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Angular momentum

  1. Mar 13, 2005 #1
    The problem is here.

    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:
    since the rod has negligible mass, it will not have a moment of inertia.

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


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    Staff Emeritus
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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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