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What is the magnitude of the angular momentum of the bar?

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    A rigid, uniform bar with mass m and length b rotates about the axis passing through the midpoint of the bar perpendicular to the bar. The linear speed of the end points of the bar is v . What is the magnitude of the angular momentum of the bar?

    2. Relevant equations

    L=Iw
    I=1/12(ML^(2) This L is the length.
    v=wr


    3. The attempt at a solution

    L=1/12*m*B^(2)*w

    v=wr
    it says that the linear speed of the bar at the end points is v
    When I first read that I automatically thought that I would have to double v=wr because it has two endpoints. However, that is wrong. Why is it that I am only supposed to use a single v=wr?
     
  2. jcsd
  3. Nov 10, 2012 #2

    tiny-tim

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    Hi Toranc3! :smile:
    v = ωr applies to the speed v of any point on the object at distance r from the centre of rotation

    so v = ω(L/2) for one end, and v = ω(L/2) for the other end :wink:
     
  4. Nov 10, 2012 #3
    Hi tiny tim, thanks for replying.
    So going back to angular momentum equation L=Iw, the omega applies only to a certain point of the rotation. Therefore when I substitute v=wr I am only replacing v/r=w at that certain point. So that is why I do not double it. Would it be correct to say that? The final answer is L=1/6mbv, forgot to add that in.
     
  5. Nov 10, 2012 #4

    tiny-tim

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    (try using the Quick Symbols box next to the Reply box :wink:)
    no, ω is the same for every point of a rigid body

    v is what varies :wink:

    (ie the speed, v, is proportional to the distance, r, from the fixed point)
     
  6. Nov 10, 2012 #5
    Well I am still not so sure why but I am going to put this error that I did in my head so I will not make the same error again. Thanks though
     
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