Answer
$$
v_C=1.52 \mathrm{~m} / \mathrm{s}
$$
Work Step by Step
$$
\begin{aligned}
& T_1+V_1=T_2+V_2 \\
& {[0+0+0]+[0]=\frac{1}{2}\left[\frac{1}{2}(3)(0.03)^2\right] \omega_A^2+\frac{1}{2}\left[\frac{1}{2}(10)(0.1)^2\right] \omega_B^2+\frac{1}{2}(2)\left(v_C\right)^2-2(9.81)(0.5)} \\
& v_C=\omega_B(0.1)=0.03 \omega_A
\end{aligned}
$$
Thus,
$$
\begin{aligned}
& \omega_B=10 v_C \\
& \omega_A=33.33 v_C
\end{aligned}
$$
Substituting and solving yields,
$$
v_C=1.52 \mathrm{~m} / \mathrm{s}
$$
