Chapter 14 Rotational Motion Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 14 Rotational Motion Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
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 w_{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 w (t) will vary with time t as [IITJEE (Screening) 2002]
w(t)
(a) w_{O}
w(t)

(b)
O
w(t)
(c) wO
(d)
w(t)
wO
t t t t
Solution: (b) The angular momentum (L) of the system is conserved i.e. L = Iw = constant
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, w
change in w will be nonlinear function of time.
will first increase and then decrease. Also the
Problem 35. The position of a particle is given by :
r = (ˆi + 2ˆj – kˆ)
and momentum
P = (3ˆi + 4ˆj – 2kˆ) . The angular
momentum is perpendicular to [CPMT 2000]
 Xaxis
 Yaxis
 Zaxis
 Line at equal angles to all the three axes
Solution: (a)
ˆi
® ® ®
= = 1
ˆj kˆ
2 – 1
= 0ˆi – ˆj – 2kˆ = ˆj – 2kˆ
and the X– axis is given by i + 0ˆj + 0kˆ
L r ´ p
3 4 – 2
Dot product of these two vectors is zero i.e. angular momentum is perpendicular to Xaxis.
Problem 36. Two discs of moment of inertia I_{1} and I_{2} and angular speeds
w_{1} and w _{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 [RPMT 2000]
(a)
I1w_{1} + I 2w _{2}
2(I1 + I 2 )
(b)
(I1 + I 2 )(w_{1} + w _{2} )^{2}
2
(c)
(I1w1 + I 2w 2 )2
2(I1 + I 2 )
 None of these
Solution: (c) By the law of conservation of angular momentum
Angular velocity of system w = I1w1 + I2w2
I1 + I2
I1w_{1} + I2w_{2} = (I1 + I2 )w
1 1 æ I w + I w ö2
(I w + I w )^{2}
Rotational kinetic energy =
(I1 + I2 )w 2 = (I1 + I2 )ç 1 1 2 2 ÷
= 1 1 2 2 .
2 2 è I_{1} + I_{2} ø
2(I1 + I2)
Problem 37. A smooth uniform rod of length L and mass M has two identical beads of negligible size, 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 w _{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 [IITJEE 1998]
(a)
(b)
(c)
(d)
w _{0}
Mw_{0} M + 12m
Mw_{0} M + 2m
Mw_{0} M + 6m
Solution: (d) Since there are no external forces therefore the angular momentum of the system remains constant.

æ 2 ö
Initially when the beads are at the centre of the rod angular momentum L1 = ç ÷w _{0}
…..(i)

ç 12
æ æ L ö ^{2}
÷
ø
æ L ö ^{2}
ML^{2} ö


When the beads reach the ends of the rod then angular momentum = çm ç
÷ + m ç ÷ +
÷w‘
…..(ii)
Equating (i) and (ii)
ML^{2}
æ 2

w_{0} = ç +
2 ö


÷w‘
Þ w‘ =
Mw _{o} .
ç è 2 ø
è 2 ø
12 ÷

12 ç 2
12 ÷
M + 6m
Problem 38. Moment of inertia of uniform rod of mass M and length L about an axis through its centre and perpendicular
to its length is given by
ML2 . Now consider one such rod pivoted at its centre, free to rotate in a vertical
12
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
 v L
2v L
3v 2L
6v L

Solution: (c) Initial angular momentum of the system = Angular momentum of bullet before collision = Mvæ L ö……………. (i)
2
let the rod rotates with angular velocity w.
æ ML2 ö
è ø
æ L ö ^{2}
Final angular momentum of the system = ç ÷w + M ç ÷ w
……(ii)


ç 12 ÷
è 2 ø
L æ ML^{2} ML^{2} ö
By equation (i) and (ii)
Mv = ç + ÷w or w = 3v / 2L




ç 12 ÷
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 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
 12 J–s
 2 Js
(d) 1.12 Js
Solution: (a) Angular momentum of the cylinder before collision L = Iw = 1 MR^{2}w = 1 (2)(0.2)^{2} ´ 3 = 0.12 Js.
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
Þ mvr + I1w = (I1 + I 2 )w‘
Þ mvr + I w = æ 1 Mr ^{2} + mr ^{2} öw‘

_{1} ç ÷
è ø
Þ 0.5 ´ 5 ´ 0.2 + 0.12 = æ 1 2(0.2)^{2} + (0.5)(0.2)^{2} öw‘
\ w ‘ = 10.3
rad/sec.

