## Engineering Mechanics: Statics & Dynamics (14th Edition)

$\omega_{BC}=2.12rad/s \circlearrowleft$ $v_C=24.59m/s \downarrow$
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: $v_C=24.59m/s \downarrow$