Pa −F · Pa + A · Pa , ω,y = . −AB + F 2 −AB + F 2
ωx, =
B−F F−A Pa, ω,y = Pa, ωz, = 0. 2 AB − F AB − F 2
Hence
Let φ be the angle which the instantaneous axis of rotation makes with the yaxis. Then ωx, = ω,y ,, ,
tan φ = , = =
Ma 2 4
B−F AB−F 2 F−A AB−F 2
Pa Pa
=
B−F F−A
5Ma 2 (6π ) 5Ma 2 2 5 − 8 − Ma 6π 4 3π 5 1 5 1 − 6π − 6π 4 4 ( ) = 21 5 8 − 45 + 6π − 45 − 3π 6π
−
854
7 Motion in Three Dimensions
=
1 4
−
5 6π 7 − 45 + 2π
=
3π − 10 10 − 3π = . −15π + 42 15π − 42
7.2.3 Example A uniform cube has its centre fixed, and is free to turn about it. It is struck by a blow along one of its edges. Find the instantaneous axis. Sol: Let M be the mass of the cube (Fig. 7.33). Here, A=B=C =M
2Ma 2 a2 + a2 = , 3 3
D = E = F = 0.
By 7.1.34, ( ) ( ) ( ) L = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz , ( , ) ( , ) ( , ) M = B ω y − ω y − D ωz − ωz − F ω x − ω x , ( , ) ( , ) ( , ) N = C ωz − ωz − E ω x − ω x − D ω y − ω y . Here, they take the form: ( ) ( ) ( ) ⎫ ⎪ 0 = A ωx, − 0 − F ω,y − 0 − E ωz, − 0 ) ( , ) ( , )⎬ ( , −P · a = B ω y − 0 − D ωz − 0 − F ωx − 0 ) ( ) ( ) ⎪ ( P · a = C ωz, − 0 − E ωx, − 0 − D ω,y − 0 ⎭ or Fig. 7.33 instantaneous axis
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
855
⎧ ⎪ ⎨
Aωx, − Fω,y − Eωz, = 0, Bω,y − Dωz, − Fωx, = −P · a, ⎪ ⎩ Cω, − Eω, − Dω, = P · a. z x y where, A=B=C =
2Ma 2 , 3
D = E = F = 0.
Thus, equations are ⎧ ⎪ ⎨ ⎪ ⎩
2Ma 2 , ωx = 0 3 2Ma 2 , ω y = −P · a 3 2Ma 2 , ωz = P · a 3
or ⎧ , ⎨ ωx = 0 ω, = −3P 2Ma ⎩ ,y 3P ωz = 2Ma . Thus, instantaneous axis acts along the line y z x = = . 0 −1 1
7.2.4 Example A uniform solid ellipsoid is fixed at its centre, and is free to turn about it. It is struck at a given point of its surface by a blow whose direction is normal to the ellipsoid. Find the equation to its instantaneous axis. Sol: Let M be the mass of the ellipsoid (Fig. 7.34). Here, A=M
b2 + c2 , 5
By Sect. 7.1.34,
B=M
c2 + a 2 a 2 + b2 , C=M , 5 5
D = E = F = 0.
856
7 Motion in Three Dimensions
Ellipsoid 2 2
2
( , , )
2
2
+
+
2
=1
S. D. Normal at ( , , ) Direction ratios (
2
,
2
, 2)
Fig. 7.34
) ( ) ( ) ( L = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz , ( , ) ( , ) ( , ) M = B ω y − ω y − D ωz − ωz − F ω x − ω x , ( , ) ( , ) ( , ) N = C ωz − ωz − E ω x − ω x − D ω y − ω y , where (L , M, N ) = (ξ, η, ζ ) × (−P(l, m, n)). Here, they take the form: (−P)(ηn − ζ m) = M
) ( ) ( ) b2 + c2 ( , ωx − 0 − 0 ω,y − 0 − 0 ωz, − 0 , etc. 5
Thus ωx, =
5 ) P(ζ m − ηn), etc. ( 2 M b + c2
So, the direction cosines of instantaneous axis are proportional to ζ m − ηn , etc b2 + c2 that is, the direction cosines of instantaneous axis are proportional to ζ
(η)
( ) − η cζ2 , etc b2 + c2
b2
that is, the direction cosines of instantaneous axis are proportional to
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
857
b2 − c2 ( ) ηζ, etc b2 c2 b2 + c2 that is, the direction cosines of instantaneous axis are proportional to ( ) a 2 b2 − c2 ηζ, etc b2 + c2 that is, the direction cosines of instantaneous axis are proportional to ( ) a 2 b2 − c2 1 , etc. b2 + c2 ξ
7.2.5 Example A disc, in the form of a portion of a parabola is bounded by its latus rectum and its axis (see Fig. 7.35). Calculate A, B, C, D, E, F. Let ρ be the mass/area. Here )) √ 2 (√ 3 5 4 4ax 2 , ρ 4ax · d x ( 3 ) = 8a3 ρ a52 = 16ρa 15 2 x=0 )) ( ( x=a 7 √ { √ 4 B = (M.I. about y − axis) = . ρ 4ax · d x x 2 = 2 aρ a72 = 4ρa 7
A = (M.I. about x − axis) =
x=a { (
2
x=0
F=
˜
4
+ C = (M.I. about z − axis) = A + B = 16ρa 15 D =⎛ 0, E = 0,⎞ (ρ · (d x · dy))x · y = ρ
lamina
=
ρ 2
( a2 ·
y=2a {
⎝
y=0 (2a)2 2
−
x=a {
64a 6 6(16a 2 )
y2 4a
)
x=
xd x ⎠ ydy =
=
ρa 4 2
(
2−
2 3
)
4ρa 4 7
ρ 2
=
=
172ρa 4 , 105
( y=2a {
a − 2
y=0 2ρa 4 . 3
Fig. 7.35 2
=4
2
A
( 2 )2 ) y 4a
ydy
858
7 Motion in Three Dimensions
7.2.6 Example A disc, in the form of a portion of a parabola bounded by its latus rectum and its axis, has its vertex A fixed, and is struck by a blow through the end of its latus rectum perpendicular to its plane. Show that the disc starts revolving about a line through A to the axis (Fig. 7.36). inclined at tan−1 14 25 By Sect. 7.2.5, A=
16ρa 4 , 15
B=
4ρa 4 172ρa 4 , C= , 7 105
D = 0,
E = 0,
F=
2ρa 4 . 3
By Sect. 7.1.34, ( ) ( ) ( ) L = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz , ( ) ( ) ( ) M = B ω,y − ω y − D ωz, − ωz − F ωx, − ωx , ) ( , ) ( , ) ( , N = C ωz − ωz − E ω x − ω x − D ω y − ω y , where (L , M, N ) = (a, 2a, 0) × P(0, 0, 1). Here, they take the form: ( ) ( ) ( )⎫ P · 2a = A ωx, − 0 − F ω,y − 0 − 0 ωz, − 0 ⎪ ) ( ) ( )⎬ ( −P · a = B ω,y − 0 − 0 ωz, − 0 − F ωx, − 0 ) ( ) ( ) ⎪ ( ⎭ 0 = C ωz, − 0 − 0 ωx, − 0 − 0 ω,y − 0 or Fig. 7.36 2
=4
2
A
Instantaneous axis
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
859
⎫ P · 2a = Aωx, − Fω,y ⎪ ⎬ −P · a = Bω,y − Fωx, . ⎪ ⎭ ωz, = 0 or ⎫ Aωx, − Fω,y − P · 2a = 0 ⎪ ⎬ Fωx, − Bω,y − P · a = 0 . ⎪ ⎭ ωz, = 0 Hence ωx, =
F Pa − B P2a −F P2a + A Pa , ω,y = . 2 −AB + F −AB + F 2
Thus φ = tan
−1
ω,y ωx,
=
−F P2a+A Pa 2 tan−1 F−AB+F Pa−B P2a −AB+F 2
−4 + 16 3 15 = tan−1 2 8 − 3 7 2 14 tan−1 55 = tan−1 . ◼ 25 7
= tan−1 =
= tan
−1
−2F + A = tan−1 F − 2B
−2
(
2ρa 4 3
2ρa 4 3
)
−2
+ (
16ρa 4 15
4ρa 4 7
)
−4 15 −10 21
7.2.7 Example A uniform triangular lamina ABC is free to turn in any way about A which is fixed. A blow is given to it at B perpendicular to its plane. Show that the lamina begins to turn about AD, where D is a point on BC such that C D = 13 C B (Fig. 7.37). Sol: Here ) x2 + 2x3 y2 + 2y3 , . D 3 3 (
Let M be the mass of the lamina. Let us take the three particles, each of mass sides of ABC, as the system II.
M 3
placed at the midpoints of the
860
7 Motion in Three Dimensions
(fixed) /3 ( 3,
/3
3)
/3
( 2,
2)
Fig. 7.37
By Sect. 3.3.16, the system of the given problem is equimomental to the system II. Thus (
( )( ( )( ) ) ) M ( x2 )2 M M x2 + x3 2 x3 2 + + 3 2 3 2 3 2 ) M( = 2(x2 )2 + 2(x3 )2 + 2x2 x3 12 ) M( = (x2 )2 + (x3 )2 + x2 x3 , 6 ( )( ( )( ) ) ( )( ) M M y2 2 y2 + y3 2 y3 2 M + + A ≡ (M.I. about x − axis) = 3 2 3 2 3 2 ) M( = 2(y2 )2 + 2(y3 )2 + 2y2 y3 12 ) M( = (y2 )2 + (y3 )2 + y2 y3 , 6 ) M( C = A+B = (x2 )2 + (y2 )2 + (x3 )2 + (y3 )2 + x2 x3 + y2 y3 , 6 D = 0, E = 0, ( )( ) ( )( ( )( x2 y2 ) M x2 + x3 y2 + y3 M x3 y3 ) M · + · + · F= 3 2 2 3 2 2 3 2 2 M = (2x2 y2 + 2x3 y3 + x2 y3 + x3 y2 ). 12
B ≡ (M.I. about y − axis) =
By Sect. 7.1.34,
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
861
) ( ) ( ) ( L = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz , ( , ) ( , ) ( , ) M = B ω y − ω y − D ωz − ωz − F ω x − ω x , ( ) ( ) ( ) N = C ωz, − ωz − E ωx, − ωx − D ω,y − ω y , where (L , M, N ) = (x2 , y2 , 0) × P(0, 0, 1). Here, they take the form: ( ) ( ) ( ) ⎫ P · y2 = A ωx, − 0 − F ω,y − 0 − 0 ωz, − 0 ⎪ ( ) ( ) ( )⎬ −P · x2 = B ω,y − 0 − 0 ωz, − 0 − F ωx, − 0 ( ) ( ) ( ) ⎪ ⎭ 0 = C ωz, − 0 − 0 ωx, − 0 − 0 ω,y − 0 or ⎫ ⎪ P y2 = Aωx, − Fω,y ⎬ −P x2 = Bω,y − Fωx, . ⎪ ⎭ ωz, = 0 or ⎫ Aωx, − Fω,y − P y2 = 0 ⎪ ⎬ Fωx, − Bω,y − P x2 = 0 . ⎪ ⎭ ωz, = 0 Hence ωx, =
F P x2 − B P y2 −F P y2 + A P x2 , ω,y = . −AB + F 2 −AB + F 2
It suffices to show that ω,y ωx,
=
y2 +2y3 3 x2 +2x3 3
that is, ( ) M −y2 12 (2x2 y2 + 2x3 y3 + x2 y3 + x3 y2 ) + x2 M6 (y2 )2 + (y3 )2 + y2 y3 y2 + 2y3 ( ) = M x2 + 2x3 x2 12 (2x2 y2 + 2x3 y3 + x2 y3 + x3 y2 ) − y2 M6 (x2 )2 + (x3 )2 + x2 x3 or,
862
7 Motion in Three Dimensions
( ) −y2 (2x2 y2 + 2x3 y3 + x2 y3 + x3 y2 ) + 2x2 (y2 )2 + (y3 )2 + y2 y3 y2 + 2y3 ( ) = 2 2 x2 + 2x3 x2 (2x2 y2 + 2x3 y3 + x2 y3 + x3 y2 ) − 2y2 (x2 ) + (x3 ) + x2 x3 or, ⎛
⎛
⎞
⎞
−⎝2x2 (y2 )2 +2x3 y2 y3 + x2 y2 y3 +x3 (y2 )2 ⎠ + 2⎝x2 (y2 )2 +x2 (y3 )2 + x2 y2 y3 ⎠ , ,, , , ,, , , ,, , , ,, , 1 2 1 2 ⎛ ⎛ ⎞ ⎞ ⎝2(x2 )2 y2 +2x2 x3 y3 + (x2 )2 y3 + x2 x3 y2 ⎠ − 2⎝(x2 )2 y2 +(x3 )2 y2 + x2 x3 y2 ⎠ , ,, , , ,, , , ,, , , ,, , 1
2
1
2
y2 + 2y3 = x2 + 2x3 or, ( ) ( ) − 2x3 y2 y3 + x3 (y2 )2 + 2x2 (y3 )2 + x2 y2 y3 y2 + 2y3 ( ) ( ) = . x2 + 2x3 2x2 x3 y3 + (x2 )2 y3 − 2(x3 )2 y2 + x2 x3 y2 ⎛ ⎞ ⎛ ⎞ −⎝2x3 y2 y3 + x3 (y2 )2 ⎠ + ⎝2x2 (y3 )2 + x2 y2 y3 ⎠ , ,, , , ,, , , ,, , , ,, , 1 2 1 2 ⎞ ⎛ ⎞ LHS = ⎛ ⎝2x2 x3 y3 + (x2 )2 y3 ⎠ − ⎝2(x3 )2 y2 + x2 x3 y2 ⎠ , ,, , , ,, , , ,, , , ,, , 1
2
1
2
2y3 (x2 y3 − x3 y2 ) + y2 (x2 y3 − x3 y2 ) 2y3 + y2 = RHS. ◼ = = 2x3 (x2 y3 − x3 y2 ) + x2 (x2 y3 − x3 y2 ) 2x3 + x2
7.2.8 Note Suppose that there is a moving rigid body. Let G be the centre of mass of the body. Let M be the mass of the body. Let the coordinate axes O x, O y, Oz are fixed in space. Clearly, the position of G can be treated as a point of the body (Fig. 7.38). By Sect. 2.2.5, M
d2x = (sum of components of impressed forces in the direction of x − axis), etc. dt 2
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
863
Fig. 7.38 Moving body at time
̅
At any instant t, let ωx be the angular velocity about the line through G parallel to xaxis, ω y be the angular velocity about the line through G parallel to yaxis, and ωz be the angular velocity about the line through G parallel to zaxis. By 7.1.31, at any instant t, the moment of momentum of the body about the line through G parallel to xaxis = Aωx − Fω y − Eωz . By Sects. 2.2.5 and 2.2.7, of momentum of the body about the line through G parallel to x − axis) = (moment of the impressed forces about the line through G parallel to x − axis).
