Acceleration of center of mass

  • #1

Homework Statement


"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



Homework 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)

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.
 

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Answers and Replies

  • #3
If the angular velocity of OA is fixed, and point O is fixed, how does anything to the right of A affect the movement of OA? Are you sure you have stated the question correctly?
And why did you mark the previous thread https://www.physicsforums.com/threads/acceleration-of-center-of-mass.912120/ as solved, then open a new one for the same question?
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
 
  • #4
haruspex
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if the reaction forces at A are worked around,
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.
 
  • #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.
 
  • #6
haruspex
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so the linear acceleration of the center of mass would be zero?
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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