Lagrangian Problem


by forty
Tags: lagrangian
forty
forty is offline
#1
Jun2-09, 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 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
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Cyosis
Cyosis is offline
#2
Jun3-09, 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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