d dt (moment
So ) d ( Aωx − Fω y − Eωz dt = (moment of the impressed forces about the line through G parallel to x − axis), etc.
In short, the general equations of a rigid body in three dimensions are Σ ⎫ 2 X⎪ M ddt x2 = Σ ⎬ 2 M ddt 2y = Y , Σ ⎪ 2 M ddt 2z = Z⎭ and (
)
d Aωx − Fω y − Eωz dt( ) d Bω y − Dωz − Fωx dt ( ) d Cωz − Eωx − Dω y dt
⎫ =L ⎬ =M . ⎭ =N
864
7 Motion in Three Dimensions
7.2.9 Note Suppose that there is a moving rigid body. Let G be the centre of mass of the body. Let M be the mass of the body. Let the coordinate axes O x, O y, Oz are fixed in space. Suppose that, just before the action of the impulsive forces, u, v, w are the component velocities of G along the coordinate axes, and ωx , ω y , ωz are the component angular velocities about the lines through G along the coordinate axes. Suppose that, just after the action of the impulsive forces, u , , v, , w, are the component velocities of G along the coordinate axes, and ωx, , ω,y , ωz, are the component angular velocities about the lines through G along the coordinate axes. Clearly, Mu , − Mu = X 1 , etc. Also, by Sect. 7.1.34, (moment of impulsive forces about the line through G along x − axis) ( ) ( ) ( ) = A ωx, − ωx − F ω,y − ω y − E ωz, − ωz , etc.
7.2.10 Example A homogeneous billiard ball, spinning about any axis, moves on a billiard table which is not rough enough to always prevent sliding. Show that the path of the centre is at first an arc of a parabola and then a straight line (Fig. 7.39). Sol: −→ Let us take the origin O as the initial position of the point of contact, and O x as the initial direction of sliding. Let u, v, 0 be the initial velocities of centre of mass G along the coordinate axes. Let Ωx , Ω y , Ωz be the initial angular velocities of the ball about G X, GY, G Z . Here, the initial direction of sliding is along xaxis, so at t = 0, we have (Fig. 7.40) ( ( ) ) (· · · , 0, 0) = (vel. of O relative to G) = − (u, v, 0) + Ωx , Ω y , Ωz × (0, 0, −a) , ,, , ) ( = −u + Ω y a, −v − Ωx a, 0 , so −v − Ωx a = 0.
(7.4)
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
865
(axis fixed in space) Ω
Ω Initially vel. =
Ω
vel. =
(axis fixed in space) ,
(axis fixed in space)
Fig. 7.39
ω
ω
At time ω ( ̅, ̅ , )
,
( ̅ , ̅ , 0)
Fig. 7.40
Here,
866
7 Motion in Three Dimensions
A=B=C =M
2a 2 , 5
D = E = F = 0.
By Sect. 7.2.7, the equations of the ball are Σ ⎫ 2 X⎪ M ddt x2 = Σ ⎬ d2 y M dt 2 = Y , Σ ⎪ 2 M ddt 2z = Z⎭ and (
)
d Aωx − Fω y − Eωz dt( ) d Bω y − Dωz − Fωx dt ( ) d Cωz − Eωx − Dω y dt
⎫ =L ⎬ =M . ⎭ =N
They take the form: ⎫ 2 M ddt x2 = −Fx ⎪ ⎪ ⎪ 2 ⎬ M ddt 2y = −Fy , 2 d (a) ⎪ ⎪ = R − Mg 0 = M0 = M ⎪ ⎭ 2 , , dt ,,
(
)
⎫
2 d M 2a5 ωx − 0ω y − 0ωz = −Fy · a ⎪ ⎪ ⎪ dt ( ⎬ ) 2 2a d ω − 0ω − 0ω · a M = F y z x x dt ⎪ ( 5 2 ) ⎪ ⎪ d 2a ⎭ ω − 0ω − 0ω M = 0 z x y dt 5
or, ⎫ 2 M ddt x2 = −Fx ⎪ ⎬ 2 M ddt 2y = −Fy , ⎪ R = Mg ⎭
⎫ dωx = −Fy ⎬ M 2a 5 dt dω y = Fx . M 2a 5 dt ⎭ dωz = 0 dt
) ( Since the resultant friction −Fx , −Fy , 0 is opposite to the velocity at A, and the velocity at A(x, y, 0) is ) ) ( ˙ d(a) + ωx , ω y , ωz × (x − x, y − y, 0 − a) ˙ y, x, dt ( ( ) ( ) ( )) ˙ ˙ = x, y, 0 + ωx , ω y , ωz × (0, 0, −a) = x˙ − ω y a, y˙ + ωx a, 0 , (
we have −Fy −Fx = , ˙x − ω y a ˙y + ωx a and hence
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point d2 y dt 2 d2 x dt 2
=
867
Fy y˙ + ωx a . = Fx x˙ − ω y a ,, , ,
Also / ( )2 ( ) (Fx )2 + Fy =  −Fx , −Fy , 0  = μ · R = μ · Mg. ,, , , Thus ( )2 (Fx )2 + Fy = μ2 M 2 g 2 . Since dω y d2 x = 2a dt 2 5 dt 2 d y dωx = − 2a 5 dt dt 2
} ,
we have y˙ + ωx a = x˙ − ω y a
d2 y dt 2 d2 x dt 2
,
=
x − dω dt
,,
dω y dt
=
,
d2 y dt 2 d2 x dt 2
x + a dω dt
−a
dω y dt
(by the law of ratio and proportion).
Hence ˙ d ( y+ω x a) y˙ + ωx a = ˙ dt = d (x−ω y a ) x˙ − ω y a dt ,, , ,
( ) d y˙ + ωx a ( ). d x˙ − ω y a
Thus ( ) ) ( d y˙ + ωx a d x˙ − ω y a . = y˙ + ωx a x˙ − ω y a It follows that ( ) ( ) ln x˙ − ω y a = ln y˙ + ωx a + C. Hence,  ( ) (  ) (  ) ln u − Ω y a = ln x˙ t=0 − ω y t=0 a = ln y˙ t=0 + ωx t=0 a + C = ln(v + Ωx a) + C ,, , , ) ( )) ( ( ˙ (v+Ωx a)(x−ω y a) ˙ ˙ = ln(v + Ωx a) + ln x − ω y a − ln y + ωx a = ln . ˙ ( y+ω x a)
868
7 Motion in Three Dimensions
Thus ( ) (v + Ωx a) x˙ − ω y a ( ) ln u − Ω y a = ln y˙ + ωx a (
)
or Fy y˙ + ωx a 0 (v + Ωx a) )=( ) (by(7.1)). = =( ˙ Fx u − Ωy a u − Ωy a x − ωy a , ,, , Thus Fy = 0. Next since ( )2 (Fx )2 + Fy = μ2 M 2 g 2 , we have Fx = μMg. Now since } 2 M ddt x2 = −Fx , 2 M ddt 2y = −Fy we have d2 x = −μg dt 2 2 d y =0 dt 2
} .
It follows that x˙ = u − μgt,
1 y˙ = v, x = ut − μgt 2 , 2
y = vt. (∗)
Thus x =u
( )2 y y 1 − μg v 2 v
This shows that the path of the centre G(x, y, a) is a parabola whose axis lies along the negative direction of the xaxis. ◼.
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
869
Next ⎫ dωx M 2a = −Fy ⎬ 5 dt dω y M 2a = Fx 5 dt ⎭ dωz = 0 dt becomes ⎫ dωx =0 ⎬ M 2a 5 dt dω y = μMg M 2a 5 dt ⎭ dωz = 0 dt or ⎫ dωx =0 ⎬ dt dω y 5 = 2a μg . dt ⎭ dωz =0 dt It follows that ωx = Ωx , ω y = Ω y +
5 μgt, ωz = Ωz . 2a
The velocity of point of contact at time t is ) ( ( ) ˙ d(a) + ωx , ω y , ωz × (x − x, y − y, 0 − a) ˙ y, = x, dt ) ( ) ( ) ( ˙ ˙ = x, y, 0 + ωx , ω y , ωz × (0, 0, −a) = x˙ − ω y a, y˙ + ωx a, 0 ) ) ( ( 5 μgt a, v + Ωx a, 0 = (u − μgt) − Ω y + 2a ) ) ( ( ) ( 7 μg t, 0, 0 . = u − Ωy a − 2 It vanishes at the time ) ( 2 u − Ωy a . t= 7μg After that rolling commences: Let us take the origin O as the initial position of the point of contact as rolling −→ commences, and O x as the initial direction of rolling as rolling commences. Now the initial velocities of centre of mass G along the coordinate axes are (0, 0, 0), and the initial angular velocities of the ball about G X, GY, G Z are
870
7 Motion in Three Dimensions
( ( )) 5 2 u − Ωy a Ωx , Ω y + , Ωz 2a 7 that is, Ωx ,
2 5u Ωy + , Ωz . 7 7a
So in the previous results, we can take 0 for u, 0 for v, and 27 Ω y + 5u for Ω y . 7a From (∗), y(= 0t = 0) is zero, so path of the particle is a straight line. ◼
7.2.11 Example If a homogeneous sphere rolls on a fixed rough plane under the action of any forces, whose resultant passes through the centre of the sphere, show that the motion is the same as if the plane were smooth and the forces reduced to fivesevenths of their given value (Fig. 7.41). Sol: ( ) Let X , , Y , , Z , be the resultant force passing through G. Here, A=B=C =M
2a 2 , 5
D = E = F = 0.
