Physics Notes for NEET Chapter 7 - PDF

Chapter 60 7 CONTENTS Introduction 7.2 Centre of mass 7.3 Angular displacement 7.4 Angular velocity 7.5 Angular acceleration 7.6 Equations of linear motion and rotational motion 7.7 Moment of inertia 7.8 Radius of gyration 7.9 Theorem of parallel axes 7.10 Theorem of perpendicular axes 7.11 Moment of inertia of two point masses about their centre of mass 7.12 Analogy between translatory motion and rotational motion 7.13 Moment of inertia of some standard bodies about different axes 7.14 Torque 7.15 Couple 7.16 Translatory and rotatory equilibrium 7.17 Angular momentum 7.18 Law of conservation of angular momentum U U D YG 7.20 7.21 Rolling without slipping 7.22 Rolling on an inclined plane 7.23 Rolling sliding and falling of a body 7.24 Velocity, acceleration and time for different bodies 7.25 Motion of connected mass 7.26 Time period of compound pendulum Work, Energy and Power for rotating body In 1988, Greg Louganis of the United States, arguably the greatest Olympic diver in history, cracked his head on the springboard while attempting a reverse 2.5 pike. After receiving stitches, Louganis won gold medal in diving. Slipping, Spinning and Rolling ST 7.19 ID E3 7.1 Sample Problems Practice Problems (Basic and Advance Level) Answer Sheet of Practice Problems Greg Louganis jumping from the spring board exhibits somersaults in air before touching the water surface. After leaving the spring board, he curls his body by rolling the arms and the legs inwards. Due to this, his moment of inertia decreases and he spins in mid air with large angular speed. As he is about to touch the water surface, he stretches out his limbs. This increases his moment of inertia and he enters the water at a gentle speed. 60 108 Rotational Motion E3 7.1 Introduction. Translation is motion along a straight line but rotation is the motion of wheels, gears, motors, planets, the hands of a clock, the rotor of jet U ID engines and the blades of helicopters. First figure shows a skater gliding across the ice in a straight line with constant speed. Her motion is called translation but second figure shows her spinning at a constant rate about a vertical axis. Here motion is called rotation. D YG Up to now we have studied translatory motion of a point mass. In this chapter we will study the rotatory motion of rigid body about a fixed axis. (1) Rigid body : A rigid body is a body that can rotate with all the parts locked together and without any change in its shape. (2) System : A collection of any number of particles interacting with one another and are under consideration during analysis of a situation are said to form a system. U (3) Internal forces : All the forces exerted by various particles of the system on one another are called internal forces. These forces are alone enable the particles to form a well defined system. Internal forces between two particles are mutual (equal and opposite). (4) External forces : To move or stop an object of finite size, we have to apply a force on ST the object from outside. This force exerted on a given system is called an external force. 7.2 Centre of Mass. Centre of mass of a system (body) is a point that moves as though all the mass were concentrated there and all external forces were applied there. (1) Position vector of centre of mass for n particle system : If a system consists of n particles of masses m1 , m 2 , m 3......m n , whose positions vectors are y r1 , r2 , r3........ rn respectively then position vector of centre of mass C.M. m r  m 2 r2  m 3 r3 ............ m n rn r  1 1 m 1  m 2  m 3 ............ m n m1 r1 m2 r r2 m3 r3 x Rotational Motion 109 Hence the centre of mass of n particles is a weighted average of the position vectors of n particles making up the system. m r  m 2 r2 (2) Position vector of centre of mass for two particle system : r  1 1 m1  m 2 60 and the centre of mass lies between the particles on the line joining them. r r If two masses are equal i.e. m1  m 2 , then position vector of centre of mass r  1 2 2 (3) Important points about centre of mass E3 (i) The position of centre of mass is independent of the co-ordinate system chosen. ID (ii) The position of centre of mass depends upon the shape of the body and distribution of mass. Example : The centre of mass of a circular disc is within the material of the body while that of a circular ring is outside the material of the body. U (iii) In symmetrical bodies in which the distribution of mass is homogenous, the centre of mass coincides with the geometrical centre or centre of symmetry of the body. D YG (iv) Position of centre of mass for different bodies Body Position of centre of mass (a) Uniform hollow sphere Centre of sphere (b) Uniform solid sphere Centre of sphere (c) Uniform circular ring Centre of ring (d) Uniform circular disc Centre of disc U S. No. Uniform rod Centre of rod (f) A plane lamina (Square, Rectangle, Parallelogram) Point of inter section of diagonals (g) Triangular plane lamina Point of inter section of medians (h) Rectangular or cubical block Points of inter section of diagonals (i) Hollow cylinder Middle point of the axis of cylinder (j) Solid cylinder Middle point of the axis of cylinder (k) Cone or pyramid On the axis of the cone at point distance ST (e) 3h from the vertex where h is the height 4 of cone 110 Rotational Motion (v) The centre of mass changes its position only under the translatory motion. There is no effect of rotatory motion on centre of mass of the body. (vi) If the origin is at the centre of mass, then the sum of the moments of the masses of the 60 system about the centre of mass is zero i.e.  m i ri  0. (vii) If a system of particles of masses m 1 , m 2 , m 3 ,...... move with velocities v 1 , v 2 , v 3 ,......  m iv i.  mi E3 then the velocity of centre of mass v cm  (viii) If a system of particles of masses m1 , m 2 , m 3 ,...... move with accelerations a1 , a 2 , a 3 ,...... then the acceleration of centre of mass A cm  ID  m i ai mi (ix) If r is a position vector of centre of mass of a system U then velocity of centre of mass v cm   dr d  m1 r1  m 2 r2  m 3 r 3 ......     dt dt  m1  m 2  m 3 ......    D YG (x) Acceleration of centre of mass A cm   d v cm d 2 r d 2  m1 r1  m 2 r2 .......     dt dt 2 dt 2  m1  m 2  m 3 .......    (xi) Force on a rigid body F  M A cm  M d2 r dt 2 (xii) For an isolated system external force on the body is zero d    v cm   0  v cm  constant. dt   U FM i.e., centre of mass of an isolated system moves with uniform velocity along a straight-line ST path. Problem 1. Sample problems based on centre of mass The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is 1.1 Å. Given, mass of carbon atom is 12 a.m.u. and mass of oxygen atom is 16 a.m.u., calculate the position of the center of mass of the carbon monoxide molecule (a) 6.3 Å from the carbon atom (b) 1 Å from the oxygen atom (c) 0.63 Å from the carbon atom (d) 0.12 Å from the oxygen atom Solution : (c) Let carbon atom is at the origin and the oxygen atom is placed at x-axis m1  12 , m 2  16 , r 1  0ˆi  0 ˆj and r 2  1.1ˆi  0 ˆj y m1 C O m2 x Rotational Motion 111 m r  m 2 r2 16  1.1 r  1 1  î m1  m 2 28 r  0. 63 ˆi i.e. 0.63 Å from carbon atom. Solution : (a) Velocity of centre of mass v cm  Problem 3. 