1. circular motion s
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1111 CIRCULAR MOTIONCIRCULAR MOTIONCIRCULAR MOTIONCIRCULAR MOTION
CET WORKSHEETS SOLUTIONS
Circular Motion
Distance and Displacement1. b) displacement
Displacement may be zero because
final position of the particle maycoincide with its initial position.
2. c) It is always positive
Displacement may be positive,negative or zero as it is a vector
quantity.
3. d)
This is because distance travelled cannever be negative.
4. b)
Relative speed = 0.When velocity of A = velocity of B∴displacement-time graphs of A andB must have same slope (but other than
zero).
5. b) 2r
BC = displacement2 2 2
BC AB AC = +
2 2 2 BC r r r ∴ = + =
6. a) / 2, 2π
Distance =
1.414 2
2 2 2 2
r ×= = =
π π π π
1.414 2
2 2 2 2
r ×= = =
π π π π
Displacement = 2 2 1.414r = ×
= 2 2 2× =
7. d) 1.14 m
AB = displacement = diameter of circle
AB = 2m
Distance = ADB =2
2
r r =
π π
= 1 3.14m× =π
Difference in displacement and
distance= ADB – AB = 3.14 – 2 = 1.14 m
8. d) both ‘a’ and ‘b’
B = initial position
A = 1/4th positionC = 3/4th position
2 2 2 AB OA OB∴ = + 2 2 2
AB r r ∴ = +
2 22 AB r ∴ = 2 2 AB r ∴ = .
Similarly, for position C.9. b) moves with velocity v tangential to
orbit
The gravitational force is the centripetalforce, which keeps the satellite bound to
Earth. As it disappears, satellite no longer
continues C.M.10. a) zero
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Circular Motion
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Scalar product ( ). sin cosr v x=
θ
But, 090∝=
cos 0∴ ∝=
∴scalar product = 0.
Circular Motion
11. b) 0 0 0 M L T Dimensionless
12. a) vector
δθ in U.C.M. is vector.
13. c) 2π
14. c) perpendicular to plane of circle
15. c) tangent outward16. c) tangent outward17. a) same as that of angular displacement
Direction of angular displacement andangular velocity vector is same.
18. d) changes in direction
19. c) i and iii are correct
20. a) 00
Angular velocity is perpendicular to
plane and passing through centre. It isoutward when motion is anticlockwiseand inward when motion is clockwise
so it will change only when direction
of motion changes. So it is 00 .
21. b) π
Position vector and centripetalacceleration are antiparallel to each other.Therefore angle between them is π .
22. c) angular acceleration23. c) the angular acceleration is zero24. d) passing through origin with a slope ω
ω is constant so V ∝ r graph between
v and r is a straight line passingthrough the origin with slope ω .
25. b) parallel to x (distance) axisAs ω is constant the straight line
graph is parallel to x-axis.
26. c) zeroTangential acceleration is zero in
U.C.M. as in U.C.M. there is only
radial acceleration.
27. b)20 / m s
The tangential acceleration is zero in
U.C.M.28. d) both velocity and acceleration change
In U.C.M. direction of velocity andacceleration change from point topoint.
29. c) periodic and non simple harmonicIn U.C.M. motion of particle isperiodic i.e. repeats in equal interval of
time. It is non simple harmonic as it isalong the circle and not in a straightline.
30. c) kinetic energyK.E. depends upon speed and not on
velocity i.e. independent of directionof motion.
31. c) will move linearly along the tangent tothe circle
Since nucleus is removed, centripetalforce (i.e. force of attraction)disappears. Hence, due to tangential
velocity of electron, it moves linearlyalong tangent.
32. b) horizontally to the South
The particle leaves the groundhorizontally to the south.
33. b) revolution of Earth round the Sun
Revolution of Earth round the Sun.34. c) angular momentum is constant and
linear momentum is changing
Velocity is changing continuously,linear momentum is changing
( ).P m V = and . , L I = ω ω is constant
therefore, angular momentum is
constant.
