Mass on the inner surface of a cone langrangian

  • #1
Hi everyone

Homework Statement

I have a cone, upside down and a mass m in it, also a homogenous gravitiy field. I found the Euler-Lagrange equation already, which led to 2 equations of motion. Now I want to find a stationary path ( acceleration=0)

Homework Equations


The Attempt at a Solution

The solution for the 2 ELE looks like:

[tex] \ddot r(1+cot^{2}( \alpha))-r \dot \phi^{2}+g*cot( \alpha)=0[/tex]

[tex] r \ddot \phi +2 \dot r \dot \phi =0[/tex]

I shall only find ONE path. My guess would be to take the second equation in order to get a linear differential equation. If the acceleration equals 0, it simplifies to:

[tex] \dot r \dot \phi =0[/tex]

But I'm a bit confused on how to solve this now? Any hints? Or did I do some mistakes?

Thanks for your help

edit: If that's correct thus far, it's just a constant I guess, but do I know sth about it?

2nd edit: if that's right phi is constant so the mass is just falling downwards?!
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  • #2
In your 2nd equation you have dropped an "r".

What you did is that you set the angular acceleration to zero.
This either means that phi is constant and the mass is just falling downwards, or it means that [itex]\dot r[/itex] is constant, meaning the mass is making a horizontal circle.

I suspect that what you wanted is not an acceleration that is zero, but just the stationary path the mass will follow.

As an observation, the generalized momentum for phi (this is basically the angular momentum in this case) is conserved:
[tex]p_\phi = {\partial \mathcal{L} \over \partial \dot\phi} = r^2\dot\phi = constant[/tex]
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  • #3
I suspect that what you wanted is not an acceleration that is zero, but just the stationary path the mass will follow.

Thanks for your reply

That's what I wanted. And I thought , by stationary, they mean :"no force". That's why I thought acceleration = 0.
So that's wrong I guess. But what do I have to do then?
  • #4
You need to solve the set of differential equations in such a way that you get a relation between r and phi (and eliminate t).
The result is probably a conic section (in polar form), but it's a bit of work to proof that.

You can compare it to the orbit of a planet.
That's an ellipse with the center of mass in one of the focus points.

Note that you already have conserved angular momentum, which is exactly what you have for a planetary orbit as well.
  • #5
Huh, ok I guess that's impossible for me, as we haven't learned how to solve non linear differential equations.
  • #6
In that case it will have to suffice that angular momentum is conserved, implying you get an ellipse (or another conic section).
  • #7
I'm trying to figure out why it's an ellipse but I don't get it. Ok you said the angular momentum is constant, so phi is a cyclic coordinate, I understand it so far. So we have no centrifugal(?) force? But why is it an ellipse now? Why not a circle.

Thanks for helping
  • #8
The cone exerts an centripetal force that keeps the mass within the cone.

The path of the mass depends on the initial speed and direction.
It can be a circle, in which case angular velocity is constant and radial velocity is zero.
This means the total radial acceleration is zero and the angular acceleration is zero too.
  • #9
Ok, thanks again. I think I understood it a bit better now.

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