Question

A hollow spherical ball rolls on a table without slipping. Ratio of its rotational kinetic energy to its total kinetic energy is

• A. $5:2$
• B. $2:7$
• C. $2:5$
• D. $7:2$

Answer 1

The correct option is C $2:5$

In case of pure rolling, body will have $K.E_{\text{Trans}}$ as well as $K.E_{\text{Rot}}$.
$\Rightarrow v=R\omega ...\left(i\right)$ for pure rolling,

where $v\to v_{\text{CM}}$ of spherical ball.
Moment of inertia of hollow sphere about axis passing through its $COM$ is:
$I_{\text{CM}}=\frac{2}{3}m{r}^2$
$K.E_{\text{Trans}}=\frac{1}{2}m{v}^2$
$K.E_{\text{Rot}}=\frac{1}{2}{I}_{\text{CM}}{\omega }^2$
Hence total kinetic energy of ball is:
$K.E_{\text{Total}}=K.E_{\text{Trans}}+K.E_{\text{Rot}}...\left(ii\right)$
From Eq $\left(i\right)$,
$K.E_{\text{Rot}}=\frac{1}{2}\left(\frac{2}{3}m{r}^2\right){\omega }^2=\frac{1}{3}m(r\omega {)}^2$
$\therefore K.E_{\text{Rot}}=\frac{1}{3}m{v}^2$
From Eq. $\left(ii\right)$,
$K.E_{\text{Total}}=\frac{1}{2}m{v}^2+\frac{1}{3}m{v}^2=\frac{5}{6}m{v}^2$
$\therefore \frac{{K.E}_{\text{Rot}}}{{K.E}_{\text{Total}}}=\frac{\frac{1}{3}m{v}^2}{\frac{5}{6}m{v}^2}=\frac{2}{5}$
$\Rightarrow {K.E}_{\text{Rot}}:{K.E}_{\text{Trans}}=2:5$