Solutions to Irodovs Problems in General Physics - Volume I (Abhay Kumar Singh)

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<p>in General Physics</p><p>Vol I</p><p>Solutions to Irodov's Problems</p><p>in General Physics</p><p>Vol I</p><p>Solutions to Irodov's Problems</p><p>Abhay Kumar Singh</p><p>Copyright © 2010 by Wiley India Pvt. Ltd., 4435-36/7, Ansari Road, Daryaganj, New Delhi-110002.</p><p>All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by</p><p>any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the</p><p>publisher.</p><p>Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and</p><p>the author make no representation or warranties with respect to the accuracy or completeness of the contents of this</p><p>book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are</p><p>no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or</p><p>extended by sales representatives or written sales materials. The accuracy and completeness of the information provided</p><p>herein and the opinions stated herein are not guaranteed or warranted to produce any particular results, and the advice</p><p>and strategies contained herein may not be suitable for every individual. Neither Wiley India nor the author shall be liable</p><p>for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or</p><p>other damages.</p><p>Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely,</p><p>Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author</p><p>shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the</p><p>book. This publication is designed to provide accurate and authoritative information with regard to the subject matter</p><p>covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services.</p><p>Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade</p><p>names of their respective holders. Wiley is not associated with any product or vendor mentioned in this book.</p><p>Other Wiley Editorial Offices:</p><p>John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA</p><p>Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany</p><p>John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia</p><p>John Wiley & Sons (Asia) Pte Ltd, 2 Clementi Loop #02-01, Jin Xing Distripark, Singapore 129809</p><p>John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1</p><p>Third Edition: 2010</p><p>ISBN: 978-81-265-2076-3</p><p>www.wileyindia.com</p><p>Printed at:</p><p>Solutions to Irodov's Problems in General Physics (Vol I)</p><p>ISBN: 978-81-265- -8030 9 (ebk)</p><p>Dedicated to</p><p>My teacher</p><p>Prof. (Dr.) J. Thakur</p><p>(Ex V.C. and HOD Department of Physics,</p><p>Patna University)</p><p>PREFACE TO THE THIRD EDITION</p><p>It is my privilege to bring to you the third edition of SOLUTIONS TO IRODOV’S PROBLEMS</p><p>IN GENERAL PHYSICS. This edition is a vastly improved version of the two previous</p><p>editions, with nearly 25 newly added alternate solutions, nearly 40 modified solutions,</p><p>better diagrams to illustrate the solutions, and crisper explanations. I have also changed</p><p>some solutions altogether, with the view that each problem has many methods of solving,</p><p>but only one, or maximum two of them are precise and fitting.</p><p>To ensure that all solutions in this edition are concise as well as compact, I have</p><p>tried to solve each problem in minimum possible steps. Besides this, I did not want to</p><p>spoon-feed the students, but to make them think and apply their intelligence. So, I have</p><p>tried to present the solution in a manner which will challenge their knowledge of concepts</p><p>of physics. I chose this method because I recognize that the students of modern India need</p><p>to be armed with a strong knowledge of concepts to be able to compete with the best</p><p>physicists in the world. And they will only reach this standard if they are proficient at prob-</p><p>lem-solving and not afraid of challenges.</p><p>Similar to the previous editions, the first volume includes the first three parts, contain-</p><p>ing 1052 problems, and the second volume contains the remaining three parts with 826</p><p>problems.</p><p>I conclude here with the belief that nothing succeeds like honest hard work and</p><p>conviction in one’s capability. I extend my deep gratitude and heartfelt wishes to all those</p><p>who have directly or indirectly contributed to not only the creation of this edition, but also</p><p>those who have shown faith in the authenticity and usefulness of the solutions. I would like</p><p>to mention especially the editorial team at Wiley India, who were dedicated throughout to</p><p>making this edition the best and the most comprehensive among all editions of Solutions</p><p>to Irodov’s Problems. Their ideas and attention to detail have given me the feeling of</p><p>satisfaction at a task well done.</p><p>Finally, my best wishes to all my students and students throughout the country who will</p><p>be helped by this book.</p><p>You can email me at: authoraks@gmail.com</p><p>Abhay Kumar Singh</p><p>May 2010</p><p>Mata Khudi lane</p><p>Mahendru</p><p>Patna-6</p><p>PREFACE TO THE FIRST EDITION</p><p>When you envisage to write a book of solutions to problems, one pertinent question crops</p><p>up in the mind that – why solution! Is this to prove one’s erudition? My only defense against</p><p>this is that the solution is a challenge to save the scientific man hours by channelizing</p><p>thoughts in a right direction.</p><p>The book entitled “Problems in General Physics” authored by I.E. Irodov (a noted</p><p>Russian physicist and mathematician) contains 1877 intriguing problems divided into six</p><p>chapters.</p><p>After the acceptance of my first book “Problems in Physics”, published by Wiley Eastern</p><p>Limited, I have got the courage to acknowledge the fact that good and honest ultimately</p><p>win in the market place. This stimulation provided me insight to come up with my second</p><p>attempt – “Solutions to I.E. Irodov’s Problems in General Physics.”</p><p>The first volume encompasses solutions of first three chapters containing 1052 prob-</p><p>lems. Although a large number of problems can be solved by different methods, I have</p><p>adopted standard methods and in many of the problems with helping hints for other meth-</p><p>ods.</p><p>In the solutions of chapter three, the emf of a cell is represented by (xi) in contrast to</p><p>the notation used in figures and in the problem book, due to some printing difficulty.</p><p>I am thankful to my students Mr. Omprakash, Miss Neera and Miss Punam for their valu-</p><p>able co-operation even in my hard days while authoring the present book. I am also thank-</p><p>ful to my younger sister Prof. Ranju Singh, my younger brother Mr. Ratan Kumar Singh, my</p><p>junior friend Miss Anupama Bharti, other well wishers and friends for their emotional</p><p>support. At last and above all I am grateful to my Ma and Pappaji for their blessings and</p><p>encouragement.</p><p>Abhay Kumar Singh</p><p>CONTENTS</p><p>Part One: Physical Fundamentals of Mechanics 1</p><p>1.1 Kinematics 1</p><p>1.2 The Fundamental Equation of Dynamics 39</p><p>1.3 Law of Conservation of Energy, Momentum and Angular Momentum 76</p><p>1.4 Universal Gravitation 123</p><p>1.5 Dynamics of a Solid Body 140</p><p>1.6 Elastic Deformations of a Solid Body 170</p><p>1.7 Hydrodynamics 187</p><p>1.8 Relativistic Mechanics 202</p><p>Part Two: Thermodynamics and Molecular Physics 223</p><p>2.1 Equation of the Gas State. Processes 223</p><p>2.2 The First Law of Thermodynamics. Heat Capacity 236</p><p>2.3 Kinetic Theory of Gases. Boltzmann’s Law and Maxwell’s Distribution 257</p><p>2.4 The Second Law of Thermodynamics. Entropy 273</p><p>2.5 Liquids. Capillary Effects 289</p><p>2.6 Phase Transformations 297</p><p>2.7 Transport Phenomena 308</p><p>Part Three: Electrodynamics 321</p><p>3.1 Constant Electric Field in Vacuum 321</p><p>3.2 Conductors and Dielectrics in an Electric</p><p>+ mw)</p><p>fr1 + fr2 … k (N1 + N2)</p><p>(fr1 + fr2) = m ( g - w)</p><p>fr2 = mg - T</p><p>N2 = mw</p><p>52 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>T</p><p>fr2</p><p>mg</p><p>T</p><p>N1</p><p>fr1</p><p>N2</p><p>w</p><p>mg</p><p>mg</p><p>N</p><p>x y</p><p>fr</p><p>w</p><p>-</p><p>- -</p><p>1.81 Let us draw the force diagram of each body, and on this basis we observe that the</p><p>prism moves towards right (say) with an acceleration and the bar 2 of mass</p><p>moves down the plane with respect to 1, say with acceleration , then</p><p>(see figure).</p><p>Let us write Newton’s second law of both bodies in projection form along positive</p><p>and axes as shown in the figure.</p><p>or (1)</p><p>and (2)</p><p>Solving Eqs. (1) and (2), we get</p><p>1.82 To analyse the kinematical relations between the bodies, sketch the force diagram of each</p><p>body as shown in the figure.</p><p>On the basis of force diagram, it</p><p>is obvious that the wedge M</p><p>will move towards right and the</p><p>block will move down along</p><p>the wedge. As the length of the</p><p>thread is constant, the distance</p><p>travelled by the block on the</p><p>wedge must be equal to the dis-</p><p>tance travelled by the wedge on the floor. Hence . As and do not</p><p>change their directions and acceleration that’s why and and</p><p>(say) and accordingly the diagram of kinematical dependence is shown in</p><p>figure.</p><p>As , so from triangle law of vector addition</p><p>(1)</p><p>From , (for the wedge),</p><p>(2)</p><p>For the bar let us fix coordinate system in the frame of ground. Newton’s law in</p><p>in projection form along x and y axes (see figure) gives</p><p>x ym</p><p>T - T cos a + N sin a = M w</p><p>Fx = mwx</p><p>wm = 2w2</p><p>M + w2</p><p>M - 2 wmM wM cos a = w22(1 - cos a)</p><p>wm = wmM + wM</p><p>wmM = w</p><p>c cvMwM</p><p>c cvmM wmM</p><p>vMvmMdsmM = dsM</p><p>w1 =</p><p>m2g sin a cos a</p><p>m1 + m2 sin</p><p>2 a</p><p>=</p><p>g sin a cos a</p><p>(m1/m2) + sin2</p><p>a</p><p>N sin a = m1w1</p><p>m2g cos a - N = m2w1 sin a</p><p>m2g cos a - N = m2w2(y2)</p><p>= m2 [w21(y2)</p><p>+ w1(y 2)</p><p>] = m2 [0 + w1 sin a]</p><p>x1</p><p>y2</p><p>w2 = w21 + w1</p><p>w21</p><p>m2w1</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 53</p><p>x2 w1</p><p>w2</p><p>w21</p><p>mg</p><p>N</p><p>y2 N</p><p>x1</p><p>mg</p><p>N Mg</p><p>T</p><p>T T</p><p>N</p><p>x</p><p>y</p><p>x1</p><p>N'</p><p>wm</p><p>w</p><p>wM</p><p>M</p><p>m</p><p>-</p><p>- -</p><p>(3)</p><p>(4)</p><p>Solving the above equations simultaneously, we get</p><p>Note: We can study the motion of the block m in the frame of wedge also, alternately</p><p>we may solve this problem using conservation of mechanical energy.</p><p>1.83 Let us sketch the diagram for the motion of the particle of mass m along the circle of</p><p>radius R as shown in the figure.</p><p>For a particle, modulus of change in linear momentum .</p><p>(a) In this case (see figure), and the time taken in describing quarter of the circle,</p><p>Hence,</p><p>(b) In this case so, . Thus</p><p>Hence,</p><p>1.84 While moving in a loop, normal reaction exerted by the flyer on the loop at different</p><p>points and uncompensated weight if any contribute to the weight of flyer at those points.</p><p>(a) When the aircraft is at the lowermost point, Newton’s second law of motion in</p><p>projection form, gives</p><p>or</p><p>(b)When it is at the upper most point, again from , we get</p><p>N – =</p><p>mv2</p><p>R</p><p>- mg = 0.7 kN</p><p>N – + mg =</p><p>mv2</p><p>R</p><p>Fn = mwn</p><p>N = mg +</p><p>mv 2</p><p>R</p><p>= 2.09 kN</p><p>N - mg =</p><p>mv2</p><p>R</p><p>Fn = mwn</p><p>| 6F7 | =</p><p>| ¢p |</p><p>¢t</p><p>=</p><p>m| ¢v |</p><p>¢t</p><p>=</p><p>m| v(t) |</p><p>t</p><p>= mwt</p><p>| ¢v | = | v(t) | = v (t) = wttvf = v(t)vi = 0</p><p>=</p><p>22 mv</p><p>pR/2v</p><p>=</p><p>2 22 mv2</p><p>pR</p><p>| 6 F 7 | =</p><p>| ¢ p |</p><p>¢t</p><p>=</p><p>m | ¢v |</p><p>¢t</p><p>¢t =</p><p>pR</p><p>2v</p><p>| ¢v | =22v</p><p>| ¢p | = m | ¢v |</p><p>w =</p><p>mg sin a</p><p>M + 2m (1 - cos a)</p><p>mg cos a - N = mwm(y) = m [wmM (y) + wm (y)] = m [0 + w sin a]</p><p>= m [wmM + wM cos (p - a)] = mw (1 - cos a)</p><p>mg sin a - T = mwm (x) = m 3wmM (x) + wM (x)4</p><p>54 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>R</p><p>vi</p><p>vf</p><p>O</p><p>O N'</p><p>mg</p><p>mg</p><p>mg</p><p>N</p><p>N"</p><p>.</p><p>(c) When the aircraft is at the middle point of the loop, again from</p><p>The uncompensated weight is . Thus effective weight</p><p>acts obliquely.</p><p>1.85 (a) Let us depict the forces acting on the small sphere m, (at an arbitrary position</p><p>when the thread makes an angle from the vertical) and write equation</p><p>via projection on the unit vectors and . From , we have</p><p>(1)</p><p>Note: Eq. (1) can be easily obtained by the conservation of mechanical energy.</p><p>From</p><p>Using Eq. (1) we have</p><p>(2)</p><p>Again from equation</p><p>(3)mg sin u = mwt or wt = g sin u</p><p>Ft = mwt</p><p>T = 3mg cos u</p><p>T - mg cos u =</p><p>mv2</p><p>l</p><p>Fn = mwn</p><p>v2</p><p>l</p><p>= 2 g cos u = wn</p><p>v 2</p><p>2</p><p>= g l cos u</p><p>3</p><p>v</p><p>0</p><p>vdv = -g l3</p><p>u</p><p>p/2</p><p>sin u d u</p><p>vdv = -gl sin ud u</p><p>= m</p><p>vdv</p><p>ds</p><p>= m</p><p>vdv</p><p>l ( - d u)</p><p>mg sin u = m</p><p>dv</p><p>dt</p><p>Ft = mwtnt</p><p>F = mwu</p><p>= 1.56 kN= 2N ¿2 + m2g</p><p>2mg</p><p>N ¿ =</p><p>mv2</p><p>R</p><p>= 1.4 kN</p><p>Fn = mwn</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 55</p><p>l</p><p>T</p><p>n</p><p>mg</p><p>t</p><p>Hence,</p><p>or</p><p>Integrating both the sides we get</p><p>or</p><p>(as vertical is reference line of angular position)</p><p>,</p><p>Hence, (using Eqs. 1 and 3)</p><p>.</p><p>(b) Vertical component of velocity,</p><p>So, (using Eq. 1)</p><p>For maximum</p><p>which yields</p><p>Therefore from Eq. (2),</p><p>(c) We have thus</p><p>But in accordance with the problem</p><p>So,</p><p>or</p><p>or or,</p><p>1.86 The ball has only normal acceleration at the lowest position and only tangential</p><p>acceleration at either of the extreme positions, Let v be the speed of the ball at its</p><p>lowest position and l be length of the thread, then according to the problem</p><p>(1)</p><p>where is the maximum deflection angle.</p><p>From Newton’s law in projection form:</p><p>or</p><p>On integrating both the sides within their limits.</p><p>- gl3</p><p>a</p><p>0</p><p>sin ud u = 3</p><p>0</p><p>v</p><p>vdv</p><p>-gl sin ud u = vdv</p><p>-mg sin u = mv</p><p>dv</p><p>ldu</p><p>Ft = mwt</p><p>a</p><p>v</p><p>2</p><p>l</p><p>= g sin a</p><p>u = 54.7°cos u =</p><p>1</p><p>23</p><p>-g sin u sin u + 2g cos2u = 0</p><p>wt (y) + wn (y) = 0</p><p>wy = 0</p><p>wy = wt (y) + wn (y)w = wtt + wnn</p><p>T = 3mg</p><p>1</p><p>23</p><p>= 23 mg</p><p>cos u =</p><p>1</p><p>23</p><p>vy or v2</p><p>y ,</p><p>d (cos u sin2</p><p>u)</p><p>d u</p><p>= 0</p><p>v2</p><p>y = v</p><p>2 sin2 u = 2 gl cos u sin2 u</p><p>vy = v sin u</p><p>= g 21 + 3 cos2u</p><p>w = 2w2</p><p>t + w2</p><p>n = 2(g sin u)2 + (2g cos u)2</p><p>56 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>l</p><p>mg</p><p>v</p><p>T</p><p>α</p><p>t</p><p>n</p><p>θ</p><p>or (2)</p><p>Note: Eq. (2) can easily be obtained by the conservation of mechanical energy of the</p><p>ball in the uniform field of gravity.</p><p>From Eqs. (1) and (2) with</p><p>or</p><p>On solving we get,</p><p>1.87 Let us depict the forces acting on the body A (which are the force of gravity mg and</p><p>the normal reaction N) and write equation via projection on the unit vectors</p><p>and (see figure).</p><p>From</p><p>or</p><p>Integrating both sides for obtaining</p><p>or (1)</p><p>From</p><p>(2)</p><p>At the moment the body loses contact with the surface, and therefore Eq. (2)</p><p>becomes</p><p>(3)</p><p>where v and correspond to the moment when the body loses contact with the surface.</p><p>Solving Eqs. (1) and (3), we obtain</p><p>or and</p><p>1.88 At first draw the free body diagram of the device as, shown. The forces, acting on the</p><p>sleeve are its weight, acting vertically downward, spring force, along the length of the</p><p>spring and normal reaction by the rod, perpendicular to its length.</p><p>Let be the spring force, and be the elongation.¢lF</p><p>v = 22gR/3u = cos-1 (2/3)�48°cos u =</p><p>2</p><p>3</p><p>u</p><p>v</p><p>2 = gR cos u</p><p>N = 0,</p><p>mg cos u - N = m</p><p>v 2</p><p>R</p><p>Fn = mwn</p><p>v</p><p>2 = 2gR (1 - cos u)</p><p>3</p><p>u</p><p>0</p><p>gR sin u du¿ = 3</p><p>v</p><p>0</p><p>vdv</p><p>v (u), we get</p><p>gR sin ud u = vdv</p><p>= m</p><p>vdv</p><p>ds</p><p>= m</p><p>vdv</p><p>Rdu</p><p>mg sin u = m</p><p>dv</p><p>dt</p><p>Ft = mwt</p><p>nt</p><p>F = m w</p><p>a�53°</p><p>sin a + 2 cos a = 2</p><p>sin a = 2(1 - cos a)</p><p>u = a</p><p>v</p><p>2 = 2gl (1 - cos a)</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 57</p><p>mg</p><p>N</p><p>R t</p><p>n</p><p>From ,</p><p>(1)</p><p>where</p><p>Similarly, from</p><p>(2)</p><p>From Eqs. (1) and (2)</p><p>On putting</p><p>On solving, we get</p><p>It is independent of the direction of rotation.</p><p>1.89 According to the question, the cyclist moves along the circular path and the centripetal</p><p>force is provided by the frictional force. Thus from the equation</p><p>or (1)</p><p>For , we should have</p><p>or , so</p><p>Hence, vmax =</p><p>1</p><p>2</p><p>2k0gR</p><p>r =</p><p>R</p><p>2</p><p>1 -</p><p>2r</p><p>R</p><p>= 0</p><p>d ar -</p><p>r 2</p><p>R</p><p>b</p><p>dr</p><p>= 0vmax</p><p>k0 a1 -</p><p>r</p><p>R</p><p>b g =</p><p>v 2</p><p>r</p><p>or v2 = k0 (r - r2/R) g</p><p>ƒr =</p><p>mv</p><p>2</p><p>r</p><p>or kmg =</p><p>mv</p><p>2</p><p>r</p><p>Fn = mwn</p><p>¢l = m v2</p><p>l0</p><p>k - m v2</p><p>=</p><p>l0</p><p>(k/m v2 - 1)</p><p>k¢l sin2</p><p>u + k ¢ l cos2</p><p>u = m v2</p><p>(l0 + ¢ l )</p><p>F = k ¢ l,</p><p>m v2 (l0 + ¢l )/cos u</p><p>F (sin u/cos u ) # sin u + F cos u = m v2 r</p><p>N cos u - F sin u = 0 or N = F sin u/cos u</p><p>Ft = mwt</p><p>r cos u = (l0 + ¢l ).</p><p>N sin u + F cos u = m v2r</p><p>Fn = mwn</p><p>58 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>F</p><p>mg</p><p>N</p><p>r</p><p>q</p><p>w=</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 59</p><p>1.90 As initial velocity is zero thus,</p><p>(1)</p><p>As the speed of the car increases with time or distance. Till the moment, slid-</p><p>ing starts, the static friction provides the required centripetal accelaration the car.</p><p>Thus,</p><p>So,</p><p>or</p><p>Hence,</p><p>So, from Eq. (1), the sought distance</p><p>1.91 Since the car follows a sinusoidal curve, so the maximum velocity at which it can ride</p><p>without sliding is the point of minimum radius of curvature obviously in this case the</p><p>static friction between the car and the road is limiting. Hence, from the equation</p><p>or</p><p>So, (1)</p><p>We know that the radius of curvature for a curve at any point (x, y) is given as</p><p>(2)</p><p>For the given curve,</p><p>Substituting this value in Eq. (2), we get</p><p>R =</p><p>[1 + (a2/a2) cos2 (x/a)]3/2</p><p>(a/a2) sin (x/a)</p><p>dy</p><p>dx</p><p>=</p><p>a</p><p>a</p><p>cos a x</p><p>a</p><p>b and</p><p>d 2y</p><p>dx2</p><p>=</p><p>-a</p><p>a2</p><p>sin</p><p>x</p><p>a</p><p>R = 2 [1 + (dy>dx)2]3/2</p><p>(d 2y)>dx2</p><p>2</p><p>vmax = 2kR min</p><p>g</p><p>v … 2 kRg</p><p>Fn = mwn</p><p>mv</p><p>2</p><p>R</p><p>… kmg</p><p>s =</p><p>v2</p><p>max</p><p>2wt</p><p>=</p><p>R</p><p>2</p><p>Ba kg</p><p>wt</p><p>b - 1 = 60 m</p><p>v2</p><p>max = R2(k2g2 - w2</p><p>t )</p><p>v</p><p>2 … R (k2g2 - w2</p><p>t )</p><p>w2 … k2g2 or wt</p><p>2 + a v2</p><p>R</p><p>b2</p><p>… k2g2</p><p>fr = mw, but ƒr … kmg</p><p>wt 7 0</p><p>v2 = 2w</p><p>4</p><p>s = f (s)</p><p>For the minimum R,</p><p>and therefore, corresponding radius of curvature</p><p>(3)</p><p>Hence, from Eqs. (1) and (3)</p><p>1.92 The sought tensile stress acts on each element of the chain. Hence divide the chain</p><p>into small, similar elements so that each element may be assumed as a particle. We</p><p>consider one such element of mass , which subtends angle at the centre. The</p><p>chain moves along a circle of known radius R with known angular speed and</p><p>certain forces act on it. We have to find one of these forces.</p><p>v</p><p>d adm</p><p>vmax = a2kg /a</p><p>Rmin =</p><p>a2</p><p>a</p><p>x</p><p>a</p><p>=</p><p>p</p><p>2</p><p>60 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>x</p><p>O</p><p>dmg</p><p>dm</p><p>dN</p><p>z</p><p>y</p><p>R</p><p>y</p><p>TT</p><p>O</p><p>dm</p><p>x</p><p>d</p><p>From Newton’s second law in projection form, , we get</p><p>and from , we get</p><p>Then putting and and solving, we get</p><p>1.93 Let us consider a small element of the thread and draw free body diagram for this</p><p>element.</p><p>(a) Applying Newton’s second law of motion in projection form, for this</p><p>element,</p><p>Fn = mwn</p><p>T =</p><p>m (v2 R + g cot u)</p><p>2p</p><p>sin (da/2) = da/2dm = mda/2p</p><p>dN sin u = (dm) g</p><p>FZ = mwz</p><p>2T sin (da/2) - dN cos u = dm v2R</p><p>Fx = mwx</p><p>or [neglecting the term</p><p>or (1)</p><p>Also, (2)</p><p>From Eqs. (1) and (2)</p><p>Integrating from to ,</p><p>(3)</p><p>So,</p><p>(b) When , the blocks will move with same value of acceleration</p><p>(say ) and clearly moves downward. From Newton’s second law in projec-</p><p>tion form (downward for and upward for ), we get</p><p>(4)</p><p>and (5)</p><p>Also (6)</p><p>Simultaneous solution of Eqs. (4), (5) and (6) yields</p><p>1.94 The force with which the cylinder wall acts on the particle will provide centripetal</p><p>force necessary for the motion of the particle, and since there is no acceleration act-</p><p>ing in the horizontal direction, horizontal component of the velocity will remain con-</p><p>stant throughout the motion.</p><p>So, vx = v0 cos a</p><p>w =</p><p>(m2 - h0m1)g</p><p>(m2 + h0m1)</p><p>=</p><p>(h - h0)</p><p>(h + h0)</p><p>g aas m2</p><p>m1</p><p>= hb</p><p>T2</p><p>T1</p><p>= h0</p><p>T2 - m1g = m1w</p><p>m2g - T2 = m2w</p><p>m1m2</p><p>m2w</p><p>m2</p><p>/m1 = h 7 h0</p><p>¢as T2</p><p>T1</p><p>=</p><p>mgg</p><p>m1g</p><p>=</p><p>m2</p><p>m1</p><p>= h0≤</p><p>k =</p><p>1</p><p>p</p><p>ln</p><p>T2</p><p>T1</p><p>=</p><p>1</p><p>p</p><p>ln h0</p><p>ln</p><p>T2</p><p>T1</p><p>= kp</p><p>u = pu = 0u</p><p>kT du = dT or</p><p>dT</p><p>T</p><p>= kd u</p><p>dfr = kdN = (T + dT) - T = dT</p><p>aas sin</p><p>du</p><p>2</p><p>L</p><p>du</p><p>2</p><p>bT du = dN</p><p>(dT sin du/2)]2T sin (du/2) = dN</p><p>(T + dT ) sin (du/2) + T sin (du/2) - dN = dmv2R = 0</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 61</p><p>dN</p><p>T+dT</p><p>T</p><p>T2</p><p>m2</p><p>m1</p><p>T1</p><p>d</p><p>kdN</p><p>we get</p><p>Using, , for the particle of mass m</p><p>which is the required normal force.</p><p>1.95 Obviously the radius vector describing the position of the particle rela-</p><p>tive to the origin of the coordinate is,</p><p>Differentiating twice with respect to time</p><p>(1)</p><p>Thus, (using Eq. 1)</p><p>1.96 (a) We have (1)</p><p>(b) Using the solution of problem 1.28 (b), the total time of motion,</p><p>Hence, using</p><p>1.97 From the equation of the given time dependence force at , the</p><p>force vanishes.</p><p>(a) Thus,</p><p>or</p><p>But, p = m v so, v =</p><p>at3</p><p>6m</p><p>p = 3</p><p>t</p><p>0</p><p>a t (t - t) dt =</p><p>at3</p><p>6</p><p>¢p = p = 3</p><p>t</p><p>0</p><p>Fdt</p><p>t = tF = a t (t - t)</p><p>=</p><p>-2m (v0</p><p># g)</p><p>g</p><p>| ¢p | = mg t</p><p>t = t in Eq. (1)</p><p>t = -</p><p>2(v0</p><p># g)</p><p>g2</p><p>¢p = 3Fdt = 3</p><p>t</p><p>0</p><p>m g dt = m gt</p><p>F = m w = -m v2r</p><p>w =</p><p>d</p><p>2r</p><p>dt</p><p>2</p><p>= - �2 (a sin v t i + b cos v t j) = -v2r</p><p>r = xi + y j = a sin v t i + b cos v t j</p><p>N =</p><p>mv2</p><p>x</p><p>R</p><p>=</p><p>mv2</p><p>0 cos2 a</p><p>R</p><p>Fn = mwn</p><p>62 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>x</p><p>v0</p><p>(b) Again from the equation</p><p>or</p><p>Integrating within the limits for</p><p>or</p><p>Thus,</p><p>Hence, distance covered during the time interval is given by</p><p>1.98 We have</p><p>or</p><p>On integrating,</p><p>(where C is integration constant)</p><p>When</p><p>Hence,</p><p>As so,</p><p>Thus, .</p><p>(see figure in the answer sheet)</p><p>s = 3</p><p>t</p><p>0</p><p>vdt =</p><p>F0t</p><p>mv</p><p>-</p><p>F0 sin vt</p><p>mv2</p><p>=</p><p>F0</p><p>mv2</p><p>(vt - sin vt)</p><p>v =</p><p>F0</p><p>mv</p><p>(1 - cos vt)</p><p>cos vt … 1</p><p>v =</p><p>-F0</p><p>mv</p><p>cos v t +</p><p>F0</p><p>mv</p><p>t = 0, v = 0, so C =</p><p>F0</p><p>mv</p><p>mv =</p><p>-F0</p><p>v</p><p>cos vt + C</p><p>m</p><p>dv</p><p>dt</p><p>= F0 sin vt or md v = F0 sin vt dt</p><p>F = F0 sin vt</p><p>s = 3</p><p>t</p><p>0</p><p>vdt = 3</p><p>t</p><p>0</p><p>at 2</p><p>m</p><p>a t</p><p>2</p><p>-</p><p>t</p><p>3</p><p>bdt =</p><p>a</p><p>m</p><p>t4</p><p>12</p><p>t = t</p><p>v =</p><p>at2</p><p>m</p><p>a t</p><p>2</p><p>-</p><p>t</p><p>3</p><p>b for t … t</p><p>v =</p><p>a</p><p>m</p><p>a tt2</p><p>2</p><p>-</p><p>t3</p><p>3</p><p>b =</p><p>at2</p><p>m</p><p>a t</p><p>2</p><p>-</p><p>t</p><p>3</p><p>b</p><p>3</p><p>t</p><p>0</p><p>a (t t - t 2) dt = m 3</p><p>v</p><p>0</p><p>d v</p><p>v (t)</p><p>a (t t - t2) dt = md v</p><p>at (t - t) = m</p><p>d v</p><p>dt</p><p>F = mw</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 63</p><p>1.99 According to the problem, the force acting on the particle of mass is, .</p><p>So,</p><p>Integrating, within the limits</p><p>(1)</p><p>It is clear from Eq. (1), that after starting at , the particle comes to rest for the</p><p>first time at .</p><p>From Eq. (1)</p><p>for (2)</p><p>Thus during the time interval , the sought distance</p><p>From Eq. (1)</p><p>1.100 (a) From the problem</p><p>Thus,</p><p>or</p><p>On integrating ln</p><p>But at</p><p>So, ln</p><p>Thus, for .</p><p>(b) We have</p><p>Integrating within the given limits to obtain , we getv (s)</p><p>m</p><p>dv</p><p>dt</p><p>= - rv so dv =</p><p>- r</p><p>m</p><p>ds</p><p>t : q, v = 0</p><p>v</p><p>v0</p><p>= -</p><p>r</p><p>m</p><p>t or v =v0e</p><p>- (r /m)t</p><p>t = 0, v = v0 so, C = ln v0</p><p>v = -</p><p>r</p><p>m</p><p>t + C</p><p>dv</p><p>v</p><p>= -</p><p>r</p><p>m</p><p>dt</p><p>m</p><p>dv</p><p>dt</p><p>= - rv [as dv c Tv]</p><p>F = - r v so, m</p><p>dv</p><p>dt</p><p>= - r v</p><p>vmax =</p><p>F0</p><p>mv</p><p>as | sin vt | … 1</p><p>s =</p><p>F0</p><p>mw</p><p>3</p><p>p/v</p><p>0</p><p>sin vt dt =</p><p>2F</p><p>mv2</p><p>t = p/v</p><p>t …</p><p>p</p><p>v</p><p>v = | v | =</p><p>F0</p><p>mv</p><p>sin vt</p><p>t = p>v t = 0</p><p>3</p><p>v</p><p>0</p><p>d v =</p><p>F0</p><p>m</p><p>3</p><p>t</p><p>0</p><p>cos vt dt or v =</p><p>F0</p><p>mv</p><p>sin vt</p><p>m</p><p>dv</p><p>dt</p><p>= F0 cos vt or d v =</p><p>F0</p><p>m</p><p>cos vt dt</p><p>F = F0 cos vtm</p><p>64 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>,</p><p>.</p><p>or, (1)</p><p>Thus for</p><p>(c) We have</p><p>or</p><p>So,</p><p>Now, average velocity over this time interval</p><p>�</p><p>1.101 According to the problem</p><p>Integrating, within the limits</p><p>(1)</p><p>To find the value of , rewrite</p><p>On integrating</p><p>3</p><p>v</p><p>v0</p><p>dv</p><p>v</p><p>= -</p><p>k</p><p>m</p><p>3</p><p>h</p><p>0</p><p>ds</p><p>mv</p><p>dv</p><p>ds</p><p>= -kv2 or dv</p><p>v</p><p>= -</p><p>k</p><p>m</p><p>ds</p><p>k</p><p>3</p><p>v</p><p>v0</p><p>dv</p><p>v2</p><p>= -</p><p>k</p><p>m</p><p>3</p><p>t</p><p>0</p><p>dt or t =</p><p>m</p><p>k</p><p>(v0 - v)</p><p>v0v</p><p>m</p><p>dv</p><p>dt</p><p>= -kv2 or, m</p><p>dv</p><p>v2</p><p>= -kdt</p><p>3</p><p>m</p><p>r ln h</p><p>0</p><p>v0e</p><p>-</p><p>r t</p><p>m dt</p><p>m r ln</p><p>h</p><p>=</p><p>v0(h - 1)</p><p>h ln h</p><p>6v7 = 3</p><p>vdt</p><p>3dt</p><p>t =</p><p>-m ln (1/h)</p><p>r</p><p>=</p><p>m ln h</p><p>r</p><p>3</p><p>v0/h</p><p>0</p><p>dv</p><p>v</p><p>=</p><p>- r</p><p>m</p><p>3</p><p>t</p><p>0</p><p>dt or, ln</p><p>v0</p><p>hv0</p><p>= -</p><p>r</p><p>m</p><p>t</p><p>mdv</p><p>dt</p><p>= - rv or,</p><p>dv</p><p>v</p><p>=</p><p>- r</p><p>m</p><p>dt</p><p>v = 0, s = stotal =</p><p>mv0</p><p>r</p><p>v = v0 -</p><p>r s</p><p>m</p><p>3</p><p>v</p><p>v0</p><p>dv = -</p><p>r</p><p>m</p><p>3</p><p>s</p><p>0</p><p>ds</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 65</p><p>/</p><p>So, (2)</p><p>Putting the value of from Eq. (2) in Eq. (1), we get</p><p>1.102 From Newton’s second law for the bar in projection form, , along x direc-</p><p>tion, we get</p><p>or</p><p>or</p><p>or</p><p>So, (1)</p><p>From Eq. (1),</p><p>at either</p><p>As the motion of the bar is unidirectional it stops after going through a distance of</p><p>.</p><p>From Eq. (1), for ,</p><p>or</p><p>Hence, the maximum velocity will be at the distance, .</p><p>Putting this value of x in Eq. (1), the maximum velocity,</p><p>1.103 Since the applied force is proportional to the time and the frictional force also ex-</p><p>ists, the motion does not start just after applying the force. The body starts its mo-</p><p>tion when F equals the limiting friction. Let the motion start after time , then t0</p><p>vmax = A</p><p>g sin a tan a</p><p>a</p><p>x = tan a/a</p><p>x =</p><p>1</p><p>a</p><p>tan a</p><p>d</p><p>dx</p><p>(x sin a -</p><p>x2</p><p>2</p><p>a cos a) = 0</p><p>vmax</p><p>2 /a tan a</p><p>x = 0, or x =</p><p>2</p><p>a</p><p>tan av = 0</p><p>v</p><p>2</p><p>2</p><p>= g (x sin a -</p><p>x2</p><p>2</p><p>a cos a)</p><p>3</p><p>v</p><p>0</p><p>vdv = g 3</p><p>x</p><p>0</p><p>(sin a - x cos a) dx</p><p>vdv = (g sin</p><p>a - ax g cos a) dx</p><p>v</p><p>dv</p><p>dx</p><p>= g sin a - ax g cos a (as k = ax)</p><p>mg sin a - kmg cos a = mw</p><p>Fx = mwx</p><p>t =</p><p>h (v0 - v)</p><p>v0v ln</p><p>v0</p><p>v</p><p>k</p><p>k =</p><p>m</p><p>h</p><p>ln</p><p>v0</p><p>v</p><p>66 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>mg</p><p>N</p><p>x</p><p>fr</p><p>So, for , the body remains at rest and for obviously</p><p>Integrating, and noting at , we have for</p><p>Thus,</p><p>1.104 While going upward, from Newton’s second law, in vertical direction</p><p>At the maximum height , the speed , so</p><p>Integrating and solving, we get</p><p>(1)</p><p>When the body falls downward, the net force acting on the body in downward di-</p><p>rection equals .</p><p>Hence net acceleration, in downward direction, according to second law of motion</p><p>Thus,</p><p>Integrating and putting the value of h from Eq. (1), we get</p><p>v ¿ =</p><p>v0</p><p>21 + k v2</p><p>0/mg</p><p>3</p><p>v ¿</p><p>0</p><p>vdv</p><p>g - kv2/m</p><p>= 3</p><p>h</p><p>0</p><p>ds</p><p>vdv</p><p>ds</p><p>= g -</p><p>kv2</p><p>m</p><p>or,</p><p>vdv</p><p>g - kv</p><p>2 /m</p><p>= ds</p><p>(mg - kv2)</p><p>h =</p><p>m</p><p>2k</p><p>ln a1 + +</p><p>k v2</p><p>0</p><p>mg</p><p>b</p><p>3</p><p>0</p><p>v0</p><p>vdv</p><p>g + (kv 2/m)</p><p>= -3</p><p>h</p><p>0</p><p>ds</p><p>v = 0h</p><p>m</p><p>vdv</p><p>ds</p><p>= - (mg + kv 2) or</p><p>vdv</p><p>g + kv2 m</p><p>= -ds</p><p>s = 3vdt =</p><p>a</p><p>2m</p><p>L</p><p>t</p><p>t0</p><p>(t - t0)</p><p>2 dt =</p><p>a</p><p>6m</p><p>(t - t0)</p><p>3</p><p>3</p><p>v</p><p>0</p><p>mdv = a3</p><p>t</p><p>0</p><p>(t - t0) dt or v =</p><p>a</p><p>2m</p><p>(t - t0)</p><p>2</p><p>t 7 t0t = t0v = 0</p><p>mdv</p><p>dt</p><p>= a (t - t0) or, mdv = a (t - t0) dt</p><p>t 7 t0t … t0</p><p>F = at0 = kmg or t0 =</p><p>kmg</p><p>a</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 67</p><p>/( )</p><p>1.105 Let us fix co-ordinate system to the given plane, taking x-axis in the direction</p><p>along which the force vector was oriented at the moment , then the fundamen-</p><p>tal equation of dynamics expressed via the projection on x and y axes gives,</p><p>and</p><p>(a) Using the condition , we obtain</p><p>and</p><p>Hence,</p><p>(b) It is seen from this that the velocity turns into zero after the time interval ,</p><p>which can be found from therelation, . Consequently, the sought dis-</p><p>tance is</p><p>Average velocity,</p><p>So,</p><p>1.106 The acceleration of the disk along the plane is determined by the projection of the</p><p>force of gravity on this plane and the friction force . In</p><p>our case, and therefore</p><p>Let us find the projection of acceleration on the direction of</p><p>the tangent to the trajectory and on the -axis</p><p>mwx = Fx - fr cos w = mg sin a (1 - cos w)</p><p>mwt = Fx cos w - fr = mg sin a (cos w - 1)</p><p>x</p><p>fr = Fx = mg sin a</p><p>k = tan a</p><p>fr = kmg cos aFx = mg sin a</p><p>6v7 = 3</p><p>2p/v</p><p>0</p><p>2F</p><p>mv</p><p>sin avt</p><p>2</p><p>b dt/(2pv) =</p><p>4F</p><p>pm v</p><p>6v7 = 3</p><p>v dt</p><p>3dt</p><p>s = 3</p><p>¢t</p><p>0</p><p>v dt =</p><p>8F</p><p>mv2</p><p>v ¢t</p><p>2 = p</p><p>¢tv</p><p>v = 2v2</p><p>x + v2</p><p>y = a 2F</p><p>mv</p><p>b 2 sin avt</p><p>2</p><p>b 2vy =</p><p>F</p><p>mv</p><p>(1 - cos vt)</p><p>vx =</p><p>F</p><p>mv</p><p>sin v tv (0) = 0</p><p>F sin vt = m</p><p>dvy</p><p>dt</p><p>F cos vt = m</p><p>dvx</p><p>dt</p><p>t = 0</p><p>x -y</p><p>68 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>F</p><p>y</p><p>x</p><p>θ=ωt</p><p>O</p><p>fr</p><p>x</p><p>v</p><p>It is seen from this that , which means that the velocity v and its projec-</p><p>tion differ only by a constant value C which does not change with time, i.e.,</p><p>, where . The constant C is found from the initial condi-</p><p>tion , since initially. Finally, we obtain</p><p>In the course of time and . (Motion then is unaccelerated.)</p><p>1.107 Let us consider an element of length at an angle</p><p>from the vertical diameter. As the speed of this element</p><p>is zero at initial instant of time, its centripetal accelera-</p><p>tion is zero, and hence, , where</p><p>is the linear mass density of the chain. Let T and</p><p>be the tension at the upper and the lower ends of ds,</p><p>then we have from</p><p>or</p><p>If we sum the above equation for all elements, the term , because there is</p><p>no tension at the free ends, so</p><p>Hence,</p><p>As a at initial moment, so</p><p>1.108 In the problem, we require the velocity of the body, relative to the sphere, which</p><p>itself moves with an acceleration in horizontal direction (say towards left).</p><p>Hence it is advisable to solve the problem in the frame of sphere (a translating</p><p>frame here).</p><p>w0</p><p>w = | wt | =</p><p>gR</p><p>l</p><p>a1 - cos</p><p>l</p><p>R</p><p>b</p><p>wn =</p><p>wt =</p><p>gR</p><p>l</p><p>a1 - cos</p><p>l</p><p>R</p><p>b</p><p>lgR 3</p><p>l /R</p><p>0</p><p>sin w d w = lwt3ds = llwt</p><p>3dT = 0</p><p>dT + l Rd wg sin w = ldswt</p><p>(T + dT) + l ds g sin w - T = ldswt</p><p>Ft = mwt</p><p>T + dT</p><p>ldN - lds g cos w = 0</p><p>wds</p><p>v : v0/2w: 0</p><p>v =</p><p>v0</p><p>(1 + cos w)</p><p>C = v0</p><p>w = p 2 Ux = 0v = v0</p><p>vx = v cos wv = - vx + C</p><p>vx</p><p>wt = -wx</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 69</p><p>dN</p><p>T+dT</p><p>T</p><p>dmg</p><p>d</p><p>/</p><p>,</p><p>At an arbitrary moment, when the body is at an angle</p><p>with the vertical, we sketch the force diagram for the body</p><p>and write the second law of motion in projection form</p><p>as, (1)</p><p>At the break off point, and let</p><p>so Eq. (1) becomes,</p><p>(2)</p><p>From</p><p>or</p><p>Integrating</p><p>(3)</p><p>Note that Eq. (3) can also be obtained by the work-energy theorem (in the</p><p>frame of sphere). Therefore,</p><p>[here is the work done by the pseudoforce</p><p>or</p><p>Solving Eqs. (2) and (3), we get</p><p>Hence, u0</p><p>2</p><p>w0 = g</p><p>= 17°</p><p>awhere k =</p><p>w0</p><p>g</p><p>bv0 = A</p><p>2gR</p><p>3</p><p>and u0 = cos-1 c2 + k 25 + 9k2</p><p>3 (1 + k2)</p><p>d</p><p>v2</p><p>0</p><p>2R</p><p>= g(1 - cos u0) + w0 sin u0</p><p>(-m w0)]mw0 R sin u0</p><p>mgR (1 - cos u0) + mw0 R sin u0 =</p><p>1</p><p>2</p><p>mv2</p><p>0</p><p>A = ¢T</p><p>v2</p><p>0</p><p>2R</p><p>= g 11 - cos u02 + w0 sin u0</p><p>3</p><p>v0</p><p>0</p><p>v dv = 3</p><p>u0</p><p>0</p><p>R (g sin u + w0 cos u) du</p><p>vdv = R (gsin u + w0 cos u) d u</p><p>mg sin u - mw0 cos u = m</p><p>vdv</p><p>ds</p><p>= m</p><p>v dv</p><p>Rdu</p><p>Ft = mwt</p><p>v2</p><p>0</p><p>R</p><p>= g cos u0 - w0 sin u0</p><p>v = v0</p><p>N = 0, u = u0</p><p>mg cos u - N - mw0 sin u =</p><p>mv2</p><p>R</p><p>Fn = mwn</p><p>u</p><p>70 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>N</p><p>t</p><p>mw0</p><p>mg</p><p>n</p><p>w0</p><p>as,</p><p>,</p><p>1.109 This is not central force problem unless the path is a circle about the said point.</p><p>Rather here (tangential force) vanishes. Thus, the equation of motion becomes</p><p>and</p><p>We can consider the latter equation as the equilibrium under two forces. When the</p><p>motion is perturbed, we write and the net force acting on the particle is</p><p>This opposes the displacement x, if is an outward directed centrifugal</p><p>force while is the inward directed external force.)</p><p>1.110 Let us observe the behaviour of the sleeve in the fra-</p><p>me fixed to the rotating rod bent into the shape of</p><p>half circle. We resolve all the forces into tangential</p><p>and normal components, then the net downward tange-</p><p>ntial force on the sleeve is</p><p>This vanishes for and for</p><p>which is real if .</p><p>If , then is always positive for small values of and</p><p>hence the net tangential force near opposes any displacement away from it,</p><p>and is then stable.</p><p>If is negative for small near and, is then</p><p>unstable.</p><p>However is stable because the force tends to bring the sleeve near the equi-</p><p>librium position .</p><p>If , the two positions coincide and becomes a stable equilibrium point. v2 R = g</p><p>u = u0</p><p>u= u0</p><p>u= 0u = 0uv2</p><p>R 7 g , then (1 - v2R/g) cosu</p><p>u = 0</p><p>u = 0</p><p>u(1 - v2R/g cos u )v2 R 6 g</p><p>v2R 7 g</p><p>u = u0 = cos-1( g/v2R)u = 0</p><p>mg sin ua1 -</p><p>v2R</p><p>g</p><p>cos ub</p><p>-A/r n</p><p>(m</p><p>v</p><p>2</p><p>0/rn 6 1.</p><p>-</p><p>A</p><p>(r0 + x)n</p><p>+</p><p>m v2</p><p>0</p><p>r0 + x</p><p>=</p><p>-A</p><p>rn</p><p>0</p><p>a1 -</p><p>nx</p><p>r0</p><p>b</p><p>+</p><p>m v2</p><p>0</p><p>r0</p><p>a1 -</p><p>x</p><p>r0</p><p>b</p><p>= -</p><p>m v2</p><p>0</p><p>r 2</p><p>0</p><p>(1 - n) x</p><p>r = r0 + x</p><p>m v2</p><p>0</p><p>r</p><p>=</p><p>A</p><p>rn</p><p>1for r = r02</p><p>vt = v0 = constant</p><p>Ft</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 71</p><p>N</p><p>O</p><p>O'</p><p>mg</p><p>mw2Rsinq</p><p>,</p><p>[ ]</p><p>][ (</p><p>( )</p><p>1.111 Define the axes as shown with z along the local vertical, x due east and y due no-</p><p>rth. (We assume we are in the northern hemisphere.) Then the Coriolis force has</p><p>the components as</p><p>Since is small when the direction in which the gun is</p><p>fired is due north. Thus the equations of motion (ne-</p><p>glecting centrifugal forces) are</p><p>Integrating we get</p><p>Finally,</p><p>Now, in the present case. So,</p><p>1.112 For the observer attached in the frame of disks</p><p>So,</p><p>So,</p><p>1.113 The sleeve is free to slide along the rod . Thus the centrifugal force acts on it in</p><p>the outward direction along the river.</p><p>So, we have</p><p>where</p><p>But, w ¿ =</p><p>vdv</p><p>dr</p><p>=</p><p>d</p><p>dr</p><p>a1</p><p>2</p><p>v2b</p><p>v¿ =</p><p>dr</p><p>dt</p><p>mv¿ = mv2r</p><p>AB</p><p>| N | = 2(2m v¿v)2 + (mg)2 + (mv2r)2</p><p>N = - (Fcor + mg + Fcf)</p><p>N + Fcor + mg + Fcf = 0</p><p>= 7 cm (to the east)</p><p>x = vv sin w a s</p><p>v</p><p>b2</p><p>= v sin u</p><p>s2</p><p>v</p><p>v 77 gt</p><p>x = vvt2 sin w +</p><p>1</p><p>3</p><p>vgt3 cos w</p><p>z = -gt and x = 2vv sin wt + vgt2 cos w</p><p>y = v (constant)</p><p>x = 2m v(vy sin w - vz cos w), y</p><p>$ = 0 and z</p><p>$ = -g</p><p>vx</p><p>= 2m v (vy cos u - vz sin u)i</p><p>= 2mv [vy cos u - vz sin u) i - vx cos u j + vx cos u k]</p><p>Fcor = -2m (v * v)</p><p>72 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>=90°–</p><p>y-North</p><p>z-Vertical</p><p>x-East</p><p>So,</p><p>or</p><p>being</p><p>the initial velocity when . The Coriolis force is then</p><p>1.114 The disk OBAC is rotating with angular velocity about the axis passing</p><p>through the edge point O. The equation of motion in rotating frame is</p><p>where is the resultant inertial force (pseudo force) which is the</p><p>vector sum of centrifugal and Coriolis forces.</p><p>(a) .</p><p>Thus,</p><p>where n is the inward drawn unit vector to the centre, from the point in question, here A.</p><p>Thus,</p><p>So,</p><p>(b) At B,</p><p>(from question)</p><p>From geometry,</p><p>(see figure)</p><p>So,</p><p>1.115 The equation of motion in the rotating coordinate system is</p><p>Now,</p><p>and v = v cos u er - v sin u e</p><p>u</p><p>v¿ = R u</p><p>.</p><p>e</p><p>u</p><p>+ R sin u w</p><p>.</p><p>e</p><p>w</p><p>m w¿ = F + m v2 R + 2m (v¿ * v)</p><p>| Fin | = mv224R2 - r2</p><p>r = 2R cos u</p><p>Fin = mv2r + 2mv2R</p><p>v ¿ = vr = constant</p><p>Fin = mv2r + 2mv¿v2 (-R)</p><p>w =</p><p>v¿2</p><p>r</p><p>=</p><p>v¿2</p><p>R</p><p>= v2R</p><p>v¿ = vR</p><p>0 = -2m v2R n + 2mv ¿vn</p><p>At A, Fin vanishes</p><p>Fin</p><p>mw¿ = F + mv2 R + 2m(v¿ * �) = F + Fin</p><p>OO¿v</p><p>2mv2v2</p><p>0 + v2r2 = 2mv2r 21 + v2</p><p>0>v2r2) = 2 # 83 N (on substituting values)</p><p>r = 0v0</p><p>v2 = v2</p><p>0 + v2r2</p><p>1</p><p>2</p><p>v2 =</p><p>1</p><p>2</p><p>v2r2 + constant</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 73</p><p>B</p><p>O'</p><p>O A</p><p>C</p><p>C</p><p>AO</p><p>O'</p><p>A</p><p>O'</p><p>R</p><p>r</p><p>B</p><p>2mv'</p><p>m 2r</p><p>v'</p><p>n</p><p>q</p><p>N</p><p>Fcf</p><p>mg</p><p>Now, on the sphere</p><p>Thus the equations of motion are</p><p>From the third equation, we get,</p><p>A result that is easy to understand by considering the motion in non-rotating frame.</p><p>Eliminating we get,</p><p>Integrating the last equation</p><p>(1)</p><p>Hence,</p><p>So the body must fly off for , exactly as if the sphere were non-</p><p>rotating.</p><p>Now, at this point</p><p>Putting the value of from Eq. (1)</p><p>we get, . (on substituting values)Fcor =</p><p>2mv2R</p><p>3</p><p>A5 +</p><p>8g</p><p>3v2R</p><p>= 17 N</p><p>u = u0</p><p>u</p><p>Fcor = 2m 2(v2R)2 sin2</p><p>u cos</p><p>2u + (v2R</p><p>2)2 sin</p><p>4 u + (vR)2cos</p><p>2 uu</p><p>#</p><p>2</p><p>F cf = centrifugal force = m v2R sin u0 = A</p><p>5</p><p>9</p><p>mv2R</p><p>u = u0 = cos-1</p><p>2</p><p>3</p><p>N = (3 cos u - 2) mg</p><p>1</p><p>2</p><p>mR u</p><p>#</p><p>2 = mg (1 - cos u)</p><p>mR u</p><p>..</p><p>= mg sin u</p><p>mR u</p><p>#</p><p>2 = mg cos - Nw</p><p>#</p><p>w</p><p># = -v</p><p>m (R sin u w</p><p>.. + 2R cos u u</p><p>.</p><p>w</p><p>#</p><p>) = -2 mvR u</p><p># #</p><p>cos u</p><p>+ mv2R sin u cos u + 2mvR sin u cos u w</p><p>#</p><p>m (R u</p><p># #</p><p>- R sin u cos u w</p><p># 2) = mg sin u</p><p>m (-R u</p><p>#</p><p>2 - R sin2u</p><p>w</p><p>#</p><p>#2) = N -mg cos u + mv2 R sin2 u + 2mvR sin2 uw</p><p>#</p><p>+ (R sin uw</p><p>..</p><p>+ 2R cos u u</p><p>.</p><p>w</p><p>#</p><p>) e</p><p>w</p><p>+ (R u</p><p>..</p><p>- R sin u</p><p>#</p><p>cos u w</p><p>#</p><p>2) e</p><p>u</p><p>w¿ = (-R u</p><p>#</p><p>2 - R sin u2 w</p><p># 2)er</p><p>= er (vR sin2</p><p>u w</p><p>#</p><p>) + [vR sin u cos u w</p><p>#</p><p>]e</p><p>u</p><p>- (vR u</p><p># cos u)ew</p><p>3 er e</p><p>u</p><p>e</p><p>w</p><p>0 R u</p><p>.</p><p>(R sin u)w</p><p>.</p><p>v cos u -v sin u 0</p><p>31</p><p>2m</p><p>Fcor =</p><p>74 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>.</p><p>1.116 (a) When the train is moving along a meridian only the Coriolis force has a lateral</p><p>component and its magnitude (see the previous problem) is,</p><p>(here we have put</p><p>So,</p><p>(b) The resultant of the inertial forces acting on the train is,</p><p>This vanishes if</p><p>Thus,</p><p>(we write for the latitude here)</p><p>Thus the train must move from the east to west along</p><p>the parallel with a speed,</p><p>.</p><p>1.117 We go to the equation given in problem 1.111. Here so we can take ;</p><p>thus, we get for the motion in the plane.</p><p>and</p><p>Integrating,</p><p>So</p><p>There is thus a displacement to the east of</p><p>2</p><p>3</p><p>* 500 *</p><p>2p</p><p>8.64</p><p>* 10-4 * 1 * A</p><p>2 * 500</p><p>9.8</p><p>= 26 cm</p><p>x =</p><p>1</p><p>3</p><p>vg cos wt</p><p>3 =</p><p>1</p><p>3</p><p>vg cos w a2h</p><p>g</p><p>b3/2</p><p>=</p><p>2vh</p><p>3</p><p>cos wA</p><p>2h</p><p>g</p><p>x</p><p># = vg cos wt2</p><p>z = -</p><p>1</p><p>2</p><p>gt2</p><p>z</p><p># = -g</p><p>x</p><p>$ = -2vvz cos w</p><p>x z</p><p>#</p><p>y = 0vy = 0</p><p>1</p><p>2</p><p>vR cos l =</p><p>1</p><p>4</p><p>*</p><p>2p</p><p>8.64</p><p>10-4 * 6.37 * 106 = 115.8 m/s L 420 km/h</p><p>60th</p><p>l</p><p>v = v</p><p>w</p><p>e</p><p>w</p><p>,v</p><p>w</p><p>= -</p><p>1</p><p>2</p><p>vR sin u = -</p><p>1</p><p>2</p><p>vR cos l</p><p>u</p><p>#</p><p>= 0, w</p><p># = -</p><p>1</p><p>2</p><p>v</p><p>+ (mv2 R sin2u + 2m vR sin2 u w</p><p>#</p><p>) er</p><p>Fin = -2mvR u</p><p>#</p><p>cos u e</p><p>w</p><p>+ (mv2 R sin u cos u + 2m vR sin u cos u w)</p><p>#</p><p>e</p><p>u</p><p>Flateral = 2 * 2000 * 103 *</p><p>2p</p><p>86400</p><p>*</p><p>54000</p><p>3600</p><p>*</p><p>23</p><p>2</p><p>R u</p><p># : v)2mvv cos u = 2mv sin l</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 75</p><p>O</p><p>Fcor</p><p>v</p><p>(we write for the latitude) l= 3.77 kN</p><p>76 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.3 Law of Conservation of Energy, Momentum</p><p>and Angular Momentum</p><p>1.118 As F is constant so the sought work done</p><p>or</p><p>1.119 As locomotive is in unidirectional motion, its acceleration</p><p>Hence, force acting on the locomotive</p><p>Let,</p><p>Hence the sought work,</p><p>1.120 We have (1)</p><p>Differentiating Eq. (1) with respect to time</p><p>(2)</p><p>Hence net acceleration of the particle</p><p>Hence the sought, force</p><p>1.121 Let make an angle with the horizontal at an arbi-</p><p>trary instant of time (see figure). Newton’s second law</p><p>in projection form along the direction of the force</p><p>gives</p><p>(because there is no acceler-</p><p>ation of the body)</p><p>F = kmg cos u + mg sin u</p><p>uF,</p><p>F = mw = 2as11 + (s/R)2</p><p>w = 2w 2</p><p>t + w 2</p><p>n = Ca2as</p><p>m</p><p>b2</p><p>+ a2as 2</p><p>mR</p><p>b2</p><p>=</p><p>2as</p><p>m</p><p>11 + (s>R) 2</p><p>2vwt =</p><p>4as</p><p>m</p><p>v or wt =</p><p>2 as</p><p>m</p><p>T =</p><p>1</p><p>2</p><p>mv 2 = as 2 or v 2 =</p><p>2 as 2</p><p>m</p><p>A = Fs =</p><p>ma 2</p><p>2</p><p>(a 2t 2)</p><p>4</p><p>=</p><p>ma 4t 2</p><p>8</p><p>s =</p><p>1</p><p>2</p><p>wt 2 =</p><p>1</p><p>2</p><p>a 2</p><p>2</p><p>t 2 =</p><p>a 2</p><p>4</p><p>t 2</p><p>v = 0 at t = 0, then the distance covered during the first t seconds</p><p>F = mw =</p><p>ma 2</p><p>2</p><p>w =</p><p>dv</p><p>dt</p><p>=</p><p>1</p><p>2</p><p>dv 2</p><p>ds</p><p>=</p><p>a 2</p><p>2</p><p>A = (3i + 4j ) # [(2i - 3j ) - (i + 2j)] = (3i + 4 j ) # (i - 5j) = 17 J</p><p>A = F # ¢r = F # (r2 - r1)</p><p>mg</p><p>N</p><p>fr</p><p>y</p><p>l</p><p>x</p><p>F</p><p>h</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 77</p><p>As , the differential work done by the force F</p><p>Hence,</p><p>1.122 Let s be the distance covered by the disk along the incline, from the equation of in-</p><p>crement of mechanical energy of the disk in the field of gravity:</p><p>or (1)</p><p>Hence the sought work</p><p>(using Eq. 1)</p><p>On putting the values</p><p>1.123 Let x be the compression in the spring when the bar is about to shift. Therefore</p><p>at this moment spring force on is equal to the limiting friction between the bar</p><p>and horizontal floor. Hence</p><p>(1)</p><p>For the block from work-energy theorem:</p><p>for minimum force. (A here includes the work done in stretching the</p><p>spring.)</p><p>So, (2)</p><p>From Eqs. (1) and (2), F = kg am1 +</p><p>m2</p><p>2</p><p>b</p><p>Fx -</p><p>1</p><p>2</p><p>kx 2 - kmgx = 0 or k</p><p>x</p><p>2</p><p>= F - km1g</p><p>A = ¢T = 0</p><p>m1</p><p>kx = km2g [where k is the spring constant (say)]</p><p>m2</p><p>m2</p><p>m2</p><p>Aƒr = -0.05 J</p><p>Aƒr = -</p><p>klmg</p><p>1 - k cot a</p><p>Aƒr = -kmg [s cos a + l ]</p><p>s =</p><p>kl</p><p>sin a - k cos a</p><p>0 + (-mgs sin a) = -kmg cos a s - kmgl</p><p>¢T + ¢U - Aƒr</p><p>= kmgl + mgh = mg (kl + h)</p><p>A = kmg3</p><p>l</p><p>0</p><p>dx + mg 3</p><p>h</p><p>0</p><p>dy</p><p>= kmg dx + mg dy</p><p>= kmg ds (cos u) + mg ds sin u</p><p>dA = F # dr = Fds (where ds = | dr | )</p><p>Fc cdr</p><p>2</p><p>N2</p><p>fr2</p><p>mg</p><p>mg</p><p>fr1</p><p>1N'1</p><p>s</p><p>l</p><p>78 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.124 From the initial condition of the problem, the limiting friction between the chain</p><p>lying on the horizontal table equals the weight of the over hanging part of the chain,</p><p>i.e., (where is the linear mass density of the chain).</p><p>So, (1)</p><p>Let (at an arbitrary moment of time) the length of the chain on the table be x. So the</p><p>net friction force between the chain and the table, at this moment</p><p>(2)</p><p>The differential work done by the friction forces</p><p>(3)</p><p>(Note that here we have written , because ds is essentially a positive term</p><p>and as the length of the chain decreases with time, dx is negative.)</p><p>Hence, the sought work done</p><p>1.125 The velocity of the body, t seconds after the beginning of the motion becomes</p><p>The power developed by the gravity at that moment, is</p><p>As is a constant force, so the average power</p><p>where is the net displacement of the body during time of flight.</p><p>As,</p><p>1.126 We have</p><p>t is defined to start from the beginning of motion from rest.</p><p>So, wt =</p><p>dv</p><p>dt</p><p>= 1aR</p><p>wn = v 2</p><p>/R = at 2 or v = 1aR t</p><p>m g � ¢r, so 6P 7 = 0</p><p>¢r</p><p>6P 7 =</p><p>A</p><p>t</p><p>=</p><p>mg # ¢r</p><p>t</p><p>mg</p><p>P = mg # v = m (g # v0 + g 2t) = mg ( gt - v0 sin a)</p><p>(mg)v = v0 + gt.</p><p>A = 3</p><p>0</p><p>(1-h)l</p><p>lg</p><p>h</p><p>1 - h</p><p>xdx = - (1 - h)h</p><p>mgl</p><p>2</p><p>= -1.3 J</p><p>ds = -dx</p><p>= -klxg (-dx) = lg a h</p><p>1 - h</p><p>bxdx</p><p>dA = fr # d r = - fr ds</p><p>ƒr = kN = klxg</p><p>k =</p><p>h</p><p>1 - h</p><p>llhlg = kl(1 - h)lg</p><p>x</p><p>fr</p><p>xg</p><p>N</p><p>(l−x)g</p><p>where</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 79</p><p>Instantaneous power, where t and n are unit</p><p>vectors along the direction of tangent (velocity) and normal, respectively.</p><p>So,</p><p>Hence the sought average power</p><p>Hence,</p><p>1.127 Let the body m acquire the horizontal velocity along positive x-axis at the point O.</p><p>(a) Velocity of the body t seconds after</p><p>the beginning of the motion,</p><p>(1)</p><p>Instantaneous power</p><p>From Eq. (1), the time of motion</p><p>Hence sought average power during the time of motion</p><p>From</p><p>or</p><p>To find let us integrate the above equation</p><p>3</p><p>v</p><p>v0</p><p>vxdvx = -ag 3</p><p>x</p><p>0</p><p>xdx or, v 2 = v 2</p><p>0 - agx 2</p><p>v (x),</p><p>vxdvx = -kgdx = -agxdx</p><p>-kmg = mwx = mvx</p><p>dvx</p><p>dx</p><p>Fx = mwx</p><p>6P 7 =</p><p>3</p><p>t</p><p>0</p><p>-kmg (v0 - kgt)dt</p><p>t</p><p>= -</p><p>kmg v0</p><p>2</p><p>= -2 W (on substituting values)</p><p>t = v0</p><p>>kg</p><p>P = F # v = (-kmg i) # (v0 - kgt) i = -kmg (v0 - kgt)</p><p>v = v0 + wt = (v0 - kg t) i</p><p>v0</p><p>6P 7 =</p><p>maRt 2</p><p>2t</p><p>=</p><p>maRt</p><p>2</p><p>6P 7 =</p><p>3</p><p>t</p><p>0</p><p>P dt</p><p>3</p><p>t</p><p>0</p><p>dt</p><p>=</p><p>3</p><p>t</p><p>0</p><p>ma Rt dt</p><p>t</p><p>P = mwt 1aR t = maRt</p><p>P = F # v = m (wt t + wnn) # (1aR t t),</p><p>.</p><p>(b)</p><p>80 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Now, (2)</p><p>For maximum power,</p><p>which yields</p><p>Putting this value of x in Eq. (2), we get</p><p>1.128 Centrifugal force of inertia is directed outward along radial line, thus the sought work</p><p>1.129 Since the springs are connected in series, the combination may be treated as a single</p><p>spring of spring constant.</p><p>So,</p><p>From the equation of increment of mechanical energy,</p><p>1.130 First, let us find the total height of ascent. At the beginning and at the end of the path,</p><p>velocity of the body is equal to zero, and therefore the increment of the kinetic energy</p><p>of the body is also equal to zero. On the other hand, according to work-energy theorem,</p><p>is equal to the algebraic sum of the work A performed by all the forces, i.e. by the</p><p>force F and gravity, over this path. However, since then Taking into ac-</p><p>count that the upward direction is assumed to coincide with the positive direction of</p><p>the y-axis, we can write</p><p>(when )h = 1>a = mg3</p><p>h</p><p>0</p><p>(1 - 2 ay) dy = mgh (1 - ah) = 0</p><p>A = 3</p><p>h</p><p>0</p><p>(F + mg) # dr = 3</p><p>h</p><p>0</p><p>(Fy - mg) dy</p><p>A = 0.¢T = 0</p><p>¢T</p><p>0 +</p><p>1</p><p>2</p><p>k¢l 2 = Aext or Aext =</p><p>1</p><p>2</p><p>a k1k2</p><p>k1 + k2</p><p>b ¢l 2</p><p>¢T + ¢Uspring = Aext</p><p>k =</p><p>k1k2</p><p>k1 + k2</p><p>A = 3</p><p>r2</p><p>r1</p><p>mv 2 rdr =</p><p>1</p><p>2</p><p>mv 2 1r</p><p>2</p><p>2 - r</p><p>2</p><p>1</p><p>) = 0.20 J (on substituting values)</p><p>Pmax = -</p><p>1</p><p>2</p><p>mv 2</p><p>01ag</p><p>x =</p><p>v0</p><p>12ag</p><p>d</p><p>dt</p><p>A2v 2</p><p>0 x 2 - lgx 4 B = 0</p><p>P = F # v = -ma xg2v 2</p><p>0 - a gx 2</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 81</p><p>The work performed by the force F over the first half of the ascent is</p><p>The corresponding increment of the potential energy is</p><p>1.131 From the equation , we get</p><p>(a) We have at the particle is in equilibrium position, i.e, .</p><p>So,</p><p>To check whether the position is steady (the position of stable equilibrium), we</p><p>have to satisfy</p><p>We have</p><p>Putting the value of , we get</p><p>(as and are positive constants)</p><p>So,</p><p>which indicates that the potential energy of the system is minimum, hence this</p><p>position is steady.</p><p>(b) We have</p><p>For F to be maximum,</p><p>So, and</p><p>As is negative, the force is attractive.Fr</p><p>Fr (max) =</p><p>-b 3</p><p>27a 2</p><p>r =</p><p>3a</p><p>b</p><p>dFr</p><p>dr</p><p>= 0</p><p>Fr = -</p><p>dU</p><p>dr</p><p>= c - 2a</p><p>r 3</p><p>+</p><p>b</p><p>r 2</p><p>d</p><p>d 2U</p><p>dr 2</p><p>=</p><p>b 2</p><p>8a 3</p><p>7 0</p><p>ba</p><p>d 2U</p><p>dr 2</p><p>=</p><p>b 4</p><p>8a 3</p><p>r = r0 = 2a /b</p><p>d 2U</p><p>dr 2</p><p>= c6a</p><p>r 4</p><p>-</p><p>2b</p><p>r 3</p><p>d</p><p>d 2U</p><p>dr 2</p><p>7 0</p><p>r0 =</p><p>2a</p><p>b</p><p>Fr = 0r = r0,</p><p>Fr = c - 2a</p><p>r 3</p><p>+</p><p>b</p><p>r 2</p><p>dFr = -</p><p>dU</p><p>dr</p><p>¢U =</p><p>mgh</p><p>2</p><p>=</p><p>mg</p><p>2a</p><p>AF = 3</p><p>h>2</p><p>0</p><p>Fydy = 2mg 3</p><p>h>2</p><p>0</p><p>(1 - ay) dy =</p><p>3 mg</p><p>4a</p><p>82 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.132 (a) We have</p><p>So,</p><p>For a central force,</p><p>Here,</p><p>Hence the force is not a central force.</p><p>(b) As</p><p>So,</p><p>So,</p><p>According to the problem</p><p>or</p><p>or</p><p>Therefore, the surface for which F is constant is an ellipse.</p><p>For an equipotential surface U is constant.</p><p>So,</p><p>or</p><p>Hence, the equipotential surface is also an ellipse.</p><p>1.133 Let us calculate the work performed by the forces of each field over the path from</p><p>a certain point to another point</p><p>(i)</p><p>(ii) dA = F # d r = (ax i + by j) # d r = ax dx + bydy</p><p>dA = F # d r = ay i # d r = ay dx or, A = a 3</p><p>x2</p><p>x1</p><p>ydx</p><p>2 (x2, y2)1 (x1, y1)</p><p>x 2</p><p>2b2</p><p>+</p><p>y 2</p><p>2a 2</p><p>=</p><p>C0</p><p>ab</p><p>= K0 (constant)</p><p>ax 2 + by 2 = C0 (constant)</p><p>x 2</p><p>b 2</p><p>+</p><p>y 2</p><p>a 2</p><p>=</p><p>C 2</p><p>2a 2b 2</p><p>= k (say)</p><p>a 2x 2 + b 2y 2 =</p><p>C 2</p><p>2</p><p>F = 22a 2x 2 + b 2y 2 = C (constant)</p><p>F = 2F 2</p><p>x + F 2</p><p>y = 24a2x 2 + 4b 2y 2</p><p>Fx =</p><p>0U</p><p>0x</p><p>= -2 ax and Fy =</p><p>-0U</p><p>0y</p><p>= -2by</p><p>U = ax 2 + by 2</p><p>= -2 bxy k - 2axy (k) Z 0</p><p>r * F = (x i + y j) * (-2a x i - 2by j)</p><p>r * F = 0.</p><p>F = 2ax i - 2by i and F = 21a2x 2 + b 2y 2</p><p>Fx = -</p><p>0U</p><p>0x</p><p>= -2ax and Fy =</p><p>-0U</p><p>0y</p><p>= -2by</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 83</p><p>Hence,</p><p>In the first case, the integral depends on the function of type i.e., on the shape</p><p>of the path. Consequently, the first field of force is not potential. In the second case,</p><p>both the integrals do not depend on the shape of the path. They are defined only by</p><p>the coordinate, of the initial and final points of the path, therefore the second field of</p><p>force is potential.</p><p>1.134 Let s be the sought distance, then from the equation of increment of mechanical energy.</p><p>or</p><p>Hence,</p><p>1.135 Velocity of the body at height , horizontally (from the figure</p><p>given in the problem book). Time taken in falling through the distance h is given by</p><p>(as initial vertical component of the velocity is zero)</p><p>Now,</p><p>For ,</p><p>Putting this value of h in the expression obtained for s, we get</p><p>1.136 To complete a smooth vertical track of radius R,</p><p>the minimum height at which a particle starts,</p><p>must be equal to Let us first prove this.</p><p>In the figure there is a smooth track and a small</p><p>body is released at height h. We have to find</p><p>minimum value for h so that the body can com-</p><p>plete smooth vertical track of radius R. As the</p><p>5</p><p>2 R.</p><p>smax = H</p><p>d</p><p>ds</p><p>(Hh - h 2) = 0, which yields h =</p><p>H</p><p>2</p><p>smax</p><p>s = vh</p><p>t = 22g (H + h) * A</p><p>2h</p><p>g</p><p>= 24(Hh - h 2)</p><p>t = A</p><p>2h</p><p>g</p><p>h, vh = 22g (H - h)</p><p>Aƒr = -kmg cos as =</p><p>-km v 2</p><p>0</p><p>2(k + tan a)</p><p>s =</p><p>v 2</p><p>0 >2g1sin a + k cos a2</p><p>a0 -</p><p>1</p><p>2</p><p>mv 2</p><p>0 b + (mg s sin a) = -kmg cos as</p><p>¢T + ¢Ugr = Afr</p><p>y (x),</p><p>A = 3</p><p>x2</p><p>x1</p><p>axdx + 3</p><p>y2</p><p>y1</p><p>bydy</p><p>h</p><p>R</p><p>mg</p><p>Nv</p><p>84 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>track is smooth and the body is under the action of only two forces, normal reac-</p><p>tion and gravitational pull. The mechanical energy of the block in the gravitational</p><p>field is conversed, because normal reaction does zero work where as the gravitation-</p><p>al force is conservative. From the conservation of mechanical energy between the</p><p>point of release and the upper most point of the circular track</p><p>which yields (1)</p><p>To complete the vertical circular track the condition is that the normal reaction at the</p><p>upper most point of the circular track</p><p>From Newtons second law in projection form towards the centre of the track</p><p>As</p><p>So,</p><p>or (2)</p><p>From Eqs. (1) and (2)</p><p>Hence,</p><p>Thus in our problem, the body could not reach the upper most point of the vertical</p><p>track of radius R/2. Let the particle A</p><p>leave the track at some point with</p><p>speed v (see figure). Now from ene-</p><p>rgy conservation for the body A in the</p><p>field of gravity</p><p>or (3)v 2 = gh (1 - sin u)</p><p>mg ch -</p><p>h</p><p>2</p><p>(1 + sin u) d =</p><p>1</p><p>2</p><p>mv 2</p><p>h Ú</p><p>5</p><p>2</p><p>R</p><p>2g (h - 2R) Ú gR</p><p>v 2 Ú gR</p><p>m a v 2</p><p>R</p><p>- gb Ú 0</p><p>N Ú 0 at upper most point of the circular track</p><p>N = m a v 2</p><p>R</p><p>- gb</p><p>N + mg = m</p><p>v 2</p><p>R</p><p>Fn = mwn (where wn is the normal acceleration)</p><p>N Ú 0.</p><p>v 2 = 2g (h - 2R)</p><p>a1</p><p>2</p><p>mv 2 - 0b - mg (h - 2R) = 0</p><p>¢T + ¢Ugr = 0</p><p>h</p><p>h/2</p><p>mg</p><p>v</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 85</p><p>At the break off point the normal reaction so from Newton’s second law at</p><p>this point</p><p>or</p><p>So, (4)</p><p>From Eqs. (3) and (4), and</p><p>After leaving the track, the body A comes in air and further goes up</p><p>and at maximum height of it’s trajectory in air and it’s velocity (say</p><p>becomes horizontal (Fig.). Hence, the sought velocity of A at this point.</p><p>1.137 Let the point of suspension be shifted with velocity in the horizontal direction</p><p>towards left, then in the rest frame of point of suspension the ball starts with same</p><p>velocity horizontally towards right. Let us work in this frame. From Newton’s sec-</p><p>ond law in projection form towards the point of suspension at the upper most point</p><p>(say B):</p><p>(1)</p><p>Condition required to complete the vertical</p><p>circle is that (2)</p><p>But,</p><p>So, (3)</p><p>From Eqs. (1), (2) and (3)</p><p>Thus</p><p>From the equation at point C</p><p>(4)T ¿ =</p><p>m v 2</p><p>C</p><p>l</p><p>Fn = mwn</p><p>vA (min) = 15gl</p><p>T =</p><p>m 1v</p><p>2</p><p>A - 4gl 2</p><p>l</p><p>- mg Ú 0 or vA Ú25gl</p><p>v</p><p>2</p><p>B = v</p><p>2</p><p>A - 4gl</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>A = mg 12l2 +</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>B</p><p>T Ú 0.</p><p>mg + T =</p><p>mv 2</p><p>B</p><p>l</p><p>or T =</p><p>mv 2</p><p>B</p><p>l</p><p>- mg</p><p>vA</p><p>v ¿ = v cos ap</p><p>2</p><p>- ub = v sin u =</p><p>2</p><p>3</p><p>A</p><p>gh</p><p>3</p><p>v ¿)</p><p>v = A</p><p>gh</p><p>3</p><p>sin u =</p><p>2</p><p>3</p><p>v 2 =</p><p>gh</p><p>2</p><p>sin u</p><p>mg sin u =</p><p>mv 2</p><p>(h>2)</p><p>Fn = mwn</p><p>N = 0,</p><p>v</p><p>θ</p><p>v'</p><p>π/2−θ</p><p>mg</p><p>T</p><p>BvB</p><p>T'</p><p>C</p><p>vC</p><p>l</p><p>A</p><p>vA</p><p>86 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Again from energy conservation</p><p>(5)</p><p>From Eqs. (4) and (5)</p><p>1.138 Since the tension is always perpendicular to the velocity vector, the work done by the</p><p>tension force will be zero. Hence, according to the work energy theorem, the kinetic</p><p>energy or velocity of the disk will remain constant during it’s motion. Hence, the</p><p>sought time , where s is the total distance traversed by the small disk during</p><p>it’s motion.</p><p>Now, at an arbitrary position (see figure)</p><p>So,</p><p>or</p><p>Hence, the required time</p><p>Note: It should be clearly understood that the only uncompensated force acting on</p><p>the disk A in this case is the tension , of the thread. It is easy to see that there is no</p><p>point here, relative to which the moment of force T is invariable in the process of</p><p>motion. Hence conservation of angular momentum is not applicable here.</p><p>1.139 Suppose that is the elongation of the rubber cord. Then, from energy conservation</p><p>or</p><p>or</p><p>or ¢l =</p><p>mg � A(mg)2 + 4 *</p><p>k</p><p>2</p><p>mgl</p><p>2 *</p><p>k</p><p>2</p><p>=</p><p>mg</p><p>k</p><p>B1 + A1 �</p><p>2kl</p><p>mg</p><p>R</p><p>1</p><p>2</p><p>k¢l 2 - mg ¢l - mgl = 0</p><p>-mg (l + ¢l) +</p><p>1</p><p>2</p><p>k ¢l 2 = 0</p><p>¢Ugr + ¢Uel = 0 (as ¢T = 0)</p><p>¢l</p><p>T</p><p>t =</p><p>l 2</p><p>0</p><p>2Rv</p><p>0</p><p>s =</p><p>l 2</p><p>0</p><p>R</p><p>-</p><p>Rl 2</p><p>0</p><p>2R 2</p><p>=</p><p>l 2</p><p>0</p><p>2R</p><p>s = 3</p><p>l>R</p><p>0</p><p>(l0 - R u) d u</p><p>ds = (l0 - R u) d u</p><p>t = s /v0</p><p>T = 3mg</p><p>1</p><p>2</p><p>mv 2</p><p>A =</p><p>1</p><p>2</p><p>mv 2</p><p>C + mgl</p><p>l0</p><p>dθ</p><p>(l0–</p><p>Rθ)dθ</p><p>ds</p><p>θ</p><p>B A</p><p>v0</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 87</p><p>Since the value of is certainly greater than 1, hence negative sign is avoided.</p><p>So,</p><p>1.140 When the thread PA is burnt, obviously the speed of the bars will be equal at any</p><p>instant of time until it breaks off. Let v be the speed of each block and be the angle,</p><p>which the elongated spring makes with the vertical at the moment when the bar A</p><p>breaks off the plane. At this stage the elongation in the spring is</p><p>(1)</p><p>Since the problem is concerned with position and there are no forces other than con-</p><p>servative forces, the mechanical energy of the system (both bars spring) in the</p><p>field of gravity is conserved, i.e., .</p><p>So, (2)</p><p>From Newton’s second law in projection form along vertical direction:</p><p>But, at the moment of break off,</p><p>Hence,</p><p>or (3)</p><p>Taking , simultaneous solution of Eqs. (2) and (3) yields</p><p>1.141 Obviously the elongation in the cord, at the moment the sliding</p><p>first starts and at the moment horizontal projection of spring force equals the limit-</p><p>ing friction.</p><p>So, (1)</p><p>(where is the elastic constant)</p><p>From Newton’s law in projection form along vertical</p><p>direction:</p><p>or (2)N = mg - k¢l cos u</p><p>k¢l cos u + N = mg</p><p>k</p><p>k¢l sin u = kN</p><p>¢l = l0 (sec u - 1),</p><p>v = C</p><p>19gl0</p><p>32</p><p>= 1.7 m/s</p><p>k =</p><p>5mg</p><p>l0</p><p>cos u =</p><p>kl0 - mg</p><p>kl0</p><p>kl0 (sec u - 1) cos u = mg</p><p>N = 0.</p><p>mg = N + k l0 (sec u - 1) cos u</p><p>2a1</p><p>2</p><p>mv 2b +</p><p>1</p><p>2</p><p>kl 2</p><p>0 (sec u - 1) 2 - mgl0 tan u = 0</p><p>¢T + ¢U = 0</p><p>+</p><p>¢l = l0 sec u - l0 = l0 (sec u - 1)</p><p>u</p><p>¢l =</p><p>mg</p><p>k</p><p>a1 + A1 +</p><p>2kl</p><p>mg</p><p>bA1 +</p><p>2kl</p><p>mg</p><p>N</p><p>mg</p><p>T</p><p>k l0(secq–1)</p><p>fr</p><p>N</p><p>l</p><p>mg</p><p>88 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>From Eqs. (1) and (2),</p><p>or</p><p>From the equation of the increment of mechanical energy:</p><p>or</p><p>or</p><p>Thus,</p><p>1.142 Let the deformation in the spring be when the rod AB has attained the angular</p><p>velocity From the second law of motion in projection form</p><p>From the energy equation,</p><p>On solving</p><p>1.143 We know that acceleration of centre of mass of the system is given by the expression.</p><p>Since</p><p>(1)</p><p>Now from Newton’s second law for bodies and</p><p>respectively,</p><p>(2) T + m1g = m1w1</p><p>m2m1F = m w,</p><p>wC =</p><p>(m1 - m2) w1</p><p>m1 + m2</p><p>w1 = -w2</p><p>wC =</p><p>m1w1 + m2w2</p><p>m1 + m2</p><p>Aext =</p><p>k</p><p>2</p><p>l 2</p><p>0 h (1 + h)</p><p>(1 - h) 2</p><p>, where h =</p><p>mv 2</p><p>k</p><p>=</p><p>1</p><p>2</p><p>mv2 a l0 +</p><p>mv2l0</p><p>k - mv2</p><p>b2</p><p>+</p><p>1</p><p>2</p><p>k a mv2l</p><p>2</p><p>0</p><p>k - mv2</p><p>b2</p><p>=</p><p>1</p><p>2</p><p>mv2 (l0 + ¢l ) 2 +</p><p>1</p><p>2</p><p>k¢l 2</p><p>Aext =</p><p>1</p><p>2</p><p>mv 2 +</p><p>1</p><p>2</p><p>k¢l 2</p><p>k¢l = mv2 (l0 + ¢l ) or ¢l =</p><p>mv2l0</p><p>k - mv2</p><p>Fn = mwn.v.</p><p>¢l,</p><p>Afr =</p><p>(kmgl0 (sec u - 1)</p><p>2 (sin u - k cos u)</p><p>= 0.09 J (on substituting values)</p><p>kmg ¢l 2</p><p>2¢l (sin u + k cos u)</p><p>= Aƒr</p><p>a1</p><p>2</p><p>k¢l</p><p>2b = Aƒr</p><p>¢U + ¢T = Aƒr</p><p>k =</p><p>kmg</p><p>¢l sin u + k¢l cos u</p><p>k¢l sin u = k (mg - k¢l cos u)</p><p>m2</p><p>m1</p><p>m g2</p><p>m g1</p><p>T</p><p>T</p><p>w</p><p>w</p><p>,</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 89</p><p>and (3)</p><p>Solving Eqs. (2) and (3) (4)</p><p>Thus, from Eqs. (1) and (4),</p><p>1.144 As the closed system consisting of two particles and is initial-</p><p>ly at rest, the centre of mass of the system will remain at rest. Further</p><p>as the centre of mass of the system divides the line join-</p><p>ing and at all the moments of time in the ratio 1:2. In addi-</p><p>tion to it, the total linear momentum of the system at all the times is</p><p>zero. So, and therefore the velocities of and are also</p><p>directed in opposite sense. Bearing in mind all these things, the</p><p>sought trajectory is as shown in the figure.