35. a) is zero at the poles and maximum atequator
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Circular Motion
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Earth is rotating about an axis passingthrough its poles. So particles whichare on poles remain their so distance
will be zero so speed is zero but at
equator distance covered is maximumso speed is maximum.
36. a) 1 : 1It is U.C.M. so, frequency ratio is 1:1.
37. c)40
Hzπ
572 / 72 20 /
18v km hr m s= = × =
v r = ω
2
20 200080
2525 10
v
r −= = = =
×ω
2 80n =π 80 40
2n Hz= =
π π
38. d) –41.453 10 / rad s×
2 2 3.14
12 60 60T
×= =
× ×
π ω
6.28 / sec
43200rad =
–41.453 10 / rad s= ×ω
39. a) 0.1
2 2 180
60 60T = = ××
π π
ω π
10.1deg/
10s= =
40. b) 1:60
2 / 60 60 1
2 / 60 60
m
s
×= =
ω π
ω π
41. d) Smaller than that of hour hand of clock
.
2 2
12 60 60hr hand
T = =
× ×
π π ω
1 /
21600rad s=
2 2
24 60 60 Earth
T = =
× ×
π π ω
1 /
43200rad s=
Earth hr.handω < ω
42. a) 0120
t θ ω =
02 220 180 120
60 3
π θ = × = × =
Or
5 min ... 030
20 min … 020 30120
5
×=
43. d) 0.35 rad
Angle =0.35
0.351
arcrad
radius= =
44. d) 1:1
( )1 11 2
2 2
2 /
2 /
T T T
T = =
ω π
ω π
1 2=ω ω
1
2
1
1=
ω
ω
45. d) 21.046 10 / m s−×
2 210 10
60v r −= = × ×
π ω
2 210 1.046 10 / 3
m s− −= × = ×π
46. a) 35 m/s
0.5 70 35 / v r m s= = × =ω
47. a) 4.8 / m sπ
.2v r r n= =ω π
2 24060 10 2 3.14
60
−= × × × ×
2
480 3.14 10
−
= × × 4.8 / v m s= ×π
48. d) 4.0 / m sπ
240 6010 2
2 60v r −= = × × ×ω π
mid point =40
2 2
r =
240 10 0.4 / m s−= × =π π
49. d) 10 : 9v r ω =
1 1 1
2 2 2
v r
v r =
ω
ω 1 2
1 2
2
T T T
=
= ∴ =
π ω
ω ω
300 30 10
270 27 9= × = =
ω
ω
50. d) 2 secπ
2
5=
π θ
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Circular Motion
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300 1
1500 5
v
r = = =ω
Now,t
= θ
ω
22 sec
15
5
t ∴ = = =
×
θ π π
ω
51. a) 22 28.966 10 / m s×
( )2
62
10
2.18 10
0.53 10
va
r −
×= =
×
22 28.966 10 / a m s= × .
52. c) 212.3 / m s
2 2 5 5
4 2
r
v T
π π
π
×
= = =
( )2 22
25 / 2 25
5 4 5r
va
r
π π = = = ×
×
259.68 12.3 /
4m s= × = .
53. d) Remain samea v ω = ×
1 22
va aω ω = × =
1a a=
54. b) apply brakesTo avoid accident velocity should bereduced so it is better to apply to
brakes.
