Lagrangian of a Heavy Symmetrical Top - Inertial or Non-inertial Frame?

Question

I'm having some confusion with the analysis of a symmetrical top (specifically, a heavy top, but this is not very important for the question).

Following Landau and Lifshitz's Mechanics, on page 110 they present Euler's angles, and then proceed to represent the angular velocity of a rigid body with respect to the so-called moving axes:

$$\vec{\Omega}=(\dot\phi \sin\theta \sin\psi+\dot\theta\cos\psi)\hat x+(\dot\phi\sin\theta\cos\psi-\dot\theta\sin\psi)\hat y+(\dot\phi\cos\theta+\dot\psi)\hat{z}.\tag{35.1}$$

Then, when going on analyzing a symmetrical top, for example on Problem 1 on page 112, they write the Lagrangian in the form:

$$\mathcal{L}=\frac{1}{2}I_1(\Omega_1^2+\Omega_2^2)+\frac{1}{2}I_3\Omega_3^2.\tag{32.8/35.2}$$

Now, I understand the definition of the angles and the projection of the angular velocity onto the moving frame. The confusion arises when they write the Lagrangian - if the velocity is written relative to the moving frame - which is a non-inertial frame, why don't we need to add terms accounting for non-inertial forces, as done on page 126, $\S$ 39?

Answer 1

The Euler angles are being measured with repect to the fixed intertial frame axes. L&L need the angular velocity components with respect to the top's axes because the intertia tensor is diagonal in that frame. They could equally as well have written the angular velocity components $\omega_x$, $\omega_y$, $\omega_z$, along the space-fixed axes, but used the (now non-diagonal and $\theta,\phi, \psi$ dependent) intertia tensor in that frame. After much algebra they would have ended up with the same kinetic energy as a function of $\dot \theta,\dot \phi, \dot \psi$. Their kinetic energy is still the energy in the inertial frame therefore.