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The text is taken from this exercise, but I'm trying to answer to now to further questions, and don't know if my procedure is correct, so I'd need a check.

Consider a particle $P$ of mass $m$ which is constrained to a semi-circle of radius R of an equation $x^2+z^2=R^2$, where $z<0$, under the action of the gravity.

The Lagrangian wrt the $x$ coordinate is $$\mathcal{L}(x,\dot x)=\frac{1}{2}m \frac{1+ x^2}{R^2-x^2}\dot{x}^2-mg\sqrt{R^2-x^2}.$$

(i) Consider $\theta \in [-\pi/2,\pi/2]$ measured w.r.t the $z$-axis. Determine a change of coordinates such that $x=x(\theta)$;

(ii) Write Lagrange equations using the $\theta$ coordinate;

(iii) Determine the equilibria of the system and study their stability.


My solution

(i) As change of coordinates, I'd say that the choice is $x=R \sin(\theta)$, where $\theta \in [-\pi/2,+\pi/2]$.

In this way, $z=-\sqrt{R^2-R^2 \sin(\theta)}=-R \cos(\theta)$.

Performing this change, the kinetic energy $T(\theta,\dot \theta)$ is

\begin{align} T=\frac{1}{2} m(R^2 \sin^2(\theta) \dot \theta^2 + R^2 \cos^2(\theta) \dot \theta^2)=\frac{1}{2}mR^2 \dot \theta^2. \end{align}

The potential energy becomes then \begin{align} V(\theta)=mgRz(\theta)=-mgR \cos(\theta). \end{align}

Therefore the Lagrangian is

\begin{align} \mathcal{L}(\theta,\dot{\theta})=\frac{1}{2}R^2 \dot \theta^2+mgR\cos(\theta). \end{align}

(ii) Lagrange equation is \begin{align} R^2 \ddot \theta +mgR \sin(\theta)=0. \end{align}

(iii) The equilibria are the critical points of the potential energy $V(\theta)$, such that $$V'(\theta)=mgR \sin(\theta)=0 \longrightarrow \theta=k \pi, \quad k \in \mathbb{Z}.$$

Checking the second derivative at those points, and distinguish the case $k \text{ even}$, $k \text{ odd}$, I get

\begin{align} V''(2k \pi)=mgR>0,\\ V''((2k+1)\pi)=-mgR<0. \end{align}

Therefore, when $k$ is even, the equilibria are stable, otherwise, when $k$ is odd, they're unstable.

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closed as off-topic by Kyle Kanos, John Rennie, GiorgioP, user191954, ACuriousMind Apr 27 at 10:27

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    $\begingroup$ It looks all right. You forgot $m$ in the kinetic part of the Lagrangian, though. Your Lagrange equation should then become $R\ddot{\theta}+g\sin(\theta) = 0$. $\endgroup$ – denklo Apr 26 at 8:18
  • $\begingroup$ Also notice that the movement along the circle is only constraint to $z<0$. Thus the $k$ even case is somewhat ill-defined. $\endgroup$ – denklo Apr 26 at 8:20
  • $\begingroup$ You're right, I forgot an $m$! In which sense the case $k$ even is ill-defined? I can't figure it out $\endgroup$ – VoB Apr 26 at 8:22
  • $\begingroup$ Oh okay, since I am in $[-pi/2, + \pi/2]$, saying $2k \pi$ is not a defined angle in my interval, you're right $\endgroup$ – VoB Apr 26 at 8:23
  • $\begingroup$ You got it. :-) $\endgroup$ – denklo Apr 26 at 8:25