Chapter 5 Motion in a Plane (two dimension) Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 5 Motion in a Plane (two dimension) Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes
Circular motion is another example of motion in two dimensions. To create circular motion in a body it must be given some initial velocity and a force must then act on the body which is always
directed at right angles to instantaneous velocity.
Since this force is always at right angles to the displacement due to the initial velocity therefore no work is done by the force on the particle. Hence, its kinetic energy and thus speed is unaffected. But due to simultaneous action of the force and the velocity the particle follows resultant path, which in this case is a circle. Circular motion can be classified into two types – Uniform circular motion and nonuniform circular motion.
Variables of Circular Motion.
 Displacement and distance : When particle moves in a circular path describing an angle q during time
t (as shown in the figure) from the position A to the position B, we see that the magnitude of the position vector r




(that is equal to the radius of the circle) remains constant. i.e., r r and the direction of the position vector changes from time to time.
(i) Displacement : The change of position vector or the displacement Dr
position B is given by referring the figure.
D r = r2 – r1
of the particle from position A to the

r r
= D = – r
Dr =
Putting r
r r2
= r = r
1
we obtain
1 2
Dr =
Þ Dr = =
q
D r = 2r sin 2
(ii) Distance : The distanced covered by the particle during the time t is given as
d = length of the arc AB = r q
(iii) Ratio of distance and displacement :
d = rq
= q cosec (q / 2)
Dr 2r sinq / 2 2
Problem 80. A particle is rotating in a circle of radius r. The distance traversed by it in completing half circle would be
 r (b) pr
(c)
2pr
(d) Zero
Solution : (b) Distance travelled by particle = Semicircumference = pr.
Problem 81. An athlete completes one round of a circular track of radius 10 m in 40 sec. The distance covered by him in 2 min 20 sec is [Kerala PMT 2002]
(a) 70 m (b) 140 m (c) 110 m (d) 220 m
Solution : (d)
No. of revolution (n) = Total time of motion = 140 sec = 3.5
Time period 40 sec
Distance covered by an athlete in revolution = n(2p r) = 3.5 (2p r) = 3.5 ´ 2 ´ 22 ´ 10 = 220 m.
7
Problem 82. A wheel covers a distance of 9.5 km in 2000 revolutions. The diameter of the wheel is [RPMT 1999; BHU 2000]
(a) 15 m (b) 7.5 m (c) 1.5 m (d) 7.5 m
Solution : (c) Distance = n(2p r) Þ
9.5 ´ 10^{3} = 2000 ´ (pD)
Þ D = 9.5 ´ 10^{3}
2000 ´ p
= 1.5 m.
 Angular displacement (q) : The angle turned by a body moving on a circle from some reference line is called angular
 Dimension = [M^{0}L^{0}T^{0}] (as q = arc / radius).
(ii) Units = Radian or Degree. It is some times also specified in terms of fraction or multiple of revolution.
 2p rad = 360^{o} = 1Revolution
(iv) Angular displacement is a axial vector quantity.
Its direction depends upon the sense of rotation of the object and can be given by Right Hand Rule; which states that if the curvature of the fingers of right hand represents the sense of rotation of
the object, then the thumb, held perpendicular to the curvature of the fingers, represents the direction of angular displacement vector.
 Relation between linear displacement and angular displacement s = q ´ r
or s = rq
Problem 83. A flywheel rotates at a constant speed of 3000 rpm. The angle described by the shaft in radian in one second is
(a) 2 p (b) 30 p (c) 100 p (d) 3000 p
Solution : (c) Angular speed = 3000 rpm = 50 rps = 50 ´ 2p rad / sec = 100p rad / sec
i.e. angle described by the shaft in one second is 100p rad.
Problem 84. A particle completes 1.5 revolutions in a circular path of radius 2 cm. The angular displacement of the particle will be – (in radian)
 6p
 3p
 2p
(d) p
Solution : (b) 1 revolution mean the angular displacement of 2p rad \1.5 revolution means 1.5 ´ 2p = 3p rad .
(3) Angular velocity (w) : Angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement.
 Angular velocity w = angletraced =
Lt Dq = dq
time taken
Dt®0 Dt dt
\ w = dq
dt
(ii) Dimension : [M^{0}L^{0}T^{–1}]
 Units : Radians per second (s^{–1}) or Degree per second.
(iv) Angular velocity is an axial vector.
 Relation between angular velocity and linear velocity v = w ´ r
Its direction is the same as that of Dq. For anticlockwise rotation of the point object on the circular path, the direction of w, according to Right hand rule is along the axis of circular path directed upwards. For clockwise rotation of the point object on the circular path, the direction of w is along the axis of circular path directed downwards.
Note : @It is important to note that nothing actually moves in the direction of the angular velocity vector w . The direction of w simply represents that the rotational motion is taking place in a plane perpendicular to it.
(vi) For uniform circular motion w remains constant where as for nonuniform motion w varies with respect to time.
