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Lagrangian dynamics problem.

  1. Oct 28, 2012 #1
    A point of mass m, affected by gravity, is obliged to be in a vertical plan on a parabola with equation z = a.r^2

    a is a constant and r is the distance between the point of mass m and the OZ vertical axis. Write the Lagrange equations in the cases that the plan of the parabola is :

    a) is fixed

    b) it rotates with angular speed ω about the OZ axis.



    http://www.google.pt/imgres?q=parabola+lagrangian+mechanics&um=1&hl=pt-PT&sa=N&biw=1097&bih=521&tbm=isch&tbnid=AbK_S7_Po3-jSM:&imgrefurl=http://stochastix.wordpress.com/2007/12/11/a-bead-sliding-on-a-rotating-parabola/&docid=vdbKHvoDOKa0bM&imgurl=http://stochastix.files.wordpress.com/2007/12/parabola.jpg%253Fw%253D450&w=300&h=300&ei=LjCNUJuvPMHDhAfL34CoCw&zoom=1&iact=hc&vpx=200&vpy=117&dur=1276&hovh=225&hovw=225&tx=132&ty=131&sig=102710367222874968480&page=1&tbnh=129&tbnw=129&start=0&ndsp=16&ved=1t:429,r:12,s:0,i:104


    I just need to know the equations of the position and speed.

    in a) i considered:

    x = r cosθ ||||| x' = r'cosθ - rθ'sinθ
    y = r sinθ ||||| y' = r'sinθ + rθ'cosθ
    z = a.r^2 ||||| z' = a.r^2'


    in b)

    x = r cos(wt) ||||| x' = r'cos(wt) - rwsin(wt)
    y = r sin(wt) ||||| y' = r'sin(wt)+ rwcos(wt)
    z = a.r^2 |||||| z' = a.r^2'


    Is that right?
     
  2. jcsd
  3. Oct 28, 2012 #2

    TSny

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    For part (a) θ is fixed, hence θ' = ? I would think that without loss of generality you could let θ be zero.

    Part (b) looks good to me.
     
  4. Oct 28, 2012 #3

    considering that θ = 0, then part a) is not going to have the y component :)


    And other question:

    z' = a.r^2' or z' = 2.a.r'.r ? is that the same?
     
  5. Oct 28, 2012 #4

    TSny

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    Correct!
     
  6. Oct 28, 2012 #5
    Thanks for all your the help!
     
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