Pendulum Potential energy equation

  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. Doc Al

    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. [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. Doc Al

    Staff: Mentor

    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. so:
    [itex]\Delta y=[/itex](center of mass level-reference level)
    not the other way around , am i correct?
     
  7. Doc Al

    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. Now i understand , thank you very much for you kindly help,I'm very appreciative
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?