Question

If a solid sphere is rolling without slipping on a horizontal plane, then the ratio of its rotational and total kinetic energies is

• A. 2:5
• B. 2:7
• C. 4:3
• D. 1:2

Hint:

Rolling without slipping: $v = r\omega$. Total KE = Rotational KE + Translational KE.

Answer 1

The Correct Option is B

Rotational kinetic energy $K_R = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a solid sphere, $I = \frac{2}{5}mr^2$, where $m$ is the mass and $r$ is the radius. Since the sphere rolls without slipping, $v = r\omega$, where $v$ is the linear velocity. $K_R = \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v^2}{r^2}) = \frac{1}{5}mv^2$. Translational kinetic energy $K_T = \frac{1}{2}mv^2$. Total kinetic energy $K = K_R + K_T = \frac{1}{5}mv^2 + \frac{1}{2}mv^2 = \frac{7}{10}mv^2$. The ratio of rotational to total kinetic energy is $\frac{K_R}{K} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1/5}{7/10} = \frac{2}{7}$.