e

ROTATORY MOTION – ASSIGNMENT-09-04-2022

SINGLE CORRECT TYPE
1. A metal disc of radius R can rotate about the vertical axis passing through its centre. The top surface
of the disc is uniformly covered with dust particles. The disc is rotated with gradually increasing speed.
At what value of the angular speed   of the disc will the 75% of the top surface become dust free.
Assume that that the coefficient of friction between the dust particles and the metal disc is 0.5. Assume
no interaction amongst the dust particles.

S
IC
YS
g 2g 3g 3g

PH
A) B) C) D)
R R 2R 4R
Key: A F
1
Sol: th of disc surface will remain covered with dust.
O
4
R
L

This means all particles beyond r  will fly away.

O

2
N  mg
O


H

x
f  m 2 x
SC

f
But f   N

 m 2 x   mg
N

R
x dust particle should fly away
IA

2
D

R
x friction should be able to provide necessary centripetal force.
2
IN

g g
2  
R/2 R
g
 
R
2. Two solid bodies rotate about stationary and mutually perpendicular intersecting axes with constant
angular velocities 1  2 rad / s and 2  4.5 rad / s . The angular acceleration of one body relative to
the other body is:
A) 2 rad / s 2 B) 4.5 rad / s 2 C) 9 rad / s 2 D) 2.25 rad / s 2
Key: C
r
Sol: 1 rotate about xy 1  1kˆ  2kˆ
r
2 rotate about yz 2  2 kˆ  4.5iˆ
r r r
21  2  1  4.5iˆ  2kˆ
r
21  (4.5) 2  (2) 2  4.92 rad/sec

r r
 21   2  1 
d (2iˆ) d 1k
r

ˆ 
dt dt
 diˆ
 ˆ d 2   dkˆ ˆ d 1 
  2  i    1 k 

S
 dt dt   dt dt 


IC
d 2 d
But 0 1

YS
dt dt

diˆ dkˆ
 2  1

PH
dt dt

dkˆ
But 0
dt
F
O
r diˆ
 21  2
dt
L
O

y ĵ
B
O
H

ld
SC

dθ
x
O A
iˆ
N
IA

l

R
D

 .R  l
IN

dR d dl
 R 
dt dt dt
d ˆ diˆ
j
dt dt
diˆ
1 ˆj 
dt
r
 21  21 ˆj  2  4.5 ˆj

 9 ˆj rad/sec2
3. A uniform solid cylinder of mass M and radius R can freely rotate around its axis O. There is a elastic
string of relaxed length L and stiffness K attached to the cylinder and a static wall. Initially, the string
is relaxed. As the cylinder starts rotating, the string will wind the cylinder. The surface of cylinder is
very rough, so that the string does not slip on the cylinder's surface. The minimum initial angular speed
of the cylinder ω 0 , so that it can rotate to angle 2π is (Assume Hooke's law to be valid).

S
IC
YS
8 2 K K  2K 4 2 K
A) B) C) D)

PH
M M M M
Key: A

1  MR 2  2 1
F
 0  K  R 
2
Sol: 
O
2 2  2

2  KR 2 2 
L

  2
O

0
MR 2
O

8 2 k
0 
H

M
SC

4. A smooth L shaped rod ABC is rotating in a horizontal smooth surface plane with angular velocity ω
with respect to a vertical axis passing through the hinge. At the instant shown, a particle of mass ‘m’
N

collides with the rod on BC part of it near end C as shown in the figure. Just before collision, its
IA

velocity makes 53o with BC part of the rod. If particle collides with the rod perfectly inelastically, then
what should be value of Vo (in terms of ω and L) such that after collision, rod comes to rest?
D
IN

m1
B L C
m

Vo
M,L
53o
hinge

250 L 250 L 250 L 250 L

A) B) C) D)
12 21 14 41
Key: A
M,L
m

Vo
M,L
53o

Sol:
Angular momentum conservation
 mL2 mL2  L2  

S

   m L   
2

IC
 3 12  4  


YS
 4V 
 m 0 L
 5 

PH
5  4
  mL2   mV0 L
3  5
F
250 L
O
V0 
12
L

5. A horizontal ring of mass M and radius R can rotate freely about a fixed smooth vertical axis passing
O

through its centre. Two beads each of mass M/8 can move radially outwards along two massless
O

