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DYNAMICS - EXAMPLE


Sliding Piston
  Example

 

Two bars link a vertically sliding piston, C, to a pinned point, A. If bar AB rotates at a rate of 3 rad/s and accelerates at a rate of 1 m/s2, what is the angular acceleration of bar BC?

 

   
    Solution


Velocity and Acceleration Diagrams

 

Velocity information about bar BC must first be found before accelerations can be determined. The key object in this system is rod BC. Velocity of points B can be found, and velocity direction of C can be found.

Velocity
First, find the velocity of point B by relating its velocity to fixed point A,
     vB = vA + vB/A = 0 + ωAB × rAB
 
         = 3k × 0.40i = 1.2j m/s

The velocity direction of C is known, thus
     vc = vcj m/s

Combining gives,
     vB = vC + vB/C = vC + ωBC× rBC
     1.2j = vCj + ωBCk × (-0.3 sin60i + 0.3 cos60j)
     1.2j = vCj + 0.2598ωBCj + 0.15ωBCi
     i terms:  ωBC = 0

Acceleration
Now the acceleration terms can be determined since the velocity information has been calculated. The acceleration of point B is,
     aB = aA + aA/B
         = aA + ωAB2 rAB + αAB × rAB
         = 0 + (3)2 (0.4i) +1k × 0.4i
         = 3.6i + 0.4j

Acceleration of point C,
     aC = acj m/s2

Combining,
     aB = aC + aB/C
     aB = aC + ωBC2 rBC + αBC × rBC
     3.6i + 0.4j = aCj + 0 +
                  + aBCk × (-0.3 sin60i + 0.3 cos60j)
     3.6i + 0.4j = aCj + 0.2598 aBCj + 0.15αBCi
     i terms:   -3.6 = -0.15 αBC

αBC = 24 rad/s2

     
   
 
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