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DYNAMICS - EXAMPLE


Two Connected Bars
  Example

 

Two rods are pinned to a fixed point at one end, and connect to each other at the other end using a slider joint. Point C can freely move along rod DE but is pinned to rod AB at C. Rod AB rotates clockwise with angular velocity ωAB = 4 rad/s and angular acceleration αAB = 5 rad/s2 when θ = 45o. Find the angular acceleration of rod DE at this instant. The collar at C is pin connected to AB and slides over rod DE.

   
    Solution

 

Since the distance between point D and C varies, two frames of reference is useful to solve the problem. Attach the local x-y coordinate to the rotating AE rod, and the global X-Y coordinate to the non-moving background.

Velocities:
Before accelerations can be found, the velocity of point C, vC/D, and angular velocity of ΩDE (= ΩDC) needs to be determined.
     vC = vD + Ω × rC/D + ( vC/D)rel
     vC = vD + ΩDC× rC/D + vC/D
     vD = 0
     ωAB = 4k
     rC/D = 0.5 m i
     ΩDC = ωDC

Since distance from A and C does not change, velocity of C can be determined from bar AB
     vC = ωAB × rC/A = -4k × (0.5i + 0.5j)
          = 2i - 2j m/s

Putting all terms together gives,
     2i - 2j = 0 + (-ωDC k) × (0.5i) + vC/D reli
     2i - 2j = -0.5ωDC j + vC/D rel i

Comparing the i and j terms give
     vC/D = 2 m/s   and   ωDC = 4 rad/s

Accelerations:
     aC = aD + dΩDC/dt × rC/D + ΩDC× (ΩDC × rC/D)
              + 2ΩDC × vC/D + aC/D
     aC = aD + αDE × rC/D + ωDE× ( ωDE × rC/D)
              + 2ωDE × vC/D + aC/D
where,
     aD = 0
     rC/D = 0.5i m
     αDC = (-αDCk) × 0.5i = -0.5αDEj m/s2
     ωDC × ( ωDC × rC/D) = (- 4)2 (0.5 i) = -8i rad/s
     2ωDC × vC/D = 2 (-4k) × 2i = 16j m/s2
     aC/D = aC/Di

Acceleration of C can be determined from bar AC
     aC = αAC × rC/A - ω2 rC/A
         = -5k × (0.5i + 0.5j) - (4)2 (0.5i + 0.5j)
         = -5.5i - 10.5j m/s2

-5.5i - 10.5j = 0 + (-αDEk)× 0.5i - (- 4)2 (0.5 i)
       + 2 (-4k)×2i + (aC/D)i

Putting all terms together gives,
     -5.5i - 10.5j = 0 - 0.5αDCj - 8i - 16j + aC/Di

Comparing the j terms
     -10.5 = -16 - 0.5αDE
     αDC = -11 rad/s2

αDC = 11k rad/s2 (counter-clockwise)

     
   
 
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