55. d) 21.256 / rad s−
( )2 12 12 n n
t t
π ω ω α
−−= =
( )2 0 2 2 2
10 10
π π − × ×= = −
2 2 3.14 6.28
5 5 5
π ×= − = − = −
2
1.256 / rad s= −
56. b)2 /
2rad s
π
2 1
t
ω ω α
−=
2 48 20 28θ π π π = − =
2 1 28 28
4 4
4
t t
t
θ θ π π − −
= =
27 5 2 /
4 4 2rad s
π π π π −= = =
57. b) 0.25 sec
21.5 3 2t t ω = − + d
dt
ω α =
1.5 3 2 0 1.5 6t t α = − × + = −
0 = 1.5 – 6t
6t = 1.5
1.50.25sec
6t = =
58. d) 800 : 1Centripetal acceleration
2 0.2 40 40r a r ω = = × ×
2320 / m s= .Tangential acceleration
2 1t a r r
t
ω ω α
− = =
40 2 380.2 0.2
19 19
− = × = ×
20.4 / m s=
320 800
0.4 1
r
t
a
a∴ = =
: 800 :1r t a a =
59. b) 24 / m s 2 2r t a a a= +
( )22 30 30 30
300 300r
va
r
×= = =
23 / m s= 27 / t a m s=
( )2
23 7 9 7a = + = +
16= 2
4 / a m s= 60. a) 75 rad
24 / ,rad sα = 1 5 / ,rad sω = t = 5 s
21
1 15 5 4 25
2 2t t θ ω α = + = × + ×
25 50= +
75rad θ =
61. b) the frictional force of the wall balances
his weight
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Circular Motion
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mg = µ N
mg = 2mr µ ω
2 g
r ω
µ =
2 g
r ω
µ =
µ = coefficient of friction
N = normal reaction
62. a) same at all points63. b) opposite to the direction of angular
velocity
64. c) 014 19′
tanv rg θ =
518 10 9.8 tan
18θ × = × ×
25 98 tanθ = ×
tan 25 98θ = ÷
tan 0.2551θ = 014.31θ ∴ =
014 19′≈
65. d) 2 1tan 4tanθ θ =
2 tanv rg θ =
21 1
222
tan
tan
v rg
rgv
θ
θ =
21 12
22
tan
tan
v
v
θ
θ =
1
2
1 tan
4 tan
θ
θ =
2 1tan 4tanθ θ ∴ =
66. c) 1.4 m/s
2 2 0.1 9.8v rg= = × ×
1.96 1.4 / m s= =
67. c) 20.6 mHere velocity of circular motion
0.8 2 4 6.4 / v r m sω π π = = × × = K.E. = P.E.
2
2
mvmgh=
2 6.4 6.420.6
2 2 9.8
vh m
g
π π ×∴ = = =
×
68. b) null vector
69. b) 1 : 3
1 1 2 2r r ω ω =
i.e. 2 1
1 2
r
r
ω
ω =
70. a)29
/ 4
rad sπ
−
2 20 2ω ω α θ = +
0 9 2 2α π = + ×
29 /
4rad sα
π
−=
71. c) rCentripital force = force friction
2 21 1 2 2mr mg mr ω µ ω ∴ = =
2 21 1 2 2r r ω ω ∴ =
( ) ( )221 2 14 2r r ω ω =
2 21 2 14 4r r ω ω =
2r r =
Centripetal Force and Centrifugal Force
72. a)2
1
r
r
Centripetal force is2mv
F r
=
i.e.1
F
r
∝
1 2
2 1
F r
F r ∴ =
73. a) centripetal forceCentripetal force is necessary for
circular motion.
74. c) zeroWork done by the force in uniformcircular motion is zero.
Work done = . cosF s Fs θ =
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Circular Motion
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Centripetal force is alwaysperpendicular to velocity hence todisplacement. So angle between F and
s is
0
90 and cos 90 = 0. Hence workdone = 0.75. b) centrifugal force balances the force of
gravityIn this case magnitudes of centrifugal
force and gravitational force are equal
and opposite in direction. Hencecentrifugal force balance the force of
gravity.
76. b) due to the lack of proper centripetalforceFor circular motion, centripetal force is
necessary. If a car is thrown out of
road, it is due to lack of centripetalforce.
77. c) friction between the coin and therecordThe centripetal force is provided by
friction between the coin and therecord.
78. d) the centripetal force does not suffer
any change in magnitudespeed is unchanged so magnitude ofcentripetal force remains same but
direction is changed as direction ofvelocity is reversed.
79. c) generate required centripetal forceturning means motion on a curvedpath, which requires centripetal force.
