Chapter 6 Motion in a Plane (two dimension) Part 4 – Physics free study material by TEACHING CARE online tuition and coaching classes
Solution : (a)
v = 72 km = 20 m/s,
h
r = 10 m,
m = 500 kg
(given)
Reaction at lowest point
NonUniform Circular Motion.
R = mg + mv 2
r
= 500 ´ 10 + 500 ´ (20)^{2}
10
= 25000 N = 25 KN
If the speed of the particle in a horizontal circular motion changes with respect to time, then its motion is said to be nonuniform circular motion.
Consider a particle describing a circular path of radius r with centre at O. Let at an instant the particle be at P
and u be its linear velocity and w be its angular velocity.
Then,
ur = wr ´ r
…..(i)
Differentiating both sides of w.r.t. time t we have
® ® ® ®

du = dw ´ r + wr dr
…..(ii) Here,
dv r

a,
(Resultant acceleration)
dt dt dt dt
®
a = ar ´ r
+ wr ´ur
dw = ar
dt
®
(Angular acceleration)
a = at
 ac
.….(iii)
dr = ur
dt
(Linear velocity)
Thus the resultant acceleration of the particle at P has two component accelerations
 Tangential acceleration : at = a ´ r
It acts along the tangent to the circular path at P in the plane of circular path.
According to right hand rule since a and r are perpendicular to each other, therefore, the magnitude of tangential acceleration is given by
at =a ´ r =a r sin 90^{o} = a r.
 Centripetal (Radial) acceleration : ac = w ´ v
It is also called centripetal acceleration of the particle at P. It acts along the radius of the particle at P.
According to right hand rule since w and u are perpendicular to each other, therefore, the magnitude of centripetal acceleration is given by
r r r _{o} = 2 2
ac =w ´u =wu sin 90 wu =w(w r) =w r = u / r
 Tangential and centripetal acceleration in different motions
Note : @Here a_{t} governs the magnitude of v while a c
its direction of motion.
 Force : In nonuniform circular motion the particle simultaneously possesses two forces
Centripetal force : Fc
= mac
= mv ^{2}
r
= mrw ^{2}
Tangential force : Ft = mat
Net force :
Fnet = ma = m


Note : @ In nonuniform circular motion work done by centripetal force will be zero since Fc v
@ In non uniform circular motion work done by tangential of force will not be zero since F_{t} ¹ 0
@ Rate of work done by net force in nonuniform circular = rate of work done by tangential force
i.e. P = dW = r r
dt Ft .v
Problem 142. The kinetic energy k
of a particle moving along a circle of radius R depends on the distance covered. It is
given as K.E. = as^{2} where a is a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 2002]
(a)
2a s 2
(b)
æ
2asç1 +
2 ö1 / 2


÷
2
(c)
2as
(d)
2a R2
ç ÷
è ø


Solution : (b) In nonuniform circular motion two forces will work on a particle Fc and Ft
So the net force FNet
= ….(i)
Centripetal force Fc
= mv ^{2}
R
= 2as^{2}
R
….(ii) [As kinetic energy
1 mv^{2} = as ^{2} given]
2
Again from :
1 mv ^{2} = as ^{2} Þ
2
v 2 = 2as ^{2}
m
Þ v = s
F_{t}
Tangential acceleration a
= dv = dv . ds Þ a
= d és
2a ù .v
at = v


2a = s 2a m m
^{t} dt
2a = 2as
m m
ds dt
t ds ê m ú
and Ft
= mat
= 2as
….(iii)
é s2 ù1 / 2
Now substituting value of F_{c} and F_{t} in equation (i)
\ FNet =
= 2as ê1 + ú

êë úû
Problem 143. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k ^{2}rt ^{2} , where k is a constant. The power delivered to the particle by the forces acting on it is [IITJEE 1994]
(a)
2pmk ^{2}r ^{2}t
(b)
mk ^{2}r ^{2}t
(c)
mk ^{4}r ^{2}t ^{5}
3
(d) Zero
Solution : (b)
a = k ^{2}r t ^{2} Þ v2 = k ^{2}r t ^{2} Þ v^{2} = k ^{2} r ^{2} t ^{2}
c r
Þ v = k r t
Tangential acceleration at = dv = k r
dt
As centripetal force does not work in circular motion.
So power delivered by tangential force P = Ftv = matv =m(kr) krt = mk^{2}r ^{2}t
Problem 144. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum,
its acceleration vector a is correctly shown in [IITJEE Screening 2002]
(a) (b) (c) (d)
Solution : (c)
ac = centripetal acceleration , a
at = tangential acceleration ,
aN = net acceleration = Resultant of ac and at
a_{N} =
Problem 145. The speed of a particle moving in a circle of radius 0.1m is v = 1.0t
where t
a t
is time in second. The resultant
acceleration of the particle at t = 5s will be
(a)
10 m / s ^{2}
(b)
100 m / s ^{2}
(c)
250 m / s ^{2}
(d)
500 m / s ^{2}
Solution : (c)
v = 1.0 t
Þ a = dv = 1m / s ^{2}

