Answer
$\omega_{AB}=3.72rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
$\vec{v_B}=\vec{v_c}+\vec{\omega_{BC}\times \vec{r_{BC}}}~~~$[eq(1)]
As $\vec{v_C}=\vec{\omega_{CD}\times \vec{r_{CD}}}$
$\implies \vec{v_C}=5\hat k\times (-4cos45\hat i+4sin45\hat j)=-14.142\hat j-14.142\hat i$
Similarly, $\vec{r_{B/C}}=8sin30\hat i+8cos30\hat j=4\hat i+6.928\hat j$
The velocity of $B$ can be defined as
$\vec{v_B}=\vec{\omega_{AB}}\times \vec{r_{B/A}}=-\omega_{AB}\hat k\times (6\hat i)=-6\omega_{AB}\hat j$
We plug in the known values in eq(1) to obtain:
$-6\omega_{AB}\hat j=(-14.142\hat j-14.142\hat i)+(\omega_{BC})\hat k\times (4\hat i+6.928\hat j)$
$\implies -6\omega_{AB}\hat j=(-14.142-6.928\omega_{BC})\hat i+(-14.142+4\omega_{BC})\hat j$
Comparing $i$ components on both sides, we obtain:
$0=-14.142-6.928\omega_{BC}$
$\implies \omega_{BC}=-2.0412rad/s$
Now comparing $j$ components, we obtain:
$-6\omega_{AB}=-14.142+4\omega_{BC}=-14.142+4(-2.0412)$
This simplifies to:
$\omega_{AB}=3.72rad/s$
