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)

$\frac{2}{5}$

Answer 2

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