</p><p>1.145 First of all, it is clear that the chain does not move in the ver-</p><p>tical direction during the uniform rotation. This means that</p><p>the vertical component of the tension T balances gravity. As</p><p>for the horizontal component of the tension T, it is constant</p><p>in magnitude and permanently directed toward the rotation</p><p>axis. It follows from this that the centre of mass of the chain,</p><p>the point C, travels along horizontal circle of radius (say).</p><p>Therefore we have</p><p>Thus</p><p>and</p><p>1.146 (a) Let us draw free body diagram and write Newton’s</p><p>second law in terms of projection along vertical and</p><p>horizontal directions, respectively,</p><p>(1)</p><p>(2)</p><p>From Eqs. (1) and (2)</p><p>ƒr cos a -</p><p>sin a</p><p>cos a</p><p>(-ƒr sin a + mg) = mv2l</p><p>ƒr cos a - N sin a = mv 2l</p><p>N cos a - mg + fr sin a = 0</p><p>T =</p><p>mg</p><p>cos u</p><p>= 5 N (on substituting values)</p><p>r =</p><p>g tan u</p><p>v2</p><p>= 0.8 cm</p><p>T cos u = mg and T sin u = mv2 r</p><p>r</p><p>m2m1p1 = -p2</p><p>m2m1</p><p>m2 = m1>2,</p><p>m2m1</p><p>wC =</p><p>(m1 - m2)</p><p>2 g</p><p>( m1 + m2)</p><p>2</p><p>w1 =</p><p>(m1 - m2) g</p><p>m1 + m2</p><p>T + m2g = m2w2 = -m2w1</p><p>m2</p><p>m1</p><p>C</p><p>T</p><p>C</p><p>mg</p><p>m 2ω ρ</p><p>l</p><p>N</p><p>fr</p><p>mg</p><p>90 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>So, (3)</p><p>(b) For rolling, without sliding,</p><p>but,</p><p>(using Eq. 3)</p><p>Rearranging, we get</p><p>Thus,</p><p>1.147 (a) Total kinetic energy in frame is</p><p>This is minimum with respect to variation in when</p><p>or</p><p>Hence, it is the frame of centre of mass in which kinetic energy of a system is</p><p>minimum.</p><p>(b) Linear momentum of the particle 1 in the or C frame</p><p>or</p><p>Similarly,</p><p>So, | ~p1 | = | p~2 | = p~ = m vrel</p><p>where vrel = | v1 - v2 |</p><p>p</p><p>'</p><p>2 = m1v2 - v12</p><p>p</p><p>'</p><p>1 = m1v1 - v22, where m =</p><p>m1m2</p><p>m1 + m2</p><p>= reduced mass</p><p>p</p><p>'</p><p>1 = m11v1 - vC2 =</p><p>m1m2</p><p>m1 + m2</p><p>1v1 - v22</p><p>K ¿</p><p>V =</p><p>m1v1 + m2v2</p><p>m1 + m2</p><p>= vC</p><p>dT</p><p>dV</p><p>= 0, i. e., m11v1 - V2 + m21v2 - V2 = 0</p><p>V,</p><p>T =</p><p>1</p><p>2</p><p>m1 (v1 - V ) 2 +</p><p>1</p><p>2</p><p>m2 (v2 - V)</p><p>2</p><p>K ¿</p><p>v … 2g (k - tan a)>(1 + k tan a)l = 2 rad>s mv2 l (cos a + k sin a) … (k mg cos a - mg sin a )</p><p>mg asin a +</p><p>v2l</p><p>g</p><p>cos ab … k (mg cos a - m v2 l sin a)</p><p>N = mg cos a - mv2l sin a</p><p>ƒr … k N</p><p>= 6 N (on substituting values)</p><p>ƒr = mg asin a +</p><p>v2l</p><p>g</p><p>cos ab</p><p>,</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 91</p><p>Now the total kinetic energy of the system in the frame is</p><p>Hence,</p><p>1.148 To find the relationship between the values of the mechanical energy of a system in</p><p>the K and C reference frames, let us begin with the kinetic energy T of the system.</p><p>The velocity of the i-th particle in the K frame may be represented as .</p><p>Now we can write</p><p>Since in the C frame , the previous expression takes the form</p><p>(since according to the problem (1)</p><p>Since the internal potential energy U of a system depends only on its configuration,</p><p>the magnitude U is the same in all reference frames. Adding U to the left and right</p><p>hand sides</p><p>of Eq. (1), we obtain the sought relationship,</p><p>1.149 As initially . From the solution of problem 1.147(b)</p><p>As</p><p>Thus,</p><p>1.150 Velocities of masses and after t seconds are, respectively,</p><p>v¿1 = v1 + gt and v¿2 = v2 + gt</p><p>m2m1</p><p>T</p><p>'</p><p>=</p><p>1</p><p>2</p><p>m1m2</p><p>m1 + m2</p><p>(v 2</p><p>1 + v 2</p><p>2)</p><p>v1 � v2</p><p>T</p><p>'</p><p>=</p><p>1</p><p>2</p><p>m | v1 - v2 |</p><p>U = U</p><p>'</p><p>= 0, so, E</p><p>'</p><p>= T</p><p>'</p><p>E = E</p><p>'</p><p>+</p><p>1</p><p>2</p><p>mV 2</p><p>vC = V )T = T</p><p>'</p><p>+</p><p>1</p><p>2</p><p>m v</p><p>'</p><p>2</p><p>C = T</p><p>'</p><p>+</p><p>1</p><p>2</p><p>mV 2</p><p>ami v</p><p>'</p><p>i = 0</p><p>= a 1</p><p>2</p><p>mi v</p><p>'</p><p>2</p><p>1 + vCami vi</p><p>' + a 1</p><p>2</p><p>mi</p><p>v 2</p><p>C</p><p>T = a 1</p><p>2</p><p>mi</p><p>v</p><p>2</p><p>i = a 1</p><p>2</p><p>mi1v'i + vC2 # 1v'i + vC2</p><p>vi = v</p><p>'</p><p>i + vC</p><p>T~ =</p><p>1</p><p>2</p><p>m v</p><p>2</p><p>rel</p><p>=</p><p>1</p><p>2</p><p>m | v1 - v2 |</p><p>2</p><p>T~ = T~1 + T~2 =</p><p>p~ 2</p><p>2m1</p><p>+</p><p>p ~2</p><p>2m2</p><p>=</p><p>p~ 2</p><p>2m</p><p>C</p><p>92 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Hence the final momentum of the system</p><p>Radius vector, is given by</p><p>1.151 After releasing, bar 2 acquires velocity obtained by the energy conservation:</p><p>Thus, the sought velocity of C.M.</p><p>1.152 Let us consider both blocks and spring as the physical system. The centre of mass</p><p>of the system moves with acceleration towards right. Let us work in</p><p>the frame of centre of mass. As this frame is a non-inertial frame (accelerated with re-</p><p>spect to the ground) we have to apply a pseudo force a towards left on the block</p><p>and towards left on the block .</p><p>As the centre of mass is at rest in this</p><p>frame, the blocks move in opposite direc-</p><p>tions and come to instantaneous rest at</p><p>some instant. The elongation of the</p><p>spring will be maximum or minimum at this instant. Assume that the block is</p><p>displaced by the distance and the block through a distance from the ini-</p><p>tial positions.</p><p>From the equation of increment of mechanical energy in C.M. frame</p><p>where also includes the work done by the pseudo forces.</p><p>Here,</p><p>Wext = aF -</p><p>m2F</p><p>m1 + m2</p><p>bx2 +</p><p>m1F</p><p>m1 + m2</p><p>x1 =</p><p>m1 F (x1 + x2)</p><p>m1 + m2</p><p>¢T</p><p>'</p><p>= 0, ¢U =</p><p>1</p><p>2</p><p>k (x1 + x2)</p><p>2 and</p><p>A</p><p>ext</p><p>¢T</p><p>'</p><p>+ ¢U = Aext</p><p>x2m2x1</p><p>m1</p><p>m2m2am1</p><p>m1</p><p>a = F /(m1 + m2)</p><p>vC =</p><p>0 + m2x 1k>m2</p><p>m1 + m2</p><p>=</p><p>x2m2k</p><p>m1 + m2</p><p>1</p><p>2</p><p>m2v</p><p>2</p><p>2 =</p><p>1</p><p>2</p><p>kx 2 or v2 = x1k>m2</p><p>v2,</p><p>=</p><p>(m1v1 + m2v2)t</p><p>(m1 + m2)</p><p>+</p><p>1</p><p>2</p><p>gt 2 = v0t +</p><p>1</p><p>2</p><p>gt 2, where v0 =</p><p>m1v1 + m2v2</p><p>m1 + m2</p><p>rC = vC t +</p><p>1</p><p>2</p><p>wC</p><p>t 2</p><p>= p0 + mgt (where, p0 = m2v1 + m2v2 and m = m1 + m2)</p><p>p = m1v1 + m2v2 = m1v1 + m1v2 + (m1 + m2) gt</p><p>m1 Fm2m1a</p><p>m2a</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 93</p><p>or</p><p>So,</p><p>Hence, the maximum and minimum separations between the blocks are</p><p>.</p><p>Alternate:</p><p>Let us link the inertial frame with</p><p>horizontal floor and point the coordi-</p><p>nate axis as shown in figure.</p><p>At an arbitrary instant of time the separation between the blocks is l and the x</p><p>coordinate of the block is From Newton’s second law for the block ,</p><p>(1)</p><p>Similarly for block</p><p>(2)</p><p>Multiplying Eq. (2) by and Eq. (1) by and further subtracting Eq. (1) from</p><p>Eq. (2), we get</p><p>or (3)</p><p>or</p><p>or (4)</p><p>Putting</p><p>and comparing with differential equation we get</p><p>(5)l - l0 =</p><p>A</p><p>v2</p><p>+ B sin (vt + d)</p><p>d 2x</p><p>dt 2</p><p>+ v2x = A,</p><p>k (m1 + m2)</p><p>m1m2</p><p>= v2 and</p><p>F</p><p>m2</p><p>= A</p><p>d 2 (l - l0)</p><p>dt 2</p><p>+</p><p>k (m1 + m2) (l - l0</p><p>)</p><p>m1m2</p><p>=</p><p>F</p><p>m2</p><p>m1m2</p><p>d 2 (l - l0</p><p>)</p><p>dt 2</p><p>= m1F - (m1 + m2) k (l - l0</p><p>)</p><p>m1m2</p><p>d 2l</p><p>dt 2</p><p>= m1F - (m1 + m2) k (l - l0</p><p>)</p><p>m1m2</p><p>d 2 (x + l )</p><p>dt 2</p><p>- m1m2</p><p>d 2</p><p>x1</p><p>dt 2</p><p>= m1F = (m1 + m2) k (l - l0</p><p>)</p><p>m2m1</p><p>m2</p><p>d 2 (x1 + l )</p><p>dt 2</p><p>= F - k (l - l0)</p><p>m2,</p><p>m1</p><p>d 2x1</p><p>dt 2</p><p>= k (l - l0</p><p>)</p><p>m1x1.m1</p><p>l0 +</p><p>2m1F</p><p>k (m1 + m2)</p><p>and l0, respectively</p><p>x1 + x2 = 0 or x1 + x2 =</p><p>2m1F</p><p>k 1m1 + m22</p><p>1</p><p>2</p><p>k (x1 + x2)</p><p>2 =</p><p>m1(x1 + x2)F</p><p>m1 + m2</p><p>m1 Fm2</p><p>x</p><p>lx1</p><p>O</p><p>94 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>But at time</p><p>So,</p><p>Therefore, Eq. (5) becomes</p><p>(6)</p><p>But at</p><p>Hence, Eq. (6) becomes</p><p>(7)</p><p>But will be maximum, when</p><p>Therefore, from Eq. (7)</p><p>or</p><p>Similarly will be minimum when cos</p><p>Therefore, from Eq. (7)</p><p>1.153 (a) The initial compression in the spring must be such that after burning of the</p><p>thread, the upper cube rises to a height that produces a tension in the spring</p><p>that is atleast equal to the weight of the lower cube. Actually, the spring will first</p><p>go from its compressed state to its natural length and then get elongated beyond</p><p>this natural length. Let l be the maximum elongation produced under these cir-</p><p>cumstances.</p><p>Then (1)</p><p>Now, from energy conservation,</p><p>(2)</p><p>(because at maximum elongation of the spring, the speed of upper cube becomes</p><p>zero).</p><p>1</p><p>2</p><p>k¢l 2 = mg (¢l + l ) +</p><p>1</p><p>2</p><p>kl 2</p><p>kl = mg</p><p>¢l</p><p>l - l0 = 0 or l = l0</p><p>v t = 1(l - l0)</p><p>l = l0 +</p><p>2Fm1</p><p>k (m1 + m2)</p><p>l - l0 =</p><p>2A</p><p>v2</p><p>=</p><p>2Fm1m2</p><p>km2 (m1 + m2)</p><p>cos v t = -1(l - l0)</p><p>l - l0 =</p><p>A</p><p>v2</p><p>(1 - cos vt)</p><p>l - l0 = 0, so, B =</p><p>-A</p><p>v2</p><p>t = 0,</p><p>l - l0 =</p><p>A</p><p>vu2</p><p>+ B cos vt</p><p>B v cos (vt + d) |t =0 = 0 and hence d = p>2.</p><p>t = 0, l - l0 = 0 and</p><p>d (l - l0</p><p>)</p><p>dt</p><p>= 0</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 95</p><p>From Eqs. (1) and (2),</p><p>Therefore, acceptable solution of equals .</p><p>(b) Let v be the velocity of upper cube at the position (say, at 2)</p><p>when the lower block breaks off the floor, then from energy</p><p>conservation,</p><p>and</p><p>or (3)</p><p>The displacement of C.M. (of two blocks systems) in the interval in which</p><p>upper block reaches position 2 is</p><p>When the upper block reaches position 2, the lower block just leaves the</p><p>floor with zero initial velocity. Now treat two blocks system like a single</p><p>particle of mass projected vertically upward with the velocity of C.M.</p><p>So, the further vertical displacement of C.M. of two block system</p><p>(using Eq. 3)</p><p>Hence, the net displacement of the C.M. of the system, in upward direction</p><p>1.154 Due to ejection of mass from a moving system (which moves due to inertia) in a di-</p><p>rection perpendicular to it, the velocity of moving system does not change. The mo-</p><p>mentum change being adjusted by the forces on the rails. Hence in our problem ve-</p><p>locities of buggies change only due to the entrance of the man coming from the</p><p>other buggy. From conservation of linear momentum for the system (buggy man</p><p>coming from buggy 2) in the direction of motion of buggy 1,</p><p>(1)Mv1 - mv2 = 0</p><p>1 +</p><p>¢yC = ¢yC1</p><p>+ ¢yC2</p><p>=</p><p>8mg</p><p>k</p><p>¢yC2</p><p>=</p><p>vC</p><p>2</p><p>2g</p><p>=</p><p>(v>2) 2</p><p>2g</p><p>=</p><p>4mg</p><p>k</p><p>vC =</p><p>mv + 0</p><p>2m</p><p>=</p><p>v</p><p>2</p><p>2m</p><p>¢yC1</p><p>=</p><p>m + m (7mg /k + mg/k)</p><p>2m</p><p>=</p><p>4 mg</p><p>k</p><p>v 2 = 32</p><p>mg 2</p><p>k</p><p>¢l = 7</p><p>mg</p><p>k</p><p>bawhere l =</p><p>mg</p><p>k</p><p>1</p><p>2</p><p>mv 2 =</p><p>1</p><p>2</p><p>k (¢l 2 - l 2) - mg (l + ¢l )</p><p>3mg>k¢l</p><p>¢l 2 -</p><p>2mg¢l</p><p>k</p><p>-</p><p>3m 2g 2</p><p>k 2</p><p>= 0 or ¢l =</p><p>3mg</p><p>k</p><p>,</p><p>- mg</p><p>k</p><p>l</p><p>l</p><p>B</p><p>C v</p><p>.0</p><p>96 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Similarly from conservation of linear momentum for the system (buggy man</p><p>coming from buggy 1) in the direction of motion of buggy 2,</p><p>(2)</p><p>Solving Eqs. (1) and (2), we get</p><p>As</p><p>So,</p><p>1.155 From momentum conservation, for the system (rear buggy man)</p><p>(1)</p><p>From momentum conservation, for the system (front buggy man coming from rear</p><p>buggy)</p><p>So,</p><p>Putting the value of from Eq. (1), we get</p><p>1.156 (i) Let be the velocity of the buggy after both men jump off simultaneously. For the</p><p>closed system (two men buggy), from the conservation of linear momentum,</p><p>or (1)</p><p>(ii) Let v be the velocity of buggy with man, when one man jumps off the buggy. For</p><p>the closed system (buggy with one man other man) from the conservation of</p><p>linear momentum:</p><p>(2)</p><p>Let be the sought velocity of the buggy when the second man jumps off the</p><p>buggy; then from conservation of linear momentum of the system (buggy</p><p>one man):</p><p>(3)(M + m)v = Mv2 + m (u + v2)</p><p>+</p><p>v2</p><p>0 = (M + m) v + m (u + v)</p><p>+</p><p>v1 =</p><p>-2m u</p><p>M + 2m</p><p>M v1 + 2m (u + v1) = 0</p><p>+</p><p>v1</p><p>vF = v0 +</p><p>mM</p><p>(M + m) 2</p><p>u</p><p>vR</p><p>vF =</p><p>M v0</p><p>M + m</p><p>+</p><p>m</p><p>M + m</p><p>(u + vR)</p><p>M v0 + m (u + vR</p><p>) = (M + m) vF</p><p>+</p><p>(M + m) v0 = m (u + vR) + M vR</p><p>v1 =</p><p>-m v</p><p>(M - m)</p><p>and v2 =</p><p>M v</p><p>(M - m)</p><p>v1</p><p>c Tv and v2</p><p>c cv</p><p>v1 =</p><p>mv</p><p>M - m</p><p>and v2 =</p><p>Mv</p><p>M - m</p><p>Mv2 - mv2 = (M + m)v</p><p>2 +</p><p>+</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 97</p><p>Solving Eqs. (2) and (3) we get</p><p>(4)</p><p>From Eqs. (1) and (4)</p><p>Hence,</p><p>1.157 The descending part of the chain is in free fall, it has speed</p><p>at the instant when the chain descended a distance y.</p><p>The length of the</p><p>chain which lands on the floor, during the differential time interval dt</p><p>following this instant is vdt.</p><p>For the incoming chain element on the floor, from (where</p><p>y-axis is directed down)</p><p>or</p><p>Hence, the force exerted on the falling chain equals and is directed upward.</p><p>Therefore from third law the force exerted by the falling chain on the table at the</p><p>same instant of time becomes and is directed downward.</p><p>Since a length of chain of weight already lies on the table and the table is at rest,</p><p>the total force on the floor is or the weight of a length 3y of</p><p>chain.</p><p>1.158 Velocity of the ball, with which it hits the slab, .</p><p>After first impact (upward) but according to the problem</p><p>, so (1)</p><p>and momentum imparted to this slab equals</p><p>Similarly, velocity of the ball after second impact is</p><p>and momentum imparted is</p><p>Again, momentum imparted during third impact is</p><p>and so on m (1 + e)e 2v,</p><p>m (v ¿ + v –) = m (1 + e)ev</p><p>v – = ev ¿ = e 2v</p><p>mv - (-mv ¿) = mv (1 + e)</p><p>e =</p><p>1</p><p>h</p><p>v ¿ =</p><p>v</p><p>h</p><p>v ¿ = ev</p><p>v = 12gh</p><p>(2lyg) + (lyg) = (3lyg)</p><p>(lyg)</p><p>lv 2</p><p>lv 2</p><p>Fy = -lv 2 = -2lgy</p><p>0 - (l vdt) v = Fydt</p><p>dpy = Fy dt</p><p>v =12gy</p><p>v2 7 v1</p><p>v2</p><p>v1</p><p>= 1 +</p><p>m</p><p>2(M + m)</p><p>7 1</p><p>v2 =</p><p>m (2M + 3m)u</p><p>(M + m)(M + 2m)</p><p>}</p><p>vdy</p><p>y</p><p>98 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Hence, net momentum imparted</p><p>(using Eq. 1)</p><p>1.159 (a) Since the resistance of water is negligibly small the resultant of all external forces</p><p>acting on the system “a man and a raft” is equal to zero. This means that the po-</p><p>sition of the C.M. of the given system does not change in the process of motion.</p><p>i.e.</p><p>or</p><p>Thus,</p><p>(b) As net external force on “man-raft” system is equal to zero, therefore the mo-</p><p>mentum of this system does not change.</p><p>So,</p><p>or, (1)</p><p>Thus, the sought force on the raft</p><p>Note: We may get the result of part (a), if we integrate Eq. (1) over the time of motion</p><p>of man or raft.</p><p>1.160 The displacement of the C.M. of the system, man ( ), ladder and the coun-</p><p>terweight is described by radius vector</p><p>(1)</p><p>But,</p><p>and, (2) ¢rm = ¢rm (M - m) + ¢r(M - m)</p><p>¢rm = - ¢r(M - m)</p><p>¢rC =</p><p>©mi ¢ri</p><p>©mi</p><p>=</p><p>M¢rM + (M –m) ¢r(M - m) + m ¢rm</p><p>2M</p><p>(M ),</p><p>(M - m)m</p><p>M</p><p>d v2(t)</p><p>dt</p><p>= -</p><p>Mm</p><p>m + M</p><p>d v¿(t)</p><p>dt</p><p>v2(t) = -</p><p>m v ¿(t)</p><p>m + M</p><p>0 = m [v¿ (t) + v2</p><p>(t)] + M v2(t)</p><p>m( l¿ + l ) + M l = 0, or l = -</p><p>m l¿</p><p>m + M</p><p>m (¢rmM + ¢rM) + M¢rM = 0</p><p>rC = constant or ¢rC = 0, a mi</p><p>¢ri = 0</p><p>= 0.2 kg m/s (on substituting values)</p><p>= 12gh</p><p>1+ 1 h</p><p>1 -1 h</p><p>= m12gh (h + 1)>(h - 1)</p><p>= mv</p><p>(1 + e)</p><p>1 - e</p><p>(from summation of G.P.)</p><p>= mv (1 + e) (1 + e + e 2 + Á )</p><p>= mv (1 + e) + mve ¿ (1 + e) + Á</p><p>m (M – m) + m</p><p>/</p><p>/(</p><p>( )</p><p>) n</p><p>,</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 99</p><p>Using Eq. (2) in Eq. (1) we get</p><p>Alternate:</p><p>Initially all the bodies of the system are at rest, and therefore the increments of lin-</p><p>ear momentum of the bodies in their motion are equal to the momentum themselves.</p><p>The rope tension is the same both on the left and on the right-hand side, and con-</p><p>sequently the momentum of the counter-balancing mass and the ladder with the</p><p>man are equal at any instant of time, i.e.,</p><p>where and are the velocities of the mass, the man, and the ladder, respec-</p><p>tively. Taking into account that and , where is the man’s</p><p>velocity relative to the ladder, we obtain</p><p>On the other hand, the momentum of the whole system</p><p>(1)</p><p>where is the velocity of the centre of inertia of the system. With allowance made</p><p>for Eq. (1) we get</p><p>Finally, the sought displacement is</p><p>1.161 Velocity of cannon as well as that of shell equals down the inclined plane</p><p>taken as the positive x-axis. From the linear impulse momentum theorem in projec-</p><p>tion form along x-axis for the system (cannon shell)</p><p>or</p><p>1.162 From conservation of momentum, for the system bullet body along the initial</p><p>direction of bullet</p><p>(1)</p><p>where is the initial velocity of the bullet and v is combined velocity, after the</p><p>collision.</p><p>v0</p><p>mv0 = (m + M) v or v =</p><p>mv0</p><p>m + M</p><p>+</p><p>¢t =</p><p>p cos a - M 12gl sin a</p><p>Mg sin a</p><p>p cos a - M 12gl sin a = Mg sin a¢ t (as mass of the shell is negligible)</p><p>¢px = Fx¢t</p><p>+</p><p>22gl sin a</p><p>¢rC = 3VC dt = (m>2M ) 3v¿dt = (m>2M ) ¢r¿</p><p>VC = v1 = (m>2M)v ¿</p><p>VC</p><p>p = p1 + p2 = 2p1 or 2M VC = 2M v1</p><p>v1 = (m>2M)v ¿</p><p>v ¿v = v2 + v ¿v2 = -v1</p><p>v2v,v1,</p><p>M v1 = m v + (M - m)v2</p><p>p1 = p2,( p2)</p><p>( p1)</p><p>¢rC =</p><p>m l</p><p>2M</p><p>or</p><p>,</p><p>i.e.,</p><p>“ ”</p><p>100 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>From conservation of mechanical energy of the system (bullet body) in the field</p><p>of gravity:</p><p>(2)</p><p>(a) From Eqs. (1) and (2),</p><p>As</p><p>(b) Fraction of initial kinetic energy turned into heat</p><p>(using Eq. 1)</p><p>1.163 When the disk breaks off the body M, its velocity towards right (along x-axis)</p><p>equals the velocity of the body M, and let the disk’s velocity in upward direction</p><p>(along y-axis) at that moment be .</p><p>From conservation of momentum, along x-axis for the system (disk body)</p><p>(1)</p><p>And from energy conservation, for the same system in the field of gravity:</p><p>where is the height of break-off point from initial level.</p><p>So, (using Eq. 1)</p><p>or</p><p>Also, if is the height of the disk from the break-off point, then,</p><p>So, 2g (h– + h¿) = v 2 -</p><p>mv 2</p><p>(M + m)</p><p>v ¿2y = 2gh–</p><p>h–</p><p>v ¿ 2y = v 2 -</p><p>mv 2</p><p>(m + m)</p><p>- 2gh¿</p><p>1</p><p>2</p><p>mv 2 =</p><p>1</p><p>2</p><p>(m + M)</p><p>m 2v 2</p><p>(M + m)</p><p>+</p><p>1</p><p>2</p><p>mv ¿2y + mgh¿</p><p>h¿</p><p>1</p><p>2</p><p>mv</p><p>2 =</p><p>1</p><p>2</p><p>(m + M )v ¿</p><p>2</p><p>x +</p><p>1</p><p>2</p><p>mv ¿</p><p>2</p><p>y + mgh¿</p><p>mv = (m + M) v ¿x or v ¿x =</p><p>mv</p><p>m + M</p><p>+</p><p>v ¿y</p><p>= 1 -</p><p>m</p><p>(m + M)</p><p>�a1 -</p><p>m</p><p>M</p><p>b</p><p>=</p><p>Ti - Tƒ</p><p>Ti</p><p>= 1 -</p><p>Tƒ</p><p>Ti</p><p>= 1 -</p><p>1</p><p>2</p><p>(m + M)v 2</p><p>1</p><p>2</p><p>mv0</p><p>2</p><p>v0 �</p><p>2M</p><p>m</p><p>2gl sin (u>2)m M,</p><p>v0 =</p><p>(m + M )</p><p>m</p><p>12gl (1 - cos u) =</p><p>2(M + m)</p><p>m</p><p>1gl sin (u>2)</p><p>1</p><p>2</p><p>(m + M )v 2 = (M + m)gl (1 - cos u), or, v 2 = 2gl (1 - cos u)</p><p>+</p><p>W</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 101</p><p>Hence, the total height, raised from the initial level is</p><p>1.164 (a) When the disk slides and comes to the plank, it has a velocity equal to</p><p>Due to friction between the disk and the plank the disk slows down</p><p>and after some time the disk moves with the plank with velocity (say).</p><p>From the momentum conservation for the system (disk plank) along horizon-</p><p>tal towards right:</p><p>Now from the equation of the increment of total mechanical energy of a system:</p><p>or</p><p>so,</p><p>Hence,</p><p>(b) We look at the problem from a frame in which the hill is moving (together with</p><p>the disk on it) to the right with speed u. Then in this frame the speed of the disk</p><p>when it just gets onto the plank is, by the law of addition of velocities,</p><p>. Similarly the common speed of the plank and the disk when</p><p>they move together is</p><p>Then as above</p><p>-</p><p>1</p><p>2</p><p>(m + M )u2 -</p><p>1</p><p>2</p><p>m2u12gh - mgh</p><p>=</p><p>1</p><p>2</p><p>(m + M)eu2 +</p><p>2m</p><p>m + M</p><p>u 12gh +</p><p>m2</p><p>(m + M)2</p><p>2gh f</p><p>Afr =</p><p>1</p><p>2</p><p>(m + M) v 2 -</p><p>1</p><p>2</p><p>mv 2 -</p><p>1</p><p>2</p><p>Mu 2</p><p>v = u +</p><p>m</p><p>m + M</p><p>12gh</p><p>v = u + 12gh</p><p>awhere m =</p><p>mM</p><p>m + M</p><p>= reduced massb</p><p>Afr = - a mM</p><p>m + M</p><p>b gh = -mgh</p><p>1</p><p>2</p><p>v 2 c m2</p><p>M + m</p><p>- m d = Afr</p><p>1</p><p>2</p><p>(M + m)</p><p>m2</p><p>v 2</p><p>(m + M)2</p><p>-</p><p>1</p><p>2</p><p>mv 2 = Afr</p><p>1</p><p>2</p><p>(M + m)v ¿2 -</p><p>1</p><p>2</p><p>mv 2 = Afr</p><p>mv = (m + M) v ¿ or v ¿ =</p><p>m v</p><p>m + M</p><p>+</p><p>v ¿</p><p>v = 12gh.</p><p>h¿ + h– =</p><p>Mv 2</p><p>2g (M + m)</p><p>.</p><p>102 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>We see that is independent of u and is in fact just as in (a). Thus the</p><p>result obtained does not depend on the choice of reference frame.</p><p>Do note however that it will be incorrect to apply “conservation of energy” for-</p><p>mula in the frame in which the hill is moving. The energy carried by the hill is</p><p>not negligible in this frame. See also the next problem.</p><p>1.165 In a frame moving relative to the Earth, one has to include the kinetic energy of the</p><p>Earth as well as Earth’s acceleration to be able to apply conservation of energy to the</p><p>problem. In a reference frame falling to the Earth with velocity the stone is initial-</p><p>ly going up with velocity and so is the Earth. The final velocity of the stone is</p><p>and that of the Earth is is the mass of the Earth), from</p><p>Newton’s third law, where time of fall. From conservation of energy</p><p>Hence,</p><p>Neglecting in comparison with 1, we get</p><p>The point in this</p><p>is that in the Earth’s rest frame the effect of Earth’s acceleration is</p><p>of order and can be neglected but in a frame moving with respect to the Earth</p><p>the effect of Earth’s acceleration must be kept because it is of order one (i.e., large).</p><p>1.166 From conservation of momentum for the closed system “both colliding particles”</p><p>or</p><p>Hence,</p><p>1.167 For perfectly inelastic collision, in the C.M. frame, final kinetic energy of the colliding</p><p>system (both spheres) becomes zero. Hence initial kinetic energy of the system in</p><p>C.M. frame completely turns into the internal energy (Q) of the formed body.</p><p>Hence,</p><p>Now from energy conservation ¢T = -Q = -</p><p>1</p><p>2</p><p>m| v1 - v2 | 2</p><p>Q = Ti =</p><p>1</p><p>2</p><p>m| v1 - v2 | 2</p><p>| v | = 11 + 4 + 16 m/s = 4.6 m/s</p><p>v =</p><p>m1v1 + m2v2</p><p>m1 + m2</p><p>=</p><p>1(3i - 2 j ) + 2(4 j - 6k)</p><p>3</p><p>= i + 2j - 4k</p><p>m1v1 + m2v2 = (m1 + m2) v</p><p>m>M</p><p>v 2</p><p>0 = 2gh or v0 = 12gh</p><p>m /M</p><p>1</p><p>2</p><p>v 2</p><p>0 am +</p><p>m 2</p><p>M</p><p>b = mgh</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>0 +</p><p>1</p><p>2</p><p>Mv</p><p>2</p><p>0 + mgh =</p><p>1</p><p>2</p><p>M av0 +</p><p>m</p><p>M</p><p>v0b2</p><p>t =</p><p>v0 + m /M gt (M0 = v0 - gt</p><p>v0</p><p>v0,</p><p>-mghAfr</p><p>( )</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 103</p><p>In laboratory frame the same result is obtained as</p><p>1.168 (a) From conservation of linear momentum</p><p>As (1)</p><p>From conservation of kinetic energy (2)</p><p>Using Eq. (1) in Eq. (2) we get</p><p>which yields (3)</p><p>So, sought fraction of kinetic energy</p><p>Hence,</p><p>(b) For two particles closed system momentum of each particle in their C.M. frame are</p><p>always equal and opposite. In C.M. frame the kinetic energy of the two particle</p><p>system where is the reduced mass. In perfectly elastic collision, the</p><p>kinetic energy of the system is conserved.</p><p>So,</p><p>Being head on collision, both the particles have to keep their motion along the</p><p>same straight line before and after the collision. On collision, momentum vector of</p><p>each particle has to change due to the reaction force</p><p>by the other particle. So only choice left to each of the</p><p>particle is to reverse the direction of its momentum</p><p>after the collision keeping the magnitude constant, i.e.,</p><p>, where i = 1, 2.p</p><p>' ¿i = -p</p><p>'</p><p>i</p><p>p</p><p>'</p><p>¿2</p><p>2m</p><p>=</p><p>p</p><p>'</p><p>2</p><p>2m</p><p>, which gives p</p><p>'</p><p>¿ = p</p><p>'</p><p>mT~ = p~2/2m</p><p>h = 1 - am2 - m1</p><p>m2 + m1</p><p>b =</p><p>2m1</p><p>m1 + m2</p><p>h = 1 -</p><p>T1¿</p><p>T1</p><p>= 1 -</p><p>p¿1</p><p>2>2m1</p><p>p1</p><p>2>2m1</p><p>= 1 -</p><p>p¿1</p><p>2</p><p>p1</p><p>2</p><p>p¿1</p><p>2</p><p>p1</p><p>2</p><p>= am2 - m1</p><p>m2 + m1</p><p>b</p><p>p1</p><p>2</p><p>2m1</p><p>=</p><p>p¿1 2</p><p>2m1</p><p>+</p><p>(p1</p><p>2 + p¿1 2)</p><p>2m2</p><p>p1</p><p>2</p><p>2m1</p><p>=</p><p>p¿1 2</p><p>2m1</p><p>+</p><p>p¿2 2</p><p>2m2</p><p>p¿1 � p1 so, p1</p><p>2 + p¿1</p><p>2 = p¿2</p><p>2</p><p>p1 = p1¿ + p2¿ or, p1 - p1¿ = p2¿</p><p>= -</p><p>1</p><p>2</p><p>m| v1 - v2 | 2</p><p>¢T</p><p>'</p><p>=</p><p>1</p><p>2</p><p>(m1v1 + m2v2)</p><p>2</p><p>m1 + m2</p><p>-</p><p>1</p><p>2</p><p>m1| v1 | 2 +</p><p>1</p><p>2</p><p>m2| v2 |2</p><p>p1∼</p><p>p′1∼ p′2∼</p><p>p2</p><p>before</p><p>after</p><p>∼</p><p>104 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>The same can be said about the velocity of each particle in the C.M. frame</p><p>but</p><p>Hence the velocity of i-th particle after collision.</p><p>So velocity of particle 1 just after the collision from above relation is</p><p>Therefore,</p><p>Thus sought fraction</p><p>1.169 (a) Being perfectly elastic head on collision, the velocity of i-th particle after collision</p><p>So, (1)</p><p>(2)</p><p>According to the problem</p><p>so,</p><p>Hence,</p><p>(b) From conservation of linear momentum in the direction perpendicular to initial</p><p>motion direction of striking particle 1 gives</p><p>So, (1)</p><p>From conservation of linear momentum</p><p>so, (2) p 1</p><p>2 + p¿1</p><p>2 - 2p1 p¿1 cos 30° = p¿2</p><p>2</p><p>p1 = p1¿ + p2¿ or, p1 - p1¿ = p2¿</p><p>p¿1 = p¿2</p><p>p¿1 sin 30° = p¿2 sin 30°</p><p>m1</p><p>m2</p><p>=</p><p>1</p><p>3</p><p>2a m1v1</p><p>m1 + m2</p><p>b =</p><p>(m2 - m1) v1</p><p>m1 + m2</p><p>which gives 2m1 = (m2 - m1)</p><p>v¿2 = -v¿1</p><p>v¿2 = 2a m1v1</p><p>m1 + m2</p><p>b</p><p>v¿1 = 2a m1v1</p><p>m1 + m2</p><p>b - v1 =</p><p>(m1 - m2) v1</p><p>m1 + m2</p><p>and</p><p>v ¿i = 2VC - vi (where i = 1, 2) (see previous problem)</p><p>= 1 -</p><p>T1¿</p><p>T1</p><p>= 1 -</p><p>v ¿1</p><p>2</p><p>v1</p><p>2</p><p>= 1 -</p><p>(m1 - m2)</p><p>2</p><p>(m1 + m2)</p><p>2</p><p>=</p><p>4m1m2</p><p>(m1 + m2)</p><p>2</p><p>v ¿1</p><p>2</p><p>v1</p><p>2</p><p>=</p><p>(m1 - m2)</p><p>2</p><p>(m1 + m2)</p><p>2</p><p>v¿1 = 2 VC - v1 = 2a m1v1</p><p>m1 + m2</p><p>b - v1 =</p><p>(m1 - m2) v1</p><p>m1 + m2</p><p>v¿i = 2 VC - vi (where i = 1, 2)</p><p>v¿i = VC + v</p><p>'¿i = VC - v</p><p>'</p><p>i = VC - (vi - VC</p><p>) = 2VC - vi</p><p>v</p><p>'¿i = -v</p><p>'</p><p>i</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 105</p><p>On using Eq. (1) in Eq. (2) we get on manipulation</p><p>(3)</p><p>From conservation of kinetic energy</p><p>On using Eq. (1) and Eq. (3), we get on solving</p><p>1.170 At the moment of maximum deformation the velocity of colliding bodies along</p><p>their common normal n (here the line joining the mass centres of the balls) must be</p><p>equal so</p><p>(1)</p><p>As the balls are smooth so, velocity of each ball along</p><p>their common tangent t remains constant.</p><p>So, (2)</p><p>Now, from conservation of linear momentum</p><p>(3)</p><p>From energy conservation, gain in internal potential energy is due to loss of kinetic</p><p>energy of the system. Initial kinetic energy of the system is the kinetic energy of</p><p>striking ball only so, the sought fraction</p><p>But,</p><p>On using Eqs. (1), (2) and (3), we get</p><p>So,</p><p>Hence, sought fraction h = 1 -</p><p>3</p><p>4</p><p>=</p><p>1</p><p>4</p><p>T ¿system</p><p>T1</p><p>=</p><p>1</p><p>2</p><p>m</p><p>3v1</p><p>2</p><p>4</p><p>n1</p><p>2</p><p>mv1</p><p>2 =</p><p>3</p><p>4</p><p>T ¿system =</p><p>1</p><p>2</p><p>m</p><p>3v1</p><p>2</p><p>4</p><p>T ¿system =</p><p>1</p><p>2</p><p>m [v¿1n</p><p>2 + v¿1t</p><p>2] +</p><p>1</p><p>2</p><p>m [v ¿2n</p><p>2 + v¿2t</p><p>2]</p><p>h = 1 - T ¿system/ T1</p><p>T1</p><p>mv1 cos 45° = 2m v¿n so, v¿n =</p><p>v 1</p><p>2 12</p><p>= v2t = 0v¿1t = v1t = v1 sin 45° and v¿2t</p><p>v¿1n = v¿2n = v¿n (say)</p><p>m1</p><p>m2</p><p>= 2</p><p>p1</p><p>2</p><p>2m1</p><p>=</p><p>p ¿1</p><p>2</p><p>2m1</p><p>+</p><p>p ¿2</p><p>2</p><p>2m2</p><p>p¿1 =</p><p>p1</p><p>2 cos 30°</p><p>=</p><p>p113</p><p>n</p><p>t</p><p>45°</p><p>v1</p><p>.</p><p>106 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.171 From the conservation of linear momentum of the shell just before and after its frag-</p><p>mentation</p><p>(1)</p><p>where and are the velocities of its fragments.</p><p>From the energy conservation</p><p>(2)</p><p>Now (3)</p><p>where velocity of the C.M. of the fragments the velocity of the shell.</p><p>Obviously in the C.M. frame the linear momentum of a system is equal to zero so</p><p>(4)</p><p>Using Eqs. (3) and (4) in Eq. (2), we get</p><p>or (5)</p><p>If we have had used then Eq. 5 would contain instead of and</p><p>so on. The problem being symmetrical we can look for the maximum of any one.</p><p>Obviously it will be the same for each.</p><p>For</p><p>So,</p><p>Hence,</p><p>Thus, owing to the symmetry</p><p>Alternate:</p><p>The maximum velocity can be attained by that fragment (say 1) which has experienced</p><p>the reaction force by the rest of two fragment (2 and 3) in the initial motion direction</p><p>of the shell. Symmetry of the problem tells us that the velocities of fragments 2 and 3</p><p>are same and so is the reaction force exerted by the fragment 2 and 3 on fragment 1.</p><p>Let is the velocity of each of the fragments 2 and 3 and of fragment 1 after the</p><p>collision.</p><p>v1¿v¿</p><p>v1(max) = v2(max) = v3(max) = v (1 + 12(h - 1)2 = 1 km/s</p><p>v2(max) = | v + v</p><p>'</p><p>2 |max = v + 12(h - 1)v = v 11 + 12(h - 1)2 = 1 km/s</p><p>v</p><p>'</p><p>2 … v A</p><p>6(h - 1)</p><p>4 - cos2u</p><p>or v</p><p>'</p><p>2(max) = 12(h - 1) v</p><p>v</p><p>' 2</p><p>2 cos 2 u Ú 8 (2 v</p><p>'</p><p>2</p><p>2 + 3 (1 - h) v 2) or 6 (h - 1)v 2 Ú (4 - cos 2 u) v</p><p>'</p><p>2</p><p>2</p><p>v</p><p>'</p><p>2v</p><p>'</p><p>3v</p><p>'</p><p>2 = -v</p><p>'</p><p>1 - v</p><p>'</p><p>3,</p><p>2 v</p><p>'</p><p>2</p><p>1 + 2 v</p><p>'</p><p>1v</p><p>'</p><p>2 cos u + 2 v</p><p>'</p><p>2</p><p>2 + 3 (1 - h)v 2 = 0</p><p>3hv2 = (v + v</p><p>'</p><p>1)</p><p>2 + (v + v</p><p>'</p><p>2)</p><p>2 + (v - v</p><p>'</p><p>1 - v</p><p>'</p><p>2)</p><p>2 = 3v2 + 2v</p><p>'</p><p>2</p><p>1 + 2v</p><p>'</p><p>2</p><p>1 + 2 v</p><p>'</p><p>1</p><p># v</p><p>'</p><p>2</p><p>v</p><p>'</p><p>1 + v</p><p>'</p><p>2 + v</p><p>'</p><p>3 = 0</p><p>vC = v =</p><p>v</p><p>' or viC = vi - vC = vi - v</p><p>3hv 2 = v 2</p><p>1 + v 2</p><p>2 + v 2</p><p>3</p><p>v3v1, v2</p><p>3v = v1 + v2 + v3</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 107</p><p>From conservation of linear momentum</p><p>(1)</p><p>From conservation of energy</p><p>So, (2)</p><p>Solving Eqs. (1) and (2), we get</p><p>km/s (on substituting values)</p><p>1.172 From the conservation of momentum</p><p>(1)</p><p>From the given condition,</p><p>or (2)</p><p>From Eqs. (1) and (2)</p><p>or</p><p>So,</p><p>Positive sign gives the velocity of the second particle which lies ahead. The negative</p><p>sign is correct for So, will continue moving</p><p>in the same direction.</p><p>Note that if .</p><p>Alternate:</p><p>As internal energy is independent of reference frame, so loss in kinetic energy of the</p><p>system in the C.M. frame is the same.</p><p>So,</p><p>or a1 -</p><p>T</p><p>'</p><p>¿</p><p>T</p><p>' b =</p><p>T</p><p>T</p><p>' a1 -</p><p>T ¿</p><p>T</p><p>b T</p><p>'</p><p>- T</p><p>'</p><p>¿ = T - T ¿</p><p>h = 0v ¿1 = 0</p><p>v ¿1 = 1 /2 v1 11 - 11 - 2h2 = 5 m/sv1.</p><p>=</p><p>1</p><p>2</p><p>Cv1 � 1v1</p><p>2 - 2h v1</p><p>2 D =</p><p>1</p><p>2</p><p>v1 A1 � 11 - 2h Bv ¿1 =</p><p>2v1 � 14v1</p><p>2 - 8h v1</p><p>2</p><p>4</p><p>2v ¿ 2</p><p>1 - 2v ¿1 v1 + hv1</p><p>2 = 0</p><p>v ¿ 2</p><p>1 + (v1 - v ¿1) 2 = (1 - h) v1</p><p>2</p><p>v ¿21 + v ¿ 2</p><p>2 = (1 - h) v1</p><p>2</p><p>h =</p><p>T - T ¿</p><p>T</p><p>= 1 -</p><p>T ¿</p><p>T</p><p>= 1 -</p><p>v ¿21 + v ¿ 2</p><p>2</p><p>v1</p><p>2</p><p>mv1 = mv ¿1 + mv ¿2 or, v1 = v ¿1 + v ¿2</p><p>v1¿ = v 11 + 12(h - 1)2 = 1</p><p>h 3v 2 = v1¿ 2 + v 2</p><p>h</p><p>2</p><p>(3m) v 2 =</p><p>1</p><p>2</p><p>mv1¿ 2 + a1</p><p>2</p><p>mv ¿ 2b</p><p>3v = v1¿ - 2v ¿</p><p>3mv = mv1¿ - 2 (mv¿)</p><p>108 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Using and , where is mass of</p><p>each particle, we get</p><p>Now, according to the problem</p><p>which yields</p><p>Being head-on collision, final momentum in C.M. frame of any particle should be re-</p><p>versed, so</p><p>(where</p><p>or</p><p>But,</p><p>Thus,</p><p>Using (where is the velocity of striking particle)</p><p>1.173 From conservation of linear momentum in the initial motion direction of striking par-</p><p>ticle 1 and perpendicular to it gives</p><p>Thus,</p><p>Putting the values of and we get % of change in kinetic energy .= -40%m /M ,u</p><p>= D 1</p><p>2</p><p>mv1</p><p>2 tan 2</p><p>u +</p><p>1</p><p>2</p><p>M a mv1</p><p>M cos u</p><p>b2</p><p>1</p><p>2</p><p>mv1</p><p>2</p><p>- 1T * 100 = a tan2</p><p>u +</p><p>m</p><p>M</p><p>sec 2</p><p>u - 1b * 100</p><p>= aT ¿system</p><p>Tsystem</p><p>- 1b * 100</p><p>Percentage change in the kinetic energy of the system =</p><p>T ¿system - Tsystem</p><p>Tsystem</p><p>* 100</p><p>v1 ¿ = v1 tan u</p><p>mv1 = Mv ¿2 cos u and mv ¿1 = Mv ¿2 sin u , respectively</p><p>vi =</p><p>v</p><p>2</p><p>(1 - 11 - 2h)</p><p>vvC =</p><p>m v</p><p>2m</p><p>=</p><p>v</p><p>2</p><p>vi ¿ = vc</p><p>[1 + 11 - 2h] - vi11 - 2h</p><p>vi¿ = v</p><p>'</p><p>i¿ + vC = -v</p><p>'</p><p>i1(1 - 2h) + vC = - (vi - vC)1(1 - 2h) + vC</p><p>v</p><p>'</p><p>i¿ = -v</p><p>'</p><p>i 1(1 - 2h),</p><p>i = 1, 2)p</p><p>'</p><p>i ¿ = p</p><p>'</p><p>i 1(1 - 2h)</p><p>p</p><p>'</p><p>¿ = p</p><p>'1(1 - 2h)</p><p>p</p><p>'</p><p>2</p><p>2m</p><p>-</p><p>p</p><p>'</p><p>¿2</p><p>2m</p><p>= 2h</p><p>p</p><p>'2</p><p>2m</p><p>T</p><p>'</p><p>- T</p><p>'</p><p>¿ = 2hT</p><p>'</p><p>a1 -</p><p>T ¿</p><p>T</p><p>' b = 2h</p><p>mT</p><p>'</p><p>=</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>rel =</p><p>1</p><p>2</p><p>am</p><p>2</p><p>b v</p><p>2a1 -</p><p>T ¿</p><p>T</p><p>' b = h, T =</p><p>1</p><p>2</p><p>mv 2,</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 109</p><p>1.174 (a) Let the particles and move with velocities and , respectively. On the</p><p>basis of solution of problem 1.147(b):</p><p>As</p><p>so,</p><p>(b) Again from solution of problem 1.147(b):</p><p>So,</p><p>1.175 From conservation of momentum</p><p>So, where is the angle between and</p><p>From conservation of energy</p><p>Eliminating we get</p><p>This quadratic equation for has a real solution in terms of and only if</p><p>or</p><p>This clearly implies (since only sign makes sense) that</p><p>Alternate:</p><p>The solution of the problem becomes standard in the frame of C.M., which is moving</p><p>with velocity in the frame of laboratory. In the frame of C.M., the</p><p>momenta of the particles must always be equal and opposite.</p><p>So, p</p><p>'</p><p>1 = p</p><p>'</p><p>2,</p><p>vC = m1v1 /(m1 + m2)</p><p>sin u1 max =</p><p>m2</p><p>m1</p><p>+</p><p>sin 2</p><p>u1 …</p><p>m 2</p><p>2</p><p>m 2</p><p>1</p><p>or, sin u1 Ú -</p><p>m2</p><p>m1</p><p>4 cos 2</p><p>u1 Ú 4a1 -</p><p>m 2</p><p>2</p><p>m 2</p><p>1</p><p>b cos u1p1p1¿</p><p>0 = p1 ¿2 a1 +</p><p>m2</p><p>m1</p><p>b - 2p1¿p1 cos u1 + p1</p><p>2 a1 -</p><p>m2</p><p>m1</p><p>b</p><p>p2¿</p><p>p1</p><p>2</p><p>2m1</p><p>=</p><p>p1¿ 2</p><p>2m1</p><p>+</p><p>p2¿ 2</p><p>2m2</p><p>p1¿p1 up 2</p><p>1 - 2 p1p1¿ cos u1 + p1 ¿ 2 = p2 ¿ 2,</p><p>p1 = p¿1 + p2 or, p1 - p¿1 = p2</p><p>T</p><p>'</p><p>=</p><p>1</p><p>2</p><p>m 1v 2</p><p>1 + v 2</p><p>2 2</p><p>T</p><p>'</p><p>=</p><p>1</p><p>2</p><p>mv 2</p><p>rel =</p><p>1</p><p>2</p><p>m| v1 - v2 |</p><p>2</p><p>p</p><p>' = m2v 2</p><p>1 + v 2</p><p>2 where m =</p><p>m1m2</p><p>m1 + m2</p><p>v1 � v2</p><p>p</p><p>' = mvrel = m| v1 - v2 |</p><p>v2v1m2m1</p><p>.</p><p>110 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>but (where is the reduced</p><p>mass of the system).</p><p>The initial kinetic energy of the system in the frame of C.M.</p><p>As the collision is perfectly elastic and the reduced mass the system is constant.</p><p>Hence,</p><p>In the frame of laboratyry from the conservation</p><p>of linear momentum</p><p>Now let us draw the so-called vector diagram (see figure) of momenta. First we de-</p><p>pict the vector as the section and the and each of which represents ac-</p><p>cording to</p><p>and</p><p>Note that this diagram is valid regardless of the angle The point C (centre of</p><p>mass) therefore can be located only on the circle of radius having it’s centre at</p><p>the point O, which divides the section AB into two parts in the ratio AO :</p><p>: . In our case, this circle passes through the point B, the end point of vector</p><p>since the section This circle is the locus of all possible</p><p>locations of the apex C of the momenta triangle ABC, whose side AC and CB rep-</p><p>resent the possible momentum of particles and after the collision in the</p><p>frame of laboratory.</p><p>Depending on the particle mass ratio the point A, the beginning of the</p><p>vector can be located inside the given circle, on it, or outside it. In our case point</p><p>A will lie outside the circle and it is also another intersting fact that the particle</p><p>can be scattered by the same angle where it possesses the momentum AC or AD</p><p>(figure) The same is true for particle The maximum scattering in the laboratory</p><p>corresponds to the case when the velocity vector in the frame of laboratory</p><p>become a tangent ( ) to the circle (see figure).</p><p>It then follows that</p><p>It is clear that is limited to for for the case point A</p><p>will be inside the sphere, and all angles of scattering from to are permitted.p>2m2 7 m1,m1 = m2,p>2umNote:</p><p>=</p><p>m2</p><p>m1</p><p>sin u1 max =</p><p>OC ¿</p><p>AO</p><p>=</p><p>p</p><p>'</p><p>m1vC</p><p>=</p><p>(m1m2>m1 + m2) v1</p><p>(m 2</p><p>1>m1 + m2) v1</p><p>AC ¿</p><p>u1 max</p><p>m2.</p><p>u1,</p><p>m1</p><p>p1</p><p>(m1 = m2)</p><p>p2¿p1¿</p><p>OB = m2vC = mv1 = p</p><p>'</p><p>.p</p><p>'</p><p>1</p><p>m2m1</p><p>OB =</p><p>p</p><p>'</p><p>u</p><p>'</p><p>.</p><p>p2¿ = p</p><p>'</p><p>2¿ + m2vC</p><p>p1¿ = p</p><p>'</p><p>1¿ + m1vC</p><p>p2p1ABp1</p><p>p1¿ + p2¿ = p1 = (m1 + m2</p><p>) vC</p><p>p</p><p>' = p</p><p>'¿</p><p>T</p><p>'</p><p>i = T</p><p>'</p><p>ƒ or</p><p>p</p><p>'2</p><p>2m</p><p>=</p><p>p</p><p>'2</p><p>2m</p><p>T</p><p>'</p><p>i = p</p><p>'</p><p>2 2m</p><p>m = m1m2 /(m1 + m2</p><p>)| p</p><p>'</p><p>1 | = | p</p><p>'</p><p>2 | = p</p><p>' = m vrel = m v1</p><p>A B</p><p>CC'</p><p>D</p><p>Om1vc(1max)</p><p>q1</p><p>q</p><p>q</p><p>q~</p><p>m2vc</p><p>m1 > m2</p><p>/ .</p><p>0</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 111</p><p>1.176 From the symmetry of the problem, the velocity of the disk A will be directed either</p><p>in the initial direction or opposite to it just after the impact. Let the velocity of the</p><p>disk A after the collision be and be directed towards right after the collision. It is</p><p>also clear from the symmetry of the problem that the disks B and C have equal speed</p><p>(say ) in the directions shown. From the condition of the problem,</p><p>(1)</p><p>For the three disk system, from the conservation of linear momentum in the symme-</p><p>try direction (towards right)</p><p>(2)</p><p>From the definition of the coefficient of restitution, we have for disks A and B</p><p>C</p><p>But for perfectly elastic collision.</p><p>So, (3)</p><p>From Eqs. (2) and (3)</p><p>(using Eq. 1)</p><p>Hence, we have,</p><p>Therefore, the disk A will recoil if and stop if</p><p>Note:One can write the equations of momentum conservation along the direction</p><p>perpendicular to the initial direction of disk A and the conservation of kinetic ener-</p><p>gy instead of the equation of restitution.</p><p>1.177 (a) Let a molecule come with velocity to strike another stationary molecule an just</p><p>after collision their velocities become and respectively. As the mass of each</p><p>molecule is same, conservation of linear momentum and conservation of kinetic</p><p>energy for the system (both molecules), respectively, gives</p><p>and v 2</p><p>1 = v ¿21 + v ¿22</p><p>v1 = v ¿1 + v ¿2</p><p>v 2¿v1¿</p><p>v1</p><p>h=12.h 6 12</p><p>v ¿ =</p><p>v (h 2 - 2)</p><p>6 - h 2</p><p>=</p><p>v (h2 - 2)</p><p>6 - h 2</p><p>v ¿ =</p><p>v (1 - 2 sin2 u)</p><p>(1 + 2 sin2 u)</p><p>v sin u = v– - v ¿ sin u</p><p>e = 1,</p><p>e =</p><p>v – - v ¿ sin u</p><p>v sin u - 0</p><p>mv = 2mv – sin u + mv ¿ or v = 2v – sin u + v ¿</p><p>cos u =</p><p>h</p><p>d</p><p>2</p><p>d</p><p>=</p><p>h</p><p>2</p><p>so, sin u = 24 - h2>2</p><p>v –</p><p>v ¿</p><p>v</p><p>A</p><p>B</p><p>C</p><p>d</p><p>v'</p><p>v"</p><p>v"</p><p>d</p><p>,</p><p>,</p><p>112 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>From the property of vector addition it is obvious from the obtained equations that</p><p>(b) Due to the loss of kinetic energy in inelastic collision so,</p><p>and therefore angle of divergence</p><p>1.178 Suppose that at time the rocket has the mass m and the velocity relative to the</p><p>reference frame, employed. Now consider the inertial frame moving with the veloc-</p><p>ity that the rocket has at the given moment. In this reference frame, the momentum</p><p>increment that the rocket and ejected gas system acquires during time dt is,</p><p>or</p><p>or</p><p>1.179 According to the question, and so the equation for this system</p><p>becomes,</p><p>As</p><p>Integrating within the limits</p><p>Thus,</p><p>As so in vector form</p><p>1.180 According to the question, F (external force)</p><p>So,</p><p>As,</p><p>So, in scalar form</p><p>or</p><p>wdt</p><p>u</p><p>= -</p><p>dm</p><p>m</p><p>mdv = -udm</p><p>d vc T u</p><p>m</p><p>d v</p><p>dt</p><p>=</p><p>dm</p><p>dt</p><p>u</p><p>= 0</p><p>v = -u ln m0 /md vc T u,</p><p>v = u ln</p><p>m0</p><p>m</p><p>1</p><p>u</p><p>3</p><p>v</p><p>0</p><p>dv = -3</p><p>0</p><p>m0</p><p>dm</p><p>m</p><p>or</p><p>v</p><p>u</p><p>= ln</p><p>m0</p><p>m</p><p>d vc T u, so, m dv = -udm.</p><p>m</p><p>d v</p><p>dt</p><p>=</p><p>dm</p><p>dt</p><p>u</p><p>m = -dm>dtF = 0</p><p>m w = F - mu</p><p>m</p><p>d v</p><p>dt</p><p>= F - mu</p><p>d p = md v + mdt u</p><p>Field 349</p><p>3.3 Electric Capacitance. Energy of an Electric Field 369</p><p>3.4 Electric Current 392</p><p>3.5 Constant Magnetic Field. Magnetics 425</p><p>3.6 Electromagnetic Induction. Maxwell’s Equations 454</p><p>3.7 Motion of Charged Particles in Electric and Magnetic Fields 486</p><p>Preface vii</p><p>PHYSICAL FUNDAMENTALS</p><p>OF MECHANICS</p><p>1.1 Kinematics</p><p>1.1 Let be the stream velocity and the velocity of motorboat with respect to water.</p><p>The motorboat reached point B while going downstream with velocity and</p><p>then returned with velocity and passed the raft at point C. Let t be the time</p><p>for the raft (which flows with stream with velocity ) to move from point A to C, dur-</p><p>ing which the motorboat moves from A to B and then from B to C.</p><p>Therefore,</p><p>On solving, we get</p><p>(on substituting values)</p><p>Alternate:</p><p>In the frame of reference moving with the stream velocity the raft becomes stationary.</p><p>The boat travels for time and turns and then travels back to reach the raft. As the</p><p>raft is still at its original position, hence it would take exactly further time (the same</p><p>time interval) for the boat to reach the raft. The boat thus takes time to reach the</p><p>raft in the frame of stream. In Newtonian mechanics time interval between two events</p><p>is frame independent, so is also the time interval in the frame of river bank. In the</p><p>mean time the raft has travelled l. Hence the speed of the river is</p><p>1.2 Let s be the total distance traversed by the point and t1 the time taken to cover half the</p><p>distance. Further let 2t be the time to cover the rest half of the distance.</p><p>Therefore,</p><p>and</p><p>s</p><p>2</p><p>= (v1 + v2) t or 2t =</p><p>s</p><p>v1 + v2</p><p>s</p><p>2</p><p>= v0t1</p><p>or t1 =</p><p>s</p><p>2v0</p><p>v0 =</p><p>l</p><p>2t</p><p>= 3 km/h</p><p>v0</p><p>2t</p><p>2t</p><p>t</p><p>t</p><p>v0 =</p><p>l</p><p>2t</p><p>= 3 km/h</p><p>t =</p><p>l</p><p>v0</p><p>= t +</p><p>(v0 + v ¿)t - l</p><p>(v ¿ – v0)</p><p>v0</p><p>(v ¿ – v0</p><p>)</p><p>(v0 + v ¿)</p><p>v ¿v0</p><p>PART1</p><p>Cl</p><p>B</p><p>( ')v + v τ0</p><p>v0</p><p>A</p><p>Hence the sought average velocity is given by</p><p>1.3 As the car starts from rest and finally comes to a stop, and the rate of acceleration and</p><p>deceleration are equal, the distances covered as well as the times taken are same in</p><p>these phases of motion.</p><p>Let be the time for which the car moves uniformly. Then the acceleration/deceleration</p><p>time is each. So,</p><p>or</p><p>Hence,</p><p>1.4 (a) Sought average velocity</p><p>(b) For the maximum velocity, should be</p><p>maximum. From the figure is maximum</p><p>for all points on the line ab thus the sought max-</p><p>imum velocity becomes average velocity for the</p><p>line ab and is</p><p>(c) Time should be such that corresponding to it the slope should coincide</p><p>with the chord Oc, to satisfy the relationship . From figure the tangent ds>dt = s>t0 ds>dtt0</p><p>=</p><p>100 cm</p><p>4 s</p><p>= 25 cm/s</p><p>ds>dt</p><p>ds>dt</p><p>6v7 =</p><p>s</p><p>t</p><p>=</p><p>200 cm</p><p>20 s</p><p>= 10 cm/s</p><p>¢t = tA1 -</p><p>46v7</p><p>wt</p><p>= 15 s</p><p>¢t</p><p>2 = t2 -</p><p>4 6v7 t</p><p>w</p><p>6v7 t = 2e 1</p><p>2</p><p>w</p><p>(t - ¢t)2</p><p>4</p><p>f + w</p><p>(t - ¢t)</p><p>2</p><p>¢t</p><p>(t - ¢t )/2</p><p>¢t</p><p>6v7 =</p><p>s</p><p>t1 + 2t</p><p>=</p><p>s</p><p>[s /2v0] + [s / [(v1 + v2 )]</p><p>=</p><p>2v0(v1 + v2 )</p><p>v1 + v2 + 2v0</p><p>2 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>2010 14 t (s)o</p><p>1</p><p>a</p><p>b</p><p>2</p><p>s (m)</p><p>c</p><p>at point c passes through the origin and thus corresponding time</p><p>1.5 Let the particles collide at the point A (see figure), whose position</p><p>vector is r3 (say). If t be the time taken by each particle to reach</p><p>point A, then from triangle law of vector addition:</p><p>So, (1) r1 - r2 = (v2 - v1)t</p><p>r3 = r1 + v1t = r2 + v2t</p><p>t = t0 = 16 s.</p><p>x</p><p>t</p><p>y Av1</p><p>v2</p><p>t</p><p>r1</p><p>r2</p><p>r3</p><p>O</p><p>Therefore, (2)</p><p>From Eqs. (1) and (2)</p><p>Thus, which is the sought relationship.</p><p>1.6 We have</p><p>In the vector diagram for above relation, let us drop a dotted perpendicular from the</p><p>tip of v onto the line of action of v0. Using the property of right angle triangle,</p><p>and</p><p>On substitution</p><p>1.7 Let us suppose one of the swimmers (say 1) crosses the river along AB, which is</p><p>obviously the shortest path. Time taken to cross the river by the swimmer 1</p><p>(where is the width of the river) (1)</p><p>For the other swimmer (say 2), who follows the quickest path, the time taken to cross</p><p>the river,</p><p>(2)t2 =</p><p>d</p><p>v ¿</p><p>AB = dt1 =</p><p>d</p><p>2v ¿2 - v 0</p><p>2</p><p>w¿ = 19.1°</p><p>tan w¿ =</p><p>v sin w</p><p>v cos w + v0</p><p>= 39.7 km/h (on substituting values)</p><p>v ¿ = 2(v cos w + v0)</p><p>2 + (v sin w)2</p><p>= 2v 0</p><p>2 + v</p><p>2 + 2v0 v cos w</p><p>v ¿ = v - v0</p><p>r1 - r2</p><p>| r1 - r2 |</p><p>=</p><p>v2 - v1</p><p>| v2 - v1 |</p><p>,</p><p>r1 - r2 = (v2 - v1) =</p><p>| r1 - r2 |</p><p>| v2 - v1|</p><p>t =</p><p>| r1 - r2 |</p><p>| v2 - v1 |</p><p>1.1 KINEMATICS 3</p><p>v0</p><p>v</p><p>v'</p><p>ϕ′ϕ</p><p>d</p><p>A</p><p>v2</p><p>v0</p><p>v'</p><p>B Cx</p><p>d</p><p>A</p><p>v1</p><p>v0</p><p>v'</p><p>B</p><p>In the time t2, drifting of the swimmer 2, becomes</p><p>(using Eq. 2) (3)x = v0t2 =</p><p>v0</p><p>v ¿</p><p>d</p><p>If t3 be the time for swimmer 2 to walk the distance x to come C to B (see figure), then</p><p>(using Eq. 3) (4)</p><p>According to the problem</p><p>or</p><p>Thus,</p><p>1.8 Let l be the distance covered by the boat A along the river as well as by the boat B</p><p>across the river. Let v0 be the stream velocity and v � the velocity of each boat with</p><p>respect to water. Therefore time taken by the boat A in its journey</p><p>and for the boat B</p><p>Hence,</p><p>On substitution,</p><p>1.9 Let v0 be the stream velocity and v � the velocity of boat with respect to water. As</p><p>some drifting of the boat is inevitable.</p><p>Let make an angle with flow direction (see figure), then the time taken to cross the</p><p>river</p><p>(where d is the width of the river)</p><p>In this time interval, the drifting of the boat</p><p>= 1v ¿cos u + v02 d</p><p>v ¿sin u</p><p>= 1cot u + h cosec u2dx = 1v ¿cos u + v02t</p><p>t =</p><p>d</p><p>v ¿sin u</p><p>uv¿</p><p>v0>v ¿ = n = 2 7 0,</p><p>tA</p><p>tB</p><p>= 1.8</p><p>tA</p><p>tB</p><p>=</p><p>v ¿</p><p>2v ¿2 - v 0</p><p>2</p><p>=</p><p>h</p><p>2h2 - 1</p><p>awhere h =</p><p>v ¿</p><p>v0</p><p>b</p><p>tB =</p><p>l</p><p>2v ¿2 - v</p><p>2</p><p>0</p><p>+</p><p>l</p><p>2v ¿2 - v</p><p>2</p><p>0</p><p>=</p><p>2l</p><p>2v ¿2 - v</p><p>2</p><p>0</p><p>tA =</p><p>l</p><p>v ¿ + v0</p><p>+</p><p>l</p><p>v ¿ - v0</p><p>=</p><p>2lv ¿</p><p>v ¿2 - v</p><p>2</p><p>0</p><p>u =</p><p>v0</p><p>(1 - v</p><p>2</p><p>0</p><p>/v¿2)-1/ 2 - 1</p><p>= 3 km/h</p><p>v0</p><p>u</p><p>=</p><p>v ¿</p><p>v ¿2 - v</p><p>2</p><p>0</p><p>- 1Q</p><p>d</p><p>2v ¿2 - v 0</p><p>2</p><p>=</p><p>d</p><p>v ¿</p><p>+</p><p>v0d</p><p>v ¿u</p><p>t1 = t2 + t3</p><p>t3 =</p><p>x</p><p>u</p><p>=</p><p>v0d</p><p>v ¿u</p><p>4 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>For xmin (minimum drifting)</p><p>Hence,</p><p>Alternate:</p><p>Let be the stream velocity, the velocity of boat with respect to water and be</p><p>the resultant velocity of boat, i.e.,</p><p>The angle from the direction of stream at which</p><p>boat can be rowed lies in the interval 0 to</p><p>Now let us draw the vector dia-</p><p>gram of velocity vectors. In the figure, a semicircle</p><p>has been drawn whose radius is . The tip of vec-</p><p>tor lies on this semicircle. One can observe that</p><p>for minimum drifting, angle should be maximum</p><p>and the line AB representing vector v becomes tan-</p><p>gent to the semicircle so that line OB representing</p><p>becomes perpendicular to it. The minimum drifting is CD.</p><p>From the figure (1)</p><p>(see figure)</p><p>Hence,</p><p>In the triangle OCD</p><p>(2)</p><p>From Eqs. (1) and (2)</p><p>or</p><p>So,</p><p>Hence, xmin = d2n</p><p>2 - 1</p><p>OC = 2(OD)2 - (OC)2 = OC 1n</p><p>2 - 1</p><p>OD = n (OC)</p><p>OC</p><p>OD</p><p>=</p><p>1</p><p>n</p><p>sin a =</p><p>OC</p><p>OD</p><p>= 120° (on substituting values) u = sin-1 (1>n) + p>2u =</p><p>p</p><p>2</p><p>+ a</p><p>sin a =</p><p>OB</p><p>AO</p><p>=</p><p>v ¿</p><p>v0</p><p>=</p><p>1</p><p>n</p><p>v¿</p><p>a</p><p>v¿</p><p>v ¿</p><p>p (0 … u … p).</p><p>u</p><p>v = v0 + v¿.</p><p>vv ¿v0</p><p>u = 120°</p><p>cos u = -</p><p>1</p><p>n</p><p>= -</p><p>1</p><p>2</p><p>d</p><p>du</p><p>1cot u + h cosec u2 = 0, which yields</p><p>1.1 KINEMATICS 5</p><p>d</p><p>O</p><p>v</p><p>v0</p><p>v'</p><p>A Bx</p><p>x</p><p>y</p><p>q</p><p>a q</p><p>a</p><p>A</p><p>O</p><p>B</p><p>C</p><p>D</p><p>x</p><p>y</p><p>v</p><p>v'</p><p>v0</p><p>1.10 Let us suppose one of the bodies, say 1, is thrown upwards and the other body</p><p>(say 2) at an angle to the horizontal. Indicate y-axis vertically upward and x-axis</p><p>along horizontal. Let after the time interval t, the particles 1 and 2 have coordinates</p><p>(0, y1) and (x2, y2)</p><p>For body 1,</p><p>and for body 2,</p><p>Thus the sought separation between the particles 1 and 2 after the time interval t</p><p>Alternate:</p><p>The solution of this problem becomes interesting in the frame attached</p><p>with one of the bodies. Let the body thrown straight up be 1 and the</p><p>other body be 2, then for the body 1 in the frame of 2 from the kine-</p><p>matical equation for constant acceleration (since both are moving</p><p>under constant acceleration) is</p><p>So,</p><p>or</p><p>But,</p><p>So, from properties of triangle</p><p>Hence, the sought distance</p><p>1.11 Let the velocities of the particles become mutually perpendicular after</p><p>time t. Then their velocities become</p><p>v¿1 = v1 + gt ; v¿2 = v2 + g t</p><p>(say v¿1 and v¿2)</p><p>| r12 | = v0t 22 (1 - sin u0) = 22 m</p><p>0(12) = 2v</p><p>2</p><p>0 + v</p><p>2</p><p>0 - 2v0v0 cos (p>2 - u0)</p><p>| v01 | = | v02 | = v0</p><p>| r12 | = | v0(12) | t</p><p>r12 = v0(12)</p><p>= F dt</p><p>v,t,</p><p>6 90°.v1</p><p># v2 7 0</p><p>v 2</p><p>1 7 v ¿21 + v ¿22</p><p>v1 � v2 or v1</p><p># v2 = 0</p><p>.</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 113</p><p>Integrating within the limit for</p><p>Hence,</p><p>1.181 As from the equation of dynamics of a body with variable mass,</p><p>(1)</p><p>Now and since (because is constant),</p><p>where is the angle by which the spaceship turns in time dt.</p><p>So,</p><p>or</p><p>1.182 We have</p><p>Integrating</p><p>As so, from the equation of variable mass system</p><p>or</p><p>Hence,</p><p>1.183 Let the car be moving in a reference frame to which the hopper is fixed and at any</p><p>instant of time, let its mass be m and velocity v.</p><p>Then from the general equation, for variable mass system</p><p>m</p><p>d v</p><p>dt</p><p>= F + u</p><p>dm</p><p>dt</p><p>v =</p><p>F</p><p>m</p><p>lna m0</p><p>m0 - mt</p><p>b</p><p>3</p><p>v</p><p>0</p><p>d v = F3</p><p>t</p><p>0</p><p>dt</p><p>(m0 - mt)</p><p>(m0 - mt)</p><p>d v</p><p>dt</p><p>= F or</p><p>d v</p><p>dt</p><p>= w =</p><p>F</p><p>(m0 - mt)</p><p>u = 0</p><p>3</p><p>m</p><p>m0</p><p>dm = -m3</p><p>t</p><p>0</p><p>dt or, m = m0 - mt</p><p>dm</p><p>dt</p><p>= -m or dm = -mdt</p><p>a = -</p><p>u</p><p>v0</p><p>3</p><p>m</p><p>m0</p><p>dm</p><p>m</p><p>=</p><p>u</p><p>v0</p><p>ln am0</p><p>m</p><p>b</p><p>- u</p><p>dm</p><p>m</p><p>= v0da or da = -</p><p>u</p><p>v0</p><p>dm</p><p>m</p><p>da</p><p>v0u � v, we must have |d v | = v0d advc Tu</p><p>m</p><p>d v</p><p>dt</p><p>= u</p><p>dm</p><p>dt</p><p>or d v = u</p><p>dm</p><p>m</p><p>F = 0,</p><p>m = m0 e</p><p>– (wt>u)</p><p>wt</p><p>u</p><p>= -3</p><p>m</p><p>m0</p><p>dm</p><p>m</p><p>or,</p><p>v</p><p>u</p><p>= - ln</p><p>m</p><p>m0</p><p>m (t)</p><p>114 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>We write the equation, for our system as,</p><p>So,</p><p>and</p><p>But</p><p>so,</p><p>Thus, the sought acceleration is</p><p>1.184 Let the length of the chain inside the smooth horizontal tube at an arbitrary instant</p><p>be x. From the equation,</p><p>as , for the chain inside the tube</p><p>(1)</p><p>Similarly for the overhanging part,</p><p>Thus,</p><p>or (2)</p><p>From Eqs. (1) and (2)</p><p>or</p><p>(As the length of the chain inside the tube decreases with time, ds = -dx.)</p><p>(x + h) v</p><p>dv</p><p>(-dx)</p><p>= gh</p><p>l(x + h) w = l hg or, (x + h) v</p><p>dv</p><p>ds</p><p>= hg</p><p>l hw = lhg - T</p><p>mw = F</p><p>u = 0</p><p>lxw = T , where l =</p><p>m</p><p>l</p><p>u = 0, Fc cw</p><p>m w = F + u</p><p>dm</p><p>dt</p><p>w =</p><p>d v</p><p>dt</p><p>=</p><p>F</p><p>m0a1 +</p><p>mt</p><p>m0</p><p>b2</p><p>v =</p><p>Ft</p><p>m0a1 +</p><p>mt</p><p>m0</p><p>b</p><p>m = m0 + mt</p><p>v =</p><p>Ft</p><p>m</p><p>on integration</p><p>d</p><p>dt</p><p>(m v) = F</p><p>m</p><p>d v</p><p>dt</p><p>= F - v</p><p>dm</p><p>dt</p><p>as, u = -v</p><p>x</p><p>h hg</p><p>T</p><p>TA</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 115</p><p>or</p><p>Integrating,</p><p>or</p><p>1.185 The angular momentum of the particle relative to point O is given as function of a</p><p>time as .</p><p>So, force moment relative to point O, is given by</p><p>As force moment so vector N coincides with vector b.</p><p>Let us depict the vectors N and M at an arbitrary instant of time</p><p>t, when angle between M and N is (see figure). It is obvious</p><p>from the figure that tan therefore , the time</p><p>.</p><p>Thus, , when vector N forms angle with vector M.</p><p>Alternate:</p><p>Let the angle between M and N, at</p><p>Then</p><p>So,</p><p>Hence, N = 2b A</p><p>a</p><p>b</p><p>2b 2t0</p><p>4 = a 2 + b 2t0</p><p>4 or t0 = A</p><p>a</p><p>b</p><p>(as t0 cannot be negative)</p><p>=</p><p>2b 2t 3</p><p>0</p><p>2a 2 + b 2t0</p><p>4 2bt0</p><p>=</p><p>bt 0</p><p>2</p><p>2a 2 + b 2t0</p><p>4</p><p>1</p><p>12</p><p>=</p><p>M # N</p><p>| M | | N |</p><p>=</p><p>(a + bt 2</p><p>0)</p><p># (2bt0)</p><p>2a 2 + b 2t 0</p><p>4 2bt0</p><p>t = t0.a = 45°</p><p>a = 45°N = 2b 1a /b</p><p>t = t0 = 1a /b</p><p>a = 45°a = a /bt 2</p><p>a</p><p>N = 2bt</p><p>N =</p><p>d M</p><p>dt</p><p>= 2bt</p><p>M = a + bt 2</p><p>v 2</p><p>2</p><p>= gh ln a l</p><p>h</p><p>b or v = A2gh ln a l</p><p>h</p><p>b</p><p>3</p><p>v</p><p>0</p><p>vdv = -gh 3</p><p>0</p><p>(l-h)</p><p>dx</p><p>x + h</p><p>vdv = -gh</p><p>dx</p><p>x + h</p><p>N</p><p>Ma</p><p>bt 2</p><p>α</p><p>,</p><p>,</p><p>116 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.186</p><p>Thus</p><p>Thus, angular momentum at maximum height, i.e., at</p><p>Alternate:</p><p>1.187 (a) The disk experiences gravity, the force of reaction of the horizontal surface, and</p><p>the force R of reaction of the wall at the moment of the impact against it. The</p><p>first two forces counter-balance each</p><p>other, leaving only the force R. It’s</p><p>moment relative to any point of the</p><p>line along which the vector R acts or</p><p>along normal to the wall is equal to</p><p>zero and therefore the angular mo-</p><p>mentum of the disk relative to any of</p><p>these points does not change in the</p><p>given process.</p><p>(b) During the course of collision with wall the position of disk is same and is</p><p>equal to Obviously the increment in momentum</p><p>Here, | ¢M | = 2mv l cos a</p><p>¢M = r00¿ * ¢p = 2mv l cos a n</p><p>r00¿</p><p>.</p><p>= 3</p><p>t</p><p>0</p><p>c av0t +</p><p>1</p><p>2</p><p>gt 2b * mg ddt = (v0 * mg)</p><p>t 2</p><p>2</p><p>M(0) = 0 so , M(t) = 3</p><p>t</p><p>0</p><p>N dt = 3</p><p>t</p><p>0</p><p>(r * mg)</p><p>= 37 kg m 2/s</p><p>M a t</p><p>2</p><p>b = amv0</p><p>3</p><p>2g bsin 2 a cos a</p><p>t</p><p>t</p><p>2</p><p>=</p><p>v0 sin a</p><p>g</p><p>M(t) =</p><p>mv0</p><p>gt 2 cos a</p><p>2</p><p>=</p><p>1</p><p>2</p><p>mv0</p><p>gt 2 cos a(–k)</p><p>= mv0 gt 2 sin ap</p><p>2</p><p>+ ab ( - k) +</p><p>1</p><p>2</p><p>mv0gt 2 sin ap</p><p>2</p><p>+ ab (k)</p><p>M(t) = r * p = av0t +</p><p>1</p><p>2</p><p>gt 2b * m (v0 + gt)</p><p>O</p><p>y</p><p>x</p><p>mg</p><p>r</p><p>v0</p><p>α</p><p>y</p><p>O</p><p>O'</p><p>R</p><p>A</p><p>n</p><p>l</p><p>directed normally</p><p>upward</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 117</p><p>1.188 (a) The ball is under the influence of forces T and mg at all the moments of time,</p><p>while moving along a horizontal circle. Obviously the vertical component of T</p><p>balances mg and so the net moment of these two about any point becomes</p><p>zero. The horizontal component of T, which provides the centripetal accelera-</p><p>tion to ball is already directed toward the centre (C ) of the horizontal circle,</p><p>thus its moment about the point C equals zero at all the moments of time.</p><p>Hence the net moment of the force acting on the ball about point C equals</p><p>zero and that’s why the angular momentum of the ball is conserved about the</p><p>horizontal circle.</p><p>(b) Let be the angle which the thread forms with the vertical</p><p>Now from equation of particle dynamics</p><p>Hence on solving,</p><p>(1)</p><p>As is constant in magnitude, so from the figure</p><p>where</p><p>Thus,</p><p>(using Eq. 1)</p><p>1.189 During the free fall, time , the reference point O moves in horizontal di-</p><p>rection by the distance . In the translating frame as , so</p><p>So, (using the value of and taking into account</p><p>Hence, | ¢M | = mVh</p><p>V� g)t | Mƒ | =</p><p>1</p><p>2</p><p>m gV a2h</p><p>g</p><p>b</p><p>=</p><p>1</p><p>2</p><p>m (g * V) t 2</p><p>= a1</p><p>2</p><p>gt 2 - Vtb * m (gt - V)</p><p>¢M = Mƒ = rf * m vƒ</p><p>Mi(O) = 0Vt</p><p>t = 22h /g</p><p>=</p><p>2mgl</p><p>v C1 - a g</p><p>v 2t</p><p>b2</p><p>| ¢M | = 2mvl cos a = 2 mvl 2 sin a cos a</p><p>= | rbo * m v | = mvl (as rbo � v)</p><p>M = | Mi | = | Mƒ |</p><p>| ¢M | = 2M cos a</p><p>| M |</p><p>cos a =</p><p>g</p><p>v 2l</p><p>T cos a = mg and T sin a = mv 2l sin a</p><p>a</p><p>C</p><p>T</p><p>O</p><p>M</p><p>Mi</p><p>Mf</p><p>mg</p><p>.</p><p>118 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.190 The Coriolis force is . Here is along the -axis (vertical). The mov-</p><p>ing disk is moving with velocity which is constant, the motion is along the x-axis,</p><p>then the Coriolis force is along y-axis and has the magnitude . At time t, the</p><p>distance of the centre of moving disk from O is (along x-axis). Thus the torque</p><p>N due to the Coriolis force is along the z-axis. Hence, equating this</p><p>to , we get</p><p>The constant is irrelevant and may be put equal to zero if the disk is originally set</p><p>in motion from the point O.</p><p>This discussion is approximate. The Coriolis force which causes the disk to swerve</p><p>from straight line motion and thus causes deviation from the above formula will be</p><p>substantial for large t.</p><p>1.191 If radial velocity of the particle then the total energy of the particle at any in-</p><p>stant is</p><p>(1)</p><p>where the second term is the kinetic energy of angular motion about the centre O.</p><p>Then the extreme values of r are determined by and solving the resulting</p><p>quadratic equation</p><p>we get</p><p>From this we see that</p><p>So,</p><p>Hence, m =</p><p>2kr1</p><p>2</p><p>v 2</p><p>2</p><p>1</p><p>2</p><p>mv2</p><p>2 + kr2</p><p>2 = k1r1</p><p>2 + r2</p><p>22</p><p>E = k1r 2</p><p>1 + r 2</p><p>12</p><p>r 2 =</p><p>E � 2E 2 - 2M</p><p>2k /m</p><p>2k</p><p>k (r 2) 2 - Er 2 +</p><p>M 2</p><p>2m</p><p>= 0</p><p>r</p><p># = 0</p><p>1</p><p>2</p><p>m r</p><p># 2 +</p><p>M 2</p><p>2mr 2</p><p>+ kr 2 = E</p><p>r</p><p># =</p><p>dM</p><p>dt</p><p>= 2mv 2</p><p>0 vt</p><p>or</p><p>M = mv 2</p><p>0 vt 2 + constant</p><p>dM /dt</p><p>N = 2m v0v</p><p># v0t</p><p>v0t</p><p>2mv0v</p><p>v0</p><p>zV2m (v ¿ * V)</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 119</p><p>Note: Eq. (1) can be derived from the standard expression for kinetic energy and</p><p>angular momentum in plane polar coordinates:</p><p>1.192 The swinging sphere experiences two forces: the gravitational force and the tension</p><p>of the thread. Now, it is clear from the condition, given in the problem, that the mo-</p><p>ment of these forces about the vertical axis, passing through the point of suspension</p><p>. Consequently, the angular momentum of the sphere relative to the given</p><p>axis (z) is constant.</p><p>Thus, (1)</p><p>where m is the mass of sphere and v is its velocity in the position, when the thread</p><p>forms an angle with the vertical. Mechanical</p><p>energy is also conserved, as the</p><p>sphere is under the influence of only one other force, i.e. tension, which does not</p><p>perform any work, as it is always perpendicular to the velocity.</p><p>So, (2)</p><p>From Eqs. (1) and (2), we get</p><p>1.193 Forces, acting on the mass m are shown in the figure. As the net torque</p><p>of these two forces about any fixed point must be equal to zero. Tension T, acting</p><p>on the mass m is a central force, which is always directed towards the centre O.</p><p>Hence the moment of force T is also zero about the point O and therefore the an-</p><p>gular momentum of the particle m is conserved about O.</p><p>So,</p><p>or</p><p>From</p><p>As the thread is pulled with constant speed</p><p>so, (where dot stands</p><p>for time derivative).</p><p>Thus T = F = mr u</p><p>#</p><p>2</p><p>r</p><p># = constant and r</p><p>$ = 0</p><p>- T = m ( r</p><p>$ - r u</p><p>#</p><p>2)</p><p>Fr = mar</p><p>u</p><p># v =</p><p>v0r</p><p>2</p><p>0</p><p>r 2</p><p>mr 2u</p><p>#</p><p>= mr 2</p><p>0 v0</p><p>N = m g,</p><p>v0 = 22gl>cos u</p><p>1</p><p>2</p><p>mv 2</p><p>0 + mgl cos u =</p><p>1</p><p>2</p><p>mv 2</p><p>p>2</p><p>mv0 (l sin u) = mvl</p><p>MZNZ = 0</p><p>T =</p><p>1</p><p>2</p><p>mr</p><p># 2 +</p><p>1</p><p>2</p><p>mr 2 u</p><p>#</p><p>2 and M = mr 2u</p><p>#</p><p>, respectively</p><p>T</p><p>N</p><p>mg</p><p>F</p><p>O</p><p>=</p><p>120 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Hence, the sought tension is</p><p>1.194 On the given system the weight of the body m is the only force whose moment is</p><p>effective about the axis of pulley. Let us take the sense of of the pulley at an ar-</p><p>bitrary instant as the positive sense of axis of rotation (z-axis).</p><p>As</p><p>So,</p><p>1.195 Let the point of contact of sphere at initial moment</p><p>be at . At an arbitrary moment, the forces</p><p>acting on the sphere are shown in the figure.</p><p>As and both ( and ) pass</p><p>through the same line, their net torque about any</p><p>point is zero. The force of static friction passes</p><p>through the point O, thus its moment about point</p><p>O becomes zero. Hence is the only force</p><p>which has effective torque about point O, and is given by normal-</p><p>ly emerging from the plane of figure.