Physics - Rotation Question with Solution | TestHub

Moment of inertia of a solid cylinder about its geometrical axis is $I_c=\frac{1}{2}M{R}^2$.

Moment of inertia of a solid sphere about its diameter is $I_s=\frac{2}{5}M{R}^2$.

Let the angular speed of the cylinder be ${\omega }_c=\omega$. Then the angular speed of the sphere is ${\omega }_s=\frac{\omega }{2}$.

Rotational kinetic energy of the cylinder is:
$K_c=\frac{1}{2}{I}_c{\omega }_c^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2=\frac{1}{4}M{R}^2{\omega }^2$

Rotational kinetic energy of the sphere is:
$K_s=\frac{1}{2}{I}_s{\omega }_s^2=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{\omega }{2}\right)}^2=\frac{1}{5}M{R}^2\left(\frac{{\omega }^2}{4}\right)=\frac{1}{20}M{R}^2{\omega }^2$

The ratio of the kinetic energy of rotation of the sphere to that of the cylinder is:
$\frac{K_s}{K_c}=\frac{\frac{1}{20}M{R}^2{\omega }^2}{\frac{1}{4}M{R}^2{\omega }^2}=\frac{4}{20}=\frac{1}{5}$

The required ratio is $1:5$.

Answer: $1:5$