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_k R)/(2 g)`
B. `(omega_0 g)/(2 mu_k R)`
C. `(2 omega_0 R)/(mu_k g)`
D. `(omega_0 R)/(2 mu_k g)`

Answer

D

Explanation

(d) Acceleration produced in the centre of mass due friction.
`a = (f)/(M) = (mu_k Mg)/(M) = mu_k g`
where `M` is the mass of the ring
Angular retardation produced by the torque due friction
`prop = (tau)/(I) = (f R)/(I) = (mu_k M g R)/(I)`
As `v = u + at`
:. `v = 0 + mu k gt (because u = 0)` (Using (i))
As `omega = omega_0 + prop t`
:. `omega = omega_0 - (mu_k M gR_1)/(I)` (Using (ii))
m For rolling without slipping
`v = R omega`
:. `(v)/(R) = omega_0 - (mu_k Mg T_1)/(I) t`
`(mu_k gt)/(R) = omega_0 - (mu_k M gR)/(I) t rArr (mu_k gt)/(R) [1 + (MR^2)/(1)] =omega`
`(mu_k gt)/(R) = (omega_0)/(1+ (MR^2)/(I)) rArr t = (R omega_0)/(mu_k g((1 + MR^2)/(I)))`
For ring, `I = MR^2`
:. `t = (R omega_0)/(mu_k g(1 + (MR^2)/(MR^2)))=(R omega_0)/(2 mu_k g)`.