NCERT Solutions for Class 11 Physics Chapter 7

Physics essentially deals with what matter constitutes, its motion and behaviour, and energy and force. While this subject can seem daunting to many students, regular practice and thorough knowledge of basic concepts right from Class 11 can help in studying Physics in a much easier way.

Physics class 11 Chapter 7 can be a tough subject for many students. Hence, getting help from experts becomes necessary while preparing for board exams. Moreover, having a strong base and grip on the topics is very important for students who want to pursue physics or engineering or subjects in any other related field in college. It will help them to understand new topics better and have an edge over the other students.

This is why students must refer to Class 11 Physics Chapter 7 NCERT Solutions that are prepared by the teachers, who are experts in teaching these subjects for years. The main motive of the solutions is to make conceptual understanding of the subject easier for the students and make it seem less daunting for them. 


NCERT Solutions for Class 11 Physics Chapter 7 (System of Particles and Rotational Motion)

The chapter will cover the motion of the centre of mass of a system of particles and the usefulness of this concept in understanding the motion of extended bodies. It explains the theoretical and mathematical parts of the concept. You will learn about the unchanging shape of an ideal rigid body in the NCERT solutions Class 11 Physics rotational motion. 

The chapter will focus on the angular velocity relation with linear velocity, torque and angular momentum. It includes the kind of motion a rigid body can have and how the centre of mass of a system of particles proceeds.

With the right resources, students can easily crack this chapter. At Extramarks, the solutions for Physics Chapter 7 class 11 have been prepared by the teachers and subject matter experts, who make sure that the answers are easy-to-understand and learn.


NCERT Solutions for Class 11 Physics Chapter 7 – (Free Download)

Physics is a lovely subject but can also be quite difficult to master. It forms the base for many scientific subjects that students take later in life. Class 11 Physics introduces students to topics like thermodynamics, laws of motion, etc. These topics are extremely important for the students to prepare from the exam point of view. They have to prepare these subjects well as they are further elaborated upon in higher classes and college. Many times, these topics are also asked in competitive exams. 

It is essential that you have a good grasp of a subject like Physics as it includes plenty of formulae and fundamental concepts that form the base for your CBSE syllabus. We provide the CBSE Class 11 Physics NCERT solutions which are planned out to aid students to perform better by making the subject easier for each student. Having the solutions handy will also allow you to revise before the exam as all the syllabus is merged into one place for the students to quickly go through. 


NCERT Solution for Class 11 Physics Chapter 7

Class 11 is an important class for students as it prepares you for Class 12 and board exams. The marks scored in Class 11 also are important in getting admission into a good college. For this, you need to have a strong grip on the basic concepts of Physics. This can only be achieved with regular practice, in-depth knowledge, and a problem-solving approach to any difficult topics. 

The Class 11 Physics NCERT solutions will help you understand the various difficult topics and master the subject. The solutions contain all the chapters laid out and answered by our panel of expert teachers. With many years of expertise, they mark out important questions and give you tips on how to solve them. With step-by-step solutions, you can also clear any doubts you may have had. 


Chapter 7 System of Particles and Rotational Motion

In the usual case scenario, the body is considered to be inflexible wherein the distances between different particles of the body do not change even after the acting of force on it. Only rotational motion is possible in the case of a rigid body fixed at one point or along a line. It is highly possible for this kind of body to either have pure translation or a combination of translation and rotation. One should have an understanding of the external forces of the body to determine the motion of the centre of mass of a system without knowledge of the internal forces of the system. Newton’s Second Law for finite-sized bodies (or systems of particles) is based on Newton’s Second Law and also Newton’s Third Law for particles.


Key Features of NCERT Solutions for Class 11 Physics Chapter 7

NCERT solutions physics class 11 chapter 11 will help students to solve questions that they might get stuck at. The solutions are explained and narrated by experienced teachers to make it easier for students to practise and revise.

  • The solution set covers all the necessary topics covered in the NCERT class 11 textbook to ensure students gain enough knowledge about everything. 
  • The solutions are written in easy language to ensure clarity so that you can score better in the exams.
  • Time management will be one of the best benefits for students if they refer to NCERT Solutions
  • It becomes convenient for students to revise the solutions before appearing for the exams. 

Q.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?


The centre of mass (C.M.) of a body is a point where the mass of a body is supposed to be concentrated. A sphere, cylinder, ring and a cube are symmetrical objects and have their centre of mass at their geometric centres. C.O.M. of a body need not necessarily lie inside the body, for example; hollow sphere and a ring.

Q.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.


The situation of the problem can be shown by the given figure.Here, seperation between H and Cl atoms=1.27ÅLet mass of H atom=mMass of Cl atom=35.5 mLet the distance of the centre of mass of the system from the Cl atom=xDistance of the centre of mass from the H atom =(1.27x)Consider that the centre of mass of the given molecule lies at the origin.m(1.27x)+35.5mxm+35.5m=0                                  m(1.27x)+35.5mx=01.27x=35.5xx=1.27(35.51)=0.037ÅHere, the negative sign shows that the centre of mass lies at the left of the molecule. The centre of mass of the HCl molecule lies at a distance 0.037 Å from the Cl atom.

