The ClassicalMechanics Chinmoy Taraphdar Asian Books Private Limitedr r The Classical Mechanics Chinmoy Taraphdar Lecturer, Dept. of Physics BankuraChristianCollege Bankura, West Bengal /4Slall?',,,,/U 7JJ'loal,l.t . . l H l l l l . ~7/28, Mahavir Lane, Vardan House, Ansari Road, Daryaganj, N. Delhi-02 Registered and Editorial Office 7/28,Mahavir Lane, Vardan House, Ansari Road,Darya Ganj, Delhi-l 10 002 E-mail:asian@asiabbooksindia.com World Wide Web: http://www.asianbooksindia.com Phones:23287577,23282098,23271887,23259161Fax:911123262021 Sales Offices Bangalore103, Swiss Complex No.33, Race Course Road, Bangalore-560 001 Ph.: 22200438 Fax: 918022256583 E-mail: asianblr@airtelbroadband.in ChennaiPalani Murugan Building No.21, West Cott Road, Royapettah, Chennai-600 014 Ph.:28486927,28486928 E-mail: asianmds@vsnl.net Delhi7/28, Mahavir Lane, Vardan House, Ansari Road, Darya Ganj, Delhi -110 002 Phones:23287577,23282098,23271887,23259161 E-mail:asian@asiabbooksindia.com Guwahati6,G.N.B.Road, Panbazar, Guwahati, Assam-781001 Ph.: 0361-2513020, 2635729 E-mail:asianghyl@sancharnet.in Hyderabad3-5-11Ol/11B I1ndFloor, Opp. Blood Bank, Narayanguda, Hyderabad-500 029 Phones: 24754941, 24750951Fax: 91-40-24751152 Email: hydasian@hd2.vsnl.net.in KolkatalOA, Hospital Street, Calcutta-700 072 Ph.: 22153040 Fax: 91-33-22159899 E-mail: calasian@cal.vsnl.com MumbaiShop No.3 &4, Ground Floor, Shilpin Centre 40, G.D. Ambekar Marg, Sewree Wadala Estate, Wadala, Mumbai-400031 Ph.:91-22-22619322, 22623572 Fax: 91-22-24159899 PuneShop No.5-8, G.F.Shaan Brahma Complex, Budhwar Peth, Pune-02 Publisher Ph.:020-24497208 Fax: 91-20-24497207 E-mail: asianpune@asianbooksindia.com I st Published 2007 ISBN978-81-8412-039-4 All Rights Reserved.No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any formor byanymeans, electronic,mechanical, photocopying, recording and/or otherwise, without the prior written permission of the publisher. Published by Kamal Jagasia for AsianBooks Pvt.Ltd., 7128,Mahavir Lane, VardanHouse, Ansari Road, Darya Ganj, New Delhi-I 10 002. Typeset at : Arpit Printographers, Delhi-32 Printed at .'Gopalji Enterprises, Delhi. Preface This book, organised into ten chapters, is written tocover the syllabus of classical mechanicsforthestudentsof physicsatthegraduateand postgraduatelevel.I hopethattheclear,lucidandcomprehensivecoverageof thisbookwillhelp students togaina thorough grounding of thesubject. Thebeginningpartof thisbookexplainsseveralchaptersonthebasisof Newtonianmechanicsandthenthetextexplainsthegeneralisedco-ordinates andLagrangian mechanicsalong with Hamiltonian mechanicsbasically forthe holonomicsystem.Theconcludingchapterdealswiththecanonical transformations by which thesolution tothe particular problem can be obtained trivially.Someproblemshavebeenworkedouttoaidinunderstandingthe underlying theory attheend of eachchapter. Finally,theexpressionsof acknowledgements.Iamindebtedtomy colleagues,studentswhohavekindlygivenmevaluablecommentsand suggestions.Iacknowledgewithadeepsenseof gratitudemyindebtednessto theauthorswhosestandardworksinthefieldIhavefreelyconsultedtomy benefit. I also acknowledge my indebtedness tomy wife'Anamika', my daughter 'Sreetama' and my son 'Jyotirmoy' for their helpat every stage of the preparation of themanuscript.My specialthanksareduetoallconcerned of 'Asian Books PrivateLimited',especiallytoMs.PurobiBiswas,ProductionManager.Mr. SubhadipKhan,theBranchManagerof Kolkataofficefortheirkindhelpin bringingoutthevolumeinitsadmirableformandbearingwithmeatevery stage withunfailing patience andgood humour. It isferventlyhopedthatthebookwillbeof valuetothestudentsand teachersalike.Commentsandsuggestionsforimprovementstothetext willbe thankfullyacknowledged. ChinmoyTaraphdar "This page is Intentionally Left Blank" Contents Preface Chapter 1.Vector 1.1.Fundamental Concept of Scalar and Vectors,1 1.2.Unit Vectors and General representation of a vector,1 1.3.Multiplication and Division of Vectors by Scalar, 2 1.4.Collinear Vectors, 3 1.5.Linear Dependence or Independence of Vectors, 3 1.6.Addition and Subtraction of two Vectors, 4 1.7.Addition of More Than Two Vectors, 5 1.8.Position Vector and Its Representation in Co-ordinate System, 5 1.9.Condition of Co-planarity of Vectors, 7 1.10.Rotational Invarience of Vector in Reference Frame, 8 1.11.Product of Two Vectors, 8 1.12.Scalar Tripple Product,10 1.13.Vector Tripple Product,11 1.14.Pseudo Vectors and Pseudo Scalars,11 1.15.Vector Derivatives (Ordinary), 13 1.16.Vector Derivatives (Partial) and Vector Operators, 13 1.17.Laplacian and D'Alambertian Operator,18 1.18.Vector Integration,19 1.19.Gauss's Divergence Theorem, 20 1.20.Green's Theorem, 21 1.21.Stoke's Theorem, 22 1.23.Reciprocal Vectors, 24 1.23.Scalar and Vector Field, 24 1.25.Elementary Idea about Vector Space, 24 1.25.Linear Operator in Vector Space, 25 Summary, 26 Worked Out Examples, 29 Exercises, 34 Chapter 2.Linear Motion 2.1.Introduction,37 2.2.Kinematics,37 2.3.Basic Definitions of Required Parameters, 37 2.4.Velocity and Acceleration in Several Co-ordinate System, 40 2.5.Tangential and Normal Component of Velocity and Acceleration,41 2.6.Radialand Transverse Component of Velocity and Acceleration, 42 2.7.Newton's Laws of Motion, 44 2.8.Accelerated Linear Motion, 45 (iii) 1-36 37-64 (vi) 2.9.Graphical Treatment of Linear Motion, 45 2.10.Conservation of Linear Motion, 46 2.11.Time Integral of Force (Impulse), 46 2.12.Work,47 2.13.Power, 48 2.14.Energy:KineticandPotential, 48 2.15.Conservative Force, 49 2.16.Conservation of Energy,49 2.17.Center of Massand Its Motion, 50 2.18.The Two Body Problem, 51 2.19.Application of the Principle of Linear Motion, 52 2.20.Mechanics of Variable Mass, 53 Summary, 55 Worked Out Examples, 58 Exercises, 64 Chapter 3.Rotational Motion: Rigid Body Rotation65-94 3.1.Introduction,65 3.2.Angular velocity and Angular Momentum, 65 3.3.Angular Acceleration, 66 3.4.Moment of Inertia and Torque, 67 3.5.Centrifugal force,70 3.6.RotationalKineticEnergy,71 3.7.Angular Momentum forRigid Body Rotation,72 3.8.KineticEnergy forRigidBody Rotation,74 3.9.Axes theorem for Moment of Inertia, 75 3.10.Calculation of Moment of Inertia in different cases, 76 3.11.Momental Ellipsoid or Ellipsoid of Inertia,83 3.13.Moment andproduct of Inertia andEllipsoid of inertia of some, symmetrical bodies,84 3.12.Moment of Inertia Tensor,87 3.14.Routh's Rule,87 3.15.Euler's Angles, 88 Summary,89 Worked Out Examples, 91 Exercises, 93 Chapter 4.Reference Frame95-111 4.1.Introduction,95 4.2.Non Inertial Frame and Pseudo Force, 95 4.3.Effect of rotation of earth onacceleration due togravity,99 4.4.Effect of Coriolis Force on a particle moving on the surface of earth,101 4.5.Effect of Coriolis force on a particle falling freelyunder gravity,103 4.6.Principle of Foucault's Pendulum,104 4.7.Flow of River on Earth Surface,106 Summary, 106 Worked out Examples,107 Exercises,110 (vii) Chapter 5.Central Force112-124 5.1.Introduction,112 5.2.Definition and Characteristics of Central force,112 5.3.Conservation of Angular Momentum under Central Force,113 5.4.Conservation of energy under central force,113 5.5.Equation of motion under attractive central force,115 5.6.Application of central forcetheorytogravitation Deduction of Keplar's law,116 5.7.Energy conservation forplanetory motion,118 5.8.Stability of Orbits,120 Summary, 120 Worked out Examples,121 Exercises,124 Chapter 6.Theory of Collision125-137 6.1.Introduction,125 6.2.Characteristicsof Collision,125 6.3.Lab Frame and Center of Mass Frame,126 6.4.Direct or Linear Collision,127 6.5.Characteristic of Direct Collision,129 6.6.Maximum Energytransfer due tohead on elastic collision,130 6.7.ObliqueCollision,131 Summary, 132 Worked Out Examples,133 Exercises,137 Chapter 7.Conservation Principle and Constrained Motion138-154 7.1.Characteristicsof Conservation Principle,138 7.2.Mechanics of a single particle andsystem of particles,139 7.3.Conservation of linear momentum,140 7.4.Conservation of Angular Momentum,141 7.5.Conservation of Energy,142 7.6.ConstrainedMotion,145 7.7.Generalised Co-ordinates and other Generalised Parameters,146 7.8.Limitation of Newton's Law,151 Summary, 151 Worked out examples,153 Exercises,154 Chapter 8.Variational Principle and Lagrangian Mechanics155-193 8.1.Introduction,155 8.2.Forces of Constraint,155 8.3.VirtualDisplacement,156 8.4.Principle of Virtual Work,156 8.5.D' Alembert's Principle, 157 8.6.Lagrange's equations for a holonomic System,158 8.7.Lagrange's equation fora conservative,non-holonomicsystem,160 8.8.IntroductiontoCalculusof variations,161 8.9.VariationalTechnique formanyindependent variables: Euler-Lagrange'sdifferentialequation,165 ( viii) 8.10.Hamilton's Variational Principle,166 8.11.Derivation of Hamilton's principle from Lagrange's equation,167 8.12.Derivation of Lagrange's equations from Hamilton's principle,168 8.13.Derivation of Lagrange's equation from D' Alambert's principle,169 8.14.Derivation of Hamilton's Principle from D' Alambert's Principle,171 8.15.Cyclic or Ignorable Co-ordinates,172 8.16.Conservation Theorems,172 8.17.Gauge Function for Lagrangian,175 8.18.Invarience of Lagrange's equations under Generalised Co-ordinate, transformations,177 8.19.Concept of Symmetry: Homogeneity and Isotropy,178 8.20.Invarience of Lagrange's equation under Galilean Transformation,179 8.21.Application's of Lagrange's equation of motion in several mechanical systems,180 Summary, 186 Worked Out Examples,187 Exercises,193 Chapter 9.Hamiltonian Formulation in Mechanics194-209 9.1.Introduction,194 9.2.Hamiltonian of the System,194 9.3.Concept of Phase Space,195 9.5.Hamilton's Canonical Equations in different Co-ordinate System,197 9.6.Hamilton's Canonical equations from Hamilton's Intergral Principle,199 9.7.Physical Significance of Hamiltonian of the System, 201 9.8.Advantage of Hamiltonian Approach,201 9.9.Principle of Least Action, 201 9.10.Difference between Hamilton's Principle and the principle of least, action,203 9.11.Application of Hamilton's Canonical Equations,203 Summary, 206 Worked Out Examples, 206 Exercises, 209 Chapter 10.Canonical Transformations210-232 10.1.Introduction toCanonical Transformations,210 10.2.Hamilton-JocobiMethod,215 10.3.Application of Hamilton-Jacobi method tothe particle fallingfreely,217 10.4.Han,iIton's Characteristics function,219 10.5.Action and Angle Variables,220 10.6.Application of Action Angle Variables to Harmonic Oscillator Problem, 221 10.7.Poisson's Bracket, 222 10.8.Poisson's Theorem, 223 10.9.Jacobi's Identity,224 10.10.Lagrange's Brackets, 225 10.11.Liouville's Theorem,226 Summary, 227 Worked Out Examples, 229 Exercises. 231 Chapter-I Vector 1.1.Fundamental Conceptof ScalarandVectors: Any physical quantity having both direction and magnitude iscalled vector quantity andthesequantitymustobeysomefundamentallawsof additionand subtraction. Butthequantityhavingonlymagnitude,butnodirectioniscalledscalar quantity.Basically,themagnitudeormeasureof ascalarquantityisquite independentof anyco-ordinatesystem,butthemeasureof vectorquantity depends on the framechoosen. Allscalarquantitiesare,however,subjecttotheordinaryalgebraicallaws of additionand multiplication: viz. (i)a+ b=b+ a ax b=bxa(Commutativelaw) (ii)(a+ b)+ c =a + (b+ c) (axb)xc =ax(bxc)(Associative law) (iii)ax (b+ c)=ax b + ax c (a+ b)xc=aXc + bx c(Distributivelaw) On theother hand vectorsaresubject tothetriangular or parallelogram law of addition(orsubtraction). 