Re: [Phys-l] solving an energy equation

[Carl Mungan <mungan@usna.edu>]

> Got it.  But if it is a one DOF problem, is there any possibility of 
> doing a "Lagrangian-in-spirit" analysis where you simply numerically 
> the component of Newton's second law along the surface?

Yes, another list member pointed that out offline and you're quite 
right: it does in fact reproduce Lagrange's equation, as one would 
expect.

But I'm still bothered by the question of whether something is lost 
in the energy approach. Let's go back to the example of a ball thrown 
vertically upward with initial speed v0.

conservation of energy => (dy/dt)^2 = v0^2 - 2gy

This has a sign ambiguity in that dy/dt is a double-valued function of y.

It would appear something is missing. However, try differentiating 
the energy equation:

2(dy/dt)(a) = -2g(dy/dt)

If you cancel 2(dy/dt) from both signs, you recover N2L, which has NO 
sign ambiguity in it.

To put it another way, Mathematica integrates (d^2y/dt^2) = -g with 
no problem, but gets stumped trying to integrate the energy equation. 
If anyone would like to try it, here's the code, where I chose g = v0 
= 1:

sol1 = NDSolve[{y''[t] == -1, y[0] == 0, y'[0] == 1}, y, {t, 0, 2}]
Plot[y[t] /. sol1, {t, 0, 2}]

sol2 = NDSolve[{(x'[t])^2 == 1 - 2x[t], x[0] == 0}, x, {t, 0, 2}]
Plot[x[t] /. sol2, {t, 0, 2}]

The first pair of lines produces the expected parabola, the second 
pair produces a bunch of error messages. If you replace the second 
pair instead with the following, you do get a correct half parabola:

sol2 = NDSolve[{x'[t] == Sqrt[1 - 2x[t]], x[0] == 0}, x, {t, 0, 1}]
Plot[x[t] /. sol2, {t, 0, 1}]

--
Carl E. Mungan, Asst Prof of Physics  410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-5002
mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/