Q.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?


Internal forces do not affect the motion of the bodies on which they act. No external force is not involved in the system, so the child’s motion will not change the velocity of the centre of mass of the system.

Q.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.


Consider two vectors a and b represented by OK and OM respectively, inclined at an angle θ, as shown in the given figure.From ΔOMN, we have:sinθ=MNOM=MN|b|MN=|b|sinθBy definition,|a×b|=|a||b|sinθ=OK.MN×22=2×Area of ΔOMKArea of ΔOMK=12|a×b|                                     Hence, proved.

Q.5 Show that a. (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.


Consider a parallelepiped with origin O and sides a, b and c respectively.Volume of the given parallelepiped = abc              Here, OA=aOB=bOC=cLet unit vector perpendicular to the plane containing b^ and c^ =n^n^ and a are along the same direction.b×c=bcsinθn^=bcsin90on^=bcn^a.b×^ = abc cos0° = abca.b×c=Volume of the parallelepiped


Find the components along the x, y, z axes of the angular momentum l of a particle, whose positionvector is r with components x, y, z and momentum is p with components p x , p y and p z . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaac H8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFf ea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaa baqaaiaacaGaaeqabaWabeqaeaaakqaabeqaaKaaGjaabAeacaqGPb GaaeOBaiaabsgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaaboga caqGVbGaaeyBaiaabchacaqGVbGaaeOBaiaabwgacaqGUbGaaeiDai aabohacaqGGaGaaeyyaiaabYgacaqGVbGaaeOBaiaabEgacaqGGaGa aeiDaiaabIgacaqGLbGaaeiiaiaabIhacaqGSaGaaeiiaiaabMhaca qGSaGaaeiiaiaabQhacaqGGaGaaeyyaiaabIhacaqGLbGaae4Caiaa bccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae yyaiaab6gacaqGNbGaaeyDaiaabYgacaqGHbGaaeOCaiaabccacaqG TbGaae4Baiaab2gacaqGLbGaaeOBaiaabshacaqG1bGaaeyBaiaabc cacaqGSbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGaaeiC aiaabggacaqGYbGaaeiDaiaabMgacaqGJbGaaeiBaiaabwgacaqGSa aakeaajaaycaqG3bGaaeiAaiaab+gacaqGZbGaaeyzaiaabccacaqG WbGaae4BaiaabohacaqGPbGaaeiDaiaabMgacaqGVbGaaeOBaiaays W7caqG2bGaaeyzaiaabogacaqG0bGaae4BaiaabkhacaqGGaGaaeyA aiaabohacaqGGaGcdaWhcaqcaawaaiaabkhaaiaawEniaiaabccaca qG3bGaaeyAaiaabshacaqGObGaaeiiaiaabogacaqGVbGaaeyBaiaa bchacaqGVbGaaeOBaiaabwgacaqGUbGaaeiDaiaabohacaqGGaGaae iEaiaabYcacaqGGaGaaeyEaiaabYcacaqGGaGaaeOEaiaabccacaqG HbGaaeOBaiaabsgacaqGGaGaaeyBaiaab+gacaqGTbGaaeyzaiaab6 gacaqG0bGaaeyDaiaab2gacaqGGaGaaeyAaiaabohacaqGGaGcdaWh caqcaawaaiaabchaaiaawEniaiaabccacaqG3bGaaeyAaiaabshaca qGObaakeaajaaycaqGJbGaae4Baiaab2gacaqGWbGaae4Baiaab6ga caqGLbGaaeOBaiaabshacaqGZbGaaeiiaiaabchakmaaBaaajeayba GaaCiEaaqabaqcaaMaaiilaiaabccacaqGWbGcdaWgaaqcbawaaiaa hMhaaeqaaKaaGjaabggacaqGUbGaaeizaiaabccacaqGWbGcdaWgaa qcbawaaiaahQhaaeqaaKaaGjaac6cacaqGGaGaae4uaiaabIgacaqG VbGaae4DaiaabccacaqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabM gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGaaeyy aiaabkhacaqG0bGaaeyAaiaabogacaqGSbGaaeyzaiaabccacaqGTb Gaae4BaiaabAhacaqGLbGaae4CaiaabccacaqGVbGaaeOBaiaabYga caqG5bGaaeiiaiaabMgacaqGUbGaaeiiaiaabshacaqGObGaaeyzai aabccacaqG4bGaaeylaiaabMhacaqGGaGaaeiCaiaabYgacaqGHbGa aeOBaiaabwgaaOqaaKaaGjaabshacaqGObGaaeyzaiaabccacaqGHb GaaeOBaiaabEgacaqG1bGaaeiBaiaabggacaqGYbGaaeiiaiaab2ga caqGVbGaaeyBaiaabwgacaqGUbGaaeiDaiaabwhacaqGTbGaaeiiai aabIgacaqGHbGaae4CaiaabccacaqGVbGaaeOBaiaabYgacaqG5bGa aeiiaiaabggacaqGGaGaaeOEaiaab2cacaqGJbGaae4Baiaab2gaca qGWbGaae4Baiaab6gacaqGLbGaaeOBaiaabshacaGGUaaaaaa@304A@