1.2.Unit VectorsandGeneral representation of a vector: Avectorisanalyticallyrepresentedbyaletter withanarrowoverit,i.e,Aanditsmagnitudeis p o * Basically,PhysicalquantitiesareTensor.Anytensoristhatquantitywhichmay changeunderco-ordinatetransformatIOn.Thetensorof rank'n'isthatquantity whichtransformsthrough'n'no.of cu-efficientmatrices.Actuallyavectorisa tensor of rank1 whereasascalarisatensor of rankO. 2TheClassical Mechanics denotedby IAIorA.Graphically,avectorisrepresentedbyanarrow0 P defining thedirection,themagnitude of thevector being indicated by thelength of thearrow.Heretheend 0' of thearrowiscalled theoriginorinitialpoint of thevector,andthehead piscalledtheterminal point or terminus. Avectorhavingunitmagnitudeiscalledunitvector.Thisisingeneral denoted by n and so ini=Iunit. Thesignificanceof anyunitvectoristhatitisrequiredtoconstruct*any vectoralonganydirection.Practically,vectorcanbeconstructedbyarranging requiredno.of unitvectorsorfractionof itoneafteranotherinachoosen direction.Thusa vector can mathematically be represented by simply thescalar multiplicationof itsmagnitudewithdirectionalunit vectoralong it. i.e,A =IAIn orA=An. Thusunitvectorisinthatsensesignificant becauseitgivesdirectionof givenvectorandif it bereversedthenoppositevectorwithsame magnitudecan be obtanied.LetA =A n isa given vector.If -Abeitsoppositevector then -A=A(-n)whenIAI=I-AI=A Nowif agivenvector beaddedwithitsoppositevectorinvectoraddition method then resultant vector incalled zero vector or null vector. Mathematically. A+(-A)=6. Thisinrepresentedsimplybya'dot'andonlyinvectoralgebra,itis significant. 1.3.MultiplicationandDivision of VectorsbyScalar Anyvectorcansimplybemultipliedordividedbyscalarquantity.Incase of multiplication**,thenewvectorwillhavemagnitudemagnifiedbythat * Let,A =5whichhasmagnitude5unitsanda specifICdirection.If wenowarrange5unitvectors oneafteranotherinthatspecificdirectionthenwe willget A=5n=AnwhenA =5unit - -+-SimilarlyforA = 3.7, A =3n + (.7n) = 3.7n **Thescalarmultiplicationof vectorbyscalarquantityobeysthefollowingrules: rnAAm(Commutativelaw). rn(nA)(rnn)A(Associativelaw) (Ill + n)ArnA+ nA(Distributivelaw) Vector 3 quantityincomparetothemagnitudeof originalvector.Also,insimilarway, fordivision by scalar quantity,themagnitudeof newvector will bereduced by thatquantity.(Herewetakethatthescalarhasmagnitude>l).Butinboth case,thenewvectorwillhavethesamedirectiontooriginalvector.Sofora vector A and scalar p, H =pA, where, B = pA > A and - 1 _1 also,B'=-AwhereB,=-A 1) pp 1 Since- = q = another scalar,so,wehave p _1 - -B'=-A =qA p andB' =qA. So,divisionof vectorbyscalarisonetypeof multiplicationbyanother scalar. 1.4.Collinear Vectors: The term collinear is synonymous with parallel.So vectors having thesame or parallel lines of action are said to be collinear. So collinear vectors are parallel toeach other and they are'like vectors'.The new vectorsH orH' , asobtained throughmultiplicationordivisionbyscalarquantitywithanoriginalvector A (asdiscussed earlier), are collinear vectorswithA. 1.5. Linear Dependence or Independence of Vectors: Asetof vectors Ai(i=1,2,.... ,n)aresaidtobelinearlydependent,if fora set of scalarsai (i =1,2,.... , n)wherenotallai' .sarezero, - - -alAI + a2A2 + .... + anAn=O. i.e. L>iAi=0. But if all0../ sarezero,thevectors Aiaresaid tobelinearly dependent. *Thecrossproductof twocollinearvectoriszero.(seecrossproduct) **LetthreevectorA=(2,1,1),13=(1,2,2),(:=(1,1,1). ThenforaA + P13+ y(:= 0, weget,20,+ 13+ Y=0 0,+ 213+ y =0 0,+ 213+ y =O. Bysolving,wegeta=-k,13=-k,y=3k.Since,kisarbitrary,thereexista,13,y whicharenotallzero.Sothevectorsarelinearlydependent. 4TheClassical Mechanics Inanotherway,if A, 13,e arethreevectorsanda,p,"Iarethreescalar quantities,thenfor, aA + P13+ "Ie= ifa=P = "I=0,thenA,13, e areindependent. But fora, p,"I,not allzero,these vectors A, 13,e aredependent. Althoughthedefinitionof linear dependenceor linear independencerefers toasetof vectors,weshallalsostatethattheindividualvectors AI> A2, .... An arelinearly dependent or independent. 1.6.Addition and Subtraction of two Vectors: Anyvectorcanbeaddedwithoneormorethanonevectorandthenew vector obtained by such addition, iscalled resultant vector.This isdenoted by R and soR= LAi. For two vector addition, the resultant vector Rcan be obtained by triangular or parallelogram lawor by component* resolution method. LetAand13are twovectors which areat anangleawitheach other.IfR betheirresultant vector then, R=A + 13. Thiscanbeobtainedfrom TriangularorParallelogramlaw. Bythislaw,theresultanthas magnitude R =IA+ 131=(A2+B2+2ABcos a)1I2. Also,if,ro= A - 13= A + ( - B)be subtractionif twovectorswhichis basically the vector addition of 1st vector withoppositevectorof 2ndvector,then bysamelaw,wegetthemagnitudeof this subtraction as,ro=IA - 131= (A2+B2 - 2ABcosa)I/2**. I 11t - a ~ I-8/ / I R ~ ~ ~~ * Here,componentmeanseffectivepartof vectoralongchoosendirection. **Byreplacingaby(7t- a)inadditionformula,wegetcorrespondingsubtraction formula.Hereweshouldrememberthat 1-131 = 1131 =B.Inclassicalmechanics,for theestimationof relativevelocityof twomovingparticlesor,forobtainingsimilar quantities,theknowledgeof vectorsubtractionisrequired. Vector5 Again,if 8and~ betherespectiveanglemadebytheresultantand subtractionvector withgivenvectorA, then, B sin aBsina tan 8 =and A+B cos a tan~ =A - B cos a. 1.7. Addition of More Than Two Vectors: PolygonRuleandPerpendicular ResolutionMethod: Any no.of given vectorscanbeaddedby anyof twomethods.OneisPolygonruleandotheris Perpendicular resolutionmethod. Incaseof Polygonrule,if aPolygonhas'n'armswith(n- 1)arms representing (n- 1)vectorsindirectionandmagnitudeinthesamesense,then thentharmof thesamepolygongivestheresultantvectorof allgivenvector indirectionandmagnitudeinoppositesense.Sobythisrule, R =IAiand basically thisrulefollowstriangular ruleof vectoraddition.Ontheother hand, incase of perpendicular resolution method,if each of given vectors beresolved intotwomutuallyperpendicularcomponentsalong xor ydirection,(say),then themagnitude of resultant of these vectorswillbe: R=(RT+~ 112 where,Rl=IAi cos 8i, R2=IAi sin 8i Here,8i istheanglemadebyAi with+veXaxis.Also,if thisresultant makesangle~ withsame +veX axis,then, tan=(R2/)=(IAI sinlIl 1 lRl"A cos8. ~ II 1.8.PositionVectorandItsRepresentationinCo-ordinateSystem: Thepositionvectorof anypointP,Y takeninanyframe(2or 3dimensionorin anydimensionof configurationspace)is1 obtained by thevector joining origin tothat point.Thispositionvectorisdenotedby r ~andsoforany point P,r=OP= position vectorof P W.r.t.theorigin'0' A er (x,~ I p(r,8) I I I I I I I I I I I I If thispoint'P'hascartesianco-~ L - - - - - - - ~ - - - - ~ - - ~ Xordinates(x,y)in2dim.Thenr=xi + yJ A i where,iand J hasunit vector along xand yaxis,respectively. 6 TheClassical Mechanics Now,inplanepolar if (r,8) be theco-ordinates or the same point P then, x=r cos 8, y=r sin 8 where i=r cos8 - e sin 8

J A r )=r sin 8 + ecos 8.__L-____________whererand e arerespectiveunit vectors along increasing rand Qdirections. Inthisplanepolarco-ordinatesystem (2-dimension). ;:=r cos 8 (r cos 8 - esin 8) + r sin 8 (r sin 8 + ecos 8) =rrwhere, ris unit vector along;: and it has nocomponent in8- Z direction . In3 -dimensionalcartesian coordinate system, if (x,y,z) be the coordinateof thepointP,thus

positionvector OP =;: p (x,y,z) =xi + y) + zkwhere, i,), kare respectiveunitvectorsalongx,y, and z directions.

Here,insphericalpolar coordinationsystemif (r,8,==,,- ==-- fdt L.jit2- tl/,. 6.Relative velocity is the velocity of one moving particle w.r.t.another particle inmotion or at rest. i.e.Vru=V2- vI Similarly, relative acceleration isthe acceleration of one accelerated particle w.r.t.another accelerated particle. - - -i.e.fru==i2- fi . AAA 7.For particle moving inthreedimension,the velocityv =vxi+ vyi + vzk ==xi + yJ+ iu(cartesian) or,v ==vr;+ vee + vee ==i-;+ ree + r sin Oe(spherical polar) or,v ==vpP + vcP+ ik(cylindericalpolar). 8.For any moving particle inthreedimension,theacceleration, J ==f) + fyJ + fJ ==xi + yJ+ zk(cartesian) or,J ==fr.p+fee+/z.z ==(r - re2); + (re + 08) e+ zz. (cylindericalpolar) or,J ==1,.; + fee + flp ==(i- - re2 - rsin2 94>2); + (re + 2;.8- rsin 9cos 94>2) e + (rsin 9 + 2i-4>sin 9 + 20r4>cos9)cP(sphericalpolar). 9.Tangential component of velocity and acceleration dsdvd2s VI==dt,1; ==dt==dt2 Linear Motion Nonnal component of velocity and acceleration v2 vN=O,jN=-. P 10.Radialand Transverse component of velocity are vr =r, va=,e 11.Radial and Transverse component of acceleration are . 2... J,. =r - rB,fa =rB+ 2Ye. 12.For accelerated motion of particle,equation of motion is d2-M_r="F.=-dt 2 IFeffective I 13.For conservation of linear momentum, L Pi=constant whenFapplied=O. 14.Timeintegralof forceor impulse,is j=rt2 Fdt=P2- PI=/).P=changein momentum. Jtl 15.Work done by the forceis W=fF.dr W=-ve for work done by the force = +veforworkdone against the force. 16.Power of asystem is mathematically given by, dw P =dt =F.v 17.Energy of asystem isgiven by E=K.E+ P.E K.E =Lmi(Vj'Vj) j P.E = W= f F.dr . 18For conservativeforceF,fldr =0 andVx F = 0when,F =-V10 - vI88= v2 cos 8. 10 + vIcos(90 + 8) But kmlhr 1 cos45=Ji 102 + 122_v2 _____=-z=>V22 2 x 10x 12 and Sin 9 =(vi - 44)/20 v2, 9~ 1020'. 244 - 120Ji , v2 =8.38 5.Amaniswalkingonahorizontalroadat3k.rnItr.Therainappearsto him tocome down vertically at therate of 3 kmIhr.Calculate the actual velocity and thedirection of the rainfall. Here,VI= velocity of man,VI= 3kmIhr Vo=Relativevelocity of rain w.r.t.man, Vo=3kmIhr v2=Trueor actualvelocity of rainfall, which makes angle 8 with vertical. . .32 =32 + v/ + 2x3xv2 cos(90 + 9) =>v2 - 6 sin 8 =0:.6 sin 8 =v2 v2sin(90 + 9) Again,tan90= 00=3 + v2cos(90 + 8) -vt v+ 2 60The Classical Mechanics :. v2 sin 8 =3 :.sin2 8= 112:::::>8= 45,v2 = 6sin 45= 3.J2kmIhr. 6.AprojectileshotfromthegroundhasrangeRand themaximum height it reachesisit.Find themagnitudeand thedirection of itsinitial velocity. Ans.Letvbetheinitialvelocityforprojectionand8istheangleof projection.I I For maximum height reached H,I v2 sin2 8 H=---. If Tbe thetotal 2g timeof flight,v 08=gT/2. . .T =2 v sin 8/g. ( 2v sin8 Soforrange,R,R=v cos 8.T =v cos 8.2 g I IH I I I R------+) v sin 28 g ecos-I (). Thisisangle ofprojection. 2gH2gH Again,v2==1- (R2g/8H)..v= 16gH2 8H _R 2 g velocity of projection 7.A particlehastotalenergy Eand theforceonit isduetopotentialfield v(x).Show that thetime taken by the particle togofrom x\to x2 is t2- t\=[2E - v(x)t/2 dx. 1 Ans.We have,total energy E ="2 mv2+ v(x). ..v =[E - v(x)]=g [E -\ :=[E - V(X)]Y :::::>5:2 dx\={2 rl2 dt I[E - v(x)]2V-;;;JI1 Linear Motion 61 8.A particle isprojected vertically with velocity v.What height it will reach fromground? If it reaches upto height 'h'then 12rR+h aMm "2mu =W =JR --;:z-drwhere,Ristheradiusof earth. 