$A$ disc of radius $R$ is rotating with an angular velocity $\omega_0$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $\mu_k$.
$(a)$ What was the velocity of its centre of mass before being brought in contact with the table?
$(b)$ What happens to the linear velocity of a point on its rim when placed in contact with the table?
$(c)$ What happens to the linear speed of the centre of mass when the disc is placed in contact with the table?
$(d)$ Which force is responsible for the effects in $(b)$ and $(c)$?
$(e)$ What condition should be satisfied for rolling to begin?
$(f)$ Calculate the time taken for the rolling to begin.

(N/A) The disc was in pure rotational motion before being brought in contact with the table,hence,$u_{CM} = 0$.
$(b)$ The linear velocity of a point on the rim decreases when the disc is placed in contact with the table due to the opposing force of kinetic friction.
$(c)$ The centre of mass acquires a linear velocity in the direction of motion when the rotating disc is placed in contact with the table due to the force of friction.
$(d)$ Kinetic friction is responsible for these effects.
$(e)$ Rolling without slipping begins when the condition $v_{CM} = \omega R$ is satisfied.
$(f)$ The acceleration produced in the centre of mass due to friction is $a_{CM} = \frac{f}{m} = \frac{\mu_k N}{m} = \frac{\mu_k mg}{m} = \mu_k g$.
The angular retardation produced by the torque due to friction is $\alpha = \frac{\tau}{I} = \frac{\mu_k mgR}{I}$ (since $\tau = fR = \mu_k mgR$).
Using $v_{CM} = u_{CM} + a_{CM}t$ (with $u_{CM} = 0$),we get $v_{CM} = \mu_k gt$.
Using $\omega = \omega_0 - \alpha t$,we get $\omega = \omega_0 - \frac{\mu_k mgR}{I}t$.
For rolling without slipping,$\frac{v_{CM}}{R} = \omega$.
Substituting the expressions: $\frac{\mu_k gt}{R} = \omega_0 - \frac{\mu_k mgR}{I}t$.
Solving for $t$: $t = \frac{R\omega_0}{\mu_k g(1 + \frac{mR^2}{I})}$.