</p><p>As</p><p>Hence,</p><p>1.196 Let position vectors of the particles of the system be and with respect to the</p><p>points O and respectively, then we have</p><p>(1)</p><p>(where is the radius vector of with respect to ).</p><p>Now, the angular momentum of the system relative to the point O can be written</p><p>as follows:</p><p>(using Eq. 1)</p><p>or (2) M = M¿ + (r0 * p) (where p = ©p</p><p>M = ©ri * pi = ©( ri¿ * pi ) + ©3(r0 * pi</p><p>)</p><p>OO¿r0</p><p>ri = ri¿ + r0</p><p>O¿</p><p>ri¿ri</p><p>M (t) = Nt = mg R sin at</p><p>M(t = 0) = 0 so ¢M = M(t) = 3N dt</p><p>| N | = mgR sin a</p><p>sin amg</p><p>mg cos aNN = mg cos a</p><p>O(t = 0)</p><p>Mz(t) = 3</p><p>t</p><p>0</p><p>mgR dt = mgRt</p><p>Mz(0) = 0 so ¢Mz = Mz(t) =3Nzdt</p><p>V</p><p>F = mr u</p><p>#</p><p>2 = mr av0r</p><p>2</p><p>0</p><p>r 2</p><p>b2</p><p>=</p><p>m v 2</p><p>0r 4</p><p>0</p><p>r 3</p><p>fr</p><p>O</p><p>x</p><p>r</p><p>N</p><p>mg sin mg cosaa a</p><p>i</p><p>)</p><p>1.3 LAW OF CONSERVATION OF ENERGY, MOMENTUM AND ANGULAR MOMENTUM 121</p><p>From Eq. (2), if the total linear momentum of the system, , then its an-</p><p>gular momentum does not depend on the choice of the point .</p><p>Note In the C.M. frame, the system of particles, as whole is at rest.</p><p>1.197 On the basis of solution of problem 1.196. we have concluded that; “in the C.M.</p><p>frame, the angular momentum of system of particles is independent of the choice of</p><p>the point, relative to which it is determined” and in accordance with the problem</p><p>this is denoted by M.</p><p>We denote the angular momentum of the system of particles, relative to the point O,</p><p>by M. Since the internal and proper angular momentum , in the C.M. frame, does</p><p>not depend on the choice of the point , this point may be taken coincident with</p><p>the point O of the K-frame, at a given moment of time. Then at that moment, the ra-</p><p>dius vectors of all the particles, in both reference frames, are equal and the</p><p>velocities are related by the equation</p><p>where is the velocity of C.M. frame, relative to the K-frame. Consequently, we may</p><p>write</p><p>or</p><p>or</p><p>1.198 Taking striking ball as particle 1 and upper ball or the dumbbell as particle 2 the</p><p>velovity acquired by particle 2 on collsion, from the formula</p><p>To find the proper angular momentum of the dumbbell, let</p><p>us denote upper ball of the dumbbell as particle 1 and</p><p>lower ball as particle 2.</p><p>Then proper angular momentum vector is</p><p>M</p><p>'</p><p>= r1C * m1v1C + r2C * m2v2C = r1C * p</p><p>'</p><p>1 + r2C * p</p><p>'</p><p>2</p><p>v2¿ = 2a m v0</p><p>m + m /2</p><p>b - 0 =</p><p>4</p><p>3</p><p>v0</p><p>vi¿ = 2vC - vi (see solution of problem 1.168)</p><p>M = M</p><p>'</p><p>+ (rC * m vC</p><p>) = M</p><p>'</p><p>+ (rC * p)</p><p>M = M</p><p>'</p><p>+ m(rC * vC</p><p>), as ©mi</p><p>ri = m rC</p><p>where m = ©mi</p><p>M = ©mi(ri * vi</p><p>) = ©mi (ri * v</p><p>'</p><p>i) + ©mi(ri * vC</p><p>)</p><p>vC</p><p>vi = v</p><p>'</p><p>i + vC</p><p>(ri¿ = ri</p><p>)</p><p>O¿</p><p>M</p><p>'</p><p>O</p><p>p = 0</p><p>v0</p><p>m m /2</p><p>m /2</p><p>1</p><p>C</p><p>2</p><p>y</p><p>x</p><p>( )</p><p>122 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>In C.M. frame net linear momentum of any system is zero, so,</p><p>So,</p><p>We have</p><p>Hence,</p><p>Hence,</p><p>1.199 In the C.M. frame of the system (both the disks spring), the linear momentum of</p><p>the disks are related by the relation, at all the moments of time.</p><p>Where,</p><p>And the total kinetic energy of the system,</p><p>[see solution of problem 1.147(b)]</p><p>Bearing in mind that at the moment of maximum deformation of the spring, the pr-</p><p>ojection of along the length of the spring becomes zero, i.e., .</p><p>The conservation of mechanical energy of the considered system in the C.M. frame gives</p><p>(1)</p><p>Now from the conservation of angular momentum of the system about the C.M.,</p><p>or (2)</p><p>Using Eq. (2) in Eq. (1),</p><p>or</p><p>or</p><p>As x Z 0, thus x =</p><p>mv 2</p><p>0</p><p>kl0</p><p>mv 2</p><p>0 x</p><p>l0</p><p>= kx 2 [neglecting x 2> l0</p><p>2 ]</p><p>1</p><p>2</p><p>mv 2</p><p>0 c1 - a1 -</p><p>2x</p><p>l0</p><p>+</p><p>x 2</p><p>l 2</p><p>0</p><p>b d = kx 2</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>0 c1 - a1 -</p><p>x</p><p>l0</p><p>b2 d = kx 2</p><p>vrel (y) =</p><p>v0l0</p><p>(l0 + x)</p><p>= v0 a1 +</p><p>x</p><p>l0</p><p>b -1</p><p>= v0a1 -</p><p>x</p><p>l0</p><p>b as x V l0</p><p>1</p><p>2</p><p>a l0</p><p>2</p><p>b am</p><p>2</p><p>v0b = 2 a l0 + x</p><p>2</p><p>b m</p><p>2</p><p>vrel (y)</p><p>1</p><p>2</p><p>am</p><p>2</p><p>b v 2</p><p>0 =</p><p>1</p><p>2</p><p>kx 2 +</p><p>1</p><p>2</p><p>am</p><p>2</p><p>b v 2</p><p>rel (y)</p><p>vrel (x) = 0vrel</p><p>T =</p><p>1</p><p>2</p><p>mv 2</p><p>rel</p><p>p</p><p>'</p><p>1 = p</p><p>'</p><p>2 = p</p><p>' = mvrel</p><p>p</p><p>'</p><p>1 = -p</p><p>'</p><p>2,</p><p>+</p><p>M</p><p>'</p><p>=</p><p>mv0 l</p><p>3</p><p>M</p><p>'</p><p>= [r12 * p</p><p>'</p><p>1] = c l j *</p><p>mv0</p><p>3</p><p>i d =</p><p>mv0 l</p><p>3</p><p>(-k)</p><p>=</p><p>m</p><p>4</p><p>a4 v0</p><p>3</p><p>i - 0b =</p><p>mv0</p><p>3</p><p>i</p><p>p</p><p>'</p><p>1 = m (v1 - v2) (see solution of problem 1.147)</p><p>M</p><p>'</p><p>= (r1C - r2C) * p1</p><p>' = r12 * p</p><p>'</p><p>1</p><p>p</p><p>'</p><p>1 = -p</p><p>'</p><p>2.</p><p>( )</p><p>1.4 Universal Gravitation</p><p>1.200 We have (here is the mass of the Sun)</p><p>Thus</p><p>So, days</p><p>(The answer is incorrectly written in terms of the planetary mass M in answer sheet.)</p><p>1.201 For any planet</p><p>So,</p><p>(a) Thus,</p><p>So,</p><p>(b)</p><p>So,</p><p>where years, mass of the Sun.</p><p>Putting the values, we get km/s</p><p>= 2.15 * 10-4 km/s2</p><p>= a 2p</p><p>T</p><p>b4>3</p><p>(gMs)</p><p>1/3</p><p>Acceleration =</p><p>V 2</p><p>J</p><p>RJ</p><p>= a 2pgMS</p><p>T</p><p>b2>3</p><p>* a 2p</p><p>T1gMS</p><p>b2>3</p><p>VJ = 12.97</p><p>MS =T = 12</p><p>V</p><p>2</p><p>J =</p><p>1gMS22>3 12p22>3</p><p>T 2>3 or VJ = a2pgMS</p><p>T</p><p>b2>3</p><p>V</p><p>2</p><p>J =</p><p>gMS</p><p>RJ</p><p>and RJ = aT</p><p>1gMS</p><p>2p</p><p>b2>3</p><p>RJ</p><p>RE</p><p>= a TJ</p><p>TE</p><p>b2>3</p><p>= (12)2>3 = 5.24</p><p>TJ</p><p>TE</p><p>= a RJ</p><p>RE</p><p>b3>2</p><p>T =</p><p>2p</p><p>v</p><p>= 2pR 3>2>2gMS</p><p>MR v2 =</p><p>gMMS</p><p>R 2</p><p>or v = A</p><p>gMS</p><p>R 3</p><p>T =</p><p>2pgMS</p><p>v 3</p><p>=</p><p>2p * 6.67 * 10-11 * 1.97 * 1030</p><p>(34.9 * 103)3</p><p>= 1.94 * 107 s = 225</p><p>v =</p><p>v</p><p>r</p><p>=</p><p>v</p><p>gMS</p><p>>v 2</p><p>=</p><p>v 3</p><p>gMS</p><p>MS</p><p>Mv 2</p><p>r</p><p>=</p><p>gMMS</p><p>r 2</p><p>or r =</p><p>gMS</p><p>v 2</p><p>1.4 UNIVERSAL GRAVITATION 123</p><p>1.202 Semi-major axis It is sufficient to consider the motion be</p><p>along</p><p>a</p><p>circle</p><p>of semi-major axis for T does not depend on eccentricity.</p><p>Hence,</p><p>(again is the mass of the Sun)</p><p>1.203 We can think of the body as moving in a very elongated orbit of maximum distance</p><p>R and minimum distance 0 so semi major axis . Hence if is the time of fall</p><p>then</p><p>or</p><p>1.204 If the distances are scaled down, decreases by a factor</p><p>and so does . Hence T does not change.</p><p>1.205 The double star can be replaced by a single star of mass moving</p><p>about the centre of mass subjected to the force . Then</p><p>So,</p><p>or</p><p>1.206 (a) The gravitational potential due to at the point of location of</p><p>V2 = 3</p><p>q</p><p>r</p><p>G .dr = 3</p><p>q</p><p>r</p><p>-</p><p>gm1</p><p>x2</p><p>dx = -</p><p>gm1</p><p>r</p><p>m2m1</p><p>r = a T</p><p>2p</p><p>b2>3</p><p>1gM21>3 = 32gM (T>2 p) 2</p><p>r 3>2 =</p><p>T</p><p>2p</p><p>1gM</p><p>T =</p><p>2pr 3>2</p><p>Agm1m2</p><p>m1m2</p><p>m1 + m2</p><p>=</p><p>2pr 3>2</p><p>1gM</p><p>gm1m2>r 2</p><p>m1m2/(m1 + m2)</p><p>Msh3>2 R</p><p>3>2T = 2p R 3>2>1r Ms</p><p>t = T>422 = 365>412 = 64.5 days</p><p>a 2t</p><p>T</p><p>b2</p><p>= aR/2</p><p>R</p><p>b3 or t2 =</p><p>T 2</p><p>32</p><p>t= R>2</p><p>MS= p 21r + R23>2gMS</p><p>T =</p><p>2p (r + R )/2</p><p>3>2</p><p>1gMS</p><p>(r + R)/2</p><p>= (r + R)>2.</p><p>124 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>n</p><p>[ ]</p><p>.</p><p>So,</p><p>Similarly,</p><p>Hence,</p><p>(b) Choose</p><p>the location of the point mass as the origin. Then the potential energy dU</p><p>of an element of mass of the rod in the field of the point mass is</p><p>where x is the distance between the element and the point. (Note that the rod</p><p>and the point mass are on a straight</p><p>line.) Then if a is the distance of the</p><p>nearer end of the rod from the point</p><p>mass,</p><p>The force of interaction is</p><p>Minus sign means attraction.</p><p>1.207 As the planet is under central force (gravitational interaction), its angular momentum</p><p>is conserved about the Sun (which is situated at one of the focii of the ellipse).</p><p>So, (1)</p><p>From the conservation of mechanical energy of the system (Sun 	 planet),</p><p>or (using Eq. 1)</p><p>Thus, (2)</p><p>Hence, M = mv</p><p>2r 2 = m22gMSr 1r 2>(r 1 + r</p><p>2)</p><p>v</p><p>2 = 22g MS r 1>r 2 (r</p><p>1 + r</p><p>2)</p><p>-</p><p>gMS</p><p>r1</p><p>+</p><p>1</p><p>2</p><p>v 2</p><p>2</p><p>r 2</p><p>2</p><p>r 2</p><p>2</p><p>= - agMS</p><p>r2</p><p>b +</p><p>1</p><p>2</p><p>v 2</p><p>2</p><p>-</p><p>gMSm</p><p>r1</p><p>+</p><p>1</p><p>2</p><p>mv 2</p><p>1 = -</p><p>gMSm</p><p>r2</p><p>+</p><p>1</p><p>2</p><p>mv 2</p><p>2</p><p>mv1r1 = mv2r2 or v 2</p><p>1 =</p><p>v 2</p><p>2 r 2</p><p>1</p><p>r 2</p><p>1</p><p>F = -</p><p>0U</p><p>0a</p><p>= g</p><p>mM</p><p>l</p><p>*</p><p>1</p><p>1 +</p><p>l</p><p>a</p><p>a -</p><p>l</p><p>a2</p><p>b = -</p><p>gmM</p><p>a 1a + l 2</p><p>U = -g</p><p>mM</p><p>l</p><p>3</p><p>a+ l</p><p>a</p><p>dx</p><p>x</p><p>= -gm</p><p>M</p><p>l</p><p>ln a1 +</p><p>l</p><p>a</p><p>b</p><p>dU = -gm</p><p>M</p><p>l</p><p>dx</p><p>1</p><p>x</p><p>dM = M /l dx</p><p>U12 = U21 = U = -</p><p>gm1m2</p><p>r</p><p>U12 = -</p><p>gm1m2</p><p>r</p><p>U21 = m2V2 = -</p><p>gm1m2</p><p>r</p><p>1.4 UNIVERSAL GRAVITATION 125</p><p>x</p><p>r m2m1</p><p>dx</p><p>x</p><p>l</p><p>dx</p><p>x</p><p>x</p><p>m</p><p>1.208 From the previous problem, if are the maximum and minimum distances from</p><p>the Sun to the planet and are the corresponding velocities, then, say,</p><p>(using Eq. 2 of problem 1.207)</p><p>where</p><p>The same result can also be obtained directly by writing an equation analogous to</p><p>Eq. (1) of problem 1.191</p><p>(Here M is angular momentum of the planet and m is its mass.) For extreme posi-</p><p>tion and we get the quadratic equation</p><p>The sum of the two roots of this equation is</p><p>Thus,</p><p>1.209 From the conservation of angular momentum about the Sun.</p><p>(1)</p><p>From conservation of mechanical energy,</p><p>or (using Eq. 1)</p><p>or</p><p>So, r1 =</p><p>-2gMS � C4g</p><p>2M 2</p><p>S + 4 1v 2</p><p>0r</p><p>2</p><p>0 sin</p><p>2 a2 av</p><p>2</p><p>0 -</p><p>2gMS</p><p>r</p><p>0</p><p>b</p><p>2 av</p><p>2</p><p>0 -</p><p>2gMS</p><p>r 0</p><p>b</p><p>av</p><p>2</p><p>0 -</p><p>2gMS</p><p>r 0</p><p>br</p><p>2</p><p>1 + 2gMS</p><p>r1 - v 2</p><p>0 r 2</p><p>0 sin a = 0</p><p>v</p><p>2</p><p>0</p><p>2</p><p>-</p><p>gMS</p><p>r0</p><p>=</p><p>v</p><p>2</p><p>0 r</p><p>2</p><p>0 sin</p><p>2 a</p><p>2r</p><p>2</p><p>1</p><p>-</p><p>gMS</p><p>r</p><p>1</p><p>1</p><p>2</p><p>mv 2</p><p>0 -</p><p>gMSm</p><p>r0</p><p>=</p><p>1</p><p>2</p><p>mv 2</p><p>1 -</p><p>gMSm</p><p>r</p><p>1</p><p>mv0r0 sin a = mv</p><p>1r 1 = mv</p><p>2 or v</p><p>1r 1 = v</p><p>2r 2 = v0r 2 sin a</p><p>E = -</p><p>gmMS</p><p>2a</p><p>= constant</p><p>r</p><p>1 + r</p><p>2 = -</p><p>gmMS</p><p>E</p><p>= 2</p><p>Er 2 + gm MS</p><p>r -</p><p>M 2</p><p>2m</p><p>= 0</p><p>r</p><p>. = 0</p><p>E =</p><p>1</p><p>2</p><p>mr 2 -</p><p>M 2</p><p>2mr 2</p><p>-</p><p>gmMS</p><p>r</p><p>2a = major axis = r</p><p>1 + r</p><p>2.</p><p>=</p><p>gmMS</p><p>r</p><p>1 + r2</p><p># r</p><p>1</p><p>r</p><p>2</p><p>-</p><p>gmMS</p><p>r2</p><p>=</p><p>-gmMS</p><p>r</p><p>1 + r</p><p>2</p><p>=</p><p>gmMS</p><p>2a</p><p>E =</p><p>1</p><p>2</p><p>mv 2</p><p>2 -</p><p>gmMS</p><p>r2</p><p>v1, v2</p><p>r</p><p>1, r 2</p><p>126 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>where .</p><p>1.210 At the minimum separation with the Sun, the cosmic body’s velocity is perpendicu-</p><p>lar to its position vector relative to the Sun. If be the sought minimum distance,</p><p>from conservation of angular momentum about the Sun (C),</p><p>(1)</p><p>From conservation of mechanical energy of the system (Sun 1 cosmic body),</p><p>So, (using Eq. 1)</p><p>or</p><p>So,</p><p>Hence, taking positive root</p><p>1.211 (a) Suppose that the sphere has a radius equal to a. We may imagine that the sphere</p><p>is made up of concentric thin spherical shells (layers) with radii ranging from 0</p><p>to a, and each spherical layer is made up of elementary bands (rings). Let us first</p><p>calculate potential due to an elementary band of a spherical layer at the point of</p><p>location of the point mass m (say point P) (see figure). As all the points of the</p><p>band are located at the distance l from the point P, so,</p><p>(where is mass of the band) (1)</p><p>(2) = adM</p><p>2</p><p>b sin udu</p><p>0M = a dM</p><p>4pa2</p><p>b (2pa sin u) (adu)</p><p>0 M 0w = -</p><p>g0 M</p><p>l</p><p>rmin = 1gMS</p><p>>v</p><p>2</p><p>02 321 + 1lv</p><p>2</p><p>0 >gMS22 - 14</p><p>rmin =</p><p>-2gMS � 24g2m2 + 4v</p><p>2</p><p>0 v</p><p>2</p><p>0 l</p><p>2</p><p>2v</p><p>2</p><p>0</p><p>=</p><p>-gMS � 2g</p><p>2MS</p><p>2 + v</p><p>4</p><p>0 l</p><p>2</p><p>v</p><p>2</p><p>0</p><p>v</p><p>2</p><p>0 r</p><p>2</p><p>min + 2gMSrmin - v</p><p>2</p><p>0 l</p><p>2 = 0</p><p>v</p><p>2</p><p>0</p><p>2</p><p>= -</p><p>gMS</p><p>rmin</p><p>+</p><p>v</p><p>2</p><p>0</p><p>2r</p><p>2</p><p>min</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>0 = -</p><p>gMSm</p><p>rmin</p><p>+</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>mv0l = mvrmin or v =</p><p>v0l</p><p>rmin</p><p>rmin</p><p>h = v</p><p>2</p><p>0 r0 >gMS</p><p>=</p><p>1 � B1 -</p><p>v</p><p>2</p><p>0 r</p><p>2</p><p>0 sin</p><p>2 a</p><p>gMS</p><p>a 2</p><p>r0</p><p>-</p><p>v</p><p>2</p><p>0</p><p>rMS</p><p>b</p><p>a 2</p><p>r0</p><p>-</p><p>v0</p><p>gMS</p><p>b =</p><p>r0 C1 � 21 - (2 - h) sin2 a D</p><p>(2 - h)</p><p>1.4 UNIVERSAL GRAVITATION 127</p><p>and (3)</p><p>Differentiating Eq. (3), we get</p><p>(4)</p><p>Hence using above equations</p><p>(5)</p><p>Now integrating Eq. (5) over the whole</p><p>spherical layer</p><p>So, (6)</p><p>Eq. (6) demonstrates that the potential produced by a thin uniform spherical</p><p>layer outside the layer is such as if the whole mass of the layer were concentrat-</p><p>ed at its centre. Hence the potential due to the sphere at point P is given by</p><p>(7)</p><p>This expression is similar to that of Eq. (6).</p><p>Hence the sought potential energy of gravitational interaction of the particle m</p><p>and the sphere,</p><p>(b) Using the equation</p><p>(using Eq. 7)</p><p>So, (8)</p><p>1.212 (The problem has already a clear hint in the answer sheet of the problem book. Here</p><p>we adopt a different method.)</p><p>Let m be the mass of the spherical layer, which is imagined to be made up of rings.</p><p>At a point inside the spherical layer at distance r from the centre, the gravitational</p><p>potential due to a ring element of radius a equals</p><p>dw = -</p><p>gm</p><p>2ar</p><p>dl (see Eq. (5) of solution of problem 1.211)</p><p>G = -</p><p>gM</p><p>r</p><p>3</p><p>r and F = m G = -</p><p>gmM</p><p>r</p><p>3</p><p>r</p><p>Gr = -</p><p>gM</p><p>r</p><p>2</p><p>Gr = -</p><p>0w</p><p>0r</p><p>U = mw = -</p><p>gMm</p><p>r</p><p>w = Ldw -</p><p>g</p><p>r</p><p>LdM = -</p><p>gM</p><p>r</p><p>dw = -</p><p>gdM</p><p>r</p><p>dw = L 0w = -</p><p>gdM</p><p>2ar</p><p>3</p><p>r +a</p><p>r -a</p><p>dl</p><p>0w = - agdM</p><p>2ar</p><p>b dl</p><p>ldl = ar sin ud u</p><p>l</p><p>2 = a</p><p>2 + r</p><p>2 - 2 ar cos u</p><p>128 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>d</p><p>O</p><p>a</p><p>r</p><p>l</p><p>P</p><p>m</p><p>So,</p><p>Hence,</p><p>Hence, gravitational field strength as well as field force becomes</p><p>zero inside a thin spherical layer.</p><p>1.213 One can imagine that the uniform hemisphere is made up of thin hemispherical lay-</p><p>ers of radii ranging from 0 to R. Let us consider such a layer (see figure). Potential at</p><p>point O due to this layer is</p><p>(This is because all points of each hemispherical shell are equidistance from O.)</p><p>Hence,</p><p>Hence, the work done by the gravitational field</p><p>force on the particle of mass m, to remove it to in-</p><p>finity is given by the formula</p><p>Hence the sought work,</p><p>(The work done by the external agent is – A .)</p><p>1.214 In the solution of problem 1.211, we have obtained and G due to a uniform sphere,</p><p>at a distance r from it’s centre outside it. We have from Eqs. (7) and (8) of problem 1.211,</p><p>(A)</p><p>In accordance with the Eq. (1) of the solution of 1.212, potential due to a spherical</p><p>shell of radius a, at any point, inside it becomes</p><p>(B)</p><p>For a point (say P) which lies inside the uniform solid sphere, the potential at that</p><p>point may be represented as a sum</p><p>winside = w1 + w2</p><p>w</p><p>w =</p><p>gM</p><p>a</p><p>= constant and Gr = -</p><p>0w</p><p>0r</p><p>= 0</p><p>w = -</p><p>gM</p><p>r</p><p>and G = -</p><p>gM</p><p>r</p><p>3</p><p>r</p><p>w</p><p>A</p><p>0: q = -</p><p>3gmM</p><p>2R</p><p>A = mw, since w = 0 at infinity</p><p>w = Ldw = -</p><p>3gM</p><p>R</p><p>3 3</p><p>R</p><p>0</p><p>rdr = -</p><p>3gM</p><p>2R</p><p>dw = -</p><p>gdm</p><p>r</p><p>= -</p><p>3gM</p><p>r</p><p>3</p><p>rdr, where dm =</p><p>M</p><p>(2>3) pR</p><p>3</p><p>a4pr</p><p>2</p><p>2</p><p>b dr</p><p>Gr = -</p><p>0w</p><p>0 r</p><p>= 0</p><p>w = Ldw = -</p><p>gm</p><p>2ar3</p><p>a+ r</p><p>a- r</p><p>dl = -</p><p>gm</p><p>a</p><p>1.4 UNIVERSAL GRAVITATION 129</p><p>dq</p><p>O</p><p>a</p><p>r</p><p>l</p><p>r</p><p>dr</p><p>M</p><p>m</p><p>r</p><p>O</p><p>where is the potential of a solid sphere having radius r and is the potential of</p><p>the layer of radii r and R. In accordance with Eq. (A)</p><p>The potential produced by the layer (thick shell) is the same at all points inside</p><p>it. The potential is easiest to calculate, for the point positioned at the layer’s cen-</p><p>tre. Using Eq. (B)</p><p>where</p><p>is the mass of a thin layer between the radii r and</p><p>Thus, (C)</p><p>From Eq. (B)</p><p>or</p><p>where is the density of the sphere</p><p>The plots and for a uniform sphere of radius R are shown in the figure</p><p>of answer sheet.</p><p>Alternate:</p><p>Like Gauss’s theorem of electrostatics, one can derive Gauss’s theorem for gravitation</p><p>in the form G.dS For calculation of G at a point inside the sphere</p><p>at a distance r from its centre, let us consider a Gaussian surface of radius r. Then</p><p>Hence, G = -</p><p>gM</p><p>R</p><p>3</p><p>r = -g</p><p>4</p><p>3</p><p>pr r aas r =</p><p>M</p><p>(4>3)pR</p><p>3</p><p>b</p><p>Gr4pr</p><p>2 = -4pg aM</p><p>r</p><p>3</p><p>br</p><p>3 or Gr = -</p><p>gM</p><p>R</p><p>3</p><p>r</p><p>= -4p gmenclosed.</p><p>G (r)w (r)</p><p>r</p><p>=</p><p>M</p><p>4 /3 pR</p><p>3</p><p>G = -</p><p>gM</p><p>R</p><p>3</p><p>r = -g</p><p>4</p><p>3</p><p>prr</p><p>Gr =</p><p>gMr</p><p>R</p><p>3</p><p>Gr =</p><p>- 0w</p><p>0r</p><p>winside = w1 + w2 = agM</p><p>2R</p><p>b a3 -</p><p>r</p><p>2</p><p>R</p><p>2</p><p>b</p><p>r + dr.</p><p>dM =</p><p>M</p><p>(4>3) pR</p><p>3</p><p>4pr</p><p>2dr = a3M</p><p>R</p><p>3</p><p>b r</p><p>2dr</p><p>w2 = -g3</p><p>R</p><p>r</p><p>dM</p><p>r</p><p>= -</p><p>3</p><p>2</p><p>gM</p><p>R</p><p>3</p><p>(R</p><p>2 - r</p><p>2)</p><p>w2</p><p>w2</p><p>w1 = -</p><p>g</p><p>r</p><p>a M</p><p>(4>3) pR</p><p>3</p><p>4</p><p>3</p><p>pr</p><p>3b =</p><p>gM</p><p>R</p><p>3</p><p>r</p><p>2</p><p>w2w1</p><p>130 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>.</p><p>So,</p><p>Integrating and summing up, we get</p><p>From Gauss’ theorem for a point outside it,</p><p>Thus,</p><p>1.215 Treating the cavity as negative mass of density in a uniform sphere density</p><p>and using the superposition principle, the sought field strength is</p><p>or</p><p>where and are the position vectors of an arbitrary</p><p>point P inside the cavity with respect to centre of sphere</p><p>and cavity, respectively.</p><p>Thus,</p><p>1.216 We partition the solid sphere into thin spherical layers and consider a layer of thick-</p><p>ness dr lying at a distance r from the centre of the ball. Each spherical layer presses</p><p>on the layers within it. The considered layer is attracted to the part of the sphere lying</p><p>within it (the outer part does not act on the layer). Hence for the considered layer</p><p>or</p><p>(where is the mean density of sphere)</p><p>or dp =</p><p>4</p><p>3</p><p>pgr2 rdr</p><p>r</p><p>dp 4 pr</p><p>2 =</p><p>g (4 /3 p r</p><p>3 r) (4pr</p><p>2drr)</p><p>r</p><p>2</p><p>dp 4pr</p><p>2 = dF</p><p>G = -</p><p>4</p><p>3</p><p>pgr (r+ - r–) = -</p><p>4</p><p>3</p><p>pgrl</p><p>r-r+</p><p>G = -</p><p>4</p><p>3</p><p>pgrr+ + -</p><p>4</p><p>3</p><p>gp (-r) r-</p><p>G = G1 + G2</p><p>+r-r</p><p>w (r) = 3</p><p>q</p><p>r</p><p>Grdr = -</p><p>gM</p><p>r</p><p>Gr4pr</p><p>2 = -4pg M or Gr = -</p><p>gM</p><p>r</p><p>2</p><p>w = -</p><p>gM</p><p>2R</p><p>a3 -</p><p>r</p><p>2</p><p>R</p><p>2</p><p>b</p><p>w = 3</p><p>q</p><p>r</p><p>Grdr = 3</p><p>R</p><p>r</p><p>-</p><p>gM</p><p>R</p><p>3</p><p>rdr + 3</p><p>q</p><p>R</p><p>-</p><p>gM</p><p>r</p><p>2</p><p>dr</p><p>1.4 UNIVERSAL GRAVITATION 131</p><p>r_</p><p>Pr+</p><p>l</p><p>R</p><p>r</p><p>dr</p><p>Thus,</p><p>(the pressure must vanish at )</p><p>or</p><p>Putting , we have the pressure at sphere’s centre, and treating it as the Earth</p><p>where mean density is equal to and we</p><p>have,</p><p>1.217 (a) Since the potential at each point of a spherical surface (shell) is constant and is</p><p>equal to , [as we have in Eq. (1) of solution of problem 1.212].</p><p>We obtain in accordance with the equation</p><p>(The factor is needed otherwise contribution of different mass elements is</p><p>counted twice.)</p><p>(b) In this case the potential inside the sphere depends only on r (see Eq. (C) of the</p><p>solution of problem 1.214)</p><p>Here dm is the mass of an elementary spherical layer confined between the radii</p><p>r and given by</p><p>=</p><p>1</p><p>23</p><p>R</p><p>0</p><p>a 3m</p><p>R</p><p>3</p><p>b r</p><p>2 dr -</p><p>3gm</p><p>2R</p><p>a1 -</p><p>r</p><p>2</p><p>3R</p><p>2</p><p>b</p><p>U =</p><p>1</p><p>2</p><p>3dm w</p><p>dm = (4pr</p><p>2 drr) = a3m</p><p>R</p><p>3</p><p>b r</p><p>2dr</p><p>r + dr</p><p>w = -</p><p>3gM</p><p>2R</p><p>a1 -</p><p>r</p><p>2</p><p>3R</p><p>2</p><p>b</p><p>1/2</p><p>=</p><p>1</p><p>2</p><p>a -</p><p>gm</p><p>R</p><p>b m = -</p><p>gm2</p><p>2R</p><p>U =</p><p>1</p><p>23dm w =</p><p>1</p><p>2</p><p>w3dm</p><p>w = -gm /R</p><p>p = 1.73 * 1011 Pa or 1.72 * 106 atm.</p><p>R = 64 * 102 kmr = 5.5 * 103 kg/m3</p><p>r = 0</p><p>p =</p><p>3</p><p>8</p><p>(1 - (r</p><p>2>R</p><p>2)) gM</p><p>2>pR</p><p>4, putting r = M>(4>3) pR</p><p>3</p><p>r = R</p><p>p = 3</p><p>R</p><p>r</p><p>dp =</p><p>2p</p><p>3</p><p>gr2 (R</p><p>2 - r</p><p>2)</p><p>132 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>After integrating, we get</p><p>1.218 Let circular frequency of the satellite in the outer orbit,</p><p>circular frequency of the satellite in the inner orbit.</p><p>So, relative angular velocity , where sign is to be taken when the satel-</p><p>lites are moving in the same sense and sign if they are moving in opposite sense.</p><p>Hence, time between closest approaches</p><p>where is 0 in the first case and 2 in the second case.</p><p>1.219</p><p>and</p><p>Then,</p><p>1.220 Let h be the sought height in the first case, so</p><p>or</p><p>99</p><p>100</p><p>= a1 +</p><p>h</p><p>R</p><p>b -2</p><p>=</p><p>gM</p><p>R</p><p>2 (1 + h/R )2</p><p>=</p><p>g</p><p>(1 + h /R )2</p><p>99</p><p>100</p><p>g =</p><p>gM</p><p>(R + h)2</p><p>v1 : v2 : v3 = 1 : 0.0034 : 0.0006</p><p>v3 =</p><p>gMS</p><p>R</p><p>2</p><p>mean</p><p>=</p><p>6.67 * 10-11 * 1.97 * 1030</p><p>(149.50 * 106 * 103)2</p><p>= 5.9 * 10-3 m/s2</p><p>v2 = v2R = a 2p</p><p>T</p><p>b2</p><p>R = a 2 * 22</p><p>24 * 3600 * 7</p><p>b2</p><p>6.37 * 106 = 0.034 m/s2</p><p>v1 =</p><p>gM</p><p>R</p><p>2</p><p>=</p><p>6.67 * 10-11 * 5.96 * 1024</p><p>(6.37 * 106)2</p><p>= 9.8 m/s2</p><p>d</p><p>2p</p><p>v0 � v</p><p>=</p><p>2p</p><p>2gME >r</p><p>3>2 1</p><p>3¢r /2r + d</p><p>= b4.5 days (d = 0)</p><p>0.80 hour (d = 2)</p><p>+ve</p><p>-ve= v0 � v</p><p>v0 = A</p><p>gME</p><p>(r - ¢r)3</p><p>=</p><p>v = A</p><p>gME</p><p>r</p><p>3</p><p>=</p><p>U = -</p><p>3</p><p>5</p><p>gm2</p><p>R</p><p>1.4 UNIVERSAL GRAVITATION 133</p><p>From the statement of the problem, it is obvious that in this case</p><p>Thus,</p><p>In the other case, if be the sought height, then</p><p>From the language of the problem, in this case is not very small in comparison with</p><p>R. Therefore in this case we cannot use the approximation adopted in the previous case.</p><p>Here,</p><p>So,</p><p>As �ve sign is not acceptable</p><p>1.221 Let the mass of the body be m and let it go upto a height h.</p><p>From conservation of mechanical energy of the system</p><p>Using , in above equation and on solving we get</p><p>1.222 Gravitational pull provides the required centripetal acceleration to the satelite. Thus</p><p>if h be the sought distance, we have</p><p>or</p><p>Hence,</p><p>1.223 A satellite that hovers above the Earth’s equator and co-rotates with it moving from</p><p>west to east with the diurnal angular velocity of the Earth appears stationary to an ob-</p><p>server on the Earth. It is called geostationary. For this calculation we may neglect the</p><p>annual motion of the Earth as well as all other influences. Then, by Newton’s law,</p><p>gMm</p><p>r</p><p>2</p><p>= m a2p</p><p>T</p><p>b2</p><p>r</p><p>h =</p><p>gR</p><p>2 - Rv</p><p>2</p><p>v</p><p>2</p><p>= R c gR</p><p>v</p><p>2</p><p>- 1 d</p><p>Rv</p><p>2 + hv</p><p>2 = gR</p><p>2 as g =</p><p>gM</p><p>R</p><p>2</p><p>mv</p><p>2</p><p>(R + h)</p><p>=</p><p>gmM</p><p>(R + h)2</p><p>or (R + h) v</p><p>2 = gM</p><p>h =</p><p>Rv</p><p>2</p><p>0</p><p>2gR - v</p><p>2</p><p>0</p><p>gM /R</p><p>2 = g</p><p>-</p><p>gMm</p><p>R</p><p>+</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>0 =</p><p>-gMm</p><p>(R + h)</p><p>+ 0</p><p>h¿ = (12 - 1) R = (12 - 1) 6400 km = 2650 km</p><p>h¿</p><p>R</p><p>= � 12 - 1</p><p>a1 +</p><p>h¿</p><p>R</p><p>b2</p><p>= 2</p><p>h¿</p><p>g</p><p>2</p><p>= g a1 +</p><p>h¿</p><p>R</p><p>b -2 or 1</p><p>2</p><p>= a1 +</p><p>h¿</p><p>R</p><p>b -2</p><p>h¿</p><p>99</p><p>100</p><p>= a1 -</p><p>2h</p><p>R</p><p>b or h =</p><p>R</p><p>200</p><p>= a6400</p><p>200</p><p>b km = 32 km</p><p>h V R</p><p>134 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>where mass of the Earth, seconds period of daily rotation of the</p><p>Earth and distance of the satellite from the centre of the Earth.</p><p>Then,</p><p>Substitution of kg, gives .</p><p>Thus instantaneous velocity with respect to an inertial frame fixed to the centre of</p><p>the Earth at that moment will be</p><p>and the acceleration will be the centripetal acceleration</p><p>1.224 We know from the previous problem that a satellite moving west to east at distance</p><p>km from the centre of the Earth will be revolving round the Earth</p><p>with an angular velocity faster than the Earth’s diurnal angular velocity. Let</p><p>angular velocity of the satellite,</p><p>angular velocity of the Earth.</p><p>Then</p><p>is the relative angular velocity with respect to Earth. Now by Newton’s law</p><p>So,</p><p>Substitution gives</p><p>1.225 The velocity of the satellite in the inertial space fixed frame is east to west.</p><p>With respect to the Earth fixed frame, from the , the velocity is</p><p>v¿ =</p><p>2pR</p><p>T</p><p>+ A</p><p>gM</p><p>R</p><p>= 7.03 km/s</p><p>vœ = v - (� * r)</p><p>2gM /R</p><p>m = 6.27 * 1024 kg</p><p>=</p><p>4 p2R</p><p>3</p><p>gT</p><p>2</p><p>a1 +</p><p>T</p><p>t</p><p>b2</p><p>M =</p><p>R</p><p>3</p><p>g</p><p>a 2p</p><p>t</p><p>+</p><p>2p</p><p>T</p><p>b2</p><p>gM</p><p>R</p><p>2</p><p>= v2R</p><p>v - v0 = 2p /t</p><p>v0 = 2p /T =</p><p>v =</p><p>R = 2.00 * 104</p><p>a 2p</p><p>T</p><p>b2</p><p>r = 0.223 m/s</p><p>2</p><p>a 2p</p><p>T</p><p>b r = 3.07 km/s</p><p>r = 4.220 * 104 kmM = 5.96 * 1024</p><p>r = A3 gM a T</p><p>2p</p><p>b2</p><p>r =</p><p>=T = 86400M =</p><p>1.4 UNIVERSAL GRAVITATION 135</p><p>Here M is the mass of the Earth and T is its period of rotation about its own axis. If</p><p>the satellite were moving from west to east, velocity would be</p><p>To find the acceleration we note the formula</p><p>Here is directed towards the centre of the Earth.</p><p>Thus, (toward the Earth’s rotation axis)</p><p>1.226 From the well known relationship between the velocities of a particle with respect</p><p>to a space fixed frame (K) and rotating frame ,</p><p>we get,</p><p>Thus kinetic energy of the satellite in the Earth’s frame</p><p>(1)</p><p>Obviously when the satellite moves in opposite sense compared to the rotation of</p><p>the Earth, its velocity relative to the same frame would be</p><p>and kinetic energy</p><p>(2)</p><p>From Eqs. (1) and (2)</p><p>(3)T ¿ =</p><p>1v + 2pR /T 221v - 2pR /T 22</p><p>T 2¿ =</p><p>1</p><p>2</p><p>mv ¿22 =</p><p>1</p><p>2</p><p>m av +</p><p>2pR</p><p>T</p><p>b2</p><p>v2¿ = v + a2p</p><p>T</p><p>b R</p><p>T ¿1 =</p><p>1</p><p>2</p><p>mv ¿</p><p>2</p><p>1 =</p><p>1</p><p>2</p><p>m av -</p><p>2pR</p><p>T</p><p>b2</p><p>v ¿1 = v - a2p</p><p>T</p><p>b R</p><p>v = vœ + (� * r)</p><p>(K ¿)</p><p>=</p><p>gM</p><p>R</p><p>2</p><p>+</p><p>2p</p><p>T</p><p>c2pR</p><p>T</p><p>+ 2 A</p><p>gM</p><p>R</p><p>d = 4.94 m/s2 on substitution</p><p>w¿ =</p><p>gM</p><p>R</p><p>2</p><p>+ 2 a 2pR</p><p>T</p><p>+ A</p><p>gM</p><p>R</p><p>b 2p</p><p>T</p><p>- a2p</p><p>T</p><p>b2</p><p>R</p><p>F = gMm /R</p><p>3 R and v � � and v¿ * �</p><p>m w¿ = F + 2 m (v¿ * �) + mv2R</p><p>-</p><p>2pR</p><p>T</p><p>+ A</p><p>gm</p><p>R</p><p>136 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Now from Newton’s second law</p><p>(4)</p><p>Using Eqs. (3) and</p><p>(4)</p><p>1.227 For a satellite in a circular orbit about any massive body, the following relation holds</p><p>between kinetic, potential and total energy</p><p>(1)</p><p>Thus, since total mechanical energy must decrease due to resistance of the cosmic</p><p>dust, the kinetic energy will increase and the satellite will . We see then, by work</p><p>energy theorem</p><p>So,</p><p>Now from Newton’s law at an arbitrary radius r from the Moon’s centre,</p><p>(where M is the mass of the moon)</p><p>Then, (where Moon’s radius)</p><p>So,</p><p>or</p><p>(where g is Moon’s gravity)</p><p>The averaging implied by Eq. (1) (for non-circular orbits)makes the result approxi-</p><p>mate.</p><p>1.228 From Newton’s second law</p><p>(1)</p><p>gMm</p><p>R</p><p>2</p><p>=</p><p>mv</p><p>2</p><p>0</p><p>R</p><p>or v0 = A</p><p>gM</p><p>R</p><p>= 1.67 km/s</p><p>=</p><p>m</p><p>a1gR</p><p>(1h - 1)</p><p>t =</p><p>m</p><p>a</p><p>a 1</p><p>vi</p><p>-</p><p>1</p><p>vf</p><p>b =</p><p>m</p><p>a2gM /R</p><p>(1h - 1)</p><p>3</p><p>vf</p><p>vi</p><p>dv</p><p>v</p><p>2</p><p>=</p><p>am</p><p>m</p><p>3</p><p>t</p><p>0</p><p>dt =</p><p>at</p><p>m</p><p>R = vi = A</p><p>gM</p><p>hR</p><p>and vf = A</p><p>gM</p><p>R</p><p>v</p><p>2</p><p>r</p><p>=</p><p>gM</p><p>r</p><p>2</p><p>or v = A</p><p>gM</p><p>r</p><p>mvdv = av</p><p>2vdt or adt</p><p>m</p><p>=</p><p>dv</p><p>v</p><p>2</p><p>dT = -dE = -dAfr</p><p>T = -E, U = 2E</p><p>T2¿</p><p>T1¿</p><p>=</p><p>11gR + 2pR /T 2211gR - 2pR /T 22 = 1.27 nearly (using Appendices)</p><p>gMm</p><p>R</p><p>2</p><p>=</p><p>mv</p><p>2</p><p>R</p><p>or v = A</p><p>gM</p><p>R</p><p>= 1gR</p><p>1.4 UNIVERSAL GRAVITATION 137</p><p>‘ fall’’‘</p><p>From conservation of mechanical energy</p><p>(2)</p><p>In Eqs. (1) and (2), M and R are the mass of the Moon and its radius. In Eq. (1) if</p><p>M and R represent the mass of the Earth and its radius, then, using appendices, we</p><p>can easily get orbital and escape velocities for Earth as</p><p>1.229 In a parabolic orbit,</p><p>So,</p><p>where mass of the Moon, R � its radius and this is just the escape velocity. On</p><p>the other hand in orbit,</p><p>Thus,</p><p>1.230 From solution of problem 1.228 for the Earth surface</p><p>Thus the sought additional velocity</p><p>This ‘kick in velocity must be given along the direction of motion of the satellite</p><p>in its orbit.</p><p>1.231 Let r be the sought distance then</p><p>or</p><p>1.232 Between the Earth and the Moon, the potential energy (P.E.) of the spaceship will have</p><p>a maximum at the point where the attractions of the Earth and the moon balance each</p><p>other. This maximum P.E. is approximately zero. We can also neglect the contribution</p><p>1hr = (nR - r) or r =</p><p>nR</p><p>1h + 1</p><p>= 3.8 * 104 km</p><p>ghM</p><p>(nR - r)2</p><p>=</p><p>gM</p><p>r</p><p>2</p><p>or hr</p><p>2 = (nR - r)2</p><p>¢v = ve - v0 = A</p><p>gM</p><p>R</p><p>(12 - 1) = 1gR (12 - 1)</p><p>v0 = A</p><p>gM</p><p>R</p><p>and ve = A</p><p>2gM</p><p>R</p><p>¢v = (1 - 12) A</p><p>gM</p><p>R</p><p>= -0.70 km/s</p><p>mvf</p><p>2 R =</p><p>gMm</p><p>R</p><p>2</p><p>or vf = A</p><p>gM</p><p>R</p><p>M =</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>i -</p><p>gMm</p><p>R</p><p>= 0 or vi = 12 A</p><p>gM</p><p>R</p><p>E = 0</p><p>v0 = 7.9 km/s and ve = 11.2 km/s</p><p>1</p><p>2</p><p>mve</p><p>2 -</p><p>gMm</p><p>R</p><p>= 0 or ve = A</p><p>2gM</p><p>R</p><p>= 2.37 km/s</p><p>138 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>‘ ’’</p><p>of either body to the P.E. of the spaceship sufficiently near the other body. Then the</p><p>minimum energy that must be imparted to the spaceship to cross the maximum of the</p><p>P.E. is clearly (using E to denote the Earth)</p><p>(1)</p><p>With this energy the spaceship will cross over the hump in the P.E. and coast down</p><p>the hill of P.E. towards the moon and crash land on it. What the problem seeks is the</p><p>minimum energy required for soft-landing. That requires the use of rockets to lower-</p><p>ing and braking of the spaceship, and since the kinetic energy of the gases ejected</p><p>from the rocket will always be positive, the total energy required for soft-landing is</p><p>greater than that required for crash-landing. To calculate this energy we assume the</p><p>rockets are used fairly close to the moon when the spaceship has nearly attained its</p><p>terminal velocity on the moon where is the mass of the moon and</p><p>is its radius. In general and since the speed of the ejected gases is not less</p><p>than the speed of the rocket, and momentum transferred to the gases must equal the</p><p>momentum of the spaceship the energy E of the gas ejected is not less than the</p><p>kinetic energy of spaceship</p><p>(2)</p><p>Adding Eqs. (1) and (2) we get the minimum work done on the ejected gases to</p><p>bring about the soft landing.</p><p>On substitution we get kJ.</p><p>1.233 Assume first that the attraction of the Earth can be neglected. Then the minimum ve-</p><p>locity, that must be imparted the body to escape from the Sun’s pull, is, as in solu-</p><p>tio of problem 1.230, equal to</p><p>where radius of the Earth’s orbit, mass of the Sun.</p><p>In the actual case near the Earth, the pull of the Sun is small and does not change</p><p>much over distances, which are several times the radius of the Earth. The velocity</p><p>in question is that which overcomes the Earth’s pull with sufficient velocity to es-</p><p>cape the Sun’s pull.</p><p>Thus,</p><p>where radius of the Earth and mass of the Earth.</p><p>Writing , we get</p><p>.v3 = 22 v</p><p>2</p><p>2 + (12 - 1)2v</p><p>2</p><p>1 = 16.6 km/s</p><p>v1</p><p>2 = gME >R ME =R =</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>3 -</p><p>gME</p><p>R</p><p>=</p><p>1</p><p>2</p><p>m (12 - 1)2v</p><p>2</p><p>1</p><p>v3</p><p>MS =v</p><p>2</p><p>1 = gMS</p><p>>r, r =</p><p>(12 - 1) v1</p><p>1.3 * 108</p><p>Amin = gm aME</p><p>RE</p><p>+</p><p>M0</p><p>R0</p><p>b</p><p>g M0m</p><p>R0</p><p>dE = vdp</p><p>R0M022gM0/R0</p><p>gMEm</p><p>RE</p><p>1.4 UNIVERSAL GRAVITATION 139</p><p>140 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.5 Dynamics of a Solid Body</p><p>1.234 Since, motion of the rod is purely translational, net torque about the C.M. of the rod</p><p>should be equal to zero.</p><p>Thus, (1)</p><p>For the translational motion of rod,</p><p>(2)</p><p>From Eqs. (1) and (2)</p><p>1.235 Sought moment</p><p>and arm of the force</p><p>1.236 Relative to point O, the net moment of force is given by</p><p>(1)</p><p>Resultant of the external force</p><p>(2)</p><p>(as ), so the sought arm l of the force F is given by</p><p>.</p><p>1.237 For coplanar forces, about any point in the same plane,</p><p>Thus length of the arm, l =</p><p>Nnet</p><p>Fnet</p><p>or Nnet = r * Fnet</p><p>(where Fnet = ©Fi = resultant force)©ri * Fi = r * Fnet</p><p>l =</p><p>N</p><p>F</p><p>=</p><p>ab - AB</p><p>2A2 + B</p><p>2</p><p>N � FN # F = 0</p><p>F = F1 + F2 = Aj + B i</p><p>= ab k + AB (-k) = (ab - AB)k</p><p>N = r1 * F1 + r2 * F2 = (a i * A j) + (b j * B i)</p><p>l =</p><p>N</p><p>F</p><p>=</p><p>aB - Ab</p><p>2A2 + B</p><p>2</p><p>= aB k + Ab (-k) = (aB - Ab) k</p><p>N = r * F = (a i + b j) * (A i + B j)</p><p>a</p><p>l>2 =</p><p>mwC</p><p>F2</p><p>or l =</p><p>2 aF2</p><p>mwC</p><p>= 1 m</p><p>F2 - F1 = mwC or 1 -</p><p>F1</p><p>F2</p><p>=</p><p>mwC</p><p>F2</p><p>F1</p><p>l</p><p>2</p><p>= F2a l</p><p>2</p><p>- ab or F1</p><p>F2</p><p>= 1 -</p><p>a</p><p>l>2</p><p>1.5 DYNAMICS OF A SOLID BODY 141</p><p>Here obviously and it is directed toward right along AC. Take the origin</p><p>at C. Then about C,</p><p>, directed normally into the plane of figure</p><p>(Here side of the square.)</p><p>Thus, , directed into the plane of the figure.</p><p>Hence,</p><p>Thus the point of application of force is at the mid point of the side BC.</p><p>1.238 (a) Consider a strip of length dx at a perpendicular distance x from the axis about</p><p>which we have to find the moment of inertia of the rod. The elemental mass of</p><p>the rod equals</p><p>Moment of inertia of this element about the axis</p><p>Thus, moment of inertia of the rod, as a whole about the given axis</p><p>(b) Let us imagine the plane of plate as x–y plane taking the origin at the intersec-</p><p>tion point of the sides of the plate (see figure).</p><p>Obviously,</p><p>Similarly, Iy =</p><p>mb</p><p>2</p><p>3</p><p>=</p><p>ma2</p><p>3</p><p>= 3</p><p>a</p><p>0</p><p>a m</p><p>ab</p><p>b dyby</p><p>2</p><p>Ix = 3dm y</p><p>2</p><p>I = 3</p><p>l</p><p>0</p><p>m</p><p>l</p><p>x</p><p>2</p><p>dx =</p><p>ml</p><p>2</p><p>3</p><p>dI = dmx</p><p>2 =</p><p>m</p><p>l</p><p>dx # x</p><p>2</p><p>dm =</p><p>m</p><p>l</p><p>dx</p><p>l =</p><p>F (a>12)</p><p>2F</p><p>=</p><p>a</p><p>212</p><p>=</p><p>a</p><p>2</p><p>sin 45°</p><p>N = F</p><p>a</p><p>12</p><p>a =</p><p>N = a22 aF +</p><p>aF</p><p>22</p><p>F - 22 aFb</p><p>| Fnet | = 2F</p><p>O</p><p>x</p><p>dx</p><p>a</p><p>dx</p><p>y</p><p>x b</p><p>a</p><p>dx</p><p>y</p><p>b</p><p>O</p><p>O x</p><p>x</p><p>y</p><p>142 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Hence from perpendicular axis theorem</p><p>which is the sought moment of inertia.</p><p>1.239 (a) Consider an elementary disk of thickness dx. Moment of inertia of this element</p><p>about the z-axis, passing through its C.M. is</p><p>where density of the material of the plate and area of cross section of</p><p>the plate. Thus, the sought moment of inertia</p><p>Putting all the values we get .</p><p>(b) Consider an element disk of radius r and thickness dx at a distance x from the</p><p>point O. Then and volume of the disk</p><p>Hence, its mass</p><p>where density of the cone .</p><p>Moment of inertia of this element, about the axis OA,</p><p>Thus the sought moment of inertia</p><p>=</p><p>prR</p><p>4 # h</p><p>5</p><p>10h</p><p>4</p><p>aas tan a =</p><p>R</p><p>h</p><p>b</p><p>I =</p><p>pr</p><p>2</p><p>tan4</p><p>a3</p><p>h</p><p>0</p><p>x</p><p>4 dx</p><p>=</p><p>pr</p><p>2</p><p>x</p><p>4 tan4</p><p>a dx</p><p>= (px</p><p>2 tan2 a dx)</p><p>x</p><p>2 tan2 a</p><p>2</p><p>dI = dm</p><p>r</p><p>2</p><p>2</p><p>= m</p><p>1</p><p>3</p><p>pR</p><p>2hr =</p><p>dm = px</p><p>2 tan a dx # r</p><p>= px</p><p>2 tan2</p><p>a dx</p><p>r = x tan a</p><p>Iz = 2.8 g>m2</p><p>=</p><p>p</p><p>2</p><p>rbR</p><p>4 (as S = pR</p><p>2)</p><p>Iz =</p><p>rSR</p><p>2</p><p>2 3</p><p>b</p><p>0</p><p>dx =</p><p>R</p><p>2</p><p>2</p><p>rSb</p><p>S =r =</p><p>dIz =</p><p>(dm)R</p><p>2</p><p>2</p><p>= r Sdx</p><p>R</p><p>2</p><p>2</p><p>Iz = Ix + Iy =</p><p>m</p><p>3</p><p>(a2 + b</p><p>2)</p><p>b dx</p><p>x</p><p>RO</p><p>x</p><p>RA</p><p>dx</p><p>r</p><p>h</p><p>n</p><p>1.5 DYNAMICS OF A SOLID BODY 143</p><p>Hence,</p><p>1.240 (a) Let us consider a lamina of an arbitrary shape and indicate by 1,2 and 3, three</p><p>axes coinciding with x, y and z-axes and the plane of lamina as plane.</p><p>Now, moment of inertia of a point mass about x-axis,</p><p>Thus moment of inertia of the lamina about this axis,</p><p>Similarly,</p><p>and</p><p>Thus,</p><p>(b) Let us take the plane of the disk as plane and</p><p>origin to the centre of the disk (see figure). From the</p><p>symmetry . Let us consider a ring element of</p><p>radius r and thickness dr, then the moment of inertia</p><p>or the ring element about the y-axis is given by</p><p>Thus the moment of inertia of the disk about z-axis</p><p>But we have</p><p>Thus,</p><p>1.241 For simplicity let us use a mathematical trick. We consider the portion of the given disk</p><p>as the superposition of two complete disks (without holes), one of positive density and</p><p>radius R and other of negative density but of same magnitude and radius .R>2</p><p>Iz =</p><p>Iz</p><p>2</p><p>=</p><p>mR</p><p>2</p><p>4</p><p>Ix = Ix + Iy = 2Ix</p><p>Iz =</p><p>2m</p><p>R</p><p>2 3</p><p>R</p><p>0</p><p>r</p><p>3 dr =</p><p>mR</p><p>2</p><p>2</p><p>dIz = dmr</p><p>2 =</p><p>m</p><p>pR</p><p>2</p><p>(2prdr)r</p><p>2</p><p>Ix = Iy</p><p>x -y</p><p>Iz = Ix + Iy or I3 = I1 + I2</p><p>= 3dm (x</p><p>2 + y</p><p>2) as r = 2x</p><p>2 + y</p><p>2</p><p>Iz = 3dmr</p><p>2</p><p>Iy = 3dmx</p><p>2</p><p>Ix = 3dmy</p><p>2</p><p>dIx = dm y</p><p>2</p><p>x -y</p><p>I =</p><p>3mR 2</p><p>10</p><p>aputting r =</p><p>3m</p><p>pR</p><p>2h</p><p>b</p><p>z</p><p>O</p><p>y</p><p>dr</p><p>x</p><p>O</p><p>r (x,y)</p><p>y z(3)</p><p>y(2)</p><p>x(1)</p><p>144 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>As (area) (mass), the respective masses of the consid-</p><p>ered disks are and respectively, and</p><p>these masses can be imagined to be situated at their re-</p><p>spective centres (C.M.). Let us take point O as origin</p><p>and point x-axis towards right. Obviously the C.M. of</p><p>the shaded position of given shape lies on the x-axis.</p><p>Hence the C.M. (C) of the shaded portion is given by</p><p>Thus C.M. of the shape is at a distance R/6 from point O toward x-axis.</p><p>Using parallel axis theorem and bearing in mind that the moment of inertia of a</p><p>complete homogeneous disk of radius m0 and radius r0 equals the moment</p><p>of inertia of the small disk of mass and radius about the axis passing</p><p>through point C and perpendicular to the plane of the disk is given by</p><p>Similarly,</p><p>Thus, the sought moment of inertia,</p><p>1.242 Moment of inertia of the shaded portion, about the axis pass-</p><p>ing through its centre,</p><p>=</p><p>2</p><p>5</p><p># 4</p><p>3</p><p>pr(R</p><p>5 - r</p><p>5)</p><p>I =</p><p>2</p><p>5</p><p>a4</p><p>3</p><p>p R</p><p>3rbR</p><p>2 -</p><p>2</p><p>5</p><p>a4</p><p>3</p><p>pr</p><p>3rbr</p><p>2</p><p>= 0.15 kg # m2</p><p>IC = I1C + I2C =</p><p>37</p><p>72</p><p>mR</p><p>2</p><p>=</p><p>2</p><p>3</p><p>mR</p><p>2 +</p><p>mR</p><p>2</p><p>27</p><p>I1C =</p><p>1</p><p>2</p><p>a4m</p><p>3</p><p>bR</p><p>2 + a4m</p><p>3</p><p>b aR</p><p>6</p><p>b2</p><p>= -</p><p>mR</p><p>2</p><p>24</p><p>-</p><p>4</p><p>27</p><p>mR</p><p>2</p><p>I2C =</p><p>1</p><p>2</p><p>a -</p><p>m</p><p>3</p><p>b aR</p><p>2</p><p>b2</p><p>+ a -</p><p>m</p><p>3</p><p>b aR</p><p>2</p><p>+</p><p>R</p><p>6</p><p>b2</p><p>R>2(-m>3)</p><p>1>2m0r 0</p><p>2;</p><p>xC =</p><p>(-m>3)(-R>2) + (4m>3)0a -</p><p>m</p><p>3</p><p>b +</p><p>4m</p><p>3</p><p>=</p><p>R</p><p>6</p><p>(-m/3),(4m</p><p>r</p><p>1</p><p>2 O x</p><p>C</p><p>r</p><p>R</p><p>/3)</p><p>1.5 DYNAMICS OF A SOLID BODY 145</p><p>Now, if the shaded portion becomes a shell, which is the required shape</p><p>to calculate the moment of inertia.</p><p>Now,</p><p>Neglecting higher terms, we get</p><p>1.243 (a) Net force which is effective on the system (cylinder is the weight</p><p>of the body m in a uniform gravitational field, which is a constant. Thus the ini-</p><p>tial acceleration of the body m is also constant.</p><p>From the conservation of mechanical energy of the said system in the uniform</p><p>field of gravity at time</p><p>or</p><p>or at all times) (1)</p><p>But</p><p>Hence using it in Eq. (1), we get</p><p>From the kinematic relationship,</p><p>Thus the sought angular velocity of the cylinder</p><p>(b) Sought kinetic energy</p><p>(on substituting value of )v =</p><p>1</p><p>2</p><p>mg</p><p>2a1 +</p><p>M</p><p>2m</p><p>b</p><p>=</p><p>1</p><p>4</p><p>(2m + M ) R</p><p>2v2</p><p>T (t) =</p><p>1</p><p>2</p><p>mv</p><p>2 +</p><p>1</p><p>2</p><p>ml</p><p>2</p><p>2</p><p>v2</p><p>v(t) = bt =</p><p>2mg</p><p>(2m + M ) R</p><p>t =</p><p>gt</p><p>(1 + M>2m) R</p><p>b =</p><p>w</p><p>R</p><p>=</p><p>2mg</p><p>(2m + M) R</p><p>1</p><p>4</p><p>(2m + M) 2w ¢h - mg ¢h = 0 or w =</p><p>2mg</p><p>(2m + M)</p><p>v</p><p>2 = 2w¢h</p><p>1</p><p>4</p><p>(2m + M)v</p><p>2 - mg ¢h = 0 (as v = vR</p><p>1</p><p>2</p><p>mv</p><p>2 +</p><p>1</p><p>2</p><p>MR</p><p>2</p><p>2</p><p>v2 - mg ¢h = 0</p><p>t = ¢t : ¢T + ¢Ugr 0</p><p>M + body m)</p><p>=</p><p>2</p><p>3</p><p>(4pr</p><p>2 drr) r</p><p>2 =</p><p>2</p><p>3</p><p>mr</p><p>2</p><p>=</p><p>2</p><p>5</p><p># 4</p><p>3</p><p>pr (r</p><p>5 + 5r</p><p>4 dr + ... - r</p><p>5)</p><p>I =</p><p>2</p><p>5</p><p># 4</p><p>3</p><p>pr {(r + dr)5 - r</p><p>5}</p><p>R = r + dr,</p><p>=</p><p>146 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.244 For equilibrium of the disk and axle</p><p>As the disk unwinds, it has an angular acceleration</p><p>given by</p><p>The corresponding linear acceleration is</p><p>Since the disk remains stationary under the combined action of this acceleration and</p><p>the acceleration of the bar which is transmitted to the axle, we must have</p><p>1.245 Let the rod be deviated through an angle from its initial position at an arbitrary in-</p><p>stant of time, measured relative to the initial position in the positive direction. From</p><p>the equation of the increment of the mechanical energy of the system</p><p>or</p><p>or</p><p>Thus,</p><p>1.246 First of all, let us sketch a free diagram of each body. Since the cylinder is rotating</p><p>and massive, the tension will be different in both the sections of threads. From</p><p>Newton’s law in projection form for the bodies and and noting that</p><p>(as the thread is not slipping), we have</p><p>and (1)</p><p>Now from the equation of rotational dynamics of a solid about sta-</p><p>tionary axis of rotation, i.e.,</p><p>(for the cylinder)</p><p>or (2) (T1 - T2)R = Ib =</p><p>mR</p><p>2b</p><p>2</p><p>Nz = Ibz</p><p>T2 - m2g = m2w</p><p>m1g - T1 = m1w = m1bR</p><p>(m1 7 m2):w1 = w2 = w = bR</p><p>m2m1</p><p>v = A</p><p>6F sin w</p><p>Ml</p><p>1</p><p>2</p><p>Ml</p><p>2</p><p>3</p><p>v2 = 3</p><p>w</p><p>0</p><p>Fl cos w dw = Fl sin w</p><p>1</p><p>2</p><p>I v2 = 3Nz</p><p>d w</p><p>¢T = Aext</p><p>w</p><p>w = mgr</p><p>2/I</p><p>(–w)</p><p>rb = w =</p><p>mgr</p><p>2</p><p>I</p><p>2Tr = Ib or b =</p><p>2Tr</p><p>I</p><p>=</p><p>mgr</p><p>I</p><p>b</p><p>2T = mg or T = mg>2</p><p>mg</p><p>TT</p><p>m2m1</p><p>m g2m g1</p><p>w</p><p>T2T1</p><p>w</p><p>.</p><p>1.5 DYNAMICS OF A SOLID BODY 147</p><p>Simultaneous solution of the above equations yields:</p><p>1.247 As the system is under constant forces, the acceleration of body</p><p>and is constant. In addition to it the velocities and acceleration of bodies and</p><p>are equal in magnitude (say v and w) because the length of the thread is constant.</p><p>From the equation of increment of mechanical energy, i.e., at time t</p><p>when block is distance h below from initial position corresponding to</p><p>(1)</p><p>(as angular velocity for no slipping of thread).</p><p>But</p><p>So using it in Eq. (1), we get</p><p>(2)</p><p>Thus the work done by the friction force on</p><p>(using Eq. 2)</p><p>1.248 In the problem, the rigid body is in translation equilibrium but there is an angular</p><p>retardation. We first sketch the free body diagram of the cylinder. Obviously the fric-</p><p>tion forces, acting on the cylinder, are kinetic. From the condition of translational</p><p>equilibrium for the cylinder,</p><p>Hence,</p><p>For pure rotation of the cylinder about its rotation</p><p>axis,</p><p>or -kN1R - kN2R =</p><p>mR</p><p>2</p><p>2</p><p>bz</p><p>Nz = Ibz</p><p>N1 =</p><p>mg</p><p>1 + k</p><p>2</p><p>; N2 = k</p><p>mg</p><p>1 + k</p><p>2</p><p>mg = N1 + kN2; N2 = kN1</p><p>= -</p><p>km1(m2 - km1)g</p><p>2t</p><p>2</p><p>m + 2(m1 + m2)</p><p>Afr = -km1gh = -km1g a1</p><p>2</p><p>wt</p><p>2b</p><p>m1</p><p>w =</p><p>2(m2 - km1)g</p><p>m + 2(m1 + m2)</p><p>v</p><p>2 = 2wh</p><p>v = V/R</p><p>1</p><p>2</p><p>(m1 + m2)v</p><p>2 +</p><p>1</p><p>2</p><p>amR</p><p>2</p><p>2</p><p>b v</p><p>2</p><p>R</p><p>2</p><p>- m2gh = -km1gh</p><p>t = 0,m1</p><p>¢T + ¢U = Afr</p><p>,</p><p>m2</p><p>m1m2</p><p>m1(m + m1 + m2)</p><p>b =</p><p>(m1 - m2)g</p><p>R am1 + m2 + m/2b and T1</p><p>T2</p><p>=</p><p>m1(m + 4m2)</p><p>m2(m + 4m1)</p><p>mg</p><p>C</p><p>N1</p><p>kN1</p><p>N2kN 2</p><p>148 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>or</p><p>or</p><p>Now, from the kinematical equation,</p><p>we have, (because )</p><p>Hence, the sought number of turns,</p><p>1.249 It is the moment of friction force which brings the disk to rest. The force of friction</p><p>is applied to each section of the disk, and since these sections lie at different distances</p><p>from the axis, the moments of the forces of friction differ from section to section.</p><p>To find where z is the axis of rotation of the disk let us</p><p>partition the disk into thin rings (see figure). The force of</p><p>friction acting on the considered element</p><p>(where is the density of the disk).</p><p>The moment of this force of friction is</p><p>Integrating with respect to r from zero to R, we get</p><p>For the rotation of the disk about the stationary axis z, from the equation</p><p>Thus, from the angular kinematical equation</p><p>1.250 According to the question,</p><p>I</p><p>dv</p><p>dt</p><p>= -k1v or I</p><p>dv</p><p>1v = -kdt</p><p>0 = v0 + a -</p><p>4kg</p><p>3R</p><p>b t or t =</p><p>3R v0</p><p>4kg</p><p>vz = v0z + bz t</p><p>-</p><p>2</p><p>3</p><p>pk sgR</p><p>3 =</p><p>(pR</p><p>2s)R</p><p>2</p><p>2</p><p>bz or bz = -</p><p>4kg</p><p>3R</p><p>Nz = Ibz</p><p>Nz = -2p k sg3</p><p>R</p><p>0</p><p>r</p><p>2dr = -</p><p>2</p><p>3</p><p>pk sgR</p><p>3</p><p>dNz = - rdfr = -2pk</p><p>sgr</p><p>2dr</p><p>s</p><p>dfr = k (2p</p><p>rdrs)g,</p><p>Nz,</p><p>n =</p><p>¢w</p><p>2p</p><p>=</p><p>v0</p><p>2 (1 + k</p><p>2</p><p>)R</p><p>8pk (1 + k)g</p><p>v = 0 ¢w =</p><p>v2</p><p>0</p><p>(1 + k</p><p>2)R</p><p>4k (1 + k)g</p><p>v2 = v0</p><p>2 + 2bz</p><p>¢w</p><p>bz = -</p><p>2k (1 + k)g</p><p>(1 + k</p><p>2)R</p><p>-</p><p>kmgR (1 + k)</p><p>1 + k</p><p>2</p><p>=</p><p>mR</p><p>2</p><p>2</p><p>bz</p><p>z</p><p>O</p><p>dr</p><p>r</p><p>R</p><p>1.5 DYNAMICS OF A SOLID BODY 149</p><p>Integrating,</p><p>or</p><p>Let the flywheel stop at then from Eq. (1),</p><p>Hence sought average angular velocity</p><p>1.251 Let us use the equation relative to the axis through O. (1)</p><p>For this purpose, let us find the angular momentum of the system Mz about the given</p><p>rotation axis and the corresponding torque Nz . The angular momentum is</p><p>So, (2)</p><p>The downward pull of gravity on the overhanging part is the only external force,</p><p>which exerts a torque about the z-axis, passing through O and is given by,</p><p>Hence, from the equation</p><p>aMR</p><p>2</p><p>2</p><p>+ mR</p><p>2bbz =</p><p>m</p><p>l</p><p>x gR</p><p>dMz</p><p>dt</p><p>= Nz</p><p>Nz = am</p><p>l</p><p>bxgR</p><p>dMz</p><p>dt</p><p>= aMR</p><p>2</p><p>2</p><p>+ mR</p><p>2bbz</p><p>where I =</p><p>m0</p><p>2</p><p>R</p><p>2 and v = vR (no cord slipping).</p><p>Mz = I v + mvR = am0</p><p>2</p><p>+ mbR</p><p>2 v</p><p>dMz /dt = Nz,</p><p><v> =</p><p>3</p><p>2I2v0>k</p><p>0</p><p>1k2t2/4I 2 - 2v0 kt/I + v02dt</p><p>3</p><p>2I2v0>k</p><p>0</p><p>dt</p><p>=</p><p>v0</p><p>3</p><p>t0 = 2I2v0 /kt = t0</p><p>v =</p><p>k</p><p>2t</p><p>2</p><p>4I</p><p>2</p><p>-</p><p>1v0 kt</p><p>I</p><p>+ v0 (noting that at t = 0, v = v0</p><p>)</p><p>1v = -</p><p>kt</p><p>2I</p><p>+ 1v0</p><p>,</p><p>150 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Thus,</p><p>Note: We may solve this problem using conservation of mechanical energy of the sys-</p><p>tem (cylinder 	 thread) in the uniform field of gravity.</p><p>1.252 (a) Let us indicate the forces acting on the sphere and their points of application.</p><p>Choose positive direction of x and (rotation angle) along the incline in down-</p><p>ward direction and in the sense of (for unidirectional rotation) respectively.</p><p>Now from equations of dynamics of rigid body, i.e.,</p><p>we get</p><p>(1)</p><p>and (2)</p><p>But, (3)</p><p>In addition, the absence of slipping provides the kinematical relationship between</p><p>the accelerations as.</p><p>(4)</p><p>The simultaneous solution of all the four equations yields</p><p>.</p><p>(b) Solving Eqs. (1) and (2) [of part (a)], we get</p><p>As the sphere starts at along positive x-axis, for pure rolling</p><p>(5)</p><p>Hence, the sought kinetic energy is</p><p>1.253 (a) Let us indicate the forces and their points of application for the cylinder.</p><p>Choosing the positive direction for x and as shown in the figure, we write the</p><p>equation of motion of the cylinder axis and the equation of moments in the C.M.</p><p>frame relative to that axis, i.e., from equation</p><p>mg - 2T = mwC</p><p>2RT =</p><p>mR</p><p>2</p><p>2</p><p>b</p><p>Fx = mwC and Nz = IC bz</p><p>w</p><p>=</p><p>7</p><p>10</p><p>ma5</p><p>7</p><p>g sin atb2</p><p>=</p><p>5</p><p>14</p><p>mg</p><p>2 sin2at</p><p>2</p><p>T =</p><p>1</p><p>2</p><p>mv</p><p>2</p><p>C</p><p>+</p><p>1</p><p>2</p><p>2</p><p>5</p><p>mR</p><p>2v2 =</p><p>7</p><p>10</p><p>mv C</p><p>2 (as v = vC</p><p>>R)</p><p>vC</p><p>(t) = wC</p><p>t =</p><p>5</p><p>7</p><p>g sin at</p><p>t = 0</p><p>wC =</p><p>5</p><p>7</p><p>g sin a</p><p>k cos a Ú</p><p>2</p><p>7</p><p>sin a or k Ú</p><p>2</p><p>7</p><p>tan a</p><p>w = bR</p><p>fr … kmg cos a</p><p>frR =</p><p>2</p><p>5</p><p>mR</p><p>2b</p><p>mg sin a - fr = mw</p><p>Fx = mwC x and NC z = IC bz</p><p>v</p><p>w</p><p>bz =</p><p>2mgx</p><p>lR (M + 2m)</p><p>7 0</p><p>N fr</p><p>mgx</p><p>j</p><p>a</p><p>we get</p><p>and</p><p>1.5 DYNAMICS OF A SOLID BODY 151</p><p>As there is no slipping of thread on the cylinder</p><p>From these three equations</p><p>(b) We have</p><p>So,</p><p>1.254 Let us depict the forces and their points of application corresponding to the cylinder</p><p>attached with the elevator. Newton’s second law for solid in vector form in the frame</p><p>of elevator, gives</p><p>(1)</p><p>The equation of moment in the C.M. frame relative to the cylinder axis, i.e., from</p><p>, is</p><p>(as thread does not slip on the cylinder,</p><p>or</p><p>From Eq. (1), , and so in vector form,</p><p>(2)</p><p>Solving Eqs. (1) and (2), we get</p><p>and sought force</p><p>1.255 Let us depict the forces and their points of application for the spool. Choosing the</p><p>positive direction for x and as shown in the figure, we apply and</p><p>and get .mg sin a - T = mw, Tr = IbNC z = IC</p><p>bz</p><p>Fx = mwxw</p><p>F = 2T =</p><p>1</p><p>3</p><p>m (g - w0)</p><p>w =</p><p>2</p><p>3</p><p>(g - w0)</p><p>T = -</p><p>m w</p><p>4</p><p>Tc Tw</p><p>T =</p><p>mw¿</p><p>4</p><p>w¿ = bR )</p><p>2TR =</p><p>mR</p><p>2</p><p>2</p><p>b =</p><p>mR</p><p>2</p><p>2</p><p>w ¿</p><p>R</p><p>Nz = IC</p><p>bz</p><p>2T + m g + m (-w0) = m wœ</p><p>= m g # a2</p><p>3</p><p>gtb =</p><p>2</p><p>3</p><p>mg</p><p>2t</p><p>P = F # v = F # (wC</p><p>t)</p><p>wC =</p><p>2</p><p>3</p><p>g 7 0 or in vector from wC =</p><p>2</p><p>3</p><p>g</p><p>b =</p><p>2</p><p>3</p><p>g</p><p>R</p><p>= 5 * 102 rad /s2</p><p>T =</p><p>mg</p><p>6</p><p>= 13 N, b =</p><p>2</p><p>3</p><p>g</p><p>R</p><p>wC = bR</p><p>mg</p><p>T</p><p>T x</p><p>j</p><p>T</p><p>mg</p><p>w0</p><p>mw0</p><p>T</p><p>j</p><p>152 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>“Notice that if a point of a solid in plane motion is</p><p>connected with a thread, the projection of velocity</p><p>vector of the solid’s point of contact along the length</p><p>of the thread equals the velocity of the other end of</p><p>the thread (if it is not slacked)”.</p><p>Thus in our problem, but hence point</p><p>P is the instantaneous centre of rotation</p><p>of zero velocity for the spool. Therefore and</p><p>subsequently</p><p>Solving the equations simultaneously, we get</p><p>1.256 Let us sketch the force diagram for solid cylinder and apply Newton’s second law in</p><p>projection form along x- and y-axes (see figure). Then,</p><p>(1)</p><p>and</p><p>or (2)</p><p>Now choosing positive direction of as shown in</p><p>the figure and using we get</p><p>(3)</p><p>(as for pure rolling</p><p>In addition,</p><p>(4)</p><p>Solving the equations, we get</p><p>or</p><p>and</p><p>=</p><p>k</p><p>m</p><p>[mg + Fmax] =</p><p>k</p><p>m</p><p>cmg +</p><p>3kmg</p><p>2 - 3k</p><p>d =</p><p>2kg</p><p>2 - 3k</p><p>wC (max) =</p><p>k (N1 + N2)</p><p>m</p><p>Fmax =</p><p>3k mg</p><p>2 - 3k</p><p>F …</p><p>3k mg</p><p>2 - 3k</p><p>fr1 + fr2 … k (N1 + N2)</p><p>wC = bR ).</p><p>=</p><p>mR</p><p>2</p><p>2</p><p>b =</p><p>mR</p><p>2</p><p>2</p><p>wC</p><p>R</p><p>FR - ( fr1 + fr2) R</p><p>NCz = IC</p><p>bz,</p><p>w</p><p>N1 + N2 = mg + F</p><p>N1 + N2 - mg - F = 0</p><p>fr1 + fr2 = mwC</p><p>w =</p><p>g sin a</p><p>1 + I/mr</p><p>2</p><p>= 1.6 m /s2</p><p>wC = br.</p><p>vC = vr</p><p>v0 = 0,vP = v0</p><p>O</p><p>T</p><p>r</p><p>P</p><p>w</p><p>mg</p><p>N</p><p>φ</p><p>fr1</p><p>fr2</p><p>N2</p><p>N1</p><p>mg F</p><p>j</p><p>1.5 DYNAMICS OF A SOLID BODY 153</p><p>1.257 (a) Let us choose the positive direction of the rotation angle , such that and</p><p>have identical signs (see figure). Equation of motion, and</p><p>gives</p><p>In the absence of the slipping of the</p><p>spool, .</p><p>From the three equations we get</p><p>(where cos ).</p><p>(b) As static friction ( fr) does not work on</p><p>the spool, from the equation of the incre-</p><p>ment of mechanical energy</p><p>Note: At there is no rolling and for i.e. the sp-</p><p>ool will move towards negative x-axis and rotate in anticlockwise sense.</p><p>1.258 For the cylinder, from equation about its stationary axis of rotation,</p><p>(1)</p><p>For the rotation of the lower cylinder, from equation ,</p><p>or</p><p>Now for the translational motion of lower cylinder, from</p><p>equation</p><p>(2)mg - 2T = mwC</p><p>Fx = mwC x</p><p>,</p><p>b ¿ =</p><p>4T</p><p>mr</p><p>= b</p><p>2Tr =</p><p>mr</p><p>2</p><p>2</p><p>b ¿</p><p>NCz = IC bz</p><p>2Tr =</p><p>mr</p><p>2</p><p>2</p><p>b or b =</p><p>4T</p><p>mr</p><p>Nz = I z</p><p>cos a 6 r>R, wcx 6 0,cos a = r>R,</p><p>=</p><p>F</p><p>2acos a - r/Rb2</p><p>t</p><p>2</p><p>2m (1 + g)</p><p>=</p><p>1</p><p>2</p><p>m (1 + g)2wCx =</p><p>1</p><p>2</p><p>m (1 + g) 2wC a12 wCt</p><p>2b</p><p>Aext =</p><p>1</p><p>2</p><p>mv C</p><p>2 +</p><p>1</p><p>2</p><p>g mR</p><p>2</p><p>v C</p><p>2</p><p>R</p><p>2</p><p>=</p><p>1</p><p>2</p><p>m (1 + g)v C</p><p>2</p><p>Aext = ¢T.</p><p>a 7 r/R</p><p>wCx = wC =</p><p>F [cos a - (r>R)]</p><p>m (1 + g)</p><p>wC x = bz</p><p>R</p><p>F cos a - fr = mwCx and frR - Fr = IC bz = gmR</p><p>2bz</p><p>NCz = IC</p><p>bzFx = mwCx</p><p>bzwCxw</p><p>mg</p><p>N</p><p>r</p><p>C</p><p>R</p><p>x</p><p>fr</p><p>F</p><p>a</p><p>T</p><p>T</p><p>mg</p><p>j</p><p>j'</p><p>,</p><p>154 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>As there is no slipping of threads on the cylinders,</p><p>(3)</p><p>Simultaneous solution of Eqs. (1), (2) and (3) yields</p><p>1.259 Let us depict the forces acting on the pulley and weight A, and indicate positive</p><p>direction for x and as shown in the figure. For the cylinder from the equation,</p><p>we get</p><p>(1)</p><p>and (2)</p><p>For the weight A from the equation</p><p>(3)</p><p>As there is no slipping of the threads on the pulleys,</p><p>(4)</p><p>Simultaneous solution of above four equations gives</p><p>1.260 (a) For the translational motion of the system</p><p>(m1 	 m2), from the equation: Fx�mwCx , we get</p><p>(1)</p><p>Now for the rotational motion of cylinder from</p><p>the equation: , we get</p><p>(2)</p><p>But,</p><p>So, (3)</p><p>(b) From the equation of increment of mechanical energy:</p><p>Here, ¢T = T (t) so, T (t) = Aext</p><p>¢T = Aext</p><p>wK =</p><p>F</p><p>m1 + m2</p><p>+</p><p>2F</p><p>m1</p><p>=</p><p>F (3m1 + 2m2)</p><p>m1(m1 + m2)</p><p>wK = wC + br</p><p>Fr =</p><p>m1r</p><p>2</p><p>2</p><p>b or br =</p><p>2F</p><p>m1</p><p>NCx = IC bz</p><p>F = (m1 + m2) wC or wC = F>(m1 + m2)</p><p>wA =</p><p>3 (M + 3m) g</p><p>M + 9m + 1>R 2</p><p>wA = wC + 2 bR = wC + 2wC = 3wC</p><p>mg - TA = mwA</p><p>Fx = mwx</p><p>2TR + TA(2R) = I b =</p><p>IwC</p><p>R</p><p>Mg + TA - 2T = MwC</p><p>Fx = m wx and NC z = IC</p><p>bz,</p><p>w</p><p>T =</p><p>mg</p><p>10</p><p>wC = b ¿r + br = 2br</p><p>T</p><p>T</p><p>Mg</p><p>mg</p><p>TA</p><p>TA</p><p>R</p><p>2R</p><p>B</p><p>A</p><p>K</p><p>F</p><p>x</p><p>We get,</p><p>1.5 DYNAMICS OF A SOLID BODY 155</p><p>As force F is constant and is directed along x-axis the sought work done is given by</p><p>(where x is the displacement of the point of application of the force F</p><p>during time</p><p>interval t) or</p><p>(using Eq. 3)</p><p>Alternate:</p><p>1.261 Choosing the positive direction for x and as shown in</p><p>figure, let us we write the equation of motion for the sphere</p><p>and</p><p>(where w2 is the acceleration of the C.M. of sphere).</p><p>For the plank, from equation , we have</p><p>In addition, the condition for the absence of slipping of the sphere yields the kine-</p><p>matic relation between the accelerations</p><p>Simultaneous solution of the four equations yields</p><p>1.262 (a) Let us depict the forces acting on the cylinder and their point of applications for</p><p>the cylinder and indicate positive direction of x and as shown in the figure.</p><p>From the equations for the plane motion of a solid ,</p><p>(1)</p><p>(2)</p><p>Let the cylinder start pure rolling at after releasing</p><p>on the horizontal floor at .t = 0</p><p>t = t0</p><p>-kmg R =</p><p>mR</p><p>2</p><p>2</p><p>bz or bz = -2</p><p>kg</p><p>R</p><p>kmg = mwC x or wC x = kg</p><p>Fx = mwC x and NC z = IC bz</p><p>w</p><p>w1 =</p><p>F</p><p>m1 + 2>7m2</p><p>and w2 =</p><p>2</p><p>7</p><p>w1</p><p>w1 = w2 + br</p><p>F - fr = m1w1</p><p>Fx = mwx</p><p>(fr) r =</p><p>2</p><p>5</p><p>m2r</p><p>2b</p><p>fr = m2w2;</p><p>NC z = IC bzFx = mwC x</p><p>w</p><p>=</p><p>1</p><p>2</p><p>(m1 + m2)a Ft</p><p>(m1 + m2)</p><p>b2</p><p>+</p><p>1</p><p>2</p><p>m1r</p><p>2</p><p>2</p><p>a 2Ft</p><p>m1r</p><p>b2</p><p>=</p><p>F</p><p>2t</p><p>2(3m1 + 2m2)</p><p>2m1 (m1 + m2)</p><p>T (t) = Ttranslation (t) + Trotation (t)</p><p>F a1</p><p>2</p><p>wKt</p><p>2b =</p><p>F</p><p>2t</p><p>2(3m1 + 2m2)</p><p>2m1(m1 + m2)</p><p>= T (t)</p><p>Aext = Fx</p><p>fr</p><p>Ffr</p><p>m2</p><p>C x</p><p>m1</p><p>frR</p><p>mgC</p><p>x</p><p>0</p><p>156 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>From the angular kinematical equation,</p><p>or (3)</p><p>From the equation of the linear kinematics,</p><p>or (4)</p><p>But at the moment when pure rolling starts .</p><p>So,</p><p>Thus,</p><p>(b) As the cylinder picks up speed till it starts rolling, the point of contact has purely</p><p>translatory movement equal to in the forward direction but there is also a</p><p>backward movement of the point of contact of magnitude</p><p>Because of slipping the net displacement is backwards. The total work done is</p><p>then,</p><p>The same result can also be obtained by the work-energy theorem,</p><p>1.263 Let us write the equation of motion for the centre of the sphere at the moment of</p><p>breaking-off</p><p>where v is the velocity of the centre of the sphere at that mo-</p><p>ment, and is the corresponding angle (see figure). The veloc-</p><p>ity v can be found from the energy conservation law</p><p>mgh =</p><p>1</p><p>2</p><p>mv</p><p>2 +</p><p>1</p><p>2</p><p>I v2</p><p>u</p><p>mv</p><p>2</p><p>R + r</p><p>= mg cos u</p><p>Afr = ¢T.</p><p>= kmg</p><p>v0</p><p>R</p><p>3kg</p><p>cv0R</p><p>6</p><p>+</p><p>v0R</p><p>3</p><p>- v0R d = -</p><p>mv0</p><p>2R</p><p>2</p><p>6</p><p>= kmg c1</p><p>2</p><p>kg t 0</p><p>2 -</p><p>1</p><p>2</p><p>a -</p><p>2kg</p><p>R</p><p>b t 0</p><p>2 R - v0t0R d</p><p>Afr = kmg c1</p><p>2</p><p>wC</p><p>t 0</p><p>2 - (v0t0 +</p><p>1</p><p>2</p><p>bt 0</p><p>2 )R d</p><p>(v0t0 - 1>2 bt 0</p><p>2) R.</p><p>1>2 wCt 0</p><p>2</p><p>t0 =</p><p>v0R</p><p>3 kg</p><p>kgt0 = av0 - 2</p><p>kg</p><p>R</p><p>t0bR</p><p>vC = vRt = t0,</p><p>vC = 0 + kgt 0</p><p>vC x = voC x + wC xt</p><p>v = v0 - 2</p><p>kg</p><p>R</p><p>t</p><p>vz = voz + bz</p><p>t,</p><p>h</p><p>mg</p><p>R</p><p>Cr</p><p>O</p><p>1.5 DYNAMICS OF A SOLID BODY 157</p><p>where I is the moment of inertia of the sphere relative to the axis passing through</p><p>the sphere’s centre. i.e. In addition,</p><p>From these four equations we obtain</p><p>1.264 Since the cylinder moves without sliding, the centre of the cylinder rotates about the</p><p>point O, while passing through the common edge of the planes. In other words, the</p><p>point O becomes the foot of the instantaneous axis of rotation of the cylinder.</p><p>At any instant during this motion the velocity of the C.M. is</p><p>When the angle (as shown in the figure) is , we have</p><p>(where N is the normal reaction of the edge)</p><p>or (1)</p><p>From the energy conservation law,</p><p>But,</p><p>(from the parallel axis theorem).</p><p>Thus,</p><p>From Eqs. (1) and (2)</p><p>The angle in this equation is clearly smaller than or equal to , so putting , we</p><p>get</p><p>(where N0 is the corresponding reaction.)</p><p>Note that No jumping occurs during this turning if Hence, must</p><p>be less than</p><p>� 1.0 m/s (on substituting values)</p><p>vmax = A</p><p>gR</p><p>3</p><p>(7 cos a - 4)</p><p>v0N0 7 0.N Ú N0.</p><p>v 0</p><p>2 =</p><p>gR</p><p>3</p><p>(7 cos a - 4) -</p><p>N0R</p><p>M</p><p>b = aab</p><p>v 0</p><p>2 =</p><p>gR</p><p>3</p><p>(7 cos b - 4) -</p><p>NR</p><p>m</p><p>v 1</p><p>2 = v 0</p><p>2 +</p><p>4</p><p>3</p><p>g R (1 - cos b)</p><p>I0 =</p><p>mR</p><p>2</p><p>2</p><p>+ mR</p><p>2 =</p><p>3</p><p>2</p><p>mR</p><p>2</p><p>1</p><p>2</p><p>I0</p><p>v 1</p><p>2</p><p>R</p><p>2</p><p>-</p><p>1</p><p>2</p><p>I0</p><p>v 0</p><p>2</p><p>R</p><p>2</p><p>= mg R (1 - cos b)</p><p>v 1</p><p>2 = gR cos b -</p><p>NR</p><p>m</p><p>mv 1</p><p>2</p><p>R</p><p>= mg cos b - N</p><p>b</p><p>v1</p><p>v = 210g (R + r)>17r</p><p>2</p><p>v = vr h = (R + r) (1 + cos u)</p><p>I = 2>5 mr</p><p>2.