Circular Motion JEE Advanced Important Questions

Circular Motion JEE Advanced Important Questions

A large sphere of radius R is moving with velocity v horizontally. A small sphere ‘B’ of mass m starts sliding
from top of the sphere downwards. Let at point C it looses contact with the large sphere. Let the velocity of
‘B’ with respect to large sphere is u, then which of the following statement follows.

A bead of mass ‘m’ is released from rest at A move along the fixed smooth circular
track as shown in figure. The ratio of magnitudes of centripetal force and normal reaction
by the track on the bead at any point P.

A string is wrapped around a cylinder of radius R. If the cylinder is released from rest,
the velocity of the cylinder after it has moves h distance is.

A point P moves along a circle of radius r with constant speed v. Its angular velocity about any fixed point on
the circle will be.

Find the magnitude and direction of the force acting on a particle of mass m during its motion in the xy plane
according to the law x = a sin t and y = b cos t, where a, b and  are constants.

An athlete completes one round of a circular track of radius R in 40 second. What will
be his displacement at the end of 2 minute 20 second.

In the aboveProblem, what is the ratio of the displacement to the distance, when the athlete has covered 3/
4 th of the circular track?

A particle P is moving in a circle of radius ‘r’ with a uniform speed v. C is the centre of the circle and AB is a
diameter. When passing through B the angular velocity of P about A and C are in the ratio.

Two beads A and B of equal mass m are connected by a light inextensible cord. They are constrained to
move on a frictionless ring in vertical plane. The blocks are released from rest as shown in figure. The tension
in the cord just after the release is.

A plastic circular disc of radius R is placed on a thin oil film, spread over a flat horizontal surface. The torque
required to spin the disc about its central vertical axis with a constant angular velocity is proportional to.

Inside a hollow uniform sphere of mass M, a uniform rod of length R 2 is released from
the state of rest. The mass of the rod is same as that of the sphere. If the inner radius of the
hollow sphere is R then find out its horizontal displacement of sphere with respect to earth
in the time in which the rod becomes horizontal.

A circular uniform hoop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also of
mass m and moving with a velocity v, strikes the hoop and gets embedded in it. The thickness of the hoop is
much smaller than R. The angular velocity with which the system rotates after the bullet strikes the hoop is.

1. (46)
SINGLE CORRECT QUESTION
1.A large sphere of radius R is moving with velocity v horizontally. A small sphere ‘B’ of mass m starts sliding
from top of the sphere downwards. Let at point C it looses contact with the large sphere. Let the velocity of
‘B’ with respect to large sphere is u, then which of the following statement follows
(A)
2 2
u v 2uvsin
gsin
R
  
  (B)
2
u
gsin
R
 

C
R
u
v
B
(C)
2 2
u v 2uvcos
gsin
R
  
  (D)
2 2
u v
gsin
R

 
2.A bead of mass ‘m’ is released from rest at A move along the fixed smooth circular
track as shown in figure. The ratio of magnitudes of centripetal force and normal reaction
by the track on the bead at any point P
0
described by the angle ‘’ (0) would
(A) increases with  (B) decreases with 
R

P
0
A
(C) remains constant (D) first increases with  then decreases.
3.A particle ‘A’ moves along a circle of radius
1
R m,
2
 so that its radius vector r

relative to the point O
rotates with the constant angular velocity  = 0.4 rad/s. The acceleration of the particle is
(A) 4
2
R
(B) 2
2
R
(C) 
2
R
R
A
B

r

0
(A) 3
2
R
4.A cyclist rides along the circumference of a circular horizontal track of radius R. The coefficient of friction
0
r
1
R
 
    
 
where 
0
is a constant, and r is the distance from the centre of the circle. The maximum
velocity of the cyclist is
(A)
0
gR (B)
0
gR
2

(C)
0
gR
2

(D)
0
g
R

5.A string is wrapped around a cylinder of radius R. If the cylinder is released from rest,
the velocity of the cylinder after it has moves h distance is
(A) 2gh (B) gh
(C)
gh
3
(D)
4gh
3
6.A point P moves along a circle of radius r with constant speed v. Its angular velocity about any fixed point on
the circle will be
(A)
v
r
(B)
v
2r
(C)
2
r
v
(D)
2
v
2r
4 CIRCULAR MOTIONclass 11 physics chapter 2 notes

