The Lagrangian of a single particle of mass $m$ is

$$ L=\dfrac{1}{2}m\dot{\textbf{r}}^2 -\dfrac{1}{2}kr^2+m\dot{\textbf{r}} \cdot (\vec{\omega} \times \textbf{r}) \tag{1}$$

where $\vec{\omega}=\omega \textbf{e}_3$.

Hint: the generalized coordinates are $(r,\theta)$ because the particle moves in $(x,y)$ plane for all time $t>0$. The motion equations are

$\hspace{6cm}\cases{r=a \cos(\Omega t)+\dfrac{\dot{r}(0)}{\Omega}\sin(\Omega t), \\ \theta=\theta_0 -\omega t,}$

where $r(0)=a$ and $k,\omega>0$. The relation between the two frequencies is

$\hspace{6cm}\Omega \equiv \sqrt{\dfrac{k}{m}+\omega^2}.$

The constants of the motion are the Hamiltonian $H$ and the polar generalized momentum $p_\theta$.

Here some trayectories in $\mathbb{R}^2$ for different parameters ($0\leq t \leq 4\pi)$:

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The question is what does the last term in the above Lagrangian mean: (energy dimension)

$\hspace{6cm} E=m\dot{\textbf{r}} \cdot (\vec{\omega}\times \textbf{r}) $


1 Answer 1


This term is (minus) the generalized velocity-dependent potential for the Coriolis force, cf. my Phys.SE answer here.

  • $\begingroup$ Thank you @Qmechanic for your answer, which has just been full responded indeed. But because of it, I have more questions about it, for example: What was the non-inertial frame in the $(x,y)$-plane? In the mass? How is it orientated? $\endgroup$ Aug 9, 2016 at 0:19
  • $\begingroup$ I think that the Lagrangian is expressed in a non-inertial reference frame with the origin on the equilibrium point of the spring for each time $t>0$ and the two axes parallel to $(x,y)$ axes of the inertial reference frame from the beginning. These arrangements simplify the harmonic potential at the expense of adding a non-inertial term: the generalized velocity-dependent potential for the Coriolis force, in order to simplify the motion equations. You can see that on the pictures: the mass is not really moving around the center but instead around the equilibrium point. $\endgroup$ Aug 9, 2016 at 1:30

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