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Homework Help: A Slider-Crank & Angular Velocities/Acceleration Problem

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data
    I have posted the problem in an attachment.

    2. Relevant equations
    Aa = Ab + Aab (acceleration at a is equal to the the acceleration of b plus the acceleration of A with respect to B)
    Aa = ωob x [ωob x Rob] + [αab x Rab]
    Va = Vb + Vab

    3. The attempt at a solution
    First, I got the relative positions (Rob and Rab) of O with respect to B (-0.1 j) and A with respect to B (-0.229i - 0.1j) I feel pretty good about this being correct

    Now, when I plug in to the 2nd equation I have in the "relevant equations" part and do a lot of cross products, I get the following:
    Aa = -1000j - 0.229αab j + 0.1αab i

    Now I am unsure how to proceed from here. I still have not found any of the three items that I'm asked to find and kind of stuck on where to go next. Any help is much appreciated.

    Attached Files:

  2. jcsd
  3. Oct 16, 2013 #2


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    As with your https://www.physicsforums.com/showthread.php?t=716711, finding the instantaneous geometry is of limited value. You need equations in which angles and the distance AO are treated as variables. Then you can differentiate to find the relationships between velocities and angular velocities etc. and as I wrote on that thread, another approach is to consider components of velocities along the rods. The two ends of a rod must have the same velocitiy in that direction.
    Btw, I believe the given answers in this question are wrong. To get the rotation rate of AB to be zero I need angle ABO to be a right angle, not angle AOB.
  4. Oct 16, 2013 #3
    I did this question and I found the answers to all three of them. Instantaneous rotation rate of AB comes to be zero both conceptually and quantitatively.

    My other two answers differed by a factor of 100 (lower), but that could be my mistake!!!
  5. Oct 16, 2013 #4


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    Yes, I realised later I was wrong about that, but only just got a chance to get back online.
  6. Oct 16, 2013 #5
    no problem!!!:wink:
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