frictionless rods fixed on the ring as shown in the figure. System is isolated. The beads and the
H

dielectric ring are charged. Initially, when the ring is at rest and beads are also at rest near point O, the
SC

ring is suddenly given angular velocity ω and left. At some instant, the angular speed of the system is
8
N

and one of the masses is at 3R/5 from point O. Then, at this instance, the distance of the other
9
IA

mass from O will be

D
IN

A) 2R/3 B) R/3 C) 3R/5 D) 4R/5

Key: D
Sol: conserving angular momentum
2
 M  3R   8w  2  8w  M 2  8w 
MR w   
2
    MR   x  
 8  5   9   9  8  9 
R 2 8R 2 x 2
R  2
 
25 9 9
R 2 R 2 x2
 
9 25 9
16R 2
 x2

S
25

IC
4R
x
5

YS
6. A heavy disc with radius R is rolling down hanging on two non-stretched string wound around the disc

PH
very tightly. The free ends of the string are attached to a fixed horizontal support. The strings are always
tensed during the motion. At some instant, the angular speed of the disc is  , and the angle between the
F
strings is  .
O
L


O
O


H
SC

R
(A) Speed of enter of mass of disc is
N

cos( / 2)
IA

R
(B) Speed of CM of disc is
D

sin( / 2)
IN

(C) The instantaneous axis of the disc lies at the circumference of the disc

(D) The instantaneous axis of the disc is at infinity

Key: A
Sol: The velocities of point A and B being perpendicular to the strings; the instantaneous axis of rotation
must be at D, instantaneous point of intersection of the line of the two strings.
Therefore,
 
B


 OB R
VCM  .OD  
 
cos cos
2 2

VP

S
R/cos
π-α 2 α

IC
IAOR α/2

YS
VR

PH
ωR
VCM  ωX 0 
α
F
cos
O
2
MULTIPLE CORRECT TYPE
L
O

7. A ring of radius R is constrained to perform pure rolling on the outer surface of a fixed pipe of radius
4R. The centre of the ring has a constant speed v as shown in figure.
O
H

D
SC

A 

B
N

4R
IA

C
D
IN

A) The acceleration of the point (B) on the ring which is in contact with the surface of the pipe is
4v 2 / 5R
B) The acceleration of the point (B) on the ring which is in contact with the surface of the pipe is
v2 / R
C) The acceleration of the point (D) on the ring which is farthest from the centre of the pipe at the
6v 2
given moment is
5R
D) Acceleration of B w.r.t A (centre of ring) is v 2 / R
Key: ACD
 v


v 4R

R

Sol:
ur
aA 
v2
5R
 $j 
r r r
 
a B  a B / A  a A  2 R  $j 
v2
 
 ˆj

S
5R

IC
4v 2 $
5R
 
j

YS
r r r
 
a D  a D / A  a A  2 R  $j 
v2
5R
 
 ˆj

PH

6v 2 $
5R
 
j F
ur v2 $
 
O
a B /A  j
R
L

8. Ending at the rim of the table there is a slope of angle of elevation of α, from which a uniform density
O

rectangular block of length ℓ and of height d is sliding down. The length of the block (‘x’ as shown
O

in figure) moved further from the end of the table until it tilts is,
H
SC

d l

N


IA

x
D

l
A) , if friction between the slope and the block is negligible.
IN

l  d
B) , if the coefficient of kinetic friction between the slope and the block is μ (0<μ<tanα,
2
and μd<ℓ).

1
C)  l  d tan   , if either surface is rough or smooth.
2

D) is independent of angle of elevation of the slope.

Key: ABD
Sol: Forces acting on the body:
 N  mg cos 
f   N  m  g cos 

d l
x
2 2
N
S

mg

The bar does not tilt until the torque of the forces acting on it is zero with respect to its center of mass.