The three rectangular components of angular momentum are given as:lx=ypzzpyly=zpxxpzlz=xpyypxLinear momentum of the particle is given as:p=pxi+pyj+pzkPosition vector of the particle is given as:r=xi+yj+zkAngular momentum, l=r×p=(xi+yj+zk)×(pxi+pyj+pzk)=|i  j  kx  y  zpx py pz|                                                      lxi+lyj+lzk=i(ypzzpy)j(xpzzpx)+k(xpyypx)On comparison of coefficients of i,j and k, we obtain:lx=ypzzpyly=xpzzpxlz=xpyypx}iAs, the particle travels in the xy plane, the z-component of the position vector and linear momentum vector becomes zero, i.e.,z=pz=0Therefore, equation (i) becomes:                        lx=0ly=0lz=xpyypx}iiWhen the particle is restricted to move in the x-y plane, the direction of angular momentum is along the z-direction.

Q.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.


Here, mass of each particle=mVelocity of each particle=vLet, at a definite instant two particles be at points S and V, as shown in the given figure.Angular momentum of the system about point S is given as:LS=mv×0+mv×dLS=mvdiAngular momentum of the system about point V is given as:LV=mv×d+mv×0LV=mvdiConsider a point T at a distance y from  point V, i.e.,TV=yST=dyAngular momentum of the system about point R is given as:LT=mv×(dy)+mv×y=mvdmvy+mvy=mvd(iii)Comparing equations (i), (ii), and (iii), we obtain:LS=LV=LTivFrom equation (iv), we conclude that the angular momentum of a system does not depend on the point about which it is taken.

Q.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long.
Calculate the distance d of the centre of gravity of the bar from its left end.


The free body diagram of the bar is shown in the given figure.Here, length of the bar, l = 2 mLet T1 and T2 be the tensions produced in the left and right strings respectively.At translational equilibrium, we have:T1sin36.9o=T2sin53.1oT1T2=sin53.1osin36.9o=0.8000.600=43T1=43T2For rotational equilibrium about the centre of gravity,we obtain:T1cos36.9o×d=T2cos53.1o(2d)T1×0.800d=T2×0.600(2d)43×T2×0.800d=T2[0.600×20.600d]1.067d+0.6d=1.2d=1.21.67=0.72mThe C.G. (centre of gravity) of the given bar lies at a distance 0.72 m from its left end.

Q.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.


Here, mass of car, m = 1800 kgSeparation between front and back axles, d = 1.8 mSeperation between the C.G. and the back axle=1.05 mThe several forces acting on the car are shown in the given figure.Here, Rf and Rb are the forces applied by the level ground on the front and back wheels respectively.For translational equilibrium, we have:Rf+Rb=mg=1800 × 9.8=17640 N (i)At rotational equilibrium, about C.G.,we have:Rf(1.05)=Rb(1.81.05)RfRb=0.751.05=57Rb=1.4Rf... (i)By solving equations (i) and (ii), we obtain:             1.4Rf+Rf=176402.4Rf=17640Rf=176402.4=7350 NRb=1.4Rf=10290 NForce applied on each front wheel = 73502=3675 NForce applied on each back wheel=102902=5145 N

Q.10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.


(a)Here, moment of inertia (M.I.) of a sphere about its diameter  =  25MR2As per the theorem of parallel axes, the M.I. of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of itsmass and the square of the distance between the two parallel axes.M.I. about a tangent to the sphere =  25MR2+MR2=75MR2

(b) Here, M.I. of a disc about its diameter=14MR2 As per the theorem of perpendicular axis, the M.I. of a planar body about an axisnormal to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axisand lying in the plane of the body.M.I. of the disc about its centre=14MR2+14MR2=12MR2This situation is shown in the figure(b). According to the theorem of parallel axes,the M.I.about an axis perpendicular to thedisc and passing through a point on its edge=12MR2+MR2=32MR2

Q.11 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is
free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?


If m be the mass of the hollow cylinder and r is the radius of the solid sphere.

M.I. of hollow cylinder about its standard axis of symmetry,I1=MR2M.I. of solid sphere about an axis through its centre,I2=25MR2From the relation, τ =Here, α,τ and I represent the angular acceleration, torque and moment of inertia respectively.Torque for the hollow cylinder,τ1=I1α1Torque for the solid sphere,τ2=I2α2Since an equal torque is exerted on both the bodies,τ1=τ2α2α1=I1I2=mr225mr2=52α2>α1(i)From the relation, ω=ω0+αtHere, ω0 = Initial angular velocityt = Time of rotationω = Final angular velocityFor equal values of ω0 and t, we obtain:ωα(ii)From equations (i) and (ii), we obtain:ω2 > ω1The angular velocity of the solid sphere will be more than the angular velocity of the hollow cylinder.