12aMmh2aMh "2mv=R(R+h)R(R+h)' aM2gRh Brg=R2 v2=(R+h)' R + h_2gR _R_(2gR_ 1) h- v2 - h- v2. - h =1( - 1)Thisistheheight reached 9.Aparticleisallowedtofallundergravityfromheighth.If theair dampingonitbeproportionaltoinstantaneousvelocityof theparticle,find thevelocity attendattimet. Here,forinstanteneousvelocity 'v', thisair damping F a.vF =-Rv. Here,R =resistive constant. dv Since,theequator of motion of thefallingparticle ism dt=mg - Rv. f=f dt =s: g v=lotdt g--v mm ( RJ g--v logegm=_R y--v =-Rt/m mge Thisisinstantaneousvelocity of fallingparticle. 10.GivenforceF =xy i - iJ ' findthework doneinmovingaparticle from(0.0)to(2,1). For displacementfrom(0,0)to(2,1),ut,Y= t,x= 2t So,trunsfrom0to1 62TheClassical Mechanics anddy= dt,dx= 2dt. workdone=JF.dr=Jxydx - Jidy rl2f 2r)2 = Jo 2t.2dt - tdt=Jo 3tdt= 1 unit 11.Twoparticlesof massm)andm2 aremovingrespectivewithvelocity v)and v2Show that the total kineticenergy of thesystem is when,v = m)v)+ m2v2 (m)+ m2) Here,totalenergy of thesystem is 1212 E =- m)v)+ -m2v2//) 221' m)v)+ m2v2 But let,U=v)- v2'V=+ m)m2 m1v1 + m2v2 =(ml + m2)v m)(u + v2)+ m2v2 =(ml + m2)v. m)u v=v---:...-2m)+ m2 m2u similarly,v1 =v + m)+m2 So,fromequation(l), thetotalenergy is E 1 ()21m)m22 =- In)+ m2v+ - () u 22m) + m2 12.Twoparticlesof massmand2mapproachestoeachother duetotheir mutualattractiveforce.Find velocity of their center of mass. Ans.Since,noexternalforceactsonthesystem,arefrommomentum conservation principle, mVI =2mv2. where,VIand v2 are their resp.velocities. Linear Motion mixi- m2x2 and for,center of mass position, Xc= ml+m2 the velocity of center of mass, dxc ml vI- m2 v2mVI- 2mv2 v=- ==0 cdtml+ m2ml+ 1n2 13.Find acceleration and tension of string for At wood's machine. 63 Ans.Fortensionof stringTandaccelerationof thesystem'f, m2g- T= m2f T-m2g= mJ (m2- md2 f=(ml+m2)T=m2g-m2f =>T =2mlm2g (ml + m2) Now, we can hence have, [(ml+ m2)2- 4mlm2 Jg2 2 (ml+ m2) Thisistherelation between acceleration f and tension T. Here, are allthat forAt wood machine,! increases forT decreases. ~ fT 14.A rocket risesfromLaunching pad withanexhaust speed of 2kg sec-I. Forwhatvalueof theratioMJMy cantherocketescapethegravitationalpull of theearth?HereMoistheinitialtotalmarsof therocketandMyisthemars of thevehicle. Ans.For rocket velocity,v=Vo+ u loge(: ) . Now,Vo=0(say). v = escape velocity = 12km/sec. u=2 km/sec. 12=210ge (~ ~ . where,Mo= initialmars M =Myfinalmars. Mo M= (e)6= 403.43. y Thisisthe ratioof ( ~ ~ ). 64TheClassical Mechanics EXERCISES 1.Define thefollowing parameters. Displacement,Speed,Velocity,Averagevelocity,Instantaneousvelocity, Instantaneous acceleration, Average acceleration, Momentum. 2.What isrelativevelocity?Find itslimit of magnitude. 3.Twobodiesareallowedtofallfreelyfromsameheightin5secinterval. Find their averagevelocity and acceleration after further10sec. 4.Draw velocity-time and acceleration-time curve in thefollowing cases: (i)Particleismovingwith uniform velocity. (ii)Particleismoving withuniform acceleration. (iii)Acceleration of particleincreasesuniformly with time. (iv)Accelerationof particle decreases uniformly with time. S.Givethesignificance of velocity timegraph. 6.Rainisfallingverticallyin downward directionand water isaccumulating in a container. Now, if wind flows horizontally, then do you think that water willaccumulate in that container at the same rate. 7.Atrainstartsfromastationatrestandthenmoveswithuniform acceleration./;along time tl. After that it moves with uniform velocity along timet2and afterthatit fmally moveswithretardation/2 along timet3and comesatrestin another station.Drawvelocity-timeandacceleration-time graph and alsofind out the distance between that two stations. 8.Amaniswalking onahorizontalroadat3kglhr.Therainappearstohim tocome down vertically at the rate of 3 kmIhr.Calculate the actual velocity and thedirectionof therain fall. 9.Threemasses2gm,3gmand5gmaresituatedatthecornersof an equilateraltrianglewithsides5cmlong.Findthepositionof centerof mars. 10.A rocket hasa mars of 1100 kg and contains 900 kg of fuel.The maximum exhaustspeedof gasesequatedfromitis104km/sec.If therocketrises vertically,thenwhat should betheminimumrateof fuelconsumptionfor just liftingitoff thelaunching pad. -:0:-Chapter-3 RotationalMotionRigidBodyRotation 3.1 Introduction For particle motion along any curved path, theinstantaneous position vector of particlewillchangeitsdirectionandmagnitudeateveryinstantof motion. Nowforcircularmotionthemagnitudeof positionwillremaininvarientand duetorotationof particlealonganycircularpath,theinstantaneousvelocity willbedirectedtangentiallyateverypositionof particleandforanycurved motion,except circular motion,thisinstantaneous velocity will betangential. * In thischapter,wewilldealthemotionthroughseveralcircular or angular parameters,like,angularvelocity,angularmomentum ...etcandthenwewill extendthewholemechanismtotherotationof systemof particle,i.e.,torigid bodyrotation. Soour main objectivity isthat in thischapter we willtry tostudy the whole rotational theory in general view and try todraw theseveral conclusions tomake the chapter easy. 3.2Angular velocity and Angular Momentum: Forpointparticlerotationalonganycircularpath,theangularvelocityis definedbytherateof changeof angularposition,i.e.therateof angular displacement.Thisisbasicallyavectorquantityandisdenotedbyw Lt~ =de tl.t -t 0!1.tdt. Mathematically,itisgiven by,w * Foranycurvedmotion,theinstantaneousvelocityhastwocomponents.Oneis radialcomponentvr =r andotheristransverseorcrossradialcomponentva=r e. So,v =v/ + veS=rr + res.Whenradialmeansalongrandcrossradial directionmeansnormaltor,i.ealongtransversedirection.Nowforcircular motion.r= constant,r = 0,andvelocityisonlytangential,v = va=r e = rw. 65 66 Thisisactually,an axialvector*andis directed along the axis of rotation. For point particlerotation,thisangular velocityw is normaltotheplaneof circle,i.e.theplane of v andrwhen,v isinstanteneous linear velocityandrisinstantaneousposition vector. Itcanbeshownthatv = because,theinstantaneous displacement forcircular motionis wxr; small Now, Hence s=ex r. v=wx rwhen,w de dt r xv=rx(wxF)=(r.r)w-(r.w)r = r2w(:r1-w,;:.w=O). 1 ~ ~ )w=zrxv. r TheClassical Mechanics So,w is normal to the plane of r andv and it isalong the axis of rotation. Ontheother hand,forrotationof point mass.'m', theangularmomentum is the moment oflinear momentum. This is also an axial vector, which isdirected along theaxisof rotation, andit isnowgiven by, [,= r x (mv)= m(r xv)= m[rx ( wx r)] Now we have [,=m[(r.r)w - (r. w)F]=mr2w So,[,=Iw.(some,r.w=0forr 1- w) where, for a point mass rotating along a circle of radius r,I = mr2= moment of inertia**of that point massaboutaxisof rotation. 3.3 Angular Acceleration: For rotationalmotionof theparticle,theangular acceleration isdefinedby therateof changeof angular velocity.Thisisavector quantity andisdenoted * Inourpreviouschapter,wehavediscussedaboutaxialvectorwhichisactually pseudovectorandisobtainedbythecrossproductof twopolarvectors.Basically, thisaxialvectordoesnotchangesignundercoordinateinversion(i.e.mirror reflection). **Thismomentof inertiagivesinertiaof rotation.Forpointparticlerotation,itis treatedasscalarquantity,butwewillseeverysoonthatforrigidbodyrotationit willbecomea2ndranktensorhaving9components. Rotational Motion:Rigid Body Rotation67 ~ dw by,-;which ismathematically given by,,-;=Lt~VI.VI.!'J.t~ 0ut- dt. Whenmagnitudeof velocityof rotatingparticlealongtangentialdirection changeswithtime,thenbody willhavealinear accelerationwhichforcircular motionisgiven byj=ii x r . andjhasmagnitude f =ar forii .1 rand ii is directed in perpendicular direction tothe plane of circle, i.e.parallel totheaxisof rotation. _de Since we have,W=dt _d2ed(de) we can write,a=dt2 =dtdt whichisanother mathematical representationforangular acceleration. 3.4 Moment of InertiaandTorque: Radiusof Gyration:Whenaparticlerotatesalongacircularpath,its rotationalinertiaisrepresentedbyarotationalparameter,called,momentof inertia.Thisisverysimilarwith'mass'forlinearmotionandgivesrotational inertia of therotating body.It isa scalar quantity and isin general denoted by I. Whenparticleof massm(pointmass)rotatesalongcircleof radiusr,this momentof inertiaisestimatedby theproductof massandsquareof radiusof thecircular path,wM i.e.I=mr2. But forarigidbody rotationor fortherotationof a system of particlesabout an axisof rotation,themoment of inertiawillbethesumof allmomentof inertiasfor allpoint particles, existing within thesystem. Mathematically, N 1=Im;r/ ;=1 N r, If M=Imibethetotalmassof thewholesystemthenforrigidbody i= 1 rotation,thismoment of inertiacan be writtenas I=Imir/=M.!c2. i Where,k isanotherrotationalparameter,called'swingradius'or'radius 68TheClassical Mechanics of gyration'. * Here, k ~ = Thisradiusof gyrationforrigidbodyrotationisdefinedbysuchnormal distanceof animaginary point fromaxisof rotation,when thetotalmassof the systemissupposedtobeconcentratedatthatimaginary point.Withrespectto thatradiusof gyration,wehavethemoment of inertia. Sinceforcontinuousmassdistribution. I=Lm/i2 =tr2dm, ia Whenlimitsof integration dependsonthegeometricstructureof body and modeof rotation,theradiusof gyrationisnowgiven by 1 k =[ ~ f: r2dm y. Nowforpointparticleorrigidbodyrotationaboutsomeaxisof rotation, the'torque'issuchparameter,whichisrotationalhomologof 'force'.Thisis defined by moment of forceandisthat parameter,which,on applicationon the body,gives angular acceleration mathematically,it isgiven by 1=r xF =r x(mJ) =r xm(axr)=m[rx(a xr)] Forpointparticlerotation, 1 =m[(r.r)a - (r.a)r]=mr2a t=lei ** (when,r.l..y,, ,, andsimilarly forIzz'whicharecalledprincipalmomentof inertiaabout x, Y and z axis respectively. On theother hand, I =" -m,x,y'I=" -m,y,z, .xyL.J""yzL.J111 ii andsimilarly forIzx'which arecalled product of inertia. * Here,equations(2),(3)and (4)can be arranged together as, * Wewillseeinthenextsectionof thischapter,howcanthisbeobtainedinseveral symmetricalcase. 74TheClassical Mechanics or,LIro. This isangular momentum for rigid body rotation. Here,Iiscalled moment of inertiatensor*of rank-2.Thishas9components,3of whichareprincipal moment of inertia and remaining 6are product of inertias.This2nd rank tensor (IxxIxyIxz) isgiven by,I=IyxIwIyz IzxIzyIzz 3.8KineticEnergy forRigidBodyRotation: Whenrigidbodyrotationaboutagivenaxisistaken,itskineticenergyis thesumof kineticenergiesof allpointparticles.Thiskineticenergyisgiven by =Lx(roI =Lmj {(r/) ( ro;+ ro;+ w;) - (Xjro x+ yjro y+ Zjro z ) 2 } I "" 1{2 (222)(222222 =L.. - mjrirox + ro y + ro z- Xjro x + Yjro y + Zjro z+ XjYjro xro y + .2 I + YjXjro yro x + YjZjro yro z + ZjYjro zro y+ XjZjro xro z + ZjXjro zro x)} So,we get,thekineticenergyforrotating body "" 1(22)2"" 1(22)2"" 1(22)2 T=L.. - mjri- Xjro x+ L.. - mjrj- Yjro y+ L.. - mjrj- Zjro z+ .2.2.2 III * Thecharacteristicsof thistensorwillbediscussedinnextsection. Rotational Motion:Rigid Body Rotation75 where,lxx'Iyy'Izz are principal moment of inertiasand Ixy'Iyz'...areproduct of inertias. 