A uniform disc of mass `m` and radius `R` is projected horizontally with velocity `v_(0)` on a rough horizontal floor so that it starts off with a purely sliding motion at `t=0`. After `t_(0)` seconds, it acquires pure rolling motion as shown in the figure.
(a) Calculate the velocity of the center of mass of the disc at `t_(0)`.
Assuming that the coefficent of friction to be `mu`, calculate `t_(0)`.

Explanation

This is the case of impure rolling. As `v_(0) gt R omega`, friction will act in backward direction to the given linear retardation and angular acceleration so that linear velocity decreases and angular velocity increases so that velocity after time `t` , `v` equals `Romega` (case of pure rolling)

(`a`) By consevation of angular momentum about `A`
`mv_(0)R=mvR+I_(c.m.)omega=mvR+(1)/(2)mR^(2)((v)/(R))`
`=(3)/(2)mvR`
`v=(2v_(0))/(3)`
(`b`) In impure rolling, friction is `f=f_(max)=muN=mumg`
Retardation `a=(f)/(m)=(mumg)/(m)=mug`
`v=v_(0)-at_(0)implies (2v_(0))/(3)=v_(0)-mug t_(0)impliest_(0)=(v_(0))/(3mug)`
After attainment of pure rolling, friction vanishes and the body rolls without slipping forever.