ROTATIONAL MOTION@iitjeehelps PDF
ROTATIONAL MOTION@iitjeehelps PDF
ROTATIONAL MOTION@iitjeehelps PDF
xyz
•
www.crackjee.xyz
Head Office : B-32, Shivalik Main Road, Malviya Nagar, New Delhi-110017
• Sales Office : B-48, Shivalik Main Road, Malviya Nagar, New Delhi-110017
Tel. : 011-26691021 / 26691713
DISHA PUBLICATION
ALL RIGHTS RESERVED
© Copyright Author
No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the
publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have
tried and made our best efforts to provide accurate up-to-date information in this book.
1.5 Couple 09
Particle
A particle is defined as an object whose mass is finite but its size is negligible small.
Rigid body
A body is said to be rigid if it does not undergo any change in size and shape, however
large external force act on it.
Translatory motion
Fig . 1.1. Translatory motion.
When different particles of a body undergo same displacement, the motion of the body
is called translatory motion.
Here AA ' BB ' CC '
Rotatory Motion
When different particles of a body undergo same angular displacement, the motion of
the body is called rotatory motion. But different particles have different linear displace-
ments.
If angular displacement traversed by the particles A, B and C on the body in time t is
then
A = B C
t
Fig. 1.2. Rotatory motion.
AA ' 0
while linear velocity vA = 0
t t
BB ' rB
vB = rB
t t
CC ' rC
and vC = rC
t t
It is clear that, rA < rB < rC
so vA < vB < vC
Fig. 1.3
Combined translation and rotation
The combination of above two motions results combined translation and rotation.
t
d = dt
0 0
t
= t0
0
or 0 = (t 0)
or = 0 t …(1)
www.crackjee.xyz
ROTATIONAL MECHANICS 3
Second equation of motion
By definition, we have
d
=
dt
or d = dt
If at t = 0, 0 , then angular displacement in time t is given by
t t
d = dt ( 0 t)dt
0 0 0
t
1 2
or = 0t t
0 2 0
1 2
or = 0t t … (2)
2
Third equation of motion
The angular acceleration can be expressed as
d d d
= = .
dt d dt
d SI units :
or =
d SI unit of angular displacement is radian.
or d = d SI unit of angular velocity is rad/s.
Integrating above equation , we get SI unit of angular acceleration is rad/s2.
d d
=
0 0
2
or 2 = 0
0
or 2 2 =
0 2
or 2 = 2 … (3)
0 2
For uniformly retarded motion, these equations become ;
= 0 t
1 2
= 0t t
2
and 2 = 2 .
0 2
Sol.
Given that = 12 t
d
i.e. = 12 t
dt
or d = (12 t ) dt
Integrating above equation, we get
Fig. 1.5
d = (12 t )dt When A moves 20 m, its angular displacement is given by
s = r
t2 s 20
or = 12t C or = = 20 rad
2 r 1
At t = 4 s, = 60 rad/s
Given 2 rad / s 2 and 0 = 0.
2
4 By second equation of motion, we have
60 = 12 4 C
2 1 2
or C = 20 = 0t t
2
t2 1
= 12t 20 … (i) 20 = 0 2 t2
2 2
or t = 4.47 s Ans.
62 Angular velocity of pulley at this time
At t = 6 s, = 12 6 20
2 = 0 t
= 74 rad/s Ans. = 0 + 2 × 4.47 = 8.94 rad/s Ans.
www.crackjee.xyz
ROTATIONAL MECHANICS 5
Now velocity of A, v A = rA 8.94 1 Since the load A starts moving with constant acceleration a, its velocity at
= 8.94 m/s time t,
and v B = rB = 8.94 × 0.75 v = at.
= 6.70 m/s Ans.
Let 1 be the angular velocity of the pulley and the first gear wheel.
Ex. 4 The angular rotation in radians of an accelerated flywheel Peripheral velocity of the pulley = 1r = at
is given by 9t 3 / 32 . Find the linear velocity and acceleration
1 = at/r
of a point at a distance of 0.75 m from the axis of rotation at the Peripheral speed of the first gear wheel 1r1.
instant when its tangential acceleration and normal accelerations
are equal. Since the two gear wheels have the same peripheral velocities, so
Sol. 2r 2 = 1r 1
We have, = 9t 3 / 32 or 2 = 1r1 / r2
= at r1/(rr2)
d 9
Angular velocity = 3t 2 Let be the angular displacement of the second gear at time t.
dt 32
2 = d /dt = at r1/(rr2)
= 27t2 /32
or d = at r1 / (rr2)dt
and angular acceleration
Integrating above equation, we get
d
= = at2 r1/(2rr2) Ans.
dt
= 27t / 16 Ex. 6 Figure shows a compound wheel which rolls without
Tangential acceleration of a point at a radius r is given by slipping. If the velocity of the centre O is 1.25 m/s. Find the velocities
at = r of the points A, B and C.
and the normal acceleration is given by an 2
r
Since the tangential and normal accelerations at a distance r = 0.75 m are
Sol.
equal, so
2
r = r
2
or =
2
27t 27t 2
or =
16 32
which gives t = 4/3 s
2
27t 2 27 4
Linear velocity = r= r 0.75
32 32 3
= 9/8 m/s Ans.
2 2 2 2
2 27t 27 4
Tangential acceleration = r r 0.75 Fig. 1.7
32 32 3
Since the wheel rolls without slipping on the supporting surface MN,
= 7/16 m/s2 Ans.
the point of contact D is the instantaneous centre of rotation. Let be
Ex. 5 In a system of gears load A rotates a pulley of radius r
and a gear wheel of radius r1. This gear wheel is geared with a the angular velocity about the instantaneous centre D.
second gear wheel of radius r2. If the load starts from rest and Velocity of O =v0 = . DO
moves down with a constant acceleration, find the equation of
Velocity of A = vA = . DA
second gear wheel.
Velocity of B = vB = . DB
Sol.
Velocity of C = vC = . DC
We have DO = 0.15 m, DA = 0.45 m
and DC = 0.15 m
v0 = 1.25 m/s
1.25 : vA : vB : vC = (0.15) : (0.45) : (0.335) : (0.15)
vA = 1.25 × 0.45 / 0.15 = 3.75 m/s
vB = 1.25 × 0.335 / 0.15 = 2.80 m/s
vC = 1.25 × 0.15 / 0.15 = 1.25 m/s Ans.
Fig. 1.6
www.crackjee.xyz
6 MECHANICS, THERMODYNAMICS & WAVES
1.3 INSTANTANEOUS AXIS OF ROTATION
At any instant it is possible to locate a point in the plane which has zero velocity and
hence plane motion of other points may be looked as pure rotation about this axis. Such
point is called instantaneous centre and the axis passing through this point and right
angles to the plane of motion is called instantaneous axis of rotation.
More about instantaneous axis of rotation
Motion of such an object may be looked as pure rotation about a point has zero velocity
to simplify the study. Such a point is called Instantaneous centre and the axis passing
through this point and perpendicular to the plane of motion is called Instantaneous Axis
of Rotation (IAOR).
In the figure shown I1 and I2 are the instantaneous centres of rotation of rod at two
different instants.
To understand instantaneous centre consider a rigid body which has plane motion. Let
A be a point having velocity vA at the instant considered. Now locate a point I on
Fig. 1.8 perpendicular to the direction of vA at A at a distance rA. The motion of A can be split into
translation of I and rotation about I. Thus we can write
vA = vI rA ,
vA vA
if we take rA , then vA = vI
vI = 0
vA
Thus point ‘I’ is selected at a distance along the perpendicular to the direction of A,
the plane motion of point A can be reduced to pure rotation about I. Hence I is the
instantaneous centre.
If B is any other point on the rigid body then its velocity will be given by
vB = vI rB
Fig. 1.9 or vB = rB (v I 0)
Method of locating instantaneous centre
Instantaneous centre can be located by any of the following two methods:
(i) If the angular velocity and linear velocity vA are known, instantaneous centre
can be located at a distance v A / along the perpendicular to the direction of vA at
A, as discussed earlier.
(ii) If the linear velocities of two points of rigid body are known, say vA and vB, drop
perpendiculars to them at A and B. The intersection point is the instantaneous
centre (see Fig 1.8).
45° 60°
A C
Fig. 1.10
At a certain instant, the point A is moving towards the mid-point of
BC with a velocity of 5 m/s and B is moving at a perpendicular
direction to AC. Find the velocity of C. Fig. 1.10 (a)
www.crackjee.xyz
ROTATIONAL MECHANICS 7
Let the rod makes an angle with the horizontal. 3. Now draw perpendiculars at A and B on directions of motion
OBD OBC /2 of vA and vB.
4. Let these perpendicular meet at O, which is the instantaneous
centre of the link AB and BC.
CIB =
5. Now join OC and draw a line at right angle to OC indicating
the direction of motion of the point C.
BC = r cot
2 vB
r F
or Moment arm =
F
iˆ ˆj kˆ
= x y z
Fx Fy Fz
Fig. 1.14
Sol. Fig. 1.13
Method I :
(b) Calculate moment of system of forces about an axis
passing through B and perpendicular to the plane of the forces.
(a) Fr sin kˆ (b) Fr sin ( kˆ ) Sol. (a) Moment of force of 20 N and 10 cos 45° N is zero about O
10 1 sin 60 kˆ N-m 10 1 sin 30 ( kˆ ) N-m because their moment arm is zero.
5 3 kˆ N-m 5 kˆ N-m The net moment of force
Method II : 10 1 kˆ (10 2 sin 45 ) 1 ( kˆ)
3ˆ 1ˆ m 10 2 N 40 N
r 1cos 30 iˆ 1sin 30 ˆj i j 10kˆ 10kˆ 0 Ans.
2 2 1m
(b) Moment of forces about B, D C
F 10 ˆj N = 20 × 0 + 40 × 0
B
3ˆ 1ˆ 50 N
r F i j 10 jˆ 5 3kˆ N-m + 10 2 × 10 + 50 × 0 + 10 × 1 1m
2 2
= 10 N-m clockwise.
0.4 m
Ex. 9 (a) A wheel of radius 1m is acted by the forces shown in
A
20 N
Fig. 1.14. Find resultant moment of force about an axis passing B
10 N
through centre of the wheel. Fig. 1.15
www.crackjee.xyz
ROTATIONAL MECHANICS 9
1.5 COUPLE
Two equal and opposite forces whose lines of action are different constitutes a couple.
The moment of couple can be found by taking moments of both the forces about any
axis perpendicular to plane of forces and adding them algebraically. Thus moment of
couple = F AO F OB
= F (AO + OB )
= Fd
or = Force × perpendicular distance between the lines of action of forces Fig. 1.16
1.6 MOMENT OF INERTIA OR ROTATIONAL INERTIA
Moment of inertia of a body is a measure of its ability to resist change in state of rotation.
It plays the same role in rotational motion whatever inertia plays in translational motion.
Mathematically moment of inertia of a particle of mass m about an axis is defined as;
I = mr2
where r is the distance of particle from axis of rotation. Fig. 1.17
It is called a tensor quantity. Its SI unit is kg-m2.
Moment of inertia of system of particles
Consider a system of n particles as shown in Fig. 1.18. Its moment of inertia is given by
I = m1r12 + m2r22 + ......+ mnrn2
n
or mi ri 2
I =
i 1 Fig. 1.18
Moment of inertia of a rigid body
Consider a body rotating about an axis as shown in Fig. 1.18. Choose small element of
mass dm at a distance r from the axis, its moment of inertia about the axis of rotation (here
z-axis)
dI = dmr2
The moment of inertia of whole body can be obtained by
I = dm r 2
Limits of integration depends on the shape of the body.
The moment of inertia of a body depends on the following factors:
(i) Mass of the body.
(ii) Size and shape of the body. Fig. 1.19
(iii) Position and orientation of axis of rotation.
Note:
1. Moment of inertia of a particle about an axis passing through the
particle itself will be zero.
2. Any body can have infinite numbers of moment of inertia.
Radius of gyration
Consider a system of particles or a rigid body is rotating about an axis as shown in
figure. Its moment of inertia about the axis is given by ;
2 2 2
I = m1r1 m2 r2 ...... mn rn
Now suppose the whole mass of the system M is concentrated at a point and placed at
a distance k from the axis, then the moment of inertia of the equivalent system can be
defined as , I = Mk2
If Mk2 = m1r12 + m2r22 + .......+ mnrn2 = I,
then k is called radius of gyration and can be written as
I Fig. 1.20
k =
M
www.crackjee.xyz
10 MECHANICS, THERMODYNAMICS & WAVES
Thus radius of gyration of any system about its axis of rotation may be defined as the
distance from the axis of rotation, if square of it multiplied by total mass of the system gives
moment of inertia of the system about the axis.
1.7 THEOREMS OF MOMENT OF INERTIA
The value of moment of inertia depends on axis of rotation, so each time we requrie
mathematical process to get moment of inertia. The calculation of getting moment of
inertia can be made easier by deriving the following theorems of moment of inertia.
1. Parallel axis theorem
Consider a body, whose moment of inertia about an axis passing through C.M. is
Icm . Let m is the mass of the particle at a distance x from the axis of rotation, then
the moment of inertia of whole body about C. M. axis,
I cm = mx 2 …(i)
Choose an axis A parallel to the C.M. axis, and the separation between the axes is
d. The moment of inertia of the body about axis A,
IA = m( d x) 2
= m(d 2 x2 2 xd )
I A = I cm Md 2 … (1)
Thus the moment of inertia of a body about any axis is equal to its moment of inertia
about a parallel axis through its centre of mass plus the product of the mass of the
body and the square of the perpendicular distance between the two axes.
Note: The parallel axis may lie inside or outside the body...
2. Perpendicular axis theorem
Consider a body lying in the xy-plane. It can be assumed to made of large number
of particles. Consider one such particle of mass m at a distance r from the origin of
axis. In terms of cartesian coordinate,
2
r2 = x y2 … (i)
The moment of inertia of the particle about x-axis
Ix = my2
The moment of inertia of the body about x-axis is
Ix = my 2 … (ii)
Moment of inertia of the body about y-axis is
Iy = mx 2 … (iii)
Fig. 1.22 Moment of inertia of the body about z-axis
Iz = mr 2
= m( x 2 y2 )
= mx 2 my 2
www.crackjee.xyz
ROTATIONAL MECHANICS 11
From equations (ii) and (iii), we get
I z = Ix I y … (2)
Thus the moment of inertia of a body lying in a plane about an axis perpendicular
to its plane is equal to the sum of the moment of inertia of the body about any two
mutually perpendicular axes in its plane and intersecting each other at the point
where the perpendicular axis passes through the body.
Note: Intersection of axes need not be the centre of mass of the body..
Perpendicular axis theorem can be used for planar body but not for cone etc.
Moment of inertia of a thin uniform rod
Consider a thin uniform rod of mass m and length L, which is rotating about an axis
passing through its C.M. and perpendicular to its length. For the thin rod its moment of
inertia about an axis passing through its length will be zero.
Consider a small element of length dx at a distance x from its C.M. The mass of the
element
M
dm = dx
L
Moment of inertia of the element dI = dm x2
Moment of inertia of whole rod
L/2
Fig. 1.23
I = 2 dm x 2
0
L/2
M
= 2 dx x 2
L
0
L/2
2 M x3 ML2
= =
L 3 12
0
ML2
=
3
Fig. 1.25
Radius of gyration about the end of the rod,
I ML2 / 3 L
k =
M M 3
www.crackjee.xyz
12 MECHANICS, THERMODYNAMICS & WAVES
Some special cases :
(i) M.I. of thin rod about an axis passing through end of rod, and it is inclined an angle
with the axis.
The mass of the element
M
dm = dx
L
The distance of the element from the axis
r = x sin
The M.I. of the element about the axis
dI = (dm)r2
The M.I. of the whole rod
Fig. 1.26
L
dm r 2
I =
0
L
M
= dx ( x sin )2
L
0
L
Fig. 1.27 M 2
= L sin x 2 dx
0
ML2
= sin 2
3
(ii) The rod of mass M and length L is bent into L-shape. Its moment of inertia about
the end
( M / 2)( L / 2) 2 M
I end = ( L / 2)2
3 2
ML2
=
6
Fig. 1.28
ML2
=
6
ML2
(iii) Iz = Md 2
12
ML2
(iv) I = 0 ML2
3
4
= ML2
3
Fig. 1.29
www.crackjee.xyz
ROTATIONAL MECHANICS 13
Ex. 10 Consider a system of two particles of masses m1 and m1m2
m2. The separation between them is r. Find M.I. of the system where , is called reduced or effective mass of the
m1 m2
about an axis passing through their C.M. and perpendicular to
the line joining them. Also find radius of gyration about centre system.
of mass.
Radius of gyration
Sol.
m1 0 m2 r m2 r m1r
Here r1 and r2 r r1
m1 m2 m1 m2 m1 m2
Fig. 1.31
a/2
M y3 Ma 2
= 2
a 3 12
0
Ma2
or Ix =
12
(ii) Similarly moment of inertia of lamina about y-axis
Fig. 1.32
2
I y = Mb
12
(iii) Moment of inertia of the lamina about z-axis: By using perpendicular axis theorem
I z = Ix Iy
M 2
or Iz = (a b2 )
12
www.crackjee.xyz
14 MECHANICS, THERMODYNAMICS & WAVES
Some special cases
(i) For square lamina, a = b
Ma 2 Ma 2
Ix , Iy
12 12
I z = Ix Iy
M 2 Ma2
= (a a 2 )
12 6
Fig. 1.33
(ii) From the figure Fig. 1.30 shown,
Io = I1 + I2 …(i)
here I1 = I2
Also Io = I3 + I4 …(ii)
here I3 = I4
From equations (i) and (ii), we get
I1 I 2 = I3 I 4
or 2I1 = 2I3
Ma 2
Fig. 1.34 Thus we have I1 I2 I3 I4
12
(iii) Now consider two perpendicular axes as shown in figure Fig 1.34
Io = I5 + I6 … (iii)
As I5 = I6
Ma 2
I1 I2 I3 I4 I5 I6
12
(iv) If we take isosceles triangle lamina of mass M, its M.I. about its diagonal
M
dm = d
2 R
Moment of inertia of this element about the axis (called geometrical axis) shown in
Fig. 1.36
2
dI = (dm) R
Moment of inertia of whole ring
2 R
Fig. 1.36
(dm) R 2
I =
0
2 R
MR 2
= 2 R d
0
MR 2 2 R MR 2
= 0
(2 R 0)
2 R 2 R
= MR2
www.crackjee.xyz
ROTATIONAL MECHANICS 15
Other cases :
(i) M.I. about the tangent parallel to the geometrical axis: By parallel axis theorem
IT = I Md 2
= MR2 + MR2 = 2MR2
MR 2
ID =
2
(iii) M.I. about tangent parallel to the diameter: By parallel axis theorem,
MR 2
I = MR 2
2 Fig. 1.38
3
= MR 2
2
M 2 2
I = 2 (a b ) Fig. 1.39
M
dm = (2 rdr )
R2
Moment of inertia of the ring element about the axis shown
dI = (dm) r2
Moment of inertia of whole disc
R
Fig. 1.40
I = (dm)r 2
0
R
M
= (2 rdr )r 2
2
R
0
R
2M r 4
=
R2 4 0
MR 2
=
2
www.crackjee.xyz
16 MECHANICS, THERMODYNAMICS & WAVES
Other cases :
(i) M.I. about the tangent parallel to geometrical axis : By parallel axis theorem
IT = I Md 2
MR 2
= MR 2
2
3
= MR 2
2
(ii) M.I. about diameter of the disc ;By perpendicular axis theorem
MR 2
ID ID =
Fig. 1.41 2
MR 2
ID =
4
(iii) M.I. about the tangent parallel to diameter ; By parallel axis theorem
I = ID Md 2
MR 2 5 2
= MR 2 = MR
4 4
(iv) M.I. of a small part of disc of mass M about the axis shown in Fig. 1.42
MR 2
I =
Fig. 1.42 2
(v) M.I. of annular disc :
Choose an elemental ring of radius r and thickness dr,
mass of the element
M
dm = (2 rdr )
2
( R2 R12 )
M.I. of the elemental ring
dl = (dm)r 2
M.I. of the annular disc about the axis shown
R2
I = (dm)r 2
Fig. 1.43 R1
R2
M
= (2 rdr )r 2
2
( R2 R12 )
R1
R2
2M
= r 3 dr
( R2 2 R12 )
R1
R2
2M r4
=
( R22 R12 ) 4 R
1
M
=
2
[ R24 R14 ]
2( R2 R12 )
M
= 2
( R22 R12 )( R22 R12 )
2( R2 R12 )
M ( R2 2 R12 )
or I =
2
www.crackjee.xyz
ROTATIONAL MECHANICS 17
Moment of inertia of a thick rod or cylinder
Consider a solid circular cylinder of mass M and cross-sectional radius R. The length of
the cylinder is L.
(i) M.I. of the cylinder about geometrical axis. Choose an element of cylinder in the
form of a pipe of length L and radius r and thickness dr.
The mass of the element
M M
dm = 2
(2 rdr ) L =
(2 rdr )
R L R2
M.I. of the elemental pipe about the axis shown in figure Fig. 1.44
dI = (dm)r2
M.I. of the whole cylinder
R
I = ( dm) r 2
0
R
M Fig. 1.44
= (2 rdr )r 2
2
0
R
R
2M
= r 3dr
R2 0
2M R4 MR 2
=
R2 4 2
(ii) M.I. of the cylinder about equitorial axis :
Choose an element disc of thickness dx at a distance x from axis of rotation.
The mass of the element
M
dm = dx
L
The M.I. of the elemental disc about the axis of rotation
dI = M.I. of disc about diameter + (dm) x2
(dm) R 2
= (dm) x 2
4
M.I of the cylinder
L/2
(dm) R 2
I = 2 (dm) x 2
4
0
L/2 L/2
MR 2 2M
= 2 dx x 2 dx
4L L
0 0
L/2
MR 2 L / 2 2 M x3
= 2L x 0 L 3
0
MR 2 ML2
=
4 12
www.crackjee.xyz
18 MECHANICS, THERMODYNAMICS & WAVES
Special cases :
(i) If M.I. of the cylinder about both the axes calculated are equal, then
MR 2 MR 2 ML2
=
2 4 12
L = 3R
(ii) M. I. of cylinder about the tangent :
Fig. 1.46
By parallel axis theorem
IT = I Md 2
MR 2
= MR 2
2
3
= MR 2
2
M.I. of thin circular pipe
I1 = MR 2
ML2 MR 2
I2 =
12 2
Fig. 1.47 Moment of inertia of sphere
Consider a sphere of mass M and radius R rotating about any of its diameter. Choose an
element in the form of a disc of radius r and thickness dx. The mass of the element
M 3M
dm = r 2 dx r 2 dx
4 3 4 R3
R
3
Here r 2 ( R 2 x 2 )
M.I. of the element about the axis shown
(dm)r 2
dI =
2
M.I. of the whole sphere
3M
R R r 2 dx r 2
(dm)r 2 4R 3
I = 2 2
2 2
0 0
R
3M
I = r 4 dx
3
4R
Fig. 1.48 0
R
3M
= (R2 x 2 )2 dx
4 R3
0
R
3M
= ( R4 x4 2 R 2 x 2 )dx
4 R3
0
R
3M 4 x5 2 R 2 x3
= R x
4 R3 5 3
0
5 5
3M R 2R
= 3
R5
4R 5 3
2
= MR 2
5
www.crackjee.xyz
ROTATIONAL MECHANICS 19
Other cases :
(i) M.I. about any tangent of the sphere :
By parallel axis theorem
IT = Icm + Md2
2
= MR 2 MR 2
5
7
= MR 2
5
(ii) Radius of gyration of the sphere about any tangent
7 2
Mk 2 = 5 MR
7 Fig. 1.49
k = R
5
Moment of inertia of thin circular shell
Consider a shell of mass M and radius R rotating about one of its diameter.
