Question

A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy and translational kinetic energy is:

• A. $\frac{2}{9}$
• B. $\frac{2}{5}$
• C. $\frac{2}{7}$
• D. $\frac{7}{2}$

Answer 1

Answer: (B)

Moment of inertia of sphere
$I=\frac{2}{5}M{R}^2$ ...(1)
and for pure rolling $v=R\omega$...(2)
$\frac{\text{ Rotational KE }}{\text{ Translational }KE}=\frac{\frac{1}{2}I{\omega }^2}{\frac{1}{2}m{v}^2}$
from (1) and (2)
$=\frac{\frac{1}{2}\times \frac{2}{5}M{R}^2{\omega }^2}{\frac{1}{2}M{R}^2{\omega }^2}=\frac{2}{5}$