Q.12 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s–1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder?
What is the magnitude of angular momentum of the cylinder about its axis?


Here, mass of the cylinder, m=20 kg                   Angular speed of the cylinder, ω=100 rad s1Radius of the cylinder, r=0.25 mM.I. of the solid cylinder is given as: I=mr22=12×20×(0.25)2=0.625kgm2Kinetic energy of rotation=12Iω2=12×0.625×(100)2=3125JAngular momentum of the cylinder, L=Iω= 0.625×100Js= 62.5 Js

Q.13 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?


Here, initial angular velocity of child, ω1=40 rev min-1Let final angular velocity of child=ω2Let I1 and I2 be initial and final M.I. of child respectively.Initial and final M.I. are related as, I2=25I1As no external force acts on the child, Angular momentum, L= constant.For the two situations, we have: I2ω2=I1ω1ω2=I1I2ω1=I125I1×40=52×40=100revmin1(b)Here, final K.E. of rotation of child, EF=12I2ω22Initial K.E. of rotation of child, EI=12I1ω12EFEI=12I2ω2212I1ω12=25I1I1(100)2(40)2=52=2.5EF=2.5EI K.E. of rotation of child increases.This is due to the internal energy spent by the child in folding his hands.

Q.14 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.


Here, mass of the hollow cylinder, m=3 kg              Radius of the hollow cylinder, r=40 cm=0.4 mForce exerted, F=30 NThe moment of inertia of the hollow cylinder about its axis is given as: I=mr2= 3×(0.4)2=0.48 kg m2Torque is given as:τ=F×r=30×0.4=12 NmIf α is the angular acceleration produced, torque is given as: τ=Iαα=τI=120.48=25rads2Linear acceleration, a=rα=0.4 × 25=10 ms2

Q.15 To maintain a rotor at a uniform angular speed of 200 rad s–1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.


Here,angular speed of the rotor, ω=200 rads -1 Requiredtorque, τ=180 Nm The power of the rotor ( P ) is related to torque and angular speed as,P=τω P=180×200=36×1 0 3 =36 kW Thepower required by the engine = 36 kW MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWabeqaeaaakqaabeqaaiaabIeacaqGLbGaaeOCaiaabwgacaqGSaGaaGjbVlaabggacaqGUbGaae4zaiaabwhacaqGSbGaaeyyaiaabkhacaqGGaGaae4CaiaabchacaqGLbGaaeyzaiaabsgacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqGVbGaaeiDaiaab+gacaqGYbGaaeilaiaabccacqaHjpWDcqGH9aqpcaqGYaGaaGimaiaaicdacaqGGaGaaeOCaiaabggacaqGKbGaae4CamaaCaaaleqabaGaaeylaiaabgdaaaaakeaacaqGsbGaaeyzaiaabghacaqG1bGaaeyAaiaabkhacaqGLbGaaeizaiaaysW7caqG0bGaae4BaiaabkhacaqGXbGaaeyDaiaabwgacaqGSaGaaeiiaiabes8a0jabg2da9iaabgdacaqG4aGaaGimaiaabccacaqGobGaaeyBaaqaaiaabsfacaqGObGaaeyzaiaabccacaqGWbGaae4BaiaabEhacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaab+gacaqG0bGaae4BaiaabkhacaqGGaWaaeWaaeaacaqGqbaacaGLOaGaayzkaaGaaeiiaiaabMgacaqGZbGaaeiiaiaabkhacaqGLbGaaeiBaiaabggacaqG0bGaaeyzaiaabsgacaqGGaGaaeiDaiaab+gacaqGGaGaaeiDaiaab+gacaqGYbGaaeyCaiaabwhacaqGLbGaaeiiaiaabggacaqGUbGaaeizaiaabccaaeaacaqGHbGaaeOBaiaabEgacaqG1bGaaeiBaiaabggacaqGYbGaaeiiaiaabohacaqGWbGaaeyzaiaabwgacaqGKbGaaeiiaiaabggacaqGZbGaaeilaiaaysW7caqGqbGaeyypa0JaeqiXdqNaeqyYdCNaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7aeaacqGH0icxcaWGqbGaeyypa0JaaeymaiaabIdacaaIWaGaey41aqRaaeOmaiaaicdacaaIWaGaeyypa0Jaae4maiaabAdacqGHxdaTcaqGXaGaaGimamaaCaaaleqabaGaae4maaaakiabg2da9iaabodacaqG2aGaaeiiaiaabUgacaqGxbaabaGaeyinIWLaaeivaiaabIgacaqGLbGaaGjbVlaabchacaqGVbGaae4DaiaabwgacaqGYbGaaeiiaiaabkhacaqGLbGaaeyCaiaabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabkgacaqG5bGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGLbGaaeOBaiaabEgacaqGPbGaaeOBaiaabwgacaqGGaGaaeypaiaabccacaqGZaGaaeOnaiaabccacaqGRbGaae4vaaaaaa@2656@

Q.16 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.


