How to deal with no-slip non-holonomic constraints in Lagrangian? I'm solving a dynamical system of a ball of mass $m$ and radius $R$ rolling on a rotating platform ("turntable") for which I found the Lagrangian to be:
$$L=\frac{1}{2} m (\dot{x} - \Omega y)^2 + \frac{1}{2} m (\dot{y}+\Omega x)^2 + \frac{1}{2} I (\omega_x^2 + \omega_y^2 + \Omega^2) - mgR$$
Where $x,y$ are the positions of a rolling sphere; $\omega_x, \omega_y$ are its angular velocities; $\Omega$ is the CONSTANT applied angular velocity of the turntable.
The two non-holonomic constraints are the no-slip conditions of the ball which are :
\begin{align*}
\dot{x} &= R \omega_y \\
\dot{y} &= -R \omega_x 
\end{align*}
Now I'm confused as to how to proceed. The generalized coordinates, I would assume are $x$ and $y$ because I can substitute the constraints into the Lagrangian and have the Lagrangian as a function of the generalized coordinates, i.e.:
$$L = L(x,y,\dot{x},\dot{y})$$
And then proceed to apply the Lagrange equations of motion and so on. Is that a correct way of dealing with these non-holonomic constraints?
 A: For a given Lagrangian, the procedure for using the constraints is fairly straightforward.  For your L, you need to change your $\omega x$ and $\omega y$ to $\dot{\theta }$ and $\dot{\phi }$ because you need coordinates rather than velocities.  You now have four generalized coordinates and two constraints.
Your constraints:
$\text{Constraint1}=x''(t)-R \phi ''(t)$
$\text{Constraint2}=y''(t)+R \theta ''(t)$
Calculate the constraint accelerations
$\text{ax1}=\frac{\partial \text{Constraint1}}{\partial x''(t)}$
$\text{ay1}=\frac{\partial \text{Constraint1}}{\partial y''(t)}$
$\text{a$\theta $1}=\frac{\partial \text{Constraint1}}{\partial \theta ''(t)}$
$\text{a$\phi $1}=\frac{\partial \text{Constraint1}}{\partial \phi ''(t)}$
$\text{ax2}=\frac{\partial \text{Constraint2}}{\partial x''(t)}$
$\text{ay2}=\frac{\partial \text{Constraint2}}{\partial y''(t)}$
$\text{a$\theta $2}=\frac{\partial \text{Constraint2}}{\partial \theta ''(t)}$
$\text{a$\phi $2}=\frac{\partial \text{Constraint2}}{\partial \phi ''(t)}$
With two constraints, we need two Lagrangian multiplyers, call them $\lambda 1$ and $\lambda 2$
The equations of motions are then:
$\text{xeq}=\frac{\partial }{\partial t}\frac{\partial L}{\partial x'(t)}-\frac{\partial L}{\partial
   x(t)}=\text{$\lambda $1}\ \text{ax1}+\text{$\lambda $2}\ \text{ax2}$
$\text{yeq}=\frac{\partial }{\partial t}\frac{\partial L}{\partial y'(t)}-\frac{\partial L}{\partial
   y(t)}=\text{$\lambda $1}\ \text{ay1}+\text{$\lambda $2}\ \text{ay2}$
$\text{$\theta $eq}=\frac{\partial }{\partial t}\frac{\partial L}{\partial \theta
   '(t)}-\frac{\partial L}{\partial \theta (t)}=\text{$\lambda $1}\ \text{a$\theta
   $1}+\text{$\lambda $2}\ \text{a$\theta $2}$
$\text{$\phi $eq}=\frac{\partial }{\partial t}\frac{\partial L}{\partial \phi '(t)}-\frac{\partial
   L}{\partial \phi (t)}=\text{$\lambda $1}\ \text{a$\phi $1}+\text{$\lambda $2}\ \text{a$\phi $2}$
With theses four equations and getting two more by setting the constraints equal to zero, you can solve for the accelerations along with $\lambda 1$ and
$\lambda 2$.  
If you don't need the reactions, you can just substitute for $\omega x$ and $\omega y$ using your constraints.  You will then have the two generalized coordinates $x$ and $y$ with no need for Lagrange multiplyers.
Since you have just asked what procedure to use for the constraints, I have not attempted to check your Lagrangian, although from reading your problem, it seems to me you don't have an object with moment of inertia $I$ spinning with an angular velocity $\Omega$, so you might check that.
