# Constrained mass point subject to elastic force and weight

1. Feb 11, 2014

### Ocirne94

Hi all, is my solution correct? I was rejected because of this...

1. The problem statement, all variables and given/known data
Consider a mass point (mass = m) constrained to move on the surface of a sphere (radius = r). The point is subject to its own weight's force and to the elastic force of a spring (elastic constant = k, rest length = 0) which at the other end is fixed to the sphere's north pole.

Write the Lagrangian and the Hamiltonian of the system.
Write the Lagrange's and Hamilton's equations of motion.
Find the constants of motion.
Give a qualitative description of the point's movement

2. Relevant equations
None given.

3. The attempt at a solution
There are 2 degrees of freedom. I choose spherical coordinates theta and phi [but is this correct? The point can reach the poles, where those coordinates aren't defined anymore].
The kinetic energy is
$T = \frac{m}{2}\cdot (r^2 \dot\theta^{2} + r^{2}sin(\theta)^{2}\dot\phi^{2})$
The potential energy is
$V = \kappa\cdot r^{2} (1-cos\theta) + mgr(1+cos\theta)$
The Lagrangian is simply
$L = T-V$
and, since there isn't any explicit dependence on time, the Hamiltonian is simply
$H = T+V$, but expressed as a function of the momenta $p_\theta$ and $p_\phi$. I computed it as $p_\theta \cdot \dot\theta + p_\phi \cdot \dot\phi - L$

$p_\theta = \frac{\partial L}{\partial \dot\theta}=mr^2\dot\theta$
$p_\phi = \frac{\partial L}{\partial \dot\phi}=\dot\phi r^{2} sin(\theta)^{2}m$

Then Lagrange's equations are only computations (I hope I haven't mistaken the derivatives), and so are the Hamilton's.

$p_\phi$ is a constant of motion; the total energy (H or E) is, too. There aren't other constants of motion.

Then I have drawn the chart of V and I have used it to trace a qualitative phase portrait, and I have made basic observations on equilibrium points (one, unstable, when the point is at the south pole; one, stable, when it is at the north pole; and a circumference (a parallel) depending on the mass and the elastic constant.

And now?

Ocirne

2. Feb 11, 2014

### haruspex

How do you get (1-cos(θ))? And shouldn't there be a factor 1/2 on that term?

3. Feb 11, 2014

### Ocirne94

I get it from geometry: the spring's square length is

$r^2(1-cos\theta)^{2} + (sin\theta^{2})$

which becomes

$r^2 + r^2cos\theta^2-2r^2cos\theta+r^2sin\theta^2$

the 2 gets simplified with the 1/2 of the elastic potential formula.
This (I forgot to say) setting potential=0 at the south pole of the sphere.

Last edited: Feb 11, 2014
4. Feb 11, 2014

### haruspex

Ah yes, of course.
Everything else looks reasonable to me. Maybe more is wanted on the qualitative description. In general, it will oscillate above and below a latitude corresponding to a stable horizontal orbit, yes? Might it be SHM, in terms of a suitable function of time?