Answer
$v_C=1.34m/s$
Work Step by Step
The required can velocity be determined as follows:
$\frac{0.5}{sin45}=\frac{r_{B/IC}}{sin60}$
This simplifies to:
$r_{B/IC}=0.61237m$
and $v_B=\omega_{AB}r_{AB}$
$\implies v_B=6\times 0.2=1.2m/s$
The angular velocity of $BC$ can be determined as
$\omega_{BC}=\frac{v_B}{r_{B/IC}}=\frac{1.2}{0.61237}=1.959rad/s$
Similarly, $\frac{0.5}{sin45}=\frac{r_{C/IC}}{sin75}$
This simplifies to:
$r_{C/IC}=0.683m$
Now $v_C=\omega_{BC}r_{IC/IC}$
We plug in the known values to obtain:
$v_C=(1.959)(0.683)=1.34m/s$