80. b) to provide centripetal force
81. b) 134 10 N −× 2F mr ω =
( )2
27 71.6 10 0.1 5 10−= × × × ×
28 141.6 10 25 10−= × × × 13
4 10F N −= ×
82. a) 431.7 N
2 2 3000 10060 60
nπ π ω π ×= = =
mass of each link (m) =2.5
100
Radius of path (r) =1.1
2π
( )22 2.5 1.1
100100 2
F mr ω π π
= = × ×
2.57 50 137.5 431.7 N π π = × = =
83. b)2
2
MLω
mass is concentrated at centre of tube
i.e. at distance2
L from one end
22
2
mLF mr
ω ω = =
84. b) half that of initial value2T mr ω =
2
T r
mω =
( )1
1 2 2 21
2 2
42
T T T r
m mmω ω ω = = =
××
212
T mω
= ×
1
1
2r r = ×
85. a) 5 rad/s2Fmr ω
2 . . 50 5025
1 2 2
F B T
mr mr ω = = = = =
×
5 / secrad ω =
86. c) sec5 2
π
Breaking Tension = Centripetal Force2mr ω =
22
60 0.3 1T
π = × ×
2
2
460 0.3
T
π = ×
2 22 4 0.3
60 50T
π π ×= =
sec50 5 2
T π π
= =
87. a) 4.72 Hz2T mr ω =
( )2100
45 9.8 5 21000
nπ × = × ×
2 2145 9.8 5 4
10nπ × = × ×
24.5 9.8 2 9.8n× = ×
245 2n=
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Circular Motion
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245
2n=
222.5n =
n = 4.72 revolution/sec.88. d) 9 times
2mv
F r
= or 2F v∝
22 2
21 1
F v
F v=
2
22 1
1
vF F
v
=
2
2 1
3vF F
v
=
( )2
2 13F v F =
22 1 19 9F v F F = =
89. b) 68 10 / m s×
Centripetal force = electrostatic force2mv
F r
=
2 .F r v
m=
7 10
31
9 10 0.64 10
9 10
Fr v
m
− −
−
× × ×= =
×
664 10= × 68 10 / secv m= ×
90. c) 0Centrifugal force and position vectorhave same direction. Therefore angle
between centrifugal force & positionvector is 0.
91. b) 2 2 Mk r t
The centripetal acceleration is2 2 2 / ca k rt v r = =
2 2 2v k r t ∴ =
v krt ∴ =
The tangential acceleration is
( ) 1t
dv d a krt kr kr
dt dt = = = × =
The tangential force on the particle is,
t t F M a M kr = =
Work is done on the particle only by
tangential force, as radial force are
perpendicular to v.Thus, power delivered to the particle
is, ( )( ) 2 2t t P F v M kr krt Mk r t = = =
92. c) it is in the accelerated frame ofreference
Force in accelerated frame of referenceis called as pseudo force.
93. d)1 1 2 M L T −
94. b) centrifugal force
The cream is separated by centrifugal
force.95. c) centrifugal force
Water is removed by centrifugation96. b) can not be an inertial frame because
the Earth is revolving round the sun
97. c) an oblate spheroid98. a) 28 m/sec
2 316 6S t t t = + −
216 6 2 3ds
v t t dt
= = + × −
0 12 6dv
a t dt
= = + −
If velocity is maximum (constant) so
acceleration will be zero.
0 12 6dv
t dt
= = −
12 2sec6
t = =
So 2max 16 12 3v t t = + −
( )2
16 12 2 3 2 16 24 12= + × − = + −
max 28 / secv m=
99. d) 8r2T mr ω =
2
T r
mω ∴ =
11
1
T r
mω
=
1 2 2 2
2 2 4 8
2
T T T r
m mm
ω ω ω
×= = =
1 8r r =
100. a) the car will skid2mv
T r
=
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Circular Motion
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2300 max500
50
v×=
2 500
max 83.36v = =
max 9.12 / v m s=
Upto this maximum speed car will notskid but here speed is more than thissafe speed so car will skid.
