solid sphere of radius `r` is gently placed on a rough horizontal ground with an initial angular speed `omega_(0)` and no linear velocity. If the coefficient of friction is `mu`, find the time `t` when the slipping stops. in addition state the linear velocity `v` and angular velocity `omega` at the end of slipping

Explanation

ltbr Let `m` be the mass of the sphere
since, it is a case of backward slipping, force of friction is in forward direction limiting friction will act in this case
linear acceleration `a=(f)/(m)=(mumg)/(m)=mug`
angular retardation `alpha=(tau)/(I)=(f.r)/((2)/(5)mr^(2))=(5)/(2)(mug)/(r)`
slipping is ceased when `v=romega`
or `(at)=r(omega_(0)-alphat)`
or `mug t=r(omega_(0)-(5)/(2)(mug t)/(r))` or `(7)/(2)mug t=romega_(0)`
`thereforet=(2)/(7)(romega_(0))/(mug)`
`v=at=mug t=(2)/(7)romega_(0)`
and `omega=(v)/(r)=(2)/(7)omega_(0)`
Alternate solution
Net torque on the sphere about the bottommost point is zero as friction is passing through that point therefore, angular mometum of the sphere will remain conserved about the bottommost point
`L_(i)=L_(f)`
`therforeIomega_(0)=Iomega+mrv`
or `(2)/(5)mr^(2)omega_(0)=(2)/(5)mr^(2)omega+mr(omegar)`
`thereforeomega=(2)/(7)omega_(0)` and `v=romega=(2)/(7)romega_(0)`