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. `(7v_(0))/(2)`
B. `(5v_(0))/(4)`
C. `(3v_(0))/(4)`
D. `(v_(0))/(4)`

Answer

D

Explanation

`v=v_(0)-(f)/(m)t&(v)/(R)=-(v_(0))/(2R)+(f)/(mR)t`
For pure rolling `v=Romega` and `alpha=(T)/(l)=(fR)/(MR^(2))`
`v_(0)-v=v+(v_(0))/(2)`
`v=(v_(0))/(4)`