101. c)1
2
Mg
mlπ
Mg = Centripetal force2 Mg mlω =
2 Mg
mlω =
, Mg
mlω ∴ = 2
Mgn
mlπ =
1
2
Mgn
mlπ ∴ =
102. a) uniform circular motion103. c) P has greater acceleration then Q
2P pa r ω = 2
Q Qa r ω =
As, P Qr r > and a r ∝ (ω is constant)
104. a)2
Fr
2mvF
r =
21.
2 2
Fr K E mv= =
105. b) Zero
sinC C r F rF uτ θ = × =
sinrF τ θ =
But r
and centripetal force are along
same line and in opposite direction0180θ ∴ =
0τ =
106. c)2
2v
r
Direction of centripetal acceleration is
opposite after completing half circle.
107. b) 216 / m s 2 2
2 1 2 1F F mr mr ω ω − = −
( )2 21 148 4 4r ω ω = −
2112 3r ω =
Initial centripetal acceleration2 21 4 / r m sω =
Final centripetal acceleration22r r ω = =
214r ω =
216 / m s=
108. c) 20 N
21
2kE mV =
21
2
mV r
r = ×
1
2kE F r ∴ = × ×
1200 20
2F ∴ = × ×
20F N ∴ =
Horizontal Curve Road109. c) the frictional force produced between
the wheels and the road
The necessary centripetal force isprovided by frictional force between
wheels and road.
110. a)2mv
mg
r
< µ
i.e. the force of friction must be greater
than the centrifugal force
111. d) v rg= µ
2mv
rgr
= µ
2v rg= µ
v rg∴ = µ
112. c) equal v
v rg= µ
It does not depend on mass of vehicleso maximum speed is common for all
vehicle.113. a) the inner wheel leaves the ground first
The car overturns when the reaction of
the inner wheel is zero. So inner wheel
leave the ground first.
114. c) 14Maximum speed is given by
v rg= µ
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Circular Motion
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0.2 100 9.8 196 14 / v m s= × × = =
115. a) / 2µ
Maximum spee is
v rg= µ
i.e. v ∝ µ
2v∴ ∝ µ
2 2 25 1 1
25 2 2
w w
d d
v
v
µ = = = =
µ
2 2
d w
µ µ∴µ = =
116. d) 0.4 m/s
0.16 0.1 10v rg= µ = × ×
0.16 0.4 / m s= =
117. b) 180 mv = 108 km/hr
5108
18v = ×
( )22 108 5 / 18
1800.5 10
vr m
g
×= = −
µ ×
118. d) / g r ω = µ
C.P.F. = force of friction2
mr mgω = µ
2 g
r µω =
g
r
µ∴ω =
119. c) 0.5 HzForce of friction = C.P.F.
2mg mr µ = ω
2 g
r
µω =
g
r
µω =
1
2
gn
r
µ=
π
Frequency does not depend upon mass
so frequency remains same i.e. 0.5 Hz.
120. a) 10.15 cm
( ) ( )2 2
0.4 10 601
602 2 1
gr n
n
µ × = = = =
π π ×
4 10.1015
4 3.14 3.14 9.85m= = =
× ×
10.15r cm=
121. b) 2961 N2
T mr ω = 2 24 2 1.5 4 9.86 25mr nπ = = × × × ×
T = 2961 N
122. d) 7 rpm2T mr ω =
2T ω ∴ ∝ i.e. 2
T n∝
21 1
22 2
T n
T n∴ =
2 2 21 12 1
2 1
25 50
T T n n
T T
∴ = = × =
2 7 . . .n r p m∴ ≈
123. b) 0.5 %2F mlω =
2
2
4F ml
T
π =
2 22
2 2
4 4ml mlT
F F
π π ′= =
2
2
T l
lT
′ ′∴ =
but 1 1011 1% 1100 100
l l l′ = + = + =
101
100
l
l
′=
2
2
101 10.05,
100 10
T T
T T
′ ′∴ = =
10.051 1
10
T
T
′∴ − = −
0.05
10
T T
T
′ −=
∴percentage increase in time period
0.05100 100 0.5%
10
T T
T
′ −× = × =
Angle of Banking124. c) the vertical component of normal
reaction of car125. d) sin N θ
126. a)
2v l
Rg
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Circular Motion
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sin h
lθ =
2
tan v
Rgθ =
θ is small
2h v
l Rg=
2v lh
Rg=
127. b) outward direction
Possibility of vehicle to skid is inoutward direction as centripetal force
required to keep vehicle in circularmotion is inward so, when vehicle skidit is in outward direction.