Choose an element in the form of ring of radius r and angular width d .
Here r = R sin .
The mass of the element
M
dm = (2 r )( Rd )
4 R2
M.I. of the element about the axis shown in figure Fig. 1.47
dI = (dm)r 2
M.I. of the whole shell
R
I = 2 (dm)r 2
0
R
M
= (2 r ) Rd r2
4 R2 Fig. 1.50
R
M
= r 3d
2R
0
M
= ( R sin )3 d
2R
0
MR 2
= sin3 d
2
0
MR 2
= sin 2 (sin d )
2
0
MR 2
= sin 2 ( d cos )
2
0
MR 2
= (1 cos 2 )(d cos )
2
0
www.crackjee.xyz
20 MECHANICS, THERMODYNAMICS & WAVES
MR 2 cos3
= cos
2 3
0
MR 2 cos3 cos3 0
= cos cos 0
2 3 3
2
= MR 2
3
Moment of inertia of a triangular plate
Consider a plate of uniform thickness t and base b and height h. If is the density of
material of plate, then mass of the plate
bh bht
M = t
2 2
Choose an element of plate of width b and thickness dy. From the similar triangles ABC
and AB C
b b'
h y
b
or b' = y
h
The mass of the element
dM = × volume of the element
Fig. 1.51
= b '(dy ) t
by
= (dy )t
h
Moment of inertia of the element about base of the plate (BC)
dI = (dM )(h y )2
Moment of inertia of whole plate about the base
h
I = dI (dM )(h y )2
0
h
by
or I = t dy (h y )2
h
0
h h
bt 2 bt
= y (h y ) dy = (h 2 y2 2hy ) ydy
h h
0 0
h
bt
= (h2 y y3 2hy 2 )dy
h
0
h
bt h 2 y 2 y4 2hy 3
= h 2 4 3
0
bt h4 h4 2h4 bth3
= h 2 4 3 =
12
bht Mh2
As M , so I =
2 6
www.crackjee.xyz
ROTATIONAL MECHANICS 21
Moment of inertia of a hollow sphere
Consider a hollow sphere of mass M and inner and outer radii R1 and R2 respectively.
We can imagine the sphere to be made up of a number of thin, concentric spherical
shells. Take one such shell of radius r and thickness dr. Its mass
2
dM = (4 r dr )
Moment of inertia of the shell about its diameter
2
dI = (dM )r 2
3
2
= (4 r 2 dr )r 2
3
8
= r 4 dr
3 Fig. 1.52
Moment of inertia of the sphere
R2
I = dI
R1
R2
8
= r 4 dr
3
R1
R2
8 r5
=
3 5
R1
8
= ( R25 R15 )
15
Mass of the hollow sphere
M = × volume
4
= ( R23 R13 )
3
4 2 ( R25 R15 )
I = ( R23 R13 )
3 5 ( R23 R13 )
2 ( R25 R15 )
or I = 5M 3
( R2 R13 )
Ex. 11 Calculate moment of inertia of a system of (2N + 1)
N
particles, separated by ‘a’ lying along a straight line about an axis
n 2 = N ( N 1)(2 N 1)
passing through the centre. 6
n 1
Sol.
2 N ( N 1)(2 N 1)
L I = 2ma
The total length, L = 2Na a 6
2N After substituting values of a and m, we get
If m is the mass of each particle and M is the total mass of the system,
then ML2 1
I = 1 Ans.
12 N
M
M = (2 N 1) m m
(2 N 1) ML2
For N , I =
The moment of inertia of the system, 12
N N
I = 2 ma 2 n 2 2 ma 2 n2
n 1 n 1
www.crackjee.xyz
22 MECHANICS, THERMODYNAMICS & WAVES
Moment of inertia of a cone
Consider a cone of mass M and base radius R . Suppose the height of cone is h. Choose
an element in the form of a disc of radius x and thickness dy as shown in Fig. 1.53. From
the similar triangles, we have
x R Ry
= x
y h h
The mass of the element
M 3M
dm = ( x 2 dy ) x 2 dy
2 2
R h R h
3
M.I. of the element about the axis shown
(dm) x 2
dI =
2
h
(dm) x 2
M.I. of the whole cone =
Fig. 1.53 2
0
3M
h
2
x 2 dy x 2
= R h
2
0
h 4
3M Ry
= dy
2
2R h h
0
h
3MR 2
= y 4 dy
5
2h
0
h
3MR 2 y 5 3
= MR 2
2h5 5 0
10
Ex. 12 Three identical thin rods, each of mass m and length Ex. 13 From a circular disc of radius R and mass 9M, a small
are joint to form an equilateral triangle. Find moment of inertia of disc of radius R/3 is removed from the disc, as shown in Fig. 1.55.
the triangle about one of its sides. Find the moment of inertia of the remaining disc about an axis
perpendicular to the plane of the disc and passing through the
point O.
Sol. Given, total mass of the disc = 9M
Fig. 1.54
Sol.
The given system of rods for its moment of inertia is equivalent to the
system shown in Fig.1.54(b). Thus the moment of inertia about the axis
given
m( sin 60 )2 m( sin 60 )2 Fig. 1.55
I = 0 M.I. of the disc about a perpendicular axis through O is
3 3
1
m 2 I = (mass) R 2
= 2 sin 2 60 2
3
1 9
2 2 3 m 2 = 9M R2 = MR 2
= m Ans. 2 2
3 4 2
www.crackjee.xyz
ROTATIONAL MECHANICS 23
Mass of the small disc removed Sol.
9M Point P must be the centre of mass of the system. Let is the mass per
2
m = 2
R /3 M unit length, then mass of AB, m1 and mass of CD , m2 2
R
The M.I. of small disc about an axis perpendicular to O,
I ' = I cm Md 2
2
1 2 2R
= M R/3 M
2 3
MR 2
=
2
M.I. of the remaining disc about an axis perpendicular to O
I net = I I'
Fig. 1.56
9 MR 2 Suppose y is the distance of C.M. from C,
= MR 2
2 2 m1 y1 m2 y2
= 4MR2 Ans. y = m1 m2
Ex. 14 A T-shaped object with dimensions shown in Fig. 1.56, ( ) 2 (2 )
=
is lying on a smooth floor. A force F is applied at the point P ( ) (2 )
parallel to AB, such that the object has only the translational 4
motion without rotation. Find the location of P with respect to C. = Ans.
3
L = (dI )
Fig. 1.59 and (dI ) is the moment of inertia of whole body, say it is I,
so L = I
(iv) Angular momentum of a rigid body due to translation and rotation both
Method I : Consider a body of mass m is rotating with angular velocity about C.
M. axis and translating with a linear velocity v as shown in the Fig. 1.61.
The angular momentum of the body
L = Ltranslation Lrotation
or L = mv0 y( kˆ ) I ( kˆ )
= (mv0 y I )kˆ kg-m2/s
Fig. 1.60
where I is the moment of inertia of the body about the
perpendicular axis through O '.
Method II : Consider the situation described in method I. Choose a particle of mass
m at position r having velcoity v in C.M. frame. Its position and velocity in
ground frame at O will be (r + r0 ) and (v + v0 ) respectively..
Angular momentum of the whole body about O
L = m( r ro ) (v vo )
Here, ro and vo are the positon vector and velocity vector of the C.M.
L m( r v) m(ro vo )
or L = Lcm M (ro vo )
Here Lcm represents the angular momentum of the body about C.M. frame, and
M (ro vo ) equals the angular momentum of the body if it is assumed to be
concentrated at C.M. translating with velocity vo . In fiugre of method I,
ro ( xiˆ yˆj ) , vo vo iˆ
(b) L I ( I cm Md 2 ) kˆ
mR 2
Md 2 kˆ kg m2 /s
2
(a) (b)
Fig. 1.63 (c) L mvd ( kˆ ) I ( kˆ )
mR 2 ( kˆ ) kg-m 2 /s
mvd
2
(d) L mvd ( kˆ ) I (kˆ )
mR 2 ( kˆ ) kg m 2 /s
mvd
2
(c) (d)
Fig. 1.64
2 2
(5.98 10 24 )(6.37 106 ) 2
5 24 60 60
= 7.1 × 1023 kg–m2/s
(ii) Angular momentum is associated with the orbital motion of earth about the sun
LOrbital = I 2 2
2 2
= ( Mr )
T Fig. 1.65
[Assuming earth as point mass in comparision to the distance between earth and
sun]
Here M = 5.98 × 1024kg,
r = mean earth-sun distance
= 1.50 × 1011 m
and T = 1year = 365 × 24 × 60 × 60 s
The directions of Lrotation and Lorbit are shown in Fig. 1.65. The resultant angular
momentum of earth will be the vector sum of these two angular momentum.
Thus we can write Lnet L2rotation L2orbital 2 Lrotation Lorbital cos 23.5
www.crackjee.xyz
26 MECHANICS, THERMODYNAMICS & WAVES
Geometrical meaning of angular momentum
Consider a particle of mass m moving with velocity v in x-y plane. Let r and ( r r)
be the position vectors of the particle at instant t and (t t ) respectively. The
displacement of the particle in time t , PQ r v t
The area vector of the triangle OPQ,
1
A = 2 (r r)
1
or A = 2 (r v t )
Fig. 1.66 A 1
or m = (r mv )
t 2
A L
or m =
t 2
A
or 2m
L = t
A
The quantity is the area covered by the position vector r per unit time and is called
t
areal velocity.
Ex. 16 A particle of mass m is moving with velocity v along a Ex. 17 Two particles, each of mass m and speed v, travel in
opposite directions along parallel lines separated by a distance d,
line y = x + 5 . Find the angular momentum of the particle about a
show that the vector angular momentum of the two particle system
perpendicular axis passing through origin O. is the same whatever be the point about which the angular
momentum is taken.
Sol.
The linear momentum of the particle p = mv. Compare given equation of
Sol.
the line of motion with y = mx + c, we have m = 1 or = 45° and c = 5 Suppose the two particles are moving parallel to the y-axis as shown in
Fig. 1.68
unit. It represents a particle moving as shown in figure. The angular
momentum about O
LO = mv cos 45 5
5
= mv unit
2
Fig. 1.68
Total angular momentum of the system
L = L1 L2
= r1 p1 r2 p2
1 2
K = (mr 2 ) 2
( mr 2 )
2 2
where mr 2 is the moment of inertia of the body about axis of rotation, say it is I. Fig. 1.70
Thus kinetic energy of rotating body
1 2
K Rot = I
2
Kinetic energy due to translation and rotation both
Consider a rotating body, which is translating with a velocity v as shown in Fig. 1.71.
The total K.E. of the body
K = KTranslation KRotation
1 2 1 2
= mv I
2 2
Fig. 1.71
www.crackjee.xyz
28 MECHANICS, THERMODYNAMICS & WAVES
1.10 ROTATIONAL WORK
Consider a body subjected to a force F. Suppose body undergoes an angular displacement
due to the torque of force F. The work done by the torque
W = F × displacement PQ
= F× s
Here, s = r
W = F×r
or W =
In case of variable torque, we can write
2
d
WRotation =
1
Fig. 1.72
Power delivered by torque
We know that W =
Dividing both sides by t , we get
W
=
t t
W
or Power P =
t
Newton’s second law for rotating rigid body
We have , torque = r F … (i)
Angular momentum , L = r p … (ii)
Differentiating both sides of equation (ii) w.r.t. time t, we get
dL d (r p)
=
dt dt
dr dp
= p r
dt dt
= v p r F
= (v mv )
= 0 [ v v 0]
dL
ext =
dt
Thus the rate of change of angular momentum is equal to the external torque. This
equation is the rotational analogue of Newton’s second law for linear motion,
dp
i.e., Fext =
dt
Also, we have L= I
dL d d
ext = (I ) = I
dt dt dt
or ext = I
www.crackjee.xyz
ROTATIONAL MECHANICS 29
1.11 ANGULAR IMPULSE
Consider a rod hinged at its one end. It is acted by an external force F for time t as
shown in Fig. 173. The rod starts rotating about the hinge due to the torque Fr .
The product of torque with the duration of its exertion is called angular impulse. Thus
angular impulse
J = t
Its direction is along the direction of . Here in the case discussed the direction of J is
along positive z-axis. Its SI unit is kg-m2/s. By Newtons second law, we have
dL
ext =
dt
or ext dt = dL Fig. 1.73
The product ext (dt ) is angular impulse and dL is the change in angular momentum.
Thus we can write
dL
ext = .
dt
If no net external torque acts on the system, this equation becomes
dL
= 0
dt
or L = constant. (For isolated system)
This equation represents the law of conservation of angular momentum.
= Fr sin
Examples based on conservation of angular momentum
1. Planetary motion around sun : (circular orbit)
In planetary motion, the gravitational force (centripetal force) always passes through
the axis of rotation, so its moment of force is zero. And therefore angular momentum
of the orbiting planet remains constant. If v1 and v2 are the speeds of planet when Fig. 1.74
it is at distances r1 and r2 respectively, then
mv1r1 = mv2r2
or v 1r 1 = v2r2
2. Fig1.75 shows a student sitting on a stool that can rotate freely about a vertical
axis.
The student, who is rotating initially with angular speed i, holds his outstretched
hands. His angular momentum vector L lies along the vertical axis as shown in
figure (a). The student now pulls his arms; this decreases his moment of inertia
from its initial value Ii to If (If < Ii). His angular speed increases from i to f. As
no net external torque acts on the system along the axis of rotation, so the angular
momentum of the system about axis remain constant. Thus we have
Ii i If f . As If < Ii, f< i
Fig. 1.75
www.crackjee.xyz
30 MECHANICS, THERMODYNAMICS & WAVES
3. Consider a device shown in Fig. 1.76 in which a small body attached with the
string is rotating initially in a circle of radius ri with angular speed i . The free end
of the string is pulled down and so body moves on a circle of smaller radius
r f (r f ri ) . As the torque of force F is zero about the axis of rotation, so the
angular momentum of the rotating body remain constant.
Therefore,
Ii i = I f f .
The increase in K.E. of the body
1 2 1
= I I 2.
2 f f 2 i i
Fig. 1.76
4. Consider another similar device in which the radius of path decreases by wrapping
the string over the pipe as shown in Fig 1.77. Here the tension in the string T which
is acting at a distance R (radius of pipe), constitutes a torque TR along the
axis of rotation (along the line of L ), so the angular momentum of the rotating
body will change. Kinetic energy of the body remains constant.
Fig. 1.77
Ex. 19 A uniform rod of length 2a is held with one end resting Ex. 20 Fig. 1.79 shows a student, sitting on a stool that can
on a smooth horizontal table making an angle with the vertical.
rotate freely about a vertical axis. The student, initially at rest, is
Show that when the rod is released its angular velocity when it
makes an angle with the vertical is given by holding a bicycle wheel whose rim is loaded with lead and whose
1/ 2
moment of inertia is I about its central axis. The wheel is rotating
6a(cos cos ) at an angular speed from an overhead perspective, the rotation
. i
a(1 3sin 2 ) is counter clockwise. The axis of the wheel points vertical, and the
angular momentum L i of the wheel points vertically upward. The
Sol. If is the agnular velocity about an axis through IAOR,
then velocity of C.M. will be (asin ). In the process,loss in P.E. student now inverts the wheel; as a result, the student and stool
will equal to gain in K.E, therefore rotate about the stool axis. With what angular speed and direction
1 1 does the student then rotate ? (The moment of inertia of the student
2
mga(cos cos ) = I m( a sin )2 + stool + wheel system about the stool axis is I0).
2 2
Since there is no force acting in the horizontal direction, therefore no Sol.
translation of c.m. in that direction.
There is no net torque acting on the system (student + stool + wheel), so
1/ 2 the angular momentum of the system about vertical axis remains constant.
6a (cos cos )
After simplifying, we get = 1 .
a(1 3sin ) Let L1 and L2 are the change in angular momentum of (student +
L1 L2 = 0
L1 = L2
Fig. 1.78
www.crackjee.xyz
ROTATIONAL MECHANICS 31
Ex. 21 Consider a disc of mass M and radius R is rotating with
angular velocity about its geometrical axis as shown in Fig. 1.80.
A small object of mass m falls gently on the edge of the disc and
stick to it. Find then the angular velocity of the disc.
Sol.
Weight of the object constitutes a torque which is perpendicular to axis
of rotation (or angular momentum), so angular momentum of the system
remain constant along its initial direction. Thus we have
Fig. 1.79
As L2 = Li Li
= 2 Li Fig. 1.80
L1 = 2 Li Li = L f
As initial angular momentum of (student + stool) is zero, so or Ii = If f
i
2 Li = I 0
MR 2 MR2
2 Li or = mR 2 f
2 2
or =
I0
The positive result tells that the angular velocity of rotation is M
f = Ans.
counterclockwise. M 2m
Analogy between translation and rotational motion
Translation Rotation
Power F v Power
www.crackjee.xyz
32 MECHANICS, THERMODYNAMICS & WAVES
Ex. 22 A flexible rope is wrapped several times around a solid
2 gh
cylinder of mass M and radius R, which rotates with no friction,
or v = M Ans.
about a fixed horizontal axis, as shown in Fig. 1.81. The free end of 1
the rope is tied to a mass m which is released from rest a distance 2m
h above the floor. Find its speed and angular velocity of the cylinder Angular velocity of cylinder
just as mass m strikes the floor.
v 2 gh 1
=
Sol. R
1
M R
Dynamical method: 2m
Energy method: Only conservative forces are acting on the system, so
loss in P.E. = gain in K.E.
1 2 1 2
or mgh = mv I
2 2
MR 2 v
The M.I. of the cylinder I , and .
2 R
Using these relations, we have
2
1 2 1 MR 2 v
mgh = mv
2 2 2 R
2 gh
v =
M
1
2m
Fig. 1.81
Let acceleration of translation of mass m is a and angular acceleration of
cylinder is . As there is no slipping of rope over the cylinder, so
Note: In this type of problems when only velocity is asked,
energy method is short and easy in calculation.
a
=
R Ex. 23 A uniform rod AB is hinged at one end A. The rod is
For the translational motion of the block; kept in the horizontal position by a massless string tied to point B
mg – T = ma …(i)
as shown in Fig. 1.82. Find the reaction of the hinge on the end A of
For rotation of cylinder ;
the rod at the instant when string is cut.
TR = I
a
or TR = I R
Ia
or T = R2 … (ii)
Solving equations (i) and (ii), we get
mg
a =
I
m
R2
I
Here m is the inertia of translation and is the inertia due to rotation. Fig. 1.82
R2
I
Sol.
m is called total inertia of motion. When string is cut, the weight of the rod constitutes torque about the
R2 hinge, so
MR 2
As I = A = mg 2 … (i)
2
According to Newton’s second law
mg g
a =
MR 2 M A = I … (ii)
m 1
2m where is the angular acceleration of the rod about the end A.
2R 2
From equations (i) and (ii)
The velocity of block when it strikes the floor : By using third equation
of motion,
I = mg
v2 = 0 + 2ah 2
g
= 2 h mg
M 2
1 or =
2m I
www.crackjee.xyz
ROTATIONAL MECHANICS 33
h = (1 cos )
2
Fig. 1.84
m1g T1 = m1a … (i)
and T2 m2 g = m2 a … (ii)
For the rotation of pulley :
T1R – T2 R = I
For no slipping between pulley and cord
a
=
R
a
T1R T2 R = I
R
Fig. 1.83 Ia
In the process, decrease in P.E. is equal to the increase in rotational K.E. or T1 T2 = … (iii)
of the rod, so R2
1 Solving above equations, we get
2
mgh = I
2 ( m1 m2 ) g MR 2
a = , here I
I 2
1 m 2 2 m1 m2 2
or mg (1 cos ) = R
2 2 3
I
3g m1 2 m2 g
= (1 cos )
and T1 = m1 ( g a ) R2
I
The velocity of C.M. of the rod m1 m2
R2
vcm = r
3g I
= (1 cos ) m2 2 m1 g
2 R2
T2 = m2 ( g a ) Ans.
I
3 g (1 cos ) m1 m2
= Ans. R2
4
www.crackjee.xyz
34 MECHANICS, THERMODYNAMICS & WAVES
Ex. 27 In Fig. 1.86(a) mass m1 slides without friction on the
Note: In using the equation net I , one should write horizontal surface, the frictionless pulley is in the form of a cylinder
of mass M and radius R, and string turns the pulley without slipping.
greater smaller I Find the acceleration of each mass, and tension in each part of the
Here is along direction of greater torque. string.
Short cut : Student can use a short -cut method for those devices in
which all the connected bodies have same acceleration magnitude.
Unblanced load
Acceleration, a
Total inertia
Unbalanced load
Inertia of translation + inertia of rotation
For tension T mdown ( g a)
or T mup ( g a)
Fig. 1.86
Ex. 26 In the device shown in Fig. 1.85, find the acceleration of
the blocks. Sol.
Positive direction of angular acceleration is taken along the
Sol. corresponding acceleration a1 which is taken positive in the direction of
Acceleration magnitude of the block net force. Also due to inertia of the pulley the tension in two parts of the
Unbalanced load string will be different. Let it is T1 and T2 as shown in Fig. 1.85 (b).
a The equation of motion for masses m1 and m2 are
Inertia of translation + inertia of rotation
T 1 = m1a … (i)
Here inertia of translation (m1 m2 ),
and m2 g T2 = m2 a … (ii)
The equation of motion of pulley is
T2 R T1R = I
(N2 constitutes no torque about the axis of rotation)
a
As there is no slipping of string over pulley, so .
R
a
T2 R T1R = I
R
Ia
or T2 T1 = … (iii)
R2
Fig. 1.85 Solving above equations, we get
I1 I2
m2 g MR 2
and inertia of rotation a = , here I
I 2
R12 R22 m1 m2
R2
(m1 m2 ) g
a = m1m2 g
I1 I2 and T1 =
( m1 m2 ) I
R12 R22 m1 m2
R2
T1 = m1 ( g a)
M
and T2 = m2 ( g a) m1 m2 g
2
T2 = Ans.
a I
To get T3 : T1R1 T3 R1 I1 Ans. m1 m2
R1 R2
1.12 ROTATION ABOUT A MOVING AXIS
Analysis of motion which includes both translation and rotation of a body required two
separate equations of motion for the body, one for translation, Fnet = ma and other for
rotation, I . In addition to these equations it often requires the relation between
these two motions i.e., a R.
www.crackjee.xyz
ROTATIONAL MECHANICS 35
Ex. 28 A string is wrapped around a solid cylinder and then the
end of the string is held stationary while the cylinder is released
from rest. Find the acceleration of the cylinder and tension in the
string.
Sol.