60    The velocities of three particles of masses 20g, 30g and 50 g are 10 i , 10 j, and 10 k respectively. The velocity of the centre of mass of the three particles is [EAMCET 2001]             (a) 2 i  3 j  5k (b) 10(i  j  k ) (c) 20 i  30 j  5k (d) 2 i  30 j  50k m1v1  m 2v 2  m 3 v 3 20  10 ˆi  30  10 ˆj  50  10 kˆ   2ˆi  3ˆj  5kˆ. m1  m 2  m 3 100 E3 Problem 2. Masses 8, 2, 4, 2 kg are placed at the corners A, B, C, D respectively of a square ABCD of diagonal 80 cm. The distance of centre of mass from A will be (b) 30 cm (c) 40 cm (d) 60 cm ID (a) 20 cm Solution : (b) Let corner A of square ABCD is at the origin and the mass 8 kg is placed at this corner (given in problem) Diagonal of square d  a 2  80 cm  a  40 2 cm U m1  8 kg, m 2  2kg, m 3  4 kg, m 4  2kg Let r 1 , r 2 , r 3 , r 4 are the position vectors of respective masses C D (0, a) D YG r 1  0ˆi  0 ˆj , r2  aˆi  0 ˆj , r3  aˆi  aˆj , r4  0ˆi  aˆj y From the formula of centre of mass m r  m 2 r2  m 3 r3  m 4 r4 r  1 1  15 2i  15 2ˆj m1  m 2  m 3  m 4 2kg 4kg C.M. 2kg (0, 8kg 0) A 402 (a, a) (a, B0) x  co-ordinates of centre of mass  (15 2 , 15 2 ) and co-ordination of the corner  (0, 0) From the formula of distance between two points (x1 , y1 ) and (x 2 , y 2 ) U distance  (x 2  x1 )2  (y 2  y1 )2 = Problem 4. The coordinates of the positions of particles of mass 7, 4 and 10 gm are (1, 5,  3), (2, 5,7) and (3, 3,  1) cm respectively. The position of the centre of mass of the system would be ST  15 85 1  , ,  cm (a)    7 17 7  Solution: (c) (15 2  0)2  (15 2  0) 2 = 900 = 30 cm  15 85 1  ,  cm (b)  , 17 7   7  15 85 1  ,  cm (c)  ,  7 21 7   15 85 7  ,  cm (d)  ,  7 21 3  m1  7 gm , m 2  4 gm , m 3  10 gm and r1  (ˆi  5 ˆj  3kˆ ), r2  (2i  5 j  7 k ), r3  (3ˆi  3ˆj  kˆ ) Position vector of center mass r   r  7(ˆi  5 ˆj  3kˆ )  4 (2ˆi  5 ˆj  7 kˆ )  10 (3ˆi  3 ˆj  kˆ ) (45 ˆi  85 ˆj  3kˆ )  7  4  10 21 15 ˆ 85 ˆ 1 ˆ  15 85  1  , i j  k. So coordinates of centre of mass  , . 7 21 7  7 21 7  7.3 Angular Displacement. It is the angle described by the position vector r about the axis of rotation. Q S  r P 112 Rotational Motion Angular displacement ( )  Linear displaceme nt (s) Radius (r) (1) Unit : radian (2) Dimension : [M 0 L0 T 0 ] 60 (3) Vector form S    r i.e., angular displacement is a vector quantity whose direction is given by right hand rule. It is also known as axial vector. For anti-clockwise sense of rotation direction of  is perpendicular to the plane, outward and along the axis of rotation and vice-versa. E3 (4) 2 radian  360   1 revolution. (5) If a body rotates about a fixed axis then all the particles will have same angular ID displacement (although linear displacement will differ from particle to particle in accordance with the distance of particles from the axis of rotation). 7.4 Angular Velocity. The angular displacement per unit time is defined as angular velocity.  where  is the angular displacement. t  d (1) Instantaneous angular velocity   lim  Q t 0 t dt D YG U If a particle moves from P to Q in time t ,   (2) Average angular velocity  av  total angular displaceme nt  2   1  t 2  t1 total time  P (3) Unit : Radian/sec (4) Dimension : [M 0 L0 T 1 ] which is same as that of frequency. (5) Vector form v    r U [where v = linear velocity, r = radius vector]  is a axial vector, whose direction is normal to the rotational plane and its direction is ST given by right hand screw rule. (6)   2  2n [where T = time period, n = frequency] T (7) The magnitude of an angular velocity is called the angular speed which is also represented by . 7.5 Angular Acceleration. The rate of change of angular velocity is defined as angular acceleration. If particle has angular velocity  1 at time t 1 and angular velocity  2 at time t 2 then, Angular acceleration    2  1 t 2  t1 Rotational Motion 113  d  d 2    (1) Instantaneous angular acceleration   lim. t 0 t dt dt 2 (2) Unit : rad/sec 2 (4) If   0 , circular or rotational motion is said to be uniform. (5) Average angular acceleration  av   2  1 t 2  t1. 60 (3) Dimension : [M 0 L0 T 2 ]. E3 (6) Relation between angular acceleration and linear acceleration a    r. (7) It is an axial vector whose direction is along the change in direction of angular velocity ID i.e. normal to the rotational plane, outward or inward along the axis of rotation (depends upon the sense of rotation). 7.6 Equations of Linear Motion and Rotational Motion. Linear Motion If linear acceleration a = constant, If angular acceleration  = constant then U If angular acceleration is 0,  = constant and ( 1   2 ) t 2 (u  v) t 2 (i)   (ii) a  v u t (ii)   (iii) v  u  at (iii)  2   1  t (iv) s  ut  (iv)   1 t  t 2 1 2 at 2 (v) v 2  u 2  2as 1 a(2n  1) 2 ST (vi) (3)   t (i) s  U (2) If linear acceleration is 0, u = constant and s = u t. D YG (1) Rotational Motion s nth  u   2  1 t 1 2 (v)  22  12  2 (vi)  nth  1  (2n  1)  2 If acceleration is not constant, the above If acceleration is not constant, the above equation will not be applicable. In this case equation will not be applicable. In this case (i) v  dx dt (i)   d dt (ii) a  dv d 2 x  2 dt dt (ii)   d  d 2  2 dt dt (iii) vdv  a ds (iii) d    d 114 Rotational Motion Sample problems based on angular displacement, velocity and acceleration The angular velocity of seconds hand of a watch will be (a)  60 (b) rad / sec  30 Solution : (b) We know that second's 2     rad/sec t 60 30 hand completes its revolution  (2 ) in 60 sec  The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the accelerator for 10 sec it starts rotating at 4500 revolutions per minute. The angular acceleration of the wheel is [MP PET 2001] (a) 30 radians/sec2 (b) 1880 degrees/sec2 E3 Problem 6. (d) 30  rad / sec (c) 60  rad / sec rad / sec 60 Problem 5. (c) 40 radians/sec2 (d) Solution: (d) Angular acceleration () = rate of change of angular speed Angular displacement ( ) of a flywheel varies with time as   at  bt 2  ct 3 then angular acceleration is given by U Problem 7. ID  4500  1200  3300 2   2 60 2 (n2  n1 )   60  360 degree  1980 degree / sec 2.    10 t 10 2 sec 2 (c) a  2b  6 t (b) 2b  6 t (a) a  2bt  3ct 2 (d) 2b  6 ct d d  2 (at  bt 2  ct 3 )  2b  6 ct 2 dt dt 2 D YG Solution: (d) Angular acceleration   2 Problem 8. A wheel completes 2000 rotations in covering a distance of 9.5 km. The diameter of the wheel is [RPMT 1999] (a) 1. 5 m (b) 1.5 cm (c) 7.5 m (d) 7.5 cm Solution: (a) Distance covered by wheel in 1 rotation = 2r  D (Where D= 2r = diameter of wheel)  Distance covered in 2000 rotation = 2000 D = 9.5  10 3 m (given) U  D  1.5 meter A wheel is at rest. Its angular velocity increases uniformly and becomes 60 rad/sec after 5 sec. The total angular displacement is ST Problem 9. (a) 600 rad (b) 75 rad Solution: (d) Angular acceleration   Now from   1 t   2  1 t (c) 300 rad  (d) 150 rad 60  0  12 rad / sec 2 5 1 1  t 2 = 0  (12 )(5)2  150 rad. 2 2 Problem 10. A wheel initially at rest, is rotated with a uniform angular acceleration. The wheel rotates through an angle  1 in first one second and through an additional angle  2 in the next one second. The ratio (a) 4 2 is 1 (b) 2 (c) 3 (d) 1 Rotational Motion 115 Solution: (c) Angular displacement in first one second  1  1   (1) 2  2 2......(i) 1 2 [From   1t   t 2 ] Now again we will consider motion from the rest and angular displacement in total two seconds 1  (2) 2  2 2......(ii) Solving (i) and (ii) we get  1   2 and  2  3 2  60 1   2  2 3. 1 Problem 11. As a part of a maintenance inspection the compressor of a jet engine is made to spin E3 300 250 0 0 200 0 1500 100 050 0 0 ID (in rev per min) according to the graph as shown. The number of revolutions made by the compressor during the test is 1 2 3 4 5 U t (in min) (b) 16570 (c) 12750 (d) 11250 D YG (a) 9000 Solution: (d) Number of revolution = Area between the graph and time axis = Area of trapezium = 1  (2.5  5)  3000 = 11250 revolution. 2 Problem 12. Figure shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If x and y be the distances travelled by A and B in the same time interval, then U (a) x  2y (b) x  y (c) y  2 x ST A B (d) None of these Solution: (c) Linear displacement (S) = Radius (r) × Angular displacement ()  S  r (if   constant) Distance travelled by mass A (x ) Radius of pulley concerned with mass A (r) 1   Distance travelled by mass B (y ) Radius of pulley concerned with mass B (2r) 2  y  2x. Problem 13. If the position vector of a particle is r  (3ˆi  4 ˆj) meter and its angular velocity is   (ˆj  2kˆ ) rad/sec then its linear velocity is (in m/s) (a) (8ˆi  6 ˆj  3kˆ ) (b) (3ˆi  6 ˆj  8 kˆ ) (c)  (3ˆi  6ˆj  6kˆ ) (d) (6ˆi  8 ˆj  3kˆ ) 116 Rotational Motion ˆi ˆj kˆ Solution: (a) v    r = (3ˆi  4 ˆj  0kˆ )  (0ˆi  ˆj  2kˆ )  3 4 0  8ˆi  6 ˆj  3kˆ 0 1 2 7.7 Moment of Inertia. 