Q: An annular disk of mass M , inner radius a and outer radius b is placed on a horizontal surface with coefficient of friction μ , as shown in the figure. At some time, an impulse J 0 x ∧ &nb

As we know, J → = ∆ P → ⇒ J 0 = M V . . . i About center of the disk, we can also write J 0 × h = I C M ω ⇒ J 0 × h = M 2 a 2 + b 2 ω . . . i i If h = h m ⇒ ω = V b pure rolling From equation(ii), w

Q: A thin circular coin of mass 5 gm and radius 4 3 cm is initially in a horizontal x y -plane. The coin is tossed vertically up ( + z direction) by applying an impulse of

By impulse - momentum theorem: J = M V C M ⇒ V C M = J M = π 2 100 × 5 1000 = 2 π m s - 1 ⇒ Total time of journey = 2 V C M g = 2 g × 2 π ⇒ ∆ t = 2 π 5 s Also, by angular impulse - momentum theorem: J × R 2 = I ω ⇒ J × R 2 = M R 2 4 ω ⇒ ω = J × R 2 M R 2 4 = J M R × 2 = π 2 100 × 2 5 1000 × 4 3 × 1 100 = 150 2 π

Q: A flat surface of a thin uniform disk A of radius R is glued to a horizontal table. Another thin uniform disk B of mass M and with the same radius R rolls without slippin

The centre of mass of B has angular velocity ω about the centre of A . Therefore, the velocity of the centre of B is V B = ω 2 R . The point of contact of A and B is at rest therefore, the angular velocity of B will be, ω B = v B R = 2 ω Now the angular momentum of B with respect to centre of A L = M v B r + I C M ω B ⇒ L = M ω 2 R 2 R + M R 2 2 2 ω L = 5 M R 2

Question 1

Consider a large disk of radius R and two smaller disks, each of radius r = R/50, lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with a

Accepted Answer

Let the radius of the large disk be R and the radius of the small disks be r. We are given r = R/50, which implies R = 50r. The centers of the small disks move along a circular path of radius R + r = 50r + r = 51r. When the two small disks are in contact, the distance between their centers is 2r. The angular separation between their centers is given by the arc length (which is approximately the chord length for small angles): (R + r) = 2r (51r) = 2r = 2 51 rad The small disks roll without slippi

Question 2

A solid cylinder of radius R rolls without slipping with a center of mass speed v_0 = gR 3 on a horizontal surface with a vertical edge, as shown in the figure. Here, g is the acceleration due to the

Accepted Answer

Let the mass of the cylinder be m and its radius be R. When the cylinder is rolling on the horizontal surface, its kinetic energy is: K_i = 1 2 mv_0^2 + 1 2 I_ cm _0^2 Since it rolls without slipping, _0 = v_0 R and I_ cm = 1 2 mR^2. K_i = 1 2 mv_0^2 + 1 2 ( 1 2 mR^2 ) ( v_0 R )^2 = 3 4 mv_0^2 Given v_0 = gR 3 , we have v_0^2 = gR 3 . K_i = 3 4 m ( gR 3 ) = 1 4 mgR Let the corner be the reference level for potential energy. The initial total mechanical energy of the cylinder just as it reaches t

Question 3

List-I shows four planar structures made of uniform solid rods each of mass m and length l. In the List-II the possible moment of inertia of these structures about an axis OCO', which lies in the plan

Accepted Answer

The moment of inertia of a uniform rod of mass m and length l about an axis passing through one end and making an angle with the rod is given by I = m l^2 3 ^2 . For structure (P): The axis OO' makes an angle of 45^ with both rods CA and CB. I_P = I_ CA + I_ CB = m l^2 3 ^2(45^ ) + m l^2 3 ^2(45^ ) = 2 m l^2 3 1 2 = m l^2 3 . This matches with (5). For structure (Q): The axis OO' makes an angle of 60^ with rods CA and CB. Rod AB is parallel to the axis at a distance d = l (60^ ) = 3 2 l. I_Q = I

