1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrangian of a pendulum system

  1. Feb 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A long light inflexible rod is free to rotate in a vertical plane about a fixed point O. A particle of mass m
    is fixed to the rod at a point P a distance ℓ from O. A second particle of mass m is free to move along the rod, and is attracted to the point O by an elastic force of strength k.

    Find any equilibrium points. (You may assume the rod is sufficiently long that the moving particle does not reach either end of the rod).

    2. Relevant equations
    The potential due to the elastic force is ##V = kx^2##

    3. The attempt at a solution
    Let ##X## be the distance of the second particle from the origin O. Then the Lagrangian I got is
    ##L = \frac{1}{2}m[\dot{X}^2 + (X^2 +l^2)\dot{\theta}^2] - mg[X(1-cos\theta) +l(1-cos\theta)] - kX^2##
    This gives the equations of motion for ##X(t)## and ##\theta (t)## as
    ##\ddot{X} = X(\dot{\theta}^2 - 2k) -g(1-cos\theta)##
    ##\ddot{\theta}(X^2+l^2) + 2\dot{X}\dot{\theta} = -gsin\theta(X + l)##.

    Equilibrium points occur when the second derivatives are both zero, so I have to solve
    ##0 = X(\dot{\theta^2} - 2k) -g(1-cos\theta)##
    ##2\dot{X}\dot{\theta} = -gsin\theta(X + l)##
    And now i'm lost. If I try to solve these I will end up with the fixed points ##X## and ##\theta## as functions of ##t##, which doesn't seem right to me.

    Any help would be appreciated, maybe the Lagrangian I got is incorrect.
     
  2. jcsd
  3. Feb 27, 2015 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    m has disappeared from the equations of motion

    I would expect an equilibrium at ##\dot X =0,\ \dot\theta = 0,\ \theta = 0, X = {mg\over 2k}##, so there must be something with ##L##. And sure enough: for ##\theta=0## V doesn't depend on X !
     
  4. Mar 11, 2015 #3
    Hi there, I am also struggling on this question? I get the same Lagrangian except the potential of the spring is 1/2*k*x^2. Can anyone explain why this L is wrong?
     
  5. Mar 11, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    As BvU hinted: the gravitational potential energy of the second mass is not correctly set up in the Lagrangian. For the special case where θ = 0, the gravitational PE of the second mass should certainly depend on the variable X. But in the Lagrangian in post #1, this is not so.

    You need to be careful about where you are defining zero gravitational PE for the two masses.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Lagrangian of a pendulum system
  1. Lagrangian of a system (Replies: 9)

  2. Lagrangian of pendulum (Replies: 2)

Loading...