128. c) independent of the mass of vehicle
max tanV rg θ =
129. a) 1000 N
Centripetal force = 200 N
( )1sin 0.2θ −
=
sin 0.2θ =
N sin θ = Centripetal force
200
sin 0.2
Centripetal force N
θ
= =
= 1000 N
130. b) 14 m/s
1tan 20 3 9.8
3v rg θ = = × ×
196 14 / m s= =
131. c) 25 m/s
Radius of track =are length
angle
tanv rg θ =
( )0 1159.23 9.8 tan 21 49= × ×
159.23 9.8 0.4003 624.669= × × =
v = 24.99 = 25 m/s
132. d)1 1
tan5
−
( )22 10 100 1
tan50 10 500 5
v
rgθ = = = =
×
1 1tan
5θ
− =
133. c) 1tan 0.02−
2 1sin
100 50
h
lθ = = =
sin 0.02θ =
θ is small so sinθ θ =
0.02θ =
tan 0.02θ = 1tan 0.02
−=θ
134. b) 10 cm2h v
l rg=
( )2
20 400 1
400 10 400 10 10
lh m
×= = =
× ×
10010
10cm cm= =
135. a) 10.8 km2 tanv rg θ =
( )22
0
150
tan tan12
vr
g gθ = =
×
3150 15010.8 10 10.8
9.8 0.2126m km
×= = × =
×
136. a)
22
tan
v
g
π
θ
If length of road is L
2 L r π =
2
tan v
rgθ =
2
tan
vr
g θ ∴ =
2 222
tan tan
v v L
g g
π π
θ θ
= × =
137. b) 14 m/s
0.5 40 9.8 14 / v rg m s µ = = × × =
Conical Pendulum
138. b) speed of revolution is almost infinite139. a) equal to that of simple pendulum of
same length l cos θ
140. c) 2.5 N
35
44n Hz=
2sinT mr θ ω ′ =
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Circular Motion
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2 2sin sin 4T m l nθ θ π ′ = × × × 2 24T ml nπ ′ = × × ×
( )
21 35
10 1 4 3.14 44
− = × × × ×
135 354 9.86 10
44 44
−×= × × ×
×
112078.510
484
−= ×
124.95 10−= ×
= 2.49 N≈ 2.5 N
141. b) 1.4 sec
cos 1 cos602 2 3.14
9.8
lT
g
θ π
×= = ×
16.28
2 9.8=
×
6.28 0.2258= ×
= 1.4 sec
142. d)2l
gπ
cos cos602 2
l lT
g g
θ π π = =
12
2
l
g
π ×
=
4
2
l
gπ
×=
2l
gπ =
Motion in Vertical Circle143. b) different at different points on circle144. c) neither kinetic energy nor potential
energy is constant145. c) near the neck
Near the neck as due to centrifugal
force soda water will move away fromcentre and lighter bubbles collect nearthe neck
146. c) 2 15v v=
1v rg=
2 15 5 5v rg rg v= = × = ×
147. a) mg + T
Tension at lowest point
2
cosmv
T mg R
θ = +
0180θ =
cos 1θ = −
T = F = mg
net F T mg= +
148. a) minimum
149. a)when the stone is at the bottom of thecircle
At top2
A
mvT mg
r = =
At bottom2
B
mvT mg
r = +
B AT T >
String breaks at bottom B.
