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Homework Help: Dynamics - Pulley System with Rotating Rod

  1. Oct 22, 2015 #1
    1. The problem statement, all variables and given/known data
    At the instant shown, the rod R is rotating about its centre of rotation with ω=3.8rad/s.


    The pulley, with mP=8.7kg and RP=0.2m, may be modelled as a uniform disc.

    The rod, with mR=4.1kg and L=0.8m, may be modelled as a thin beam rotating about one end.

    g=9.8m/s ².

    What is the magnitude of the acceleration of point B at this instant?

    2. Relevant equations
    ΣF=ma (N2) ΣM=Iα (Eulers equation)

    3. The attempt at a solution
    IRod at centre of rotation=(1/3)ML2

    I defined upwards and anticlockwise to be positive and thus derived the following equations:
    ΣMP at centre=RpTA-RpTB=IPαP
    ΣMRod at end=-LTB+(1/2)LMRg=IRαR
    where TA= Tension force acting between A and pulley and TB=Tension force acting between rod and pulley

    I then found these constraints on aB in terms of aAPR
    assuming that aB is acting upwards

    Then, by subbing aB into the three original equations, I got the following system of equations:

    However, when I solve this system of linear equations I get the wrong answer. I have a feeling this is because I ignored the angular velocity of the rod but I can't see that would affect the acceleration of B.

    Attached Files:

  2. jcsd
  3. Oct 23, 2015 #2


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    If you ignored everything (gravity and tension) except for the fact that the rod is rotating. Would B have zero acceleration?
  4. Oct 23, 2015 #3
    The question does not specify. All the information in the problem statement is all the information that the question gives.
  5. Oct 23, 2015 #4


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    It is not a question about the problem, it is a question to you.
  6. Oct 23, 2015 #5
    Ah ok. So the end of the rod would be accelerating towards the pivot point as well.
    I took this into consideration and got the right answer. I guess I should have studied the end of the rod more closely. Thank you so much for your help :)
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