Complete analysis of a point on a surface with Lagrangian

Question

I have to study the dynamic of a point with mass $m$ on the constraint defined by $\Gamma\equiv y^2+z^2-1=0$.

1. Write Lagrangian function and Lagrange's equations.

The point $P\in\Gamma$ is parametrised by $P(x,\theta)=(x,\sin\theta,\cos\theta)\implies\dot P=(\dot x,\cos(\theta)\dot\theta,-\sin(\theta)\dot\theta)$.
The form of the kinetic energy is $T=\dfrac{1}{2}m\Vert\dot P\Vert^2=\dfrac{1}{2}m(\dot x^2+\dot\theta^2)$, while the potential energy is $mg\sin\theta.$
Hence $L=\dfrac{1}{2}m(\dot x^2+\dot\theta^2)-mg\sin\theta$ and Lagrange's equations are: $$\begin{cases} \ddot x=0\\\ddot\theta+g\cos\theta=0 \end{cases}$$

2. Find conserved quantities and their generalised momentum. What's their physical meaning?

Since $L$ is a function of $\dot x,\theta$ and $\dot\theta$, we know that $p=\dfrac{\partial L}{\partial \dot x}=m\dot x$ is the genralised momentum, hence $\dot x=\dfrac{p}{m}$ so we can rewrite the total energy $E=T+V$ as $E^*=\dfrac{1}{2}m((p/m)^2+\dot\theta^2)+mg\sin\theta$ in order to reprent the phase portrait of the system.

$\textbf{first question}$: I still can't really understand why we write the energy in terms of the generalised momentum. My idea is that $p$ is not dependent on time so, we can represent the phase portrait considering $p$ as a constant parameter.

$\textbf{second question}$: for the representation of the phase portrait, can I only analyze the function $V^*$? I mean, should I first consider the entire surface $E^*$ and its intersections with planes parralel to $(q,\dot q)$ (the energy levels $z=\overline E$) plane or I can directly consider the graph of the reduced potential energy?

3. Describe the type of trajectories of the point $P$ on $\Gamma$.
   My idea was to analyse the frequency of small oscillations in a neighbourhood of the points of minimum of $V$, in order to understand if the trajectories are periodic or quasi-periodic. So I'd make a local analyisis of the trajectories.

4. Supposing $x\in [-L,L], P(0)=(0,1,0)$ and $\dot P(0)=(\dot x_0,0,0)$, with $\dot x_0>0$ find, in terms of $\dot x_0$ the necessary time for $P$ to leave $\Gamma$.

First time I see this kind of request so I'm not sure about what's asking :-(

Answer 1

The Problem:

So OP is dealing with the following problem. A point particle with mass $m$ is confined to the submanifold $$\mathcal{M} := \{(x,y,z)\in\mathbb R^3\ |\ y^2 + z^2 = 1\} \subset \mathbb R ^3 $$ which represents a cylinder which is infinitely long and its symmetry axis is the $x$-axis. So the configuration space can be parameterized by the generalized coordinates - $(x,\theta) \in \mathbb R \times(0,2\pi)$. Hence a point $\vec r$ on the submanifold $\mathcal M \subset \mathbb R^3$ is given by $$ \vec r (x, \theta)= \begin{pmatrix}x\\\lambda\sin\theta\\\lambda\cos\theta\end{pmatrix}, $$ where $\lambda = 1$m for dimensional convenience. To express the lagrange function $L$ by the generalized coordinates we calculate $$ \dot{\vec{r}}^2 = \dot x ^2 + \lambda^2\ \dot \theta^2 $$ with which we arrive at the Lagrangian $$ L(x, \theta,\dot \theta) = \frac m 2 (\dot x ^2 + \lambda^2 \dot \theta ^2) - mg\lambda \sin\theta $$ since we deal with a constant gravitational force in negative $y$-direction. (Since OP did not add a drawing of the Cartesian system used I stick to his*her Lagrangian. Although usually I would have the force in negative $z$-direction.) Hence we arrive at the equations of motion $$ \begin{align} \ddot x(t) &= 0\\ \ddot \theta (t) &= - \frac g \lambda \cos(\theta(t)) . \end{align} $$ Generalized Momenta:

OP now investigates the generalized momentum for $x$. The generalized momentum is given by $$ p_x = \frac{\partial L}{\partial \dot x} = m \dot x $$ which for physical trajectories (hence solutions of the EoM) is a constant of motion, which is seen using the EoM (hence the Euler-Lagrange-Equation) $$ \dot p_x = \frac{\text{d}}{\text{d} t} \frac{\partial L}{\partial \dot x} = \frac{\partial L}{\partial x} = 0\ ^1 $$

Now OP writes the total energy of the System as a function of only the generalized momentum $p_x$, which is not the Hamiltonian for the system! What you could use it for though, is to use this new energy to integrate the EoMs, since one could use separation of variables. But solving the EoM of $\theta(t)$ analytically for this problem is only more or less possible.

If you want to arrive at the Hamiltonian of the System you also need to use the generalized momentum for the $\theta$ degree of freedom. Hence you would need to use $$ p_\theta = \frac{\partial L}{\partial \dot \theta} = m \lambda \dot \theta $$ and now you can write down the Hamiltonian $$ H(p_x, p_\theta, \theta) = \frac{p_x ^2}{2m} + \frac{p_\theta ^2}{2m} + mg\lambda \sin{\theta}, $$ which is a function on the phase-space of the system. And now you could of course use the fact that $p_x$ is a constant of motion and reduce the phase-space of the problem to $2$ dimensions by treating $p_x$ as a parameter and start to analyze the "now simpler" system. I hope this helps as a motivation. For a more "rigorous" discussion of the topic in more geometrical terms I would recommend reading chapters 3, 4, 7, 8 and 9 in "Mathematical Methods of Classcial Mechanics" by V. I. Arnolds

Questions 3. and 4.:

The possible trajectories of the system can be discussed intuitively in the following way, and you can also perform some easy calculations for this matter. The $x$-coordinate is kinda boring, since $\dot x (t) = \dot x (0)\ \forall t$, which can be seen by integration of its degree of freedom.

So the interesting part of the trajectory would be $\theta(t)$. So here we have different possibilities: What if $\theta(0) = \frac 3 2 \pi$ and $\dot\theta(0) = 0$? What if the energy in the $\theta$ degree of freedom is high enough to "spiral" on the cylinder? What if the energy in the $\theta$ degree of freedom is less then that? (Here you could expand the problem around the "stable" $^2$ fix point of the system.)

The last question seems to reduce the cylinder to a finite length, such that $(x, \theta) \in [-L, L] \times (0, 2\pi)$, dependent on the initial conditions you could now calculate how long it takes for your point particle to leave the now finite cylinder. (So find $t_1$ such that $|x(t_1)| > L$)

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$^1$ : Sometimes the approx symbol "$\approx$" is used on this site to indicate "equality for solutions of the EoM" so one would write $\dot p_x \approx 0$.

$^2$ : Actually there is no stable fix point for this system in the sense of Lyapunov stability, since small pertubations from the "stable" fix point do not decrease but oscillate.