A disc of radius `R` and mass `m` is projected on to a horizontal floor with a backward spin such that its centre of mass speed is `v_(0)` and angular velocity is `omega_(0)`. What must be the minimum value of `omega_(0)` so that the disc eventually returns back?

Explanation

Initally there is forward slipping, therefore, friction is backwards and maximum.
Let velocity becomes zero in time `t_(1)` and angular velocity becomes zero in time `t_(2)`
then `0=v_(0)-at_(1)`
or `t_(1)=(v_(0))/(a)=(v_(0))/(mug)` ..(i)
and `0=omega_(0)-alphatt_(2)` ltbr. or `t_(2)=(omega_(0))/(alpha)`
here `alpha=(mumgR)/((1)/(2)mR^(2))=(2mug)/(R)`
`thereforet_(2)=(omega_(0)R)/(2mug)` ...(ii)
disk will return back when
`t_(2)gtt_(1)` or `(omega_(0)R)/(2mug)gt(v_(0))/(mug)`
or `omega_(0)gt(2v_(0))/(R)`