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Rotational dynamics

  1. Apr 10, 2008 #1
    [SOLVED] rotational dynamics

    1. The problem statement, all variables and given/known data
    A thin uniform rod has length 2.0 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest as angle theta=40 above the horizontal. Use the principle of conservation of energy to determine the angular speed of the rod as it passes through the horizontal position.

    2. Relevant equations

    3. The attempt at a solution

    I used mgh=.5mv^2 and found v=6.26

    Since w=v/r I found the angular speed to be 6.26/2=3.1 rad/sec

    I think this is how the problem is worked out...I'm just not sure. Thanks!
  2. jcsd
  3. Apr 10, 2008 #2

    Doc Al

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    Staff: Mentor

    What about the rotational KE?
  4. Apr 10, 2008 #3
    Would the rotational KE be .5Iwf^2?

    Where I=(1/12)ML^2 and then you just add that on to the final kinetic energy and solve for vf?
  5. Apr 10, 2008 #4

    Doc Al

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    Staff: Mentor

    You could find the total KE by adding the KE of the center of mass to the rotational KE about the center of mass. But much easier to treat the rod as purely rotating about one end.
  6. Apr 10, 2008 #5
    ok thanks I got it!
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