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.


1 Answer 1


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 $$\tan(\theta)=\frac{l_y-y_{CM}+y_0}{l_x-x_{CM}+x_0}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.