60 Moment of inertia plays the same role in rotational motion as mass plays in linear motion. It is the property of a body due to which it opposes any change in its state of rest or of uniform rotation. E3 (1) Moment of inertia of a particle I  mr 2 ; where r is the perpendicular distance of particle from rotational axis. (2) Moment of inertia of a body made up of number of particles (discrete distribution) I  m 1 r12  m 2 r22  m 3 r32 ....... U  dI  dm r 2 i.e., I  r 2 dm ID (3) Moment of inertia of a continuous distribution of mass, treating the element of mass dm at position r as particle r1 m2 m1 r2 D YG r m r3 r dm m3 (4) Dimension : [ML2 T 0 ] (5) S.I. unit : kgm2. (6) Moment of inertia depends on mass, distribution of mass and on the position of axis of U rotation. (7) Moment of inertia does not depend on angular velocity, angular acceleration, torque, ST angular momentum and rotational kinetic energy. (8) It is not a vector as direction (clockwise or anti-clockwise) is not to be specified and also not a scalar as it has different values in different directions. Actually it is a tensor quantity. (9) In case of a hollow and solid body of same mass, radius and shape for a given axis, moment of inertia of hollow body is greater than that for the solid body because it depends upon the mass distribution. 7.8 Radius of Gyration. Radius of gyration of a body about a given axis is the perpendicular distance of a point from the axis, where if whole mass of the body were concentrated, the body shall have the same moment of inertia as it has with the actual distribution of mass. Rotational Motion 117 When square of radius of gyration is multiplied with the mass of the body gives the moment of inertia of the body about the given axis. I. M I  Mk 2 or k  m m m From the formula of discrete distribution I  mr 1  mr 2  mr 3 .......  mr n2 2 2 If m1 = m2 = m3 =....... = m then I  m(r1  r2  r3 .......... rn2 ) 2 2 2 ID........(ii) By equating (i) and (ii) m r4 m........(i) From the definition of Radius of gyration, I  Mk 2 E3 2 r1 60 Here k is called radius of gyration. r2 r3 r5 k M 2 nmk 2 2  m(r1  r2  r3 ..........  rn2 ) 2 2 [As M  nm ] 2 2 D YG k 2 r1  r2  r3 ...........  rn2 n 2  2 U Mk 2  m(r1  r2  r3 ............  rn2 ) Hence radius of gyration of a body about a given axis is equal to root mean square distance of the constituent particles of the body from the given axis. (1) Radius of gyration (k ) depends on shape and size of the body, position and configuration of the axis of rotation, distribution of mass of the body w.r.t. the axis of rotation. U (2) Radius of gyration (k ) does not depends on the mass of body. (3) Dimension [M 0 L1 T 0 ]. ST (4) S.I. unit : Meter. (5) Significance of radius of gyration : Through this concept a real body (particularly irregular) is replaced by a point mass for dealing its rotational motion. Example : In case of a disc rotating about an axis through its centre of mass and perpendicular to its plane k (1 2)MR 2 I R   M M 2 So instead of disc we can assume a point mass M at a distance (R / 2 ) from the axis of rotation for dealing the rotational motion of the disc. Note :  For a given body inertia is constant whereas moment of inertia is variable. 118 Rotational Motion 7.9 Theorem of Parallel Axes. Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of the body about an axis parallel to given axis and passing through centre of mass of the body Ig IG I 60 and Ma 2 where M is the mass of the body and a is the perpendicular distance between the two axes. I  I g  Ma 2 a G E3 Example: Moment of inertia of a disc about an axis through its 1 centre and perpendicular to the plane is MR 2 , so moment of inertia 2 about an axis through its tangent and perpendicular to the plane will be  I 1 MR 2  MR 2 2 3 MR 2 2 I R IG G U I ID I  Ig  Ma2 7.10 Theorem of Perpendicular Axes. D YG According to this theorem the sum of moment of inertia of a plane lamina about two mutually perpendicular axes lying in its plane is equal to its moment of inertia about an axis perpendicular to the plane of lamina and passing through the point of intersection of first two axes. ST U Iz  I x  Iy Z X Y Example : Moment of inertia of a disc about an axis through its centre of mass and 1 perpendicular to its plane is MR 2 , so if the disc is in x–y plane then by theorem of 2 Z ID perpendicular axes X I I I i.e. ID O z x y   1 MR 2  2 ID 2 1 ID  MR 2 4 [As ring is symmetrical body so I x  I y  I D ] Y Rotational Motion 119 Note :  In case of symmetrical two-dimensional bodies as moment of inertia for all axes passing through the centre of mass and in the plane of body will be same so the two axes in the plane of body need not be perpendicular to each other. 60 7.11 Moment of Inertia of Two Point Masses About Their Centre of Mass. Let m 1 and m 2 be two masses distant r from each-other and r1 and r2 be the distances of their centre of mass from m 1 and m 2 respectively, then (1) r1  r2  r E3 (2) m 1 r1  m 2 r2 (3) r1  Centre of mass m1 r1 m2 r and m1  m 2 r2  m1 r m1  m 2 r2 ID (4) I  m 1 r12  m 2 r22 m2  m m  m1m 2 (5) I   1 2  r 2   r 2 [where   is known as reduced mass   m 1 and   m 2.] m1  m 2 m1  m 2  derived from above formula. U (6) In diatomic molecules like H 2 , HCl etc. moment of inertia about their centre of mass is D YG 7.12 Analogy Between Tranlatory Motion and Rotational Motion. Translatory motion Mass Linear momentum (m ) Moment Inertia P  mv Angular Momentum L  I Torque   I P  2mE F  ma U Force ST Kinetic energy Rotatory motion of (I) L 2 IE E 1 mv 2 2 E 1 2 I 2 E P2 2m E L2 2I 7.13 Moment of Inertia of Some Standard Bodies About Different Axes. Body Axis of Rotation Figure Moment of inertia k k2/R2 120 Rotational Motion Axis of Body Ring About an axis passing through C.G. and U Disc About a tangential axis perpendicular to its own plane R 2 1 2 3 MR 2 2 3 R 2 3 2 2MR 2 2R 2 1 MR 2 2 R 2 1 2 1 MR 2 4 R 2 1 4 60 1 E3 1 MR 2 2 ID Ring k2/R2 R MR 2 About its diameter About a tangential axis in its own plane k U Ring inertia About an axis passing through C.G. and perpendicular to its plane D YG Ring Rotation Moment of Figure ST perpendicular to its plane About its Diameter Disc Rotational Motion 121 Passing through the centre and perpendicular to the plane 5 4 3 R 2 3 2 M 2 [R1  R 22 ] 2 – – 3 MR 2 2 R2 R1 D YG Annular disc inner radius = R1 and outer radius = R2 5 R 2 k2/R2 60 Disc About a tangential axis perpendicular to its own plane 5 MR 2 4 k E3 Disc About a tangential axis in its own plane inertia ID Rotation Moment of Figure U Axis of Body Diameter M 2 [R1  R 22 ] 4 – – Annular disc Tangential and Parallel to the diameter M [5 R 12  R 22 ] 4 – – M [3 R 12  R 22 ] 2 – – ST U Annular disc Annular disc Tangential and perpendicular to the plane 122 Rotational Motion Axis of Body Rotation Moment of Figure inertia k k2/R2 R 2 1 2 3 R 2 3 2 R Solid cylinder 3 MR 2 2 (Generator) passing through its C.G. and D YG perpendicular to its own axis  L2 R 2  M   4   12 L2 R 2  12 4  L2 R 2  M   4   3 L2 R 2  3 4 ST U Solid cylinder About the diameter of one of faces of the cylinder 60 E3 Tangential About an axis 1 MR 2 2 L ID Solid cylinder About its own axis U Solid cylinder Cylindrical shell About its own axis MR2 R 1 Rotational Motion 123 Rotation Tangential About the diameter of one of faces of the cylinder Cylindrical shell E3 R2 Axis of cylinder U Hollow cylinder with inner radius = R1 and outer radius L2 R 2  12 2  L2 R 2  M   2   3 L2 R 2  3 2 R1 M 2 (R1  R 22 ) 2 ST = R2 Hollow cylinder with inner radius = R1 and outer radius = R2 Tangential 2R  L2 R 2  M   2   12 ID About an axis passing through its C.G. and perpendicular to its own axis k 2MR2 (Generator) D YG Cylindrical shell inertia U Cylindrical shell Moment of Figure k2/R2 60 Axis of Body M 2 (R 1  3 R 22 ) 2 2 124 Rotational Motion Axis of Body Rotation inertia About its diametric axis About its diametric axis Spherical shell 2 5 7 R 5 7 5 2 MR 2 3 2 R 3 2 3 5 MR 2 3 5 R 3 5 3 60 E3 Spherical shell 7 MR 2 5 About a tangential axis U Hollow sphere of inner radius R1 and outer radius R2 About its diametric axis 2  R2 5  R15  M  5  R2 3  R13  ST Hollow sphere k2/R2 2 R 5 ID About a tangential axis D YG Solid sphere k 2 MR 2 5 U Solid Sphere Moment of Figure 2 M [R 25  R15 ] Tangential 5(R 23 L  R13 )  MR 22 Rotational Motion 125 Axis of Body Rotation inertia About on axis passing through its centre of mass and ML 2 12 perpendicular to the rod. About an axis passing through its edge and E3 Long thin rod ML 2 3 ID perpendicular to the rod Rectangular M 2 [l  b 2 ] 12 D YG mass and perpendicular to the plane U Passing through the centre of lamina of length