By Sect. 7.2.7, the equations of the sphere are Fig. 7.41
ω
ω ′
At time
′
ω
( ̅, ̅ , ) ′
,
( ̅ , ̅ , 0)
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
871
Σ ⎫ 2 M ddt x2 = X⎪ Σ ⎬ d2 y M dt 2 = Y , Σ ⎪ 2 M ddt 2z = Z⎭ and (
)
d Aωx − Fω y − Eωz dt( ) d Bω y − Dωz − Fωx dt ( ) d Cωz − Eωx − Dω y dt
⎫ =L ⎬ =M . ⎭ =N
They take the form: ⎫ 2 M ddt x2 = X , − Fx ⎪ ⎪ ⎪ d2 y ⎬ , M dt 2 = Y − Fy , 2 d (a) ⎪ , ⎪ 0 = M0 = M = Z + R ⎪ ⎭ 2 , dt ,, ,
(
)
⎫
2 d M 2a5 ωx − 0ω y − 0ωz = −Fy · a ⎪ ⎪ ⎪ dt ( ⎬ ) d 2a 2 ω − 0ω − 0ω · a M = F y z x x dt 5 ⎪ ( ) ⎪ ⎪ d 2a 2 ⎭ ω − 0ω − 0ω M = 0 z x y dt 5
or, ⎫ 2 M ddt x2 = X , − Fx ⎪ ⎬ 2 M ddt 2y = Y , − Fy , ⎪ ⎭ R = −Z ,
⎫ dωx = −Fy ⎬ M 2a 5 dt dω y = Fx . M 2a 5 dt ⎭ dωz = 0 dt
Since the sphere rolls, the velocity at A(x, y, 0) is zero. Now since the velocity at A(x, y, 0) is (
) ) ( ˙x, y, ˙ d(a) + ωx , ω y , ωz × (x − x, y − y, 0 − a) dt ( ( ) ( ) ( )) ˙ y, ˙ 0 + ωx , ω y , ωz × (0, 0, −a) = x˙ − ω y a, y˙ + ωx a, 0 , = x, we have x˙ = ω y a,
y˙ = −ωx a.
Since ( ( ) ) d2x 2 d aω y 2a dω y 2 d x˙ 2 d2x , , , M 2 =X −M , =X −M = X, − M =X −M 5 dt , 5 dt 5 dt 5 dt 2 , dt ,, we have M
d2x 2 d2x , = X − M , dt 2 5 dt 2
872
7 Motion in Three Dimensions
or M
5 d2x = X ,. 2 dt 7
Since ( ) 2 d(aωx ) 2 d − y˙ 2 d2 y 2a dωx d2 y , , , =Y +M = Y, − M M 2 =Y +M =Y +M 5 dt 5 dt 5 dt 2 5 dt , , dt ,, we have M
2 d2 y d2 y , = Y − M dt 2 5 dt 2
or M
5 d2 y = Y ,. ◼ 2 dt 7
7.2.12 Example A sphere is projected obliquely up a perfectly rough plane. Show that the equation of the path of the point of contact of the sphere and the plane is y = x tan β −
5 gx 2 sin α , 14 V 2 cos2 β
where α is the inclination of the plane to the horizon, and V is the initial velocity at an angle β to the horizontal line in the plane. (Fig. 7.42) Sol: Here, the following 3 external forces are acting: 1. Normal reaction R(0, 0, 1) at the point (x, y, 0), sin α, cos α) at the point (x, y, a) 2. Gravitational pull −Mg(0, ( ) 3. Friction −Fx , −Fy , 0 at the point (x, y, 0). Here, the net force ) ( = R(0, 0, 1) − Mg(0, sin α, cos α) + −Fx , −Fy , 0 ) ( = −Fx , −Mg sin α − Fy , R − Mg cos α , And the net torque about G(x, y, a)
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
873
ω ω ,
Line of greatest slope ω ( ̅ , ̅ , 0)
Horizontal line
Fig. 7.42
= ((x, y, 0) − (x, y, a)) × R(0, 0, 1) + ((x, y, 0) − (x, y, a)) ( ) ( ) ( ) × −Fx , −Fy , 0 = (0, 0, −a) × −Fx , −Fy , R = a −Fy , Fx , 0 . Also A=B=C =M
2a 2 , 5
D = E = F = 0.
So equations of motion are d 2 (a) = R − Mg cos α , M y¨ = −Mg sin α − Fy , 0 = M 2 , dt , ,, ( ) ( ( ) ) ) ( 2a 2 d ω y 2a 2 d(ωz ) 2a 2 d(ωx ) M M = −Fy · a, = Fx · a, = 0. M 5 dt 5 dt 5 dt M x¨ = −Fx ,
Thus M x¨ = −Fx , M y¨ = −Mg sin α − Fy , ( ) ( ) 2a d ω y M = Fx . 5 dt
(
) 2a d(ωx ) M = −Fy , 5 dt
It follows that 2a x¨ = − 5
( ) d ωy , dt
2a d(ωx ) y¨ = −g sin α + . 5 dt
Since the sphere rolls, the velocity at A(x, y, 0) is zero. Now since the velocity at A(x, y, 0) is
874
7 Motion in Three Dimensions
(
) ( ) ˙ y, ˙ d(a) + ωx , ω y , ωz × (x − x, y − y, 0 − a) x, dt ) ( ) ( )) ( ( ˙ ˙ = x, y, 0 + ωx , ω y , ωz × (0, 0, −a) = x˙ − ω y a, y˙ + ωx a, 0 ,
we have x˙ = ω y a,
y˙ = −ωx a.
It follows that ( ) x¨ = −
x˙ a
2a d
( ,
y¨ = −g sin α +
x¨ = 0,
5 y¨ = − g sin α. 7
5
dt
2a d 5
− y˙ a
)
dt
and hence
It follows that x = (V cos β)t,
y = (V sin β)t −
( ) 1 5 g sin α t 2 . 2 7
Hence y = (V sin β)
( )( )2 1 5 x x − g sin α V cos β 2 7 V cos β
or y = tan β x −
5 g sin α (x)2 . ◼ 14 V 2 cos2 β
7.2.13 Example A homogeneous sphere is projected, so as to roll, in any direction along the surface of a rough plane inclined at α to the horizontal. Show that the coefficient of friction must be > 27 tan α. Sol: Here, the following 3 external forces are acting: 1. Normal reaction R(0, 0, 1) at the point (x, y, 0), 2. Gravitational pull −Mg(0, sin α, cos α) at the point (x, y, a)
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
875
) ( 3. Friction −Fx , −Fy , 0 at the point (x, y, 0). Here, the net force ) ( = R(0, 0, 1) − Mg(0, sin α, cos α) + −Fx , −Fy , 0 ) ( = −Fx , −Mg sin α − Fy , R − Mg cos α , and the net torque about G(x, y, a) ( ) = ((x, y, 0) − (x, y, a)) × R(0, 0, 1) + ((x, y, 0) − (x, y, a)) × −Fx , −Fy , 0 ) ( ) ( = (0, 0, −a) × −Fx , −Fy , R = a −Fy , Fx , 0 . See Fig. 7.43. Also A=B=C =M
2a 2 , 5
D = E = F = 0.
So, equations of motion are M x¨ = −Fx , (
M 2a5
2
)
d(ωx ) dt
d 2 (a) = R − Mg cos α , M y¨ = −Mg sin α − Fy , 0 = M dt 2 , , ( ( ,,2 ) ) 2 d (ω y ) d(ωz ) 2a = −Fy · a, M 2a5 = F · a, M = 0. x 5 dt dt
Thus
ω ω ,
Line of greatest slope ω
Horizontal line
Fig. 7.43
( ̅ , ̅ , 0)
876
7 Motion in Three Dimensions M x¨ = −Fx ,
M y¨ = −Mg sin α − Fy ,
( M
2a 5
)
(
d(ωx ) = −Fy , dt
M
2a 5
) ( ) d ωy = Fx . dt
It follows that 2a x¨ = − 5
( ) d ωy , dt
2a d(ωx ) y¨ = −g sin α + . 5 dt
Since the sphere rolls, the velocity at A(x, y, 0) is zero. Now since the velocity at A(x, y, 0) is (
) ) ( ˙ y, ˙ d(a) + ωx , ω y , ωz × (x − x, y − y, 0 − a) x, dt ( ( ) ( ) ( )) ˙ ˙ = x, y, 0 + ωx , ω y , ωz × (0, 0, −a) = x˙ − ω y a, y˙ + ωx a, 0 , we have x˙ = ω y a,
y˙ = −ωx a.
It follows that (
( ) x¨ = −
x˙ a
2a d
dt
5
,
y¨ = −g sin α +
2a d
− y˙ a
)
dt
5
and hence −Fx = 0,, = ,x¨ ,, M
−Mg sin α − Fy 5 = y¨ = − g sin α . M 7 ,, , ,
Thus Fx = 0,
2 Fy = − Mg sin α. 7
Hence /
( )2 (Fx )2 + Fy R
/ =
( )2 (0)2 + − 27 Mg sin α Mg cos α
=
2 tan α. ◼ 7
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
877
7.2.14 Example A perfectly rough sphere, of mass M and radius a, is rotating with angular velocity Ω about an axis at right angles to the direction of motion of its centre. It impinges directly on another rough sphere of mass m which is at rest. Show that after separation the component velocities of the two spheres at right angles to the original direction of motion of the first sphere are respectively 2 M 2 m aΩ and aΩ. 7m+M 7m+M Sol: For the first sphere, suppose that, just after the action of the impulsive forces, u , , v, , w, are the component velocities of G along the coordinate axes, and ωx , , ω y , , ωz , are the component angular velocities about the lines through G along the coordinate axes. For the second sphere, suppose that, just after the action of the impulsive forces, u ,, , v,, , w,, are the component velocities of G 1 along the coordinate axes, and ωx ,, , ω y ,, , ωz ,, are the component angular velocities about the lines through G 1 along the coordinate axes (Fig. 7.44). By Sect. 7.2.8, equations of impulsive motion of the first sphere, of mass M, is , , Mw, − M0 Mu ( =2 0, ) − Mu )= −R, ( Mv 2− )(M0 = −F, )( ) ) 2 ( 2a 2a , , 0 = M 5 ωx − 0 , 0 = M 5 ω y − 0 , −F · a = M 2a5 ωz, − Ω .
(
Thus,
Mv, = −F, w, = 0,
(
) ) 2a ( , ωz − Ω , 5 , Mu − Mu = −R.
ωx, = 0, ω,y = 0, −F =
M
′
Ω
′′
Second sphere
First sphere 1
Fig. 7.2.13 ,
, ′′ ′
Fig. 7.44
878
7 Motion in Three Dimensions
By Sect. 7.2.8, equations of impulsive motion of the second sphere, of mass m, is (
0 = m 2b5
2
,, = R, ( mv,, ) − m0 = F, mw,, − m0(= 0, ) )(mu − m0 ) ) ) 2 ( 2 ( ωx ,, − 0 , 0 = m 2b5 ω y ,, − 0 , −F · b = m 2b5 ωz ,, − 0 .