dt
and a
= v^{2} = (5)^{2}
= 250 m / s ^{2}
[At
t = 5 sec, v = 5 m / s ]
c r
\ aN =
0.1
= Þ a_{N}
= 250 m / s ^{2} (approx.)
Problem 146. A particle moving along the circular path with a speed v and its speed increases by ‘g’ in one second. If the radius of the circular path be r, then the net acceleration of the particle is
(a)
v2 + g
(b)
v2 + g ^{2}
(c)
1


év4 + g 2 ù 2
(d)
1
év2 + gù 2
r r ^{2}
v2
ê r 2 ú
ê ú

r úû
Solution : (c)
at = g
(given) and ac = r and aN = = =
Problem 147. A car is moving with speed 30 m/sec on a circular path of radius 500 m. Its speed is increasing at the rate of 2m/sec^{2}. What is the acceleration of the car [Roorkee 1982; RPET 1996; MH CET 2002; MP PMT 2003]
(a) 2 m/s^{2} (b) 2.7 m/s^{2} (c) 1.8 m/s^{2} (d) 9.8 m/s^{2}
Solution : (b) a_{t} = 2m/s^{2} and ac
= v 2
r
= 30 ´ 30 = 1.8 m/s ^{2}
500
\ a = =
= 2.7m/s ^{2} .
Problem 148. For a particle in circular motion the centripetal acceleration is [CPMT 1998]
 Less than its tangential acceleration (b) Equal to its tangential acceleration
(c) More than its tangential acceleration (d) May be more or less than its tangential acceleration
Solution : (d)
Problem 149. A particle is moving along a circular path of radius 3 meter in such a way that the distance travelled measured
along the circumference is given by S = t 2 + t 3 . The acceleration of particle when t = 2 sec is
2 3
(a) 1.3 m/s^{2} (b) 13 m/s^{2} (c) 3 m/s^{2} (d) 10 m/s^{2}
Solution : (b)
s = t 2 + t 3 Þ v = ds = t + t ^{2} and a
= dv = d (t + t ^{2})
= 1 + 2t
2 3 dt
^{t} dt dt
At t = 2 sec, v = 6 m/s and at
= 5 m / s ^{2} , ac
= v 2
r
= 36 = 12 m/s ^{2}
3
a _{N} = = = 13m / s ^{2} .
Equations of Circular Motion.
Where
w_{1} = Initial angular velocity of particle
w_{2} = Final angular velocity of particle
a = Angular acceleration of particle
q = Angle covered by the particle in time t
q_{n} = Angle covered by the particle in n^{th} second
Problem 150. The angular velocity of a particle is given by ceases to be zero will be
w = 1.5 t – 3t ^{2} + 2 , the time when its angular acceleration
(a)
25 sec
(b)
0.25 sec
(c)
12 sec
(d)
1.2 sec
Solution : (b)
w = 1.5 t – 3t ^{2} + 2 and a = dw
dt
= 1.5 – 6t
Þ 0 = 1.5 – 6t \ t = 1.5 = 0.25 sec
6
Problem 151. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the
first
2 sec , it rotates through an angle q_{1} . In the next
2 sec , it rotates through an additional angle q _{2} . The
ratio of
q1 / q 2 is [AIIMS 1982]
(a) 1 (b) 2 (c) 3 (d) 5
Solution : (c) From equation of motionq = w_{1}t + 1 a t ^{2}
2
q_{1} = 0 + 1 a (2)^{2} = 2a …..(i) [As w_{1} = 0,
2
t = 2 sec, q = q_{1} ]
For second condition
q_{1} + q _{2} = 0 + 1 a (4)^{2}
2
q_{1} + q _{2} = 8a
….(ii)
[As w_{1} = 0,
t = 2 + 2 = 4 sec, q = q_{1} + q _{2} ]
From (i) and (ii) q_{1} = 2a, q _{2} = 6a \ q 2 = 3
q_{1}
Problem 152. If the equation for the displacement of a particle moving on a circular path is given by (q ) = 2t ^{3} + 0.5 , where
q is in radians and t in seconds, then the angular velocity of the particle after 2 sec from its start is
[AIIMS 1998]
(a)
8 rad / sec
(b)
12 rad / sec
(c)
24 rad / sec
(d)
36 rad / sec
Solution : (c)
q = 2t ^{3} + 0.5
and
w = dq
dt
= 6t ^{2}
at t = 2 sec, w = 6 (2)^{2} = 24 rad / sec
Problem 153. A grinding wheel attained a velocity of 20 rad/sec in 5 sec starting from rest. Find the number of revolutions made by the wheel
(a)
p rev/ sec (b)
25
1 rev/sec (c)
p
25 rev/sec (d) None of these
p
Solution : (c)
w_{1} = 0,
w _{2} = 20 rad / sec, t = 5 sec
a = w _{2} – w_{1}
t
= 20 – 0 = 4 rad / sec ^{2}
5
From the equation q = w_{1}t + 1 a t ^{2} = 0 + 1 (4).(5)^{2} = 50 rad
2 2
2p rad means 1 revolution. \ 50 Radian means
50 or
2p
25 rev.
p
Problem 154. A grind stone starts from rest and has a constant angular acceleration of 4.0 rad/sec^{2}. The angular displacement and angular velocity, after 4 sec. will respectively be
 32 rad, 16 rad/sec (b) 16 rad, 32 rad/sec (c) 64 rad, 32 rad/sec (d) 32 rad, 64 rad/sec
Solution : (a)
w_{1} = 0, a = 4 rad / sec ^{2} , t = 4 sec
Angular displacement q = w_{1}t + 1 a t ^{2} = 0 + 1 4 (4)^{2} = 32 rad.
2 2
\ Final angular w _{2} = w_{1} + a t = 0 + 4 ´ 4 = 16 rad / sec
Problem 155. An electric fan is rotating at a speed of 600 rev/minute. When the power supply is stopped, it stops after 60 revolutions. The time taken to stop is
(a) 12 s (b) 30 s (c) 45 s (d) 60 s
Solution : (a)
w_{1} = 600
rev / min = 10 rev / sec,
w _{2} = 0 and q = 60 rev
From the equation w ^{2} = w ^{2} – 2aq Þ 0 = (10)^{2} – 2a 60 \a = 100 = 5
^{2} ^{1} 120 6
Again w _{2} = w_{1} – a t Þ 0 = w_{1} – a t
t = w1
a
= 10 ´ 6 = 12 sec .
5
Motion in Vertical Circle.
This is an example of nonuniform circular motion. In this motion body is under the influence of gravity of earth. When body moves from lowest point to highest point. Its speed decrease and becomes minimum at highest point. Total mechanical energy of the body remains conserved and KE converts into PE and vice versa.
 Velocity at any point on vertical loop : If u is the initial velocity imparted to body at lowest point then. Velocity of body at height h is given by
v = = [As h = l – l cosq = l (1 – cosq)]
where l in the length of the string
 Tension at any point on vertical loop : Tension at general point P, According to Newton’s second law of
Net force towards centre = centripetal force
T – mg cosq
= mv ^{2}
l
or T = mg cosq + mv 2
l
