Question

A ball of radius $R$ rolls without slipping. Find the fraction of total energy associated with its rotational energy, if the radius of the gyration of the ball about an axis passing through its center of mass is $K$.

• 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 $\frac{K^2}{K^2+R^2}$

Kinetic energy of rotation is
$K_{rot}=\frac{1}{2}l{\omega }^2=\frac{1}{2}M{K}^2\frac{V^2}{R^2}$
where, $k$ is radius of gyration.
Kinetic energy of translation is $K_{trans}=\frac{1}{2}M{v}^2$
Thus, total energy,
$E=K_{rot}+K_{trans}$
$=\frac{1}{2}M{K}^2\frac{v^2}{R^2}+\frac{1}{2}M{v}^2$
$=\frac{1}{2}M{v}^2(\frac{K^2}{R^2}+1)$
$=\frac{1}{2}\frac{M{v}^2}{R^2}(K^2+R^2)$
Hence, $\frac{K_{rot}}{Totalenergy,E}=\frac{\frac{1}{2}M{K}^2\frac{v^2}{R^2}}{\frac{1}{2}\frac{M{v}^2}{R^2}(K^2+R^2)}$
$=\frac{K^2}{K^2+R^2}$