ç ÷
è ø
Work, Energy and Power for Rotating Body.
(1) Work : If the body is initially at rest and angular displacement is dq due to torque then work done on the body.
W = òt dq [Analogue to work in translatory motion W = ò F dx ]
(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.
 Power : Rate of change of kinetic energy is defined as power
P = d (K
) = d é 1 Iw ^{2} ù = Iw dw
= Iwa = Iaw = tw
dt ^{R}
dt êë 2 úû dt
® ®
In vector form Power = t × w
Slipping, Spinning and Rolling.
[Analogue to power in translatory motion
® ®
P = F× v ]
(1) Slipping : When the body slides on a surface without rotation then its motion is called slipping motion.
In this condition friction between the body and surface F = 0 .
Body possess only translatory kinetic energy KT
= 1 mv ^{2} .
2
Example : Motion of a ball on a frictionless surface.
(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 = 1 Iw ^{2} .
R 2
or KR
= 1 mK ^{2} v 2
2 R ^{2}
= 1 mv
2
æ 2 ö

ç ÷
R 2
è ø
æ K 2 ö
i.e., Rotatory kinetic energy = ç ÷ times translatory kinetic energy.
ç R 2 ÷
è ø
Here K 2 is a constant for different bodies. Value of K 2 = 1 (ring),
K ^{2} = 1
(disc) and
K ^{2} = 1
(solid sphere)
R 2 R 2
R ^{2} 2
R ^{2} 2
(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.
Example : (i) Motion of a wheel of cycle on a road.
(ii) Motion of football rolling on a surface.
In this condition friction between the body and surface F ¹ 0 .
Body possesses both translational and rotational kinetic energy. Net kinetic energy = (Translatory + Rotatory) kinetic energy.
K = K + K = 1 mv ^{2} + 1 Iw ^{2} = 1 mv ^{2} + 1 mv ^{2} K 2
N T R
1 2 æ
2 2 2
K 2 ö
2 R ^{2}
\ KN
= 2 mv
ç1 + ÷
R 2
è ø
Rolling Without Slipping.
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 w.
By the law of conservation of energy
K = 1 mv ^{2} + 1 Iw ^{2}
[\ As v = Rw ]
^{N} 2 2
= 1 mR ^{2}w ^{2} +
2
1 Iw ^{2}
2
= 1 w ^{2}[mR ^{2} + I]
2
= 1 w ^{2} [I + mR ^{2} ] = 1 I w ^{2}
[As I
= I + mR ^{2} ]
2 2 ^{P} ^{P}
By theorem of parallel axis, where I = moment of inertia of rolling body about its centre ‘O’ and I_{P} = moment of inertia of rolling body about point of contact ‘P’.
 Linear velocity of different points in rolling : In case of rolling, all points of a rigid body have same angular speed but different linear
Let A, B, C and D are four points then their velocities are shown in the following figure.
 Energy distribution table for different rolling bodies :
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 [MP PET 2003]
(a) 10 J (b) 100 J (c) 500 J (d) 250 J
Solution: (d) Rotational kinetic energy
1 Iw ^{2} = 1 æ 1 MR^{2} ö w ^{2} = 1 æ 1 ´ 10 ´ (0.5)^{2} ö (20)2 = 250 J
ç ÷
2 2 è 2 ø
ç ÷
2 è 2 ø
Problem 42. An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min. What torque does it deliver [CBSE PMT 2000]
(a) 350 Nm (b) 440 Nm (c) 531 Nm (d) 628 Nm
Solution: (c)
P = tw
Þ t = 100 ´ 103 = 531 N– m
2p 1800
60
Problem 43. A body of moment of inertia of 3 kgm^{2} 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 [MH CET (Med.) 1999]
(a) 8 m/s (b) 0.5 m/s (c) 2 m/s (d) 1 m/s
Solution: (d) Rotational kinetic energy of the body = 1 Iw ^{2} and translatory kinetic energy = 1 mv ^{2}
2 2
According to problem = 1 Iw ^{2} = 1 mv ^{2} Þ 1 ´ 3 ´ (2)^{2} = 1 ´ 12 ´ v^{2} Þ v = 1 m/s .
2 2 2 2
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
(a)
:


