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More Help With Lagrangian Mechanics

  1. Sep 11, 2012 #1
    So in my internet readings on Lagrangian mechanics I started researching applications with non-potential and/or non-conservative forces and came across this page:


    This page is fascinating but I'm having a bit of difficulty understanding a piece of the first example. Can some one explain to me the constraint function they came up with? Its labeled as equation 727. I feel like I'm missing something obvious but I just can't figure it out. Thank you!
  2. jcsd
  3. Sep 11, 2012 #2
    Isn't this the no slip condition?

    That is in words

    The distance (χ) down the plane = the angle turned through by the cylinder times the radius
  4. Sep 11, 2012 #3


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    Precisely: If the cylinder is rolling without slipping the distance, travelled of a fixed point from the initial time, where we set [itex]\phi=0[/itex] is given by the circumference along the cylinder boundary, i.e., it's [itex]a \phi[/itex], where [itex]a[/itex] is the cylinder's radius. The same distance the cylinder's center axis has travelled, i.e., we must have the constraint
    [tex]x=a \phi,[/tex]
    where we have chosen [itex]x=0[/itex] as an initial condition. The constraint function is then given (up to a sign and an overall multiplicative constant, which both are irrelevant for the solution of the problem) thus reads in this case as given by Eq. (727).

    All the scripts of Fitzpatrick's are just mavelous by the way!:smile:
  5. Sep 11, 2012 #4
    Sigh.....I knew it was something obvious, thanks a lot guys!
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