3.9AxestheoremforMoment ~ Inertia: Themoment of inertiaforrotationof rigid body about anyaxisparallelor perpendicular totheaxispassingthroughcenter of gravity,canbeobtained by twoaxes theorem, which are (i)Perpendicular Axis theorem (ii)Parallel Axis theorem Let us discuss this twotheorems. (i)PerpendicularAxistheorem: (a)Fortwodimensionalbody:Foratwo dimensionallaminated body,if lx,Iyand Iz be the respective momentof inertia forrotation about x, yand z axis respectively, then by the perpendicular axistheorem. Iz =Ix+ Iy when thebody initially exist inxy plane Ix =Iy+ Iz when the body initially exist inyz plane Iy =Ix+ Iz when thebody initially exist is xz plane Because,Let the body exist in xy plane. Here,Ix= Emy2(Em= total mass) I= Emx2 and 1= Em? yz z y whenthepointPhasco-ordinates(x,y)andthemassissupposedtobe concentrated at P. Now since,? =x2 + y2 Iz = Em(x2 + y2)= Emx2 + Emy 1=1+1=1+1. zyxxy Thisisperpendicular axistheorem. 76TheClassical Mechanics (b)For three dimensional body: In this case, by perpendicular axis theorem, if I,I,I. beprincipalmomentof inertiaof athreedimensionalbody,when xy rotating about x, yand z axisrespectively,thenzp Ix+ Iv+ Iz =2Lm(Op)2. wheriPisimaginarypointatwhichtotalmass issupposed to be concentrated, and'0' isarbitrarily choosenorigin. Here,bydefinition, y It=Lm(y2+ z2). I=Lm(x2 + z2). Y I, =Lm(x2+ y2).0x Because,Ix==~ m P C 2 ) =Lm(AC2+ PA2)==Lm(y2+ z2)andsimilarlyfor Iv'Iz .Since,OP2==x2 + y2+ z2 ..Ix+ Iy+ Iz =2Lm(x2 + r + z2)=2Lm(OP)2. (ii)Parallel Axis theorem : Thistheoremstatesthatif ABandCDbetwoparallelaxessuchthatthe formeraxispassesthroughthecenterof gravityof thebodythenformoment of inertia lABand ICDabout ABand CDaxisrespectively, ICD=lAB+ Ma2(M =massof thebody).C Where,'a'isthenormalseparationof two parallel axis AB and CD. Here,forpoint Pon thebody,if ithasnormal distance x from the axis AB, then I=Lmx2 AB andICD=Lm(x + a)2 Now,ICD=Lm(x + a)2 =Lm(x2 + 2ax + a2) =Lm(x2) + 2aLmx + a2 Lm ICD==lAB+2aLmx +Ma2. But inrotationalequilibrium*,Lmx=O. So,ICD=lAB+ Ma2. This isparallelaxistheorem. D 3.10Calculation of Moment of Inertia in different cases: A a 8 (a)CircularRing:Consideracircularringof radius'a'andmassM.It is rotating about an axis passing through its center and perpendicular tothe plane of circle.If A be massper unit lengthof thering,then 'L.nlX *Here,wehavethepositionof centerof massx=='L.m.;Nowif theaxispasses throughthecenterof massthen x=O. For thatcase,'L.nlX=O. i.e.themomentof masswillbezero. Rotational Motion:Rigid Body Rotation M A =-21ta' Now,forasmallelementarysegment'dl', havingmassdm=Adl,themoment of inertiaiselI =dm.a2. So,forwholering,themomentof inertia*I= f dI=f dm.a2 =fA dl.a2 ='A.a2 f dl M2 =--.a .2n:a=Ma2 21ta I=Ma2 (b)Circular Disc: Here,we takea circular disc of mass'M'andradius'a'.Itisrotatingaboutan axispassingthoughcenterof itandnormaltothe planeof thedisc.If eJbethemassperunitareaof the disc, then eJ=Ml1ta2 Herethedisccanbe dividedintoalargeno.of co-axial circular rings,oneof which hasradius xand width dx. If dmbethemassof thiselementarydisc,then dm=21tXdx.eJ. w Sothe moment of inertia of that elementary ring about thesaid axis,is elI= dm.x2 = 21tXdx.eJ.x2 = 21teJ.x3dx. 77 So,themoment of inertiaforthewholediscabouttheaxispassing though itscenter and normal toitsplane is I=felI=1: 21teJ.x3dx a1 1M412 =21teJ.- =2-1t--2a=- Ma 441ta2 12 1=-Ma 2 Thisisthemoment of inertiaof circular disc,rotating about thegivenaxis. (c)HollowSphere: Herewetaketherotation of ahollowsphereof mass * Here,if wewanttofindoutthemomentof inertiaof theringaboutitsdiameter thentakingtheringonxyplane,wehavefromperpendicularaxistheorem,Iz =Ix ~ .Butforsymmetryof rotationaboutxor yaxis 1=1.1=.!..I=.!..Ma2 xy'"x2z2 78 Mandradius'a',aboutitsanydiametric axis.Themassper unit area of thesphere M is0"=-42If thisspherebedivided 7ta intoalargeno.of co-axialcircularrings. One of which has radius asin e and width ad8,themassof thiselementary ring is dm= 27ta sin 8.and 8.a = 27ta2asin 8 d8. TheClassical Mechanics Now,themoment of inertia of thiselementary ring isdI= dm.(asin 8)2 = 27ta2.a=sin8 dv.a2 sn28 dI= 27ta4a.sin38 d8. So,themoment of inertia of thewholesphereisnowgiven by a I=fdI=27ta40".rersin2 8 d8 J9 =0 =27ta4a. Jolt (1- cos2 8) sine de 4 f+l(2) =27taa-I1- zdz(when, z = cos9) _4M (1)4M 4 - 27ta.--2.21- - =27ta.--2 .-. 47ta347ta3 22 1= -Ma 3 Thisistherequired result. (d)SolidSphere:Forasolidsphereof mass M,radius'a', and densityp =1i7ta3 . Ifwetaketherotationaboutitsany diametricaxis,thendividingthewholesphere intoalargeno.of co-axialcircular disc,weget an elementary disc of mass dm = 7tYdx.p. when, y= radiusof elementary disc,anddxisitswidth. Nowfromthefigure, y2=(a2 - x2)where,xisnormalpositionof theelementarydiscw.r.t.the center of sphere.So,weget, dm=7t(a2 - x2)dx.p. Now,themoment of inertia of thisdisc about the said axisis Rotational Motion:Rigid/Body Rotation 121(22) dI= 2,.dm.y = 2,dma-x. 1(22)2 =2,.7ta-xp.dx. Sothe moment of inertiaof thewhole sphere,is f 1f+a(22)2 I=dI=2,7tP-aa- xdx [ ]+a 142X3 x5 I- -7tp.ax-2a.-+-- 235 -a (15 -10 + 3)53M I=15tta. 47ta3 22 1=-Ma 5 Thisisthe required result. (e)RectangularLamina:WetakeaY rectangular laminated disc of length 'a' and breadth 'b'.SoifM be itsmass,thenmass M per unit length cr=db. If we now want to calculate its moment of inertiaaboutanyof itstwoadjacent sides,then placing it in xy plane with xand x yaxis along itstwo adjacent sides,(a Let us now findout itsmoment of inertia about y-axis. 79 1 b ) x Taking a rectangular narrow strip of massdm=bdxcr,its moment of inertia dl=dm.x2 =bx2dxcr. Sothemoment of inertiaof thewhole foilabout y-axis,is Iy=fdI= f;bx2.dx.cr.= bcria3. 1M312 I= -b.-.a= -Ma y3ab3 80TheClassical Mechanics Similarly,themoment of inertia*of thesamefoilabout x-axisis I=1Mb2 x3 (j)EllipticFoil:Consideran elliptic foil,having semimajor axis'a' andsemimirror axis'b' and massM. Wecansimplyfindoutthemoment of inertiaof thisfoilaboutanyof its axes.If crbethemassperunitarea of thefoil,then M cr= nab P(x,y) x Letusnowfindoutmomentof inertiaof thisfoilaboutitsminoraxis, alongwhichy-axisistaken.If wenowtakearectangularnarrowstrip,ata distance xfrom yaxiscorresponding totheterminal point P(x, y) on theellipse, then mass of thisstripisdm=(2y.dx).cr. g2bl22 But forellipse, y=b1 - 2"=- '\ja- x aa 2bcr 22 dm=-- a- x.dx a Now themoment of inertiaof thislinear stripabout minor axis,is 2bcrx2 22 dl = dm.x2 =a- x.dx a Sothemoment of inertia of thewholefoilabout thatsaid axisis Let, for f 2bcrf+a2I22 1=dl=--x'\ja-xdx a-a X= asin 8.:.dx = acose de. x=a,sine =l,e=nI2. x= -a sine = -1, e= -nI2. 2bcrJ+1t/2 I=--.a2 sin2 e.acose.acose de a-1t/2 * Here,if wetaketherotationof foilabout z-axis,thenI, + Iy= M( a2 + b2) (Byperpendicularaxistheorem) Here,if wewanttofindmomentof inertiaof thesamefoilaboutaxispassing thoughcenterof foilandparalleltoz-axis,weget, a2 1(22)12 I=I=I+M- =>I=-M a+b--Ma zZ434 Rotational Motion:Rigid Body Rotation =2ber.a3 -.!..f+1t/2(4sin2 8 cos2 8)d8 4-1t12 11 f+1t/2 I=- bera3 . - (1- cos 8 48) d8 22-1t/2 (I).= =-.!..Ma2 mmor4nab4 Thisismoment of inertiaof thisellipticfoilaboutitsminor axis. Similarly,themoment of inertiaof thisfoilaboutitsmajor axis*,is =-.!..Mb2 (I)major4 (g)RightCircularCylinder:(A)About itsownaxis: Here,if we divide the whole cylinder into a large no.of circular disc, one of which has mass'm' when M =Lm =mass of whole cylinder, then,themomentof inertiaof thewholecylinderabout its own axis is ,,12I2 I=LJ 2. rna=2. Ma. (B)Aboutanaxis perpendiculartoitsownaxisand passing throughcenter o/gravity o/the cylinder: Ac 81 Here,wetakeanaxisABwhichpasses thoughthecenter of gravityof thecylinder. Ifwenowconsiderasmallelementary circulardiscatadistancexfromthecenter of gravityandparalleltotheaxisAB,then foranother parallel axisCD,passing through thecenterof gravityof thiselementary disc, \ I I 1 1 I I 1\ta J.l____ j)__ x B 1 I I themoment of inertia of thisdiscabout axisCD,isdIeD=dm.a2 . ** When,dmisthemassof theelementarydisc.and'a'isradiusof the cylinder. * Similarly,fromperpendicularaxistheorem,if zaxisbetakenperpendiculartothe foil,passingthoughthecenterof foil,then I22 I - I+ I- - M(a+ b) majormmor4 **Itisfollowedfromperpendicularaxistheorem. 82TheClassical Mechanics M Nowif P =--2- bethedensity of thecylinder(where,ListotalLength 7taL of cylinder)then,dm=7ta2.dx. p. 122 dlco ="41tapdx.a =1.1ta4pdx 4. Now by parallel axistheorem,the moment of inertia of this elementary disc about AB axis is dIAB =dIco + dm.x2 1422 dI=-1ta.pdx + 1tapdx.x AB4 So,themoment of inertia of the whole cylinder about thesaid axisis I=JdI=1.1tpa4 J+L/2dx +1ta2p J+L/2 x2dx ABAB4-L!2-L!2 1421[L3) I=-1tpa.L + 1tap.-.-AB434. 1M4123M =-7t--a L + -7ta L.--47ta2L121ta2L 1212 I=-Ma+-ML AB412 I=1. M (a2 +.!. L2) AB43 Thisisthemomentof inertiaof arightcircular cylinder about an axis passing through itscenter of gravity and normal totheaxisof cylinder. (h)For a Right Circular Cone about itsown axis: Wenowconsiderarightcircularconeof massM,base radius aand height M3M p=-.!. 7ta2L- 1ta2L 3 Herewetakeanelementarycirculardiscof radius 'y',atadistancexfromthevertexof thecone,then momentof inertiaof thatelementarydiscaboutaxisof cone is Rotational Motion" Rigid Body Rotation 1z dI=-dm.y 2 where,dmismass of the elementary disc and dm=(nYdx).p. But 1zz dI=-7tPYdx.y 2 14 =27tPYdx. xL =-::::} Ya 1a4 4 dI=-np.-.x dx 2L4. Sothe moment of inertia of the wholecone about itsown axisis f 1a4 lL I=dl=-np.-.x4dx 2L40 1a4 L5 143M =-np.-.- =-na L.--z-2L4510naL 3z 1= -.Ma 10 Thisis therequired result. 3.11Momental Ellipsoidor Ellipsoidof Inertia: Letusnowconsidertherotationof any rigidbodyaboutanyaxis'OA'having direction cosines(t,m,n). If Pbeapoint withinthebody atwhich themassof thebodyissupposedtobe concentrated,thenforperpendicularPM, drawnfromPontheaxisOA,weget,Op2 =x2+ y2+ z2. and,OM=Ix my + nz. where, (x, y, z)are coordinates ofP. Now fromthefig,themomentof inertiax of thisrotating body forrotationaboutaxisOA,is lOA=I =:Lm'(PM)2.where,:Lm' =Totalmass 1= :Lm' (OP2- OM2) =:Lm'[(xl + Y + z2)- (Ix+ my + nzf] z 83 y 84TheClassical Mechanics 1= Lm'[x2(1- Z2)+ y(1- m2) + z2(1- n2) -21m xy - 2mnyz - 2/nxz] =Lm'[x2(m2 + n2) + y(n2 + [2)+ z2([2+ m2) -2 Imxy- 2mnyz-2nlxz] =Lm'(I + z2)[2+ Lm'(x2 + z2)m2 + Lm'(x2' + y)n2 + 2 1mL(-mxy) + 2mnL(-myz) + 2nl'ir--m'zx I =Iz2+ Im2 + In2 + 21mI+ 2 mnI+ 2nl I. xxyyzz'xyyzzx Sinceforseveralaxis,(t,m,n)arevariables,soletsuchaxisforwhich, body hasconstant moment of inertiaforrotation,I = X,m= Y,n = Z We now get I =IX2+ Iy2 + IZ2+ 21XY+ 21YZ + 21ZX xxyyzzxyyzzx :::::>Ax2 + By + Cz2+ Dxy + Eyz + Fzx=1 IxxIyy When,A=-1-' B =-1- ... 1 Izz 2Ixy2Iyz2Izx C=-I ,D=-I- E=-I-F=-I-' and all of theseareconstants. Now theequations (1)isan equation of anellipsoid, * whichisanellipsoid of inertiaormomentalellipsoid.Moreclearly,wecansaythatif severalaxes of rotationarepossibletorigidbodyrotation,forwhichmomentof inertiaof the body remainsinvariant, then in this condition, the locus of all direction cosines forseveral axes,will be ellipsoid. Thiselliposid isknown asellipsoid of inertia. 