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Lagrangian of a double pendulum system (with a spring)

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the Lagrangian for the double pendulum system given below, where the length of the massless, frictionless and non-extendable wire attaching [itex]m_1[/itex] is [itex]l[/itex]. [itex]m_2[/itex] is attached to [itex]m_1[/itex] through a massless spring of constant [itex]k[/itex] and length [itex]r[/itex]. The spring may only stretch in the [itex]m_1[/itex]-[itex]m_2[/itex] direction and its unstretched length is [itex]l_0[/itex]. [itex]\theta[/itex] and [itex]\phi[/itex] are the angles of [itex]m_1[/itex] and [itex]m_2[/itex] with respect to the red [itex]y[/itex]-axis.

    double_pendulum_with_spring.png



    2. Relevant equations

    Lagrangian:
    [tex]L = T - U[/tex]

    Kinetic Energy of i-th mass:

    [tex]T_i = \frac{1}{2}m_i \left(\dot{x}_i^2 + \dot{y}_i^2 \right) [/tex]

    Potential Energy of i-th mass:
    [tex]U_i = m_i g y_i [/tex]

    Potential Energy of the spring:
    [tex]U_{spr} = \frac{1}{2}kr^2 [/tex]

    3. The attempt at a solution

    Since I'm still learning this topic, I need someone to help me by verifying if my work is correct, as there's no solutions manual.

    [tex]L = T - U = \left \{ \frac{1}{2}m_1 \left(\dot{x}_1^2 + \dot{y}_1^2 \right) - m_1gy_1 \right \} + \left \{ \frac{1}{2}m_2 \left(\dot{x}_2^2 + \dot{y}_2^2 \right) - m_2gy_2 \right \} - \frac{1}{2}k r^2 [/tex]

    I'm considering [itex]\left(r, \theta, \phi \right)[/itex] as my generalized coordinates

    [tex]\left\{\begin{matrix}
    x_1& =& l \sin \theta &\\
    x_2& =& l \sin \theta + r \sin \phi &\\
    y_1 &=& l \cos \theta &\\
    y_2 &=& l \cos \theta + r \cos \phi&
    \end{matrix}\right.[/tex]

    Plugging those in the original lagrangian yields

    [tex]\begin{align*} L = \left \{ \frac{1}{2}m_1 \left(l^2\dot{\theta}^2 \cos^2 \theta + l^2\dot{\theta}^2 \sin^2 \theta \right) - m_1gl \cos \theta \right \}\\
    + \left \{ \frac{1}{2}m_2 \left(l^2\dot{\theta}^2 \cos^2 \theta + r^2\dot{\phi}^2 \cos^2 \phi + 2lr \dot{\theta} \dot{\phi} \cos \theta \cos \phi + l^2 \dot{\theta}^2 \sin^2 \theta + r^2 \dot{\phi}^2 \sin^2 \phi + 2lr \dot{\theta} \dot{\phi} \sin \theta \sin \phi \right)\\
    - m_2g \left( l \cos \theta + r \cos \theta \right) \right \} - \frac{1}{2}k r^2 \end{align*}[/tex]

    which summarizes to

    [tex]L = \left \{ \frac{1}{2}m_1 \left(l^2\dot{\theta}^2 \right) - m_1gl \cos \theta \right \} + \left \{ \frac{1}{2}m_2 \left(l^2\dot{\theta}^2 + r^2\dot{\phi}^2 + 2lr \dot{\theta} \dot{\phi} \cos \left(\theta-\phi \right)\right) - m_2g \left( l \cos \theta + r \cos \phi\right) \right \} - \frac{1}{2}k r^2 [/tex]

    Is this correct? Something tells me it isn't, is there another way of representing [itex]r[/itex]? Also, I haven't used [itex]l_0[/itex] which was given in the problem statement, and I feel like it should be there in the lagrangian somewhere.
     

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    Last edited: Oct 15, 2016
  2. jcsd
  3. Oct 16, 2016 #2

    kuruman

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    Science Advisor
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    Gold Member

    What is the difference between ##l_0## and ##r##? I would assume that ##r## is the stretched or compressed length of the spring in which case the elastic potential energy should be ... ?
     
  4. Oct 17, 2016 #3
    Sorry, the comment above answers your question.
     
    Last edited: Oct 17, 2016
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