Let M and R be the mass and radius of the
cylinder. ‘a’ is the acceleration of C.M. of the
cylinder.
MR 2 [vman ]ground = [v ]
Mg man disk [vdisk ]ground
= I
MR 2 2 or [vman ]ground = v r
M 2
2R Thus angular momentum of the man m( v r )r .
2g
= Ans.
3
MR 2 2g
Ia 2 3
and T =
R2 R2
Mg
= Ans. Fig. 1.89
3
Ex. 29 In figure the mirror of weight W is initially at the same And angular momentum of the disk I disk
MR 2
level as a monkey of equal weight. The fixed pulley is massless and 2
has a radius R. If the monkey starts running up the rope, can he get By conservation of angular momentum, we have
away from the mirror? MR 2
0 = m( v r )r
Sol. 2
Both sides of the pulley, the weight suspended are equal, so net torque After solving, we get
acting on the centre of the pulley will be zero. mvr
= Ans.
2 MR 2
mr
2
www.crackjee.xyz
36 MECHANICS, THERMODYNAMICS & WAVES
Ex. 31 A thin rod is held resting on the ground with its length and for pulley
inclined at an angle to the horizontal. The coefficient of friction 4000 × 0.5 – 2000 × 0.75 = I … (iii)
between the rod and the ground is µ. Show that when the rod let go, where I = m pulley × k2 = (800 / 9.81) × 0.62
= 29.36 kg-m2
it will start slipping on the ground, if
Solving above equations, we get
3sin cos
µ
1 3sin
. = 2.03 rad/s 2 , TA = 3585. 58 N, TB= 2310.82 N Ans.
Sol. By Newton’s second law
Ex. 33 In Fig. 1.92 the steel balls A and B have a mass of 500 g
mg – R = ma cos … (i)
each, and are rotating about the vertical axis with an angular
For not slipping, µR > ma sin velocity of 4 rad/s at a distance of 15 cm from the axis. Collar C is
a sin now forced down until the balls are at a distance of 5 cm from the
Solving above equations, we get µR … (ii) axis. How much work must be done to move the collar down ?
g a cos
Sol.
a sin
and for slipping, µ … (iii)
g a cos
Fig. 1.90
Fig. 1.92
In this process angular momentum remain constant
a can be calculated as
I1 1 = I 2 2
a
I = mg L cos where … (iv) where I1 = 2m (0.15)2, 1 4 rad / s , I2 = 2m(0.05)2
L
Substituting these values in above equation, we get
3sin cos I1 1
After solving equation, we get µ < 36 rad / s
1 3sin 2 = I2
Ex. 32 The composite pulley shown in Fig. 1.91 weighs 800 N Work done on the collar
and has a radius of gyration of 0.6 m. The 2000 N and 4000 N blocks
1 2 1 2
are attached to the pulley by inextensible strings as shown in figure. W = I2 2 I1 1
2 2
Determine the angular acceleration of the pulley and the tension
in the strings. Neglecting the weight of the string. 1 1
or W = 2m(0.05)2 36 2 2m(0.15) 2 4
2 2
Sol. where m = 500/1000 = 0.5 kg
= 1.44 J Ans.
Fig. 1.91
Since the moment of 4000 N is more than that of 2000 N block about the
axis of rotation of pulley, therefore pulley rotates in anticlockwise
direction. Let aA be the acceleration of block A, aB that of B and is the
angular acceleration of the pulley, then
aA = × 0.5, aB = × 0.75
Equations of motion are Fig. 1.93
4000 – TA = (4000/9.81)aA …(i)
mv2 2
TB – 2000 = (2000 / 9.81) aB … (ii) 600 = … (i)
r
www.crackjee.xyz
ROTATIONAL MECHANICS 37
In the process of shorting of the cord, angular momentum remains clutch disconnected, A is brought upto an angular velocity 0.
constant. The accelerating torque is then removed from A and it is coupled
mv1r = mv2 r2 … (ii) to disk B by the clutch (Bearing friction may be neglected). It is
found that 2000 J of heat are developed in the clutch when the
or m 4 0.5 = m v2 r
connection is made. What are the original kinetic energy of disk
2 A?
or v2 =
r Sol. Let IA = I and IB = 2I
Substituting the value of v2 in equation (i), we get In the connection of disk A with disk B, the angular momentum
r = 0.2988 m Ans. remain constant
Ex. 35 A block of mass M rests on a turntable that is rotating or I 0 0 = ( I 2I )
at constant angular velocity . A smooth cord runs from the block
through a hole in the centre of the table down to a hanging block of
mass m. The coefficient of friction between the first block and the
turntable is . (see Fig. 195). Find the largest and the smallest
values of the radius r for which the first block will remain at rest
relative to the turntable.
Sol.
For minimum value of r = r1, the tendency of motion of the block Fig. 1.95
M is towards the centre, and therefore frictional force will act which gives = 0
away from the centre. 3
For the equilibrium of M, we have The change in K.E. evolved as heat i.e.,
1 2 1 2
I 0 (3I ) = 2000
2 2
Substituting the value of in the above equation, we get
2
1 2 1 0
I 0 3I = 2000
2 2 3
1 2
or I0 = 2000
3
1
I 02 = 3000J
2
Fig. 1.94 1
where I 02 in the initial energy of the disk A.
T = M 2 r1 N … (i) 2
where N = Mg Ex. 37 A particle A moves along a circle of radius R = 50 cm so
For the equilibrium of block m we have that its radius vector r relative to the point O rotates with the
T = mg … (ii) constant angular velocity 0.40 rad/s. Find the modulus of the
From (i) and (ii), we get velocity of the particle, and the modulus and direction of its total
acceleration.
mg = M 2
r1 Mg Sol. In OAB, by sine rule, we can write
or r1 = ( mg Mg ) /( M ) 2
R r
=
For maximum value of r = r2, the tendency of motion of block M sin sin(180 2 )
will be away from the centre, therefore frictional force acts towards
centre.
Doing the similar treatment for this case,
T Mg = M 2
r2
and T = mg
After solving for r2, we get
r2 = ( mg 2 Ans.
Mg ) /( M )
Ex. 36 Disks A and B are mounted on a shaft SS and may be
connected or disconnected by a clutch C, as in Fig. 1.95. The
Fig. 1.96
moment of inertia of disk A is one half that of disk B. With the or r = 2 R cos … (i)
www.crackjee.xyz
38 MECHANICS, THERMODYNAMICS & WAVES
r can be written as ; Ex. 39 A disc of mass M has a radius R can rotate freely about
a horizontal shaft O which is located at a distance r from the
r = r cos iˆ r sin ˆj
centre of mass of the disc C. Assume that the disc is released from
= (2 R cos )cos ˆi (2 R cos )sin ˆj the position shown in the Fig. 142. Find minimum value of r for
which the angular acceleration of the disc is maximum.
= 2 R cos 2 iˆ R sin 2 ˆj … (ii)
Sol. The moment of inertia of the disc about axis of rotation which
The velocity of the particle passes through O;
dr d
v = dt 2 R cos 2 iˆ R sin 2 ˆj I o = I cm Mr 2
dt
d ˆ d MR 2
2 R sin 2 i 2 R cos 2 jˆ = Mr 2
= dt dt 2
The angular acceleration of the disc is given by
= 2 R(sin 2 iˆ cos 2 ˆj )
= Mg r Fig. 1.98
v = 2 R
I MR 2 2
= 2 0.40 (0.5) 0.4 m/s Mr
2
Here is constant, so the tangential acceleration of the particle is zero.
Thus the magnitude of total acceleration 2 gr
v 2 (2 R ) 2 = 2
a = 4 2R R 2r 2
R R
= 4 × (0.40)2 × 0.5 = 0.32 m/s2 Ans. d
The angular acceleration to be maximum, 0
Ex. 38 A cubical block of side a moves down on a rough inclined dr
plane of inclination with constant velocity. Find the position of d 2 gr
normal reaction. or = 0
dr R 2 r 2
2
Sol. As the block moves down with constant velocity, the net
force on the block in the direction of motion is zero. Thus we have, (R 2 2r 2 ) 2g – 2gr (0 2 2r ) 0
mg sin f = 0
R
or f = mg sin …(i) r = 2
Ans.
Fig. 1.97 mv
V = M
Also, the block is not rotating, so net torque on it must be zero. The net
torque can be zero only if normal reaction on the block will pass some and mvd = I
distance away, from the C.M. (see figure). Thus for rotational equilibrium
of the block, mvd
=
I
a
f Nx = 0 Since collision is elastic, therefore
2
1 2 1 1 2
fa mv = MV 2 I Fig. 1.99
or x = 2 2 2
2N Substituting the values of v and in the above equation, we get
From equation (i), we have f mg sin
mg sin a
x = 2 mg cos 2 2
1 2 1 mv 1 ML2 mvd
mv = M
a tan 2 2 M 2 12 mL2
=
2 12
Thus normal reaction exerted by inclined plane on the block will pass
a tan ML2
through a distance from C.M.to keep the block in rotational m = Ans.
2 L2 + 12d 2
equilibrium.
Fig. 1.100
www.crackjee.xyz
ROTATIONAL MECHANICS 39
Ex. 41 A thin uniform rod of length L is initially at rest with Sol.
respect to an inertial frame of reference. The rod is tapped at one (a) Let vc and are the velocity of C.M. and angular velocity just
end perpendicular to its length. How far the centre of mass translate after collision.
while the rod complete one revolution about its centre of mass. Using conservation of linear momentum, we have
Neglect gravitational effect. mv0 = 0 Mvc … (i)
Sol. and by conservation of angular momentum
Let t is the duration of impact, then L
mv0 = I
2
L ML2
or mv0 = … (ii)
2 12
F t = 2
P 1 1 mv0 1 ML2 6mv 0
or mv02 = M vc 2
or F t = ( Mvc – 0) … (i) 2 2 M 2 12 ML
Also angular impulse is equal to change in angular momentum
m 1 v0
which gives and vc Ans.
F M 4 4
or t = I
2
(b)
= I( 0) … (ii)
Dividing equation (i) by (ii), we get
vc
=
6
Let t be the time in which rod rotates an angle 2 , then
vc t
=
t 6 Fig. 1.102
Velocity of point P immediately after collision be zero, let it is at
xc a distance y from C.M.
or =
6
vc y = 0
xc mv0 6mv0
or = xc Ans. y=0
2 6 3 or M ML
Ex. 42 A rod AB of mass M and length L is lying on a horizontal L
frictionless surface. A particle of mass m travelling along the surface which gives y =
6
hits the end A of the rod with a velocity v 0 in a direction
perpendicular to AB. The collision is completely elastic. After the L L 2L
AP = 2 6 3
collision the particle comes to rest.
m L
(a) Find the ratio . (c) Angle rotated by rod in time 3v
M 0
(b) A point P on the rod is at rest immediately after the collision. 6mv0 L
Find the distance AP. = t
ML 3v0
L
(c) Find the linear speed of a point P at time 3v after the
0 = Ans.
2
collision.
www.crackjee.xyz
40 MECHANICS, THERMODYNAMICS & WAVES
1 1 2
The rod turns through , in this interval of time. The velocity of or P 1.5 = Mg (0.485) MvC2 I where
2 2 2
point P in y-direction will be
ML2 18 3M
I = M
6 mv0 L 12 12 2
vy = y
ML 6 Substituting the all known values in above equation, we get
v0 1/2
= 0.498 P
4 = – 0.16 Ans.
Mg
The resultant velocity of point P
Ex. 44 A uniform bar of length 6a and mass 8m lies on a
v = vc 2 vy2 smooth horizontal table. Two point masses m and 2m moving in
the same horizontal plane with speed 2v and v respectively strike
2 2 the bar and stick to the bar after collision. Calculate
v0 v0 (a) the velocity of centre of mass after collision,
=
4 4 (b) angular velocity of rotation about the centre of mass,
(c) total energy.
v0 Sol.
= Ans.
2 2 (a) In the process of collision the linear momentum remains constant
Ex. 43 A constant force P is exerted on a rod of mass M. The rod 2mv m 2v 0 = (2m m 8m)vcm
is supported by frictionless wall. If the rod starts from a position of or vcm = 0 Ans.
rest when 45 , as shown in Fig 1.154, what is its angular (b) Angular momentum of the system also remain constant
speed when the end A has moved a distance 1.5 m? 2mv.a (2 mv).2a 0 I … (i)
Sol.
Fig. 1.104
where I is the moment of inertia of the system after collision.
I = M.I. of the rod + M.I. of the mass 2m + M.I. of the mass m
(8m)(6a) 2
or I = + 2m. a2 + m. (2a)2
12
= 30 ma2
Substituting this value in equation (i), we get
= v/5a Ans.
(c) Total energy after collision
1 2 1
E = 2I 30ma 2 ( v / 5a ) 2
Fig. 1.103 2
or E = 3mv2/5 Ans.
The length of the rod = 32 32 18 m
Ex. 45 Two 2.00 kg walls are attached to the ends of a thin rod
When the end A moves a distance x = 1.5 m , let end B moves distance y, of negligible mass, 50 cm long. The rod is free to rotate in vertical
which can be obtained as : plane without friction about a horizontal axis through its centre.
(3 – x)2 + (3 + y)2 = 18 While the rod is horizontal, a 50 g putty wad was dropped onto one
Substituting x = 1.5 m, y will be equal to 0.97 m . Thus the distances of of the balls with a speed of 3.0 m/s and sticks to it.
ends A and B from axis of rotation are 1.5 m and 3.97 m respectively. (a) What is the angular speed of the system just after the putty
Let vA and vB are the velocities of the ends A and B respectively, then wad hits ?
(b) What is the ratio of the kinetic energy of the entire system
vB 1.5 and v A 3.97 after the collision to that of putty wad just before ?
The velocity of the centre C, (c) Through what angle the system rotate until momentarily
2 2 stops?
vA vB
vC2 = Sol.
2 2
(a) Let M is the mass of each ball and m is the mass of the putty wad
given :
or vC2 = 4.52 2
M = 2.00 kg m = 0.05 kg, Length of the rod, L = 0.05 m
In this process centre will rise = y/2 = 0.97/2 = 0.485 m Using the principle of conservation of angular momentum, we have
Using work-energy theorem, we have
L
Work done by the force = increase in P.E. of the rod + gain in translational mv 0 = I … (i)
K.E. + gain in rotational K.E. 2
www.crackjee.xyz
ROTATIONAL MECHANICS 41
Ex. 46 Four thin rods, each with mass M and length d = 1.0 m,
are rigidly connected in the form of a plus sign; the entire assembly
rotates in a horizontal plane around a vertical axle at the centre,
with initial (clockwise) angular velocity i = –2.0 rad/s (see Fig.
1.107). A mud ball with mass m and initial speed vi = 12 m/s is
thrown at and sticks to the end of one rod. Let M = 3m. What is the
Fig. 1.105 final angular velocity f of the plus sign + mud ball system if the
2 2 2 initial path of the mud ball is each of the four paths shown in
L L L
where I = M 2 M
2
m
2
figure: path 1 (contact is made when the ball’s velocity is
perpendicular to the rod), path 2 (radial contact), path 3
2 (perpendicular contact), and path 4 (contact is made at 60° to the
L
= (2M m) perpendicular)?
2
Substituting this value in above equation, we get
=
mvL
I
2mv Sol.
2 (2 M m) L
= 0.148 rad/s Ans.
1 2
(b) The initial K.E. = mv
2
Fig. 1.107
During the collision, the total angular momentum of the system remains
constant.
or Li = L f
rod
I rod i Lmud
i = (I I mud ) f … (i)
If = I rod I mud
Fig. 1.106 Md 2
= 4 md 2
1 3
2
Final K.E. = I
2
4(3m)d 2
1 2 = md 2 5md 2
I 3
2
Their ratio f =
1 2 = (4md 2 i Lmud ) /(5md 2 )
f … (ii)
mv
2
Angular momentum about axle taken as +ve, if it rotates counterclockwise,
After substituting the values of L and , we get and -ve if rotates clockwise.
m Lmud :
f = 0.0123 Ans.
(2 M m) path 1 :L1 = – mdv path 2 : L2 = 0
(c) As the rod rotates, its mechanical energy remain constant. If one path 3 : L3 = mdv path 4 : L4 = (mvcos 60°)d
on the balls is lowered a distance d the other is raised the same Here v = 12 m/s substituting these values in equation (ii) along with
distance, and the sum of the P.E. of the balls does not change. Now v = 12 m/s, we get f
using principle of conservation of mechanical energy, we have
path 1 : – 4.0 rad/s path 2 : – 1.6 rad/s
mechanical energy at A = mechanical energy at B
path 3 : 0.80 rad/s path 4 : – 0.40 rad/s Ans.
L 1 2 L
or mg I = mg (1 cos )
Ex. 47 A rigid body is made of three identical thin rods, each
2 2 2
with length fastened together in the form of letter H (see Fig. 1.
2 2
L 1 L 2mv L 159). The body is free to rotate about a horizontal axis that runs
or mg (2M m) mg (1 cos )
2 2 2 2M m 2 along the length of one of the length of the H. The body is allowed
After substituting the all known values in above equation, we get to fall from rest from a position in which the plane of the H is
= 92.6 horizontal. What is the angular speed of the body when the plane
of the H is vertical ?
The total angle of the swing is = 90° + 92.6° = 182.6° Ans.
www.crackjee.xyz
42 MECHANICS, THERMODYNAMICS & WAVES
Sol. Using principle of conservation of mechanical energy, we have Ex. 49 A homogeneous rod AB of length = 1.8 m and mass M
is pivoted at the centre O in such a way that it can rotate freely in
mg 1 2 the vertical plane. The rod is initially in the horizontal position.
mg = I
2 2 An insect S of the same mass M falls vertically with speed v on the
point C, midway between the points O and B. Immediately after
falling, the insect moves towards the end B such that the rod rotates
with a constant angular velocity .
(a) Determine the angular velocity in terms of v and L.
(b) If the insect reaches the end B when the rod has turned
through an angle of 90°, determine v.
Sol.
Fig. 1.108
where is the angular speed of the rod when it becomes vertical, and
m 2 2
I m , assuming mass of each rod is m.
3 Fig. 1.110
Substituting value of I in the above equation , we get (a) In the process the angular momentum of the system remains
constant.
9g L
= Ans. Mv 0 = I … (i)
4 4
where I is the moment of inertia of the system which is
Ex. 48 A uniform spherical shell of mass M and radius R rotates
2
about a vertical axis on frictionless bearing as shown in Fig. 1.109. ML2 L 7 ML2
M =
A massless cord passes around the equator of the shell, over a 12 4 48
pulley of rotational inertia I and radius r, and is attached to a small
Substituting this value in above equation, we get
object of mass m that is otherwise free to fall under the influence
of gravity. There is no friction of pulley’s axle; the cord does not 12v
= … (ii) Ans.
slip on the pulley. What is the speed of the object after it has fallen 7L
a distance h from rest ? Use work-energy considerations. (b) Let the insect is at a distance x from the centre of the rod after time
Sol. Using a conservation of energy principle, we have ML2
Fall in P.E. of the block = Gain in K.E. of block + rotational K.E. of t, the moment of inertia of the system will be Mx 2 , and
12
the pulley + rotational K.E. of the shell.
1 2 1 1 2 MR 2 ML2
2
'2 therefore angular momentum at this instant Mx 2 .
or mgh = 2 mv 2
I
2 3 12
Torque exerted by the weight of the insect at that moment
mgx.
dL
Since we have =
dt
d ML2
Therefore, Mx 2 Mgx … (iii)
dt 12
dx
or M .2 x = Mgx
dt
Fig. 1.109 dx g
where v / r and ' v / R or =v … (iv)
dt 2
After substituting these values in above equation, and solving, we get
( / 2)
The time taken by rod to rotate through rad , it is
2
mgh 2
v = Ans.
m I M given that the rod rotates with constant angular velocity. The time
+ 2+
2 2r 3 taken by the insect to reach the end B = (L/4)/v.
www.crackjee.xyz
ROTATIONAL MECHANICS 43
It is given that both times are equal , therefore Ex. 51 Two uniform thin rods A and B of length 0.6 m each and
( / 2) ( L / 4) of masses 0.01 kg and 0.02 kg respectively are rigidly jointed, end
=
v to end. The combination is pivoted at the lighter end P as shown in
( / 2) ( L / 4) the Fig. 1.112, such that it can freely rotate about the point P in a
or = vertical plane. A small object of mass 0.05 kg, moving horizontally
(g / 2 )
After solving above equation, we get hits the lower end of the combination and strikes to it. What should
be the velocity of the object so that the system could just be raised
= ( g / L)
to the horizontal position?
Substituting this value in equation (ii), we get
7 L g L
Sol.
v = 7 where L = 1.8 m
12 L 12 In the process of collision, the angular momentum of the system remain
= 4.4 m/s Ans. constant.
Ex. 50 In the system shown in Fig. 1.164. The masses of the mv(2L) + 0 = I … (i)
bodies are known to be m1 and m2, the coefficient of friction between
2
the body m1 and the horizontal plane is equal to , and a pulley of mA L2 mB L2 3L 2
mass m is assumed to be a uniform disc. The thread does not slip where I= mB m 2L
3 12 2
over the pulley. At the moment t = 0 the body m2 starts descending.
Assuming the mass of the thread and the friction in the axle of the
pulley to be negligible, find the work performed by the friction
forces acting on the body m 1 over the first t second after the
beginning of motion.
Sol. Let a is the acceleration of the blocks.
Equation of motion of the blocks are
Fig. 1.111
T1 N = m1a … (i)
m2g – T2 = m 2a … (ii) Fig. 1.112
and for the rotation of the pulley
T2R – T1R = I = I(a/R) … (iii) where mA 0.01 kg , mB 0.02 kg , m = 0.05 and L = 0.6 m
2 Substituting the values in above equation, we get
mR
where I =
2
Solving above equations, we get 2v
=
3
( m2 µm1 ) g
a = Now using conservation of energy principle, we have
m
( m1 m2 )
2
1 2
The distance travelled by the block in t second I = Mgycm … (ii)
2
1 2
s = at where M = (0.01 + 0.02 + 0.05) = 0.08 kg and ycm can be obtained as
2
Work done by the friction
1 2 0.01 0.3 0.02 (0.6 0.3) 0.05 (0.6 0.6)
W = fr s N at ycm
2 0.01 0.02 0.05
1 = 1.0125
( m1 g ) (m µm1 ) g
= 2 2 t2
m Substituting the value of ycm in equation (ii), we get
( m1 m2 )
2 v = 6.3 m/s Ans.
m1 g 2 t 2 ( m2 - µm1 )
= Ans.
(2 m1 + 2 m2 + m )
www.crackjee.xyz
44 MECHANICS, THERMODYNAMICS & WAVES
1.13 ROLLING MOTION
Consider a wheel moving along a straight track, the centre of the wheel moves forward
in pure translation. A point on the rim of the wheel, however traces out a complex curve
called cycloid. We can analyse the motion of a rolling wheel as a combination of translation
and pure rotation.
Fig. 1.13
In one complete rotation of wheel, the C.M. of the wheel moves a distance s 2 R . In
this motion there is no relative motion between point of contact of moving body and the
surface. If T is the time to complete a rotation, then we have
s 2 R
=
T T
or vcm = R
Thus in pure rolling, the velocity of the point of contact is zero and the velocity of centre
of mass is vcm = R. If the wheel moves through a distance greater than 2 R in one full
rotation, then vcm > R and the wheel slips forward. This type of motion occurs when
you apply sudden breaks to the car. The car stops after a long distance but the wheels
rotate only a little during the period.