Problem 85. A scooter is going round a circular road of radius 100 m at a speed of 10 m/s. The angular speed of the scooter will be [Pb. PMT 2002]
 01 rad/s (b) 0.1 rad/s (c) 1 rad/s (d) 10 rad/s
Solution : (b)
w = v = 10 = 0.1rad / sec
r 100
Problem 86. The ratio of angular velocity of rotation of minute hand of a clock with the angular velocity of rotation of the earth about its own axis is
(a) 12 (b) 6 (c) 24 (d) None of these
Solution : (c)
w = 2p rad
and w
= 2p Rad = 2p rad
\ w Minute hand
= 24 : 1
Minute hand
60 min
Earth
24 hr
24 ´ 60 min
w Earth

Problem 87. A particle P is moving in a circle of radius ‘a’ 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 [NCERT 1982] (a) 1 : 1 (b) 1 : 2 (c) 2 : 1
Solution : (b) Angular velocity of P about A
w A = v
2a
Angular velocity of P about C
w _{C} = v
a
\ w _{A}
w _{C}
= 1 : 2
Problem 88. The length of the seconds hand of a watch is 10 mm. What is the change in the angular speed of the watch after 15 seconds
 Zero (b)
(10p / 2) mms ^{–}^{1}
(c)
(20 / p ) mms^{–}^{1}
(d) 10
mms ^{–}^{1}
Solution : (a) Angular speed of seconds hand of watch is constant and equal to angular speed will be zero.
2p rad
60 sec
= p rad / sec . So change in
30
(4) Change in velocity : We want to know the magnitude and direction of the change in velocity of the particle which is performing uniform circular motion as it moves from A to B during time t as shown in figure. The change in velocity vector is given as
Dv = v2 – v1
r r r
or D = –
Þ Dv =
v v2 v1
For uniform circular motion v1 = v2 = v
q
So Dv =
= 2v sin 2
The direction of Dv is shown in figure that can be given as
f = 180^{o} – q
2
= (90^{o} – q / 2)
Note : @Relation between linear velocity and angular velocity.
In vector form v = w ´ r
Problem 89. If a particle moves in a circle describing equal angles in equal times, its velocity vector
[CPMT 1972, 74; JIPMER 1997]
(a) Remains constant (b) Changes in magnitude
(c) Changes in direction (d) Changes both in magnitude and direction
Solution : (c) In uniform circular motion velocity vector changes in direction but its magnitude always remains constant.
v1 =v2 =v3 =v4 = constant
Problem 90. A body is whirled in a horizontal circle of radius 20 cm. It has angular velocity of 10 rad/s. What is its linear velocity at any point on circular path [CBSE PMT 1996; JIPMER 2000]
(a) 10 m/s (b) 2 m/s (c) 20 m/s (d) m/s
Solution : (b) v = r w = 0.2 ´ 10 = 2 m/s
Problem 91. The linear velocity of a point on the equator is nearly (radius of the earth is 6400 km)
(a) 800 km/hr (b) 1600 km/hr (c) 3200 km/hr (d) 6400 km/hr
Solution : (b)
v = r w = 6400 km ´ 2p rad
= 1675 km / hr = 1600 km / hr
24 hr
Problem 92. A particle moves along a circle with a uniform speed v. After it has made an angle of 60^{o} its speed will be
(a)
v (b)
v (c)
2
v (d) v
3
[RPMT 1998]
Solution : (d) Uniform speed means speed of the particle remains always constant.
Problem 93. A particle is moving along a circular path of radius 2 m and with uniform speed of 5 ms^{–1}. What will be the change in velocity when the particle completes half of the revolution
 Zero (b) 10 ms^{–1} (c)
10 2 ms ^{–}^{1}
(d)
10 /
2 ms ^{–}^{1}
æ q ö æ 180 ^{o} ö
Solution : (b)
Dv = 2v sinç ÷ = 2 ´ 5 sinç ÷


è 2 ø ç ÷
= 2 ´ 5 sin 90 ^{o} = 10 m / s

Problem 94. What is the value of linear velocity, if wr = 3ˆi – 4ˆj + kˆ and r = 5ˆi – 6ˆj + 6kˆ
[Pb. PMT 2000]
(a) 6iˆ+ 2ˆj – 3kˆ
ˆi Jˆ
kˆ
18ˆi + 13ˆj – 2kˆ (c) 4ˆi – 13ˆj + 6kˆ
(d)
6ˆi – 2ˆj + 8kˆ
Solution : (b)
r r r


v = r = 3
5
r
– 4 1
– 6 6
ˆ
ˆ ˆ ˆ ˆ
v = (24 + 6)i – (18 – 5) J + (18 + 20)k = 18 i + 13 J – 2k
Problem 95. A particle comes round circle of radius 1 m once. The time taken by it is 10 sec. The average velocity of motion is [JIPMER 1999]
(a) 0.2 p m/s (b) 2 p m/s (c) 2 m/s (d) Zero
Solution : (d) In complete revolution total displacement becomes zero. So the average velocity will be zero.
Problem 96. Two particles of mass M and m are moving in a circle of radii R and r. If their timeperiods are same, what will be the ratio of their linear velocities [CBSE PMT 2001]
 MR : mr (b) M : m (c) R : r (d) 1 : 1
Solution : (c)
v1 = r1w1 . Time periods are equal i.e. w = w
\ v1
= r1 = R
v2 r2w _{2}
^{1} ^{2} v2 r2 r
(5) Time period (T) : In circular motion, the time period is defined as the time taken by the object to complete one revolution on its circular path.
 Units :
(ii) Dimension : [M^{0}L^{0}T]
 Time period of second’s hand of watch = 60 second.
(iv) Time period of minute’s hand of watch = 60 minute
 Time period of hour’s hand of watch = 12 hour
(6) Frequency (n) : In circular motion, the frequency is defined as the number of revolutions completed by the object on its circular path in a unit time.
