A disc of circumference s is at rest at a point A on a horizontal surface when a constant horizontal force begins to act on its centre. Between A and B there is sufficient friction to prevent slipping and the surface is smooth to the right of B, AB = s. The disc moves from A to B in time T. To the right of B, 

Between A and B ,

Let P = external force, F = force of friction between A and B, a₁ = acceleration between A and B, a₂ = acceleration beyond B

$\mathrm{P}-\mathrm{F}={\mathrm{ma}}_1$

$$\begin{gathered} \mathrm{Torque} \mathrm{\tau }=\mathrm{I\alpha } \\ \mathrm{rF}=\mathrm{I\alpha } \end{gathered}$$

$\mathrm{P}={\mathrm{ma}}_2$

${\mathrm{a}}_2>{\mathrm{a}}_1$

As no torque acts to the right of B, angular acceleration disappear.

Between A and B:

Let $\mathrm{\alpha }$ = angular acceleration between A and B.

Angular velocity at B , ${\omega }_B=\alpha T$

For one rotation,2$\pi =\frac{1}{2}\alpha T^2$

To the right of B:

Angular velocity at $\mathrm{B}={\mathrm{\omega }}_{\mathrm{B}}=\mathrm{\alpha T}$

For one rotation to the right of B

$\mathrm{\theta }=2\mathrm{\pi }={\mathrm{\omega }}_{\mathrm{B}}\mathrm{t}$

$\mathrm{t}=\frac{2\mathrm{\pi }}{\mathrm{\alpha T}}=\frac{\frac{1}{2}{\mathrm{\alpha T}}^2}{\mathrm{\alpha T}}=\frac{\mathrm{T}}{2}$