Answer
$$
\omega=22.7 \mathrm{rad} / \mathrm{s}
$$
Work Step by Step
For the weight
$$
\begin{aligned}
& T_1+V_1=T_2+V_2 \\
& 0+10(2)=\frac{1}{2}\left(\frac{10}{32.2}\right) v_2^2 \\
& v_2=11.35 \mathrm{ft} / \mathrm{s} \\
& \left(H_A\right)_2=\left(H_A\right)_3 \\
& m v_2(0.5)+0=m v_3(0.5)+I_A \omega \\
& \left(\frac{10}{322}\right)(11.35)(0.5)+0=\left(\frac{10}{32.2}\right) v_3(0.5)+\left[\frac{1}{2}\left(\frac{20}{32.2}\right)(0.5)^2\right] \omega \\
& (+\downarrow) \quad e=\frac{0.5 \omega-v_3}{v_2-0} \quad 1=\frac{0.5 \omega-v_3}{11.35-0} \quad 11.35=0.5 \omega-v_3
\end{aligned}
$$
Solving Eqs.[1] and [2] yields:
$$
\begin{aligned}
& \omega=22.7 \mathrm{rad} / \mathrm{s} \\
& v_3=0
\end{aligned}
$$
