Two bars are pinned at one end, and connect with sliding collar at the other end. If the angular velocity of bar AC is clockwise at 3 rad/s, what is the angular velocity of bar BD? Point C slides smoothly along bar BD but is pinned to bar AC 15 cm from point A. .

Angular Velocity

Since the distance BC is not constant, this problem is well suited for using two frames of reference. Fix the local frame of reference, x-y, on bar BCD so that the motion of C can be easily modeled relative to B, (r_C/B)_rel. Then the velocity of C is,
     v_C = v_B + Ω × r_C/B + (v_C/B)_rel

This equation includes the unknown angular velocity of bar BD, Ω = Ω_BC = Ω_BD, that is required, and can be determined. The global frame of reference is fixed to the background and is not moving.

The velocity of B is zero (it is pinned).
     v_B = 0

The velocity of C with respect to B, v_C/B, is
     v_C/B = -v_C/Bi = -v_C/Bcos45I - v_C/Bsin45J m/s

The velocity of point C, based on the motion of bar AC, in the global coordinate system is
     v_C = -ω_ACr_AC J = -3(0.15)J = -0.45J m/s

The relative position of C with respect to B is
    r_C/B = 0.15/cos45i = 0.15I + 0.15J m

Putting all the terms together, gives
    -0.45J = -v_C/Bcos45I - v_C/Bsin45J + 0 +
                    + Ω_BDK × (0.15I + 0.15J)

     -0.45J = -0.7071v_C/BI - 0.7071v_C/BJ +
              + 0.15Ω_BDJ - 0.15Ω_BDI

Summing the I terms gives,
     0.7071v_C/B= -0.15 Ω_BD
     v_C/B= -0.2121Ω_BD

Summing J terms:
     -0.45 = -0.7071 (-0.2121Ω_BD) + 0.15Ω_BD
     Ω_BD = -1.50K rad/s

Ω_BD = 1.50 rad/s