# Lagrangian Problem

by forty
Tags: lagrangian
 P: 136 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
 HW Helper P: 1,495 The arc length is given by $s(t)=\int_a^b \sqrt{1+f'(x)^2}dx$, notice the square. Therefore $ds=\sqrt{1+f'(x)^2}dx$. Secondly y=f(x), dy/dx=f'(x). Can you take it from here?