Let mass per unit area of the disc=mLet radius of the original disc=RMass of the original disc, M=πR2mThe disc with the removed portion is shown in the given figure.Here, radius of the smaller disc=R2Mass of the smaller disc, M’=π(R2)2m=14πR2m=M4Let O and O¢ be the respective centres of the original disc and the disc removed from the original. According to the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O,while that of the smaller disc is supposed to be concentrated at O¢.Given, OO=R2After the smaller disc has been removed from the original, the remainingportion is assumed to be a system of two masses.These masses are:M (concentrated at O), andM(=M4)concentrated at O¢(The negative sign shows that this portion has been cut off from the original disc.)Let distance of centre of mass of the remaining portion = xRelation between the centres of masses of two masses is given as:x=m1r1+m2r2m1+m2Using the above relation for the given system, we have,x=M×0M×(R2)M+(M)=M4×R2MM4=MR8×43M=R6The negative sign shows that the centre of mass of disc gets displaced toward the left of point O.

Q.17 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?


Let, mass of the meter stick= mHere, mass of each coin, m=5 gAs the coins are placed 12 cm away from the end P, the centre of mass gets displaced by 5 cm from point R towards the end P.The centre of mass is positioned at the seperation of 45 cm from point P.For the equilibrium state about the 45 cm mark, we have:10×g(4512)=mg(5045)m=10×335=66gMass of the metre stick, m’ =66 g.

Q.18 A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
(a) Will it reach the bottom with the same speed in each case?
(b) Will it take longer to roll down one plane than the other?
(c) If so, which one and why?


Let, mass of the sphere=mLet height of the inclined planes=hLet velocity of the sphere at the bottom of the plane=vAt the top of the inclined plane, Total energy of the sphere=Potential energy=mghAccording to the principle of conservation of energy, we have:12mv2+122=mghIn case of a solid sphere, M.I. about its centre is given as:I=25mr212mv2+12(25mr2)ω2=mghLinear and angular velocity are related as:=v12mv2+15mv2=mghv=107ghSince h is same in each case, v must also be same. Sphere will reach the bottom with same speed in each case.

Q.19 A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?


Here, radius of the hoop, r=2 mMass of the hoop, m=100 kgVelocity of the hoop, v=20 cms-1=0.2 ms-1Moment of inertia of the hoop about its centre, I=mr2(Total energy of the hoop = Translational KE + Rotational KE)=12mv2+122=12mv2+12(mr2)ω2  (I=mr2)=12mv2+12mv2   (=v)=mv2  Work required to be done to stop the hoop = Total energy of the hoop.Required work, W=mv2 =100×(0.2)2=4 J

Q.20 The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms.
Suppose the mean speed of such a molecule in a gas is 500 ms-1 and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.


Here, mass of the oxygen molecule, m =5.30×1026 kgMoment of inertia of oxygen molecule, I=1.94×1046 kg m2Velocity of the oxygen molecule, v=500 ms-1Let distance of separation between the two atoms of the oxygen molecule=2rMass of each oxygen atom=m2Moment of inertia I is given as:I=m2r2+m2r2=mr2(i)r=Im=1.94 × 10465.30 × 1026=0.61×1010m                Given, K.E. of rotation=23×K.E. of translation122=23×12mv212(mr2)ω2=13mv2  (I=mr2)ω=23vr=23×5000.61×1010=6.75×1012rads1

Q.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?


A solid cylinder rolling up an inclination is shown in the given figure.Here, Angle of inclination, θ=30°Initial velocity of the solid cylinder, v=5 ms-1Height attained by the cylinder=h(a) Energy of the cylinder at point P:K.E.Rot=K.E.TransEnergy of the cylinder at point Q=mgh According to the law of conservation of energy, we have:12mv2+122=mgh(i)Moment of inertia of the solid cylinder,I=12mr2(ii)Substituting equation (ii) in equation (i), we obtain:12mv2+12(12mr2)ω2=mgh(iii)12mv2+14mv2=mgh   (=v)34mv2=mghh=3v24g=3×(5)24×9.8=1.913mLet s be the distance up the inclined plane, then using the relation,sinθ=hss=hsinθ=1.913sin30o=3.826mIn ΔPQR:sinθ=QRPQsin300=hPQPQ=1.910.5=3.82mThe cylinder will move 3.82 m up the inclined plane.(b) For radius of gyration K, the velocity of the cylinder at the instance it rolls back to the bottom is given as:v=(2gh1+K2R2)In case of the solid cylinder, K2=R22 v= ( 2gABsinθ 1+ 1 2 ) 1 2 = ( 4 3 gABsinθ ) 1 2 The time taken to comeback to the bottom isgivenas: t= AB v = AB ( 4 3 gABsinθ ) 1 2 = ( 3AB 4gsinθ ) 1 2 Total time taken by the cylinder to comeback to the bottom =( 2×0.764 )s =1.53 s. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EA92@

Q.22  As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 ms-2)(Hint: Consider the equilibrium of each side of the ladder separately.)


