Question

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its center 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. $\frac{K^2}{K^2 +R^2}$
• B. $\frac{R^2}{K^2 +R^2}$
• C. $\frac{K^2 +R^2}{R^2}$
• D. $\frac{K^2}{R^2}$

Answer 1

Answer: (A)

Total energy
$=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}$
$=\frac{1}{2} m v^{2}\left(1+K^{2} / R^{2}\right)$
Required fraction $=\frac{K^{2} / R^{2}}{1+K^{2} / R^{2}}=\frac{K^{2}}{R^{2}+K^{2}}$.