The radius of a wheel is R and its radius of gyration about its axis passing through its center and perpendicualr to its plane is K. If the wheel is roling without slipping. Then the ratio of tis rotational kinetic energy to its translational kinetic energy is

Options

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

Explanation

a) (Rotational KE)/(Translational KE) = ((1/2)lomega^(2))/(1/2Mv^(2)) = K^(2)/R^(2)` [`therefore` l = `MK^(2)`, v =`Romega]