T = m [u ^{2} – gl(2 – 3 cosq )]
l
[As v = ]
 Velocity and tension in a vertical loop at different positions
It is clear from the table that :
TA > TB > TC
and T_{B} = T_{D}
TA – TB = 3mg,
and
TA – TC TB – TC
= 6mg
= 3mg
(4) Various conditions for vertical motion :
Note : @K.E. of a body moving in horizontal circle is same throughout the path but the K.E. of the body moving in vertical circle is different at different places.
@ If body of mass m is tied to a string of length l and is projected with a horizontal velocity u then :
Height at which the velocity vanishes is h = u 2
2g
Height at which the tension vanishes is h = u 2 + gl
3g
(5) Critical condition for vertical looping : If the tension at C is zero, then body will just complete revolution in the vertical circle. This state of body is known as critical state. The speed of body in critical state is called as critical speed.
From the above table T_{C} =
mu 2 – 5mg = 0 Þ u =
l
It means to complete the vertical circle the body must be projected with minimum velocity of lowest point.
(6) Various quantities for a critical condition in a vertical loop at different positions :
at the
 Motion of a block on frictionless hemisphere : A small block of mass m slides down from the top of a frictionless hemisphere of radius r. The component of the force of gravity (mg cosq) provides required centripetal force but at point B it’s circular motion ceases and the block lose contact with the surface of the
For point B, by equating the forces, mg cosq = mv2
r
For point A and B, by law of conservation of energy Total energy at point A = Total energy at point B K.E._{(}_{A}_{)} + P.E._{(}_{A}_{)} = K.E._{(}_{B}_{)} + P.E._{(}_{B}_{)}
1
…..(i)
0 + mgr =
mv^{2} + mgh
2
Þ v =………………………….. (ii)
and from the given figure h = r cosq…………………………………. (iii)
By substituting the value of v and h from eq^{n} (ii) and (iii) in eq^{n} (i)
mg æ h ö = m ( )2
ç r ÷ r
è ø
Þ h = 2(r – h)
Þ h = 2 r
3
i.e. the block lose contact at the height of 2 r
3
from the ground.
and angle from the vertical can be given by cosq = h = 2 \ q = cos^{1} 2 .
r 3 3
Problem 156. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [IITJEE (Screening) 2001]
(a) (b) (c) (d)
Solution : (a) Normal reaction at the highest point of the path R = mv 2 – mg
r
For maximum R, value of the radius of curvature (r ) should be minimum and it is minimum in first condition.
Problem 157. A stone tied to string is rotated in a vertical circle. The minimum speed with which the string has to be rotated
[EAMCET (Engg.) 1998; CBSE PMT 1999]
 Decreases with increasing mass of the stone (b) Is independent of the mass of the stone
(c) Decreases with increasing in length of the string (d) Is independent of the length of the string
Solution : (b)
v = for lowest point of vertical loop.
v µ m^{0} i.e. it does not depends on the mass of the body.
Problem 158. A mass m is revolving in a vertical circle at the end of a string of length 20 cms. By how much does the tension of the string at the lowest point exceed the tension at the topmost point
 2 mg (b) 4 mg (c) 6 mg (d) 8 mg
Solution : (c)
TLowest point – THighest point = 6 mg (Always)
Problem 159. In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released
from rest when the string is horizontal. The string breaks when it makes an angle q with the vertical
(a)
q = cos^{1}(1 / 3)
q = 60^{o}
q = cos^{1}(2 / 3)
q = 0^{o}
Solution : (c) Let the string breaks at point B.
mv ^{2}
Tension = mg cosq + ^{B} = Breaking strength
r
mv ^{2}
= mg cosq + ^{B} = 2 mg
r
….(i)
If the bob is released from rest (from point A) then velocity acquired by it at point B
vB =
v_{B} = ....(ii) [As h= r cos q ] By substituting this value in equation (i)
mg cosq + m (2g r cosq ) = 2 mg
r
or 3mg cosq = 2 mg Þ cosq = 2 \ q = cos
3
_{1}æ 2 ö
ç 3 ÷
è ø
Problem 160. A toy car rolls down the inclined plane as shown in the fig. It goes around the loop at the bottom. What is the
relation between H and h
(a)
H = 2
h
(b)
H = 3
h
(c)
H = 4
h
(d)
H = 5
h
Solution : (d) When car rolls down the inclined plane from height H, then velocity acquired by it at the lowest point
v = ….(i)
and for looping of loop, velocity at the lowest point should be v = ….(ii)
From eq^{n} (i) and (ii) v = =
\ H = 5r
2
….(iii)
From the figure H = h + 2r Þ
r = H – h
2
Substituting the value of r in equation (iii) we get H = 5 é H – hù Þ H = 5
2 êë 2 úû h
Problem 161. The mass of the bob of a simple pendulum of length L is m . If the bob is left from its horizontal position then the speed of the bob and the tension in the thread in the lowest position of the bob will be respectively.
(a)
and 3mg
(b)
3mg
and
(c)
2mg
and
(d)
2gl
and 3mg
Solution : (a) By the conservation of energy
Potential energy at point A = Kinetic energy at point B
mg l = 1 mv^{2}
2
Þ v =
2
and tension = mg + mv Þ T = mg + m (2gl) Þ T = 3 mg
l l
Problem 162. A stone of mass m is tied to a string and is moved in a vertical circle of radius r making n revolutions per minute. The total tension in the string when the stone is at its lowest point is [Kerala (Engg.) 2001]
(a)
m {g + (p ^{2}n^{2}r) / 900}
 m (g + p nr ^{2})
(c)
m (g + p nr)
(d)
m (g + n^{2}r ^{2})
Solution : (a) Tension at lowest point T = mg + mw ^{2}r = mg + m4p ^{2}n^{2}r
If n is revolution per minute then T = mg + m4p ^{2} n2
r = mg + mp 2n2r =
é + p ^{2}n^{2}r ù
3600
900
mêg


