Answer
(a) $\frac{2}{5}$
(b) same
Work Step by Step
(a) We know that rotational kinetic energy is given as
$K.E_r=\frac{1}{2}I\omega^2$
$\implies K.E_r=\frac{1}{2}(\frac{2}{3}MR^2)(\frac{v}{R})^2$
$K.E_r=\frac{1}{3}Mv^2$
The total kinetic energy is given as
$K.E_{total}=K.E_t+K.E_r$
$\implies K.E_{total}=\frac{1}{2}Mv^2+\frac{1}{3}Mv^2$
$K.E_{total}=\frac{5}{6}Mv^2$
Now the required ratio is given as
$\frac{K.E_r}{K.E_{total}}=\frac{\frac{1}{3}Mv^2}{{\frac{5}{6}Mv^2}}$
$\frac{K.E_r}{K.E_{total}}=\frac{2}{5}$
(b) We can see that the ratio in part(a) does not depend on the speed. Hence, if the speed is doubled, then the answer to part (a) will stay the same.
