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 is $R$, 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

The correct option is A $\frac{k^2}{k^2+R^2}$

$RE=\frac{1}{2}m{K}^2{\omega }^2$
$TE=\frac{1}{2}m{R}^2{\omega }^2+\frac{1}{2}m{k}^2{w}^2$

$RE/TE=K^2/(R^2+K^2)$