Question 4

Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point C such that it can rotate freely around C in the XY plane, as shown in the figure. Initially, when the disk i

Accepted Answer

Moment of inertia of the disk about the pivot C is I_C = 1 2 MR^2 + MR^2 = 3 2 MR^2. Substituting M = 1 kg and R = 0.2 m, we get I_C = 3 2 (1)(0.2)^2 = 0.06 kg m^2. The initial angular momentum of the particle about C is L_i = m u y_ , where y_ is the perpendicular distance from C to the initial line of motion. L_i = m u (R + R 45^ ) = (0.02)(100)(0.2) (1 + 1 2 ) = 0.4 + 0.2 2 kg m^2/s. This initial angular momentum is in the clockwise direction. The final angular momentum of the particle about

Question 5

Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point C such that it can rotate freely around C in the XY plane, as shown in the figure. Initially, when the disk i

Accepted Answer

Let the center of the uniform circular disk be the origin O(0,0). The disk is pivoted at its top point C(0, R). Radius of disk R = 0.2 m, Mass M = 1 kg. Mass of particle m = 20 g = 0.02 kg. Initial velocity of particle v _i = -100 i ms^ -1 . Final velocity of particle v _f = -90 j ms^ -1 . The particle hits the disk at point P. From the figure, the line OP makes an angle of 45^ with the negative y-axis. Coordinates of P with respect to O are (R 45^ , -R 45^ ). Position vector of P with respect t

Question 6

A disc of mass M and radius R is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Anoth

Accepted Answer

On applying conservation of angular momentum about axis of larger disc. aligned & 1 2 M ( R 2 )^2 - M ( ^ R ) R - MR ^2 2 ^ =0 \ & 8 = 3 ^ 2 \ & ^ = 12 Hence, n =12 aligned

Question 7

A thin uniform rod of length L and certain mass is kept on a frictionless horizontal table with a massless string of length L fixed to one end (top view is shown in the figure). The other end of the s

Accepted Answer

Linear impulse F d t= momentum aligned & = m ( V _ cm -0 ) \ & P = m ( r _ cm ) \ & = m ( L + L 2 ) aligned P = m ( 3 ~L 2 ) ...(i) Angular impulse dt = angular momentum aligned & r Fdt = L \ & aligned & r Fdt = I ( -0), and I is moment of inertia about axis of rotation. \ & aligned & ( L + L 2 + x ) P = ( I _ cm + md ^2 ) \ &= ( mL ^2 12 + m ( L + L 2 )^2 ) aligned \ & ( 3 ~L 2 + x ) P = mL ^2 ( 1 12 + ( 3 2 )^2 ) aligned aligned ( 3 L 2 +x ) P= mL ^2 ( 7 3 ) ...(ii) Divide eq.-(i) & (ii)

Question 8

An annular disk of mass M , inner radius a and outer radius b is placed on a horizontal surface with coefficient of friction &#956; , as shown in the figure. At some time, an impulse J 0 x &#8743; &nb

Accepted Answer

As we know,  J &#8594; = &#8710; P &#8594; &#8658; &#160; J 0 = M V &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . &#160; i About center of the disk, we can also write J 0 &#215; h = I C M &#969; &#8658; &#160; J 0 &#215; h = M 2 a 2 + b 2 &#969; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . &#160; i i If  h = h m &#8658; &#969; = V b pure&#160;rolling From equation(ii), w

Question 9

A thin circular coin of mass 5 &#160; gm and radius 4 3 &#160; cm  is initially in a horizontal x y -plane. The coin is tossed vertically up ( + z direction) by applying an impulse of