3.13Moment andproduct of Inertiaand Ellipsoidof inertia of some symmetrical bodies: Wewillnowfindout principalmomentof inertiaand productofinertiaandalso ellipsoidof inertiaforsome symmetrical bodies. (i)AThinRectangular Foil:Herewetakearectangular foilorlaminain xy planehaving length a,breadth band mass per z M unit axiscr=ab' ~ - - - - a- - - - ~ )* Wehavethegeneralequationof theellipse, x2 i2xy -+-+-cosS = sin28 =>A'x2 + BY + C'xy= D' a2 b2 ab Similarly,forellipsoid,theequationshouldbe, A'x2 + BY + C'z2+ D'xy +Eyz + F'zx+ a'. =>Ax2 + By + Cz2+ Dxy + Eyz+ Fzx=1. Thisisequationof anellipsoid. Rotational Motion:Rigid BodyRotation Assuming uniformmassdistribution, Wehave,theprincipal moment of inertia Ixx=Imdyl + zl)=Sf (l + z2) dm i =Sf (l + z2) crdxdy= Sf lcrdxdy(.,'z = 0) J+a12 2J+b/2Ma2 I=cr.ydydx= --xx-a12-b/212 Similarly, Mb2 M(22) 1=--.1 =-a +b Y.Y12zz12 (By perpendicular axistheorem). Alsotheproduct of inertia IX)!= IyX= I-mjXjYi= - JJ crxy dxdy i f+b/2f+al2 =-cr.xdxydy= -b/2-a12 and some z = 0,Iyz = Izy= 0, I=I=0. zxxz 85 Sofor this thin rectangular foilwith'0' at the centre, the ellipsoid of inertia isgiven by, Ix2+Iy2+Iz2=1 xxY.Yzz +b2)z2=1 =>121212 =>= [a2x2 + b2l+ (a2 + b2)z2]=1 which istheequation of a spheroid. (ii)Circular disc:Here,weconsider a uniform circular disc of radius' a', massper unitareacr=M! 1ta2. Assuminguniformmassdistributionandusing cylinderical co-ordinates x=rcos e, y=rsine,z= z and taking any elementary area dA =dxdy=dr.rde which has mass dm=rdrde.cr We may write Izz =Sf (x2 + l) rdrdecr=cr f: r3 drfo;;e a4 Ma2 =cr.-.21t= --42 86TheClassical Mechanics which isour expected result Also, a4 Ma2 =(j-'lt =-- =1 44y.y We can alsoshow that,in this case of circular disc,Ixy= Iyz = Izx= O. Soellipsoid of inertia isIxx x2 + Iyyy2+ Izz z2=1 Ma2 (22)Ma2 2 =>-- x+y+--z=1 =>x2+y2+zz2=4 a2 which is equation of an oblate spheroid. (iii)Thinsphericalshell: Wenowconsiderathinsphericalshellof radius'r', thicknessdr,mass per unit volume ',---------- ,-' --- -M p= 4'ltr2dr Assuminguniformmassdistributionand using spherical polar coordinates, We now have, x=r sin S cos S y=rsin S sin S z=rcos S. ---------\ ,- \".,,- r\\.......... \\, \I ,0I:".' .... __I1.--_..... -------------j--, ,I ,I II II /"./ V Izz =f(x2+ i)pdv=H(r sinS COSx = -wgtcos8 dt3 d2z12 z=-=-g =>z=zo --gt dt2 2 when,t = 0,z = zO'x =0, y=0istheinitial position,with initial velocities i( 0) = 0 = i( 0). Since,thetimeof fallfromheight 'h'is t =f 'thedeflectionof aparticletowardstheeastwhenitisdropped fromrestis 18h ( 3 )1/2 8="3wcos82 This is called 'eastward effect' due tothe effect of coriolis forceon a freely fallingbody.Asforexample,if theparticlebedroppedfromaheightof 100 104TheClassical Mechanics mfromrestatlatitudee =45,itwillbedeflectedbyabout1.55x1O-2m toward east in northen hemisphere. 4.6Principleof Foucault'sPendulum The French scientist Leon Foucault showed that thesmall effect of coriolis forcemay be remarkable fortheoscillation of apendulum.Henoticedthat the rightangledcoriolisdeflectionononeswingof thependulumcouldnotbe undoneinthereturnswing.Theeffectof corio lisforceof terrestrialorigin moved from the domain of theory and outdoor observations tothat of observation inalaboratoryexperiment. Theequation of motionfora pendulum*including thecoriolistermis 2-dr2- ( -2 + kr=-2 Wxv) dt where, k = =stiffness constant for pendulum and' l' istheeffective lengthof pendulum. Now taking cartesian components of thisequation, weget x+k2x2(Ywz Foriand z negligible in compare toX,y and X,y . Thus the above equations are the coupled equations of motion in xand y. Tosolve theabove coupled equations, let us choose a complex variable u= x+ iy(i= .J=l) . Then,aftersomealgebraicmanipulation,wefinallyget, or,it + 2iwzu+ eu0 Thegeneral solution of thisequation isnow givenby, * Forpendulummotioninsimpleharmonicmannerunderrestoringforceonly,the equationof motionbecomes, d2- d2- k-- _r_ror m/=-k r=>m-=-kor =>-+-=0 odt2 dPm kogd2;:= 0d2;:2-But-;;; = Iforpendulumoscillation:.dt2 I=>dt2 + kr= 0whenno otherforceact Reference Frame where, , / ",;r"-, /\ /\ I\ I\ I\ I\ JJ II II II IJ IJ \, \, \I \I \/ \/ \, "" ~rotation of plane of oscillation In clock wise sense for Foucaults pendulum. y uexp(-iwzt)[c\eik't +C2e-ik't] ~ X +/. k' + .(.) -iw t u=Xly =Xo+ lyOeZ 105 x where,Xoand YoarethesolutionswhenCorio lisforceisabsentand u':::;k for k> > wz' Sofromthissolution,wecanconcludethattheplaneof oscillationof the pendulumrotateswithanangularvelocity-w), inoppositetothesenseof rotation of the earth. The period for a complete rotation of the plane of oscillation is 21t21t Twz =wsinElat latitudeEl. Soatthepoles,T24hr,andattheequator 106TheClassical Mechanics Sothatnorotationof theplaneof oscillationisobservedattheequator. Foucaulthaddemonstratedthetruthof theabovefactsthroughhishistoric pendulumexperimentin1851.Bymeasuringtheperiodof oscillation,T, Foucault measured theperiod of earth rotation.This wasthefirstexperimental proof that the earth isinfact rotating w.r.t.the inertial framein which Newton's Lawsarevalid,withanangularvelocitywhichispreciselythesameasthat inferred framethe apparent durinal rotation of thesun, moon and the star sphere. 4.7Flow of River onEarthSurface Basically, rivers flow approximately in a horizontal plane for which a slight downwardslopeinthedirectionof flowisrequiredtomaintainthespeedof theflow.Sothecomponentof gravitationalforcealongthedirectionof river flowisprimarilyimportantdependingonthemagnitudeof theslopeof the down stream. But there is absolutely nocomponent of geff(including the effect of centrifugalacceleration)actingalongthebreadthof ariver.Hence,the component of corio lisacceleration, however small,can act freely on the moving water acrossthedirection of thestream. Tofindtheeffectof suchtransversecoriolisaccelerationontheflowof river,letuschoosethedirectionof riverflowalong x-axis.y-axisistakenin transverse direction to x-axis on horizontal plane and z -axis in vertical direction. Theflowisnowconsidered at latitude'e' of earth surface,making anglee(in theanticloth wise sense)w.r.t.thegeographical north direction.Therefore,the earth'sangularvelocityvectorw w.r.t.thesaid referenceframeisnowgiven by, w=wsinek + wcose coscpi- wcose sincp} ..w=w(sinek+cos9coscpt -cos9sincp}).andthevelocityof flowis v= vialong x-direction.So thecoriolis acceleration is lie=-2(wxv)=-2vw(sin9}+cos9sincpk) Thus,a corio lisforcewill be experienced by thewater inthe riversflowing inanydirection,causingadirectiontowardstherightof theflowdirectionin thenorthenhemisphereandtotheleftof theflowdirectioninthesouthern hemisphere.Asaresult,thecorrespondingbanksof theriverwillbedenuded, which isactually observed. SUMMARY 1.Frameisacertainregionw.r.t.restormotionof particleorsystemof particles isconsidered. 2.Frame has two types which are inertial and noninertial frame.Inertial frame isatrest or inuniformmotionwhereasnoninertialframeisaccelerated frame. Reference Frame107 3.Pseudo force will appear in non inertial frame and in that case, the effective forceinnoninertialS '-frameis F'= F + (- Fp ).,F = Real or actualforce. - Fp= Pseudo forceappeared. 4.For rotation of the frame with angular velocityW,(;).=(;)+wX FIxedrot 5.Centrifugal forcedeveloped in rotatingframeis -w x(w x r). For earth's rotation, it reduces the effective gravitational acceleration resulting its effective value as geff=g-wx(wxr) 6.Coriolis force developed in a moving body w.r.t.the rotating frame on earth surface is-2m( wxv). 7.Themaximum valueof centrifugalaccelerationat equator is ~ = w2R= 3.4 x10-2 mls2 = 0.3%of g. 8.The maximum effect of transverse coriolis acceleration at any pole of earth is ac = 2w sin 90 v = 2w v =1.458XlO-4v. 9.Angular deflection of particle moving in time t at latitude eof earth surface due totransverse coriolis acceleration isa=wsin e.t 10.If any latitude eon northen hemisphere of earth, the components of angular speed 'w'of earth rotation are,Wx= Wy= Wcos e, Wz = w sin e. 11.The deflection of particle when dropped from certain height 'h'above earth 18h ( 3 )1/2 surface,duetocoriolisaccelerationis8= "3wcosegandthis deflection occurs towards east. 12.For Foucault's pendulum, the time period for each oscillation at any latitude eon earth surface isT=~ . wsine \Vorked Out Examples Example1: Anobjectisthrowndownwardwithinitialspeedvo.Prove that after time t,it isdeflected east of the vertical byanamount 2 wvot2 cose +-wgt3 cose 3 108TheClassical Mechanics Ans: Here, the effective component of angular velocityw for earth rotation along x,)',zaxisof the frameS' taken on earth surface,are Wx= 0,Wy= Wcos 9,Wz = Wsin 9 Now, for velocity in time t,if X,y, z be the components thenx =0,y =0, z =Vo+ gt. Again the instantaneous transverse coriolis acceleration x= x =d ~ = 12(w xv)1= 2wcos9( Vo+ gt) dt 2wvo cos 9 + 2wg cos 9t. wvot2 cose +w g t 3 cose 3 This isthedeflection towards east of vertical. Example2:Show that if theparticlemovesnear theearth's surface,then theequation of motion are given by x2wsin9y ji-2(wsin9 x + wcos9z) z- g + 2w cos 9 y forlatitude'9' on earthsurface. Ans:We have forany frameS'(xyz)on the earth surface, AAA W=wxi+wy}+wzk AAA =w) +wy}+wzk wx v=(-wsin9y)i +(wsinex+wcos9z)}-(wcoseY)k But the equation of motion isgiven by, d2r - =g-2(wx v) dt2 =-gu + 2wsine yi - 2(w sine x + xcose z)} + 2wcosey k x =2wsin9 y ji =-2(wsinex+wcosi) i=- g + 2w cos 9 y Example 3: An object isthrownvertically upward at latitude 9with speed va- Prove that when it returns,it will be at a distance west ward fromits starting 43cos9 point equal to3WVo --2-g Reference Frame Ans:Here,by followingEx.1. thedeflectionis x =WVocose t2 +.!.wgt3 cose. 3 But t is thetimeof ascending whichisactually the timeof fall. 93 so,Vo=gt...x="3wgt cose Vo Also,t =-, andthen, g 4v3 the totaldeflection x=-. wg-i} cos e 3g 43cose x=3WVO-2-' g 109 Example4:Anobjectattheequatoristhrownverticallyupwardwitha speed of 60mph.Howfarfromitsorginal position willitland? Ans:Here thedeflectionisgiven by 43cose x=3WVO-2-g Butw =7.29x10-5 rad/sec Vo=60 mph =2.682x103 cm/sec q =0at equator g=980 cm/sec x=l.952 Ri2cm. Example5:Howlongwouldittakefortheplaneof oscillationof a Foucault pendulum tomake one complete revolution if the pendulum islocated at aplane of colatitude 45. Ans:Herelatitudee=90 - 45=45. SoforFoucault's pendulum,thetime takenforone complete revolution of the plane of oscillation. T ==2n= 33.84 hr. w sine7.29 x 10-5 x sin45 Example6:Provethatif anobjectisthrowndownwardfromheighth withinitialvelocityVoabovetheearthsurfacethenitwillhittheearthata point east of the vertical at adistance.(atlatitude e) w;;;e (V + 2gh_Vo )2(v + 2gh+2vo). Ans:Herebyfollowingthesimilarproblemof Ex.1.wehavetheeast 110TheClassical Mechanics warddeflection x=WVoCOSe(2+ .!wg cose (3 3 But forinstant T,the downwardinstanteneousvelocity v =Vo+ gt 1 (= -(v- vo) g Butv2=v5+ 2gh 1 =~ [ ~ v J + 2gh- Va] and WCOSe12(3) X=gl +Vo 3 w; ~ e (v5+ 2gh- Vo r(vJ+ 2gh + 2vo ) Thisiseastward linear deflectionof that falling body. Example 7:A train havinga maximum speed equalmovingVoismoving roundacurvewithradiusof curvature'p'.Provethatif thereisnolateral deflection through out the outer tract, then the track should be at a height above theinner track isgiven byaV5/v + p2g2where'a'isthedistance between tracks. ADS:Here,if xbethehightdifferencebetweenouterandinnertrack, then xv2 tane=_=_0_ apletT ButfetT = effective accn.=~IJ + /f 2 Vo IN= normalaccn.=-p IT=Tangential accn;.= g EXERCISES 1.What doyou mean by a frame?Giveitssignificance. Reference Frame111 2.Discusstheappearance of Pseudoforceina non inertial accelerated frame intranslationalmotion. 