The situation of the problem can be shown in the given figure.

Here, N B =Force applied on the ladder by the floor atpoint B N C =Force applied on the ladder by the floor atpoint C T=Tension producedin the rope BA=CA=1.6 m DE=0. 5 m BF=1.2 m Mass of the weight, m=40 kg Draw a normal from A on the floor BC,which intersects DE at themidpoint H. Since,ΔABI and ΔAIC are similar BI=IC I is the mid – point of BC. DE || BC BC=2×DE =1 m AF=BABF =0.4 m ( i ) SinceD is the mid – point of AB AD= 1 2 ×BA =0.8m(ii) From equations ( i ) and ( ii ), we obtain: FE=0.4 m F is the mid – point of AD. FG || DH and F is the middlepoint of AD. G will also be the middlepoint of AH. AsΔAFG and ΔADH are similar MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5113@ FGDH = AFADFGDH=0.40.8=12FG=12DH=12×0.25=0.125mIn ΔADH:AH=AD2-DH2=(0.8)2(0.25)2=0.76mIn case of translational equilibrium of the ladder, the upward force must be same as that of the downward force.NC+NB=mg=392 N (iii)In case of rotational equilibrium of the ladder, the total moment about A  is given as:NB×BI+mg×FG+NC×CI+T×AGT×AG=0NB×0.5+40×9.8×0.125+NC×(0.5)=0(NCNB)×0.5=49NCNB=98N...(iv)By adding equation (iii) and (iv), we obtain:NC=245NNB=147NIn case of rotational equilibrium of side AB, assume the moment about A.NB×BI+mg×FG+T×AG=0245×0.5+40+9.8×0.125+T×0.76=00.76T=122.549T=96.7N

Q.23 A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?


(a)Here, M.I. of the man-platform system,I=7.6 kgm2when the man stretches his hands to a distance of 90 cm, M.I., Ia=2×mr2=2×5×(0.9)2=8.1 kgm2Initial M.I. of the system, I1=I+Ia=7.6 +8.1=15.7 kgm2Angular speed, ω1=30rpmAngular momentum, L1=I1ω1=15.7×30=471 kgm2s1Moment of inertia when the man bends his hands to a distance of 20 cm, Ib=2×mr2=2×5×(0.2)2=0.4 kgm2Final moment of inertia,I2=I+Ib=7.6 +0.4 =8kgm2Let, final angular speed =ω2Final angular momentum, L2=I2ω2=8ω2 (ii)According to the law of conservation of angular momentum, we have:Final angular momentum = Initial angular momentumI2ω2=I1ω1ω2=4718=58.875 rev/min(b)No, K.E. is not conserved in this process. The fact is that, as the M.I. decreases, the K.E. increases. The work done by the man to bend his hands closer to his body provides the additional K.E.

Q.24 A bullet of mass 10 g and speed 500 ms-1 is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.. (Hint: The moment of inertia of the door about the vertical axis at one end is ML2/3.)


Here, mass of the bullet, m=10 g=10×103 kgSpeed of the bullet, v=500 ms-1Thickness of the door, L=1 mRadius of door, r=12mMass of door, M=12 kgAngular momentum provided by the bullet on the door, α=mvr=(10 × 103)×500×12=2.5 kgm2s-1M.I. of the door, I=ML23=12×123=4kgm2Since, angular momentum is given as:L=             ω=LI=2.54=0.625rads1Angular speed of the door, ω=0.625rads1

Q.25 Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident.
(a) What is the angular speed of the two-disc system?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω1 ≠ ω2.


aHere, two discs of moments of inertia I1 and I2 about their respectiveaxes are brought into contact.Total initial angular momentum of the two discs, L1=I1ω1+I2ω2As the two discs are joined, their M.I. get added up.M.I. of the system of two discs is given as: I=I1+I2Let the angular speed of the system=ω Total final angular momentum of the system,L2=I1+I2ωAccording to the law of conservation of angular momentum, we have:Final angular momentum = Initial angular momentum  I1+I2ω=I1ω1+I2ω2ω=I1ω1+I2ω2I1+I2  (i)bLet, kinetic energy of disc I be12I1ω12Kinetic energy of disc II be 12I2ω22Total initial kinetic energy,Ei=12I1ω12+12I2ω22As the discs are joined together, their M.I. get added up.Moment of inertia of the system, I=I1+I2Let angular speed of the system=ωFinal kinetic energy, Ef=12I1+I2ω2Using equation (i), we getEf=12I1+I2I1ω1+I2ω22I1+I22=I1ω1+I2ω222I1+I2Now, EiEf=12I1ω12+12I2ω22I1ω1+I2ω222I1+I2Thus, EiEf=I1I2ω1ω222I1+I2EiEf>0Therefore, Ef<EiThe loss of K.E. takes place in the process due to the frictional forcethat comes into play when the two discs come in contact with each other.

