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

$(\omega_B)_{max}=8.49~rad/s$ $v_C=0.6~m/s$
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: $v_C=6\sqrt{2}\times 50\sqrt{2}$ $\implies v_C=0.6m/s$