A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc

At point $\mathrm{G}$ :

${\mathrm{V}}_{\mathrm{G}}=\mathrm{V}-\mathrm{\omega R}$

${\mathrm{V}}_{\mathrm{G}}=\mathrm{V}-\mathrm{V} \left(\because \mathrm{V}=\mathrm{\omega R}\right)$

${\mathrm{V}}_{\mathrm{G}}=0$

So, the velocity at lowest point ${\mathrm{V}}_{\mathrm{G}}=0$

Acceleration at point $\mathrm{G}:$

${\mathrm{a}}_{\mathrm{G}}={\mathrm{a}}_{\mathrm{com}}+\mathrm{\alpha }$                        $\left(\begin{array}{c}\mathrm{Here} ; \mathrm{\alpha }=\mathrm{Angular} \mathrm{acceleration}\\ {\mathrm{a}}_{\mathrm{COM}}=\hspace{0.17em}\mathrm{Acceleration} \mathrm{of} \mathrm{center} \mathrm{of} \mathrm{mass}.\end{array}\right)$

${\mathrm{a}}_{\mathrm{G}}=\frac{{\mathrm{V}}^2}{\mathrm{R}}+0$

${\mathrm{a}}_{\mathrm{G}}=\frac{{\mathrm{V}}^2}{\mathrm{R}}$   $$\begin{gathered} \left(\because \mathrm{\omega }= \mathrm{constant}\right) \\ So, \alpha =0 \end{gathered}$$