l and breadth b Rectangular lamina Tangential perpendicular to the plane and at the mid-point of breadth M [4 l 2  b 2 ] 12 Tangential perpendicular to the plane and M 2 [l  4 b 2 ] 12 U Rectangular lamina at the mid-point of length ST Rectangular parallelopiped length l, breadth b, thickness t Passing through centre of mass and parallel to (i) Length (x) (ii) breadth (z) k k2/R2 L 60 Long thin rod Moment of Figure ii b iii (i) M[b 2  t 2 ] 12 (ii) M[l 2  t 2 ] 12 i t l (iii) (iii) thickness (y) i ii iii M[b 2  l 2 ] 12 12 L 3 126 Rotational Motion Rotation Solid cone of Equilateral (i) length (x) (i) M 2 [3l  b 2  t 2 ] 12 (ii) M 2 [l  3b 2  t 2 ] 12 (iii) M 2 2 [l  b  3 t 2 ] 12 (ii) breadth (y) Passing through CM and perpendicular to the plane Axis joining the vertex and centre of the base M 2 [a  b 2 ] 4 3 MR 2 10 Passing through ST Right angled triangular lamina of sides a, b, c Ma 2 6 CM and perpendicular to the plane U triangular lamina with side a Along the edges k E3 (iii) thickness(z) D YG radius R and height h Tangential and parallel to ID Elliptical disc of semimajor axis = a and semiminor axis =b inertia U Rectangular parallelepiped length l, breath b, thickness t Moment of Figure k2/R2 60 Axis of Body (1) Mb 2 6 (2) Ma 2 6 (3) M 6  a 2b 2   2 2  a  b  Rotational Motion 125 Sample problem based on moment of inertia Problem 14. Five particles of mass = 2 kg are attached to the rim of a circular disc of radius 0.1 m and 60 negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is (a) 1 kg m2 (b) 0.1 kg m2 (c) 2 kg m2 (d) 0.2 kg m2 Solution: (b) We will not consider the moment of inertia of disc because it doesn't have any mass so moment of inertia of five particle system I  5 mr 2  5  2  (0.1)2  0.1 kg-m 2. Problem 15. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of moment of inertia IX and IY is (a) IY = 64IX (b) IY = 32IX Solution: (a) Moment of Inertia of disc I = t. 4 Then the relation between the E3 radius 4R is made from an iron plate of thickness [AIEEE 2003] (c) IY = 16IX (d) IY = IX 1 1 1 MR 2  (R 2 t )R 2  t R 4 2 2 2 ID [As M  V   = R 2 t where t  thickness,  = density]  Ix Iy       4 [If  = constant] 1 (4 ) 4  64 4 [Given Ry  4 R x , ty  tx ] 4 D YG  Ix ty  Ry   t x  R x U Iy I y  64 I x Problem 16. Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia U about an axis perpendicular to its plane and passing through a point on its rim will be (a) 5 I (b) 6 I (c) 3 I (d) 4 I 1 Solution: (b) Moment of inertia of disc about a diameter = MR 2  I (given)  MR2  4 I 4 Now moment of inertia of disc about an axis perpendicular to its plane and passing through a point on its rim = 3 3 MR 2  (4 I)  6 I. 2 2 ST Problem 17. Four thin rods of same mass M and same length l, form a square as shown in figure. Moment of inertia of this system about an axis through centre O and perpendicular to its plane is [MP PMT 2002] 4 (a) Ml 2 l 3 A B Ml 2 (b) 3 P l Ml 2 (c) 6 2 (d) Ml 2 3 Solution: (a) Moment of inertia of rod AB about point P  O D C l 1 Ml 2 12 l 126 Rotational Motion M.I. of rod AB about point O  Ml 2 l  M  12 2 2  1 Ml 2 3 [by the theorem of parallel axis] and the system consists of 4 rods of similar type so by the symmetry ISystem  4 Ml 2. 3 Problem 18. Three rings each of mass M and radius R are arranged as shown in the figure. The moment 60 of inertia of the system about YY will be Y (a) 3 MR2 3 MR 2 2 1 (d) 3 2 (c) 5 MR2 E3 (b) Y 7 MR 2 2 I  I1  I2  I3 Solution: (d) M.I of system about YY '  1 3 mR 2  mR 2 2 2  7 3 mR 2  mR 2 2 2 U I ID where I1 = moment of inertia of ring about diameter, I2 = I3 = M.I. of inertia of ring about a tangent in a plane Problem 19. Let l be the moment of inertia of an uniform square plate about an axis AB that passes D YG through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle  with AB. The moment of inertia of the plate about the axis CD is then equal to (b) l sin 2  (a) l [IIT-JEE 1998] (c) l cos 2  (d) l cos 2  2 Solution: (a) Let IZ is the moment of inertia of square plate about the axis which is passing through the centre and perpendicular to the plane. A IZ  I AB  I A'B'  ICD  IC 'D' [By the theorem of perpendicular axis] C C  U IZ  2 IAB  2 IA' B'  2 ICD  2 IC ' D' A [As AB, A' B' and CD, C' D' are symmetric axis] Hence ICD  IAB  l B D ST D B Problem 20. Three rods each of length L and mass M are placed along X, Y and Z-axes in such a way that one end of each of the rod is at the origin. The moment of inertia of this system about Z axis is [AMU 1995] (a) 2 ML2 3 (b) 4 ML2 3 (c) 5 ML2 3 ML2 3 (d) Solution: (a) Moment of inertia of the system about z-axis can be find out by calculating the moment of inertia of individual rod about z-axis z I1  I 2  ML 3 2 3 because z-axis is the edge of rod 1 and 2 x 2 y 1 Rotational Motion 127 and I3  0 because rod in lying on z-axis  Isystem  I1  I2  I3  ML2 ML2 2 ML2.  0  3 3 3 Problem 21. Three point masses each of mass m are placed at the corners of an equilateral triangle of (b) 3ma 2 (a) ma 2 (c) 3 ma 2 4 2 ma 2 3 C m I  I A  I B  IC  0  0  mx 2 a 2    3 ma 2  4  ID a 3  m  2  (d) E3 Solution: (c) The moment of inertia of system about AB side of triangle 60 side a. Then the moment of inertia of this system about an axis passing along one side of the triangle is [AIIMS 1995] a x m m A a B Ml 2 6 D YG (a) U Problem 22. Two identical rods each of mass M. and length l are joined in crossed position as shown in figure. The moment of inertia of this system about a bisector would be B1 B2 2 (b) Ml 12 (c) Ml 2 3 (d) Ml 2 4 U Solution: (b) Moment of inertia of system about an axes which is perpendicular to plane of rods and passing through the common centre of rods I z  Ml 2 Ml 2 Ml 2   12 12 6 ST Again from perpendicular axes theorem I z  I B1  I B2  2 I B1  2 I B2   I B1  I B2  Ml 2 6 [As I B1  I B2 ] Ml 2. 12 Problem 23. The moment of inertia of a rod of length l about an axis passing through its centre of mass and perpendicular to rod is I. The moment of inertia of hexagonal shape formed by six such rods, about an axis passing through its centre of mass and perpendicular to its plane will be (a) 16I (b) 40 I (c) 60 I (d) 80 I Solution: (c) Moment of inertia of rod AB about its centre and perpendicular to the length = ml 2  12 I ml 2 =I  12 128 Rotational Motion Now moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of hexagon Irod= ml 2  mx 2 12 [From the theorem of parallel A axes] x l 2  3  ml 2 5 ml 2  m l   2  12 6   B l l Now the moment of inertia of system Isystem = 6  Irod  6  60 O 5 ml 2  5ml 2 6 E3 [As ml 2  12 I ] Isystem = 5 (12 I) = 60 I Problem 24. The moment of inertia of HCl molecule about an axis passing through its centre of mass and (a) 0.61  10 47 kg. m 2 ID perpendicular to the line joining the H  and Cl  ions will be, if the interatomic distance is 1Å (b) 1.61  10 47 kg. m 2 (c) 0.061  10 47 kg. m 2 (d) 0 Solution: (b) If r1 and r2 are the respective distances of particles m1 and m 2 from the centre of mass U then m1r1  m 2 r2  1  x  35.5  (L  x )  x  35.5 (1  x )  x  0.973 Å and L  x  0.027 Å D YG L–x H Moment of inertia of the system about centre of mass I  m1 x 2  m 2 (Lm1 x )2x C.M. Cl m2 I  1amu  (0.973 Å)2  35.5 amu  (0.027 Å)2 Substituting 1 a.m.u. = 1.67  10–27 kg and 1 Å = 10–10 m I  1.62  10 47 kg m 2 U Problem 25. Four masses are joined to a light circular frame as shown in the figure. The radius of ST gyration of this system about an axis passing through the centre of the circular frame and perpendicular to its plane would be A 3m (a) a / 2 2m a O (b) a / 2 2m m A (c) a (d) 2a Solution: (c) Since the circular frame is massless so we will consider moment of inertia of four masses only. I  ma 2  2ma 2  3ma 2  2ma 2  8ma 2.....(i) Now from the definition of radius of gyration I  8mk 2.....