Answer
(a) $1/3$
(b) $2/7$
(c) $2/5$
(d) $5/13$
Work Step by Step
Using the formulas for the moments of inertia of all the shapes:
(a) $$ K_{rotational} = \frac{I\omega^2}{2} = \frac{MR^2\omega^2}{4} = \frac{Mv_{cm}^2}{4} $$
$$ K_{total} = \frac{Mv_{cm}^2}{4} + \frac{Mv_{cm}^2}{2} = \frac{3Mv_{cm}^2}{4} $$
$$ \frac{K_{rotational}}{K_{total}} = \frac{1}{3} $$
(b) $$ K_{rotational} = \frac{I\omega^2}{2} = \frac{MR^2\omega^2}{5} = \frac{Mv_{cm}^2}{5} $$
$$ K_{total} = \frac{Mv_{cm}^2}{5} + \frac{Mv_{cm}^2}{2} = \frac{7Mv_{cm}^2}{10} $$
$$ \frac{K_{rotational}}{K_{total}} = \frac{2}{7} $$
(c) $$ K_{rotational} = \frac{I\omega^2}{2} = \frac{MR^2\omega^2}{3} = \frac{Mv_{cm}^2}{3} $$
$$ K_{total} = \frac{Mv_{cm}^2}{3} + \frac{Mv_{cm}^2}{2} = \frac{5Mv_{cm}^2}{6} $$
$$ \frac{K_{rotational}}{K_{total}} = \frac{2}{5} $$
(d) $$ I = \frac{M \Big( R^2 + \big( \frac{R}{2} \big)^2 \Big)}{2} = \frac{5MR^2}{8} $$
$$ K_{rotational} = \frac{I\omega^2}{2} = \frac{5MR^2\omega^2}{16} = \frac{5Mv_{cm}^2}{16} $$
$$ K_{total} = \frac{5Mv_{cm}^2}{16} + \frac{Mv_{cm}^2}{2} = \frac{13Mv_{cm}^2}{16} $$
$$ \frac{K_{rotational}}{K_{total}} = \frac{5}{13} $$
