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

Answer

D

Explanation

Taking `tau` about CM

`mumgR=MR^(2) alpha`
`mug=Ralpha`
`alpha=(mug)/R`
`omega=omega_(0)=(mug)/Rf=(omega_(0))/2`
`(omega_(0))/2=(mug)/Rfimpliesf=(omega_(0)R)/(2mug)`