2. (47)
7.Find the magnitude and direction of the force acting on a particle of mass m during its motion in the xy plane
according to the law x = a sin t and y = b cos t, where a, b and  are constants.
(A) m
2 2 2
x y, radially inwards (B) m
2
(x
2
+ y
2
), radially outwards
(C) m(x + y), tangentially (D)
2 2
x y radially inwards
8.An athlete completes one round of a circular track of radius R in 40 second. What will
be his displacement at the end of 2 minute 20 second
(A) Zero (B) 2 R
(C) 2R (D) 7R
9.In the aboveProblem, what is the ratio of the displacement to the distance, when the athlete has covered 3/
4 th of the circular track?
(A)
2 2
3
(B)
2
3
(C)
2
3
(D) None of these
10.A particle P is moving in a circle of radius ‘r’ with a uniform speed v. C is the centre of the circle and AB is a
diameter. When passing through B the angular velocity of P about A and C are in the ratio
(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 4 : 1
11.Two beads A and B of equal mass m are connected by a light inextensible cord. They are constrained to
move on a frictionless ring in vertical plane. The blocks are released from rest as shown in figure. The tension
in the cord just after the release is
(A)
mg
4
(B) 2 mg
(C)
mg
2
(D)
mg
2
12.A plastic circular disc of radius R is placed on a thin oil film, spread over a flat horizontal surface. The torque
required to spin the disc about its central vertical axis with a constant angular velocity is proportional to
(A) R
2
(B) R
3
(C) R
4
(D) R
6
13.Inside a hollow uniform sphere of mass M, a uniform rod of length R 2 is released from
the state of rest. The mass of the rod is same as that of the sphere. If the inner radius of the
hollow sphere is R then find out its horizontal displacement of sphere with respect to earth
in the time in which the rod becomes horizontal.
(A) R/2 (B) R/4 (C) R / 2 2 (D) None
14.A circular uniform hoop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also of
mass m and moving with a velocity v, strikes the hoop and gets embedded in it. The thickness of the hoop is
much smaller than R. The angular velocity with which the system rotates after the bullet strikes the hoop is
(A) V/(4R) (B) V/(3R) (C) 2V/(3R) (D) 3V/(4R)class 12th physics chapter 1 notes

3. (48)
15.A solid uniform sphere is rolling without slipping on a frictionless surface, shown in figure with a translational
velocity v m/s. If it is to climb on the inclined surface then v should be:
(A) >
10
gh
7
(B) > 2gh (C) 2gh (D)
10
7
gh
16.If the relative velocity between the pair of point A
1
B
1
, A
2
B
2
and A
3
B
3
are V
1
, V
2
and V
3
respectively then
(A) V
1
= V
2
= V
3
(B) V
1
> V
2
> V
3
(C) V
1
< V
2
< V
3
(D) Data's are insufficient to decide
17.A small bead of mass m moving with velocity v gets threaded on a stationary semicircular ring of mass
m and radius R kept on a smooth horizontal table. The ring can freely rotate about its centre. The bead
comes to rest relative to the ring. What will be the final angular velocity of the system?
(A) v/R
(B) 2v/R
(C) v/2R
(D) 3v/R
18.A monkey jumps from ball A onto ball B which are suspended from inextensible light string
each of length L. The mass of each ball & monkey is same. What should be the minimum
relative velocity of jump w.r.t. ball, if both the balls manage to complete the circle?
(A) 5gL (B) 20gL (C) 4 5gL (D) none
19.A satellite revolving in a circular equatorial orbit from west to east appears over a certain point on the equator
every 8 hours. Therefore it's period is
(A) 16 hr (B) 8 hr (C) 6 hr (D) 32 hr
20.A point moves on a circle of radius 2 meter and its speed depends on the distance covered as v = S.
Then the time taken by the particle in making the full circle is
(A) 2 sec. (B) 3 sec. (C) 4 sec. (D) none
21.Two particles tied to different light strings are whirled in a horizontal circle as shown in figure. The ratio of
length of the strings (L
2
/L
1
) so that they complete their circular path with equal time period is
(A)
3
2
(B)
2
3
(C) 1 (D) none of theseNCERT Maths class 10 exemplar

4. (49)
22.A small block of mass m is released from A inside the frictionless circular groove of radius 2 m on an inclined
plane as shown in figure. The contact force between the block and inclined plane at point B is
(A)28mg (B) 2.5 mg (C)
2
28
mg (D) 18.5 mg
23.

A body moves in a circle of radius R having centre at origin, with constant angular velocity  in the x-y plane
as shown in the figure. Another body moves parallel to y-axis with constant velocity (R/2). At time t = 0,
both particles are at (R, 0). The time t, when first body has velocity only along positive x-axis w.r.t. the
second body is
(A) /(6
(B) 5/(3
(C) 5/(6
(D) /(2
24.A small ball is attached with a string of length l which is fixed at point O on an
inclined plane. What minimum velocity should be given (at the lowest point) to
the ball along the incline so that it may complete a circle on inclined plane?
(plane is smooth and initially particle was resting on the inclined plane.)
(A) 5gl (B)
5gl
2
(C)
5 3gl
2
(D) None of these
25.The small sphere at P is given a downward velocity v
0
and swings in a vertical plane at the end of a rope of
l = 1 m attached to a support at O. The rope breaks at angle 30° from horizontal, knowing that it can
withstand a maximum tension equal to three times the weight of the sphere. Then the value of v
0
will be: (g =

2
m/s
2
)
(A) /2 m/s (B) 2/3 m/s
(C)
3
2
 m/s (D) None of these
26.Two racing cars of masses m
1
and m
2
are moving in circles of radii r
1
and r
2
respectively. Their speeds are
such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to the
second car is
(A) 1 : 1 (B) m
1
: m
2
(C) r
1
: r
2
(D) m
1
r
1
: m
2
r
2
27.A car is moving with constant acceleration ‘a’ there is a conical pendulum in the car which just touch the roof
while performing circular motion (of conical pendulum). Calculate time period of periodic motion.
(A)
L
2
g
 (B)
g


