Lagrangian Problem

  1. I'm not sure if this is in the right section, if it isn't can someone please move it :)

    Lagrangian mechanics has me completely stumped. Just doesn't seem to make any sense to me. So lets see how this goes.

    A best of mass m is threaded onto a frictionless wire and allowed to move under the pull of a constant gravitational acceleration g. the wire is bent into a curve y=f(x) in the x-y plane, with gravity pointing in the -y direction.

    (a) Let s(t) be the arc length along the bead's trajectory. Show that ds2 = dx2 +dy2

    From calculus i remember this being integral(a->b) of (1 + f'(x))1/2 dx

    a = x, b = x + dx

    how do I solve this :S

    (b) treating s(t) as a generalized coordinate, argue that the Lagrangian is given by

    L = (1/2)ms'2 - mgf[x(s)]

    Well if s(t) is it's position then s' is it's velocity so KE = (1/2)ms'2 and f[x(s)] is just its height so mgf[x(s)] is the PE. L = KE - PE

    (c) Argue that there exists a constant of the motion E such that

    E = (1/2)s'2 + gf[x(s)]

    What is E physically.

    Well E is the energy per unit mass. This exists due to the Lagrangians independence of time?

    (d) With the help of a diagram explain under what conditions the motion is periodic.

    PE > KE?

    (e) Show that the period is given by

    T = 21/2.integral(s1->s2) ds/((E-gf[x(s)])1/2)

    where s1 and s2 satisfy E=gf[x(s1)] and E=gf[x(s2)]

    To do this do I have to find the equation of motion with respect to s?

    (There is more, but i think ill stop here!)

    Sorry for being so vague but this stuff really does my head in. Any help or pointers would be greatly appreciated.

    Thanks
     
  2. jcsd
  3. Cyosis

    Cyosis 1,495
    Homework Helper

    The arc length is given by [itex]s(t)=\int_a^b \sqrt{1+f'(x)^2}dx[/itex], notice the square. Therefore [itex]ds=\sqrt{1+f'(x)^2}dx[/itex]. Secondly y=f(x), dy/dx=f'(x). Can you take it from here?
     
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?