Answer
$v_C=2.45m/s$
$\omega_{BC}=7.81rad/s$
Work Step by Step
The required can be determined velocity and angular velocity follows:
$v_B=\omega_{AB}r_{AB}=4(0.5)=2m/s$
We know that in Cartesian coordinates
$v_B=(-2cos30\hat i+2sin30 \hat j)m/s$
and $v_C=-v_C cos45\hat i-v_C sin45\hat j$
As $v_C=v_B+\omega\times r_{C/B}$
We plug in the known values to obtain:
$-v_Ccos45\hat i-v_Csin45\hat j=(-2cos30\hat i+2sin30\hat j)+(\omega_{BC}\hat k)(-0.35\hat i)$
This can be rearranged as:
$-v_Ccos45\hat i-v_Csin45\hat j=-2cos30\hat i+(2sin30-0.35\omega_{BC})\hat j$
Comparing $i$ components on both sides, we obtain:
$-v_c cos45=_2cos30$
$\implies v_C=2.45m/s$
and comparing $j$ components on both sides, we obtain:
$-2.45 sin45=2sin30-0.35\omega_{BC}$
This simplifies to:
$\omega_{BC}=7.81rad/s$
