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Lagrangian Mechanics

  1. Mar 25, 2012 #1
    First, to make sure i have this right, lagrangian mechanics, when describing a dynamic system, is the time derivative of the positional partial derivatives (position and velocity) of the lagrangian of the system, which is the difference between the kinetic and potential energy of the system. (set equal to external forces on the system, Q)

    L = T-U

    is this right?

    So my question is:

    is the definition of a lagrangian, T-U, valid only for constant mass systems or can it still be used for variable mass systems? what about the derivatives of the lagrangian, would i need to find the partial derivative of the lagrangian with respect to mass? and finally, im assuming the answers to these questions are very different if we are talking about a time varying mass vs a mass that varies with postion, yes?

    Any help would be much appreciated. As im sure you can tell, i am very lost at the moment

    As background for what im working on, it's a mass on a spring (assuming attached to a rigid/ non-moving body, so for now 1 degree of freedom) with the mass decresing with respect to time. Finally as the mass decreases there is a perodic downward force applied to the mass decreasing in amplitude proportional to the decrease in mass.
     
    Last edited by a moderator: Mar 29, 2012
  2. jcsd
  3. Mar 25, 2012 #2
    Hmmm, interesting question. My initial response was that it should still work as long as all the mass is accounted for, but then I saw your specific scenario. If mass was simply moving from body A to body B, both in the problem, I think it would work. If mass was a function of position, it would be differentiated in the partial derivatives. If it was a function of just time, it would be be differentiated in the total time derivative, so I don't see why it wouldn't work.

    However, if mass was simply disappearing it seems like you would just have energy disappearing, which strikes me as something that would mess things up (since a big part of the lagrangian is the trade in energy between T and U).

    But then your situation is even more interesting, because you actually are putting more T in, in the form of that force. Hmmm, lemme think about this.
     
  4. Mar 29, 2012 #3
    Well i ended up proceeding as i would in a constant mass problem and then just differentiating mass w/respect to time in the total time derivative. Which gave me the following diff equation:

    [itex]\ddot{m}\ddot{x} + \dot{m}\dot{x} + kx + mg = F(t) [/itex]

    substituting in F(t),

    [itex]\ddot{m}\ddot{x} + \dot{m}\dot{x} + kx + mg = g(NM-\frac{tω}{2\pi}M)[\frac{1}{2}cos(ωt)+1][/itex]

    Now to see if i remember how to use maple :eek:
     
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