Lagrangian Constraint Problem

1. Feb 22, 2004

Ed Quanta

So a bead slides down a frictionless parabolic wire of shape y=ax^2. I have to express the Lagrangian in terms of x and y. Then I have to use the constraint equation to express this solely in terms of x. Then I have to find the equations of motion, and simplify them for small oscillations.

How do I get from

L=1/2m(x')^2 + 1/2m(y')^2 - mgy

to

L=1/2m(x')^2(1 + 4a^2x^2) -mgax^2 ???

And in this specific example I am having trouble seeing what the constraint force or constraint potential is. It seems to that we know that the force of gravity is present and acting in the negative y direction, and we are told there is no force of friction, so what is the constraint?

2. Feb 22, 2004

Kalimaa23

You have two degrees of freedom, x and y, and one constraint, so you are left with one degree of freedom. Your bead has to roll on a parabola with equation y=ax^2, so (surprise, surprise), the constraint is y=ax^2. Simply substitute y and you'll get the correct lagrangrian.

3. Feb 22, 2004

Ed Quanta

Yes, but since y'=2ax, how do you come up with the term 1/2m(x')^2 *4ax^2? In other words, how does a term of x' end up in there? And I am not talking about the term 1/2m(x')^2 *1 which is also present in the Lagrangian.

4. Feb 22, 2004

Hurkyl

Staff Emeritus
I don't think that's true...

5. Feb 23, 2004

pmb_phy

You've made an error in taking the derivative. Yo didn't take the derivative with respect to time, you took the derivative with respect to x. The correct derivative is obtained using the chaing rule. The result is

y' = dy/dt = 2ax dx/dt = 2ax x'

6. Feb 23, 2004

Ed Quanta

Stupid stupid me