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Find the Lagrangian of a System of Particles

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Ok, so in this system, there are two point particles of mass M connected by massless levers of length L. The pair of masses pivots about the upper point and rotates about the axis at an angular frequency ω. The lower mass is constrained to slide on the vertical axis. The system is illustrated below:
    Uk4c9oI.png
    Find the Lagrangian in terms of Θ.
    2. Relevant equations
    I know that the Lagrangian is represented by the difference in kinetic and potential energies,
    L = T - U
    T = ½Mv^2,
    and U should be derived from the gravitational force only,
    U = Mgh,
    where h is the vertical distance from the origin.

    3. The attempt at a solution
    I have a fundamental lack of understanding regarding finding the kinetic energy of this system. It has been suggested that I view this in cylindrical coordinates, where
    x=LsinΘ
    y=LcosΘ
    z= z.
    I intuit that the origin of my cylindrical coordinate system should reside at the pivot point above the two masses. Could someone please illuminate the subtleties in finding T? I understand conceptually the Lagrangian and the parts of the problem to follow.
     
  2. jcsd
  3. Oct 27, 2014 #2

    BvU

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    Hi CM, and welcome to PF :)

    Does U increase if z (h) increases ?

    For T you have to express v for each of the masses in terms of ##\theta## and ##\omega##. x and y you already have, so ##\dot x## and ##\dot y## aren't too difficult. You will have to do something about z, though. I can see that if ##\theta## is given, z is determined; i.e. there is a way to express z in terms of ##\theta##
     
  4. Oct 27, 2014 #3
    I worked out this problem in the mean time. You are correct!
    I see that the potential increases with z, which can be determined to be 2Lcosθ.
    The gravitational potential energy is thus U = 3Lcosθ*mg when we account for g's effect on each mass.
    The kinetic energy I have determined as well.

    Then, L = T - U, and voila, we have a Lagrangian! I can now find the equation of motion in the generalized coordinate θ.
    Thanks for your help; hope to see you here again :)
     
  5. Oct 28, 2014 #4

    BvU

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    However, your expression for U decreases when ##\theta## increases from zero. I would expect it to increase; wouldn't you ?
     
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