Thus ( ) 2b ωz ,, , mv,, = F, w,, = 0, mu ,, = −R. ωx ,, = 0, ω y ,, = 0, −F = m 5 It suffices to show that −v, =
2 M 2 m aΩ, v,, = aΩ, 7m+M 7m+M
that is, F 2 m = aΩ, M 7m+M
F 2 M = aΩ. m 7m+M
It suffices to show that F=
2 mM aΩ. 7m+M
From question, the first sphere impinges directly on the second sphere. So (y − comp. of vel. of A rel. to the centre of first sphere) = (y − comp. of vel. of A rel. to the centre of second sphere), and hence −7F −F + Ωa = + 2M M
(
) ) ( , + ω, · a = v ,, − ω ,, · b = F − −F b = 7F . + Ω a = v z z 2m , ,, , m M 2a m 2b 5 5 −F
Thus −7F 7F + Ωa = 2M 2m or F=
2 mM aΩ. ◼ 7m+M
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
879
7.2.15 Example A homogeneous sphere spinning about its vertical axis moves on a smooth horizontal table and impinges directly on a perfectly rough vertical cushion. Show that the kinetic energy of the sphere is diminished by the impact in the ratio ( ) ( ) 2e2 5 + 7 tan2 θ : 10 + 49e2 tan2 θ , where e is the coefficient of restitution of the ball and θ is the angle of reflection (Fig. 7.45). Sol: By Sect. 7.2.8, equations of impulsive motion of the sphere, of mass M, is ( ) ( ) M) v, sin θ − M(0)(= F, ) M −v, cos θ − M(v)(= −R,) ) ) ) ( ( 2 2 2 ( 0 = M 2a5 ωx, − 0 , 0 = M 2a5 ω,y − 0 , −F · a = M 2a5 ωz, − ω . (
Thus, ) ( ) ( M v, sin θ = F, M v, cos θ + v = R, ωx, = 0, ω,y = 0, ( ) ( , ) ) 2a ( M v sin θ = F = M ω − ωz, . 5 ,, , , Also, v, cos θ = ev, v, sin θ = a.ωz, (because, A is at rest). Since Fig. 7.45
′
,
cos
′ ′
sin
880
7 Motion in Three Dimensions
( ) ) ) 2a 2a ( v, sin θ 2( v, sin θ = ω − ωz, = ω− = aω − v, sin θ , 5 5 a 5 , ,, , we have ) ( , 7 a.ωz, = 7v θ = 2aω, . , sin ,, Thus, 2ω . 7
ωz, = Next since ,
v sin θ = ,, ,
) 2ω , =a 7 , (
a.ωz,
we have v, =
2aω . 7 sin θ
Since v, cos θ v= = , ,, e ,
(
2aω 7 sin θ
)
cos θ
e
=
2aω cot θ , 7e
we have v=
2aω cot θ . 7e
Now ( , )2 1 ( 2a 2 )( , )2 v + 2 M 5 ωz final kinetic energy ( ) = 1 initial kinetic energy 2 + 1 M 2a 2 ω2 Mv 2 2 5 ( , )2 ( 2a 2 )( , )2 ( 2aω )2 ( 2a 2 )( 2ω )2 + 5 v + 5 ωz 7 sin θ 7 ( 2) ( 2) = = ( ) 2 2aω cot θ v2 + 2a5 ω2 + 2a5 ω2 7e ( ) ( 2 )2 ( 2 )( 2 )2 4 8 1 + cot 2 θ + 5×49 + 5 7 49 7 sin θ = = ( 2 cot θ )2 2 4 cot 2 θ + 25 + 49e2 1 M 2
7e
5
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
=
) ( 4 1 + cot 2 θ + 4 e2
( 2 =
5
cot 2 θ +
2×49 5
7 + 5 cot 2 θ
2 cot 2 e2 ( 2
θ+
8 5
)
49 5
=
=
881
) ( 2 1 + cot 2 θ +
2 cot 2 θ + e2 ( ) 2 7 tan2 θ + 5 5 2 e2
) 2e 7 tan2 θ + 5 = . ◼ 10 + 49e2 tan2 θ
+
4 5
49 5
49 tan2 θ 5
7.2.16 Example A sphere, of radius a, rotating with angular velocity ω about an axis inclined at an angle β to the vertical, and moving in the vertical plane containing that axis with velocity u in a direction making an angle α with the horizon, strikes a perfectly rough horizontal plane. Find the resulting motion, and show that the vertical plane containing the new direction of motion makes an angle tan
−1
(
2aω sin β 5u cos α
)
with the original plane (Fig. 7.46). Sol: By Sect. 7.2.8, equations of impulsive motion of the sphere is Fig. 7.46 cos
, sin cos
sin
882
7 Motion in Three Dimensions
Mvx, − M(0) = −Fx ,
Mv,y − M(u cos α) = −Fy ,
Mvz, − M(u sin α) = R, ( ) ) 2a 2 ( , − Fy · a = M ωx − 0 , 5 ( ) ) 2a 2 ( , 0= M ωz − ω cos β . 5
( Fx · a =
M
) ) 2a 2 ( , ω y − ω sin β , 5
Thus Mvx, = −Fx , Mv,y − M(u cos α) = −Fy , Mvz, − M(u sin α) = R, ) , )( , ( ) ( −Fy = M 2a ωx , Fx = M 2a ω y − ω sin β , ωz, = ω cos β. 5 5 Also, vx, = aω,y , −v,y = a.ωx, (because, A is at rest). Since ) ( −M aω,y = −Mvx, = Fx = ,
( M
) ) 2a ( , ω y − ω sin β , 5 ,, ,
we have −ω,y =
) 2( , ω − ω sin β 5 y
or vx, 2 = ω,y = ω sin β . a 7 , ,, , Thus vx, =
2a ω sin β. 7
Since ( ) ( ) 2a M −a.ωx, − M(u cos α) = Mv ,y − M(u cos α) = −Fy = M ωx, 5 , ,, , we have −a.ωx, − u cos α =
2a , ω 5 x
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
883
or −v,y a
−5 = ωx, = u cos α . 7a , ,, ,
Thus v,y =
5 u cos α. 7
It suffices to show that ( )) ( v, 2aω sin β tan tan−1 = x, = 5u cos α vy ,, , ,
2a ω sin β 7 5 u cos α 7
=
2aω sin β . ◼ 5u cos α
7.2.17 Example A ball, moving horizontally with velocity u and spinning about a vertical axis with angular velocity ω, impinges directly on an equall ball at rest. Show that the maximum deviation of the first ball from its initial direction of motion produced by the impact is tan−1
μ(1 + e) , 1−e
where μ is the coefficient of friction, and e is the coefficient of restitution between the balls. Also show that the least value of ω which will produce this deviation is (Fig. 7.47) 7μu (1 + e). 2a Sol: For the first ball, suppose that, just after the action of the impulsive forces, u , , v, , w, are the component velocities of G along the coordinate axes, and ωx , , ω y , , ωz , are the component angular velocities about the lines through G along the coordinate axes. For the second ball, suppose that, just after the action of the impulsive forces, U, V , W are the component velocities of G 1 along the coordinate axes, and ωx ,, , ω y ,, , ωz ,, are the component angular velocities about the lines through G 1 along the coordinate axes. By 7.2.8, equations of impulsive motion of the first ball are
884
7 Motion in Three Dimensions ′ ′′
Second ball First ball ′ 1 ′ ′
,
′ ′′
Fig. 7.47 , , Mu Mw, − M0 ( =2 0, ) − Mu )= −R, ( Mv 2−)(M0 = −F, )( ) ) 2 ( 2a 2a , , 0 = M 5 ωx − 0 , 0 = M 5 ω y − 0 , −F · a = M 2a5 ωz, − ω .
(
Thus, ωx,
= 0,
ω,y
( = 0, −F =
) ) 2a ( , −F −R M ωz − ω , v , = , w, = 0, u , − u = . 5 M M
Also ( ) −7F −F −5F + aω = +a + ω = v, + aωz, = V − aωz ,, . u − U = −e(u − 0), 2M M 2Ma , ,, , ,
Thus u , − U = −eu,
−7F + aω = V − aωz ,, . 2M
By Sect. 7.2.8, equations of impulsive motion of the second ball are
0=
(
2 M 2a5
= R, ( M V − M W − M0( = 0, ) )(MU − M0 )(M0 = F, ) ) ) 2 ( 2a 2 ,, ,, ωx − 0 , 0 = M 5 ω y − 0 , −F · a = M 2a5 ωz ,, − 0 .
Thus ,,
,,
ωx = 0, ω y = 0, −F = (Fig. 7.48). It follows that
(
) 2a M ωz ,, , 5
M V = F, W = 0, U =
R . M
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
885
Second sphere First sphere ′ 1 ′
,
, ′ ′′
Fig. 7.48
R R − (1 + e)u = U − (1 + e)u = (U − eu) − u = u , − u = − . M M , ,, , and hence R=
Mu(1 + e) . 2
Since R u, − u = − =− M , ,, ,
Mu(1+e) 2
M
=−
u(1 + e) 2
we have u, =
u(1 − e) . 2
Since −
u(1 + e) −R u(1 − e) −MU = − u = u, − u = = −U, = 2 2 M , ,, M,
we have U= Since
u(1 + e) . 2
886
7 Motion in Three Dimensions
) ( −7F 5F F 5F 7F −5F + aω = V − aωz ,, = V − a =V+ = + = , 2M 2a M 2M M 2M 2M , ,, , we have F=
M aω. 7
Now since ( ) M 2a − aω = −F = M ωz ,, , 7 5 ,, , , we have ωz ,, = −
5 ω. 14
Since M ,M V,,= F, = 7 aω, we have V =
1 aω. 7
Since ) ( − M7 aω −F 1 = − aω, = v = M 7 , ,, M , ,
we have 1 v, = − aω. 7 Here, ( ) Mu(1 + e) M aω = F ≤ μR = μ , , ,, , 7 2 so ω 7(1 + e) ≤μ· . u 2a
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point
887
Here, the deviation of the first ball from its initial direction of motion produced by the impact    , ( )  − 1 aω    2a ω   −1 7 = tan  ,  = tan  u(1−e)  = tan−1   u 7(1 − e) u 2 ( ) 2a 7(1 + e) μ(1 + e) μ· = tan−1 . ◼ ≤ tan−1 7(1 − e) 2a 1−e −1  v
Since tan−1
(
2a ω 7(1 − e) u
)
≤ tan−1
μ(1 + e) , 1−e
we have ω≤
7μu (1 + e). 2a
7.2.18 Example Show that the loss of kinetic energy at the impact of two perfectly rough inelastic uniform spheres, of masses M and M , , which are moving before impact with their centres in one plane, is ) ( M M, ) 2u 2 + 7v2 , , 14 M + M (
where u and v are the relative velocities before impact of the points of contact tangentially, in the plane of motion of the centres, and normally. Sol: In the Fig. 7.49, the plane of the paper is the plane of motion of the centres of the spheres. For the first sphere, suppose that, just before the action of the impulsive forces, U, V are the component velocities of G along the coordinate axes, and ω is the angular velocity about the line through G perpendicular to the plane. Suppose that, just after the action of the impulsive forces, u , , v, are the component velocities of G along the coordinate axes, and ω, is the angular velocity about the line through G perpendicular to the plane.
888
7 Motion in Three Dimensions
Ω Ω′ Second sphere
′ ′ 1
′
First sphere
1
′ 1
1
′ 1
,
′
,
Fig. 7.49
For the second sphere, suppose that, just before the action of the impulsive forces, U1 , V1 are the component velocities of G 1 along the coordinate axes, and Ω is the angular velocity about the line through G 1 perpendicular to the plane. Suppose that, just after the action of the impulsive forces, u 1 , , v1 , are the component velocities of G 1 along the coordinate axes, and Ω, is the angular velocity about the line through G 1 perpendicular to the plane. By Sect. 7.2.8, equations of impulsive motion of the first sphere are Mu , − MU = −R,
Mv , − M(−V ) = F,
( F ·a =
M
) ) 2a 2 ( , ω −ω . 5
By Sect. 7.2.8, equations of impulsive motion of the second sphere are M , u ,1 − M , U1 = R,
M , v1, − M , (−V1 ) = −F,
( F ·b =
M,
) ) 2b2 ( , Ω −Ω . 5
Here, the geometrical equations are 7F 2M
(F ) ( 5F ) + (−V + aω) = M − V + a 2Ma + ω = v, + aω, = v1, − bΩ, ,, , , ( ) ( 5F ) −7F = −F − b − V + Ω = + − bΩ) (−V 1 1 M, 2M , b 2M ,
and u , − u ,1 = −0(U − U1 ) = 0. ,, , , Thus F =− =−
(−V + aω) − (−V1 − bΩ) 7 7 2M + 2M , , MM
=−
2 M M, ((−V + aω) − (−V1 − bΩ)) 7 M + M,
2 (relative velocities before impact of the points of contact tangentially) 7 M + M,
7.2 Examples on 3D Motion of a Rigid Body About a Fixed Point =−
2 M M ,u . 7 M + M,
Hence F =−
2 M M ,u . 7 M + M,
Next since ( ) ) ( M , u , − M , U1 = M , u ,1 − M , U1 = R = − Mu , − MU , , ,, , we have −
R MU + M , U1 −R + MU +U = = u, = , , M M M ,, + M , ,
and hence
=
, , (U −U1 ) +M , U1 U −M , U1 R = U − MUM+M = M M+M = MM+M , , , M , , M (before impact vel. of A rel. to G in the normal direction) M,v = . M+M , M+M ,
Thus R=
M M ,v . M + M,
Now the change in kinetic energy ) ( ( ) )( ) 1 ( )( ) a )( ) 1 ( , 2a ( , = M u − U u , + U + M v, − V v, + V + M ω − ω ω, + ω 2 2 2 5 ( ) ( ) )( , ) 1 ,( , )( , ) b )( ) 1 ,( , 2b ( , + M u 1 − U1 u 1 + U1 + M v1 − V1 v1 + V1 + M, Ω − Ω Ω, + Ω 2 2 2 5 ) ( ) 1 ( , ) a ( , ) ( , 1 = (−R) u + U + F v − V + F ω + ω 2 2 2 ) ( ) 1 ) b ( , ) ( , 1 ( , + R u 1 + U1 + (−F) v1 − V1 + F Ω + Ω 2 2 2 ) ( , )) 1 ( )) 1 (( , ) ( ) 1 (( , = R u 1 + U1 − u + U + F v − V − v1, − V1 + F aω, + aω + bΩ, + bΩ 2 2 2 ) ( )) 1 ( )) 1 (( ) ( ) 1 (( = R u , + U1 − u , + U + F v , − V − v1, − V1 + F aω, + aω + bΩ, + bΩ 2 2 2 ) ) ( ) 1 (( 1 = R(U1 − U ) + F v , + aω, − (V − aω) − v1, − bΩ, + (V1 + bΩ) 2 2 ) ) ( ) 1 (( 1 = R(U1 − U ) + F v , + aω, − v1, − bΩ, + ((−V + aω) − (−V1 − bΩ)) 2 2 ) ( ) ) −1 1 (( = R(U − U1 ) + F v , + aω, − v1, − bΩ, + u 2 2
889
890
7 Motion in Three Dimensions ) ( ) ) 1 (( −1 Rv + F v , + aω, − v1, − bΩ, + u 2 2 ) ( ) ) −1 1 (( 1 −1 Rv + F v1, − bΩ, − v1, − bΩ, + u = Rv + Fu = 2 2 2 2 ( ( ) ) ) −1 M M , v 2 M M ,u 1 −M M , ( 2 ) 2u + 7v 2 . ◼ = − v + u= ( 2 M + M, 2 7 M + M, 14 M + M , =
Chapter 8
Conservation Principles of A.M. and Energy
In any theory, conservation principles are regarded as “god’s principles”. First of all they make calculation simple. Secondly, in Newtonian rigid body dynamics, physicists regard conservation principle of angular momentum and conservation principle of energy more fundamental than Newton’s laws of motion itself. It can be shown that nature works in an “economical” manner. Although Lagrange’s equations in generalized coordinates is a topic of analytical dynamics (a sequel book under preparation), we have derived it here—for the simple reason that a treatment of motion of a top, requires an application of Lagrange’s equations. The question of stability of a top, that defies gravity, is also dealt with in this chapter. Gyroscope works on these principles.
