Systems of Particles and Rotational Motion
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Question : 30
Total: 33
A solid disc and a ring, both of radius 10 cm are placed on a horizontaltable simultaneously, with initial angular speed equal to 10 π rad s – 1 . µ k = 0.2 .
Solution:
Here, radius of both the ring and solid disc
R = 10 c m = 0.1 m , µ k = 0.2
Moment of inertia of the solid disc=
M R 2
Initial angular velocityω 0 = 10 π rad s – 1
Initial velocity of centre of mass is zero. Frictional force causes the C.M. to accelerate
F = m a ; m a = µ k m g
a = µ k g . . . ( i )
Now,v = u + a t
v = µ k g t ( ∵ u = 0 ) . . . ( i i )
Torque due to friction causes retardation in the initial angular speedω 0
µ k m g × R = − I α
α =
. . . ( i i i )
As ω = ω 0 + α t
∴ ω = ω 0 −
. . . ( i v )
Rolling begins, whenv = R ω
From (ii) and (iv),
µ k g t = R ω 0 −
. . . ( v )
For a ring, I = m R 2
∴ µ k g t = R ω 0 −
µ k g t = R ω 0 − µ k g t 2 µ k g t = R ω 0
t 1 =
. . . ( v i )
For a disc,I =
m R 2
∴ From equation (v)µ k g t = R ω 0 −
µ k g t = R ω 0 − 2 µ k g t
3 µ k g t = R ω 0 . . . ( v i i )
t 2 =
PuttingR = 0.1 m , ω 0 = 10 π rad s − 1 ,
µ k = 0.2
g = 9.8 m s − 2
t 1 =
= 0.8 s
t 2 =
= 0.53 s
It is clear thatt 2 < t 1
Moment of inertia of the solid disc
Initial angular velocity
Initial velocity of centre of mass is zero. Frictional force causes the C.M. to accelerate
Now,
Torque due to friction causes retardation in the initial angular speed
Rolling begins, when
From (ii) and (iv),
For a ring
For a disc,
∴ From equation (v)
Putting
It is clear that
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