Accepted Answer

By impulse - momentum theorem: J = M V C M &#8658; &#160; V C M = J M = &#960; 2 100 &#215; 5 1000 = 2 &#960; &#160; m &#160; s - 1 &#8658;  Total time of journey  = 2 V C M g = 2 g &#215; 2 &#960; &#8658; &#160; &#8710; t = 2 &#960; 5 &#160; s Also, by angular impulse - momentum theorem: J &#215; R 2 = I &#969; &#8658; J &#215; R 2 = M R 2 4 &#969; &#8658; &#160; &#969; = J &#215; R 2 M R 2 4 = J M R &#215; 2 = &#960; 2 100 &#215; 2 5 1000 &#215; 4 3 &#215; 1 100 = 150 2 &#960; &#160;

Question 10

A flat surface of a thin uniform disk A  of radius  R is glued to a horizontal table. Another thin uniform disk  B of mass M  and with the same radius R  rolls without slippin

Accepted Answer

The centre of mass of  B  has angular velocity  &#969;  about the centre of  A . Therefore, the velocity of the centre of  B  is V B = &#969; 2 R . The point of contact of  A  and  B  is at rest therefore, the angular velocity of  B  will be, &#969; B = v B R = 2 &#969; Now the angular momentum of B  with respect to centre of  A L = M v B r + I C M &#969; B &#8658; L = M &#969; 2 R 2 R + M R 2 2 2 &#969; L = 5 M R 2

Question 11

At time t = 0 , a disk of radius 1 &#160; m starts to roll without slipping on a horizontal plane with an angular acceleration of &#945; = 2 3 &#160; rad &#160; s - 2 . A small stone is stuck to the d

Accepted Answer

Using second equation of motion, The angle rotated by disc in t = &#960; &#160; s is  &#952; = &#969; 0 t + 1 2 &#945; t 2 &#8658; &#952; = 1 2 &#215; 2 3 &#960; 2 = &#960; 3 &#160; rad Now using first equation of motion, the angular velocity of disc is  &#969; = &#969; 0 + &#945; t &#8658; &#969; = 0 + 2 &#960; 3 = 2 &#960; 3 &#160; rad &#160; s - 1 And, velocity of disk about centre of mass is  v cm = &#969; R = 2 &#960; 3 &#215; 1 = 2 &#960; 3 &#160; m &#160; s - 1 At the momen

Question 12

A solid sphere of mass 1 &#160; kg and radius 1 &#160; m rolls without slipping on a fixed inclined plane with an angle of inclination &#952; = 30 &#176; from the horizontal. Two forces of magnitude 1

Accepted Answer

The forces acting on the solid sphere is shown below. Here,  N  is the normal force acting on sphere and weight  m g  is acting downwards.    Taking torque about contact point. &#964; &#8594; = m g R sin 30 &#8855; + 1 &#215; 1 &#8857; = 10 &#215; 1 &#215; 1 2 - 1       (Taking &#8855;  as positive) Then, we have  &#8658; 5 - 1 = I sphere&#160;about&#160;tangent &#945; Using parallel axis theorem, &#964; = 2 5 m R 2 + m R 2 &#945; = 7 5 m R

Question 13

A pendulum consists of a bob of mass m=0.1 ~kg and a massless inextensible string of length L=1.0 ~m . It is suspended from a fixed point at height H=0.9 ~m above a frictionless horizontal floor. Init

Accepted Answer

The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is, Angular momentum about top point L = m V &#215; 0 . 9 L = 0 . 2 &#215; 0 . 9 &#8658; L = 0 . 18 &#160; kg &#160; m - 2 &#160; s Hence,  J = 0 . 18 .

