Answer
$v_C=28.98m/s \rightarrow$
$\omega_{BC}=10.6 rad/s \circlearrowleft$
Work Step by Step
We can determine the required velocity and angular velocity as follows:
$v_B=\omega_{AB}\times r_{AB}$
$\implies v_B=(5\hat k)\times (-0.3cos 45\hat i-0.3sin45\hat j)$
$\implies v_B=(\frac{15\sqrt{2}}{2}\hat i-\frac{15{\sqrt{2}}}{2}\hat j)m/s$
We know that
$v_C=v_B+\omega_{BC}\times r_{C/B}$
$\implies v_{C \hat i}=(\frac{15\sqrt{2}}{2}\hat i-\frac{15\sqrt{2}}{2}\hat j)+(\omega_{BC}\hat k)\times (2sin30\hat i-2cos30\hat j)$
$\implies v_{C \hat i}=(\frac{15\sqrt{2}}{2}+\omega_{BC} \sqrt{3})\hat i+(\omega_{BC}-\frac{15\sqrt{2}}{2})\hat j$
Comparing $j$ components on both sides, we obtain:
$0=\omega_{BC}-\frac{15\sqrt{2}}{2}$
$\implies \omega_{BC}=\frac{15\sqrt{2}}{2}=10.6 rad/s \circlearrowleft$
and comparing $i$ components on both sides, we obtain:
$v_C=\frac{15\sqrt{2}}{2}+\frac{15\sqrt{2}}{2}\sqrt{3}$
This simplifies to:
$v_C=28.98m/s \rightarrow$
