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

Total kinetic energy
=K.E of translation + K.E of rotation
`=1/2 Mv^(2)+1/2Iomega^(2)`
`=1/2Mv^(2) + 1/2Mk^(2) v^(2)/R^(2) (therefore I=Mk^(2)` and `v=omegaR)`
`=1/2Mv^(2)(1+k^(2)/R^(2))`
`therefore ("K.E of rotation")/("Total K.E.") = (1/2 Mk^(2)v^(2)/R^(2))/(1/2Mv^(2)(1+k^(2)/R^(2))`
`=(k^(2)/R^(2))/(1+k^(2)/R^(2))= k^(2)/(k^(2)+R^(2))`