8.1 Conservation of Angular Momentum 8.1.1 Note Suppose that there is a system of particles. Suppose that m is a particle of the system whose rectangular coordinates are (x, y, z) at time t. Let X, Y, Z be the components parallel to the axes of the external forces acting on the particle m. By Sect. 2.2.2
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 R. Sinha, Advanced Newtonian Rigid Dynamics, https://doi.org/10.1007/9789819920228_8
891
892
8 Conservation Principles of A.M. and Energy
d (total momentum of the system measured parallel to the axis of x) dt ( ( )) Σ ( d2x ) Σ dx d Σ m m 2 = X = = dt system dt dt system ︸ ︷︷ ︸ ( ) the sum of the resolved parts of the external forces = , etc measured parallel to the axis of x and d (total moment of momentum of the system about the axis of x) dt ( )) ( dy d Σ dz −z = m y dt system dt dt ) Σ ( ) Σ Σ ( d2z dz dy d d2 y = y −z = m m y 2 −z 2 = (y Z − zY ) dt dt dt dt dt system system ︸ ︷︷ ︸ = (the sum of the moments of external forces about the axis of x), etc. It follows that Σ 1. If X = 0, then the total momentum of the system measured parallel to the axis of x remains cstant throughout the motion, etc. (Principle of conservation ofΣ linear momentum) 2. If (y Z − zY ) = 0, then the total moment of momentum of the system about the axis of x remains constant throughout the motion, etc. (Principle of conservation of angular momentum (or, moment of momentum))
8.1.2 Example A bead, of mass m, slides on a circular wire, of mass M and radius a, the wire turns ' freely about a vertical diameter. If ω and ω are the angular velocities of the wire when the bead is, respectively, at the ends of a horizontal and vertical diameter, show that M + 2m ω' = . ω M Sol: By system, we shall mean “circular wire + bead”, and by axis we shall mean − → vertical diameter A A' . In Fig. 8.1, R denotes the reaction force experienced by wire due to bead. Now by Newton’s third law, the reaction force experienced by bead due
8.1 Conservation of Angular Momentum
893
− → to wire is − R . It follows that the sum of the moments of external forces about the axis ) ( = moment of force Mg acting at G about the axis AG A' + (moment of force mg acting at bead about the axis) ( ) the moment of vertical force mg acting at =0+ = 0 + 0 = 0, bead about the vertical axis AG A' and hence, by the conservation principle of angular momentum, throughout the motion, total moment of momentum of “circular wire + bead” about the vertical axis AG A' is constant. Hence ) ( total moment of momentum about the vertical axis AG A' of “circular wire + bead” when the bead is horizontal ( ) total moment of momentum about the vertical axis AG A' of = “circular wire + bead” when the bead is vertical Remark: Here bead m has two velocities: (1) linear velocity due to motion along the ring (2) angular velocity due to rotation about the fixed axis AG A' . Observe that moment of momentum, about the vertical axis AG A' , of bead due to motion (1) is zero. Again observe that moment of momentum, about the vertical axis AG A' , of bead due to motion (2) ( ) = M.I.of bead about the vertical axis AG A' (angular velocity of ring)
Fig. 8.1 Linear velocity −⃗
, angular velocity
Bead, ⃗
′
894
8 Conservation Principles of A.M. and Energy
When the bead is horizontal, total moment about the vertical axis AG A' of “circular wire + bead”
of
momentum
( ) = moment of momentum about the vertical axis AG A' of circular wire ( ) + moment of momentum about the vertical axis AG A' of bead ( ) ( ) Ma 2 = ω + moment of momentum about the vertical axis AG A' of bead 2 ) ( ( ( ) ) Ma 2 ω + 0 + ma 2 ω , = 2 and when the bead is vertical, total moment about the vertical axis AG A' of “circular wire + bead”
of
momentum
( ) = moment of momentum about the vertical axis AG A' of circular)ire ( of momentum about the vertical axis AG A' of bead ( + )moment ( ) 2 = Ma ω' + moment of momentum about the vertical axis AG A' of bead 2 ( 2) ) ) ( 2) ' ( ( = Ma ω' + 0 + m02 ω' = Ma ω. 2 2 So (
) ) ( ( ) Ma 2 Ma 2 ω + ma 2 ω = ω' . 2 2
or Mω + 2mω = Mω' or M + 2m ω' = . ω M
8.1.3 Note We shall try to show that the same results of Sect. 8.1.1 are true in the case of impulsive forces. Suppose that the duration of the impulse is a small time τ. Next since
8.1 Conservation of Angular Momentum
895
⎛ ⎞ ( ( 2 ) Σ ) Σ Σ d⎝ dx ⎠ d x m m 2 = X, = dt system of bodies dt dt system of bodies ︸ ︷︷ ︸ we have Σ
(
)
t=τ
∫ X dt
t=0
t=τ (Σ
= ∫
t=0
⎞ ( ) t=τ d x ⎠ m  dt  system of bodies t=0 ︷︷ ︸
⎛ ) X dt = ⎝
︸
Σ
= (total inear momentum parallel to the axis of x)t=τ t=0 = (change in total linear momentum parallel to the axis of x). Thus ) Σ (t=τ ∫ X dt (change in total linear momentum parallel to the axis of x) = t=0 ︸ ︷︷ ︸ Σ = (impulse of force X parallel to the axis of x), etc. Conclusion 1. If the axis is such that the sum of impulses parallel to the axis vanishes, then the total linear momentum parallel to the axis just before the action of the impulsive forces is equal to the total linear momentum parallel to that axis just after their action. (Principle of conservation of linear momentum for impulsive forces). Next since ) ) ( dy m y dz − z dt dt system of bodies ( 2 ) Σ ) Σ d z d2 y − z dy m y − z = = (y Z − zY ) dt dt 2 dt 2 system of bodies ︸ ︷︷ ︸ (
d dt
Σ
=
system of bodies
( d m dt y dz dt
Σ
we have ⎛ ( ) ) t=τ (Σ Σ t=τ ∫ (y Z − zY )dt = ∫ (y Z − zY ) dt = ⎝ t=0
t=0
︸
⎞ ) t=τ dy ⎠ dz −z m y  dt dt  system of bodies t=0 ︷︷ ︸ Σ
(
= (total moment of momentum about the axis of x)t=τ t=0 = (change in total moment of momentum about the axis of x).
Thus
896
8 Conservation Principles of A.M. and Energy
) Σ (t=τ ∫ (y Z − zY )dt (change in total A.M. about the axis of x) = t=0 ︸ ︷︷ ︸ Σ = (moment of impulses about the axis of x), etc. Conclusion 2. If the axis is such that the sum of moments of impulsive forces about the axis vanishes, then the total A.M. about that axis just before the action of the impulsive forces is equal to the total A.M. about that axis just after their action. (Principle of conservation of A.M. for impulsive forces).
8.1.4 Example A rod, of length 2a, is moving on a smooth table with a velocity v perpendicular to its length and impinges on a small inelastic obstacle at a distance c from its centre. . When the end leaves the obstacle, show that the angular velocity of the rod is 3cv 4a 2 Sol: By the conservation principle of A.M. about the chosen axis, we have 2
velocity as the end leaves the obstacle) M 4a) 3 (angular
( = ︸
M
a2 + Ma 2 (angular velocity as the end leaves the obstacle) + 0 = (Mv)c 3 ︷︷ ︸
we have 3vc ∎ (angular velocity as the end leaves the obstacle) = vc 4a = 4a . 3 By the conservation principle of A.M. about the chosen axis for impulsive forces, we have ) ( a2 M + Mc2 (angular velocity just after leaving the obstacle) + 0 = (Mv)c, 3 ︸ ︷︷ ︸ and hence (angular velocity just after leaving the obstacle) =
Fig. 8.2 ′
cv a2 3
+ c2
.
Chosen axis
′
End that leaves the obstacle
8.1 Conservation of Angular Momentum
897
8.1.5 Example A uniform circular plate is turning on its own plane about a point A on its circumference with uniform angular velocity ω. Suddenly A is released and another point B of the circumference is fixed. Show that the angular velocity about B is ω (1 + 2cosα), 3 where α is the angle that AB subtends at the centre. Sol: Let M be the mass of the circular plate, and a its radius. Let G be the centre of the circular plate. Just before fixing at B, the velocity of G is aω along the line K through G perpendicular to AG. So, perpendicular distance of line K from B = a cos α. Here, as the point B is fixed, an impulsive force P is in action at B. Suppose that L is the line through B perpendicular to the plane. Since the impulsive force P intersects the line L , its moment about L vanishes, and hence we can apply the conservation of A.M. for impulsive forces about B. Thus (
) ) ( A.M. of circular plate about A.M. of circular plate about . = B just after impact B just before impact
(8.1)
By Sect. 4.2.15, (
) ( ) Σ A.M. about centre of A.M. of lamina about any = point O fixed in space mass G of particle m lamina ) ( A.M. about O of a particle of mass M placed . + at G and moving with it So, in our case, Fig. 8.3 , vel. =
line cos
898
8 Conservation Principles of A.M. and Energy
(A.M. of circular plate about B just before impact) ( ) a2 = M ω + (M(aω))(perpendicular distance of line K from B) 2 ( ) a2 = M ω + (M(aω))(a cos α) 2 Ma 2 ω = (2 cos θ + 1). 2 Next ( (A.M.of circular plate about B just after impact) =
M
) 3a 2 ' a2 + Ma 2 ω' = M ω 2 2
Thus, from (1), (
or
) Ma 2 ω 3a 2 ' ω = M (2cosθ + 1) 2 2
ω' = 13 (2 cos θ + 1)ω.
∎
8.1.6 Example A uniform square lamina, of mass M and side 2a, is moving freely about a diagonal with uniform angular velocity ω when one of the corners not in that diagonal is fixed. Show that the new angular velocity is ω7 , and that the impulse of the force on the fixed point is Sol:
√ 2 Maω. 7
Fig. 8.4
,2
′
8.1 Conservation of Angular Momentum
899
Here, we can apply the conservation of A.M. for impulsive forces about the line D X . Thus ) ( (√ ) a2 a2 ω + (M0) 2a = M ω= M 3 3 ( ) ( ) A.M. of lamina about line A.M. of lamina about line = D X just before impact D X just after impact ︸ ︷︷ ︸ (√ )2 ) a2 + M 2a (angular velocity about D X after fixing) 3 7a 2 =M (angular velocity about D X after fixing). 3 (
=
M
Thus, (angular velocity about D X after fixing) =
ω . 7
Now since D X and D B are perpendicular lines in the plane, it suffices to show that (A.M.of lamina about line D B just after fixing) = 0. Clearly, we can apply the conservation of A.M. for impulsive forces about the line D B. Thus ( ) ( ) A.M. of lamina about line A.M. of lamina about line = . D B just before fixing D B just after fixing From symmetry, ∎ (A.M.of lamina about line D B just before fixing) = 0. Observe that, before fixing of D, the centre of mass G was at rest, and after the fixing it is moving with velocity ( √ ω) . DG · (angular velocity about D X after fixing) = 2a · 7 ) (√ So the change in linear momentum of G is M 2a · ω7 . Thus ) (√ P = M 2a · ω7 .
