In polar coordinates,
$\begin{align*}x &= r \cos \theta \\y &= r \sin \theta\end{align*}$
and
$\begin{align*}\dot{x} &= \dot{r} \cos \theta - r \dot{\theta}\sin\theta \\\dot{y} &= \dot{r} \sin \theta + r \dot{\th...$
so, the kinetic energy
$\begin{align*}T &= \frac{1}{2} m \left(\dot{x}^2 + \dot{y}^2\right) \\&= \frac{1}{2}m \left(\dot{r}^2 + r^2\dot{\thet...$
The potential energy is
$\begin{align*}V = \frac{1}{2}\left(r - s \right)^2\end{align*}$
So, the Lagrangian is
$\begin{align*}L &= T - V \\&= \boxed{\frac{1}{2}m \left(\dot{r}^2 + r^2\dot{\theta}^2 \right) - \frac{1}{2}\left(r -...$
Therefore, answer (D) is correct.

Solution to 2008 Problem 82