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

Options

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

Answer

A

Explanation

We have, `(K_(R))/(K_(T))=(I)/(mR^(2))`, if `v=R omega` in case of pure rolling `= (K^(2))/(R^(2))`
`:. K_(R)=((K^(2))/(K^(2)+R^(2)))K_("Total") or (K_(R))/(K_("Total"))=(K^(2))/(K^(2)+R^(2))`.