(C)
2
g


(D) None
a
Class 12 English book

5. (50)
28.A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. The centripetal
force acting on the particle is 10 N. It's kinetic energy is
(A) 0.1 J (B) 0.2 J (C) 2.0 J (D) 1.0 J
29.A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components
of its acceleration are equal. If its speed at t = 0 is
0
v the time taken to complete the first revolution is
(A)
0
R
v
(B)
0
v
R
(C)
2
0
R
(1 e )
v
 

(D)
2
0
R
(e )
v
 
.
30.A spool is being applied by a horizontal force F as shown in the figure. The spool is placed on rough
horizontal surface. The direction of frictional force acting on spool is
(A) Forward direction
(B) Zero if R = 2r and I = 2 mr
2
(C) Backward direction if R = r and   2
F
R
r

(D) Data is insufficient
31.The bar shown in the figure is made of a single piece of material. It is fixed at one end. It consists of two
segments of equal length
2
L
0
but different cross-sectional area A and 2A. What is the change in length of the
entire system under the action of an axial force F. Consider the shape of joint to remain circular.
(A)
Ay4
FL3
(B)
Ay8
FL3
(C)
Ay2
FL3
(D) None of these
L/2 L/2
A
F
A=area of cross section
2A
32.A particle of mass m and charge q is attached to a light rod of length L. The rod can
rotate freely in the plane of paper about the other end, which is hinged at P. The entire
assembly lies in a uniform electric field E also acting in the plane of paper as shown.
The rod is released from rest when it makes an angle  with the electric field direction.
Determine the speed of the particle when the rod is parallel to the electric field.

E
m,q
L

P
(A)
2/1
m
)cos1(qEL2





 
(B)
2/1
m
)sin1(qEL2





 
(C)
2/1
m2
)cos1(qEL





 
(D) None of these
33.A large insulating thick sheet of thickness 2d carries a uniform charge per unit volume . A particle of mass m,
carrying a charge q having a sign opposite to that of the sheet, is released from the surface of the sheet. The
sheet does not offer any mechanical resistance to the motion of the particle. Find the oscillation frequency 
of the particle inside the sheet.
(A)  =
2
1
0
m
q


(B)  =
3
1
0
m
q


(C)  =
4
1
0
m
q


(D) None of these
34.Figure shows a radar screen, with the dots denoting respective positions of Indian SUKHOI. A, B, C and
PAKI F-16-E. All are flying with constant velocity in horizontal plane. SUKHOI-A reports to ground control
that E is moving due north with velocity 160 m/s. At same time SUKHOI-B reports that PAKI F-16-E is
moving due east at 120 m/s. Through what minimum angle SUKHOI-C (originally moving in north east) turns

6. (51)
so that it is alligned in direction of motion of E.
(A) 8° clockwise
(B) 8° anticlockwise
(C) 16° clockwise
N
S
EW
y
x
45°
B
CA
E
Radar Screen
Plane Direction
SUKHOI A East
SUKHOI B North
SUKHOI C North East
PAKI F-16 E Unknown direction
(D) 16° anticlockwise
35.A motor car is travelling at 60 m/s on a circular road of radius 1200m. It increases its speed at the rate of 4
m/s
2
. The acceleration of the car is (at initial time)
(A) 3 m/s
2
(B) 4 m/s
2
(C) 5m/s
2
(D) 6 m/s
2
(E) 7m/s
2
36.A particle of mass m is attached to one end of string of length  while the other end is fixed to a point height
h (h < ) above the smooth horizontal table. The particle is made to revolve in a circle on the table so as to
make n revolutions per second. The value of n if the particle is in contact with the table will be-
(A)
h
g
2
1

(B)

g
2
1

(C)
h2
g
2
1

(D) None of these
37.A particle is moving in the vertical plane. It is attached at one end of a string of length and whose other end is
fixed. The velocity at lowest point is u. The tension in string is
T

and velocity of particle is v

at any position.
Then which of the following quantity will remains constant.
(A) Kinetic energy(B) Gravitational P.E.(C) T

·v

(D) T

×v

ASSERTION AND REASON
38.Statement-1 : A particle moving at constant speed and constant magnitude of radial acceleration must
be undergoing uniform circular motion.
Statement-2 : In uniform circular motion speed cannot change as their is no tangential acceleration.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
39.Abhi and John are in a rotor at rest relative to wall to rotor.
Statement-1 : Centrifugal force on Abhi in reference frame of John mr
2
radially onward as shown in figure.
Statement-2 : Angular velocity of Abhi with respect to John is same as angular velocity of Abhi with respect
to axis of rotation.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for
statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation
for statement-1.
(C) Statement-1 is true, statement-2 is false.
JOHNABHI
r
(D) Statement-1 is false, statement-2 is true.