When the wheel moves a distance shorter than 2 R in full rotation, vcm < R and the
wheel slips backward. It happens when you derive the car on muddy rod, then wheels
rotate more that the forward motion of the car.
Equation of cycloid
Suppose the motion of the wheel is along the positive x-direction in xy-plane with ox
along horizontal and oy along the vertical, the origin O is at the centre of the wheel. Let
R be the radius of the wheel, – , its angular velocity (being clockwise) and – , the angle
that a point P on its axis makes with the x-axis at O at t = 0. In time t the point is given by
x = R cos(– t – ) R cos( t ),
y = R sin( t ) R sin( t )
For an observer on the ground, the velocity of C.M. of the wheel is R . Hence after time
t, the position of the point is given by the coordinates
Fig. 1.14
x = vcm t R cos( t ),
= Rt R cos( t ),
y = R R sin( t )
Velocity of a point on the rolling body
Consider a point P on a rolling body, the velocity of point P is the vector sum of velocity
due to translation and due to rotation. Thus
v p = v translation v rotation
= vcm r
or vp = 2
vcm 2
vcm 2vcm vcm cos
Fig. 1.15
or v p = 2vcm cos .
2
www.crackjee.xyz
ROTATIONAL MECHANICS 45
Ex. 52 Point on the periphery of the rolling body which has
We have v p = 2vcm cos
velocity equal to the velocity of centre of mass of body. 2
1
or cos =
2 2
Fig. 1.116
or = 120°
Fig. 1.117
It is clear from the figure (c) that the point of contact of the wheel (point P) is stationary,
i.e., there is no relative motion between point of contact of body and the surface. And
the top most point is moving at speed 2vcm, faster than any other point on the wheel.
Note:
1. If point of contact of surface is moving with velocity u with respect to ground,
then Fig. 1.118
vcm R u
2. For accelerated surface (see figure 1.118 and 1.119)
acm R a
The kinetic energy of rolling wheel
Let us calculate the kinetic energy of the rolling wheel as measured by the stationary Fig. 1.119
observer.
K.E. of rolling wheel = K.E. associated with translational motion + K.E. associated with
the rotation motion
1 2 1 2
or K rolling = Mvcm I cm
Body I cm Rolling K.E.
2 2
2
For pure rolling, vcm = R Ring MR MR 2 2
1 1 MR 2 3
M( R)2 Icm 2 Disc MR2 2
Krolling = 2 4
2 2
2 7
1 Sphere MR2 MR2 2
(MR 2 I cm ) 2 5 10
=
2
As we know (Icm+ MR2) is the moment of inertia of the wheel about point of contact P,
say it is Ip. Thus,
1 2
K rolling = Ip
2
www.crackjee.xyz
46 MECHANICS, THERMODYNAMICS & WAVES
Angular momentum of rolling wheel
Angular momentum of rolling wheel about an axis passing through point of contact P
and perpendicular to plane of wheel :
L = Ltranslation Lrotation
= m( R vcm ) Icm
or 2
L = m R I cm
Fig. 1.120 or L = ( I cm mR 2 ) Ip
vs = ( 2 R) 2 R 2vcm
Fig. 1.121
Acceleration of a point on a rolling wheel
Consider a rolling wheel; each point on the periphery of the wheel is rotating in a circle
of radius R, due to which centripetal acceleration of each point on the wheel is 2 R. As
the wheel is moving with constant speed, so acceleration due to translation motion of
each point is zero.
Fig. 1.122
Ex. 53 Consider a wheel rolls without slipping and its centre moves with constant
acceleration a. Find the accelerations of points O, P, Q and S when linear velocity of centre
of wheel is v.
Sol.
Fig. 1.123
Fig. 1.124
www.crackjee.xyz
ROTATIONAL MECHANICS 47
R = a 2
2 R cos / 2
R = v =
1/ 2 0
From figure (d),
Resultant acceleration of point O, = 4 R[ cos ( cos 0)]
aO = aiˆ = 8R Ans.
Ex. 55 A rod AB of length 3m which remains in the same vertical
a P = aiˆ aiˆ + 2 Rˆj = 2
Rjˆ plane has its ends A and B constrained to remain in contact with a
horizontal floor and a vertical wall respectively as shown in Fig.
aQ = aiˆ aiˆ 2
Rˆj 127. Determine the velocity and acceleration of the end B at the
position shown in figure, if the point A has a velocity of 2 m/s and an
= 2aiˆ 2
Rˆj acceleration of 1.6 m/s2 rightward.
aS = aiˆ 2
Riˆ aˆj
Sol.
2
= (a R) iˆ ajˆ
Ex. 54 Determine distance travelled by any point P on the
rolling body in one revolution.
Sol.
Fig. 1.127
Velocity of end A :
Method I
The motion of B can be looked as ; translation of end B with velocity vA
and rotation of B about A with tangential velocity vBA. If be the angular
Fig. 1.125 velocity of B about A, then vBA 3 3 . Since the velocity vector
The speed of the point P at any angular position as shown in Fig.
of end B moves downward, so resultant of v A and vBA must be equal to
1.126
vB (see figure).
vp = 2vcm cos
2 vA
From figure (b), vB = tan 60
180
= 2vcm cos
2 vA 2
vB =
tan 60 tan 60
= 2vcm cos 90 /2
= 1.16 m/s Ans.
Fig. 1.126
vA 2
= 2vcm sin Also vBA = 2.3 m / s
2 sin 60 sin 60
The distance travelled by point P in small time dt Method II
d = v p dt
= 2vcm sin ( dt )
2
d d
We have = dt
dt
Fig. 1.128
d Velocity of end A, v A is horizontal, while velocity of the end B, vB is
d = 2vcm sin
2
vertical downward. Drop perpendiculars to the directions of v A and vB
The distance travelled in one complete revolution at points A and B respectively. The intersection point I be the instantaneous
2 centre.
d
= 2vcm sin For the end A, vA = ( IA)
2
0 vA 2
= = 0.77 rad/s
2 IA 3sin 60
2vcm
= sin d For the end B, vB = ( IB ) 0.77 3cos 60
2
0 = 1.16 m/s Ans.
www.crackjee.xyz
48 MECHANICS, THERMODYNAMICS & WAVES
Acceleration of end B :
The acceleration of the end B has three components.
(i) Due to translation of end A, aA = 1.6 m/s2.
(ii) Due to rotation of end B about A, i.e., tangential acceleration
at = rAB 3 3 m / s2
Fig. 1.132
The resultant acceleration of point A,
Fig. 1.130 aA = (a0 at ) 2 an2
Sol.
Velocity : Translational velocity of centre of mass O = (4 4) 2 252 26.25 m/s 2 Ans.
v0 = r 5 1 5 m/s Acceleration of point B
The motion of point A and B may be looked as translational of O and rB = 0.6 m
rotation about O.
For point A, v A = vO vAO at = 4 0.6 2.4 m / s 2
2
or v A = vO (OA) 5 5 1 vBO 32
an = 15 m / s 2
= 10 m/s Ans. rB 0.6
www.crackjee.xyz
ROTATIONAL MECHANICS 49
x and y component of acceleration of point B, aB
ax = aO at cos30 an sin 30
= 4 + 2.4 cos 30° – 15 sin 30°
= –1.4 m/s2
ay = – (at sin 30° + an cos 30°)
= – (2.4 sin 30° + 15 cos 30°)
= – 14.2 m/s2
The resultant acceleration of point B,
aB = ax2 a 2y
= (1.4)2 (14.2)2
Fig. 1.133
= 14.27 m/s2 Ans.
Note:
F f fR
From equation (i), we have acm and from equation (ii),
m I
As the maximum value of frictional force f can be f flim sN , so angular
f lim R
acceleration a can not be greater than a certain limit, i.e., max . In case, if
I
F is large, then acm becomes larger than amax R. The wheel will have linear speed greater
than angular speed and therefore wheel will not have pure rolling motion.
www.crackjee.xyz
50 MECHANICS, THERMODYNAMICS & WAVES
The minimum value of coefficient of friction required for
rolling of a body on horizontal surface for given value of F
1
We have got f = F
mR 2
1
I
The coefficient of friction (static)
f F 1
min =
N Mg mR 2
1
I
F
Ring or hoop I = mR2
2mg
mR 2 F
Disc or cylinder I
2 3mg
2 2F
Sphere I mR 2
5 7mg
3. Let a wheel is thrown on a rough surface with initial velocity v0. The tendency of
point of contact P is to slide forward, so frictional force acts in backward direction
of motion of wheel. Due to the frictional force, the linear speed of centre of wheel
decreases and angular speed increases due to the torque exerted by frictional
force. A condition is reached when v = R, and then wheel will start in pure rolling
motion. Thereafter friction stops acting.
Retardation for translation motion,
f N mg
a = g …(i)
m m m
Acceleration for rotational motion,
fR
= … (ii)
I I
Let the wheel starts pure rolling after time t, then
v = v0 at … (iii)
and = 0 t … (iv)
When wheel starts rolling,
v = R … (v)
Work done by friction K.E.
Fig. 1.136 Kf Ki
1 2 1 2 1
mv
= I – mv02 … (vi)
2 2 2
We can solve above equations to get unknowns.
4. Let a rotating wheel, 0 is placed on a rough surface. The tendency of point
of contact P is to slide backward, so the frictional force will act in forward direction.
Because of this frictional force, the centre of wheel starts accelerating, while torque
of frictional force decreases the angular speed. After some time wheel starts rolling.
Thereafter friction stop acting.
www.crackjee.xyz
ROTATIONAL MECHANICS 51
Acceleration for translational motion ,
fr N mg
a = … (i)
m m m
= g
Retardation for rotational motion,
fR
= … (ii)
I I
Let wheel starts pure rolling after time t, then we have
v = 0 + at … (iii)
Fig. 1.137
= 0 t … (iv)
When wheel starts rolling , we have
v = R … (v)
Work done by friction K.E.
= Kf – Ki
1 2 1 2 1 2
mv I – I 0 … (vi)
2 2 2
We can solve above equations to get unknowns.
(a)
Fig. 1.139
If a be the acceleration of C.M. of the cylinder, then the acceleration
of the top most point of the cylinder is 2a, and hence acceleration
of the hanging block is 2a.
For the motion of the cylinder :
Fig. 1.138
The accelerations magnitude of the block and the centre of mass T f = Ma … (i)
of the cylinder are equal, Let it is a.
For the motion of the cylinder : and TR fR = I … (ii)
T f = Ma … (i) a
For pure rolling, = … (iii)
and fR = I … (ii) R
a For the motion of the block :
For pure rolling, … (iii)
R mg T = m(2a ) … (iv)
For the motion of the block :
mg – T = ma … (iv) After solving above equations, we get
After solving above equations, we get
2 mg
mg a = I
a = 4m M
I
m M R2
R2
www.crackjee.xyz
52 MECHANICS, THERMODYNAMICS & WAVES
1.14 ACCELERATED PURE ROLLING
1. Rolling motion of spool
Take an example of a spool in which a tangential force F is acting on the axel of
spool. In first case, it is applied tangentially on the top of the axel and in second
case it is applied tangentially at the bottom of the axel.
Case I. In this case sliding tendency of point of contact P is backward
F+ f = macm … (i)
Fr - f R = Ia … (ii)
Fig. 1.140 - I acm
For pure rolling a = … (iii)
R
After simplifying above equation we get,
æ rö
F ç1 + ÷
è Rø
acm = I
m+ 2
R
Case II. In this case sliding tendency of point of contact P is forward
F- f = macm … (i)
Fig. 1.141 - II
f R – Fr = Ia … (ii)
acm
For pure rolling motion, a = …(iii)
R
After simplifying above equation we get,
æ rö
F ç1 – ÷
è Rø
acm = I
m+ 2
R
2. Let a wheel vcm = w R lands on a smooth inclined plane. The centre of the wheel
will accelerate due to a force mgsin q, but the angular speed of wheel will remain
Fig. 1.142 constant as no torque is acting on the wheel about axis of rotation. So, the motion
of wheel will no remain pure rolling. Thus the body can have pure rolling only on
rough inclined surface.
3. Let a wheel is placed on rough inclined plane. The tendency of contact point P is
to slide down the inclined due to a net force down the plane. The friction will act up
the plane at the point of contact of the wheel. This frictional force constitute a
torque fR. Due to which wheel starts rotating in addition to translation. Here role of
friction is to transfer some part of tanslational K.E. into rotational K.E. The
mechanical energy of pure rolling wheel remains constant.
Rolling on rough inclined plane :
Dynamical method
For translational motion of wheel
mg sin q - f = macm … (i)
For rotational motion of wheel
Fig. 1.143 fR = Ia … (ii)
acm
For pure rolling motion, a =
R
Iacm
\ fR = … (iii)
R
Iacm
or f = … (iv)
R2
www.crackjee.xyz
ROTATIONAL MECHANICS 53
Solving equations (i) and (iv), we get
mg sin q é g sin q ù
acm = or acm = ê ú
æ I ö I
çè m + 2 ÷ø ê1 + ú
R ë mR 2 û
Unbalanced load
Short- cut method : acm = {Inertia of translation + inertia of rotation}
mg sin q
=
(m + I / R 2 )
Velocity of C.M. after falling a height h
v2 = 0 + 2acm ( s )
é g sin q ù h
2ê
I ú ´ sin q
= ê1 + ú
ë mR 2 û
2 gh
or v =
I
1+
mR 2
The minimum frictional force and coefficient of friction required to cause pure rolling of
Iacm
a body can be obtained from, f min = .
R2
Acceleration f min
Moment of Iacm m min =
Body æ g sin q ö f min = N
inertia acm =ç R2
è 1 + I / mR 2 ø÷ N = mg cos q
g mg sin q tan q
Ring MR 2 sin q
2 2 2
MR 2 2 mg sin q tan q
Disc/Cylinder g sin q
2 3 3 3
2 5 2 2 tan q
Sphere MR 2 g sin q mg sin q
5 7 7 7
Ex. 58 A plank of mass m1 with uniform sphere of mass m2 For the motion of sphere : If f is the frictional force, then
placed on it rests on a smooth horizontal plane. A constant horizontal f = m2ac … (ii)
force F is applied to the plank. With what acceleration will the and f r = Ia
plank and the centre of the sphere move provided there is no sliding
between the plank and the sphere? 2
or fr = m2 r 2a … (iii)
Sol. 5
For the motion of the plank :
F - f = m1a p … (iv)
After solving equations, we get
é 2F ù
ac = ê 2m + 7m ú
ë 2 1û
é 7F ù
and ap = ê ú Ans.
Fig. 1.144 ë 2m2 + 7m1 û
If ac is the acceleration of the centre of the sphere, and ap is the acceleration
of the plank (or acceleration of point of contact of the sphere), then Ex. 59 Two solid bodies rotate about stationary mutually
perpendicular intersecting axes with constant angular velocities
a p = ac + a r … (i)
ω1 = 3.0 rad/s and w2 = 4.0 rad/s. Find the angular velocity and
Here r is the radius of the sphere. angular acceleration of one body relative to the other.
www.crackjee.xyz
54 MECHANICS, THERMODYNAMICS & WAVES
Sol. We know that Ex. 61 A circular wooden hoop of mass m and radius R rests flat
on a frictionless surface. A bullet, also of mass m, and moving with
r r r
w12 = w1 -w 2 a velocity v strikes the hoop and gets embedded in it. The thickness
of hoop is much smaller than R. Find the angular velocity with
\ w12 = w12 + w 2 2 which the system rotates after the bullet strikes the hoop
Fig. 1.151 Ex. 65 Two thin circular disks of mass 2 kg and radius 10 cm
w1 = 0 + a t each are joined by a rigid massless rod of length 20 cm. The axis of
the rod is along the perpendicular to the planes of the disk through
5mg their centres as shown in the Fig.1.153. This object is kept on a
w1 = 0 + t … (i)
2R truck in such a way that the axis of the object is horizontal and
and v1 = v0 – at perpendicular to the direction of the motion of the truck. Its friction
with the floor of the truck is large enough so that the object can roll
3v
= - ( m g )t … (ii) on the truck without slipping. Take x-axis as the direction of motion
5
of the truck and z-axis as the vertically upwards direction. If the
After getting pure rolling
truck has an acceleration of 9 m/s2, calculate :
v1 = w1R … (iii) (i) The force of friction on each disk.
3v æ 5 m gt ö (ii) The magnitude and the direction of the frictional torque
- m gt = ç R
\
5 è 2 R ÷ø acting on each disk about the centre of mass O of the object.
Express the torque in the vector form in terms of unit vectors
6v i, j and k in the x, y and z directions.
or t =
35
www.crackjee.xyz
56 MECHANICS, THERMODYNAMICS & WAVES
Sol. and N2 = µN1 … (ii)
(i) In the reference frame of the truck, frictional force will pull the Solving above equations, we get
disks in the direction of motion of the truck, while pseudo force mg
(F = ma), where a = 9 m/s2 acts in backward direction of motion N1 =
(1 + µ2 )
of the truck. Consider the forces on the either wheel.
For the translation of the disk µmg
and N2 =
(1 + µ2 )
Method 1 : Dynamical method
The net retarding torque on the cylinder = (µN1 + µN2) × R
therefore we have, (µN1 + µN2) × R = Ia
mR 2
Substituting the values of N1, N2 and I = , we get
2
2 m g ( µ + 1)
Fig. 1.153
a = R ( m 2 + 1)
F – fr = ma cm … (i)
and for rotation of the disk, we have The time taken by the cylinder to come in rest can be obtained as
fr × R = Ia
0 = w0 - at ,
or fr × R = I acm / R … (ii)
mR 2 w0 w0 R( m 2 + 1)
where \ t = =
I = a 2m g ( m + 1)
2
Now using second equation of motion
mR 2
Solving above equations, and substituting I = , m = 2 kg and 1 2
2 q = w 0t - a t
2
R = 0.10 m,
we get acm = 6 m/s2 After substituting the values of a and t in above equation, we get
and fr = 6 N or fr = (6i) N w 02 R( m 2 + 1)
(ii) The position vectors of point of contacts are (taking c. m. of the q = [4µg ( µ + 1)]
system as the origin)
r
r = (0iˆ - 0.1 ˆj - 0.1kˆ) Number of turns n =
q
=
(1 + μ 2 )ω 02 R
Ans.
r 2p [8πgm (μ + 1)]
and r2 = (0iˆ + 0.1 ˆj - 0.1kˆ)
Method II : Energy method
The torque on the disks are Using work-energy theorem,
r r r DK = W f
t1 = r1 ´ f r = (0iˆ - 0.1 ˆj - 0.1kˆ) ´ (6iˆ)
1
= (-0.6jˆ + 0.6k)N
ˆ -m or I w 02 = ( m N1 + m N 2 ) ´ (2p Rn)
2
r r r Substituting all the known values in above equation, we get
t 2 = r2 ´ fr = (0iˆ + 0.1 ˆj - 0.1kˆ) ´ (6iˆ)
(1 + µ 2 )ω02 R
= (– 0.6jˆ – 0.6k)N
ˆ -m n = Ans.
[8πgµ(μ + 1)]
Ex. 66 A uniform cylinder of radius R is spinned about its axis
Ex. 67 A uniform solid cylinder of radius R = 15 cm rolls over
to the angular velocity w0 and then placed into a corner Fig. 1.165.
The coefficient of friction between the corner walls and the cylinder a horizontal plane passing into an inclined plane forming an angle
is equal to µ. How many turns the cylinder accomplish before it a = 30° with the horizontal shown in the Fig. 1.155. Find the
stops? maximum value of the v0 which still permits the cylinder to roll
onto the inclined plane section without a jump. The sliding is
Sol.
assumed to be absent.
For the vertical and horizontal equilibrium of the cylinder, we have
N1 + µN2 = mg … (i) Sol.
1 1 1 2 1 2
mv02 + I w 2 \ mg ( R + r )(1 - cos q ) = mv + I w
K1 = 2 2
2 2
1 1
2 mg ( R + r )(1 - cos q ) = m(w r )2 + I w 2 … (i)
1 1 æ mR2 ö æ v0 ö 2 2
= mv02 + ç ÷ çè ÷ø
2 2è 2 ø R At the break-off
mv 2
3 mg cos q = N +
= mv02 … (i) (R + r)
4
Substituting N = 0 and v = wr, we have
When the cylinder passes on to the inclined plane its centre of mass
descends through a distance h = R (1 – cos a). m(w r ) 2
mg cos q = … (ii)
If v is the velocity of its centre of mass now, (R + r)
Solving equations (i) and (ii), we get
3 2
then rolling kinetic energy = mv .
4 10g( R + r )
w = Ans.
From energy conservation, we have 17 r 2
gR
v0 = (7 cos a - 4) = 1.0 m / s Ans.
3
7m 2 2r 2
K.E. = v [1 + 2 ] Ans.
10 7R
Ex. 70 A uniform solid cylinder of mass M and radius R rolls
on a rough inclined plane with its axis perpendicular to the line of
the greatest slope. As the cylinder rolls it winds up a light string
which passes over a small mass m, the part of the string between
pulley and the cylinder being parallel to the line of greatest slope.
If q is the inclination of the plane with the horizontal, calculate
the tension in the string.
Fig. 1.156 Sol.
Let it happens at a vertical height h below the top. Therefore we have, Method 1: Dynamic method
Let the acceleration of c.m. of the cylinder is a . The acceleration of a
1 2 1 2 point on its tangent will be 2a, so the acceleration of the mass m will be
mgh = mv + I w
2 2 2a.
www.crackjee.xyz
58 MECHANICS, THERMODYNAMICS & WAVES
Equation of motion for the translation of the cylinder
Fig. 1.159
Sol.
Fig. 1.158 Given, velocity of centre = v0
Mg sin q - T - f r = Ma … (i)
v0
and for its rotation about axis passing through its c.m. Angular velocity about the centre =
2r
fr R – T R = I a
Here initial velocity of rotation w 0 < v0 / r , the sphere slips in forward
MR 2 direction. Frictional force decelerates it to decrease its translational velocity
or fr R – T R = a … (ii)
2 v0 to a value v, which corresponds to pure rolling . Frictional force
Equation of motion for mass m increases angular velocity w0 to a value w, which corresponds to pure
T – mg = m(2a) … (iii) rolling and satisfies the relation v = wr.
Solving above equations, we get
fr
2Mg sin q - 4mg Deceleration of the centre of mass of the sphere a =
a = M
8m + 3M
\ v = v0 – at
(3 + 4sinθ)mMg
and T = Ans. æ fr ö
8m + 3 M or v = v0 - çè ÷ø t … (i)
M
Method II : Energy method
Let v be the velocity of the c.m. of the cylinder when it travelled a
t
distance y along the plane. The velocity and the distance travelled by the and angular acceleration about centre a =
I
mass m in this interval will be 2v and 2y respectively.