 Units : s^{–1} or hertz (Hz).
(ii) Dimension : [M^{0}L^{0}T^{–1}]
Note : @Relation between time period and frequency : If n is the frequency of revolution of an object in circular motion, then the object completes n revolutions in 1 second. Therefore, the object will complete one revolution in 1/n second.
\T = 1 / n
@ Relation between angular velocity, frequency and time period : Consider a point object describing a uniform circular motion with frequency n and time period T. When the object completes one revolution, the angle traced at its axis of circular motion is 2p radians. It means, when time t = T,
q = 2p
radians. Hence, angular velocity w = q
t
= 2p
T
= 2pn
(Q T = 1/n)
w = 2p
T
= 2pn
@ If two particles are moving on same circle or different coplanar concentric circles in same direction with different uniform angular speeds w_{A} and w_{B} respectively, the angular velocity of B relative to A will be
w rel = w B – w A
So the time taken by one to complete one revolution around O with respect to the other (i.e., time in which B complete one revolution around O with respect to the other (i.e., time in which B completes one more or less revolution around O than A)
T = 2p
= 2p =
T1T2
éas T = 2p ù
w rel
w _{2} – w_{1}
T1 – T2
ëê w úû
Special case : If w _{B} = w _{A},w _{rel} = 0
and so T = ¥., particles will maintain their position relative to
each other. This is what actually happens in case of geostationary satellite (w_{1} = w_{2} = constant)
(7) Angular acceleration (a) : Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity.
 If Dw be the change in angular velocity of the object in time interval t and t + Dt, while moving on a circular path, then angular acceleration of the object will be
a =
 Units : rad. s^{–2}
Lt Dw
Dt®0 Dt
= dw
dt
= d ^{2}q
dt ^{2}
(iii) Dimension : [M^{0}L^{0}T^{–2}]
 Relation between linear acceleration and angular acceleration a = a ´ r
(v) For uniform circular motion since w is constant so a = dw = 0
dt
(vi) For nonuniform circular motion a ¹ 0
Note : @Relation between linear (tangential) acceleration and angular acceleration a = a ´ r
@ For uniform circular motion angular acceleration is zero, so tangential acceleration also is equal to zero.
@ For nonuniform circular motion a ¹ 0 (because a ¹ 0).
Problem 97. A body is revolving with a uniform speed v in a circle of radius r. The angular acceleration of the body is
(a)
v (b) Zero
r
(c)
v2 along the radius and towards the centre (d)
r
v2 along the radius and away from the centre
r
Solution : (b) In uniform circular motion w constant so a = dw = 0
dt
Problem 98. The linear acceleration of a car is 10m/s^{2}. If the wheels of the car have a diameter of 1m, the angular acceleration of the wheels will be
 10 rad/sec^{2} (b) 20 rad/sec^{2} (c) 1 rad/sec^{2} (d) 2 rad/sec^{2}
Solution : (b)
Angular acceleration = linear acceleration
radius
= 10
0.5
= 20 rad / sec ^{2}
Problem 99. The angular speed of a motor increases from 600 rpm to 1200 rpm in 10 s. What is the angular acceleration of the motor
(a)
600 rad sec ^{–}^{2}
60p rad sec ^{–}^{2}
60 rad sec ^{–}^{2}
2p rad sec ^{–}^{2}
Solution : (d)
a = w _{2} – w_{1} = 2p (n_{2} – n_{1} ) = 2p (1200 – 600) rad = 2p rad / sec 2
t t 10 ´ 60
sec ^{2}
Centripetal Acceleration.
(1) Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
 It always acts on the object along the radius towards the centre of the circular
(3) Magnitude of centripetal acceleration
a = v 2
r
= w ^{2}r = 4pn^{2}r = 4p 2 r
T 2
(4) Direction of centripetal acceleration : It is always the same as that of Du . When Dt
decreases, Dq also decreases. Due to which Du becomes more and more perpendicular to
u . When D t ® 0, Du becomes perpendicular to the velocity vector. As the velocity vector of the particle at an
instant acts along the tangent to the circular path, therefore Du and hence the centripetal acceleration vector acts
along the radius of the circular path at that point and is directed towards the centre of the circular path.
Problem 100. If a cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration of the cycle will be
[Pb. PMT 2001]
(a)
p ^{2}m / s ^{2}
(b)
2p ^{2}m / s ^{2}
(c)
4p ^{2}m / s ^{2}
(d)
8p m/s ^{2}
Solution : (c) Given r = 4 m and T = 2 seconds.
