Question

A solid sphere of radius $r$ is rolling on a horizontal surface. The ratio between the rotational kinetic energy and total kinetic energy is:

• A. $\frac{5}{7}$
• B. $\frac{2}{7}$
• C. $\frac{1}{2}$
• D. $\frac{1}{7}$

Answer 1

The correct option is B $\frac{2}{7}$

Moment of inertia of solid sphere about an axis passing through its centre is given by $I=\frac{2}{5}{MR}^2$

Velocity of centre of mass is given by $v=R\omega$

Hence, translational kinetic energy is given by, ${KE}_L=\frac{1}{2}{Mv}^2$

$\therefore K{E}_L=\frac{1}{2}M{R}^2{\omega }^2$

Rotational kinetic energy is given by $K{E}_R=\frac{1}{2}\times \frac{2}{5}M{R}^2{\omega }^2$

Total kinetic energy, $K{E}_T=K{E}_L+K{E}_R=\frac{1}{2}\times \frac{7}{5}M{R}^2{\omega }^2$

Ratio of rotational kinetic energy to total kinetic energy is $\frac{K{E}_R}{K{E}_T}=\frac{2}{7}$