# Lagrangian of the Euler equations - why are Lin constraints required?

The following equation describes the motion of a rigid body rotation, such as a gyroscope:

$$\frac{d\textbf{L}}{dt} ={\bf{\tau}}= \textbf{r}\times m\textbf{g}= {\omega}\times \textbf{L}$$

where $$L$$ is the angular momentum, $$\tau$$ is the torque, $$r$$ is the position vector, $$g$$ is gravity, and $$\omega$$ is the angular velocity.

In thread 1, it is shown that the relation $$\frac{dL}{dt} = \omega\times L$$ can be derived from the following Lagrangian:

$$S[\omega, {\bf p}, {\bf r}]= \int \left(\frac 12 I_1\omega_1^2+\frac 12 I_2\omega_2^2+\frac 12 I_3\omega_3^2+ {\bf p}\cdot (\dot {\bf r}+ \omega \times {\bf r})\right)dt$$

where $${\bf p}$$ here is a Lagrange multiplier for the Lin constraint $$\dot {\bf r}+ \omega \times {\bf r} =0$$.

My questions are -

1) Why does this equation cannot be derived from a simple (Kinetic - Potential) energy Lagrangian? Why do we need an additional constraint? I tried reading some papers about Lin constraints in Fluid mechanics, in which they are required to derive Navier-Stokes equations in Euler's specification. But what is the intuition behind this here?

2) If the constraint is imposed in order to keep a rotational reference frame, how, I would guess that (as mentioned in 2), we need an additional potential energy, because of the effect of gravity, but it seems like this is not represented in this Lagrangian. What is the reason for this? (an equivalent question is how to account for the torque $$r\times mg$$ in the Largrangian? )

I will be glad to receive some intuition about Lin constraints in this particular case (any further details, or good references, will be gratefully appreciated).

Thanks