1 2 æ k 2 ö
 :
3 _{2}
 :
é k 2 1 ù
 :
Solution: (a)
Kdisc
= mvd ç1 + ÷

2 è ø
= 4 mvd
ê As =
êë R ^{2} 2
for discú
úû


1 æ k 2 ö 2 é k 2 ù
Kring
= mvr ç1 +
2 è
2 ÷ = mvr
R ø
ê As = 1
êë R ^{2}
for ringú
úû
According to problem K = K
Þ 3 mv ^{2} = mv ^{2} Þ vd = .
disc
ring 4 d r vr
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 4s . If the moment of inertia of the wheel about its axis is 0.20 kgm^{2}, then the work done by the torque in two seconds will be
(a) 10 J (b) 20 J (c) 30 J (d) 40 J
Solution: (c)
w_{1} = 20
rad/sec, w _{2}
= 0, t = 4sec. So angular retardation a = w1 – w 2
t
= 20 = 5rad / sec ^{2}
4
Now angular speed after 2 sec
w_{2} = w_{1} – at = 20 – 5 ´ 2 = 10 rad/sec
Work done by torque in 2 sec = loss in kinetic energy = 1 I (w ^{2} – w ^{2} ) = 1 (0.20)((20)^{2} – (10)^{2})
2 1 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% (b) 800% (c) 200% (d) 100%
L^{2} E
æ L ö2
æ 3L ö ^{2}
Solution: (b) As
E = Þ
2 = ç 2 ÷
= ç ^{1} ÷
[As L2 = L1 + 200%.L1 = 3L_{1}]
2I E_{1}
è L1 ø
è L1 ø
Þ E2 = 9E1 = E1 + 800% of E1
Problem 47. A ring, a solid sphere and a thin disc of different masses rotate with the same kinetic energy. Equal torques are applied to stop them. Which will make the least number of rotations before coming to rest
 Disc (b) Ring
(c) Solid sphere (d) All will make same number of rotations
Solution: (d) As W = tq = Energy Þ q = Energy = 2np
t
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
power will be
®

w = 2i + 3 j + 4k
and a torque
®

t = i + 2 j + 3k
acts on it. The rotational
(a) 20 W (b) 15 W (c) W (d) W
Solution: (a) Power (P) = t .w
= (i + 2ˆj + 3kˆ).(2ˆi + 3ˆj + 4kˆ) = 2 + 6 + 12 = 20 W
Problem 49. A flywheel of moment of inertia 0.32 kg–m^{2} is rotated steadily at 120 rad / sec by a kinetic energy of the flywheel is
50 W electric motor. The
(a)
4608 J
(b)
1152 J
(c)
2304 J
(d)
6912 J
Solution: (c) Kinetic energy K
= 1 Iw ^{2} = 1 (0.32)(120)^{2} = 2304 J.
^{R} 2 2
Rolling on an Inclined Plane.
When a body of mass m and radius R rolls down on inclined plane of height ‘h’ and angle of inclination q , it loses potential energy. However it acquires both linear and angular speeds and hence, gain kinetic energy of translation and that of rotation.
1 2 æ
k 2 ö
By conservation of mechanical energy mgh = 2 mv
ç1 + ÷
R 2
è ø
 Velocity at the lowest point : v =
 Acceleration in motion : From equation v ^{2} = u ^{2} + 2aS
By substituting u = 0, S =
a = g sinq
1 + k 2
R 2
h
sinq
and v =
we get
 Time of descent : From equation v = u + at
By substituting u = 0 and value of v and a from above expressions
t = 1
sinq
From the above expressions it is clear that,
æ k 2 ö
v µ 1 ;
a µ 1 ; t µ
1 + k 2
R 2
Note : @ Here factor ç ÷ is a measure of moment of inertia of a body and its value is constant for
ç R 2 ÷
è ø
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 on
k 2 . Lesser
R 2
the moment of inertia of the rolling body lesser will be the value of velocity and acceleration and lesser will be the time of descent.
k 2 . So greater will be its
R 2
@ If a solid and hollow body of same shape are allowed to roll down on inclined plane then as
æ k 2 ö æ k 2 ö
ç ÷ < ç ÷ , solid body will reach the bottom first with greater velocity.
ç R 2 ÷ ç R 2 ÷
è ø _{S} è ø _{H}
@ If a ring, cylinder, disc and sphere runs a race by rolling on an inclined plane then as
æ k 2 ö æ k 2 ö
ç ÷ = minimum while ç ÷ = maximum , the sphere will reach the bottom first with
ç R 2 ÷ ç R 2 ÷
è øsphere è øRing
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. velocity µ q ° , time of decent µ q ^{–}^{1} and acceleration µ q .
Rolling Sliding and Falling of a Body.
Velocity, Acceleration and Time for Different Bodies.
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]
(a)
(b)
(c)
(d)
Solution: (b) Velocity at the bottom (v) = = = .
Problem 51. A sphere rolls down on an inclined plane of inclination q. What is the acceleration as the sphere reaches bottom [Orissa JEE 2003]
(a)
5 g sinq
7
(b)
3 g sin q
5
(c)
2 g sin q
7
(d)
2 g sin q
5
Solution: (a) Acceleration (a) = g sinq
1 + K 2
R 2
= g sinq 1 + 2
5
= 5 g sinq .
7
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 [RPMT 1999]
 Ring, disc, sphere (b) Sphere, disc, ring (c) Disc, ring, sphere (d) Sphere, ring, disc
Solution: (b) Time of descent µ
moment of inertia µ k 2

