Answer
$$
\omega=2.71 \mathrm{rad} / \mathrm{s}
$$
Work Step by Step
$$
\begin{aligned}
& T_1+V_1=T_2+V_2 \\
& 0+0=\frac{1}{2}\left[\frac{1}{12}\left(\frac{180}{32.2}\right)(8)^2\right] \omega^2+\frac{1}{2}\left(\frac{180}{32.2}\right) v_G^2-180\left(4 \sin 30^{\circ}\right) \\
& r_{I C-G}=\sqrt{(1)^2+(4.3301)^2-2(1)(4.3301) \cos 30^{\circ}} \\
& r_{I C-G}=3.50 \mathrm{~m}
\end{aligned}
$$
Thus,
$$
v_G=3.50 \omega
$$
Substitute into Eq. (1) and solving,
$$
\omega=2.71 \mathrm{rad} / \mathrm{s}
$$