S
d l  d l  l  d
f  N   x   0  m g cos     x  mg cos   x 

IC
2 2  2 2  2

YS
9. A solid sphere of mass m is at rest on a smooth horizontal surface and there is a very small cylindrical
hole in the solid sphere upto centre of the solid sphere. A small particle of mass m enters in the hole

PH
with velocity v0 and sticks with the sphere at the centre of solid sphere. Size of hole is so small that a
point mass can just move inside it. (collision is such that sphere does not lose contact with the surface
F
after collision). Surface on the left side of point Q is smooth and right side is rough as shown in the
O
figure. Then (Assume moment of inertial of sphere ICM = 2mR2 / 5)
L

Vo
m
O

60o solid sphere

O
H

P smooth Q rough
SC

A) linear momentum conserved in horizontal direction

N

B) angular momentum conserved about point P which is point of· ground

IA

C) after collision when pure rolling starts on the rough surface the angular velocity of solid sphere is
D

5 v0
IN

24 R

v0
D) just after collision, velocity centre of mass of the system is
4

Key: ABCD
 mv 0

solid sphere
60o

Sol:

mv0 v

solid sphere v1 mg
60o rough

P N  mg
fk

S
Along horizontal direction,  momentum  just before collision   momentum Just after collision

IC
YS
mv 0
 2mv x
2

PH
v0
vx  … (1)
4
When pure rolling starts
F
2
O
2mvx R  2mvR  mR 2
5
L

v  R
O

Vo 2  v
R  2mvR  mR 2  
O

2m
4 5  R
H

2
 2mvR  mvR
SC

5
mVo R 12mvR

N

2 5
IA

5Vo
ω=
24R
D

10. A uniform rod of mass m and (length l is released from rest in the vertical position   00  on a
IN

rough (Surface is sufficiently rough to prevent just sliding) square corner A, as shown in figure, then
choose correct options.

A
Rough
 A) If the rod just begins to slip when   37 0 with vertical, then the co-efficient of static friction 
between the rod and the corner is 0.3

B) If the end of the rod is notched so that It cannot slip, then the angle  at which contact between the
rod and the corner ceases is 530

C) If the rod just begins to slip when   37 0 , then the coefficient of static friction between the rod
3
and the corner is
4

D) If the end of the rod is notched so that it cannot slip, then the angle  at which contact between the
rod and the corner ceases is 37 0

S
IC
Key: AB

YS
m, l N
B



PH
fs   l / 2   v0 

A m1 F
O
Sol:
L

l l  1  ml 2  2
O

mg  mg   cos      
1
2 2  2 3 
O

mg cos   N  m 2 (l / 2) 
2
H
SC

l ml 2
 mg sin     
3
2 3
l
mg sin   f s  m    
4
N

2
IA

fs   N 

5
D

1  ml 2  2
IN

l
mg 1  cos     
2 2  3 

l  4  1  ml 2  2
mg  1     
2 5  2  3 

mg l 1  ml 2  2
  
10 2  3 

3g 3g
  …. (1)
5l 5l
 l ml 2
 mg sin    
2 3

3 g sin 
 …. (2)
2l

m 2 l m 2 l
mg cos   N  N    mg cos 
2 2

m 2 l
 mg cos  
2

4 ml  3 g 
N  mg   

S
5 2  5l 

IC
1 3mg 
  4mg 
2 

YS
5

1  8mg  3mg 
 

PH
5 2 


1 mg
F
N  5mg 
10 2
O
mg
N
L

2
O
O

l
mg sin   m   f s
2
H
SC

  3 
 3g   
3 l   5  
mg  m    fs
5 2  2l 
N

 
 
IA

3mg  3
1    f s
D

5   4
IN

3mg
 fs
20

fs   N
3mg mg

20 2
  0.3

When   53o
 l 1  ml 2  2
mg 1  cos   
 
2 2  3 

3 g l  3 
1  2
l 2  5 

6g

5l

m 2 l
N  mg cos  
2

3 m6 g  x
N  mg 

S
5 2  5l

IC
N 0

YS
11. A uniform solid cylinder of mass M and radius R, rolling without sliding on a rough horizontal plane,

PH
is pulled at its axis with a horizontal velocity of V0. By means of a string of length 2R attached to its
axis, the cylinder is dragging a thin plate of mass m = 2M lying on the plane. If the pulling force is
removed and the coefficient of friction between the cylinder and ground is sufficient enough for rolling
F
O
always. (µ = 0.4, V0 = 2m/sec, R = 0.5 m, g = 10 m/sec2). Choose the CORRECT option (s) before
the cylinder comes to rest
L
O

y
O
H

M
SC

V0
2R
m
R
N

o x
IA

A) The net force on the cylinder is along positive x direction.