3.Discuss theappearanceof Pseudoforceina noninertial rotating frame. 4.Discusstheeffectof transversecoriolisforceonaparticlewhenfalling froma certain height aboveearth surface. S.Discuss the effect of centrifugal acceleration on the gravitational acceleration of earth surface. 6.Discuss the appeamce of cyclone. 7.Discussthe effect of Coriolisforceon riverflow. 8.Di8cussthe priaciple of foucaultspendulum. 9.Showthattheangulardeviationof a plumb-lineframethroughverticalat a point on the earth surface at a latitude e is ro w2 sin e cos e 22where,r 0istheradiusof theearth. g-rowcose 10.Supposethatthemassmof aconicalpendulumof lengthImovesina I 1/4 horizontalcircle of radius a.Prove that (a)Tht speed isali(P - a2) mgl and(b)Thetension inthestring isI vP _a2 11.If anobjectisdroppedtotheearth'ssurfaceprovethatitspathisa semicubical parabola. 12.Howlongwouldittaketheplaneof oscillationof aFoucaultpendulum tomake one complete revolution if the pendulum is located at 45 Latitude or atthenorth pole. 13.Explainphysicallywhythetrueverticalandapparentverticalwould coincide at the equator and alsothenorth and south poles. -:0:-Chapter-S CentralForce 5.1Introduction Thediscussionof 'Centralforce'isveryrichestdiscussioninclassical mechanics,sincetheknowledgeaboutcentralforceisimportantfordealing with several inverse square force,specially, in case of plane tory motion.Several forces,like Newton's force of gravitation, Van' dar Waal' s forcebetween'atoms and moleculesina gas,or even theYukawaforcebetween thenucleousinthe nucleons of atoms are examples of central forces.The idea about it is in general, applicableindiscussionontides,dynamicalmanifulationsof theorbitsof spaceships inthespace age,thegeometry of theorbits of plannets... etc.In this chapter, wewill restrict ourself indiscussion of central forcecharacteristics and inseveralapplicationsof it.Basically,thechapter willgothrough thedepthof central forcephenomena. 5.2DefinitionandCharacteristicsof Centralforce Theforcefield,eitherattractiveorrepulsivewhichisalwaysinradial direction towardsor away from the center (origin)of theforce,iscalled central force. Mathematically, itisrepresented as FCentral=F= f(r).c. = f(r)r . r Where,'+'signindicatesrepulsiveforceand'-' signgivestheattractive natureof force,and theforceisalwaysdirected along(r)radialdirection. The basic characteristics of theforceare, (i)Theforcemayberepulsiveorattractive,whichisdirectedtowards or away fromthecentre of force. (ii)Thisforceobeysinversesquarelawof force,andthiskindof force fieldisderivablefroma scalar potentialfield,cp(r)such that 112 Central Force113 f(r)r = - vr8 + 2rre= 0 we have ;(rZe) o rZeconstant =h (say). Thisisanoutcomeforconservative centralforce. So,wehavefromequation(1) .f(r) ;: _re2 =--m 1(.2h2)f "2 m r+ ?=f(r )dr + constant ...(1) ... (2) Central Force =>~ m(,:2+ r2(2) + {-f j(r) dr}=constant =>E =constant. Thisgivesenergy conservation under centralforce. 5.5Equation of motionunderattractivecentral force 115 Forattractivecentralforce,wehavetheradialformof centralforce F(r) =- j(r)rwhich givestheequations of motionforparticlesmotionunder such central forceas m(r _re2) m(,B+ ue) - f(r) o Then equation (2)givestheconstant of motion r2e =Constant =h ...(1) ...(2) Buttosimplifytheformof 1slequation,letusnowmakeachangein 1 variableas,r= - . U Also, But Also, ,: dr dt 1. --u u2 drdue1 edu du . de .=--;;de ~ = h U 2r2 d(dr) _d(h dU) dtdtdtde Sowehave fromequation(1), - ~ ~ m 116TheClassical Mechanics This isequation of motion under attractivecentral force. 5.6Applicationof centralforcetheorytogravitation:Deductionof Keplar's law Actually,thegravitational forcebetween twomass pointsm]and m2 obeys inversesquarelawof forceandinthatcasetheforceisattractivecentralforce innature. Sothegravitational centralforceis F{r)= Gm]m2 where,f(r)=2 r Nowforplanetory motionundersuchattractivecentralforceforaplanet of massm arround thesun of mass M,theequation of motion will be d2u1 ~ --+u=+---de2 mh2u2 1 where,u =- ,(r, e) =position of planet r and 2'j 2 re=eu=h So,forplanetory motionundergravitationalforce,theequationof motion should be, d2u GMmu2 GM --+u+ mh2u2 =+--=+f.! de2 h2 GM where, f.! -2- = Constant h d2u' .. --+u' 0...(1) de2 foru'u - f.!. Wecannowapplythisequationof motion(1)todeduceKeplar'slawof planetorymotion. For planetory motion,Keplar gavethree basiclaws which are Central Force117 (i)Law-I:Everyplanetwillrotatearroundsuninellipticorbitwith sunat any of itsfocie. (ii)Law-II:Fororbitalmotionof planet,itsarealvelocitywillremain constant. (iii)Law-III:Forrevolutionof planetaroundsun,thesquareof time periodof revolutionwillbeproportionaltothecubeif semimajor axisof thatellipticorbit. We now obtainthesolution of equation(1), u'A cos e=>u - 11= A cos e.(A= Const.) u11+ A cose GM =>-2-+Acose rh r 1+(~ c o s eI=1 +Ecos e ...(2) r Thisispolarequationof theplanetorbitwhere'l'issemilatusrectum andEis eccentricity of theorbit. It isnow obvious that theeccentricity E= Ah2 -- --a .-- =--a h2 GMGM 4n2 T2= __ .a3 =>T2ex.a3 GM i.e.thesquareof timeperiodwillbeproportionaltocubeof semimajoraxis of the elliptic orbit.Thus Keplar' s3rd law isestablished. 5.7Energyconservation forplanetorymotion For planetory motion,thetotalenergy isnowgiven by EEk+ Ep But(m=mass of planet) Nowinpresenceof gravitationalforcefieldofferedbysuntotheplanet, the potential energy is Now considering we have, W = + S'GMm dr =_GMm r2r 00 E 1(.22e 2)GMm -mr+r---2r 1e r=- andr2e = - = h uu2 lIE U=- = - + - cos e rII Where, I issemi latusrectum of the planetory orbit andEiseccentricity. 1E. --r--sinee .. r2 I => r (.: r2e= h) I Central Force119 ( 22)1E2.2hGMm E- -m-h SIn8 + - ---- 2Pr2r { 2 2()2}() 1E'"h.221E1E E=-m--SIn 8+h-+-cos8-GMm-+-cos8 2PIIII ElE'"h.2 8h2EhcoshE2 8 =-m--SIn+-++--cos {2222822} 2PPPP GMm ---(1+ E cos8) I =-m--+-(1+2 Ecos8)---(1+ Ecos8) 1{E2h2 h2 }GMm 2PPI =-m- 1+E+--cos8------Ecos8 1{h2 (2)2h2E}GMmGMm 2PPII But taking our previousnotations GMm and =--2-GM =h h2 1 I = GM= ;- , we can writedown GMmE I and then we have,thetotalenergy of planet is mh2 (2)mh2 E mh2E2 E=-- 1+E'"+--cos8------cos 8 2PPIP mh2 (2) E=2Pl+E--1-1 But also, =I mh2 mh2 E= 2PP mh2 E= = constant (E 0 r S'l'b.~ ov(r)C'fI ~lllceat eqm 1num, lorceTr =entnugalorce ..(n+ 1) aT'= 12br-31 =>n+ 1)a= + 2b rn - 3 Now, 2b-(n+3) a= --r n+1 v(r) v-(n+3)n+12b_2 = --rr= --r n+ 1n+ 1 [OVeff ] -- -4-4( or2r=ro=6bro=2bro3 + n) Now foranycircular orbit withr =rothestability condition is o2Veff( v) 2>0 or n + 3>0 =>n > -3. and forcircular orbit n = -3. SUMMARY 1.Central forcehas thestandard formF(r) = j(r); . Central Force121 2.Thisforcecanbederivablefromascal:!rpotentialenergy = i.e.,F = - \7lj)(r) 3.Magnitudeof centralforce f( r)couldbeuniquelyrepresentedasapower inris, +C() f(r) =Iknr-n n=-C() 4.Angular momentum will be conserved for planner motion of particle under centralforce. 5.Only radialdependence of force preserves the isotropy of space. 6.Conservative nature of forcerestricts the time independence of the force. 7.Energy conservation holds for conservative central force. S.The equation of motion of particle under attractive central forceis d2u --+u = ---de2 mh2u2 12"j 2 where,u = -, h = re = e u. r 9.The equation of planetory orbit under central forceis I - = 1 + ecos e r h2 Ah2 where,I = --, e = --; A= constant GMGM ande< 1 10.The areal velocity of planet under gravitational central forceis I2' V - - re = h/2 = constant A- 2. 11.For planetorymotion,T2aa3,T= Timeperiod.a= Semimajoraxisof planetory ellipticorbit mh2 12.Totalenergyof planetE=- -2-(1- e2)=const ( - ve). 21 13.For stability of orbit under effective central forcepotential a 1"'+1,n > -3 and forcircular orbit, n=-3. Worked out Examples e 1.Aparticledescribestheparabolar = asec2 '2suchthatthecrossradial 122TheClassical Mechanics d2r component of the velocity isconstant.Show that-2 isconstant. dt Ans:Hereforthecrossradialcomponent,r2e = constant = k (say) But .. o r=asec2-2 i-F d2r dt2 o00.1 a.2 sec- sec - tan -0.-2222 asec2 =r8 222 k2 2 dr= dt2 222a k2 - =constant 2a 2.Aparticledescribesacircularorbitgivenbyr=2acos0underthe influenceof anattractivecentralforcedirectedtowardsapointonthecircle. Show that theforcevaries astheinverse fifthpower of the distance. ADS:Here the equation of the circle r=2acos 0 1secO Putting r =- we haveu =--u2a dusecOtanO dO2a d2usec20+secOtan20 d02 2a Central Force f(r)= ('1 :,fr)as r i. e.,theforcevariesastheinverse fifthpower of thedistance. 3. A particle of massm movesina centralforcefieldbyF:;,;-r r4 123 Show that if Ebe the totalenergy supplied totheparticle,thenitsspeedis givenby 1 v= (_k + 2E)2 mr2m ADs:Here,F IFI

r3 v(r) 2r2 Sotheenergy will beEEk+ vCr) E v= .!. m(r2 + r2{)2) _ 22r2 12k -my--22r2 I e:+ m:2Y So the centralforc.eisa conservative force. ADs: We know that if a forceF be conservative,thenV x F = 0 or, fF.dr=0 c Now forcentralforce,Ff(r)!'" r V xFvx(V x[ (xi + yJ +Zk)] 124 But TheClassical Mechanics ijk 000 -oxByoz x --j(r) r l.. f(r) r :.. f(r) r = of _l.. of) +of _:.. 8f)+k(l.. of 81) rByrozrozroxroxrBy ofof orxof - =---=--oxoroxror of_l.. ofof =:.. of By- ror'ozror VXF==i 1. of (zy - yz) + j 1. of (xz - zx) + k1. of (yx - xy) r orrorr or =0 :.The central forceisconservative EXERCISES 1.What is central force?Give its characteristics. 2.Consider themotionof aparticleunder centralforce_.!.. Show thatits 2 areal velocity isconstant.r 3.Find thedifferentialequation of orbitformotion of particle under central force. 4.Findtheequivalentonedimensionalproblemforthemotionof particle under centralforce. S.Discuss Keplar's problem under inverse square law of force. 6.Discuss thecondition forstability of orbit. 7.A particle of mass m moves under the action of central force whose potential v(r)=kmrZ(r> 0),then, (i)For what energy and angular momentum,will theorbit become circle of radius'a'about theorigin. (ii)Calculate theperiod forcircular motion. (iii)If theproblembeslightlydisturbedfromcircular.motionwhatwill be the period for small radialoscillation about r=a. 8.Show that thegravitationalforceisconservative. 9.AparticlemovesinaplaneundertheforceF(r)=ar,whereristhe distancefromthecenter of force.Find the path of theparticle. -:0:-Chapter-6 TheoryofCollision 6.1Introduction Collisionisaprocessthroughwhichtwomovingparticleoronemoving andonerestparticlecomesincontactandalsotransfertheirmomentum mutually keeping total momentum conserve. For twoelastic bodies, this collision iscommonlyknownas'impact'.Since, u ~ i n g collisionbetweentwobodies, noexternalforceactsonthesystemfromoutside,thetotalmomentumof the system willremainconservethrough theimpact. Basically, during collision or impact between two elastic bodies, the property of elasticityplaysitsroleandthusthetime*duringwhichthedeformation (slight)inshapetakesplacethroughcollision,iscalled'resistitution time'and within thistime theshapeisrestored. During collision between twoobjects, say twospherical bodies,they exerts mutualforcestoeach other along thecommonnormaltotheir point of contact andthiscommonnormaliscalled'lineof impact'.Thislineistheline joining thecentersof twoimpingingspheres.Forsuchlineof impact,wecanclassify wheather thecollision isdirect or oblique collision because, collision takes place suchthatthedirectionof motionof bothcollidingobjectsbeforeandafter collisionbealongthelineof impact,thenitisdirectcollision.Otherwise,for obliquecollision,thedirectionof motionisnotalongthelineof impact.But foranycollision,particlecomesincloserandtheyexchangetheirmomentum with each other keeping totalmomentum constant. 