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Lagrangian mechanics: Bar connected to a spring

  1. May 28, 2016 #1

    Mass 1 can slide on a vertical rod under the influence of a constant gravitational force and and is connected to the rod via a spring with the spring konstant k and rest length 0. A mass 2 is connected to mass 1 via a rod of length L (forms a 90 degree angel with the first rod, and is otherwise rotatable).

    1. Find the lagrangian
    2. Find the equations of motion with the euler-lagrange equations
    3. Find the solutions to the equations of motions
    2. Relevant equations
    L = T-V
    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot z} - \frac{\partial L}{\partial z}[/tex]
    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot \varphi} - \frac{\partial L}{\partial \varphi}[/tex]

    3. The attempt at a solution
    This is the first time I have to deal with lagrangian mechanics.
    I thought we have 2 free coordinates. We have z for the up and down motion and [tex]\varphi[/tex] for the rotation on the x-y Plane of mass 2.
    The potential energy is: [tex]V = V_{Spring} + V_{grav1} + V_{grav2} = \frac{1}{2} k z^2 + m_1 g z + m_2 g z[/tex]
    The kinetic energy: [tex]T = T_1 + T_2 = \frac{1}{2} m_1 \dot z^2 + \frac{1}{2} m_2 (\dot L^2 + L^2 \dot \varphi ^2 + \dot z^2)[/tex]
    [tex]\dot L^2[/tex] must be 0 cause the length of the 2nd rod doesn't change over time
    Also I thought, since mass 2 is moving on the edge of a cylinder, I would need to use the kinetic energy in cylinder coordinates for mass 2

    I plugged my lagrangian into my 2 euler-lagrange equations and got this:
    for z : [tex](m_1 + m_2) \ddot z = -kz + (m_1+m_2) g[/tex]
    for [tex]\varphi: \frac{d}{dt}(m_2 l^2 \dot \varphi) = 0 [/tex]

    I am not sure about r. This looks like the equation of motion of a damped harmonic oscillator. It is not clear from the exercise if the spring is damped or not, so I am not sure if that is correct.
    The 2nd equation just means, that angular momentum doesn't change with respect to time? Is there anything I could use from the conversation of angular momentum to solve the first equation or am I done here?

    Last edited: May 28, 2016
  2. jcsd
  3. May 28, 2016 #2


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    No damping (usually that's a term ##\beta \dot z##). Jus a mass on a spring.
    Correct !
    There is nothing in common, so: no. The two degrees of freedom are totally independent.
  4. May 28, 2016 #3
    Thank you very much!! I'll try to solve it now

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