Picture a rigid square with one of its vertices attached to the end of a massless rigid rod whose other end is attached to a point fixed in space. The motion is restricted to the plane containing the square. If I want the system to be described in terms of the square's degrees of freedom, i.e. its center of mass coordinates and its rotational angle, the Lagrangian of the system is: $$ {\cal L}=\frac{m}{2}\lvert \dot {\vec r}_{CM}\rvert^2+\frac{I}{2}\dot \theta^2+\lambda\left(\lvert\vec r_{CM}+\vec \rho(\theta)-\vec r_0\rvert^2-l^2\right) $$ where $m$ is the mass of the square, $\vec r_{CM}$ is the location of its center of mass, $I$ is its moment of inertia, and $\theta$ is its rotational angle, $\lambda$ is a Lagrange multiplier, $\vec \rho(\theta)$ is the vector pointing to the linked vertex from the square's center of mass, $\vec r_0$ is the location of the fixed point, and $l$ is the fixed length of the rod.
Deriving the Euler-Lagrange equation for the rotational angle yields
$$ \frac{d}{dt}\frac{\partial \cal L}{\partial \dot \theta}=\frac{\partial \cal L}{\partial \theta}\\ I\ddot \theta=2\lambda\left(\vec r_{CM}+\vec \rho(\theta)-\vec r_0\right)\cdot\frac{d\vec \rho(\theta)}{d\theta}\\ =2\lambda\left(\vec r_{CM}+\vec \rho(\theta)-\vec r_0\right)\cdot\left(\hat e_z\times\vec \rho(\theta)\right)\\ =2\lambda\left[\vec \rho(\theta)\times\left(\vec r_{CM}+\vec \rho(\theta)-\vec r_0\right)\right]\cdot\hat e_z $$
To me, this makes total sense since, basically, it's saying that the angular acceleration is proportional to the torque between the rigid rod and $\vec \rho(\theta)$.
On the other hand, for the center of mass' location:
$$ \frac{d}{dt}\frac{\partial \cal L}{\partial \dot {\vec r}_{CM}}=\frac{\partial \cal L}{\partial \vec r_{CM}}\\ m \ddot {\vec r}_{CM}=2\lambda\left(\vec r_{CM}+\vec \rho(\theta)-\vec r_0\right)\cdot\frac{d\vec r_{CM}}{d\vec r_{CM}}\\ =2\lambda\left(\vec r_{CM}+\vec \rho(\theta)-\vec r_0\right) $$
This result doesn't make sense. It's saying that the whole effect of the constraint is devoted to the translation of the square disregarding the rotation.
The equation of motion I would expect is
$$ m \ddot {\vec r}_{CM}=2\lambda\left[\left(\vec r_{CM}+\vec \rho(\theta)-\vec r_0\right)\cdot\hat\rho(\theta)\right]\hat\rho(\theta) $$
Thus the effect of the constraint is distributed between the torque and a perpendicular component driving a translational motion of the square.
What am I missing? Why am I not getting the expected EOM from the Lagrangian?
Any help will be much appreciated.