êë
900 ú
Problem 163. A particle is kept at rest at the top of a sphere of diameter 42m. When disturbed slightly, it slides down. At what height h from the bottom, the particle will leave the sphere [IMSBHU 2003] (a) 14 m (b) 28 m (c) 35 m (d) 7 m
Solution : (c) Let the particle leave the sphere at height ‘h’ from the bottom
We know for given condition
x = 2 r
3
and h = r + x = r + 2 r = 5 r = 5 ´ 21 = 35 m
[As r= 21 m]
3 3 3
Problem 164. A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position (Take g = 10 m/sec^{2}) [AIIMS 1987]
(a)
4 m / sec
(b)
6.25 m / sec
(c)
16m / sec
 None of these
Solution : (a) v = = = = 4 m / s
Problem 165. The ratio of velocities at points A, B and C in vertical circular motion is
(a)
1 : 9 : 25
(b)
1 : 2 : 3
(c)
1 : 3 : 5
(d) 1 : :
Solution : (d)
v A : vB : vc = : :
= 1 : :
Problem 166. The minimum speed for a particle at the lowest point of a vertical circle of radius R, to describe the circle is ‘v’.
If the radius of the circle is reduced to onefourth its value, the corresponding minimum speed will be
[EAMCET (Engg.) 1999]
 v
4
v (c) 2v (d) 4v
2
Solution : (b) v =
\ v µ
So v2 = =
v1
= 1 Þ v = v / 2