6.2Characteristicsof Collision Newton's Collision Rule Any typeof collision hasthefollowing basic characterswhich are-(j)Momentum Conservationtakesplace during collision. (ii)Kineticenergy(total)maybeconservedorenergylossmayoccur duringcollision. 125 126TheClassical Mechanics (iii)Momentumtransfertakesplaceduringcollisionandthroughsuch transfer impulsiveaction takesplace during collision. (iv)Collisionmaybeperfectlyelastic,orperfectlyinelasticenergy conservationtakesplacealongwithmomentumconservation.For perfectlysemielasticcollision,kineticenergylossoccursandthe particlemovestogetheraftercollison,andforsemielasticcollision, although energy loss takes place, both particles moves separately after collision. Duringanycollision,either directorobliquecollision,Newton'scollision ruleisthatthemagnitudeof relativevelocityof twocollidingobjectafter collision will beproportional tothat beforecollision. Thisconstantof proportionalityiscalledcoefficientof resistitution*(e) andmathematically itisgiven by where,UIand U2arerespectivevelocitiesof twoparticlesbeforecollisionand VIand V2arethat after collision. Here,by conservation principle, e 1 and,fore =1,collisionissaid tobe elastic,fore =0,collision isinelastic. and,for < e u2'v2 > vi'wehavefromNewton's collision rule, v2 - v\=e(u\- u2)...(2) Multiplying equation(2)by m2 andthensubstracting from(1)weget, m\v\ + m2 v\m\u\+ m2u2 - em2u\+ em2u2 =>v\(m\+ m2)=u\(m\ - em2)+ m2u2 (1+ e) 128TheClassical Mechanics v= I 1[(ml-em2)ul+(1 + e)m2u2] ml+m2 ...(3) Similarly,multiplying equation(2)ml and thenadding with(1),weget, (ml + m2)v2 mlul(1+ e)+ (m2- eml)u2 v2 =1[(m2- emdu2 + (1 + e)mlud nil+m2. ...(4) Thistwoequations(3)and(4)respectivelygivesthevelocitiesVIandv2 of twoimpinging particlesafter directimpact. Forthisgeneraldirectimpact,if wewanttofindoutthelossof kinetic energy duetoimpact,then We have ~ =lossinK.E ~ -[(mIVI + m2v2)2+ m\m2(vl- V2)2J} 11[22 2J ()mjm2(uIU2)-m\m2(UI-u2)e 2m\+m2 [Useing equation (2)and eauation(1)1 1mlm2(2)()2 -'l-eU\-U2 2ml+m2 ...(5) ThisisthegeneralexpressionforlossinK.Efordirectcollisionandwe should have, for perfectly in elasticcollision,e =0 nllUI+ f1'12U21mlm2()2 V== V2,andloss,~ =-- --- Uj- u2andthistwo Inil+ m22nil+ nil Theory of Collision129 particleswillmovetogetheraftercollisionalso,forperfectlyelasticcollision, e=l then, VI(1)[(ml-m2)ul+2m2u2] m1 +m2 v2 =(1) [2m)u)+ (m2- mdu2] m)+mz and then theloss,ilE =O. i.e.forelasticcollision,particleswillmoveseparately along thesameline of impact after collision without any lossin kineticenergy. 6.5Characteristic of Direct Collision We havefrom our previous discussion that fordirect collision between two bodiesmovinginthesamedirectionbeforeand aftercollisionalongtheirline of impact, the velocities after collision. V= I V= 2 Wheresymbolshastheir usualmeaning andalso,thelossinK.Eforsuch collision.ilE=..!..m)mZ(l-eZ)(u)-uZ)2.(O:S;e:S;1) 2m)+mz With this results, we can have the following characteristics for direct impact, which are, (i)Form)=m2,VI=u2'v2 =ul fore=1i.e.forelasticcollision betweentwobodieshavingsamemass,thevelocityexchangetakes place between them. (ii)For elastic collision,e =1,ilE =0 i.e.nolossin K.E occurs forelastic collision between two particles. (iii)Forelasticcollision,e=1,if u2 =0i.e.if 2nd particleisatrest initially,then V= I 1- mz m)-mzm) u)=---u) m)+mz1+ mz m) (a)form)= m2 VI= 0,v2 =u) 2m)u)2u) andVz= ---'-...!....- = ---m)+mzl+_m_z m) i.e.the1stbody isstopped and the2nd onetakesoff with thevelocity whichthe1stoneoriginally had.Both the momentum and thekinetic 130 TheClassical Mechanics energy of thefirstarecompletely transferredtothe2nd. (b)form2 mp VI::::;- ul and v2 ::::;o. i.e.whenalightbody collideswithamuchheavier body at rest,the velocity of light body isapproximately reversed and the heavier body remainsapproximatelyatrest.viz,aballdroppedonearthrebounds with reverse velocity when the collision iselastic. (c)form2 mp vI::::;ul and u2 ::::;2uI i.e.whenheavybodycollideelasticallywhichremainspractically unchanged,but thelight body rebounds approximately with twice the velocityof theheavybody.By theseconsiderations,wecan explain the'slow down'of thefast nutronsin anuclear reactor, taking proton of hydrogeneousmaterial(paraffin)has nearly thesame massas that of aneutron inagood moderator. 6.6MaximumEnergy transferduetoheadonelasticcollision Consider a particle of mass ml which collides with another mass m2,initially at rest.For such head on collision,if theinitial velocity ul of massml changes tothevelocity VIthen,foru2 =o. IS We have, andthefractionallossinkineticenergyof themassml forsuchcollision Now let L\E E 4x Herethetransferinenergy willbe maximumif I2 Ef ="2mlVI=0 isL\E=I4x=I E=>(l+x)2 Theory of Collision131 Thisisthecondition formaximum energy transfer forheadon collision. 6.7ObliqueCollision Here we consider the oblique collisionof twomassesml and m2 withinitialvelocities UIand U2beforecollisionat resp.angles(IIand(I2withline of impact and the velocities after collisionarevIand v2withthe samelineof impact. Nowfrommomentumconservationalongthehorizontallineof impact, weget, mlul cos (II+ m2u2 cos (I2=mivi cosPI+ m2v2 cosP2 ...(1) Butsincethetransversecomponentsof momentaremainsunchanged,we have, mlul sin (II= mivi sinPI;m2u2 sin (I2= m2v2 sinP2 Also,fromNewton's collision rule, (v2cosP2 - vIcosPI) =e(ul cos (II- u2 cos (I2) By solving these equations, wenow get, (ml- em2)uI COS(II+ m2(1 + e)u2COS(I2 vIcos PI=(ml+ m2) (ml- em2)uI COS(I2+ m2(1 + e)u2 COS(I2 or, vI(ml+ +m2)cosPI and, (m2- emdu2 COS(I2+ ml(l+e)uI COS(II v= 2(ml+ m2)cosP2 ...(2) ...(3) ...(4) ...(5) Sofor such oblique impact,if we want toobtain thelossof kinetic energy, we get 1 [{2(2.2)2(2.2)} =2"mlulcos(II+ sm(II+ mlu2cos(I2+ sm(I2 {ml V?( cos2 PI+ sin2 PI) + m 2 V ~ cos2 P2+ sin2 P2)} ] 1 [2222(22 A22 A)] ~ E =2"mlulcos(II+ m2u2 cos(I2- mivicos1-'1+ m2v2 cos1-'2 (Using equation(2)) 132TheClassical Mechanics 1[222222 =()mlulcosUI+m2u2 cosu2 2ml+m2 +mlm2(ufcoS2UI +uiCOS2(2)+mlm2(vf cos2 +vi cos2 = = E=(mlm2 )(1-e2)(UICOSUI-U2COSU2)2 2ml +m2 This isthe energy loss for oblique impact. So for perfect inelastic oblique collision,thislosswill be 1mlm2()2 ilE=UICOSUI-U2COSU2(fore =0) 2ml+m2 SUMMARY 1.For anycollision, momentum conservation takesplace. ...(6) 2.Theratioof magnitudeof relativevelocityaftercollisiontothatbefore collision iscalled coefficient of resistitution (e) which is1 or 0 or 0 < e < 1 according as the collision is elastic, inelastic or sernielastic. 3.For anycollision between twoparticleshaving initialvelocitiesiiiand ii2 before collision andvIand v2after collision, liil -ii21= IV2- vII 4.For direct collision,we have frommomentum conservation mliil + m2ii2= ml VI+ m2 v2. 5.The velocities of particles after direct collision;are Theory of Collision 6.The lossin kinetic energy fordirect collision, 1mjm2(2)()2 llE=1- eUj- U2. 2mj+m2 7.For u2 = 0,Vj= 0,v2 = uj forml = m2 vI~ -uI, v2 = form2 mj vI~ upVz 2uI for m2 ml 8.For Uz = 0,themaximum energytransfer takes place for ml = m2 9.For oblique collision, mlul cosal + m2uZ cosaz = mivi cos~ I + m2 v2 cos~ z(u2 cos~ 2 - vIcos~ I = e(ul cosal - Uz cosaz) when, (mj- em2)Uj cosaj - U2cosa2 vIcos~ I =(mj+ m2) (m2- emdU2 cosa2 + (1+ e)mjUj cosaj V2cos~ 2 =(mj+ m2) Energy lossfor oblique collision llE = Mmjm2/(mj + m2)](1- e2)(u2 cosa2 - Ujcosaj)2 Worked Out Examples 133 1.Twomasses 5gm and10gm approachestoeach other with respectively velocity 5rnIsand10rnIsec.They makeshead oncollisionandafter collision, the1stparticlecomesatrest.Whatwillbethevelocityof 2nd particleafter collision. ADS:Here,given that ml = 5gm, m2 =10gm ul = 5 rnIsecu2 = -10 rnIsec After collision,Vj= sofrommomentum conservation mjul + m2v2 = mivi + m2v2 =>5x5 - 10x10=5x+ 10v2 75 ..vI=-10= -7.5rnIsec. So,aftercollisionthe2nd particlemoveswithvelocity7.5rnIsecinits originaldirection. 2. A particle collides elastically with another particle of same mass initially at rest.So that after collision,theymovesat right anglestoeach other. 134 Ans:Let both particles has the same mass'm'. Here,ul = U,u2 = 0,e= 1 Sofrommomentum conservation, mumVI+mv2 =:}UvI+ V2 The Classical Mechanics But,if aftercollision,thistwoparticlesbescatteredatanangleawith each other, then I _IV2+ vll e- 1 == -'-----------'--UU 2+22-u2 vIv2- vlv2 cosa-Also,forelastic collision,we have fromenergy conServation I2 -mu 2 I2I2 -mvl+-mv2 22 ...(1) v?+ u2 ...(2) Sofromthistwoequations,u2 - 2vI v2 cosa=u2 2vlv2 cos a= or,cos a= 0,a= '!t/2 i.e.they will moveat rightangle toeach other after collision. 3.Showthatif twoequalmassescollideselastically,theywillexchange velocities. Ans:Let,both hassame mass mandthey has respectivevelocities ul and u2 before collision and has velocities vIand v2 after collision. sofrommomentum conservation, mUI + mU2 =mVI + mV2 ul + u2 =vI+ v2 Again for elasticcollision,e =I . .v2 - vI=ul - v2 By solving thistwoequations(1)and (2),we get vI=u2'v2 =ul Sothey exchange velocity throughcollision. ...(1) ...(2) 4.Show that for perfectly elastic collision of particles, the total kinetic energy willremain conserved. Ans:Letml andm2 aretwomassesandul andu2 aretherespective velocities before collision and VIand v2 are that after collision. Soforelasticcollision,(taken tobedirect or head on collision) v2 - vI=ul - u2 ...(2)(,'e =1) andfrommomentum conservation, ...(2) Theory of Collision we can have, (ml- m2 )UI+ 2m2u2 ml+m2 v= 2 ml+m2 sothetotal kineticenergy after collision, 1212 E=-mivi+-m2v2 'f22 1212 =-mlul+-m2u2 22 = Ej = Total kineticenergy beforecollision. 135 5.Consider thedirectcollision of twoparticlesof massml andm2 having initial velocities U Iand u2 before collision. If e be the coefficient of resistitution, thenshowthatthemomentumwhichistransferredfromthe1stparticletothe 2nd particle is mlm2(l+e)(ul-U2) ml+m2 Ans: For thisdirect collision, mlul + m2u2 =mlv}+ m2v2 ...(1) andv2 - VI=e(ul - u2)...(2) Where VIand v2 are their respective velocities after collision.Sothe change in momentum duetosuch impact is I'!.p= m2v2 - m2u2 = m2(v2 - u2) [Because by solving equation (1)and (2),weget m}(l+e)(m2-eml) v=u\+U2 2(m\+ m2)(ml + m2) (m\-em2)m2(1+e) v=~ U2 I(m\+ m2)(ml + m2) So,thismomentum transfer 136TheClassical Mechanics :.f'o..p=mlm2(1 + e)(ul- U2) ml+m2 6.AballwhichisdroppedfromaheightHontoafloorreboundstoa height h < H.Determine thecoefficient of resistitution Ans:If theparticleafterfallingfromheightH,collideswiththefloor withvelocityuandreboundswithinstantaneousvelocityvaftercollisionthen v thecofficient of resistitution,e =-u Hereu2 =2gH,y2=2gh e = ~7.A billiard ballstrikesanother identical billiard ball obliquely at anangle 45withtheirlineof centersatthetimeof impact.If thecoefficientof 1 resistitutionis"2' findtheangleat whichthe1slballwill'bounceoff'. Ans: For such obliqueimpact, initially,ul = U,al = a= 45. u2 =0 If vIandv2 be theirrespectivevelocitiesafterimpact thenfortheirangles 131 and132 withthelineof impact, mucosa=mVI cos 131 + mV2 cos132 U vIcos 131 + v2 cos132 =Ucos 45=..fi Again,v2 cos132 - vIcos 131 = e(u cos 45) = .!.u.