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

$\omega_{BC}=2.83 rad/s \circlearrowleft$ $\omega_{AB}=2.83 rad/s \circlearrowright$
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: $-2\omega_{AB}=-5.6568$ $\implies \omega_{AB}=2.83 rad/s \circlearrowright$