Nonlinear Pendulum

Question

The question I have is mostly on stability analysis but the problem is:


Consider a nonlinear pendulum. Using a linearized stability analysis, show that the inverted position is unstable. What is the exponential behavior of the angle in the neighborhood of this unstable equilibrium position.

I don't understand what inverted position means, what position was it in to begin with? I found this thing on linearized stability analysis but I'm not totally sure if that applies to this problem.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The pendulum equation of motion is given by $$ mL^{2}\ddot{\theta} = mgL\sin\pars{\theta}\quad\imp\quad \ddot{\theta} - {g \over L}\,\sin\pars{\theta} = 0 $$ where $m$: mass, $L$: length, $g$: gravity acceleration $\ds{\approx 9.8\ {{\rm m} \over {\rm sec}^{2}}}$. $\theta$ is the angle from the vertical line where $\theta = 0$ $\pars{~\it\mbox{inverted pendulum}~}$ is the highest point and $\theta = \pi$ is the lowest point.

1. When $\theta \approx 0$, the equation of motion becomes $$ \ddot{\theta} - {g \over L}\,\theta \approx 0$$ which has solutions $\theta\pars{t} \sim \exp\pars{\pm\root{g \over l}t}$ which renders the motion unstable.
2. When $\theta \approx \pi$, the equation of motion becomes $$ \ddot{\theta} + {g \over L}\,\pars{\theta - \pi}\approx 0$$ which has solutions $\theta\pars{t} - \pi \sim \exp\pars{\pm\ic\root{g \over l}t}$ which oscillates around the lowest point $\ds{\pars{~\mbox{with angular frequency}\ \root{g \over L}~}}$ and it means the motion is stable..

Answer 2

The phrase "inverted position" means that the pendulum is pointing straight up.

Intuitively, we know that this is unstable since any perturbation away from perfectly vertical will cause the pendulum to swing downwards. To show this, you'll compute the linearization of the system when the pendulum is pointing straight up and you'll find that it's unstable (and probably that the distance from vertical grows exponentially).