Answer
$\omega_S=57.5rad/s \circlearrowleft$
$\omega_{OA}=10.6 rad/s\circlearrowleft$
Work Step by Step
We can determine the required angular velocity as follows:
$\frac{0.5}{0.1-x}=\frac{0.75}{x}$
This simplifies to:
$x=0.01304m$
We know that
$\omega_S=\frac{r_{O/H}}{x}\omega_H$
We plug in the known values to obtain:
$\omega_S=\frac{0.15}{0.01304}(5)$
This simplifies to:
$\omega_S=57.5rad/s \circlearrowleft$
We know that
$v_A=\omega_S(r_A-x)$
$\implies v_A=57.5(0.05-0.01304)=2.125m/s$
and $v_A=\omega_{OA} r_{OA}$
This can be rearranged as:
$\omega_{OA}=\frac{2.125}{0.2}=10.6 rad/s\circlearrowleft$
