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Homework Help: Angular momentum

  1. Mar 13, 2005 #1
    The problem is http://home.earthlink.net/~urban-xrisis/clip002.jpg [Broken].

    I got a different answer from what my book tells me and I want to know why. Angular momentum is:
    [tex]L=I \omega[/tex]
    So... the moment of inertia for the block would be:
    [tex]I=((M+m)l^2)[/tex]
    since the rod has negligible mass, it will not have a moment of inertia.
    [tex]\omega=v/r=v/l[/tex]
    [tex]L=\frac{(M+m)l^2v}{l}[/tex]
    [tex]L=vlM+vlm[/tex]

    my text gives an answer of [tex]L=mvl[/tex]
    I dont understand what I misunderstood
     
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Mar 13, 2005 #2

    SpaceTiger

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    Try again without assuming that the block/bullet combination are moving at the same speed as the original bullet. You'll need to consider conservation of linear momentum.
     
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