\ a = 4p 2 r = 4p 2 4 = 4p ^{2} m / s ^{2}
c T 2
(2)^{2}
Problem 101. A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stone [Pb. PMT 2000]
(a) 493 cm/sec^{2} (b) 720 cm/sec^{2} (c) 860 cm/sec^{2} (d) 990 cm/sec^{2}
Solution : (a) Time period = Total time = 20 = 2 sec
No. of revolution 10
\a = 4p 2 .r = 4p 2 ´ (1 / 2)m / s ^{2} = 4.93 m / s^{2} = 493 cm / s ^{2}
c T 2
(2)^{2}
Problem 102. A particle moves with a constant speed v along a circular path of radius r and completes the circle in time T. What is the acceleration of the particle [Orissa JEE 2002]
 mg
2pv T
pr ^{2}
T
pv ^{2}
T
Solution : (b)
a = v 2 = w ^{2}r = vw = v æ 2p ö = 2p v

_{c} ç ÷
r è ø T
Problem 103. If the speed of revolution of a particle on the circumference of a circle and the speed gained in falling through a distance equal to half the radius are equal, then the centripetal acceleration will be
(a) g
2
(b)
g (c)
4
g (d) g
3
Solution : (d) Speed gain by body falling through a distance h is equal to v = =
[As h = r
2
given]
Þ v = Þ
v 2 = g r
Problem 104. Two cars going round curve with speeds one at 90 km/h and other at 15 km/h. Each car experiences same acceleration. The radii of curves are in the ratio of [EAMCET (Med.) 1998]
(a) 4 : 1 (b) 2 : 1 (c) 16 : 1 (d) 36 : 1
v2 v2
Solution : (d) Centripetal acceleration = ^{1} = ^{2} (given)
r1 r2
r æ v ö 2 æ 90 ö ^{2} 36
\ ^{ } ^{1} = ç 1 ÷ = ç ÷ =
r2 è v2 ø
è 15 ø 1
Problem 105. A wheel of radius 0.20m is accelerated from rest with an angular acceleration of 1 rad / s ^{2} . After a rotation of
90^{o} the radial acceleration of a particle on its rim will be
(a)
p m / s ^{2}
(b)
0.5 p m / s ^{2}
(c)
2.0p m / s ^{2}
(d)
0.2 p m / s ^{2}
Solution : (d) From the equation of motion
Angular speed acquired by the wheel, w ^{2} = w ^{2} + 2aq = 0 + 2 ´ 1´ p Þ w ^{2} = p
2 1 2 2
Now radial acceleration w ^{2}r = p ´ 0.2 = 0.2p m / s ^{2}
Centripetal Force.
According to Newton’s first law of motion, whenever a body moves in a straight line with uniform velocity, no force is required to maintain this velocity. But when a body moves along a circular path with uniform speed, its direction changes continuously i.e. velocity keeps on changing on account of a change in direction. According to Newton’s second law of motion, a change in the direction of motion of the body can take place only if some external force acts on the body.
Due to inertia, at every point of the circular path; the body tends to move along the tangent to the circular path at that point (in figure). Since every body has directional inertia, a velocity
cannot change by itself and as such we have to apply a force. But this force should be such that it changes the direction of velocity and not its magnitude. This is possible only if the force acts perpendicular to the direction of velocity. Because the velocity is along the tangent, this force must be along the radius (because the radius of a circle at any point is perpendicular to the tangent at that point). Further, as this force is to move the body in a circular path, it must acts towards the centre. This centreseeking force is called the centripetal force.
Hence, centripetal force is that force which is required to move a body in a circular path with uniform speed.
The force acts on the body along the radius and towards centre.
 Formulae for centripetal force :
F = mv^{2}
r
= mw ^{2}r = m 4p ^{2}n^{2}r = m4p 2r
T 2
 Centripetal force in different situation
Situation  Centripetal Force 
A particle tied to a string and whirled in a horizontal circle  Tension in the string 
Vehicle taking a turn on a level road  Frictional force exerted by the road on the tyres 
A vehicle on a speed breaker  Weight of the body or a component of weight 
Revolution of earth around the sun  Gravitational force exerted by the sun 
Electron revolving around the nucleus in an atom  Coulomb attraction exerted by the protons in the nucleus 
A charged particle describing a circular path in a magnetic  Magnetic force exerted by the agent that sets up the magnetic 
field  field 
Centrifugal Force.
It is an imaginary force due to incorporated effects of inertia. When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangential at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body with the same velocity), the body appears to be stationary before it is released. When the body is released, it appears to B, as if it has been thrown off along the radius away from the centre by some force. In reality no force is actually seen to act on the body. In absence of any real force the body tends to continue its motion in a straight line due to its inertia. The observer A easily relates this events to be due to inertia but since the inertia of both the observer B and the body is same, the observer B can not relate the above happening to inertia. When the centripetal force ceases to act on the body, the body leaves its circular path and continues to moves in its straightline motion but to observer B it appears that a real force has actually acted on the body and is responsible for throwing the body radially outwords. This imaginary force is given a name to explain the effects on inertia to the observer who is sharing the circular motion of the body. This inertial force is called centrifugal force. Thus centrifugal force is a fictitious force which has significance only in a rotating frame of reference.
Problem 106. A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed of 10
r.p.m. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is [MP PMT 2001]
(a) 5 r.p.m. (b) 10 r.p.m. (c) 20 r.p.m. (d) 14 r.p.m.
Solution : (a) Tension in the string T = mw ^{2}r = m4p ^{2}n^{2}r
T µ n^{2} or n µ [As m and r are constant]
\ n2 = = Þ n = n1 = 10 = 5 rpm
n_{1} ^{2} 2 2
Problem 107. A cylindrical vessel partially filled with water is rotated about its vertical central axis. It’s surface will
[RPET 2000]
(a) Rise equally (b) Rise from the sides (c) Rise from the middle (d) Lowered equally
Solution : (b) Due to the centrifugal force.
