Answer
$$ \omega=0.190 \mathrm{rad} / \mathrm{s}
$$
Work Step by Step
$$
\begin{aligned}
& (↻+) \quad\left(H_A\right)_1=\left(H_A\right)_2 \\
& 0+0=\left[\frac{1}{12}\left(\frac{200}{32.2}\right)(4)^2+\frac{200}{32.2}(11)^2\right] \omega-\left[\left(\frac{150}{32.2}\right)(5-10 \omega)\right](10) \\
& \omega=0.190 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
