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Angular speed around a bar/clay system's center of mass after impact

  1. Apr 13, 2005 #1
    On a frictionless table, a glob of clay of mass 0.740 kg strikes a bar of mass 1.740 kg perpendicularly at a point 0.140 m from the center of the bar and sticks to it. If the bar is 0.660 m long and the clay is moving at 9.600 m/s before the impact, what is the final speed of the center of mass?

    I have solved this part. The next part is the one I'm stuck on.

    At what angular speed does the bar/clay system rotate about its center of mass after the impact in radians/second?

    I have tried conservation of energy 1/2m1v1 = 1/2m2v2 + 1/2 Iw^2 but my answer is wrong.

    Final equation. w = (m1v1-m2v2)/(1/2ML^2)

    Could someone tell me where I went wrong?
     
  2. jcsd
  3. Apr 14, 2005 #2
    Might just be a typo, for energy, I believe its [tex] \frac{1}{2}mv^2 [/tex] But you might already have that, think you are on the right track with both angular energy and translational though.
     
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