Choosing coordinates in Lagrangian Mechanics Consider the problem of a hoop rolling down an inclined plane, with the plane sliding (frictionless) in a horizontal motion.
I don't know how to choose the generalized coordinates for this system. In the instructor's solution, he uses two rather obscure coordinates, which he calls $(\xi,S)$, as it follows from the diagram:

Where I've also included the transformations from cartesian coordinates. I don't understand how does one even think about using those coordinates, especially $\xi$. Why not just put the block in the origin? And for $S$, why not just use $l$?
Is that because of the hoop? Say, if was just a block, would it be much different in that regard?
 A: 
I don't know how to choose the generalized coordinates for this system.

The nice thing about the Lagrangian method is that you can use any coordinates you like. The only real requirement is that you can figure out how to write the Lagrangian in those coordinates. It is nice, but not required, if the coordinates have some immediate physical meaning and if the action is particularly simple in those coordinates.

Why not just put the block in the origin?

You can start with the block at the origin simply by setting $\xi(0)=0$. But we need to write the KE of the block in terms of $\xi$, so the coordinate cannot be permanently attached to the block.

And for S, why not just use l?

$l$ is a constant, so you cannot use it to express either the KE or the PE of the circle. Remember, the whole point of choosing coordinates is to write down the Lagrangian.
A: 
you have two generalized coordinates $~\xi~$ and $~s~$.
to apply Newton second law or Euler- Lagrange , the position of the
center of mass, must be given in inertial system (black coordinate system )
starting with the position of CM in the red coordinate system you obtain
\begin{align*}
  &\begin{bmatrix}
    x_{red} \\
      y_{red} \\
  \end{bmatrix}=
  \begin{bmatrix}
   s \\
    r \\
  \end{bmatrix}
\end{align*}
from here "rotate" to obtain the blue  coordinates
\begin{align*}
&\begin{bmatrix}
    x_{\rm blue} \\
      y_{\rm blue} \\
  \end{bmatrix}=
\begin{bmatrix}
  \cos(\alpha) & \sin(\alpha) \\
  -\sin(\alpha) & \cos(\alpha) \\
\end{bmatrix}
  \begin{bmatrix}
    x_{red} \\
      y_{red} \\
  \end{bmatrix}
\end{align*}
from here to the green coordinate system
\begin{align*}
& x_{\rm green}=x_{\rm blue}\\
&y_{\rm green}= y_{\rm blue}+l\,\sin(\alpha)
\end{align*}
and to the black one
\begin{align*}
&x_{\rm black}=x_{\rm green}+\xi\\
&y_{\rm black}= y_{\rm green} 
\end{align*}
$~x_{\rm black}~,y_{\rm black}~$ are the results of equation (1)