</p><p>mg</p><p>N</p><p>O</p><p>and</p><p>(2)</p><p>.</p><p>158 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.265 Clearly the tendency of bouncing of the hoop will be maximum when the small body</p><p>A will be at the highest point of the hoop during its rolling motion. Let the velocity</p><p>of C.M. of the hoop equal v at this position. The static friction does no work on the</p><p>hoop, so from conservation of mechanical energy:</p><p>or</p><p>or (1)</p><p>From the equation Fn � mwn for body A at final position, as shown in Fig. (b):</p><p>(2)</p><p>As the hoop has no acceleration in vertical</p><p>direction, so for the hoop,</p><p>(3)</p><p>From Eqs. (2) and (3),</p><p>(4)</p><p>As the hoop does not bounce,</p><p>(5)</p><p>So from Eqs. (1), (4) and (5),</p><p>Hence,</p><p>1.266 Since the lower part of the belt is in contact with the rigid floor, velocity of this part</p><p>becomes zero. The crawler moves with velocity v, hence the velocity of upper part of</p><p>the belt becomes 2v by the rolling condition and kinetic energy of upper part is</p><p>given by</p><p>which is also the sought kinetic energy, assuming that the length of the belt is much</p><p>larger than the radius of the wheels.</p><p>1.267 The sphere has two types of motion, one is the rotation about its own axis and the</p><p>other is motion in a circle of radius R. Hence the sought kinetic energy is,</p><p>(1)T =</p><p>1</p><p>2</p><p>I1 v1</p><p>2 +</p><p>1</p><p>2</p><p>I2 v2</p><p>2</p><p>1</p><p>2</p><p>am</p><p>2</p><p>b(2v)2 = mv</p><p>2</p><p>v0 … 28 gR</p><p>8 gR - v 0</p><p>2</p><p>3R</p><p>Ú 0 or 8gR Ú v 0</p><p>2</p><p>N Ú 0</p><p>N = 2mg -</p><p>mv</p><p>2</p><p>R</p><p>N + N ¿ = mg</p><p>mg + N ¿ = mv2R = m a v</p><p>R</p><p>b2</p><p>R</p><p>3v</p><p>2 = v 0</p><p>2 - 2gR</p><p>0 +</p><p>1</p><p>2</p><p>mv 0</p><p>2 +</p><p>1</p><p>2</p><p>mR</p><p>2a v0</p><p>R</p><p>b2</p><p>- mgR =</p><p>1</p><p>2</p><p>m (2v)2 +</p><p>1</p><p>2</p><p>mv</p><p>2 +</p><p>1</p><p>2</p><p>mR</p><p>2 a v</p><p>R</p><p>b2</p><p>+ mgR</p><p>E1 = E2</p><p>(a)</p><p>A</p><p>R</p><p>v0</p><p>(b) N</p><p>mg</p><p>v</p><p>N' mg</p><p>N'</p><p>1.5 DYNAMICS OF A SOLID BODY 159</p><p>where I1 is the moment of inertia about its own axis, and I2 is the moment of iner-</p><p>tia about the vertical axis, passing through O.</p><p>But, (using parallel axis theorem) (2)</p><p>In addition</p><p>(3)</p><p>Using Eqs. (2) and (3) in Eq. (1), we get</p><p>1.268 For a point mass of mass dm, looked at from C rotating frame, the equation is</p><p>where r� � radius vector in the rotating frame with respect to rotation axis and</p><p>v� � velocity in the same frame. The total centrifugal force is clearly</p><p>where RC is the radius vector of the C.M. of the body with respect to rotation axis. Also</p><p>where we have used the definitions,</p><p>.</p><p>1.269 Consider a small element of length dx at a distance x from the</p><p>point C, which is rotating in a circle of radius .</p><p>Now, mass of the element</p><p>So, centrifugal force acting on this element</p><p>and moment of this force about C is</p><p>and hence, total moment</p><p>N = 23</p><p>l>2</p><p>0</p><p>mv2</p><p>2l</p><p>sin 2ux</p><p>2dx =</p><p>1</p><p>24</p><p>mv2l 2 sin 2u</p><p>=</p><p>mv2</p><p>2l</p><p>sin 2ux</p><p>2dx</p><p>| dN | = am</p><p>l</p><p>bdx v2x sin u # x cos u</p><p>= am</p><p>l</p><p>bdx v2 x sin u</p><p>= am</p><p>l</p><p>bdx</p><p>r = x sin u</p><p>m RC = adm rœ and m v¿C = adm v¿</p><p>Fcor = 2m vœ</p><p>C * v</p><p>Fcf = adm v2rœ = mv2 RC</p><p>dm wœ = f + dm v2rœ + 2dm (vœ * �)</p><p>T ¿ =</p><p>7</p><p>10</p><p>mv</p><p>2a1 +</p><p>2r</p><p>2</p><p>7R</p><p>2</p><p>b</p><p>v1 =</p><p>v</p><p>r</p><p>and v2 =</p><p>v</p><p>R</p><p>I1 =</p><p>2</p><p>5</p><p>mr</p><p>2 and I2 =</p><p>2</p><p>5</p><p>mr</p><p>2 + mR</p><p>2</p><p>O</p><p>O</p><p>x</p><p>dx</p><p>A</p><p>C</p><p>r</p><p>B</p><p>160 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.270 Let us consider the system in a frame rotating with the rod. In this frame, the rod is</p><p>at rest and experiences not only the gravitational force mg and the reaction force R,</p><p>but also the centrifugal Fcf .</p><p>In the considered frame, from the condition of equilibrium i.e. N0z� 0</p><p>or (1)</p><p>where Ncf is the moment of centrifugal force about O. To calculate Ncf, let us consider</p><p>an element of length dx, situated at a distance x from the point O. This element</p><p>is subjected to a horizontal pseudo force The moment of this</p><p>pseudo force about the axis of rotation through the point O is</p><p>So,</p><p>(2)</p><p>It follows from Eqs. (1) and (2) that,</p><p>(3)</p><p>1.271 When the cube is given an initial velocity on</p><p>the table in some direction (as shown in</p><p>figure) it acquires an angular momentum</p><p>about an axis on the table perpendicular to</p><p>the initial velocity and (say) just below the</p><p>centre of gravity (C.G.) This angular momen-</p><p>tum will disappear when the cube stops and</p><p>this can only be due to a torque. Frictional</p><p>forces cannot do this by themselves because</p><p>they act in the plane containing the axis.</p><p>But if the forces of normal reaction act ec-</p><p>centrically (as shown in figure), their torque</p><p>cos u = a 3g</p><p>2v2l</p><p>b or u = cos-1</p><p>a 3g</p><p>2v2l</p><p>b</p><p>=</p><p>mv2l</p><p>2</p><p>3</p><p>sin u cos u</p><p>Ncf = 3</p><p>l</p><p>0</p><p>mv2</p><p>l</p><p>sin u cos u x</p><p>2dx</p><p>=</p><p>mv2</p><p>l</p><p>sin u cos ux</p><p>2dx</p><p>dNcf = am</p><p>l</p><p>bdx v2 x sin ux cos u</p><p>(m /l )dx v2x sin u.</p><p>Ncf = mg</p><p>l</p><p>2</p><p>sin u</p><p>C</p><p>mg</p><p>Fcf</p><p>0</p><p>Axis to the</p><p>initial velocity on</p><p>the table</p><p>Initial</p><p>velocity</p><p>NG</p><p>Initial</p><p>angular momentum</p><p>x</p><p>, ,</p><p>1.5 DYNAMICS OF A SOLID BODY 161</p><p>can bring about the vanishing of the angular momentum. We can calculate the dis-</p><p>tance x between the point of application of the normal reaction and the C.G. of</p><p>the cube as follows. Take the moment about C.G. of all the forces. This must van-</p><p>ish because the cube does not turn on the table.</p><p>Then if the force of friction is fr</p><p>But and so</p><p>1.272 As the rod is smooth, in the process of motion of the given system, the kinetic en-</p><p>ergy and the angular momentum relative to rotation axis do not vary. Hence, it fol-</p><p>lows that</p><p>and</p><p>From these equations we obtain</p><p>1.273 As surface is smooth, for further motion the frame attached with the C.M. is inertial.</p><p>Due to hitting of the ball, the angular impulse received by the rod about the C.M. is</p><p>equal to . If is the angular velocity acquired by the rod, we have</p><p>(1)</p><p>In the frame of C.M., the rod is rotating about an axis passing through its mid point</p><p>with the angular velocity . Hence the force exerted by one half on the other</p><p>mass of one half acceleration of C.M. of that part, in the frame of C.M.</p><p>=</p><p>m</p><p>2</p><p>av2</p><p>l</p><p>4</p><p>b = m</p><p>v2l</p><p>8</p><p>=</p><p>9p2</p><p>2ml</p><p>= 9 N</p><p>*</p><p>=v</p><p>ml</p><p>2</p><p>12</p><p>v =</p><p>pl</p><p>2</p><p>or v =</p><p>6p</p><p>ml</p><p>vp l /2</p><p>v¿ =</p><p>v0l</p><p>21 + 3m/M</p><p>v =</p><p>v0</p><p>(1 + 3m /M )</p><p>Ml</p><p>2</p><p>3</p><p>v0 =</p><p>Ml</p><p>2</p><p>3</p><p>v + ml</p><p>2v</p><p>1</p><p>2</p><p>Ml</p><p>2</p><p>3</p><p>v0</p><p>2 =</p><p>1</p><p>2</p><p>m (v2l</p><p>2 + v ¿2) +</p><p>1</p><p>2</p><p>Ml</p><p>2</p><p>3</p><p>v2</p><p>¢x =</p><p>ka</p><p>2</p><p>fr = kmg,N = mg</p><p>fr</p><p>a</p><p>2</p><p>= N¢x</p><p>¢</p><p>(where is the final angular velocity of the rod) v</p><p>and</p><p>162 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.274 (a) In the process of motion of the given system the kinetic energy and the angular</p><p>momentum relative to rotation axis do not vary. Hence it follows that</p><p>and</p><p>From these equations we obtain</p><p>As so in vector form</p><p>(b) Obviously the sought force provides the centripetal acceleration to the C.M. of the</p><p>rod and is</p><p>1.275 (a) About the axis of rotation of the rod, the angular momentum of the system is</p><p>conserved. Thus if the velocity of the flying bullet is v,</p><p>(1)</p><p>Now from the conservation of mechanical energy of the system (rod with bullet)</p><p>in the uniform field of gravity,</p><p>(2)</p><p>[because C.M. of rod raises by the height</p><p>Solving Eqs. (1) and (2), we get</p><p>v = aM</p><p>m</p><p>bA2</p><p>3</p><p>gl sin</p><p>a</p><p>2</p><p>and v = A</p><p>6g</p><p>l</p><p>sin</p><p>a</p><p>2</p><p>l</p><p>2</p><p>(1 - cos a)].</p><p>1</p><p>2</p><p>aml</p><p>2 +</p><p>Ml</p><p>2</p><p>3</p><p>b v2 = (M + m) g</p><p>l</p><p>2</p><p>(1 - cos a)</p><p>v =</p><p>mv</p><p>m + (M>3)l</p><p>=</p><p>3mv</p><p>Ml</p><p>as m V M</p><p>mvl = aml</p><p>2 +</p><p>Ml</p><p>2</p><p>3</p><p>b v</p><p>= Mv2</p><p>l</p><p>2</p><p>=</p><p>8Mv</p><p>2</p><p>l (1 + 4M>3m)2</p><p>Fn = mwcn</p><p>v¿ = a3m - 4M</p><p>3m + 4M</p><p>b v</p><p>v¿ c cv,</p><p>v ¿ = a3m - 4m</p><p>3m + 4M</p><p>bv and v =</p><p>4v</p><p>l (1 + 4m>3M)</p><p>mv</p><p>l</p><p>2</p><p>= mv ¿</p><p>l</p><p>2</p><p>+</p><p>Ml</p><p>2</p><p>3</p><p>v</p><p>1</p><p>2</p><p>mv</p><p>2 =</p><p>1</p><p>2</p><p>mv ¿2 +</p><p>1</p><p>2</p><p>aMl</p><p>2</p><p>3</p><p>bv3</p><p>( )</p><p>1.5 DYNAMICS OF A SOLID BODY 163</p><p>(b) Sought</p><p>where is the velocity of the bullet and equals the velocity of C.M. of</p><p>the rod after the impact. Putting the value of v and , we get</p><p>This is caused by the reaction at the hinge on the upper end.</p><p>(c) Let the rod start swinging with angular velocity in this case. Then, like in</p><p>part (a),</p><p>Final momentum is</p><p>So,</p><p>This vanishes for</p><p>1.276 (a) As force F on the body is radial so its angular momentum about the axis becomes</p><p>zero and the angular momentum of system about the given axis is conserved.</p><p>Thus</p><p>(b) From the equation of the increment of the mechanical energy of the system</p><p>Putting the value of from part (a) and solving we get</p><p>1.277 (a) Among the external forces are, the reaction force by the support at the centre of</p><p>the disk, weight of disk, and weight of the man, acting on the system (disk 	 man).</p><p>Aext =</p><p>mv0</p><p>2R</p><p>2</p><p>2</p><p>a1 +</p><p>2m</p><p>M</p><p>bv</p><p>1</p><p>2</p><p>MR</p><p>2</p><p>2</p><p>v2 -</p><p>1</p><p>2</p><p>aMR</p><p>2</p><p>2</p><p>+ mR</p><p>2bv0</p><p>2 = Aext</p><p>¢T = Aext</p><p>MR</p><p>2</p><p>2</p><p>v0 + m v0R</p><p>2 =</p><p>MR</p><p>2</p><p>2</p><p>v or v = v0a1 +</p><p>2m</p><p>M</p><p>b</p><p>x L 2>3l .</p><p>¢p = pf - pi L mv a3x</p><p>2l</p><p>- 1b</p><p>pf = mx v¿ + 3</p><p>t</p><p>0</p><p>y v¿</p><p>M</p><p>l</p><p>dy L</p><p>M</p><p>2</p><p>v¿l L</p><p>3</p><p>2</p><p>mv</p><p>x</p><p>l</p><p>mvx = aMl</p><p>2</p><p>3</p><p>+ mx</p><p>2bv¿ or v¿ L</p><p>3mvx</p><p>Ml</p><p>2</p><p>v¿</p><p>¢p =</p><p>1</p><p>2</p><p>mv = M A</p><p>gl</p><p>6</p><p>sin</p><p>a</p><p>2</p><p>v</p><p>v (l>2)vl</p><p>¢p = cm (vl ) + M av</p><p>l</p><p>2</p><p>b d - mv</p><p>164 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>First two forces pass through the centre O of the disk, so no torque about O,</p><p>that’s why about the vertical axis of rotations OZ is also zero. The weight of the</p><p>man is collinear to axis OZ, so no torque about this axis.</p><p>Let the man m1 stand at point A (say) at the edge of the disk. Let Ox be a fixed</p><p>line and such that at t � 0 the line joining O to A coincides with it. At time t, let</p><p>�AOx � � and �AOP � where P denotes the position of the man on the disk</p><p>(relative to disk). The angle xOP increases at the rate therefore the ve-</p><p>locity of the man at right angles to OP is R and its angular momentum</p><p>about the rotation axis Oz of the disk is</p><p>The disk has angular momentum about the Oz axis which becomes (m2R</p><p>2/2) in</p><p>the opposite sense to conserve the angular momentum of the (disk man) sys-</p><p>tem, which is zero.</p><p>Therefore,</p><p>or</p><p>On integrating</p><p>or (1)</p><p>This gives the total angle of rotation of the disk.</p><p>(b) From Eq. (1)</p><p>Differentiating with respect to time</p><p>d</p><p>2u</p><p>dt</p><p>2</p><p>= - a m1</p><p>m1 + m2>2 b 1</p><p>R</p><p>dv¿(t)</p><p>dt</p><p>du</p><p>dt</p><p>= - a m1</p><p>m1 + m2>2 bdw¿</p><p>dt</p><p>= - a m1</p><p>m1 + m2>2 b v¿(t)</p><p>R</p><p>u = a m1</p><p>m1 + m2>2 bw¿</p><p>3</p><p>u</p><p>0</p><p>d u = -3</p><p>w¿</p><p>0</p><p>£ m1</p><p>m1 +</p><p>m2</p><p>2</p><p>≥d w¿</p><p>d u = J</p><p>m1</p><p>m1 + am2</p><p>2</p><p>b Kdw¿</p><p>m1R</p><p>2w</p><p>.</p><p>¿ = am2R</p><p>2</p><p>2</p><p>+ m1R</p><p>2b</p><p>u</p><p>m2R</p><p>2</p><p>2</p><p>u</p><p>.</p><p>- m1R</p><p>2 (w</p><p>. - u</p><p>.</p><p>) = 0</p><p>+</p><p>.</p><p>u</p><p>mR</p><p>2 (</p><p>.</p><p>w¿ -</p><p>.</p><p>u).</p><p>(</p><p>.</p><p>w -</p><p>.</p><p>u)</p><p>.</p><p>w¿ -</p><p>.</p><p>u,</p><p>w¿</p><p>A</p><p>z</p><p>x</p><p>P</p><p>O j'</p><p>¿</p><p>1.5 DYNAMICS OF A SOLID BODY 165</p><p>Thus, the sought force moment from the equation</p><p>Hence,</p><p>1.278 (a) From the law of conservation of angular momentum of the system relative to ver-</p><p>tical axis z, it follows that</p><p>Hence, (1)</p><p>Note that for the corresponding vector coincides with the positive di-</p><p>rection to the z-axis, and vice versa. As both disks rotate about the same verti-</p><p>cal axis z, thus in vector form</p><p>However, the problem makes sense only if</p><p>(b) From the equation of increment of mechanical energy of a system:</p><p>Using Eq. (1)</p><p>1.279 For the closed system (disk 	 rod), the angular momentum is conserved about any</p><p>axis. Thus from the conservation of angular momentum of the system about the ro-</p><p>tation axis of rod passing through its C.M. gives:</p><p>(1)</p><p>For the closed system linear momentum is also conserved. Hence,</p><p>(2)</p><p>From Eqs. (1) and (2) we get</p><p>vC =</p><p>lv</p><p>3</p><p>and v - v ¿ = hvC</p><p>mv = mv ¿ + h mvC</p><p>mv</p><p>l</p><p>2</p><p>= mv¿</p><p>l</p><p>2</p><p>+</p><p>hml</p><p>2</p><p>12</p><p>v</p><p>Afr =</p><p>I1I2</p><p>2(I1 + I2)</p><p>(v1z - v2z)</p><p>2</p><p>Afr =</p><p>1</p><p>2</p><p>(I1 + I2) vz</p><p>2 -</p><p>1</p><p>2</p><p>I1 v 1</p><p>2 +</p><p>1</p><p>2</p><p>I2 v2z</p><p>2</p><p>Afr = ¢T</p><p>�1</p><p>cc �2 or �1</p><p>c T �2</p><p>� = (I1�1 + I2�2)>(I1 + I2)</p><p>�vz 7 0,</p><p>vz = (I1v1z + I2v2z )>(I1 + I2)</p><p>I1 v1z + I2v2z = (I1 + I2)vz</p><p>Nz = -</p><p>m1m2R</p><p>2m1 + m2</p><p>dv ¿(t)</p><p>dt</p><p>Nz =</p><p>m2R</p><p>2</p><p>2</p><p>d</p><p>2u</p><p>dt</p><p>2</p><p>= -</p><p>m2R</p><p>2</p><p>2</p><p>a m1</p><p>m1 + m2>2 b 1</p><p>R</p><p>dv¿(t)</p><p>dt</p><p>Nz = Ibz is</p><p>(where is the final velocity of the disk and angular velocity of the rod). vv ¿</p><p>(where vC is the velocity of C.M. of the rod).</p><p>.</p><p>166 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Applying conservation of kinetic energy, as the collision is elastic</p><p>(3)</p><p>or</p><p>Then</p><p>Vectorially, noting that we have taken parallel to</p><p>So,</p><p>Note: Instead of Eq. 3 one can also write the equation of coefficient of restitution.</p><p>1.280 See the diagram in the book (Fig. 1.72).</p><p>(a) When the shaft is turned through 90�, the platform must start turning with</p><p>angular velocity so that the angular momentum remains constant. Here,</p><p>The work performed by the motor is therefore,</p><p>If the shaft is turned through 180�, angular velocity of the sphere changes sign.</p><p>Thus from conservation of angular momentum,</p><p>Here � is the complete angular momentum of the sphere, i.e., we assume</p><p>that the angular velocity of the sphere is just . Then</p><p>and the work done must be</p><p>(b) In the first part of (a), the angular momentum vector of the sphere is processing</p><p>with angular velocity . Thus torque needed is</p><p>I0v0æ =</p><p>I 0</p><p>2 v0</p><p>2</p><p>I + I0</p><p>æ</p><p>1</p><p>2</p><p>I æ2 +</p><p>1</p><p>2</p><p>I0</p><p>v0</p><p>2 -</p><p>1</p><p>2</p><p>I0</p><p>v0</p><p>2 =</p><p>2I</p><p>2</p><p>0</p><p>v0</p><p>2</p><p>I</p><p>æ = 2I0</p><p>v0</p><p>I</p><p>- v0</p><p>I0 v0</p><p>I æ - I0v0 = I0v0</p><p>1</p><p>2</p><p>(I + I0)æ2 =</p><p>1</p><p>2</p><p>I</p><p>2</p><p>0 v2</p><p>0</p><p>I + I0</p><p>(I + I0)æ = I0 v0 or æ =</p><p>I0v0</p><p>I + I0</p><p>æ</p><p>BB¿</p><p>v¿ = 0 for h = 4 and v¿ Tcv for h 7 4</p><p>v¿ = a4 - h</p><p>4 + h</p><p>b v</p><p>vv¿</p><p>v ¿ =</p><p>4 - h</p><p>4 + h</p><p>v and v =</p><p>12v</p><p>(4 + h)l</p><p>v</p><p>2 - v ¿2 = 4hv C</p><p>2 and hence v + v ¿ = 4vC</p><p>1</p><p>2</p><p>mv</p><p>2 =</p><p>1</p><p>2</p><p>mv ¿</p><p>2 +</p><p>1</p><p>2</p><p>hmv C</p><p>2 +</p><p>1</p><p>2</p><p>hml</p><p>2</p><p>12</p><p>v</p><p>2</p><p>l/2</p><p>C</p><p>1.5 DYNAMICS OF A SOLID BODY 167</p><p>1.281 The total centrifugal force can be calculated by,</p><p>Then for equilibrium,</p><p>and</p><p>Thus T1 vanishes, when</p><p>Then</p><p>1.282 See the diagram in the book (Fig. 1.72).</p><p>(a) The angular velocity about OO� can be resolved into a com-</p><p>ponent parallel to the rod and a component sin perpendi-</p><p>cular to the rod through C. Then component parallel to the rod</p><p>does not contribute to the angular momentum.</p><p>Also,</p><p>This can be obtained directly also.</p><p>(b) The modulus of M does not change but the modulus of the change of M is</p><p>�</p><p>(c) Here,</p><p>Now,</p><p>as M processes with angular velocity .</p><p>1.283 (a) Here is along the symmetry axis. It has two</p><p>components, the part is constant and the part precesses.</p><p>Then,</p><p>M</p><p>�</p><p>= I v sin uI v cos u</p><p>M = Iv</p><p>v</p><p>` d M</p><p>dt</p><p>` = I v sin u cos u</p><p>vdt</p><p>dt</p><p>=</p><p>1</p><p>24</p><p>ml</p><p>2v2 sin2 u</p><p>M</p><p>�</p><p>= M cos u = I v sin u cos u</p><p>2M sin (90° - u) =</p><p>1</p><p>12</p><p>ml</p><p>2 v sin 2 u</p><p>Mz = M sin u =</p><p>1</p><p>12</p><p>ml</p><p>2 v sin2</p><p>u</p><p>M = I v sin u =</p><p>1</p><p>12</p><p>ml</p><p>2 v sin u</p><p>uv</p><p>v</p><p>T2 = mg</p><p>l0</p><p>l</p><p>= 25 N</p><p>v2 =</p><p>2g</p><p>l</p><p>, v = A</p><p>2g</p><p>t</p><p>= 6 rad>s</p><p>T2 + T1 =</p><p>1</p><p>2</p><p>ml0v</p><p>2</p><p>(T2 - T1)</p><p>l</p><p>2</p><p>= mg</p><p>l0</p><p>2</p><p>3</p><p>l0</p><p>0</p><p>m</p><p>l0</p><p>v2</p><p>xdx =</p><p>1</p><p>2</p><p>ml0</p><p>v2</p><p>mgmg</p><p>O</p><p>O'</p><p>B</p><p>T1</p><p>T2</p><p>A</p><p>90−</p><p>90−</p><p>90−</p><p>90−</p><p>Mz</p><p>Mz</p><p>wdt</p><p>MI</p><p>MI</p><p>| ¢M ||</p><p>168 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>or</p><p>(b) This force is the centripetal force due to precession. It acts inward and has the</p><p>magnitude</p><p>is the distance of the i th element from the axis. This is the force that the table</p><p>will exert on the top. See the diagram in the answer sheet.</p><p>1.284 See the diagram in the book (Fig. 1.73).</p><p>The moment of inertia of the disk about its symmetry axis is . If the angular</p><p>velocity of the disk is , then the angular moment is . The precession fre-</p><p>quency being we have</p><p>This must equal the effective gravitational torques (g being replaced by</p><p>in the elevator). Thus,</p><p>1.285 The effective g is inclined at angle tan–1 with the vertical. Then</p><p>with reference to the new “vertical” we proceed as in solution of problem 1.283.</p><p>Thus,</p><p>v¿ =</p><p>ml2g</p><p>2 + w</p><p>2</p><p>Iv</p><p>= 0.8 rad/s</p><p>w/g2g2 + w</p><p>2</p><p>v =</p><p>(g + w)l</p><p>pR</p><p>2n</p><p>= 300 rad/s</p><p>g + w</p><p>m (g + w) l,</p><p>` d M</p><p>dt</p><p>` =</p><p>1</p><p>2</p><p>mR</p><p>2 v * 2pn</p><p>2pn,</p><p>1>2 m R2vv</p><p>1>2 mR</p><p>2</p><p>ri</p><p>| F | = ` amiv¿2ri ` = m v¿2 l sin u = 12 mN</p><p>v¿ = precession frequency =</p><p>mgl</p><p>Iv</p><p>= 0.7 rad/s</p><p>` dM</p><p>dt</p><p>` = I v sin u v¿ = mgl sin u</p><p>M</p><p>M</p><p>^̂</p><p>^</p><p>mg</p><p>1.5 DYNAMICS OF A SOLID BODY 169</p><p>The vector forms an angle with the normal vertical.</p><p>1.286 The moment of inertia of the sphere is and hence the value of angular</p><p>momentum is Since it processes at speed the torque required is</p><p>So, (the force F � must be vertical)</p><p>1.287 The moment of inertia is and angular momentum is The axle</p><p>oscillates about a horizontal axis making an instantaneous angle</p><p>This means that there is a variable precession with a rate of precession The</p><p>maximum value of this is When the angle between the axle and the axis is</p><p>at its maximum value, a torque acts on it which is equal to</p><p>The corresponding gyroscopic force will be</p><p>1.288 The revolutions per minute of the flywheel being n, the angular momentum of the</p><p>flywheel is The rate of precession is .</p><p>Thus,</p><p>1.289 As in the previous problem, a couple must come in play. This can be done</p><p>if a force, acts on the rails in opposite directions in addition to the centrifu-</p><p>gal and other forces. The force on the outer rail is increased and that on the inner</p><p>rail decreased. The additional force in this case has the magnitude 1.4 kN.</p><p>2pInv/Rl</p><p>2pInv>R</p><p>N = 2pINv>R = 5.97 kNm</p><p>v/Rl * 2pn.</p><p>pmr</p><p>2v w</p><p>lT</p><p>= 90 N</p><p>1</p><p>2</p><p>mr</p><p>2v</p><p>2pw</p><p>T</p><p>=</p><p>pmr</p><p>2v w</p><p>T</p><p>I vÆ</p><p>2pwm</p><p>/T.</p><p>dw/dt.</p><p>w = wm sin</p><p>2pt</p><p>T</p><p>1>2 mr</p><p>2v.1>2 mr</p><p>2</p><p>F ¿ =</p><p>2</p><p>5</p><p>mR</p><p>2vv¿>l = 300 N</p><p>2</p><p>5</p><p>mR</p><p>2vv¿ = F ¿l</p><p>v¿2>5 mR</p><p>2 v.</p><p>2>5 mR</p><p>2</p><p>u = tan</p><p>-1</p><p>v</p><p>g</p><p>= 6°� ¿</p><p>1.6 Elastic Deformations of a Solid Body</p><p>1.290 Variation of length with temperature is give by</p><p>(1)</p><p>But,</p><p>Thus, , which is the sought stress of pressure.</p><p>Putting the value of and E from Appendix and taking , we get</p><p>1.291 (a) Consider a transverse section of the tube and concentrate on an element which</p><p>subtends angle at the centre. The forces acting on a portion of length on</p><p>the element are:</p><p>(1) Tensile forces side ways of magnitude</p><p>The resultant of these is radically towards</p><p>the centre.</p><p>(2) The force due to fluid pressure</p><p>Since these balance, we get</p><p>where is the maximum tensile force.</p><p>Putting the values, we get atm.</p><p>(b) Consider an element of area about z-axis</p><p>chosen arbitrarily. There are tangential tensile forces all around the ring of the</p><p>cap. Their resultant is</p><p>Hence, in the limit,</p><p>or pm =</p><p>2 sm¢r</p><p>r</p><p>= 39.5 atm</p><p>pmp a r¢u</p><p>2</p><p>b2</p><p>= smp a r¢u</p><p>2</p><p>b¢r ¢u</p><p>s c2p ar ¢u</p><p>2</p><p>b ¢r d sin</p><p>¢u</p><p>2</p><p>dS = p (r ¢u>r)2</p><p>pmax = 19.7</p><p>sm</p><p>pmax L sm ¢r>r= pr ¢</p><p>¢l</p><p>2 s¢r ¢l sin ¢</p><p>>2 L s¢rl ¢</p><p>s¢r¢l.</p><p>¢l¢w</p><p>s = 2.2 * 103 atm</p><p>¢t = 100°Ca</p><p>s = a ¢tE</p><p>e =</p><p>s</p><p>E</p><p>lt = l0 (1 + a¢t) or ¢l</p><p>l0</p><p>= a¢t = e</p><p>2</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 170</p><p>.</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 171</p><p>1.292 Let us consider an element of rod at a distance x from its rotation axis (see figure).</p><p>From Newton’s second law in projection form directed towards the rotation axis,</p><p>On integrating</p><p>But at</p><p>Thus,</p><p>Hence</p><p>Thus (at mid point)</p><p>Condition required for the problem is</p><p>So,</p><p>Hence the sought number of rps</p><p>(using the table rps)</p><p>1.293 Let us consider an element of the ring (see figure). From Newton’s law</p><p>for this element, we get</p><p>(see solution of problem 1.93 or 1.92)</p><p>So,</p><p>Condition for the problem is</p><p>or �2</p><p>max =</p><p>2p2</p><p>smr</p><p>pr</p><p>2(2prr)</p><p>=</p><p>sm</p><p>rr</p><p>2</p><p>T</p><p>pr</p><p>2</p><p>… sm or</p><p>m�2r</p><p>2p2r</p><p>2</p><p>… sm</p><p>T =</p><p>m</p><p>2p</p><p>�2r</p><p>Td u = a m</p><p>2p</p><p>dub �2r</p><p>Fn = mwn</p><p>n = 0.8 * 103 n =</p><p>�</p><p>2p</p><p>=</p><p>1</p><p>pl</p><p>A</p><p>2 sm</p><p>r</p><p>mv2l</p><p>8</p><p>= S sm or v =</p><p>2</p><p>l</p><p>B</p><p>2 sm</p><p>r</p><p>Tmax = S sm</p><p>Tmax =</p><p>m�2l</p><p>8</p><p>T =</p><p>m�2</p><p>2</p><p>a l</p><p>4</p><p>-</p><p>x</p><p>2</p><p>l</p><p>b</p><p>0 =</p><p>mv2</p><p>2</p><p>l 2</p><p>4</p><p>+ C or C = -</p><p>mv2l</p><p>8</p><p>x = �</p><p>l</p><p>2</p><p>or free end, T = 0</p><p>-T =</p><p>m�2</p><p>l</p><p>x2</p><p>2</p><p>+ C (constant)</p><p>-dT = (dm) �2x =</p><p>m</p><p>l</p><p>�2x dx</p><p>x dx</p><p>T</p><p>T</p><p>dq</p><p>r</p><p>172 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Thus, sought number of rps is</p><p>Using the table of appendices rps.</p><p>1.294 Let the point O descend by the distance x (see figure). From the condition of equi-</p><p>librium of point O,</p><p>(1)</p><p>Now, (2)</p><p>In addition to this,</p><p>(3)</p><p>From Eqs. (1), (2) and (3)</p><p>or</p><p>1.295 Let us consider an element of the rod at a distance x from the free end (see figure).</p><p>For the considered element are internal restoring forces which produce</p><p>elongation and dT provide the acceleration to the element. For the element, from</p><p>Newton’s law</p><p>As free end has zero tension, on integrating the above expression,</p><p>3</p><p>T</p><p>0</p><p>dT =</p><p>F0</p><p>l</p><p>3</p><p>x</p><p>0</p><p>dx or T =</p><p>F0</p><p>l</p><p>x</p><p>dT = (dm) w = am</p><p>l</p><p>dxb F0</p><p>m</p><p>=</p><p>F0</p><p>l</p><p>dx</p><p>T - T ¿</p><p>x = l a mg</p><p>2pEd</p><p>2</p><p>b1>3</p><p>= 2.5 cm</p><p>x -</p><p>x</p><p>21 + 12x>l22 =</p><p>mgl</p><p>pEd</p><p>2</p><p>e =</p><p>2(l>2)2 + x</p><p>2 - l>2</p><p>l>2 = B1 + a2x</p><p>l</p><p>b2</p><p>- 1</p><p>(Here s is stress and e is strain.)</p><p>T</p><p>p(d>2)2</p><p>= s = eE or T = eEp</p><p>d</p><p>2</p><p>4</p><p>2T sinu = mg or T =</p><p>mg</p><p>2sin u</p><p>=</p><p>mg</p><p>2x</p><p>2(l>2)2 + x2</p><p>n = 23</p><p>n =</p><p>�max</p><p>2p</p><p>=</p><p>1</p><p>2pr</p><p>A</p><p>sm</p><p>r</p><p>x</p><p>l/2 l/2</p><p>TT</p><p>mg</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 173</p><p>Elongation in the considered element of length dx is given by</p><p>Thus, total elongation</p><p>Hence,</p><p>t (because w12 = 0 and r0(12) = 0)</p><p>r12 = r0(12) + v0(12) t +</p><p>1</p><p>2</p><p>w12t</p><p>2</p><p>= 22 m, on putting the values of v0, t and u</p><p>= v0 t 22(1 - sin u)</p><p>l = 2x 2</p><p>2 + ( y1 - y2 )2 = 2(v0 cos ut)2 + {v0t (1 - sin u)2}</p><p>x2 = v0 cos u t and y2 = v0</p><p>sin u t -</p><p>1</p><p>2</p><p>gt 2</p><p>y1 = v0t -</p><p>1</p><p>2</p><p>gt2</p><p>u</p><p>6 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>0</p><p>v0(12)</p><p>v02</p><p>v01</p><p>v0</p><p>x</p><p>y</p><p>(x2, y2)</p><p>v0</p><p>l(0, y1)</p><p>v| |</p><p>As</p><p>or</p><p>or</p><p>Hence,</p><p>In the frame attached with 2 for the particle 1</p><p>Both the particles are initially at the same position and have same acceleration g, so</p><p>1.12 From the symmetry of the problem all the three points are</p><p>always located at the vertices of equilateral triangles of varying</p><p>side length and finally meet at the centroid of the initial</p><p>equilateral triangle. Symmetry of the problem tells us that the ve-</p><p>locity with which point 1 approaches 2 is equal to the velocity</p><p>by which 2 approaches 3 and 3 approaches 1.</p><p>Method 1:</p><p>The rate at which 1 approaches 2 or the separation between 1 and 2 decreases with</p><p>time is the projection of v12 towards r21. At an arbitrary instant when the equilateral</p><p>triangle has edge length l � a, the velocity with which 1 approaches 2 becomes</p><p>On integrating: (where t is the sought time)</p><p>or</p><p>Method 2:</p><p>If we concentrate on the motion of any one point say 3, one can observe that at all</p><p>the time its velocity makes an angle 30� with 3O and is equal to v cos 30° = 13/2 v.</p><p>a =</p><p>3</p><p>2</p><p>vt so, t =</p><p>2a</p><p>3v</p><p>3</p><p>0</p><p>a</p><p>-dl =</p><p>3v</p><p>2</p><p>3</p><p>t</p><p>0</p><p>dt</p><p>-dl</p><p>dt</p><p>= v - v cos a2p</p><p>3</p><p>b</p><p>= 2.5 m, on putting the values of v1, v2 and g</p><p>=</p><p>v1 + v2</p><p>g</p><p>1v1v2 (using value of t</p><p>| r | = | v0 |t = (v1 + v2) t</p><p>w = 0, and v0 = | v1 - v2 |r0 = 0,</p><p>r = r0 + v0t +</p><p>1</p><p>2</p><p>w t 2</p><p>t =</p><p>2v1v2</p><p>g</p><p>-v1v2 + g2t</p><p>2 = 0</p><p>(v1 + gt) # (v2 + gt) = 0</p><p>v¿1 � v¿2 so, v¿1 # v¿2 = 0</p><p>1.1 KINEMATICS 7</p><p>0</p><p>a</p><p>a a</p><p>1</p><p>3</p><p>v</p><p>2p/3</p><p>v</p><p>v</p><p>l</p><p>2</p><p>v1tv t2</p><p>v1</p><p>v2</p><p>v2′ v1′</p><p>. Thus the sought distance is</p><p>)</p><p>Initially 3O equals so the sought time</p><p>(In the figure, velocity of point 3 has been resolved into</p><p>two rectangular components, one pointing towards the</p><p>centre of the triangle and other perpendicular to it.)</p><p>1.13 Let us locate the points A and B at an arbitrary instant of</p><p>time (see figure).</p><p>If A and B are separated by the distance s at this moment,</p><p>then the points converge or point A approaches B with</p><p>velocity where angle a varies with</p><p>time. On integrating,</p><p>(where T is the sought time)</p><p>or</p><p>(1)</p><p>As both A and B cover the same distance in x-direction during the sought time inter-</p><p>val, so the other condition which is required, can be obtained by the equation,</p><p>(2)</p><p>Solving Eqs. (1) and (2), we get</p><p>One can see that if , or u � v, point A cannot catch B.</p><p>1.14 In the reference frame fixed to the train, the distance between the two events is</p><p>obviously equal to l. Suppose the train starts moving at time in the positive</p><p>direction and take the origin at the head-light of the train at . Then the</p><p>coordinate of first event in the earth’s frame is</p><p>x1 =</p><p>1</p><p>2</p><p>wt 2</p><p>t = 0(x = 0)</p><p>t = 0</p><p>u = v</p><p>T =</p><p>ul</p><p>v</p><p>2 - u2</p><p>uT = 3</p><p>T</p><p>0</p><p>v cos a dt</p><p>= vT - u3</p><p>T</p><p>0</p><p>cos a dt</p><p>l = 3</p><p>T</p><p>0</p><p>(v - u cos a) dt</p><p>-3</p><p>0</p><p>l</p><p>ds = 3</p><p>T</p><p>0</p><p>(v - u cos a) dt</p><p>-ds>dt = v - u cos a</p><p>t =</p><p>a>13</p><p>(13>2)v</p><p>=</p><p>2a</p><p>3v</p><p>a>13,</p><p>8 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1</p><p>3</p><p>v</p><p>v</p><p>3</p><p>2 v v/2</p><p>2</p><p>Oa</p><p>l</p><p>A</p><p>v</p><p>B</p><p>u</p><p>and similarly the coordinate of the second event is</p><p>The distance between the two events is obviously</p><p>in the reference frame fixed on the Earth.</p><p>For the two events to occur at the same point in the reference frame K, moving with</p><p>constant velocity V relative to the earth, the distance traveled by the frame in the time</p><p>interval must be equal to the above distance.</p><p>Thus,</p><p>So,</p><p>The frame K must clearly be moving in a direction opposite to the train so that if (for</p><p>example) the origin of the frame coincides with the point x1 on the earth at time t, it</p><p>coincides with the point x2 at time</p><p>1.15 (a) As time interval between two events is reference independent in Newtonian</p><p>mechanics, so it is better to solve the problem in the elevator�s frame having the</p><p>observer at its floor. (see figure). Let us denote separation between floor and ceil-</p><p>ing by and the acceleration of the elevator by</p><p>From the kinematical formula</p><p>Here,</p><p>and</p><p>So,</p><p>or t = A</p><p>2h</p><p>g + w</p><p>= 0.7 s (on substituting values)</p><p>0 = h +</p><p>1</p><p>2</p><p>[- ( g + w)]t</p><p>2</p><p>= (–g) - (w) = - ( g + w)</p><p>wy = wbolt(y) - wele(y)</p><p>y = 0, y0 = +h, v0y = 0</p><p>y = y0 + v0y t +</p><p>1</p><p>2</p><p>wy t</p><p>2</p><p>w = 1.2 m/s2h = 2.7 m</p><p>t + t.</p><p>V =</p><p>l</p><p>t</p><p>- w (t + t>2) = 4.03 m/s</p><p>Vt = l - w t (t + t>2)</p><p>t</p><p>x1 - x2 = l - w t (t + t>2) = 0.242 km</p><p>x2 =</p><p>1</p><p>2</p><p>w (t + t)2 - l</p><p>1.1 KINEMATICS 9</p><p>O</p><p>y</p><p>h w</p><p>.</p><p>(b) At the moment the bolt loses contact with the elevator, it has already acquired the</p><p>velocity equal to elevator, given by;</p><p>In the reference frame attached with the elevator shaft (ground) and pointing the</p><p>y-axis upward, we have for the displacement of the bolt,</p><p>or</p><p>Hence the bolt comes down or displaces downward relative to the point, when it</p><p>loses contact with the elevator by the amount 0.7 m (see figure).</p><p>Obviously the total distance covered by the bolt during its free fall time</p><p>1.16 After time the separation between the particles</p><p>For l and hence l 2 to be minimum</p><p>or</p><p>which gives</p><p>On using the value of in the expression of</p><p>.</p><p>Alternate:</p><p>Let the particles 1 and 2 be at points B and A respectively at t � 0, at the distances l1</p><p>and l2 from intersection point O.</p><p>Let us fix the inertial frame with the particle 2. Now the particle 1 moves relative to this</p><p>reference frame with a relative velocity and its trajectory is the straight</p><p>line BP. Obviously, the minimum distance between the particles is equal to the length</p><p>of the perpendicular AP dropped from point A on to the straight line BP (Fig. (a)).</p><p>v12 = v1 - v2,</p><p>lmin =</p><p>| v1l2 - v2l1 |</p><p>1v</p><p>2</p><p>1 + v</p><p>2</p><p>2</p><p>lt,</p><p>t =</p><p>(v1l 1 - v2l2)</p><p>2v</p><p>2</p><p>1 + v 2</p><p>2</p><p>d</p><p>dt</p><p>{(l1 + v1t)</p><p>2 + (l2 - v2t)</p><p>2} = 0</p><p>dl</p><p>2</p><p>dt</p><p>= 0</p><p>l = 2(l1 - v1t)</p><p>2 + (l2 - v2t)</p><p>2</p><p>t 7 0,</p><p>s = | ¢y | + 2a v 0</p><p>2</p><p>2g</p><p>b = 0.7 m +</p><p>(2.4)2</p><p>(9.8)</p><p>m = 1.3 m</p><p>= -0.7 m</p><p>¢y = (2.4)(0.7) +</p><p>1</p><p>2</p><p>(-9.8)(0.7)2</p><p>= v0t +</p><p>1</p><p>2</p><p>(-g)t</p><p>2</p><p>¢y = v0y</p><p>t +</p><p>1</p><p>2</p><p>wy</p><p>t</p><p>2</p><p>v0 = (1.2) (2) = 2.4 m/s</p><p>10 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>f</p><p>i</p><p>y</p><p>v0</p><p>From Fig. (b),</p><p>(1)</p><p>From Fig. (a), the shortest distance</p><p>or</p><p>The sought time</p><p>1.17 Let the car turns off the highway at a distance x from the point D.</p><p>So, CD � x, and if the speed of the car in the field is v, then the</p><p>time taken by the car to cover the distance AC � (AD � x) on the</p><p>highway</p><p>(1)</p><p>and the time taken to travel the distance CB in the field</p><p>(2)</p><p>So, the total time elapsed to move the car from point A to B</p><p>t = t1 + t2 =</p><p>AD - x</p><p>hv</p><p>+</p><p>2l</p><p>2 + x 2</p><p>v</p><p>t2 =</p><p>2l 2 + x</p><p>2</p><p>v</p><p>t1 =</p><p>AD - x</p><p>hv</p><p>=</p><p>l2 cos u + l1 sin u</p><p>2v 1</p><p>2 + v 2</p><p>2</p><p>=</p><p>l2v2 + l1v1</p><p>v 1</p><p>2 + v 2</p><p>2</p><p>(using Eq. 1)</p><p>t =</p><p>BP</p><p>| v12 |</p><p>=</p><p>BM + MP</p><p>| v12 |</p><p>=</p><p>l1 cosec u + (l2 - l1 cot u) cos u</p><p>2v 1</p><p>2 + v 2</p><p>2</p><p>AP = a l2 - l1</p><p>v2</p><p>v1</p><p>b</p><p>v1</p><p>2v 1</p><p>2 + v</p><p>2</p><p>2</p><p>=</p><p>| v1l2 - v2l1 |</p><p>2v 1</p><p>2 + v 2</p><p>2</p><p>(using Eq. 1)</p><p>AP = AM sin u = (OA - OM ) sin u = (l2 - l1 cot u) sin u</p><p>and sin u =</p><p>v1</p><p>2v 1</p><p>2 + v</p><p>2</p><p>2</p><p>So, cos u =</p><p>v2</p><p>2v 1</p><p>2 + v 2</p><p>2</p><p>v12 = 2v</p><p>2</p><p>1 + v</p><p>2</p><p>2, tan u =</p><p>v1</p><p>v2</p><p>,</p><p>1.1 KINEMATICS 11</p><p>B</p><p>x</p><p>l</p><p>DC</p><p>l2 + x 2</p><p>A</p><p>P</p><p>A</p><p>M</p><p>O</p><p>v2</p><p>l2</p><p>v12</p><p>v2</p><p>v1</p><p>l1</p><p>v12</p><p>v1</p><p>B</p><p>(a) (b)</p><p>q</p><p>q</p><p>q</p><p>q</p><p>For t to be minimum</p><p>or</p><p>or</p><p>1.18 From question so the motion starts from</p><p>origin, thus from the plot of we have</p><p>and for</p><p>As at ,</p><p>So,</p><p>Now,</p><p>Now, ax =</p><p>dvx</p><p>dt</p><p>= L</p><p>1 ; 0 … t 6 1</p><p>0 ; 1 6 t … 3</p><p>-1 ; 3 6 t … 6</p><p>2 ; 6 6 t … 7</p><p>s (t) = 3</p><p>t</p><p>0</p><p>| vx |dt = h</p><p>t</p><p>2</p><p>2</p><p>; 0 … t … 1</p><p>t - 1</p><p>2</p><p>; 1 6 t … 3</p><p>4t - t</p><p>2</p><p>2</p><p>- 5 ; 3 6 t … 4</p><p>t</p><p>2</p><p>2</p><p>- 4t + 11 ; 4 6 t … 6</p><p>14t - t</p><p>2 - 43 ; 6 6 t … 7</p><p>x (t) = 3</p><p>t</p><p>0</p><p>vx</p><p>dt = f</p><p>t</p><p>2</p><p>2</p><p>; 0 … t … 1</p><p>t - 1</p><p>2</p><p>; 1 6 t … 3</p><p>4t - t</p><p>2</p><p>2</p><p>+ 5 ; 3 6 t … 6</p><p>t</p><p>2 - 14t + 49 ; 6 6 t … 7</p><p>t = 0, x = 0</p><p>t 7 7.vx = 0</p><p>vx = L</p><p>t ; 0 … t … 1</p><p>1 ; 1 6 t … 3</p><p>4 - t ; 3 6 t … 6</p><p>2t - 14 ; 6 6 t … 7</p><p>vx(t)</p><p>vx (0) = 0,</p><p>h 2x 2 = l 2 + x 2 or x =</p><p>l</p><p>2h 2 - 1</p><p>1</p><p>the sought strain is</p><p>1.296 Let us consider an element of the rod at a distance r from its rotation axis. As the</p><p>element rotates in a horizontal circle of radius r, we have from Newton’s second law</p><p>in projection form directed toward the axis of rotation</p><p>or</p><p>At the free end tension becomes zero. Integrating the above expression we get</p><p>Thus,</p><p>Elongation in elemental length dr is given by</p><p>(where S is the cross sectional area of the rod and T is the tension in the rod at the</p><p>considered element)</p><p>or</p><p>Thus, the sought elongation</p><p>j = 3dj =</p><p>m�2l</p><p>2SE 3</p><p>l</p><p>0</p><p>a1 -</p><p>r</p><p>2</p><p>l</p><p>2</p><p>b dr</p><p>0j =</p><p>m�2l</p><p>2SE</p><p>a1 -</p><p>r</p><p>2</p><p>l</p><p>2</p><p>b dr</p><p>0j =</p><p>s(r)</p><p>E</p><p>dr =</p><p>T</p><p>SE</p><p>dr</p><p>T =</p><p>m�2</p><p>l</p><p>a l</p><p>2 - r</p><p>2</p><p>2</p><p>b =</p><p>m�2l</p><p>2</p><p>a1 -</p><p>r</p><p>2</p><p>l</p><p>2</p><p>b</p><p>-3</p><p>0</p><p>T</p><p>dT =</p><p>m</p><p>l</p><p>�2 3</p><p>l</p><p>r</p><p>rdr</p><p>-dT = am</p><p>l</p><p>drb �2r =</p><p>m</p><p>l</p><p>�2rdr</p><p>T - (T + dT ) = (dm)�2r</p><p>s =</p><p>j</p><p>l</p><p>=</p><p>F0</p><p>2SE</p><p>j =</p><p>F0</p><p>SEl</p><p>3</p><p>l</p><p>x</p><p>xdx =</p><p>F0l</p><p>2SE</p><p>0j =</p><p>s</p><p>E</p><p>(x) dx =</p><p>T</p><p>SE</p><p>dx =</p><p>Fxo xdx</p><p>SE l</p><p>x</p><p>dx</p><p>T+dTT</p><p>r</p><p>dr</p><p>dTT</p><p>174 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>or</p><p>(where is the density of the copper)</p><p>1.297 Volume of a solid cylinder</p><p>So, (1)</p><p>But longitudinal strain and accompanying lateral strain are related as</p><p>(2)</p><p>Using Eq. (2) in Eq. (1), we get</p><p>(3)</p><p>But</p><p>(Because the increment in the length of cylinder is negative.)</p><p>So, (where is the Poisson’s ratio for copper)</p><p>Thus,</p><p>(Negative sign means that the volume of the cylinder has decreased.)</p><p>1.298 (a) As free end has zero tension, thus the tension in the rod at a vertical distance y</p><p>from its lower end</p><p>(1)</p><p>Let be the elongation of the element of length dy, then</p><p>= T</p><p>SE</p><p>dy =</p><p>mgydy</p><p>SlE</p><p>=</p><p>rgydy</p><p>E</p><p>(where r is the density of copper)</p><p>0l =</p><p>s(y)</p><p>E</p><p>dy</p><p>0l</p><p>T =</p><p>m</p><p>l</p><p>gy</p><p>= -1.6 mm3</p><p>¢V =</p><p>-Fl</p><p>E</p><p>(1 - 2m)</p><p>m</p><p>¢V</p><p>V</p><p>=</p><p>-F</p><p>pr</p><p>2E</p><p>(1 - 2m)</p><p>¢l</p><p>¢l</p><p>l</p><p>=</p><p>-F>pr</p><p>2</p><p>E</p><p>¢V</p><p>V</p><p>=</p><p>¢l</p><p>l</p><p>(1 - 2m)</p><p>¢r</p><p>r</p><p>= -m</p><p>¢l</p><p>l</p><p>¢r/r¢l/l</p><p>¢V</p><p>V</p><p>=</p><p>p2r¢rl</p><p>pr</p><p>2l</p><p>+</p><p>pr</p><p>2¢l</p><p>pr</p><p>2l</p><p>=</p><p>2¢r</p><p>r</p><p>+</p><p>¢l</p><p>l</p><p>V = pr</p><p>2l</p><p>r =</p><p>1</p><p>3</p><p>r�2l</p><p>3</p><p>E</p><p>j =</p><p>m�2l</p><p>2SE</p><p>2l</p><p>3</p><p>=</p><p>(Sl r)</p><p>3SE</p><p>�2l</p><p>3</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 175</p><p>Thus, the sought elongation is</p><p>(2)</p><p>(b) If the longitudinal (tensile) strain is , the accompanying lateral (compres-</p><p>sive) strain is given by</p><p>(3)</p><p>Then, since , we have</p><p>(using Eq. 3)</p><p>(where is given in part (a) and is the Poisson ratio for copper).</p><p>1.299 (a) Since the hydrostatic pressure is uniform, the stress on each face of the bar is the</p><p>same (see figure (a)). The change in length of the bar can be thought of as the</p><p>sum of changes in length that would occur in the three independent cases sh-</p><p>own in the figures (b), (c) and (d).</p><p>m¢l/l</p><p>= (1 - 2m)</p><p>¢l</p><p>l</p><p>¢V</p><p>V</p><p>=</p><p>2¢r</p><p>r</p><p>+</p><p>¢l</p><p>l</p><p>V = pr</p><p>2l</p><p>e¿ =</p><p>¢r</p><p>r</p><p>= -me</p><p>e = ¢l>l</p><p>¢l = 30l = rg3</p><p>l</p><p>0</p><p>ydy</p><p>E</p><p>=</p><p>1</p><p>2</p><p>rgl</p><p>2>E</p><p>p</p><p>p</p><p>pp</p><p>(a)</p><p>pp</p><p>(b)</p><p>p</p><p>(d)</p><p>p</p><p>p</p><p>(c)</p><p>If we push on the ends of the block with a pressure p, the compressional strain</p><p>is and it is negative,</p><p>¢l1</p><p>l</p><p>= -</p><p>p</p><p>E</p><p>p>E,</p><p>If we push on the two sides of the block with pressure p, the compressional st-</p><p>rain is again , but now we want the lengthwise strain. We can get that from</p><p>the sideways strain multiplied by . The sideways strain is</p><p>So,</p><p>If the we push on the top of the block, the compressional strain in once more</p><p>, and the corresponding strain in the sideways direction is again . We</p><p>get</p><p>Combining the results of the three problems i.e., taking</p><p>we get</p><p>(1)</p><p>The problem is, of course, symmetrical in all three directions; it follows that</p><p>(2)</p><p>The change in the volume under hydrostatic pressure is also of some interest.</p><p>Since , we can write, for small displacements.</p><p>Using Eqs. (1) and (2), we have</p><p>(3)</p><p>(b) But volume strain</p><p>(4)</p><p>From Eqs. (3) and (4), we have</p><p>Hence the compressibility</p><p>b =</p><p>1</p><p>K</p><p>=</p><p>3 (1 - 2m)</p><p>E</p><p>K =</p><p>E</p><p>3 (1 - 2m)</p><p>¢V</p><p>V</p><p>= -</p><p>p</p><p>K</p><p>(where K is called the bulk modulus)</p><p>¢V</p><p>V</p><p>= -3</p><p>p</p><p>E</p><p>(1 - 2m)</p><p>¢V</p><p>V</p><p>=</p><p>¢l</p><p>l</p><p>+</p><p>¢w</p><p>w</p><p>+</p><p>¢h</p><p>h</p><p>V = lwh</p><p>¢w</p><p>w</p><p>=</p><p>¢h</p><p>h</p><p>= -</p><p>p</p><p>E</p><p>(1 - 2m)</p><p>¢l</p><p>l</p><p>= -</p><p>p</p><p>E</p><p>(1 - 2m)</p><p>¢l = ¢l1 + ¢l2 + ¢l3,</p><p>¢l3</p><p>l</p><p>= + m</p><p>p</p><p>E</p><p>-mp>Ep>E</p><p>¢l2</p><p>l</p><p>= + m</p><p>p</p><p>E</p><p>¢w</p><p>w</p><p>= -</p><p>p</p><p>E</p><p>-m</p><p>p>Y</p><p>176 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 177</p><p>1.300 A beam clamped at one end and supporting an applied load at the free end is called</p><p>a cantilever. The theory of cantilevers is discussed in the advanced text book on me-</p><p>chanics. The key result is that elastic forces in the beam generate a couple, whose</p><p>moment, called the moment of resistances, balances the external bending moment</p><p>due to weight of the beam, load etc. The moment of resistance, also called internal</p><p>bending moment (I.B.M) is given by</p><p>Here R is the radius of curvature of the beam at the</p><p>representative point (x, y ). I is called the geometrical</p><p>moment of inertia of the cross section relative to the axis</p><p>passing through the natural layer which remains un-</p><p>scratched Fig. (a).</p><p>(1)</p><p>The section of the beam beyond P exerts the bending mo-</p><p>ment N (x) and we have,</p><p>(2)</p><p>If there is no load other than that due to the weight of</p><p>the beam, then</p><p>(where density of steel).</p><p>Hence, at</p><p>Here, width of the beam perpendicular to paper.</p><p>Also,</p><p>Hence,</p><p>R =</p><p>1</p><p>6</p><p>Eh</p><p>2</p><p>rgl</p><p>2</p><p>= 0.121 km</p><p>a I</p><p>R</p><p>b</p><p>0</p><p>=</p><p>6rgl</p><p>2</p><p>Eh2</p><p>I = 3</p><p>h/2</p><p>-h/2</p><p>z</p><p>2 bdz =</p><p>bh3</p><p>12</p><p>b =</p><p>a I</p><p>R</p><p>b</p><p>0</p><p>=</p><p>rgl</p><p>2bh</p><p>2EI</p><p>x = 0</p><p>r =</p><p>N (x) =</p><p>1</p><p>2</p><p>rg(l - x)2 bh</p><p>EI</p><p>R</p><p>= N (x)</p><p>I = 3z</p><p>2 dS</p><p>I.B.M. = EI/R</p><p>ds</p><p>z</p><p>(a)</p><p>(b)</p><p>h</p><p>y</p><p>p(x,y)</p><p>x</p><p>178 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.301 We use the equation given above and use the result that when y is small</p><p>(a) Here is a constant. Then integration gives</p><p>But,</p><p>Integrating again,</p><p>where we have used for to set the constant of integration at zero.</p><p>This is the equation of a parabola. The sag of the free end is</p><p>(b) In this case because the load F at the extremity is balanced by</p><p>a similar force at F directed upward and they constitute a couple. Then</p><p>Integrating,</p><p>As before . Integrating again, using for ,</p><p>Here for a square cross section</p><p>1.302 One can think of it as analogous to the previous</p><p>case but with a beam of length 2 loaded upward</p><p>by a force F/2.</p><p>Thus,</p><p>on using the last result of the previous problem.</p><p>l =</p><p>Fl</p><p>3</p><p>48 EI</p><p>l></p><p>I = 3</p><p>a>2</p><p>-a>2z</p><p>2 adz =</p><p>a</p><p>4</p><p>12</p><p>y =</p><p>F 1lx2>2 - x3>62</p><p>EI</p><p>ahere l =</p><p>Fl</p><p>3</p><p>3EI</p><p>bx = 0y = 0C1 = 0</p><p>dy</p><p>dx</p><p>=</p><p>F (lx - x</p><p>2>2)</p><p>El</p><p>+ C1</p><p>d</p><p>2y</p><p>dx</p><p>2</p><p>=</p><p>F (l - x)</p><p>EI</p><p>N (x) = F (l - x)</p><p>l = y (x = l ) =</p><p>N0l</p><p>2</p><p>2EI</p><p>x = 0y = 0</p><p>y =</p><p>N0x</p><p>2</p><p>2EI</p><p>a dy</p><p>dx</p><p>b = 0, for x = 0, so C1 = 0.</p><p>dy</p><p>dx</p><p>=</p><p>N0x</p><p>EI</p><p>+ C1</p><p>N (x) = N0</p><p>1</p><p>R</p><p>=</p><p>d</p><p>2y</p><p>dx</p><p>2</p><p>; thus,</p><p>d</p><p>2y</p><p>dx</p><p>2</p><p>=</p><p>N (x)</p><p>EI</p><p>F/2 F/2</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 179</p><p>1.303 (a) In this case</p><p>Also</p><p>Then,</p><p>Integrating,</p><p>Using</p><p>Integrating again,</p><p>Thus,</p><p>As before, EI where N (x) is the bending moment due to section PB.</p><p>Thus, bending moment is clearly</p><p>(here is weight of the beam per unit length)</p><p>Now integrating,</p><p>or since</p><p>Integrating again,</p><p>As for . From this we find</p><p>l = y (x = l ) =</p><p>5wl</p><p>4>24</p><p>EI</p><p>=</p><p>5rgl</p><p>4</p><p>2Eh2</p><p>x = 0, c1 = 0y = 0</p><p>EIy = w ax</p><p>4</p><p>24</p><p>-</p><p>x</p><p>3l</p><p>6</p><p>b +</p><p>wl</p><p>3x</p><p>3</p><p>+ c1</p><p>dy</p><p>dx</p><p>= 0 for x = l, c0 =</p><p>wl</p><p>3</p><p>3</p><p>El</p><p>dy</p><p>dx</p><p>= w ax</p><p>3</p><p>6</p><p>-</p><p>x</p><p>2l</p><p>2</p><p>b + c0</p><p>w = r gb h = w ax</p><p>2</p><p>2</p><p>- xlb</p><p>= w a2l</p><p>2 - 2xl +</p><p>x</p><p>2</p><p>2</p><p>b –wl (2l - x)</p><p>N = 3</p><p>2l</p><p>x</p><p>w dj(j - x) - wl (2l - x)</p><p>d</p><p>2y</p><p>dx</p><p>2</p><p>= N (x)(b)</p><p>l =</p><p>6rgl</p><p>4</p><p>Eh2</p><p>a 1</p><p>2</p><p>-</p><p>1</p><p>3</p><p>+</p><p>1</p><p>12</p><p>b =</p><p>6rgl</p><p>4</p><p>Eh2</p><p>3</p><p>12</p><p>=</p><p>3rgl</p><p>4</p><p>2Eh2</p><p>y =</p><p>6rg</p><p>Eh2</p><p>a l</p><p>2x</p><p>2</p><p>2</p><p>-</p><p>lx</p><p>3</p><p>3</p><p>+</p><p>x</p><p>4</p><p>12</p><p>b</p><p>dy</p><p>dx</p><p>= 0 for x = 0</p><p>dy</p><p>dx</p><p>=</p><p>6rg</p><p>Eh2</p><p>a l</p><p>2x - lx</p><p>2 +</p><p>x</p><p>3</p><p>3</p><p>b</p><p>Ebh2</p><p>12</p><p>d</p><p>2y</p><p>dx</p><p>2</p><p>=</p><p>r gbh</p><p>2</p><p>(l</p><p>2 - 2lx + x</p><p>2)</p><p>I =</p><p>bh3</p><p>12</p><p>N (x) =</p><p>1</p><p>2</p><p>rgbh (l - x)2 (where b = width of the girder).</p><p>2l</p><p>A By</p><p>P</p><p>x</p><p>180 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.304 The deflection of the plate can be noticed by going to a co-rotating frame. In this</p><p>frame each element of the plate experiences a pseudo force proportional to its mass.</p><p>These forces have a moment which constitutes the bending moment of the problem.</p><p>To calculate this moment we note that the acceleration of an element at a distance</p><p>from the axis is and the moment of the forces exerted by the section</p><p>between x and l is</p><p>From the fundamental equation</p><p>The moment of inertia</p><p>Note that the neutral surface (i.e. the surface which contains lines which are neither</p><p>stretched nor compressed) is a vertical plane here and z is perpendicular to it.</p><p>Since</p><p>Thus,</p><p>1.305 (a) Consider a hollow cylinder of length l, outer radius , inner radius r, fixed</p><p>at one end and twisted at the other by means of a couple of moment N. The an-</p><p>gular displacement , at a distance l from the fixed end, is proportional to both</p><p>l and N. Consider an element of length dx at the twisted end. It is moved by an</p><p>angle as shown. A vertical section is also shown and the twisting of the par-</p><p>allelopipe of length l and area under the action of the twisting couple can¢r dx</p><p>w</p><p>w</p><p>r + ¢r</p><p>l = y (x = l ) =</p><p>9rbl</p><p>5</p><p>5Eh2</p><p>c2 = 0 because y = 0 for x = 0</p><p>y =</p><p>4rb</p><p>Eh2</p><p>a l</p><p>3x</p><p>2</p><p>2</p><p>-</p><p>x</p><p>5</p><p>20</p><p>b + c2</p><p>dy</p><p>dx</p><p>= 0, for x = 0, c1 = 0. Integrating again,</p><p>dy</p><p>dx</p><p>=</p><p>4rb</p><p>Eh2</p><p>a l</p><p>3x -</p><p>x</p><p>4</p><p>4</p><p>b + c1</p><p>d</p><p>2y</p><p>dx</p><p>2</p><p>=</p><p>4rb</p><p>Eh2</p><p>(l</p><p>3 - x</p><p>3)</p><p>Integrating</p><p>I = 3</p><p>+h>2</p><p>-h>2 z</p><p>2Idz =</p><p>lh3</p><p>12</p><p>EI</p><p>d</p><p>2y</p><p>dx</p><p>2</p><p>=</p><p>1</p><p>3</p><p>rlh b(l</p><p>3 - x</p><p>3)</p><p>N = rlhb3</p><p>x</p><p>l</p><p>j2dj =</p><p>1</p><p>3</p><p>rlh b (l</p><p>3 - x</p><p>3)</p><p>a = jbj</p><p>( )</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 181</p><p>be discussed by elementary means. If f is the tangential force generated, then</p><p>shearing stress is dx and this must equal</p><p>Hence,</p><p>The force f has moment fr about the axis and so the total moment is</p><p>(b) For a solid cylinder we must integrate over r. Thus</p><p>N = 3</p><p>r</p><p>0</p><p>2pr</p><p>3 dr</p><p>G</p><p>l</p><p>=</p><p>pr</p><p>4 G</p><p>2l</p><p>N = G¢r</p><p>l</p><p>r</p><p>2 3dx =</p><p>2pr</p><p>3¢r</p><p>l</p><p>G</p><p>f = G¢r dx</p><p>r</p><p>l</p><p>G u = G</p><p>r</p><p>l</p><p>, since u =</p><p>r</p><p>l</p><p>ƒ/¢r</p><p>dx</p><p>dr</p><p>f</p><p>F</p><p>dx</p><p>q</p><p>1.306 Clearly</p><p>Using</p><p>= 0.5033 * 103 N # m L 0.5 kNm</p><p>N =</p><p>p * 8.1 * p</p><p>32 * 3 * 90</p><p>(625 - 81) * 102 Nm</p><p>= 2.0° =</p><p>p</p><p>90</p><p>radians, l = 3 m</p><p>d2 = 5 * 10-2 m, d1 = 3 * 10-2 m</p><p>G = 81 GPa = 8.1 * 1010 Nm-2</p><p>N = 3</p><p>d2>2</p><p>d1>2</p><p>2pr</p><p>3 dr</p><p>G</p><p>l</p><p>=</p><p>p</p><p>32l</p><p>G</p><p>(d 2</p><p>4 - d 1</p><p>4)</p><p>182 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.307 The maximum power that can be transmitted by means of a shaft rotating about its</p><p>axis is clearly where N is moment of the couple producing the maximum per-</p><p>missible torsion, . Thus,</p><p>1.308 Consider an elementary ring of width dr at a distant r from the axis. The part outside</p><p>exerts couple on this ring while the part inside exerts a couple N in the</p><p>opposite direction. We have for equilibrium</p><p>where dI is the moment of inertia of the elementary ring, is the angular accelera-</p><p>tion and minus sign is needed because the couple N(r) decreases, with distance,</p><p>vanishing at the outer radius, . Now</p><p>Thus,</p><p>On integration, N =</p><p>1</p><p>2</p><p>mb</p><p>r</p><p>2</p><p>2 - r</p><p>2</p><p>1</p><p>r</p><p>4</p><p>2 - r</p><p>4</p><p>1</p><p>dN =</p><p>2mb</p><p>r</p><p>2</p><p>2 - r</p><p>2</p><p>1</p><p>r</p><p>3 dr</p><p>dI =</p><p>m</p><p>p r 2</p><p>2 - r</p><p>2</p><p>1</p><p>2pr dr r</p><p>2</p><p>N (r2) = 0</p><p>b</p><p>dN</p><p>dr</p><p>dr = -dIb</p><p>N +</p><p>dN</p><p>dr</p><p>dr</p><p>P =</p><p>pr</p><p>4G</p><p>2l</p><p># � = 16.9 kW</p><p>N�</p><p>dr</p><p>r</p><p>N + dN</p><p>dr</p><p>dr</p><p>1.309 We assume that the deformation is wholly due to external load, neglecting the effect</p><p>of the weight of the rod (see next problem). Then a well known formula says, elas-</p><p>tic energy per unit volume</p><p>This gives kJ for the total deformation energy.</p><p>1</p><p>2</p><p>m</p><p>r</p><p>Ee2 L 0.04</p><p>=</p><p>1</p><p>2</p><p>stress * strain =</p><p>1</p><p>2</p><p>se</p><p>(</p><p>(</p><p>(</p><p>)</p><p>)</p><p>)</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 183</p><p>1.310 When a rod is deformed by its own weight the stress increases as one moves up, the</p><p>stretching force being the weight of the portion below the element considered.</p><p>The stress on the element dx is</p><p>The extension of the element is</p><p>Integrating we get the extension of the whole rod as</p><p>The elastic energy of the element is</p><p>Integrating</p><p>1.311 The work done to make a loop out of a steel band appears as the elastic energy of</p><p>the loop and may be calculated from the same.</p><p>If the length of the band is the radius of the loop Now consider an element</p><p>ABCD of the loop. The elastic energy of this element can be calculated by the same</p><p>sort of arguments as used to derive the formula for internal bending moment.</p><p>Consider a fiber at a distance z from the neutral surface PQ. This fiber experiences</p><p>a force p and undergoes an extension ds where while</p><p>Thus strain</p><p>If is the cross sectional area of the fiber, the elastic energy associated with it is</p><p>Summing over all the fibres we get</p><p>For the whole loop this gives, using</p><p>El p</p><p>R</p><p>=</p><p>2EI p2</p><p>l</p><p>3d w = 2 p</p><p>EI</p><p>2R</p><p>aaZ</p><p>2 =</p><p>E ld</p><p>2R</p><p>1</p><p>2</p><p>E aZ</p><p>R</p><p>b2</p><p>Rd</p><p>a</p><p>a</p><p>ds /s = Z /R #</p><p>PQ = s = Rd</p><p>.d = Zd</p><p>,</p><p>R = l>2p.l ,</p><p>¢U =</p><p>1</p><p>6</p><p>pr</p><p>2r2g2l</p><p>3</p><p>E</p><p>=</p><p>2</p><p>3</p><p>pr</p><p>2lE a¢l</p><p>l</p><p>b2</p><p>1</p><p>2</p><p>rg (l - x)</p><p>rg ( l - x)</p><p>E</p><p>pr</p><p>2dx</p><p>¢l =</p><p>1</p><p>2</p><p>rgl</p><p>2</p><p>E</p><p>¢dx = d¢x = rg(l - x) dx>E</p><p>rpr</p><p>2 (l - x) g>pr</p><p>2 = rg (l - x) x</p><p>dx</p><p>l–x</p><p>x</p><p>dx</p><p>l–x</p><p>R</p><p>R</p><p>P</p><p>A D</p><p>pp</p><p>B C</p><p>d</p><p>Q</p><p>ds</p><p>184 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Now,</p><p>So the energy is</p><p>Alternate:</p><p>Suppose that the steel band was made into a hoop of radius R, then length of the</p><p>hoop</p><p>Consider an infinitesimally thin section of radius and thickness in the hoop. The</p><p>length of this section of the hoop is . Hence the longitudinal strain correspon-</p><p>ding to this section is</p><p>So elastic energy density is</p><p>Thus the sought work done is nothing but elastic energy stored.</p><p>(on integrating)</p><p>1.312 A rod should be treated as a solid cylinder. Let the upper end of the cylindrical rod be</p><p>fixed and let a couple be applied to the lower end in a plane perpendicular to its length</p><p>(with its axis coinciding with that of the cylinder) twisting it through an angle (radi-</p><p>ans). This is an example of “pure shear”. In equilibrium position the twisting couple is</p><p>equal and opposite to the restoring couple. Let us calculate the value of this couple.</p><p>Imagine the cylinder to consist of a large number of co-axial, hollow cylinders, and con-</p><p>sider one such hollow cylinder of radius and radial thickness Fig. (a). Each ra-</p><p>dius of the lower end is turned through the same angle , but the displacement is the</p><p>greatest at the rim, decreasing as the centre is approached, where it is reduced to zero.</p><p>Or</p><p>B O</p><p>l</p><p>(a) (b)</p><p>B'</p><p>O'</p><p>B'</p><p>B</p><p>A</p><p>dr</p><p>r</p><p>r q</p><p>f</p><p>u</p><p>drv,</p><p>u</p><p>U = 1udV = L</p><p>R + d2</p><p>R - d2</p><p>1</p><p>2</p><p>E a x</p><p>R</p><p>- 1b2</p><p>2prdr h =</p><p>1 p2Ehd3</p><p>l</p><p>u =</p><p>1</p><p>2</p><p>E e2 =</p><p>1</p><p>2</p><p>E a r</p><p>R</p><p>- 1b2</p><p>e =</p><p>2pr - 2rR</p><p>2pR</p><p>=</p><p>r</p><p>R</p><p>- 1</p><p>2pr</p><p>drr</p><p>l = 2pR.</p><p>1</p><p>6</p><p>p2Ehd3</p><p>l</p><p>= 0.08 kJ</p><p>I = 3</p><p>d>2</p><p>-d>2z</p><p>2h dZ =</p><p>hd3</p><p>12</p><p>1.6 ELASTIC DEFORMATIONS OF A SOLID BODY 185</p><p>Let AB Fig. (b) be a line, parallel to the axis, before the cylinder is twisted. On twist-</p><p>ing, since the point B shifts to , the line AB takes up the position . The angle</p><p>through which this hollow cylinder is sheared is, therefore, , say. Then</p><p>clearly,</p><p>(see Figs. a and b)</p><p>so, angle of shear or shear strain</p><p>Obviously, will have the maximum value where is the greatest, i.e., the maxi-</p><p>mum strain is on the outermost part of the cylinder, and the least, on the innermost.</p><p>In other words, the shearing stress is not uniform all through.</p><p>Thus, although the angle of shear is the same for any one hollow cylinder, it is dif-</p><p>ferent for different cylinders, being the greatest for the outermost and the least for</p><p>the innermost one.</p><p>Since</p><p>We have</p><p>Now, face area of this hollow cylinder</p><p>Therefore total shearing force this on area</p><p>Therefore, moment of this force about the axis Fig. (b) of the cylinder is</p><p>equal to</p><p>Integrating this expression between the limits, and , we have total twi-</p><p>sting couple on the cylinder,</p><p>=</p><p>2pGu</p><p>l L</p><p>r</p><p>0</p><p>r3dr =</p><p>2pG u</p><p>l</p><p>cr4</p><p>4</p><p>d r</p><p>0</p><p>=</p><p>2pGur 4</p><p>2l</p><p>=</p><p>pGur 4</p><p>2l</p><p>= Cu</p><p>N (u) = 1</p><p>r</p><p>0</p><p>2pG 2p G</p><p>u</p><p>l</p><p>r3dr</p><p>r = rr = 0</p><p>2pG ur2dr r</p><p>l</p><p>=</p><p>2pG ur3dr</p><p>l</p><p>OO ¿</p><p>= 2pr dr *</p><p>Gru</p><p>t</p><p>= 2pG</p><p>u</p><p>l</p><p>r2dr</p><p>= 2pr dr</p><p>F = Gf =</p><p>Gru</p><p>l</p><p>G =</p><p>shearing stress</p><p>strain or angle of shear</p><p>=</p><p>F</p><p>f</p><p>rf</p><p>f =</p><p>ru</p><p>l</p><p>Also BB¿ = ru,</p><p>BB¿ = lf</p><p>BAB ¿ = f</p><p>AB ¿B ¿</p><p>.</p><p>186 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>where is twisting couple per unit twist of the cylinder or torsional rigid-</p><p>ity of the cylinder.</p><p>Thus, when the rod is twisted through an angle</p><p>, a couple</p><p>appears to resist this. Work done in twisting the rod by an angle or elastic ener-</p><p>gy stored is then</p><p>(on substituting values)</p><p>1.313 The energy between radii r and is given by differentiation of</p><p>(see solution of problem 1.312)</p><p>So,</p><p>Its density is</p><p>Alternate:</p><p>The sought energy density</p><p>But, (see solution of problem 1.312)</p><p>So,</p><p>1.314 The energy density is as usual Stress is the pressure . Strain</p><p>is by definition of . Thus</p><p>(on substituting values)u =</p><p>1</p><p>2</p><p>b (rgh)2 = 23.5 kJ/m3</p><p>bb * rgh</p><p>rgh1>2 stress * stress.</p><p>u =</p><p>1</p><p>2</p><p>G</p><p>2r</p><p>2</p><p>l</p><p>2</p><p>f = ar</p><p>l</p><p>b</p><p>u =</p><p>1</p><p>2</p><p>Gf2 (where f is angle of shear or shear strain)</p><p>u =</p><p>dU</p><p>dV</p><p>=</p><p>pr</p><p>3dr</p><p>2prdr l</p><p>G</p><p>2</p><p>l</p><p>=</p><p>1</p><p>2</p><p>G</p><p>2r</p><p>2</p><p>l</p><p>2</p><p>dU =</p><p>pr</p><p>3dr</p><p>l</p><p>G</p><p>2</p><p>U =</p><p>pr</p><p>4G</p><p>4l</p><p>2</p><p>r + dr</p><p>U = L</p><p>0</p><p>N (u) d(u) =</p><p>pr4G</p><p>4l</p><p>2 = 7 J</p><p>N (u) =</p><p>pr 4G</p><p>2l</p><p>u</p><p>u</p><p>C = pGr</p><p>4>2l</p><p>1.7 HYDRODYNAMICS 187</p><p>1.7 Hydrodynamics</p><p>1.315 Between points 1 and 2, fluid particles are in nearly circular motion and therefore</p><p>have centripetal acceleration. The force for this acceleration, like for any other</p><p>situation in an ideal fluid, can only come from the pressure variation along the line</p><p>joining 1 and 2. This requires that pressure at 1 should be greater than the pressure</p><p>at 2 so that the fluid particles can have required acceleration. If there is no turbu-</p><p>lence, the motion can be taken as irrotational. Then by considering</p><p>along the circuit shown, we infer that</p><p>The portion of the circuit near 1 and 2 are streamlines while the other two arms are</p><p>at right angle to streamlines.</p><p>(By electrostatic analogy, the density of streamlines is proportional to the velocity.)</p><p>1.316 From the conservation of mass</p><p>(1)</p><p>But as shown in the figure of the problem book, therefore</p><p>As every streamline is horizontal between 1 and 2, Bernoulli’s theorem becomes</p><p>As the difference in height of the water column is , therefore</p><p>(2)</p><p>From Bernoulli’s theorem between points 1 and 2 of a streamline</p><p>or p2 - p1 =</p><p>1</p><p>2</p><p>r (v</p><p>2</p><p>1 - v</p><p>2</p><p>2)</p><p>p1 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>1 = p2 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>2</p><p>p2 - p1 = rg¢h</p><p>¢h</p><p>p1 6 p2 as v1 7 v2</p><p>p +</p><p>1</p><p>2</p><p>rv</p><p>2 = constant</p><p>which gives</p><p>v1 7 v2</p><p>S1 6 S2</p><p>v1S1 = v2S2</p><p>v2 7 v1</p><p>Cv # d l = 0</p><p>2</p><p>1</p><p>.</p><p>188 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>or (using Eq. 2) (3)</p><p>Using Eqs. (1) in (3), we get</p><p>Hence the sought volume of water flowing per second</p><p>1.317 Applying Bernoulli’s theorem for points A and B,</p><p>or</p><p>So,</p><p>Thus, rate of flow of gas,</p><p>The gas flows over the tube past it at B. But at A the gas becomes stationary as the</p><p>gas will move into the tube which already contains gas.</p><p>In applying Bernoulli’s theorem we should remember that</p><p>is constant along a streamline. In the present case, we are really applying Bernoulli’s</p><p>theorem somewhat indirectly. The streamline at A is not the streamline at B.</p><p>Nevertheless the result is correct. To be convinced of this, we need only apply</p><p>Bernoulli’s theorem to the streamline that goes through A by comparing the situa-</p><p>tion at A with that above B on the same level. In steady conditions, this agrees with</p><p>the result derived because there cannot be a transverse pressure differential.</p><p>1.318 Since, the density of water is greater than that of kerosene oil,</p><p>it will collect at the bottom. Now, pressure due to water level</p><p>equals and pressure due to kerosene oil level equals</p><p>. So, net pressure becomes .</p><p>From Bernoulli’s theorem, this pressure energy will be con-</p><p>verted into kinetic energy while flowing through the hole A.</p><p>h1r1g + h2r2gh2r2g</p><p>h1r1g</p><p>p</p><p>r</p><p>+</p><p>1</p><p>2</p><p>v</p><p>2 + gz</p><p>Q = Sv = S C</p><p>2¢hr0g</p><p>r</p><p>v = C</p><p>2¢hr0g</p><p>r</p><p>1</p><p>2</p><p>rv</p><p>2 = pA - pB = ¢hr0</p><p>g</p><p>pA = pB +</p><p>1</p><p>2</p><p>rv</p><p>2 as vA = 0</p><p>Q = v1S1 = S1S2 A</p><p>2g¢h</p><p>S</p><p>2</p><p>2 - S</p><p>2</p><p>1</p><p>v1 = S2 A</p><p>2g¢h</p><p>S</p><p>2</p><p>2 - S</p><p>2</p><p>1</p><p>rg¢h =</p><p>1</p><p>2</p><p>r (v</p><p>2</p><p>1 - v</p><p>2</p><p>2)</p><p>A</p><p>B</p><p>h</p><p>h2</p><p>h1</p><p>A</p><p>Kerosene oil</p><p>Water</p><p>1.7 HYDRODYNAMICS 189</p><p>i.e.,</p><p>Hence,</p><p>1.319 Let H be the total height of the water column and the hole is made at a height h</p><p>from the bottom.</p><p>Then from Bernoulli’s theorem</p><p>or</p><p>which is directed horizontally.</p><p>For the horizontal range,</p><p>Now, for maximum l</p><p>which yields</p><p>1.320 Let the velocity of the water jet near the orifice be Then applying Bernoulli’s</p><p>theorem,</p><p>or (1)</p><p>Here the pressure term on both sides is the same and is equal to the atmospheric</p><p>pressure.</p><p>Now, if water rises up to a height h, then at this height, whole of its kinetic energy</p><p>will be converted into potential energy. So,</p><p>(using Eq. 1) =</p><p>v</p><p>2</p><p>2g</p><p>- h0 = 20 cm</p><p>1</p><p>2</p><p>rv¿2 = rgh or h =</p><p>v¿2</p><p>2g</p><p>v¿ = 2v</p><p>2 - 2gh0</p><p>1</p><p>2</p><p>rv</p><p>2 = h0rg +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>v ¿.</p><p>h =</p><p>H</p><p>2</p><p>= 25 cm</p><p>d (Hh - h2)</p><p>dh</p><p>= 0</p><p>= 2 2(Hh - h2)</p><p>= 22g (H - h) # A</p><p>2h</p><p>g</p><p>l = vt</p><p>v = 1(H - h) 2g</p><p>1</p><p>2</p><p>rv</p><p>2 = (H - h)rg</p><p>v = C2 ah1 + h2</p><p>r2</p><p>r1</p><p>bg = 3 m/s</p><p>h1r1g + h2r2g =</p><p>1</p><p>2</p><p>r1v</p><p>2</p><p>H h</p><p>v</p><p>l</p><p>190 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.321 Water flows through the small clearance into the orifice. Let d be the clearance, then</p><p>from the equation of continuity</p><p>or (1)</p><p>where and are respectively the inward radial velocities of the fluid at 1, 2 and</p><p>3. Now by Bernoulli’s theorem just before 2 and just after it in the clearance</p><p>(2)</p><p>Applying the same theorem at 3 and 1, we find that this also equals</p><p>(3)</p><p>(since the pressure in the orifice is ).</p><p>From Eqs. (2) and (3) we also have</p><p>(4)</p><p>and</p><p>(using Eqs. 1 and 4)</p><p>1.322 For diagram of piston confirm that</p><p>(1)</p><p>as f is constant and is constant, p is also constant.</p><p>From Bernouli’s equation from just inside and outside points of the orifice we get</p><p>or (2)</p><p>If l is the length of liquid column at an arbitrary instant</p><p>of time. Then from mass conservation</p><p>(3)</p><p>(where S is the piston area).</p><p>a- dl</p><p>dt</p><p>b S = vefflux</p><p>(p - p0) =</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>efflux</p><p>p = p0 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>efflux</p><p>p0</p><p>ƒ = (p - p0) S</p><p>= p0 + hrg a1 - aR1</p><p>r</p><p>b2b</p><p>p = p0 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>1 a1 - a v</p><p>v1</p><p>b2b v1 = 12gh</p><p>p0</p><p>p +</p><p>1</p><p>2</p><p>rv</p><p>2 = p0 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>1</p><p>p0 + hrg = p2 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>2</p><p>vv1,v2</p><p>v1R1 = vr = v2R2</p><p>(2pR1d)v1 = (2prd) v = (2pR2d) v2</p><p>R2</p><p>R1</p><p>r</p><p>ho</p><p>23 1</p><p>B</p><p>S</p><p>V</p><p>l</p><p>A</p><p>1.7 HYDRODYNAMICS 191</p><p>From Eq. (2) for constant p, we have .</p><p>So,</p><p>(where V and have meaning in accordance with the piston).</p><p>So, (4)</p><p>Now the differential work done by the constant for is given by</p><p>As both p and are constants, so, total work</p><p>(using Eqs. 2 and 3)</p><p>(using Eq. 4)</p><p>1.323 Water jet coming out from the orifice of area s at an arbitrary instant of time when</p><p>the height of water column in the cylindrical vessel is</p><p>As the rate of out-flux should be equal to the rate with which volume of the water</p><p>in the vessel decreases</p><p>Hence, t = -</p><p>S</p><p>s12h</p><p>3</p><p>0</p><p>h</p><p>dh</p><p>2H</p><p>=</p><p>S</p><p>sA</p><p>2h</p><p>g</p><p>-S</p><p>dH</p><p>dt</p><p>= s 22gH</p><p>vefflux = 22gH</p><p>H.