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Acceleration of center of mass

  1. Apr 21, 2017 #1
    1. The problem statement, all variables and given/known data
    "Represent (linear) acceleration of center of mass of link OA in terms of variables shown." OmegaOA is given to be some constant value, I have assigned it 'omegaOA'.
    dynamicscompassignment2.PNG


    2. Relevant equations
    Already derived equations to previous parts of the problem, fairly certain these are correct. Relative variables are neglected, since this is a single body problem. 'a' is linear acceleration, alpha is angular acceleration.
    omegaAB = (omegaOA*b*cos(theta))/sqrt(d^2 - (b*sin(theta) + h)^2)
    alphaOA = 0
    alphaAB = [-(omegaOA^2)*b*sin(theta) - (omegaAB^2)*(b*sin(theta)+h)]/sqrt(d^2 + (b*sin(theta) + h)^2)
    alphaOA = (a/r)

    3. The attempt at a solution
    If the weight (mg) of the rod was the only force acting on the rod, the problem is very simple.

    I*alpha = torque where I = (1/12)*m*length^2
    Mass cancels and the problem is easy to solve. However, I am guessing there are force vectors at A that need to be taken into account. Any help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Apr 21, 2017 #2

    haruspex

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  4. Apr 22, 2017 #3
    Sorry about the repost, I had stated the problem poorly the first time, so I figured it would save the time of those replying to make a more precise and accurate post of the question.
    And the 'relevant equations' that I have listed are just there because I'm trying to give as much info as possible. And if the reaction forces at A are worked around, there are still support forces at O (according to my analysis, could be wrong). I.e. a simple moment analysis appears to lead to a confusing situation, with either forces at A or O coming up
     
  5. Apr 23, 2017 #4

    haruspex

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    Why do you care about forces at all? Point O is fixed, the rotation rate of OA is fixed. That tells you everything about the motion of all points on OA.
     
  6. Apr 24, 2017 #5
    That makes sense..so the linear acceleration of the center of mass would be zero? I would assume that all points along the link OA would have an acceleration of zero.
     
  7. Apr 24, 2017 #6

    haruspex

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    No. It has a tangential acceleration of zero.
    Linear acceleration of a point is its total acceleration, tangential plus radial (vectorially). Linear acceleration of a rigid body is the linear acceleration of its mass centre.
     
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