Answer
$\omega_{AB}=2 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We know that
$\vec{v_C}=\vec{v_B}+\vec{\omega_{BC}}\times \vec{r_{C/B}}$..eq(1)
As $\vec{v_C}=-4\hat j$
and $\vec{r_{C/B}}=3cos30\hat i+3sin 30\hat j=2.597\hat i+1.5\hat j$
similarly, $\vec{v_B}=-(\omega_{AB}r_{AB})\hat j=-2\omega_{AB}\hat j$
We plug in the known values in eq(1) to obtain:
$-4\hat j=-2\omega_{AB}\hat j+\omega_{BC} \hat k\times (2.598\hat i+1.5\hat j)$
$\implies -4\hat j=-1.5\omega_{BC}\hat i+(-2\omega_{AB}+2.598\omega_{BC})\hat j$
Comparing the $i$ component, we obtain:
$0=-1.5\omega_{BC}$
$\implies \omega_{BC}=0 rad/s$
and comparing the $j$ component, we obtain:
$-4=-2\omega_{AB}+2.598\omega_{BC}$
$\implies -4=-2\omega_{AB}+0$
This simplifies to:
$\omega_{AB}=2 rad/s$
