State of motion of the sphere at t = 0 is described in figure, Centre of the sphere is moving towards right with a speed u and angular velocity of the sphere is u/2r in the anticlockwise sense. Here r is the radius of sphere. We can understand that it is a case of slipping and after some time sphere starts pure rolling due to friction from the surface. We can assume direction towards right as positive and thus clockwise sense should be treated positive for rotational motion. Assuming coefficient of friction between sphere and the surface. Mass of sphere is m.

At what time sphere stops rotating for a moment?

Options

A. `(u)/(5mug)`
B. `(2u)/(5mug)`
C. `(3u)/(7mug)`
D. `(2u)/(7mug)`

Answer

A

Explanation

Here we can see that point of contact has net motion in the forward direction hence it is a case of forward slipping and force of friction acts in the backward direction as shown in figure.

In the vertical direction we can write the following equation:
`N=mg`
Since it is a case of slipping hence friction acts at its maximum.
`f= muN= mumg`
Let a is linear acceleration of sphere and a is angular acceleration then we can write the following equations:
`- mu m g = ma rArr a=- mug " "...(i)`
`(mu g) r=(2)/(5) alpha rArr alpha=(5 mug)/(2r) " "...(ii)`
`omega= omega_(0)+alpha t`
`rArr 0-(u)/(2r)+(5mug)/(2r)t`
`rArr t=(u)/(5 mug)`
Hence option (a) is correct.