Question 14

A pendulum consists of a bob of mass m=0.1 ~kg and a massless inextensible string of length L=1.0 ~m . It is suspended from a fixed point at height H=0.9 ~m above a frictionless horizontal floor. Init

Accepted Answer

The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is, Angular momentum about top point L = m V &#215; 0 . 9 L = 0 . 2 &#215; 0 . 9 &#8658; L = 0 . 18 &#160; kg &#160; m - 2 &#160; s Now after the bob slides for some distance, the string becomes taut and the bob lifts off the floor so when string get taut an impulse act by string on particle. Due to impulse, component of  V  along string get change and becomes zero but comp

Question 15

A horizontal force  F  is applied at the centre of mass of a cylindrical object of mass  m  and radius  R , perpendicular to its axis as shown in the figure. The coefficient o

Accepted Answer

Let moment of inertia  = I For pure rolling,  a = &#945; R and  F &#8722; f = m a &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . 1 and  f R = I &#945; or  f R = I a R or  f = I a R 2 Therefore,  F &#8722; I a R 2 = m a &#8756; &#160; &#160; &#160; F = m + I R 2 a &#8658;  For the same value of  F , value of a depends on  I . &#8658;  For a solid cylinder I = m R 2 2  (to calculate maximum  a ) f = &#956; m g &#8756; &#160;

Question 16

A particle of mass  M = 0 . 2 &#160; kg  is initially at rest in the  x y -plane at a point  x = &#8722; l , &#160; y = &#8722; h , where  l = 10 &#160; m  and  h =

Accepted Answer

At  t = 2 &#160; s : displacement along + X  axis = 1 2 &#215; 10 &#215; t 2 = 1 2 &#215; 10 &#215; 4 = 20 &#160; m Hence, particle arrives at  10 , - 1 At  x = l , &#160; y = &#8722; h , &#964; &#8594; = r &#8594; &#215; F &#8594; = 10 i ^ &#8722; j ^ &#215; 0 .2 &#215; 10 i ^ = 2 k ^ &#160; N &#160; m L &#8594; = r &#8594; &#215; p &#8594; = 10 i ^ &#8722; j ^ &#215; 0 .2 &#215; v i ^ and  v at &#160; t = 2 s = 0 + 10 &#215; 2 = 20 &#160; m &#160; s - 1 &#8756; &#160;

Question 17

A thin rod of mass  M  and length  a  is free to rotate in horizontal plane about a fixed vertical axis passing through point  O . A thin circular disc of mass  M  a

Accepted Answer

Angular momentum of the system about  O , L &#8594; = L &#8594; Rod + L &#8594; Disc &#8658; L &#8594; = L &#8594; Rod + L &#8594; C M Disc + r &#8594; C M &#215; M D i s c V &#8594; C M &#8658; L = M a 2 &#937; 3 + M a 2 8 &#937; + M 3 a 4 &#937; 3 a 4 &#8658; L = M a 2 3 &#937; + M a 2 8 &#937; + 9 M a 2 16 &#937; &#8658; L = 16 M a 2 &#937; + 6 M a 2 &#937; + 27 M a 2 &#937; 48 &#8658; L = 49 48 M a 2 &#937; Compare it with  n M a 2 48 &#937; , we get  n = 49 Hence,

Question 18

A rod of mass m and length L , pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets

Accepted Answer

The net torque on the system (rod+bullet) will be zero about hinge point. Therefore, we can apply the principle of angular momentum conservation about hinge point. From angular momentum conservation, m v x = m L 2 3 + m x 2 &#969; &#8658; &#969; = 3 v x L 2 + 3 x 2 = 3 v L 2 x + 3 x For maximum angular velocity, d &#969; d x = 0 &#8658; d d x L 2 x + 3 x = 0 &#160; &#8658; - L 2 x 2 + 3 = 0 &#160; &#8658; x = L 3 &#8658; &#969; max = 3 v 3 L 2 L + 3 L = 3 v 2 L

Question 19

A football of radius R is kept on a hole of radius r ( r R ) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle θ from the horizontal as shown

Accepted Answer

m g sin θ R 2 − r 2 = m g cos θ r ⇒ tan θ = r R 2 − r 2 ⇒ sin θ = r R

Question 20

A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the

Accepted Answer

Angular displacement θ = S 0 10 = 5 Displacement of point B is, S = r . θ = 15 × 5 = 75 cm Displacement of roller = 50 cm Relative displacement = 75 - 50 = 25 cm

Rotational Motion — Previous Year Questions & Solutions