(ii) Rotational Motion 129 comparing (i) and (ii) radius of gyration k  a. Problem 26. Four spheres, each of mass M and radius r are situated at the four corners of square of side R. The moment of inertia of the system about an axis perpendicular to the plane of square and passing through its centre will be 60 (b) 5 M (4 r 2  5 R 2 ) 2 r 2 M (4 r 2  5 R 2 ) 5 R O 2 (c) M (4 r 2  5 r 2 ) 5 (d) 5 M (4 r 2  5 r 2 ) 2 2 Mr 2 5 ID Solution: (b) M. I. of sphere A about its diameter I O '  E3 (a) Now M.I. of sphere A about an axis perpendicular to the plane of square and passing through its centre will be 2 U  R  2 MR 2   Mr 2  IO  IO '  M   5 2  2 [by the theorem of parallel axis] O A  D YG 2 MR 2  Moment of inertia of system (i.e. four sphere)= 4 IO  4  Mr 2   2   5  2 M 4r2  5 R 2 5  B R/ 2 O C D Problem 27. The moment of inertia of a solid sphere of density  and radius R about its diameter is 105 5 R  176 (b) 105 2 R  176 U (a) (c) 176 5 R  105 Solution: (c) Moment of inertia of sphere about it diameter I  24 2  MR 2   R 3   R 2 5 3 5   176 2 R  105 [As 4 R 3  ] 3 ST M  V = (d) I= 8  22 5 176 5 8 5 R  R  R   15  7 105 15 Problem 28. Two circular discs A and B are of equal masses and thickness but made of metals with densities d A and d B (d A  d B ). If their moments of inertia about an axis passing through centres and normal to the circular faces be I A and IB , then (a) I A  IB (b) I A  IB (c) I A  IB (d) I A    IB Solution : (c) Moment of inertia of circular disc about an axis passing through centre and normal to the circular face 130 Rotational Motion  I 1  M  1  MR 2  M  2 2   t  M2 2 t I or So, in the problem [As M  V   R 2 t  R 2  IA d B  IB d A 1  M t ] If mass and thickness are constant. 60 I  I A  IB [As d A  d B ] 7.14 Torque. E3 If a pivoted, hinged or suspended body tends to rotate under the action of a force, it is said to be acted upon by a torque. or The turning effect of a force about the axis of rotation is called moment of force or torque due to the force. Rotatio n P r r O F (B) D YG (A) F U O ID Rotatio n If the particle rotating in xy plane about the origin under the Y F sin  effect of force F and at any instant the position vector of the particle is r then,  F  r  F cos  P  U Torque  = r  F 90o   r F sin  X d ST [where  is the angle between the direction of r and F ] (1) Torque is an axial vector. i.e., its direction is always perpendicular to the plane containing vector r and F in accordance with right hand screw rule. For a given figure the sense of rotation is anti-clockwise so the direction of torque is perpendicular to the plane, outward through the axis of rotation. (2) Rectangular components of force Fr  F cos   radial component of force , As or F  F sin   transverse component of force   r F sin    r F = (position vector)  (transverse component of force) Rotational Motion 131 Thus the magnitude of torque is given by the product of transverse component of force and its perpendicular distance from the axis of rotation i.e., Torque is due to transverse component of force only. (3) As   r F sin     F(r sin  )  Fd or [As d  r sin  from the figure ] 60 i.e. Torque = Force  Perpendicular distance of line of action of force from the axis of rotation. Torque is also called as moment of force and d is called moment or lever arm. E3 (4) Maximum and minimum torque : As   r  F or   r F sin   m aximum  rF When sin   max  1  minimum  0 When sin   min  0 i.e.   0 or 180  i.e.,   90  F is perpendicular to r ID F is collinear to r (5) For a given force and angle, magnitude of torque depends on r. The more is the value of r, the more will be the torque and easier to rotate the body. U Example : (i) Handles are provided near the free edge of the Planck of the door. (ii) The handle of screw driver is taken thick. D YG (iii) In villages handle of flourmill is placed near the circumference. (iv) The handle of hand-pump is kept long. (v) The arm of wrench used for opening the tap, is kept long. (6) Unit : Newton-metre (M.K.S.) and Dyne-cm (C.G.S.) (7) Dimension : [ML2T 2 ]. (8) If a body is acted upon by more than one force, the total torque is the vector sum of each torque. U    1   2   3 ........ (9) A body is said to be in rotational equilibrium if resultant torque acting on it is zero i.e. ST   0. (10) In case of beam balance or see-saw the system will be in rotational equilibrium if, R  1   2  0 or F1 l1  F2 l 2  0  F1 l1  F2 l 2 l1 l2 However if,  1   2 , L.H.S. will move downwards and if  1   2. R.H.S. will move downward. and the system will not F1 F2 be in rotational equilibrium. (11) On tilting, a body will restore its initial position due to torque of weight about the point O till the line of action of weight passes through its base on tilting, a body will topple due to torque of weight about O, if the line of action of weight does not pass through the base. Tilt Torque G R O G Tilt R O Torqu e 60 132 Rotational Motion (12) Torque is the cause of rotatory motion and in rotational motion it plays same role as force plays in translatory motion i.e., torque is rotational analogue of force. This all is evident E3 from the following correspondences between rotatory and translatory motion. Rotatory Motion Translatory Motion   I   d P    7.15 Couple.  F ds P  F v dP F dt D YG dL   dt W ID  U W Fm a A special combination of forces even when the entire body is free to move can rotate it. This combination of forces is called a couple. (1) A couple is defined as combination of two equal but oppositely directed force not acting along the same line. The effect of couple is known by its moment of couple or torque by a couple ST U τ  r F. F r F (2) Generally both couple and torque carry equal meaning. The basic difference between torque and couple is the fact that in case of couple both the forces are externally applied while in case of torque one force is externally applied and the other is reactionary. (3) Work done by torque in twisting the wire W  1 C 2. 2 Where   C ; C is known as twisting coefficient or couple per unit twist. 7.16 Translatory and Rotatory Equilibrium. Rotational Motion 133 Forces are equal and does not act along the same line. F  F  0 and    0 Rotation i.e. spinning.  F  0 and    0 same F2 F1 same line. skidding. F1 Forces are unequal and does not act along the Translation i.e. slipping or E3 the Body will remain stationary if initially it was at rest. F Forces are unequal and act along line.  F  0 and    0 F F 60 Forces are equal and act along the same line.  F  0 and    0 ID F2 Rotation and translation both i.e. rolling. Sample problems based on torque and couple Problem 29. A force of (2ˆi  4 ˆj  2kˆ ) N acts at a point (3ˆi  2ˆj  4 kˆ ) metre from the origin. The magnitude of (a) Zero (b) 24.4 N-m (c) 0.244 N-m (d) 2.444 N-m F  (2ˆi  4 ˆj  2kˆ ) N and r  (3 i  2  4 kˆ ) meter D YG Solution: (b) U torque is ˆi ˆj kˆ Torque   r  F  3 2  4    12 ˆi  14 ˆj  16 kˆ and |  |  (12)2  (14 )2  (16)2 = 24.4 2 4 2 N-m Problem 30. The resultant of the system in the figure is a force of 8 N parallel to the given force through U R. The value of PR equals to R (a) 1 4 RQ P Q ST (b) 3 8 RQ 5N (c) 3 5 RQ 3N (d) 2 5 RQ Solution: (c) By taking moment of forces about point R, 5  PR  3  RQ  0  PR  3 RQ. 5 Problem 31. A horizontal heavy uniform bar of weight W is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to (a) W (b) W 2 (c) 3W 4 (d) W 4 134 Rotational Motion  Weight (W) = Mg Solution: (d) Let the mass of the rod is M Initially for the equilibrium F  F  Mg  F  Mg / 2 F When one man withdraws, the torque on the rod B l 2 Mg Ml 2 l   Mg 3 2 60  A [As I = Ml 2/ 3]  Angular acceleration   3 g 2 l and linear acceleration a  l 3g  2 4 F A E3   I  Mg F Mg B B Now if the new normal force at A is F ' then Mg  F'  Ma Mg W 3 Mg.   4 4 4 ID  F'  Mg  Ma  Mg  7.17 Angular Momentum. U The turning momentum of particle about the axis of rotation is called the angular momentum of the particle. D YG or The moment of linear momentum of a body with respect to any axis of rotation is known as angular momentum. If P is the linear momentum of particle and r its position vector from the point of rotation then angular momentum. L r P L  r P L  r P sin  nˆ U Angular momentum is an axial vector i.e. always directed perpendicular to the plane of rotation and along the axis of rotation. ST (1) S.I. Unit : kg-m2-s–1 or J-sec. (2) Dimension : [ML2 T -1 ] and it is similar to Planck’s constant (h). (3) In cartesian co-ordinates if r  xˆi  yˆj  zkˆ and P  Px ˆi  Py ˆj  Pz kˆ ˆi Then L  r  P  x Px ˆj y Py kˆ z = (yPz  z Py )ˆi  (xPz  zPx ) ˆj  (xPy  yPx ) kˆ Pz (4) As it is clear from the figure radial component of momentum P r  P cos  Y P P sin   P cos  P r  X Rotational Motion 135 Transverse component of momentum P   P sin  So magnitude of angular momentum L  r P sin  60 L  r P  Angular momentum = Position vector × Transverse component of angular momentum E3 i.e., The radial component of linear momentum has no role to play in angular momentum. (5) Magnitude of angular momentum L  P (r sin  )  L  Pd [As d  r sin  from the figure.]  