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DYNAMICS - EXAMPLE


Two Connected Bars
  Example

 

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. .

 

   
    Solution


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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