2
Problem 167. A body slides down a frictionless track which ends in a circular loop of diameter D, then the minimum height h
of the body in term of D so that it may just complete the loop, is [AIIMS 2000]
(a)
h = 5D
2
(b)
h = 5D
4
h = 3D
4
h = D
4
Solution : (b) We know h = 5 r
= 5 æ D ö = 5D
[For critical condition of vertical looping]
ç ÷
2 2 è 2 ø 4
Problem 168. A can filled with water is revolved in a vertical circle of radius 4m and the water just does not fall down. The time period of revolution will be [CPMT 1985; RPET 1999]
 1 sec (b) 10 sec (c) 8 sec (d) 4 sec
Solution : (d) At highest point mg = mw ^{2}r Þ g = 4p 2 r Þ 10 = 4p 2 4
Þ T ^{2} = 16
\ T = 4 sec
T 2 T 2
Problem 169. A particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles 30^{o} and 60^{o} from vertical (lowest position) are T_{1} and T_{2} respectively, Then [Orissa JEE 2002]
(a)
T1 = T2
T1 > T2
T1 < T2
T1 ³ T2
Solution : (b)
T = mg cosq + mv2
r
As q increases T decreases So T > T
1 2
Problem 170. A mass of 2kg
is tied to the end of a string of length 1m. It is, then, whirled in a vertical circle with a constant
speed of 5 ms^{1} . Given that
70 N
g = 10 ms^{2} . At which of the following locations of tension in the string will be
(a) At the top (b) At the bottom
 When the string is horizontal (d) At none of the above locations
Solution : (b) Centrifugal force F = mv2 = 2 ´(5)2 = 50 Newton
r 1
Weight = mg = 2 ´ 10 = 20 Newton
Tension = 70 N (sum of above two forces)
i.e. the mass is at the bottom of the vertical circular path
Problem 171. With what angular velocity should a 20 m long cord be rotated such that tension in it, while reaching the highest point, is zero
(a) 0.5 rad/sec (b) 0.2 rad/sec (c) 7.5 rad/sec (d) 0.7 rad/sec
Solution : (d) w = = = = 0.7 rad / sec
Problem 172. A body of mass of 100g
is attached to a 1m long string and it is revolving in a vertical circle. When the string
makes an angle of 60^{o} with the vertical then its speed is 2 m / s . The tension in the string at q = 60^{o} will be
(a)
89 N
(b)
0.89 N
(c)
8.9 N
(d)
0.089 N
Solution : (b)
T = mg cosq + mv2
r
= 0.1 ´ 9.8 ´ cos 60 + 0.1 ´ (2)2
1
= 0.49 + 0.4 = 0.89 Newton
Problem 173. A body of mass
2kg
is moving in a vertical circle of radius
2m . The work done when it moves from the
lowest point to the highest point is
(a)
80 J
(b)
40 J
(c)
20 J
 0
Solution : (a) work done = change in potential energy = 2 mgr = 2 ´ 2 ´ 10 ´ 2 = 80 J
Problem 174. A body of mass m
is tied to one end of a string of length l and revolves vertically in a circular path. At the
lowest point of circle, what must be the K.E. of the body so as to complete the circle [RPMT 1996]
(a)
5 mgl
(b)
4 mgl
(c)
2.5 mgl
(d)
2 mgl
Solution : (c) Minimum velocity at lowest point to complete vertical loop =
So minimum kinetic energy = 1 m(v ^{2} )
2
= 1 m( 2
5gl )^{2} = 5 mgl = 2.5 mgl
2
Conical Pendulum.
This is the example of uniform circular motion in horizontal plane.
A bob of mass m attached to a light and inextensible string rotates in a horizontal circle of radius r with constant angular speed w about the vertical. The string makes angle q with vertical and appears tracing the surface of a cone. So this arrangement is called conical pendulum.
The force acting on the bob are tension and weight of the bob.
mv ^{2}
From the figure
T sinq =
r
….(i)
and
T cosq
= mg
….(ii)
 Tension in the string : T = mg
T = mg = mgl
cosq
[As cosq = h = ]
l l
v 2
(2) Angle of string from the vertical : tanq =
rg
(3) Linear velocity of the bob : v =
 Angular velocity of the bob : w = = =
(5) Time period of revolution : TP
= 2p
= 2p
= 2p
= 2p
Problem 175. A point mass m is suspended from a light thread of length l, fixed at O, is whirled in a horizontal circle at constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are [AMU (Med.) 2001]
 T
 T
 T
 T
F F F F
W W W
Solution : (c) Centrifugal force (F) works radially outward, Weight (w) works downward
Tension (T) work along the string and towards the point of suspension
Problem 176. A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2/p revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is [BHU 2002]
(a) ML (b) 2 ML (c) 4 ML (d) 16 ML
Solution : (d)
T sinq = Mw ^{2}R
T sinq = Mw ^{2} L sinq
From (i) and (ii)
….. (i)
….. (ii)
2 2 2
_{2}æ 2 ö^{2}
T = Mw
L = M 4p n L
= M 4p
ç p ÷
L = 16ML
è ø
Problem 177. A string of length 1m is fixed at one end and a mass of 100gm is attached at the other end. The string makes
2 / p rev/sec around a vertical axis through the fixed point. The angle of inclination of the string with the
vertical is ( g = 10 m / sec ^{2} )
(a)
tan^{1} 5
8
(b)
tan^{1} 8
5
(c)
cos ^{1} 8
5
(d)
cos ^{1} 5
8
Solution : (d) For the critical condition, in equilibrium
T sinq = mw ^{2}r
w ^{2}r
and
T cosq = mg
\ tanq =
g
Þ 4p ^{2}n^{2}r =
g
4p ^{2} (2 / p )^{2} .1 = 8
10 5
Problem 178. If the frequency of the rotating platform is f and the distance of a boy from the centre is r, which is the area swept out per second by line connecting the boy to the centre
(a)
prf
(b)
2prf
(c)
pr ^{2} f
(d)
2pr ^{2} f
Solution : (c) Area swept by line in complete revolution = pr ^{2}
If frequency of rotating platform is f per second, then Area swept will be p r ^{2} f
per second.
Problem 179. Figure below shows a body of mass M moving with uniform speed v along a circle of radius R. What is the change in speed in going from P_{1} to P_{2}
(a) Zero (b)
 v /
 2v
Solution : (a) In uniform circular motion speed remain constant. \change in speed is zero.
Problem 180. In the above problem, what is change in velocity in going from P_{1} to P_{2}
(a) Zero (b) (c) v / (d) 2 v
Solution : (b) Change in velocity = 2v sin(q / 2) = 2v sinæ 90 ö = 2v sin 45 = 2v = 2 v

