Answer
$\omega_{BC}=2.83 rad/s \circlearrowleft$
$\omega_{AB}=2.83 rad/s \circlearrowright$
Work Step by Step
The required angular velocity can be determined as follows:
We know that
$\vec{v_B}=\vec{v_C}+\vec{\omega_{BC}}\times \vec{r_{B/C}}~~~$[eq(1)]
The velocity of $C$ is given as
$\vec{v_C}=8sin45\hat i-8sin45\hat j=5.6568\hat i-5.6568\hat j$
and $\vec{r_{B/C}=2\hat j}$
Similarly $\vec{v_B}=\vec{\omega_{AB}}\times \vec{r_{AB}}$
$\vec{v_B}=(-\omega_{AB}\hat k)\times (2\hat i)=-2\omega_{AB}\hat j$
We plug in the known values in eq(1) to obtain:
$-2\omega_{AB}\hat j=(5.6568\hat i-5.6568\hat j)+(\omega_{BC}\hat k)\times (2\hat j)$
$-2\omega_{AB}\hat j=(5.6568-2\omega_{BC})\hat i-5.6568\hat j~~$[eq(2)]
Comparing $i$ components on both sides, we obtain:
$0=5.6568-2\omega_{BC}$
This simplifies to:
$\omega_{BC}=2.83 rad/s \circlearrowleft$
Now, comparing the $j$ components on both sides of eq(2), we obtain:
$-2\omega_{AB}=-5.6568$
$\implies \omega_{AB}=2.83 rad/s \circlearrowright$
