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 rotation will be.

Options

A. `(k^2 + R^2)/(R^2)`
B. `(k^2)/(R^2)`
C. `(k^2)/(k^2 + R^2)`
D. `(R^2)/(k^2 + R^2)`

Answer

C

Explanation

( c) force of friction,
`= K.E. of translation + K.E. of rotation`
=`(1)/(2) Mv^2 + (1)/(2) I omega^2`
=`(1)/(2) Mv^2 + (1)/(2) Mk^2 (v^2)/(R^2) (because I = Mk^2 and v = omega R)`
=`(1)/(2) Mv^2 (1 + (k^2)/(R^2))`
`(K.E of rotation)/(Total K.E) = ((1)/(2) Mv^2 (v^2)/(R^2))/((1)/(2) Mv^2 (1 + (k^2)/(R^2)))`
=`((k^2)/(R^2))/(1 + (k^2)/(R^2)) = (k^2)/(k^2 + R^2)`.