Take a spherical pendulum with bob mass $m$, rod length $\ell$ and physical coordinates $\theta$, $\phi$ (spherical angles) and $h$ (the hinge height with respect to the coordinate origin). The rod is massless and infinitely stiff. The derivation of the equations of motion of such a system using Lagrangian dynamics is outlined here. Note that the hinge height $h$ is an ignorable coordinate and thus plays no role in the equations of motion.

However, I'd like to extend this system by placing the spherical pendulum in a non-inertial reference frame that is both rotating and accelerating (linearly and angularly), and change the pendulum such that it has damping at the hinge with ratio $\gamma$, and known functions of time for $\dot{m}$, $\dot{\ell}$ and $\dot{h}$, as well as $\ddot{m}$, $\ddot{\ell}$ and $\ddot{h}$ (these are all $\neq 0$).

My Lagrangian dynamics is rather rusty (to put it mildly), so when I started writing out the equations for the kinetic energy $T$ and potential $V$, I immediately got stuck on the following questions:

  • Can the potential $V$ still be written as $-mg\ell\cos\theta$, or should the potential take into account the non-inertialness of the frame?
  • In fact, is this pendulum system actually still conservative (there is dissipation due to the damping after all)? How should I go about it if it is not?
  • Can I somehow do the required vector products & additions afterwards? I mean: can I first derive the equations of motion as if the pendulum's reference frame is inertial, and then get to the final form by adding the linear, centripetal, Coriolis and Euler terms? Or do these need to be included right from the start somehow?

Most examples I find or have available seem too trivial for this sort of problem...And this problem is simply beyond anything I've ever done in the past. Some assistance and/or guidance and/or links to similar problems would be greatly appreciated.

Note that I've asked a similar question before, but the solution I have there is rather unsatisfactory; it considers the constraint force (tension in the string), which it turns out, is not really necessary. It is also quite evil in numerical terms, because of the time-dependent discontinuities in the forces applied to the bob. So I'm going with a rod now, which has negligible length changes (save for the $\dot{\ell}$ and $\ddot{\ell}$ terms).

Although I could do it in the "Newtonian" way by carefully considering all the forces, etc., I'd really like to learn (again) how to do this cleanly by using the Lagrangian formulation.


1 Answer 1


Not sure if you want us to explicitly help you solve this new potential energy and kinetic energy, or if you just want us to address the bullet points. I'll start out by just addressing them and you can comment if it's not enough =).

  • The non-inertialness will need to be taken into account. This is an extremely difficult topic though, unfortunately. If you don't have much experience dealing with non-inertial reference frames, this is going to be a bear to learn. Whenever I do a non-inertial reference frame problem, I feel like I need to take into account a half dozen extra things! Here is a great link to get started. http://www.dartmouth.edu/~phys44/lectures/Chap_6.pdf. This pdf takes you through everything. Get a cup of coffee and sit down for a couple hours and pore over the details if you really want to understand all this. The Lagrangian formulation in a non-inertial reference frame starts on page 4.
  • Lagrangians are rather annoying to use for non-conservative systems (your system is non-conservative since the hinge is dissapating energy into heat, therefore leaving your system since you're not accounting for heat, nor would you want to). You have to literally "inject in" a potential energy to account for it by knowing how it the disappative force will act. See http://en.wikiversity.org/wiki/Advanced_Classical_Mechanics/Dissipative_Forces for some really good examples, one includes a pendulum even!
  • To your final bullet point, I believe the answer is...don't think about it this way. Solving in a non-inertial reference frame kind of encompasses a different approach entirely. Many of your old terms change, and the formalism for solving it takes into account everything at the same time...I think it's bad practice to solve a Lagrangian in an inertial reference frame and then try adding terms in later...possible, but unadvised.
  • $\begingroup$ Isn't the Langrangian a scalar? So shouldn't it be constant in all reference frames? $\endgroup$
    – resgh
    Aug 24, 2013 at 13:18
  • $\begingroup$ EDIT: I shouldn't have said constant, I mean same. $\endgroup$
    – resgh
    Aug 25, 2013 at 3:08
  • 1
    $\begingroup$ The Lagrangian will look superficially different in different coordinates systems, despite being a 'scalar' with respect to some transformations (symmetries). $\endgroup$ Aug 25, 2013 at 10:13
  • $\begingroup$ @EdwardHughes Is the difference only in the form of the expression when the Langrangian is expressed in different coordinate systems, and is the value of the Langrangian the same? $\endgroup$
    – resgh
    Aug 25, 2013 at 11:43
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    $\begingroup$ @namehere: indeed the value of the Lagrangian doesn't change (for any given point in phase space). But if you rewrite your coordinates for phase space then the Lagrangian will look different in form. $\endgroup$ Aug 25, 2013 at 11:46

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