Answer
$K = 0.41~J$
Work Step by Step
We can find the rotational inertia:
$I = 2M~(L~sin~\theta)^2+2M~(L~sin~\theta)^2$
$I = 4M~(L~sin~\theta)^2$
$I = (4)(1.6~kg)[(0.60~m)~sin~30^{\circ}]^2$
$I = 0.576~kg~m^2$
We can find the rotational kinetic energy:
$K = \frac{1}{2}~I~\omega^2$
$K = (\frac{1}{2})~(0.576~kg~m^2)~(1.2~rad/s)^2$
$K = 0.41~J$
