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

$\omega_{AB}=1.24rad/s$
We can determine the required angular velocity as follows: According to the law of sines $\frac{4}{sin 45}=\frac{r_{IC/B}}{sin30}=\frac{r_{IC/C}}{sin 105}$ This simplifies to: $r_{IC/B}=\frac{4sin30}{sin45}=2.828in$ and $r_{IC/C}=\frac{4sin105}{sin45}=5.464in$ We know that $v_C=\omega_{BC}(r_{IC/C})$ We plug in the known values to obtain: $12=5.464\omega_{BC}$ This simplifies to: $\omega_{BC}=2.1962rad/s$ Similarly, $v_B=\omega_{BC}(r_{IC/B})$ We plug in the known values to obtain: $v_B=(2.1962)(2.828)=6.211in/s$ Now $v_B=\omega_{AB}r_{AB}$ We plug in the known values to obtain: $6.211=5\omega_{AB}$ This simplifies to: $\omega_{AB}=1.24rad/s$