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_(o)mu_(k)R)/2g`
B. `(omega_(o)g)/2mu_(k)R`
C. `(2omega_(o)R)/mu_(k)g`
D. `(omega_(o)R)/2mu_(k)g`

Answer

D

Explanation

Acceleration produced in the center of mass due to friction.
`a=f/M = (mukMg)/(M) = mu_(k)g`
Where M is the mass of the ring …………….(i)
Angular retardation produced by the torque due to friction
`alpha = tau/I = (fR)/(I) = (mu_(k)MgR)/(I)`...............(ii)
As `v=u=at`
`therefore v=0+mu_(k)gt` `(therefore mu=0)` Using (i)
As `omega = omega_(0) + alphat`
`therefore omega= omega_(0) - (mu_(k)MgR)/(I) t` (using ii)
For rolling without slipping
v=R`omega`
`therefore v/R=omega_(0)-(mu_(k)MgR)/(I) t rArr (mu_(k)"gt")/(R) [1+(MR^(2))/(I)]=omega_(0)`
`(mu_(k)"gt")/(R) = (mu_(0))/(1+(MR^(2))/(I)) rArr (Romega_(0))/(mu_(k)g(1+(MR^(2))/(I))`
For ring, `I=MR^(2)`
`therefore t=(Romega_(0))/(mu_(k)g(1+(MR^(2))/(MR^(2))))=(Romega_(0))/(2mu_(k)g)`