This could be solved using relative motion equations (vectors) as explained previous section, General Plane Motion: Velocities. However, for many velocity plane motion problems, using the instantaneous center of zero velocity (IC) can simplify the problem.
The IC point will be at the intersection of the two lines perpendicular to the velocities of point B and D. For this instant, the triangle OBD is rotating about point O.
The distances DO and BO are,
rOD = rBD tan45 = 0.3 (1) = 0.3 m
rOB = rBD / cos45 = 0.3 / 0.7071 = 0.4243 m
The velocity of point B is,
vB = ωAB rAB = 50 (0.08) = 4 m/s
Thus, the angular velocity of BD is,
ωBD = ωBDO = vB/rBO = 4 / 0.4243 = 9.428 rad/s
The velocity of point D can now be determined,
vB = ωBD rOD = 9.428 (0.3) = 2.828 mr/s
Finally, the angular velocity of CD is
ωCD = vB / rCD = 2.828 / 0.075 = 37.71
ωCD = 37.71 rad/s