ç ÷
è ø
Problem 181. In the above problem, what is the change in angular velocity in going from P_{1} to P_{2}
(a) Zero (b)
2 v / R
 v / 2 R
(d)
2v / R
Solution : (a) Angular velocity remains constant, so change in angular velocity = Zero.
Problem 182. A particle of mass m
is fixed to one end of a light spring of force constant k and unstretched length l . The
system is rotated about the other end of the spring with an angular velocity w , in gravity free space. The increase in length of the spring will be
 mw2l
k
mw ^{2}l
k – mw ^{2}
mw ^{2}l k + mw ^{2}
 None of these
Solution : (b) In the given condition elastic force will provides the required centripetal force
k x = mw ^{2}r
k x = mw ^{2} (l + x) Þ k x = mw ^{2}l + mw ^{2} x Þ x(k – mw ^{2} ) = mw ^{2}l
\ x =
mw ^{2}l k – mw ^{2}
Problem 183. A uniform rod of mass m
and length l rotates in a horizontal plane with an angular velocity w about a
vertical axis passing through one end. The tension in the rod at a distance x from the axis is
(a)
1 mw ^{2}x
(b)
1 mw 2 x ^{2}
(c)
1 mw 2 æ x ö
(d)
1 mw 2 [l ^{2} – x ^{2}]
2 2 l
ç ÷


