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Relate rotational kinetic energy to potential energy

  1. Apr 15, 2017 #1
    1. The problem statement, all variables and given/known data
    This problem is from the 2015 AP Physics C Mechanics free response, question 3 part b.
    https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap15_frq_physics_c-m.pdf
    upload_2017-4-15_2-49-11.png

    2. Relevant equations
    K = 1/2Iω2
    U = mgh

    3. The attempt at a solution
    The potential energy of the bar when it's horizontal gets transferred to kinetic energy when vertical, so I originally had the equation mgL = 1/2(1/3ML22
    However, the scoring guidelines say that the change in height should be L/2, not L, resulting in a potential energy of mgL/2. Could someone explain why the change in height should be L/2 and not L?
     
    Last edited by a moderator: Apr 15, 2017
  2. jcsd
  3. Apr 15, 2017 #2

    vela

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    The change in vertical position of the end of the bar is L. Is that the point that's important in the problem?
     
  4. Apr 15, 2017 #3
    It asks for the velocity of the "free end of the rod", so I think that it means the end of the bar. But why does the answer sheet say L/2 then?
     
    Last edited: Apr 15, 2017
  5. Apr 15, 2017 #4

    vela

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    Is that relevant to the change in potential energy of the rod? Why not use the position of the other end or a point 1/3 the way in from one end? I'm trying to get you to think about what point is important when you talk about the potential energy of the rod.
     
  6. Apr 15, 2017 #5
    So I have to take the entire bar into account and not just the end...kind of like taking the "average" change in height for all pieces of the rod?
     
  7. Apr 15, 2017 #6

    vela

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    Yes. More precisely, the average position of the mass of the bar, i.e., the center of mass.
     
  8. Apr 15, 2017 #7
    Okay, that makes sense. Thanks for the help!
     
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