D
IN

B) The net force on the cylinder is along negative x direction.

C) Frictional force on the cylinder is along negative x direction.

C) Frictional force on the plate is along positive x direction.

Key: ACD
Sol: F2 F1   m  M  a s
1 a
F1R  MR 2 s
2 R
 1
F1  Ma s
2
F2    mg  T sin  
T Cos  F1  Ma s

W as
v T
T

F1 F2

Solving these equation

S
F2  F1  3Ma s 



IC
1

YS
F1  Ma s 


2
7

PH
F2  Ma s 


2
7
Ma s = μ(mg-Tsinα)
F
2
O
7
μTsinα = 2μMg- Ma s
2
L
O

1
T cos  = Ma s  Ma s
2
O

3Ma s
H

T cos α =
2
SC

4g
as   2.08m / sec 2
3 1   tan    4
N

V02 4
IA

S   0.96m
2as 2  2.08
D

u
t  0.96
IN

a
12. A dumbbell consists of two equal particles of mass ‘m’ (connected with massless rod) separated by a
distance ‘2a’. It is initially attached along a radial line to a disk that is rotating at  rad / s about a
vertical axis (see figure). The center of the dumbbell is at a distance ‘R’ from the center of the disk.
Choose the CORRECT statement(s) about the motion of dumbbell after it is suddenly free. Assume
all the motion occurs on a horizontal frictionless plane.
 2a

A) Motion of dumbbell will be pure translational

B) Angular velocity of dumbbell remain same i.e.,  .

S
C) Kinetic energy of dumbbell will be given by mR 2 2

IC
D) Kinetic energy of dumbbell will be given by m  R 2  a 2   2

YS
Key: BD

PH
1 1
KE  m 2  R  a   m 2  R  a 
2 2
Sol:
2 2 F
13. Slender bar A is rigidly connected to a light rod BC in Case 1 and two light inextensible cords in Case
O
2 as shown. The thickness of bar A is negligible compared to L.
For Option (A) and (B) : In both case (1) and case (2) A is released from rest at an angle θ=θ o . when
L
O

θ=0 o
O

For Option (C) and (D) : Consider system to be at rest initially in vertical position. If bullet D strikes
H

A with a speed v0 initially directed horizontally and becomes embedded in it

SC
N
IA
D
IN

Mark the CORRECT statement(s) : -

A) The kinetic energy will be the same in both the cases
B) The speed of the centers of gravity will be the same
C) The speed of the center of gravity of A immediately after the impact in case-1 will be larger.
D) The speed of the center of gravity of A immediately after the impact in case-2 will be larger.
Key: AD
Sol: For (A) since loss in potential energy in both case is same.
hence kinetic energy at θ 0  0 will be same in both case.

For (D) in case (2) slender bar will have more speed because it has only translational kinetic energy.
14. A horizontal square platform of mass m and side a is free to rotate about a vertical axis passing through
its centre O. The platform is stationary and a person of the same mass (m) is standing on the platform
at point A. The person now starts walking along the edge from A to B (see figure). The speed v of the
person with respect to the platform is constant. Taking v=5m/s and a=1m. During the motion.

S
A

IC
YS
A) Angular velocity of platform when the person is at A is 3.75 rad/s.