7. (52)
COMPREHENSION
Passage for Q. no. (40 to 41)
A child is swinging a toy airplane on a string. The airplane moving with constant speed describes a horizontal
(in X-Y plane) circle of radius 2 m and takes 2 s to complete one orbit. At t = 0, the airplane is directly
infront of the child and moving in the +Y direction. The child is facing in the positive X-direction. The
approximate plot of X-components of
40.velocity of the airplane as a function of time over one complete circle is
(A) (B)
(C) (D)
41.centripetal acceleration of the airplane as a function of time over one complete circle is
(A) (B)
(C) (D)
Passage for Q. no. (42 to 43)
A uniform bar AB of mass m and length L is placed horizontally at rest on a smooth table.
Another uniform bar PQ of mass 2m but of same length L is moving horizontally with
velocity V on same table. When bar PQ reaches near bar AB the end P is attached to end
B and combined rod moves forward and also rotates.
42.The angular velocity of composite rod will be
(A)  =
4v
11L
(B) 
11v
4L
(C) 
8v
11L
(D)  =
v
L
43.The velocity of end A just after the rod joined together
(A)
2
v
11
 (B)
2v
3
 (C)
28
v
33
 (D) none of these

8. (53)
Passage for Q. no. (44 to 46)
Two beads of mass 2m and m, connected by a rod of length  and of negligible mass are free to move
in a smooth vertical circular wire frame of radius  as shown. Initially the system is held in horizontal
position (Refer figure)

2m m
44.The velocity that should be given to mass 2m (when rod is in horizontal position) in counter-clockwise
direction so that the rod just becomes vertical is :
(A)
5g
3

(B)
3 3 1
g
3
 
 
 
 (C)
3
g
2
 (D) None of these
45.If the rod is replaced by a massless string of length and the system is released when the string is
horizontal then :
(A) Mass 2m will arrive earlier at the bottom.
(B) Mass m will arrive earlier at the bottom.
(C) Both the masses will arrive together but with different speeds.
(D) Both the masses will arrive together with same speeds.
46.The string is now replaced by a spring of spring constant k and natural length . Mass 2m is fixed at the
bottom of the frame. The mass m which has the other end ofthe spring attached to it is brought near the
mass 2m and released as shown in figure. The maximum angle  that the spring will substend at the centre
will be : (Take k = 10 N/m,  = 1 m, m = 1 kg and  = r)
2m m
fixed
attached
to spring
2m
m
fixed

(A) 60° (B) 30° (C) 90° (D) None of these
MULTIPLE CORRECT QUESTION
47.A car is of mass m moving along a circular track of radius r with a speed which increases linearly with time t
as v = kt, where k is a constant. Then
(A) the instantaneous power delivered by the centripetal force is mk
3
t
3
/r.
(B) the power delivered by the centripetal force is zero.
(C) the instantaneous power delivered by the tangential force is mk
2
t.
(D) the power delivered by the tangential force is zero.
48.An object follows a curved path. The following quantities may remain constant during the motion.
(A) speed (B) velocity
(C) acceleration (D) magnitude of acceleration

9. (54)
49.A point object P of mass m is slipping down on a smooth hemispherical body of mass M & radius R. The
point object is tied to a wall with light inextensible string as shown. At a certain instant the speed of hemisphere
is V & its acceleration a (as shown in figure). Then speed V
p
& acceleration a
p
of the particle has value
(neglect friction)
(A) V
p
= V sin 30 (B) V
p
= V
(C) a
p
= a (D) a
p
=
1/ 2
2
22
V a 3 a
R 2 2
 
    
     
    
 
50.A smooth track in the form of a quarter circle of radius 6 m lies in the vertical plane. A particle moves from P
1
to P
2
under the action of forces
1 2
F , F
 
and
3
F

. Force
1
F

is always toward P
2
and is always 20 N in magnitude.
Force
2
F

always acts horizontally and is always 30 N in magnitude. Force
3
F

always acts tangentially to the
track and is of magnitude 15 N. Select the correct alternative(s)
(A) work done by
1
F

is 120 J
(B) work done by
2
F

is 180 J
(C) work done by
3
F

is 45 
(D)
1
F

is conservative in nature
51.Little Jai is sitting on a seat of merry-go-round moving with constant angular
velocity. At t = 0, Jai is at position A shown in figure.
Top view of
merry-go-round
x
A
v at t = 0
y
Which of the graphs shown in figure are correct.
(A)
time
F
y
F
y
is the y-component of the force keeping Jai moving in a circle.
(B) time
x
x is the x component of Jai’s position.
(C)
time

 is the angle that Jai’s position vector makes with x-axis.
(D) time
v
x
v
x
is the x component of Jai’s velocity.

