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?

Answer

5

Explanation

Initially 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) - at_(2)` or `t_(2) = omega_(0)/alpha`
Here, `alpha = (mu m g R)/(1/2 mR^(2)) = (2 mu g)/R`
`therefore t_(2) = (omega_(0) R)/(2 mu g)`...........(ii)
Disk will return back when `t_(2) gt t_(1)` or `(omega_(0)R)/(2mug) gt v_(0)/(mu g)` or `omega_(0) gt (2v_(0))/R`