Answer
$K = 1.1\times 10^{-3}~J$
Work Step by Step
We can consider the two rods as a single rod with a length $2d$
We can find the rotational inertia:
$I = md^2+m(2d)^2+\frac{1}{3}(2M)(2d)^2$
$I = 5md^2+\frac{8}{3}Md^2$
$I = (5)(0.85~kg)(0.056~m)^2+\frac{8}{3}(1.2~kg)(0.056~m)^2$
$I = 0.0234~kg~m^2$
We can find the rotational kinetic energy:
$K = \frac{1}{2}I\omega^2$
$K = \frac{1}{2}(0.0234~kg~m^2)(0.30~rad/s)^2$
$K = 1.1\times 10^{-3}~J$