</p><p>=</p><p>1</p><p>2</p><p>r</p><p>V</p><p>3</p><p>S2t 2</p><p>=</p><p>1</p><p>2</p><p>ra V</p><p>St</p><p>b3</p><p>St</p><p>=</p><p>1</p><p>2</p><p>rv</p><p>3</p><p>efflux St</p><p>= a 1</p><p>2</p><p>rv</p><p>2</p><p>effluxb S a veff s</p><p>S</p><p>b t</p><p>W = 3dA = (p - p0) S a -</p><p>dl</p><p>dt</p><p>b t</p><p>(-dl /dt)</p><p>dA = F a -</p><p>dl</p><p>dt</p><p>b dt = (p - p0) S a -dl</p><p>dt</p><p>b dt</p><p>F = (p - p0) S</p><p>vefflux =</p><p>V</p><p>St</p><p>+</p><p>V = vefflux</p><p>St</p><p>vefflux = constant also</p><p>ve</p><p>192 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.324 In a rotating frame (with constant angular velocity), the Eulerian equation is</p><p>In the frame of rotating tube the liquid in the “column” is practically static because</p><p>the orifice is sufficiently small. Thus the Eulerian equation in projection form along</p><p>r (which is the position vector of an arbitrary liquid element of length dr relative to</p><p>the rotation axis) reduces to</p><p>or</p><p>So,</p><p>Thus, (1)</p><p>Hence the pressure at the end B just before the orifice is</p><p>(2)</p><p>Then applying Bernoulli’s theorem at the orifice for the points just inside and out-</p><p>side of the end B</p><p>So,</p><p>1.325 The Euler’s equation is</p><p>(where z is vertically upwards)</p><p>Now, (1)</p><p>But (2)(v # ¥) v = ¥a1</p><p>2</p><p>v</p><p>2b - v * Curl v</p><p>d v</p><p>dt</p><p>=</p><p>0v</p><p>0t</p><p>+ (v # ¥) v</p><p>r</p><p>d v</p><p>dt</p><p>= f - ¥p = - ¥(p + rgz)</p><p>v = vh A</p><p>2l</p><p>h</p><p>- 1</p><p>p0 +</p><p>1</p><p>2</p><p>rv2 (2lh - h2) = p0 +</p><p>1</p><p>2</p><p>rv</p><p>2 (where v is the sought velocity)</p><p>p (l ) = p0 +</p><p>rv2</p><p>2</p><p>(2lh - h 2)</p><p>p (r) = p0 +</p><p>rv2</p><p>2</p><p>[r 2 - ( l - h) 2]</p><p>3</p><p>p</p><p>p0</p><p>dp = rv2</p><p>3</p><p>r</p><p>(l-h)</p><p>rdr</p><p>dp = rv2rdr</p><p>-dp</p><p>dr</p><p>+ rv2r = 0</p><p>- §p + rg + 2r(v¿</p><p>* �) + rv2 r = r</p><p>d v¿</p><p>dt</p><p>h</p><p>drr</p><p>l</p><p>BA</p><p>O</p><p>O</p><p>1.7 HYDRODYNAMICS 193</p><p>If we consider the steady (i.e. ) flow of an incompressible fluid then</p><p>and as the motion in irrotational Curl</p><p>So from Eqs. (1) and (2)</p><p>or</p><p>1.326 Let the velocity of water flowing through A be and that through B be then dis-</p><p>charging rate through and similarly through .</p><p>Now, force of reaction at A is</p><p>Hence, the net force is</p><p>(1)</p><p>Applying Bernoulli’s theorem to the liquid flowing out of A we get</p><p>And similarly at B</p><p>Hence,</p><p>Thus,</p><p>1.327 Consider an element of height at a distance y from the top. The velocity of the</p><p>fluid coming out of the element is</p><p>The force of reaction due to this is as in the previous</p><p>problem.</p><p>dF = rdAv</p><p>2 = r(bdy) 2gy,dF</p><p>v = 22gy</p><p>dy</p><p>F = 2rgS ¢h = 0.50 N</p><p>1v</p><p>2</p><p>B - v</p><p>2</p><p>A2</p><p>r</p><p>2</p><p>= ¢hrg</p><p>r0 + rg (h + ¢h) = r0 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>B</p><p>r0 + rgh = r0 +</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>A</p><p>F = rS 1v</p><p>2</p><p>A - v</p><p>2</p><p>B2 as FA c TFB</p><p>FA = rQAvA = rSv</p><p>2</p><p>B</p><p>B = SvBA = QA = SvA</p><p>vB,vA</p><p>p +</p><p>1</p><p>2</p><p>rv</p><p>2 + rgz = constant</p><p>§ap +</p><p>1</p><p>2</p><p>rv</p><p>2 + rgzb = 0</p><p>r§a 1</p><p>2</p><p>v2b = - § (p + rgz)</p><p>v = 0r = constant</p><p>0 v>0t = 0</p><p>h</p><p>B</p><p>S</p><p>h</p><p>S</p><p>A</p><p>194 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Integrating</p><p>(The slit runs from a depth to a depth h from the top.)</p><p>1.328 Let the velocity of water flowing through the tube at a certain instant of time be u,</p><p>then where Q is the rate of flow of water and is the cross section</p><p>area of the tube.</p><p>From impulse momentum theorem, for the stream of</p><p>water striking the tube corner, in x-direction in the time</p><p>interval ,</p><p>Similarly,</p><p>Therefore, the force exerted on the water stream by the</p><p>tube,</p><p>According to third law, the reaction force on the tube’s wall by the stream equals</p><p>Hence, the sought moment of force about O becomes</p><p>and</p><p>1.329 Let us take an arbitrary cross section of radius of the narrowing tube. If is the in-</p><p>side pressure at the location of taken ring element of radius and width and is</p><p>outside atmosphere pressure, from symmetry of the problem, net force exerted on the</p><p>taken ring element due to the inside and outside pressure difference.</p><p>Hence the sought net force on the tube</p><p>(1)F = Fx = LdF sin u = L(p - p0) 2prdr</p><p>= (p - p0) 2p rdr (because dl sin u = dr)</p><p>= dF sin u = (p - p0) 2prdl sin u</p><p>p0dl,r</p><p>pr</p><p>| N | =</p><p>rQ 2l</p><p>pr 2</p><p>= 0.70 Nm</p><p>N = - li * (rQui - Qu j) = rQul k =</p><p>rQ 2</p><p>pr2</p><p>l k</p><p>rQu i - rQu j</p><p>(-F)</p><p>F = - rQ u i + rQu j</p><p>Fy = rQu</p><p>Fxdt = -rQ udt or Fx = -rQu</p><p>dt</p><p>pr</p><p>2u = Q >pr</p><p>2,</p><p>h - l</p><p>= 5 N (on substituting values)</p><p>= rgb [h2 - (h - l)2] = r gbl (2h - l )</p><p>F = r gb 3</p><p>h</p><p>h-1</p><p>2ydy</p><p>l</p><p>O</p><p>x</p><p>y</p><p>1.7 HYDRODYNAMICS 195</p><p>From Bernoulli’s equation</p><p>,</p><p>where (2)</p><p>and v is the speed of water at the location of taken ring element.</p><p>From conservation of mass</p><p>or (3)</p><p>(where is the radius at open end).</p><p>Using Eq. (3) in Eq. (2), we get</p><p>(4)</p><p>Using Eq. (4) in Eq. (1), we get</p><p>(5)</p><p>After integration and using , we get</p><p>(on substituting the values)</p><p>Note: If we try to calculate F from the momentum change of the liquid flowing out we will</p><p>be wrong even as regards the sign of the force. There is of course the effect of pressure</p><p>at S and s but quantitative derivation of F from Newton’s law is difficult.</p><p>1.330 The Euler’s equation is</p><p>in the space fixed frame where downward. We assume incompressible</p><p>fluid so is constant.</p><p>Then where z is the height vertically upwards from some fixed origin.</p><p>We go to the rotating frame where the equation becomes</p><p>r</p><p>d v</p><p>dt</p><p>= - § (p + rgz) + rv2r + 2r (v * �)</p><p>f = - § (rgz)</p><p>r</p><p>f = -rgk</p><p>r</p><p>d v</p><p>dt</p><p>= f - §p</p><p>F =</p><p>rgh (S - s)2</p><p>S</p><p>= 6 N</p><p>pr2</p><p>2 = S and pr1</p><p>2 = s</p><p>F = 3</p><p>r2</p><p>r1</p><p>rgh a1 -</p><p>r1</p><p>2</p><p>r</p><p>4</p><p>b 2prdr</p><p>=</p><p>1</p><p>2</p><p>rv</p><p>2</p><p>efflux a1 -</p><p>r1</p><p>2</p><p>r4</p><p>b = rgh a1 -</p><p>r1</p><p>2</p><p>r4</p><p>b</p><p>p - p0 =</p><p>1</p><p>2</p><p>r av</p><p>2</p><p>efflux - v</p><p>2</p><p>efflux</p><p>r 1</p><p>4</p><p>r</p><p>4</p><p>b</p><p>r1</p><p>v = vefflux</p><p>r1</p><p>2</p><p>r2</p><p>rvpr2 = rvefflux pr1</p><p>2</p><p>vefflux = 22gh</p><p>p - p0 =</p><p>1</p><p>2</p><p>r (v</p><p>2</p><p>efflux - v</p><p>2)</p><p>S</p><p>r</p><p>dF</p><p>dF</p><p>s</p><p>196 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>The additional terms on the right are the well known coriolis and centrifugal forces.</p><p>In the frame rotating with the liquid , so</p><p>or</p><p>On the free surface ; thus</p><p>If we choose the origin at point (i.e. , the axis) of the free surface then</p><p>“constant” and</p><p>(the parabolic of revolution)</p><p>At the bottom</p><p>So,</p><p>If on the axis at the bottom, then</p><p>1.331 When the disk rotates the fluid in contact with it co-rotates, but the fluid in contact</p><p>with the walls of the cavity does not rotate. A velocity gradient is then set up lead-</p><p>ing to viscous forces.</p><p>At a distance r from the axis, the linear velocity is so there is a velocity gradient</p><p>both in the upper and lower clearance. The corresponding force on the element</p><p>whose radial width is is</p><p>The torque due to this force is</p><p>and the net torque considering both the upper and lower clearance is</p><p>23</p><p>R</p><p>0</p><p>h</p><p>2pr</p><p>3 dr</p><p>v</p><p>h</p><p>= pR</p><p>4h</p><p>v</p><p>h</p><p>h2prdr</p><p>vr</p><p>h</p><p>r</p><p>h2prdr</p><p>vr</p><p>h</p><p>a from the formula F = h A</p><p>dv</p><p>dx</p><p>b</p><p>dr</p><p>vr>h vr</p><p>p = p0 +</p><p>1</p><p>2</p><p>rv2r</p><p>2</p><p>p = p0</p><p>p =</p><p>1</p><p>2</p><p>rv2r</p><p>2 + constant</p><p>z = constant.</p><p>z =</p><p>v2</p><p>2g</p><p>r</p><p>2</p><p>= 0</p><p>r = 0</p><p>z =</p><p>v2</p><p>2g</p><p>r</p><p>2 + constant</p><p>p = constant</p><p>p + rgz -</p><p>1</p><p>2</p><p>rv2r</p><p>2 = constant</p><p>§ap + rgz -</p><p>1</p><p>2</p><p>rv</p><p>2r</p><p>2b = 0</p><p>v = 0</p><p>1.7 HYDRODYNAMICS 197</p><p>So power developed is</p><p>(on substituting values)</p><p>(As instructed, end effects, i.e., rotation of fluid in the clearance , has been</p><p>neglected.)</p><p>1.332 Let us consider a coaxial cylinder of radius r and thickness , then force of friction</p><p>or viscous force on this elemental layer is given by</p><p>This force must be constant from layer so that steady motion may be possible.</p><p>So, (1)</p><p>Integrating,</p><p>or (2)</p><p>Putting , we get</p><p>Dividing Eqs. (2) by (3), we get</p><p>Note: The force F is supplied by the agency which tries to carry the inner cylinder</p><p>with velocity .</p><p>1.333 (a) Let us consider an elemental cylinder of radius r and thickness then from</p><p>Newton’s formula</p><p>(1)</p><p>and moment of this force acting on the element,</p><p>or (2)2pl hdv = N</p><p>dr</p><p>r</p><p>3</p><p>N = 2pr</p><p>2l h</p><p>dv</p><p>dr</p><p>r = 2pr</p><p>3l h</p><p>dv</p><p>dr</p><p>F = 2prl hr</p><p>dv</p><p>dr</p><p>= 2pl hr</p><p>2</p><p>dv</p><p>dr</p><p>dr</p><p>v0</p><p>v = v0</p><p>ln r /R2</p><p>ln R1/R2</p><p>F ln</p><p>R1</p><p>R2</p><p>= 2pl hv0</p><p>r = R1</p><p>F lna r</p><p>R2</p><p>b = 2pl hv</p><p>F3</p><p>r</p><p>R2</p><p>dr</p><p>r</p><p>= 2pl h3</p><p>v</p><p>0</p><p>dv</p><p>Fdr</p><p>r</p><p>= 2pl h3</p><p>v</p><p>0</p><p>dv</p><p>F = 2prlh</p><p>dv</p><p>dr</p><p>dr</p><p>r Ú R</p><p>P =</p><p>pR</p><p>4v2h</p><p>h</p><p>= 9.05 W</p><p>dr</p><p>rR2 R1</p><p>dr</p><p>r</p><p>R2</p><p>R1</p><p>2</p><p>(3)</p><p>198 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>As in the previous problem N is constant when conditions are steady</p><p>Integrating,</p><p>or (3)</p><p>Putting , we get</p><p>(4)</p><p>From Eqs. (3) and (4)</p><p>(b) From Eq. (4)</p><p>1.334 (a) Let be the volume flowing per second through the cylindrical shell of thick-</p><p>ness dr then,</p><p>dV</p><p>N1 =</p><p>N</p><p>l</p><p>= 4 phv2</p><p>R</p><p>2</p><p>1 R</p><p>2</p><p>2</p><p>R</p><p>2</p><p>2 - R</p><p>2</p><p>1</p><p>v = v2</p><p>R</p><p>2</p><p>1 R</p><p>2</p><p>2</p><p>R</p><p>2</p><p>2 - R</p><p>2</p><p>1</p><p>c 1</p><p>R</p><p>2</p><p>1</p><p>-</p><p>1</p><p>r</p><p>2</p><p>d</p><p>2pl hv2 =</p><p>N</p><p>2</p><p>c 1</p><p>R</p><p>2</p><p>1</p><p>-</p><p>1</p><p>R</p><p>2</p><p>2</p><p>dr = R2, v = v2</p><p>2pl hv =</p><p>N</p><p>2</p><p>c 1</p><p>R</p><p>2</p><p>1</p><p>-</p><p>1</p><p>r</p><p>2</p><p>d</p><p>2pl h 3</p><p>v</p><p>0</p><p>dv = N 3</p><p>r</p><p>R1</p><p>dr</p><p>r</p><p>3</p><p>dr</p><p>r</p><p>2</p><p>R</p><p>r</p><p>v</p><p>The total volume is</p><p>(b) Let be the kinetic energy within the above cylindrical shell. Then,</p><p>=</p><p>1</p><p>2</p><p>(2plr)r dr v</p><p>2</p><p>0 a1 -</p><p>r</p><p>2</p><p>R</p><p>2</p><p>b2</p><p>= plrv</p><p>2</p><p>0 cr -</p><p>2r</p><p>3</p><p>R</p><p>2</p><p>+</p><p>r</p><p>5</p><p>R</p><p>4</p><p>d dr</p><p>dT =</p><p>1</p><p>2</p><p>(dm) v</p><p>2 =</p><p>1</p><p>2</p><p>(2prldrr) v</p><p>2</p><p>dE</p><p>V = 2pv0 3</p><p>R</p><p>0</p><p>ar -</p><p>r</p><p>3</p><p>R</p><p>2</p><p>b dr = 2pv0</p><p>R</p><p>2</p><p>4</p><p>=</p><p>p</p><p>2</p><p>R</p><p>2v0</p><p>dV = - (2prdr) v0 a1 -</p><p>r</p><p>2</p><p>R</p><p>2</p><p>bdr = 2pv0ar -</p><p>r</p><p>3</p><p>R</p><p>2</p><p>bdr</p><p>1.7 HYDRODYNAMICS 199</p><p>Hence, total energy of the fluid,</p><p>(c) Here frictional force is the shearing force on the tube, exerted by the fluid, which</p><p>equals .</p><p>Given,</p><p>So,</p><p>At</p><p>Then, viscous force is given by</p><p>(d) Taking a cylindrical shell of thickness and radius r, viscous force,</p><p>Let be the pressure difference, then net force on the element is</p><p>But, since the flow is steady,</p><p>or</p><p>1.335 The loss of pressure head in travelling a distance l is seen from the middle section to</p><p>be . Since in our problem and</p><p>, we see that pressure head of 5 cm remains in compensated and must</p><p>be converted into kinetic energy of the liquid flowing out. Thus,</p><p>Thus, v = 1 2g¢h � 1 m/s</p><p>rv</p><p>2</p><p>2</p><p>= rg¢h (where ¢h = h3 - h2 )</p><p>5 + h2 - h1</p><p>h3 - h2 = 15 cm =h2 - h1 = h1h2 - h1 = 10 cm</p><p>¢p =</p><p>-2plhr dv>dr</p><p>pr</p><p>2</p><p>=</p><p>-2plhr 1-2v0 r>R</p><p>22</p><p>pr</p><p>2</p><p>=</p><p>4hv0l</p><p>R</p><p>2</p><p>Fnet = 0</p><p>¢ppr</p><p>2 + 2phlr</p><p>dv</p><p>dr</p><p>¢p</p><p>F = -h(2prl )</p><p>dv</p><p>dr</p><p>dr</p><p>= -2pR hl a -</p><p>2v0</p><p>R</p><p>b = 4phv0l</p><p>F = -h (2pRl ) adv</p><p>dr</p><p>b</p><p>r = R</p><p>dv</p><p>dr</p><p>= -</p><p>2v0</p><p>R</p><p>r = R</p><p>dv</p><p>dr</p><p>= -2v0</p><p>r</p><p>R</p><p>2</p><p>v = v0 a1 -</p><p>r</p><p>2</p><p>R</p><p>2</p><p>b- hS dv>dt</p><p>T = plrv</p><p>2</p><p>03</p><p>R</p><p>0</p><p>ar -</p><p>2r</p><p>3</p><p>R</p><p>2</p><p>+</p><p>r</p><p>5</p><p>R</p><p>4</p><p>b dr =</p><p>pR</p><p>2rl v</p><p>2</p><p>0</p><p>6</p><p>,</p><p>200 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.336 We know that, Reynold’s number is defined as, , where v is the ve-</p><p>locity, l is the characteristic length, and the coefficient of viscosity. In the case of</p><p>circular cross-section the characteristic length is the diameter of cross-section d, and</p><p>v is taken as average velocity of flow of liquid.</p><p>Now, (Reynold’s number at from the pipe end) , where is the</p><p>velocity at distance</p><p>Similarly,</p><p>So,</p><p>From equation of continuity,</p><p>or</p><p>Thus,</p><p>1.337 We know that Reynold’s number for turbulent flow is greater than that for laminar</p><p>flow.</p><p>Now,</p><p>But,</p><p>So m/s (on substituting values)</p><p>1.338 We have</p><p>and v is given by</p><p>( density of lead, density of glycerine)</p><p>v =</p><p>2</p><p>9h</p><p>(r - r0)gr</p><p>2 =</p><p>1</p><p>18h</p><p>(r - r0) gd</p><p>2</p><p>r 0=r =</p><p>6phrv =</p><p>4p</p><p>3</p><p>r</p><p>2 (r - r0) g</p><p>R =</p><p>vr0d</p><p>h</p><p>v2min</p><p>=</p><p>r1v1r1 h2</p><p>r2r2h1</p><p>= 5 m</p><p>(Re</p><p>)t Ú (Re</p><p>)l</p><p>(Re</p><p>)l =</p><p>rvd</p><p>h</p><p>=</p><p>2r1v1r1</p><p>h1</p><p>and (Re)t =</p><p>2r2v2r2</p><p>h</p><p>Re2</p><p>Re1</p><p>= e</p><p>a</p><p>¢</p><p>x = 5</p><p>d1v1</p><p>d2v2</p><p>=</p><p>r2</p><p>r1</p><p>=</p><p>r0e</p><p>-ax2</p><p>r0e</p><p>-ax1</p><p>= e-a¢x (as x2 - x1 = ¢x)</p><p>pr 21v1 = pr</p><p>2</p><p>2 v2</p><p>or d1v1r1 = d2v2r2</p><p>A1v1 = A2v2</p><p>Re1</p><p>Re2</p><p>=</p><p>d1v1</p><p>d2v2</p><p>Re2</p><p>=</p><p>rd2v2</p><p>h</p><p>x1</p><p>v1=</p><p>rd1v1</p><p>mh</p><p>x1Re1</p><p>h</p><p>Re = rvl/h(Re)</p><p>where</p><p>.</p><p>1.7 HYDRODYNAMICS 201</p><p>Thus,</p><p>and mm (on substituting values)</p><p>1.339</p><p>or</p><p>or</p><p>or</p><p>or</p><p>Since</p><p>or,</p><p>Thus,</p><p>The steady state velocity is</p><p>v differs from by n where</p><p>or ln n</p><p>Thus,</p><p>(We have neglected buoyancy in olive oil.)</p><p>(on substituting values) = 0.20 s</p><p>t =</p><p>-rd</p><p>2</p><p>18 h</p><p>lnn</p><p>1</p><p>k</p><p>= -</p><p>4p>3 r</p><p>3P</p><p>6phr</p><p>= -</p><p>4r</p><p>2r</p><p>18h</p><p>= -</p><p>d</p><p>2r</p><p>18h</p><p>t =</p><p>1</p><p>k</p><p>e-kt = n</p><p>g /k</p><p>g /k</p><p>v =</p><p>g</p><p>k</p><p>(1 - e-kt)</p><p>C = -</p><p>g</p><p>k</p><p>v = 0 for t = =</p><p>g</p><p>k</p><p>+ C</p><p>ve</p><p>kt =</p><p>g</p><p>k</p><p>e</p><p>kt + C or v =</p><p>g</p><p>k</p><p>+ Ce-kt (where C is constant)</p><p>e</p><p>kt</p><p>dv</p><p>dt</p><p>+ ke</p><p>ktv = ge</p><p>kt or d</p><p>dt</p><p>e</p><p>kt v = ge</p><p>kt</p><p>dv</p><p>dt</p><p>+ kv = g where k =</p><p>6phr</p><p>m</p><p>dv</p><p>dt</p><p>+</p><p>6phr</p><p>m</p><p>v = g</p><p>m</p><p>dv</p><p>dt</p><p>= mg - 6phrv</p><p>d = [9h2/r0(r - r0)g]1/3 = 5.2</p><p>1</p><p>2</p><p>=</p><p>1</p><p>18h2</p><p>(r - r0) gr0d</p><p>3</p><p>0, so 0</p><p>.</p><p>( )</p><p>a a</p><p>,</p><p>202 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.8 Relativistic Mechanics</p><p>1.340 From the formula for length contraction</p><p>So, or</p><p>1.341 (a) In the frame in which the triangle is at rest, the space coordinates of the vertices</p><p>are (000), , all measured at the same time In</p><p>the moving frame the corresponding coordinates at time are</p><p>and</p><p>The perimeter P is then</p><p>(b) The coordinates in the first frame are shown at time t. The coordinates in the</p><p>moving frame are</p><p>The perimeter P is then</p><p>= a111 - b2 + 14 - b22 awhere b =</p><p>v</p><p>c</p><p>b</p><p>P = a11 - b2 +</p><p>a</p><p>2</p><p>[1 - b2 + 3]1>2 * 2</p><p>A : (vt¿, 0, 0), B : aa</p><p>2</p><p>11 - b2 + vt¿, a</p><p>13</p><p>2</p><p>, 0b , C : (a11 - b2 + vt¿, 0, 0)</p><p>A</p><p>C</p><p>B</p><p>A</p><p>B</p><p>C(0,0,0) (a,0,0)</p><p>a</p><p>2</p><p>‚a 3</p><p>2</p><p>‚0</p><p>P = a + 2aa3</p><p>4</p><p>(1 - b2) +</p><p>1</p><p>4</p><p>b1>2</p><p>= a C1 + 24 - 3b2 D</p><p>C : aa</p><p>2</p><p>3311 - b2 + vt¿, -</p><p>a</p><p>2</p><p>, 0b</p><p>A : (vt¿, 0, 0), B : aa</p><p>2</p><p>23 21 - b2 + vt¿,</p><p>a</p><p>2</p><p>, 0b</p><p>t ¿</p><p>t.aa</p><p>23</p><p>2</p><p>, +</p><p>a</p><p>2</p><p>, 0b aa</p><p>23</p><p>2</p><p>, -</p><p>a</p><p>2</p><p>, 0b</p><p>v = c 2h(2 - h)1 -</p><p>v2</p><p>c2</p><p>= (1 - h)2</p><p>a l0 - l0</p><p>A1 -</p><p>v2</p><p>c2</p><p>b = hl0</p><p>1.8 RELATIVISTIC MECHANICS 203</p><p>1.342 In the rest frame, the coordinates of the ends of the rod in terms of proper length</p><p>at time t. In the laboratory frame the coordinates at time are</p><p>Therefore, we can write</p><p>and</p><p>Hence,</p><p>or</p><p>(on substituting values)</p><p>1.343 In the frame K in which the cone is at rest, the coordinates of are (0, 0, 0) and of</p><p>are In the frame which is moving with velocity along the axis</p><p>of the cone, the coordinates of and at time are</p><p>Thus the taper angle in the frame is</p><p>or (on substituting values)</p><p>The lateral surface area is</p><p>(on substituting values)</p><p>Here, is the lateral surface area in the rest frame and</p><p>1.344 Because of time dilation, a moving clock reads less time. We write,</p><p>Thus, 1 -</p><p>2 ¢t</p><p>t</p><p>+ a¢t</p><p>t</p><p>b2</p><p>= 1 - b2</p><p>t - ¢t = t11 - b2 awhere b =</p><p>v</p><p>c</p><p>b</p><p>h¿ = h11 - b2 awhere b =</p><p>v</p><p>c</p><p>bS0 = ph2</p><p>sec u tan u</p><p>= 3.3 m2</p><p>= S011 - b2cos2 u</p><p>= ph2 (1 - b2)</p><p>tan u</p><p>11 - b2A1 +</p><p>tan2 u</p><p>1 - b2</p><p>S = ph¿2 sec u¿ tan u¿</p><p>u¿ = 59°tan u¿ =</p><p>tan u</p><p>11 - b2</p><p>= a y¿B - y ¿A</p><p>x ¿B - xA</p><p>bK¿</p><p>A : (-vt¿, 0, 0), B : (h - 11 - b2 - vt¿, h tan u, 0)</p><p>t¿BA</p><p>vK¿,(h, h tan u, 0).B</p><p>A</p><p>= 1.08 m</p><p>awhere b =</p><p>v</p><p>c</p><p>bl0 = A</p><p>1 - b2</p><p>sin2 u</p><p>1 - b2</p><p>l</p><p>2</p><p>0 = (l</p><p>2) a cos2 u + (1 - b2) sin2 u</p><p>1 - b2</p><p>b</p><p>l sin u = l0</p><p>sin u0l cos u0 = l0 cos u011 - b2</p><p>A : (vt¿, 0, 0), B : (l0 cos u011 - b2 + vt¿, l0sin u0, 0)</p><p>t¿</p><p>A : (0, 0, 0), B : (l0 cos u0, l0</p><p>sin u0, 0)</p><p>l0</p><p>B</p><p>A</p><p>or</p><p>(on substituting values)</p><p>In the frame K, the length l of the rod is related to the time of flight by</p><p>In the reference frame fixed to the rod (frame ) the proper length of the rod is</p><p>given by</p><p>But,</p><p>Thus,</p><p>So, or</p><p>and (on substituting values)</p><p>The distance traveled in the laboratory frame of reference is where is the</p><p>velocity of the particle. But by time dilation</p><p>So,</p><p>Thus the distance traversed is</p><p>(on substituting values)</p><p>(a) If is the proper life time of the mount, the life time in the moving frame is</p><p>and hence</p><p>Thus,</p><p>(on substituting values)</p><p>(b) The words “from the muon’s stand point” are not part of any standard</p><p>terminology.</p><p>= 1.4 ms</p><p>t0 =</p><p>l</p><p>v</p><p>11 - v2>c2</p><p>l =</p><p>vt0</p><p>11 - v</p><p>2>c</p><p>2</p><p>t0</p><p>11 - v</p><p>2>c</p><p>2</p><p>t0</p><p>= 5 m</p><p>s = c¢t11 - (¢t0>¢t)2</p><p>v = c 11 - (¢t0>¢t)2</p><p>¢t =</p><p>¢t0</p><p>11 - v</p><p>2>c</p><p>2</p><p>vv¢t</p><p>l0 = c 1(¢t¿)2 - (¢t)2 = c ¢t¿ A1 - a ¢t</p><p>¢t¿</p><p>b = 4.5 m</p><p>v = c A1 - a ¢t</p><p>¢t¿</p><p>b2</p><p>1 - b2 = a ¢t</p><p>¢t¿</p><p>b2</p><p>v ¢t¿ =</p><p>v ¢t</p><p>11 - b2</p><p>l0 =</p><p>l</p><p>11 - b2</p><p>=</p><p>v ¢t</p><p>11 - b2</p><p>awhere b =</p><p>v</p><p>c</p><p>bl0 = v¢t¿</p><p>l0k¿</p><p>l = v¢t</p><p>¢t</p><p>= 0.6 * 108 m/s</p><p>v = c A</p><p>¢t</p><p>t</p><p>a2 -</p><p>¢t</p><p>t</p><p>b</p><p>204 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.345</p><p>1.346</p><p>1.347</p><p>1.8 RELATIVISTIC MECHANICS 205</p><p>1.348 In the frame K in which the particles are at rest, their positions are A and B whose</p><p>coordinates may be taken as,</p><p>In the frame with respect to which K is moving with a</p><p>velocity the coordinates of A and B at time in</p><p>the moving frame are</p><p>Suppose B hits a stationary target in after time while A hits it after time</p><p>Then,</p><p>So,</p><p>(on substituting values)</p><p>1.349 In the reference frame fixed to the ruler the rod is moving with a velocity and</p><p>suffers Lorentz contraction. If is the proper length of the rod, its measured</p><p>length will be</p><p>In the reference frame fixed to the rod, the ruler suffers Lorentz contraction and we</p><p>must have</p><p>and</p><p>(on substituting values) v = c A1 -</p><p>¢x1</p><p>¢x2</p><p>= 2.2 * 108 m/s</p><p>¢x1 = l011 - b2 awhere b =</p><p>v</p><p>c</p><p>b</p><p>l0</p><p>v</p><p>= 17 m</p><p>l0 =</p><p>v¢t</p><p>11 - v</p><p>2>c2</p><p>l011 - b2 + vt¿B = v (t¿B + ¢t)</p><p>tB + ¢t.t¿BK¿</p><p>A = (vt¿, 0, 0), B = (l011 - b2 + vt¿, 0, 0) awhere b =</p><p>v</p><p>c</p><p>b</p><p>t¿v,</p><p>K¿</p><p>A : (0, 0, 0), B = (l0, 0, 0)</p><p>A B</p><p>K</p><p>¢x211 - b2 = l0</p><p>Thus, l0 =1¢x1¢x2 = 6.0 m (on substituting values)</p><p>1 - b2 =</p><p>¢x1</p><p>¢x2</p><p>or</p><p>1.350 The coordinates of the ends of the rods in the frame fixed to the left rod are shown</p><p>in the figure. The points B and D coincide when</p><p>The points A and E coincide when</p><p>Thus,</p><p>or</p><p>From this</p><p>1.351 In the rest frame of the particles, the events corresponding to the decay of the</p><p>particles are</p><p>In the reference frame K, the corresponding coordinates are by Lorentz transforma-</p><p>tion</p><p>Now, by Lorentz Fitzgerald contraction formula.</p><p>Thus the time lag of the decay time of B is</p><p>(on substituting values)</p><p>B decays later. (B is the forward particle in the direction of motion.)</p><p>= 20 ms</p><p>¢t =</p><p>vl0</p><p>c</p><p>211 - b2</p><p>=</p><p>vl</p><p>c</p><p>2(1 - b2)</p><p>=</p><p>vl</p><p>c</p><p>2 - v</p><p>2</p><p>l011 - b2 = l</p><p>A : (0, 0, 0, 0), B : a vl0</p><p>c211 - b2</p><p>,</p><p>l0</p><p>11 - b2</p><p>, 0, 0b</p><p>A : (0, 0, 0, 0) and B : (0, l0, 0, 0)</p><p>K0,</p><p>v =</p><p>2c2¢t>l0</p><p>1 + c2¢t2>l02 =</p><p>2l0>¢t</p><p>1 + (l0>c¢t)2</p><p>a v¢t</p><p>l0</p><p>- 1b2</p><p>= 1 - b2 = 1 -</p><p>v2</p><p>c2</p><p>¢t = t1 - t0 =</p><p>l0</p><p>v</p><p>11 + 11 - b22</p><p>t1 =</p><p>c1 + lo11 - b2</p><p>v</p><p>0 = c1 + l011 - b2 - vt1</p><p>l0 = c1 - vt0 or t0 =</p><p>c1 - l0</p><p>v</p><p>206 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>(c1+l0 1−β2−vt,0,0)</p><p>EA B D</p><p>(0,0,0) (l0,0,0) (c1–vt,0,0)</p><p>,</p><p>1.8 RELATIVISTIC MECHANICS 207</p><p>1.352 (a) In the reference frame K with respect to which the rod is moving with velocity</p><p>the coordinates of A and B are</p><p>Thus,</p><p>So,</p><p>(b)</p><p>Thus,</p><p>i.e.,</p><p>or</p><p>1.353 At the instant the picture is taken the coordinates of in the rest frame</p><p>of AB are</p><p>(a) In this frame the coordinates of at other times are is</p><p>opposite to at time t (B) � In the frame in which is at rest, the time</p><p>corresponding to this is given by Lorentz transformation.</p><p>(b) Similarly in the rest frame of A , B, the coordinates of A at other times are</p><p>A¿ : a t, - l0 21 - v</p><p>2/c</p><p>2 + vt, 0, 0b</p><p>t0 (B¿) =</p><p>1</p><p>21 - v</p><p>2/c</p><p>2</p><p>a l0</p><p>v</p><p>-</p><p>vl0</p><p>c</p><p>2</p><p>b =</p><p>l0</p><p>v</p><p>21 - v</p><p>2/c2</p><p>B¿, A¿l0</p><p>/v.B</p><p>B¿ : (t, vt, 0, 0). So B¿B¿</p><p>A¿: (0, or - l021 - v</p><p>2>c</p><p>2, 0, 0)</p><p>B¿: (0, 0, 0, 0)</p><p>B : 10, l0, 0, 02A : 10, 0, 0, 02</p><p>A, B, A¿, B¿</p><p>tB - tA =</p><p>l0</p><p>v 11 + 11 - v2/c22</p><p>tA - tB =</p><p>l0</p><p>v</p><p>A1 - 21 - v 2/c2</p><p>Bv 1tA - tB2 = 1�1 - 21 - v</p><p>2/c</p><p>2)l0</p><p>1Since xA - tB</p><p>can be either + l0 or - l0 .2�l0 - v 1tA - tB2 = l = l021 - v</p><p>2/c</p><p>2</p><p>l0 =</p><p>xA - xB - v 1tA - tB2</p><p>21 - v</p><p>2/c2</p><p>l = xA - xB - v 1tA - tB2 = l011 - �2</p><p>B : t, xB + v 1t - tB2, 0, 0</p><p>A : t, xA + v 1t - tA2, 0, 0</p><p>v,</p><p>AB</p><p>v</p><p>A' B'</p><p>A B</p><p>is opposite to at time</p><p>The corresponding time in the frame in which are rest is</p><p>1.354 By Lorentz transformation</p><p>So at time</p><p>If so we get the diagram given below “in terms of</p><p>the K-clocks”.</p><p>The situation in terms of the K clock is reversed.</p><p>1.355 Suppose is the locus of points in the frame at which the readings of the clocks</p><p>of both reference system are permanently identical, then by Lorentz transformation</p><p>On differentiating</p><p>Let . Then,</p><p>= c</p><p>cos h u - 1</p><p>sin h u</p><p>= c A</p><p>cos h u - 1</p><p>cos h u + 1</p><p>= c tan h</p><p>u</p><p>2</p><p>… V</p><p>x (t) =</p><p>c</p><p>tan h u</p><p>11 - 21 - tanh2u2 = c</p><p>cos h u</p><p>sin h u</p><p>a1 -</p><p>1</p><p>cos h u</p><p>b</p><p>b = tan h u, 0 … u 6 q</p><p>=</p><p>c</p><p>b</p><p>11 - 21 - b22 awhere b =</p><p>V</p><p>c</p><p>b</p><p>x (t) =</p><p>c</p><p>2</p><p>V</p><p>a1 - 21 - V 2>c2</p><p>b</p><p>t ¿ =</p><p>1</p><p>21 - V</p><p>2>c</p><p>2</p><p>a t -</p><p>Vx (t)</p><p>c</p><p>2</p><p>b = t</p><p>Kx (t)</p><p>K :</p><p>K :</p><p>x 7 0,t¿ 6 0 and if x 6 0,t¿ 7 0,</p><p>t = 0, t¿ =</p><p>vx</p><p>c2</p><p>1</p><p>21 - v2>c2</p><p>t¿ =</p><p>1</p><p>21 - v</p><p>2/c</p><p>2</p><p>a t -</p><p>vx</p><p>c</p><p>2</p><p>b</p><p>t (A¿) = gt (A) =</p><p>l 0</p><p>v</p><p>A¿, B¿</p><p>t (A) =</p><p>l0</p><p>v</p><p>21 - v</p><p>2/c</p><p>2AA¿</p><p>208 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>¿</p><p>( is a monotonically increasing function of .)utan h u</p><p>1.8 RELATIVISTIC MECHANICS 209</p><p>1.356 We can take the coordinates of the two events to be</p><p>For B to be the effect and A to be cause we must have</p><p>In the moving frame, the coordinates of A and B become</p><p>Since,</p><p>we must have</p><p>1.357 (a) The four-dimensional interval between A and B (assuming ) is:</p><p>Therefore the time interval between these two</p><p>events in the reference frame in which the</p><p>events occurred at the same place is</p><p>or</p><p>(b) The four dimensional interval between A and C (assuming ) is:</p><p>So the distance between the two events in the frame in which they are simulta-</p><p>neous is</p><p>1.358 By the velocity addition formula</p><p>v ¿y =</p><p>vy 21 - V 2>c 2</p><p>1 - vxV>c2</p><p>vx¿ =</p><p>vx - V</p><p>1 - Vvx >c 2</p><p>4 units = 4 m.</p><p>32 - 52 = -16</p><p>¢y = ¢z = 0</p><p>t¿B - t¿A =</p><p>4</p><p>c</p><p>=</p><p>4</p><p>3</p><p>* 10-8 s</p><p>c (t¿B - t¿A) = 216 = 4 m</p><p>52 - 32 = 16 units</p><p>¢y = ¢z = 0</p><p>¢t¿ 7</p><p>| a | ¿</p><p>c</p><p>(¢t¿)2 -</p><p>a¿2</p><p>c</p><p>2</p><p>= g2 c a¢t -</p><p>aV</p><p>c</p><p>2</p><p>b2</p><p>-</p><p>1</p><p>c</p><p>2</p><p>(a - V ¢t)2 d = (¢t)2 -</p><p>a</p><p>2</p><p>c</p><p>2</p><p>7 0</p><p>A : (0, 0, 0, 0); B: c g a¢t -</p><p>aV</p><p>c2</p><p>b , g (a - V ¢t), 0, 0 d ¢where g =</p><p>1</p><p>21 - 1V 2>c22≤</p><p>¢t 7</p><p>| a |</p><p>c</p><p>.</p><p>A : (0, 0, 0, 0); B : (¢t, a, 0, 0)</p><p>B</p><p>C</p><p>A</p><p>0 1 3 4 5 6 7</p><p>1</p><p>2</p><p>3</p><p>4</p><p>5</p><p>6</p><p>7</p><p>ct</p><p>x</p><p>2</p><p>.</p><p>and .</p><p>1.359 (a) By definition, the velocity of approach is</p><p>in the reference frame K</p><p>(b) The relative velocity is obtained by the transformation law</p><p>1.360 The velocity of one of the rods in the reference frame fixed to the other rod is</p><p>The length of the moving rod in this frame is</p><p>1.361 The approach velocity is defined by</p><p>in the laboratory frame.</p><p>So,</p><p>On the other hand, the relative velocity can be obtained by using the velocity addition</p><p>formula and has the components</p><p>B -v1, v2 C1 - a v 2</p><p>1</p><p>c 2</p><p>b R so Vr = Av 2</p><p>1 + v 2</p><p>2 -</p><p>v1v2</p><p>c2</p><p>Vapproach = 2v 2</p><p>1 + v 2</p><p>2</p><p>Vapproach =</p><p>d r1</p><p>dt</p><p>-</p><p>d r2</p><p>dt</p><p>= v1 - v2</p><p>l = l0 C1 -</p><p>4v</p><p>2>c 2</p><p>(1 + b2)2</p><p>= l0</p><p>1 - b2</p><p>1 + b2</p><p>V =</p><p>v + v</p><p>1 + v 2>c 2</p><p>=</p><p>2v</p><p>1 + b2</p><p>= 0.19c (where c is the velocity of light )</p><p>vr =</p><p>v1 - (–v2)</p><p>1 -</p><p>v1 (-v2)</p><p>c 2</p><p>=</p><p>v1 + v2</p><p>1 +</p><p>v1v2</p><p>c2</p><p>vapproach =</p><p>dx1</p><p>dt</p><p>-</p><p>dx2</p><p>dt</p><p>= v1 - (-v2) = v1 + v2 = 1.25c</p><p>v ¿ = 2v ¿2x + v ¿2y =</p><p>2(vx - V )2 + v 2</p><p>y (1 - V 2>c 2)</p><p>1 - vxV>c 2</p><p>210 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>,</p><p>.</p><p>1.8 RELATIVISTIC MECHANICS 211</p><p>1.362 The components of the velocity of the unstable particle in the frame K are</p><p>So the velocity relative to K is</p><p>The life time in this frame dilates to</p><p>and the distance traversed is</p><p>1.363 In the frame the components of the velocity of the particle are</p><p>Hence,</p><p>1.364 In the coordinates of A and B are</p><p>After performing Lorentz transformation to the frame K we get</p><p>By translating , we can write the coordinates of B as</p><p>x = gl a1 -</p><p>V 2</p><p>c 2</p><p>b + Vt¿g = l A1 -</p><p>v ¿2</p><p>c 2</p><p>+ Vt ¿g</p><p>B : t = gt ¿t ¿: t ¿ -</p><p>Vl</p><p>c 2</p><p>z = 0 z = 0</p><p>y = v ¿t ¿ y = -v ¿t¿</p><p>x = g Vt ¿ x = g (l + Vt ¿)</p><p>A : t = gt ¿ B : t = g a t¿ +</p><p>Vl</p><p>c 2</p><p>b</p><p>A : (t¿, 0, -v¿ t¿, 0); B : (t¿, l, - v¿ t¿, 0)</p><p>K ¿</p><p>tan u¿ =</p><p>v ¿y</p><p>v ¿x</p><p>=</p><p>v sin u</p><p>v cos u - V</p><p>21 - V 2>c 2</p><p>v¿y =</p><p>v sin u 21 - V 2>c 2</p><p>1 -</p><p>vVcos u</p><p>c 2</p><p>v¿x =</p><p>v cos u - V</p><p>1 -</p><p>vV cos u</p><p>c 2</p><p>K ¿</p><p>s = ¢t0</p><p>2V 2 + v¿ 2 - (v¿2V 2)>c 2</p><p>21 - V 2>c 2 21 - v¿2>c 2</p><p>¢t0nA1 -</p><p>V 2</p><p>c 2</p><p>-</p><p>v¿2</p><p>c 2</p><p>+</p><p>v¿2V 2</p><p>c 4</p><p>AV 2 + v¿2 -</p><p>v¿2V 2</p><p>c2</p><p>aV, v¿ A1 -</p><p>V 2</p><p>c 2</p><p>, 0b</p><p>v'</p><p>y'y</p><p>v</p><p>v</p><p>xθ</p><p>K'K</p><p>;</p><p>;</p><p>;</p><p>;</p><p>.</p><p>Thus,</p><p>Hence,</p><p>1.365</p><p>(a) In K the velocities at time t and are respectively and along x-</p><p>axis which is parallel to the vector V. In the frame moving with velocity V</p><p>with respect to K, the velocities are, respectively,</p><p>The latter velocity is written as</p><p>Also by Lorentz transformation</p><p>Thus the acceleration in the frame is</p><p>(b) In the K frame, the velocities of the particle at the time t and are, respec-</p><p>tively and where V is along -axis.x(0, v + wdt, 0),(0, v, 0)</p><p>t + dt</p><p>w¿ =</p><p>dv¿</p><p>dt¿</p><p>=</p><p>w11 - vV>c223 a1 -</p><p>V 2</p><p>c 2</p><p>b3>2</p><p>K ¿</p><p>dt¿ =</p><p>dt - Vdx>c2</p><p>21 - V 2>c 2</p><p>= dt</p><p>1 - vV/c 2</p><p>21 - V 2>c 2</p><p>=</p><p>v - V</p><p>1 - vV>c 2</p><p>+</p><p>wdt 11 - V 2>c2211 - vV>c222</p><p>v - V</p><p>1 - vV>c2</p><p>+</p><p>wdt</p><p>1 - vV>c2</p><p>+</p><p>v - V</p><p>(1 - vV>c2)</p><p>wV</p><p>c</p><p>2</p><p>dt</p><p>v - V</p><p>1 - vV>c2</p><p>and v + wdt - V</p><p>1 - (v + wdt) V>c</p><p>2</p><p>K¿</p><p>v + wdtvt + dt</p><p>-</p><p>t</p><p>v</p><p>-</p><p>t + dt</p><p>v + wdt</p><p>K</p><p>tan u¿ =</p><p>v¿V</p><p>c 221 - 1V> c22</p><p>¢x = l C1 - aV 2</p><p>c 2</p><p>b , ¢y =</p><p>v¿Vl</p><p>c 2</p><p>y = -v¿ a t¿ -</p><p>Vl</p><p>c2</p><p>b , z = 0</p><p>212 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>v</p><p>K'K</p><p>v' B</p><p>A</p><p>1.8 RELATIVISTIC MECHANICS 213</p><p>In the frame the velocities are</p><p>and</p><p>Thus the acceleration</p><p>We have used</p><p>1.366 In the instantaneous rest frame and</p><p>So,</p><p>is constant by assumption. Thus, integration gives</p><p>Integrating once again</p><p>The percentage difference of rocket velocity from velocity of light is given by</p><p>1.367 The boost time in the reference frame fixed to the rocket is related to the time</p><p>elapsed on the Earth by</p><p>t0 = 3</p><p>t</p><p>0</p><p>A1 -</p><p>v 2</p><p>c 2</p><p>dt = 3</p><p>t</p><p>0</p><p>D1 -</p><p>1w¿t>c22</p><p>1 + 1w¿t>c2T</p><p>2</p><p>dt</p><p>tt0</p><p>c - v</p><p>c</p><p>=</p><p>1</p><p>2</p><p>a c</p><p>w ¿t</p><p>b2</p><p>= 0.47%</p><p>= 0.91 light year</p><p>x =</p><p>c 2</p><p>w¿</p><p>¢B1 + aw¿t</p><p>c</p><p>b2</p><p>- 1≤</p><p>v =</p><p>w ¿t</p><p>21 + 1w ¿t>c22</p><p>w¿</p><p>=</p><p>dv11 - V 2>c223>2 = w¿dt</p><p>w¿ =</p><p>w11 - V 2>c 223>2 (from solution problem 1.365(a))</p><p>v = V</p><p>dt¿ =</p><p>dt</p><p>21 - V 2>c 2</p><p>w¿ =</p><p>wdt 2(1 - V 2>c 2)</p><p>dt¿</p><p>= w a1 -</p><p>V 2</p><p>c 2</p><p>b along the y-axis</p><p>1-V, (v + wdt) 21 - V 2>c 2, 02 respectively.</p><p>1-V, v 21 - V 2>c 2, 02K¿</p><p>,</p><p>. aa</p><p>1.368</p><p>For</p><p>1.369 We define the density in the frame K is such a way that is the rest mass</p><p>of the element. That is , where is the proper density,</p><p>are the dimensions of the element in the rest frame . Now,</p><p>if the frame K is moving with velocity relative to the frame . Thus</p><p>Defining by</p><p>We get</p><p>or</p><p>1.370 We have</p><p>or</p><p>or v =</p><p>cp</p><p>2p 2 + m2c 2</p><p>=</p><p>c</p><p>21 + (moC> 22</p><p>1 -</p><p>v 2</p><p>c 2</p><p>=</p><p>m2</p><p>0c</p><p>2</p><p>m2</p><p>0c</p><p>2 + p 2</p><p>= 1 -</p><p>p 2</p><p>p 2 + m2</p><p>0c</p><p>2</p><p>m0v</p><p>21 - v2>c2</p><p>= p or m0</p><p>21 - v 2>c2</p><p>= Am0</p><p>2 +</p><p>p 2</p><p>c 2</p><p>= 0.6c (where c is the velocity of light)</p><p>v = c A</p><p>h(2 + h)</p><p>(1 + h)2</p><p>=</p><p>c 2h (2 + h)</p><p>1 + h</p><p>1 + h =</p><p>1</p><p>21 - v2>c2</p><p>or v 2</p><p>c 2</p><p>= 1 -</p><p>1</p><p>(1 + h)2</p><p>=</p><p>h(2 + h)</p><p>1(1 + h)2</p><p>r = r0 (1 + h)h</p><p>r =</p><p>r0</p><p>21 - v 2>c2</p><p>K0v</p><p>dy = dy0, dz = dzo, dx = dx0 21 - v2>c2</p><p>K0dx0, dy0, dz0</p><p>r0r dxdydz = r0dx0dy0dz0dm0</p><p>r dx dy dzr</p><p>L 70 (on substituting values)</p><p>b L 1,</p><p>m</p><p>m0</p><p>L</p><p>1</p><p>22 (1 - b)</p><p>=</p><p>1</p><p>22h</p><p>m =</p><p>m0</p><p>21 - b2</p><p>= 3.5 months (on substituting values)</p><p>= 3</p><p>t</p><p>0</p><p>dt</p><p>21 + 1w¿t>c2 =</p><p>c</p><p>w¿</p><p>3</p><p>w ¿t c</p><p>0</p><p>dj</p><p>21 + j2</p><p>=</p><p>c</p><p>w¿</p><p>ln cw ¿t</p><p>c</p><p>+ B1 + aw ¿t</p><p>c</p><p>b2 d</p><p>214 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>/</p><p>o p</p><p>1.8 RELATIVISTIC MECHANICS 215</p><p>So, (on substituting values)</p><p>1.371 By definition of ,</p><p>or</p><p>1.372 The work done is equal to change in kinetic energy which is different in the two</p><p>cases. Classically, i.e., in non-relativistic mechanics, the change in kinetic energy is</p><p>Relativistically, it is</p><p>1.373</p><p>or</p><p>or</p><p>1.374 Relativistically,</p><p>So,</p><p>Thus, -brel = c 2T</p><p>m0c</p><p>2</p><p>- 3</p><p>T 2</p><p>m2</p><p>0c</p><p>4</p><p>d1>2 = A</p><p>2T</p><p>m0c</p><p>2</p><p>a1 -</p><p>3</p><p>4</p><p>T</p><p>m0c</p><p>2</p><p>b</p><p>b</p><p>2</p><p>rel �</p><p>2T</p><p>m0c</p><p>2</p><p>-</p><p>3</p><p>4</p><p>1b 2</p><p>rel22 =</p><p>2 T</p><p>m0c</p><p>2</p><p>-</p><p>3</p><p>4</p><p>a 2T</p><p>m0c</p><p>2</p><p>b2</p><p>T</p><p>m0c</p><p>2</p><p>= a 1</p><p>21 - b 2</p><p>- 1b =</p><p>1</p><p>2</p><p>b 2 +</p><p>3</p><p>8</p><p>b4</p><p>v</p><p>c</p><p>=</p><p>23</p><p>2</p><p>, i.e. v = c</p><p>23</p><p>2</p><p>= 2.6 * 108 m/s</p><p>A1 -</p><p>v</p><p>2</p><p>c</p><p>2</p><p>=</p><p>1</p><p>2</p><p>or 1 -</p><p>v</p><p>2</p><p>c 2</p><p>=</p><p>1</p><p>4</p><p>m0c</p><p>2</p><p>A1 -</p><p>v 2</p><p>c2</p><p>= 2m0c</p><p>2</p><p>= 0.416 m0c</p><p>2 = 0.42 m0 c</p><p>2</p><p>m0c</p><p>2</p><p>21 - (0.8)2</p><p>-</p><p>m0c</p><p>2</p><p>21 - (0.6)2</p><p>=</p><p>m0c</p><p>2</p><p>0.6</p><p>-</p><p>m0c</p><p>2</p><p>0.8</p><p>= m0c</p><p>2 (1.666 - 1.250)</p><p>1</p><p>2</p><p>m0c</p><p>2 ((0.8)2 - (0.6)2) =</p><p>1</p><p>2</p><p>m0c</p><p>2 0.28 = 0.14 m0c</p><p>2</p><p>v = c A1 -</p><p>1</p><p>h2</p><p>=</p><p>c</p><p>h</p><p>2h2 - 1 =</p><p>1</p><p>2</p><p>c23 (on substituting for h)</p><p>m0v</p><p>21 - v2>c2</p><p>= hm0v 1 -</p><p>v</p><p>2</p><p>c</p><p>2</p><p>=</p><p>1</p><p>h2</p><p>h</p><p>c - v</p><p>c</p><p>= c1 - a1 + am0c</p><p>p</p><p>b2b -1>2 d * 100 % = 0.44%</p><p>or</p><p>,</p><p>But classically,</p><p>Hence if,</p><p>with an error less than .</p><p>1.375 From the formula</p><p>we find</p><p>or GeV/c</p><p>1.376 Let the total force exerted by the beam on the target surface be F and the power lib-</p><p>erated there be P. Then, using the result of the previous problem, we see</p><p>since, being the number of particles striking the target per second. Also,</p><p>These will be, respectively, equal to the pressure and power developed per unit area</p><p>of the target if I is current density.</p><p>1.377 In the frame fixed to the sphere, the momentum transferred to the elastically scat-</p><p>tered particle is</p><p>The density of the moving element is n</p><p>1</p><p>21 - v 2>c2</p><p>(from solution of problem 1.369)</p><p>2mv</p><p>21 - v2>c2</p><p>P = N a m0c</p><p>2</p><p>21 - v 2>c 2</p><p>- m0c</p><p>2b =</p><p>I</p><p>e</p><p>T</p><p>I = Ne, N</p><p>F = Np =</p><p>N</p><p>c</p><p>2T (T + 2m0c</p><p>2) =</p><p>I</p><p>ec</p><p>2T (T + 2m0c</p><p>2)</p><p>T (2 m0c</p><p>2 + T) = c 2p 2 i.e. p =</p><p>1</p><p>c</p><p>2T (2m0c</p><p>2 + T ) = 1.09</p><p>E 2 = c 2p 2 + m 2</p><p>0c 4 or (m0c</p><p>2 + T ) 2 = c 2p 2 + m 2</p><p>0c 4</p><p>E =</p><p>m0c</p><p>2</p><p>21 - v</p><p>2/c</p><p>2</p><p>, p =</p><p>m0v</p><p>21 - v 2>c 2</p><p>e</p><p>b</p><p>T</p><p>m0c</p><p>2</p><p>6</p><p>4</p><p>3</p><p>e L 0.013, then the velocity is given by the classical formula</p><p>bcl = A</p><p>2T</p><p>m0c</p><p>2</p><p>so brel - bcl</p><p>bcl</p><p>=</p><p>3</p><p>4</p><p>T</p><p>m0c</p><p>2</p><p>= e</p><p>216 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>,</p><p>1.8 RELATIVISTIC MECHANICS 217</p><p>and the momentum transferred per unit time per unit area is the pressure, given by</p><p>In the frame fixed to the gas, when the sphere hits a stationary particle, the latter</p><p>recoils with a velocity</p><p>The momentum transferred is</p><p>and the pressure is</p><p>1.378 The equation of motion is</p><p>Integrating,</p><p>or</p><p>or using at , we get,</p><p>1.379</p><p>or</p><p>v</p><p>21 - v2>c2</p><p>=</p><p>c 2t</p><p>a</p><p>x = 2a 2 + c 2t 2</p><p>so, x</p><p>#</p><p>= v =</p><p>c 2t</p><p>a 2 + c 2t 2</p><p>x = Bc 2t 2 + am0c</p><p>2</p><p>F</p><p>b2</p><p>-</p><p>m0c</p><p>2</p><p>F</p><p>t = 0x = 0</p><p>x = 3</p><p>Fct dt</p><p>2F 2t 2 + m2</p><p>0c</p><p>2</p><p>=</p><p>c</p><p>F</p><p>3</p><p>jdj</p><p>2j 2 + (m0c)</p><p>2</p><p>=</p><p>c</p><p>F</p><p>2F 2t 2 + m 2</p><p>0 c</p><p>2 + constant</p><p>b 2</p><p>1 - b 2</p><p>= a Ft</p><p>m0c</p><p>b2</p><p>or b2 =</p><p>(Ft)2</p><p>(Ft)2 + (m0c)</p><p>2</p><p>or v =</p><p>Fct</p><p>2(m0c)</p><p>2 + (Ft)2</p><p>v>c</p><p>21 - v2>c2</p><p>=</p><p>b</p><p>21 - b2</p><p>=</p><p>Ft</p><p>m0 c</p><p>using v = 0 for t = 0</p><p>d</p><p>dt</p><p>a m0v</p><p>21 - v2>c2</p><p>b = F</p><p>2mv</p><p>1 - v 2>c2</p><p># n # v =</p><p>2mnv 2</p><p>1 - v 2>c2</p><p>m # 2v</p><p>1 + v 2>c 2</p><p>A1 -</p><p>4v 2>c 2</p><p>(1 - v 2>c 2)2</p><p>=</p><p>2mv</p><p>1 - v 2>c2</p><p>=</p><p>v + v</p><p>1 + v 2>c2</p><p>=</p><p>2v</p><p>1 + v 2>c2</p><p>p =</p><p>2mv</p><p>21 - v2>c2</p><p>n</p><p>1</p><p>21 - v 2>c2</p><p># v</p><p>2mnv2</p><p>1 - v 2>c2</p><p>( )</p><p>Since,</p><p>Thus,</p><p>1.380</p><p>Thus,</p><p>1.381 By definition,</p><p>where is the invariant interval</p><p>Thus,</p><p>1.382 For a photon moving in the direction</p><p>In the moving frame,</p><p>Note that</p><p>1.383 As before</p><p>px = m0c</p><p>dx</p><p>ds</p><p>E = m0c</p><p>3</p><p>dt</p><p>ds</p><p>e¿ =</p><p>e</p><p>2</p><p>if, 1</p><p>4</p><p>=</p><p>1 - b</p><p>1 + b</p><p>or b =</p><p>3</p><p>5</p><p>, V =</p><p>3c</p><p>5</p><p>e¿ =</p><p>1</p><p>21 - b 2</p><p>ae - V</p><p>e</p><p>c</p><p>b = e A</p><p>1 - V>c</p><p>1 + V>c</p><p>py = pz = 0</p><p>e = cpx</p><p>x</p><p>E ¿ = m0c</p><p>3</p><p>dt¿</p><p>ds</p><p>- c 3 m0g</p><p>1dt - Vdx >c22</p><p>ds</p><p>=</p><p>E - Vpx</p><p>A1 -</p><p>V 2</p><p>c 2</p><p>px¿ = cm0</p><p>dx¿</p><p>ds</p><p>= cm0g</p><p>(dx - Vdt)</p><p>ds</p><p>=</p><p>px - VE>c 2</p><p>21 - V 2>c2</p><p>(dy = dz = 0).ds 2 = c 2 dt 2 - dx 2</p><p>E = m0</p><p>c 2</p><p>21 - vx</p><p>2>c2</p><p>=</p><p>m0c</p><p>3dt</p><p>ds</p><p>, px = m0</p><p>vx</p><p>21 - v2>c2</p><p>=</p><p>cm0dx</p><p>dx</p><p>F7 = m0</p><p>w</p><p>(1 - b 2)3>2, w = v</p><p>#</p><p>, w7 v</p><p>F</p><p>�</p><p>= m0w 21 - b2, w = v</p><p>#</p><p>, w</p><p>�</p><p>v</p><p>F =</p><p>d</p><p>dt</p><p>a m0v</p><p>21 - v 2>c2</p><p>b = m0</p><p>v</p><p>#</p><p>21 - v 2>c2</p><p>+ m0</p><p>v</p><p>c 2</p><p>v # v</p><p>#</p><p>111 - v2>c223>2</p><p>d</p><p>dt</p><p>a m0v</p><p>21 - v 2>c2</p><p>b =</p><p>m0c</p><p>2</p><p>a</p><p>= F</p><p>218 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>and</p><p>1.8 RELATIVISTIC MECHANICS 219</p><p>Similarly</p><p>Then</p><p>1.384 (b) In the C.M. frame, the total momentum is zero, Thus</p><p>where we have used the result of problem 1.375.</p><p>(a) Then</p><p>Total energy in the C.M. frame is</p><p>So,</p><p>Also</p><p>1.385</p><p>Also cp = 2T (T + 2m0c</p><p>2), v =</p><p>c 2p</p><p>E</p><p>= c A</p><p>T</p><p>T + 2m0c</p><p>2</p><p>2(2m0c</p><p>2 + T ) 2 - T (2m0c</p><p>2 + T ) = 22m0c</p><p>2 (2m0c</p><p>2 + T ) = c 22m0 (2m0c</p><p>2 + T )</p><p>M0c</p><p>2 = 2E 2 - c 2p 2</p><p>or p</p><p>'</p><p>= A</p><p>1</p><p>2</p><p>m0T = 940 MeV>c</p><p>2 2c 2p</p><p>'</p><p>2 + mc 4 = 22m2</p><p>0c</p><p>2(T + 2m0c</p><p>2) 4c 2p</p><p>'</p><p>2 = 2m0c</p><p>2T</p><p>T</p><p>'</p><p>= 2m0c</p><p>2 aA1 +</p><p>T</p><p>2m0c</p><p>2</p><p>- 1b = 777 MeV</p><p>2m0c</p><p>2</p><p>21 - V 2>c 2</p><p>= 2m0c</p><p>2 C</p><p>T + 2m0c</p><p>2</p><p>2m0c</p><p>2</p><p>= 22m0c</p><p>2 ( T + 2m0c</p><p>2) = T</p><p>'</p><p>+ 2m0c</p><p>2</p><p>1</p><p>21 - V 2>c 2</p><p>=</p><p>1</p><p>A1 -</p><p>T</p><p>T + 2m0c</p><p>2</p><p>= C</p><p>T + 2m0c</p><p>2</p><p>2m0c</p><p>2</p><p>V = cA</p><p>T</p><p>T + 2m0c</p><p>2</p><p>= 2.12 * 108 m/s</p><p>V</p><p>c</p><p>=</p><p>cp1x</p><p>E1 + E2</p><p>=</p><p>2T (T + 2m0c</p><p>2)</p><p>T + 2m0c</p><p>2</p><p>= A</p><p>T</p><p>T + 2m0c</p><p>2</p><p>= m 2</p><p>0c 4</p><p>(c 2dt 2 - dx</p><p>2 - dy 2 - dz</p><p>2)</p><p>ds 2</p><p>= m 2</p><p>0c 4 is invariant</p><p>E 2 - c 2p 2 = E 2 - c 2 1p 2</p><p>x + p 2</p><p>y + p 2</p><p>z2 py = m0c</p><p>dy</p><p>ds</p><p>, pz = m0c</p><p>dz</p><p>ds</p><p>Q</p><p>1.386 Let kinetic energy of a proton striking another stationary particle of the same</p><p>rest mass. Then, combined kinetic energy in the C.M. frame</p><p>1.387 We have</p><p>Hence,</p><p>The R.H.S. is an invariant. We can evaluate it in any frame. Choose the C.M. frame</p><p>of the particles 2 and 3.</p><p>In this frame,</p><p>Thus,</p><p>or</p><p>1.388 The velocity of ejected gases is relative to the rocket. In an Earth centered frame</p><p>it is</p><p>in the direction of the rocket. The momentum conservation equation then reads</p><p>or</p><p>Here, is the mass of the ejected gases. So,</p><p>mdv -</p><p>-u + uv>c2</p><p>1 - uv2>c2</p><p>dm = 0 or mdv + u a1 -</p><p>v 2</p><p>c 2</p><p>b dm = 0</p><p>-dm</p><p>mdv - a v - u</p><p>1 - uv>c2</p><p>- vb dm = 0</p><p>(m + dm) (v + dv) +</p><p>v - u</p><p>1 - uv>c2</p><p>(-dm) = mv</p><p>v - u</p><p>1 - vu>c2</p><p>u</p><p>2m0c</p><p>2E1 … Em 2</p><p>0 + m 2</p><p>1 - (m2 + m3)</p><p>2Fc 4 or E1 …</p><p>m 2</p><p>0 + m 2</p><p>1 - (m2 + m3)</p><p>2</p><p>2m0</p><p>c</p><p>2</p><p>(m 2</p><p>0 + m 2</p><p>1) c</p><p>4 - 2m0c</p><p>2E1 = (m2 + m3)</p><p>2c 4</p><p>R.H.S. = (E ¿2 + E ¿3) 2 = (m2 + m3)</p><p>2c 4</p><p>L.H.S. = (m 2</p><p>0 c</p><p>4 - E1)</p><p>2 - c 2p 2</p><p>1 = (m 2</p><p>0 + m 2</p><p>1) c</p><p>4 - 2m0c</p><p>2E1</p><p>(m0c</p><p>2 - E1)</p><p>2 - c 2 p 2</p><p>1 = (E</p><p>2 + E3)</p><p>2 - (p2 + p3)</p><p>2c 2</p><p>E1 + E2 + E3 = m0c</p><p>2, p1 + p2 + p3 = 0</p><p>T ¿</p><p>2m0c</p><p>2</p><p>=</p><p>T (2m0c</p><p>2 + T)</p><p>m2</p><p>0c</p><p>4</p><p>T ¿ =</p><p>2T (T + 2m0c</p><p>2)</p><p>m0c</p><p>2</p><p>= 2m0c</p><p>2aA1 +</p><p>T ¿</p><p>2m0c</p><p>2</p><p>- 1b = 2T, a T</p><p>m0c</p><p>2</p><p>+ 1b2</p><p>= 1 +</p><p>T ¿</p><p>2m0c</p><p>2</p><p>T ¿ =</p><p>220 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>or</p><p>1.8 RELATIVISTIC MECHANICS 221</p><p>(neglecting since u is non-relativistic).</p><p>Integrating, using</p><p>The constant since initially.</p><p>Thus,</p><p>1 - b</p><p>1 + b</p><p>= a m</p><p>m0</p><p>bu>c</p><p>or b =</p><p>1 - 1m>mo2u>c</p><p>1 + 1m>mo2u>c</p><p>b = 0=</p><p>u</p><p>c</p><p>ln m0,</p><p>b =</p><p>v</p><p>c</p><p>, we get</p><p>3</p><p>db</p><p>1 - b 2</p><p>+</p><p>u</p><p>c</p><p>3</p><p>dm</p><p>m</p><p>= 0</p><p>ln</p><p>1 + b</p><p>1 - b</p><p>+</p><p>u</p><p>c</p><p>ln m = constant</p><p>1 - uv/c 2,</p><p>or</p><p>THERMODYNAMICS AND</p><p>MOLECULAR PHYSICS</p><p>2.1 Equation of the Gas State. Processes</p><p>2.1 Let and be the masses of the gas in the vessel before and after the gas is</p><p>released. Hence mass of the gas released,</p><p>Now from ideal gas equation</p><p>as V and T are same before and after the release of the gas.</p><p>So,</p><p>or (1)</p><p>We also know (2)</p><p>(where standard atmospheric pressure and</p><p>From Eqs. (1) and (2), we get</p><p>2.2 Let V is volume of each vessel. For the vessel which contained the ideal gas,</p><p>When heated, some gas, passes into the evacuated vessel till pressure difference be-</p><p>comes Let and be the pressure on the two sides of the valve. Then</p><p>and p¿2V = n¿2 RT2 = (n1 - n¿1)RT2</p><p>p¿1V = n¿1RT2</p><p>p¿2p¿1¢p.</p><p>p1V = n1RT1</p><p>¢m = rV</p><p>¢p</p><p>p0</p><p>= 1.3 * 30 * 0.78</p><p>1</p><p>= 30 g</p><p>T0 = 273 K).p0 =</p><p>p = r</p><p>R</p><p>M</p><p>T so, M</p><p>RT0</p><p>=</p><p>r</p><p>p0</p><p>¢m =</p><p>(p1 - p2)VM</p><p>RT0</p><p>=</p><p>¢pVM</p><p>RT0</p><p>(p1 - p2)V = (m1 - m2)</p><p>R</p><p>M</p><p>T0 = ¢m</p><p>R</p><p>M</p><p>T0</p><p>p1V = m1</p><p>R</p><p>M T0 and p2V = m2</p><p>R</p><p>MT0</p><p>¢m = m1 – m2</p><p>m2m1</p><p>PART2</p><p>Putting the values of and from the first two equations</p><p>But,</p><p>So,</p><p>or</p><p>2.3 Let the mixture contain and moles of H2 and He, respectively. If molecular weights</p><p>of H2 and He are and , then their respective masses in the mixture are</p><p>Therefore, for the total mass of the mixture we get,</p><p>(1)</p><p>Also, if is the total number of moles of the mixture in the vessels, then we know,</p><p>(2)</p><p>Solving Eqs. (1) and (2) for and we get</p><p>Therefore, we get</p><p>and</p><p>or</p><p>One can also express the above result in terms of the effective molecular weight M of</p><p>the mixture, defined as</p><p>Thus,</p><p>m1</p><p>m2</p><p>=</p><p>M1</p><p>M2</p><p># M2 - M</p><p>M - M1</p><p>=</p><p>1 - M>M2</p><p>M>M1 - 1</p><p>M = m</p><p>n</p><p>= m RT</p><p>pV</p><p>m1</p><p>m2</p><p>=</p><p>M1</p><p>M2</p><p>(nM2 - m)</p><p>(m - nM1)</p><p>m2 = M2</p><p>(m - nM1)</p><p>M2 - M1</p><p>m1 = M1</p><p>(nM2 - m)</p><p>M2 - M1</p><p>n1 =</p><p>(nM2 - m)</p><p>M2 - M1</p><p>, n2 =</p><p>m - nM1</p><p>M2 - M1</p><p>n2,n1</p><p>n = n1 + n2</p><p>n</p><p>m = m1 + m2 or m = n1M1 + n2M2</p><p>m1 = n1M1 and m2 = n2M2</p><p>M2M1</p><p>n2n1</p><p>p¿2 = 1</p><p>2</p><p>ap1T2</p><p>T1</p><p>- ¢pb = 0.08 atm</p><p>=</p><p>p1T2</p><p>T1</p><p>- p¿2 - ¢p</p><p>p¿2 = a p1</p><p>T1</p><p>-</p><p>p¿2 + ¢p</p><p>T2</p><p>bT2</p><p>p¿1 - p¿2 = ¢p</p><p>p¿2V = ap1V</p><p>RT1</p><p>-</p><p>p¿1V</p><p>RT2</p><p>bRT2 or p¿2 = a p1</p><p>T1</p><p>-</p><p>p¿1</p><p>T2</p><p>bT2</p><p>n2n1</p><p>224 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>Using the data and table, we get</p><p>2.4 We know, for the mixture, N2 and CO2 (being regarded as ideal gases, their mixture</p><p>too behaves like an ideal gas) where, is the total number of moles of</p><p>of the gases (mixture) present, is the pressure and V is the volume of the vessel.</p><p>If and are number of moles of N2 and CO2 respectively present in the mixture,</p><p>then .</p><p>Now number of moles of N2 and CO2 is, by definition, given by</p><p>where, is the mass of N2 (molecular weight in the mixture and is the</p><p>mass of CO2 (molecular weight in the mixture.</p><p>Therefore density of the mixture is given by</p><p>2.5 (a) The mixture contains and moles of O2, N2 and CO2 respectively. Then</p><p>the total number of moles of the mixture</p><p>We know, ideal gas equation for the mixture</p><p>or</p><p>(b) Mass of oxygen (O2) present in the mixture:</p><p>Mass of nitrogen (N2) present in the mixture:</p><p>Mass of carbon dioxide (CO2) present in the mixture:</p><p>So, mass of the mixture:</p><p>m = m1 + m2 + m3 = n1M1 + n2M2 + n3M3</p><p>m3 = n3M3</p><p>m2 = n2M2</p><p>m1 = n1M1</p><p>p =</p><p>(n1 + n2 + n3)RT</p><p>V</p><p>= 1.968 atm (on substituting values)</p><p>pV = nRT or p = nRT</p><p>V</p><p>n = n1 + n2 + n3</p><p>n3n1, n2</p><p>= 1.5 kg/m3 (on substituting values)</p><p>=</p><p>p0</p><p>RT</p><p>m1 + m2</p><p>n1 + n2</p><p>=</p><p>p0(m1 + m2)M1M2</p><p>RT (m1M2 + m2M1)</p><p>r =</p><p>m1 + m2</p><p>V</p><p>=</p><p>m1 + m2</p><p>(nRT> 0)</p><p>= M2)</p><p>m2= M1)m1</p><p>n1 =</p><p>m1</p><p>M1</p><p>and n2 =</p><p>m2</p><p>M2</p><p>n = n1 + n2</p><p>n2n1</p><p>n , p0V = nRT</p><p>M = 3.0 g and m1</p><p>m2</p><p>= 0.50</p><p>2.1 EQUATION OF THE GAS STATE. PROCESSES 225</p><p>,</p><p>,</p><p>,</p><p>p0</p><p>p</p><p>Molecular mass of the mixture</p><p>2.6 Let and be the pressure in the upper and lower part of the cylinder respectively</p><p>at temperature . At the equilibrium position for the piston:</p><p>or</p><p>But, (where is the initial volume of the lower part)</p><p>So, (1)</p><p>Let be the sought temperature and at this temperature the volume of the lower part be-</p><p>comes then according to the problem the volume of the upper part becomes .</p><p>Hence, (2)</p><p>From Eqs. (1) and (2),</p><p>(3)</p><p>As, the total volume must be constant, so</p><p>Putting the value of in Eq. (3), we get</p><p>=</p><p>T0(h</p><p>2 - 1)h¿</p><p>(h¿2 - 1)h</p><p>= 0.42 kK</p><p>T =</p><p>T011 - 1>h2V0</p><p>(1 + h)</p><p>(1 + h¿)</p><p>V0</p><p>11 - 1>h¿2</p><p>V ¿</p><p>V0(1 + h) = V ¿(1 + h¿) or V ¿ =</p><p>V0(1 + h)</p><p>1 + h¿</p><p>RT0</p><p>V0</p><p>a1 -</p><p>1</p><p>h</p><p>b =</p><p>RT ¿</p><p>V ¿</p><p>a1 -</p><p>1</p><p>h¿</p><p>b or T ¿ =</p><p>T011 - 1>h2V ¿</p><p>V011 - 1>h¿2</p><p>mg</p><p>S</p><p>= RT</p><p>V ¿ a1 - 1</p><p>h¿ b</p><p>h¿V V ¿,</p><p>T ¿</p><p>RT0</p><p>hV0</p><p>+</p><p>mg</p><p>S</p><p>=</p><p>RT0</p><p>V0</p><p>Q</p><p>mg</p><p>S</p><p>=</p><p>RT0</p><p>V0</p><p>a1 -</p><p>1</p><p>h</p><p>b</p><p>V0p1 =</p><p>RT0</p><p>hV0</p><p>p1 +</p><p>mg</p><p>S</p><p>= p2 (where m is the mass of the piston)</p><p>p1S + mg = p2S</p><p>T0</p><p>p2p1</p><p>=</p><p>n1M1 + n2M2 + n3M3</p><p>n1 + n2 + n3</p><p>= 36.7 g/mol (on substituting values)</p><p>M = mass of the mixture</p><p>total number of moles</p><p>226 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>,</p><p>2.7 Let be the density after the first stroke. The mass remains constant.</p><p>So,</p><p>Similarly, if is the density after second stroke</p><p>or</p><p>In this way after nth stroke,</p><p>Since pressure density,</p><p>(because temperature is constant)</p><p>It is required by to be ,</p><p>So,</p><p>Hence,</p><p>2.8 From the ideal gas equation</p><p>(1)</p><p>In each stroke, volume v of the gas is ejected, where v is given by</p><p>In case of continuous ejection, if corresponds to mass of gas in the vessel at time</p><p>t, then mN is the mass at time where is the time in which volume v of the</p><p>gas has come out.</p><p>The rate of evacuation is therefore , i.e.,</p><p>In the limit we get</p><p>(2)C = V</p><p>m dm</p><p>dt</p><p>¢t: 0,</p><p>C = v</p><p>¢t</p><p>= - V</p><p>m (t + ¢t)</p><p># m (t + ¢t) - m (t)</p><p>¢t</p><p>v >¢t</p><p>¢t,t + ¢t,</p><p>mN -1</p><p>v = V</p><p>mN</p><p>[mN -1 - mN</p><p>]</p><p>dp</p><p>dt</p><p>= RT</p><p>MV</p><p>dm</p><p>dt</p><p>p = m</p><p>M</p><p>RT</p><p>V</p><p>n =</p><p>ln h</p><p>ln 11 + ¢V >V2</p><p>1</p><p>h</p><p>= a V</p><p>V + ¢V</p><p>bn</p><p>or h = aV + ¢V</p><p>V</p><p>bn</p><p>1>hpn</p><p>>p0</p><p>pn = a V</p><p>V + ¢V</p><p>bn</p><p>p0</p><p>r</p><p>rn = a V</p><p>V + ¢V</p><p>bn</p><p>r0</p><p>r2 = a V</p><p>V + ¢V</p><p>br1 = a V</p><p>V + ¢V</p><p>b2</p><p>r0</p><p>Vr1 = (V + ¢V ) r2</p><p>r2</p><p>Vr = (V + ¢V ) r1 Q r1 =</p><p>Vr</p><p>(V + ¢V )</p><p>r1</p><p>2.1 EQUATION OF THE GAS STATE. PROCESSES 227</p><p>From Eqs. (1) and (2)</p><p>Integrating we get,</p><p>or ln</p><p>Thus,</p><p>2.9 Let be the instantaneous density, then instantaneous mass . In a short interval</p><p>dt the volume is increased by Cdt.</p><p>So, (because mass remains constant in a short interval dt)</p><p>So,</p><p>Since pressure density,</p><p>or</p><p>or</p><p>2.10 The physical system consists of one mole of gas confined in the smooth vertical tube.</p><p>Let and be the masses of upper and lower pistons and and be their</p><p>respective areas.</p><p>For the lower piston</p><p>or (1)</p><p>Similarly for the upper piston</p><p>or (2) T = (p - p0)S1 - m1g</p><p>p0S1 + T + m1g = pS1</p><p>T = (p - p0) S2 + m2g</p><p>pS2 + m2g = p0S2 + T</p><p>S2S1m2m1</p><p>t = V</p><p>C</p><p>ln</p><p>p1</p><p>p2</p><p>= V</p><p>C</p><p>ln 1</p><p>h</p><p>= 1.0 min</p><p>3</p><p>p2</p><p>p1</p><p>-</p><p>dp</p><p>p = C</p><p>V</p><p>t</p><p>dp</p><p>p = - C</p><p>V</p><p>dt</p><p>r</p><p>dr</p><p>r</p><p>= -</p><p>C</p><p>V</p><p>dt</p><p>Vr = (V + Cdt) (r + dr)</p><p>= Vrr</p><p>p = p0e</p><p>- Ct>V</p><p>p</p><p>p0</p><p>= -</p><p>C</p><p>V</p><p>t</p><p>3</p><p>p0</p><p>p</p><p>dp</p><p>p</p><p>= -</p><p>C</p><p>V3</p><p>0</p><p>t</p><p>dt</p><p>dp</p><p>dt</p><p>= -</p><p>C</p><p>V</p><p>mRT</p><p>MV</p><p>= -</p><p>C</p><p>V</p><p>p or dp</p><p>p</p><p>= -</p><p>C</p><p>V</p><p>dt</p><p>228 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>T</p><p>T</p><p>p</p><p>p0</p><p>p0</p><p>.</p><p>From Eqs. (1) and (2)</p><p>or</p><p>So,</p><p>From the gas law, or .</p><p>So,</p><p>Hence,</p><p>2.11 (a) Given that (as for one mole of gas).</p><p>Thus, (1)</p><p>For</p><p>which yields,</p><p>Hence, using this value of p in Eq. (1), we get</p><p>(b) Given that</p><p>So, (2)</p><p>For , the condition is which yields .</p><p>Hence using this value of p in Eq. (2), we get</p><p>2.12 Given that (as for one mole of gas).</p><p>So, (1)p = 1a RT (T - T0)</p><p>-1>2</p><p>V = RT>pT = T0 + aV 2 = T0 + a (R 2T</p><p>2>p2)</p><p>Tmax =</p><p>p0</p><p>ebR</p><p>p = p0</p><p>>edT >dp = 0,Tmax</p><p>bRT</p><p>p = ln</p><p>p0</p><p>p and T =</p><p>p</p><p>bR</p><p>ln</p><p>p0</p><p>p</p><p>p = p0e</p><p>-bV = p0e</p><p>-bRT>p</p><p>Tmax = 1</p><p>R 1a # 2</p><p>3</p><p>p0 Ap0 - 2</p><p>3</p><p>p0 = 2</p><p>3</p><p>ap0</p><p>R</p><p>bC p0</p><p>3a</p><p>p = 2>3 p0</p><p>d</p><p>dp</p><p>(p0</p><p>p</p><p>2 - p</p><p>3) = 0Tmax,</p><p>T = 1</p><p>R 1a p 1p0 - p = 1</p><p>R 1a 2p0</p><p>p 2 - p 3</p><p>V = RT>pp = p0 - aV</p><p>2 = p0 - a1RT>p22 = 0.9 K</p><p>¢T =</p><p>1</p><p>R</p><p>(p0¢S + mg)l</p><p>ap0 +</p><p>mg</p><p>¢S</p><p>b¢Sl = R¢T</p><p>p¢V = nR¢T (because p is constant)pV = nRT</p><p>p =</p><p>mg</p><p>¢S</p><p>+ p0 = constant</p><p>(p - p0) ¢S = mg</p><p>(p - p0)(S1 - S2) = (m1 + m2) g</p><p>2.1 EQUATION OF THE GAS STATE. PROCESSES 229</p><p>For Pmin,</p><p>(2)</p><p>From Eqs. (1) and (2), we get</p><p>2.13 Consider a thin layer at a height h and thickness dh. Let p and be the pres-</p><p>sures on the two sides of the layer. The mass of the layer is . Equating vertical</p><p>downward force to the upward force acting on the layer, we get</p><p>So, (1)</p><p>But,</p><p>So,</p><p>or (using Eq. 1)</p><p>So,</p><p>That means, temperature of air drops by at a height of 1 km above bottom.</p><p>2.14 From the previous problem we have,</p><p>(1)</p><p>But, from (where C is a constant), we get</p><p>(2)</p><p>We have from gas law,</p><p>So, (using Eq. 2)</p><p>Thus, (3)</p><p>But,</p><p>dT</p><p>dh</p><p>=</p><p>dT</p><p>dr</p><p># dr</p><p>dp</p><p># dp</p><p>dh</p><p>dT</p><p>dr</p><p>=</p><p>M</p><p>R</p><p># C (n</p><p>- 1) rn–2</p><p>T =</p><p>M</p><p>R</p><p>Crn -1</p><p>p = r</p><p>R</p><p>M</p><p>T or Crn = r</p><p>R</p><p>M</p><p># T</p><p>dp >dr = Cnrn -1</p><p>p = Crn</p><p>dp</p><p>dh</p><p>= -rg</p><p>34 K</p><p>dT</p><p>dh</p><p>= -</p><p>gM</p><p>R</p><p>= -34 K>km</p><p>-</p><p>rR</p><p>M</p><p>dT = rgdh</p><p>dp =</p><p>rR</p><p>M</p><p>dT</p><p>p =</p><p>r</p><p>M</p><p>RT</p><p>dp</p><p>dh</p><p>= -rg</p><p>Sdhrg + (p + dp)S = pS</p><p>Sdhr</p><p>dp + p</p><p>pmin = 1a R 2T0 (2T0 - T0</p><p>)-1>2 = 2R 1aT0</p><p>T = 2T0</p><p>dp</p><p>dT</p><p>= 0, which gives</p><p>230 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>dh</p><p>(p+dp)S</p><p>pS</p><p>Sdh</p><p>h</p><p>Substituting from Eqs. (1), (2) and (3) we get</p><p>2.15 We have,</p><p>Thus,</p><p>Integrating, we get</p><p>or (where p0 is the pressure at the surface of the Earth)</p><p>So,</p><p>Under standard conditions,</p><p>Pressure at a height of 5 km</p><p>Pressure in a mine at a depth of 5 km</p><p>2.16 We have and from gas law .</p><p>Thus,</p><p>So,</p><p>Integrating within limits</p><p>or</p><p>So,</p><p>(a) Given that T = 273 K and r0</p><p>>r = e.</p><p>r = r0e</p><p>-Mgh>RT and h = -</p><p>RT</p><p>Mg</p><p>ln</p><p>r</p><p>r0</p><p>ln</p><p>r</p><p>r0</p><p>=</p><p>gM</p><p>RT</p><p>h</p><p>3</p><p>r</p><p>r0</p><p>dr</p><p>r</p><p>= 3</p><p>h</p><p>0</p><p>gM</p><p>RT</p><p>dh</p><p>dr</p><p>r</p><p>=</p><p>gM</p><p>RT</p><p>dh</p><p>dp =</p><p>dr</p><p>M</p><p>RT (at constant temperature)</p><p>p = ( r>M )RTdp = -rgdh</p><p>= 1 * e-28 * 9.81 * (-5000)>8.314 * 273 = 2 atm</p><p>= 1 * e-28 * 9.81 * 5000>8.314 * 273 = 0.5 atm</p><p>p0 = 1 atm, T = 273 K</p><p>p = p0e</p><p>-Mgh>RT</p><p>ln</p><p>p</p><p>p0</p><p>= -</p><p>Mg</p><p>RT</p><p>h</p><p>3</p><p>p</p><p>p0</p><p>dp</p><p>p</p><p>= -</p><p>Mg</p><p>RT3</p><p>h</p><p>0</p><p>dh</p><p>dp</p><p>p</p><p>= -</p><p>Mg</p><p>RT</p><p>dh</p><p>dp = -rgdh and from gas law r = (M>RT )p</p><p>dT</p><p>dh</p><p>=</p><p>M</p><p>R</p><p>C (n - 1) rn -2</p><p>1</p><p>Cnrn -1</p><p>(-rg) =</p><p>-Mg (n - 1)</p><p>nR</p><p>2.1 EQUATION OF THE GAS STATE. PROCESSES 231</p><p>.</p><p>Thus,</p><p>(b) Given that and .</p><p>Thus,</p><p>2.17 From the Barometric formula, we have</p><p>and from gas law</p><p>So, at constant temperature from these two equations</p><p>(1)</p><p>Eq. (1) shows that density varies with height in the same manner as pressure. Let us</p><p>consider the mass element of the gas contained in the column.</p><p>Hence the sought mass,</p><p>2.18 As the gravitational field is constant the centre of gravity and the centre of mass</p><p>(C.M.) are same. The location of C.M.</p><p>But from Barometric formula and gas law r = r0e</p><p>-Mgh>RT</p><p>h =</p><p>3</p><p>q</p><p>0</p><p>h dm</p><p>3</p><p>q</p><p>0</p><p>dm</p><p>=</p><p>3</p><p>q</p><p>0</p><p>h r dh</p><p>3</p><p>q</p><p>0</p><p>rdh</p><p>m =</p><p>Mp0S</p><p>RT</p><p>3</p><p>h</p><p>0</p><p>e - Mgh>RT dh =</p><p>p0S</p><p>g</p><p>(1 - e-Mgh>RT )</p><p>dm = r (Sdh) =</p><p>Mp0</p><p>RT</p><p>e-Mgh>RT Sdh</p><p>r =</p><p>Mp0</p><p>RT</p><p>e-Mgh>RT = r0e</p><p>-Mgh>RT</p><p>r =</p><p>pM</p><p>RT</p><p>p = p0e</p><p>-Mgh>RT</p><p>h = -</p><p>RT</p><p>Mg</p><p>ln</p><p>r</p><p>r0</p><p>= 0.09 km (on substituting values)</p><p>(r0 - r)>r0 = 0.01 or r>r0 = 0.99T = 273 K</p><p>=</p><p>RT</p><p>Mg</p><p>= 8 km.</p><p>h = -</p><p>RT</p><p>Mg</p><p>ln e-1</p><p>232 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>So,</p><p>2.19 (a) We know that the variation of pressure with height of a fluid is given by</p><p>But from gas law</p><p>From these two equations</p><p>(1)</p><p>or</p><p>Integrating we get,</p><p>or</p><p>Hence, .</p><p>(b) Proceed up to Eq. (1) of part (a), and then put and proceed fur-</p><p>ther in the same fashion to get</p><p>2.20 Let us consider the mass element of the gas (thin layer) in the</p><p>cylinder at a distance r from its open end as shown in the figure.</p><p>Using Newton’s second law for the element</p><p>(p + dp) S - pS = ( r S dr) �2r</p><p>Fn = mwn</p><p>p =</p><p>p0</p><p>(1 + ah)Mg>aRT0</p><p>T = T0(1 + ah)</p><p>p = p0(1 - ah)Mg>aRT0 aobviously h 6</p><p>1</p><p>a</p><p>b</p><p>ln</p><p>p</p><p>p0</p><p>= ln (1 - ah) Mg>aRT0</p><p>3</p><p>p</p><p>p0</p><p>dp</p><p>p</p><p>=</p><p>-Mg</p><p>RT0 3</p><p>h</p><p>0</p><p>dh</p><p>(1 - ah)</p><p>dp</p><p>p</p><p>=</p><p>-Mgdh</p><p>RT0(1 - ah)</p><p>dp = -</p><p>pMg</p><p>RT</p><p>dh</p><p>p =</p><p>r</p><p>M</p><p>RT or r =</p><p>pM</p><p>RT</p><p>dp = -rgdh</p><p>h =</p><p>3</p><p>q</p><p>0</p><p>h(e-Mgh>RT )dh</p><p>3</p><p>q</p><p>0</p><p>(e-Mgh>RT )dh</p><p>=</p><p>RT</p><p>Mg</p><p>2.1 EQUATION OF THE GAS STATE. PROCESSES 233</p><p>dr</p><p>r</p><p>w</p><p>or</p><p>So,</p><p>or</p><p>Thus,</p><p>2.21 For an ideal gas law</p><p>On substitution, we get</p><p>From Van der Waal gas equation</p><p>or</p><p>2.22 (a) (where is the molar volume)</p><p>(The pressure is less for a Van der Waal gas than for an ideal gas.)</p><p>or</p><p>or</p><p>=</p><p>1.35 * 1.1 * (1 - 0.039)</p><p>0.082 * (0.139)</p><p>L 125 K</p><p>T =</p><p>a (1 + h)(VM - b)</p><p>RVM</p><p>(hVM + b)</p><p>a(1 + h)</p><p>V 2</p><p>M</p><p>= RT c -1</p><p>VM</p><p>+</p><p>-1 + h</p><p>VM - b</p><p>d = RT</p><p>hVM + b</p><p>VM</p><p>(VM - b)</p><p>VMp = c RT</p><p>VM - b</p><p>-</p><p>a</p><p>V 2</p><p>M</p><p>d (1 + h) =</p><p>RT</p><p>VM</p><p>,</p><p>=</p><p>rRT</p><p>M - rb</p><p>-</p><p>ar2</p><p>M 2</p><p>= 79.2 atm</p><p>p =</p><p>nRT</p><p>V - nb</p><p>-</p><p>an2</p><p>V 2</p><p>=</p><p>mRT>M</p><p>V - mb>M -</p><p>am2</p><p>V</p><p>2M 2</p><p>ap +</p><p>n2a</p><p>V</p><p>2</p><p>b(V - nb) = n RT (where V = nVM )</p><p>p = 0.082 * 300 *</p><p>500</p><p>44</p><p>atm = 279.5 atm</p><p>p =</p><p>r</p><p>M</p><p>RT</p><p>ln</p><p>p</p><p>p0</p><p>=</p><p>Mv2</p><p>2RT</p><p>r 2 or p = p0e</p><p>Mv2r 2>2RT</p><p>3</p><p>p</p><p>p0</p><p>dp</p><p>p</p><p>=</p><p>Mv2</p><p>RT 3</p><p>r</p><p>0</p><p>rdr</p><p>dp</p><p>p</p><p>=</p><p>Mv2</p><p>RT</p><p>rdr</p><p>dp = rv2rdr =</p><p>pM</p><p>RT</p><p>v2rdr</p><p>234 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>(b) The corresponding pressure is</p><p>and</p><p>So,</p><p>or</p><p>Also,</p><p>or</p><p>Using we get</p><p>and .</p><p>or</p><p>2.25 For an ideal gas, .</p><p>Now,</p><p>= k0e 1 -</p><p>2b</p><p>V</p><p>+</p><p>2a</p><p>RTV</p><p>f (to leading order in a, b)</p><p>k =</p><p>(V - b)2</p><p>RTV</p><p>e 1 -</p><p>2a (V - b)2</p><p>RTV 3</p><p>f -1</p><p>= k0a1 -</p><p>b</p><p>V</p><p>b2e1 -</p><p>2a</p><p>RTV</p><p>a1 -</p><p>b</p><p>V</p><p>b2 f -1</p><p>k0 = V>RT</p><p>= cRTV 3 - 2a (V - b)2</p><p>V 2(V - b)2</p><p>d -1</p><p>=</p><p>V 2(V - b)2</p><p>[RTV 3 - 2a (V - b)2]</p><p>k =</p><p>-1</p><p>V</p><p>a0V</p><p>0p</p><p>b</p><p>T</p><p>p =</p><p>RT</p><p>V - b</p><p>-</p><p>a</p><p>V 2</p><p>- V a 0p</p><p>0V</p><p>b</p><p>T</p><p>=</p><p>RTV</p><p>(V - b)2</p><p>-</p><p>2a</p><p>V</p><p>2</p><p>b = 0.045 l/moll2>mol2,</p><p>atma = 1.87p2 = 110 atm,V = 0.250,T2 = 350 K,T1 = 300 K, p1 = 90 atm,</p><p>a = V 2</p><p>T1 p2 - p1T2</p><p>T2 - T1</p><p>a</p><p>V 2</p><p>=</p><p>T1 ( p2 - p1)</p><p>T2 - T1</p><p>- p1 =</p><p>T1 p2 - p1 T2</p><p>T2 - T1</p><p>Q</p><p>p1 = T1</p><p>p2 - p1</p><p>T2 - T1</p><p>-</p><p>a</p><p>V</p><p>2</p><p>V - b =</p><p>R (T2 - T1)</p><p>p2 - p1</p><p>Q b = V -</p><p>R (T2 - T1)</p><p>p2 - p1</p><p>p2 - p1 =</p><p>R (T2 - T1)</p><p>V - b</p><p>p2 = RT2</p><p>1</p><p>V - b</p><p>-</p><p>a</p><p>V 2</p><p>p1 = RT1</p><p>1</p><p>V - b</p><p>-</p><p>a</p><p>V 2</p><p>=</p><p>1.35</p><p>1</p><p>*</p><p>0.961</p><p>0.139</p><p>L 9.3 atm</p><p>=</p><p>a</p><p>V 2</p><p>M</p><p>(VM + hVM - hVM - b)</p><p>(hVM + b)</p><p>=</p><p>a</p><p>V 2</p><p>M</p><p>(VM - b)</p><p>(VM + b)</p><p>p =</p><p>RT</p><p>VM - b</p><p>-</p><p>a</p><p>V 2</p><p>M</p><p>=</p><p>a (1 + h)</p><p>VM</p><p>(hVM + b)</p><p>-</p><p>a</p><p>V 2</p><p>M</p><p>2.1 EQUATION OF THE GAS STATE. PROCESSES 235</p><p>2.23</p><p>2.24</p><p>Since,</p><p>Now,</p><p>If a and b do not vary much with temperature, then the effect at high temperature is</p><p>clearly determined by b and its effect is repulsive so compressibility is less.</p><p>2.2 The First Law of Thermodynamics. Heat Capacity</p><p>2.26 Internal energy of air, treating it as an ideal gas is given by</p><p>(1)</p><p>(using ( and ).</p><p>Thus at constant pressure because the volume of the room is a constant.</p><p>Putting the value of and V in Eq. (1), we get</p><p>2.27 From energy conservation</p><p>or (1)</p><p>Using, (from previous problem) (2)</p><p>We get</p><p>Hence from Eqs. (1) and (2),</p><p>2.28 On opening the valve, the air will flow from the vessel at higher pressure to the ves-</p><p>sel at lower pressure till both vessels have the same air pressure. If this air pressure</p><p>is p, the total volume of the air in the two vessels will be Also if and</p><p>be the number of moles of air initially in the two vessels, we have</p><p>(1)</p><p>After the air is mixed, the total number of moles are and the mixture is at tem-</p><p>perature T.</p><p>Hence, (2)</p><p>Let us look at the two portions of air as one single system. Since this system is con-</p><p>tained in a thermally insulated vessel, no heat exchange is involved in the process.</p><p>That is, total heat transfer for the combined system Q = 0.</p><p>p (V1 + V2) = (n1 + n2)RT</p><p>(v1 + v2)</p><p>p1V1 = n1RT1 and p2V2 = n2RT2</p><p>n2n1(V1 + V2).</p><p>¢T =</p><p>Mv 2 (g - 1)</p><p>2R</p><p>¢U =</p><p>nR</p><p>g - 1</p><p>¢T</p><p>U =</p><p>nRT</p><p>g - 1</p><p>¢U = (1>2)nMv</p><p>2</p><p>Ui +</p><p>1</p><p>2</p><p>(nM ) v</p><p>2 = Uƒ</p><p>U = 10 M J.p = patm</p><p>U = constant,</p><p>Cp >CV = gCV = R>(g - 1), since Cp - CV = R</p><p>U =</p><p>m</p><p>M</p><p>CVT =</p><p>m</p><p>M</p><p>R</p><p>g - 1</p><p>T =</p><p>pV</p><p>g - 1</p><p>k 7 k0 if</p><p>2a</p><p>RTV</p><p>7</p><p>2b</p><p>V</p><p>or T 6</p><p>a</p><p>bR</p><p>236 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>Moreover, this combined system does not perform mechanical work either. The walls of</p><p>the containers are rigid and there are no pistons, etc to be pushed, looking at the total system,</p><p>we know</p><p>Hence, internal energy of the combined system does not change in the process.