Bar AB of a classic three bar linkage system has an angular velocity of 5 rad/s. The lengths of the bars AB and DC are each 3 ft. What is the angular velocity of bar BC?

This is a three step process. First find the velocity of point B. Then find the direction velocity direction of point C. Finally, the angular velocity of bar BC can be determined the information known about point B and C.

Velocity of point B,
     v_B = ω × r_B/A
         = 5k × 3(cos 45i + sin 45j)
         = -10.61i + 10.61j ft/s

Velocity of point C,
     v_C = ω_CD × r_C/D
         = ω_CDk × 3(-cos 60i + sin 60j)
         = -1.5ω_CDj - 2.598ω_CDi ft/s

To understand the motion of Bar BC, the position vector, r_C/B needs to be identified.
     r_C/B =(6 - 3cos45 - 3cos60)i + (3sin60 - 3cos45)j
           = 2.379i + 0.4768j

Angular velocity of bar BC is found by relating the velocity of point C to point B,
     v_C = v_B + v_C/B = v_B + ω_BC × r_C/B
     -1.5ω_CDj - 2.598ω_CDi = -10.61i + 10.61j +
                 ω_BCk × (2.379i + 0.4768j)
     -2.598 ω_CDi -1.5 w_CDj = -10.61i + 10.61j +
                 2.379ω_BCj - 0.4768ω_BCi

Solving i and j terms,
     -2.598ω_CD = -10.61 - 0.4768ω_BC
     -1.5ω_CD = 10.61 + 2.379ω_BC

Solving equations for ω_BC,

ω_BC = -6.305 rad/s