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

$\omega_{AB}=3.46 rad/s$ $\omega_{BC}=2.31rad/s$
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 $\vec{v_C}=6\hat j$ and $\vec{r_{B/C}}=-3cos30\hat j+3sin30\hat j=-2.598\hat i+1.5\hat j$ The velocity of $B$ is $\vec{\omega_B}=\vec{\omega_{AB}}\times \vec r_{AB}$ $\implies \vec{v_B}=(\omega_{AB})\hat k\times (1\hat j)=-\omega_{AB}\hat i$ We plug in the known values in eq(1) to obtain: $-\omega_{AB}\hat i=6\hat j+(\omega_{BC})\hat k(-2.598\hat i+1.5\hat j)$ $\implies \omega_{AB}\hat i=-1.5\omega_{BC}\hat i+(6-2.598\omega_{BC})\hat j~$[eq(2)] Comparing $i$ components on both sides of eq(2) $-\omega_{AB}=-1.5\omega_{BC}$ $\implies \omega_{AB}=1.5\omega_{BC}~$[eq(3)] Comparing $j$ components of eq(2) on both sides $0=6-2.598\omega_{BC}$ This simplifies to: $\omega_{BC}=2.3095rad/s$ We plug in this value in eq(3) to obtain: $\omega_{AB}=1.5(2.3095)=3.464 rad/s$