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)

$\frac{K^2}{K^2 +R^2}$

Answer 2

In rolling without slipping, total energy of ball is the sum of its translational and rotational energy.
Kinetic energy of rotation
$K_{rot}=\frac{1}{2}I\omega^2=\frac{1}{2}MK^2\frac{v^2}{R^2}$
where $K$ is radius of gyration.
Kinetic energy of translation,
$K_{\text{trans}}=\frac{1}{2}Mv^2$
Thus, total energy
$E=K_{rot}+K_{\text{trans}}$
$=\frac{1}{2}MK^2\frac{V^2}{R^2}+\frac{1}{2}Mv^2$
$=\frac{1}{2}Mv^2\left(\frac{K62}{R^2}+1\right)$
$=\frac{1}{2}\frac{Mv^2}{R^2}(K^2+R^2)$
Hence $\frac{K_{rot}}{K_{trans}}=\frac{\frac{1}{2}MK^2\frac{v^2}{R^2}}{\frac{1}{2}\frac{Mv^2}{R^2}(K^2+R^2)}$
$=\frac{K^2}{K^2+R^2}$