I'm trying to solve the following exercise, but I'm totally new to Lagrangian mechanics and I don't have any solution

Let $P$ be a particle of mass $m$ in the space which is under the action of the gravity (w.r.t the $z$-axis) and of an elastic force w.r.t a fixed point $O$, and $K>0$ elastic constant

(i) Choose Lagrangian coordinated and estabilish the configuration manifold $Q$.

(ii) Write Lagrange equations

My solution

(i) I would choose cylindrical coordinates,because of the geometry of the system: \begin{align} x&=r \cos(\theta), \\ y &=r \sin(\theta), \\ z&=z . \end{align}

As configuration manifold honestly I don't know how to proceed: I'd say that it is, euristically, $Q=(\vec{P-O} )\times S^2 $, where $S^2$ is the 2-sphere.

(ii) With the above configurations, I start computing the kinetic energy $T$, which is, \begin{align} T(r,\theta,z,\dot r, \dot \theta,\dot z)=\frac{1}{2}m (\dot r^2+r^2 \dot \theta^2+ \dot z^2) \end{align}

Since gravity acts on the system, then $V(r,\theta,z)=+mgz$

I have also to take into account the elastic force, where the potential energy is $V_{el}=\frac{1}{2}K || \vec{P-O}||^2$.

But $\vec{P-O}=(r \cos(\theta),r \sin(\theta),z)$, and hence

\begin{align} V_{el}(r,\theta,z)=\frac{1}{2}K(r^2+z^2) \end{align}

Finally, the Lagrangian is

\begin{align} \mathcal{L}=\frac{1}{2}m (\dot r^2+r^2 \dot \theta^2+ \dot z^2) -mgz-\frac{1}{2}K(r^2+z^2) \end{align}

Of course, deriving Lagrang. equations is not a problem. What I need to know is if my approach is correct or if I am missing something important


Following the suggestion in the comment, I notice that probably spherical coordinates are better.

The configuration manifold $Q$ should be $Q=\mathbb{R}^+ \times S^2$.

From some computations, I get

\begin{align} T=\frac{1}{2}m(\dot r^2+r^2 \dot \theta^2+r^2 \dot \phi^2 \sin^2(\theta)) \end{align}

and for the potential energies I get

\begin{align} V=mg r \cos(\theta) \end{align}

for the elastic potential energy I need to compute $|| \vec{P-O}||^2$, where $\vec{P-O}=(r \sin (\theta)\cos(\phi),r \sin(\theta) \sin(\phi),r \cos(\theta))$, and hence

\begin{align} V_{el}=\frac{1}{2}Kr^2 \end{align}

and hence

\begin{align} \mathcal{L}=\frac{1}{2}m(\dot r^2+r^2 \dot \theta^2+r^2 \dot \phi^2 \sin^2(\theta)) -mg r \cos(\theta) -\frac{1}{2}Kr^2 \end{align}

Could it be okay now?


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  • $\begingroup$ Out of curiosity, can you explain why you choose cylindrical coordinates? My first choice would have been spehrical coordinates, I think.. Or in other words, what kind of geometry do you see here? $\endgroup$ – Sito Apr 24 at 21:45
  • $\begingroup$ Uhm right, they're definitely the right choice. In this case the configuration manifold would be $Q= S^2$, right? $\endgroup$ – VoB Apr 24 at 21:50
  • $\begingroup$ Please wait, I'm not saying that they are.. I'm unsure about this myself, but on first glance I don't see anything constraining the motion of the system to a cylinder so I'm not sure if cylindrical coordinates are the best choice... Regarding the configuration space, well if your dealing with spherical coordinates I'd assume the configuration manifold is something along the lines of $Q=\mathbb{R}^+\times S^2$, but again, don't quote me on that... $\endgroup$ – Sito Apr 24 at 21:52
  • $\begingroup$ Actually I think that your observation make the point: there is no cylindrical simmetry in this problem. For the configuration manifold... yes, I forgot to write $\mathbb{R}^+$, and you're right, at least I think $\endgroup$ – VoB Apr 24 at 21:59
  • 1
    $\begingroup$ The Lagrangian has only quadratic and linear terms, so in Cartesian coordinates the Euler-Lagrange equations will be linear: \begin{align} &\ddot{x} = - \, \frac{K}{m} \, x\\ &\ddot{y} = - \, \frac{K}{m} \, y\\ &\ddot{z} = - \, \frac{K}{m} \, z - g\\ \end{align} In fact you have three decoupled linear osciallotrs. These Cartesian coordinates are more convenient then spherical or other curvilinear coordinates $\endgroup$ – Futurologist Apr 26 at 20:05