1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Classical mechanics

  1. Feb 14, 2009 #1
    Hello all, I'm having trouble with the following problem:

    Pb: A chain with constant density and infinite length is slipping down from the table without friction. Determine the position of the tip of the chain at time t.

    I know there are a few ways to approaching this problem, namely from newtons equations, or lagranges equations, but I am quite rusty with this, so any suggestions would help a lot, thanks.
  2. jcsd
  3. Feb 14, 2009 #2

    do you know the Euler–Lagrange equations ?
    You have to find an expression for the potential and kinetic energy,
    the difference is the Lagrange-function. Put this function in the Langrange equations
    and you get a second order diff.-equation.

    kind regards
  4. Feb 14, 2009 #3


    User Avatar
    Science Advisor
    Homework Helper

    Hi johnson12! :smile:

    Infinite length? … presumably only in one direction? :wink:

    Use conservation of energy.
  5. Feb 14, 2009 #4
    are you referring to this equation:

    [tex]\frac{\partial L}{\partial x_{i}}[/tex] - [tex]\frac{d}{dt}[/tex][tex]\frac{\partial L}{\partial \dot{x_{i}}}[/tex] = 0, i=1,2,3

    where L = T - U is the lagrange function,

    how can I use this to model my problem?
    (ps. im apologize if my physics is wrong, unfortunately im a math major).
  6. Feb 14, 2009 #5
    Yes, but we need only one variable x_1 = y for example. Now you have to find an expression
    for the kinetic energy T which is simple and for the potential energy U which is simple
    as well. U depends of course at your point of reference. Make a sketch.
  7. Feb 15, 2009 #6
    I get that [tex]T(\dot{x}) = \frac{1}{2}m \dot{x}^{2}[/tex]
    [tex] U(x) = mgx [/tex],

    [tex]\frac{\partial L}{\partial x}= - m g [/tex]
    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{x}}= m\ddot{x}[/tex]
    Lagranges equation implies [tex]m\ddot{x} + mg = 0[/tex]

    but I'm a little confused, if the chain is of infinite length, would it then have infinite mass?

    so how can the chain move?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook