Fraction of Rotational Energy for Rolling Ball – Rankers Physics

Fraction of Rotational Energy for Rolling Ball

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: (2003)

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

Solution:

Rotational kinetic energy $E_{rot} = \frac{1}{2}MK^2\omega^2$ and total energy $E = \frac{1}{2}M(K^2+R^2)\omega^2$. The fraction is $E_{rot}/E_{total} = \frac{K^2}{K^2+R^2}$.

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