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Potential energy of a spring

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the potential energy of this system without gravity. And after that i have to find the point of equilibrium.
    http://img534.imageshack.us/img534/2667/img0121km.jpg [Broken]

    2. Relevant equations

    [tex] V=\frac{1}{2} K d^2 [/tex]

    3. The attempt at a solution

    [tex] V=\frac{1}{2} K d^2 = \frac{1}{2} K \left( l^2 +4l^2 - 4l^2 \cos{ (\phi - \theta)} +4l^2 \sin^2{\theta} \right) [/tex]

    is it correct ? i think it isn't becouse i find strange result in the next step.
    Thanks
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 24, 2011 #2

    kuruman

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    I do not quite understand what the picture shows. Are the lines labeled l and 2l massless rods pivoted at O? Are the two angles shown in the picture assumed given?

    It might be a good idea to state the problem as it was given to you.
     
  4. Aug 24, 2011 #3
    sorry i left out some details. Firts of all the subject is Lagrangian Mechanics, after that, the lines l and 2l is massless rods pivoted ad O and the 2 angles are the Lagrangian coordinate.
     
  5. Aug 24, 2011 #4

    kuruman

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    When you write the potential as V = (1/2)kd2, the "d" in the expression is the additional stretching (or compression) of the spring from the equilibrium position. Your potential is written as if "d" stands for the length of each spring.

    It seems to me that first you need to consider what the assembly looks like at equilibrium, then find by how much each spring stretches (or is compressed) when the assembly looks like the figure.
     
  6. Aug 25, 2011 #5
    ok, so you are tell me that, for each spring
    [tex]V=\frac{1}{2}k (d-d_0)^2[/tex]
    is it true ?
    i had considered this solution but the expression of potential became so complicated that i tought it was wrong.
    now i retry to do it, thanks.
     
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