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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.

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Here is how it should have gone:

$$\frac{d}{dt}\frac{\partial \cal L}{\partial \dot{\vec r}_{CM}}=\frac{\partial \cal L}{\partial {\vec r}_{CM}}\\ m\ddot{\vec r}_{CM}=2\lambda \left(\vec r_{CM}+\vec\rho(\theta)-\vec r_0\right)\cdot\left(\frac{\partial \vec r_{CM}}{\partial \vec r_{CM}}+\frac{\partial \vec \rho(\theta)}{\partial \vec r_{CM}}\right)\\ =2\lambda \left(\vec r_{CM}+\vec\rho(\theta)-\vec r_0\right)\cdot\left({\bf 1}+\frac{d \vec \rho(\theta)}{d \theta}\frac{\partial \theta}{\partial \vec r_{CM}}\right)\\ =2\lambda \left(\vec r_{CM}+\vec\rho(\theta)-\vec r_0\right)\cdot\left\{{\bf 1}+\left[\hat e_z \times\vec \rho(\theta)\right]\otimes \left[-\hat e_z \times\vec \rho(\theta)\right]\right\}\\ =2\lambda \left(\vec r_{CM}+\vec\rho(\theta)-\vec r_0\right)\cdot\left[\vec \rho(\theta)\otimes\vec \rho(\theta)\right]\\ =2\lambda \left[\left(\vec r_{CM}+\vec\rho(\theta)-\vec r_0\right)\cdot\vec \rho(\theta)\right]\vec \rho(\theta)$$

Where $$ \vec\rho(\theta)=\rho\left[\cos(\theta)\hat e_x+\sin(\theta)\hat e_y\right]$$ and $$\tan(\theta)=\frac{l_y-y_{CM}+y_0}{l_x-x_{CM}+x_0}$$

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