Pendulum Potential energy equation

  1. Feb 5, 2013 #1
    hello every one
    i have this pendulum:
    [​IMG]
    i need to stabilize the pendulum in the inverted position , i need to know the potential energy for the pendulum , i read several articles in each one i have a different equation :
    [itex]V=mgl_{p}cos\alpha[/itex]
    [itex]V=-mgl_{p}cos\alpha[/itex]
    [itex]V=mgl_{p}(1-cos\alpha)[/itex]
    [itex]V=mgl_{p}(cos\alpha-1)[/itex]
    Now i'am really confused which equation is the correct one?
    Please help me
     
  2. jcsd
  3. Feb 5, 2013 #2

    Doc Al

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    Staff: Mentor

    What are you taking as your reference level? (Where PE = 0.) The pivot? The gravitational PE is given by mgΔy, where Δy is measured from your chosen reference level. Using that you should be able to pick the correct formula.
     
  4. Feb 5, 2013 #3
    [itex]P.E=0[/itex] at the vertical position ([itex]\alpha=0[/itex] from the vertical line in the picture above), now which formula should i use :
    [itex]P.E=mgl_{p}(1-cos\alpha)[/itex]
    or
    [itex]P.E=mgl_{p}(cos\alpha-1)[/itex]
     
  5. Feb 5, 2013 #4

    Doc Al

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    Using the vertical position as your reference, Δy will be negative for any nonzero angle. The second of those is the one you want.
     
  6. Feb 5, 2013 #5
    so:
    [itex]\Delta y=[/itex](center of mass level-reference level)
    not the other way around , am i correct?
     
  7. Feb 5, 2013 #6

    Doc Al

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    Staff: Mentor

    Δy is measured from the PE = 0 reference point. Another way to write the PE is mgy, where y is the vertical position of the center of mass and y = 0 is the PE = 0 point.

    Since you are choosing the reference level to be where the pendulum is vertical (the angle is zero) and thus the mass is at its highest point, all values for y and thus PE for non-zero angles will be negative since they are below that point.
     
  8. Feb 5, 2013 #7
    Now i understand , thank you very much for you kindly help,I'm very appreciative
     
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