10. (55)
52.A small sphere of mass m is connected by a string to a nail at O and moves in a circle of radius r on the smooth
plane inclined at an angle  with the horizontal. If the sphere has a velocity u at the top position A. Mark the
correct options.
(A) Minimum velocity at A so that string does not get slack instantaneously is gr
5
3
.
(B) Tension at B if sphere has required velocity in option A is
5
11
mg.
(C) Tension at C in situation of option B is
5
23
mg
C
B
A
O
90°
=37°
u
(D) None of these
53.A solid sphere is given a angular velocity  and kept on a rough fixed incline plane. The choose the correct
statement.
/ / / / / / / / / / / / / / / / / / /


o
(A) If  = tan  then sphere will be in linear equilibrium for some time and after that pure rolling down the
plane will start.
(B) If  = tan  then sphere will move up the plane and frictional force acting all the time will be 2 mg sin.
(C) If m =
2
tan
there will be never pure rolling (consider inclined plane to be long enough.)
(D) If incline plane is not fixed and it is on smooth horizontal surface then linear momentum of the system
(wedge to sphere) can be conserved in horizontal.
54.In which of the following examples of motion, can the body be considered approximately a point object:
(A) A railway carriage moving without jerks between two stations
(B) A sparrow sitting on top of a man cycling smoothly on a circular track
(C) A spinning cricket ball that turns sharply on hitting the ground
(D) A tumbling beaker that has slipped off the edge of a table
55.As shown in figure BEF is a fixed vertical circular tube. A block of mass m starts moving in the tube at point
B with velocity V towards E. It is just able to complete the vertical circle, then
(A) velocity at B must be Rg3.
(B) velocity at F must be Rg2.
C
R
F
E
60°
B
(C) Normal reaction at point F is 2mg.
(D) The normal reaction at point E is 6 mg.

11. (56)
MATCH THE COLUMN
56.A narrow tube exists in a vertical plane. A small sphere is given a horizontal velocity u at its lowest
position (Refer diagram). Match the values of velocity u in Column-I to the properties of upward motion
indicated in Column-II.
Inner Surface
R
u
Outer Surface
Column-I Column-II
(A) u 2gR (p) Sphere is always in contact with inner
surface of tube during its upward motion.
(B) u 3gR (q) Sphere is always in contact with outer
surface of tube during its upward motion.
(C) u 4gR (r) Sphere is in contact with outer surface of
tube first then with inner surface during its
upward motion.
(D) u 5gR (s) Sphere is in contact with inner surface of
tube first then with outer surface during its
upward motion.
57.A particle of mass ‘m’ moves in a circular path. The particle starts from rest at t = 0. Speed of the particle
is given as v = kt
2
where ‘k’ is a constant. Now match the following:
List-I List-II
(A). Acceleration of particle at time ‘t’ (P) 2mk
2
t
3
(B) Work done by resultant force from (Q)
2 31
mk t
2
t – 0 time ‘t’
(C) Instantaneous power at time (R)
2
2
kt
kt 4
R
 

 
 
‘t’ (due to resultant force)
(D) Average power upto time ‘t’ (due to (S)
2 41
mk t
2
resultant force)

12. (57)
SUBJECTIVE QUESTION
58.Prove that a motor car moving over a convex bridge is lighter than the same car resting on the same bridge.
59.A small block of mass 2m initially rests at the bottom of a circular, vertical track, which
has a radius of R. The contact surface between the mass and the loop is frictionless. A
bullet of mass m strikes the block horizontally with initial speed v
0
and remain embedded
in the block as the block and the bullet circle the loop. Determine each of the following
in terms of m, v
0
, R and g.
(a) The speed of the masses immediately after the impact.
(b) The minimum initial speed of the bullet if the block and the bullet are successfully to execute a complete
ride on the loop.
60.A particle moves along a circle of constant radius with radial acceleration changing with time as a
r
= k t
n
where k is constant and n > 1. How does the power developed by the net force on the particle vary with
time?
61.An annular wheel (M.I. = 32 kgm
2
) hinged at its centre is rotating with initial
angular velocity 10 rad/s in anticlockwise direction. If the inner radius is 5 cm, the
outer radius is 20 cm and the wheel is acted upon by the constant forces shown in
the figure, then what will be the angular velocity of the wheel after 10 sec. (Assume
that the lever arm of all forces about centre remains constant)
62.A hollow sphere of radius R = 0.5m rotates about a vertical axis through its centre with an
angular velocity of  = 5 rad/s. Inside the sphere a small block is moving together with the
sphere at height of R/2 (see figure) (g=10m/s
2
). What will be the least coefficient of
friction for fulfill this condition?
63.A particle is revolving with a constant angular acceleration  in a circular path of radius r. Find the time when
the centripetal acceleration will be numerically equal to the tangential acceleration.
64.A particle is moving along a vertical circle of radius r = 20 m with a constant vertical
circle of radius r = 20 m with a constant speed v = 31.4 m/s as shown in figure. Straight
line ABC is horizontal and passes through the centre of the circle. A shell is fired from
point A at the instant when the particle is at C.
If distance AB is 20 3 m and the shell collide with the particle at B, then prove
2
(2n 1)
tan
3

  where  is
the angle of projection of particle and n is an integer. Further, show that smallest value of  is 30°.