~2v2 u =2J2 2vIcos 131 =~ 1 - ~ ) =2:n Againfornormalmomentum about thelineof impact, musina=mVI sin 131 ...(1) Theory of Collision U vIsinPI= u sin 45=fi tanPI= 2.=>PI=tan-I(2) 137 ...(2) Thisistheangleatwhichthe1slballwill'bounce off after collision. EXERCISES 1.What are the basiccharacteristics of collision? 2.State and explain Newton's collision rule. 3.Showthatforinelasticdirectcollisionbetweentwoparticles,thelossin 1 -mlm2 kineticenergyis~ E =2(VI- V2)2. ml+m2 4.Find thevelocities of twoparticles after oblique collision between thistwo particles.Alsofindlossinkineticenergy inthisregard. 5.Discuss collision between twoparticles incenter of massframe. 6.Agunfiresabulletof massm withhorizontalvelocityv intoablock of mass m which rests on a horizontal frictionlessplane.If the bullet becomes embedded intheblock,findthelossin kineticenergy inthisimpact. 7.Repeatthepreviousproblem(6)if theblockismovingawayfromthe gunwithvelocity vO' 8.A massml travelling withspeedv on a horizontalplanehitsanother mass m2 whichisatrest.If thecoefficient of resistitutionisE,provethat there mlm2(1- ~ v 2isa lossof kineticenergy equal to2() ml+m2 9.Aballisdroppedfromaheight' h'aboveahorizontalplaneontoan inclinedplaneorangleawhichisrestingonthehorizontalplane.Prove thatif thecofficientof resistitutionise,thentheballwillnexthitthe inclined at a point which isat a distance 4e(1+ e)sin abelow theoriginal pointof impact. -:0:-Chapter-7 ConservationPrincipleand ConstrainedMotion 7.1Characteristicsof ConservationPrinciple In classicalmechanics,wedealseveralproblemswithmotionof asingle particleorasystemof particles.Forsuchmotion,wehavetofaceaterm 'conservation'manytimes.Theconservationof aparameter,involvewiththe motion,isbasicallytheconstantnessof thatparameterwithrespecttotime. Clearly,the parameter which isaconserved quantity, will not change with time throughout.Wealsoobservethattheparameterconservationdemandsoneor morethanoneconditionswhicharebasically constrainedof motionand these conditions restricts the conservation of that parameter. With thedealing with several conserved parameter, we can now say that, (i)Conservationprinciplesareindependentonthetrajectoriesof the particles or system,but it dependsonsome restrictions. (ii)Conservationof anyparametermeansitsconstantnessandall conserved quantitiesare constant of motion. (iii)Conservationlawshaveanintimaterelationwithinvariance.Their failureincertaincasesmayresultinthediscoveryof newandnot yet understood phenomenon. (iv)Fromconservationlawwecanfindequationof motionevenwhen theeffective forceactionon thesystem isnot totally known. (v)Conservation lawsassure usmany timesthatsomeaspectsof motion areimpossible and must beleft out. (vi)Conservation phenomenon tellsus agreat deal about the motioneven if forcesaffecting motion arenot knownin advance. Inour physicalworld,thereexistano.of conservationprinciplesor laws, someexactandsomeapproximate.Theseprinciplesarerelatedwithenergy, linear momentum,angular momentum change ...etc. and various other quantities. Wewilldiscussallsuchconservationsinourpresentchapterwhicharevery 138 Conservation Principle and Constrained Motion139 powerful toolsinour presentchapter whichareverypowerful toolsinsolving several mechanical problems. 7.2Mechanicsof a singleparticle andsystemof particles Themotionof asingleparticleorasystemof particlescanprimarily be studiedby applyingNewton's2nd lawof motionwhichestimatetheeffective forceactingonthatparticleor system.Tostudy Newtonian mechanicsforthe motion of particle,itismandatory torealizethenatureand interaction of force on it.Togetasharp knowledgeabout thisforthemotionof particleor forthe motion of system of particles,we should havethefollowing discussion-For asingleparticlemotion,wehavefromNewton's2nd law,theapplied ..- djJd_dVd2; forceor theeffectiveforce1SF = - = -(my)= m-d = m---:x dtdttdt Where,p isinstantaneousmomentumof theparticle'm'isitsmassand v isinstantaneousvelocitywhichhasmagnitudemuchlessthanthatof light wave invacuum.Somotion isrestrictedintheregionof Newtonian mechanics whichisfarapartfromrelativisticregion.Here,;,isinstantaneousposition of particle duringitsmotion w.r.t somestationary point. But if we now consider the system ofN particles,then from Newton's law, - d2;,-(e)"-the effectiveforceon ith particle,Fj = mj -+ = Fj +Fij. dtj*j Where, isinstantaneouspositionof ithparticlehavingmass'm;'.Here, F}e)isexternalforceactingonithparticleandLFijisthesumof internal forcesonithparticlebyall Jthparticles(j #- i).Soforanyconstituentof the system,theeffectiveforceistheresultantof externalforceandallpossible internal forcesgiven by the neighboursparticles. Buthereconsideringallparticlestogetherforthewholesystem,thenet force, F= LFj = LFj(e) + LLFij jj*i andF = LFj(e) + LLFij = Lmj= d:jj(j*j)idtdtj - -But we must have,Fij=-Fji - 1 L(- -) ""p..=- P.+F=0 lj2I}Jl jjij i. e.,thenetinternalforceformotionof allparticlesinthesystemiszero, and asa result,formotion of asystem of particles, 140TheClassical Mechanics F IF/e)= d22 lIm,r;]. jdtj If we now take the average of all radius vectors for all constituents particles, then we get theinstantaneous position of center of mass of the system, Mrcwhere,M=Imj=Totalmass, F~ F / e =M(~ ]Whichisasingleparticleequation.Sothemotionof asystem of particles caningeneralbereplacedby themotionof thecenterof massof thatsystem under the net forcewhich is the sum of all external forces,applied on that system. 7.3Conservationof linearmomentum The principle of linear momentum conservation states that if the net effective forceonasingleparticleor system of particlesbezerothenthemomentum of that particleor thevector sum of momentum of allparticlesof thesystemwill beconstantorwillremaininvariantthroughoutthemotionof thatparticleor thatsystem. This conservation can easily be understood from Newton's law and for single particlemotion, _dVd(_) F=m- = - mv dtdt SoforF=0,mv=constant or conserved. Also,forsystem of particles, "F-(e)_"d2r;_"dVj -!!...-"-F=L.Jj- L.J mjdt2- L.J mjdt- dt L.J m, Vj IIII sofor, F :, (mi Vi)~ 0 =>L miV'~ Constant Thisgivestheconservationof linear momentum.Soforeffectiveforceto Conservation Principle and Constrained Motion141 bezeroforthemotion,thenetmomentumwillbecomeaconservedquantity or constant of motion. 7.4Conservationof AngularMomentum For asingleparticlerotationorforrigidbody rotationif nettorqueacting ontheparticleor thesystembeequaltozerothentheangularmomentum will beconserved.Thisisangularmomentumconservation.Wecannowdiscuss such conservation principle for single particle motion or formotion of the system of particles,inthefollowingway-For singleparticlerotation,theangular momentumisthemoment of linear momentum whichisgiven by L=F x Ii =F x (mv) =m(Fx v) Sothetorqueacting on it,is andfor - F- - djJd(- -) t=rx= r xdt=dtrxp -t dL dt 0, ( dF- fidF-) .:- xp= 0,or-lip dtdt dL - = 0L = Constant dt' Thisisangularmomentumconservationforsingleparticlerotation.Also forsystem of particles,thetorqueon ithparticle. - .r.::.- d(-)- -r..x "p.= r.x - mv= r..x F IIIdtIIII But, So, But we have,forallconstituent particles in the system L ~ xFij I#i ~ ~ ~ xFij+FjxFji) J"#l Here,~ - Fj = ~ j andthisIFijlgivestheseparationbetweenithand jth particle.Since,theinternalforcebetweenithand jthparticleFijactsalong 142TheClassical Mechanics :. - 0) x Fij=0 and thetotalinternaltorquex Fij=0

Sotheeffective torqueonith particleiseffectively thetorqueforexternal forceF(e)acting on that particle, I - -(e) Ie. x Fj = x Fi and thenet torque on thewholesystem 1:= "i:. xdPj x P')= dL Idtdt IIdt II ( d"l;_.d"i_) for, - x P = 0, SInce, _I Ilpi dtdt dLL = I("l; x Pj) = Constant Now,for:r= 0,- = 0 , dt Thisisconservation of angular momentum for system of particles by which if net torque for all external forcesacting on the system be zero, the total angular momentum forthewhole system willbecomeconstant. 7.5Conservation of Energy Forasingleparticlemotionorformotionof asystemof particles,if conservativeforceactsonthatparticleoronthatsystem,thenitstotalenergy isthesum of kineticandpotential energy willbeconstant. NowaforceF issaidtobeconservativeif thecloselineintegralof it along a closepath'C' becomes zero.i.e.,if F be conservative, fF.dr =0 c andinthat casewehavefromStoke's theoremV x F =0theconservative natureof forcedemandsthatthisforcecanbeexpressedasa -ve gradientof somescalar function,v(r),called scalar potential. I.e.F=-Vv(r) Under action of such conservative force,we have for a single particle system, Conservation Principle and Constrained Motion 22d-Work done,W= f F.dr = f m.di 11 When particle goesfromstate1 tostate2. 222 Wf mv.av =f v) =f 111 1212 - mv2 --mvl 22 143 But also,with respect toconservative forceF , the work donein thiscase, 22 WfF.dr = - f VY.dr II W 1212 - mv2 --mvl YI - Y2 and in thiscase, we finally have, 22 1212 = 22 =(KE)I+ (PE)I)(KE)2+ (PE)2 i.e.thesumof kineticandpotentialenergywillbeconstantineverystate of the particle.Thisisenergy conservation. Similarly,forN-particlesystem thetotal potential energy when, Fint_Vjyint I Here,onethingisveryclearthatif themutualinteractingforcesbe Newtonian,yIn!mustbeafunctionof therelativedistancebetweentwo interacting particles,i. e.,forithparticle. ) yintIyr =+ yr) j 144TheClassical Mechanics = -rjl) i/F-i 1 Hereclearly,thefactor"2comesbecauseof thefactthatwhilesumming themutualpotentialenergies,apairof particle(i,j)appearstwice,oncefor ithdueto Jthandotherfor jthdueithparticle,wherebotharethesameand should appear only once. So,forithparticleof thesystem, But p.=p(e)+ "pi.nt IIfF-i F-(e)d- "F- d-i.ri+ij'1"; joF-i -(e)d- "F- d-F,.ri +y'ri

d(l ) F-(e)d- "F-d-__m .f..2dt=i.ri+ij'1"; dt2IIThusforwhole system, L = +jdt2iiBut also,= +Pji.d0) IrF-Ii=-0) ,Conservation Principle and Constrained Motion Also, Wethen finally have, d(""l.2]_ "V(V(e)).. ( )dt dt- -jj. d'i-2 'VI}v,;.dry i',1*' " 1 .2=,,(e)1""' .2,2 ,., I,,J*' "..!.m.f,2+"v(e) +..!. ""v.1,nt 2I'2 y=Constant ,,,J*' T+V where,T Constant ,,1.2 totalK.E= 2miri I 145 "V(e)l""Vint andV=totalpotentialenergy = j+ 2 ij ,,J This isenergy conservation for system of particles under conservative forces actingonit. 7.6ConstrainedMotion Degreeof Motion,Constraints:Themotionof aparticleorsystemof particles under some conditions or restrictions iscalled constrained motion.For much motion,we must havesomeknowledgeabout thedegreeof freedomand constraints. Theminimumno.of independentvariableswhicharerequiredtospecify themotionof particleorsystemof particlesasawhole,iscalleddegreeof freedom for that particle or system.It can be shown frummathematical version, that if motion of a system ofN-particles areinvolve with 'm'no.of restrictions, then degree of freedom of that system isd"" 3N - m.Wewill analyse the reason behinditlateron,butatthisjunction,whatwewanttosaythatif wetakeall exampleof gasmolecules,thenformono atomicmolecule,N 1.m=0, d= 3. For diatomicmolecule,N=2,m ==I,d =5 ...etc. Nowthemathematicalrestrictionswhichcanbeexpressedthroughthe equationsorinequations,arecalledconstraints.Sotheconstrainedmotionis restricted by the'