Decrease in P. E. of cylinder = Increase in (rolling K. E. of the cylinder
f r .r 5 fr
+ translational K.E. of the mass m + P.E. of the mass) or a = =
2 2 Mr
(2 Mr / 5)
1 2 1 2 1 2
Mg ( y sin q ) = Mv + I w + m(2v ) + mg (2 y )
2 2 2 w = w0 + at
v MR 2 5 fr
where w=
R
and I =
2
or w = w0 + (2Mr ) t
Substituting these values in above equation and solving for v, we get
v0 5 fr
4 g ´ ( M sin q - 2 m) or w = + t … (ii)
v 2 = 2r (2Mr )
8m + 3M
When pure rolling occurs v = w r … (iii)
Using third equation of motion v2 = u 2 + 2ay , where u = 0 Solving above equations, we get
Substituting this value in equation (iii), we can get T. Ex. 72 A billiard ball, initially at rest, is given a sharp impulse
by a cue. The cue is held horizontally a distance h above the centre
Ex. 71 A sphere of mass M and radius r shown in Fig. 1.159
line as shown in Fig. 1.160. The ball leaves the cue with a speed v0
slips on a rough horizontal plane. At some instant, it has
and because of its forward energy eventually acquires a final speed
translational velocity v0 and rotational velocity about the centre
(v0/2r). Find the translational velocity after the sphere starts pure æ 9v 0 ö
of ç ÷ , show that h = (4/5) R where R is the radius of the ball.
rolling. è 7 ø
www.crackjee.xyz
ROTATIONAL MECHANICS 59
t fr R 5m g
where a = I = =
(2mR 2 / 5) 2R
q 2 - q1 a
Dq = qn = w 0 + 2 ( 2 n - 1)
where Dq can be taken as positive for anticlockwise and
6. Relationship between linear and angular variables : If a
negative for clockwise rotation.
body rotates through an angle q , the point at a radial distance
3. Angular velocity : If a body rotates through an angular
r from axis of rotation moves along an arc with length s is given
displacement Dq in a time interval Dt , its angular velocity by
r
ur Dq s = qr
w av = Also, v = wr
Dt
The instantaneous angular velocity at = ar
r
ur dq n2
w = and an = = w 2r
dt r
Angular velocity is a vector quantity whose direction is given by 7. Moment of inertia or rotational inertia : M.I. of a particle
right hand rule. Its SI units is rad/s. of mass m at a distance r from axis of rotation is given by
4. Angular acceleration : If the angular velocity of a body changes I = mr 2
by Dw in time interval Dt , then the average angular acceleration
of the body is given by M.I. of a body can be defined as I = Smi ri2
ur
ur
a av =
Dw
Dt
For a system of discrete particles I=
ò r dm
2
www.crackjee.xyz
60 MECHANICS, THERMODYNAMICS & WAVES
8. Parallel-axes theorem : The parallel-axes theorem relates the 10. Radius of gyration : Radius of gyration of system of particles
moment of inertia of a body about any axis to that about a parallel about an axis is given by
axis passing through centre of mass of the body :
I
K = M
2
I = I cm + Md
where I = Smi ri2 and M = Smi
9. Perpendicular-axes theorem : If I x and I y are the moment Radius of gyration of a body of mass M and moment of inertia
of inertia of the planer body about the x and y-axis respectively,
then moment of inertia about z axis is given by about an axis is given by K = I
Iz = Ix + I y M
ML2 ML2
I= I=
12 3
Thin rod about axis through centre Thin rod about axis through one end
perpendicular to length perpendicular to length
M (a2 + b2 )
I= I = MR 2
12
Thin rectangular lamina about Ring or hoop about central axis
perpendicular axis about centre
MR 2
I=
M 2
I=
2
(
R1 + R22 )
2
Hoop about any diameter Annular ring or cylinder
MR 2
I= Circular disc about diameter
2
MR 2
Circular disc I=
4
www.crackjee.xyz
ROTATIONAL MECHANICS 61
MR 2 ML2 MR 2
I= I= +
2 12 4
Solid cylinder or disc Solid cylinder about equatorial axis
2 2
I= MR 2 I= MR 2
5 3
Sphere about any diameter Thin spherical shell about any diameter
2g
acm = sin q
3
mg
f min = sin q
3
Only one option correct 7. Figure shows a graph of the angular velocity versus time for the
1. Fly wheel is an important part of an engine : rotating disk. For a point on the rim of the disk, the only radial
(a) it gives strength to engine acceleration is represented by :
(b) it accelerates the speed of the wheel
(c) it reduces the moment of inertia
(d) it helps the engine in keeping the speed uniform
2. A body is in pure rotation. The linear speed v of a particle, the
distance r of the particle from the axis and the angular velocity
of the body are related as = v/r. Then :
(a) 1/r (b) r a c t
b d
(c) is independent of r (d) all
3. A uniform rod is kept vertically on a horizontal smooth surface at (a) ab (b) ac
O. If it is rotated slightly and released, it falls down on the (c) bc (d) cd
horizontal surface. The lower end will remain : 8. Figure shows the overhead view of a disk rotating counter
(a) at O clockwise. The angular speed of the disk is decreasing. Which of
l from O the vector correctly represents total acceleration of a point on the
(b) at a distance less than rim of the disk :
2
(c) at a distance l from O 2
2
l
(d) at a distance larger than from O
2 3 1
4. Consider the following two equations
dL
(A) L = I (B) ,
dt 4
in non inertial frame :
(a) 1 (b) 2
(a) both A and B are true (b) A is true but B is false
(c) 3 (d) 4
(c) B is true but A is false (d) both A and B are false
9. Figure shows an assembly of three small spheres of the same mass
5. Which of the following equations can be used for uniformly
that are attached to a massless rod with the indicated spacings.
accelerated rotating body :
Consider the moment of inertia I of the assembly about each
(a) =5 (b) =t–3
sphere, in turn. The sphere(s) about which moment of inertia is
(c) = 5t2 – 3 (d) = 4t2 + t + 3
greatest is :
6. Figure is a graph of the angular position of the rotating disk. The
angular velocity of the disk at t = 3 s : 1 2 3
d 2d
(rad)
(a) 1 (b) 2
(c) 3 (d) 1 and 3
10. A body is rotating uniformly about a vertical axis fixed in an inertial
frame. The resultant force on a particle of the body not on the axis
t(s) is :
1 2 3 (a) vertical
(b) horizontal and skew with the axis
(a) Zero (b) Positive (c) horizontal and intersecting the axis
(c) Negative (d) None (d) none of these
1 2 w2
F3 F2 w1
(a)
(b)
F1
O
F4
(a) w1 = w2 (b) w1 < w2
(a) F1 (b) F2 (c) w1 > w2 (d) none of the above
(c) F3 (d) F4 18. Figure shows a particle moving at constant velocity v and four
13. A body is rotating nonuniformly about a vertical axis fixed in an points with their xy coordinates. If L 1, L 2, L3 and L 4 are the
inertial frame . The resultant force on a particle of the body not on
angular momentum about the points a, b, c and d respectively,
the axis is
then :
(a) vertical
(b) horizontal and skew with the axis
(c) horizontal and intersecting the axis
(d) none of these a v
(9, 1)
14. A solid sphere, a hollow sphere and a disc , all having same mass (–2, +1) c
and radius, are placed at the top of a smooth incline and released
. The friction coefficients between the objects and the incline are
same and not sufficient to allow pure rolling. Least time will be b (–2, –2) d (3, –1)
taken in reaching the bottom by :
(a) the solid spehere (b) the hollow sphere (a) L1 < L3 (b) L1 > L4
(c) the disc (d) all will take same time (c) L1 = L3 = L4 (d) (L1 = L3) < L4 < L2
15. The density of a rod gradually decreases from one end to the
19. Figure shows three particles of the same mass and the same
other. It is pivoted at an end so that it can move about a vertical
axis through the pivot. A horizontal force F is applied on the free constant speed moving as indicated by the velocity vectors. Points
end in a direction perpedicular to the rod. The quantities, that do a, b, c and d form a square with point e at the centre. The points
not depend on which end of the rod is pivoted, are : about which angular momentum is greatest :
(a) angular acceleration
(b) angular velocity when the rod completes one rotation
a b
(c) angular momentum, when the rod completes one rotation
(d) torque of the applied force
16. A hollow sphere and a solid sphere having same mass and same e
radii are rolled down a rough inclined plane :
(a) The hollow sphere reaches the bottom first
d c
(b) The soild sphere reaches the bottom with greater speed
(c) The soild sphere reaches the bottom with greater kinetic
energy (a) a (b) a, b
(d) The two spheres will reach the bottom with same linear
(c) c, d (d) e
momentum
O x
angular momentum with respect to origin :
(a) remains constant (b) goes on increasing
(c) goes on decreasing (d) zero (a) 10k N-m (b) 5 3 j N-m
22. Four spheres each of diameter 2 r and the mass M are placed with (c) 5k N-m (d) 5j N-m
their centres on the four corners of a square of side l then the 29. Two point masses m1 and m2 joined by a masssless string of length
moment of inertia of the system about an axis along one of the r. The moment of inertia of the system about an axis passing
sides of the square is : through the centre of mass and perpendicular to the string is :
(a) M[
4 r2 + 2 l2] (b)
8 r2 + 2 l2]
M[ m1m2 r 2
5 5 (a) (m1 + m2)r 2
(b)
m1 + m2
(c)
8 Mr2 (d) M[
4 r2 + 4 l2]
5 5
23. Three point masses, each of mass m are placed at corners of an
m1r 2 m2 r 2
(c) (d)
equilateral triangle of side l then the moment of inertia of this m2 (m1 + m2 ) m1 (m1 + m2 )
system about an axis along one side of the triangle is : 30. A uniform cylinder has a length l and radius R. If moment of interia
(a) ml2 (b) 3 ml2 of this cylinder about its own geometrical axis is equal to moment
(c)
3 ml2 (d) 2 ml2 of inertia of the same cylinder about an axis passing through centre
4 3 and perpendicular to its length is :
24. A body slides down on an incline and reaches the bottom with a
(a) l=R (b) l= 3 R
velocity v. If the same body were in the form of a ring , its velocity
at the bottom would have been : 1
(c) l= R (d) l=3R
(a) v (b) v 2 3
31. A rod of mass m and length l is bent in to shape of L. Its moment
v 2
(c) (d) v of inertia about the axis shown in figure :
2 5
25. A ring rolls on a plane surface.The fraction of its total energy
associated with its rotation is:
m/2
(a) 1 (b) 1
2
1
(c) (d) 2
3
m/2
26. A soild cylinder of mass M and radius R rolls down an inclined
plane with height h without slipping. The speed of its centre of
mass when it reaches the bottom is :
4 ml 2 ml 2
(a) 2gh (b) gh (a) (b)
3 6 3
(c)
3
gh (d) 4gh ml 2
4 (c) (d) None
2
v (a) (b)
P
300
r x
O x
450 O
x
point of intersection of the rods and perpendicular to this plane is: (c) 3ml2 (d) ml2
F
(a) is always upwards
(b) is always downwards F
(c) is initially upwards and then downwards periodically
(d) none of these.
52. A disc is performing pure rolling on a smooth stationary surface
with constant velocity v. For the point of contact of the disc (a) centre of mass of spool moves towards right
(a) velocity is v, acceleration is zero (b) centre of mass of spool moves towards left
(b) velocity is zero, acceleration is zero (c) centre of mass remains at rest
(d) none of these
(c) velocity is v, acceleration is v 2 / R
58. A horizontal force F is applied at the top of an equilateral triangular
(d) velocity is zero, acceleration is v 2 / R block having mass m and side a as shown in figure. The minimum
53.. A disc of radius R rolls on a horizontal surface with linear velocity value of the coefficient of friction required to topple the block
v and angular velocity w. This is a point P on the circumference of before translation will be
the disc at an angle q which has a vertical velocity. Here q is equal
to F
p 2 1
æ v ö æ v ö
p + sin -1 ç – sin -1 ç (a) (b)
(a) è w R ÷ø (b)
2 è w R ÷ø 3 2
æ v ö æ v ö 1 1
p – cos -1 ç p + cos -1 ç
(c) è w R ÷ø (d) è w R ÷ø (c)
3
(d)
3
m2, R A
m q
u
a2 ( a2 + g) (a) Zero
(a) (b)
R R
2gR
2( a2 + g) (b)
(c) (d) None of these 3
R
60. A particle of mass m is attached to a rod of length L and it rotates
gR
in a circle with a constant angular velocity w. An observer P is (c)
rigidly fixed on the rod at a distance L/2 from the centre. The 5
acceleration of m and the pseudo force on m from the frame of
(d) it cannot come on the surface for any value of u.
reference of P must be respectively.
63. A hoop of radius 0.10m and mass 0.50 kg rolls across a table
w parallel to one edge with a speed of 0.50 m/s. Refer its motion to
P a rectangular coordinate system with the origin at the left rear
m
corner of the table. At a certain time t, a line drawn from the origin
L to the point of contact of the hoop with the table has length 1m
O and makes an angle of 30° with the X-axis (figure). What is the spin
angular momentum of the hoop with respect to the origin at this
2 L time t ?
(a) zero, zero (b) zero, mw
2
L L Z
(c) w2 , mw 2 (d) zero, mw 2 L
2 2
61. Consider the two bobs are shown in the figure. The bobs are
pivoted to the hinges through massless rods. If tA be the time 30°
taken by the bob A to reach the lowest position and tB be the time Y
taken by the bob B to reach the lowest position. (Both bobs are X vcm
released from rest from a horizontal position) then ratio tA / tB is
m m
A B
l l /2
(a) 3 (b) 5
(a) -0.25 iˆ kg m 2 /s (b) - 0.005 iˆ kg m 2 /s
1
(c) 2 (d)
2 (c) - 0.025 iˆ kg m 2 /s (d) - 0.5 iˆ kg m 2 /s
Lev el -2
Only one option correct 5. If E1 is the kinetic energy of the block and E2 is the kinetic energy
1. Figure shows four rotating disks that are sliding across a frictionless of sphere at the position B, then :
floor. ThreeforcesF, 2F or 3F act on each disk, either at the rim, (a) E1 = E2 (b) E1 < E2
at the centre, or halfway between rim and centre. Which disks are (c) E1 > E2 (d) none of the above
in equilibrium : 6. If K 1 is the translational kinetic energy of block and K 2 is the
F F translational kinetic energy of the sphere at B, then:
(a) K1 = K2 (b) K1 < K2
3F 2F (c) K1 > K2 (d) none of the above
2F 7. Two spheres of radii R and 2R roll on an inclined plane without
F
slipping. If v1, v2 and w1, w2 are their velocities of centre of mass
A B and angular velocities of rotation after moving a distance l, then :
(a) v1 = v2, w1 = w2 (b) v1 < v2, w1 < w2
F F
(c) v1 < v2, w1 = w2 (d) v1 = v2, w1 > w2
2F 8. Figure shows the motions of three identical spheres on three
F different ramps of same inclination.
F
2F
C D
(a) A, D (b) A, B
(c) C, D (d) A, C
2. Figure shows overhead views of three structures on which three
forces act. The direction of the forces are as indicated. If the
(a) (b) (c)
magnitude of the forces are adjusted properly, which structure
can be in stable equilibrium :
The rough part of the ramps are shown by bold line. If KA, KB and
KC are their kinetic energy at the bottom of the ramp, then :
(a) KA < KB (b) (KA = KB) < KC
(c) KA < (KB = KC) (d) (KA = KB) > KC
9. End of the bar AB in figure rests on a horizontal surface, while end
B is hinged. A horizontal force P of 60 N is exerted on end A.
A B C Neglect the weight of the bar. The vertical component of the force
exerted by the bar on the hinge at B is :
(a) A (b) B
(c) C (d) None
3. In figure, a block slides down a frictionless inclined plane and a B
sphere rolls without sliding down a ramp of the same angle q. The
block and sphere have the same mass, start from rest at point A, 5m
and descend to point B. If work done by gravitational force on the 4m
block is W1 and that on sphere is W2, then :
P = 60N
A
A A
3
(c) gl (d) None
4
12. A hoop of radius r rolls on a horizontal plane with constant velocity
v without slipping. The velocity of any point, t second after it
passes the top position is :
mgh
(a) v2 = 2 gh (b) v2 =
w m I
q + 2 +M
v 2 2r
mgh mgh
(c) v2 = (d) v2 =
m I
m I M + +M
+ 2+ 2 r2
2 2r 3
(a) 2v (b) 2vt
16. The angular momentum of a particle relative to a point O varies
æ vt ö æ vt ö r r
(c) 2v cos çè ÷ø (d) 2v sin çè ÷ø with time as J = a + bt2, where a and b are constant vectors, with
2r 2r r
r
13. A block of mass M rests on a turntable that is rotating at constant a perpendicular b . The moment of force ( t ) relative to the point
r
angular velocity w. A smooth cord runs from the block through a O acting on the particle when the angle between the vectors t and
hole in the centre of the table down to hanging block of mass m. J equal 45° :
The coefficient of friction between the first block and the turnable
a a
is m (see figure). The smallest value of the radius r for which the (a) (b) 2
first block will remain at rest relative to the turntable: b b
a b
(c) 2b (d) 2a
M b a
17. A rod AB of length L slides in the xy- plane. If the rod makes an
angle q with the vertical, the angular velocity of the rod will be :
y
m
A
mg - m Mg mg + m Mg L
(a) r= (b) r= q
2 2
Mw Mw
mg x
(c) r= (d) None of the above B
Mw2
14. Moment of inertia of a uniform circular disc about a diameter is I. (a) Directly proportional to the length of the rod and the linear
Its moment of inertia about an axis perpendicular to its plane and velocity of the end A at that instant.
(b) Independent of the length of the rod but will depend on the
passing through a point its rim will be :
angle q.
(a) 5I (b) 3I
(c) Independent of q but will depend on the length of the rod
(c) 6I (d) 4I
and linear velocity of the end A at that instant.
15. A uniform spherical shell of mass M and radius R rotates about a (d) Dependent upon the length of the rod, the angle q and also
vertical axis on frictionless bearing as shown in figure. A massless on the linear velocity of the end A of the rod at that instant.
cord passes around the equator of the shell over a pulley of rotational
w0 t Iy
w0 thickness . The ratio between moments of inertia is :
(a) (b) 4 Ix
(a) 1 (b) 16
t t (c) 32 (d) 64
30. A thin wire of length L and uniform linear mass density r is bent
w (t) w (t) into a circular loop with centre at O as shown. The moment of
inertia of the loop about the axis xx’ is :
x x’
t t O
5rL3 3rL3
A B (c) (d)
16 p 2 8p 2
a
31. In the device shown in figure, the cylinder in the pure rolling
a motion. At certain instant the angular speed of the cylinder is w.
The velocity of the block at that instant is
(a) wR
2R
R
(b) 2wR
Mg (a) 1 : 1 (b) 2 : 1
(a) infinitesimal (b)
4 (c) 3 : 8 (d) 8 : 11
37. A sphere is given some angular velocity about a horizontal axis
Mg
(c) (d) Mg (1 – m) through the centre, and gently placed on a plank with coefficient
2
of friction µ. The plank rests on a smooth horizontal surface. The
33. A rectangular piece of dimension l × b is cut out of central portion
initial acceleration of the sphere relative to the plank is : (mass of
of a uniform circular disc of mass m and radius r. The moment of
sphere = mass of plank)
inertia of the remaining piece about an axis perpendicular to the
(a) m g
plane of the disc and passing through its centre is :
é lb ù mé 2 lb ù 7
m êr 2 – (l 2 + b 2 )ú r – (l 2 + b 2 )ú (b) mg
(a) (b) ê 5
ë 6p r 2 û 2ë 6p r 2 û
(c) 2m g
m é 2 (l 2 + b 2 ) ù
(d) zero
2 êë ú
(c) r –
6 û 38. A force F is applied at the top of a ring placed on a rough horizontal
(d) not determinable as mass of the rectangular piece is not surface as shown in figure. Friction is sufficient to cause pure
given rolling. The frictional force acting on the ring is :
34. A small particle of mass m is given an initial high velocity in the
horizontal plane and winds its cord around the fixed vertical shaft
of radius a. All motion occurs essentially in horizontal plane. If
theangular velocity of thecord is w0 when the distance from the
particle to the tangency point is r0, then the angular velocity of F F
the cord w after it has turned through an angle q is : : (a) towards left (b) towards right
2 2
F
(c) towards left (d) zero
3
39. A cylinder rolls without slipping on a plank is device shown in
figure. The acceleration of the plank to keep the cylinder in a fixed
position during the motion is
g
(a) sin q
2
a (b) g sin q
(a) w = w0 (b) w = r w0
0 (c) 2 g sin q
æ w0 ö (d) 2 g sin q
(c) w= (d) w = w0q
ç a ÷ 40. A time varying force F = 2t is applied on a spool as shown in
ç1- q÷
è r0 ø figure. The angular momentum of the spool at time t about
bottommost point is
35. Moment of inertia of a uniform–disc of mass m about an axis
x = a is mk2, where k is the radius of gyration. What is its moment r 2t 2
of inertia about an axis x = a + b : (a)
2
( a + b) 2
(a) mk2 + m(a + b)2 – ma2 (b) mk2 + m ( R + r )2 2
2 (b) t
r
b2
(c) mk2 + m (d) mk2 + mb2 (c) ( R + r )t 2
2
(d) none of these
r
(c) cos q =
R (a) v0 = w 0 R (b) 2v0 = 5w 0 R
2r (c) 5v0 = 2w 0 R (d) 2v0 = w 0 R
(d) cos q =
R
20 cm
F F B A
20 cm
12 v 4v
(a) vcm = 0, w = (b) vcm = 0, w =
7 l l
5v v
(c) vcm = 0, w = (d) vcm = 0, w =
l 5l
(a) l/3 (b) l/6 56. A straight uniform metal rod of length 3 l is bent through angle as
shown. The bent rod is then placed on a rough horizontal table. A
(c) l/4 (d) l /12
53. Two thin rods; each of mass m and length l are joined to form L light string is attached to the vertex of the right angle. The string is
shape as shown. The moment of inertia of rods about an axis then pulled horizontally so that the rod translates at constant
passing through free end (O) of a rod end and perpendicular to velocity. Then the angle a which the side 2 l makes with string is
both the ends is
l 2l
l a
l
2 2 ml 2
(a) ml (b)
7 6
Top view
5 2
(c) ml (d) ml2
3
1 1
(a) p - tan -1 (b) p - tan -1
3 2 4
54. A block of mass kg is placed on a rough horizontal surface
10
1 1
as shown in the figure. A force of 1 N is applied at one end of (c) r - sin -1 (d) p - sin -1
2 4
the block and the block remains stationary. The normal force
v0
L-a L+a
(a) W (b) W
L+a L-a
m v
m
L - 2a L-a
(a) v (b) v/3 (c) W (d) W
L+a L + 2a
3v
(c) 2v/3 (d)
4
59. Three identical rods are hinged at point A as shown. The angle
made by rod AB with vertical is
B 90°
D
90°
C
Answer Key
57 (c) 58 (c) 59 (b) 60 (a)
Solution from page 97
www.crackjee.xyz
78 MECHANICS, THERMODYNAMICS & WAVES
C
q2
q1 a
r2 B
r1 F2 v
F1 O A
(a) (F2 r2 sin q2 – F1 r1 sin q1) out of the plane of the page.
