Equilibrium point of a system described by a Lagrangian

Question

I have the Lagrangian $L=\frac{m}{2}\Big(\dot{r}^2+r^2\dot{\theta}^2\Big) - \kappa\big(r^2+a^2-ar(1+\cos\theta)\big)+mgr\sin\theta$ describing a system. In this system a mass is displaced slightly from equilibrium so that we can write $r=R+\epsilon$ and $\theta=\theta_0+\delta$. I am also given that $\epsilon,\delta,\dot{\epsilon},\dot{\delta}$ are all much less than zero. I need to find approximate linear equations of motion for $\epsilon$ and $\delta$.

Firstly I need to find the position $r=R$, $\theta=\theta_0$ for which a particle remains at rest in this system. From the Euler-Lagrange equations and setting $\dot{r}=0, \dot{\theta}=0$ I end up with the $2$ equations:

$-2kr+ka(1+\cos\theta)+mg\sin\theta=0$, from the E-L equation for the coordinate $r$, and

$-kar\sin\theta +mgr\cos\theta=0$, for the E-L equation for $\theta$.

So from the second one I find that $\tan\theta_0=\frac{mg}{ka}$. After making use of $1+\tan^2\theta=\sec^2\theta$ and a bit of tedious algebra I get the pretty ugly expression $R=\frac{1}{2}\Big(\frac{k^2a^2+m^2g^2}{k\sqrt{1+m^2g^2}}+a\Big)$.

Are these correct? If so how do I proceed from here? I have a feeling that I am supposed to substitute in $R$, $\theta_0$ and Taylor expand, disregarding the $O(\epsilon^2), O(\delta^2)$ terms but this makes everything pretty messy so I am a bit stuck.

Answer 1

The equilibrium values $\theta_0$ and $r_0$ seem correct.

Now you have to find the quadratic approximation of $L$ in the neighbourhood of the equilibrium point $(0,0,r_0,\theta_0) $ (the first two entries are the velocities).

Let $H$ be the hessian matrix at the equilibrium point: $$ H= \begin{bmatrix} \frac{\partial^2V}{\partial \theta ^2} & \frac{\partial^2V}{\partial \theta\partial r}\\ \frac{\partial^2V}{\partial \theta\partial r} & \frac{\partial^2V}{\partial r^2} \\ \end{bmatrix}_{(\theta_0, r_0)} $$ where $V(r,\theta)=\kappa\big(r^2+a^2-ar(1+\cos\theta)\big)+mgr\sin\theta $. The explicit values can be messy, but that's it. The quadratic approximation of the Lagrangian is $$ L_0=\frac{m}{2}\Big(\dot{r}^2+r_0^2\dot{\theta}^2\Big)-\frac12\left[H_{11} (r-r_0)^2+2H_{12}(r-r_0)(\theta-\theta_0)+H_{22}(r-r_0)^2\right] $$ The corresponding linear equations are \begin{align} m\ddot r= -H_{11}(r-r_0)-H_{12}(\theta-\theta_0),\\ mr_0^2\ddot\theta=-H_{12}(r-r_0)-H_{22}(\theta-\theta_0) \end{align}

Answer 2

What is $\kappa$ ?

Assuming that $\kappa$ is a Lagrange multiplier to handle the holonomic constraint $r^2+a^2-ar(1+\cos\theta)=0$ the movement equations are

$$ \left\{ \begin{array}{rcl} -a \kappa (\cos (\theta)+1)+g m \sin (\theta)+r \left(m \theta '^2+2 \lambda \right)-m r'' & = & 0\\ r \left(g m \cos (\theta)+a \kappa \sin (\theta)-2 m r' \theta '-m \theta ''\right) & = & 0\\ 2 r'^2+2 a \sin (\theta (t)) \theta ' r'-a (\cos (\theta)+1) r''(t)+r \left(a \cos (\theta)\theta '^2+2 r''+a \sin (\theta) \theta ''\right) & = & 0\\ \end{array} \right. $$

here the last equation is the holonomic constraint derived twice regarding $t$. Now after solving for $(r'',\theta'',\kappa)$ we have

$$ \kappa = \frac{m \left(a g \sin (\theta)+r(t) (a-2 r) \theta '^2-2 g r \sin (\theta )-2 r'^2\right)}{2 a^2 (\cos (\theta)+1)-4 a r (\cos (\theta)+1)+4 r^2} $$