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Finding angular speed of a thin rod.

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A thin 1.0-meter-long rod pivoted at one end falls (rotates) from a horizantal position, starting from rest and with no friction. What is the angular speed of the rod when it is vertical?



    2. Relevant equations

    I (of the thin rod) = (1/3)ML^2



    3. The attempt at a solution


    I tried solving this problem using conservation of energy, but I'm not getting the right answer (5.4 rad/s).

    For kinetic energy, I was using (1/2)I[tex]\omega[/tex]^2; and for potential energy, I was using mgL. Then I solved for [tex]\omega[/tex]. What's wrong with that?
     
  2. jcsd
  3. Mar 9, 2008 #2

    Doc Al

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    Staff: Mentor

    In figuring out the change in PE of the rod, consider what happens to its center of mass.
     
  4. Mar 9, 2008 #3
    It accelerates?
     
  5. Mar 9, 2008 #4

    Doc Al

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    Staff: Mentor

    How does its position change? That determines the change in PE of the rod.
     
  6. Mar 9, 2008 #5
    The CoM moves from a height of L to a height of L/2 ?
     
  7. Mar 9, 2008 #6

    Doc Al

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    Staff: Mentor

    Good. So what's the change in PE of the rod?
     
  8. Mar 9, 2008 #7
    MgL - Mg(L/2)

    I got the answer now, thanks!
     
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