#### Answer

(a) The mass of the child's ball is reduced by a factor of $\frac{1}{8}$
(b) The rotational inertia of the child's bowling ball is reduced by a factor of $\frac{1}{32}$

#### Work Step by Step

(a) $V = \frac{4}{3}\pi~R^3$
If the radius of the child's ball is decreased by a factor of $\frac{1}{2}$, then the volume is decreased by a factor of $(\frac{1}{2})^3 = \frac{1}{8}$
Then the mass of the child's ball is reduced by a factor of $\frac{1}{8}$
(b) We can write an expression for the rotational inertia of the adult's bowling ball:
$I_a = \frac{2}{5}MR^2$
We can write an expression for the rotational inertia of the child's bowling ball:
$I_c = \frac{2}{5}(\frac{M}{8})(\frac{R}{2})^2$
$I_c = \frac{1}{32}\times \frac{2}{5}MR^2$
$I_c = \frac{1}{32}\times I_a$
The rotational inertia of the child's bowling ball is reduced by a factor of $\frac{1}{32}$.