Angular momentum = (Linear momentum) (Perpendicular distance of line of action of force from the axis of rotation) ID (6) Maximum and minimum angular momentum : We know L  r  P L  m [ r  v ]  m v r sin   P r sin  [As P  m v ] U  Lm aximum  mvr When sin   max  1 D YG When sin   min  0 i.e.   0 or 180  Lminimum  0 i.e.,   90  v is perpendicular to r v is parallel or anti-parallel to r (7) A particle in translatory motion always have an angular momentum unless it is a point on the line of motion because L  mvr sin  and L  1 if   0 o or 180 o  L  mvr  mr 2  [As r  v and v  r ] L  I [As mr2 = I] ST or U (8) In case of circular motion, L  r  P  m ( r  v ) = mvr sin  In vector form L  I  dL d I (9) From L  I   = I   dt dt d   and   I  ] [As dt i.e. the rate of change of angular momentum is equal to the net torque acting on the particle. [Rotational analogue of Newton's second law] (10) If a large torque acts on a particle for a small time then 'angular impulse' of torque is given by 136 Rotational Motion   J   dt   av t2 t1 dt or Angular impulse J   av t   L 60  Angular impulse = Change in angular momentum (11) The angular momentum of a system of particles is equal to the vector sum of angular momentum of each particle i.e., L  L1  L 2  L 3 .......  Ln. Ln E3 (12) According to Bohr theory angular momentum of an electron in nth orbit of atom can be taken as, h 2 ID orbit] [where n is an integer used for number of 7.18 Law of Conservation of Angular Momentum. U dL Newton’s second law for rotational motion   dt D YG dL 0 So if the net external torque on a particle (or system) is zero then dt i.e. L  L1  L 2  L 3 ....... = constant. Angular momentum of a system (may be particle or body) remains constant if resultant torque acting on it zero.  I 1  U As L  I so if   0 then I  constant ST Since angular momentum I remains constant so when I decreases, angular velocity  increases and vice-versa. Examples of law of conservation of angular momentum : (1) The angular velocity of revolution of a planet around the sun in an elliptical orbit increases when the planet come closer to the sun and vice-versa because when planet comes closer to the sun, it's moment of inertia I decreases there fore  increases. (2) A circus acrobat performs feats involving spin by bringing his arms and legs closer to his body or vice-versa. On bringing the arms and legs closer to body, his moment of inertia I decreases. Hence  increases. Rotational Motion 137 (3) A person-carrying heavy weight in his hands and standing on a rotating platform can change the speed of platform. When the person suddenly folds his arms. Its moment of inertia E3 60 decreases and in accordance the angular speed increases. (4) A diver performs somersaults by Jumping from a high diving board keeping his legs and arms out stretched first and then curling his body. (5) Effect of change in radius of earth on its time period L 2 2 MR 2   constant 5 T T  R2 U  ID L  I  constant Angular momentum of the earth [if M remains constant] D YG If R becomes half then time period will become one-fourth i.e. 24  6 hrs. 4 Sample problems based on angular momentum Problem 32. Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = Mv is imparted to the body at one of its ends, what would be its angular velocity [IIT-JEE (Screening) 2003] U (a) v/L (b) 2v/L L M M (c) v/3L ST J = Mv (d) v/4L Solution: (a) Initial angular momentum of the system about point O = Linear momentum × Perpendicular distance of linear momentum from the axis of rotation L  Mv  ....(i) 2 Final angular momentum of the system about point O 2   L 2 L   I1  I2  (I1  I2 )   M    M   ....(ii)  2     2  Applying the law of conservation of angular momentum O M 1 L/2 2 M L/2 138 Rotational Motion 2  L L Mv    2 M    2 2   v L Problem 33. A thin circular ring of mass M and radius R is rotating about its axis with a constant M M  4m (b) (M  4 m ) M (c) Solution: (a) Initial angular momentum of ring  I  MR 2 (M  4 m ) M  4m (d) M 4m E3 (a) 60 angular velocity . Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be If four object each of mass m, and kept gently to the opposite ends of two perpendicular diameters of the ring then final angular momentum = (MR 2  4 mR 2 ) ' By the conservation of angular momentum ID Initial angular momentum = Final angular momentum ST U D YG U M   MR 2  (MR 2  4 mR 2 ) '   '   .  M  4m  Rotational Motion 137 Problem 34. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its center. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity 0. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform  (t) will vary with time t as (t) [IIT-JEE (Screening) 2002] (t) (t) O O (a) (b) (c) t O (d) t t E3 t 60 (t) O Solution: (b) The angular momentum (L) of the system is conserved i.e. L = I = constant ID When the tortoise walks along a chord, it first moves closer to the centre and then away from the centre. Hence, M.I. first decreases and then increases. As a result,  will first increase and then decrease. Also the change in  will be non-linear function of time. Problem 35. The position of a particle is given by : r  (ˆi  2ˆj  kˆ ) and momentum P  (3ˆi  4 ˆj  2kˆ ). The U angular momentum is perpendicular to (a) X-axis D YG (b) Y-axis (c) Z-axis (d) Line at equal angles to all the three axes L  r p  ˆi 1 3 ˆj 2 4 kˆ 1 2  0ˆi  ˆj  2kˆ  ˆj  2kˆ and the X- axis is given by i  0 ˆj  0kˆ U Solution: (a) Dot product of these two vectors is zero i.e. angular momentum is perpendicular to X-axis. ST Problem 36. Two discs of moment of inertia I1 and I2 and angular speeds  1 and  2 are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational KE of system will be (a) I1 1  I 2  2 2(I1  I 2 ) (b) (I1  I 2 ) ( 1   2 ) 2 (I   I 2  2 ) 2 (c) 1 1 2(I1  I 2 ) 2 (d) None of these Solution: (c) By the law of conservation of angular momentum I11  I22  I1  I2  Angular velocity of system   I11  I2 2 I1  I2 138 Rotational Motion 2 Rotational kinetic energy  2   1 I1  I2   2  1 I1  I2   I11  I22   (I11  I22 ). 2 2 2(I1  I2 )  I1  I2  Problem 37. A smooth uniform rod of length L and mass M has two identical beads of negligible size, 60 each of mass m , which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity  0 about an axis perpendicular to the rod and passing through the mid point of the rod (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is [IIT-JEE 1998] (c) M 0 M  2m (d) M 0 M  6m L/2 L/2 ID M 0 M  12 m U (b) E3 (a)  0 Solution: (d) Since there are no external forces therefore the angular momentum of the system remains constant. D YG  ML 2 Initially when the beads are at the centre of the rod angular momentum L1    12    0  .....(i) When the beads reach the ends of the rod then angular momentum U 2   L 2 ML 2  L  m    m     '.....(ii)  2  2 12     Equating (i) and (ii)  mL 2 ML2  M o ML2  '   ' . 0     M  6m 12 12   2 ST Problem 38. Moment of inertia of uniform rod of mass M and length L about an axis through its centre ML 2. Now consider one such rod pivoted at its 12 centre, free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass M moving horizontally at a speed v strikes and embedded in one end of the rod. The angular velocity of the rod just after the collision will be and perpendicular to its length is given by (a) v L (b) 2v L (c) 3v 2 L (d) 6v L Solution: (c) Initial angular momentum of the system = Angular momentum of bullet before collision L  Mv  .....(i) 2 Rotational Motion 139 let the rod rotates with angular velocity .  ML2 Final angular momentum of the system    12  L  ML 2 ML 2     or   3v / 2 L 2  12 4  v M 60 Mv......