A ring of radius `R` is first rotated with an angular velocity `omega` and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is `mu`. Time after which its angular speed is reduced to half is

Options

A. `(omega_(0)muR)/(2g)`
B. `(2omega_(0)R)/(mug)`
C. `(omega_(0)R)/(2mug)`
D. `(omega_(0)g)/(2mug)`

Answer

3

Explanation

`alpha= (tau)/(I) = (mug)/(R)`
`omega_(t) =(omega_(0))/(2) = omega_(0) - alphat`
So`t=(omega_(0)R)/(2 mug)`