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Rotational Kinetic Energy of a rod

  1. Mar 23, 2008 #1
    1.A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.


    What fraction of the original kinetic energy is converted into internal energy in the collision?

    Here is my attempt:

    Ki = (1/2)mv^2

    kf = (m+M) L^2

    Now I'm a bit confused what they mean by fraction, so I initial thought I had to divide Kf/Ki , which is:

    2(m + M)L^2 / (mv^2)

    but WebAssign marked it wrong. I have one more try and I want to try the other way, Ki/Kf but I don't want to get it wrong.

    Can anyone help me and tell me if I'm doing this right or not. Thanks

  2. jcsd
  3. Mar 23, 2008 #2
    Your final kinetic energy equation is wrong. First off, it doesn't have the right units. It looks like you just used I. The final energy should be 1/2 I w^2, where the equation v = Lw holds.
  4. Mar 23, 2008 #3
    Oh okay I see how I got the final kinetic energy wrong. I did only use I. I fixed it, and now its:

    Kf = (1/2)(m + M)v^2

    Can anyone confirm this?

    And if thats correct, how do I find the fraction?
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