A sphere of mass $M$ and radius $r$ shown in the figure slips on a rough horizontal plane. At some instant, it has a translational velocity $v_0$ and rotational velocity about the centre $\left(v_0/2r\right)$. Find the translational velocity after the sphere starts pure rolling.

Given, the velocity of centre=$v_0$

Angular velcoty about the centre $=\frac{v_0}{2r}$

Here initial velocity of rotation ${\omega }_0<V_o/r$, the sphere slips in forwarding direction. Frictional force decelerates it to decrease its translational velocity $v_0$ to a value $v$, which corresponds to pure rolling. The frictional force increases angular velocity ${\omega }_0$, to a value $\omega$, which corresponds to pure rolling and satisfies the relation $v=\omega r$

Deceleration of the centre of mass of the sphere $a=\frac{f_r}{M}$

$\therefore v=v_0-at$

or $v=v_0-\left(\frac{f_r}{M}\right)t \dots \left(1\right)$

and angular acceleration about centre $\alpha =\frac{\tau }{I}$

or $\alpha =\frac{F_r.r}{\left(2M{r}^2/5\right)}=\frac{5f_r}{2Mr}$

$\omega ={\omega }_0+\alpha t$

or $\omega ={\omega }_0+\frac{5f_r}{\left(2Mr\right)}t$

or $\omega =\frac{v_0}{2r}+\frac{5f_r}{2Mr}t \dots \left(2\right)$

When pure rolling occurs $v=\omega r \dots \left(3\right)$

Solving the above equations, we get

$v=\frac{6v_0}{7}$