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Trying to find the angular speed of a rod.

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data
    A uniform rod (length = 2.0m) is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through the rod at a point of 0.5m from one end of the rod. If the rod is released from rest in a horizontal position, what is the angular speed of the rod as it rotates through its lowest position?


    2. Relevant equations
    I believe (although I might be wrong on some of them):
    [itex]\tau = I \alpha[/itex]
    [itex]K = \frac{1}{2} I \omega^{2}[/itex]
    [itex]L_{z} = I \omega[/itex]
    For a rod the moment of inertia is:
    [itex]\frac{1}{12} M L^{2}[/itex]
    And the parallel axis thereom:
    [itex]I = I_{cm} + M D^{2}[/itex]
    3. The attempt at a solution
    So far I have used the parallel axis theorem and the moment of inertia for a rod to get the moment of inertia to be 0.5833 for this rods rotation.
    I am not sure exactly what to do after that. I can figure it out if it is the moment right after release, but not sure how to do it when the rod is at the bottom.
     
  2. jcsd
  3. Mar 6, 2013 #2

    haruspex

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    Energy?
     
  4. Mar 6, 2013 #3
    I first started thinking I could take the potential energy the system has before release as (mgh) and then at the bottom i could take mgh = (1/2)I*w^2 and find omega... But I couldn't figure out how to use it right...
     
  5. Mar 6, 2013 #4

    haruspex

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    That's the right method. Just need to figure out h and I. Any ideas?
     
  6. Mar 6, 2013 #5
    I believe I have found I for the entire rod. The issue is with the height I think. I want to say the height is 1.5 but what about the 0.5 end? I try using m = 1, g = 9.8 and h = 1.5.

    1.5*9.8 = 14.7
    14.7 = (1/2)(0.5833333)(w^2)
    29.4 = 0.5833(w^2)
    50.4 = w^2
    w = 7.099

    But the answer is supposedly 4.1...
    I then tried subtracting the small 0.5 stub:
    0.5*9.8*1 (mgh)
    4.9 = (1/2)(0.5833)(w^2)
    9.8 = (0.5833)w^2
    16.8 = w^2
    w = 4.099

    Subtracting the two gives 3.00 which is not right either though.
     
  7. Mar 6, 2013 #6
    [itex]I[/itex] was found using doing the following:

    I = I_cm + m*d^2

    I = (1/12)m*L^2 + m*d^2

    d = 0.5 since there is a 0.5 shift from the center.

    I = (1/12)*1*4 + 1*(0.5^2)

    I = (0.5833333)
     
  8. Mar 6, 2013 #7
    Actually never mind I should use h = 2 not 1.5 for the first part of my original attempt and then everything else should be the same.

    1.5*9.8 = 14.7
    14.7 = (1/2)(0.5833333)(w^2)
    29.4 = 0.5833(w^2)
    50.4 = w^2
    w = 8.199

    0.5*9.8*1 (mgh)
    4.9 = (1/2)(0.5833)(w^2)
    9.8 = (0.5833)w^2
    16.8 = w^2
    w = 4.099

    8.199 - 4.099

    Thanks!

    Although I am still not sure why I should use 2...
     
  9. Mar 6, 2013 #8

    haruspex

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    No, 0.5m is right for h (it's the distance the CoM descends).
    You should have
    mgh = (1/2)m(L2/12+h22
    4.9*2 = (1/3+1/4)ω2
    ω ≈ 4.1
    Are you saying the given answer is 8.2?
     
  10. Mar 7, 2013 #9
    Oh, my bad. Haha the given answer was 4.1, for some reason I didn't even realize using h = 0.5 gives the answer approx. 4.1. My bad thanks again!
     
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