A ring of radius `R` is rotating with an angular speed `omega_0` about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is `mu_k`. The time after it starts rolling is.

Options

A. `(omega _(0) mu _(1) R)/(2g)`
B. `(omega _(a) R)/(2 mu _(g)R)`
C. `(2 omega _(0) R)/(omega _(k) g)`
D. `(omega _(0) R)/( 2 mu _(g) g)`

Explanation

Aoceleration produced in the centre of mass due to frition
` = (f)/(M) = ( mu _(k ) Mg)/( M ) = mu _(k) g " "...(i)`
Angular retardation produced by the torque due to friction `therefore v = 0 + mu _(k) g t(therefore u = 0)` (Using (i))
`therefore omega = omega _(0) -( mu _(k) g t )/( R)` (Using (ii))
For rolling without slipping
`v = R omega `
`(mu _(1) g t)/( R) = omega _(0) - (mu _(k) g t)/( R)`
`implies t = ( R omega _(0))/( 2 mu _(k)g )`