Answer
$\omega_{BC}=2.12rad/s \circlearrowleft$
$v_C=24.59m/s \downarrow$
Work Step by Step
We can determine the required velocity and angular velocity as follows:
$v_B=\omega_{AB}\times r_{AB}$
$\implies v_B=(60\hat k)\times (-0.3sin60\hat i+0.3 cos 60\hat j)=(-9\hat i-9\sqrt{3})m/s$
Similarly $v_C=v_B+\omega_{BC}\times r_{C/B}$
$\implies -v_C\hat j=(-9\hat i-9\sqrt{3}\hat j)+(\omega_{BC}\hat k)\times (-0.6 sin45\hat i-0.6cos 45\hat j)$
This can be rearranged as:
$-v_C\hat j=(0.3\sqrt{2}\omega_{BC}-9)\hat i-(0.3\sqrt{2}\omega_{BC}+9\sqrt{3})\hat j$
Comparing $i$ components on both sides, we obtain:
$0=0.3\sqrt{2}\omega_{BC}-9$
$\implies \omega_{BC}=2.12rad/s \circlearrowleft$
and comparing $j$ components, we obtain:
$-v_C=-0.3\sqrt{2}(2.12)-9\sqrt{3}$
This simplifies to:
$v_C=24.59m/s \downarrow$