(a) A is vertically upwards
(b) (F1 r1 sin q1 – F2 r2 sin q2) out of the plane of the page. (b) B may be vertically downwards
(c) (F2 r2 sin q2 – F1 r1 sin q1) in to the plane of the page. (c) C cannot be horizontal
(d) Zero (d) Some point on the rim may be horizontal leftwards.
w0
(a) The time until rolling begins. (a) the speed of the centre of mass of the cylinder is 2v.
(b) The displacement of the disc until rolling begins. (b) the speed of the centre of mass of the cylinder is zero
(c) The velocity when rolling begins. (c) the angular velocity of the cylinder is v/R
(d) The work done by the force of friction. (d) the angular velocity of the cylinder is zero
15. The angular acceleration of the toppling pole shown in figure is 19. If a cylinder is rolling down the incline with sliding
given by a = k sin q, where q is the angle between the axis of the (a) after sometime it may start pure rolling
pole and the vertical, and k is a constant. The pole starts from rest (b) after some time it will start pure rolling
at q = 0. Choose the correct options (c) it may be possible that it will never start pure rolling
a (d) none of these
20. The moment of inertia of a thin square plate ABCD of uniform
as thickness about an axis passing through the centre O and perpen-
an
dicular to the plane of the plate is
I2
l q A B I3
(a) I1 + I3
(b) I1 + 2 I2
I1
(c) I1 + I2 O
O
(a) The tangential acceleration of the upper end of the pole is lk I1 + I 2 + I 3 + I 4 D C I4
(d)
sin q 2
(b) The centripetal acceleration of the upper end of the pole is
2kl (1 – cos q) 21. A small ball is connected to a block by a light string of length l .
(c) The tangential acceleration of the upper end of the pole is Both are initially on the ground. There is sufficeint friction on the
2kl (1 – cos q) ground to prevent the block from slipping. The ball is projected
(d) The centripetal acceleration of the upper end of the pole is vertically up with a velocity u, where 2g l < u2 < 3g l . The centre
lk sin q of mass of the block + ball system is C.
16. A particle of mass m is projected with a velocity v, making an
angle 45° with the horizontal. The magnitude of the angular l
momentum of the particle about the point of projection when the
particle is at its maximum height h is
1. Statement - 1 Statement - 2
It is harder to open and shut the door if we apply force close to the Moment arm about vertical line is zero.
hinge. 8. Statement - 1
Statement - 2 The sum of all the forces acting on a particle is zero.
Torque is minimum about the hinge. Statement - 2
2. Statement - 1 The body must be in equilibrium.
Torque is equal to rate of change of angular momentum. 9. Statement - 1
Statement - 2 A rod placed on two rigid support be in equilibrium..
Angular momentum depends on moment of inertia and angular Statement - 2
velocity.
The net torque about any point must be zero.
3. Statement - 1
10. Statement - 1
The radius of gyration of a circular ring is equal to the radius of the
When tall buildings are constructed on earth, the duration of day-
ring .
night slightly increases.
Statement - 2
Statement - 2
I = mk2, where k is the radius of gyration.
Moment of inertia of earth about the axis of rotation increases.
4. Statement - 1
11. Statement - 1
A person standing on a rotating table when suddenly folds his
If the ice at poles melts, the moment of inertia of earth increases.
arms, the platform moves fast.
Statement - 2
Statement - 2
The angular momentum of the earth about axis of rotation increases.
The angular momentum in the process remain constant.
12. Statement - 1
5. Statement - 1
The total kinetic energy of a rolling ring is the twice of its
Moment of inertia of a particle is same, whatever be the axis of
translational kinetic energy.
rotation.
Statement - 2
Statement - 2
For any body rolling kinetic energy is twice of its translational
Moment of inertia depends on mass and distance of the particle
kinetic energy
from the axis.
13. Statement - 1
6. Statement - 1
The total kinetic energy of a rolling ring is the twice of its
a v translational kinetic energy.
In a rotating body, a = a r and v = wr, so = .
a w Statement - 2
Passage for (Q. 1 - 3) : 5. Frictional force acting on the cylinder just after its release is
Two disc A and B are mounted coaxially on a vertical axle. The discs
have moment of inertia I and 2I respectively about the common axis. Mg
Disc A is imparted an initial angular velocity 2 w using the entire potential (a) Mg (b)
3
energy of a spring compressed by a distance x1. Disc B is imparted an
angular velocity w by a spring having the same spring constant and 2Mg 4Mg
compressed by a distance x2. Both the discs rotate in the clockwise (c) (d)
3 7
direction.
6. Angular speed of the cylinder as the line AB becomes vertical is
1. The ratio x1/x2 is
g g
1 1 (a) (b)
R 3R
(a) (b)
2 2
(c) (d) 2 g g
2 (c) (d)
2. When disc B is brought in contact with disc A they acquire a 5R 6R
common angular velocity in time t. The average frictional torque
on one disc by the other during this peirod is Passage for (Q. 7 - 9) :
2Iw 9I w Consider a cylinder of mass M = 1kg and radius R = 1 m lying on a rought
(a) (b)
3t 2t horizontal plane. It has a plank lying on its top as shown in the figure.
9I w 3I w
(c) (d)
4t 2t F
m = 1kg
3. The loss of kinetic energy in the above process is 60°
1 2 1 2 A
(a) Iw (b) Iw
2 3
1 2 1 2 M R
(c) Iw (d) Iw
4 5
Passage for (Q. 4 - 6) :
A non uniform cylinder of mass 2 M and radius 2R is released from the B
position shown. The moment of inertia of the cylinder about longitudinal
axis passing through A is 4 MR2. Point B (distance R from the axis) is the
centre of mass of the cylinder. Friction between the cylinder and the A force F = 55 N is applied on the plank such that the plank moves and
horizontal surface is sufficient to prevent slipping. causes the cylinder to roll. The plank always remains horizontal. There is
no slipping at any point of contact.
Rw 0 5 Rw 0
(a) (b)
6µg 12µg
Rw0 Rw 0
(c) (d)
µg 4µg 16. Determine the angle q through which the cylinder rotates before it
leave contact with the edge.
12. The angular momentum of the disc about the point of contact,
when slipping ceases is equal to (a) cos -1 ( 3 / 4) (b) cos -1 (4 / 7)
w0
MR 2
w0 (c) cos -1 (5 / 7) (d) sin -1 (4 / 5)
(a) MR 2 (b)
2 12 17. Determine the speed of the centre of mass of the cylinder before
leaving contact with edge.
w0 w0
(c) MR2 (d) MR 2
4 3 (a) 3gR / 7 (b) 2gR
Passage for (Q. 13 - 15) : (c) (4gR / 7) (d) 7 / 4gR
A uniform rod AB of length 2l falls without rotation on a smooth 18. Determine the ratio of the translational to rotational kinetic energies
horizontal surface at an angle q to the horizontal. The speed of rod just of the cylinder when its centre of mass is in horizontal line with its
before collision is v0 and the collision is elastic. The magnitude of the edge.
angular velocity and magnitude of the velocity of centre of mass after (a) 3 (b) 6
collision are w and v' respectively. (c) 4 (d) 2
Passage for (Q. 19 - 21) :
A uniform flat disc of mass M and radius R rotates about a horizontal axis
B
through its centre with angular speed w0
A q
3 2
(c) mv (d) 2mv 2
2
27. The work done by friction on the ring in a displacement pR is :
(a) µmg × R (b) µmg × 2pR
1 1 (c) zero (d) none of these
(a) MR 2 w 0 , MR 2w 2 (b) mR 2w 0 , mR 2w 0
2 2 Passage for (Q. 28 - 30) :
A cylinder of mass 6 kg is suspended through two ideal strings wrapped
æM ö 2 æM ö 2 2 around it symmetrically as shown in the figure. The strings are taut and
(c) çè - m÷ø R w 0 , çè - m÷ø R w 0 the cylinder is initially at rest. Take g = 10 m/s2.
2 2
æM ö 2 1 éæ M ö 2 2ù
(d) çè - m÷ø R w 0 , 2 êçè 2 - m÷ø R w 0 ú
2 ë û
mR 2
P2 (c) a (d) data insufficient
3
O v, a
P1 33. The magnitude of acceleration of the point of disc, which is in
contact with ground is
(a) zero (b) a = Ra
31. The speed of point P1 with respect to P2 is
(c) 2a (d) none of these
R 3
(a) w+v (b) Rw
2 2
R R
(c) w (d) w
2 3
34. Match Column I with Column II and select the correct answer :
Column-I Column-II
(Quantity) (Expression)
r r
A. Angular momentum (p) r ´ (mv )
1 2
B. Impulse (q) Iw
2
C. Torque (r) r r
r ´F
r
D. Energy (s) mDv
35. Match the following columns.
Column-I Column-II
(Object) (Moment of inertia about any axis)
mR2
A. ring (p)
4
mR 2
B. sphere (q)
2
3
C. disc (r) mR 2
2
D. cylinder (s) 2 mR 2
36. Match Column I (Body rolling on a surface without slipping) with Column II (Ratio of Translational energy to Rotational energy) and
select the correct answer using the codes given below :
Column I Column II
A. Circular ring (p) 1/2
B. Circular disc (q) 1
C. Solid sphere (r) 3/2
D. Spherical shell (s) 2
(t) 5/2
F
A small particle is rotating in horizontal plane with
decreasing radius in the device shown
h F
A. R (p) The direction of static friction may be backward or
static friction may be forward or friction may be zero
disc
F
B. R (q) The direction of static friction is towards backward
disc
R
h
C. F (r) The angular acceleration will be clockwise
disc
h F
D. R (s) Acceleration of the centre mass will be along the
direction of F
disc
Answer Key 37 A ® (p, r); B ® (p, s); C ® (p, s); D ® (q, r) 38 A ® (s); B ® (p); C ® (p); D ® (q, r)
Sol. from page 104 39 A ® (p, s, r); B ® (q, r, s); C ® (q, r, s); D ® (p, r, s)
www.crackjee.xyz
86 MECHANICS, THERMODYNAMICS & WAVES
40. Match Column I with Column II and select the correct answer :
Column I Column II
A. Moment of Inertia of a solid uniform
sphere about the diameter (p) MR 2
B. Moment of inertia of a thin uniform
1
spherical shell about the tangent (q) MR2
2
5
C. Moment of inertia of a uniform disc through (r) MR2
3
centre of a mass and perpendicular to plane of the disc
2
D. Moment of inertia of disc about tangent (s) MR2
5
in the plane of disc.
5
(t) MR2
4
41. The following figures showns different bodies which are either free to rotate or translate on smooth horizontal surface. An impulse J is given
to the bodies in the direction shown in figure. Match the columns:
Column I Column II
l
A. (p) Translation
J M
Dumbell with a massless rod placed on smooth table
M
J
l
C. (r) Angular momentum about CM increases
L- shaped strip J
not fixed anywhere
hinge
l
Mg sin q
B. For solid sphere (q)
3
Mg sin q
C. For solid cylinder (r)
3.5
Mg sin q
D. For hollow spherical shell (s)
2
43. A uniform solid cylinder of mass m and radius R is placed on a rough horizontal surface where friction is sufficient to provide pure rolling.
A horizontal force of magnitude F is applied on cylinder at different positions with respect to its centre O in each of four situations of
column-I, due to which magnitude of acceleration of centre of mass of cylinder is ‘a’. Match the appropriate results in column-II for
conditions of column-I.
Column I Column II
F
R
A.
O (p) Friction force on cylinder will not be zero
//////////////////////////
F
R/2
O F
B. (q) a=
m
//////////////////////////
F O
F
C. (r) a¹
////////////////////////// m
R/2 O
D. F (s) the direction of friction force
//////////////////////////
acting on cylinder is towards left
1 2g 2
1 Ans. T = mg = 13N, a = = 5.102 rad / s 2 (b) P = mg2t
2 2 2 6 3R 3
Ans. W = M (3a + b )w = 2.6 J
2
www.crackjee.xyz
ROTATIONAL MECHANICS 89
9. A uniform rod of mass m = 5.0 kg length L = 90 cm rests on a 10. A roller of diameter 6 cm rides between two horizontal bars mov-
smooth horizontal surface. One of the ends of the rod is struck ing in opposite direction as shown in figure. Find the distance of
defining the position of the path of the instantaneous centre of
with the impulse J = 3.0 N. s in a horizontal direction perpendicular
rotation of the roller. (Assume no slip at points of contact P and Q)
to the rod. As a result the rod obtains the momentum p = 3.0 N.s.
Find the force with which one half of the rod will act on the other Q
20 m/s
in the process of motion.
9J 2
Ans. F = = 9N
2 mL
y
10 m/s
P
Ans. 2 cm
where a, b, A, B are constants, and iˆ, ˆj are the unit vectors of the
x and y -axes. Find the moment N and the arm l of the force F
(a) From what minimum height h above the bottom of the track relative to the point O.
must the marble be released in order that it is not leave the
track at the top of the loop? (The radius of the loop-the- Ans. N = (aB – bA) k̂ , where k̂ is the
loop is R; assume R >>r) aB - bA
(b) If the marble is released from the height 6R above the bottom unit vector of the z-axis; l =
of the track, what is the horizontal component of the force A2 + B 2
acting on it at point Q? 18. A force F1 = Aˆj is applied to a point whose radius vector r1 = aiˆ ,
Ans. (a) 2.7 R, (b) 50 mg/ 7
13. Figure shows two blocks, each of mass m, suspended from the while a force F2 = Biˆ is applied to the point whose radius vector
ends of a rigid weightless rod of length l1 + l2, with l1 = 20 cm and
r2 = bˆj . Both radius vectors are determined relative to the origin
l2 = 80 cm. The rod is held in horizontal position shown in figure
and then released. Calculate the accelerations of the two blocks as of co-ordinates O, iˆ and ĵ are the unit vectors of the x and y-axes,
they starts to move. a, b, A, B are constants. Find the arm l of the resultant force
relative to the point O.
[ aA - bB ]
Ans. l =
( A2 + B 2 )
19. Three forces are applied to a square plane as shown in the figure.
Find the modulus, direction and the point of application of the
resultant force, if this point is taken on the side AB.
Ans. a1 = 1.73 m/s2, a2 = 6.92 m/s2
14. A 2.0 kg object moves in a plane with velocity components vx =
30 m/s and vy = 60 m/s as it passes through the point (x, y) = (3.0,
– 4.0)m.
(a) What is its angular momentum relative to the origin at this
moment?
(b) What is its angular momentum relative to the point
(–2.0, –2.0)m at the same moment.
Ans. (a) 600 kg. m2/s (b) 720 kg.m2/s both about z-axis Ans. Fres = 2F. This force is parallel to the diagonal AC and is
applied at the midpoint of the side BC.
www.crackjee.xyz
ROTATIONAL MECHANICS 91
20. Find the moment of inertia 26. A spool with a thread would on it is placed on an inclined smooth
(a) of a thin uniform rod relative to the axis which is perpendicular plane set at an angle a = 30° to the horizontal. The free end of
to the rod and passes through its end, if the mass of the rod
the thread is attached to the wall as shown in figure. The mass of
is m and its length l;
(b) of a thin uniform rectangular plate relative to the axis passing the spool is m = 300g, its moment of inertia relative to its own axis
perpendicular to the plane of the plate through one of its I = 0.45 g- m2, the radius of the would thread layer r = 3.0 cm .
vertices, if the sides of the plate are equal to a and b, and its Find the acceleration of the spool axis.
mass is m. Ans. a =g sin a / (1 + I/mr2) = 1.6 m/s2
1 27. A uniform solid cylinder of mass m rests on two horizontal planks.
Ans. (a) L = 1 ml 2 (b) I = m( a 2 + b 2 ) A thread is wound on the cylinder. The hanging end of the thread
3 3
is pulled vertically down with a constant force shown in the figure.
21. Calculate the moment of inertia
(a) of a copper uniform disc relative to the symmetry axis Find the maximum magnitude of the force F which still does not
perpendicular to the plane of the disc. If its thickness is bring about any sliding of the cylinder, if the coefficient of friction
equal to b = 2.0 mm and its radius R = 100 mm; between the cylinder and the planks is equal to µ. What is the
(b) of a uniform solid cone relative to its symmetry axis, if the acceleration amax of the axis of the cylinder rolling down the inclined
mass of the cone is equal to m and the radius of its base to R. plane?
1 3 2µg
Ans. (a) I = prbR 2 = 2.8 g .m2 ; (b) I = mR2 Ans. Fmax = 3µmg/(2 – 3µ); amax =
2 10 (2 - 3µ)
22. A thin horizontal uniform rod AB of mass m and length l can
rotate freely about a vertical axis passing through its end A. At a
certain moment the end B starts experiencing a constant force F
which is always perpendicular to the original position of the
stationary rod and directed in a horizontal plane. Find the angular
velocity of the rod as a function of its rotation angle f counted
relative to the initial position. Ans. w = 6 F sin f / ml
23. In the arrangement shown in figure. The mass of the uniform solid
cylinder of radius R is equal to m and the masses of two bodies are 28. A spool with thread wound on it, of mass m, rests on a rough
equal to m1 and m2. The thread slipping and the friction is the axle horizontal surface. Its moment of inertia relative to its own axis is
of the cylinder are supposed to be absent. Find the angular
acceleration of the cylinder and the ratio of tensions T1/T2 of the equal to I = g mR 2 , where g is a numerical factor and R is the
vertical sections of the thread in the process of motion. outside radius of the spool. The radius of the wound thread layer
(m2 - m1 ) g T m ( m + 4 m2 ) is equal to r. The spool is pulled without sliding by the thread with
Ans. a = , 1 = 1 a constant force F directed at an angle a to the horizontal as shown
(m1 + m2 + m / 2) R T2 m2 (m + 4m1 )
in the figure. Find
24. A uniform disc of radius R is spinned to the angular velocity w and
then carefully placed on a horizontal surface. How long will the
disc be rotating on the surface if the friction coefficient is equal to
µ? The pressure exerted by the disc on the surface can be regarded
as uniform.
Ans. t = 3wR/4µg
25. A uniform cylinder of radius R and mass M can rotate freely about
a stationary horizontal axis O figure. A thin cord of length l and
mass m is wound on the cylinder in a single layer. Find the angular (a) the projection of the acceleration vector of the spool axis on
acceleration of the cylinder as function of the length x of the the x-axis.
hanging part of the cord. The wound part of the cord is supposed (b) the work performed by the force F during the first t seconds
to have its centre on gravity of the cylinder axis. after the beginning of motion
Ans. a = 2mgx / Rl( M + 2 m) Ans. (a) ax = F(cos a – r/R) / m( 1+ g);
(b) W = F2t2(cos a – r/R)2/2m(1 + g)
29. A uniform solid cylinder of mass m and radius R is set in rotation
about its axis with an angular velocity w0, then lowered with its
lateral surface onto a horizontal plane and released. The coefficient
of friction between the cylinder and the plane is equal to µ. Find
(a) how long the cylinder will move with sliding
(b) the total work performed by the sliding friction force acting
on the cylinder.
w0 R - mw 0 2 R 2
Ans. (a) t = ; (b) W =
3m g 6
www.crackjee.xyz
92 MECHANICS, THERMODYNAMICS & WAVES
30. A block of mass M is moving with a velocity v1 on a frictionless 36. A long uniform rod of length L and mass M is pivoted about a
surface as shown in figure. It passes over to a cylinder of radius R horizontal, frictionless pin through one end. The rod is released
and moment of inertia I which has a fixed axis and initially at rest. from rest in a vertical position as shown in the figure. At the
When it first make contact with the cylinder, it slips on the cylinder,
instant the rod is horizontal, find (a) the angular velocity of the rod
but the friction is large enough so that slipping ceases before it
(b) its angular acceleration (c) the x and y component of the
losses contact with the cylinder. Final it goes to the dotted position
with velocity v2. Compute v2 in terms of v1, M, I and R. acceleration of its c.m. (d) the component of reaction force at the
pivot.
v1
Ans. v2 =
é I ù
ê1 + ú
ë MR 2 û
4F 8F
Ans : ( a ) a1 = 8m + 3m , a2 = 8m + 3m
2 1 2 1
m1F 3m1 F
(a) Calculate the velocity of the centre of mass of the disc at t0. (b) f1 = , f2 =
3m1 + 8m2 8m2 + 3m1
(b) Assuming that the coefficient of friction to be m, calculate t0.
43. Figure shows two cylinders of radii r1 and r2 having moments of
Also calculate the work done by the frictional force as a
inertia I1 and I2 about their respective axes. Initially, the cylinders
function of time and the total work done by it over a time t
rotate about their axes with angular speed w1 and w2 as shown in
1 2 the figure. The cylinders are moved closer to touch each other
much longer than t0 Ans : 2v0/3, v0/3µg, , – mv0 .
6 keeping the axes parallel. The cylinders first slip over each other at
the contact but the slipping finally ceases due to the friction
41. Two heavy metallic plates are joined together at 900 to each other.
between them. Find the angular speeds of the cylinders after the
A linear sheet of mass 30 kg is hinged at the line AB joining the
slipping ceases.
two heavy metallic plates. The hinges are frictionless. The moment
of inertia of the laminar sheet about an axis parallel to AB and
passing through its centre of mass is 1.2 kg-m 2. Two rubber
obstacles P and Q are fixed one on each metallic plate at a distance
0.5 m from the line AB. This distance is chosen so that the reaction
due to the hinges on the laminar sheet is zero during the impact.
Initially the laminar sheet hits one of the obstacles with an
angular velocity 1 rad/s and turns back. If the impulse on the sheet I1w1r2 + I 2w 2 r1 Iw r +I w r
Ans. w '1 = r2 and w '2 = 1 1 2 2 2 1 r1
due to each obstacle is 6 N-s. I 2 r12 + I1r22 I 2 r12 + I1r22
44. A light rod with the balls A and B is clamped at the centre in such a
A
way that it can rotate freely about a horizontal axis through the
Q clamp. The system is kept at rest in the horizontal position. A
5m
x
particle P of the same mass m is dropped from a height h on the ball
0.
(a) Find the angular momentum and the angular speed of the
(a) Find the location of the centre of mass of the laminar sheet
system just after the collision.
from AB.
(b) What should be the minimum value of h so that the system
(b) At what angular velocity does the laminar sheet come back makes a full rotation after the collision.
after the first impact ?
mL gh 8 gh 3
(c) After how many impacts does the laminar sheet come to Ans. (a) , (b) L.
rest? 2 3L 2
Ans : (a) 0.1 m (b) 1 rad/s (c) infinite. 45. A hollow sphere is released from the top of an inclined plane of
inclination q.
42. A man pushes a cylinder of mass m1 with the help of a plank of
(a) What should be the minimum coefficient of friction between
mass m 2 as shown. There is no slipping at any contact. The
the sphere and the plane to prevent sliding ?
horizontal component of the force applied by the man is F. Find
(b) Find the kinetic energy of the sphere as it moves down a
length l on the incline if the friction coefficient is half the
value calculated in part (a).
2 7
Ans. (a) tan q (b) mgl sin q .
5 8
www.crackjee.xyz
94 MECHANICS, THERMODYNAMICS & WAVES
46. A uniform plate of length a = 0.6 m and width b = 0.4 m having body is set into rotational motion on the table about A with a
mass M = 3 kg is free to rotate about an edge. Initially the plate is constant angular velocity w.
kept horizontally and small particles each of mass M = 0.01 kg
collide elastically and perpendicular to the plate on the second
half of the plate at the rate n = 100 particles per unit time per unit
area as shown. What will be the velocity of striking particles so
that plate does not rotate ?