Problem 108. A proton of mass 1.6 × 10^{–27} kg goes round in a circular orbit of radius 0.10 m under a centripetal force of 4 × 10^{–13} N. then the frequency of revolution of the proton is about [Kerala PMT 2002]
(a) 0.08 × 10^{8} cycles per sec (b) 4 × 10^{8} cycles per sec
(c) 8 × 10^{8} cycles per sec (d) 12 × 10^{8} cycles per sec
Solution : (a)
F = 4 ´ 10 ^{–}^{13} N ; m = 1.6 ´ 10 ^{–}^{27} kg ; r = 0.1m
Centripetal force F = m4p ^{2}n^{2}r \ n =
= 8 ´ 10^{6} cycles / sec = 0.08 ´ 10^{8} cycle / sec .
Problem 109. Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v_{0}, then the ratio of tensions in the three sections of the string is [UPSEAT 2003]
(a) 3 : 5 : 7 (b) 3 : 4 : 5 (c) 7 : 11 : 6 (d) 3 : 5 : 6
Solution : (d) Let the angular speed of the thread is w

For particle ‘C’ Þ T_{3} = mw ^{2} 3l
For particle ‘B’
T2 – T3
= mw ^{2} 2l Þ T_{2}
= mw ^{2} 5l
For particle ‘C’
T1 – T2 = mw ^{2}l Þ T1 = mw ^{2} 6l
\ T_{3} : T_{2} : T_{1} = 3 : 5 : 6
Problem 110. A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle with a uniform angular velocity of 2 rad/s. The tension of the string is (in N) [KCET 1998]
(a) 2 (b) 1
3
(c) 4 (d) 1
4
Solution : (c)
T = mw ^{2}r = 1 ´ (2)^{2} ´ (1) = 4 Newton
Problem 111. A cord can bear a maximum force of 100 N without breaking. A body of mass 1 kg tied to one end of a cord of length 1 m is revolved in a horizontal plane. What is the maximum linear speed of the body so that the cord does not break
(a) 10 m/s (b) 20 m/s (c) 25 m/s (d) 30 m/s
Solution : (a) Tension in cord appears due to centrifugal force
T = mv2 and for critical condition this tension will be equal
r
to breaking force (100 N)
2

\ ^{max} = 100 Þ
r
2
max
100 ´ 1

1
Þ vmax = 10 m / s
Problem 112. A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity w_{0}. If the length of the string and angular velocity are doubled, the tension in the string which was initially T_{0} is now [AIIMS 1985]
(a) T_{0} (b) T_{0}/2 (c) 4T_{0} (d) 8T_{0}
T æ w
ö ^{2} æ l ö
T æ 2w ö ^{2} æ 2l ö
Solution : (d)
T = mw ^{2}l \
2 = ç 2 ÷
ç ^{2} ÷ Þ
2 = ç
÷ ç ÷ Þ T_{2} = 8T_{0}
T1 è w1 ø
è l1 ø
T0 è w ø
è l ø
Problem 113. A stone is rotated steadily in a horizontal circle with a period T by a string of length l. If the tension in the string is kept constant and l increases by 1%, what is the percentage change in T
(a) 1% (b) 0.5% (c) 2% (d) 0.25%
Solution : (b) Tension = m 4p 2l
T 2
\ l µ T ^{2} or T µ [Tension and mass are constant]
Percentage change in Time period = 1
2
(percentage change in length) [If % change is very small]
= 1 (1%) = 0.5%
2
Problem 114. If mass speed and radius of rotation of a body moving in a circular path are all increased by 50%, the necessary force required to maintain the body moving in the circular path will have to be increased by
(a) 225% (b) 125% (c) 150% (d) 100%
Solution : (b) Centripetal force F = mv2
r
æ m ö æ v ö ^{2}
çm + 2 ÷ çv + 2 ÷
9 mv ^{2} 9
If m, v and r are increased by 50% then let new force F‘ = è
ø è ø =
ær + r ö 4 r
= 4 F

ç ÷
è ø
Percentage increase in force
DF ´ 100 = F‘ – F ´ 100% = 500 % = 125%
F F 4
Problem 115. Two masses m
and M are connected by a light string that passes through a smooth hole O at the centre of a
table. Mass m lies on the table and M hangs vertically. m is moved round in a horizontal circle with O as the centre. If l is the length of the string from O to m then the frequency with which m should revolve so that M remains stationary is
(a)
(b) 1
p
(c)
(d) 1
p
Solution : (a) ‘m’ Mass performs uniform circular motion on the table. Let n is the frequency of revolution then centrifugal force = m 4p ^{2}n^{2}l
For equilibrium this force will be equal to weight Mg
m 4p ^{2}n^{2}l = Mg
\ n =
Problem 116. A particle of mass M moves with constant speed along a circular path of radius r under the action of a force F. Its speed is [MP PMT 2002]
(a)
Solution : (a) Centripetal force F = mv 2
r
(b)
\ v =
(c)
(d)
Problem 117. In an atom for the electron to revolve around the nucleus, the necessary centripetal force is obtained from the following force exerted by the nucleus on the electron [MP PET 2002]
(a) Nuclear force (b) Gravitational force (c) Magnetic force (d) Electrostatic force
Solution : (d)
Problem 118. A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be
[AFMC 2002]
(a) Double (b) Half (c) 4 times (d)
1 times
4
mv ^{2}
F æ v ö ^{2} æ 2v ö 2
Solution : (c) F =
\F µ v ^{2} or
2 = ç 2 ÷ = ç
÷ = 4
Þ F2 = 4 F1
r F_{1}
è v1 ø
è v ø
Problem 119. A bottle of soda water is grasped by the neck and swing briskly in a vertical circle. Near which portion of the bottle do the bubbles collect
 Near the bottom (b) In the middle of the bottle
 Near the neck (d) Uniformly distributed in the bottle
Solution : (c) Due to the lightness of the gas bubble they feel less centrifugal force so they get collect near the neck of the bottle. They collect near the centre of circular motion i.e. near the neck of the bottle.