PH
B) Angular velocity of platform increases first and then decreases again as it reaches B.
C) Angular velocity of platform decreases first and then increases again as it reaches B.
F
D) Angular velocity of platform remains constant as the person goes from A to B.
O
Key: AB
L
O
O
H

Sol:
SC

v m =v mp +v p
N

= v2 +x 2ω2 +2vωcos  90+θ 

IA

= v2 +x 2ω2 -2vxωsinθ
D

Li  0  L f
IN

m 2 

12
 a  a 2    m  v  x sin    0
2
a2 a2 va 6v
     
6 4 2 5a
15. A large, horizontal disk of radius R, shown below, starts to rotate from rest with an angular acceleration
of α . The rotation is about a vertical axis through the centre of disk. The disk contains a narrow
channel of length 2R and rectangular cross-section. Gravity acts in the vertical direction with an
acceleration of g. There is a small rectangular puck that just fits easily in the aforementioned channel,
as shown. The puck is situated a distance r from the axis of rotation.
 A) If the sides of the channel are friction less but the bottom of the channel has a static coefficient of
μ
friction μ , then puck begins to slide at t=
α

g
B) In case mentioned in option (A), t=
r 2

S
IC
C) Now, instead, the situation is that the bottom of the channel is frictionless but the walls have a static
μg

YS
coefficient of friction μ . Now puck begins to slide at t =
rα 2

PH
μ
D) In case mentioned in option (C), t =
α
Key: BD
F
O
Sol: (A) μmg=mω 2 r

μg =α 2 rt 2
L
O

μg
t=
O

α 2r
H

(C) N=mr
SC

μN=mω2 r

μmrα=mω2 r
N

μ =α 2 t 2
IA

μ
t=
D

α
IN

16. A massless spool of inner radius r and outer radius R is placed against a vertical wall and tilted split
floor as shown. A light inextensible thread is tightly wound around the spool through which a mass m
is hanging. There exists no friction at point of contact with tilted floor but there is friction between
spool and the wall with coefficient of friction  . The angle between the two surfaces is  .
 S
2 2
r  r
A) Magnitude of force on spool with wall to maintain equilibrium is mg    1   cot 2 

IC
R  R

YS
 r
B) Magnitude of force on spool with wall to maintain equilibrium is mg  1   cot 
 R

PH
cot 
C) Minimum value of  to maintain equilibrium is
R
1
F
r
O
tan 
D) Minimum value of  to maintain equilibrium is
R
L

1
O

r
O

Key: AD
H

Sol: mgr  fR .........  i 

SC

N1 sin   f  mg ....... ii 

N1 cos   N 2 .......  iii 

N
IA
D
IN

Solving, F from wall  f 2  N 22

For maximum  , f   N 2
tan 

R
1
r
17. A uniform circular ring of mass ‘M’ and radius ‘R = 10 cm’ is placed on a smooth horizontal surface
at rest. An insect of mass m(= M/2) initially sitting on the circumference of the ring starts moving with
a velocity v0 = 2 m/s relative to the ring along its circumference. Then choose the correct option (s).

R
O m

M,R

A) The angular velocity of the ring about its centre is 5 rad/s.

B) The angular velocity of the ring about its centre is 10 rad/s.

S
C) The velocity of the centre of mass of the ring relative to ground is 0.5 m/s.

IC
D) The velocity of the insect relative to ground is 1 m/s.

YS
Key: ACD
Sol: Using conservation of angular momentum of system about center of mass

PH
2
MR 20  mR 2  0
3 F
  angular velocity of the system w.r.t. axis through centre of mass.
O
0  angular velocity of the ring about its axis through point O.
L

2
2mR 20  mR 2
O

3
O

  30 … (i)
H

 
Now,   R  R   2R  v
SC

 0  0
 3  3

 0   R  v0
N

v0
0   
IA

R
D

v0 2
40  0   5rad / s
IN

R 4  0.1
  30  15rad / s

R 0.1
 velocity of centre of mass of the ring relative to ground =  15   0.5m / s
3 3
2R
 velocity of the insect relative to ground =  1m / s
3
Angular velocity of the ring about its centre = 0  5rad / s
 M, R
2R / 3

R / 3 2R / 3
m
o

R / 3

0

PARAGRAPH TYPE
Paragraph for Question Nos.18 to 19
In the figure shown, the ring of mass 4m has tiny mass m attached rigidly to its rim. At the moment shown in
the figure, suppose the geometric centre of the ring has a rightward velocity v, and the ring is rolling purely