150. c) 2 mgr
( ) ( )5 1
. . . .2 2 L H
K E K E mgr mgr − = −
42
2mgr mgr = =
151. b) 5 2 / m s
For a motor cyclist to complete the
vertical circle2mv
mgr
=
2vg
r =
2v gr =
v gr =
10 5 50 25 2 5 2 / m s= × = = × =
152. b) 2gR
In hemisphere, at the bottom position,P.E. = K.E.
21
2mgR mv=
2 2v gR∴ =
2v gR=
153. a)5
2 R
For just completing vertical circle, the
velocity required at lowest point A, is
5v gR= … (i)
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Circular Motion
12
When the ball falls, its potential energyis converted into kinetic energy
21
2
mgh mv=
2 5 5
2 2 2
v gRh R
g g= = =
154. b)5
4
Dh =
Potential energy = kinetic energy
21
2mgh mv=
2 5 5 5 5
2 2 2 2 2 4
v rg D Dh r
g g= = = = × =
155. a)5
2 mgr
Total energy 21
2mv mgh= +
2 21 10
2 2mv mv= +
( )21 5
52 2
m rg mrg= =
156. d) 45 N2mv
T mgr
= +
2 253 105
vm gr
= + = +
= 3[15] = 45 N.
157. a)2
mvmg
r =
i.e., centrifugal force balances the
weight of water in bucket.158. b) at the bottom of circle
2
cosmv
T mgr
θ = +
21 5
22.5 1 10cos2 θ
×
= + ×
22.5 12.5 10 cosθ = +
10 10cosθ =
cos 1θ ∴ = 0
Oθ ∴ =
Therefore the stone is at bottom of
circle.159. b) 5 rad/s
max 5v rg=
max 5r rgω =
max
5 5 10
2
g
r ω
×= =
max 25ω =
max 5 / rad sω ∴ =
160. b) 12.52 m/sIf the vehicle does not loose the
contact at the highest point ofoverbridge, then
Its weight in Centrifugal force
downward onitinoutward
direction direction
=
2mvmg
R=
v Rg=
(R = 15.5 + 0.5 = 16 m)
16 9.8 4 9.8 12.52 / m s= × = =
161. c) 31.24 10 / secrad −
×
Centripetal force = weight of body2
mr mgω =
2 g
Rω =
6
9.8
6.4 10
g
Rω = =
×
31.24 10 / rad s−= ×
162. c) 67.55 rpm
Centripetal force on 1m = Weight of
2m
21 2m r m gω =
2 g
r ω = ( )1 2m m=∵
1050
0.2
g
r ω = = =
2 nω π =
50 2 3.14 n= × ×
50
2 3.14n=
×
1.125n rps∴ =
67.55rpm=
163. a) 00
Tension in the string is
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Circular Motion
13
2
cos mv
T mgl
θ = +
( )2
4 4103.2 4 9.8cos 1θ
×= × +
103.2 4 9.8cos 4 16θ = × + ×
103.29.8cos 16
4θ = +
25.8 9.8cos 16θ = +
9.8cos 9.8θ =
cos 1θ = 00θ = .
164. c)3
Rh =
At point P net centripetal force is givenby2
cosmv
mg N R
θ = −
When particle will leave the circle,N = 0
2
cosmv
mg R
θ = … (i)
We know, 2 2 2v u as− =
Body is initially at rest so u = 0,2 2v as=
Putting a = g and s = h, we get2 2v gh= … (ii)
From fig, cos R h
Rθ
−= … (iii)
Putting value of cos θ and 2v in
equation (i), we get
( ) ( )2gh R hm mg
R R
−=
R – h = 2h
3
Rh =
Bridges
165. b)2mv
mg R
−
Net force = weight – centripetal force2
mvmg
R= −
166. d) 44.3 m/s
The force acting on the motorcycle athighest point is
2mv
R mgr
= −
Condition for motorcycle not loosing
contact with bridge is 0 R ≥ 2
Maxmvmg
r
≥
2 Maxmv
mgr
=
2maxv gr =
10 196 1960 Maxv gr = = × =
= 44.3 m/s.
167. d)2mv
mg R
+
168. b) decreases