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Lagrangeian problem

  1. Feb 10, 2010 #1
    lagrangeian problem....

    1. The problem statement, all variables and given/known data
    in a spring pendulum system the spring is initially in equilibrium at upright position with theta = 0 and elongation (mg)/k, (k is spring constant). total length of spring in equilibrium condition is a with respect to which we define additional deviations in terms of a variable r. write dwn the lagrangian for this system

    2. Relevant equations

    L = kinetic energy - potential energy
    kinetic energy = 0.5 mv^2
    potential = mgh

    3. The attempt at a solution
    i know that the l = kinetic energy - the potential energy...but not exactly sure how to form my equations..please help...

    i think potential energy would be 0.5(k*x^2) ?
    Last edited: Feb 10, 2010
  2. jcsd
  3. Feb 10, 2010 #2
    Re: lagrangeian problem....

    The first thing here is picturing the whole problem schematically! So graph the pendulum with given data and show it to us!

  4. Feb 10, 2010 #3
  5. Feb 10, 2010 #4
    Re: lagrangeian problem....

    ok from what i can guess

    potential energy = mgrcos(theta) - 0.5k*(r^2)
    kinetic energy = 0.5(m(dr/dt)^2) + 0.5(mr^2)(dtheta/dt)^2
  6. Feb 10, 2010 #5
    Re: lagrangeian problem....

    You got it right!

  7. Feb 10, 2010 #6
    Re: lagrangeian problem....

    ok thnx.. :)
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