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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 :
A
$\dfrac{K^2 + R^2}{R^2}$
B
$\dfrac{K^2}{R^2}$
C
$\dfrac{K^2}{K^2 + R^2}$
D
$\dfrac{R^2}{K^2 + R^2}$
Detailed Solution
For rolling without slipping, $v = R\omega$ and $I = mK^2$.
Total kinetic energy: $T.K.E. = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}mv^2\left(1 + \dfrac{K^2}{R^2}\right)$
Rotational kinetic energy: $R.K.E. = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}mK^2\dfrac{v^2}{R^2} = \dfrac{1}{2}mv^2\left(\dfrac{K^2}{R^2}\right)$
Fraction $= \dfrac{R.K.E.}{T.K.E.} = \dfrac{K^2/R^2}{1 + K^2/R^2} = \dfrac{K^2}{K^2 + R^2}$
Total kinetic energy: $T.K.E. = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}mv^2\left(1 + \dfrac{K^2}{R^2}\right)$
Rotational kinetic energy: $R.K.E. = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}mK^2\dfrac{v^2}{R^2} = \dfrac{1}{2}mv^2\left(\dfrac{K^2}{R^2}\right)$
Fraction $= \dfrac{R.K.E.}{T.K.E.} = \dfrac{K^2/R^2}{1 + K^2/R^2} = \dfrac{K^2}{K^2 + R^2}$