I tried to find the equation of this pendulum, but I think I did something wrong. I know I have to get the Bessel's equation but I can't see it.
It's a simple 2-D pendulum, without any dissipation.
The length is $l$.
\begin{gather}
x(t)= l(t) \sin\theta(t), \\
y(t)=l(t) \cos\theta(t), \\
\dot x(t)= \dot l(t) \sin\theta(t) + l(t) \dot \theta(t) \cos\theta(t) \\
\dot y(t)= \dot l(t) \cos\theta(t) – l(t) \dot \theta(t) \sin\theta
\end{gather}
and:
\begin{equation}
\dot x^2 + \dot y^2 = \dot l^2 + l^2 \dot \theta^2
\end{equation}
The potential and kinetic energy:
\begin{gather}
V=-mgy= -mgl(t) \cos\theta(t), \\
T=\frac{m}{2} (\dot l^2(t)+ l^2(t) \dot \theta^2(t))
\end{gather}
So the Langrangian is:
\begin{equation}
L=T-V= \frac{m}{2}(\dot l^2 + l^2 \dot \theta^2) + mgl \cos\theta
\end{equation}
After this, we get:
\begin{gather}
\frac{ \partial L}{\partial \theta} = -mgl \sin\theta, \\
\frac{ d}{dt} \frac{\partial L}{\partial \dot \theta}= m2l \dot l \dot \theta +ml^2 \ddot \theta.
\end{gather}
From thism we get the equation:
\begin{equation}
\ddot \theta + 2 \dot l \dot \theta = g \sin\theta.
\end{equation}
And for the radial part:
\begin{gather}
\frac{\partial L}{\partial l} = ml \dot \theta^2 + mg \cos\theta , \\
\frac{d}{dt} \frac{\partial L}{\partial \dot l}= m \ddot l.
\end{gather}
The equation is:
\begin{equation}
\dot l = l \dot \theta^2 + g \cos\theta
\end{equation}
My question is how can I get to the Bessel's equation from here?
Best Answer
The correct equation of motion for $\theta$ is \begin{equation} l\ddot \theta + 2 \dot l \dot \theta = -g \sin\theta. \end{equation} Now if you assume $l(t)=l_{0}+vt$ then you will get a Bessel Differential Equations. \begin{equation} \ddot \theta+\frac{2}{l}\dot \theta+\frac{g}{v^{2}l}\theta=0 \end{equation} The solution for which is the Bessel Function. ($\sin\theta$ approximated by $\theta$ for small angle)