Two Connected Bars


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, (rC/B)rel. Then the velocity of C is,
     vC = vB + Ω × rC/B + (vC/B)rel

This equation includes the unknown angular velocity of bar BD, Ω = ΩBC = ΩBD, 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).
     vB = 0

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

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

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

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

     -0.45J = -0.7071vC/BI - 0.7071vC/BJ +
              + 0.15ΩBDJ - 0.15ΩBDI

Summing the I terms gives,
     0.7071vC/B= -0.15 ΩBD
     vC/B= -0.2121ΩBD

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

ΩBD = 1.50 rad/s

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