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

$v_A=10.77in/s$ $21.8^{\circ}$ $v_B=14in/s$
The required velocity can be determined as follows: $r_{IC/B}=r_{O/IB}+r_{IC/O}$ $\implies r_{IC/B}=5+2=7in$ and $r_{IC/A}=\sqrt{(2)^2+(5)^2}=\sqrt{29}in$ We know that $v_A=\omega r_{IC/A}$ We plug in the known values to obtain: $v_A=2(\sqrt{29})=10.77in/s$ and $v_B=\omega r_{IC/B}$ We plug in the known values to obtain: $v_B=(2)(7)=14in/s \downarrow$ $\theta=tan^{-1}(\frac{2}{5})=21.8^{\circ}$