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Angular Momentum

  1. Oct 20, 2007 #1
    1. A mass m is moving with velocity v. it then collids with the end of a rigid rod perpendicular to its intial path. They stick together. The other end of the rod is attached to a frictionless hinge which allows it to rotate in any direction. rod mass = M length = l
    What is the angular velocity of the particle rod system after impact?

    3. Initally the only angular momentum would be L = mvl
    angular momentum is conserved so the intial L = final L
    I am a bit confused about how to find the final L?
    L=rmv
    L = Iw(for the rod) +mwl for the particle?
    I for the rod = MR^(2) / 3
     
  2. jcsd
  3. Oct 20, 2007 #2

    Doc Al

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    Staff: Mentor

    Once they collide and stick together they become one body. What's the rotational inertia of "rod + mass"?
     
  4. Oct 20, 2007 #3
    Is it just I = (m+M)R^(2)
     
  5. Oct 20, 2007 #4

    Doc Al

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    No. You already know the rotational inertia of the stick. Just add the rotational inertia of a point mass at its end.
     
  6. Oct 20, 2007 #5
    So the final angular momentum would equal
    L = [Ml^(2) / 3]w + [Ml^2]w?
    where w = vl?
     
  7. Oct 20, 2007 #6

    Doc Al

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    Staff: Mentor

    Almost. That second mass should be m, not M.
    No. (If that were true, you could just write down the answer!) Set initial angular momentum equal to final angular momentum and solve for w.
     
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