Problem 120. A body is performing circular motion. An observer O_{1} is sitting at the centre of the circle and another observer
O_{2} is sitting on the body. The centrifugal force is experienced by the observer
(a)
O_{1} only (b)
O2 only (c) Both by O1 and O2
 None of these
Solution : (b) Centrifugal force is a pseudo force, which is experienced only by that observer who is attached with the body performing circular motion.
Work done by Centripetal Force.
The work done by centripetal force is always zero as it is perpendicular to velocity and hence instantaneous displacement.
Work done = Increment in kinetic energy of revolving body Work done = 0
Also W = F . S = F × S cosq
= F×S cos 90^{o} = 0
Example : (i) When an electron revolve around the nucleus in hydrogen atom in a particular orbit, it neither absorb nor emit any energy means its energy remains constant.
(ii) When a satellite established once in a orbit around the earth and it starts revolving with particular speed, then no fuel is required for its circular motion.
Problem 121. 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
 1 Joule (b) 0.2 Joule (c) 2.0 Joule (d) 1.0 Joule
Solution : (d)
mv 2 = 10 N (given) Þ mv ^{2} = 10 ´ r
r
= 10 ´ 0.2 = 2
Kinetic energy
1 mv ^{2} = 1 (2) = 1 Joule.
2 2
Problem 122. A body of mass 100 g is rotating in a circular path of radius r with constant velocity. The work done in one complete revolution is [AFMC 1998]
(a) 100r Joule (b) (r/100) Joule (c) (100/r) Joule (d) Zero
Solution : (d) Because in uniform circular motion work done by the centripetal force is always zero.
Problem 123. A particle of mass m is describing a circular path of radius r with uniform speed. If L is the angular momentum of the particle about the axis of the circle, the kinetic energy of the particle is given by [CPMT 1995]
(a)
L^{2} / mr ^{2}
(b)
L^{2} / 2mr ^{2}
 2L^{2} / mr ^{2}
(d)
mr ^{2} L
Solution : (b) Rotational kinetic energy
I = mr ^{2} )
E = L2
2I
= L2
2mr ^{2}
(As for a particle
Skidding of Vehicle on a Level Road.
When a vehicle turns on a circular path it requires centripetal force.
If friction provides this centripetal force then vehicle can move in circular path safely if
Friction force ³ Required centripetal force
m mg ³ mv2
r
\ vsafe £
This is the maximum speed by which vehicle can turn in a circular path of radius r, where coefficient of friction between the road and tyre is m.
Problem 124. Find the maximum velocity for overturn for a car moved on a circular track of radius 100m . The coefficient of
friction between the road and tyre is 0.2
[CPMT 1996]
(a)
0.14 m / s
(b)
140 m / s
(c)
1.4 km / s
(d)
14 m / s
Solution : (d)
vmax = =
= 10
= 14 m / s
Problem 125. When the road is dry and the coefficient of friction is
m , the maximum speed of a car in a circular path is
10 m / s . If the road becomes wet and m¢ = m , what is the maximum speed permitted
2
Solution : (d)
(a)
v µ
5 m / s
Þ v 2 =
v1
(b)
=
10 m / s
=
Þ v2 =
(c)
v1
10
Þ v2
m / s
= = 5
 5 2 m / s
m / s
Problem 126. The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m with skidding is (assume g = 10 ms^{–2}) [Kerala PMT 2002]
(a)
40 ms ^{–}^{1}
(b)
20 ms ^{–}^{1}
(c)
15 ms ^{–}^{1}
(d)
10 ms ^{–}^{1}
Solution : (d)
vmax = =
= 10 m / s
Skidding of Object on a Rotating Platform.
On a rotating platform, to avoid the skidding of an object (mass m) placed at a distance r from axis of rotation, the centripetal force should be provided by force of friction.
Centripetal force = Force of friction
mw^{2}r = mmg
\ w max =
(mg / r) ,
Hence maximum angular velocity of rotation of the platform is
Bending of a Cyclist.
(mg / r) , so that object will not skid on it.
A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going round a curve. Consider a cyclist of weight mg taking a turn of radius r with velocity v. In order to provide the necessary centripetal force, the cyclist leans through angle q inwards as shown in figure.
The cyclist is under the action of the following forces :
The weight mg acting vertically downward at the centre of gravity of cycle and the cyclist.
The reaction R of the ground on cyclist. It will act along a linemaking angle q with the vertical.
The vertical component R cosq of the normal reaction R will balance the weight of the cyclist, while the horizontal component R sin q will provide the necessary centripetal force to the cyclist.
R sinq = mv 2
r
…..(i)
and R cos q = mg …..(ii) Dividing equation (i) by (ii), we have
R sinq
R cosq
= mv ^{2} r
mg
or tanq = v 2
rg
…..…(iii)
1 æ v 2 ö
Therefore, the cyclist should bend through an angle q = tan
ç ÷
ç rg ÷
è ø
It follows that the angle through which cyclist should bend will be greater, if
(i) The radius of the curve is small i.e. the curve is sharper
 The velocity of the cyclist is
Note : @For the same reasons, an ice skater or an aeroplane has to bend inwards, while taking a turn.