Q.26 (a) Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x2 + y2).
(b) Prove the theorem of parallel axes. (Hint: If the centre of mass is chosen to be the origin).


aAccording to the theorem of perpendicular axes, the MI of a planar bodyabout an axis perpendicular to its plane is equalto the sum of its M.I. abouttwo perpendicular axes concurrent with perpendicular axis and lying in theplane of the body. A physical body with centre O and a point mass m, inthe xy plane at x, y is shown in the given figure.Here, M.I. about xaxis, Ix=mx2M.I. about yaxis, Iy=my2M.I. about zaxis, Iz=mx2+y22=Ix + Iy=mx2+my2= m(x2+y2)=mx2+y22Ix + Iy =IzHence, theorem is proved.b According to the theorem of parallel axes, the MI of a body about anyaxis is equal to the sum of the MI of the body abouta parallel axis passingthrough its CM and the product of its mass and the square of the distancebetween the two parallel axes. Suppose a rigid body is made up of n particles,having masses m1, m2, m3, , mn, at normal distances r1, r2, r3, , rnrespectively from the centre of mass of the rigid body.


Prove the result that the velocity v of translation of a rolling body(like a ring, disc, cylinder or sphere ) at the bottom of an inclinedplane of a height h is given by v2=2gh1+k2R2. Using dynamical consideration(i.e. by consideration of forces and torques). Note k is the radius of gyrationof the body about its symmetry axis, and R is the radius of the body The bodystarts from rest at the top of the plane.


A body rolling on an inclined plane of height h,is shown in the given figure.Here, radius of the body=R Height of the inclined plane=h Acceleration due to gravity=g Translational velocity of the body=v Radius of gyration of the body=k Let mass of body = mTotal energy at the top of the plane, E1=mghM.I. of the rolling body is given as: I = mk2According to the principle of conservation of energy:K.E. of translation + K.E. of rotation=P.E. at the top12mv2+122=mgh12mv2+12(mk2)ω2=mgh  (I=mk2)Since, ω=vR12mv2+12mk2R2v2=mgh12mv2(1+k2R2)=mghv2=2gh(1+k2R2)Hence, proved.

Q.28 A disc rotating about its axis with angular speed ωo is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?


Here, angular speed of the disc = ωoRadius of the disc=RFrom the relation for linear velocity, v=ωoR For point A:vA=Rωo; tangential to the right For point B:vB=Rωo; tangential to the left For point C:vC=(R2)ωo; in the direction of vAThe directions of motion of points A, B, and C on the disc are shown in the given figure As the disc is placed on a frictionless table,it will not roll at all as friction is essential for the rolling of a body.

Q.29 Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?


In order to roll a disc, we need a torque tangential to the disc; provided only by frictional force in this case.
(a) Frictional force acts in the direction opposite to the direction of velocity at point B. The linear velocity at point B acts along tangentially leftward. Thus, force of friction will act tangentially rightward and the sense of frictional torque will be in normal direction to the plane of the disc and outwards before the start.
(b) Since force of friction acts opposite velocity at point B, perfect rolling will start when the velocity at that point becomes zero. The force of friction will also become zero at this stage.

Q.30 A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2.


Here, radii of the ring and the disc, r=10 cm= 0.1 mInitial angular speed, ω0=10 π rads1Coefficient of kinetic friction, μk=0.2Initial velocity of ring and disc, u=0Frictional force gives rise to motion of the two objects.According to Newton’s second law of motion:Force of friction, f = maμkmg=maHere, a=Acceleration produced m=Massa=μkg iAccording to the first equation of motion, v=u+at=0+μkgtv=μkgt iiTorque due to frictional force will act perpendicularly outwards and giverise to retardation in the initial angular speed ωo.Torque is given by the relation, τ=IαHere, α=Angular accelerationμkmgr=Iαα=μkmgrI(iii)Applying the first equation of rotational motion, we have: ω=ωo+αtω=ωo+μkmgrIt(iv)Rolling begins when linear velocity, v=rωv=rωoμkmgrtI(v)Equating equations ii and v, we obtain:μkgt=rωoμkmgrtIμkgt=rωoμkmgr2tI(vi)In case of the ring: I=mr2μkgt=rωoμkmgr2tmr2

Q.31 A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µs = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?


