Q. A hollow sphere of radius $R$ and mass $m$ is fully filled with water of mass $m$. It is rolled down a horizontal plane such that its centre of mass moves with a velocity $v$. If it purely rolls :

System of Particles and Rotational Motion Report Error

Solution:

Total kinetic energy = Rotational KE of sphere + Translational KE of sphere + Translation KE of water inside it.
$K=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow K=\left(\frac{1}{2} \times \frac{2}{3} m R^{2} \times \frac{v^{2}}{R^{2}}\right) +\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow K=\frac{1}{3} m v^{2}+m v^{2}$
$\Rightarrow K=\frac{4}{3} m v^{2}$
Angular momentum of sphere about a fixed point on ground,
$L=(I \omega+m v R)+m v R$
$\Rightarrow L=\frac{2}{3} m R^{2} \times \frac{v}{R}+2 m v R$
$\Rightarrow L=\frac{2}{3} m v R+2 m v R$
$\Rightarrow L=\frac{8 m v R}{3}$