Answer
(a) $I = 0.0225~kg~m^2$
(b) The mass of the turntable should be 0.50 kg.
Work Step by Step
(a) We can convert the angular speed to units of rad/s.
$\omega = (45~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega = 4.71~rad/s$
We can find the moment of inertia of the turntable.
$\frac{1}{2}I~\omega^2 = K$
$I = \frac{2~K}{\omega^2}$
$I = \frac{(2)(0.250~J)}{(4.71~rad/s)^2}$
$I = 0.0225~kg~m^2$
(b) We can find the mass of the turntable.
$\frac{1}{2}MR^2 = I$
$M = \frac{2I}{R^2}$
$M = \frac{(2)(0.0225~kg~m^2)}{(0.30~m)^2}$
$M = 0.50~kg$
The mass of the turntable should be 0.50 kg.
