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Lagrangian Mechanics Problem

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a vertical plane in a constant gravitational feild. Let the origin of a coordinate system be located at some point in this plane. A particle of mass m moves in the vertical plane under the influence of gravity and under the influence of an aditional force f = -Ar^(a-1) directed toward the origin (r is the distance from the origin; A and a [does not = 0 or 1] are constants). Choose appropriate generalized coordinates, and find the Langrangian equations of motion. Is the angular momentum about thet origin conserved?


    2. Relevant equations
    [tex]L = T - U[/tex]
    [tex]\frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial \dot q_i} = 0[/tex]


    3. The attempt at a solution
    Choose the coordinate system (r', y') centered at the origin, so that the r' axis is in the direction of the origin to the particle of mass m and the y' axis is perpendicular to r'.

    First, [tex]\arrow F = (-mg\cos{\theta}, -mg\sin{\theta}) + (-Ar^{a-1}, 0) = (-mg\cos{\theta} - Ar^{a-1}, -mg\sin{\theta})[/tex].

    From here we can guess the potential:
    [tex]U = mgr\cos{\theta} + \frac{A}{a}r^a + mgy\sin{\theta}[/tex].

    (The preceding steps may be wrong, but the following is what I'm not sure about):
    Notice that in our coordinate system y is always 0. So our potential is [tex]U = mgr\cos{\theta} + \frac{A}{a}r^a[/tex]. For the kinetic energy, we get [tex]T = 1/2m(\dot r^2 + \dot \theta^2r^2) <ACCIDENT = 1/2m\dot \theta^2r^2>[/tex]. So our Langrangian is [tex]L = 1/2m(\dot r^2 + \dot \theta^2r^2) - mgr\cos{\theta} - \frac{A}{a}r^a[/tex].

    The rest is easy if I did the preceding correctly. You get that angular momentum is not conserved when you find [tex]\frac{\partial L}{\partial \theta} - \frac{d}{dt}\frac{\partial L}{\partial \dot \theta} = 0[/tex].

    Thanks in advanced!
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    kuruman

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    Gold Member

    Looks OK. You should get two equations of motion. What does this mean? Are you saying r-dot is zero?

    [tex]
    T = 1/2m(\dot r^2 + \dot \theta^2r^2) = 1/2m\dot \theta^2r^2
    [/tex]
     
    Last edited: Oct 11, 2009
  4. Oct 11, 2009 #3
    That was an accident. In the Lagrangian, you can see I did not carry that part through. I will put the incorrect part in brackets. Thank you! My confidence in this material has been elevated.
     
    Last edited: Oct 11, 2009
  5. Oct 11, 2009 #4

    kuruman

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    Just checking. Lagrangians are fun once you get used to them.
     
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