Problem 127. A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 metres/sec. The combined mass
of the boy and the cycle is
(g = 9.8 m / sec ^{2} )
90kg . The angle that the cycle makes with the vertical so that it may not fall is
[MP PMT 1995]
(a)
60.25 ^{o}
(b)
63.90 ^{o}
(c)
26.12^{o}
(d)
 ^{o}
Solution : (b)
r = 20 m,
v = 20 m / s,
m = 90 kg,
g = 9.8 m / s ^{2}
(given)
1 æ v 2 ö
_{1} æ 20 ´ 20 ö _{1}
q = tan
ç ÷ = tan ç
÷ = tan
(2)
= 63.90°
ç r g ÷
è 20 ´ 10 ø
è ø
Problem 128. If a cyclist moving with a speed of 4.9 m / s
on a level road can take a sharp circular turn of radius 4m , then
coefficient of friction between the cycle tyres and road is [AIIMS 1999]
(a) 0.41 (b) 0.51 (c) 0.71 (d) 0.61
Solution : (d)
v = 4.9 m / s, r = 4 m and g = 9.8 m / s ^{2}
(given)
m = v 2
r g
= 4.9 ´ 4.9
4 ´ 9.8
= 0.61
Problem 129. A cyclist taking turn bends inwards while a car passenger taking same turn is thrown outwards. The reason is
[NCERT 1972]
 Car is heavier than cycle
 Car has four wheels while cycle has only two
 Difference in the speed of the two
 Cyclist has to counteract the centrifugal force while in the case of car only the passenger is thrown by this force
Solution : (d)
Banking of a Road.
For getting a centripetal force cyclist bend towards the centre of circular path but it is not possible in case of four wheelers.
Therefore, outer bed of the road is raised so that a vehicle moving on it gets automatically inclined towards the centre. In the figure (A) shown reaction R is resolved into two components, the component R cosq balances weight of vehicle
\ R cos q
= mg
……(i)
and the horizontal component R sin q provides necessary centripetal force as it is directed towards centre of desired circle
mv ^{2}
Thus
R sin q =
r
……(ii)
Dividing (ii) by (i), we have
tanq = v 2
r g
or tanq = w 2r = vw
…… (iii)
…… (iv) [As v = rw ]
g rg
If l = width of the road, h = height of the outer edge from the ground level then from the figure (B)
tanq = h = h
…….(v) [since q is very small]
x l
From equation (iii), (iv) and (v)
tanq = v 2
= w ^{2}r = vw = h
rg g rg l
Note : @If friction is also present between the tyres and road then
@ Maximum safe speed on a banked frictional road v =
v 2 =
rg
m + tanq 1 – m tanq
Problem 130. For traffic moving at 60 km / hr
along a circular track of radius 0.1 km , the correct angle of banking is
[MNR 1993]
(a)
(60)^{2} (b)
tan1 é (50 / 3)^{2} ù
(c)
tan1 é100 ´ 9.8 ù
(d)
tan^{1}
0.1
ê100 ´ 9.8 ú
ê (50 / 3)^{2} ú
Solution : (b)
v = 60 km / hr = 50 m / s ,
3
êë úû
r = 0.1km = 100 m, g = 9.8 m / s ^{2}
ë û
(given)
v 2 1 æ v 2 ö
_{–}_{1} é (50 / 3)2 ù
Angle of banking tanq =
or q = tan ç ÷ = tan ê ú
r g ç r g ÷ ê100 ´ 9.8 ú
è ø ë û
Problem 131. A vehicle is moving with a velocity v
on a curved road of width b and radius of curvature R . For
counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is [EAMCET 1983; MP PMT 1996]
(a)
v^{2}b Rg
(b)
rb (c)
Rg
vb2 (d)
Rg
vb R^{2} g
Solution : (a) For Banking of road tanq = v 2 and tanq = h
\ v ^{2} = h r g l
Þ h = v 2l
r g
r g
= v ^{2}b R g
l
[As l = b and r = R given]
Problem 132. The radius of curvature of a road at a certain turn is 50m . The width of the road is 10m and its outer edge is
1.5m higher than the inner edge. The safe speed for such an inclination will be
(a)
6.5 m / s
(b)
8.6 m / s
(c)
8 m / s
(d)
10 m / s
Solution : (b)
h = 1.5 m,
r = 50 m,
l = 10 m,
g = 10 m / s ^{2}
(given)
v ^{2} = h r g l
Þ v = =
= 8.6 m / s
Problem 133. Keeping the banking angle same to increase the maximum speed with which a car can travel on a curved road by 10%, the radius of curvature of road has to be changed from 20m to [EAMCET 1991]
(a)
16m
(b)
18m
(c)
24.25m
(d)
30.5 m
Solution : (c)
tanq = v 2
r g
Þ r µ v ^{2} (if q is constant)
r æ v ö 2 æ 1.1v ö ^{2}
2 = ç 2 ÷ = ç
÷ = 1.21 Þ r_{2} = 1.21 ´ r_{1} = 1.21 ´ 20 = 24.2 m
r1 è v1 ø
è v ø
Problem 134. The slope of the smooth banked horizontal road is with which a car can negotiate the curve is given by
p . If the radius of the curve be r , the maximum velocity
(a) prg (b)
(c)
p / rg
(d)
Solution : (b)
tanq = v 2
r g
Þ p = v 2
r g
\ v =
Overturning of Vehicle.
