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



Circular Motion

2

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



Circular Motion

3

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



Circular Motion

4

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



Circular Motion

5

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 =



Circular Motion

6

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=



Circular Motion

7

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

=



Circular Motion

8

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



Circular Motion

9

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



Circular Motion

10

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 ′ =



Circular Motion

11

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)



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



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

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1. Circular Motion S

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