#### Answer

(a) $I = 0.032~kg~m^2$
(b) The rotational kinetic energy is 15.8 J

#### Work Step by Step

(a) We can find $R$, the distance from each ball to the axis of rotation.
$\frac{0.20~m}{R} = cos(30^{\circ})$
$R = \frac{0.20~m}{cos(30^{\circ})}$
$R = 0.23~m$
We can find the moment of inertia of the system.
$I = (3)(M~R^2)$
$I = (3)(0.20~kg)(0.23~m)^2$
$I = 0.032~kg~m^2$
(b) We can find the angular velocity.
$\omega = (5.0~rev/s)(2\pi~rad/rev)$
$\omega = 10\pi~rad/s$
We can find the rotational kinetic energy.
$KE_{rot} = \frac{1}{2}I~\omega^2$
$KE_{rot} = \frac{1}{2}(0.032~kg~m^2)(10\pi~rad/s)^2$
$KE_{rot} = 15.8~J$
The rotational kinetic energy is 15.8 J.