Answer
$(\omega_B)_{max}=8.49~rad/s$
$v_C=0.6~m/s$
Work Step by Step
We can determine the required angular velocity and the maximum speed as follows:
We know that
$(r_A){max}=(r_B)_{max}$
$\implies (r_A)_{max}=\frac{\sqrt{(100)^2+(100)^2}}{2}=50\sqrt{2}min$
and $(r_A)_{min}=(r_B)_min=50mm$
Now $\omega_(r_A)_{max}=(\omega_B)_{max}(\omega_B)_{min}$
We plug in the known values to obtain:
$6\times 50\sqrt{2}=(\omega_B)_{max}\times 50$
This can be rearranged as:
$(\omega_B)_{max}=\frac{6\times 50\sqrt{2}}{50}=8.485rad/s$
The maximum speed is given as
$v_C=(\omega_B)_{max}r_C$
We plug in the known values to obtain:
$v_C=6\sqrt{2}\times 50\sqrt{2}$
$\implies v_C=0.6m/s$