</p><p>Initially, energy of the combined system is equal to the sum of internal energies of the</p><p>two portions of air:</p><p>(3)</p><p>Final internal energy of moles of air at temperature T is given by</p><p>(4)</p><p>Therefore, implies</p><p>From Eq. (2) therefore, final pressure is given by</p><p>This process in an example of free adiabatic expansion of ideal gas.</p><p>2.29 By the first law of thermodynamics, .</p><p>Here as the volume remains constant.</p><p>So,</p><p>From gas law,</p><p>So,</p><p>Hence amount of heat lost</p><p>2.30 By the first law of thermodynamics, .</p><p>But,</p><p>So, Q =</p><p>A</p><p>g - 1</p><p>+ A =</p><p>g # A</p><p>g - 1</p><p>=</p><p>1.4</p><p>1.4 - 1</p><p>* 2 = 7 J</p><p>¢U =</p><p>p ¢V</p><p>g - 1</p><p>=</p><p>A</p><p>g - 1</p><p>(as p is constant)</p><p>Q = ¢U + A</p><p>= - ¢U = 0.25 k J</p><p>¢U = -</p><p>p0V¢T</p><p>T0(g - 1)</p><p>= -0.25 k J</p><p>p0V = nRT0</p><p>Q = ¢U =</p><p>nR</p><p>g - 1</p><p>¢T</p><p>A = 0,</p><p>Q = ¢U + A</p><p>p =</p><p>n1 + n2</p><p>V1 + V2</p><p>RT =</p><p>R</p><p>V1 + V2</p><p>(n1T1 + n2T2) =</p><p>p1V1 + p2V2</p><p>V1 + V2</p><p>T =</p><p>n1T1 +</p><p>n2T2</p><p>n1 + n2</p><p>=</p><p>p1V1 + p2V2</p><p>(p1V1>T1) + (p2V2>T2)</p><p>= T1T2</p><p>p1V1 + p2V2</p><p>p1V1T2 + p2V2T1</p><p>Ui = Uf</p><p>Uƒ =</p><p>(n1 + n2) RT</p><p>g - 1</p><p>(n1 + n2)</p><p>Ui = U1 + U2 =</p><p>n1RT1</p><p>g - 1</p><p>+</p><p>n2RT2</p><p>g - 1</p><p>A = 0.</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 237</p><p>2.31 Under an isobaric process,</p><p>From the first law of thermodynamics,</p><p>Again increment in internal energy, (for )</p><p>Thus,</p><p>2.32 Given moles of the gas. In the first phase, under isochoric process, there-</p><p>fore from gas law if pressure is reduced n times, the temperature, i.e., new temperature</p><p>becomes</p><p>Now from first law of thermodynamics,</p><p>During the second phase (under isobaric process),</p><p>Thus from first law of thermodynamics,</p><p>Hence the total amount of heat absorbed,</p><p>2.33 Number of moles in the mixture</p><p>At a certain temperature,</p><p>Thus, CV =</p><p>n1CV1</p><p>+ n2CV2</p><p>n</p><p>=</p><p>an1 R</p><p>g1 - 1</p><p>+ n2</p><p>R</p><p>g2 - 1</p><p>b</p><p>n</p><p>U = U1 + U2 or nCV = n1CV1</p><p>+ n2CV2</p><p>n = n1 + n2</p><p>=</p><p>nRT0(n - 1)</p><p>n (g - 1)</p><p>(-1 + g) = nRT0a1 -</p><p>1</p><p>n</p><p>b = 2.5 kJ (on substituting values)</p><p>Q = Q1 + Q2 =</p><p>nRT0(1 - n)</p><p>n (g - 1)</p><p>+</p><p>nRT0(n - 1)g</p><p>n (g - 1)</p><p>=</p><p>nR aT0 -</p><p>T0</p><p>n</p><p>bg</p><p>g - 1</p><p>=</p><p>nRT0(n - 1)g</p><p>n (g - 1)</p><p>Q2 = ¢U2 + A2 =</p><p>nR¢T</p><p>g - 1</p><p>+ nR¢T</p><p>A2 = p¢V = nR ¢T.</p><p>=</p><p>nR</p><p>g - 1</p><p>aT0</p><p>n</p><p>- T0b =</p><p>nRT0 (1 - n)</p><p>n (g - 1)</p><p>Q1 = ¢U1 =</p><p>nR¢T</p><p>g - 1</p><p>T0>n.</p><p>A1 = 0,n = 2</p><p>Q - R ¢T =</p><p>R ¢T</p><p>g - 1</p><p>or g =</p><p>Q</p><p>Q - R ¢T</p><p>= 1.6</p><p>n = 1¢U =</p><p>R¢T</p><p>g - 1</p><p>¢U = Q - A = Q - R¢T = 1 k J</p><p>A = p¢V = R¢T (as n = 1) = 0.6 k J.</p><p>238 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>Similarly,</p><p>Thus,</p><p>2.34 From the previous problem,</p><p>and</p><p>Now molar mass of the mixture</p><p>Hence,</p><p>2.35 Let S be the area of the piston and F be the force exerted by the external agent.</p><p>Then, (see figure) at an arbitrary instant of time. Here p is the pressure at</p><p>the instant the volume is V. (Initially the pressure inside is p0 ). Work done by the agent,</p><p>A</p><p>= 3</p><p>hV0</p><p>V0</p><p>(p0 - p)S # dx = 3</p><p>hV0</p><p>V0</p><p>(p0 - p)dV</p><p>= 3</p><p>hV0</p><p>V0</p><p>F dx</p><p>F + pS = p0S</p><p>cV =</p><p>CV</p><p>M</p><p>= 0.42 J>gK and cp =</p><p>Cp</p><p>M</p><p>= 0.66 J>gK</p><p>(M ) =</p><p>Total mass</p><p>Total number of moles</p><p>=</p><p>20 + 7</p><p>(1>2) + (1>4)</p><p>= 36</p><p>Cp =</p><p>n1</p><p>g1R</p><p>g1 - 1</p><p>+ n2</p><p>g2R</p><p>g2 - 1</p><p>n1 + n2</p><p>= 23.85 J>mol K (on substituting values)</p><p>CV =</p><p>n1</p><p>R</p><p>g1 - 1</p><p>+ n2</p><p>R</p><p>g2 - 1</p><p>n1 + n2</p><p>= 15.2 J>mol K (on substituting values)</p><p>= 1.33 (on substituting values)</p><p>=</p><p>n1g1(g2 - 1) + n2g2(g1 - 1)</p><p>n1(g2 - 1) + n2(g1 - 1)</p><p>g =</p><p>Cp</p><p>CV</p><p>=</p><p>n1</p><p>g1</p><p>g1 - 1</p><p>R + n2</p><p>g2</p><p>g2 - 1</p><p>R</p><p>n1</p><p>R</p><p>g1 - 1</p><p>+ n2</p><p>R</p><p>g2 - 1</p><p>=</p><p>n1g1CV1</p><p>+ n2g2CV2</p><p>n</p><p>=</p><p>an1 g1R</p><p>g1 - 1</p><p>+ n2</p><p>g2R</p><p>g2 - 1</p><p>b</p><p>n</p><p>Cp =</p><p>n1Cp1</p><p>+ n2Cp2</p><p>n</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 239</p><p>pS</p><p>F</p><p>p S0</p><p>V</p><p>2.36 Let the agent move the piston to the right by x. In equilibrium position,</p><p>Work done by the agent in an infinitesimal change dx is</p><p>By applying for the two parts,</p><p>So,</p><p>When the volume of the left end is times the volume of the right end,</p><p>2.37 In the isothermal process, heat transfer to the gas is given by</p><p>In the isochoric process, . Thus heat transfer to the gas is given by</p><p>Q2 = ¢U = nCV ¢T =</p><p>nR</p><p>g - 1</p><p>¢T a for CV =</p><p>R</p><p>g - 1</p><p>bA = 0</p><p>Q1 = nRT0 ln</p><p>V2</p><p>V1</p><p>= n RT0 ln h a for h =</p><p>V2</p><p>V1</p><p>=</p><p>p1</p><p>p2</p><p>b</p><p>= -p0V0a ln 4h</p><p>(h + 1)2</p><p>b = p0V0 ln</p><p>(h + 1)2</p><p>4h</p><p>= -p0V0 c lneV 2</p><p>0 - ah - 1</p><p>h + 1</p><p>b2</p><p>V 2</p><p>0 f - ln V 2</p><p>0 d</p><p>= -p0V0 3ln1V 2</p><p>0 - V 22 - ln V 2</p><p>04</p><p>A = 3</p><p>v</p><p>0</p><p>(p2 - p1)dV = 3</p><p>V</p><p>0</p><p>2p0V0V</p><p>V 2</p><p>0 - V 2</p><p>dV = -p0V0 [ln (V 2</p><p>0 - V 2)]V0</p><p>(V0 - V) = h (V0 - V ) or V =</p><p>h - 1</p><p>h + 1</p><p>V0</p><p>h</p><p>p2 - p1 =</p><p>p0V0 2Sx</p><p>V 2</p><p>0</p><p>- S</p><p>2x</p><p>2 =</p><p>2p0V0V</p><p>V 2</p><p>0 - V 2</p><p>(where Sx = V )</p><p>p1 (V0 + Sx) = p0V0 and p2 (V0 - Sx) = p0V0</p><p>pV = constant,</p><p>Fagent</p><p># dx = (p2 - p1) Sdx = (p2 - p1)dV</p><p>p1S + Fagent = p2S or Fagent = (p2 - p1)S</p><p>= nRT (h - 1 - ln h) = RT (h - 1 - ln h) (for n = 1 mole)</p><p>= (h - 1) p0V0 - nRT ln h = (h - 1) nRT - nRT ln h</p><p>= p0(h - 1)V0 - 3</p><p>hV0</p><p>V0</p><p>pdV = p0(h - 1)V0 - 3</p><p>hV0</p><p>V0</p><p>nRT # dV</p><p>V</p><p>240 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>But,</p><p>or</p><p>So,</p><p>Thus, net heat transfer to the gas is</p><p>or</p><p>or</p><p>2.38 (a) From ideal gas law,</p><p>For isochoric process, thus represents a straight line pass-</p><p>ing through the origin and its slope is equal to k.</p><p>For isobaric process constant, thus on curve, it is a horizontal straight line</p><p>parallel to T-axis, if T is along horizontal (or x-axis).</p><p>For isothermal process, constant, thus on curve, it represents a vertical</p><p>straight line if T is taken along horizontal (or -axis).</p><p>For an adiabatic process constant.</p><p>On differentiating, we get</p><p>The approximate plots of isochoric, isobaric, isothermal, and adiabatic processes</p><p>are drawn in the answer sheet.</p><p>(b) As p is not considered variable, we have from ideal gas law</p><p>On co-ordinate system let us, take T along -axis.</p><p>For an isochoric process, constant, thus constant and obviously</p><p>represents a straight line passing through the origin of the coordinate system and</p><p>is its slope. k¿</p><p>V = k¿Tk ¿ =V =</p><p>xV-T</p><p>V =</p><p>nR</p><p>p</p><p>T = k¿T awhere k¿ =</p><p>nR</p><p>p b</p><p>dp</p><p>dT</p><p>= a g</p><p>1 - g</p><p>b ap 1-g</p><p>p-g b aT</p><p>g-1</p><p>T g</p><p>b = a g</p><p>g - 1</p><p>b p</p><p>T</p><p>(1 - g) p-gdp T</p><p>g + gp1-g # T g-1 # dT = 0</p><p>T</p><p>gp1-g =</p><p>x</p><p>p-TT =</p><p>p-Tp =</p><p>p = kT,V = constant,</p><p>p = 1nR>V2T = kT 1where k = nR>V 2</p><p>g = 1 +</p><p>h - 1</p><p>Q</p><p>nRT0</p><p>- ln h</p><p>= 1 +</p><p>6 - 1a 80 * 103</p><p>3 * 8.314 * 273</p><p>b - ln 6</p><p>= 1.4</p><p>Q</p><p>nRT0</p><p>= ln h +</p><p>h - 1</p><p>g - 1</p><p>Q Q</p><p>nRT0</p><p>- ln h =</p><p>h - 1</p><p>g - 1</p><p>Q = nRT0 ln h +</p><p>nR</p><p>g - 1</p><p>. (h - 1)T0</p><p>Q2 =</p><p>nR</p><p>g - 1</p><p>. (h - 1)T0</p><p>¢T = hT0 - T0 = (h - 1)T0</p><p>p2</p><p>p1</p><p>=</p><p>T0</p><p>T</p><p>Q T = T0</p><p>p1</p><p>p2</p><p>= hT0 a for h =</p><p>p1</p><p>p2</p><p>b</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 241</p><p>.</p><p>So,</p><p>For isothermal process constant. Thus on the stated coordinate system it repr-</p><p>esents a straight line parallel to the -axis.</p><p>For an adiabatic process, constant.</p><p>On differentiating, we get</p><p>The approximate plots of isochoric, isobaric, isothermal and adiabatic processes</p><p>are drawn in the answer sheet.</p><p>2.39 (a) According to relation in adiabatic process, (where k � constant).</p><p>Therefore,</p><p>So,</p><p>Hence,</p><p>(b) Using the solution of part (a), sought work done</p><p>� 5.61 kJ (on substituting values)</p><p>2.40 Let (p0, V0, T0) be the initial state of the gas.</p><p>We know that work done by the gas in adiabatic process is</p><p>But from the equation T V �–1 � constant, we get �T � T0 (���1 – 1).</p><p>Thus, Aadia</p><p>On the other hand, work done by the gas in isothermal process is</p><p>Thus,</p><p>2.41 Since here the piston is conducting and it is moved slowly, the temperature on the two</p><p>sides increases and is maintained at the same value.</p><p>Elementary work done by the agent � work done in compression – work done in</p><p>expansion i.e. where are pressures at any</p><p>instant of the gas on expansion and compression side, respectively.</p><p>p1 and p2dA p2dV – p1dV = (p2 – p1) dV</p><p>Aadia</p><p>A iso</p><p>=</p><p>hg-1 - 1</p><p>(g - 1) ln h</p><p>=</p><p>50.4 - 1</p><p>0.4 * ln 5</p><p>= 1.4</p><p>A iso = nRT0 ln (1>h) = -nRT0 ln h</p><p>=</p><p>-nRT0(h</p><p>g-1 - 1)</p><p>g - 1</p><p>Aadia =</p><p>-nR ¢T</p><p>g - 1</p><p>A =</p><p>nR¢T</p><p>g - 1</p><p>=</p><p>nRT0</p><p>g - 1</p><p>1h(g–1)/g -12</p><p>T = T0</p><p># hg-1>g = 290 * 10(1.4-1)>1.4 = 0.56 kK</p><p>T</p><p>g>T0</p><p>g = hg-1 1for h = p2>p12 1T2>T12g = 1p2>p12g-1</p><p>T g = kp g-1T-p</p><p>dV</p><p>dT</p><p>= - a 1</p><p>g - 1</p><p>b # V</p><p>T</p><p>(g - 1)V g-2 dVT + V g-1dT = 0</p><p>T V g-1 =</p><p>V</p><p>T =</p><p>242 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>, , = ,</p><p>So,</p><p>From the gas law and , for each section</p><p>(x is the displacement of the piston towards section 2).</p><p>So, (as Sx � V )</p><p>So,</p><p>Also, from the first law of thermodynamics</p><p>dA � �dU � �2 dT (as dQ � 0)</p><p>So, work done on the gas � �dA � 2 dT</p><p>Thus, 2 dT �</p><p>or</p><p>When the left end is � times the volume of the right end,</p><p>(V0 	 V ) � �(V0 � V ) or V � V0</p><p>On integrating,</p><p>or ln � (� – 1)</p><p>�</p><p>�</p><p>Hence,</p><p>2.42 From energy conservation as in the derivation of Bernoulli’s theorem it reads</p><p>2 	 gz 	 u 	 Qd � constant (1)</p><p>1</p><p>2</p><p>p</p><p>r</p><p>T = T0 a (h + 1)2</p><p>4h</p><p>b g-1</p><p>2</p><p>c ln V 2</p><p>0 - ln V 2</p><p>0</p><p>e 1 - ah - 1</p><p>h + 1</p><p>b2 f d =</p><p>g - 1</p><p>2</p><p>In</p><p>(h + 1)2</p><p>4h</p><p>g - 1</p><p>2</p><p>[ln 1V 2</p><p>0 - V 22 - ln V 2</p><p>0]-</p><p>g - 1</p><p>2</p><p>c - 1</p><p>2</p><p>ln 1V 2</p><p>0 - V 22 dV</p><p>0</p><p>T</p><p>T0</p><p>3</p><p>V</p><p>0</p><p>VdV</p><p>V 2</p><p>0 - V 23</p><p>T</p><p>T0</p><p>dT</p><p>T</p><p>= (g - 1)</p><p>h - 1</p><p>h + 1</p><p>dT</p><p>T</p><p>= g - 1</p><p>VdV</p><p>V 20 - V</p><p>2</p><p>nRT</p><p>2V # dV</p><p>V 20 - V</p><p>2</p><p>R</p><p>g - 1</p><p>n</p><p>R</p><p>g - 1</p><p>n</p><p>R</p><p>g - 1</p><p>n</p><p>p S1</p><p>p S2</p><p>Fagent</p><p>dA = nRT</p><p>2V</p><p>V 20 -</p><p>v</p><p>c - 1</p><p>h</p><p>+</p><p>x</p><p>2l 2 + x 2</p><p>d = 0</p><p>dt</p><p>dx</p><p>= 0</p><p>12 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1 2 3 4 5 6</p><p>ec</p><p>7</p><p>ba</p><p>d t</p><p>–2</p><p>–1</p><p>0</p><p>1</p><p>vx</p><p>1 2 3 4 5 6 7</p><p>t</p><p>x(t)</p><p>1</p><p>2</p><p>3</p><p>0</p><p>31 2 4 5 6 7</p><p>t</p><p>s(t)</p><p>1</p><p>2</p><p>3</p><p>4</p><p>5</p><p>6</p><p>1/2</p><p>5/2</p><p>7/2</p><p>0</p><p>1 2 3 4 5 6 7</p><p>t</p><p>1</p><p>2</p><p>a x</p><p>1.19 (a) Mean velocity</p><p>(1)</p><p>(b) Modulus of mean velocity vector (2)</p><p>(c) Let the point moves from i to f along the half circle (see figure) and and be</p><p>the speed at the points respectively.</p><p>We have</p><p>or</p><p>So, (3)</p><p>So, from Eqs. (1) and (3)</p><p>(4)</p><p>Now, the modulus of the mean vector of total acceleration</p><p>(see figure) (5)</p><p>Using Eq. (4) in Eq. (5), we get</p><p>(on substituting values)</p><p>1.20 (a) We have</p><p>So,</p><p>and</p><p>(b) From the equation</p><p>at and also at t = ¢t =</p><p>1</p><p>a</p><p>t = 0r = 0,</p><p>r = at (1 - at)</p><p>w =</p><p>dv</p><p>dt</p><p>= -2aa</p><p>v =</p><p>dr</p><p>dt</p><p>= a (1 - 2at)</p><p>r = at (1 - at)</p><p>| < w > | =</p><p>2pR</p><p>t2</p><p>= 10 cm/s2</p><p>| < w > | =</p><p>| ¢v |</p><p>¢t</p><p>=</p><p>| v - v0 |</p><p>t</p><p>=</p><p>v0 + v</p><p>t</p><p>v0 + v</p><p>2</p><p>=</p><p>pR</p><p>t</p><p><v > =</p><p>3</p><p>t</p><p>0</p><p>(v0 + wt</p><p>t)dt</p><p>3</p><p>t</p><p>0</p><p>dt</p><p>=</p><p>v0 + (v0 + wt</p><p>t)</p><p>2</p><p>=</p><p>v0 + v</p><p>2</p><p>v = v0 + wt</p><p>t (as wt is constant, according to the problem)</p><p>dv</p><p>dt</p><p>= wt</p><p>vv0</p><p>| <v> |=</p><p>| ¢r |</p><p>¢t</p><p>=</p><p>2R</p><p>t</p><p>= 32 cm/s</p><p>=</p><p>s</p><p>t</p><p>=</p><p>pR</p><p>t</p><p>= 50 cm/s</p><p>< v > =</p><p>Total distance covered</p><p>Time elapsed</p><p>1.1 KINEMATICS 13</p><p>R</p><p>f</p><p>vi</p><p>v0</p><p>So, the sought time</p><p>As</p><p>So,</p><p>Hence, the sought distance</p><p>Simplifying, we get,</p><p>1.21 (a) As the particle leaves the origin at</p><p>So, (1)</p><p>As</p><p>(where is directed towards the + x-axis).</p><p>So, (2)</p><p>From Eqs. (1) and (2),</p><p>(3)</p><p>Hence x coordinate of the particle at s</p><p>Similarly at</p><p>and at</p><p>x = 10 * 20 a1 -</p><p>20</p><p>2 * 5</p><p>b = -200 cm = -2 m</p><p>t = 20 s</p><p>x = 10 * 10 a1 -</p><p>10</p><p>2 * 5</p><p>b = 0</p><p>t = 10 s</p><p>x = 10 * 6 a1 -</p><p>6</p><p>2 * 5</p><p>b = 24 cm = 0.24 m</p><p>t = 6</p><p>x = 3</p><p>t</p><p>0</p><p>v0a1 -</p><p>t</p><p>t</p><p>bdt = v0t a1 -</p><p>t</p><p>2t</p><p>b</p><p>vx = v0a1 -</p><p>t</p><p>t</p><p>bv0</p><p>v = v0a1 -</p><p>t</p><p>t</p><p>b</p><p>¢x = x = 3vx dt</p><p>t = 0</p><p>s =</p><p>a</p><p>2a</p><p>s = 3v dt = 3</p><p>1>2a</p><p>0</p><p>a (1 - 2a t) dt + 3</p><p>1>a</p><p>1>2aa (2at - 1) dt</p><p>v = | v | = d a (1 - 2at) for t …</p><p>1</p><p>2a</p><p>a (2at - 1) for t 7</p><p>1</p><p>2a</p><p>v = a (1 - 2at)</p><p>¢t =</p><p>1</p><p>a</p><p>14 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>ve</p><p>(b) At the moment the particle is at a distance of 10 cm from the</p><p>Putting in Eq. (3)</p><p>So,</p><p>Now putting in Eq. (3)</p><p>On solving,</p><p>As t cannot be negative, so,</p><p>Hence the particle is at a distance of 10 cm from the origin at three moments of time:</p><p>(c) We have</p><p>So,</p><p>So, (4)</p><p>and</p><p>(5)</p><p>For s</p><p>s = 3</p><p>5</p><p>0</p><p>10a1 -</p><p>t</p><p>5</p><p>b dt + 3</p><p>8</p><p>5</p><p>10a t</p><p>5</p><p>- 1b dt</p><p>t = 8</p><p>s = 3</p><p>4</p><p>0</p><p>v0a1 -</p><p>t</p><p>t</p><p>b dt = 3</p><p>4</p><p>0</p><p>10a1 -</p><p>t</p><p>5</p><p>b dt = 24 cm</p><p>= v0t[1 + (1 - t>t)2]>2 for t 7 t</p><p>s = 3</p><p>t</p><p>0</p><p>v0a1 -</p><p>t</p><p>t</p><p>b dt + 3</p><p>t</p><p>t</p><p>v0a t</p><p>t</p><p>- 1b dt for t 7 t</p><p>s = 3</p><p>t</p><p>0</p><p>v0a1 -</p><p>t</p><p>t</p><p>b dt for t … t = v0t (1 - t/2t)</p><p>v = | v | =</p><p>v0a1 -</p><p>t</p><p>t</p><p>b for t … t</p><p>v0a t</p><p>t</p><p>- 1b for t 7 t</p><p>v = v0a1 -</p><p>t</p><p>t</p><p>bt = 5 � 215 s, 5 + 235 s</p><p>t = (5 + 235) s</p><p>t = 5 � 235 s</p><p>-10 = 10a1 -</p><p>t</p><p>10</p><p>bx = -10</p><p>t =</p><p>10 � 2100 - 40</p><p>2</p><p>= 5�215 s</p><p>10 = 10t a1 -</p><p>t</p><p>10</p><p>b or t 2 - 10t + 10 = 0</p><p>x = +10</p><p>x = �10 cm.</p><p>1.1 KINEMATICS 15</p><p>origin,</p><p>t</p><p>On integrating and simplifying, we get</p><p>cm</p><p>On the basis of Eqs. (3) and (4), x(t) and s(t) plots can be drawn as shown in the</p><p>answer sheet.</p><p>1.22 (a) As particle is in unidirectional motion, it is directed along the x-axis all the time. At</p><p>.</p><p>So, and</p><p>According to the problem</p><p>So,</p><p>and (1)</p><p>As, so,</p><p>On integrating, (2)</p><p>(b) Let t be the time to cover first s m of the path. From Eq. (2)</p><p>Hence, (3)</p><p>The mean velocity of particle</p><p>1.23 According to the problem</p><p>(as v decreases with time)</p><p>or</p><p>On integrating we get, s =</p><p>2</p><p>3a</p><p>v</p><p>3>2</p><p>0</p><p>-3</p><p>0</p><p>v0</p><p>1v dv = a3</p><p>s</p><p>0</p><p>ds</p><p>-</p><p>v dv</p><p>ds</p><p>= a1v</p><p><v > = 1v (t) dt</p><p>1dt</p><p>=</p><p>3</p><p>21s>a</p><p>0</p><p>a2</p><p>2</p><p>t dt</p><p>21s>a =</p><p>a1s</p><p>2</p><p>t =</p><p>2</p><p>a</p><p>1s</p><p>s = 3v dt = 3</p><p>t</p><p>0</p><p>a2</p><p>2</p><p>t dt =</p><p>a2</p><p>2</p><p>t</p><p>2</p><p>2</p><p>v =</p><p>a 2</p><p>2</p><p>t = v (t)</p><p>dv =</p><p>a 2</p><p>2</p><p>dtw =</p><p>dv</p><p>dt</p><p>=</p><p>a 2</p><p>2</p><p>w =</p><p>1</p><p>2</p><p>d</p><p>ds</p><p>(v 2) =</p><p>1</p><p>2</p><p>d</p><p>ds</p><p>(a 2s) =</p><p>a 2</p><p>2</p><p>v 2 = a 2s</p><p>v = a1x = a1s</p><p>w =</p><p>dv</p><p>dt</p><p>=</p><p>v dv</p><p>ds</p><p>=</p><p>1</p><p>2</p><p>d</p><p>ds</p><p>(v 2)¢x = x = s</p><p>t = 0, x = 0</p><p>s = 34</p><p>16 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Again according to the problem</p><p>or</p><p>Thus,</p><p>1.24 (a) As</p><p>So,</p><p>and therefore</p><p>which is equation of a parabola, whose graph is shown</p><p>in the figure.</p><p>(b) As</p><p>(1)</p><p>So,</p><p>Differentiating Eq. (1) with respect to time, we get</p><p>So,</p><p>(c)</p><p>or</p><p>So,</p><p>(d) The mean velocity vector</p><p>Hence, | 6v7 | = 2a2 + (-bt)2 = 2a</p><p>2 + b</p><p>2t</p><p>2</p><p>6v7 =</p><p>1vdt</p><p>1dt</p><p>=</p><p>1</p><p>t</p><p>0</p><p>(a i - 2bt j)dt</p><p>t</p><p>= a i - bt j</p><p>tan a =</p><p>a</p><p>2 bt</p><p>or a = tan-1a a</p><p>2bt</p><p>b</p><p>cos a =</p><p>2bt</p><p>2a2 + 4b</p><p>2t</p><p>2</p><p>cos a =</p><p>v # w</p><p>vw</p><p>=</p><p>(a i - 2bt j) # (-2b j)</p><p>(2a</p><p>2 + 4b</p><p>2 t</p><p>2) 2b</p><p>| w | = w = 2b</p><p>w =</p><p>d v</p><p>dt</p><p>= -2b j</p><p>v = 2a2 + (-2bt )2 = 2a2 + 4b</p><p>2t</p><p>2</p><p>v =</p><p>d r</p><p>dt</p><p>= a i - 2b j</p><p>r = at i - bt</p><p>2 j</p><p>y =</p><p>–bx 2</p><p>a 2</p><p>x = at, y = -bt</p><p>2</p><p>r = at i - bt</p><p>2 j</p><p>t =</p><p>2 1v0</p><p>a</p><p>-3</p><p>0</p><p>v0</p><p>dv</p><p>1v</p><p>= a3</p><p>t</p><p>0</p><p>dt</p><p>-</p><p>dv</p><p>dt</p><p>= a1v or -</p><p>dv</p><p>2v</p><p>= a dt</p><p>1.1 KINEMATICS 17</p><p>y</p><p>xO</p><p>1.25 (a) We have</p><p>(1)</p><p>Hence, y (x) becomes</p><p>(parabola)</p><p>(b) Differentiating Eq. (1) we get</p><p>(2)</p><p>So,</p><p>Differentiating Eq. (2) with respect to time</p><p>So, (3)</p><p>(c) From Eqs. (2) and (3) we have</p><p>So at</p><p>On simplifying,</p><p>As</p><p>1.26 Differentiating motion law: with respect to time,</p><p>.</p><p>So, (1)</p><p>and (2)</p><p>Differentiating Eq. (1) with respect to time</p><p>(3)</p><p>(a) The distance s traversed by the point during the time is given by</p><p>(using Eq. 2)</p><p>(b) Taking scalar product of v and w</p><p>We get, v # w = (av cos vt i + a v sin vt j) # (av 2 sin vt (- i) + a v 2 cos vt j)</p><p>s = 3</p><p>t</p><p>0</p><p>v dt = 3</p><p>t</p><p>0</p><p>a v dt = avt</p><p>t</p><p>w =</p><p>d v</p><p>dt</p><p>= -a v2 sin vt i + a v2 cos vt j</p><p>v = a v = constant</p><p>v = a v cos vt i + a v sin vt j</p><p>vx = a v cos vt and vy = a v sin vt</p><p>x = a sin vt, y = a (1 - cos vt),</p><p>t0 Z 0, t0 =</p><p>1</p><p>a</p><p>1 - 2at0 = �1</p><p>cos</p><p>p</p><p>4</p><p>=</p><p>1</p><p>22</p><p>=</p><p>v # w</p><p>vw</p><p>=</p><p>-a (1 - 2a t0)2aa</p><p>a21 + (1 - 2 at0)</p><p>2 2a a</p><p>t - t0,</p><p>v = a i + a (1 - 2at) j and w = 2a a j.</p><p>w = 2w x</p><p>2 + w y</p><p>2 = 2a a</p><p>wx = 0 and wy = -2a a</p><p>v = 2v x</p><p>2 + v y</p><p>2 = a 21 + (1 - 2at)2</p><p>vx = a and vy = a (1 - 2 at)</p><p>y =</p><p>ax</p><p>a</p><p>a1 -</p><p>ax</p><p>a</p><p>b = x -</p><p>a</p><p>a</p><p>x</p><p>2</p><p>x = at and y = at (1 - at)</p><p>18 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>So,</p><p>Thus, i.e., the angle between velocity vector and acceleration vector equals .</p><p>1.27 According to the problem</p><p>So, (1)</p><p>Differentiating equation of trajectory, with respect to time</p><p>(2)</p><p>So,</p><p>Again differentiating with respect to time</p><p>or (using Eq. 1)</p><p>or (using Eq. 1) (3)</p><p>Using Eq. (3) in Eq. (2) (4)</p><p>Hence, the velocity of the particle at the origin</p><p>(using Eqs. 3 and 4)</p><p>Hence,</p><p>1.28 As the body is under gravity of constant acceleration g, its velocity vector and</p><p>displacement vectors are:</p><p>(1)</p><p>and (2)¢r = r = v0t +</p><p>1</p><p>2</p><p>gt 2 (as r = 0 at t = 0)</p><p>v = v0 + gt</p><p>v = A</p><p>w</p><p>2b</p><p>(1 + a 2)</p><p>v = A adx</p><p>dt</p><p>b</p><p>x =0</p><p>2</p><p>+ ady</p><p>dt</p><p>b</p><p>x =0</p><p>2</p><p>= A</p><p>w</p><p>2b</p><p>+ a</p><p>2</p><p>w</p><p>2b</p><p>dy</p><p>dt</p><p>`</p><p>x = 0</p><p>= a A</p><p>w</p><p>2b</p><p>dx</p><p>dt</p><p>= A</p><p>w</p><p>2b</p><p>-w = a (0) - 2b adx</p><p>dt</p><p>b2</p><p>- 2bx (0)</p><p>d</p><p>2y</p><p>dt 2</p><p>=</p><p>ad 2x</p><p>dt 2</p><p>- 2b adx</p><p>dt</p><p>b2</p><p>- 2b x</p><p>d</p><p>2x</p><p>dt</p><p>2</p><p>dy</p><p>dt</p><p>`</p><p>x = 0</p><p>=</p><p>a dx</p><p>dt</p><p>`</p><p>x = 0</p><p>dy</p><p>dt</p><p>=</p><p>a d x</p><p>dt</p><p>- 2bx</p><p>dx</p><p>dt</p><p>y = ax - bx 2,</p><p>wx =</p><p>dvx</p><p>dt</p><p>= 0 and wy =</p><p>dvy</p><p>dt</p><p>= -w</p><p>w = w (- j)</p><p>p 2v � w,</p><p>v # w = -a 2v 2 sin vt cos vt + a 2v 3 sin vt cos vt = 0</p><p>1.1 KINEMATICS 19</p><p>/</p><p>So, �v � over the first t seconds</p><p>(3)</p><p>Hence from Eq. (3), �v� over the first seconds</p><p>(4)</p><p>For evaluating t, take</p><p>or</p><p>But we have at and also at (also from energy conservation).</p><p>Hence using this property in Eq. (5),</p><p>As so,</p><p>Putting this value of in Eq. (4), the average velocity over the time of flight</p><p>1.29 The body thrown in air with velocity at an angle from the horizontal lands at</p><p>point P on the Earth’s surface at same horizontal level (see figure). The point of pro-</p><p>jection is taken as origin, so and .</p><p>(a) From the equation</p><p>or</p><p>As so, time of motion</p><p>(b) At the maximum height of ascent,</p><p>so, from the equation</p><p>Hence maximum height</p><p>During the time of motion the net horizontal displacement or horizontal range,</p><p>will be obtained by the equation</p><p>¢x = v0xt +</p><p>1</p><p>2</p><p>wxt</p><p>2</p><p>H =</p><p>v 0</p><p>2 sin2</p><p>a</p><p>2g</p><p>0 = (v0 sin a)2 - 2gH</p><p>v y</p><p>2 = v 0y</p><p>2 + 2wy ¢y</p><p>vy = 0</p><p>t =</p><p>2v0 sin a</p><p>g</p><p>t Z 0,</p><p>0 = v0 sin at -</p><p>1</p><p>2</p><p>g t2</p><p>¢y = v0y t +</p><p>1</p><p>2</p><p>wyt</p><p>2</p><p>¢y = y¢x = x</p><p>av0</p><p>6 v7 = v0 - g</p><p>(v0</p><p># g)</p><p>g 2</p><p>t</p><p>V</p><p>2</p><p>dV</p><p>p2 – p1 = nRT</p><p>2Sx</p><p>V 20 - S</p><p>2x</p><p>2</p><p>= nRT # 2V</p><p>V 20 - V</p><p>2</p><p>p2 (V0 – Sx) = nRTp1 (V0 + Sx) = nRT</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 243</p><p>v</p><p>In Eq. (1), u is the internal energy per unit mass and in this case is the thermal energy</p><p>per unit mass of the gas. As the gas vessel is thermally insulated Qd � 0. Just inside</p><p>the vessel,</p><p>u � � also �</p><p>Inside the vessel and just outside p � 0, and u � 0. In general gz is not very</p><p>significant for gases.</p><p>Thus applying Eq. (1) just inside and outside the hole, we get</p><p>v2� 	 u</p><p>�</p><p>Hence,</p><p>or</p><p>Alternate:</p><p>From Eulerian equation</p><p>In projection form along a stream line for steady flow, it</p><p>becomes,</p><p>In this problem,</p><p>Experiments show, when the particles move sufficiently fast it can be assumed that</p><p>the process is adiabatic. Then the connection between the pressure and the density is</p><p>given by the relation,</p><p>where is the adiabatic exponent dependent on the nature of the gas and is the</p><p>density of the gas inside the vessel. The adiabatic relation is a consequence of the fact</p><p>that during the expansion of a gas there is no heat exchange with the surrounding,</p><p>that is there is no loss or gain of heat.</p><p>rig</p><p>p</p><p>rg</p><p>=</p><p>pi</p><p>ri</p><p>g</p><p>= constant</p><p>-dp</p><p>ds</p><p>= rv</p><p>dv</p><p>ds</p><p>(only because a becomes 90°)</p><p>- 0p</p><p>0 s</p><p>+ g cos a = rv</p><p>0v</p><p>0 s</p><p>- ¥p + rg = ra</p><p>v = A</p><p>2gRT</p><p>M (g - 1)</p><p>= 3.22 km/s</p><p>v2 =</p><p>2gRT</p><p>M(g - 1)</p><p>+</p><p>RT</p><p>M (g-1)</p><p>=</p><p>gRT</p><p>M(g-1)</p><p>RT</p><p>M</p><p>p</p><p>r</p><p>1</p><p>2</p><p>= 0</p><p>RT</p><p>M</p><p>p</p><p>r</p><p>RT</p><p>M(g - 1)</p><p>CVT</p><p>M</p><p>244 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>v</p><p>On finding the density as a function of the pressure and substituting, we obtain the</p><p>equation</p><p>This equation should be integrated along the stream tube. Since the pressure in the</p><p>vessel is pi and the pressure outside the vessel is p0 we must integrate from pi to p0</p><p>for the pressure and from zero to (the outlet velocity) for the velocity.</p><p>Thus,</p><p>On performing the integration we find the outlet velocity,</p><p>According the problem because outside is vacuum.</p><p>Hence,</p><p>Note: The velocity here is the velocity of hydrodynamic flow of the gas into vacuum.</p><p>This requires that the diameter of the hole is not too small mean free path l ). In</p><p>the opposite case the flow is called effusion. Then the above result does not</p><p>apply and kinetic theory methods are needed.</p><p>2.43 The differential work done by the gas,</p><p>So,</p><p>From the first law of thermodynamics</p><p>2.44 According to the problem: or (where a is proportionality constant).</p><p>So, (1)pdV =</p><p>anRdT</p><p>g - 1</p><p>dA = aUA r U</p><p>= nR¢T # 2 - g</p><p>g - 1</p><p>= R ¢T # 2 - g</p><p>g - 1</p><p>(for n = 1 mole)</p><p>Q = ¢U + A =</p><p>nR</p><p>g - 1</p><p>¢T - nR¢T</p><p>A = - 3</p><p>T+ ¢T</p><p>T</p><p>nRdT = -nR¢T</p><p>1as pV = nRT and V = a>T2 dA = pdV =</p><p>nRT 2</p><p>a</p><p>a- a</p><p>T 2</p><p>bdT = -nRdT</p><p>(D V l )</p><p>(D 7</p><p>v = A</p><p>2g</p><p>g - 1</p><p>pi</p><p>ri</p><p>= A</p><p>2g</p><p>g - 1</p><p>RT</p><p>M</p><p>= 3.3 km/s (on substituting values)</p><p>p0 = 0</p><p>v = C</p><p>2g</p><p>g - 1</p><p>pi</p><p>ri</p><p>c1 - a p0</p><p>pi</p><p>1>g b g - 1</p><p>g d</p><p>-3</p><p>p0</p><p>pi</p><p>p-1>g</p><p>dp =</p><p>ri</p><p>pi</p><p>1>g3</p><p>v0</p><p>0</p><p>v dv</p><p>v = v0</p><p>-p-1>gdp</p><p>ds</p><p>=</p><p>ri</p><p>p i</p><p>1>g v</p><p>dv</p><p>ds</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 245</p><p>,</p><p>,</p><p>From ideal gas law, which on differentiating gives</p><p>(2)</p><p>Thus, from Eqs. (1) and (2)</p><p>or</p><p>or</p><p>or</p><p>or</p><p>Dividing both the sides by pV, we get</p><p>On integrating we get,</p><p>or</p><p>2.45 In the polytropic process work done by the gas is given by,</p><p>(where and are initial and final temperatures of the gas like in an adiabatic process).</p><p>Also,</p><p>By the first law of thermodynamics,</p><p>According to definition of molar heat capacity when number of moles and</p><p>, then molar heat capacity. Q =¢T = 1</p><p>n = 1</p><p>= (Tƒ - Ti )nR c 1</p><p>g - 1</p><p>-</p><p>1</p><p>n - 1</p><p>d =</p><p>nR [n - g]</p><p>(n - 1) (g - 1)</p><p>¢T</p><p>=</p><p>nR</p><p>g - 1</p><p>(Tf - Ti ) +</p><p>nR</p><p>n - 1</p><p>(Ti - Tƒ )</p><p>Q = ¢U + A</p><p>¢U =</p><p>nR</p><p>g - 1</p><p>(Tƒ - Ti )</p><p>TƒTi</p><p>A =</p><p>nR CTi - Tƒ D</p><p>n - 1</p><p>ln (pV n) = ln C or pV n = C</p><p>n ln V + ln p = ln C (where C is constant)</p><p>n</p><p>dV</p><p>V</p><p>+</p><p>dp</p><p>p</p><p>= 0</p><p>npdV + Vdp = 0 awhere</p><p>k - 1</p><p>k</p><p>= n = ratiob</p><p>pdV</p><p>k - 1</p><p>k</p><p>+ Vdp = 0</p><p>pdV (k - 1) + kVdp = 0 (where k =</p><p>a</p><p>g - 1</p><p>= another constant)</p><p>pdV a a</p><p>g - 1</p><p>-1b +</p><p>a</p><p>g - 1</p><p>Vdp = 0</p><p>pdV =</p><p>a</p><p>g - 1</p><p>( pdV + Vdp)</p><p>pdV + Vdp = nRdT</p><p>pV = nRT,</p><p>246 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>Here,</p><p>2.46 Let the process be polytropic, according to the law constant.</p><p>Thus,</p><p>or</p><p>So,</p><p>In the polytropic process molar heat capacity is given by</p><p>So,</p><p>2.47 (a) Increment of internal energy for becomes</p><p>From first law of thermodynamics</p><p>(b) Sought work done,</p><p>2.48 Law of the process is or , so the process is polytropic index</p><p>.As p = aV so, pi = aV0 and pf = ah V0</p><p>n = -1.</p><p>pV - 1 =ap =aV</p><p>=</p><p>nR¢T</p><p>n - 1</p><p>= -</p><p>R¢T</p><p>n - 1</p><p>= 0.43 kJ (as n = 1 mole)</p><p>=</p><p>pƒVƒ - piVi</p><p>1 - n</p><p>=</p><p>nR (Tƒ - Ti )</p><p>1 - n</p><p>=</p><p>k</p><p>1 - n</p><p>AV 1- n</p><p>ƒ - V 1- n</p><p>i B =</p><p>Apƒ V n</p><p>ƒ V</p><p>1- n</p><p>ƒ - piV</p><p>n</p><p>i V</p><p>1- n</p><p>i B</p><p>1 - n</p><p>An = 3pdV = 3</p><p>Vƒ</p><p>Vi</p><p>k</p><p>V n</p><p>dV 1where pV n = k = piVi</p><p>n = pƒV</p><p>n</p><p>ƒ 2</p><p>Q = ¢U + A =</p><p>R¢T</p><p>g - 1</p><p>-</p><p>R¢T</p><p>n - 1</p><p>= 0.11 kJ</p><p>¢U =</p><p>nR¢T</p><p>g - 1</p><p>=</p><p>R ¢T</p><p>g - 1</p><p>= -324 J (as n = 1 mole)</p><p>¢T</p><p>Cn =</p><p>8.314</p><p>1.66 - 1</p><p>-</p><p>8.314 ln 4</p><p>ln 8 - ln 4</p><p>= -42 J> mol K</p><p>=</p><p>R</p><p>g - 1</p><p>-</p><p>R ln a</p><p>ln b ln a</p><p>awhere n =</p><p>ln b</p><p>ln a</p><p>b</p><p>Cn =</p><p>R (n - g)</p><p>(n - 1) (g - 1)</p><p>=</p><p>R</p><p>g - 1</p><p>-</p><p>R</p><p>n - 1</p><p>an = b or ln b = n ln a or n =</p><p>ln b</p><p>ln a</p><p>1pi >pƒ2 = b</p><p>pƒV</p><p>n</p><p>ƒ = pi V n</p><p>i</p><p>pV n =</p><p>Cn =</p><p>R (n - g)</p><p>(n - 1) (g - 1)</p><p>6 0 for 1 6 n 6 g</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 247</p><p>, with</p><p>(a) Increment of the internal energy is given by</p><p>Putting and</p><p>(b) Work done by the gas is given by</p><p>(c) Molar heat capacity is given by</p><p>2.49 (a) and (where is the molar heat capacity in the</p><p>process.) It is given that</p><p>So,</p><p>(b) By the first law of thermodynamics,</p><p>or</p><p>So,</p><p>or</p><p>(c) We know</p><p>But from part (a), we have Cn = -R>(g - 1).</p><p>Cn =</p><p>(n - g) R</p><p>(n - 1) (g - 1)</p><p>dT</p><p>T</p><p>+</p><p>g - 1</p><p>2</p><p>dV</p><p>V</p><p>= 0Q TV (g-1)>2 = constant</p><p>2Rv</p><p>g - 1</p><p>dT +</p><p>nRT</p><p>V</p><p>dV = 0 or 2dT</p><p>(g - 1) T</p><p>+</p><p>dV</p><p>V</p><p>= 0</p><p>2nCn dT = pdV or 2R n</p><p>g - 1</p><p>dT + pdV = 0</p><p>2dQ = dA (as dQ = -dU)</p><p>dQ = dU + dA</p><p>Cn¢T = -</p><p>R</p><p>g - 1</p><p>¢T or Cn = -</p><p>R</p><p>g - 1</p><p>Q = - ¢U.</p><p>CnQ = nCn¢T¢U =</p><p>nR</p><p>g - 1</p><p>¢T</p><p>Cn =</p><p>R (n - g)</p><p>(n - 1) (g - 1)</p><p>=</p><p>R (-1 - g)</p><p>(-1 - 1) (g - 1)</p><p>=</p><p>R</p><p>2</p><p># g + 1</p><p>g - 1</p><p>=</p><p>aV 2</p><p>0 (1 - h2)</p><p>-2</p><p>=</p><p>1</p><p>2</p><p>a V 2</p><p>0 (h2 - 1)</p><p>A =</p><p>piVi - pƒVƒ</p><p>n - 1</p><p>=</p><p>aV 2</p><p>0 - ahV0</p><p># hV0</p><p>-1 - 1</p><p>¢U =</p><p>aV 2</p><p>0 (h</p><p>2 - 1)</p><p>g - 1</p><p>pƒ = a hV0pi = aV0</p><p>¢U =</p><p>nR</p><p>g - 1</p><p>[Tƒ - Ti ] =</p><p>pƒVƒ - piVi</p><p>g - 1</p><p>248 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>, we get</p><p>,</p><p>Thus,</p><p>which yields</p><p>From part (b), we know .</p><p>So,</p><p>Work done by the gas for one mole is given by</p><p>2.50 Given (for one mole of gas).</p><p>So,</p><p>Here polytropic exponent</p><p>(a) In the polytropic process for one mole of gas,</p><p>(b) Molar heat capacity is given by</p><p>2.51 Given</p><p>or</p><p>or</p><p>So polytropic index n = 1 - a.</p><p>pV 1-a = Ra>CV = constant = a (g - 1) [as CV = R>(g - 1)]</p><p>aV a # R</p><p>CV</p><p># 1</p><p>pV</p><p>= 1 or V a-1 # p-1 =</p><p>CV</p><p>Ra</p><p>Q</p><p>nCVT = aV a or nCV</p><p>pV</p><p>nR</p><p>= aV a</p><p>U = aV a</p><p>=</p><p>R</p><p>g - 1</p><p>+ R (a - 1)</p><p>C =</p><p>R</p><p>g - 1</p><p>-</p><p>R</p><p>n - 1</p><p>=</p><p>R</p><p>g - 1</p><p>-</p><p>R</p><p>[a>(a - 1)] - 1</p><p>A =</p><p>R¢T</p><p>1 - n</p><p>=</p><p>R¢T</p><p>1 - [a>(a - 1)]</p><p>= R¢T (1 - a)</p><p>n = a>(a - 1).</p><p>p 1-aV</p><p>-a = aR -a or pV</p><p>a>(a-1) = constantQ</p><p>pT -a = a or p apV</p><p>R</p><p>b -a</p><p>= a</p><p>p = aT</p><p>a</p><p>A =</p><p>R (T0 - T )</p><p>n - 1</p><p>=</p><p>2RT0</p><p>g - 1</p><p>C1 - h(1-g>2) D</p><p>T0</p><p>T</p><p>= a V</p><p>V0</p><p>b (g-1)>2</p><p>= h (g-1)>2 (where T is the final temperature)</p><p>TV (g-1)>2 = constant</p><p>n =</p><p>1 + g</p><p>2</p><p>-</p><p>R</p><p>g - 1</p><p>=</p><p>(n - g) R</p><p>(n - 1) (g - 1)</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 249</p><p>(a) Work done by the gas is given by</p><p>Hence,</p><p>By the first law of thermodynamics</p><p>(b) Molar heat capacity is given by</p><p>2.52 (a) By the first law of thermodynamics</p><p>Molar specific heat according to definition is given by</p><p>We have</p><p>On differentiating,</p><p>So,</p><p>Hence,</p><p>(b) For the process, .</p><p>So,</p><p>or</p><p>So, C = CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>= CV + p0e</p><p>aV # R</p><p>p0e</p><p>aV</p><p>(1 + aV)</p><p>= CV +</p><p>R</p><p>1 + aV</p><p>T = (p0>R)e aV # V</p><p>p = RT>V = p0 e aV</p><p>p = p0e</p><p>aV</p><p>C = CV +</p><p>RT</p><p>V</p><p># 1</p><p>aT0e</p><p>aV</p><p>= CV +</p><p>RT0e</p><p>aV</p><p>aVT0e</p><p>aV</p><p>= CV +</p><p>R</p><p>aV</p><p>dV</p><p>dt</p><p>=</p><p>1</p><p>a T0e</p><p>aV</p><p>dT = aT0e aV # dV</p><p>T = T0e</p><p>aV</p><p>=</p><p>nCV</p><p>dT + (nRT>V )dV</p><p>ndT</p><p>= CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>C =</p><p>dQ</p><p>ndT</p><p>=</p><p>nCVdT + pdV</p><p>ndT</p><p>dQ = dU + dA = nCVdT + pdV</p><p>=</p><p>R</p><p>g - 1</p><p>+</p><p>R</p><p>a</p><p>(as n = 1 - a)</p><p>C =</p><p>R</p><p>g - 1</p><p>-</p><p>R</p><p>n - 1</p><p>=</p><p>R</p><p>g - 1</p><p>-</p><p>R</p><p>1 - a - 1</p><p>= ¢U +</p><p>¢U (g - 1)</p><p>a</p><p>= ¢U c1 +</p><p>g - 1</p><p>a</p><p>d Q = ¢U + A</p><p>A =</p><p>- ¢U (g - 1)</p><p>n - 1</p><p>=</p><p>¢U (g - 1)</p><p>a</p><p>(as n = 1 - a)</p><p>A =</p><p>-nR¢T</p><p>n - 1</p><p>and ¢U =</p><p>nR¢T</p><p>g - 1</p><p>250 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>2.53 (a) Using solution of Problem 2.52,</p><p>(for one mole of gas)</p><p>We have,</p><p>Therefore,</p><p>So,</p><p>Hence,</p><p>(b) Work done is given by</p><p>By the first law of thermodynamics,</p><p>2.54 (a) From solution of Problem 2.52, heat capacity is given by</p><p>We have T � T0</p><p>V V �</p><p>T</p><p>a</p><p>-</p><p>T0</p><p>a</p><p>Q</p><p>dV</p><p>dT</p><p>C = CV +</p><p>RT</p><p>V</p><p>=</p><p>gp0 (V2 - V1)</p><p>(g - 1)</p><p>+ a ln</p><p>V2</p><p>V1</p><p>=</p><p>p01V2 - V12</p><p>(g - 1)</p><p>+ p0(V2 - V1) + a ln</p><p>V2</p><p>V1</p><p>Q = ¢U + A</p><p>=</p><p>1</p><p>g - 1</p><p>c(p0 + aV2) V2 - ap0 +</p><p>a</p><p>V1</p><p>b V1 d =</p><p>p0(V2 - V1)</p><p>g - 1</p><p>=</p><p>R</p><p>(g - 1)R</p><p>(p2V2 - p1V1)</p><p>¢U = CV (T2 - T1) = CV ap2V2</p><p>R</p><p>-</p><p>p1V1</p><p>R</p><p>b (for one mole)</p><p>A = 3</p><p>V2</p><p>V1</p><p>ap0 +</p><p>a</p><p>V</p><p>b dV = p0(V2 - V1) + a ln</p><p>V2</p><p>V1</p><p>R =</p><p>Rg</p><p>g - 1</p><p>+</p><p>aR</p><p>p0V</p><p>C = CV + ap0 +</p><p>a</p><p>V</p><p>b # R</p><p>p0</p><p>=</p><p>R</p><p>g - 1</p><p>+ a1 +</p><p>a</p><p>p0V</p><p>b dV>dT = R>p0</p><p>RdT = p0dV</p><p>p = p0 +</p><p>a</p><p>V ¿</p><p>or RT</p><p>V</p><p>= p0 +</p><p>a</p><p>V ¿</p><p>Q RT = p0V + a</p><p>C = CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>= CV +</p><p>pdV</p><p>dT</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 251</p><p>On differentiating, we get</p><p>Hence,</p><p>(b) Given that T � T0</p><p>V. for one mole of an ideal gas, so</p><p>Now,</p><p>By the first law of thermodynamics</p><p>2.55 Heat capacity is given by</p><p>(a) Given that</p><p>C = CV + aT</p><p>C = CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>= aCp(V2 – V1) + RT0 ln</p><p>V2</p><p>V1</p><p>= aR (V2 – V1) c1 +</p><p>1</p><p>g - 1</p><p>d + RT0 ln</p><p>V2</p><p>V1</p><p>=</p><p>aR</p><p>g - 1</p><p>(V2 – V1) + RT0 ln</p><p>V2</p><p>V1</p><p>+ aR (V2 – V1)</p><p>Q = ¢U + A</p><p>= CV [T0 + aV2 – (T0 + aV1)] = aCV (V2 - V1)</p><p>¢U = CV (T2 - T1)</p><p>= RT0 ln</p><p>V2</p><p>V1</p><p>+ a(V2 – V1)R</p><p>A = 3</p><p>V2</p><p>V1</p><p>pdV = 3</p><p>V2</p><p>V1</p><p>aRT0</p><p>V</p><p>+ aRb dV (for one mole)</p><p>p =</p><p>R</p><p>V</p><p>(T0 + aV ) =</p><p>RT0</p><p>V</p><p>+ aR</p><p>T = pV>R</p><p>= CV +</p><p>RT</p><p>aV</p><p>= Cp +</p><p>RT0</p><p>aV</p><p>=</p><p>R</p><p>g - 1</p><p>+ R a T0</p><p>aV</p><p>+ 1b =</p><p>gR</p><p>g - 1</p><p>+</p><p>RT0</p><p>aV</p><p>=</p><p>R</p><p>g - 1</p><p>+</p><p>R (T0 + aV )</p><p>V</p><p># 1</p><p>a</p><p>C = CV +</p><p>RT</p><p>V</p><p># 1</p><p>a</p><p>dV</p><p>dT</p><p>=</p><p>1</p><p>a</p><p>252 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>and as</p><p>Also,</p><p>So, or</p><p>Integrating both sides, we get</p><p>(where C0 is a constant)</p><p>or</p><p>Hence,</p><p>(b) According to the problem .</p><p>But,</p><p>So,</p><p>or</p><p>Integrating both sides, we get</p><p>(where C0 is a constant)</p><p>So,</p><p>or</p><p>Hence,</p><p>(c) Given that C � CV 	 ap and C � CV</p><p>So,</p><p>or (as for one mole of gas)</p><p>or dV � adT or</p><p>So, constant or V � aT � constant T =</p><p>V</p><p>a</p><p>+</p><p>dT = dV>adV>dT = aQ</p><p>p =</p><p>RT</p><p>V</p><p>a</p><p>RT</p><p>V</p><p>=</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>ap =</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>QCV + ap = CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>dV</p><p>dT</p><p>RT</p><p>V</p><p>Te</p><p>R/bV =</p><p>1</p><p>C0</p><p>= constant</p><p>T # C0 = e- R/bV</p><p>lnT # C0 = -</p><p>R</p><p>bV</p><p>-</p><p>R</p><p>b</p><p>V -1 = ln T + ln C0 = ln T # C0</p><p>R</p><p>b</p><p>V -2 dV =</p><p>dT</p><p>T</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>= bV</p><p>C = CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>C = CV + bV</p><p>V # e</p><p>-aT>R =</p><p>1</p><p>C0</p><p>= constant</p><p>V # C0 = e</p><p>-aT>R</p><p>a</p><p>R</p><p>T = ln V + ln C0 = ln VC0</p><p>a</p><p>R</p><p>dT =</p><p>dV</p><p>V</p><p>CV + aT = CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 253</p><p>2.56 (a) By the first law of thermodynamics</p><p>or A � CdT � CV dT � (C � CV ) dT (for one mole)</p><p>Given that</p><p>So,</p><p>(b) Given that</p><p>So,</p><p>or</p><p>or</p><p>or</p><p>Integrating both sides, we get</p><p>or</p><p>or</p><p>or pV gea(g-1)/pV = RK = constant</p><p>pV</p><p>g>RK = e -a(g-1)/pV</p><p>ln cV</p><p>g-1 # pV</p><p>RK</p><p>d =</p><p>-a(g - 1)</p><p>pV</p><p>ln cV</p><p>g-1 # T</p><p>K</p><p>d =</p><p>-a(g - 1)</p><p>RT</p><p>(g - 1) lnV = -</p><p>a(g - 1)</p><p>RT</p><p>- ln T + ln K</p><p>(g - 1)</p><p>dV</p><p>V</p><p>=</p><p>a(g - 1)</p><p>RT 2</p><p>dT -</p><p>dT</p><p>T</p><p>dV</p><p>V</p><p>=</p><p>a</p><p>RT 2</p><p>dT -</p><p>1</p><p>g - 1</p><p># dT</p><p>T</p><p>R</p><p>g - 1</p><p>1</p><p>RT</p><p>+</p><p>dV</p><p>V</p><p>=</p><p>a</p><p>RT 2</p><p>dT</p><p>CV +</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>=</p><p>a</p><p>T</p><p>C = a>TC = +</p><p>dQ</p><p>dT</p><p>=</p><p>RT</p><p>V</p><p>dV</p><p>dT</p><p>+ CV</p><p>= a ln h -</p><p>RT0</p><p>g - 1</p><p>(h - 1)</p><p>= a ln h - CVT0 (h - 1)</p><p>A = 3</p><p>hT0</p><p>T0</p><p>aa</p><p>T</p><p>- CVbdT = a ln</p><p>hT0</p><p>T0</p><p>- CV(hT0 - T0)</p><p>C = a>T</p><p>A = Q - ¢U</p><p>254 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>and</p><p>2.57 The work done is</p><p>2.58 (a) The increment in the internal energy is</p><p>But from the second law of thermodynamics</p><p>On the other hand</p><p>or</p><p>So,</p><p>(b) From the first law of thermodynamics</p><p>2.59 (a) From the first law, for an adiabatic process, .</p><p>From the previous problem</p><p>So,</p><p>This equation can be integrated if we assume that CV and b are constant, then</p><p>or or T (V – b)R>CV = constantln T +</p><p>R</p><p>CV</p><p>ln (V – b) = constant</p><p>R</p><p>CV</p><p>dV</p><p>V - b</p><p>+</p><p>dT</p><p>T</p><p>= 0</p><p>0 = CV dT +</p><p>RTdV</p><p>V - b</p><p>dU = a 0U</p><p>0T</p><p>b</p><p>V</p><p>dT + a 0U</p><p>0V</p><p>b</p><p>T</p><p>dV = CV dT +</p><p>a</p><p>V 2</p><p>dV</p><p>dQ = dU + pdV = 0</p><p>Q = A + ¢U = RT ln</p><p>V2 - b</p><p>V1 - b</p><p>= 3.8 kJ (on substituting values)</p><p>¢U = a a 1</p><p>V1</p><p>-</p><p>1</p><p>V2</p><p>b = 0.11 kJ (on substituting values)</p><p>T a 0p</p><p>0T</p><p>b V =</p><p>RT</p><p>V - b</p><p>and a 0U</p><p>0V</p><p>b</p><p>T</p><p>=</p><p>a</p><p>V 2</p><p>p =</p><p>RT</p><p>V - b</p><p>-</p><p>a</p><p>V 2</p><p>a 0U</p><p>0V</p><p>b</p><p>T</p><p>= T a 0 s</p><p>0V</p><p>b</p><p>T</p><p>- p = T a 0 p</p><p>0T</p><p>b</p><p>V</p><p>- p</p><p>¢U = 3</p><p>V2</p><p>V1</p><p>a0U</p><p>0V</p><p>b</p><p>T</p><p>dV</p><p>= RT ln</p><p>V2 - b</p><p>V1 - b</p><p>+ a a 1</p><p>V2</p><p>-</p><p>1</p><p>V1</p><p>b</p><p>A = 3</p><p>V2</p><p>V1</p><p>p dV = 3</p><p>V2</p><p>V1</p><p>a RT</p><p>V - b</p><p>-</p><p>a</p><p>V2</p><p>bdV</p><p>2.2 THE FIRST LAW OF THERMODYNAMICS. HEAT CAPACITY 255</p><p>(b) We use</p><p>Now,</p><p>So along constant p,</p><p>Thus,</p><p>But,</p><p>On differentiating,</p><p>or</p><p>and</p><p>2.60 From the first law</p><p>Q � Uf � Ui 	 A � 0 (as the vessels are thermally insulated)</p><p>As this is free expansion, A � 0, so, Uf � Ui.</p><p>But,</p><p>So,</p><p>or</p><p>On substitution we get, �T � �3 K.</p><p>2.61 From the first law</p><p>Q � Uf � Ui 	 A � Uf � Ui (as A � 0 in free expansion).</p><p>¢T =</p><p>-a(g - 1)V2n</p><p>RV1(V1 + V2)</p><p>CV (Tf - Ti) = a a</p><p>V1 + V2</p><p>-</p><p>a</p><p>V1</p><p>bn =</p><p>-aV2n</p><p>V1(V1 + V2)</p><p>U = nCVT -</p><p>an2</p><p>V</p><p>Cp - CV =</p><p>R</p><p>1 -</p><p>2a(V - b)2</p><p>RT V 3</p><p>T a 0V</p><p>0T</p><p>b</p><p>P</p><p>=</p><p>RT>V - b</p><p>RT</p><p>(V - b)2</p><p>-</p><p>2a</p><p>V 3</p><p>=</p><p>V - b</p><p>1 -</p><p>2a (V - b)2</p><p>RTV 3</p><p>0 = a -</p><p>RT</p><p>(V - b)2</p><p>+</p><p>2a</p><p>V 2</p><p>b a 0V</p><p>0T</p><p>b</p><p>P</p><p>+</p><p>R</p><p>V - b</p><p>p =</p><p>RT</p><p>V - b</p><p>a</p><p>V 2</p><p>Cp - CV =</p><p>RT</p><p>V - b</p><p>a 0V</p><p>0T</p><p>b</p><p>Cp = CV +</p><p>RT</p><p>V - b</p><p>a 0V</p><p>0T</p><p>b</p><p>dQ = CV dT +</p><p>RT</p><p>V - b</p><p>dV</p><p>dU =CVdT +</p><p>a</p><p>V 2</p><p>dV</p><p>256 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>p</p><p>p</p><p>So at constant temperature,</p><p>� 0.33 kJ (on substituting values)</p><p>2.3 Kinetic Theory of Gases. Boltzmann’s Law and Maxwell’s</p><p>Distribution</p><p>2.62 From the formula</p><p>Mean distance between molecules</p><p>2.63 After dissociation each molecule becomes two N-atoms and so contributes,</p><p>degrees of freedom. Thus the number of moles becomes</p><p>and atm (on substituting values)</p><p>(Here M is the molecular weight, in grams of .)</p><p>2.64 Let number density of He atoms, number density of molecules.</p><p>Then,</p><p>(where and are masses of helium and nitrogen molecules).</p><p>Also,</p><p>From Eqs. (1) and (2), we get</p><p>n1 =</p><p>a p</p><p>kT</p><p>-</p><p>r</p><p>m2</p><p>b</p><p>a1 -</p><p>m1</p><p>m2</p><p>b = 1.6 * 10-19 cm-3</p><p>p = 1n1 + n22 kT</p><p>m2m1</p><p>r = n1m1 + n2m2</p><p>N2n2 =n1 =</p><p>N2</p><p>p =</p><p>mRT</p><p>MV</p><p>11 + h2 = 1.9</p><p>m</p><p>M</p><p>11 + h2</p><p>2 * 3N2</p><p>110-5 cm321>3 = 101>3 * 10-2 cm = 0.2 mm</p><p>= 1 * 1011 per m3 = 105 cm 3</p><p>n =</p><p>p</p><p>kT</p><p>=</p><p>4 * 10-15 * 1.01 * 105</p><p>1.38 * 1023 * 300</p><p>m 3</p><p>p = nkT</p><p>Q =</p><p>-an2</p><p>V2</p><p>- a -</p><p>an2</p><p>V1</p><p>b = an2</p><p>V2 - V1</p><p>V1</p><p># V2</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 257</p><p>-</p><p>-</p><p>(1)</p><p>(2)</p><p>2.65 Number of nitrogen molecules coming to strike the differential area ds</p><p>of the wall per unit time is given by</p><p>where number density or concentration of molecules in the beam.</p><p>The change in linear momentum of each of molecule is</p><p>(see figure)</p><p>So for all the molecules striking the wall per unit time, i.e., the force exerted by</p><p>the wall on incoming molecules is</p><p>Therefore from Newton’s law, the force exerted by molecules on the wall is</p><p>Thus the sought pressure i.e. normal force per unit differential is</p><p>2.66 From the formula</p><p>If number of degrees of freedom of the gas, then</p><p>2.67</p><p>So,</p><p>(a) For monoatomic gases</p><p>vsound</p><p>vrms</p><p>= A</p><p>5</p><p>9</p><p>= 0.75</p><p>i = 3, therefore</p><p>vsound</p><p>vrms</p><p>= A</p><p>g</p><p>3</p><p>= A</p><p>i + 2</p><p>3i</p><p>vsound = A</p><p>gp</p><p>r</p><p>= A</p><p>gRT</p><p>M</p><p>and vrms = A</p><p>3kT</p><p>m</p><p>= A</p><p>3RT</p><p>M</p><p>=</p><p>2</p><p>(</p><p>2>p) - 1</p><p>= 5 1on substituting values2</p><p>g =</p><p>C</p><p>CV</p><p>= 1 +</p><p>2</p><p>i</p><p>or i =</p><p>2</p><p>g - 1</p><p>Cp = CV + RT and CV =</p><p>i</p><p>2</p><p>RT</p><p>i =</p><p>v = A</p><p>gp</p><p>r</p><p>or g =</p><p>rv 2</p><p>p</p><p>dFnormal</p><p>ds</p><p>= 2nmv</p><p>2cos2 u</p><p>2nmv 2 cos2u dS 1-n2 N2</p><p>1nvds cosu)¢pmolecule = 2nv2cos2udS n</p><p>N2</p><p>¢p</p><p>¢pmolecule = 2mv cosu n</p><p>N2</p><p>N2n = N>V =</p><p>dN</p><p>dt</p><p>= n | 1v # dS2 | = 1nv dS cosu2</p><p>258 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>n</p><p>p</p><p>vr</p><p>(b) For rigid diatomic molecules</p><p>2.68 For a general non-collinear, non-planar molecule, mean energy</p><p>(translational) (rotational) (vibrational)</p><p>(per molecule)</p><p>For linear molecules, mean</p><p>energy</p><p>(translational) (rotational) (vibrational)</p><p>(per molecule)</p><p>Translational energy corresponds to fraction and in non-linear and</p><p>linear molecules, respectively.</p><p>2.69 (a) A diatomic molecule has 3 translational, 2 rotational and one vibrational degrees</p><p>of freedom. The corresponding energy per mole is</p><p>(for translational) (for rotational) (for vibrational)</p><p>Thus,</p><p>(b) For linear N-atomic molecules, energy per mole</p><p>So,</p><p>(c) For non-collinear N-atomic molecules</p><p>R (as before from Problem 2.68)</p><p>and g =</p><p>3N - 2</p><p>3N - 3</p><p>=</p><p>N - 2>3</p><p>N - 1</p><p>CV = 31N - 12</p><p>CV = a3N -</p><p>5</p><p>2</p><p>bR and g =</p><p>6N - 3</p><p>6N - 5</p><p>= a3N -</p><p>5</p><p>2</p><p>bRT (as before)</p><p>CV =</p><p>7</p><p>2</p><p>R and g =</p><p>Cp</p><p>CV</p><p>=</p><p>9</p><p>7</p><p>=</p><p>7</p><p>2</p><p>RT</p><p>+ 1 * RT+ 2 *</p><p>1</p><p>2</p><p>RT</p><p>3</p><p>2</p><p>RT</p><p>1</p><p>2N - 5>31</p><p>2(N - 1)</p><p>= a3N -</p><p>5</p><p>2</p><p>b kT</p><p>+ (3N – 5)kT+ kT=</p><p>3</p><p>2</p><p>kT</p><p>= 13N – 32kT</p><p>+ 13N – 62kT+</p><p>3</p><p>2</p><p>kT=</p><p>3</p><p>2</p><p>kT</p><p>vsound</p><p>vrms</p><p>= A</p><p>7</p><p>15</p><p>= 0.68</p><p>i = 5, therefore</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 259</p><p>2.70 In the isobaric process, work done is</p><p>On the other hand heat transferred is .</p><p>Now R for non-collinear molecules and R for linear</p><p>molecules.</p><p>Thus,</p><p>For monoatomic gases,</p><p>2.71 Given specific heats (per unit mass), then</p><p>or g/mol (on substituting values)</p><p>Also,</p><p>or (on substituting values)</p><p>2.72 (a) Given,</p><p>(b) In the process, and constant.</p><p>So,</p><p>Thus,</p><p>or</p><p>Hence,</p><p>(where is the polytropic index).n = 1>2 i = 2 cC</p><p>R</p><p>+</p><p>11n - 12 d = 3 (monoatomic)</p><p>C = CV + 2R = a 29</p><p>8.3</p><p>bR Q CV =</p><p>12.4</p><p>8.3</p><p>R =</p><p>3</p><p>2</p><p>R</p><p>CdT = CV dT + pdV = CVdT +</p><p>RT</p><p>V</p><p>dV = CVdT +</p><p>2RT</p><p>T</p><p>dT</p><p>2</p><p>dT</p><p>T</p><p>-</p><p>dV</p><p>V</p><p>= 0</p><p>(T</p><p>2>V ) =pT = constant</p><p>i = 2aCp</p><p>R</p><p>- 1b = 5</p><p>CV =</p><p>20.7</p><p>8.3</p><p>R, g =</p><p>29</p><p>20.7</p><p>= 1.4 =</p><p>7</p><p>5</p><p>Cp = 29 J>K mol =</p><p>29</p><p>8.3</p><p>R</p><p>i =</p><p>2</p><p>(cp >cV</p><p>) - 1</p><p>=</p><p>2cV</p><p>cp - cV</p><p>= 5</p><p>g =</p><p>cp</p><p>cV</p><p>=</p><p>2</p><p>i</p><p>+ 1</p><p>M =</p><p>R</p><p>cp - cV</p><p>= 32</p><p>M1cp - cV2 = R</p><p>cp , cV</p><p>Cp = 5>2 and A>Q = 2>5</p><p>A</p><p>Q</p><p>= d 1</p><p>3N - 2</p><p>non-collinear</p><p>1</p><p>3N - (3>2)</p><p>linear</p><p>C = [3N - (3>2)]Cp = (3N - 2)</p><p>Q = C dT</p><p>A = pdV = RdT (per mole).</p><p>260 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>p</p><p>p</p><p>2.73 We know that,</p><p>(since a monoatomic gas has and a diatomic gas has )</p><p>and</p><p>Hence,</p><p>2.74 The internal energy of the molecules is</p><p>(where velocity of the vessel, number of molecules, each of mass m).</p><p>When the vessel is stopped, internal energy becomes</p><p>So, there is an increase in internal energy of . This will lead to a rise in</p><p>temperature of</p><p>Since there is no flow of heat, this change of temperature will lead to an excess pr-</p><p>essure, given by</p><p>and finally</p><p>(where molecular weight of number of degrees of freedom of</p><p>2.75 (a) From the equipartition theorem</p><p>(b) In equilibrium, the mean kinetic energy of the droplet will be equal to that of a</p><p>molecule.</p><p>So,</p><p>1</p><p>2</p><p>p</p><p>6</p><p>d 3 rv 2</p><p>rms =</p><p>3</p><p>2</p><p>kT or vrms = 3A</p><p>2kT</p><p>pd 3r</p><p>= 0.15 m/s</p><p>e =</p><p>3</p><p>2</p><p>kT = 6 * 10-21 J and vrms = A</p><p>3kT</p><p>m</p><p>= A</p><p>3RT</p><p>M</p><p>= 0.47 km/s</p><p>N2 = 5).N2, i =M =</p><p>¢p</p><p>p</p><p>=</p><p>Mv 2</p><p>iRT</p><p>= 2.2%</p><p>¢p =</p><p>R¢T</p><p>V</p><p>=</p><p>mNv 2</p><p>iV</p><p>=</p><p>mNv 2</p><p>iR</p><p>¢T =</p><p>(1>2)mNv 2</p><p>(i>2)R</p><p>¢U = (1>2)mNv 2</p><p>(1>2)mN 6u27</p><p>N =v =</p><p>U =</p><p>1</p><p>2</p><p>mN 6 1u - v22 7 =</p><p>1</p><p>2</p><p>mN 6 u2 - v 27</p><p>g =</p><p>Cp</p><p>CV</p><p>=</p><p>5 n1 + 7 n2</p><p>3 n1 + 5 n2</p><p>1</p><p>R</p><p>Cp =</p><p>3</p><p>2</p><p>n1 +</p><p>5</p><p>2</p><p>n2 + n1 + n2</p><p>CV = (5>2)RCV = (3>2)R</p><p>1</p><p>R</p><p>CV =</p><p>3</p><p>2</p><p>n1 +</p><p>5</p><p>2</p><p>n2</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 261</p><p>.</p><p>2.76 Given that,</p><p>So,</p><p>Now, in an adiabatic process</p><p>So,</p><p>The gas must be expanded times, i.e., 7.6 times.</p><p>2.77 Here,</p><p>(where mass of the gas, molecular weight of the gas).</p><p>If increases times, the temperature will have increased times. This will require</p><p>(neglecting expansion of the vessels), a heat flow of amount</p><p>2.78 The root mean square angular velocity is given by</p><p>or</p><p>2.79 Under compression, the temperature will rise, as</p><p>So,</p><p>So mean kinetic energy of rotation per molecule in the compressed state is given by</p><p>2.80 Number of collisions is given by</p><p>Now,</p><p>v ¿</p><p>v</p><p>=</p><p>n ¿</p><p>n</p><p>6v ¿ 7</p><p>6v 7</p><p>=</p><p>1</p><p>hA</p><p>T ¿</p><p>T</p><p>1</p><p>4</p><p>n 6v 7 = v</p><p>kT ¿ = kT0h</p><p>2>i = 0.72 * 10-20 J</p><p>T ¿1h-1 V022>i = T0V</p><p>2>i</p><p>0 or T ¿ = h+2>i T0</p><p>TV</p><p>g-1 = constant or TV</p><p>2>i = constant</p><p>v = A</p><p>2kT</p><p>I</p><p>= 6.3 * 1012 rad/s</p><p>1</p><p>2</p><p>I v2 = 2 *</p><p>1</p><p>2</p><p>kT 1for 2 degrees of rotations2</p><p>5</p><p>2</p><p>m</p><p>M</p><p>R 1h2 - 12 T = 10 kJ</p><p>h2hvrms</p><p>M =m =</p><p>CV =</p><p>5</p><p>2</p><p>m</p><p>M</p><p>R 1as i = 52</p><p>hi</p><p>V ¿ a 1</p><p>h2</p><p>Tb i>2</p><p>= VT</p><p>i>2 or V ¿h-i = V or V ¿ = hiV</p><p>TV</p><p>g-1 = TV</p><p>2>i = constant or VT</p><p>i>2 = constant</p><p>v ¿rms = A</p><p>3 RT</p><p>M</p><p>=</p><p>1</p><p>h</p><p>vrms =</p><p>1</p><p>hA</p><p>3RT</p><p>M</p><p>or T ¿ =</p><p>1</p><p>h 2</p><p>T</p><p>i = 5, CV = 5>2 R, g = 7>5.</p><p>262 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>(When the gas is expanded times, n decreases by a factor )</p><p>Also</p><p>So,</p><p>i.e., collisions decrease by a factor</p><p>2.81 In a polytropic process constant, where n is the polytropic index. For this</p><p>process</p><p>Then,</p><p>Now,</p><p>or</p><p>Now,</p><p>2.82 If is the polytropic index, then</p><p>Now,</p><p>Hence,</p><p>Then, C =</p><p>iR</p><p>2</p><p>+</p><p>R</p><p>2</p><p>= 3 R</p><p>1</p><p>a - 1</p><p>= -</p><p>1</p><p>2</p><p>or a = -1</p><p>v ¿</p><p>v</p><p>=</p><p>n ¿</p><p>n</p><p>6v ¿ 7</p><p>6v7</p><p>=</p><p>V</p><p>V ¿</p><p>A</p><p>T ¿</p><p>T</p><p>=</p><p>V T -1>2</p><p>V ¿T ¿-1>2 = 1</p><p>pV</p><p>a = constant, TV</p><p>a-1 = constant</p><p>a</p><p>= 1h2-(i -1)>(i -2) times =</p><p>1</p><p>2.52</p><p>times</p><p>=</p><p>1</p><p>h</p><p>a 1</p><p>h</p><p>b1>(i -2)</p><p>= a 1</p><p>h</p><p>b (i -1)>(i -2)</p><p>v ¿</p><p>v</p><p>=</p><p>n ¿</p><p>n</p><p>6v ¿ 7</p><p>6v7</p><p>=</p><p>1</p><p>hA</p><p>T ¿</p><p>T</p><p>=</p><p>1</p><p>h</p><p>a V</p><p>V ¿</p><p>b n - 1</p><p>2</p><p>1</p><p>n - 1</p><p>=</p><p>i</p><p>2</p><p>- 1 =</p><p>i - 2</p><p>2</p><p>or n =</p><p>i</p><p>i - 2</p><p>C = R so i</p><p>2</p><p>-</p><p>1</p><p>n - 1</p><p>= 1</p><p>=</p><p>i</p><p>2</p><p>RdT +</p><p>RT</p><p>V</p><p>dV =</p><p>i</p><p>2</p><p>RdT -</p><p>1</p><p>n - 1</p><p>RdT = a i</p><p>2</p><p>-</p><p>1</p><p>n - 1</p><p>bRdT</p><p>dQ = CdT = dU + pdV = CV</p><p>dT + pdV</p><p>dT</p><p>T</p><p>+ 1n - 12 dV</p><p>V</p><p>= 0</p><p>pV</p><p>n = constant or TV</p><p>n-1 = constant</p><p>pV</p><p>n =</p><p>h(i +1)>i (where i = 5 h).</p><p>v ¿</p><p>v</p><p>=</p><p>1</p><p>h</p><p>h-1>i = h-(i +1)>i</p><p>T ¿1hV 22>i = TV</p><p>2>i or T ¿ = h2>i T</p><p>hh</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 263</p><p>.</p><p>So,</p><p>2.83</p><p>2.84 (a) The formula is</p><p>Now, Prob</p><p>(b) Prob</p><p>2.85 (a)</p><p>(b) Clearly is the most probable speed at this temperature.</p><p>So, A</p><p>2kT</p><p>m</p><p>= v or T =</p><p>mv</p><p>2</p><p>2k</p><p>= 342 K</p><p>v</p><p>T =</p><p>m</p><p>k</p><p>¢ ¢v113 - 122≤2 = 384 K</p><p>vrms - vp = A23 - 22 B AkT</p><p>m</p><p>= ¢v</p><p>=</p><p>1223</p><p>22p</p><p>e-3>2dh = 0.0185 = 1.85%</p><p>=</p><p>4</p><p>2p *</p><p>3</p><p>2</p><p>e-3>2 * 2A</p><p>3</p><p>2</p><p>dh</p><p>A</p><p>3</p><p>2</p><p>- A</p><p>3</p><p>2</p><p>dh</p><p>= 3</p><p>4</p><p>p</p><p>u2e -u2</p><p>du</p><p>A</p><p>3</p><p>2</p><p>+ A</p><p>3</p><p>2</p><p>dh</p><p>= Prob ¢ ` u - A</p><p>3</p><p>2</p><p>` 6 A</p><p>3</p><p>2</p><p>dh≤</p><p>¢ ` v - vrms</p><p>vrms</p><p>` 6 dh≤ = Prob ¢ ` v</p><p>vp</p><p>-</p><p>vrms</p><p>vp</p><p>` 6 dh</p><p>vrms</p><p>vp</p><p>≤</p><p>=</p><p>4</p><p>2p e-1 * 2 dh =</p><p>8</p><p>2pe</p><p>dh = 0.0166 = 1.66%</p><p>¢ ` v - vp</p><p>vp</p><p>` 6 dh≤ = 3</p><p>1+dh</p><p>1-dh</p><p>df 1u2</p><p>df 1u2 =</p><p>4</p><p>2p u2e-u2</p><p>du 1where u = v>v 2</p><p>vavg = A</p><p>8</p><p>p</p><p>p</p><p>r</p><p>= 0.51 km>s and vrms = A</p><p>3p</p><p>r</p><p>= 0.55 km>s</p><p>vp = A</p><p>2kT</p><p>m</p><p>= A</p><p>2RT</p><p>M</p><p>= A</p><p>2p</p><p>r</p><p>= 0.45 km/s</p><p>264 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>p</p><p>2.86 (a) We have,</p><p>So,</p><p>(b)</p><p>Thus,</p><p>Now,</p><p>Thus,</p><p>2.87</p><p>2.88</p><p>On substitution, we get,</p><p>2.89</p><p>For a given range to ) is maximum when</p><p>d</p><p>dvp</p><p>dN (v)</p><p>Nv</p><p>2dv</p><p>= 0 = a -3v</p><p>-4</p><p>p +</p><p>2v</p><p>2</p><p>v</p><p>6</p><p>p</p><p>be -v</p><p>2>v</p><p>2</p><p>p</p><p>v + dvv</p><p>dN (v) =</p><p>N 4</p><p>1p</p><p>v</p><p>2dv</p><p>v</p><p>3</p><p>p</p><p>e-v</p><p>2>v</p><p>2</p><p>pH</p><p>v = 1.60 km/s.</p><p>v</p><p>2 = 3kT</p><p>ln mHe</p><p>>mH</p><p>mHe - mH</p><p>v</p><p>2</p><p>v</p><p>3</p><p>pH</p><p>e-v</p><p>2>v</p><p>2</p><p>pH</p><p>=</p><p>v</p><p>2</p><p>v</p><p>3</p><p>pHe</p><p>e-v</p><p>2>v</p><p>2</p><p>pHe</p><p>or e</p><p>v</p><p>2 mHe</p><p>2kT -</p><p>mH</p><p>2kT = amHe</p><p>mH</p><p>b3>2</p><p>T =</p><p>MN 1¢v22</p><p>2R a1 - A</p><p>MN</p><p>MO</p><p>b2</p><p>=</p><p>mN 1¢v22</p><p>2k a1 - A</p><p>mN</p><p>mO</p><p>b2</p><p>= 363 K</p><p>vpN</p><p>- vpo</p><p>= ¢v = A</p><p>2RT</p><p>MN</p><p>a1 - B</p><p>MN</p><p>MO</p><p>b</p><p>vp = A</p><p>2kT</p><p>mN</p><p>= A</p><p>2RT</p><p>MN</p><p>, vp</p><p>O</p><p>= A</p><p>2RT</p><p>MO</p><p>v = B</p><p>3kT0</p><p>m</p><p>A</p><p>ln h</p><p>1 - 1>h</p><p>e-</p><p>mv 2</p><p>2kT0</p><p>A1- 1</p><p>hB =</p><p>1</p><p>h3>2 or mv</p><p>2</p><p>2kT0</p><p>a1 -</p><p>1</p><p>h</p><p>b =</p><p>3</p><p>2</p><p>ln h</p><p>v</p><p>2</p><p>v</p><p>3</p><p>p1</p><p>e-v</p><p>2>v</p><p>2p1 =</p><p>v</p><p>2</p><p>v p</p><p>2</p><p>=</p><p>2kT0</p><p>m</p><p>, v</p><p>2</p><p>p2</p><p>=</p><p>2kT0</p><p>m</p><p>h</p><p>F 1v2 =</p><p>4</p><p>2p</p><p>v</p><p>2</p><p>v</p><p>2</p><p>p</p><p>e-v</p><p>2>v</p><p>2</p><p>p *</p><p>1</p><p>vp</p><p>a 1</p><p>vp</p><p>comes from F 1v2dv = df 1u2, du =</p><p>dv</p><p>vp</p><p>b</p><p>T =</p><p>m1v</p><p>2</p><p>1 - v</p><p>2</p><p>22</p><p>2 k ln v</p><p>2</p><p>1>v</p><p>2</p><p>2</p><p>= 330 K</p><p>v</p><p>2</p><p>1</p><p>v</p><p>2</p><p>p</p><p>e-v</p><p>2</p><p>1 >v</p><p>2</p><p>p =</p><p>v</p><p>2</p><p>2</p><p>v</p><p>2</p><p>p</p><p>e-v</p><p>2</p><p>2 >v</p><p>2</p><p>pQ a v1</p><p>v2</p><p>b2</p><p>= e</p><p>v</p><p>2</p><p>1 - v</p><p>2</p><p>2 >v</p><p>2</p><p>p or v</p><p>2</p><p>p =</p><p>2kT</p><p>m</p><p>=</p><p>v</p><p>2</p><p>1 - v</p><p>2</p><p>2</p><p>(ln v</p><p>2</p><p>1>v</p><p>2</p><p>2)</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 265</p><p>2</p><p>N</p><p>( )</p><p>dN (v)(,</p><p>or</p><p>Thus,</p><p>2.90 Given</p><p>Thus,</p><p>2.91 (by symmetry)</p><p>2.92</p><p>2.93 Here is equal to number of molecules hitting an area of the wall per second</p><p>i.e.,</p><p>= 3</p><p>q</p><p>0</p><p>dN 1vx2 vxdA</p><p>dA</p><p>=</p><p>2kT</p><p>m 3</p><p>q</p><p>0</p><p>xe-x</p><p>dx</p><p>22x</p><p>3</p><p>q</p><p>0</p><p>e-x</p><p>dx</p><p>22x</p><p>=</p><p>2kT</p><p>m</p><p>≠ a3</p><p>2</p><p>b</p><p>≠ a1</p><p>2</p><p>b =</p><p>kT</p><p>m</p><p>6v</p><p>2</p><p>x7 =</p><p>3</p><p>q</p><p>0</p><p>v</p><p>2</p><p>x e-mv 2</p><p>x >2kT dvx</p><p>3</p><p>q</p><p>0</p><p>e-mv 2</p><p>x >2kT dvx</p><p>=</p><p>B</p><p>2kT</p><p>m</p><p>≠(1)</p><p>≠a1</p><p>2</p><p>b = B</p><p>2kT</p><p>mp</p><p>=</p><p>A</p><p>2kT</p><p>m</p><p>3</p><p>q</p><p>0</p><p>1</p><p>2</p><p>e -xdx</p><p>3</p><p>q</p><p>0</p><p>e -x</p><p>dx</p><p>21x</p><p>=</p><p>A</p><p>2kT</p><p>m</p><p>3</p><p>q</p><p>0</p><p>ue-u2</p><p>du</p><p>3</p><p>q</p><p>0</p><p>e-u2</p><p>du</p><p>6 | vx |7 =</p><p>3</p><p>q</p><p>0</p><p>| vx |</p><p>e -mv</p><p>2</p><p>x >2kT dvx</p><p>3</p><p>q</p><p>0</p><p>e -mv</p><p>2</p><p>x >2kT dvx</p><p>=</p><p>3</p><p>q</p><p>0</p><p>vxe</p><p>-mv</p><p>2</p><p>x >2kT dvx</p><p>3</p><p>q</p><p>0</p><p>e -mv</p><p>2</p><p>x >2kT dvx</p><p>6vx 7 = 0</p><p>dn 1v2 = N a m</p><p>2pkT</p><p>b3>2</p><p>e- m</p><p>2kT 1v</p><p>2</p><p>x + v</p><p>2</p><p>�</p><p>2</p><p>2pv</p><p>�</p><p>dv</p><p>�</p><p>dvx</p><p>d</p><p>3v = 2pv</p><p>�</p><p>dv</p><p>�</p><p>dvx</p><p>T =</p><p>1</p><p>3</p><p>mv</p><p>2</p><p>k</p><p>v</p><p>2 =</p><p>3</p><p>2</p><p>v</p><p>2</p><p>p =</p><p>3kT</p><p>m</p><p>266 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>vdA</p><p>vdA</p><p>or</p><p>2.94 Let the number of molecules per unit volume with x component of velocity in the</p><p>range be</p><p>Then,</p><p>2.95</p><p>2.96</p><p>or</p><p>dN 1e2</p><p>de</p><p>= N a m</p><p>2pkT</p><p>b3>2</p><p>e -mv 2>2RT</p><p>4pv 2</p><p>dv</p><p>de</p><p>dN 1v2 = N a m</p><p>2pkT</p><p>b3>2</p><p>e -mv 2>2kT</p><p>4pv 2dv = dN (e) =</p><p>dN 1e2</p><p>de</p><p>de</p><p>= 2a m</p><p>2pkT</p><p>b1>2</p><p>= a 2m</p><p>pkT</p><p>b1>2</p><p>= a 16</p><p>p2</p><p>mp</p><p>8kT</p><p>b1>2</p><p>=</p><p>4</p><p>p6v7</p><p>= a m</p><p>2pkT</p><p>b3>2</p><p>4p</p><p>1</p><p>2</p><p>2kT</p><p>m 3</p><p>q</p><p>0</p><p>e- xdx</p><p>61>v7 = 3</p><p>q</p><p>0</p><p>a m</p><p>2pkT</p><p>b3>2</p><p>e-mv 2</p><p>x >2kT 4pv</p><p>2dv</p><p>1</p><p>v</p><p>=</p><p>4</p><p>2pnkT 3</p><p>q</p><p>0</p><p>xe -x</p><p>dx</p><p>22x</p><p>= nkT</p><p>= 2mn</p><p>1</p><p>2p</p><p>2kT</p><p>m 3</p><p>q</p><p>0</p><p>u2e-u2</p><p>du</p><p>= 3</p><p>q</p><p>0</p><p>2mv</p><p>2</p><p>x</p><p>n a m</p><p>2pkT</p><p>b1>2</p><p>e-mv</p><p>2</p><p>x >2kT</p><p>p = 3</p><p>q</p><p>0</p><p>2mvx</p><p># vx</p><p>dn 1vx2</p><p>dn1vx2 = na m</p><p>2pkT</p><p>b1>2</p><p>e</p><p>- mv</p><p>2</p><p>x /2kTdvx</p><p>vx to vx + dvx</p><p>awhere 6v7 = A</p><p>8kT</p><p>mp</p><p>b =</p><p>1</p><p>2</p><p>n A</p><p>2kT</p><p>mp</p><p>= n A</p><p>kT</p><p>2mp</p><p>=</p><p>1</p><p>4</p><p>n 6v7</p><p>= 3</p><p>q</p><p>0</p><p>n</p><p>2p a2kT</p><p>m</p><p>b1>2</p><p>e-u2</p><p>udu</p><p>v = 3</p><p>q</p><p>0</p><p>na m</p><p>2pkT</p><p>b1>2</p><p>e-mv 2</p><p>x >2kTvxdvx</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 267</p><p>dA</p><p>dvx</p><p>,</p><p>Now,</p><p>or</p><p>i.e.,</p><p>The most probable kinetic energy is given by</p><p>or</p><p>The corresponding velocity is</p><p>2.97 The mean kinetic energy is</p><p>Thus,</p><p>If the fraction of molecules is 0.9%.</p><p>2.98</p><p>=</p><p>2</p><p>2p 1kT 2-3>21e0kT e-e0 >kT = 2A</p><p>e0</p><p>pkT</p><p>e-e0 >k T</p><p>=</p><p>2</p><p>2p 1kT 2-3>21e03</p><p>q</p><p>e0</p><p>e-e>kTde 1e0 W kT 2</p><p>¢N</p><p>N</p><p>=</p><p>2</p><p>2p1kT 2-3>2</p><p>3</p><p>q</p><p>e0</p><p>1e e-e>kT de</p><p>dh = 1%,</p><p>=</p><p>2</p><p>2p e-3>2 a3</p><p>2</p><p>b3>2</p><p>2dh = 3 A</p><p>6</p><p>p</p><p>e-3>2 dh</p><p>dN</p><p>N</p><p>= 3</p><p>3</p><p>211+dh2kT</p><p>3</p><p>2kT 11-dh2</p><p>2</p><p>2p1kT 2-3>2 e -e>kTe1>2 de</p><p>6e7 =</p><p>3</p><p>q</p><p>0</p><p>e3>2 e-e>kTde</p><p>3</p><p>q</p><p>0</p><p>e1>2 e-e>kTde</p><p>= kT</p><p>≠15>22</p><p>≠13>22 =</p><p>3</p><p>2</p><p>kT</p><p>v = A</p><p>kT</p><p>m</p><p>Z vp</p><p>1</p><p>2</p><p>e 1>2 e - >kT</p><p>e1>2</p><p>kT</p><p>e>kT = 0 or e 1</p><p>2</p><p>kT = ep</p><p>d</p><p>de</p><p>dN 1e2</p><p>de</p><p>= 0</p><p>dN 1e2 = N</p><p>2</p><p>2p 1kT 2-3>2 e -e>kT e1>2 de</p><p>= N</p><p>2</p><p>2p 1kT 2-3>2 e-e>kT e1>2</p><p>dN 1e2</p><p>de</p><p>= N a m</p><p>2pkT</p><p>b3>2</p><p>e-e>kT 4pA</p><p>2e</p><p>m</p><p>1</p><p>m</p><p>e =</p><p>1</p><p>2</p><p>mv 2 so dv</p><p>de</p><p>=</p><p>1</p><p>mv</p><p>268 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>− − =e e</p><p>(In evaluating the integral, we have taken out as since the integral is domi-</p><p>nated by the lower limit.)</p><p>2.99 (a)</p><p>For the most probable value of the velocity</p><p>So,</p><p>This should be compared with the value for the Maxwellian distribution.</p><p>(b) In terms of energy,</p><p>From this the probable energy comes out as follows:</p><p>2.100 The number of molecules reaching a unit area of wall at angle between and</p><p>to its normal per unit time is</p><p>2.101 The number of molecules reaching per unit area of the wall with velocities in the</p><p>interval per unit time is</p><p>dv = 3</p><p>u</p><p>=</p><p>p>2</p><p>u</p><p>=</p><p>0</p><p>dn1v2 dÆ</p><p>4p</p><p>v cosu</p><p>v to v + dv</p><p>= n a 2kT</p><p>mp</p><p>b1>2</p><p>3</p><p>q</p><p>0</p><p>e-x xdx sin u cos u du = n a 2kT</p><p>mp</p><p>b1>2</p><p>sin u cos u du</p><p>= 3</p><p>q</p><p>0</p><p>n a m</p><p>2pkT</p><p>b3>2</p><p>e-mv</p><p>2>2kT v</p><p>3 dv sin u cos u du * 2p</p><p>dv = 3</p><p>v = q</p><p>v = 0</p><p>dn1v2 dÆ</p><p>4p</p><p>v cos u</p><p>u + duu</p><p>2A</p><p>m2</p><p>ae-e>kT -</p><p>e</p><p>kT</p><p>e-e>kTb = 0 or ep = kT F 1e2 = 0</p><p>= A a2e</p><p>m</p><p>b3>2</p><p>e -e>kT</p><p>1</p><p>12me</p><p>= A</p><p>2e</p><p>m2</p><p>e-e>kT</p><p>F 1e2 = Av 3 e-mv 2>2kT</p><p>dv</p><p>de</p><p>e =</p><p>1</p><p>2</p><p>mv 2</p><p>vp = A</p><p>2kT</p><p>m</p><p>vp = A</p><p>3kT</p><p>m</p><p>dF 1v2</p><p>dv</p><p>= 0 or 3 Av</p><p>2e -mv 2>2kT - Av 3</p><p>2mv</p><p>2kT</p><p>e -mv 2>2kT = 0</p><p>F 1v2 = Av 3e-mv 2>2k T</p><p>1e01e</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 269</p><p>v</p><p>Q</p><p>2.102 If the force exerted is F, then by the law of variation of concentration with height</p><p>So,</p><p>2.103 Here,</p><p>Hence,</p><p>2.104</p><p>2.105</p><p>They are equal at a height where</p><p>or</p><p>2.106 At a temperature T the concentration varies with height according to</p><p>This means that particles per unit area of the base of cylinder are</p><p>3</p><p>q</p><p>0</p><p>n1z-2 d z-</p><p>n1z-2 = n0e</p><p>-mgz>kT</p><p>n1z-2</p><p>h =</p><p>kT</p><p>g</p><p>ln n1 - ln n2</p><p>m1 - m2</p><p>n1>n2 = e</p><p>gh (m1 - m2)>kT h</p><p>n1(h) = n1e</p><p>-m1gh>kt n21h2 = n2e</p><p>-m2gh> kt</p><p>h</p><p>h0</p><p>= e</p><p>(-MN2</p><p>+ MH2</p><p>)gh>RT = 1.39</p><p>h =</p><p>concentration of H2</p><p>concentration of N2</p><p>= h0</p><p>e -MN2</p><p>gh>RT</p><p>e -MH2</p><p>gh>RT</p><p>= h0e</p><p>(-MN2</p><p>+ MH2</p><p>)gh>RT</p><p>Na =</p><p>6 * 8.31 * 290 * ln2</p><p>p * 64 * 9.8 * 200 * 4</p><p>* 1026 = 6.36 * 1023 mol-1</p><p>= 0.2 * 103 kg>m3</p><p>and</p><p>T = 290 K, h = 2, h = 4 * 10-5 m, d = 4 * 10-7m, g = 9.8 m>s2,</p><p>¢r</p><p>F =</p><p>p</p><p>6</p><p>d</p><p>3¢rg =</p><p>RT lnh</p><p>Nah</p><p>or Na =</p><p>6RT lnh</p><p>pd 3g¢rh</p><p>h = eF¢h>kT or F =</p><p>kT ln h</p><p>¢h</p><p>= 9 * 10-20 N</p><p>n 1Z2 = n0 e</p><p>-FZ>kT</p><p>= np a m</p><p>2pkT</p><p>b3>2</p><p>e-mv 2>2kTv</p><p>3dv</p><p>= 3</p><p>u</p><p>=</p><p>p>2</p><p>u</p><p>=</p><p>0</p><p>na m</p><p>2pkT</p><p>b3>2</p><p>e-mv 2>2kT v</p><p>3 dv sin u cos u du * 2p</p><p>270 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>R = >K.8.31 J</p><p>Given</p><p>We know that</p><p>and</p><p>,</p><p>Clearly this cannot change. Thus pressure at the bottom of the cylin-</p><p>der, must not change with change of temperature.</p><p>2.107</p><p>When there are many kinds of molecules, this formula holds for each kind and the</p><p>average energy is given by</p><p>where fractional concentration of each kind of molecule at the ground level.</p><p>2.108 The constant acceleration is equivalent to a pseudo force wherein a concentration</p><p>gradient is set up. Then</p><p>or</p><p>2.109 In a centrifuge rotating with angular velocity about an axis, there is a centrifugal</p><p>acceleration where r is the radial distance from the axis. In a fluid if there are sus-</p><p>pended colloidal particles they experience an additional force. If m is the mass of</p><p>each particle then its volume is and the excess force on this particle is</p><p>outward, corresponding to a potential energy</p><p>This gives rise to a concentration variation, given by</p><p>Thus,</p><p>(where is the molecular weight).