Answer
$v_A=10.77in/s$ $21.8^{\circ}$
$v_B=14in/s$
Work Step by Step
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:
$v_B=(2)(7)=14in/s \downarrow$
$\theta=tan^{-1}(\frac{2}{5})=21.8^{\circ}$
