Q. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be
A) $\frac{K^{2}}{R^{2}}$
B) $\frac{K^{2}}{{K^{2} + R}^{2}}$
C) $\frac{R^{2}}{{K^{2} + R}^{2}}$
D) $\frac{{K^{2} + R}^{2}}{R^{2}}$
Explanation
(b) $\frac{Rotational\ KE}{Total\ KE} = \frac{\frac{1}{2}mv^{2}\left( \frac{K^{2}}{R^{2}} \right)}{\frac{1}{2}mv^{2}\left( 1 + \frac{K^{2}}{R^{2}} \right)} = \frac{K^{2}}{{K^{2} + R}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.