(ii) E3 By equation (i) and (ii) 2    M  L    2  Problem 39. A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with angular velocity 3 rad / s. A particle of mass 0.5 kg and moving with a velocity 5 m/s ID strikes the cylinder and sticks to it as shown in figure. The angular momentum of the cylinder before collision will be (a) 0.12 J-s U (b) 12 J-s 3 rad/s (c) 1.2 J-s D YG (d) 1.12 J-s Solution: (a) Angular momentum of the cylinder before collision L  I  1 1 MR 2  (2) (0.2)2  3 = 0.12 J-s. 2 2 Problem 40. In the above problem the angular velocity of the system after the particle sticks to it will be (a) 0.3 rad/s (b) 5.3 rad/s (c) 10.3 rad/s (d) 89.3 rad/s Solution: (c) Initial angular momentum of bullet + initial angular momentum of cylinder = Final angular momentum of (bullet + cylinder) system U  mvr  I1  (I1  I2 ) ' ST 1   mvr  I1 =  Mr 2  mr 2  ' 2  1   0.5  5  0.2  0.12   2(0.2)2  (0.5)(0.2)2  ' 2    '  10.3 rad/sec. 7.19 Work, Energy and Power for Rotating Body. (1) Work : If the body is initially at rest and angular displacement is d  due to torque then work done on the body.  W   d  [Analogue to work in translatory motion W  F dx ] 140 Rotational Motion (2) Kinetic energy : The energy, which a body has by virtue of its rotational motion is called rotational kinetic energy. A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place. 1 2 I 2 KR  1 L 2 1 mv 2 KT  1 Pv 2 KT  P2 2m 2 L2 2I ID KR  KT  E3 KR  Analogue to translatory kinetic energy 60 Rotational kinetic energy (3) Power : Rate of change of kinetic energy is defined as power d d 1 d   I  I   (K R )   I 2   I dt dt  2 dt  U P In vector form Power     D YG [Analogue to power in translatory motion P  F v ] 7.20 Slipping, Spinning and Rolling. (1) Slipping : When the body slides on a surface without rotation then its motion is called slipping motion. =0 In this condition friction between the body and surface F  0. v 1 Body possess only translatory kinetic energy K T  mv 2. 2 U Example : Motion of a ball on a frictionless surface. ST (2) Spinning : When the body rotates in such a manner that its axis of rotation does not move then its motion is called spinning motion. In this condition axis of rotation of a body is fixed.  Example : Motion of blades of a fan. In spinning, body possess only rotatory kinetic energy K R  or KR  1 mK 2 2 v2 R2   K2 1 mv 2  2 2 R  K2 i.e., Rotatory kinetic energy =  2 R 1 2 I. 2       times translatory kinetic energy.   Rotational Motion 141 K2 K2 1 K2 1 K2 is a constant for different bodies. Value of (ring), (disc) and    1 2 R2 R2 R2 R2 2 (solid sphere) Here Example : (i) Motion of a wheel of cycle on a road. 60 (3) Rolling : If in case of rotational motion of a body about a fixed axis, the axis of rotation also moves, the motion is called combined translatory and rotatory.  (ii) Motion of football rolling on a surface. v E3 In this condition friction between the body and surface F  0. Body possesses both translational and rotational kinetic energy. Net kinetic energy = (Translatory + Rotatory) kinetic energy. KN   1 K2  mv 2  1  2  2 R   7.21 Rolling Without Slipping. U  1 1 1 1 K2 mv 2  I 2  mv 2  mv 2 2 2 2 2 2 R ID K N  KT  KR  D YG In case of combined translatory and rotatory motion if the object rolls across a surface in such a way that there is no relative motion of object and surface at the point of contact, the motion is called rolling without slipping. Friction is responsible for this type of motion but work done or dissipation of energy against friction is zero as there is no relative motion between body and surface at the point of contact. Rolling motion of a body may be treated as a pure rotation about an axis through point of contact with same angular velocity .  By the law of conservation of energy 1 1 [ As v  R  ] mv 2  I 2 v 2 2 O 1 1  mR 2 2  I 2 2 2 P 1 2 2 =  [mR  I] 2 1 1 =  2 [I  mR 2 ]  I P  2 [As I P  I  mR 2 ] 2 2 By theorem of parallel axis, where I = moment of inertia of rolling body about its centre ‘O’ and IP = moment of inertia of rolling body about point of contact ‘P’. ST U KN  (1) Linear velocity of different points in rolling : In case of rolling, all points of a rigid body have same angular speed but different linear speed. Let A, B, C and D are four points then their velocities are shown in the following figure. B B v C D v B v v v=0 v D D C v 2v v C 2 v 142 Rotational Motion + (2) Energy distribution table for different rolling bodies : Translatory (KT) Rotatory (KR) 1 mv 2 2 1 K2 mv 2 2 2 R 1 mv 2 2 1 mv 2 2 1 2 2 5 1 mv 2 2 1 mv 2 4 1 mv 2 2 1 mv 2 5 7 mv 10 2 3 1 mv 2 2 1 mv 2 3 5 mv 2 6 Ring Cylindrical shell 1 Disc solid cylinder Solid sphere 1 2 mv 2    K2 R2 1 mv 2 D YG Hollow sphere Total (KN) KT (%) KN KR (%) KN 1 (50%) 2 1 (50%) 2 2 (66.6%) 3 5 (71.5%) 7 1 (33.3%) 3 2 (28.5%) 7 3 (60%) 5 2 (40 %) 5 E3 R 2     ID K2 3 mv 2 4 U Body 60 = 2 Sample problems based on kinetic energy, work and power Problem 41. A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with an angular velocity of 20 rad/s. Its kinetic energy is (a) 10 J (b) 100 J U Solution: (d) Rotational kinetic energy (c) 500 J (d) 250 J 1 1 1 1 1   I 2   MR 2   2    10  (0. 5) 2  20 2  250 J 22 2 22   Problem 42. An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min. What torque does it deliver ST (a) 350 N-m Solution: (c) P      [CBSE PMT 2000] (b) 440 N-m (c) 531 N-m (d) 628 N-m 100  10  531 N- m 1800 2 60 3 Problem 43. A body of moment of inertia of 3 kg-m2 rotating with an angular velocity of 2 rad/sec has the same kinetic energy as a mass of 12 kg moving with a velocity of (a) 8 m/s (b) 0.5 m/s (c) 2 m/s Solution: (d) Rotational kinetic energy of the body = According to problem  1 1 I 2  mv 2 2 2  (d) 1 m/s 1 1 I 2 and translatory kinetic energy  mv 2 2 2 1 1  3  (2)2   12  v 2  v  1 m/s. 2 2 Rotational Motion 143 Problem 44. A disc and a ring of same mass are rolling and if their kinetic energies are equal, then the ratio of their velocities will be (c) 3 : 4 K disc   3 1 k2  mv d2 1  2   mv d2   4 2 R    k2 1  As 2  2  R  for disc   K ring   1 k2  mv r 1  2   mv r2  2 R    k2  As 2  1  R  for ring   According to problem K disc  K ring  (d) 3: 2 v 3 mv d2  mv r2  d  4 vr 2: 3 4. 3 E3 Solution: (a) (b) 4 : 3 60 (a) Problem 45. A wheel is rotating with an angular speed of 20 rad / sec. It is stopped to rest by applying a constant torque in 4 s. If the moment of inertia of the wheel about its axis is 0.20 kg-m2, then the work done by the torque in two seconds will be Solution: (c) (b) 20 J (c) 30 J ID (a) 10 J  1  20 rad/sec,  2  0, t  4 sec. So angular retardation   1   2 t  20  5 rad / sec 2 4 2  1  t  20  5  2 = 10 rad/sec U Now angular speed after 2 sec (d) 40 J D YG Work done by torque in 2 sec = loss in kinetic energy =    1 1 I 12   22  (0.20 )((20 )2  (10 )2 ) 2 2 1  0.2  300 = 30 J. 2 Problem 46. If the angular momentum of a rotating body is increased by 200%, then its kinetic energy of rotation will be increased by (a) 400% E L2 2I 2   3L  E 2  L2       1  E1  L1   L1  (c) 200% (d) 100% 2 [As L2  L1  200 %.L1 = 3L1] U Solution: (b) As (b) 800%  E2  9 E1  E1  800 % of E1 Problem 47. A ring, a solid sphere and a thin disc of different masses rotate with the same kinetic ST energy. Equal torques are applied to stop them. Which will make the least number of rotations before coming to rest (a) Disc (b) Ring (c) Solid sphere (d) All will make same number of rotations Solution: (d) As W   = Energy    Energy   2n  So, if energy and torque are same then all the bodies will make same number of rotation. Problem 48. The angular velocity of a body is   2ˆi  3ˆj  4 kˆ and a torque   ˆi  2ˆj  3kˆ acts on it. The rotational power will be (a) 20 W (b) 15 W (c) 17 W Solution: (a) Power (P )  .   (i  2ˆj  3kˆ ). (2ˆi  3ˆj  4 kˆ )  2  6  12 = 20 W (d) 14 W 144 Rotational Motion Problem 49. A flywheel of moment of inertia 0.32 kg-m2 is rotated steadily at 120 rad / sec by a 50 W electric motor. The kinetic energy of the flywheel is (b) 1152 J Solution: (c) Kinetic energy K R  (c) 2304 J (d) 6912 J 1 1 I 2  (0.32 ) (120 )2 = 2304 J. 2 2 60 (a) 4608 J 7.22 Rolling on an Inclined Plane. By conservation of mechanical energy mgh  E3 When a body of mass m and radius R rolls down on inclined plane of height ‘h’ and angle of inclination  , it loses potential energy. However it acquires both linear and angular speeds and hence, gain kinetic energy of translation and that of rotation. B S h  1 k2  mv 2  1  2  2 R   Translatio n ID 2 gh (1) Velocity at the lowest point : v  k2 1 2 R Rotation (2) Acceleration in motion : From equation v 2  u 2  2aS 2 gh we get k2 1 2 R C D YG U h and v  By substituting u  0, S  sin   a g sin  k2 1 2 R (3) Time of descent : From equation v  u  at By substituting u = 0 and value of v and a from above expressions 1 sin  U t 2h  k2  1  2  g  R  ST From the above expressions it is clear that, v  Note 1 1 k2 R2 ; a 1 k2 ; t  1  k2 R2 1 2 R  k2  :  Here factor  2  is a measure of moment of inertia of a body and its value is R  constant for given shape of the body and it does not depend on the mass and radius of a body.  