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

$v_B=4ft/s$ $v_C=6ft/s$
We can determine the required linear velocity as follows: From the given diagram $r_{IC/A}=\frac{8}{2}=4ft$ Now, $v_C=\omega r_{IC/C}$ We plug in the known values to obtain: $v_C=2\times 3=6ft/s$ and $v_B=\omega{r_{IC/B}}$ We plug in the known values to obtain: $v_B=2\times 2=4ft/s$