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Is this conservation of momentum?

  1. Nov 14, 2004 #1
    The problem is:
    A uniform thin rod of length = .5m and mass M = 4.0kg can rotate in a horizontal plane about a vertical axis through its center (I = ML^2/12). The rod is at rest when a bullet of mass m = 3.0g traveling in the horizontal plance of the rod is fired into one end of the rod. As viewed from above, the direction fo the bullets velocity makes an angle of theta=60 with the rod. If the bullet lodges into the rod and the angular velocity of the rod is 10rad/s immediately after the collision, what is the bullet's speed just before impact?
    This is a review question for a test I have monday, the answer is1290m/s, but I cant get that.

    Here's what I did (which is obviously incorrect):
    I = (1/12)ML^2 + m(L/2)^2

    conservation of momentum????
    mv = Iw
    v = Iw/m
    = (1/12)ML^2 + m(L/2)^2 / m
    = wrong.

    Is conservation of momentum the right tool to be using to solve this? ...how does theta play into it? Any tips will be *greatly* appreciated.
     
  2. jcsd
  3. Nov 14, 2004 #2

    Doc Al

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    Staff: Mentor

    This is correct for the rotational inertia of the system (rod + bullet) after the collision.

    Two problems here:
    (1) You are mixing up linear and angular momentum; they aren't the same thing---the units don't even match.
    (2) In this problem, angular momentum is conserved. What's the angular momentum of the bullet just before it hits the rod?

    Conservation of angular momentum is.
    It will allow you to determine the angular momentum of the bullet.
     
  4. Nov 14, 2004 #3

    ehild

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    Homework Helper
    Gold Member

    No. Use conservation of angular momentum instead. The initial angular momentum is that of the bullet and it is mv(L/2)sin(theta) with respect to the rotation axis.

    ehild
     
  5. Nov 14, 2004 #4
    Great! Thanks guys. You helped a bunch...now I know what I need to go back and study :-) Have a great sunday!
     
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