Equations of motion with Euler-Lagrange equations I try to determine the equations of motion of the system defined by the lagrangian:
$$L=\frac{1}{2}m(\overset{.}{x}^2+\overset{.}{y}^2)+\frac{1}{2}I\overset{.}{\theta}^2.$$
I found that the system operates under the following constraint: $$\omega=\sin \theta dx-\cos\theta dy .$$
To do this I think I can use the Euler-Lagrange equations to solve:
$$\frac{d}{dt}\frac{\partial L}{\partial \overset{.}{z}} - \frac{\partial L}{\partial z}=\lambda\omega$$
where $z=(x,y,\theta)$.
But I don't understand how to solve these. I also wonder if I can transform this relation into the following way
$$\frac{d}{dt}\frac{\partial \mathscr{L}}{\partial \overset{.}{z}} - \frac{\partial \mathscr{L}}{\partial z}=\lambda\omega$$
with $\mathscr{L}=L-\lambda\omega$.
Can someone give me some hints?
Have a nice day!
 A: I'm guessing the configuration space is $\Bbb R^2 \times \Bbb S^1$ and you're modelling the motion of an ice skate with the non-holonomic constraint $\ker \omega$. The perfect reaction force will be a multiple of $\omega$ itself (by a dimension count), say $\mathcal{R}=\lambda \omega$. There is no potential energy, so the external force of the full system vanishes. So the equations of motion become just $$\mu\left(\frac{D\dot{c}}{dt}\right)=\lambda(c(t))\omega_{c(t)}.$$But the left side equals $$\left(\frac{\partial L}{\partial x} -\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\right)\,dx+\cdots \mbox{similar terms with $y$ and $\theta$},$$and you know what the right side is. Equate the coefficients of $dx$, $dy$ and $d\theta$ on both sides to get a $3\times 3$ differential system. You won't even need to find $\lambda$. See section 5.4 of the book by Godinho & Natario for more details.
A: *

*OP's differential constraint is a semi-holonomic constraint, which can equivalently be written as
$$f~\equiv~\dot{x}\sin \theta  -\dot{y} \cos\theta~=~0.\tag{1}$$


*The Lagrange equations become
$$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^j}-\frac{\partial L}{\partial q^j}~=~ \lambda \frac{\partial f}{\partial \dot{q}^j} ,\tag{2} $$
cf. e.g. this Phys.SE post. Eq. (2) is equivalent to the last equation in Ivo Terek's answer (v1).


*Warning: Note that the action $S[q,\lambda]=\int\! dt ( L-\lambda f)$ leads to wrong EOMs different from the correct eq. (2)!


*Also note that OP's various EOMs (v3) are inconsistent since their LHS are 0-forms, while their RHS are 1-forms.