</p><p>m</p><p>k</p><p>=</p><p>M</p><p>R</p><p>M = NAm</p><p>n1r22</p><p>n1r12 = h = exp a +</p><p>M</p><p>2 r RT</p><p>1r - r02�21r 2</p><p>2 - r 2</p><p>12b</p><p>n1r2 = n0 exp a +</p><p>m</p><p>2 r kT</p><p>1r - r02 �2r 2b</p><p>- (m>2r) r - r0 v2r</p><p>2.m>r 1r- r02v2r</p><p>m/r</p><p>v2</p><p>r</p><p>�</p><p>w = -</p><p>RT ln11 - h2</p><p>MAl</p><p>L</p><p>hRT</p><p>MAl</p><p>L 70 g</p><p>e -MAwl>RT = 1 - h</p><p>fi</p><p>r</p><p>6U 7 = aFikT</p><p>a fi</p><p>= kT</p><p>6U 7 =</p><p>3</p><p>q</p><p>0</p><p>mgz e-mgz>ktdz</p><p>3</p><p>q</p><p>0</p><p>e-mgz>kTdz</p><p>= kT</p><p>3</p><p>q</p><p>0</p><p>xe-xdx</p><p>3</p><p>q</p><p>0</p><p>e-xdx</p><p>= kT</p><p>≠(2)</p><p>≠(1)</p><p>= kT</p><p>n0kT = p0 =</p><p>= 3</p><p>q</p><p>0</p><p>n0e</p><p>-mgz>kT dz- =</p><p>n0kT</p><p>mg</p><p>2.3 KINETIC THEORY OF GASES. BOLTZMANN’S LAW AND MAXWELL’S DISTRIBUTION 271</p><p>,</p><p>of ( )</p><p>and</p><p>Thus,</p><p>2.110 The potential energy associated with each molecule is and there is a</p><p>concentration variation, given by</p><p>Thus,</p><p>Using cm, erg/K, K,</p><p>rad/s</p><p>2.111 Here,</p><p>(a) The number of molecules located at the distance between r and is</p><p>(b) is given by</p><p>(c) The fraction of molecules lying between r and is</p><p>Thus,</p><p>(d)</p><p>So,</p><p>When T decreases times, will increase times. h3>2n102 = n0h</p><p>n1r2 = N a a</p><p>pkT</p><p>b1>2</p><p>exp a -ar 2</p><p>kT</p><p>b</p><p>dN = N a a</p><p>pkT</p><p>b3>2</p><p>4pr</p><p>2dr exp a -ar 2</p><p>kT</p><p>b</p><p>dN</p><p>N</p><p>= a a</p><p>pkT</p><p>b3>2</p><p>4pr</p><p>2dr exp a -ar 2</p><p>kT</p><p>b</p><p>= a kT</p><p>a</p><p>b3>2</p><p>2p≠ a3</p><p>2</p><p>b = apkT</p><p>a</p><p>b3>2</p><p>3</p><p>q</p><p>0</p><p>4pr 2dr exp a -</p><p>ar 2</p><p>kT</p><p>b = a kT</p><p>a</p><p>b3>2</p><p>4p3</p><p>q</p><p>0</p><p>x</p><p>dx</p><p>22x</p><p>exp 1-x2</p><p>dN</p><p>N</p><p>=</p><p>4pr 2dr n0 exp 1-ar 2>kT 2</p><p>3</p><p>q</p><p>0</p><p>4pr 2drn0 exp 1ar 2>kT 2</p><p>r + dr</p><p>d</p><p>dr</p><p>r 2 n 1r2 = 0 or 2r -</p><p>2ar 3</p><p>kT</p><p>= 0 or rp = A</p><p>kT</p><p>a</p><p>rp</p><p>4 pr 2dr n1r2 = 4pn0exp a -</p><p>ar 2</p><p>kT</p><p>br</p><p>2dr</p><p>r + dr</p><p>n1r2 = n0exp a -</p><p>ar 2</p><p>kT</p><p>b</p><p>� = 280</p><p>T = 300R = 8.31 * 107M = 12 + 32 = 44 g, l = 100</p><p>h = exp aMv2l 2</p><p>2RT</p><p>b or v = A</p><p>2RT</p><p>Ml 2</p><p>ln h</p><p>n 1r2 = n0 exp am�2r 2</p><p>2kT</p><p>b = n0 exp aM�2r 2</p><p>2RT</p><p>b</p><p>- 1>2 m v2r 2</p><p>M =</p><p>2 r RT ln h1r - r02�21r 2</p><p>2 - r 2</p><p>12</p><p>272 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>( )</p><p>we get,</p><p>the concentration variation is</p><p>2.112 (a) Let</p><p>So,</p><p>So,</p><p>(b) The most probable value of U is given by</p><p>From 2.111 (b), the potential energy at the most probable distance is kT.</p><p>2.4 The Second Law of Thermodynamics. Entropy</p><p>2.113 The efficiency is given by</p><p>Now, in the two cases the efficiencies are</p><p>increased)</p><p>decreased)</p><p>Thus,</p><p>2.114 (a) For</p><p>Define n</p><p>p4V</p><p>g</p><p>4 = p1V</p><p>g</p><p>1 so V 1-g</p><p>4 = V 1-g</p><p>1 n1–g or V4 = nV1</p><p>p4V4 = p3V3 = p2V2n</p><p>1-g = p1V1n</p><p>1-g</p><p>p3 = p2n</p><p>-g</p><p>V3 =</p><p>n V2.</p><p>p2V</p><p>g</p><p>2 = p3V</p><p>g</p><p>3 , p1V</p><p>g</p><p>1 = p4V</p><p>g</p><p>4</p><p>p1V1 = p2V2, p3V3 = p4V4</p><p>H2, g = 7>5. From the figure</p><p>hh 6 n1</p><p>hl =</p><p>T1 - T2 + ¢T</p><p>T1</p><p>(T2</p><p>hh =</p><p>T1 + ¢T - T2</p><p>T1 + ¢T</p><p>(T1</p><p>h =</p><p>T1 - T2</p><p>T1</p><p>(for T1 7 T2)</p><p>d</p><p>dU</p><p>adN</p><p>dU</p><p>b = 0 a 1</p><p>21U</p><p>-</p><p>U 1>2</p><p>kT</p><p>b exp a -U</p><p>kT</p><p>b or Up =</p><p>1</p><p>2</p><p>kT</p><p>= 2pn0 a</p><p>-3>2 U 1>2 exp a U</p><p>kT</p><p>bdU</p><p>dN = n04p</p><p>U</p><p>a</p><p>dU</p><p>21aU</p><p>exp a U</p><p>kT</p><p>b</p><p>dr = A</p><p>1</p><p>a</p><p>dU</p><p>21U</p><p>=</p><p>dU</p><p>21aU</p><p>U = ar</p><p>2 or r = A</p><p>U</p><p>a</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 273</p><p>adiabatic</p><p>p V1 1</p><p>Q1</p><p>p V2 2</p><p>p V3 3Q'2</p><p>p4 4,V</p><p>by</p><p>Then</p><p>So,</p><p>since, ,</p><p>and</p><p>Also,</p><p>Finally</p><p>(b) Define</p><p>So we get the formulae here by in the previous case.</p><p>2.115 Used as a refrigerator, the refrigerating efficiency of a heat engine is</p><p>given by</p><p>(where is the efficiency of the heat engine).</p><p>2.116 Given</p><p>Heat taken at the upper temperature</p><p>Now,</p><p>or</p><p>Similarly,</p><p>Thus, heat ejected at the lower temperature</p><p>= –RT3 ln aT1</p><p>T2</p><p>b1>g-1 1</p><p>n2</p><p>aT1</p><p>T2</p><p>b -1>g-1</p><p>= 2RT3 ln n</p><p>= –RT3 ln aT1</p><p>T2</p><p>b1>g-1V1</p><p>V4</p><p>= -RT3 in aT1</p><p>T2</p><p>b1>g-1 V2</p><p>n2V3</p><p>RT3 ln</p><p>V6</p><p>V5</p><p>Q2 =</p><p>V6 = aT1</p><p>T3</p><p>b1>g-1</p><p>V1</p><p>V5 = aT2</p><p>T3</p><p>b1>g-1</p><p>V4,</p><p>V3 = aT1</p><p>T2</p><p>b1>g-1</p><p>V2</p><p>T1V</p><p>g-1</p><p>2 = T2V</p><p>g-1</p><p>3</p><p>= RT1 ln n + RT2 ln n = R (T1 + T2) ln n</p><p>Q1 =</p><p>V2 = nV1,V4 = nV3</p><p>h</p><p>e =</p><p>Q¿2</p><p>A</p><p>=</p><p>Q¿2</p><p>Q1 - Q¿2</p><p>=</p><p>Q¿2 >Q1</p><p>1 -</p><p>Q ¿2</p><p>Q1</p><p>=</p><p>1 - h</p><p>h</p><p>= 9</p><p>h = 1 - n(1/g)-1 = 1 - n-(2/7) = 0.18</p><p>n n1/g</p><p>p2V</p><p>g</p><p>2 =</p><p>p2</p><p>n</p><p>V3</p><p>g or V3 = n1/gV2</p><p>n by p3 = p2>n</p><p>h = 1 -</p><p>Q ¿2</p><p>Q1</p><p>= 1 - n1-g = 0.242</p><p>Q1 = p2V2 ln</p><p>V2</p><p>V1</p><p>, Q ¿2 = p3V3 ln</p><p>V3</p><p>V4</p><p>n</p><p>1-g = p2V2 ln</p><p>V3</p><p>V4</p><p>274 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>A=Q –1 Q'2</p><p>Q1</p><p>Q'2</p><p>p V1, 1</p><p>p V4, 4</p><p>p V5, 5p V6, 6</p><p>p V2, 2</p><p>p V3, 3</p><p>T1</p><p>T2</p><p>T3</p><p>Then,</p><p>substituting by</p><p>-</p><p>Then,</p><p>Thus,</p><p>2.117</p><p>So,</p><p>On the other hand,</p><p>Thus,</p><p>So,</p><p>2.118</p><p>So,</p><p>Now,</p><p>So,</p><p>2.119 Since the absolute temperature of the gas rises times both in the</p><p>isochoric heating and the isobaric expansion, we have</p><p>where,</p><p>and</p><p>Heat rejected is</p><p>where, Q¿21 = CVT1 (n - 1) Q¿22 = C T1 a1 -</p><p>1</p><p>n</p><p>b Q¿2 = Q¿21 + Q¿22</p><p>Q12 = CVT1a1 -</p><p>1</p><p>n</p><p>b Q11 = CP 1n - 12T1</p><p>Q1 = Q11 + Q12Heat taken is</p><p>p1 = np2 and V2 = nV1</p><p>n</p><p>h = 1 -</p><p>1</p><p>n</p><p># n1>g = 1 - n(1>g)-1</p><p>p2V</p><p>g</p><p>4 = p1V</p><p>g</p><p>1 or V4 = n1>gV1</p><p>p1 = np2 or V3 = n1/gV2</p><p>h = 1 -</p><p>p2 1V3 - V42</p><p>p1</p><p>1V2 - V12</p><p>Q1 =</p><p>C</p><p>R</p><p>p1</p><p>1V2 - V12</p><p>Q ¿2 =</p><p>C</p><p>R</p><p>p2 1V3 - V42</p><p>h = 1-n1-g (where g = 7>5 for N2</p><p>h = 0.602 = 60.2%</p><p>p1 = p2n</p><p>g p4 = p3n</p><p>g</p><p>p1V</p><p>g</p><p>1 = p2V</p><p>g</p><p>2</p><p>, p3V</p><p>g</p><p>2 = p4V</p><p>g</p><p>1 also V2 = nV1</p><p>h = 1-</p><p>V2</p><p>1 p2 - p32</p><p>V1 1p1 - p42</p><p>Q1 =</p><p>CV</p><p>R</p><p>V11p1 - p42</p><p>Q ¿2 = CV 1T2-T32 =</p><p>CV</p><p>R</p><p>V2 1p2 - p32</p><p>h = 1-</p><p>2T3</p><p>T1 + T2</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 275</p><p>p V1, 1</p><p>p V2, 2</p><p>p V3, 2</p><p>p V4, 1</p><p>Q1</p><p>Q'2</p><p>p V1, 1 p V1, 2</p><p>p V3, 2p V2, 4</p><p>Q1</p><p>Q'2</p><p>p V2, 2</p><p>p V1, 2p V1, 1</p><p>p V2, 1</p><p>Q12 Q'21</p><p>Q'22</p><p>Q11</p><p>T1 nT1</p><p>(T1/n) T1</p><p>p</p><p>,</p><p>and</p><p>Therefore,</p><p>)</p><p>p</p><p>Since,</p><p>and</p><p>and</p><p>Since,</p><p>p</p><p>Thus,</p><p>2.120 (a) Here,</p><p>But,</p><p>Thus,</p><p>(b) Here,</p><p>and</p><p>i.e.,</p><p>Also,</p><p>or</p><p>Thus,</p><p>2.121 Here the isothermal process proceeds at the maximum</p><p>temperature instead of at the minimum temperature of</p><p>the cycle as in Problem 2.120.</p><p>(a) Here</p><p>p2V</p><p>g</p><p>1 = p0V</p><p>g</p><p>0 or p1 V</p><p>g</p><p>1 = np0V</p><p>g</p><p>0</p><p>p1V1 = p0V0, p2 =</p><p>p1</p><p>n</p><p>h = 1 -</p><p>ln n</p><p>n - 1</p><p>Q ¿2 = RT0 ln ng>(g-1) =</p><p>R g</p><p>g - 1</p><p>T0 ln n = C T0 ln n</p><p>Q ¿2 = RT0 ln</p><p>V0</p><p>V1</p><p>Q1 = C T0 (n - 1),</p><p>ngV g-1</p><p>1 = V g-1</p><p>1 or V1 = ng>g-1</p><p>V0</p><p>p1(n V1)</p><p>g = p0V</p><p>g</p><p>0</p><p>V2 = nV1, p1V1 = p0V0</p><p>h = 1 -</p><p>ln n</p><p>n - 1</p><p>aon using CV =</p><p>R</p><p>g - 1</p><p>b</p><p>Q¿2 = RT0 ln n-1>g-1</p><p>RT0</p><p>g - 1</p><p>ln n</p><p>nV g-1</p><p>1 = V g-1</p><p>0 or V1 = V0n</p><p>-1>g-1</p><p>Q¿2 = RT0 ln</p><p>V0</p><p>V1</p><p>, Q1 = CVT0 (n - 1)</p><p>np1V</p><p>g</p><p>1 = p0V</p><p>g</p><p>0</p><p>p2 = np1, p1V1 = p0V0,</p><p>= a1 -</p><p>h - 1 + g(1 - 1>n)</p><p>g(n - 1) + (1 - 1>n)</p><p>b = 1 -</p><p>1 +</p><p>g</p><p>n</p><p>g +</p><p>1</p><p>n</p><p>= 1 -</p><p>n + g</p><p>1 + ng</p><p>h = 1 -</p><p>Q ¿2</p><p>Q1</p><p>= 1 -</p><p>CV</p><p>(n - 1) + Cp</p><p>(1 - 1>n)</p><p>Cp(n - 1) + CV</p><p>(1 - 1>n)</p><p>276 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>Q1</p><p>Q'2</p><p>Temp nT0</p><p>Temp T0</p><p>isothermal</p><p>p V1, 1</p><p>adiabatic</p><p>p V0, 0</p><p>isothermal</p><p>adiabatic</p><p>Q1</p><p>Q'2</p><p>p V1, 1</p><p>p V1, , nT2 0</p><p>T0</p><p>p V0, 0</p><p>T0</p><p>isothermal</p><p>adiabatic</p><p>p V0, 0</p><p>T0 p V1, ,1 0T</p><p>p V1, ,1 0T /n</p><p>Q1</p><p>Q'2</p><p>isochoric</p><p>(a)</p><p>p</p><p>p</p><p>and</p><p>Also,</p><p>and</p><p>i.e.,</p><p>Thus,</p><p>(b) Here</p><p>Thus,</p><p>2.122 The section from to is a polytropic process of index . We shall</p><p>assume that the corresponding specific heat C is positive.</p><p>Here,</p><p>Now, constant</p><p>So,</p><p>Then,</p><p>We have</p><p>Now,</p><p>Thus, h = 1 -</p><p>n - 1</p><p>n ln n</p><p>Q¿2 = CT0 a1 -</p><p>1</p><p>n</p><p>b , Q1 = RT0 ln</p><p>V1</p><p>V0</p><p>= RT0 a 1</p><p>g - 1</p><p>-</p><p>1</p><p>a - 1</p><p>b ln n = CT0 ln n</p><p>V a-1</p><p>1 =</p><p>1</p><p>n</p><p>V a-1</p><p>2 or V1 = n-1>(a-1) V2 = n1>(g-1)-1>(a -1) V0</p><p>p1V</p><p>a</p><p>1 = p2V</p><p>a</p><p>2 or V g-1</p><p>0 =</p><p>1</p><p>n</p><p>V g-1</p><p>2 or V2 = V0n</p><p>1>(g-1)</p><p>p0V0 = p1V1 = np2V2, p0V</p><p>g</p><p>0 = p2V</p><p>g</p><p>2</p><p>p1V1 = RT0, p2V2 =</p><p>RT0</p><p>n</p><p>=</p><p>p1V1</p><p>n</p><p>C = CV -</p><p>R</p><p>a - 1</p><p>= R a 1</p><p>g - 1</p><p>-</p><p>1</p><p>a - 1</p><p>b</p><p>pdV =</p><p>RT</p><p>V</p><p>dV = -</p><p>R</p><p>a - 1</p><p>dT</p><p>pV</p><p>a = constant or TV</p><p>a-1 =</p><p>dQ = CdT = CVdT + pdV</p><p>aT0>n)V2,(p2,T0)V1,(p1,</p><p>h = 1 -</p><p>n - 1</p><p>n ln n</p><p>Q¿2 = C T0 a1 -</p><p>1</p><p>n</p><p>b , Q1 = RT0 ln</p><p>V1</p><p>V0</p><p>=</p><p>R g</p><p>g - 1</p><p>T0 ln n = C T0 ln n</p><p>p0V</p><p>g</p><p>0 = p1V</p><p>g</p><p>2 = p1n</p><p>-g</p><p>V g</p><p>1 = V g-1</p><p>0 n-g V g-1</p><p>1 or V1 = n(g>g-1) V0</p><p>V2 =</p><p>V1</p><p>n</p><p>, p0V0 = p1V1</p><p>h = 1 -</p><p>Q¿2</p><p>Q1</p><p>= 1 -</p><p>n - 1</p><p>n ln n</p><p>Q¿2 = CVT0 a1 -</p><p>1</p><p>n</p><p>b , Q1 = RT0 ln</p><p>V1</p><p>V0</p><p>=</p><p>RT0</p><p>g - 1</p><p>ln n = CVT0 ln n</p><p>V g-1</p><p>1 = nV g-1</p><p>0 or V1 = n1>g-1 V0</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 277</p><p>adiabatic</p><p>(b)</p><p>p V1, ,2 0T /n</p><p>isothermal</p><p>Q1</p><p>Q'2 p V1, ,1 0T</p><p>p V0, ,0 0T</p><p>adiabatic</p><p>Q1</p><p>p V2 , ,2 0T /n</p><p>p V1 , ,1 0T</p><p>p V0 , ,0 0T</p><p>Q'2</p><p>isothermal</p><p>Polytropic of</p><p>index</p><p>pp</p><p>,</p><p>2.123 (a) Here,</p><p>Along the adiabatic line</p><p>So,</p><p>Thus,</p><p>(b) Here</p><p>Along the adiabatic line constant.</p><p>So,</p><p>Thus,</p><p>2.124 (a)</p><p>So,</p><p>(b)</p><p>So,</p><p>2.125 We have</p><p>and</p><p>as well as</p><p>and Q2¿ = Q2– + Q2‡</p><p>Q1 = Q1¿ + Q1¿</p><p>Q2¿ = RT0 ln , Q1– = CVT0 (t - 1)</p><p>Q¿1 = tRT0 ln , Q¿2 = CVT0(t - 1)</p><p>h = 1 -</p><p>Q¿2</p><p>Q1</p><p>= 1 -</p><p>n - 1 + (g - 1) ln n</p><p>g(n - 1)</p><p>Q ¿2 = Q –</p><p>2 + Q ‡2</p><p>Q1 = CpT0 (n - 1), Q –2 = CVT0 (n - 1),</p><p>Q ‡2 = RT0 ln n</p><p>= 1 -</p><p>g</p><p>1 +</p><p>R</p><p>CV</p><p>n ln n</p><p>n - 1</p><p>= 1 -</p><p>g (n - 1)</p><p>n - 1 + (g - 1)n lnn</p><p>h = 1 -</p><p>Q¿2</p><p>Q1</p><p>= 1 -</p><p>C (1 - 1>n)</p><p>CV</p><p>(1 - 1>n) + R ln n</p><p>Q ¿2 = CpT0a1 -</p><p>1</p><p>n</p><p>b , Q –1 = RT0 ln n</p><p>Q</p><p>¿1 = CVT0</p><p>a1 -</p><p>1</p><p>n</p><p>b</p><p>Q1 = Q</p><p>¿1 + Q</p><p>–1</p><p>h = 1 -</p><p>ng - 1</p><p>gng-1(n - 1)</p><p>T0V</p><p>g-1</p><p>0 = T1aV0</p><p>n</p><p>bg-1</p><p>or T1 = ng-1 T0</p><p>TV</p><p>g-1 =</p><p>Q ¿2 = CV</p><p>(nT1 - T0), Q1 = C # T1(n - 1)</p><p>h = 1 -</p><p>g (n - 1)</p><p>ng-1</p><p>Q1 = CV</p><p>T1</p><p>n</p><p>(ng - 1)</p><p>T0V</p><p>g-1</p><p>0 = T11n V02g-1 or T0 = T1n</p><p>g-1</p><p>Q ¿2 = Cp aT1 -</p><p>T1</p><p>n</p><p>b = C T1a1 -</p><p>1</p><p>n</p><p>b , Q1 = CV</p><p>aT0 -</p><p>T1</p><p>n</p><p>b</p><p>278 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>Q'2</p><p>Q1</p><p>p V1 , ,1 1T /n</p><p>p V0 , ,0 0T</p><p>p V1 , ,n T0 1</p><p>adiabatic</p><p>(a)</p><p>Q2'</p><p>p V1 , ,0 1nT</p><p>p V0 , ,0 0T</p><p>(b)</p><p>adiabatic</p><p>Q1</p><p>p V1 , / ,0 0n T</p><p>Q "1Q1</p><p>p V0/ , ,n T /n0 0</p><p>p V0, ,0 0T</p><p>p V0/ ,n,n T0 0</p><p>adiabatic</p><p>Q2</p><p>(a)</p><p>Q2''</p><p>np V0, ,0 0nT n</p><p>p V0, ,0 0T</p><p>Q "'=RT ln n2 0</p><p>Q1</p><p>n n T, / ,0 0p V0</p><p>(b)</p><p>T0</p><p>V0</p><p>v 0V</p><p>Q2'''</p><p>Q2''Q1''</p><p>T0</p><p>Q1'</p><p>p</p><p>p</p><p>p</p><p>and</p><p>Here,</p><p>Here,</p><p>Now,</p><p>Now,</p><p>and</p><p>ν</p><p>ν</p><p>So,</p><p>2.126 Here,</p><p>In addition, we have</p><p>and</p><p>So,</p><p>2.127 Because of the linearity of the section BC, whose equation is</p><p>We have</p><p>Here,</p><p>Thus,</p><p>Along BC, the specific heat C is given by</p><p>Thus,</p><p>Finally h = 1 -</p><p>Q ¿2</p><p>Q1</p><p>= 1 - 2</p><p>1t + g</p><p>1t + 1</p><p>1</p><p>g + 1</p><p>=</p><p>(g - 1) 11t - 12</p><p>(g + 1)11t + 12</p><p>Q1 =</p><p>1</p><p>2</p><p>RT0</p><p>g + 1</p><p>g - 1</p><p>t - 1</p><p>1t</p><p>CdT = CV</p><p>dT + pdV = CV</p><p>dT + d a1</p><p>2</p><p>aV 2b = aCV +</p><p>1</p><p>2</p><p>Rb dT</p><p>Q¿2 = Q2– + Q2‡ =</p><p>RT0</p><p>g - 1</p><p>11t - 12 a1 +</p><p>g</p><p>1t b</p><p>Q‡2 = CpT0a1 -</p><p>1</p><p>1t = Cp</p><p>T0</p><p>1t11t - 12</p><p>Q–2 = CVT0 11t - 12,</p><p>t</p><p>= v or = 1t</p><p>p</p><p>p0</p><p>=</p><p>V</p><p>V0</p><p>* (K p = aV )</p><p>= 1 -</p><p>t - 1 + a1 -</p><p>1</p><p>g</p><p>b ln n</p><p>t - 1 + a1 -</p><p>1</p><p>g</p><p>bt ln n</p><p>=</p><p>(t - 1) ln n</p><p>t ln n +</p><p>g(t - 1)</p><p>g - 1</p><p>h = 1 -</p><p>Q ¿2</p><p>Q1</p><p>= 1 -</p><p>Cp(t - 1) + R ln n</p><p>Cp</p><p>(t - 1) + tR ln n</p><p>Q2¿ = Q2– + Q2‡</p><p>Q1 = Q1¿ + Q1–</p><p>Q2– = CpT0 (t - 1), Q2‡ = RT0 ln n</p><p>Q1– = CpT0(t - 1), Q1¿ = tRT0 ln n</p><p>= 1 -</p><p>t - 1</p><p>g - 1</p><p>+ ln</p><p>t - 1</p><p>g - 1</p><p>+ t ln</p><p>=</p><p>(t - 1) ln</p><p>t ln +</p><p>t - 1</p><p>g - 1</p><p>= 1 -</p><p>CV</p><p>(t - 1) + R ln</p><p>CV</p><p>(t - 1) + t R ln</p><p>h = 1 +</p><p>Q¿2</p><p>Q1</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 279</p><p>Q2'''</p><p>nT0</p><p>V0</p><p>Q2'' P0</p><p>T0</p><p>Q1''</p><p>nV0</p><p>T0</p><p>V0</p><p>Q1'</p><p>n V0</p><p>Q1</p><p>Q2'''</p><p>p V0, ,0 0TAB</p><p>C</p><p>Q2''</p><p>p V0, /0 0V,T /V</p><p>p V0/ ,0</p><p>0 0V , T /V</p><p>and</p><p>ν</p><p>ν</p><p>ν</p><p>ν ν</p><p>ν</p><p>ν ν</p><p>ν</p><p>2.128 We write Claussius inequality in the form</p><p>where is the heat transferred to the system but is heat rejected by the system,</p><p>both are positive and this explains the minus sign before</p><p>In this inequality and we can write</p><p>Thus,</p><p>or</p><p>2.129 We consider an infinitesimal carnot cycle with isothermal process at temperatures</p><p>and T. Let be the work done in the cycle and be the heat received</p><p>at</p><p>the higher temperature. Then by Carnot’s theorem</p><p>On the other hand,</p><p>while,</p><p>Hence,</p><p>2.130 (a) In an isochoric process, the entropy change will be</p><p>For carbon dioxide</p><p>¢S = 19.2 J/K mol</p><p>g = 1.30, so</p><p>¢S = 3</p><p>Tƒ</p><p>Ti</p><p>CVdT</p><p>T</p><p>= CV ln</p><p>Tƒ</p><p>Ti</p><p>= CV ln n =</p><p>R ln n</p><p>g - 1</p><p>a 0U</p><p>0V</p><p>b</p><p>T</p><p>+ p = T a 0p</p><p>0T</p><p>b</p><p>V</p><p>dQ1 = dU12 + pdV = c a 0U</p><p>0V</p><p>b</p><p>T</p><p>+ p d dV</p><p>dA = dpdV = a 0p</p><p>0T</p><p>b</p><p>V</p><p>dTdV</p><p>dA</p><p>dQ1</p><p>=</p><p>dT</p><p>T</p><p>dQ,dAT + dT</p><p>h = 1 -</p><p>Q ¿2</p><p>Q1</p><p>6 1 -</p><p>Tmin</p><p>Tmax</p><p>= hcarnot</p><p>Q1</p><p>Tmax</p><p>6</p><p>Q¿2</p><p>Tmin</p><p>or</p><p>Tmin</p><p>Tmax</p><p>6</p><p>Q¿2</p><p>Q1</p><p>3</p><p>d1Q</p><p>Tmax</p><p>- 3</p><p>d2Q</p><p>Tmin</p><p>6 0</p><p>Tmax 7 T 7 Tmin</p><p>d</p><p>2Q</p><p>d</p><p>2Qd1Q</p><p>3</p><p>d1Q</p><p>T</p><p>- 3</p><p>d2 Q</p><p>T</p><p>… 0</p><p>280 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>1 2</p><p>4 3p</p><p>p+dp</p><p>V V+dV</p><p>Q1</p><p>isotherm</p><p>.</p><p>(b) For an isobaric process, entropy change will be</p><p>2.131 In an isothermal expansion, entropy change is</p><p>So,</p><p>2.132 The entropy change depends on the final and initial states only, so we can calculate</p><p>it directly along the isotherm, it is</p><p>volume is n times the initial volume.</p><p>2.133 If the initial temperature is and volume is then in an adiabatic expansion,</p><p>So,</p><p>being the volume at the end of the adiabatic process. There is no entropy change</p><p>in this process. Next the gas is compressed isobarically and the net entropy change is</p><p>But,</p><p>So,</p><p>2.134 The entropy change depends on the initial and final state only so can be calculated for</p><p>any process. We choose to evaluate the entropy change along the pair of lines shown</p><p>in the figure. Then</p><p>= (-CV ln b + Cp ln a) n =</p><p>nR</p><p>g - 1</p><p>(g ln a - ln b)</p><p>L -11 J>K</p><p>¢S = 3</p><p>T0>b</p><p>T0</p><p>CV</p><p>dT</p><p>T</p><p>+ 3</p><p>aT0>b</p><p>T0>b nCp</p><p>dT</p><p>T</p><p>¢S = am</p><p>M</p><p>Cpb ln</p><p>1</p><p>n</p><p>= -</p><p>m</p><p>M</p><p>Cp ln n = -</p><p>m</p><p>M</p><p>Rg</p><p>g - 1</p><p>ln n = -9.7 J>K</p><p>V1</p><p>T1</p><p>=</p><p>V0</p><p>Tƒ</p><p>or Tƒ = T1</p><p>V0</p><p>V1</p><p>= T0n</p><p>-g</p><p>¢S = am</p><p>M</p><p>Cpb ln</p><p>Tƒ</p><p>T1</p><p>V1</p><p>T = T0n</p><p>1-g = T1 (where n = V1>V0</p><p>)</p><p>TV g-1 = T0V</p><p>g-1</p><p>0</p><p>V0T0</p><p>¢S = 2 R ln n = 20 J>K, assuming that the final</p><p>Vƒ</p><p>Vi</p><p>= e ¢S> R = 2.0 times</p><p>¢S = R ln</p><p>Vƒ</p><p>Vi</p><p>= 25 J>K mol</p><p>¢S = Cp ln</p><p>Tƒ</p><p>Ti</p><p>= Cp ln n =</p><p>gR ln n</p><p>g - 1</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 281</p><p>p V0, ,0 0T</p><p>p0 ,V0,</p><p>T0 p0 , V0 ,</p><p>T0</p><p>ν</p><p>ν</p><p>2.135 To calculate the required entropy difference we only have to calculate the entropy differ-</p><p>ence for a process in which the state of the gas in vessel 1 is changed to that in vessel 2.</p><p>(using</p><p>2.136 For the polytropic process with index n constant.</p><p>Along this process (as in solution of Problem 2.122),</p><p>So,</p><p>2.137 The process in question may be written as</p><p>(where is a constant and are some reference values).</p><p>For this process the specific heat is</p><p>Along the line volume increases times then so does the pressure. The temperature</p><p>must then increase times. Thus,</p><p>n = 2, g =</p><p>5</p><p>3</p><p>a = 2)</p><p>¢S = 46.1 J/K.</p><p>¢S = 3</p><p>a2T0</p><p>T0</p><p>nC</p><p>dT</p><p>T</p><p>=</p><p>nR</p><p>2</p><p>g + 1</p><p>g - 1</p><p>ln a2 = nR</p><p>g + 1</p><p>g - 1</p><p>ln a</p><p>a2</p><p>a</p><p>C = CV +</p><p>1</p><p>2</p><p>R = R a 1</p><p>g - 1</p><p>+</p><p>1</p><p>2</p><p>b =</p><p>1</p><p>2</p><p>R</p><p>g + 1</p><p>g - 1</p><p>(from solution of Problem 2.127),</p><p>p0, V0a</p><p>p</p><p>p0</p><p>= a</p><p>V</p><p>V0</p><p>¢S = 3</p><p>tT0</p><p>T0</p><p>C</p><p>dT</p><p>T</p><p>=</p><p>n - g</p><p>(g - 1)(n - 1)</p><p>R ln t</p><p>C = R a 1</p><p>g - 1</p><p>-</p><p>1</p><p>n - 1</p><p>b =</p><p>n - g1g - 121n - 12 # R</p><p>pV n =</p><p>¢S = 0.85 J>K</p><p>b = 1.5 n = 1.2).g = 5>3, a = 2,</p><p>= naR ln a -</p><p>R</p><p>g - 1</p><p>lnbb = n R aln a -</p><p>ln b</p><p>g - 1</p><p>b = n(Cp ln a - CV ln ab)</p><p>¢S = n± 3</p><p>T1>ab</p><p>T1</p><p>CV</p><p>dT</p><p>T</p><p>+ 3</p><p>T1>b</p><p>T1>abC</p><p>dT</p><p>T</p><p>≤</p><p>282 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>p V1, ,1 1T</p><p>p0 ,V0 ,</p><p>T0</p><p>=</p><p>p1 ,</p><p>T1aV1,</p><p>a a</p><p>p</p><p>,</p><p>and</p><p>and(using</p><p>2.138 Let be a reference point on the line and let be any other point.</p><p>Then,</p><p>The entropy difference</p><p>For an extremum of</p><p>or</p><p>or</p><p>This gives a maximum of because</p><p>Note that a maximum of is a maximum of</p><p>2.139 For the process,</p><p>and specific heat</p><p>On the other hand for an ideal gas.</p><p>Thus,</p><p>or ln</p><p>Using</p><p>2.140 For a Van der Waal gas ap +</p><p>a</p><p>V</p><p>2</p><p>b1V - b2 = RT</p><p>T = T0 +</p><p>R</p><p>a</p><p>ln</p><p>V</p><p>V0</p><p>T = T0 when V = V0</p><p>V + constant = T</p><p>R</p><p>a</p><p>dV</p><p>V</p><p>= dT or</p><p>R</p><p>a</p><p>pdV =</p><p>RT</p><p>V</p><p>dV = aTdT</p><p>dQ = CdT = CV</p><p>dT + pdV</p><p>C = T</p><p>dS</p><p>dT</p><p>= aT + CV</p><p>S = aT + CV ln T</p><p>S (p , V ).¢S</p><p>02¢S</p><p>0 V 2</p><p>6 0¢S</p><p>g1p0-aV2 - aV = 0 or V = Vm =</p><p>gp0</p><p>a1g + 12</p><p>Cp1p0 - aV2-aVCV = 0</p><p>0¢S</p><p>0V</p><p>=</p><p>-aCV</p><p>p0 - aV</p><p>+</p><p>Cp</p><p>V1</p><p>= 0</p><p>¢S</p><p>= CV ln</p><p>p0 - aV</p><p>p1</p><p>+ Cp ln</p><p>V</p><p>V1</p><p>= CV ln</p><p>p</p><p>p1</p><p>+ Cp ln</p><p>V</p><p>V1</p><p>¢S = S1p, V) - S 1p1, V12</p><p>p = p0 - aV</p><p>1p, V2p1, V1</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 283</p><p>p V1,</p><p>p V,</p><p>p V1, 1</p><p>we get</p><p>The entropy change along an isotherm can be calculated from</p><p>It follows from solution of Problem 2.129 that</p><p>(assuming a, b to be known constants).</p><p>Thus,</p><p>2.141 We use,</p><p>(assuming to be known constants).</p><p>2.142 We can take .</p><p>Then,</p><p>2.143</p><p>(on substituting values)</p><p>2.144 Here,</p><p>Then,</p><p>Clearly, C 6 0 if n 6 0</p><p>C = T</p><p>1</p><p>n</p><p>T</p><p>1</p><p>n -1</p><p>a1>n =</p><p>S</p><p>n</p><p>T = aS</p><p>n or S = aT</p><p>a</p><p>b 1</p><p>n</p><p>= 2.0 kJ</p><p>¢S = 3</p><p>T2</p><p>T1</p><p>CdT</p><p>T</p><p>= 3</p><p>T2</p><p>T1</p><p>m(a + bT)</p><p>T</p><p>dT = mb 1T2 - T12 + ma ln</p><p>T2</p><p>T1</p><p>S = 3</p><p>T</p><p>0</p><p>C</p><p>dT</p><p>T</p><p>= 3</p><p>T</p><p>0</p><p>a T</p><p>2dT =</p><p>1</p><p>3</p><p>aT</p><p>3</p><p>S: 0 as T: 0</p><p>CV, a, b</p><p>= 3</p><p>T2</p><p>T1</p><p>CVdT</p><p>T</p><p>+ 3</p><p>V2</p><p>V1</p><p>R</p><p>V - b</p><p>dV = CV ln</p><p>T2</p><p>T1</p><p>+ R ln</p><p>V2 - b</p><p>V1 - b</p><p>¢S = 3</p><p>V2,T2</p><p>V1,T1</p><p>dS (V,T ) = 3</p><p>T2</p><p>T1</p><p>a 0S</p><p>0T</p><p>b</p><p>V1</p><p>dT + 3</p><p>V2</p><p>V1</p><p>a 0S</p><p>0V</p><p>b</p><p>T = T2</p><p>dV</p><p>¢S = R ln</p><p>V2 - b</p><p>V1 - b</p><p>a 0S</p><p>0V</p><p>b</p><p>T</p><p>= a 0p</p><p>0T</p><p>b</p><p>V</p><p>=</p><p>R</p><p>V - b</p><p>¢S = 3</p><p>V2</p><p>V1</p><p>a 0S</p><p>0V</p><p>b</p><p>T</p><p>dV</p><p>284 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>2.145 We know,</p><p>(assuming C to be a known constant).</p><p>Then,</p><p>2.146 (a)</p><p>(b)</p><p>(c)</p><p>Since for an ideal gas CV is constant and (as U does not depend</p><p>on V ).</p><p>2.147 (a) We have from the definition</p><p>Thus, using</p><p>(b) Here,</p><p>h = 1 -</p><p>2T1</p><p>T1 + T0</p><p>=</p><p>T0 - T1</p><p>T0 + T1</p><p>=</p><p>n - 1</p><p>n + 1</p><p>Q¿2 = T1 1S1 - S02 Q1 =</p><p>1</p><p>2</p><p>1S1 - S021T1 + T02</p><p>h = 1 -</p><p>T0 + T1</p><p>2T0</p><p>= 1 -</p><p>1 + 1>n</p><p>2</p><p>=</p><p>n - 1</p><p>2n</p><p>T1 =</p><p>T0</p><p>n</p><p>Q ¿2 =</p><p>1</p><p>2</p><p>1T0 + T1 21S1 - S02</p><p>Q1 = T0 1S1 - S02</p><p>Q = 3TdS = area under the curve</p><p>¢U = CV (T2 - T1)</p><p>= a ln</p><p>T1</p><p>T2</p><p>+ CV (T1 - T2)</p><p>W = ¢Q - ¢U</p><p>Q = 3</p><p>T2</p><p>T1</p><p>CdT = a ln</p><p>T1</p><p>T2</p><p>C = T</p><p>dS</p><p>dT</p><p>= -</p><p>a</p><p>T</p><p>T = T0 exp a S - S0</p><p>C</p><p>b</p><p>S - S0 = 3</p><p>T</p><p>T0</p><p>CdT</p><p>T</p><p>= C ln</p><p>T</p><p>T0</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 285</p><p>S0</p><p>T0</p><p>C<0</p><p>C>0</p><p>S</p><p>T</p><p>S</p><p>T S0,T0 S1,T0</p><p>S1,T1</p><p>Q1</p><p>Q</p><p>(a)</p><p>2'</p><p>S</p><p>T S0 ,T0</p><p>S1,T1</p><p>Q1</p><p>S0 ,T1 Q2'</p><p>(b)</p><p>We get</p><p>2.148 In this case, known as free expansion, no work is done and no heat is exchanged. So</p><p>internal energy must remain unchanged, i.e., For an ideal gas this implies co-</p><p>nstant temperature, i.e., The process is irreversible but the entropy change</p><p>can be calculated by considering a reversible isothermal process. Then, as before</p><p>2.149 The process consists of two parts. The first part is a free expansion in which . The</p><p>second part is an adiabatic compression in which work done results in change of</p><p>internal energy. Obviously,</p><p>Now in the first part</p><p>(because there is no change of temperature).</p><p>In the second part,</p><p>p</p><p>Thus,</p><p>The entropy change</p><p>and as the process is reversible adiabatic. Thus,</p><p>2.150 In all adiabatic processes, by virtue of first law of thermodynamics</p><p>Thus, Uƒ = Ui - A</p><p>Q = Uƒ - Ui + A = 0</p><p>¢S = R ln 2.¢SII = 0¢SI = R ln 2</p><p>¢S = ¢SI + ¢SII.</p><p>¢U = UF - Ui =</p><p>RT0</p><p>g - 1</p><p>12 g-1 - 12</p><p>=</p><p>RT0</p><p>g - 1</p><p>11 - 2 g-12</p><p>= 2 g-1p0V</p><p>g</p><p>0 V</p><p>-g+1</p><p>0</p><p>2 g-1 - 1</p><p>g - 1</p><p>3</p><p>V0</p><p>2V0</p><p>pdV = 3</p><p>V0</p><p>2V0</p><p>2 g-1p0V</p><p>g</p><p>0</p><p>V g</p><p>dV = c2 g-1p0V</p><p>g</p><p>0</p><p>-g + 1</p><p>V</p><p>1-g dV0</p><p>2V0</p><p>V g =</p><p>1</p><p>2</p><p>p0 12V02 g = 2 g-1p0V</p><p>g</p><p>0</p><p>pƒ =</p><p>1</p><p>2</p><p>p0 Vƒ = 2V0</p><p>0 = UF - Uƒ + 3</p><p>V0</p><p>Vƒ</p><p>pdV Vƒ = 2V0</p><p>Uƒ =Ui</p><p>¢S = 3</p><p>dQ</p><p>T</p><p>= 3</p><p>V2</p><p>V1</p><p>pdV</p><p>T</p><p>= nR ln n = 20.1 J/K</p><p>Tƒ = Ti .</p><p>Uƒ= Ui .</p><p>286 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>and</p><p>an</p><p>For a slow process,</p><p>(where for a quasistatic adiabatic process</p><p>On the other hand for a fast process, the external work done is In fact</p><p>for free expansion.</p><p>Thus (slow) (fast).</p><p>Since U depends on temperature only, if consequently</p><p>ideal gas equation</p><p>on.</p><p>2.151 Let</p><p>Since the temperature is the same, the required entropy change can be calculated by</p><p>considering isothermal expansion of the gas in either parts into the whole vessel.</p><p>Thus,</p><p>2.152 Let specific heat of copper and specific heat of water..</p><p>Then,</p><p>can be found from,</p><p>Using</p><p>2.153 For an ideal gas the internal energy depends on temperature only. We can consider</p><p>the process in question to be one of simultaneous free expansion, then the total en-</p><p>ergy U = U1 + U2.</p><p>T0 L 300 K and ¢S = 28.4 - 24.5 � 3.9 J>K c1 = 0.39 J>g K, c2 = 4.18 J>g K,</p><p>c2m2 1T0 - 2802 = m1c1 1370 - T02 or T0 =</p><p>280m2c2 + 370m1c1</p><p>c2m2 + m1c1</p><p>T0</p><p>= m2c2 ln</p><p>T0</p><p>280</p><p>- m1c1 ln</p><p>370</p><p>T0</p><p>¢S = 3</p><p>T0</p><p>7+273</p><p>c2m2dT</p><p>T</p><p>- 3</p><p>97 + 273</p><p>T0</p><p>m1c1dT</p><p>T</p><p>c2c1</p><p>= 5.1 J>K= 1R ln (1 + n) + 2R ln a1 + n</p><p>n</p><p>b</p><p>= 1R ln</p><p>V1 + V2</p><p>V1</p><p>+ 2R ln</p><p>V1 + V2</p><p>V2</p><p>¢S = ¢SI + ¢SII</p><p>V1 = V0, V2 = h</p><p>v0.</p><p>pV = RT</p><p>p–ƒ 7 p¿ƒT ¿ƒ 6 T –ƒ ,</p><p>6U –ƒU ¿ƒA– = 0</p><p>A– 6 A¿.</p><p>pV</p><p>g = constant)A¿ = 3</p><p>V</p><p>V0</p><p>pdV</p><p>2.4 THE SECOND LAW OF THERMODYNAMICS. ENTROPY 287</p><p>from the</p><p>Thus the pressure will be higher after the fast expansi</p><p>be the be the</p><p>.</p><p>ν ν</p><p>ν ν</p><p>Since and is the final</p><p>temperature.</p><p>The entropy change is obtained by considering isochoric process because in effect,</p><p>the gas remains confined to its vessel.</p><p>So,</p><p>Since,</p><p>2.154 (a) Each atom has probability to be in either compartment. Thus</p><p>(b) Typical atomic velocity at room temperature is cm/s so it takes an atom</p><p>s to cross the vessel. This is the relevant time scale for our problem. Let</p><p>s, then in time T there will be crossing or arrangements of the</p><p>atoms. This will be large enough to produce the given arrangement if</p><p>2.155 The statistical weight is</p><p>The probability distribution is</p><p>2.156 The probabilities that the half A contains n molecules is</p><p>2.157 The probability of one molecule being confined to the marked volume is</p><p>We can choose this molecule in many ways. The probability that n molecules get</p><p>confined to the marked volume is clearly</p><p>NCn</p><p>P</p><p>n 11 - P2N - n =</p><p>N !</p><p>n !1N - n2 ! P</p><p>n 11 - P2N - n</p><p>1NC1</p><p>P =</p><p>V</p><p>V0</p><p>NCn</p><p>* 2-N =</p><p>N!</p><p>n ! (N - n) !</p><p>2-N;</p><p>1</p><p>32</p><p>,</p><p>5</p><p>32</p><p>,</p><p>10</p><p>32</p><p>,</p><p>5</p><p>32</p><p>,</p><p>1</p><p>32</p><p>, respectively.</p><p>NCN>22-N = 252 * 2-10 = 24.6%</p><p>NCN>2 =</p><p>N !</p><p>N>2 !</p><p>N</p><p>2</p><p>!</p><p>=</p><p>10 * 9 * 8 * 7 * 6</p><p>8 * 4 * 3 * 2</p><p>= 252</p><p>t</p><p>t</p><p>2-N L 1 or N L</p><p>ln t>t</p><p>ln 2</p><p>L 75</p><p>t>TT = 10-5</p><p>10-5</p><p>' 105</p><p>P = 2-N.1>2</p><p>1T1 + T222 = 1T1 - T222 + 4T1T2,</p><p>¢S 7 0</p><p>¢S = 3</p><p>1T1 +T22>2</p><p>T1</p><p>CV</p><p>dT</p><p>T</p><p>- 3</p><p>T2</p><p>1T1 +T22>2CV</p><p>dT</p><p>T</p><p>= CV ln</p><p>1T1 + T222</p><p>4T1T2</p><p>1T1 + T22>2U1 = CVT1, U2 = CVT2 U = 2CV (T1 + T2)>2</p><p>288 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>so,</p><p>We get</p><p>2.158 In a sphere of diameter d, the number of molecules are</p><p>where Loschmidt’s number number of molecules per unit volume (1 cc) under NTP.</p><p>The relative fluctuation in this number is</p><p>or</p><p>The average number of molecules in this sphere is</p><p>2.159 For a monoatomic gas per mole.</p><p>The entropy change in the process is</p><p>Now from the Boltzmann equation, .</p><p>So,</p><p>Thus the statistical weight increases by this factor.</p><p>2.5 Liquids. Capillary Effects</p><p>2.160 (a)</p><p>(b) The soap bubble has two surfaces.</p><p>So,</p><p>=</p><p>8 * 45</p><p>3 * 10-3</p><p>= 1.2 * 10-3 atm</p><p>¢p = 2a a 1</p><p>d>2 +</p><p>1</p><p>d>2 b =</p><p>8a</p><p>d</p><p>=</p><p>4 * 490 * 10-3</p><p>1.5 * 10-6</p><p>N</p><p>m2</p><p>= 1.307 * 106</p><p>N</p><p>m2</p><p>= 13 atm</p><p>¢p = a a 1</p><p>d>2 +</p><p>1</p><p>d>2 b =</p><p>4a</p><p>d</p><p>Æ</p><p>Æ0</p><p>= e 1S - S02>k = a1 +</p><p>¢T</p><p>T0</p><p>b 3NA</p><p>2</p><p>= a1 +</p><p>1</p><p>300</p><p>b 3 * 6</p><p>2 *1023</p><p>= 101.31*1021</p><p>S = k ln Æ</p><p>¢S = S - S0 = 3</p><p>T0 + ¢T</p><p>T0</p><p>CV</p><p>dT</p><p>T</p><p>=</p><p>3</p><p>2</p><p>R lna1 +</p><p>¢T</p><p>T0</p><p>b</p><p>CV = 3>2R</p><p>1</p><p>h2</p><p>= 106</p><p>1</p><p>h2</p><p>=</p><p>p</p><p>6</p><p>d 3n0 or d 3 =</p><p>6</p><p>pn0h</p><p>2</p><p>Q d = a 6</p><p>ph2n0</p><p>b1>3</p><p>= 0.41 mm</p><p>0N</p><p>N</p><p>=</p><p>1N</p><p>N</p><p>=</p><p>1</p><p>1N</p><p>= h</p><p>=n0 =</p><p>N =</p><p>pd</p><p>3</p><p>6</p><p>n0</p><p>2.5 LIQUIDS. CAPILLARY EFFECTS 289</p><p>,</p><p>2.161 The pressure just inside the hole will be less than the outside pressure by . This</p><p>can support a height h of Hg where,</p><p>So,</p><p>2.162 By Boyle’s law</p><p>or</p><p>Thus,</p><p>2.163 The pressure has terms due to hydrostatic pressure and capillarity and they add as</p><p>2.164 By Boyle’s law</p><p>or</p><p>or</p><p>2.165 Clearly</p><p>¢h =</p><p>4a| cos u |1d2 - d12</p><p>d1d2rg</p><p>= 11 mm</p><p>¢hrg = 4a| cos u |a 1</p><p>d1</p><p>-</p><p>1</p><p>d2</p><p>b</p><p>h = cp01n3 - 12 +</p><p>4a</p><p>d</p><p>1n2 - 12 d gr = 4.98 m of water.</p><p>hgr - p0 n3 - 1 =</p><p>4a</p><p>d</p><p>1n2 - 12</p><p>ap0 + hgr +</p><p>4a</p><p>d</p><p>bp</p><p>6</p><p>d 3 = ap0 +</p><p>4a</p><p>nd</p><p>b p</p><p>6</p><p>n3d 3</p><p>= a1 +</p><p>5 * 9.8 * 103</p><p>105</p><p>+</p><p>4 * 0.73 * 10-3</p><p>4 * 10-6</p><p>* 10-5b atm = 2.22 atm</p><p>p = p0 + rgh +</p><p>4a</p><p>d</p><p>a =</p><p>1</p><p>8</p><p>p0d a1 -</p><p>h3</p><p>n</p><p>bn1h2 - 12</p><p>p0a1 -</p><p>h3</p><p>n</p><p>b =</p><p>8a</p><p>d</p><p>1h2 - 12</p><p>ap0 +</p><p>8a</p><p>d</p><p>b4p</p><p>3</p><p>ad</p><p>2</p><p>b3</p><p>= ap0</p><p>n</p><p>+</p><p>8a</p><p>hd</p><p>b</p><p>4p</p><p>3</p><p>ahd</p><p>2</p><p>b3</p><p>h =</p><p>4 * 490 * 10-3</p><p>13.6 * 103 * 9.8 * 70 * 10-6</p><p>=</p><p>200</p><p>13.6 * 70</p><p>L 0.21 m of Hg</p><p>r gh =</p><p>4a</p><p>d</p><p>or h =</p><p>4a</p><p>rgd</p><p>4a>d</p><p>290 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>n</p><p>( )</p><p>2.166 In a capillary with diameter mm, water will rise to a height</p><p>Since this is greater than the height ( mm) of the tube, a meniscus of radius</p><p>R will be formed at the top of the tube, where</p><p>2.167 Initially the pressure of air in the capillary is and it’s length is l. When submerged</p><p>under water, the pressure of air in the portion above water must be , since</p><p>the level of water inside the capillary is the same as the level outside. Thus by</p><p>Boyle’s law</p><p>or</p><p>or</p><p>2.168 We have by Boyle’s law</p><p>or</p><p>Hence,</p><p>2.169 Suppose the liquid rises to a height h. Then the total energy of the liquid in the</p><p>capillary is</p><p>Minimising E we get h =</p><p>4a</p><p>rg1d2 - d12 = 6 cm</p><p>E 1h2 =</p><p>p</p><p>4</p><p>Ad 2</p><p>2 - d 2</p><p>1 B h * rg *</p><p>h</p><p>2</p><p>- p 1d2 + d12 ah</p><p>a = argh +</p><p>p0h</p><p>l - h</p><p>b d</p><p>4 cos u</p><p>4a cos u</p><p>d</p><p>= rgh +</p><p>p0h</p><p>l - h</p><p>ap0 - rgh +</p><p>4a cosu</p><p>d</p><p>b 1l - h2 = p0 l</p><p>x =</p><p>l</p><p>1 + p0d>4a = 1.4 cm (on substituting values)</p><p>4a</p><p>d</p><p>1l - x2 = p0x</p><p>ap0 +</p><p>4a</p><p>d</p><p>b1l - x2 = p0l</p><p>p0 + 4 a>dp0</p><p>R =</p><p>2a</p><p>rgh</p><p>=</p><p>2 * 73 * 10-3</p><p>103 * 9.8 * 25 * 10-3</p><p>L 0.6 mm</p><p>= 25</p><p>=</p><p>4 * 73 * 10-3</p><p>103 * 9.8 * 0.5 * 10-3</p><p>= 59.6 mm</p><p>h =</p><p>2a</p><p>rgr</p><p>=</p><p>4a</p><p>rgd</p><p>d = 0.5</p><p>2.5 LIQUIDS. CAPILLARY EFFECTS 291</p><p>R</p><p>,</p><p>2.170 Let h be the height of the water level at a distance x from the edge. Then the total</p><p>energy of water in the wedge above the level outside is</p><p>This is minimum when</p><p>2.171 From the equation of continuity</p><p>We then apply Bernoulli’s theorem</p><p>The pressure p differs from the atmospheric pressure by capillary effects. At the</p><p>upper section, neglecting the curvature in the vertical plane.</p><p>Thus,</p><p>or</p><p>Finally, the liquid coming out per second is,</p><p>2.172 The radius of curvature of the drop is at the upper end of the drop and at the lower</p><p>end. Then the pressure inside the drop is at the top end and</p><p>at the bottom end. Hence,</p><p>p0 +</p><p>2a</p><p>R1</p><p>= p0 +</p><p>2a</p><p>R2</p><p>+ rgh or 2a1R2 - R12</p><p>R1R2</p><p>= rgh</p><p>p0 + 12a>R22p0 + 12a>R12 R2R1</p><p>V =</p><p>1</p><p>4</p><p>pd 2B</p><p>2gl - 14a>rd21n - 12</p><p>n4 - 1</p><p>= 0.9 cm3>s (on substituting values)</p><p>v = B</p><p>2gl - 14a>rd21n - 12</p><p>n4 - 1</p><p>p0 + 2a>d</p><p>r</p><p>+</p><p>1</p><p>2</p><p>v 2 + gl =</p><p>p0 + 2na>d</p><p>r</p><p>+</p><p>1</p><p>2</p><p>n4v 2</p><p>p = p0 + 2 a>d</p><p>p</p><p>r</p><p>+</p><p>1</p><p>2</p><p>v 2 + £ = constant</p><p>p</p><p>4</p><p>d 2 # v =</p><p>p</p><p>4</p><p>a d</p><p>n</p><p>b2 #V or V = n2v</p><p>h =</p><p>2a cosu</p><p>xrgdw</p><p>=Ldx</p><p>1</p><p>2</p><p>xrgdw c ah -</p><p>2a cosu</p><p>xrgdw</p><p>b2</p><p>-</p><p>4a2 cos2u</p><p>x2r2g2dw2</p><p>=Ldx</p><p>1</p><p>2</p><p>xrgdwah2 - 2</p><p>2a cos u</p><p>xrgdw</p><p>hb</p><p>E =Lxdw # dx # h # rg</p><p>h</p><p>2</p><p>- 2Ldx # h # a cos u</p><p>292 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>x</p><p>h</p><p>To a first approximation</p><p>So,</p><p>and if</p><p>2.173 We must first calculate the pressure difference inside the film from that outside. This is</p><p>given by</p><p>Here, and , where R is the radius of</p><p>the tablet and can be neglected. Thus the total force exert-</p><p>ed by mercury drop on the upper glass plate is typically,</p><p>We should put h/n for h because the tablet is compressed n times. Then since Hg</p><p>is nearly incompressible, constant, so . Thus, total force</p><p>Part of the force is needed to keep the Hg in the shape of a tablet rather than in the shape</p><p>of infinitely thin sheet. This part can be calculated being putting above. Thus</p><p>or</p><p>2.174 The pressure inside the film is less than that outside by an amount</p><p>(where and are the principal radii of curvature of the meniscus).</p><p>One of these is small, being given by while the other is large and will</p><p>be ignored.</p><p>Then,</p><p>(where</p><p>A � area of the water film between the plates)</p><p>Now,</p><p>So, [when (the angle of contact) � 0]</p><p>N (on substituting values) = 1.0</p><p>uF =</p><p>2ma</p><p>rh2</p><p>A =</p><p>m</p><p>rh</p><p>F L</p><p>2A cos u</p><p>h</p><p>a</p><p>h = 2r1 cos u</p><p>r2r1</p><p>p = a a 1</p><p>r1</p><p>+</p><p>1</p><p>r2</p><p>b</p><p>m =</p><p>2pR 2a | cos u |</p><p>hg</p><p>1n2 - 12 = 0.7 kg</p><p>mg +</p><p>2pR 2a | cos u |</p><p>h</p><p>=</p><p>2pR 2a | cos u |</p><p>h</p><p>n2</p><p>n = 1</p><p>=</p><p>2pR 2a | cos u |</p><p>h</p><p>n2</p><p>R: R1npR 2h =</p><p>2pR 2a | cos u |</p><p>h</p><p>r2 L R2r1| cos u | = h</p><p>p = aa</p><p>1</p><p>r1</p><p>+</p><p>1</p><p>r2</p><p>b</p><p>h = 2.3 mm, a = 73 mN>mR2 - R1 L 1>8 rgh3>a L 0.20 mm</p><p>R1 L R2 L h>2.</p><p>2.5 LIQUIDS. CAPILLARY EFFECTS 293</p><p>h</p><p>2.175 This is analogous to the previous problem except that, .</p><p>So,</p><p>2.176 The energy of the liquid between the plates is</p><p>This energy is minimum when, and the minimum potential energy is then</p><p>The force of attraction between the plates can be obtained from this as</p><p>(minus sign means the force is attractive)</p><p>Thus,</p><p>2.177 Suppose the radius of bubble is x at some instant. Then the pressure inside is</p><p>. The flow through the capillary is given by Poiseuille’s equation,</p><p>Integrating we get,</p><p>(where we have used the fact that and ).</p><p>The life time of the bubble corresponding to is</p><p>2.178 If the liquid rises to a height h, the energy of the liquid column becomes</p><p>E = rgpr2h # h</p><p>2</p><p>- 2prha =</p><p>1</p><p>2</p><p>rgp arh - 2</p><p>a</p><p>rg</p><p>b2</p><p>-</p><p>2pa2</p><p>rg</p><p>t =</p><p>2hlR 4</p><p>ar 4</p><p>x = 0</p><p>x = R t = 0</p><p>pr 4a</p><p>2hl</p><p>t = p1R 4 - x 42</p><p>Q =</p><p>pr 4</p><p>8hl</p><p>4a</p><p>x</p><p>= -4px</p><p>2</p><p>dx</p><p>dt</p><p>p0 + 4a>x</p><p>F =</p><p>alh</p><p>d</p><p>= 13 N</p><p>F =</p><p>-0Emin</p><p>0d</p><p>= -</p><p>2a2l</p><p>rgd 2</p><p>Emin = -</p><p>2a2l</p><p>rgd</p><p>h = 2a>rgd</p><p>=</p><p>1</p><p>2</p><p>rgld ah -</p><p>2a</p><p>rgd</p><p>b2</p><p>-</p><p>2a2l</p><p>rgd</p><p>E = ldh rg</p><p>h</p><p>2</p><p>- 2 alh =</p><p>1</p><p>2</p><p>rgldh2 - 2alh</p><p>F =</p><p>2pR 2a</p><p>h</p><p>= 0.6 kN</p><p>A = pR 2</p><p>294 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>h</p><p>This is minimum when and that is relevant height to which water must</p><p>rise.</p><p>At this point,</p><p>Since in the absence of surface tension, heat liberated must be</p><p>2.179 (a) The free energy per unit area being ,</p><p>(on substituting values)</p><p>(b) because the soap bubble has two surfaces.</p><p>Substitution gives</p><p>2.180 When two mercury drops each of diameter d merge, the resulting drop has diameter</p><p>where,</p><p>The increase in free energy is</p><p>2.181 Work must be done to stretch the soap film and compress the air inside. The former</p><p>is simply , there being two sides of the film. To get the latter</p><p>we note that the compression is isothermal and work done is</p><p>where,</p><p>or</p><p>and minus sign is needed because we are calculating work done on the system. Thus</p><p>since pV remains constant, the work done is</p><p>So, A¿ = 8p R</p><p>2 a + p V ln</p><p>p</p><p>p0</p><p>pV ln</p><p>V0</p><p>V</p><p>= p V ln</p><p>p</p><p>p0</p><p>V0 =</p><p>pV</p><p>p0</p><p>p = p0 +</p><p>4a</p><p>R</p><p>V0</p><p>p0 = ap0 +</p><p>4a</p><p>R</p><p>b # V V =</p><p>4p</p><p>3</p><p>R</p><p>3</p><p>- 3</p><p>Vf =V</p><p>Vi =V0</p><p>pdV</p><p>2a * 4pR 2 = 8pR 2 a</p><p>¢F = p 22>3 d 2 a - 2pd 2 a = 2pd 2 a 12-1>3 - 12 = –1.43 mJ</p><p>p</p><p>6</p><p>d</p><p>1</p><p>3 =</p><p>p</p><p>6</p><p>d 3 * 2 or d1 = 21>3d</p><p>d1</p><p>F = 10 mJ.</p><p>F = 2pad 2</p><p>F = pad 2 = 3 mJ</p><p>a</p><p>Q =</p><p>2pa2</p><p>rg</p><p>E = 0</p><p>Emin = -</p><p>2pa2</p><p>rg</p><p>rh = 2a>rg</p><p>2.5 LIQUIDS. CAPILLARY EFFECTS 295</p><p>,</p><p>and</p><p>and</p><p>2.182 When heat is given to a soap bubble, the temperature of the air inside rises and the</p><p>bubble expands but unless the bubble bursts, the amount of air inside does not</p><p>change. Further we shall neglect the variation of the surface tension with tempera-</p><p>ture. Then from the gas equations</p><p>Differentiating we get,</p><p>or</p><p>Now from the first law of thermodynamics</p><p>or</p><p>using</p><p>2.183 Consider an infinitesimal Carnot cycle with isotherms at and T. Let A be the</p><p>work done during the cycle. Then</p><p>(where is the change in the area of film; we are consider-</p><p>ing only one surface here).</p><p>Then by Carnot theorem</p><p>or</p><p>-</p><p>da</p><p>dT</p><p>dT ds</p><p>q ds</p><p>=</p><p>dT</p><p>T</p><p>or q = -T</p><p>da</p><p>dT</p><p>h =</p><p>A</p><p>Q1</p><p>=</p><p>dT</p><p>T</p><p>ds</p><p>A = [a1T - dT 2 - a1T 2] ds = -</p><p>da</p><p>dT</p><p>dT ds</p><p>T - dT</p><p>C = Cp +</p><p>1>2 R</p><p>1 + 3p0r>8a</p><p>C = CV + R, we get</p><p>C = CV + R</p><p>p0 + 4a>r</p><p>p08a>3r</p><p>d Q = nCdT = nCV</p><p>dT +</p><p>nRdT</p><p>p0 + 8a>3r</p><p>ap0 +</p><p>4a</p><p>r</p><p>b</p><p>dV = 4pr 2dr =</p><p>nRdT</p><p>p0 +</p><p>8a</p><p>3r</p><p>ap0 +</p><p>8a</p><p>3r</p><p>b 4pr 2dr = nRdT</p><p>ap0 +</p><p>4a</p><p>r</p><p>b 4p</p><p>3</p><p>r</p><p>3 = nRT n = constant)</p><p>296 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>T</p><p>T – dT</p><p>q</p><p>σ</p><p>p</p><p>(</p><p>2.184 As before we can calculate the heat required. It is taking into account two sides of the</p><p>soap film. So,</p><p>Thus,</p><p>Now,</p><p>So,</p><p>2.6 Phase Transformations</p><p>2.185 The condensation takes place at constant pressure and temperature and the work</p><p>done is where is the volume of the condensed vapour phase. It is</p><p>(where g is the molecular weight of water).</p><p>2.186 The specific volume of water (the liquid) will be written as . Since ,</p><p>most of the weight is due to water. Thus if is mass of the liquid and that of</p><p>the vapour then</p><p>So, g</p><p>Its volume is</p><p>2.187 The volume of the condensed vapour was originally at temperature</p><p>K. Its mass will be given by</p><p>g</p><p>(where standard atmospheric pressure).p0 =</p><p>p0 1V0 - V 2 =</p><p>m</p><p>M</p><p>RT or m =</p><p>Mp01V0 - V 2</p><p>RT</p><p>= 2</p><p>T = 373V0 - V</p><p>V ¿vap = 1.0 l</p><p>mvap =</p><p>V - mV l¿</p><p>V ¿vap - V l¿</p><p>= 20</p><p>V = mlV l¿ + mvapV ¿vap or V - mV l¿ = mvap 1V ¿vap - V l¿2 m = ml + mvap</p><p>mvapml</p><p>V ¿vap 7 7 V l¿V l¿</p><p>M = 18</p><p>p ¢V =</p><p>¢m</p><p>M</p><p>RT = 1.2 J</p><p>¢VA = p ¢V</p><p>¢U = ¢F + T¢S = 2 aa - T</p><p>da</p><p>dT</p><p>b ds ¢F = 2ads</p><p>¢S =</p><p>dq</p><p>T</p><p>= -2</p><p>da</p><p>dT</p><p>ds</p><p>dq = -T</p><p>da</p><p>dT</p><p>ds * 2</p><p>2.6 PHASE TRANSFORMATIONS 297</p><p>,</p><p>2.188 We let specific volume of liquid. specific volume of vapour.</p><p>Let original volume of the vapour.</p><p>Then,</p><p>So,</p><p>In the case when the final volume of the substance corresponds to the midpoint of</p><p>a horizontal portion of the isothermal line in the pV diagram, the final volume must</p><p>be per unit mass of the substance. Of this, the volume of the liquid is</p><p>per unit total mass of the substance.</p><p>Thus,</p><p>2.189 From the first law of thermodynamics</p><p>(where q is the specific latent heat of vaporization).</p><p>So, increment in entropy is given by</p><p>(on substituting values)</p><p>Now,</p><p>Thus,</p><p>For water, this gives</p><p>2.190 Some of the heat used in heating water to the boiling temperature</p><p>The remaining heat (here specific heat of</p><p>water, K) is used to create vapour. If the piston rises to a height h then</p><p>the volume of vapour will be Sh (neglecting water). Its mass will be</p><p>Heat of vapourization will be</p><p>. To this must be added the work done in creating the saturated vapour .</p><p>Thus, Q - m c¢T L p0</p><p>S h a1 +</p><p>qM</p><p>RT</p><p>b or h L</p><p>Q - mc¢T</p><p>p0S a1 +</p><p>qM</p><p>RT</p><p>b = 20 cm</p><p>= p0Sh</p><p>p0Sh Mq</p><p>RT</p><p>p0Sh</p><p>RT</p><p>* M</p><p>L</p><p>¢T = 100</p><p>c == Q - mc¢TT = 100°C = 373 K.</p><p>L 2.08 * 106 J.</p><p>¢U = m aq -</p><p>RT</p><p>M</p><p>b</p><p>A = p 1V ¿vap - V l¿2 m = m</p><p>RT</p><p>M</p><p>¢S =</p><p>Q</p><p>T</p><p>=</p><p>mq</p><p>T</p><p>= 6.0 kJ/K</p><p>¢U + A = Q = mq</p><p>h =</p><p>1</p><p>1 + N</p><p>V l¿>211 + N 2 V l¿ >2</p><p>1N - 12 ml V l¿ = V a1 -</p><p>1</p><p>n</p><p>b =</p><p>V</p><p>n</p><p>1n - 12 or h =</p><p>mlV l¿</p><p>V>n =</p><p>n - 1</p><p>N - 1</p><p>M</p><p>pV</p><p>RT</p><p>= ml + mvap =</p><p>V</p><p>NV l¿</p><p>or V</p><p>n</p><p>= 1ml + Nmvap2 V l¿</p><p>V =</p><p>V ¿vap = NV l¿ = V l¿ =</p><p>298 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>2.191 A quantity of saturated vapour must condense to heat the water to</p><p>bolling point (Here specific heat of water, water</p><p>temperature.)</p><p>The work done in lowering the piston will then be</p><p>Since work done per unit mass of the condensed vapour is .</p><p>2.192 Given</p><p>or</p><p>For water, dynes/cm, g/cc and K. Using these va-</p><p>lues and with , we get</p><p>2.193 At equilibrium the number of “liquid” molecules evaporating must equal the number</p><p>of “vapour” molecules condensing. By kinetic theory, this number is</p><p>The required mass is</p><p>where is standard atmospheric pressure , K and molecular weight of</p><p>water.</p><p>2.194 Here we must assume that is also the rate at which the tungsten filament loses</p><p>mass when in an atmosphere of its own vapour at this temperature and that (of the</p><p>previous problem) Then</p><p>where of the saturated vapor.p = pressure</p><p>p = mA</p><p>2pRT</p><p>M</p><p>= 0.9 nPa</p><p>L 1.</p><p>h</p><p>m</p><p>M =T = 373p0</p><p>= h p0A</p><p>M</p><p>2pRT</p><p>= 0.35 g/cm2</p><p>m = m * h * n * A</p><p>kT</p><p>2pm</p><p>= h nkT A</p><p>m</p><p>2pkt</p><p>h *</p><p>1</p><p>4</p><p>n 6 v 7 = h *</p><p>1</p><p>4</p><p>n * A</p><p>8</p><p>p</p><p>kT</p><p>m</p><p>d L 0.2 m m.h = 0.01</p><p>T = 300M = 18 g, rl = 1a = 73</p><p>d =</p><p>4aM</p><p>rl RTh</p><p>= h Pvap = h</p><p>1m>M2 RT</p><p>Vvap</p><p>=</p><p>hRT</p><p>M</p><p>rvap</p><p>¢P =</p><p>rvap</p><p>rl</p><p>2a</p><p>r</p><p>=</p><p>rvap</p><p>rl</p><p>*</p><p>4a</p><p>d</p><p>= h pvap</p><p>pV = RT>M</p><p>mc1T - T02</p><p>q</p><p>*</p><p>RT</p><p>M</p><p>= 25 J</p><p>T0 = 295 K = initialc =T = 373 K.</p><p>mc1T - T02>q</p><p>2.6 PHASE TRANSFORMATIONS 299</p><p>2.195 From the Van der Waal’s equation</p><p>(where</p><p>volume of one gram mole of the substances).</p><p>For water and</p><p>If molecular attraction vanishes, the equation will be</p><p>(for the same specific volume)</p><p>Thus,</p><p>2.196 The internal pressure being , the work done in condensation is</p><p>This by assumption is being the molecular weight, q being specific talent heat of</p><p>vaporization and being the molar volumes of the liquid and gas, respectively</p><p>Thus,</p><p>(where is the density of the liquid).</p><p>For water atm</p><p>2.197 The Van der Waal’s equation can be written as (for one mole)</p><p>At the critical point and vanish. Thus,</p><p>Solving these simultaneously, we get on division</p><p>V - b =</p><p>2</p><p>3</p><p>V, V = 3 b L VMCr</p><p>0 =</p><p>2RT1V - b2 3</p><p>-</p><p>6a</p><p>V 4</p><p>or RT1V - b2 3</p><p>=</p><p>3a</p><p>V 4</p><p>0 =</p><p>RT1V - b22 -</p><p>2a</p><p>V 3</p><p>or RT1V - b22 =</p><p>2a</p><p>V 3</p><p>a 02p</p><p>0V 2</p><p>b</p><p>T</p><p>a 0 p</p><p>0V</p><p>b</p><p>T</p><p>p 1V 2 =</p><p>RT</p><p>V - b</p><p>-</p><p>a</p><p>V 2</p><p>pi = 2 * 104</p><p>r</p><p>pi =</p><p>a</p><p>V</p><p>l</p><p>2</p><p>=</p><p>Mq</p><p>Vl</p><p>= rq</p><p>Vl , Vg</p><p>Mq, M</p><p>3</p><p>Vg</p><p>Vl</p><p>a</p><p>V 2</p><p>dV =</p><p>a</p><p>Vl</p><p>-</p><p>a</p><p>Vg</p><p>L</p><p>a</p><p>Vl</p><p>a>V 2</p><p>¢p =</p><p>a</p><p>V 2</p><p>m =</p><p>5.47</p><p>1.8 * 1.8</p><p>* 104 atm L 1.7 * 104 atm</p><p>p¿ =</p><p>RT</p><p>V - b</p><p>a = 5.47 atm l2/mol2V = 18 cm3 per mol = 1.8 * 10-2 l mol-1</p><p>V =</p><p>p =</p><p>RT</p><p>V - b</p><p>-</p><p>a</p><p>V 2</p><p>300 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>.</p><p>This is the critical molar volume. Putting this back, we get</p><p>Finally</p><p>From these we see that</p><p>2.198 We have,</p><p>Thus,</p><p>and</p><p>2.199 Specific volume is molar volume divided by molecular weight. Thus</p><p>2.200 We have,</p><p>or</p><p>or</p><p>where</p><p>or</p><p>When</p><p>t =</p><p>3</p><p>8</p><p>* 24 *</p><p>1</p><p>6</p><p>=</p><p>3</p><p>2</p><p>p = 12 and n = 1/2, we get</p><p>ap +</p><p>27b 2</p><p>V 2</p><p>M</p><p>b an -</p><p>1</p><p>3</p><p>b =</p><p>8</p><p>3</p><p>t or ap +</p><p>3</p><p>n2</p><p>b an -</p><p>1</p><p>3</p><p>b =</p><p>8</p><p>3</p><p>t</p><p>p =</p><p>p</p><p>pCr</p><p>, n =</p><p>VM</p><p>VMCr</p><p>, t =</p><p>T</p><p>TCr</p><p>ap +</p><p>a</p><p>pCrV</p><p>2</p><p>M</p><p>b * an -</p><p>b</p><p>VMCr</p><p>b =</p><p>8</p><p>3</p><p>t</p><p>p + a>V 2</p><p>M</p><p>pCr</p><p>*</p><p>VM - b</p><p>VMCr</p><p>=</p><p>8</p><p>3</p><p>T</p><p>TCr</p><p>ap +</p><p>a</p><p>V 2</p><p>M</p><p>b1VM - b2 = RT</p><p>V ¿Cr =</p><p>VMCr</p><p>M</p><p>=</p><p>3RTCr</p><p>8MpCr</p><p>=</p><p>3 * 0.082 * 562</p><p>8 * 78 * 47</p><p>l>g = 4.71 cm3>g</p><p>1RTCr22</p><p>pCr</p><p>=</p><p>64a</p><p>27</p><p>or a =</p><p>27</p><p>64</p><p>1RTCr22</p><p>pCr</p><p>= 3.59 atm l 2/mol</p><p>b = R</p><p>TCr</p><p>8pCr</p><p>=</p><p>0.082 * 304</p><p>73 * 8</p><p>= 0.043 l/mol</p><p>pCr</p><p>RTCr</p><p>=</p><p>a>27b2</p><p>8a>27b</p><p>=</p><p>1</p><p>8b</p><p>pCrVMCr</p><p>RTCr</p><p>=</p><p>a>9b</p><p>8a>27b</p><p>=</p><p>3</p><p>8</p><p>pcr =</p><p>RTCr</p><p>VMCr - b</p><p>-</p><p>a</p><p>V 2</p><p>MCr</p><p>=</p><p>4a</p><p>27b2</p><p>-</p><p>a</p><p>9b 2</p><p>=</p><p>a</p><p>27b 2</p><p>RTCr</p><p>4b2</p><p>=</p><p>2a</p><p>27b3</p><p>or TCr =</p><p>8a</p><p>27bR</p><p>2.6 PHASE TRANSFORMATIONS 301</p><p>a a</p><p>2.201 (a) The critical volume is the maximum volume in the liquid phase and the</p><p>minimum volume in the gaseous. Thus,</p><p>(b) The critical pressure is the maximum possible pressure in the vapour phase in</p><p>equilibrium with liquid phase. Thus,</p><p>2.202</p><p>2.203 The vessel is such that either vapour or liquid of mass m occupies it at critical point.</p><p>Then its volume will be</p><p>The corresponding volume in liquid phase at room temperature is</p><p>(where density of liquid ether at room temperature).</p><p>Thus,</p><p>(using the given data and g/l).</p><p>2.204 We apply the following relation, at to the</p><p>cycle 1234531</p><p>Here,</p><p>So, CpdV = 0</p><p>dS = CdU = 0</p><p>T CdS = CdU + CpdV</p><p>Tconstant</p><p>r = 720</p><p>h =</p><p>V</p><p>VCr</p><p>=</p><p>8MpCr</p><p>3RTCr r</p><p>L 0.254</p><p>r =</p><p>V =</p><p>m</p><p>r</p><p>VCr =</p><p>m</p><p>M</p><p>VMCr =</p><p>3</p><p>8</p><p>RTCr</p><p>pCr</p><p>m</p><p>M</p><p>rCr =</p><p>M</p><p>3b</p><p>=</p><p>44</p><p>3 * 43</p><p>g/cm3 = 0.34 g/cm3</p><p>TCr =</p><p>8</p><p>27</p><p>a</p><p>bR</p><p>=</p><p>8</p><p>27</p><p>*</p><p>3.62</p><p>0.043 * 0.082</p><p>L 304 K</p><p>pmax =</p><p>a</p><p>27b 2</p><p>=</p><p>5.47</p><p>27 * 0.03 * 0.03</p><p>= 225 atm</p><p>Vmax =</p><p>1000</p><p>18</p><p>* 3 * 0.030 = 5</p><p>VMCr</p><p>302 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>1</p><p>4</p><p>3</p><p>2</p><p>5</p><p>I</p><p>II</p><p>p</p><p>V</p><p>ll</p><p>We have,</p><p>This implies that the areas I and II are equal. This reasoning is inapplicable to the</p><p>cycle 1231, for example. This cycle is irreversible because it involves the irreversible</p><p>transition from a single phase to a two-phase state at the point 3.</p><p>2.205 When a portion of super-cool water turns into ice some heat is liberated, which</p><p>should heat it upto ice point. Neglecting the variation of specific heat of water, the</p><p>fraction of water turning into ice is clearly</p><p>(where specific heat of water and latent heat of fusion of ice).</p><p>Clearly at</p><p>2.206 From the Clausius–Clapeyron (C - C ) equation</p><p>Here, is the specific latent heat absorbed in</p><p>2.207 Here .</p><p>So,</p><p>or</p><p>2.208 From C-C equation</p><p>Assuming the saturated vapour to be ideal gas, we can write</p><p>dp</p><p>dT</p><p>=</p><p>q12</p><p>T 1V ¿2 - V ¿12 =</p><p>q12</p><p>TV ¿2</p><p>V ¿2 L</p><p>q12</p><p>T</p><p>¢T</p><p>¢p</p><p>=</p><p>2250</p><p>373</p><p>*</p><p>0.9</p><p>3.2</p><p>* 10-3 m3/g = 1.7 m3/kg</p><p>¢T =</p><p>T 1V ¿2 - V ¿12</p><p>q12</p><p>¢p</p><p>1 = liquid, 2 = steam</p><p>1</p><p>atm * cm3</p><p>J</p><p>=</p><p>105N/m2 * 10-6m3</p><p>J</p><p>= 10-1 Nm/J, ¢T = 0.0075 K</p><p>¢T =</p><p>T 1V ¿2 - V ¿12</p><p>q12</p><p>¢p = -</p><p>273 * 0.091</p><p>333</p><p>* 1 atm cm3 K>J</p><p>1: 2 and 1 = solid, 2 = liquid.q12</p><p>dT</p><p>dp</p><p>=</p><p>T 1V ¿2 - V ¿1 2</p><p>q12</p><p>t = -80°C.f = 1</p><p>q =c =</p><p>f =</p><p>c | t |</p><p>q</p><p>= 0.25</p><p>2.6 PHASE TRANSFORMATIONS 303</p><p>Thus,</p><p>and</p><p>2.209 From C-C equation, neglecting the volume of the liquid</p><p>or</p><p>Now, (for an ideal gas)</p><p>So,</p><p>2.210 From C-C equation</p><p>Integrating</p><p>So,</p><p>This is reasonable for and far below critical temperature.</p><p>2.211 As before from solution of Problem 2.206, the lowering of melting point is given by</p><p>The superheated ice will then melt in part. The fraction that will melt is</p><p>where C is the specific heat capacity of ice, T � 273 K and q is the specific heat of</p><p>melting.</p><p>h =</p><p>CT¢V ¿</p><p>q2</p><p>p = 0.03</p><p>¢T = -</p><p>C¢V ¿</p><p>q</p><p>p</p><p>|T - T0|<<T0,</p><p>p = p0 exp cMq</p><p>R</p><p>a 1</p><p>T0</p><p>-</p><p>1</p><p>T</p><p>b d</p><p>ln p = constant -</p><p>Mq</p><p>RT</p><p>dp</p><p>dT</p><p>L</p><p>q</p><p>TV ¿2</p><p>L</p><p>Mq</p><p>RT 2</p><p>p</p><p>= aMq</p><p>RT</p><p>- 1b dT</p><p>T</p><p>= a18 * 2250</p><p>8.3 * 373</p><p>- 1b *</p><p>1.5</p><p>375</p><p>L 4.85%</p><p>dm</p><p>m</p><p>=</p><p>dp</p><p>p</p><p>-</p><p>dT</p><p>T</p><p>1V is constant = specific volume)</p><p>pV =</p><p>m</p><p>M</p><p>RT or m =</p><p>MpV</p><p>RT</p><p>dp</p><p>p</p><p>=</p><p>Mq</p><p>RT</p><p>dT</p><p>T</p><p>dp</p><p>dT</p><p>L</p><p>q12</p><p>TV ¿2</p><p>L</p><p>Mq</p><p>RT 2</p><p>p 1 q = q122</p><p>p = p0 a1 +</p><p>Mq</p><p>RT 2</p><p>¢Tb L 1.04 atm</p><p>¢p =</p><p>Mq</p><p>RT 2</p><p>p0 ¢T</p><p>1</p><p>V ¿2</p><p>=</p><p>mp</p><p>RT</p><p>304 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>as</p><p>2.212 (a) The equations of the transition lines are</p><p>At the triple point they intersect. Thus,</p><p>The corresponding ptr is 5.14 atm.</p><p>In the formula log we compare b with the corresponding term in the</p><p>equation in Problem 2.210. Then</p><p>or</p><p>Finally J/g (on subtraction)</p><p>2.213</p><p>2.214</p><p>2.215 Specific heat of copper, J/gK. Suppose all the ice does not melt, then</p><p>heat rejected � 90 � 0.39 (90 � 0) � 3159 J</p><p>and heat gained by ice � 50 � 2.09 � 3 	 x � 333</p><p>This gives, x � 8.5 g</p><p>c = 0.39</p><p>=</p><p>333</p><p>273</p><p>+ 4.18 ln</p><p>373</p><p>283</p><p>L 8.56 J/K</p><p>¢S L</p><p>qm</p><p>T1</p><p>+ c ln</p><p>T2</p><p>T</p><p>+</p><p>qv</p><p>T2</p><p>= 103a4.18 ln</p><p>373</p><p>283</p><p>+</p><p>2250</p><p>373</p><p>b L 7.2 kJ/K</p><p>¢S = 3</p><p>T2</p><p>T1</p><p>mc</p><p>dT</p><p>T</p><p>+</p><p>mq</p><p>T2</p><p>L m ac ln</p><p>T2</p><p>T1</p><p>+</p><p>q</p><p>T2</p><p>b</p><p>q</p><p>solid-liquid = 213</p><p>q</p><p>liquid-gas =</p><p>2.303 * 1310 * 8.31</p><p>44</p><p>= 570 J/g</p><p>q solid-gas =</p><p>2.303 * 1800 * 8.31</p><p>44</p><p>= 783 J/g</p><p>ln p = a * 2.303 -</p><p>2.303 b</p><p>T</p><p>Q 2.303 =</p><p>Mq</p><p>R</p><p>p = a - b>T ,</p><p>2.27 =</p><p>490</p><p>Ttr</p><p>or Ttr =</p><p>490</p><p>2.27</p><p>= 216 K</p><p>= 6.78 -</p><p>1310</p><p>T</p><p>(liquid-gas)</p><p>log p = 9.05 -</p><p>1800</p><p>T</p><p>(solid-gas)</p><p>2.6 PHASE TRANSFORMATIONS 305</p><p>The hypothesis is correct and final temperature will be T � 273 K.</p><p>Hence change in entropy of copper piece</p><p>2.216 (a) Here Suppose the final temperature is Then,</p><p>heat lost by water</p><p>heat gained by ice (if all the ice melts)</p><p>In this case, for</p><p>So the final temperature will be 0�C and only some ice will melt.</p><p>Then</p><p>Finally</p><p>(b) If , then all the ice will melt as one can check and the final temper-</p><p>ature can be obtained like this</p><p>or</p><p>and</p><p>2.217 ¢S = -</p><p>mq2</p><p>T2</p><p>-mc ln</p><p>T2</p><p>T1</p><p>+</p><p>Mqice</p><p>T1</p><p>¢S =</p><p>m1q</p><p>T1</p><p>+ c am1 ln</p><p>T</p><p>T1</p><p>- m2ln</p><p>T2</p><p>T</p><p>b = 19 J/K</p><p>T =</p><p>m2T2 + m1T1 -</p><p>m1qm</p><p>c</p><p>m1 + m2</p><p>L 280 K</p><p>1m2T2 + m1T12 c - m1qm = 1m1 + m22 cT</p><p>m2c 1T2 - T2 = m1qm + m1c 1T - T12</p><p>m2ct2 > m1qm</p><p>= m2c cT2</p><p>T1</p><p>- 1 - ln</p><p>T2</p><p>T1</p><p>d = 8.8 J/K</p><p>= m2c</p><p>(T2 - T1)</p><p>T1</p><p>- m2c ln</p><p>T2</p><p>T1</p><p>¢S =</p><p>m¿1qm</p><p>T1</p><p>+ m2c ln</p><p>T1</p><p>T2</p><p>¢S = 75.3 *</p><p>333</p><p>273</p><p>+ 100 * 4.18 ln</p><p>273</p><p>333</p><p>m¿1 = 75.3 g = amount of ice that will melt</p><p>100 * 4.18 (60) = m¿1 * 333</p><p>m1 = m2m1qm = m2 * 4.18 160 - t2= m1qm + m1 c 1t - t12= m2 c 1t2 - t2 t °C.t2 = 60°C.</p><p>¢S = mc ln</p><p>T</p><p>T1</p><p>= mc ln</p><p>273</p><p>363</p><p>= -10 J/K</p><p>306 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>( )</p><p>or</p><p>where,</p><p>So,</p><p>2.218 When heat dQ is given to the vapour its temperature will change by dT, pressure by</p><p>dp and volume by dV, it being assumed that the vapour remains saturated.</p><p>Then by C-C equation,</p><p>On the other hand,</p><p>So,</p><p>Hence,</p><p>Finally</p><p>refer to unit mass here).</p><p>Thus,</p><p>For water,</p><p>So,</p><p>and</p><p>2.219 The</p><p>required entropy change can be calculated along a process in which the water is</p><p>heated from T1 to T2 and then allowed to evaporate. The entropy change for this</p><p>is</p><p>where specific latent heat of vaporization.q =</p><p>¢S = Cp ln</p><p>T2</p><p>T1</p><p>+</p><p>qM</p><p>T2</p><p>C = -4.13 J/g K = -74 J/mol K</p><p>Cp = 1.90 J/g K</p><p>Cp =</p><p>Rg</p><p>g - 1</p><p>1</p><p>M</p><p>with g = 1.32 and M = 18</p><p>C = Cp -</p><p>q</p><p>T</p><p>(Cp , CV</p><p>= CvdT + a T</p><p>M</p><p>-</p><p>q</p><p>T</p><p>bdT = CpdT -</p><p>q</p><p>T</p><p>dT</p><p>dQ = CdT = dU + pdV ¿</p><p>pdV ¿ = a R</p><p>M</p><p>-</p><p>q</p><p>T</p><p>b dT</p><p>pdV ¿ + V ¿dp =</p><p>RdT</p><p>M</p><p>pV ¿ =</p><p>RT</p><p>M</p><p>dp</p><p>dT</p><p>=</p><p>q</p><p>TV ¿</p><p>(V ¿vap 7 7V l¿) or dp =</p><p>q</p><p>TV ¿</p><p>dT</p><p>= 0.2245 + 0.2564 = 0.48 J/K</p><p>¢S = mq2 a 1</p><p>T1</p><p>-</p><p>1</p><p>T2</p><p>b + mc aT2</p><p>T1</p><p>- 1 - ln</p><p>T2</p><p>T1</p><p>bmq</p><p>ice = m 1q2 + c 1T2 - T122</p><p>2.6 PHASE TRANSFORMATIONS 307</p><p>. ( )</p><p>2.7 Transport Phenomena</p><p>2.220 (a) The fraction of gas molecules which traverses distances exceeding the mean free</p><p>path without collision is just the probability to traverse the distance with-</p><p>out collision.</p><p>Thus,</p><p>Hence the sought fraction</p><p>(b) This probability is</p><p>Hence the sought fraction</p><p>2.221 From the formula</p><p>2.222 (a) Let P (t) � probability of no collision in the interval (0, t). Then</p><p>or</p><p>where we have used</p><p>(b) The mean interval between collision is also the mean interval of no collision. Then,</p><p>2.223 (a)</p><p>=</p><p>1.38 * 10-23 * 273</p><p>22p (0.37 * 10-9)2 * 105</p><p>= 6.2 * 10-8 m</p><p>l =</p><p>1</p><p>22pd 2n</p><p>=</p><p>kT</p><p>22pd 2p</p><p>6 t7 =</p><p>3</p><p>q</p><p>0</p><p>te-atdt</p><p>3</p><p>q</p><p>0</p><p>e-atdt</p><p>=</p><p>1</p><p>a</p><p>≠(2)</p><p>≠(1)</p><p>=</p><p>1</p><p>a</p><p>P (0) = 1.</p><p>dP</p><p>dt</p><p>= -aP (t) or P (t) = e-at</p><p>P (t + dt) = P (t) (1 - adt)</p><p>1</p><p>h</p><p>= e</p><p>¢l/l or l =</p><p>¢l</p><p>ln h</p><p>h = 0.23.</p><p>P = e-1 - e-2 = 0.23</p><p>h = 0.37.</p><p>P = e-1 =</p><p>1</p><p>e</p><p>= 0.37</p><p>s = l</p><p>308 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>.</p><p>(b)</p><p>2.224 The mean distance between molecules is of the order</p><p>This is about 18.5 times smaller than the mean free path calculated in Problem</p><p>2.223(a) above.</p><p>2.225 We know that the Van der Waal’s constant b is four times the molecular volume. Thus,</p><p>Hence, (on substituting values)</p><p>2.226 The velocity of sound in is</p><p>So,</p><p>GHz (on substituting values)</p><p>2.227 (a)</p><p>(b) The corresponding molecular concentration n is obtained by dividing by kT and</p><p>is per cm and the corresponding mean dis-</p><p>tance is given by</p><p>l</p><p>n1>3 =</p><p>10-2</p><p>(0.184)1>3 * 105</p><p>= 1.8 * 10-7 m L 0.18 mm</p><p>1.84 * 1020 per m3 = 1.84 * 1014</p><p>L 0.7 Pa for O2</p><p>l 7 l if p 6</p><p>kT</p><p>12p d</p><p>2l</p><p>= 5.5</p><p>1</p><p>n</p><p>= A</p><p>gRT0</p><p>M =</p><p>RT0</p><p>22p d 2p0NA</p><p>A</p><p>gp</p><p>r</p><p>= A</p><p>gRT</p><p>M</p><p>N2</p><p>l = a kT0</p><p>22pP0</p><p>b a2pNA</p><p>3b</p><p>b2/3</p><p>= 84 nm</p><p>b = 4NA</p><p>p</p><p>6</p><p>d 3 or d = a 3d</p><p>2p NA</p><p>b1/3</p><p>a22.4 * 10-3</p><p>6.0 * 1023</p><p>b1/3</p><p>= a224</p><p>6</p><p>b1/3</p><p>* 10-9 m « 3.34 * 10-9 m</p><p>t = 1.36 * 104s = 3.8 h</p><p>l = 6.2 * 106 m</p><p>t =</p><p>l</p><p>6v7</p><p>=</p><p>6.2 * 10-8</p><p>454</p><p>s = 0.136 ns</p><p>2.7 TRANSPORT PHENOMENA 309</p><p>3</p><p>2.228 (a)</p><p>where</p><p>(b) Total number of collisions is</p><p>Note, the factor , which come because when two molecules collide we must</p><p>not count it twice.</p><p>2.229 (a)</p><p>d is a constant and n is a constant for an isochoric process, so is constant for</p><p>an isochoric process.</p><p>i.e.,</p><p>(b) (for an isobaric process)</p><p>(for an isobaric process)</p><p>2.230 (a) In an isochoric process is constant and</p><p>so v increases n times</p><p>(b) must decrease n times in an isothermal process and v must in-</p><p>crease n times because is constant in an isothermal process.</p><p>2.231 (a)</p><p>Thus,</p><p>But in an adiabatic process constant (as here).</p><p>or</p><p>Thus, v V</p><p>-6>5 T</p><p>1>2 r V</p><p>-1>5 g = 7>5 TV g-1 = constant, so TV 2>5 =</p><p>l r V and v r T 1>2</p><p>V</p><p>l r 1</p><p>n</p><p>Ú</p><p>1</p><p>N>V =</p><p>V</p><p>N</p><p>6v7</p><p>l =</p><p>kT</p><p>12pd 2p</p><p>v r 1T r 1pV r 1p r 1n</p><p>l</p><p>v =</p><p>6v7</p><p>l</p><p>r 1T</p><p>T</p><p>=</p><p>1</p><p>1T</p><p>l =</p><p>1</p><p>12pd 2</p><p>kT</p><p>P</p><p>r T</p><p>v r 1T</p><p>v =</p><p>6v7</p><p>l</p><p>= B</p><p>8RT</p><p>Mp</p><p>nl</p><p>l</p><p>l =</p><p>1</p><p>12pd 2n</p><p>1>2</p><p>1</p><p>2</p><p>nv = 1.0 * 1029 s-1 cm-3</p><p>n =</p><p>p0</p><p>kT0</p><p>and 6v7 = A</p><p>8 RT</p><p>pM</p><p>= 12pd 2 n 6v7 = 0.74 * 1010 s-1</p><p>v =</p><p>1</p><p>t</p><p>=</p><p>1</p><p>l>6v7</p><p>=</p><p>6v7</p><p>l</p><p>310 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>r</p><p>(b)</p><p>But, constant</p><p>or or</p><p>Thus,</p><p>(c)</p><p>But,</p><p>Thus,</p><p>2.232 In the polytropic process of index n constant, constant and</p><p>constant.</p><p>(a) When</p><p>(b) When</p><p>So,</p><p>(c)</p><p>2.233 (a) The number of collision between the molecules in a unit volume is</p><p>1</p><p>2</p><p>nn =</p><p>1</p><p>12</p><p>pd 2 n 26v7 r 1T</p><p>V 2</p><p>v r</p><p>p</p><p>1T</p><p>r T 1n>n -12- 11>22 = T 1n +12>21n -12</p><p>l r T 1- n>n -1 = T</p><p>-1>n -1 = T 1>1- n</p><p>l r T</p><p>p</p><p>,</p><p>p r T n>n -1</p><p>v =</p><p>6v7</p><p>l</p><p>r</p><p>p</p><p>1T</p><p>r p</p><p>1-1>2 + 1>2n = p</p><p>(n +1)>2n</p><p>l r p -1>n</p><p>l r T</p><p>p</p><p>,</p><p>T n r p n -1 or T r p</p><p>1-1>n</p><p>v r T 1>2</p><p>V</p><p>= V</p><p>1- n>2 V -1 = V</p><p>-n +1>2l r V</p><p>p 1–n T n = TV n -1 =pV n =</p><p>v r T 1>2</p><p>V</p><p>T</p><p>3</p><p>l r T -5>2 TV 2/5 = constant or V r T -5>2l r V</p><p>v =</p><p>6v7</p><p>l</p><p>r</p><p>p</p><p>1T</p><p>r p1>2+1>2g = pg+1>2g = p6>7</p><p>l r p -1>g = p-5>7</p><p>T r p 1-1>gT</p><p>p</p><p>r p-1>g</p><p>p aT</p><p>p</p><p>b g</p><p>=</p><p>l r T</p><p>p</p><p>2.7 TRANSPORT PHENOMENA 311</p><p>When,</p><p>When</p><p>r</p><p>,</p><p>Given</p><p>Given</p><p>,</p><p>This remains constant is polytropic process</p><p>Using solution of Problem 2.122, the molar specific heat for the polytropic pro-</p><p>cess, constant, is</p><p>Thus,</p><p>It can also be written as where</p><p>On substituting values J/K mol.</p><p>(b) In this case constant and so</p><p>So,</p><p>which can also be written as (where</p><p>On substituting values J/K mol.</p><p>2.234 We can assume that all molecules incident on the hole, leak out. Then,</p><p>or</p><p>Integrating,</p><p>Hence,</p><p>2.235 If the temperature of the compartment 2 is times more than that of compartment</p><p>1, it must contain times less number of molecules since pressure must be the</p><p>same when the big hole is open. If mass of the gas in 1 than the mass of the</p><p>gas is 2 must be . So immediately after the big hole is closed,</p><p>n0</p><p>1 =</p><p>M</p><p>mV</p><p>n0</p><p>2 =</p><p>M</p><p>mVh</p><p>M>h M =</p><p>1>h h</p><p>6v7 = A</p><p>8RT</p><p>pM</p><p>n = n0 e</p><p>-t>t</p><p>dn = -n</p><p>dt</p><p>4v>S6v7</p><p>= -n</p><p>dt</p><p>t</p><p>-dN = -d1nV2 =</p><p>1</p><p>4</p><p>n 6v7 S dt</p><p>C = 29</p><p>i = 5</p><p>R</p><p>2</p><p>(i + 1)</p><p>C = R a 1</p><p>g - 1</p><p>+</p><p>1</p><p>2</p><p>b = R a5</p><p>2</p><p>+</p><p>1</p><p>2</p><p>b = 3R</p><p>pV -1 = constant.</p><p>1T</p><p>V</p><p>=</p><p>C = 23</p><p>i = 5</p><p>1</p><p>4</p><p>R (1 + 2i)</p><p>C = R a 1</p><p>g - 1</p><p>+</p><p>1</p><p>4</p><p>b = R a5</p><p>2</p><p>+</p><p>1</p><p>4</p><p>b =</p><p>11</p><p>4</p><p>R</p><p>C = R a 1</p><p>g - 1</p><p>-</p><p>1</p><p>a - 1</p><p>bpV</p><p>a</p><p>=</p><p>pV -3 = constant.</p><p>312 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>,</p><p>)</p><p>).