Answer
$\omega_{CD}=57.7rad/s \circlearrowleft$
Work Step by Step
We can determine the required angular velocity as follows:
According to sine rule
$\frac{r_{IC/B}}{sin90}=\frac{r_{IC/F}}{sin60}=\frac{r_{B/F}}{sin30}$
This simplifies to:
$r_{IC/B}=\frac{r_{B/F}}{sin30}=\frac{0.3}{sin30}=0.6m$
$r_{IC/F}=0.6sin60=0.5196m$
and $\omega_{BF}=\frac{0.1}{0.6}(50)=8.333 rad/s$
Now $v_F=\omega_{BF}r_{IC/F}$
$\implies v_F=8.333(0.5196)=4.33m/s$
Similarly $v_F=\omega_{CD}r_{CF}$
This can be rearranged as:
$\omega_{CD}=\frac{v_F}{r_{CF}}$
We plug in the known values to obtain:
$\omega_{CD}=\frac{4.33}{0.075}=57.7rad/s \circlearrowleft$
