(a) Before being brought in contact with the table the disc was in only rotational motion about its axis passing through center.
So, $\left(v_{C M}=0\right)$ as the point on axis are considered at rest.


(b) When the rotating disc is placed in contact with surface of table due to force of friction linear velocity of a point on the rim will decreases.
(c) When the rotating disc is placed in the contact with the surface of table due to force of friction centre of mass acquires some linear velocity.
(d) Force of friction is responsible for the effects in (b) and (c).
(e) When rolling of disc starts on table then velocity of centre of mass $v_{C M}$ is due to reaction force due to rotation angular speed $\omega_0$ of disc of radius $R$, so $v_{C M}$ at start when just comes in contact with table is :

$$
v_{C M}=\omega_0 R
$$
where $\omega_0$ is angular speed of the disc when rolling just starts.
(f) Acceleration produced in centre of mass due to reaction force $F$ due to frictional force in disc of mass $m$ is
$$
F=m A \Rightarrow a_{C M}=\frac{F}{m}=\frac{\mu_{\mathrm{K}} m g}{m}=\mu_\kappa g
$$

Angular retardation ( $\alpha$ ) produced by the torque due to frictional friction.
$$
\begin{aligned}
\tau &=I \alpha \Rightarrow \alpha=\frac{\tau}{I}=\left(\frac{r \times F}{I}\right) \\
&=\frac{\vec{R} \times \mu_{\mathrm{K}} m g}{I}=\frac{R \mu_{\mathrm{K}} m g \sin \theta}{I}
\end{aligned}
$$
(As the angle between $R$ and $F$ is $90^{\circ}$ ).
$$
\alpha=\frac{-\mu_\kappa m g R}{I}\left[\because \tau=\left(\mu_\kappa N\right) R=\mu_\kappa m g R\right]
$$
For linear velocity, $v_{C M}=u_{C M}+a_{C M} t\left(\because u_{C M}=0\right)$
$$
v_{C M}=\mu_{\mathrm{K}} g t
$$
For rotational motion,
$$
\omega=\omega_0+\alpha t \Rightarrow \omega=\omega_0-\frac{\mu_\kappa m g R}{I} t
$$
Condition for rolling without slipping,
$$
\begin{aligned}
&v_{C M}=\omega R \frac{v_{C M}}{R}=\omega \Rightarrow \frac{v_{C M}}{R}=\omega_0-\frac{\mu_{\mathrm{K}} m g R}{I} t \\
&\frac{\mu_{\mathrm{K}} g t}{R}=\omega_0-\frac{\mu_{\mathrm{K}} m g R}{I} t \Rightarrow \frac{\mu_{\mathrm{K}} g t}{R}\left[1+\frac{m R^2}{I}\right]=\omega_0 \\
&t=\frac{R \omega_0}{\mu_\kappa g\left(1+\frac{m R^2}{I}\right)}
\end{aligned}
$$