Problem 5/57  Page 358, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A rod of length L with a known velocity at the left end and a known speed at the right end.

Find: The angular velocity of the rod.

L = 2 ft
vA = vA at an angle of q (ccw) with the link (left end)
vA = 10 ft/s
q = 30 degrees
vB = vB at an angle of b (cw) with the link (right end)
vB = 12 ft/s

The velocity two point formula can be used to relate the velocities of the two ends of the rod.  Letting +X run along the rod from the left end toward the right end, and positive Y be perpendicular to the rod so that ccw rotation is positive:
vB = vA + w X r
vB ( cos b  i - sin b  j ) = vA ( cos q  i + sin q j ) + w k X L i
vB  ( cos b  i - sin b  j ) = vA ( cos q  i + sin q j ) + w L j

Equating components:
vB  cos b  = vA   cos q 
-vB sin b = vA sin q + w L

Solving the first equation for the angle b and then solving the second equation for the angular speed:
cos b = vA cos q   / vB         b  = 43.81 degrees
w = -(1/L) ( vA sin q  + vB sin b ) = -6.65 rad/s