A uniform ring of radius `R` is given a back spin of angular velocity `V_(0)//2R` and thrown on a horizontal rough surface with velocity of centre to be `V_(0)`. The velocity of the centre of the ring when it starts pure rolling will be

Options

A. `(V_(0))/2`
B. `(V_(0))/4`
C. `(3V_(0))/4`
D. `0`

Answer

B

Explanation

Here `u=V_(0), omega_(0)=(V_(0))/(2R)`
At pure rolling
and `V/R=-(V_(0))/(2R)=(f/(mR))t`
(in pure rolling `V=Romega)`
`(alpha=tau/I=(fR)/(mR^(2))=f/(mR))`
`V_0-V=V+(V_(0))/2`
`2V=(V_(0))/2impliesV=(V_(0))/4`