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Homework Help: Spring Pendulum - Lagrangian Mechanics

  1. May 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Please see attached image :)

    2. Relevant equations
    Euler-Lagrange Equation
    [itex] \frac{\partial{L}}{\partial{q}} - \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}}} = 0[/itex]
    [itex] L = T - V[/itex]

    3. The attempt at a solution
    a. The potential energy V is the potential energy from the spring and the gravitational potential energy. The kinetic energy is the energy in the radial direction and in the theta direction.
    [tex] L = \frac{1}{2}m(\dot{r}^2+\dot{\theta}^2(r_0+r)^2) + mg(r_0+r)cos\theta - \frac{1}{2}kr^2[/tex]

    b. Use one Euler-Lagrange equation for r and one for theta and I get: [tex] m\dot{\theta}^2 + mgcos\theta - kr = \frac{d}{dt}(m\dot{r}) [/tex] [tex] -mg(r_0+r)sin\theta = \frac{d}{dt}(m\dot{\theta}(r_0+r)^2)[/tex]
    Which are the F=ma equation for the radial direction and the Torque = (d/dt) Angular momentum equation.

    c. Confused here, if angular position and velocity are to be fixed, I assume they mean [itex] \theta [/itex] and [itex] \dot{\theta}[/itex], do I consider them both to be zero, or just a constant number? And if the position is fixed doesn't that mean that the velocity should be zero? Is this a differential equation I have to solve?

    d. and e. Need c to continue.

    Any help is appreciated, thanks !

    Attached Files:

  2. jcsd
  3. May 23, 2016 #2


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    "Fixed" means "constant." The position and velocity aren't actually fixed (so ##\dot{\theta} \ne 0##), but because the timescale over which the radial motion occurs is so short, you can analyze the motion as if ##\theta## and ##\dot{\theta}## were constant. It's an approximation.

    By the way, the first term in the radial differential equation is incorrect.
  4. May 23, 2016 #3
    Thanks I must have missed a (r_0 + r) factor in the first term when I was putting it into tex, I'll give it a shot with theta and theta dot as constants.
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