13. (58)
65.A rod of length R and mass M is free to rotate about a horizontal axis passing through hinge P as shown in the
figure. First it is taken aside such that it becomes horizontal and then released.
At the lowest point the rod hits the block B of mass m and stops. If
mass of rod is 75 kg, find mass of the block if it just complete the
circle.
R
m
2
P
m
1
66.Two particles A and B are moving in a horizontal plane anticlockwise on two different concentric circles with
different constant angular velocities 2 and  respectively. Find the relative velocity of B w.r.t. A after time
t = /. (Take  = 3rad/sec, r = 2m ] (Both are moving in same sense)
2r
r
A B
X
Y
at t= 0
67.A flexible drive belt runs over a frictionless flywheel (see Figure). The mass per unit length of the drive belt
is 1 kg/m, and the tension in the drive belt is 10N. The speed of the drive belt is 2m/s. The whole system is
located on a horizontal plane. Find the normal force (in N) exerted by the belt on the flywheel.
T
T
v

14. SOLUTION [CIRCULAR MOTION]
(59)
1. (B) Centripetal acceleration is taken in frame of reference of
sphere because circular motion is in this frame
2. (C) F
centripetal
= N - mg sin 
 N = F
centripetal
+ mg sin 
2
mV
mgsin
R
  
From conservation of energy
mg R sin  =
1
2
mV
2

2
mV
2 mg sin
R
 
3. (A) Suppose the body moves from A to B in time t. Then radius
vector describes the angle . Since the angle subtended at the
centre of the circle is twice that at the circumference. Hence the
angular velocity of the body is 2. Linear velocity = 2R
Acceleration
2 2 2
2v 4R
4 R
R R

   
Put the values.
4. (B)
2
0
mv r
mg mg 1
r R
 
    
 
 
2
2
0
r
y v g r
R
 
    
 
For y to be maximum,
dy 2r
0 1
dr R
  
R
r
2

2
2
max 0 0
R R R
v g g
2 4R 4
 
     
 
0
max
gR
v
2


5. (D)
6. (B) Let O be the centre of circle and P the position of the point
at any time and A the point about which angular velocity is to be
found.
N
O
P
T
A
Draw PN perpendicular to AP.
The velocity v at P is along PT,
the tangent at P.
Also if OAP = OPAs = ,
TPN =  and therefore the resolved part of v along PN = v cos
.

Resolved part of v along PN
Angular velocity about A
AP

v cos v
2r cos 2r

 

CIRCULAR MOTION SOLUTION
7. (A) Given x = a sin t
y = b cos t
Radius vector ˆ ˆ
r xi yj 

, = a sin tˆ
i
+ b cos tˆ
j
Velocity vector
dr
v
dt



,
d
ˆ ˆ
(a sin i b cos t j)
dt
   
= a cos t ˆ
i
– b sin tˆ
j
Acceleration vector
dv
a
dt



= –a
2
sin tˆ
i
– b
2
cos tˆ
j
= – 
2
(a sin tˆ
i
+ b cos tˆ
j,
2
r 

Forces
2
F m a m r    
 
The magnitude of force
2
F ( m r)  

,
2 2 2 2
F m r m x y    
The direction of force is radially inwards.
8. (B)
9. (A)
10.(B) Angular velocity of particle P about point A,
A
AB
T 2r
 
  

A
2r
r
v
B C
P
Angular velocity of particle
P about point C,
c
BC
T r
 
  
Ratio
A
c
/ 2r 1
/ r 2

 
 

15. SOLUTION [CIRCULAR MOTION]
(60)
11.(D) Just after the release B moves downwards and A moves
horizontally leftwards with the same acceleration say a. Drawing
free body diagram of both A and B :


Tcos45 ma
or T 2ma ....(1)
 


mg Tcos45 ma
or mg ma ma
g
or a ... 2
2
  
 

Substituting this in (1) we get
mg
T
2

12.(C)
Consider an element of radius r and thickness dr.
Thickness of oil film is h.
 dA = 2r dr, v = r
 dF =
2v 2
(dA) r dr
h h

 
 dT = (dF)r =
32
r dr
h

 T =
4
R
2h

or
4
T R
13.(B) x = displacement of rod w.r.t. sphere
R 2 R
22 2
 
y = displacement of sphere w.r.t. ground
My = M(x – y) 
x R
y
2 4
 
14.(B) LMC
mV = mV
0
+ m(V
0
+ R)  V = 2V
0
+ 2
AMC mVR = m(V
0
+ R)R + mR
2

 V = V
0
+ 2 R
solving  = V/3 R
15.(A)
16.(A) So relative velocity in every case is of magnitude 2V
 V
1
= V
2
= V
3
= 2V
17.(C) P
i
= mv  At final position, both the bead and ring are
rotating about axis through O and | to the plane.
I
m
= mR
2
, I
ring
= mR
2
 I = I
m
+ I
ring
= 2mR
2
 L
final
= I = 2mR
2

V'
R
 
 
 
= 2mRV'
 P
f
=
final
L
R
= 2mV'
P
f
= P
i
 2mV' = mV  V' =
V
2
=
V'
R
=
V
2R
18.(C)
19.(C)
20.(C) v = S 
dv 1 dS
dt dt2 S
 
= · S
2 S


=
2
2

= a (tangential acceleration)
If required time is t,2R =
21
at
2
solving we get t = 4 sec.
21.(A) Tsin = m
2
L sin
Tcos = mg
& hence, for constant ''
(angular speed), L 
1
cos

2 1
1 2
L cos
L cos



=
3
· 2
2
=
3
2
22.(C) N – mgsin =
2
mv
R
2
1
mv
2
= mgH 
2
mv
R
=
2mgH
R
= 2mg
 N =
mg
2
+ 2mg =
5mg
2

