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Instantaneous center of velocity for a inverted slider crank

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the instantaneous centers of the linkage below.

    upload_2015-11-1_18-14-26.png
    2. Relevant equations

    # of ICs = n(n-1) / 2

    3. The attempt at a solution

    From the equation above, there are 4(4-1)/2 = 6 ICs in this linkage. But one thing I don't understand is how there are 4 links when 2 and 3 are welded together to make one piece? And how is the IC of point 2,3 at the center of the circle?
     
  2. jcsd
  3. Nov 1, 2015 #2

    haruspex

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    I'm not sure I understand the diagram. I'm guessing 2 represents a solid disk which pivots eccentrically about a point near its base, while the black hoop is free to slide around the disk. But as you say, the black hoop is rigidly connected to the black bar. How is the black bar connected to the block numbered 4? That looks rigid too, so I only see three pivot points, so only two links. Or can it slide through the block?
     
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