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Potential energy equations

  1. May 30, 2005 #1
    A small ring of mass M can move freely on a smooth, circular hoop, of radius R. A couple of light inextensible strings that pass over smooth pegs situated below the center of the hoop at the same horinzontal level are attached to the ring. Their other ends are attached to particles of mass m.

    Given that the ring is below the pegs and that M > m sin(x/2), the distance between the pegs is 2 R sinx, and that the hoop is fixed in a vertical plane, prove that the system has three positions of equilibrium.


    I'm having a very hard time setting up the potential energy equations -- I always end up with too many variables and constants. Can anyone help?
     
  2. jcsd
  3. May 30, 2005 #2
    Can you draw a picture in Paint or something? The setup is too complicated to visualize.
     
  4. May 30, 2005 #3
    This is my interpretation...
     

    Attached Files:

  5. May 31, 2005 #4
    You can express the potential energy in terms of one variable, the displacement of the ring along the circular hoop. For instance, the angle that the ring makes with the vertical line through the center of the hoop would make a good variable. Remember that an overall constant added to the potential energy function does not change the mechanical properties of the system--the forces, equilibria, and motion are unchanged. This fact will be helpful when you calculate the contribution to the potential energy from the two masses hanging from strings. The displacement of the ring completely determines the position of both of those masses, since I think we can assume the weight of those masses pulls the strings taut (they form straight lines).
     
  6. Jun 1, 2005 #5
    I realise that. I'm just having a very tough time tranforming those ideas into equations. :/
     
  7. Jun 1, 2005 #6
    Hmmm I didn't know that energy relationships can solve equilibrium problems. Couldn't you use forces?
     
  8. Jun 1, 2005 #7
    Sure, but it's easier to use the potential energy. A minimum of the potential energy function is an equilibrium, and corresponds to zero force.

    Pick a reference point for each mass. They can be different because of the reason mentioned above: a constant added to the potential energy does not change the motion of the system. For the ring on the hoop, I would pick the lowest point on the hoop. Then the ring's contribution to the potential energy is

    [tex]V_{\mbox{ring}} = MgR(1 - \cos \theta)[/tex]

    where theta is the displacement angle measured from the vertical. This is just Mg times its height relative to the reference point at theta = 0. That's the easy one. For each of the other two masses, a good reference point is the position of the mass when the ring is at the bottom of the hoop. Remember, the position of the ring determines the positions of the other two masses. You can then work out the change in height of each mass as the ring moves to a displacement angle theta.

    edit: A maximum or any other place where the derivative of the potential vanishes is also an equilibrium (zero force), but it must be a minimum to be stable. But I just realized the problem does not require that equilibria are stable, so look for any critical points in the appropriate range.
     
    Last edited: Jun 1, 2005
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