Question

A thin spherical shell of mass $M$ and radius $R$ as shown in the figure slips on a rough horizontal plane. At the same instant, it has translational velocity $v_0$ and rotational velocity about the centre $\frac{3v_0}{R}$ . Find the translation velocity after the shell starts pure rolling.

• A. $\frac{9}{5}{v}_0$
• B. $2v_0$
• C. $\frac{5}{9}{v}_0$
• D. $\frac{v_0}{3}$

Answer 1

The correct option is A $\frac{9}{5}{v}_0$


For translation motion,
$f=Ma---\left(1\right)$
For rotational motion,
${\tau }_0=-fR$
$\Rightarrow I\alpha =-fR$
$\Rightarrow \frac{2}{3}M{R}^2\alpha =-fR$
$\Rightarrow \alpha =\frac{-3f}{2MR}--\left(2\right)$

Let after time $t$, the thin spherical shell starts pure rolling. At that instant, let the linear and angular velocity be $v$ and $\omega$.
Then, $v-R\omega =0$
$\Rightarrow v_0+at=R({\omega }_0+\alpha t)$
$\Rightarrow v_0+\frac{f}{M}t=R[\frac{3v_0}{R}+\left(\frac{-3f}{2MR}\right)t]$
[using (1) & (2) and ${\omega }_0=\frac{3v_0}{R}$]
$\Rightarrow v=\frac{f}{M}t=3v_0-\frac{3f}{2M}t$
$\Rightarrow \frac{5f}{2M}t=2v_0$
$\Rightarrow t=2v_0\times \frac{2M}{5f}=\frac{4M{v}_0}{5f}$
Thus, velocity when it starts pure rolling
$v=u+at$
$=v_0+\frac{f}{M}\times \frac{4M{v}_0}{5f}$
$=v_0+\frac{4}{5}{v}_0$
$=\frac{9}{5}{v}_0$