∎
900
8 Conservation Principles of A.M. and Energy
8.2 Examples on Conservation of A.M. 8.2.1 Example If the Earth, supposed to be a uniform sphere, had in a certain period contracted slightly so that its radius was less by n1 th than before, show that the length of the h. day would have shortened by 48 n Sol: Let M be the mass of the earth. Suppose that its radius contracted from R ( )( ) to R − n1 R = n−1 R . By “axis” we shall mean the diameter of the Earth about n which it revolves. By the conservation of A.M., we have (
( n−1 )2 ) ) ( R 2π 2R 2 2π n , = M M 5 5 24h (24h − D) 2
and hence 24h − D = 24h Thus ( ( )2 ) ( 24 h = n2 − D = 1 − n−1 n
1 n2
(
n−1 n
)2 .
) ( ) 24 h ≈ n2 − 0 24 h =
48 n
h.
∎
8.2.2 Example A heavy circular disc is revolving in a horizontal plane about its centre which is fixed. An insect, of mass n1 th that of the disc, walks from the centre along a radius, n times the original and then flies away. Show that the final angular velocity is n+2 angular velocity of the disc. Sol: It is easy to see that the line L through G, which is perpendicular to the plane of the disc, can be taken for the purpose of applying C.P.A.M. So (
Hence n ω' = n+2 ω.
) ( ( ) ) a2 a2 M 2 ' M ω= M + a ω. 2 2 n ∎
8.2 Examples on Conservation of A.M.
901
8.2.3 Example A uniform circular board, of mass M and radius a, is placed on a perfectly smooth horizontal plane and is free to rotate about a vertical axis through its centre. A man, of ' mass M , walks round the edge of the board whose upper surface is rough enough to prevent his slipping. When he has walked completely round the board to his starting point, show that the board has turned through an angle M' · 4π. M + 2M ' Sol: By system we shall mean “circular board + man”. Observe that, at the end of walk, φ − θ = 2π. We have to show that θ (t)φ(t)−θ (t)=2π = ±
M' .4π. M + 2M '
'
Here, equation of motion of man M is (
) M ' a 2 φ¨ = F · a.
Here, equation of motion of circular board M is (
) a2 ¨ θ = −F · a. M 2
It follows that M ' φ˙ = −
M θ˙ + C. 2
≡ ( ) ,
≡
( )
Direction fixed in board At time = 0 ′
Direction fixed in space
Fig. 8.5
′
At time
Direction fixed in space
902
8 Conservation Principles of A.M. and Energy
  Since θ˙ t=0 = 0, and φ˙ t=0 = 0, we have C = 0, and hence M ' φ˙ = −
M θ˙ . 2
It follows that M 'φ = −
M θ + C1 . 2
Since θ t=0 = 0, and φt=0 = 0, we have C1 = 0, and hence M 'φ = −
M θ. 2
Now, when φ(t) − θ (t) = 2π, we have M ' (2π + θ ) = −
M θ, 2
and hence M' θ = − M+2M ' 4π .
∎
8.2.4 Example A circular ring, of mass M and radius a, lies on a smooth horizontal plane, and an insect, of mass m, resting on it starts and walk round it with uniform velocity v relative to the ring. Show that the centre of the ring describes a circle with angular velocity v m . M + 2m a Sol: By system we shall mean “ring + insect”. Since there is no external force acting on the system, the centre of mass G of the system is fixed in space. Observe ' that insect I , G, O are always collinear. Here GO =
m a, M +m
m G is fixed in space, and M+m a is a constant, the locus of O, that is, the locus of m a and centre G. ∎ centre of ring, is a circle of radius M+m This shows that the ring is rotating about an axis L through G which is perpendicular to the plane of the ring. Clearly, L is fixed in space.
8.2 Examples on Conservation of A.M.
903
Path of insect I Path of Insect
,
′
=
≡
+
I =
Point fixed in space
≡
+
direction fixed in space
≡ ( ),
≡
( ),
=
(ang. vel. of ′ ) = ̇ (ang. vel. of ) = ̇ (ang. vel. of ) =
( + )
Fig. 8.6 '
Also, the locus of insect I is a circle of radius It remains to show that
M a M+m
and centre G.
m v d(π + θ ) =± . dt M + 2m a Here, from question, ( v= ︸
) ( ) ( ) velocity velocity of velocity of insect relative = − to point I of the ring of insect point I of the ring ︷︷ ︸
so (velocity of insect) = (G I )θ˙ + v. Since G is fixed in space, and there is no external force acting on the system, we can apply C.P.A.M. about G. We get ) ( M a 2 + x 2 θ˙ + my 2 θ˙ + mvy ) ( ( )) ( = Ma 2 + M x 2 θ˙ + m y θ˙ + v y ( ) = Ma 2 + M(OG)2 θ˙ + (m(velocity of insect))(G I ) = 0, ︸ ︷︷ ︸ and hence
904
8 Conservation Principles of A.M. and Energy
θ˙ = ︸
−mvy ) ( = 2 M a + x 2 + my 2 ︷︷ ︸ =
M −m av ( M+m )
M −mv ( M+m a) ) ( 2 2 m M M a 2 +( M+m a ) +m ( M+m a)
) ( 2 m M M+ M ( M+m )2 +m ( M+m )
=
M −m av ( M+m ) Mm M+ M+m
=
M −m av ( M+m ) ) ( 2 2 m M M 1+( M+m ) +m ( M+m )
∎ =
−Mm av M(M+m)+Mm
=
−m av . M+2m
8.2.5 Example If a merrygoround is set in motion and left to itself, show that in order that a man may 1. Move with the greatest velocity, 2. Be most likely to slip, he must place himself at a distance from the centre equal to √ (1) k √n, (2) k n3 , k being the radius of gyration of the machine about its axis and n the ratio of its weight to that of the man. (Remark: Conservation of angular momentum is found to be true, even when the “system” itself is changing but during the process no “external torque” is exerted.) Sol: Let M be the mass of the man. Suppose that x is the distance of the seated man from the centre of the machine. Let Ω be the original angular velocity of the machine. It follows that the velocity of the seated man = x(angular velocity of machine ( ) after2 sitting of man) (n M)k 2 )Ω ( nk Ωx = x (n M)k 2 +M x 2 = nk 2 +x 2 . ( ) “tendency to slip” of the seated ) (The M x(angular velocity of machine after sitting of man)2
man,
that
(
)2 ( ) )2 ( (n M)k 2 Ω x nk 2 Ω ( ) = Mx = Mx = Mn 2 k 4 Ω2 ( )2 . 2 nk 2 + x 2 (n M)k 2 + M x 2 nk + x 2 1. For maximum velocity of the seated man, we have ) ( 2 ) ( 2 − x(2x) nk 2 Ωx nk 2 − x 2 d 2 1 nk + x = 0, nk Ω ( = )2 = nk Ω ( )2 d x nk 2 + x 2 nk 2 + x 2 nk 2 + x 2 ︸ ︷︷ ︸ 2
is,
8.2 Examples on Conservation of A.M.
905
√ and hence x = k n. ∎ 2. For most likely to slip, we have nk 2 +x 2 −4x 2 nk 2 −3x 2 = Mn 2 k 4 Ω2 ( 2 )2 3 (nk 2 +x 2 )3 (nk +x ) ) ( 2 2 2 2 2 x d 2 4 2 2 4 2 1(nk +x ) −x2(nk +x )(2x) = Mn k Ω = Mn k Ω ( )2 = 0, (nk 2 +x 2 )4 dx nk 2 + x 2 ︸ ︷︷ ︸
Mn 2 k 4 Ω2
and hence√ x = k n3 .
∎
8.2.6 Example A uniform circular wire, of radius a, lies on a smooth horizontal table and is movable about a fixed point O on its circumference. An insect, of mass equal to that of the wire, starts from the other end of the diameter through O and crawls along the wire with a uniform velocity v relative to the wire. Show that at the end of time t the wire has turned through an angle ( ) 1 1 −vt vt + √ tan−1 √ tan . 2a 3 3 2a Sol: Let m be the mass of the ring. Hence the mass of insect is m. By system we shall mean “ring + insect”. Observe the following: 1. 2. 3. 4. 5.
Ring revolves about O. A.M. of the system about O is a constant. Initially, A.M. of the system about (O is zero. ) ( ) = 2ma 2 φ˙ . At time t, A.M. of ring about O is ma 2 + ma 2 dφ dt At time t, the velocity of insect is the composite of two velocities: vel v relative ' ' to the point I on the ring, and vel. of I rel to O. 6. At time t, A.M. of insect about O. )) )) ( ( ( ( ( ) θ θ = − m v cos (O I ) + m 2a φ˙ cos (O I ) 2 2 ))( ) ( ( ))( ) ( ( ( ) θ θ θ θ 2a cos + m 2a φ˙ cos 2a cos = − m v cos 2 2 2 2 θ θ ˙ = 2amv cos2 + 4a 2 m cos2 φ. 2 2
906
8 Conservation Principles of A.M. and Energy Path of ,
direction fixed in space ′
Initial line
(point fixed in space) Insect, At time = 0
Fig. 8.7
Circle, fixed in 2 space, of radius
√3
Point fixed in space
=
√3
√3
tan( + )
′
vel. = (
Fig. 8.8
′)
̇
8.2 Examples on Conservation of A.M.
907
Thus ) ( θ θ −2amv cos2 + 4a 2 m cos2 φ˙ + 2ma 2 φ˙ = 0 2 2 or ( ) ( vt ) dφ 2 a 2 θ 2 θ ˙ 1 + 2 cos = −v cos −v cos +a + a φ 1 + 2 cos 2 dt 2 2 2 ( ) 2θ + a φ˙ = 0 . = −v + 2a φ˙ cos ︸ ︷︷ 2 ︸ vt 2 a
Thus vt
v cos2 2a dφ = a dt 1 + 2 cos2
vt a
2
or ( ) ) ( vt −1 1 v 1 + 2 cos2 2a v 1− φ= dt = vt vt dt 2a 1 + 2 cos2 2a 2a 1 + 2 cos2 2a ( ) ( )) ( vt 2a vt d tan 2a sec2 2a v v v = 1− dt = dt − . vt vt 2a sec2 2a +2 2a tan2 2a +3 It follows that φ=
( )) ( tan vt 2a 1 v t− + C. √ tan−1 √ 2a 2a v 3 3
Now since φt=0 = 0, C = 0. Thus ( wevt have ) tan 2a v 1 −1 √ . φ = 2a t − √3 tan 3
∎
8.2.7 Example A small insect moves along a uniform bar, of mass equal to itself and of length 2a, the ends of which are constrained to remain on the circumference of a fixed circle, whose radius is √2a3 . If the insect starts from the middle point of the bar and moves along the bar with relative velocity v, show that the bar in time t will then turn through an angle
908
8 Conservation Principles of A.M. and Energy
1 vt √ tan−1 . a 3 Sol: Let m be the mass of the bar. Hence the mass of insect is m. By system we shall mean “bar + insect”. Observe the following: 1. Bar revolves about O. 2. A.M. of the system about O is a constant. 3. Initially, A.M. of the system about ⎛ O is zero. (/ )2 ⎞ ) ( )2 ( 2 2 2a √ 4. At time t, A.M. of bar about O is ⎝m a + m − a 2 ⎠ dφ = m 2a φ˙ . 3
3
dt
3
5. At time t, the velocity of insect is the composite of two velocities: vel v relative ' ' to the point I on the bar, and vel. of I rel to O. 6. At time t, A.M. of insect about O. )2 ( ( (( ' ) ))( ' ) mva a a ˙ ˙ = − (mv) √ + m O I φ O I = − √ + m φ √ sec(θ + φ) 3 3 3 ) mva a2 ( 1 + tan2 (θ + φ) = − √ + m φ˙ 3 3 ) ( 2 a mva + (vt)2 . = − √ + m φ˙ 3 3 Thus )) ( ( 2 ) 2a 2 mva mva dφ ( 2 a 2 2 ˙ +m a + (vt) = − √ + m φ + (vt) − √ +m φ˙ = 0 . dt 3 3 3 3 ︸ ︷︷ ︸ It follows that ) ( 1 va a 1 −1 vt tan dφ = √ 2 dt = d √ a 3 a + (vt)2 3 a ︸ ︷︷ ︸ or vt 1 φ = √ tan−1 + C. a 3 Now since φt=0 = 0, w have C = 0. Thus vt 1 φ = √ tan−1 . a 3