2 è ø 2 l
Solution : (d) Let rod AB performs uniform circular motion about point A. We have to calculate the tension in the rod at a distance x from the axis of rotation. Let mass of the small segment at a distance x is dm
So dT = dmw ^{2} x = æ m ö dx .w ^{2} x = mw 2
[x d x]
ç l ÷ l
è ø
l mw 2 l
mw ^{2} é x ^{2} ù l
Integrating both sides ò dT = l ò x dx Þ T =
l ê 2 ú
x x
\ mw 2 [ 2 2 ]
ëê úû x
T = 2l l – x
Problem 184. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration a . If the coefficient of friction between the rod and the bead is m , and gravity is neglected, then the time after which
the bead starts slipping is [IITJEE (Screening) 2000]
(a)
(b)
(c)
1 (d) Infinitesimal
Solution : (a) Let the bead starts slipping after time t
For critical condition
Frictional force provides the centripetal force mw ^{2}L = m R = m m ´ at =mmLa
m (at)^{2}L = mmLa Þ t = (As w = at)
Problem 185. A smooth table is placed horizontally and an ideal spring of spring constant k = 1000 N / m
and unextended length
of 0.5m
has one end fixed to its centre. The other end is attached to a mass of 5kg
which is moving in a circle with
constant speed 20m / s . Then the tension in the spring and the extension of this spring beyond its normal length are
(a)
500 N, 0.5 m
(b)
600 N, 0.6 m
(c)
700 N, 0.7 m
(d)
800 N, 0.8 m
Solution : (a) k = 1000, m = 5 kg, l = 0.5 m, v = 20 m / s (given)
Restoring force = kx = mv2 = mv 2 Þ 1000 x = 5 (20)2 Þ x = 0.5 m
r l + x 0.5 + x
and Tension in the spring = kx = 1000 ´ 1
2
= 500 N
Problem 186. A particle describes a horizontal circle at the mouth of a funnel type vessel as shown in figure. The surface of the funnel is frictionless. The velocity v of the particle in terms of r and q will be
 v =
 v =
 v =
 v =
/ cot q
mv^{2}
Solution : (c) For uniform circular motion of a particle
and mg = R sinq
Dividing (i) by (ii)
= R cosq
r
….(i)
….(ii)
v 2 = cot q
rg
Þ v =
Problem 187. Figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a
spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that
the block presses the track with a force mg when it reaches the point P [see. Fig], where the radius of the track is horizontal
(a)
(b)
(c)
mv ^{2}
(d)
Solution : (c) For the given condition, centrifugal force at P should be equal to mg i.e. ^{P} = mg \ vP =
R
From this we can easily calculate the required velocity at the lowest point of circular track.
v ^{2} = v ^{2} – 2gR (by using formula : v ^{2} = u ^{2} – 2gh )
p L
vL = = =
It means the block should possess kinetic energy = 1 mv ^{2} = 1 m´ 3gR
2 ^{L} 2
And by the law of conservation of energy
1 kx ^{2} = 1 3m ´ g R
Þ x = .
2 2