The several forces acting on the cylinder are shown in the given figure. Acceleration of the cylinderisgivenbytherelation: a= mgsinθ m+ I r 2 = mgsinθ m+ 1 2 m r 2 r 2 = 2 3 gsin 30 o = 2 3 ×9.8×0.5 =3.27m s 2 ( a ) Accordingto Newton’s second law of motion, wehave: f net =ma mgsin 30 o f=ma f=mgsin 30 o ma =10×9.8×0.510×3.27 =4932.7 =16.3N ( b ) Atthetimeof rolling, the instantaneous point of contact with the plane comes to rest. Work done against theforce offriction = 0 ( c ) Incaseof rolling without slipping, we have: μ= 1 3 tanθ tanθ=3μ =3×0.25 =0.75 θ=ta n 1 ( 0.75 ) = 36.87 o MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2D4C@


Separation of Motion of a system of particles into motion of thecentre of mass and motion about the centre of mass:a Show pi = p’i+ miV Where pi is the momentum of the ith particle(of mass mi)  and pi= miv’i. Note vi is the velocity of the ith particlerelative to the centre of mass. Also, prove using the definition of thecentre of mass pi=0b Show K = K + 12MV2 Where K is the total kinetic energy of thesystem of particles, K is the total kinetic energy of the system whenthe particle velocities are taken with respect to the centre of massand MV22 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system).The resulthas been used in Sec. 7.14.c Show L = L+ R × MV Where L=ri × pi is the angular momentumof the system about the centre of mass with velocities taken relative to thecentre of mass. Remember ri= ri R; rest of the notation is the standardnotation used in the chapter. Note L and MR × V can be said to be angularmomenta, respectively, about and of the centre of mass of the system of particles.d Show dLdt=ri x dpdt Further, show that where dLdt=τext is the sumof all external torques acting on the system about the centre of mass.(Hint: Use the definition of centre of mass and Newton’s Third Law.Assume the internal forces between any two particles act along theline joining the particles.)


(a)Let us consider a system of i moving particles.Here, mass of the ith particle = miVelocity of the ith particle = viMomentum of the ith particle, pi=miviVelocity of the CM=VVelocity of the ith particle w.r.t. the CM of the system is given by the relation:vi= viV (1)By multiplying equation (1) with mi, we obtain:mivi=mivimiVpi=pimiVHere,pi=mivi=Momentum of the ith particle w.r.t. the CM of the systempi=pi+miVFrom the relation: pi=miviSummation of momentum of all the particles w.r.t. the CM of the system is given as:ipi=imivi=imidridtHere,ri=Position vector of ith particle w.r.t. the CMvi=dridtAccording to the definition of CM, we haveimiri=0imidridt=0ipi=0(b) As the relation for velocity of the ith particle is given as:vi=vi+Vimivi=imivi+imiV      (2)By taking the dot product of equation (2) with itself, we obtain:imivi.imivi=imi(vi+V).imi(vi+V) case of the CM of the system of particles, we M2ivi2=M2ivi2+M2V212Mivi2=12Mivi2+12MV2K=K+12MV2Here, K=12Mivi2=Total K.E. of the system of particlesK=12Mivi2=Total K.E. of the system of particles w.r.t. the CM12MV2=K.E. of the translation of the system w.r.t. the CM(c) Position vector of the ith particle w.r.t. origin= riPosition vector of the ith particle w.r.t. the CM = riPosition vector of the CM w.r.t. origin=RGiven that,ri = riRri = ri+RFrom part (a), we obtain:pi = pi + miVTaking the cross product of the above equation by ri, we obtain:iri×pi=iri×pi+iri×miVL=i(ri+R)×pi+i(ri+R)×miV=iri×pi+iR×pi+iri×miV+iR×miV=L+iR×pi+iri×miV+iR×miVHere, R×ipi=0,(iri)×MV=0,imi=M,L=L+R×MV(d) We have,L=iri×pidLdt=ddt(iri×pi) =ddt(iri)×pi+iri×ddt(pi)=ddt(imiri)×vi+iri×ddt(pi)Here, ri=Position vector w.r.t. the CM of the system of particles.imiri=0dLdt=iri×ddt(pi)From the relation:dLdt=iri×ddt(pi)=iri×middt(vi)Here, ddt(vi)= Rate of change of velocity of the ith particle w.r.t. the CM of the system.As per Newton’s third law of motion, we have:middt(vi)=External force on the ith particle=i(τi)extiri×middt(vi)=τext=External torque on the systemdLdt=τext

Q.33 Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.


(a)The given statement is falseas the force of friction and motion of CM are in opposite directions.
(b) The given statement is trueas rolling can be considered to be rotation of a body about an axis passing through the point of contact of the body with the ground. Thus, instantaneous speed of the point of contact of the body during rolling is zero.
(c) The given statement is falseas a rortating body has an instantaneous acceleration not equal to zero.
(d) The given statement is true. On perfect rolling frictional force becomes zero and therefore the work done against it also becomes zero.
(e) The given statement true. The required torque for rolling comes from the tangential force generated by friction between the body and the surface. Without friction the body will simply slip from the inclined plane.

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