8.2 Examples on Conservation of A.M.
909
8.2.8 Example A circular disc is moving with an angular velocity Ω about an axis through its centre perpendicular to its plane. An insect alights on its edge and crawls along a curve drawn on the disc in the form of lemniscates with uniform angular velocity 18 Ω, the 1 curve touches the disc. The mass of the insect being, 16 th of that of the disc, show that when the insect gets to the centre is √ 24 π −1 7 − . √ tan 3 4 7 Sol: Let M be the mass of the circular disc. Hence, the mass of insect is system we shall mean “disc + insect”. Observe the following:
M . 16
By
1. Disc revolves about O. 2. A.M. of the system about O is a constant. ( 2) 3. At time t, A.M. of disc about O is M a2 dφ . dt ( M 2 ) d(θ +φ) 4. At time t, A.M. of insect about O is 16 r . dt Thus, C.P.A.M, (
2
M a2
)
dφ dt
+
(M
r2
)( dθ
+
dφ dt
)
( 2) = A.M.of system at time t = A.M.of disc before alighting of insect = M a2 Ω. ︸ ︷︷ ︸
Fig. 8.9
16
dt
910
8 Conservation Principles of A.M. and Energy
Hence a2
) ) ( ( ) dθ dθ dφ dφ 1( dφ 1 dφ = a2 = a2Ω + a 2 cos 2θ + + r2 + dt 8 dt dt dt 8 dt dt ︸ ︷︷ ︸
or (8 + cos 2θ )
(
= (8 + cos 2θ ) dφ dt
dφ Ω dθ 8
)
cos 2θ = (8 + cos 2θ ) dφ + Ω8 cos 2θ ) (dt dφ dθ dφ d ( Ω8 t ) = 8Ω + cos 2θ dt = 8 + cos 2θ + dt dt dt ︸ ︷︷ ︸ +
Ω 8
Hence (8 + cos2θ )
Ω dφ Ω = 8Ω − cos2θ. dθ 8 8
Thus (8 + cos2θ )
dφ = 64 − cos2θ. dθ
So ) ( 64 − cos 2θ 72 dθ = −1 + dθ. dφ = 8+ 8 + cos 2θ ︸ ︷︷cos 2θ ︸ It( follows)that required angle
=−
π
θ= 4 T =1 1 1 dT π π + 72 ∫ dθ = − + 72 ∫ 2 2 4 4 θ=0 8 + cos 2θ T =0 8 + 1−T 1 + T 2 1+T
T =1 1 π = − + 72 ∫ dT 4 9 + 7T 2 T =0 ( √ 1 1 7T π tan−1 = − + 72 √ 4 3 73
∎ )T =1 √  24 7 π  . = − + √ tan−1   4 3 7 T =0
8.2.9 Example A rod O A can turn freely in a horizontal plane about the end O and lies at rest. An insect, whose mass is enethird that of the rod, alight on the end A and commences crawling along the rod with uniform velocity V , at the same instant the rod is set in
8.2 Examples on Conservation of A.M.
911
Path of insect
2 − point fixed in space direction fixed in space
2
Fig. 8.10
rotation about O in such a way that the initial velocity of A is V . Prove that, when the insect reaches O, the rod has rotated through a right angle, and that the angular velocity of the rod is then the twice the initial angular velocity. Sol: Let M be the mass of the rod. Hence, the mass of insect is M3 . By system we shall mean “rod + insect”. Observe the following: 1. Rod revolves about O. 2. A.M. of the system about O is a constant. ) ( 2 . 3. At time t, A.M. of rod about O is M a3 + Ma 2 dθ dt (M ) 2 dθ 4. At time t, A.M. of insect about O is 3 (2a − V t) dt . Thus, by C.P.A.M, M 4a3
2
dθ dt
+
M 3 (2a
( 2 ) ) ( = M a3 + Ma 2 dθ + M3 (2a − V t)2 dθ − V t)2 dθ dt dt dt
= A.M.of system at time t = A.M.of rod before alighting of insect ︸ ︷︷ ︸ ( 2 ) ) ( V V + M3 (2a − V 0)2 2a = M 4a V. = M a3 + Ma 2 2a 3 Hence
912
8 Conservation Principles of A.M. and Energy
4a 2
dθ dθ + (2a − V t)2 = 4aV . dt dt
It follows that dθ =
4aV dt. (V t − 2a)2 + 4a 2
Hence 1 θ = 4aV . V ︸
(
) 1 V t − 2a −1 V t − 2a tan + C. +C = 2 tan−1 2a 2a 2a ︷︷ ︸
Since θ t=0 = 0, we have C =
π , 2
and hence
θ = 2tan−1
π V t − 2a + . 2a 2
It follows that θ t= 2aV = π2 . Since
∎
4a 2
dθ dθ + (2a − V t)2 = 4aV , dt dt
we have 4a 2
    ) (   dθ  2a 2 dθ  2 dθ  2 dθ  = 4a + 2a − V = 4aV = 8a    dt t= 2a dt t= 2a V dt t= 2a dt t=0 V V V ︸ ︷︷ ︸
and hence  dθ  =2· dt t= 2a V

dθ  . dt t=0
∎
8.2.10 Example A particle, of mass m, moves within a rough circular tube, of mass M, lying on a smooth horizontal plane and initially the tube is at rest while the particle has an angular velocity round the tube. Show that by the time relative motion ceases, the M of the initial kinetic energy has been dissipated by friction. fraction 2m+M
8.2 Examples on Conservation of A.M.
913
Fig. 8.11
≡
+ +
,
vel. =
Rough Circular tube
≡
0
At time = 0 0
0
+ +
Path of
Ang. vel. ≡
Path of
At time Ang. vel. ≡ vel. =
′
0
+
Sol: By system we shall mean “circular tube + point mass”. Let G be the centre of mass of the system. Observe the following: mV0 1. G moves with constant velocity V1 , where V1 ≡ M+m . 2. A.M. of the system about G is a constant. aM . 3. Initially, A.M. of the system about G is (mV0 ) M+m ( ( am )2 ) 4. At time t, A.M. of the circular tube about G is Ma 2 + M M+m ω. ( ( ) ) ' aM 2 ω. 5. At time t, A.M. of the point mass about O is m M+m
Thus, by C.P.A.M, (
Ma 2 + M
(
) am 2 M+m
)
( ( ) ) ' aM 2 ω + m M+m ω
aM = (A.M.of system at time t) = (A.M.of system at time t = 0) = (mV0 ) M+m , ︸ ︷︷ ︸
and hence ( a 1+ Thus
(
m M +m
)2 ) ω+
mV0 Mm ' aω = . 2 M +m (M + m)
914
8 Conservation Principles of A.M. and Energy
) ( ' a (M + m)2 + m 2 ω + Mmaω = m(M + m)V0 . '
When the relative motion ceases, we have ω = ω, and hence ) ( a (M + m)2 + m 2 ω + Mma(ω) = m(M + m)V0 . It follows that V0 m(M + m)V0 m(M + m) ) ω= ( = 2 M + 3Mm + 2m 2 a a (M + m)2 + m 2 + Mma ︸ ︷︷ ︸ =
Thus,
m V0 V0 m(M + m) = . M + 2m a (M + m)(M + 2m) a
⎞ ⎛ ⎞⎞ 1 1 M(V1 )2 m(V1 )2 ⎟ ⎜2 ⎟⎟ 1 ⎜⎜ 2 ⎟⎟ ⎜ ⎜⎜ ( ( ( ( )2 ) ⎟ )2 ) K f − K i = ⎜⎜ ⎟+⎜ ⎟⎟ − m(V0 )2 a M 1 am ⎠ ⎝ ⎝⎝ 1 2 2 ⎠⎠ Ma 2 + M m ω2 + + (ω) 2 M +m 2 M +m ( ) 1 1 Mma 2 1 Ma 2 + ω2 − m(V0 )2 = (M + m)(V1 )2 + 2 2 M +m 2 ) ( ) ( mV0 2 1 1 Mma 2 1 Ma 2 + ω2 − m(V0 )2 = (M + m) + 2 M +m 2 M +m 2 ) ( 1 M + 2m 1 m 2 (V0 )2 1 + Ma 2 ω2 − m(V0 )2 = 2 M +m 2 M +m 2 ) ( )( ( ) M + 2m 1 m 1 −1 + Ma 2 ω2 = m(V0 )2 2 M +m M +m 2 ( )( ) ) ( M + 2m 1 V0 2 m M 1 + Ma 2 = − m(V0 )2 2 M +m M +m 2 M + 2m a ( )( ) 2 1 1 m M 1 + M = − m(V0 )2 (V0 )2 2 M +m M +m 2 M + 2m ( ) m 1 M 2 −1 + = m(V0 ) 2 M +m M + 2m ( ) 1 −M M = − m(V0 )2 = Ki . 2 M + 2m 2m + M ⎛⎛
8.2.11 Example A rod, of length 2a, is moving about one end with uniform angular velocity upon a smooth horizontal plane. Suddenly this end is loosed and a point, distant b from this , b > 4a , b = 4a . end, is fixed. Find the motion, considering the cases when b < 4a 3 3 3 Sol: Let M be the mass of the rod. Let ω be the initial angular velocity of the rod. Here, A.M. about P is unaltered by fixing at P. So
∎
8.2 Examples on Conservation of A.M.
915
Original case
Ω 2 ⁄3
second case 2 ⁄3
Fig. 8.12 ( )x=a  ) )  ( )) MΩ (a + x)3 MΩ 8a 3 4a (a + x)2   = −b Ω= − b 2a 2 −b     3 2a 3 2a 3 2 x=−a ) ) x=a M = ∫ d x ((a + x)Ω)(a + x − b) x=−a 2a )
Ma
= (A.M. about P in the original case) = (A.M. about P in the second case) ,, , , ) =
M
) ) 2 ) a2 a + M(b − a)2 ω = M + (b − a)2 ω. 3 3
Thus ω=
a a2 3
( 4a 3
−b
)
+ (b − a)2
Ω.
Case I: when b < 4a . Here ω is positive, so rod turns the same way as before. 3 . Here ω is zero, so rod is brought to rest. Case II: when b = 4a 3 . Here ω is negative, so rod turns the opposite way from Case III: when b > 4a 3 before.
8.2.12 Example A circular plate rotates about an axis through its centre perpendicular to its plane with angular velocity ω. This axis is set free and a point in the circumference of the plate fixed. Show that the resulting angular velocity is ω3 . Sol: Let M be the mass of the circular plate. Here, A.M. about P is unaltered by fixing at P. So
916
8 Conservation Principles of A.M. and Energy
Fig. 8.13
Original case
′
second case
) ( 4) ( Ma 2 −Mω −Mω θ =2π a4 a a3 ∫ ω= − 2π = dθ − cos θ 2 πa 2 4 πa 2 θ =0 3 4 ) ( ) −Mω θ =2π r =a ( 2 3 ∫ ∫ cos θr dr dθ − r = πa 2 θ =0 r=0 ( (( ) ( )) ) )( θ =2π r =a − sin θ, r cos θ − 1, M ∫ × =− ∫ (dr )(r dθ ) (r ω) πa 2 cos θ, 0 r sin θ, 0 θ =0 r =0 = (A. M. about P in the original case) = (A.M. about P in the second case) ︸ ︷︷ ︸ ( =
M
) a2 3a 2 ' + Ma 2 ω' = M ω 2 2
and hence ω' =
ω 3
.
∎
8.2.13 Example Three equal particles are attached to the corners of an equilateral triangular area ABC, whose mass is negligible, and the system is rotating in its own plane about A. A is released and the middle point of AB is suddenly fixed. Show that the angular velocity is unaltered. Sol: Let 2a be the length of each side of the equilateral triangle. Here, A.M. about P is unaltered by fixing at P. So ) (/ ) ( ) ( ) ( m 5a 2 Ω = m 2a 2 Ω + m 3a 2 Ω = m((2a)Ω)a + (m((2a)Ω)) (2a)2 − a 2 sin 2π 3 +0 = (A.M.about P in the original case) = (A.M.about P in the second case) ︸ ︷︷ ︸ ) ( ) ( ( )) ( ) ( 2 2 2 2 2 ω = m 5a ω, = ma ω + ma ω + m (2a) − a
8.2 Examples on Conservation of A.M.
917
Fig. 8.14
Ω
Original case Vel. = (2 )Ω
Vel. = (2 )Ω
second case
and hence ω = Ω.