</p><p>(</p><p>and</p><p>where mass of each molecule and are concentrations in 1 and 2. After</p><p>the big hole is closed the pressures will differ and concentration will become and</p><p>, where</p><p>On the other hand</p><p>Thus,</p><p>So,</p><p>2.236 We know</p><p>Thus changing times implies T changing times.</p><p>On the other hand</p><p>Thus D changing times means changing times.</p><p>So p must change times times (on substituting values).</p><p>2.237</p><p>(a) D will increase n times, and will remain constant if T is constant.</p><p>(b)</p><p>Thus D will increase times, will increase times, if p is constant.n1>2hn 3>2 h r 1pV</p><p>D r T 3>2</p><p>p</p><p>r</p><p>1pV2 3>2</p><p>p</p><p>= p 1>2 V 3>2</p><p>h</p><p>D r 1T</p><p>n</p><p>r V 1T h r 1T</p><p>= 2</p><p>a 3</p><p>b</p><p>b</p><p>T 3>2</p><p>p</p><p>b</p><p>D =</p><p>1</p><p>3</p><p>6v7l =</p><p>1</p><p>3A</p><p>8kT</p><p>pm</p><p>kT</p><p>12pd 2p</p><p>a2ah</p><p>h =</p><p>1</p><p>3</p><p>6v7lr =</p><p>1</p><p>3</p><p>6v7</p><p>1</p><p>12 p d 2</p><p>m a 1T</p><p>n2 = n0</p><p>2</p><p>1 + h</p><p>1 + 1h</p><p>n2 A1 + 1h B =</p><p>m</p><p>mVh</p><p>11 + h2 = n0</p><p>211 + h2</p><p>n1 6v17 = n26v27 i.e. n1 = 1h n2</p><p>n1 + n2 =</p><p>M</p><p>mVh</p><p>11 + h2n2</p><p>n1</p><p>n0</p><p>1, n</p><p>0</p><p>2m =</p><p>2.7 TRANSPORT PHENOMENA 313</p><p>and</p><p>, ,</p><p>Also,</p><p>2.238</p><p>In an adiabatic process</p><p>Now V is decreased times.</p><p>So D decreases times and increase times and 1.6 times, respectively.</p><p>2.239 (a)</p><p>Thus D remains constant in the process constant. So polytropic index</p><p>.</p><p>(b)</p><p>So remains constant in the isothermal process. constant, here.</p><p>(c) Heat conductivity and CV is a constant for the ideal gas.</p><p>Thus here also.</p><p>2.240</p><p>or</p><p>2.241</p><p>=</p><p>1</p><p>3</p><p>A</p><p>8kT</p><p>mp</p><p>1</p><p>12 p d 2n</p><p>mn</p><p>CV</p><p>M</p><p>k =</p><p>1</p><p>3</p><p>6v7lrcV</p><p>= 10-10a 2</p><p>3 * 18.9</p><p>b1>2a4 * 83.1 * 273</p><p>p3 * 0.36</p><p>b1>4</p><p>L 0.178 nm</p><p>d = a 2</p><p>3 h b1>2amkT</p><p>p3</p><p>b1>4</p><p>= a 2</p><p>3 * 18.9</p><p>* 106b1>2a4 * 8.31 * 273 * 10-3</p><p>p3 * 36 * 1046</p><p>b1>4</p><p>h =</p><p>1</p><p>3</p><p>A</p><p>8kT</p><p>pm</p><p>m</p><p>12 p d 2</p><p>=</p><p>2</p><p>3</p><p>A</p><p>mkT</p><p>p 3</p><p>1</p><p>d 2</p><p>n = 1</p><p>k = hCV</p><p>n = 1,pV =h</p><p>h r 1T r 1pV</p><p>n = 3</p><p>pV 3 =</p><p>D r V 1T r 1pV 3</p><p>L 6.3n 1>5hn 4>5</p><p>h r V 1g>2 = a 1</p><p>n</p><p>b -1>5</p><p>= n 1>5</p><p>D r V 3-g>2 = a 1</p><p>n</p><p>b3-g>2</p><p>= a 1</p><p>n</p><p>b4>5</p><p>1>nTV g-1 = constant or T r V 1-g</p><p>D r V 1T h r 1T</p><p>314 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>and</p><p>Thus,</p><p>of the process</p><p>is the specific heat capacity which is Now is the same for all monoatom-</p><p>ic gases such as He and A.</p><p>Thus,</p><p>or</p><p>2.242 In this case</p><p>or</p><p>To decrease N1 n times,</p><p>t = -</p><p>2(v0</p><p># g)</p><p>g 2</p><p>t Z 0,</p><p>v</p><p>2</p><p>0 = v</p><p>2</p><p>0 + 2(v0</p><p># g)t + g</p><p>2 t2</p><p>t = tt = 0v = v0</p><p>v</p><p>2 = v</p><p>2</p><p>0 + 2(v0</p><p># g) t + g</p><p>2t</p><p>2</p><p>v # v = (v0 + gt) # (v0 + gt) = v</p><p>2</p><p>0 + 2(v0</p><p># g)t + g</p><p>2t</p><p>2</p><p>6v7 = v0 +</p><p>g</p><p>2</p><p>t</p><p>t</p><p>6v7 =</p><p>¢r</p><p>¢t</p><p>=</p><p>r</p><p>t</p><p>= v0 +</p><p>1</p><p>2</p><p>gt</p><p>20 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>R/2</p><p>wn</p><p>x</p><p>P</p><p>H</p><p>y v0</p><p>w = g</p><p>O</p><p>α</p><p>α</p><p>or</p><p>when</p><p>(c) For the body, and are</p><p>(1)</p><p>and (2)</p><p>Hence putting the value of t from Eq. (1) into Eq. (2) we get</p><p>Note: One can use the formula of curvature radius of a trajectory , to solve part (d),</p><p>1.30 We have,</p><p>Thus,</p><p>Now,</p><p>Hence,</p><p>Now,</p><p>or</p><p>On the basis of obtained expressions or facts the sought plots can be drawn as shown</p><p>in the figure of answer sheet.</p><p>wv = wt = - g</p><p>vy</p><p>v</p><p>wn = g</p><p>vx</p><p>vt</p><p>Cwhere vx = 2vt</p><p>2 - v2</p><p>y = 2v</p><p>2 - v</p><p>2</p><p>y Dwn = 2w2 - w2</p><p>t = Ag2 - g2</p><p>v2</p><p>y</p><p>v t</p><p>2</p><p>| wt | = g</p><p>| vy |</p><p>v</p><p>= -</p><p>g</p><p>v</p><p>(v0 sin a - gt) = -g</p><p>vy</p><p>vt</p><p>wt =</p><p>dvt</p><p>dt</p><p>=</p><p>1</p><p>2vt</p><p>d</p><p>dt</p><p>(vt)</p><p>2 =</p><p>1</p><p>v</p><p>(g2t - gv0 sin a)</p><p>v2 = v</p><p>2</p><p>x + v</p><p>2</p><p>y = v2</p><p>0 - 2gt v0 sin a + g2t 2</p><p>vx = v0 cos a, vy = v0 sin a - gt</p><p>R =</p><p>[1 + (dy/dx)2]3>2</p><p>|d 2y /dx 2|</p><p>y (x)</p><p>y = v0 sin a a x</p><p>v0 cos a</p><p>b -</p><p>1</p><p>2</p><p>g a x</p><p>v0 cos a</p><p>b2</p><p>= x tan a -</p><p>gx2</p><p>2 v2</p><p>0 cos2 a</p><p>y = v0 sin at -</p><p>1</p><p>2</p><p>gt 2</p><p>x = v0 cos at</p><p>y (t)x (t)</p><p>v2</p><p>0 sin 2 a</p><p>g</p><p>=</p><p>v2</p><p>0 sin</p><p>2 a</p><p>2g</p><p>or tan a = 4, so, a = tan-1 4</p><p>R = H</p><p>R = v0cos at -</p><p>1</p><p>2</p><p>(0) t2 = v0</p><p>cos a t =</p><p>v 2</p><p>0 sin 2 a</p><p>g</p><p>1.1 KINEMATICS 21</p><p>1.31 The ball strikes the inclined plane (Ox) at point O (origin) with velocity</p><p>(1)</p><p>As the ball classically rebounds, it recalls with same velocity , at the same angle</p><p>from the normal on y axis (see figure). Let the ball strike the incline second time at P,</p><p>which is at a distance l (say) from the point O, along the incline. From the equation</p><p>where is the time of motion of ball in air while moving</p><p>from O to P.</p><p>As , so, (2)</p><p>Now from the equation.</p><p>So, (using Eq. 2)</p><p>Hence the sought distance,</p><p>(using Eq. 1)</p><p>1.32 Total time of motion</p><p>(1)</p><p>and horizontal range</p><p>(2)</p><p>From Eqs. (1) and (2)</p><p>On simplifying</p><p>Solving for , we get</p><p>t2 =</p><p>2400 �21425000</p><p>2</p><p>=</p><p>2400 � 1194</p><p>2</p><p>t2</p><p>t4 - 2400t2 + 1083750 = 1</p><p>(9.8)2t2</p><p>(480)2</p><p>+</p><p>(85)2</p><p>(4t2)2</p><p>= 1</p><p>R = v0 cos at or cos a =</p><p>R</p><p>v0t</p><p>=</p><p>5100</p><p>240 t</p><p>=</p><p>85</p><p>4t</p><p>t =</p><p>2v0 sin a</p><p>g</p><p>or sin a =</p><p>tg</p><p>2v0</p><p>=</p><p>9.8 t</p><p>2 * 240</p><p>l =</p><p>4(2gh) sin a</p><p>g</p><p>= 8h sin a</p><p>l = v0 sin aa2v0</p><p>g</p><p>b +</p><p>1</p><p>2</p><p>g sin a a2v0</p><p>g</p><p>b =</p><p>4v0</p><p>2 sin a</p><p>g</p><p>l = v0 sin at +</p><p>1</p><p>2</p><p>g sin a t2</p><p>x = v0xt +</p><p>1</p><p>2</p><p>wxt</p><p>2</p><p>t =</p><p>2v0</p><p>g</p><p>t Z 0</p><p>t</p><p>0 = v0 cos at -</p><p>1</p><p>2</p><p>g cos at2</p><p>y = v0y t +</p><p>1</p><p>2</p><p>wy t2</p><p>av0</p><p>v0 = 22gh</p><p>22 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>y</p><p>x</p><p>Pg</p><p>h</p><p>90°–</p><p>v0</p><p>α</p><p>O</p><p>Thus,</p><p>1.33 Let the shells collide at the point . If the first shell takes t s to collide with sec-</p><p>ond and be the time interval between the firings, then</p><p>(1)</p><p>and (2)</p><p>From Eq. (1) (3)</p><p>From Eqs. (2) and (3)</p><p>as</p><p>(on substituting values)</p><p>1.34 According to the problem</p><p>(a) or</p><p>Integrating (1)</p><p>Also, we have or (using Eq. 1)</p><p>So, (using Eq. 1)</p><p>(b) According to the problem</p><p>(2)</p><p>So,</p><p>Therefore, wt =</p><p>dv</p><p>dt</p><p>=</p><p>a2y</p><p>2v2</p><p>0 + ay 2</p><p>dy</p><p>dt</p><p>=</p><p>a2y</p><p>21 + (ay/v0)</p><p>2</p><p>v = 2v2</p><p>x + v2</p><p>y = 2v2</p><p>0 + a2y2</p><p>vy = v0 and vx = ay</p><p>3</p><p>x</p><p>0</p><p>dx = av03</p><p>t</p><p>0</p><p>t dt, or x =</p><p>1</p><p>2</p><p>av0t</p><p>2 =</p><p>1</p><p>2</p><p>ay2</p><p>v0</p><p>dx = ay dt = av0t dt</p><p>dx</p><p>dt</p><p>= ay</p><p>3</p><p>y</p><p>0</p><p>dy = v03</p><p>t</p><p>0</p><p>dt or y = v0t</p><p>dy = v0 dt</p><p>dy</p><p>dt</p><p>= v0</p><p>= 11 s</p><p>¢t Z 0 ¢t =</p><p>2v0 sin (u1 - u2)</p><p>g (cos u2 + cos u1)</p><p>t =</p><p>¢t cos u2</p><p>cos u2 - cos u1</p><p>y = v0 sin u2 (t - ¢t) -</p><p>1</p><p>2</p><p>g (t - ¢t)2</p><p>x = v0 cos u1 t = v0 cos u2 (t – ¢t)</p><p>¢t</p><p>P (x, y)</p><p>t = 24.55 s = 0.41 min (depending on the angle a)</p><p>t = 42.39 s = 0.71 min</p><p>1.1 KINEMATICS 23</p><p>v0</p><p>v0</p><p>P</p><p>x</p><p>y</p><p>O</p><p>(x,y)</p><p>q1q2</p><p>and</p><p>Differentiating Eq. (2) with respect to time.</p><p>So,</p><p>Hence,</p><p>1.35 (a) The velocity vector of the particle</p><p>So, (1)</p><p>From Eq. (1) (2)</p><p>and</p><p>Integrating (3)</p><p>From Eqs. (2) and (3), we get,</p><p>(4)</p><p>(b) The curvature radius of trajectory is</p><p>(5)</p><p>Let us differentiate the path equation</p><p>(6)</p><p>From Eqs. (5) and (6), the sought curvature radius is</p><p>R =</p><p>a</p><p>b</p><p>c1 + a b</p><p>a</p><p>xb2 d3>2</p><p>dy</p><p>dx</p><p>=</p><p>b</p><p>a</p><p>x and d</p><p>2y</p><p>dx</p><p>2</p><p>=</p><p>b</p><p>a</p><p>y =</p><p>b</p><p>2a</p><p>x2 with respect to x,</p><p>R =</p><p>[1 + (dy >dx)2]3>2</p><p>| d 2y > dx 2 |</p><p>y (x)</p><p>y =</p><p>b</p><p>2a</p><p>(x2)</p><p>3</p><p>y</p><p>0</p><p>dy = ab3</p><p>t</p><p>0</p><p>t dt or y =</p><p>1</p><p>2</p><p>ab t</p><p>2</p><p>dy = bx dt = bat dt</p><p>3</p><p>x</p><p>0</p><p>dx = a3</p><p>t</p><p>0</p><p>dt or x = at</p><p>dx</p><p>dt</p><p>= a and</p><p>dy</p><p>dt</p><p>= bx</p><p>v = a i + bx j</p><p>wn = 2w2 - w2</p><p>t = Aa2v2</p><p>0 -</p><p>a4y 2</p><p>1 + (ay/v0)</p><p>2</p><p>=</p><p>av0</p><p>21 + (ay/v0)</p><p>2</p><p>w = | wx | = av0</p><p>dvy</p><p>dt</p><p>= wy = 0 and dvx</p><p>dt</p><p>= wx = a</p><p>dy</p><p>dt</p><p>= av0</p><p>24 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>1.36 In accordance with the problem</p><p>But,</p><p>So,</p><p>or (because a is directed towards the x-axis)</p><p>So,</p><p>Hence,</p><p>1.37 The velocity of the particle</p><p>So, (1)</p><p>and (using ) (2)</p><p>From</p><p>So, (3)</p><p>From Eqs. (2) and (3),</p><p>Hence,</p><p>1.38 (a) According to the problem</p><p>For v (t),</p><p>Integrating this equation from</p><p>(1) -3</p><p>v</p><p>v0</p><p>dv</p><p>v</p><p>2</p><p>=</p><p>1</p><p>R3</p><p>t</p><p>0</p><p>dt or v =</p><p>v0a1 +</p><p>v0t</p><p>R</p><p>b</p><p>v0 … v … v and 0 … t … t</p><p>-dv</p><p>dt</p><p>=</p><p>v2</p><p>R</p><p>| wt | = | wn |</p><p>= 2a2 + (4pa h )2 = a21 + 16p2h2 = 0.8 m/s2</p><p>w = 2w2</p><p>t + w2</p><p>n</p><p>wn = 4pah</p><p>4ph</p><p>a</p><p>=</p><p>t 2</p><p>R</p><p>2pRh = 3</p><p>t</p><p>0</p><p>v dt = 3</p><p>t</p><p>0</p><p>at dt</p><p>s = Lv dt</p><p>v = at wn =</p><p>v 2</p><p>R</p><p>=</p><p>a2t 2</p><p>R</p><p>dv</p><p>dt</p><p>= wt = a</p><p>v = at</p><p>v 2 = 2ax or v = 22ax</p><p>3</p><p>v</p><p>0</p><p>v dv = a3</p><p>x</p><p>0</p><p>dx</p><p>v d v = a i # dr = a dx</p><p>v dv = (a # T) ds = a # ds T = a # d r</p><p>wt =</p><p>vd v</p><p>ds</p><p>or v dv = wt</p><p>ds</p><p>wt = a # T</p><p>1.1 KINEMATICS 25</p><p>.</p><p>, we get</p><p>Now for</p><p>Integrating this equation from</p><p>Hence, (2)</p><p>(b) The normal acceleration of the point</p><p>(using Eq. 2)</p><p>In accordance with the problem</p><p>So,</p><p>1.39 From the equation</p><p>and</p><p>As is a positive constant, the speed of the particle increases with time, and the tan-</p><p>gential acceleration vector and velocity vector coincides in direction.</p><p>Hence the angle between v and w is equal to angle between and w and can</p><p>be found by the formula</p><p>1.40 Differentiating the equation</p><p>So, (1)</p><p>and (2) wn =</p><p>v2</p><p>R</p><p>=</p><p>a2v2 cos2 vt</p><p>R</p><p>wt =</p><p>dv</p><p>dt</p><p>= -av2 sin vt</p><p>dl</p><p>dt</p><p>= v = av cos vt</p><p>l = a sin v t</p><p>tan a =</p><p>| wn |</p><p>| wt |</p><p>=</p><p>a2s/R</p><p>a2/2</p><p>=</p><p>2s</p><p>R</p><p>awt t</p><p>wt</p><p>wn =</p><p>v2</p><p>R</p><p>=</p><p>a2s</p><p>R</p><p>wt =</p><p>dv</p><p>dt</p><p>=</p><p>a</p><p>21s</p><p>ds</p><p>dt</p><p>=</p><p>a</p><p>21s</p><p>a 1s =</p><p>a2</p><p>2</p><p>v = a1s</p><p>w = 22wn =22</p><p>v</p><p>2</p><p>0</p><p>R</p><p>e</p><p>-2s/R = 22</p><p>v2</p><p>R</p><p>| wt | = | wn | and wt t � wnn</p><p>wn =</p><p>v2</p><p>R</p><p>=</p><p>v</p><p>2</p><p>0e</p><p>-2s/R</p><p>R</p><p>v = v0e</p><p>-s/R</p><p>3</p><p>v</p><p>v0</p><p>dv</p><p>v</p><p>= -</p><p>1</p><p>R3</p><p>s</p><p>0</p><p>ds or ln</p><p>v</p><p>v0</p><p>= -</p><p>s</p><p>R</p><p>v0 … v … v and 0 … s … s</p><p>-</p><p>v dv</p><p>ds</p><p>=</p><p>v2</p><p>R</p><p>v (s),</p><p>26 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>, we get</p><p>(a) At the point so, etc.</p><p>Hence, (on substituting values)</p><p>Similarly at so,</p><p>Hence, (on substituting values)</p><p>(b) Using in Eqs. (1) and (2) we get</p><p>and</p><p>Therefore,</p><p>On differentiating with respect to l and putting</p><p>we get,</p><p>So, (on substituting values)</p><p>The corresponding is</p><p>(on substituting values)</p><p>1.41 As and at , the point is at rest.</p><p>So, v(t) and s (t) are (1)</p><p>Let R be the curvature radius, then</p><p>(using Eq. 1)</p><p>But according to the problem</p><p>wn = bt</p><p>4</p><p>wn =</p><p>v2</p><p>R</p><p>=</p><p>a2t 2</p><p>R</p><p>=</p><p>2as</p><p>R</p><p>v = at and s =</p><p>1</p><p>2</p><p>at2</p><p>t = 0wt = a</p><p>= av2A1 - a R</p><p>2a</p><p>b2</p><p>= 2.5 m/s2</p><p>= av2A1 -</p><p>R 2</p><p>2a2</p><p>+</p><p>R2</p><p>4a2</p><p>wmin = Av4aa2 -</p><p>R2</p><p>2</p><p>b +</p><p>v4</p><p>R2</p><p>aR2</p><p>2</p><p>b2</p><p>at l = lmwmin</p><p>lm = �aA1 -</p><p>R2</p><p>2a2</p><p>= �0.37 m</p><p>l 2 = lm</p><p>2 = a2 –</p><p>R 2</p><p>2</p><p>= a2 a1 –</p><p>R 2</p><p>2a2</p><p>b</p><p>dwt</p><p>dl</p><p>= 0</p><p>wt = Av4l 2 +</p><p>v4</p><p>R</p><p>2</p><p>(a2 - l 2)2 = f (l )</p><p>wn =</p><p>v2</p><p>R</p><p>(a2 - l 2)</p><p>wt = -v2l</p><p>sin vt = l /a</p><p>w = | wt | = av2 = 3.2 m/s2</p><p>wn = 0l = � a, sin vt = � 1 and cos vt = 0,</p><p>w = wn =</p><p>a2v2</p><p>R</p><p>= 2.6 m/s2</p><p>vt = 0, pl = 0, sin vt = 0 and cos vt = �1</p><p>1.1 KINEMATICS 27</p><p>,</p><p>.</p><p>So, (using Eq. 1) (2)</p><p>Therefore, (using Eq. 2)</p><p>Hence,</p><p>1.42 (a) Let us differentiate twice the path equation with respect to time,</p><p>Since the particle moves uniformly, its acceleration at all points of the path is normal</p><p>and at the point it coincides with the direction of derivative .</p><p>Keeping in mind that at the point ,</p><p>we get</p><p>So,</p><p>Note: We can also calculate it from the formula of problem 1.35(b).</p><p>(b) Differentiating the equation of the trajectory with respect to time we see that</p><p>(1)</p><p>which implies that the vector is normal to the velocity vector</p><p>which, of course, is along the</p><p>h must be decreased n times. Now h does not depend on</p><p>pressure until the pressure is so low that the mean free path equals, say, .</p><p>Then the mean free path is fixed and h decreases with pressure. The mean free path</p><p>equals , when</p><p>Corresponding pressure is</p><p>The sought pressure is n times less, given by</p><p>The answer is qualitative and depends on the choice for the mean free path.</p><p>2.243 We neglect the moment of inertia of the gas in a shell. Then the moment of friction</p><p>forces on a unit length of the cylinder must be a constant as a function of r.</p><p>So,</p><p>and v =</p><p>N1</p><p>4ph</p><p>a 1</p><p>r 2</p><p>1</p><p>-</p><p>1</p><p>r 2</p><p>2</p><p>b or h =</p><p>N1</p><p>4pv</p><p>a 1</p><p>r 2</p><p>1</p><p>-</p><p>1</p><p>r 2</p><p>2</p><p>b</p><p>2pr3h</p><p>dv</p><p>dr</p><p>= N1 or v(r) =</p><p>N1</p><p>4ph a 1</p><p>r 2</p><p>1</p><p>-</p><p>1</p><p>r 2</p><p>b</p><p>1>2 ¢R</p><p>p =</p><p>22kT</p><p>pd2n¢R</p><p>= 70.7 *</p><p>10-23</p><p>10-18 * 10-3</p><p>L 0.71 Pa</p><p>p0 =</p><p>22kT</p><p>pd2¢R</p><p>1</p><p>22pd2n0</p><p>= ¢R (where n0 = concentration)</p><p>1>2 ¢R</p><p>1>2 ¢R</p><p>N1</p><p>2R ¢R</p><p>R 4</p><p>L 4p h v or N1 =</p><p>2p h vR3</p><p>¢R</p><p>N1</p><p>r 2</p><p>2 - r 2</p><p>1</p><p>r 2</p><p>1r</p><p>2</p><p>2</p><p>= 4 p h �</p><p>dA</p><p>dHe</p><p>= A</p><p>8.7</p><p>110</p><p>= 1.658 L 1.7</p><p>kHe</p><p>kA</p><p>= 8.7 =</p><p>1MA d 2</p><p>A1MHe d 2</p><p>He</p><p>= 110</p><p>d 2</p><p>A</p><p>d 2</p><p>He</p><p>k r 1</p><p>1Md 2</p><p>CCV >M.cV</p><p>2.7 TRANSPORT PHENOMENA 315</p><p>V</p><p>2.244 We consider two adjoining layers. The angular velocity gradient is So the mo-</p><p>ment of the frictional force is</p><p>2.245 In the ultra-rarefied gas we must determine h by taking</p><p>Then,</p><p>So,</p><p>2.246 Take an infinitesimal section of length dx and apply Poiseuille’s equation to this.</p><p>Then,</p><p>From the formula,</p><p>or</p><p>This equation implies that if the flow is isothermal, then</p><p>constant</p><p>Thus,</p><p>2.247 Let T be the temperature of the interface. Then heat flowing from left is equal to the</p><p>heat flowing into right in equilibrium.</p><p>Thus,</p><p>2.248 We have</p><p>k1</p><p>T1 - T</p><p>l1</p><p>= k2</p><p>T - T2</p><p>l2</p><p>= k</p><p>T1 - T2</p><p>l1 - l2</p><p>k1 =</p><p>T1 - T</p><p>l1</p><p>= k2 =</p><p>T - T2</p><p>l2</p><p>or T =</p><p>1k1T12>l1 + 1k2T22>l2</p><p>k1>l1 + k2>l2</p><p>m =</p><p>pa4M</p><p>16hRT</p><p>|p2</p><p>2 - p2</p><p>1|</p><p>l</p><p>p</p><p>dp</p><p>dx</p><p>=</p><p>|p2</p><p>2 - p2</p><p>1|</p><p>2l</p><p>=</p><p>dm</p><p>dt</p><p>= m = -</p><p>pa</p><p>4M</p><p>8hRT</p><p>pdp</p><p>dx</p><p>pdV =</p><p>RT</p><p>M</p><p>dm</p><p>pV = RT # m</p><p>M</p><p>dV</p><p>dt</p><p>=</p><p>-pa4</p><p>8h</p><p>0p</p><p>0x</p><p>N =</p><p>1</p><p>3</p><p>va4pA</p><p>pM</p><p>2RT</p><p>h =</p><p>1</p><p>3A</p><p>8kT</p><p>mp</p><p>*</p><p>1</p><p>2</p><p>h *</p><p>mp</p><p>kT</p><p>=</p><p>1</p><p>3A</p><p>2M</p><p>pRT</p><p>hp</p><p>l = 11>22h.</p><p>N = 3</p><p>a</p><p>0</p><p>r 2prdr hr</p><p>�</p><p>h</p><p>=</p><p>pha4�</p><p>2h</p><p>v>h.</p><p>316 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>or using the previous result</p><p>or</p><p>2.249 By definition the heat flux per unit area is</p><p>Integrating</p><p>(where T1 � temperature at the end x � 0)</p><p>So,</p><p>2.250 Suppose the chunks have temperatures T1, T2 at time t and T1 � dT1, T2 	 dT2 at</p><p>time dt 	 t.</p><p>Then,</p><p>Thus,</p><p>Hence,</p><p>2.251 (where A � constant)</p><p>Thus, T 3/2 = constant -</p><p>x</p><p>l</p><p>(T 3/2</p><p>1 - T 3/2</p><p>2 )</p><p>=</p><p>2</p><p>3</p><p>A</p><p>(T 3/2</p><p>1 - T 3/2</p><p>2 )</p><p>l</p><p>= -</p><p>2</p><p>3</p><p>A</p><p>0T 3/2</p><p>0x</p><p>Q</p><p>#</p><p>= k</p><p>0T</p><p>0x</p><p>= -A2T</p><p>0T</p><p>0x</p><p>¢T = (¢T )0e</p><p>-t/t awhere</p><p>1</p><p>t</p><p>=</p><p>k # S</p><p>l</p><p>a 1</p><p>C1</p><p>+</p><p>1</p><p>C2</p><p>b b</p><p>d¢T = –</p><p>kS</p><p>l</p><p>a 1</p><p>C1</p><p>+</p><p>1</p><p>C2</p><p>b¢T dt 1where ¢T = T1 - T22</p><p>C1dT1 = C2dT2 =</p><p>kS</p><p>l</p><p>(T1 - T2) dt</p><p>T = T1aT2</p><p>T1</p><p>bx/l and Q</p><p>#</p><p>=</p><p>a</p><p>l</p><p>ln T1/T2</p><p>ln T =</p><p>x</p><p>l</p><p>ln</p><p>T2</p><p>T1</p><p>+ ln T1</p><p>Q</p><p>#</p><p>= -K</p><p>dT</p><p>dx</p><p>= -a</p><p>d</p><p>dx</p><p>ln T = constant = +</p><p>a</p><p>ln T1/T2</p><p>l</p><p>k1</p><p>l1</p><p>k2</p><p>l2</p><p>(T1 - T2)</p><p>k1</p><p>l1</p><p>+</p><p>k2</p><p>l2</p><p>= k</p><p>T1 - T2</p><p>l1 + l2</p><p>Q k =</p><p>l1 + l2</p><p>l1</p><p>k1</p><p>+</p><p>l2</p><p>k2</p><p>k1</p><p>l1</p><p>§T1 -</p><p>k1T1</p><p>l1</p><p>+</p><p>k2T2</p><p>l2</p><p>k1</p><p>l1</p><p>+</p><p>k2</p><p>l2</p><p>¥ = k</p><p>T1 - T2</p><p>l1 + l2</p><p>2.7 TRANSPORT PHENOMENA 317</p><p>Using</p><p>where x is the distance from the plate maintained at the temperature T1.</p><p>2.252</p><p>Then from the previous problem</p><p>i � 3 here and d is the effective diameter of helium atom.</p><p>2.253 At this pressure and average temperature (� 27°C � 300 K)</p><p>The gas is ultra-thin and we write here.</p><p>Then,</p><p>where,</p><p>and</p><p>where,</p><p>2.254 At equilibrium,</p><p>So, T = B -</p><p>A</p><p>2pk</p><p>ln r</p><p>2pr k</p><p>dT</p><p>dr</p><p>= -A = constant.</p><p>6v7 = A</p><p>8RT</p><p>Mp</p><p>.</p><p>We have used T2 - T1 6 6</p><p>T2 + T1</p><p>2</p><p>here.</p><p>q =</p><p>p 6v 7</p><p>6T (g - 1)</p><p>(T2 - T1) = 22 W/m2 (on substituting values)</p><p>k =</p><p>1</p><p>3</p><p>6v7 *</p><p>1</p><p>2</p><p>l *</p><p>MP</p><p>RT</p><p>*</p><p>R</p><p>g - 1</p><p>*</p><p>1</p><p>M</p><p>=</p><p>p 6v 7</p><p>6T (g - 1)</p><p>l</p><p>q = k</p><p>dT</p><p>dx</p><p>= k</p><p>T2 - T1</p><p>l</p><p>l = (1>2) l</p><p>l =</p><p>1</p><p>22pd2</p><p>kT</p><p>p</p><p>= 2330 * 10-5 m = 23.3 mm 77 5.0 mm = l</p><p>T =</p><p>(T1 + T2)</p><p>2</p><p>q =</p><p>2iR 3>2(T 3>2</p><p>2 - T 3>2</p><p>1 )</p><p>9p3>2d 22MNAl</p><p>,</p><p>k =</p><p>1</p><p>3</p><p>A</p><p>8RT</p><p>pM</p><p>1</p><p>22pd 2n</p><p>mn</p><p>R (i>2)</p><p>M</p><p>=</p><p>R 3>2iT 3>2</p><p>3p3>2d22MNA</p><p>T = T1 c1 +</p><p>x</p><p>l</p><p>eaT2</p><p>T1</p><p>b3/2</p><p>- 1 fd2/3</p><p>T 3/2 = T 3/2</p><p>1 +</p><p>x</p><p>l</p><p>(T 3/2</p><p>2 - T 3/2</p><p>1 ) or a T</p><p>T1</p><p>b3/2</p><p>= 1 +</p><p>x</p><p>l</p><p>aaT2</p><p>T</p><p>b3/2</p><p>- 1b</p><p>T = T1 at x = 0, we get</p><p>318 PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS</p><p>,</p><p>But T � T1 when r � R1 and T � T2 when r � R2</p><p>From this we find T � T1</p><p>2.255 At equilibrium</p><p>Using T � T1 when r � R1 and T � T2 when r = R2,</p><p>2.256 The heat flux vector is and its divergence equals w. Thus,</p><p>or in cylindrical coordinates.</p><p>or</p><p>Since T is finite at r � 0, A � 0. Also T � T0 at r � R</p><p>So,</p><p>Thus,</p><p>where r is the distance from the axis of the wire also called axial radius.</p><p>2.257 Here again</p><p>So in spherical polar coordinates,</p><p>or</p><p>Again</p><p>So finally, T = T0 +</p><p>w</p><p>6k</p><p>* (R2 - r 2)</p><p>A = 0 and B = T0 +</p><p>w</p><p>6k</p><p>R2</p><p>T = B -</p><p>A</p><p>r</p><p>-</p><p>w</p><p>6k</p><p>r 2</p><p>1</p><p>r 2</p><p>0</p><p>0r</p><p>ar 2</p><p>0T</p><p>0r</p><p>b = –</p><p>w</p><p>k</p><p>or r2</p><p>0T</p><p>0r</p><p>= -</p><p>w</p><p>k</p><p>or r2</p><p>0T</p><p>0r</p><p>= -</p><p>w</p><p>3k</p><p>r 3 + A</p><p>§2T = –</p><p>w</p><p>k</p><p>T = T0 +</p><p>w</p><p>4k</p><p>(R 2 - r 2)</p><p>B = T0 +</p><p>w</p><p>4k</p><p>R 2</p><p>T = B + A ln r -</p><p>w</p><p>2k</p><p>r 2</p><p>1</p><p>r</p><p>a 0</p><p>0r</p><p>b ar</p><p>0T</p><p>0r</p><p>b = –</p><p>w</p><p>k</p><p>§2T = –</p><p>w</p><p>k</p><p>-k§T</p><p>T = T1 +</p><p>T2 - T1</p><p>(1>R2) - (1>R1)</p><p>a1</p><p>r</p><p>-</p><p>1</p><p>R1</p><p>b</p><p>T = B +</p><p>A</p><p>4pk</p><p>1</p><p>r</p><p>4pr2k</p><p>dT</p><p>dr</p><p>= -A = constant</p><p>T2 - T1</p><p>ln R2>R1</p><p>ln</p><p>r</p><p>R1</p><p>2.7 TRANSPORT PHENOMENA 319</p><p>, , .</p><p>ELECTRODYNAMICS</p><p>3.1 Constant Electric Field in Vacuum</p><p>3.1</p><p>Thus,</p><p>Similarly</p><p>For</p><p>3.2 Total number of atoms in the sphere of mass</p><p>So the total nuclear charge l =</p><p>6.023 * 10</p><p>23</p><p>63.54</p><p>* 1.6 * 10</p><p>-19 * 29</p><p>1 gram =</p><p>1</p><p>63.54</p><p>* 6.023 * 10</p><p>23</p><p>= B</p><p>6.67 * 10</p><p>–11m</p><p>3 (kg # s</p><p>2)</p><p>9 * 10</p><p>9</p><p>= 0.86 * 10</p><p>–10 C>kg</p><p>q</p><p>2</p><p>4pe0r</p><p>2</p><p>=</p><p>gm</p><p>2</p><p>r</p><p>2</p><p>or q</p><p>m</p><p>= 24pe0g</p><p>Fele = Fgr</p><p>=</p><p>(1.602 * 10-19 C)2a 1</p><p>9 * 109</p><p>b * 6.67 * 10-11 m3> (kg # s2) * (1.672 * 10-27 kg)2</p><p>= 1 * 1036</p><p>Fele</p><p>Fgr</p><p>(for proton) =</p><p>q</p><p>2</p><p>4pe0gm</p><p>2</p><p>=</p><p>(1.602 * 10</p><p>–19 C)</p><p>2a 1</p><p>9 * 10</p><p>9</p><p>b * 6.67 * 10</p><p>–11m</p><p>3>(kg # s</p><p>2) * (9.11 * 10</p><p>–31 kg)</p><p>2</p><p>= 4 * 1042</p><p>Fele</p><p>Fgr</p><p>(for electrons) =</p><p>q</p><p>2</p><p>4pe0gm</p><p>2</p><p>Fele</p><p>(for electrons) =</p><p>q</p><p>2</p><p>4pe0r</p><p>2</p><p>and Fgr =</p><p>gm</p><p>2</p><p>r</p><p>2</p><p>PART3</p><p>.</p><p>.</p><p>,</p><p>Now the charge on the sphere total nuclear charge total electronic charge</p><p>Hence, force of interaction between these two spheres,</p><p>3.3 Let the balls be deviated by an angle from the vertical when separation between</p><p>them equals x.</p><p>Applying Newton’s second law of motion for any one of the spheres, we get</p><p>(1)</p><p>and (2)</p><p>From the Eqs. (1) and (2)</p><p>(3)</p><p>But from the figure</p><p>(4)</p><p>From Eqs. (3) and (4)</p><p>Thus (5)</p><p>Differentiating Eq. (5) with respect to time</p><p>According to the problem</p><p>so, a2pe0mg</p><p>l</p><p>x</p><p>3b1>2</p><p>dq</p><p>dt</p><p>=</p><p>3pe0mg</p><p>l</p><p>x</p><p>2</p><p>a</p><p>2x</p><p>dx</p><p>dt</p><p>= v =</p><p>a</p><p>1x</p><p>aapproach velocity is</p><p>dx</p><p>dt</p><p>b</p><p>2q</p><p>dq</p><p>dt</p><p>=</p><p>2pe0mg</p><p>l</p><p>3x</p><p>2</p><p>dx</p><p>dt</p><p>q</p><p>2 =</p><p>2pe0mgx</p><p>3</p><p>l</p><p>Fele =</p><p>mg x</p><p>2l</p><p>or q</p><p>2</p><p>4pe0x</p><p>2</p><p>=</p><p>mg x</p><p>2l</p><p>tan u =</p><p>x</p><p>2Bl</p><p>2 - ax</p><p>2</p><p>b2</p><p>�</p><p>x</p><p>2l</p><p>as x 66 l</p><p>tan u =</p><p>Fele</p><p>mg</p><p>T sin u = Fele</p><p>T cos u = mg</p><p>u</p><p>F =</p><p>1</p><p>4pe0</p><p># [4.398 * 10</p><p>2]</p><p>2</p><p>12</p><p>N = 9 * 10</p><p>9 * 10</p><p>4 * 19.348 N = 1.74 * 10</p><p>15 N</p><p>=</p><p>6.023 * 10</p><p>23</p><p>63.54</p><p>* 1.6 * 10</p><p>-19 *</p><p>29 * 1</p><p>100</p><p>= 4.298 * 10</p><p>2 C</p><p>-=</p><p>322 PART THREE ELECTRODYNAMICS</p><p>x</p><p>T</p><p>mg</p><p>Fele</p><p>3.1 CONSTANT ELECTRIC FIELD IN VACUUM 323</p><p>Hence,</p><p>Note: If the plane of figure is x-y plane and point of suspension is O, Eq. (3) can be</p><p>obtained more easily using net torque equal to zero about OZ-axis i.e.</p><p>3.4 Let us choose coordinate axes as shown in the figure and fix three charges, and</p><p>having position vectors and respectively.</p><p>Now, for the equilibrium of</p><p>or</p><p>because</p><p>So,</p><p>or</p><p>Also for the equilibrium of</p><p>or</p><p>Substituting the value of , we get</p><p>3.5 When the charge is placed at the center of the ring, the wire</p><p>gets stretched. The component of extra tension towards the</p><p>center will balance the electric force due to the charge on</p><p>a differential part of the ring which subtends angle at the</p><p>center, having the charge</p><p>As is very small</p><p>2¢T sin</p><p>du</p><p>2</p><p>L Tdu</p><p>du</p><p>2¢T sin</p><p>du</p><p>2</p><p>= dFele</p><p>dq = a q</p><p>2p</p><p>b du</p><p>du</p><p>q0dFele</p><p>¢T</p><p>q0</p><p>q3 =</p><p>-q1q211q1 + 1q222</p><p>r3</p><p>q3 =</p><p>-q2</p><p>| r2 - r1 | 2</p><p>| r1 - r3 |</p><p>tangent. Thus the former vector is</p><p>along the normal and the normal component of acceleration is clearly</p><p>on using</p><p>At , so at</p><p>Differentiating Eq. (1)</p><p>b2 adx</p><p>dt</p><p>b2</p><p>+ b2x ad2x</p><p>dt2</p><p>b + a2 ady</p><p>dt</p><p>b2</p><p>+ a2y ad2y</p><p>dt</p><p>b = 0</p><p>wn = �</p><p>d2y</p><p>dt2</p><p>`</p><p>x = 0</p><p>x = 0x = 0, y = � b</p><p>wn =</p><p>w # n</p><p>| n |</p><p>bawn =</p><p>b2x</p><p>d2x</p><p>dt2</p><p>+ a2y</p><p>d2y</p><p>dt2</p><p>(b4x2 + a4y2)1/2</p><p>v =</p><p>dx</p><p>dt</p><p>i +</p><p>dy</p><p>dt</p><p>j</p><p>(b2xi + a2y j)</p><p>b2x</p><p>dx</p><p>dt</p><p>+ a2y</p><p>dy</p><p>dt</p><p>= 0</p><p>wn = 2av2 =</p><p>v2</p><p>R</p><p>or R =</p><p>1</p><p>2a</p><p>w = ` d2y</p><p>dt2</p><p>`</p><p>x = 0</p><p>= 2av2 = wn</p><p>x = 0, ` dx</p><p>dt</p><p>` = v</p><p>d2y/dt2x = 0</p><p>dy</p><p>dt</p><p>= 2ax</p><p>dx</p><p>dt</p><p>; d 2y</p><p>dt2</p><p>= 2a c adx</p><p>dt</p><p>b2</p><p>+ x</p><p>d2x</p><p>dt2</p><p>dy (x)</p><p>w = a21 + (4bs2 /a3)2</p><p>w = 2w2</p><p>t + w</p><p>2</p><p>n = 2a2 + (2as /R)2 = 2a2 + (4bs2/a2</p><p>)2</p><p>bt 4 =</p><p>a2t2</p><p>R</p><p>or R =</p><p>a2</p><p>bt2</p><p>=</p><p>a2</p><p>2bs</p><p>28 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Also from Eq. (1)</p><p>So, (since tangential velocity is constant )</p><p>Thus,</p><p>and</p><p>This gives</p><p>1.43 Angular velocity of point A, with respect to centre C of the circle or turning rate of</p><p>line CA taking the line OCX as reference line becomes</p><p>because angular speed of line OA is</p><p>The turning rate of line CA is also the turning rate of velocity</p><p>vector of point A, which is given by .</p><p>So, m/s (on substituting values)</p><p>Alternate:</p><p>Angular speed of a point relative to origin</p><p>So, angular speed of point A relative to origin O</p><p>But from sine property of triangle OAC,</p><p>So, or</p><p>From both the methods the obtained speed is constant, so</p><p>the tangential acceleration of particle A is zero and</p><p>(on substituting values)w = wn =</p><p>vA</p><p>2</p><p>R</p><p>=</p><p>(2vR)2</p><p>R</p><p>= 4v2R = 0.32 m/s2</p><p>vA</p><p>vA = 2v Rv =</p><p>vA</p><p>2R</p><p>r = 2R cos u</p><p>v =</p><p>vA cos u</p><p>r</p><p>=</p><p>component velocity of point A perpendicular to OA</p><p>Length of line OA</p><p>=</p><p>transverse velocity</p><p>magnitude of position vector</p><p>vA = vC R = (2v) R = 0.4</p><p>vA /R</p><p>v = -du/dt.</p><p>vC = -</p><p>d (2u)</p><p>dt</p><p>= 2a -d u</p><p>dt</p><p>b = 2v</p><p>R =</p><p>a2</p><p>b</p><p>| wn | =</p><p>bv2</p><p>a2</p><p>=</p><p>v 2</p><p>R</p><p>ad2y</p><p>dt2</p><p>b = �</p><p>b</p><p>a2</p><p>v 2</p><p>= v adx</p><p>dt</p><p>b = �v</p><p>dy</p><p>dt</p><p>= 0 at x = 0</p><p>1.1 KINEMATICS 29</p><p>r</p><p>A</p><p>C</p><p>x</p><p>O</p><p>2q</p><p>q</p><p>q</p><p>R</p><p>R</p><p>C</p><p>2q</p><p>q</p><p>q</p><p>q x</p><p>vAA</p><p>r</p><p>O</p><p>.</p><p>1.44 Differentiating with respect to time</p><p>(1)</p><p>For fixed axis rotation, the speed of the point A</p><p>(2)</p><p>Differentiating with respect to time</p><p>(using Eq. 1)</p><p>But, (using Eq. 2)</p><p>So,</p><p>(on substituting values)</p><p>1.45 The shell acquires a constant angular acceleration at the same time as it accelerates</p><p>linearly.</p><p>So,</p><p>(where w � linear acceleration and angular acceleration).</p><p>Then,</p><p>But,</p><p>Hence, (on substituting values)</p><p>1.46 Let us take the rotation axis as z-axis whose positive direction is associated with the pos-</p><p>itive direction of the co-ordinate , the rotation angle, in accordance with the right-hand</p><p>screw rule (see figure).</p><p>(a) Differentiating with respect to time twice, we get</p><p>(1)</p><p>(2)</p><p>From Eq. (1) the solid body comes to stop at</p><p>¢t = t = A</p><p>a</p><p>3b</p><p>d 2w</p><p>dt2</p><p>=</p><p>dvz</p><p>dt</p><p>= bz = -6bt</p><p>dw</p><p>dt</p><p>= a - 3bt2 = vz</p><p>w(t)</p><p>w</p><p>v =</p><p>2pnv</p><p>l</p><p>= 2.0 * 103 rad/s</p><p>v2 = 2wl</p><p>v = 22b 2pn = A2</p><p>w</p><p>l</p><p>(2pn)2</p><p>b =</p><p>w</p><p>l</p><p>=</p><p>b</p><p>2pn</p><p>l =</p><p>1</p><p>2</p><p>wt</p><p>2 and 2pn =</p><p>1</p><p>2</p><p>bt</p><p>2</p><p>= 0.7 m/s2</p><p>w = 2w2</p><p>t + w2</p><p>n = 2(v/t)2 + (2atv)2 =</p><p>v</p><p>t</p><p>21 + 4a2t 4</p><p>wn =</p><p>v2</p><p>R</p><p>=</p><p>v2</p><p>v/2at</p><p>= 2atv</p><p>wt =</p><p>dv</p><p>dt</p><p>= 2aR =</p><p>v</p><p>t</p><p>v = vR = 2 at R or R =</p><p>v</p><p>2at</p><p>d w</p><p>dt</p><p>= vz = 2at</p><p>w (t)</p><p>30 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>z</p><p>ϕ</p><p>1.1 KINEMATICS 31</p><p>The angular velocity , for</p><p>So,</p><p>Similarly,</p><p>So,</p><p>(b) From Eq. (2),</p><p>So,</p><p>Hence,</p><p>1.47 Angle is related with and by means of the formula</p><p>(1)</p><p>where R is the radius of the circle which an arbitrary point of the body circumscribes.</p><p>From the given equation (here as is positive for all val-</p><p>ues of t)</p><p>Integrating</p><p>So, wn = v2R = aat2</p><p>2</p><p>b2</p><p>R =</p><p>a2t 4</p><p>4</p><p>R</p><p>3</p><p>v</p><p>0</p><p>dv = a3</p><p>t</p><p>0</p><p>tdt or v =</p><p>1</p><p>2</p><p>at2</p><p>bb = dv/dtb = dv/dt = at</p><p>| wt | = bRwn = v2Rtan a =</p><p>wn</p><p>| wt |</p><p>wn| wt |a</p><p>b = | (bz )t =1a /3b</p><p>| = 213ab = 12 rad/s2</p><p>(bz )t =2a/3b = -6bA</p><p>a</p><p>3b</p><p>= -223ab</p><p>bz = - 6bt</p><p>6 b7 = L</p><p>bdt</p><p>Ldt</p><p>=</p><p>3</p><p>2a/3b</p><p>0</p><p>6 bt dt</p><p>3</p><p>2a/3b</p><p>0</p><p>dt</p><p>= 23 ab = 6 rad/s2</p><p>b = | bz | = 6bt for all values of t.</p><p>/2a/3b = 2a/3 = 4 rad/s= [at - bt</p><p>3]2a/3b</p><p>0</p><p>6v7 = 3</p><p>vdt</p><p>3dt</p><p>=</p><p>3</p><p>2a/3b</p><p>0</p><p>(a - 3bt2) dt</p><p>3</p><p>2a/3b</p><p>0</p><p>dt</p><p>0 … t … 2a/3bv = a - 3 bt2</p><p>where and</p><p>and</p><p>Putting the values of and wn in Eq. (1), we get</p><p>s (on substituting values)</p><p>1.48 In accordance with the problem,</p><p>k is proportionality constant)</p><p>or (1)</p><p>When , total time of rotation</p><p>Average angular velocity</p><p>Hence,</p><p>1.49 (a) We have</p><p>Integrating this Eq. within its limit for</p><p>or ln</p><p>Hence, (1)</p><p>(b) From the equation, and Eq. (1) or by differentiating Eq. (1)</p><p>v = v0e</p><p>-kt</p><p>v = v0 - k w</p><p>w =</p><p>v0</p><p>k</p><p>(1 - e-kt)</p><p>v0 - kw</p><p>v0</p><p>= -kt 3</p><p>t</p><p>0</p><p>dt 3</p><p>w</p><p>0</p><p>dw</p><p>v0 - a w</p><p>=</p><p>(w) t</p><p>v = v0 - a w =</p><p>d w</p><p>dt</p><p>6v 7 =</p><p>cv0t +</p><p>k2t3</p><p>12</p><p>-</p><p>k</p><p>2</p><p>2v0t</p><p>2 d22v0/k</p><p>0</p><p>22v0>k =</p><p>v0</p><p>3</p><p>6 v7 = 3</p><p>v dt</p><p>3dt</p><p>=</p><p>3</p><p>22v0/k</p><p>0</p><p>av0 +</p><p>k</p><p>2t</p><p>2</p><p>4</p><p>- kt2v0b dt</p><p>22v0 /k</p><p>t = t =</p><p>21v0</p><p>k</p><p>v = 0</p><p>-3</p><p>v</p><p>v0</p><p>d v</p><p>1v = k3</p><p>t</p><p>0</p><p>dt or 1v = 1v0 -</p><p>kt</p><p>2</p><p>-</p><p>d v</p><p>dt</p><p>= k1v</p><p>bx 6 0.</p><p>t = 7</p><p>tan a =</p><p>a2t4R/4</p><p>atR</p><p>=</p><p>at</p><p>3</p><p>4</p><p>or t = c a 4</p><p>a</p><p>b tan a d1/3</p><p>| wt |</p><p>| wt | = bR = at R</p><p>32 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Thus (where</p><p>1.50 Let us choose the positive direction of z-axis (stationary rotation axis) along the vec-</p><p>tor . In accordance with the equation</p><p>or</p><p>Integrating this equation within its limit for</p><p>or</p><p>Hence,</p><p>The plot is shown in the figure. It can be seen that as the angle grows, the vec-</p><p>tor first increases, coinciding with the direction of the vector , reaches</p><p>the maximum at , then starts decreasing and finally turns into zero at .</p><p>After that the body starts rotating in the opposite direction in a similar fashion</p><p>. As a result, the body will oscillate about the position with an am-</p><p>plitude equal to .</p><p>1.51 A rotating disk moves along the x-axis, in plane motion in plane. For the calcula-</p><p>tion of velocity only, plane motion of a solid can be imagined to be in pure rotation</p><p>about a point (say I) at a certain instant known as instantaneous centre of rotation. The</p><p>axis is directed along of the solid which passes through the point I at that instant</p><p>and is known as instantaneous axis of rotation.</p><p>Therefore the velocity vector of an arbitrary point (P ) of the solid can be represented as:</p><p>(1)</p><p>(where is normal location of point P relative to instantaneous rotation axis pass-</p><p>ing through point I ).</p><p>So instantaneous rotation axis I is at the perpendicular distance from point P.</p><p>On the basis of Eq. (1) for the centre of mass (C.M.) of the disk, velocity is</p><p>(2)</p><p>According to the problem and so to satis-</p><p>fy the Eq.(2), is directed towards . Hence point I is</p><p>at a distance , above the centre of the disk along y-</p><p>axis. Using all these facts in Eq. (2), we get</p><p>(3)vC = vy or y =</p><p>vC</p><p>v</p><p>CI = y</p><p>(- j)C I</p><p>� c c kvC c c i</p><p>vC = � * �CI</p><p>rPI = vP/v</p><p>PI</p><p>vP = � * rPI = v * �PI</p><p>�</p><p>x -y</p><p>p/2</p><p>w = w/2(vz 6 0)</p><p>w = pw = w/2</p><p>�0(vz 7 0)�</p><p>wvz (w)</p><p>vz = �22b0 sin w</p><p>3</p><p>vt</p><p>0</p><p>vz dvz = b0 sin w</p><p>vz (w)</p><p>vzdvz = bzdw = b cos w dw</p><p>dvz</p><p>dt</p><p>= bz or vz</p><p>dvz</p><p>dw</p><p>= bz</p><p>b0</p><p>1.1 KINEMATICS 33</p><p>z</p><p>/2</p><p>z</p><p>z</p><p>O</p><p>x</p><p>y</p><p>C</p><p>y</p><p>I</p><p>rCI</p><p>v</p><p>O</p><p>�</p><p>�</p><p>�</p><p>(a) From the angular kinematical equation</p><p>(4)</p><p>On the other hand</p><p>(where x is the x coordinate of the C.M.) (5)</p><p>From Eqs. (4) and (5),</p><p>Using this value of in Eq. (3) we get</p><p>(hyperbola)</p><p>(b) As centre C moves with constant acceleration w, with zero initial velocity</p><p>So, and</p><p>Therefore,</p><p>Hence, (parabola)</p><p>1.52 (a) The general plane motion of a solid can be imagined as the combination of trans-</p><p>lation with C.M. and rotation about C.M.</p><p>So, (1)</p><p>and</p><p>(2)</p><p>where is the component of r normal to the axis of rota-</p><p>tion and directed away from it. In this problem</p><p>and .</p><p>Let the point A touch the horizontal surface at , further let us locate the point</p><p>A at time t, when it makes an angle from vertical (see figure).</p><p>On the basis of Eqs. (1) and (2) the pictorial diagrams for velocity and accelera-</p><p>tion are as follows:</p><p>u</p><p>t = 0</p><p>vC = vrAC = rAC = R</p><p>= wC + v2 (- �AC) + (� * �AC)</p><p>wA = wC + wAC = wC + � * (� * rAC) + (� * rAC )</p><p>vA = vC + vAC = vC + � * rAC = vC + � * AC</p><p>y =</p><p>vC</p><p>v</p><p>=</p><p>12wx</p><p>v</p><p>vC = w A</p><p>2x</p><p>w</p><p>= 12xw</p><p>vc = wtx =</p><p>1</p><p>2</p><p>wt 2</p><p>y =</p><p>vC</p><p>v</p><p>=</p><p>v</p><p>bx/v</p><p>=</p><p>v2</p><p>bx</p><p>v</p><p>v =</p><p>bx</p><p>v</p><p>x = vt, t = x/v</p><p>v = bt</p><p>vz = v0z + bz t</p><p>34 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>r</p><p>axis of</p><p>rotation</p><p>wR C CR</p><p>A</p><p>v</p><p>A</p><p>w</p><p>v w</p><p>θ θ</p><p>w</p><p>w2R</p><p>�</p><p>�</p><p>As the rolling is without slipping along a line, so, and .</p><p>According to the problem (constant), so , and . Using</p><p>these facts and the vector w, is directed toward centre C of the</p><p>wheel:</p><p>Hence, distance covered by the point A during time interval</p><p>(on substituting values)</p><p>Note: One can easily find , assuming the body to rotate about the instantaneous cen-</p><p>tre of rotation of zero velocity (not of zero acceleration), which is the contact point of</p><p>the rolling body in this case.</p><p>1.53 As the ball rolls without slipping on the rigid surface along a line so, contact point O</p><p>has zero velocity, which is possible when the ball rotates in clock wise sense if it’s</p><p>centre moves towards right and such that , and . As motion</p><p>starts at with constant acceleration w so at time t, the velocity of mass centre C</p><p>becomes and .</p><p>(a) The contact point O becomes instantaneous centre of rotation, thus, the velocity</p><p>of any arbitrary point P (say) of the ball can be obtained by the relation</p><p>(1)</p><p>Using Eq. (1), the pictorial diagram to find the velocities of the points A and B</p><p>of the ball is shown below</p><p>vP = � * �PO</p><p>v = vC /R = wt /RvC = wt</p><p>t = 0</p><p>wC = w = bRvC = vR</p><p>vA</p><p>=</p><p>8v</p><p>v</p><p>= 8R = 4.0 m</p><p>s = 3vAdt = 3</p><p>2p/v</p><p>0</p><p>2 v sin (vt/2)dt</p><p>2p/v</p><p>= 2v sin (u/2) = 2v sin (vt /2)</p><p>= 2v</p><p>2 + v</p><p>2 + 2v</p><p>2 cos (p - u) = v 22(1 - cos u)</p><p>vA = 2v2 + (vR)2 + 2v (vR) cos(p-u)</p><p>wA = v 2/R =2.0 m/s2</p><p>b = 0wC = 0v = v/RvC = v</p><p>wC = bRvc = vR</p><p>1.1 KINEMATICS 35</p><p>A</p><p>B</p><p>ω 2R</p><p>A</p><p>C B</p><p>w</p><p>R</p><p>w</p><p>2R</p><p>wO O</p><p>wc+bRw(2R)</p><p>w2R</p><p>w2R</p><p>wR</p><p>Hence,</p><p>and vB = 22 vC = 22 wt = 7.1 cm/s</p><p>vA = 2vC = 2wt = 10 cm/s</p><p>,</p><p>(b) One can write for the any arbitrary point P</p><p>(2)</p><p>Taking into account Eq. (2), one can easily get the acceleration of the points O,</p><p>A and B on using the pictorial diagram.</p><p>So,</p><p>and</p><p>1.54 In the frame of horizontal plane the direction of velocity at a point is along the tan-</p><p>gent of the curve at that point, so normal acceleration at that point is perpendicular</p><p>to the direction of velocity vector and directed inwards. We make the pictorial dia-</p><p>gram as in solution of problem 1.53.</p><p>wB = A aw -</p><p>w2t2</p><p>R</p><p>b2</p><p>+ w</p><p>2 = 2.5 m/s2</p><p>wA = C4w2 +</p><p>w4t4</p><p>R2</p><p>= 2w C1 + awt2</p><p>2R</p><p>b2</p><p>= 5.6 cm/s</p><p>2</p><p>w0 = w2t2/R = 2.5 cm/s2</p><p>wP = wC + wPC = wC + �2 (- �PC) + (� * �PC)</p><p>36 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>A</p><p>B</p><p>2vC</p><p>A</p><p>C B w</p><p>r</p><p>w+</p><p>w</p><p>w2R</p><p>w2R</p><p>45°</p><p>2vC</p><p>C</p><p>vC</p><p>n n</p><p>45°</p><p>n</p><p>O O</p><p>r</p><p>As an arbitrary point of the cylinder follows a curve, its normal acceleration and ra-</p><p>dius of curvature are related by the well-known equation</p><p>So, for point A,</p><p>or</p><p>which gives</p><p>Similarly for point B, wB (n) =</p><p>v2</p><p>B</p><p>RB</p><p>RA = 4r (on using vC = vr)</p><p>v2r =</p><p>(2vC)</p><p>2</p><p>RA</p><p>wA =</p><p>v2</p><p>A</p><p>RA</p><p>wn =</p><p>v2</p><p>R</p><p>( )n</p><p>or</p><p>(again on using</p><p>1.55 The angular velocity is a vector as infinitesimal rotation commute. As for relative lin-</p><p>ear velocity, the relative angular velocity of the body 1 with respect to the body 2 is</p><p>clearly</p><p>As (on substituting values)</p><p>If frame rotates with angular velocity with respect to frame K, the relation be-</p><p>tween the time derivatives of any vector a, seen from different frames is:</p><p>If frame attached with the intersection point of two axes (about which solids are ro-</p><p>tating) is K and the frame attached with rotating solid with angular velocity is</p><p>Then,</p><p>However, (as the first body rotates with constant angular velocity in space)</p><p>and (the sought angular velocity)</p><p>Hence,</p><p>So, (on substituting values)</p><p>1.56 (a) We have (1)</p><p>So,</p><p>Differentiating Eq. (1) with respect to time</p><p>(2)</p><p>So, � = 2a2 + (2bt)2</p><p>� =</p><p>d�</p><p>dt</p><p>= a i + 2bt j</p><p>v = 2(at)2 + (bt</p><p>2)2, thus, v | t = 10s = 7.81 rad/s</p><p>� = at i + bt2 j</p><p>| �12 | = v1v2 = 12 rad/s2</p><p>�12 = �1 * �2</p><p>ad�1</p><p>dt</p><p>b</p><p>K ¿</p><p>= �12</p><p>ad�1</p><p>dt</p><p>b</p><p>K</p><p>= 0</p><p>ad�1</p><p>dt</p><p>b</p><p>K</p><p>= ad�1</p><p>dt</p><p>b</p><p>K ¿</p><p>+ �2 * �1</p><p>K ¿�2</p><p>d a</p><p>dt</p><p>2</p><p>K</p><p>=</p><p>d a</p><p>dt</p><p>2</p><p>K ¿</p><p>+ � * a</p><p>�K ¿</p><p>�1 � �2, so, | �12 | = 2v1</p><p>2 + v2</p><p>2 = 5 rad/s</p><p>�12 = �1 - �2</p><p>vC = vr)RB = 222</p><p>v2</p><p>C</p><p>v2r</p><p>= 222r</p><p>v2r cos 45° =</p><p>(22vC</p><p>)2</p><p>RB</p><p>1.1 KINEMATICS 37</p><p>and</p><p>(b)</p><p>Putting the values of (a) and (b) and taking t � 10 s, we get</p><p>1.57 (a) Let the axis of the cone (OC) rotate in anticlockwise sense with constant angular</p><p>velocity and the cone itself about it’s own axis (OC) in clockwise sense with</p><p>angular velocity . Then the resultant angular velocity of the cone</p><p>As the rolling is pure the magnitudes of the vectors</p><p>and can be easily found from the figure as</p><p>As , hence</p><p>(b) Vector of angular acceleration</p><p>If any vector a (say) rotates with angular velocity keeping its values constant</p><p>then . The vector keeping its magnitude constant rotates about</p><p>the axis with the angular velocity . So . Hence,</p><p>. The magnitude of the vector is equal to (as</p><p>So,</p><p>, on putting the values of v, R and a =</p><p>v2</p><p>R2</p><p>tan a = 2.3 rad/s2</p><p>b =</p><p>v</p><p>R cot a</p><p>a v</p><p>R</p><p>b</p><p>� ¿ � �0).b = v¿ v0�� = � ¿ * �0</p><p>(� ¿ * �0) d�0 /dt =� ¿OO¿</p><p>�0 da /dt = � * a</p><p>�</p><p>� =</p><p>d�</p><p>dt</p><p>=</p><p>d (� ¿ + �0)</p><p>dt</p><p>=</p><p>d�0</p><p>dt</p><p>(as � ¿ = constant)</p><p>=</p><p>v</p><p>R cos a</p><p>= 2.32 rad/s</p><p>v = 2v¿2 + v2</p><p>0 A a v</p><p>R cot a</p><p>b2</p><p>+ a v</p><p>R</p><p>b2</p><p>� � �0</p><p>v¿ =</p><p>v</p><p>R cot a</p><p>v0 =</p><p>v</p><p>R</p><p>�0� ¿</p><p>� = � + �0</p><p>�0</p><p>� ¿</p><p>a = 17°</p><p>cos a =</p><p>� # �</p><p>vb</p><p>=</p><p>(at i + bt2 j) # (ai + 2bt j)</p><p>2(at)2 + (bt</p><p>2)2 2a2 + (2bt)2</p><p>� | t = 10 s = 1.3 rad/s2</p><p>38 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>R</p><p>C</p><p>v</p><p>O'</p><p>A</p><p>O</p><p>w'w0</p><p>w</p><p>and</p><p>Alternate:</p><p>Vector is turning keeping the magnitude constant. To find , let us make a vec-</p><p>tor diagram as shown in the figure. The circle shown in the figure has the radius</p><p>and the tail of and of coincides at the centre of the circle.</p><p>Let turn by the small angle in the differential time interval</p><p>dt, so . Thus arc length</p><p>As is along radial line and is along the tangent in the</p><p>turning sense of so . In vector form</p><p>Hence,</p><p>1.58 The axis AB acquired the angular velocity</p><p>Using the facts of the solution of problem 1.57, the angular velocity of the body is</p><p>The angular acceleration,</p><p>But,</p><p>So, (because )</p><p>As,</p><p>1.2 The Fundamental Equation of Dynamics</p><p>1.59 Let R be the constant upward thrust on the aerostat of mass m, coming down with a</p><p>constant acceleration w. Applying Newton’s second law of motion for the aerostat in</p><p>projection form</p><p>(1)mg - R = mw</p><p>Fy = mwy</p><p>�0 � �0 so, � = 2(v0b0t)</p><p>2 + b2</p><p>0 = b0 21 + (v0t)</p><p>2 = 0.2 rad/s2</p><p>� ¿ = �0</p><p>t � = (�0 + � ¿ * �0) = �0 + (�0t * �0)</p><p>d�0</p><p>dt</p><p>= � ¿ * �0 and</p><p>d� ¿</p><p>dt</p><p>= �0</p><p>� =</p><p>d� ¿</p><p>dt</p><p>=</p><p>d(� ¿ + �0)</p><p>dt</p><p>=</p><p>d� ¿</p><p>dt</p><p>+</p><p>d�0</p><p>dt</p><p>= 2v2</p><p>0 + b2</p><p>0 t</p><p>2 = 0.6 rad/s</p><p>v = 2v2</p><p>0 + v¿2</p><p>� ¿ = �0t</p><p>� =</p><p>d�0</p><p>dt</p><p>= (� ¿ * �0)</p><p>d�0 = (� ¿ * �0) dt</p><p>d�0 � �0�0</p><p>d�0�0</p><p>= v0 v¿dt.= v0 dw| d�0 | Ldw = v¿dt</p><p>dw�0</p><p>�0 + d�0�0</p><p>v0</p><p>d�0�0</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 39</p><p>B</p><p>A0</p><p>''</p><p>d =w'dt</p><p>dw0 w0</p><p>w0+dw0</p><p>f</p><p>Now, if be the mass to be dumped, then using the equation</p><p>(2)</p><p>From Eqs. (1) and (2), we get</p><p>1.60 Let us write the fundamental equation of dy-</p><p>namics for all the three blocks in terms of pro-</p><p>jections, having taken the positive direction of</p><p>x and y axes as shown in the figure and using</p><p>the fact that kinematical relation between the</p><p>accelerations is such that the blocks move</p><p>with same value of acceleration (say w)</p><p>(1)</p><p>(2)</p><p>and</p><p>The simultaneous solution of Eqs. (1), (2) and (3) yields,</p><p>and</p><p>As the block moves down with acceleration w, so in vector form</p><p>1.61 (a) Let us indicate the positive direction of x-</p><p>axis along the incline (see figure). The fig-</p><p>ures show the force diagram for the blocks.</p><p>Let be the force of interaction between the</p><p>bars and they are obviously sliding down</p><p>with the same constant acceleration w.</p><p>Newton’s second law of motion in projec-</p><p>tion form along x-axis for the blocks gives</p><p>(1)</p><p>(2)m2g sin a - R - k2m2 g cos a = m2w</p><p>m1g sin a - k1m1 g cos a + R = m1w</p><p>w =</p><p>[m0 - k (m1 + m2)] g</p><p>m0 + m1 + m2</p><p>m0</p><p>T2 =</p><p>(1 + k) m0</p><p>m0 + m1 + m2</p><p>m2g</p><p>w = g</p><p>[m0 - k (m1 + m2)]</p><p>m0</p><p>+ m1 + m2</p><p>T2 - km2g = m2w</p><p>T1 - T2 - km1g = m1w</p><p>m0g - T1 = m0w</p><p>¢m =</p><p>2mw</p><p>g + w</p><p>R - (m - ¢m) g = (m - ¢m)w</p><p>Fy = mwy¢m</p><p>40 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>T1</p><p>T1</p><p>T2 T2</p><p>y</p><p>x</p><p>kN1</p><p>N2N1</p><p>kN2</p><p>m2gm1g</p><p>m0g</p><p>x</p><p>R</p><p>N2</p><p>fr2</p><p>m2g</p><p>x</p><p>RN1</p><p>fr1</p><p>m1g</p><p>(3)</p><p>R</p><p>Solving Eqs. (1) and (2) simultaneously, we get</p><p>(3)</p><p>and</p><p>(b) When the blocks just slide down the plane, , then from Eq. (3)</p><p>or</p><p>Hence,</p><p>1.62 Case 1: When the body is launched up.</p><p>Let k be the coefficient of friction, u the velocity of projection and l the distance tra-</p><p>versed along the incline. Retarding force on the block and</p><p>hence the retardation</p><p>Using the equation of particle kinematics along the incline,</p><p>or (1)</p><p>and</p><p>or (2)</p><p>Using Eq. (2) in Eq. (1) we get (3)</p><p>Case 2: When the block comes downward.</p><p>The net force on the body � mg sin � � kmg cos � and hence its acceleration</p><p>� g sin � � kg cos �.</p><p>Let, be the time required, then</p><p>(4)</p><p>From Eqs. (3) and (4)</p><p>But, (according to the question)</p><p>t</p><p>t ¿</p><p>=</p><p>1</p><p>h</p><p>t</p><p>2</p><p>t ¿2</p><p>=</p><p>sin a - k cos a</p><p>sin a + k cos a</p><p>l =</p><p>1</p><p>2</p><p>(g sin a - k g cos a)t ¿2</p><p>t ¿</p><p>l =</p><p>1</p><p>2</p><p>(g sin a + k g cos a) t</p><p>2</p><p>u = (g sin a + k g cos a) t</p><p>0 = u - (g sin a + k g cos a) t</p><p>l =</p><p>u2</p><p>2(g sin a + k g cos a)</p><p>0 = u2 - 2(g sin a + k g cos a) l</p><p>= g sin a + kg cos a.</p><p>= mg sin a + kmg cos a</p><p>tan a =</p><p>(k1m1 + k2m2)</p><p>m1 + m2</p><p>(m1 + m2) sin a = (k1m1 + k2m2) cos a</p><p>g sin a - g cos a</p><p>k1m1 + k2m2</p><p>m1 + m2</p><p>= 0</p><p>w = 0</p><p>R =</p><p>m1m2 (k1 - k2) g cos a</p><p>m1 + m2</p><p>w = g sin a - g cos a</p><p>k1m1 + k2m2</p><p>m1 + m2</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 41</p><p>Hence, on solving we get</p><p>1.63 At the initial moment, obviously the tension in the thread</p><p>connecting and equals the weight of .</p><p>(a) For the block to come down or the block to go</p><p>up, the conditions is</p><p>where is tension and is friction which in the lim-</p><p>iting case equals</p><p>Then</p><p>or</p><p>(b) Similarly in the case</p><p>or</p><p>or</p><p>(c) For this case, neither kind of motion is possible, and need not be limiting.</p><p>Hence,</p><p>1.64 From the conditions obtained in the previous problem, first we will check whether the</p><p>mass goes up or down.</p><p>Here, , (substituting the values). Hence the mass will</p><p>come down with an acceleration (say w). From the free body diagram of previous</p><p>problem,</p><p>(1)</p><p>and (2)</p><p>Adding Eqs. (1) and (2), we get,</p><p>w =</p><p>(m2/m1 - sin a - k cos a) g</p><p>(1 + m2/m1)</p><p>=</p><p>(h - sin a - k cos a)g</p><p>1 + h</p><p>m2g - m1g sin a - km1g cos a = (m1 + m2) w</p><p>T - m1g sin a - km1g cos a = m1w</p><p>m2g - T = m2w</p><p>m2m2/m1 = h 7 sin a + k cos a</p><p>m2</p><p>(k cos a + sin a) 7</p><p>m2</p><p>m1</p><p>7 (sin a - k cos a)</p><p>ƒr</p><p>m2</p><p>m1</p><p>6 (sin a - k cos a)</p><p>m1g sin a - m2g 7 km1g cos a</p><p>m1g sin a - m2g 7 frlim</p><p>m2</p><p>m1</p><p>7 (k cos a + sin a)</p><p>m2g - m1g</p><p>sin a 7 km1 g cos a</p><p>km 1</p><p>g cos a</p><p>ƒrT</p><p>m 2</p><p>g - T Ú 0 and T - m1g sin a - fr Ú 0</p><p>m1m2</p><p>m2m2m1</p><p>k =</p><p>(h2 - 1)</p><p>(h2 + 1)</p><p>tan a = 0.16</p><p>42 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>T</p><p>m1g m2g</p><p>T</p><p>fr</p><p>N</p><p>.</p><p>Substituting all the values,</p><p>As moves down with acceleration of magnitude , thus in vector</p><p>form, acceleration of is given by</p><p>1.65 Let us write Newton’s second law in projection form along positive x-axis for the</p><p>plank and the bar</p><p>(1)</p><p>At the initial moment, fr represents the static friction, and as</p><p>the force F grows so does the friction force fr, but up to its</p><p>limiting value, i.e., .</p><p>Unless this value is reached, both bodies move as a single</p><p>body with equal acceleration. But as soon as the force fr reaches the limit, the bar</p><p>starts sliding over the plank, i.e., .</p><p>Substituting here the values of and taken from Eq. (1) and taking into account that</p><p>, we obtain,</p><p>“=” corresponds to the moment</p><p>Hence,</p><p>If</p><p>and if</p><p>On this basis and plots are as shown in the figure of answer sheet.</p><p>1.66 Let us designate the -axis (see figure) and apply for body A</p><p>or</p><p>Now, from kinematical equation:</p><p>or</p><p>(using Eq. 1) = 22l/(sin 2a/2 - k cos2</p><p>a)g</p><p>t = 22l sec a/(sin a - k cos a) g</p><p>l sec a = 0 + (1/2) wt</p><p>2</p><p>w = g sin a - kg cos a</p><p>mg sin a - kmg cos a = mw</p><p>Fx = mwxx</p><p>w2(t)w1(t)</p><p>t 7 t0 w1 =</p><p>km2g</p><p>m1</p><p>= constant, w2 =</p><p>(at - k m2</p><p>g)</p><p>m2</p><p>w1 = w2 =</p><p>at</p><p>m1 + m2</p><p>t … t0 , then</p><p>t0 =</p><p>kgm2(m1 + m2)</p><p>am1</p><p>t = t0).(where the sign</p><p>(at - m2g)/m2 Ú</p><p>km2</p><p>m1</p><p>g</p><p>ƒr = km2g</p><p>w2w1</p><p>w2 Ú w1</p><p>fr = frs (max) = kN = km2g</p><p>fr = m1w1, F - fr = m2w2</p><p>w2 =</p><p>(h - sin a - k cos a)g</p><p>1 + h =</p><p>= 0.05 g</p><p>m2</p><p>w = 0.05 g 7 0m2</p><p>w = 0.048 g 0.05 g</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 43</p><p>m1</p><p>m2</p><p>fr</p><p>F</p><p>fr</p><p>m2</p><p>m1</p><p>fr</p><p>F</p><p>fr</p><p>mg</p><p>fr</p><p>N</p><p>l</p><p>x</p><p>L</p><p>then,</p><p>For</p><p>i.e.,</p><p>or</p><p>and putting the values of and l in Eq. (2) we get .</p><p>1.67 Let us fix the co-ordinate system to the wedge, taking the x-axis up along the</p><p>incline and the y-axis perpendicular to it (see figure).</p><p>Now, we draw the free body diagram for the bar.</p><p>Let us apply Newton’s second law in projection form along x and</p><p>y axes for the bar</p><p>(1)</p><p>or (2)</p><p>But, and using Eq. (2) in Eq. (1), we get</p><p>(3)</p><p>For the value of ( should be maximum</p><p>So,</p><p>Putting this value of in Eq. (3), we get</p><p>1.68 First of all let us draw the free body diagram for the small body of mass m and indi-</p><p>cate x-axis along the horizontal plane and y-axis, perpendicular to it, as shown in the</p><p>figure. Let the block break off the plane at i.e.,</p><p>So,</p><p>or (1)</p><p>From , for the body under investigation</p><p>mdvx</p><p>dt</p><p>= at cos a</p><p>Fx = mwx</p><p>t0 =</p><p>mg</p><p>a sin a</p><p>N = mg - at0 sin a = 0</p><p>N = 0.t = t0</p><p>Tmin =</p><p>mg (sin a + k cos a)</p><p>21 + k2</p><p>b</p><p>d (cos b + k sin b)</p><p>db</p><p>= 0 or tan b = k</p><p>cos b + k sin b)Tmin</p><p>T = mg sin a + kmg cos a/(cos b + k sin b)</p><p>ƒr = kN</p><p>N = mg cos a - T sin b</p><p>T sin b + N - mg cos a = 0</p><p>T cos b - mg sin a - fr = 0</p><p>x -y</p><p>tmin = 1 sa, k</p><p>tan 2a = -</p><p>1</p><p>k</p><p>Q a = 49°</p><p>2 cos 2a</p><p>2</p><p>+ 2k cos a sin a = 0</p><p>d a sin 2a</p><p>2</p><p>- k cos2</p><p>ab</p><p>d a</p><p>= 0tmin,</p><p>44 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>x</p><p>y mg</p><p>N</p><p>fr</p><p>T</p><p>Fy</p><p>x</p><p>N</p><p>mg</p><p>α</p><p>,</p><p>-</p><p>-</p><p>Integrating within the limits for</p><p>(using Eq. 1)</p><p>So, (2)</p><p>Integrating, Eq. (2) for</p><p>(3)</p><p>Using the value of from Eq. (1), into Eqs. (2) and (3)</p><p>1.69 Newton’s second law of motion in projection form, along horizontal or x-axis, i.e.,</p><p>gives</p><p>or</p><p>Integrating, over the limits for</p><p>or</p><p>This is the sought relationship.</p><p>1.70 From the Newton’s second law in projection form</p><p>For the bar,</p><p>(1)</p><p>For the motor,</p><p>(2)</p><p>Now, from the equation of kinematics in the frame of</p><p>bar or motor</p><p>(3) l =</p><p>1</p><p>2</p><p>(w + w¿) t2</p><p>T - kmg = mw ¿</p><p>T - 2 kmg = (2m) w</p><p>(using F = mg /3)= 22g sin a/3a</p><p>(as a = as)v = A</p><p>2F sin a</p><p>ma</p><p>F</p><p>m</p><p>3</p><p>x</p><p>0</p><p>cos (ax) dx =</p><p>v2</p><p>2</p><p>v (s)</p><p>F cos (as) ds = mvdv</p><p>(as a = as)F cos (as) = mv</p><p>dv</p><p>ds</p><p>Fx = mwx</p><p>v =</p><p>mg2 cos a</p><p>2 a sin2</p><p>a</p><p>and s =</p><p>m2g</p><p>3cos a</p><p>6 a2sin3 a</p><p>t = t0</p><p>s =</p><p>a cos a</p><p>2m</p><p>t3</p><p>3</p><p>s (t)</p><p>v =</p><p>ds</p><p>dt</p><p>=</p><p>a cos a</p><p>2m</p><p>t</p><p>2</p><p>m3</p><p>v</p><p>0</p><p>dvx = a cos a 3</p><p>t</p><p>0</p><p>tdt</p><p>v (t)</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 45</p><p>FN</p><p>mg</p><p>x</p><p>α</p><p>m</p><p>2m</p><p>T T</p><p>rfrf</p><p>we get</p><p>From Eqs. (1), (2) and (3) we get on eliminating T and</p><p>1.71 (a) Let us write Newton’s second law in vector form , for both the blocks (in</p><p>the frame of ground)</p><p>(1)</p><p>(2)</p><p>These two equations contain three unknown quantities</p><p>and T. The third equation is provided by the kinematic rela-</p><p>tionship between the accelerations</p><p>(3)</p><p>where w is the acceleration of the mass with respect to the</p><p>pulley or elevator car.</p><p>Summing up term wise the left hand and the right-hand sides</p><p>of these kinematical equations, we get</p><p>(4)</p><p>The simultaneous solution of Eqs. (1), (2) and (4) yields</p><p>Using this result in Eq. (3), we get,</p><p>Using the results in Eq. (3) we get,</p><p>(b) Obviously the force exerted by the pulley on the ceiling of the car</p><p>Note: One could also solve this problem in the frame of elevator car.</p><p>1.72 Let us write Newton’s second law for both, bar 1 and body 2 in terms of projection</p><p>having taken the positive direction of and as shown in the figure and assuming</p><p>that body 2 starts sliding, say, upward along the incline.</p><p>(1)</p><p>(2) m2g - T2 = m2w</p><p>T1 - m1g sin a = m1w1</p><p>x2x1</p><p>F = -2T =</p><p>4m1m2</p><p>m1 + m2</p><p>(g - w0)</p><p>w =</p><p>m1 - m2</p><p>m1 + m2</p><p>(g - w0)</p><p>w1¿ =</p><p>m1 - m2</p><p>m1 + m2</p><p>(g - w0) and T =</p><p>2m1m2</p><p>m1 + m2</p><p>(w0 - g)</p><p>w1 =</p><p>(m1 - m2) g + 2m2w0</p><p>m1 + m2</p><p>w1 + w2 = 2w0</p><p>m1</p><p>w1 = w0 + w¿, w2 = w0 - w¿</p><p>w1, w2</p><p>T + m2g = m2w2</p><p>T + m1g = m1w1</p><p>F = m w</p><p>t = 22l/(kg + 3w)</p><p>w¿</p><p>46 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>m2</p><p>m1</p><p>m2 g</p><p>m1 g</p><p>TT</p><p>w0</p><p>For the pulley, moving in vertical direction from the</p><p>equation</p><p>(as mass of the pulley )</p><p>or (3)</p><p>As the length of the threads are constant, the kinemat-</p><p>ic relationship of accelerations becomes</p><p>(4)</p><p>Simultaneous solutions of all these equations yields</p><p>As , w is directed vertically downward, and hence in vector form</p><p>1.73 Let us write Newton’s second law for masses and and moving pulley in verti-</p><p>cal direction along positive x-axis (see figure):</p><p>or</p><p>Again using Newton’s second law in projection form for</p><p>mass along positive direction (see figure), we get</p><p>The kinematic relationship between the accelerations of masses given in terms of pro-</p><p>jection on the x-axis</p><p>Simultaneous solution of the obtained five equations yields, in vector form</p><p>w1 =</p><p>[4m1m2 + m0 (m1 - m2)]g</p><p>4m1m2 + m0(m1 + m2)</p><p>w1x + w2x = 2w0</p><p>T1 = m0w0</p><p>x1m0</p><p>T1 = 2T</p><p>T1 - 2T = 0 (as m = 0)</p><p>m2g - T = m2w2x</p><p>m1g - T = m1w1x</p><p>m2m1</p><p>w =</p><p>2 (2h - sin a)</p><p>4h + 1</p><p>g</p><p>h 7 1</p><p>w =</p><p>2g (2 m2 /m1 - sin a)</p><p>(4 m2 /m1 + 1)</p><p>=</p><p>2g (2h - sin a)</p><p>4h + 1</p><p>w = 2w1</p><p>T1 = 2T2</p><p>mp = 0</p><p>2T2 - T1 = (mp) w1 = 0</p><p>Fx = mwx</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 47</p><p>m g1</p><p>N T1</p><p>T1</p><p>x1</p><p>T2</p><p>T2</p><p>T2</p><p>x2</p><p>m2g</p><p>w0</p><p>m0</p><p>T1</p><p>T</p><p>T</p><p>m g2</p><p>m g1</p><p>x</p><p>w0</p><p>T1</p><p>x1</p><p>1.74 As the thread is not tied with m, so if there were no friction between the thread</p><p>and the ball m, the tension in the thread would be zero and as a result both bod-</p><p>ies will have free fall motion. Obviously in the given problem it is the friction force</p><p>exerted by the ball on the thread, which becomes the tension in the thread. From</p><p>the condition or language of the problem and as both are directed down-</p><p>ward so, relative acceleration of and is directed downward.</p><p>Kinematical equation for the ball in the frame of rod in projection form along</p><p>upward direction gives</p><p>(1)</p><p>Newton’s second law in projection form along vertically down di-</p><p>rection for both, rod and ball gives,</p><p>(2)</p><p>(3)</p><p>Multiplying Eq. (2) by m and Eq. (3) by M and then subtracting Eq.</p><p>(3) from Eq. (2) and after using Eq. (1), we get</p><p>1.75 Suppose, the ball goes up with acceleration and the rod comes down with the</p><p>acceleration .</p><p>As the length of the thread is constant, (1)</p><p>Newton’s second law in projection form, vertically upwards for the ball</p><p>and vertically downwards for the rod respectively gives,</p><p>(2)</p><p>and (3)</p><p>but, (because pulley is massless) (4)</p><p>From Eqs. (1), (2), (3) and (4)</p><p>(upwards)</p><p>and (downwards) w2 =</p><p>2(2 - h)g</p><p>(h + 4)</p><p>w1 =</p><p>(2M - m)g</p><p>m + 4M</p><p>=</p><p>(2 - h)g</p><p>h + 4</p><p>T = 2T ¿</p><p>Mg - T ¿ = Mw2</p><p>T - mg = mw1</p><p>2w1 = w2</p><p>w2</p><p>w1</p><p>ƒr =</p><p>2l Mm</p><p>(M - m) t</p><p>2</p><p>mg - ƒr = mwm</p><p>Mg - ƒr = MwM</p><p>l =</p><p>1</p><p>2</p><p>(wM - wm) t</p><p>2</p><p>M = wM - wm</p><p>wM 7 wm</p><p>48 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>Mg</p><p>mg</p><p>fr</p><p>T =fr</p><p>wm</p><p>wM</p><p>Mg</p><p>T</p><p>T</p><p>T'</p><p>T'</p><p>mg</p><p>T'</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 49</p><p>From kinematical equation in projection form, we get</p><p>as and are in the opposite direction.</p><p>Putting the values of and , the sought time becomes</p><p>1.76 Using Newton’s second law in projection form along x-axis for the body 1 and along</p><p>negative x-axis for the body 2, respectively, we get</p><p>(1)</p><p>(2)</p><p>For the pulley lowering in downward direction from along x -axis,</p><p>Newton’s law gives</p><p>(as pulley is massless)</p><p>or (3)</p><p>As the length of the thread is constant, so</p><p>(4)</p><p>The simultaneous solution of above equations yields</p><p>(5)</p><p>Obviously during the time interval in which the body 1 comes to the horizontal floor cov-</p><p>ering the distance h, the body 2 moves upward the distance 2h. At the moment when</p><p>the body 2 is at the height 2h from the floor its velocity is given by the expression</p><p>After the body touches the floor, the thread becomes slack or the tension in the</p><p>thread zero, thus as a result body 2 is only under gravity for it’s subsequent motion.</p><p>Owing to the velocity at that moment or at the height 2h from the floor, the body</p><p>2 further goes up under gravity by the distance,</p><p>Thus the sought maximum height attained by body 2 is</p><p>h¿ =</p><p>v2</p><p>2</p><p>2g</p><p>=</p><p>4h (h - 2)</p><p>h + 4</p><p>v2</p><p>m1</p><p>v 2</p><p>2 = 2w2(2h) = 2 c2(h - 2) g</p><p>h + 4</p><p>d 2h =</p><p>8h (h - 2)g</p><p>h + 4</p><p>w2 =</p><p>2(m1 - 2m2) g</p><p>4m2 + m1</p><p>=</p><p>2(h - 2)</p><p>h + 4</p><p>g aas m1</p><p>m2</p><p>= hb</p><p>w2 = 2w1</p><p>T1 = 2T2</p><p>T1 - 2T2 = 0</p><p>T2 - m2g = m2w2</p><p>m1g - T1 = m1w1</p><p>t = 22l (h + 4)/3 (2 - h)g = 1.4 s</p><p>w2w1</p><p>w2w1</p><p>l =</p><p>1</p><p>2</p><p>(w1 + w2) t</p><p>2</p><p>T2</p><p>T2</p><p>T2</p><p>T1</p><p>T1</p><p>m g1</p><p>h</p><p>m g2</p><p>1.77 Let us draw free body diagram of each body, i.e., of rod A and of wedge B and also</p><p>draw the kinematical diagram for accelerations, after analysing the directions of</p><p>motion of A and B. The kinematic relationship of accelerations is</p><p>(1)</p><p>Let us write Newton’s second law for both bodies in terms of projections having taken</p><p>positive directions of and axes as shown in the figure:</p><p>(2)</p><p>and (3)</p><p>Simultaneous solution of Eqs. (1), (2) and (3) yields</p><p>wB =</p><p>wA</p><p>tan a</p><p>=</p><p>g</p><p>(tan a + h cot a)</p><p>wA =</p><p>mAg sin a</p><p>mA sin a + mB cot a cos a</p><p>=</p><p>g</p><p>(1 + h cot2 a)</p><p>N sin a = mBwB</p><p>m A</p><p>g - N cos a = mAwA</p><p>xy</p><p>tan a =</p><p>wA</p><p>wB</p><p>=</p><p>6hh</p><p>h + 4</p><p>= 0.6 m (on substituting values)</p><p>H = 2h + h¿ = 2h +</p><p>4h (h - 2)</p><p>(h + 4)</p><p>50 PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS</p><p>N</p><p>B</p><p>A</p><p>x</p><p>wAB</p><p>wA</p><p>m gA</p><p>N"</p><p>Ny</p><p>m gA</p><p>N'</p><p>wB</p><p>Note: We may also solve this problem using conservation of mechanical energy instead</p><p>of Newton’s second law.</p><p>1.78 Let us draw free body diagram of each body and fix the coordinate system, as shown</p><p>in the figure. After analysing the motion of M and m on the basis of force diagrams,</p><p>let us draw the kinematic diagram for accelerations (see figure).</p><p>As the length of thread is constant so,</p><p>and</p><p>- -</p><p>do not change their directions that’s why</p><p>(say)</p><p>As</p><p>So, from the triangle law of vector addition</p><p>(1)</p><p>From equation , for the wedge and the block,</p><p>(2)</p><p>and (3)</p><p>Now, from equation , for the block,</p><p>(4)</p><p>Simultaneous solution of Eqs. (2), (3) and (4) yields</p><p>Hence, using Eq. (1)</p><p>1.79 Bodies 1 and 2 will remain at rest with respect to bar A for where</p><p>is the sought minimum acceleration of the bar. Beyond these limits there will be</p><p>a relative motion between bar and the bodies. For , the tendency of</p><p>body 1 in relation to the bar A is to move towards right and is in the opposite sense</p><p>for . On the basis of above argument the static friction on 2 by A is directed</p><p>upward and on 1 by A is directed towards left for the purpose of calculating .</p><p>Let us write Newton’s second law for bodies 1 and 2 in terms of projection along</p><p>positive x-axis (see figure).</p><p>(1)T - fr1 = mw or fr1 = T - mw</p><p>wmin</p><p>w Ú wmax</p><p>0 … w … wmin</p><p>wmin</p><p>wmin … w … wmax</p><p>wm =</p><p>g 22</p><p>(2 + k + M/m)</p><p>w =</p><p>mg</p><p>(km + 2m + M)</p><p>=</p><p>g</p><p>(k + 2 + M/m)</p><p>mg - T - kN = mw</p><p>Fy = mwy</p><p>N = mw</p><p>T - N = Mw</p><p>Fx = mwx</p><p>wm = 22 w</p><p>wm = wmM + wM</p><p>| wmM | = | wM | = w</p><p>dsmM = dsM and as vmM and vM</p><p>1.2 THE FUNDAMENTAL EQUATION OF DYNAMICS 51</p><p>mg</p><p>T</p><p>x</p><p>T</p><p>T</p><p>T</p><p>N</p><p>wm</p><p>T</p><p>y</p><p>N</p><p>kN</p><p>wmM</p><p>M</p><p>m wmM</p><p>wM</p><p>wM</p><p>kN</p><p>(2)</p><p>As body 2 has no acceleration in vertical direction, so</p><p>(3)</p><p>From Eqs. (1) and (3)</p><p>(4)</p><p>But,</p><p>or (5)</p><p>From Eqs. (4) and (5)</p><p>Hence,</p><p>1.80 On the basis of the initial argument of the solution of problem 1.79, the tendency of</p><p>bar 2 with respect to 1 will be to move up along the plane.</p><p>Let us fix coordinate system in the frame of ground as shown in the figure.</p><p>From second law of motion in projection form along and x axes</p><p>or (1)</p><p>or (2)</p><p>but</p><p>So from Eqs. (1) and (2),</p><p>or</p><p>or</p><p>So, the sought maximum acceleration of the wedge is</p><p>where cot a 7 kwmax =</p><p>(k cos a + sin a) g</p><p>cos a - k sin a</p><p>=</p><p>(k cot a + 1)g</p><p>cot a - k</p><p>w … g</p><p>(k cos a + sin a)</p><p>cos a - k sin a</p><p>w (cos a - k sin a) … g (k cos a + sin a)</p><p>(w cos a - g sin a) … k (g cos a + w sin a)</p><p>ƒr … kN</p><p>ƒr = m (w cos a - g sin a)</p><p>mg sin a + ƒr = mw cos a</p><p>N = m (g cos a - w sin a)</p><p>mg cos a - N = mw sin a</p><p>y</p><p>x y</p><p>wmin =</p><p>g (1 - k)</p><p>(1 + k)</p><p>m (g - w) … mk (g + w), or w Ú</p><p>g (1 - k)</p><p>(1 + k)</p><p>fr1 + fr2 … k (mg</p>