16. SOLUTION [CIRCULAR MOTION]
(61)
Contact force = mg
2
25 3
4 2
 
 
 
 
=
28
2
mg
23.(B) 1, g
R
ˆ
V i
2



R cos =
R
2

 = 60º

Rotated
= 360º – 60º
= 300º or
5
3

; t =
5
3


24.(B) for complete circular motion
V
min
= 5gL
here g
eff
= g sin 30º =
g
2
 V
min
=
5
gL
2
25.(C)
2
1
mv mg
1 2
 = 3mg
v
1
=
5
g
2
now
1
2
mV
0
2
+ mg
1
2
=
1
2
mV
1
2
V
0
=
3g
2
=
3
2

26.(A) 
1
= 
2
= 
 
1
= 
2
= 1
27. a = g,
eff
g g 2
45º
ma
equb.
mg
So geometry is like
T 
2
m
2

mg2

45º
g2

2
m
T cos 45
2

 

Tsin 45 mg 2 
2
m 2g 2g
1
2mg

      

 
2
T 2
2g g
   
 
28.(D)
2
mv
10
r
 
21 r
mv 10 1J
2 2
  
29.(C) 30. (B) 31. (A)32. (A)
33.(A) 34. (B) 35. (C)
36.(A) Tcos = mg
h

Tsin = m
2
 sin
n =
1 g
2 h
37.(C) T · V 0
 
because T is always  to v.
Assertion and Reason
38.(B) 39. (C)
Passsage
40.(D) r = 2m
 =
2
2


= 1 rad/sec.
 = t
V
x
= –Vsint
= –Vsin t
41.(B) a
x
= –
2
V
cos t
r
 = –
2
V
cos t
r
42.(C)
cm
L 3L
2m m
2 2
y
3m
 
 
 
 

=
5
6
L
From linear momentum conservation
(2m) V = (3m) V
cm
 V
cm
=
2
3
V ..............(1)
From angular momentum conservation (about 0)
L
3
 
 
 
2mV = I
cm

cm
=
2 2
2 2
cm
mL 2 L 2
m L 2m 2m L
12 3 12 3
 
   
        
     

cm
=
8 V
11 L
43.(A) Velocity of 
A

A
= 
cm
– 
cm
7
L
6



17. SOLUTION [CIRCULAR MOTION]
(62)
=
2
3
 –
8
11 L
 
 
 
7
L
6
;
=
2 28
3 33
 

 
 
 


A
=
2
11 


44. (B) The speeds given to 2m will also be possessed by m
 KE in horizontal position gets converted in PE in vertical
position.
1
2
2mv
2
+
1
2
mv
2
= change in PE in vertical position.
 PE = 2 mg [ cos

30° –  cos

60°] + mg [ cos

30° +
2

]
2 mg
3 3
mg
2 2 2 2
   
     
      
   
2m
m

3 1
mg [ 3 1] mg
2
 
   
  
 
=
3 1
mg 3 1
2 2
 
   
  

=
3 3 1
mg
2 2
 
 
  

K.E. =
1
2
3mv
2
= mg
3 3 1
2
 
 
  
 v =
3 3 1
g
2
 
 
 
 

45.(D) Both the masses will have same acceleration all the
time.
 Their velocities and distance covered will be same.
46.(A) Length of spring at maximum = 2 cos
 Extension is x = (2 cos – )
Now initial potential energy of the spring is converted into
final PE of spring and gravitational PE.

1
2
k
2
=
1
2
k (2 cos – )
2
+ mg ( – cos )
Putting values
1
2
× 10 × 1
2
=
1
2
× 10 (2 cos – 1)
2
+ 10 (1 – cos

)
5 = 5 (2 cos

 – 1)
2
+ 10 – 10 cos
2 cos
m q
1 = (2 cos

 – 1)
2
+ 2 – 2 cos
2 cos

 – 1 = (2 cos – 1)
2
 cos =
1
2
  = 60°
Multiple Choice Questions
47.(B, C)
48.(A, C, D)
49.(B, D) Velocity of point object w.r.t. hemispherical body V '

will be same in magnitude to that of weight of hemispherical
body. Hence, for total velocity
pV

which is the resultant of V

& V'

we have V
p
= 
2 2 2
V V 2V cos 120  = V
acceleration of p = acc. of p w.r.t.
hemispherical body + acc. of hemispherical body
=
t'p
a

+
n'p
a

+ a

where
'p
a

is acceleration of 'p' w.r.t. hemispherical body &
t'p
a

&
n'p
a

the tangential & normal component in
corresponding circular motion.
Here
t'p
a

= a
&
n'p
a

=
2
pV
R
Hence a
p
=  
2
2
20P
V
a cos30 a a cos 60
R
 
   
 
=
2
2
2
V a 3 a
R 2 2
   
     
  
50.(B, C, D)
2F
w = F
2
2
OP = 30 · (6) = 180 J
3F
w =
2
1
P
3
P
Along the path
F dx

=
2 6
4
3
0
F dx


= 15 · (3) = 45J
1F
w =
2
1
P
11
P
F dx

1F
w =
/ 2
0
20 cos Rd
4 2

  
 