S
on the flat sufficient rough surface. Answer the following questions:

IC
YS
m

PH
18. The total kinetic energy of the system is
A) 3mv 2 B) 5 mv 2 C) 6 mv 2 D) 4mv 2
F
O
19. The magnitude of acceleration of the geometric centre of the ring at the moment shown in the figure
is ? Given v  1m / s and radius of ring R = 10 cm
L
O

A) 1 m / s 2 B) 2 m / s 2 C) 0 D) 5 m / s 2
O

Key: B
H


SC
N

R
mg 
IA

5m g
2R
D
IN

Sol: N


R 2R
5

R/5

R
5m g  N  5m   ................. 1
 5 
  2 R 

f  5m  R   .................  2 
 5 

R  4m 2 
N.  f .R   .R  4m .R 2   .................  3
5  5 
 5mg  mR   5mR  m2 R  24 mR
5 5
mR 24
mg   5m R   m 2R  m R 
5 5
m g  m 2R  10m R 

g  2R  10R 

S
INTEGER TYPE

IC
20. In the figure shown, a wedge is being moved on a horizontal surface with a velocity ‘v’ as shown. At

YS
the instant shown, if angular velocity of the rod of length l (about hinge) touching the wedge as shown
xv  3

PH
is . The value of x ?  take sin 37  
l  5

hinge
F
O
l
37 0
L
O
O

370
H

Key: 0.75
SC

g
N
IA

l
D
IN

530 V
0
37
Sol:
In the common normal
direction l cos 37  V cos 53
v 3

l 4
21. A uniform thin hollow spherical shell of mass m and radius R is released from rest with its diametrical
plane vertical on a very rough horizontal surface. It doesn’t slip on the floor during its subsequent
 motion. At an instant when the shell achieves its maximum value of kinetic energy, the value of normal
reaction on it from the floor is xm g , where g is acceleration due to gravity, then find the value of x ?

Key: 1.75
N

S
IC
p 

YS
mg
Sol:

PH
R
N  mg  m 2
2 F
3g
2 
O
2R
3g R
L

N  mg  m  
O

2R 2
O

3
N  mg  mg
4
H

7
SC

 mg
4
22. Two small balls are fixed at the ends of a light rigid rod of length l. The system is released from height
N

‘h’ with rod horizontal. The balls collide with the horizontal surface and rebound. The coefficient of
IA

restitution between A and ground is e1 and that between B and ground is e 2 . Find the angular speed
D

of rod (in rad/s) just after the collision. Taking g  10 m / s 2 , l  0.4 m , h  5m , e 1  0.6, e 2  0.4
IN

A B

Key: 5.00
Sol. After rebounding
l
VA  e1u  VC  
2
 l
VB  e2u  VC  
2
VA  VB   e1  e2  u

l   e1  e2  u


 e1  e2  u
l
2 gh  0.2 
 5
0.4
23. On a fixed rough inclined plane of inclination 45º a block of mass m and a cylinder of mass 2m are
released. Contact surface between block and cylinder is smooth, at all other contact surfaces the

S
g

IC
Coefficient of friction is 0.5. The acceleration of the cylinder when they move together is . Find
x 2

YS
x?
  0.5

2m
PH
F
m
O
L

45o
O
O

Key: 1.60
H

  0.5
SC

2m
M
N

NS
IA

NC N
D
IN

fC 2m fS
2m g sin 
m g sin 
2m g cos 
m g cos 

N  mg sin   f C  m  a  ........... 1

2 mg sin   N  f C  2m  a  ...........  2 

 2m  R 
2
a 
fS .R    .............  3
2 R 
 mg mg
N   Ma
2 2 2
2mg
 N  3ma
2
mg
 mg 2  4 ma
2 2
5g
a 
8 2
g
a 
1.6 2
24. A bicycle wheel (ring) of mass 1kg and radius 2m is attached at the end of a long rod of mass 1kg and

