Answer
$\omega_{BC}=6.79rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
According to the law of sines
$\frac{0.125}{sin30}=\frac{r_{IC/B}}{sin45}=\frac{r_{IC/C}}{sin 105}$
This simplifies to:
$r_{IC/B}=sin45\times 0.125\times \frac{1}{sin30}=0.1768m$
and $r_{IC/C}=sin105\times 0.125\times \frac{1}{sin30}=0.2415m$
We know that
$v_B=\omega_{AB}r_{AB}$
$\implies v_B=4(0.3)=1.2m/s$
and $v_B=\omega_{BC}(r_{IC/B})$
We plug in the known values to obtain:
$1.2=0.1768\omega_{BC}$
This simplifies to:
$\omega_{BC}=6.79rad/s$