Velocity, acceleration and time of descent (for a given inclined plane) all depends k2 on 2. Lesser the moment of inertia of the rolling body lesser will be the value of R Rotational Motion 145 60 k2. So greater will be its velocity and acceleration and lesser will be the time of R2 descent.  If a solid and hollow body of same shape are allowed to roll down on inclined plane  k2   k2  then as  2    2  , solid body will reach the bottom first with greater velocity.  R S  R H E3  If a ring, cylinder, disc and sphere runs a race by rolling on an inclined plane then  k2   k2  as  2   maximum , the sphere will reach the bottom  minimum while  2  R R    sphere  Ring first with greatest velocity while ring at last with least velocity.  Angle of inclination has no effect on velocity, but time of descent and acceleration depends on it. ID velocity    , time of decent   1 and acceleration  . 7.23 Rolling Sliding and Falling of a Body. R2 k2 Sliding R 2 k2 R2 Falling 0 Acceleration h  2 gh 1  k 2 R2 g sin  1  K2 R2 2 gh g sin  2 gh g D YG k2 Rolling Velocity U Figure 0 U  = 90o 2h  k2  1 2  g  R  1 sin  1 sin   0 Time 2h g 2h g  ST 7.24 Velocity, Acceleration and Time for Different Bodies. Body Ring or k2 R2 Velocity v  2 gh 1 R 2 a gsin θ 1 k 2 R2 Time of descent t 1 2 h  k2  1 2 sin θ g  R  1 gh 1 g sin  2 1 sin  4h g 1 or 0.5 2 4 gh 3 2 g sin  3 1 sin  3h g Hollow cylinder Disc or solid cylinder k 2 Acceleration 146 Rotational Motion 2 or 0.4 5 10 gh 7 5 g sin  7 1 sin  14 h 5 g Hollow sphere 2 or 0. 66 3 6 gh 5 3 g sin  5 1 sin  10 h 3 g 60 Solid sphere Sample problems based on rolling on an inclined plane 3 gh 4 4 gh 3 (b) 2 gh Solution: (b) Velocity at the bottom (v)  1 K 2 2 gh  1 1 2  R2 (d) 4 gh 2 gh 4 gh. 3 A sphere rolls down on an inclined plane of inclination . What is the acceleration as the sphere reaches bottom [Orissa JEE 2003] (a) 5 g sin  7 g sin  K 2  3 g sin  5 (c) 1 R2 2 g sin  7 (d) 2 g sin  5 g sin  5  g sin . 2 7 1 5 D YG Solution: (a) Acceleration (a)  (b) U Problem 51. (c) ID (a) E3 Problem 50. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom [CBSE PMT 2003] Problem 52. A ring solid sphere and a disc are rolling down from the top of the same height, then the sequence to reach on surface is (a) Ring, disc, sphere (b) Sphere, disc, ring (c) Disc, ring, sphere (d) Sphere, ring, disc Solution: (b) Time of descent  moment of inertia  U  k2     0.4 ,  R2    sphere  k2   R2     0.5 ,   disc  k2   R2  k2 R2   1   ring  t sphere  t disc  t ring. ST Problem 53. A thin uniform circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linear acceleration along the inclined plane will be (a) g 2 Solution: (c) a (b) g 3 g sin  g g sin 30 o   2 4 1 1 k 1 2 R (c) g 4 [As k2 R2 (d) 2g 3  1 and   30 o ] Problem 54. A solid sphere and a disc of same mass and radius starts rolling down a rough inclined plane, from the same height the ratio of the time taken in the two cases is (a) 15 : 14 (b) 15 : 14 (c) 14 : 15 (d) 14 : 15 Rotational Motion 147 1 Solution: (d) Time of descent t  sin  2h  k2  1 2  g  R   t shpere t disc = 2   1  k   R 2  sphere   2   1  k   R 2 disc  2 5  1 1 2 1 7 2 14   5 3 15 2 gh  k2 1 2 R  v 2  9.8  l sin  2 1 5 2  9.8  1.4  7 /5 [As h 1 k2 2 and sin   given]  , l 2 sin  5 10 R E3 Solution: (a) v  60 Problem 55. A solid sphere of mass 0. 1 kg and radius 2 cm rolls down an inclined plane 1.4 m in length (slope 1 in 10). Starting from rest its final velocity will be (a) 1.4 m / sec (b) 0.14 m / sec (c) 14 m / sec (d) 0.7 m / sec 1 10  1.4 m / s. U ID Problem 56. A solid sphere rolls down an inclined plane and its velocity at the bottom is v1. Then same sphere slides down the plane (without friction) and let its velocity at the bottom be v2. Which of the following relation is correct 5 7 (a) v1 = v2 (b) v 1  v 2 (c) v 1  v 2 (d) None of these 7 5 Solution: (d) When solid sphere rolls down an inclined plane the velocity at bottom v1  10 gh 7 D YG but, if there is no friction then it slides on inclined plane and the velocity at bottom v 2  2 gh  v1  v2 5. 7 7.25 Motion of Connected Mass. ST U A point mass is tied to one end of a string which is wound round the solid body [cylinder, pulley, disc]. When the mass is released, it falls vertically downwards and the solid body rotates unwinding the string m = mass of point-mass, M = mass of a rigid body R = radius of a rigid body, I = moment of inertia of rotating body g (1) Downwards acceleration of point mass a  I T h 1 2 mR m I   (2) Tension in string T  mg  2  I  mR  (3) Velocity of point mass v  2 gh I 1 mR 2 (4) Angular velocity of rigid body   2mgh I  mR 2 Sample problems based on motion of connected mass 148 Rotational Motion Problem 57. A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be [MP PMT 1994; DPMT 2001] 2 gh I  mr (b) 2mgh I  mr 2mgh (c) 2 I  2mr 2 Solution : (b) According to law of conservation of energy mgh  (d) 2 gh 60 (a) 1 (I  mr 2 ) 2    2 2mgh I  mr 2. E3 Problem 58. In the following figure, a body of mass m is tied at one end of a light string and this string mgRt (M  m ) (b) 2 Mgt (M  2 m ) (c) 2mgt R(M  2m ) U (a) ID is wrapped around the solid cylinder of mass M and radius R. At the moment t = 0 the system starts moving. If the friction is negligible, angular velocity at time t would be D YG 2mgt (d) R(M  2m ) g Solution : (d) We know the tangential acceleration a  1 mR 2 m g  I M 1 1 / 2 MR mR 2  2mg 2m  M [As I  1 MR 2 for 2 2 cylinder] After time t, linear velocity of mass m, v  u  at  0  So angular velocity of the cylinder   2mgt 2m  M 2mgt v . R(M  2 m ) R U Problem 59. A block of mass 2 kg hangs from the rim of a wheel of radius 0.5 m. On releasing from rest ST the block falls through 5 m height in 2 s. The moment of inertia of the wheel will be (a) 1 kg-m2 (b) 3.2 kg-m2 (c) 2.5 kg-m 2kg 2 (d) 1.5 kg-m2 Solution : (d) On releasing from rest the block falls through 5m height in 2 sec. 5 0  1 a(2)2 2 a  2.5 m / s 2 [As S  ut  1 2 at ] 2 Rotational Motion 149 g Substituting the value of a in the formula a  1 mR 2.5  10 I 1 2  (0.5)2 2  I  1. 5 kg  m 2 60  and by solving we get I 7.26 Time Period of Compound Pendulum. L l2  k 2 where L  l g E3 Time period of compound pendulum is given by, T  2 Here l = distance of centre of mass from point of suspension k = radius of gyration about the parallel axis passing through centre of mass. Figure Tangent passing through the rim and perpendicular to the plane Ring U Tangent, Perpendicular plane Tangent parallel the plane ST Disc to R D YG Tangent parallel the plane Spherical shell Tangent Solid sphere Tangent to to l k2 ID Axis of rotation L l2  k2 l T  2π L g T  2 2R g 2 2R R R2 2 3 R 2 T  2 3R 2g R R2 2 3 R 2 T  2 3R 2g R R2 4 5 R 4 T  2 5R 4g R 2 2 R 3 5 R 3 T  2 5R 3g R 2 2 R 5 7 R 5 T  2 7R 5g U Body R 150 Rotational Motion Sample problems based on time period of compound pendulum (a) 1 / 4 sec 2R 1 = 2 sec  2 g g (c) 1 sec (d) 2 sec [As diameter 2 R  1 meter given] E3 Solution: (d) T  2 (b) 1 / 2 sec 60 Problem 60. A ring whose diameter is 1 meter, oscillates simple harmonically in a vertical plane about a nail fixed at its circumference. The time period will be Problem 61. A number of holes are drilled along a diameter of a disc of radius R. To get minimum time period of oscillations the disc should be suspended from a horizontal axis passing through a hole whose distance from the centre should be R (b) 2 L g R 2  L l2  l R2 2 l R2 2l D YG Here k 2  2 2 2 l2  k 2 l where L  R (c) ID Solution: (b) T  2 R 2 U (a) For minimum time period L should be minimum  1 R2 R R2   1 .  2   1 2  0  l  2  l  2l 2 ST U  d  R 2  l 0 dl  2l  dL 0 dl (d) Zero

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