L = mvd (constant) d
As d constant mv MR 2
46. (c) I=
34. (a) The angular momentum, 2
L = mvr sin(180 30 ) M
1 R
= 2 4 3 12 kg m 2/s 2 3
MR 2 = mr 2
2 47. (a)
5 2
35. (b) vA r ( ˆj )
2R
r=
10 1(– ˆj ) –10 ˆj m / s 15
2 2
2 centripetal acceleration , a = v / R
53. (c, d) For velocity to be vertical,
1 2 1 2 v cos = v
K1 I I
2 2 or R cos = v v= R
K2 1 2 1
I (2 I )( / 2) 2
2 2 2 2 v v
cos
R
K
K2 =
2 54 (b) Li = – mvd cos and L f mvd cos
41. (a) Its moment of inertia will be equal to the moment of inertia L = 2mvd cos
MR 2 55. (c) The tendency of sliding of point of contact is downward, so
of the disc. So I . friction will act upward.
2
www.crackjee.xyz
ROTATIONAL MECHANICS 97
56. (d) mg sin 60 (T f) ma … (i)
T
and f R TR I f
60°
sin
or f – T
Ia
… (ii) mg
R2 60°
After simplifying, we get f mg / 5
57. (a) The sliding tendency of bottom most point of the spool is
backward, so friction acts rightwards. Therefore centre of 1 2
I mg sin
mass of spool moves rightwards. 2
58. (c) For topple F 2 mg sin 2 mg sin
F sin 60 mg /2 A
I 2
m
F mg / 3
2g sin
For translation N A
F = µmg
60° In the similar position
mg µmg
or µmg
3 mg 4g sin tB 1 AtA
B 2
B t A 2 tB
1
µ= 62. (d) The angular momentum of the system about O = 0
3
= 0.
59. (c) mg – T = ma2 … (i) R 63. (c) The position vector of the center of mass at the time t is
and TR = I 2
rcm iˆ (cos 30 ) ˆj (sin 30 ) kˆ (0.10)
I 2 T
or T=
R = 0.866 iˆ 0.5 ˆj 0.10 kˆ
2 and the total momentum of the hoop is
mR 2 T
or T = … (ii)
2 R R p mvcm (0.50) (0.50 ˆj) 0.25 ˆj
or 1R = ( a2 2R ) … (iii)
Thus, Lorb rcm p
mR 2
and TR =
2
1 … (iv) mg = (0.866 iˆ 0.5 ˆj 0.10 kˆ) 0.25 ˆj
From equations, we get the answer.
60. (d) In the shown frame the particle appears to be at rest. = 0.025 iˆ 0.216 kˆ kg m 2 /s
Net force on it must be zero. Therefore pseudo force must To find the spin angular momentum, note that every element
be equal and opposite to the tension. of mass of the hoop is at the same distance from the centre
of mass r' = 0.10m, and every element rotates about the
P center of mass with a velocity v (of magnitude 0.50m/s)
m
perpendicular to r . Thus,
L
O Lspin r v dm = r v ( iˆ) dm
61. (c) Consider a situation when the bob A has fallen through an
angle . = mr v iˆ 0.025 iˆ kg m 2 /s
Loss in PE = Gain in KE
Solutions Exercise1.1 Level -2
1. (d) For the equilibrium, F 0 and 0 5. (a) E1 E2 mgh
which is possible in A and C. 6. (c) K1 + 0 = mgh K1 = mgh
2. (c) In (c) the net force and net torque can be zero. and K2 + Krotational= mgh K2 = mgh – Krotational
Clearly, K K2
1
2 gh 2 gh 10
3. (a) Work done by gravitational force in both the cases is 7. (d) v gh
W1 = W2 = mgh. 1
I 2mR 2 7
1
mR 2 5mR 2
2 gh Clearly velocity of c.m. does not depend on radius of the
4. (c) v1 2 gh and v2 I
1 sphere.
mR 2
v v
1 and 2 .
Clearly, v1 v2 R 2R
www.crackjee.xyz
98 MECHANICS, HEAT, THERMODYNAMICS & WAVES
8. (d) In case (a) and (b), the sphere will move in pure rolling 13. (a) For m to be stationary
motion, while in (c), the sphere will not be in pure rolling. So T = mg
some part of mechanical energy will convert into heat by
N
friction. T µN
9. (b) Taking moment of forces about O, and put 0
Mg
Fv T
O FH
mg
4m For M : T µN = M 2r
or 60 × 4 – Fv × 3 = 0
or mg – µMg = M 2r
60 N
mg µMg
A 3m r =
M 2
Fv = 80 N
10. (c) On inclined plane, the minimum friction needed for pure MR 2
14. (c) Given I =
4
Iam
rolling, f . MR 2 3
R2 Itangent = MR 2 MR 2
2 2
2 mR 2
For cylinder a g sin , I 3
3 2 [4 I ] 6 I .
=
2
mg sin 15. (c) By conservation of energy, we have
f .
3 1 2 1 1
mgh = mv I pulley 12 I shell 2
2
Also N = mg cos 2 2 2
f tan 1 2 1 v 2 1 2 2
v
µ=
N
=
3
. = mv I MR 2
2 2 r 2 3 R
11. (c) In the process,
mgh
m
v2 = m I M
2 2 3
2r
dJ d ˆ
16. (c) = ( ai bt 2 ˆj ) 2btjˆ
dt dt
.J
cos 45° =
J
L 2
13 L
= MR 2 . or Mv
2
= 2M
24 2
5R 5R 2 v
19. (b) (A) = ; and at R = .
I I I L
' TR TR 2
(B) ; and at ' R 26. (c)
I I I x
As T 5 N , so at ' at .
F 1 2 1 F 2
20. (c) a ; s at t
M 2 2M
I = L (constant)
FR 2F
= L
I MR 2 MR or =
I
2
L
=
1 2 1 2F Ft 2 2
t t2 . MR
mx 2
2 2 MR MR 2
21. (b) We know that , J = m(v f vi ) When tortoise moves along a chord, x decreases and then
increases. So first increases and then decreases.
or 7.5 = 15(v f 0)
27. (d) I AB = I cm Md 2
vf = 0.5 m/s
1.6 Ma 2
1
mvt 2
1
15 (0.5) 2 = M (2 a )2
Energy transferred, E = 2
2 2
= 4.8 Ma2.
= 1.9 J
22. (c) By conservation of angular momentum about C.M, we have ( R / 3)2
L 28. (b) The mass of the hole removed = 9m =m
Mv = R2
I
2 I = Iwhole disc – Iremoved disc
2
2 M (2 L) (9 m) R 2 2 2
= = – m( R / 3) m
2R
12 2 2 3
3v = 4 mR2
or =
, counterclockwise
4L Ix m1R2 / 2 ( R 2 )tR 2
23. (b) By conservation of angular momentum about C.M. 29. (d) I y = m (4 R) 2 / 2 t
2 (4 R ) 2 (4 R) 2
4
C.m 1
=
64
L
30. (d) L = 2 R R
2 x x
mv
mR2
mv
R I = mR 2
= I 2 O
2
2
2 2 3 3 L
mR 2 m
R
m
R = mR 2 = ( L )
= 2 2 2
2 2
v 3 L3
or = . =
3R 8 2
www.crackjee.xyz
100 MECHANICS, HEAT, THERMODYNAMICS & WAVES
31. (c) vblock = vtangent = vcm R 1 1 2
v
= Mv 2 ( MR 2 )
= 2R R 3 R 2 2 R
32. (c) Taking moment O, we get F
= Mv 2
a a
Fa = Mg × K plank
2 2
K cylinder
Mg a/2 O
or F =
2 Mg
33. (b)
r 37. (c)
b
f
f
Acceleration of sphere
f
The mass of the rectangular piece a = µg
m
m( b ) Acceleration of plank
m = 2
r f
a' = µg
m
mr 2 m' 2 Acceleration of sphere relative to plank
Moment of inertia, I = [ b2 ]
2 12 = a a ' 2 µg .
mr 2 m b 2 38. (d) F
= – ( b2 )
2 12 r 2
m 2 b f
= r – ( 2 b2 )
2 6 r2 F+f = ma … (i)
and FR – f R = I
34. (c) r r0 – a .
In the process v remains constant, so Ia
or F–f =
R2
0r0 = r
mR2 a
or = 0r0 0 . =
r0 – a a R2
1
r or F – f = ma … (ii)
From above equations, f = 0.
39. (d)
f
35. (a) R
sin a
a =g
a
mgsin – f = m×0
a+b f = mg sin … (i)
Also fR = I
Ia = I cm ma 2
mR 2 a
or 2 = 2 or fR =
mk I cm ma 2 R
ma
I cm = mk 2 ma 2 or f = … (ii)
2
Now I ( a b) = 2
I cm m (a b ) From equations (i) and (ii), we get
a = 2g sin .
=mk 2 ma 2 m ( a b) 2 . 40. (c) Torque about bottom most point is
36. (b) Velocity of plank will be 2v. = F (R r ) 2t ( R r) .
Kinetic energy of plank, Angular momentum
1 t t
K plank = M (2v) 2 2 Mv 2
2 dt 2t ( R r )
L =
Kinetic energy of hollow cylinder,
0 0
1 1
Kcylinder = Mv 2 I 2
2 2 = t 2 (R r)
www.crackjee.xyz
ROTATIONAL MECHANICS 101
41. (a) mvr = I 10cos 30 f = ma … (i)
2 = 2a
a ma 2 a Also fR = I
or mv = m
2 6 2
Ia
or f =
3v R2
=
4a
2 a 2 4a
42. (a) mR2 = 2a … (ii)
3 R2 3 3
v0
From above equations, we get
v1 f = 2 3 N
With respect to ground it has only rotation, so Normal reaction, N = 20 + 5 = 25 N
v1 = r f
Now using conservation of angular momentum about a fixed µ = 0.08 3
N
point at the level of bottom of the sphere,
mv0r = I
47. (d) Given, I = 2 x 2 – 12 x 15
dI
2 2 v1 As moment of inertia is minimum about C.M. so 0,
= mr dx
5 r
d
2 or (2 x 2 12 x 15) 0
v0 = v1 . dx
5
or 4 x 12 = 0
43. (b)
or x = 3
J
r 48. (a) Angular impulse, J = I ( f – i )
f
Using conservation of angular momentum about a fixed point m 2
or 10 × 1 = ( f 0)
at the level of contact point of the ring, we have 3
J×r = I
2 12
= f
= (2mr 2 ) 3
J or f = 15 rad/s
or =
2mr 1 2
K = I
J 2
v . = r
2m 1 m 2 2
44. (c) Velocity of the particle perpendicular to z-axis is =
2 3
v 2(iˆ ˆj ) . Angular momentum about z-axis,
1 2 12
= m (r v) = 152
L 2 3
= 1 [(iˆ ˆj ) (2iˆ 2 ˆj )] = 75 J.
or L = –2kˆ – 2kˆ –4kˆ kg - m2/s 49. (b) mv = I
2
45. (b) For spool to be stationary , F 0
N
or F cos f 0 …(i) M 2 v
= v
Also, 3 /2
0, m
F M 3
or Fr f R 0 … (ii) r = .
From above equations, m 4
50. (c)
r f v0
cos = .
R
mg
f
10 sin 30°
46. (b) f
a = g
m
10 cos 30° N fR mg R
and = 2
I I mR 2
5
f
5µg
= .
mg = 20 N 2R
www.crackjee.xyz
102 MECHANICS, HEAT, THERMODYNAMICS & WAVES
Using equation of motion, we have 56. (d)
0 = v0 – at … (i)
2 mg
and 0 = 0– t … (ii) 3
From above equations, we get
v0 O
a 1 mg /2
= 3
0
/2
g 2R For rotational equilibrium,
= =
5 g / 2R 5 0 0
or 5v0 = 2 0 R . 2 1 1
51. (a) The tendency of sliding of contact point is rightward, so or mg (l sin ) mg cos 0
3 3 2
frictional force acts leftward.
52. (b) 1
B or tan
4
v 57. (c) K = K ring K particles
1 2 1 2 1 1 1
A = mv0 I m( 2v0 )2 m(2v0 ) 2 m( 2v0 ) 2 0
mv0 0 = 2 2 2 2 2
m 0 mv
v = v0 v0
Also , I = mR2
m 2 R
Also mv0 x = I = K = 5 mv02
12
58. mvR = I mv ' R
12v0 x
or = 2 mR2 v '
or mvR = mv ' R
For end A to be stationary 2 R
vA = v0 – v' 2v / 3
2 59. (b) Vertical line from hinge A must pass
v x through C.M. of rod system. A
v0 12 0 2 /3
or 0 = 2 2 OP /2
tan P
AP 2 /3
or x = . B
6 /2 D
3 1 3 O
2 tan tan
53. (c) d = ( / 2) 2 4 4
60. (a) FBD of rod will be as shown below
m 2 m 2
I = md 2 TB C
3 12
5 2 TB
= m N 1/2 a RAH
3
3 B A
54. (b) For equilibrium of the block C
2 W
L
1
F 0 or N 3 RAV
2
Summation of moments about A should be zero
3/2 x MA = 0 = – TB × L – (TB – W) × a = 0
Torque of all the force except N about A is found equal to
zero. Wa
TB
55. (a) By conservation of linear momentum, we have (L a)
mv – mv = ( m m m)vcm Summation of moments about B should be zero
MB = 0 = (TB – W) × (L – a) + RAV × L = 0
or vcm = 0
Now using conservation of angular momentum, we get Wa
W × (L – a) + RAV × L = 0
mv = I ( L a)
m 2
= 2m( / 2)2 RAV × L = –
Wa WL Wa
12 ( L a)
( L a)
12v
= WL ( L a ) (L a)
7 RAV W
( L a) L (L a)
www.crackjee.xyz
ROTATIONAL MECHANICS 103
Solutions Exercise-1.3
1. (a) Torque = F × moment arm. 10. (a) In this process, I = constant. When buildings are
If force is applied close to hinge, moment arm will be small, constructed, I will increase, will decrease and T will
so torque will be small. increase.
11. (c) As ice melt, it spread towards the periphery, so moment of
dL
2. (b) and L I . inertia increases. But L remains constant, because
dt 0.
12. (c) K K translational K rotation
3. (a) Mk 2 MR 2k R.
4. (a) In the process I = constant. As the person folds the hands, 1 2
mv
1 2
I
I decreases and so increases. 2 2
5. (d) I = mr2, it depends on m and r both.
6. (c) and a are different quantities, so they cannot be added. 2 1 2 1 2
1 2 1 v
mv ( mR 2 ) mv mv
Similarly and v. 2 2 R 2 2
7. (a) = × moment arm. In this case r = 0, so = 0.
As I is different for different objects, so the result is not
8. (c) The body may not be in rotational equilibrium. same for all objects.
F 1 2 1 2
13. (a) K = mv mv
2 2
F
= 2K translational
9. (b) If rod is in stable equilibrium, then net torque about any axis 14. (b) On smooth inclined plane, torque about geometric axis is
must be zero. zero. So rolling is not possible.
Solutions Exercise-1.4
Passage (Q. 1 - 3) : f = (2M )acm
1 2 1
1. (c) kx1 = I (2 ) 2 and 2 MgR f (2 R ) = I
2 2
acm
1 2 1 or 2 MgR f (2 R ) = (4 MR 2 )
and kx2 = (2 I ) 2 (2 R)
2 2
or 2MgR 4Macm R = 2MRacm
x1
x2 = 2 g
acm =
2. (a) If 3
' is the final angular velocity, then
I (2 ) 2 I ( ) = ( I 2I ) ' acm g
so = .
(2R ) 6R
4
or ' = 2 Mg
3 5. (c) f 2M g /3 .
(L f Li ) / t 3
Thus JB =
1 2
= [(2 I ) ' 2 I ] / t 6. (a) I = (2 M ) gR
2
4
= [2 I 2 I ]/ t 1
3 or (4 MR 2 ) 2
= 2 MgR
2
2I
= g
3t =
R
3. (b) Loss = ki kf Passage (Q. 7 - 9) :
1 1 1 7. (b); 8. (a); 9. (c)
= I (2 )2 (2 I ) 2
– (I 2 I ) '2 Drawing the F.B. D of the plank and the cylinder.
2 2 2
F sin
1 2 N1
= I .
3
Passage (Q. 4 - 6) :
4. (d)
F cos
R f1
f
mg
2Mg
www.crackjee.xyz
ROTATIONAL MECHANICS 105
f1 0
= 0 – 2
2
N1
0 0
Mg 3 = or, =
2 6
N2 R 0
f2 Now from (ii), t2 =
6 µg
Equations of motion are Maximum displacement of the disc in forward direction,
F cos – f1 = ma ....(1) 2
R 0 R 0 1 R 0
F sin + N1 = mg ....(2) S= µg
f1+ f2 = MA .....(3) 4 4µg 2 4µg
f1R – f2R = I .....(4)
A=R .....(5) R 2 02 1 1 R 2 20
= =
µg 16 32 32µg
1
4 55 The displacement of the disc when it starts pure rolling
4 F cos 2
a 10 m/s 2
3M 8m 3 1 8 1 2
1 R 2 20 1 R 0 5 ( R 0 )2
= µg =
1 32 µg 2 6µg 32 9 µg
3MF cos 3 1 55
f1 2 7.5 N
3M 8m 3 1 8 1 0R 5 ( 0 R) 2
t3
6 32 9 µg
1
1 55
MF cos 2 5 0R
and f 2 2.5 N or, t3 =
3M 8m 3 1 8 1 48 µg
Passage (Q. 10 - 12) :
25 0 R
10. (a) Let linear velocity of the disc will become zero after a time Total time = t1 + t2 + t3 = .
t1. Then it starts moving in backward direction and at time t2 48 µg
it comes in pure rolling. When disc starts pure rolling its 11. (b) Time after which disc starts pure rolling.
linear and angular velocities will become constant and friction R 0 R 0 5R 0
will be zero. t = t1 + t2 = =
4µg 6µg 12 µg
12. (c) Angular momentum of disc after it starts pure rolling,
L = MvR + I
R 0
v0=
4 MR 0 R MR 2 0
= 6 2 6
At time t1, 2 1 1 MR 2 0
= MR 0 = 4
R 6 12
0
0= µgt1
4
Passage (Q.1 3 -1 5) :
R 0 13. (b) The force of impact at A is vertically upward.
t1 = ... (i)
4µg 14. (d)
At time t2,
v = µg t2 v'
µg t2 = R
A
R A
t2 = ... (ii) v0
µg
vA = v' + cos
2 µg collision is elastic
= 0– (t1 t2 )
R v0 = v ' cos
2µg R 0 R 15. (b) Angular momentum L r mv
= –
0 R 4µg µg
| L | mv0 cos .
www.crackjee.xyz
106 MECHANICS, HEAT, THERMODYNAMICS & WAVES
Passage for (Q. 16 - 18) : Angular momentum
Suppose cylinder gets rotated through an angle before leaving contact. In L = I
the process 0
N MR 2
= 0
2
20. (c) In the process, mechanical energy remains constant and so
R cos
1 2
h mv = mgh
2
mg v2
h =
1 1 2g
mgh = Mv 2 + I w2
2 2
2 2 2
( 0R ) 0R
1 æç MR 2 ö÷÷æç v ÷ö
2
1 = = .
Mv 2 + çç ÷ç ÷÷÷ 2g 2g
mgR (1 – cos ) = 2 2 çèç 2 ø÷ç ÷èç R ø
21. (d) Final angular momentum of the broken disc
3
= Mv 2 L' = I' 0
4
4 MR 2
v = gR(1 - cos q) .....(i) mR 2
3 = 2
0
For the motion of the cylinder, we have
mv 2 M
mg cos – N = = m R2 0
R 2
To leave the contact, N = 0,
1 2
mv 2 Kinetic energy K' = I' 0
mg cos 2
R
v2 1 M
or cos = ...(ii) = m R2 2
0 Ans.
gR 2 2
16 . (b) After solving equations (i) and (ii), we get Passage for (Q. 22 - 24) :
4 22. (a) Let vc and are the velocity of C.M. and angular velocity
cos = .
7 just after collision.
17 . (c) From equation (i), we get Using conservation of linear momentum, we have
4 gR mv0 = 0 Mvc … (i)
v = .
7 and by conservation of angular momentum
æç 4 ö÷ 1 2 1 2 L
18 . (b) We havemgR èçç1 - ø÷÷÷ = mv + mv mv0 = I
7 2 4 2
1 2 mgR
mv = ML2
4 7 L
or mv0 = … (ii)
mgR 2 12
or Krot =
7 From equations (i) and (ii)
Total decrease in potential energy
= mgR mv0 6mv0
vc = and
mgR M ML
Ktrans = mgR –
7 Since collision is completely elastic, therefore K.E. before collision
is equal to after collision
6
= mgR . 1 1 1
7 or mv0 2 0 = Mvc 2 I 2
2 2 2
K trans
= 6.
K rot 1 1 mv0
or mv02 = M vc 2
Passage for (Q. 19 - 21) : 2 2 M
19. (a) Kinetic energy of the disc
2
1 2 1 ML2 6mv0
k = I 0
2 2 12 ML
1 MR 2 2 1
= 0 = MR 2 2
0. m 1 v0
2 2 4 which gives and vc
M 4 4
www.crackjee.xyz
ROTATIONAL MECHANICS 107
23 (b) Passage for (Q. 28 - 30) :
. T T
mg
28. (a) mg 2T = macm … (i)
mR 2 acm
and 2TR = I =
Velocity of point P immediately after collision be zero, let it is at 2 R
a distance y from C.M. macm
vc y = 0 or 2T = … (ii)
2
mv0 6mv0 From above equations, we get
or y = 0
M ML 2g 20
acm = m / s2
L 3 3
which gives y =
6 20
6
29. (c) macm 3
L L 2L T 10 N
AP = 4 4
2 6 3
2
L 1 2 1 mR 2 v
24. (d) Angle rotated by rod in time 3v 30 (b) mg 7.5 = mv
0 2 2 2 R
6 mv0 L v = 10 m/s
= t Passage for (Q. 31 - 33) :
ML 3v0
31 (c) v p1 v R / 2 and v p2 v R/2
=
2
32. (b) =I mR2
2
The rod turns through , in this interval of time. The velocity of
2
point P in y-direction will be v2
33. (d) a .
6mv0 L R
vy = y 34. A (p); B (s); C (r); D (q)
ML 6
Impulse, J P m v
v0
= 35. A (q, r, s ); B (q, r, s); C (p, q, r, s); D (p, q, r, s)
4
The resultant velocity of point P MR 2
(A) Moment of inertia of ring
v = vc 2
vy 2 2
2
2 2 (B) Moment of inertia of sphere MR 2
v0 v0 5
=
4 4
MR 2
(C) Moment of inertia of disc of cylinder
v0 4
=
2 2 36. Solution is given in the theory
Passage for (Q. 25 - 27) : 37. A (p, r); B (p, s); C (p, s); D (q, r)
(A) A planet around sun in circular orbit; its angular momentum
25. (c) If 2 vcm = v
remains constant ( = 0). As its distance from sun is same,
v so speed is also same.
vcm = (B) A planet in elliptical orbit, = 0 , so angular momentum is
2
zero. But its distance from sun changes, so speed changes.
vcm v
Angular velocity , = . (C) F r F 0 0 F
R 2R
In the process L is constant.
