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Total Angular Momentum

  1. Oct 27, 2004 #1
    This is a problem that seemed easy to me. I keep recalculating the answer, and I keep getting it wrong. I'm not sure if I'm approaching the problem incorrectly, or if i'm just making some small dumb mistake. Here's the question:

    Big Ben (see the figure below), the Parliament Building tower clock in London, has hour and minute hands with lengths of 2.65 m and 4.45 m and masses of 61.0 kg and 101 kg, respectively. Calculate the total angular momentum of these hands about the center point. Treat the hands as long, thin uniform rods.

    Ltot = Lmin + Lhour
    L=I*W
    Irod = .5*m*L^2
    Wmin = 2pi/3600
    Whour = 2pi/86400

    Lmin = .5(101)4.45^2 * 2pi/3600 = 1.745
    Lhour = .5(61)2.65^2 * 2pi/86400 = .0156
    Ltot = 1.745 + .0156 = 1.76

    1.76 is what I keep getting. Any thoughts, comments, help would be greatly appreciated. Thanks in advance.
    -Brad
     
  2. jcsd
  3. Oct 27, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The rotational inertia of a thin rod about one end is [itex]I = 1/3 m L^2[/itex].
     
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