When a car moves in a circular path with speed more than maximum speed then it overturns and it’s inner wheel leaves the ground first
Weight of the car = mg
Speed of the car = v
Radius of the circular path = r
Distance between the centre of wheels of the car = 2a
Height of the centre of gravity (G) of the car from the road level = h
Reaction on the inner wheel of the car by the ground = R_{1}
Reaction on the outer wheel of the car by the ground = R_{2}
When a car move in a circular path, horizontal force F provides the required centripetal force
i.e.,
F = mv^{2}
R
…….(i)
For rotational equilibrium, by taking the moment of forces R_{1}, R_{2} and F about G
Fh + R_{1}a = R_{2}a
…….(ii)
As there is no vertical motion so R_{1} + R_{2} = mg…………………… (iii)
By solving (i), (ii) and (iii)
R = 1
é – v ^{2}hù
…….(iv)

 2 M êg
ú
ra úû
and
R = 1
é + v ^{2}hù
…….(v)

 2 M êg
ú
ra úû
It is clear from equation (iv) that if v increases value of R_{1} decreases and for R_{1} = 0
v ^{2}h = g ra
or v =
i.e. the maximum speed of a car without overturning on a flat road is given by v =
Problem 135. The distance between two rails is 1.5m . The centre of gravity of the train at a height of 2m from the ground.
The maximum speed of the train on a circular path of radius 120m can be
(a)
10.5 m / s
(b)
42 m / s
(c)
21 m / s
(d)
84 m / s
Solution : (c) Height of centre of gravity from the ground h = 2m, Acceleration due to gravity g = 10 m / s ^{2} , Distance between two rails 2a = 1.5m, Radius of circular path r = 120 m (given)
vmax =
Þ vmax =
= 21.2 m / s
Problem 136. A car sometimes overturns while taking a turn. When it overturns, it is [AFMC 1988]
 The inner wheel which leaves the ground first
 The outer wheel which leaves the ground first
 Both the wheels leave the ground simultaneously
 Either wheel leaves the ground first
Solution : (a)
Problem 137. A car is moving on a circular path and takes a turn. If wheels respectively, then
R1 and
R_{2} be the reactions on the inner and outer
(a)
R1 = R2
(b)
R1< R2
(c)
R1 > R2
(d)
R1 ³ R2
Solution : (b) Reaction on inner wheel R
= M ég – v ^{2}hù
and Reaction on outer wheel R
= M ég + v ^{2}hù




\ R1 < R2 .
_{1} 2 ê
r a ú
2 2 ê
r a ú
Problem 138. A train
A runs from east to west and another train B of the same mass runs from west to east at the same
speed along the equator. A presses the track with a force F_{1} and B presses the track with a force F_{2}
 F1 > F2
 F1< F2
 F1 = F2
 The information is insufficient to find the relation between F_{1} and F_{2}
Solution : (a) We know that earth revolves about its own axis from west to east. Let its angular speed is w_{e}
speed of the train is w _{t}
and the angular
For train A : Net angular speed = ( w _{e} – w _{t} ) because the sense of rotation of train is opposite to that of earth
So reaction of track R1 = F1 = m g – m(w _{e} – w _{t} )^{2} R
For train B : Net angular speed = (w _{e} + w _{t} ) because the sense of rotation of train is same as that of earth
So reaction of track R2 = F2 = m g – m(w _{e} + w _{t} )^{2} R
So it is clear that F1 > F2
Motion of Charged Particle in Magnetic Field.
When a charged particle having mass m , charge q enters perpendicularly in a magnetic field B, with velocity v
then it describes a circular path of radius r.
Because magnetic force (qvB) works in the perpendicular direction of v and it provides required centripetal force Magnetic force = Centripetal force
qvB =
mv ^{2}
r
mv
\ radius of the circular path r = qB
Reaction of Road on Car.
(2) When car moves on a convex bridge
Centripetal force = mg cos q – R
= mv ^{2}
r
and reaction
R = mg cosq – mv 2
r
Problem 139. The road way bridge over a canal is in the form of an arc of a circle of radius
20m . What is the minimum
speed with which a car can cross the bridge without leaving contact with the ground at the highest point
(g = 9.8 m / s ^{2})
(a)
7 m / s
(b)
14 m / s
(c)
289 m / s
(d)
5 m / s
Solution : (b) At the highest point of the bridge for critical condition mg – mv 2 = 0 Þ
r
mv ^{2}
r
= m g
\ v_{max} = =
= = 14 m / s
Problem 140. A car moves at a constant speed on a road as shown in the figure. The normal force exerted by the road on
the car is
N A and N B
when it is at the points A and B respectively
 N A = NB
 N A > NB
 N A < NB
 All possibilities are there
Solution : (c) From the formula
N = mg – mv 2
r
\ N µ r As r_{A}
< r_{B}
\ N A < NB
Problem 141. A car while travelling at a speed of 72 km / hr . Passes through a curved portion of road in the form of an arc
of a radius 10m . If the mass of the car is 500 kg the reaction of the car at the lowest point P is
 25 kN
 50 kN
 75 kN
 None of these