 
 

1F
w = 20 (–2) R ·
/ 2
0
sin
4 2

 
  
 
 
  
 
1F
w = (–40) · 6 0 sin
4
 

 
 

18. SOLUTION [CIRCULAR MOTION]
(63)
1F
w = 120 2 J
Work done by F
1
is independent of path. So it is conservative in
nature
51.(A,C) 52. (A)
53.(A,D) 54. (A,B)
55.(A,B,C)
2
E
V = V
2
+ 2g(R – R cos 60°)
4gr = V
2
+ gR
V = 3gR
2
B
V =
2
E
V – 2gR = 2gR
N =
2
P
mV
R
= 2mg
N
E
mg =
2
E
mV
R
 N
E
= 5mg]
56.(A-q; B-r; C-r; D-q)
57.(A-r; B-s; C-p; D-q)
58.The motion of the motor car over a convex bridge AB is the
motion along the segment of a circle AB (Figure;
The centripetal force is provided by the difference of weight mg
of the car and the normal reaction R of the bridge.
mg – R =
2
mv
r
or R = mg –
2
mv
r
Clearly R < mg, i.e., the weight of the moving car is less than the
weight of the stationary car.
59.(a) v
0
/3, (b) 35gR
(a) Conserving momentum in horizontal direction
mv
0
= (2m + m)v, v =
0
v
3
Collision is perfectly inelastic
(b) Now 3 m mass will move in the circle and 3 m mass requires
5Rgminimum velocity at bottom most point to execute the
loop so
V =
0
V
3
= 5Rg, then v
0
= 3 5Rg
60. t
n–1
Given radial acceleration
a
r
= kt
n
=
2
v
R
 v
2
= Rkt
n
 v = Rk t
n/2

dv
dt
=
n
Rk
2

n
1
2
t
 
 
 
 Tangential force, F
t
= m
dv
dt
= m
n
2
Rk
n
1
2
t
 
 
 
 Power developed = v·F

=  r t t
F F ·v F v 
 
= m
n
2
Rk
n
1
2
t
 
 
  ·  
n / 2
Rk t =
n 1mnRk
t
2

 P  t
n–1
61.
net
19(0.2) – 12(0.05)
= 3.8 – 0.6
= 3.2 N mt = 32 
 = 0.1 rad/s
2
 = 
0
+ t
 = 10 + (0.1) (10)
= 11 rad/s
62.FBD of block (Ref. sphere)
here for  = 5 rad/s
mg sin60 > F
p
cos60
0
 block will have
tendency to move
down along inclined
 friction will be directed upward
x-axis Nsin60 – Ncos60 = m
2
Rsin60
.............(1)
y-axis Ncos60 – Nsin60 = mg
.............(2)
From (1) and (2)
 =
3 3
23
63.Le the speed of the particle after time t from starting be v
 The centripetal acceleration
2
2
r
v
a r
r
   & the corresponding angular speed t .

2 2 2
r
a r( t) r t    ...(i)
We known that the tangential acceleration ra
t
...(ii)
Since,
tr
aa (given)

2 2
r t r   
1
t

64.As at the time of firing of the shell, the particle was at C and the
shell collides with it at B, therefore the number of the revolutions
completed by the particle is odd multiple of half i.e., (2n – 1)/2,
where n is an integer.
Let T be the time period of the particle, then
2 r 2 3.14 20
T 4second
v 31.4
  
  
If t be the time of the flight of the shell, then
t = time of [(2n – 1)/2] revolutions of the particle

19. SOLUTION [CIRCULAR MOTION]
(64)
(2n 1)
4 2(2n 1)
2

    second
for a projectile, the time of flight is given by
2u sin
t
g


Hence,
2u sin
2(2n 1)
g

  ...(i)
The range of the projectile is given by
2
u sin 2
R
g


Hence,
2
u sin 2
20 3
g

 ...(ii)
From equation (i) and (ii)
2
(2n 1)
tan
3

 
For  to be smallest, n = 1, so
2
(2n 1)
tan
3

 
65.(0015) Let the angular velocity of rod at the time of collision be
w
According to the law of conservation of energy
For the rod at the horizontal and vertical positions, we get
m
2
gR =
2
R
gmI
2
1 22

2
gRm
2
=
2
2
2
·
3
Rm
2
1
 =  =
R
g3
Applying the law of conservation of angular momentum about
P
Let the angular speed of block about P after the collision be 
0
.
I = m
1
R
2

0

3
Rm
2
2
= m
1
R
2

0

0
=
R
g3
m3
m
1
2
Linear velocity of ball is
v
0
= 
0
R  v
0
=
gR3
m3
m
1
2
For ball to complete the circle
v
0
= gR5 =
gR3
m3
m
1
2

1
2
m
m
= 15 ]
66.(0024 )
A2 r
2 r
v = 4 r
rel
67.(0012)
T
T
N
2T – N = m
2
R
cm
= µ × R ×
2
R
V






×

R2
2T – N = 2µv
2
N = 2T – 2µv
2
= 2 × 10 – 2 × 1 × 4 = 12 N]