Question

A solid sphere of mass $m$ is kept between two planks rolls moving without slipping with the velocities as shown.


Determine the total kinetic energy of the solid sphere.

• A. $\frac{1}{2}m{v}^2$
• B. $\frac{4}{5}m{v}^2$
• C. $\frac{11}{5}m{v}^2$
• D. $\frac{13}{5}m{v}^2$

Answer 1

The correct option is C $\frac{11}{5}m{v}^2$


Let the velocity of the centre of mass of the sphere be $v_c$ and radius of the sphere $r$.
Assuming that the sphere moves in anticlockwise direction with angular speed $\omega$
The velocities at the contact points $A$ and $B$ are


At point $A$
$v_c-\omega r=v....\left(1\right)$
At point $B$
$v_c+\omega r=3v.....\left(2\right)$
Adding (1) and (2)
$2v_c=4v$
$\Rightarrow v_c=2v$

Total kinetic energy of the sphere = Rotational kinetic energy + Translational kinetic energy
Now total kinetic energy will be
$K{E}_{\text{total}}=\frac{1}{2}m{v}_c^2+\frac{1}{2}{I}_c{\omega }^2$
From (2), $\omega r=3v-v_c\Rightarrow \omega =\frac{v}{r}$
Also $I_C=\frac{2}{5}m{r}^2$
$K{E}_{\text{total}}=\frac{1}{2}m(2v{)}^2+\frac{1}{2}\times \frac{2}{5}\times m{r}^2\times \frac{v^2}{r^2}$
$K{E}_{\text{total}}=2m{v}^2+\frac{1}{5}m{v}^2$
$K{E}_{\text{total}}=\frac{11}{5}m{v}^2$

Hence, option (c) is correct.