S
length 3m, with the axis of wheel perpendicular to the rod. The rod is rotating counter clockwise as

IC
seen from above, about its other end with an angular speed   0.5 rad / s . If wheel is also found to be

YS
rotating uniformly about its axis with same angular speed  in counter clock wise sense as seen in
rotating frame of horizontal rod. The angular momentum of the system about the axis passing through

PH
x
one end (A) of rod in x kg  m 2 / s. Find ? (Neglect mass of spokes and mass of rod connecting end
2
F
of rod of length 3m and centre of ring).
O
L
O
O
H

A
SC

Key: 8
N
IA

Sol: Angular momentum of rod with respect to axis passing through O is

ml 2
D

LR / O  I   
3
IN

LB / O  L1  L2

LB / O  I11  mvl

LB /O  mr 2  m l 2

Total Angular Momentum  LR / O  LB / O

1
L  (1)(3) 2 (0.5)  (1)(0.5)(3)(3) (1)(2)2 (0.5)
3
 1.5  4.5  2  8kg  m 2 / s
25. What minimum force (in N) should be applied to the end of the lever of a double pulley system to hold
a load P = 10 kg? The length of the lever is l  1m , the radius of the larger pulley of the system
r1  20cm , the radius of the smaller pulley r2  10cm (figure). (Axle of the double pulley is fixed, take

g  10 m / s 2 ).

S
Key: 5

IC
Sol: The force required to keep the double pulley system in equilibrium can be determined from the

YS
torque balancing.
P P

PH
r2  F l  r1
2 2
P  r1  r2  10 g  0.1  5 N
F  
F
2l 2 1
O
26. A uniform solid cylinder A of mass m1 can freely rotate about a horizontal axis fixed to a mount of
L

mass m2 . A constant horizontal force F is applied to the end K of a light thread tightly wound on the
O

cylinder. The friction between the mount and the supporting horizontal plane is assumed to be absent.
O

Take m1 =m 2 = 5kg and F=10N ,The acceleration of the point K is a m/s 2 then value of a
H
SC
N
IA
D
IN

Key: 5
Sol: For whole arrangement

F
a
m1  m2
 for cylinder
m1R 2
FR  
2
Now acceleration of point K

F 2F
So ak   5
m1  m2 m1

S
27. A uniform rod AB of mass M and length 2R is moving in a vertical plane inside a hollow sphere

IC
of radius R. The sphere is rolling on a fixed horizontal surface without slipping with velocity of its

YS
centre 2v. When the end B is at the lowest position, its speed is found to be v as shown in the figure.
If the kinetic energy of the rod at this instant is 4Mv2/k, find k.

PH
Hollow sphere
F
O
A 2v
L

v
O

B
O

Key: 3
H

Hollow sphere
SC

V 

A
N
IA

V I
B
D

Sol:
IN

w.r.t sphere the ends A and B of rod move as shown in figure. I

w.r.t ground the ends A and B move as shown in figure - II
Let  be the angular velocity of rod about its COM
V 3V
y x
2
2V V
A
 2
Vcx
45o
C Vcy V
v
B 2
V
2
 V
VCy  ------(1)
2

VC . x  2V
l V
VCx   
2 2
l 3V
VCx   
2 2
2v 2v
l  or               2
2 2l
1 1
KE  mvcm 2  I c 2

S
2 2

IC
1  v2  1 ml 2 2v 2
 m   2v 2   2
2  2  2 12 l

YS
1 5 2 1 v2 1 2  5 1 
 m v  m  mv   

PH
2 2 2 6 2 2 6
1 2  15  1  4 2
 mv    mv
2
F
 6  3
O
28. A disc of radius r = 2cm attached with a horizontal axle OA, is rolling without sliding on a circular track
of radius R =10 cm as shown. In the process, the centre of the disc moves with a constant speed v = 10
L
O

cm/s. Find the angular acceleration (in rad/s2) of the disc. (in rad s-2)
O
H
SC
N
IA

Key: 5.00
D

v v v2
Sol: angular acceleration of the disc,   0  
r R rR
IN

 0.1
2

 4
 5rad / s 2
20  10