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 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. Assume ng coefficient of friction between sphere and the surface. Mass of sphere is m. At what time sphere starts pure rolling?

State of motion of the sphere at t = 0 is described in figure, Centre

1.	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 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. Assume ng coefficient of friction between sphere and the surface. Mass of sphere is m. At what time sphere starts pure rolling?
Answer» `(2u)/(7mug)` `(3u)/(5mug)` `(3u)/(7mug)` `(5u)/(7mug)` Solution :We can continue with the solution discussed so far in previous part. LET us first write linear and angular speed of the sphere at some instant of time t. `u=u+at+ rArr u= v- mug t ""...(iii)` `omega= omega_(0)+ alpha t rArr omega=-(u)/(2r)+(5mug)/(2r) t""...(iv)` For pure ROLLING we have v rohence we can write the following: `u=r omega rArr u- mu g=-(u)/(2) +(5mug)/(2)t` `rArr mu"GT"=(3u)/(7) ""...(v)` Hence time to start pure rolling can be written from equation (v) `t=(3u)/(7 mug)` Hence OPTION (C) is correct.