# 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?

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.

$$\text{Constraint1}=x''(t)-R \phi ''(t)$$

$$\text{Constraint2}=y''(t)+R \theta ''(t)$$

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.