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Question

A sphere of mass M and radius r shown in figure slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotational velocity about the centre v02r Find the translational velocity after the sphere starts pure rolling.


A
2v0
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B

52v0

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C

v0

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D

67v0

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Solution

The correct option is D

67v0


Velocity of the centre =v0 and the angular velocity about the centre= v02r Thus, v0>ω0r.The sphere slips forward and thus the friction by the plane on the sphere will act backward. As the friction is kinetic, its value is μN=μMg and the sphere will be decelerated by acm=fM.

Hence,

v(t)=v0fMt. .......(i)

This friction will also have a torque l = f r about the centre. This torque is clockwise and in the direction of ω0. Hence the angular acceleration about the centre will be

α=fr(25)Mr2=5f2Mr

and the clockwise angular velocity at time t will be

ω(t)=ω0+5f2Mrt=v02r+5f2Mrt

pure rolling starts when v(t)=rω(t)

i.e., v(t)=v02+5f2Mt ...(ii)

Eliminating from (i) and (ii),

52v(t)+v(t)=52v0+v02

or, v(t)=27×3v0=67v0.

Thus, the sphere rolls with translational velocity 6v07 in the forward direction.

Alternative: Let us consider the torque about the initial point of contact A. The force of friction passes through this point and hence its torque is zero. Thenormal force and the weight balance each other. The net torque about A is zero. Hence the angular momentum about A is conserved.

Initial angular momentum is,

L=Lcm+Mrv0=Icmω+MRv0

=(25Mr2)(v02r)+Mrv0=65Mrv0.

Suppose the translational velocity of the sphere, after it starts rolling, is v0. The angular velocity is v/r. The angular momentum about A is,

L=Lcm+Mrv

Thus, 65Mrv0=75Mrv

or, v=67v0


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