∎
8.2.14 Example A uniform square plate ABC D, of mass M and side 2a, lies on a smooth horizontal ' plane. It is struck at A by a particle of mass M moving with velocity V in the direction of AB. The particle remains attached to the plate. Determine the subsequent motion of the system, and show that its angular velocity is 3V M' . M + 4M ' 2a Sol: Here, A.M. about G is unaltered due to impact. So
918
8 Conservation Principles of A.M. and Energy
Fig. 8.15 ,2
′
vel. = (√2 ) +
⎛⎛
(√
′
′ (√2
≡
+
) ′
≡
)⎞
⎞ ) )( ( ' ) ( M M aV 1 ⎝⎝ ⎠ √ ⎠ = M ' V AG sin π = M V M + M' M + M' 4 2 '
M
2a
= (A.M. about G just before impact) = (A.M. about G just after impact) ︷︷ ︸ ︸ ((
) ) 2a 2 M + M(G O)2 + M ' (G A)2 ω' 3 ⎛⎛ ⎛ (√ ) ⎞2 ⎞ ⎛ (√ ) ⎞2 ⎞ ' 2 2a M M 2a ⎜⎜ 2a '⎝ ' ⎠ ⎟ ⎠ ⎟ + M = ⎝⎝ M + M⎝ ⎠ ⎠ω ' ' 3 M+M M+M
=
( =
M
) ( ) M M' 2a 2 2a 2 ' M + 4M ' ' = M + 2a 2 ω ω . 3 M + M' 3 M + M'
Hence 3M ' V ω' = 2a(M+4M ') .
∎
8.2.15 Example A lamina in the form of an ellipse is rotating in its own plane about one of its foci with angular velocity ω. This focus is set free and the other focus at the same instant 2−5e2 is fixed. Show that the ellipse now rotates about it with angular velocity ω 2+3e 2.
8.2 Examples on Conservation of A.M.
919
2 2
vel. = ( Original case
′ (−
, 0)
2
+
2
=1
)
(
, 0)
2
2 2
Second case
′ (−
, 0)
(
+
2
=1
, 0)
Fig. 8.16
Sol: Let M be the mass of the elliptical lamina. Here, A.M. about S is unaltered by fixing at S. So ( ) ( ) ) a 2 + a 2 1 − e2 a2 ( 2 − 5e2 ω = Mω − a 2 e2 4 4 ) ( 2 ) ( 2 a +b a 2 + b2 − a 2 e2 = −(M((ae)ω))(ae) + M ω = Mω 4 4 M
= (A.M. about S in the original case) = (A.M. about S in the second case) ︸ ︷︷ ︸ ( =
M
( ) ( ) ) ) a 2 + a 2 1 − e2 a2 ( a 2 + b2 + M(ae)2 ω' = M + (ae)2 ω' = M 2 + 3e2 ω' 4 4 4
and hence ω' =
2−5e2 ω. 2+3e2
∎
8.2.16 Example An elliptic area, of eccentricity e, is rotating with angular velocity ω about one latus rectum. Suddenly, this latus rectum is loosed and the other fixed. Show that the new 1−4e2 angular velocity is 1+4e 2 ω.
920
8 Conservation Principles of A.M. and Energy
Fig. 8.17
Sol: Let M be the mass of the elliptical lamina. Here, A.M. about the latus rectum through S is unaltered by fixing the latus rectum through S. So ) ( ) a2 a2 ( 2 ω 1 − 4e ω = −(M((ae)ω))(ae) + M M 4 4 ( ) ( ) A.M. about the latus rectum A.M. about the latus rectum = = through S in the original case through S in the second case ︸ ︷︷ ︸ ( =
M
) ) a2 a2 ( + M(ae)2 ω' = M 1 + 4e2 ω' 4 4
and hence ω' =
1−4e2 ω. 1+4e2
∎
8.2.17 Example A uniform circular disc is spinning with angular velocity ω about a diameter when a point P on its rim is suddenly fixed. If the radius vector to P make an angle α with this diameter, show that the angular velocities after fixing about the tangent and normal at P are 15 ωsinα and ωcosα. Sol: Here the required angular velocity after fixing the tangent at P
8.2 Examples on Conservation of A.M.
= = = = = = = =
921
( ) ) θ= π ( cos θ M ∫θ =−2 π ∫rr =2a dθ )) ((a sin α + r sin(θ − α))ω)(r cos θ ) =0 π a 2 ((dr )(r 2
2
M a4 + Ma 2
) ( Mω θ = π2 r=2a cos θ ( ) π a2 2 3 ∫ ∫ a sin αr cos θ + r cos θ sin(θ − α) dr dθ 2 r =0 M 5a4 θ =− π2 ) ω θ= π ( 2 (2a cos θ )4 (2a cos θ )3 πa 2 ∫ cos θ + cos θ sin(θ − α) dθ a sin α 5a 2 π 3 4 4 θ =− 2 ) ( 3 4 θ = π2
ω4 (2 cos θ ) (2 cos θ ) ∫ sin α cos θ + cos θ sin(θ − α) dθ π 5 θ =− π2 3 4 ) π ( ( ) ω 4 θ= 2 8 4 5 6 ∫ sin α cos θ + 4 cos α cos θ sin θ − sin α cos θ dθ π 5 θ =− π2 3 ( ( ) ( ( ))) 3 1 π 5 3 1 π ω4 8 sin α 2 · · · + 4 0 − sin α 2 · · · · π5 3 4 2 2 6 4 2 2 ( )( ( )) ω4 3 1 π 8 5 sin α 2 · · · −4 π5 4 2 2 3 6 ( ) ω1 3π −2 ω sin α = − sin α. π5 2 3 5
Here the required angular velocity after fixing the normal at P Fig. 8.18 ,
∎
922
8 Conservation Principles of A.M. and Energy
= = = =
( ) ) θ= π ( cos θ M ∫θ =−2 π ∫rr =2a dθ )) ((a sin α + r sin(θ − α))ω)(r sin θ ) =0 π a 2 ((dr )(r 2
2
M a4
) Mω θ = π (r =2a cos θ ( ) 2 π a2 2 3 ∫ ∫ a sin αr sin θ + r sin θ sin(θ − α) dr dθ 2 r=0 M a4 θ =− π2 ) ω θ= π ( 2 (2a cos θ )4 (2a cos θ )3 πa 2 ∫ sin θ + sin θ sin(θ − α) dθ a sin α a2 π 3 4 4 θ =− 2 ) π ( 4ω θ = 2 (2 cos θ )4 (2 cos θ )3 π
∫
θ =− π2 θ = π2
sin α
(
3
sin θ +
(
4
sin θ sin(θ − α) dθ
cos α cos4 θ sin2 θ 8 sin α cos3 θ sin θ + 4 − sin α cos5 θ sin θ θ =− π2 3 ) ) ( 4ω ( = 0 + 4 cos α cos4 θ − cos6 θ − 0 π ( ) 3 1 π 5 3 1 π 16ω cos α ·2 · · − · · · = π 4 2 2 6 4 2 2 ( ) ω cos α 5π = · 2 3π − = ω cos α. π 2 =
4ω π
∎
))
∫
dθ
8.2.18 Example A cube is rotating with angular velocity ω about a diagonal when suddenly the diagonal is let go, and one of the edges which does not meet this diagonal is fixed. Show that the resulting angular velocity about this edge is ω√ 3. 12
Fig. 8.19
,2
√3
√3 √3 Diagonal Edge
8.2 Examples on Conservation of A.M.
Sol: Required angular velocity ( =
(M.I. about diagonal)ω M.I. about edge
=
M
Ma
a 2 +a 2 3
2 +a 2 3
923
)
ω √ 3
+M (a 2 +a 2 )
=
2 ω √ 3 8a 2 3
M 2a3 M
=
ω 1√ 4 3
=
ω 12
√ 3.
8.2.19 Note Let l, m be positive numbers. Observe that ( ) x=1 y=1−x ∫ x l−1 y m−1 dy d x x l−1 y m−1 d xd y = ∫ x=0 y=0 ( ) y=1−x x=1 l−1 m−1 ∫ y dy d x = ∫ x
˜ 0≤x,0≤y,x+y≤1
x=0
x=1
= ∫ x
l−1 (1
x=0
=
y=0
1 x=1 l−1 − x)m ∫ x (1 − x)m d x dx = m m x=0
1 B(l, m + 1) (see 4.2.4) m = m1 ⎡(l)·⎡(m+1) 4.2.4) ⎡(l+(m+1)) (by = m1 ⎡(l)·(m·⎡(m)) 4.2.4) (by ⎡(l+m+1) ⎡(l)⎡(m) = ⎡(l+m+1) .
Thus ¨ x l−1 y m−1 d xd y = 0≤x,0≤y,x+y≤1
⎡(l)⎡(m) . (∗) ⎡(l + m + 1)
Next observe that ˜
x l−1 y m−1 d xd y = h l+m
0≤x,0≤y,x+y≤h
=
˜
( x )l−1 ( y )m−1 ( x ) ( y ) d h d h h h
0≤ hx ,0≤ hy , hx + hy ≤1 l+m ⎡(l)⎡(m) , (by (∗)). h ⎡(l+m+1)
Thus, ¨ x l−1 y m−1 d xd y = h l+m 0≤x,0≤y,x+y≤h
⎡(l)⎡(m) . (∗∗) ⎡(l + m + 1)
∎
924
8 Conservation Principles of A.M. and Energy
8.2.20 Note Let l, m, n be positive numbers. Observe that ˚ x l−1 y m−1 z n−1 d xd ydz = 0≤x,0≤y, ˚ 0≤z,x+y+z≤1
x l−1 y m−1 z n−1 d xd ydz ⎛
0≤x,0≤y, 0≤z,x+y≤1−z
¨
z=1⎜
= ∫ ⎝z n−1 z=0
z=1
(
= ∫
⎞
⎟ x l−1 y m−1 d xd y ⎠dz
)) ( l+m ⎡(l)⎡(m) dz (by 8.2.19(∗∗)) · (1 − z) ⎡(l + m + 1) z=1 ( ) ⎡(l)⎡(m) ∫ z n−1 · (1 − z)l+m dz = ⎡(l + m + 1) z=0 ⎡(l)⎡(m) · B(n, (l + m) + 1) = ⎡(l + m + 1) ⎡(l)⎡(m) ⎡(n)⎡((l+m)+1) = ⎡(l+m+1) · ⎡(n+((l+m)+1)) (by 4.2.4) . = ⎡(l)⎡(m)⎡(n) ⎡(l+m+n+1) 0≤x,0≤y,x+y≤1−z
z
n−1
z=0
Thus ˚ x l−1 y m−1 z n−1 d xd ydz = 0≤x,0≤y,0≤z,x+y+z≤1
⎡(l)⎡(m)⎡(n) . ⎡(l + m + n + 1)
Ts is known as the Dirichlet Theorem on triplet integral.
8.2.21 Note By Sect. 4.2.4, θ = π2
∫ sin2x−1 θ cos2y−1 θ dθ =
θ =0
On taking
1 2
for x, and
1 2
for y, we have
⎡(x) · ⎡(y) . 2⎡(x + y)
8.2 Examples on Conservation of A.M.
925
( ) ( ) ( ) ( ) θ = π2 ⎡ 21 · ⎡ 21 ⎡ 21 · ⎡ 21 π = = ∫ 1dθ = −u 1−1 du 2 2⎡(1) 2 ∫u=∞ θ =0 u=0 e u ︸ ︷︷ ︸ ( ) ( ) ( ) ( ) ( ) ( ) ⎡ 21 · ⎡ 21 ⎡ 21 · ⎡ 21 ⎡ 21 · ⎡ 21 = , = = −u u=∞ −2(0 − 1) 2 2e  −1 u=0
and hence ( ) √ 1 = π. ⎡ 2
8.2.22 Example If an octant of an ellipsoid bounded by three principal planes be rotating with uniform angular velocity ω about the axis a, and if this axis be suddenly freed and the axis b fixed, show that the new angular velocity is 2abω ). ( π a 2 + c2 Sol: Let M be the mass of the octant. Let G be the centre of mass of the octate. Let the ellipsoid be x2 y2 z2 + 2 + 2 = 1. 2 a b c Let ωx be the angular velocity about the line through G parallel to x − axis, ω y be the angular velocity about the line through G parallel to y − axis, and ωz be the angular velocity about the line through G parallel to z − axis. By Sect. 7.1.31, the moment of momentum of the octant about the line, through its centre of mass, parallel to y − axis = Bω y − Dωz − Fωx = B0 − D0 − Fω Σ Σ = −ω (dm)(x · y) = −ω (dm)(x · y) octant
(
˚ = −ω x2 a2
2 + by2
2 + cz 2