R2
æ k2 ö
æ k 2 ö
æ k 2 ö
ç ÷

ç 2 ÷
è øsphere
= 0.4 , ç
è
÷

2 ÷
ø disc
= 0.5 , ç
è
÷

2 ÷
øring
\ t sphere
< tdisc
< tring .


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 [CBSE PMT 1992; BHU 1998]
 g 2
 g 3
 g 4
2g 3
Solution: (c)
a = g sin q
1 + k 2
R2
= g sin 30 ^{o} = g
1 + 1 4
[As
k 2 = 1
R 2
and q = 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) : (c) 14 : 15 (d) :
Solution: (d) Time of descent t =
\ tshpere =
tdisc
= = =
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)
 m / sec
Solution: (a) v = =
[As
k ^{2} = 2 , l = h
and sin q = 1
given]
R^{2} 5 sin q 10
Þ v = = 1.4 m / s.
Problem 56. A solid sphere rolls down an inclined plane and its velocity at the bottom is v_{1}. Then same sphere slides down the plane (without friction) and let its velocity at the bottom be v_{2}. Which of the following relation is correct
 v_{1}
= v_{2}
v1 = 5 v2
7
v1 = 7 v 2
5
 None of these
Solution: (d) When solid sphere rolls down an inclined plane the velocity at bottom v_{1} =
but, if there is no friction then it slides on inclined plane and the velocity at bottom v_{2} =
\ v1 = .
v2
Motion of Connected Mass.
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 pointmass, M = mass of a rigid body
R = radius of a rigid body, I = moment of inertia of rotating body
 Downwards acceleration of point mass
a = g
1 + I
(2) Tension in string
T = mgé I ù
êë I + mR^{2} úû
mR^{2}
(3) Velocity of point mass v =
 Angular velocity of rigid body w =
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]
(a)
(b)
(c)
(d)
Solution : (b) According to law of conservation of energy mgh = 1 (I + mr ^{2})w ^{2} Þ w = .
2
Problem 58. In the following figure, a body of mass m is tied at one end of a light string and this string 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
(a)
(b)
(c)
(d)
mgRt
(M + m)
2Mgt
(M + 2m)
2mgt R(M – 2m)
2mgt R(M + 2m)
Solution : (d) We know the tangential acceleration a =
g
1 + I
mR ^{2}
= g
1 + 1 / 2MR ^{2}
mR ^{2}
= 2mg 2m + M
[As I = 1 MR ^{2} for cylinder]
2
After time t, linear velocity of mass m, v = u + at
= 0 + 2mgt
2m + M
So angular velocity of the cylinder w = v =
R
2mgt .
R(M + 2 m)
Problem 59. A block of mass
2 kg hangs from the rim of a wheel of radius
 m . On releasing from rest the block falls
through 5 m height in 2 s . The moment of inertia of the wheel will be
 1 kg–m^{2}
 2 kg–m^{2}
 5 kg–m^{2}
 5 kg–m^{2}
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 at ^{2} ]
2
Substituting the value of a in the formula a = g
1 + I
and by solving we get
Þ 2.5 = 10
1 + I
Þ I = 1.5kg – m^{2}
mR ^{2}
2 ´ (0.5)^{2}
Time Period of Compound Pendulum.
Time period of compound pendulum is given by, T = 2p Here l = distance of centre of mass from point of suspension
where
L = l ^{2} + k ^{2}
l
k = radius of gyration about the parallel axis passing through centre of mass.
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
(a)
1 / 4 sec
(b)
1 / 2 sec
(c)
 sec
(d)
 sec
Solution: (d)
T = 2p
= 2p
= 2 sec [As diameter 2R = 1 meter given]
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
(a)
R (b)
2
R (c)
R (d) Zero
Solution: (b)
T = 2p
where L = l 2 + k 2
l
l ^{2} + R 2
Here k ^{2} = R2
2
\ L = 2
l
= l + R 2
2l
For minimum time period L should be minimum
dL = 0
dl

Þ d æ l +

dl ç
R2 ö


2l ÷
Þ 1 + R^{2} æ – 1 ö =
1 – R2 = 0 Þ l = R .
2 ç l 2 ÷
2l ^{2} 2
è ø