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)
$=\frac{\frac{1}{2} \times \frac{2}{5} M R^{2} \omega^{2}}{\frac{1}{2} M R^{2} \omega^{2}}=\frac{2}{5}$