1 1
26. (a) K mvcm2 I 2 2
mvcm (D) In this case Fr ,
2 2
so angular momentum does not remain constant.
2
v mv 2
m F r
2 2
27. (c) In pure rolling, W f s 0
www.crackjee.xyz
108 MECHANICS, HEAT, THERMODYNAMICS & WAVES
38. A (s); B (p); C (p); D (q, r)
Also J = P , so linear momentum increase.
(A) In pure rolling on horizontal surface, friction is zero.
(B) (C) On inclined plane, the point of contact has the tendency of J = L , so angular momentum also increases.
sliding downward, so friction acts in upward direction. (B) 0 , so it experiences translation and linear momentum
(D) In case of slipping vcm R. increases
39. A (p, s, r); B (q, r, s); C (q, r, s); D (p, r, s) (D) Due to hinge strip will not move.
(A, D) The sliding tendency of point of contact of the disc may be 42. Solution is given in the theory.
forward or backward depending on relative value of F and 43. A p, r; B p, s, r; C p, r, s; D p, r, s
Fh. Assume friction to be absent and horizontal force F is applied at
(B,C) Sliding tendency of point of contact is forward so friction a distance x above centre
act in backward direction. The acceleration of C.M. will be in
the direction of F. F
a .......... (1)
40. A s, ; B r;C q;D t m
2
MR 2 mR 2
(A) MI of sphere = and Fx
5 2
2 5 2Fx
(B) MI of shell about tangent = MR 2 MR 2 MR 2 or R .......... (2)
3 3 mR
MR 2 R
(C) MI of disc = If a = R then from eq. (1) and (2) x
2 2
(D) MI of disc about tangent in its plane F
The friction force will be zero and a
MR 2
5 m
= MR 2 MR 2
4 4 R F
41. A p, q, r, s ; B p, s ; C p, q, r, s ; D q, r, s If a > R or x , friction force is towards left and a
2 m
(A), (C) Dumbell experiences a force and net torque,
so it has translation and rotation. R F
If a < R or x , friction force is towards right and a
2 m
Solutions Exercise-1.5
1. The rolling kinetic energy of the hoop is For equal angular momentum, we have
IA A = IB B
1 2 1 2
K = mv I IA B
2 2
IB A
2
1 2 1 v rA
= mv (mR 2 ) 1
2 2 R rB Ans.
3
= mv 2 (b) For equal rotational kinetic energy, we have
= 100 × (0.20)2
1 1
= 4 J. I A 2A = I B 2B
Thus work done to stop the hoop W = 4 J. Ans. 2 2
2. The torque = F r 2
IA B
= 25 × 0.2 = 5 N-m = 2
IB A
mR2 2
The moment of inertia I = rA
2 =
rB
20 (0.2)2
= = 0.4 kg-m2 1
2 . = Ans.
9
5
(a) Angular acceleration, = = 12.5 rad /s2
= 4. The moment of inertia of the body,
0.4 I I = 2 [2 Ma2] + M [b2 – a2]
(b) Work done by the pull W = F . s = 25 × 2 = 50 J = M ( 3a2 + b2 ).
(c) In case when there is no slipping between wheel and cord, Thus work done in the process,
friction does no work and so kinetic energy of the wheel 1 2
K = work done W = I
2
= 50 J. Ans.
3. (a) In the device, A rA = r 1
B B = M (3a 2 b2 ) 2
2
A rB = 2.6 J. Ans.
= .
B rA
www.crackjee.xyz
ROTATIONAL MECHANICS 109
5. (a) If IB be the moment of inertia of the wheel B, then by
conservation of angular momentum. T
IA IB B = ( I A IB )
A T
or 8 × 600 + 0 = ( 8 + IB) × 400
I B = 4 kg-m2 Ans.
(b) The energy lost in the process
= ki – kf
1 1 mg
= I A 2A (I A IB) 2
2 2 For rotation about axis of the cylider
= 5200 J Ans. 2TR = I ii
6. By Newton's second law
F2 – F1 = ma MR 2
where I =
F1 = F2 – ma 2
= 5 – 1 × 2 = 3 N.
For rotational equilibrium, taking moment of forces about centre a
and =
of mass, we get R
l l On solving above equations, we get
F1 F2 y = 0 mg
2 2
(a) T =
6
l l
3 5 0.2 = 0 2g
2 2 and =
l = 1 m. Ans. 3R
(b) The velocity of centre of mass
7. Given – v = at
2 gt
As = =
I 3
– The average power generated by gravitational force
P = Fv
d
or = k 2 gt
dt = mg
3
d t 2
k dt = mg 2t Ans.
or = 3
0
0
L mL2 6p
2| | = – kt 9. p
0 2 12 mL
kt Rod will rotate about its c.m., one half exerts centrifugal force on
or = 0
2
m 2 L
2 the other half, therefore F
kt 2 4
or = 0 ....(i)
2
For to be zero,
kt
0 = 0
2
2 0
t = ....(ii)
k
Average angular velocity
t
dt 10 Given, diameter = 6 cm
0 20 10
av = ....(iii) =
t 6 y y
After simplifying equations (i), (ii) and (iii), we get or 20y = 10 ( 6 – y)
or 20y + 10 y = 60
0
av = Ans. or 30 y = 60
3
or y = 2
8. For the translation motion of the cylinder, we have
mg – 2T = ma ...(i)
www.crackjee.xyz
110 MECHANICS, HEAT, THERMODYNAMICS & WAVES
Solutions Exercise-1.6
1. (a) If h is the height reached by the cylinder, then f mg
1 2 1 The acceleration for translation a = = = g
2 m m
mv I = mgh
2 2 = 0.2 × 9.8
= 1.96 m/s2.
1 2 1 mR 2 v
2 s h
or 2 mv = mgh fR mg R
2 2 R Angular retardation, = = =
30° I I I
3 2
or mv = mgh mR2
4 For disc I = ,
2
3v 2 3 52 mgR
h = 1.9 m =
4g 4 9.8 mR 2
h 1.9 2
The distance s = 3.8 m
sin 30 1/ 2 2 g
=
R
2
(b) Acceleration of the cylinder = g sin 30
3 2 0.2 9.8
= = 39.2 rad /s2.
0.1
2 1
or a 9.8= = 3.27 m/s2 For ring I = mR2,
3 2
mgR g
The time taken to move the distance of 3.8 m = 2 = 1.96 rad /s2
=
mR R
1 2 If t is the time taken to start pure rolling, then
3.8 at =
2 = t, .....(i)
0
t 1.5 s
v = 0 + at ....(ii)
Total time taken T = 2t = 3 s Ans. and v = R ....(iii)
2. If be the necessary retardation, then After solving above equations, we get
0 = 0 – t 0
t =
a
240
2 R
0 60
= =
t 20 Thus tdisc 10
= 0.53 s
=
1.96
2 39.2
= rad/s 2 . 0.1
5 Similarly tring = 0.80 s Ans.
The torque needed ( ) = I Obviously disc begins to roll earlier than the ring.
4. (a) In the process angular momentum remains constant, so
mR 2 Ii i = If f
=
2
Ii i
25 0.22 2 f = If
=
2 5
7.6 2 5(0.9) 2 (2 30)
= N- m =
5 7.6 2 5(0.2)2
/5 59 rpm.
If F is the required force, then F = = = N Ans. (b) Kinetic energy in the process does not remain constant.
R 0.2
Workdone by the man using his muscle power will increase
3. Given 0 = 10 rad/ s. the kinetic energy. Thus
1
kf If f
2
=
ki 1
I
2 i i
0
mg
50 N
0.5 m and TR = I
As = a/R
TR = Ia/R ......(ii)
After solving above equations, we get
= µ N × 0.5 + 5 mg
a =
= 0.6 × 50 × 0.5 + 5 I
m
= 20 N-m R2
If is the required torque, then by Newton's second law, we have
– rest = I mg
=
= rest + I MR 2 / 2
m
2
120 R2
60
= 20 + 3.75 × 9 mg
=
= 25.23 N-m. M
m
2
The force applied F =
r 20 9.8
or a = 20
25.23 20
= 2
0.5
50 N. Ans. = 6.53 m/s2
(b) To maintain the constant speed, the torque needed Ia
(a) Tension T =
R2
= rest
= 20 N. MR 2 / 2
= a
rest R2
Thus force required =
r
Ma
20 =
= 2
0.5
20 6.53
= 40 N. Ans. =
2
(c) The anguler retardation = = 65.3 N. Ans.
I (b) If v is the velocity of the bucket, then
5 v2 = 0+2al
= = 1.33 rad/s2. = 2 × 6.53 × 20
3.75
v = 16.2 m/s Ans.
www.crackjee.xyz
112 MECHANICS, HEAT, THERMODYNAMICS & WAVES
(c) If t be is the time of fall, then
1 2 1 2
1 2 mg (h – 2R) = mv I
l = at 2 2
2
2
1 2 1 2 2 v
2l = mv mr
t = 2 2 5 r
a
7
2 20 = mv 2
= 10
6.53
= 2.47 s Ans. 7
= m gR
10. If is the final angular velocity of the system, then 10
Itable – mvR = ( Itable + Iman ) ' h = 2.7 R. Ans.
or 400 × 0.2 – 60 × 1 × 2= (400 + 60 × 22) ' (b) By using conservation of mechanical energy, we have
' = 0.0625 rad /s Ans.
1 2 1 2
mgh = mgR mvQ I
11. Velocity of the particle just before striking the rod v (2 gh) 2 2
In the process of collision the angular momentum of the system
7
remain constant or mg (6R) = mgR mvQ2
…(i) 10
mv 0 I
M 2 50
where I m 2 vQ = gR
3 7
Substituting value of I in equation (i), we get The normal (horizontal) force at Q
mvQ2
FQ =
R
50
= mg. Ans.
7
13. Net torque, net = mgl2 – mgl1
If is the angular accelerations, then
net
=
I
mv
(M 2
/ 3 m 2) mg (l2 l1 )
=
Let rod deflects through an angle before momentarily stops. ml12 ml22
The rotational K. E. of the rod + particle is used to raise the
particle as well as to raise the c.g. of the rod. g (l2 l1 )
=
1 2 Mgy l12 2
l2
I mgy … (ii)
2 2 9.8(0.8 0.2)
where y = (1 – cos ). =
Solve above equations to get the value of 0.22 0.82
12. (a) If v is the min. velocity to complete the loop, then = 8.65 rad / s2
a1 = l1
mv 2 = 8.65 × 0.2
mg + N = (R >> r).
R = 1.73 m/s2
As N = 0, a2 = l2
= 8.65 × 0.8
v = gR .
= 6.92 m/s2
14. Given, v = (30iˆ 60 ˆj ) m/s
r = (3iˆ 4 ˆj ) m
v
(a) The angular mometum, L = m( r v)
h mg
N = 2[(3iˆ 4 ˆj ) (30iˆ 60 ˆj )]
= 600 k̂ kg-m2/s Ans.
(b) Here r = [(3 ( 2)iˆ {( 4 ( 2) ˆj )]
For the rolling marble, we can write = (5iˆ 2 ˆj ) m
www.crackjee.xyz
ROTATIONAL MECHANICS 113
The angular mometum,
17. Moment of force N = r×F
L = m( r v)
= ( ai b j ) ( Ai B j)
= 2[(5iˆ 2 ˆj ) (30iˆ 60 ˆj )]
= bAk aBk
= 720 k̂ kg-m2/s = ( aB bA) k .
15. (a) Moment of inertia of the sculpture
I = |r F|
Moment arm l =
2 |F|
mR 2 mL2 L
mR 2 m R
2 12 2 ( aB bA)
= 2 . Ans.
A B2
2 2
mR m(2 R) 2
= mR 2 m R R 18. Resultant force = Aj Bi .
2 12 F
= aAk bB( k )
= ( aA bB ) k
| |
Moment arm l =
Ki + Ui = Kf + Uf |F|
1 2 ( aA bB )
0 + [mgR + mg (R + R)] = I [ mgR mg ( R R)] = Ans.
2 A2 B2
After substituting the values and solving, we get 19. The given system of forces is like as shown in figure. Thus
= 12 rad/s. Fx = F + F
16. The mass of the carpet of radius R / 2 = 2F
M ( R / 2) 2
m' = B F
R2
F 2F
M
= . A C
4 x x
v D F
F
Using conservation of mechanical energy, we have
and Fy = F–F
M R 1 M 2 1 2
= 0
MgR + 0 = g v I
4 2 2 4 2
Resultant of these, R = Fx2 Fy2
M R
2 = 2F
Mv 2
1 4 v
2 Suppose the resultant force R passes from a point, a distance y
7 2
or MgR = R from A, then
8 8 2 2
F x – 2F y = 0
2
x
7 3 y =
MgR Mv 2 2
=
8 16 Thus the resultant is applied at the mid point of side BC.
20. Solution is given in the theory.
14 21. Solution is given in the theory.
v = gR . Ans.
3 22. For any angular position , the torque of the force F relation to A
= F × l cos
www.crackjee.xyz
114 MECHANICS, HEAT, THERMODYNAMICS & WAVES
1 2 R
Thus d = I 2 mg
0 2 = r 2 dr
R2
0
F
2
= µmgR
B 3
A dr
1 ml 2
2 r
Fl cos d =
0 2 3
ml 2 2
or F l sin = dF
6
= 6 F sin / ml
Ans.
23. From FBD, we havem2g – T2 = m2 a ...(i)
The angular retardation =
I
2
µmgR
3
R = mR 2
2
T1 T2
4 g
T2 =
3R
m2 Now from = 0– t, we have
0
m2g t =
T1
m1 = g
3R
m1g 3 R
= . Ans.
4 g
and T1 – m1g = m 1 a ...(ii)
For the rotation of the pulley 25. The mass of the hanging part of the cord
T2R – T1R = I ...(iii) mx
a m' =
Also = and R
R
mR2
I = ....(iv)
2
After simplifying above equations, we get
R
m1) g (m2
= Ans.
m T
m1 m 2 R
2
T
24. To get frictional torque on the disc, take an element of width dr a
distance r from the centre of the disc. The frictional torque
d = dF r
= µ (dN) r mg
mg
= µ 2 rdr r
R2 Thus m'g – T = m'a ....(i)
and TR = I
2 mgr 2 dr
= Ia
R2 or T = ...(ii)
R2
www.crackjee.xyz
ROTATIONAL MECHANICS 115
On solving equations, we get 28. If ax be the acceleration of the spool along x-axis, then
m'g F cos – f = ma x .....(i)
a = and fR–Fr = I .....(ii)
I
m'
R2 ax
where =
R
MR 2 m and I = mR2
Here I = (l x) R 2
2 l (a) After simplifying above equations, we get
After simplifying, we get
2mgx R r F
= Ans.
Rl ( M 2m)
f
26. For translational motion of the cylinder, we have
mg sin – T = ma cm .......(i) F (cos r / R)
ax =
m(1 g )
(b) The displacement of the spool in t second
T 1 2
s = axt
2
mg Thus work done W = F s cos
sin
F 2t 2 (cos r / R) 2
and for rotational motion 2m(1 g )
acm Ans.
Tr = I .......(ii)
R f
29. Acceleration for translation,a =
On solving above equations, we get m
a in = g sin Ans. mg
= N
I m
1
mR 2 = µg
0
27. From FBD, we have
N = F + mg Retardation for rotation = f
I
Frictional force mg
f = µN fR
=
= µ ( F + mg ) .....(i) I
N =
mgR
mR 2
N 2
f = 2µg / R
f f = 2f If t be the required time and v and are linear and angular velocities
N=2N after pure rolling, then
mg F = 0– t
and v = 0+at
For translational motion of the cylinder After getting pure rolling, v = R.
f = ma cm .....(ii) After solving equations, we get
For rotational motion of the cylinder
FR – f R = I .....(iii) 0R
(a) t =
acm 3 g
Here = (b) Work done by frictions W = ki – kf
R
MR 2 1 2 1 2 1 2
and I = = I mv I
2 2 0 2 2
After simplifying above equations, we get
2
m R2
2 g = –
0 Ans.
a cm =
2 3 6
30. By conservation of angular momentum, we have
3 mg M v1R = I + Mv2R
and Fmax =
2 3 MR 2
where I =
2
www.crackjee.xyz
116 MECHANICS, HEAT, THERMODYNAMICS & WAVES
v2 ML2 ML3
and = 32. Hint : mr 2
R 3
0
3
v2
Thus, M v1R = I + Mv2R
R 1 ML2 2 1 ML2 2 1 1
0 mv '2 ( mr 2 ) 2
v1 2 3 2 3 2 2
v2 =
I
1 MR 2 MR2
MR 2 ( mR 2 )
33. Hint : 2 2
0
Ans
31. Choose an element of the rod of width dx at a distance x from the W.d. = change in rotational K. E.
hinge. 34. (a) In the process angular momentum remains constant.
m If be the final angular velocity, then
Mass of the element, dm = dx. The centrifugal force on this I1 1 + I2 2 = (I1 + I2)
element I1 1 I2 2
dF = (dm) 2 (x sin ). = I1 I2 Ans.
2
I1I 2 ( 1 2)
= Ans.
2( I1 I 2 )
35. Acceleration for translation
f
a = 0
m
Fig. 6.44 mg f
Its moment of force about the hinge =
m
d = dF × x cos = µg
2
= (dm) (x sin ) (x cos ) Angular retardation =
I
m 2 2 sin 2
= dx x fR
2 =
I
m 2
= sin 2 x2 dx ....(i) mgR
2 =
2
For the moment of force of whole length of rod, integrating (i) mR 2
3
2 3 g
m
= sin 2 x 2dx =
2 2 R
0
For translational motion v2 = 0 + 2 a s
For rotational motion 2 = 2
0 2
2 2
m
= sin 2 ....(ii) where = s/R
6 After getting pure rolling, v = R
In the rotating frame, apart from other forces the centrifugal force After solving above equations and substituting the values, we get
also act. For rotational equilibrium of the rod, we have = 0. s = 0.58 m. Ans.
Taking moment of all forces about hinge and putting their algebraic 36. In the process of falling of rod
(a) decrease in P.E. = increase in rotational K.E.
sum zero, we get
L 1 2
m 2 or Mg = I
mg sin = sin 2 2 2
2 6
L 1 ( ML2 ) 2
3g or Mg =
or cos = Ans. 2 2 3
2
2
3g
=
L
www.crackjee.xyz
ROTATIONAL MECHANICS 117
(b) Torque acting on the rod about left end Velocity of combined mass (4 + 2) kg after collision:
Rx 4 × 3 + 0 = (4 + 2) v
v = 2 m/s
The velocity of c.m. is given by
Ry Mg
m1v1 m2v2
L v cm = m1 m2
= Mg
2
By Newton's second law 6 2 3 0
=
6 3
L
Mg = I
2 4
= m/s Ans.
3
L ML2
or Mg = (b) Using conservation of angular momentum about c.m. we have
2 3 4×3×y = I
or 4 × 3 × 0.5 = [ 6 × 0.52 + 3 × 12 ] ×
3g
= . 4
2L = m/s Ans.
(c) ax = 2 r 3
38. (i) Suppose a is the acceleration of the block, then the tangential
3g L acceleration of the pulley will be a. Thus angular acceleration
=
L 2 of the pulley
3g a
= =
2 R
and, ay = r M
T
3g L
= T
2L 2
3g
= °
4 30
sin
Thus Fx = Ma x mg 30°
3Mg Thus mg sin 30° – T = ma .....(i)
= and TR = I
2
For the translation of centre of mass along y-axis, we have MR 2 a
Mg – Ry = Ma y or TR =
2 R
3g Ma
Ry = M T = ....(ii)
4 2
From equations (i) and (ii), we get
Mg
Ry = . Ans. mg sin 30
4 M
a = m
37. (a) The position of c.m. 2
6 0 3 1.5 mg
y = =
6 3 2m M
= 0.5 m from top. 0.5 9.8
=
2 0.5 2
4 kg 2 kg = 1.63 m / s2
2 1.63
and T =
y 2
= 1.63 N Ans.
(ii) In process the K.E. of the system will convert into potential
c.m. energy of the block and so
1 2 1 2
mv I = mgh
2 2
2
1 2 1 MR 2
or 2 m( R) 2 2 = mg (s sin 30°)
Q
m
x
5
0.
B 1
2
c.m.
f
www.crackjee.xyz
ROTATIONAL MECHANICS 119
For first cylinder 45. (a) If acm is the acceleration of the cm of the hollow sphere, then
– f r1 t = I1 ( 1' – 1) .....(ii) mg sin – f = ma cm .....(i)
For second cylinder and fR = I
f r2 t = I2 ( 2' – 2) .....(iii) acm
After solving above equations, we get or fR = I
R
I1 1r2 I2 2 r1 acm
r2 I
1' =
I 2 r12 I1r22 f = .....(ii)
R2
After solving equations, we get
I1 1r2 I2 2 r1
and r1 Ans. g sin
2' =
I 2 r12 2
I1r2 a cm = I
1
44. The velocity of the particle before collision mR2
v = 2gh 2
(a) Angular momentum of the system after collision For hollow sphere, I = mR 2
3
= angular momentum before
collision 3
a cm = g sin
5
= L particle L rod
Iacm
L Thus f =
= mv 0 R2
2
2 3
L mR 2 g sin
= m 2 gh = 3 5
2 R2
mL gh 2
= Ans. or f = mg sin
2 5
By conservation of angular momentum, we have
f
L µ min =
mv = I N
2
2
2 mg sin
mL gh L = 5
or = 2 mL2 m
2 mg cos
2
2
8 gh = tan Ans.
= Ans. 5
3L
µ min
(b) The velocity of the particle + ball B after collision (b) Given, µ =
2
B
P tan
=
h 5
A A Now mg sin – f = ma
L/2 L/2
or mg sin – µ N = ma
tan
B or mg sin mg cos = ma
5
m 2 gh 0
v' = 4 g sin
m m a =
5
gh For rotational motion of the sphere
= . fR = I
2
2
For the full rotation to occur, ball B will reach to the highest or µ NR = mR 2
point, and so 3
Ki + Ui = Kf + Uf tan 2
or mg cos R = mR 2
1 L L 5 3
(m m)v '2 0 = 0 2 mg mg
2 2 2 3 g sin
After substituting the value of v' and simplifying, we get =
10 R
3L The linear velocity after moving a distance
h = Ans.
2
www.crackjee.xyz
120 MECHANICS, HEAT, THERMODYNAMICS & WAVES
v2 = 0+2a For the rotational equilibrium of the plate, we have
4 weight = force
= 2 g sin
5 [3g ] × 0.3 = F × 0.45
= 0.24 × 0.45
8
= g sin v = 83.3 m/s Ans.
5
Thus translational K.E., 47. (a) The effective system is shown in figure. Here
r = cos 30°
1 2
K Trans = mv
2 3
= .
4 2
= mg sin .
5
y
We can write = Fx
v a x
Fy
= v
a
r
sin
3g
8 10 R Fy
g sin
= 5 sin
4g
5
F 2m
3 8
= R g sin 2
8 5 Thus, Fy = M r