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Lagrangian Problemby forty
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#1
Jun209, 09:25 PM

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 xy plane, with gravity pointing in the y direction. (a) Let s(t) be the arc length along the bead's trajectory. Show that ds^{2} = dx^{2} +dy^{2} 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 = 2^{1/2}.integral(s_{1}>s_{2}) ds/((Egf[x(s)])^{1/2}) where s_{1} and s_{2} satisfy E=gf[x(s_{1})] and E=gf[x(s_{2})] 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
Jun309, 05:05 AM

HW Helper
P: 1,495

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?



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