How do we get $p_{\psi}=M_3$ and $p_{\phi}=M_z$ for a symmetrical top?

Question

I am working on Landau and Lifshitz's Mechanics, 3rd edition. In Section 35 Problem 1 it studies a heavy symmetrical top. We know that the Lagrangian is given by $$ L=\frac{1}{2}(I_1+\mu l^2)(\dot{\theta}^2+\dot{\phi}^2\sin^2\theta)+\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos\theta)^2-\mu gl\cos \theta. $$ Since $\psi$ and $\phi$ are cyclic coordinates, we have two integrals $$ p_{\psi}=\frac{\partial L}{\partial \dot{\psi}}=I_3(\dot{\psi}+\dot{\phi}\cos\theta)\equiv M_3 $$ and $$ p_{\phi}=\frac{\partial L}{\partial \dot{\phi}}=((I_1+\mu l^2)\sin^2\theta+I_3cos^2\theta)\dot{\phi}+I_3\dot{\psi}\cos\theta\equiv M_z. $$

I understand that $p_{\psi}$ and $p_{\phi}$ are constants. On the other hand, I understand that by physics argument, the projection of the angular momentum to the $z$ axis and the $x_3$ axis are both constants.

My question is: how do we know that $p_{\psi}=M_3$ and $p_{\phi}=M_z$? Do we get it from some general result in mechanics?

Edit: by (33.2) of Mechanics $M_3=I_3\Omega_3$ and by (35.1) $\Omega_3=\dot{\psi}+\dot{\phi}\cos\theta$. Therefor $M_3=I_3(\dot{\psi}+\dot{\phi}\cos\theta)$ is clear.

As for $M_z$, we can use (33.2) and $(35.1)$ to get the expression of the angular momentum vector $\mathbf{M}$ as $$ \mathbf{M}=(I_1+\mu l^2)(\dot \phi\sin\theta\sin\psi+\dot \theta\cos \psi)\mathbf{e_1}+(I_1+\mu l^2)(\dot \phi\sin\theta\cos\psi-\dot \theta\sin \psi)\mathbf{e_2}+I_3(\dot{\psi}+\dot{\phi}\cos\theta)\mathbf{e_3} $$

Then we can use Eulerian angles to project $\mathbf{e_1}$, $\mathbf{e_2}$, and $\mathbf{e_3}$ to the $z$-axis and obtain the projection of $\mathbf{M}$ to the $z$-axis. But the computation is complicated.

My edited question is: can we get $M_z=\frac{\partial L}{\partial \dot\phi}$ in a simple way?

Answer 1

The transformation matrix between body fixed system and inertial system is

$$S=S_z(\varphi)\,S_x(\theta)\,S_z(\psi)$$

form here you obtain the angular velocity vector (components in body system)

$$\vec\omega= \left[ \begin {array}{c} \sin \left( \theta \right) \sin \left( \psi \right) \dot\varphi +\cos \left( \psi \right) \dot\theta \\ \sin \left( \theta \right) \cos \left( \psi \right) \dot\varphi -\sin \left( \psi \right) \dot\theta \\ \dot\psi +\dot\varphi \,\cos \left( \theta \right) \end {array} \right] $$

the components of the angular momentum vector in body system are:

$$\vec L=\begin{bmatrix} I_1 & 0 & 0 \\ 0 & I_1 & 0 \\ 0 & 0 & I_3 \\ \end{bmatrix}\,\vec\omega$$

thus the components of angular momentum vector in inertial system are:

$$\begin{bmatrix} L_x \\ L_y \\ L_z \\ \end{bmatrix}=S\,\vec L$$

from here you obtain that

$$L_z=(I_1\,\sin^2(\theta)+I_3\,\cos^2(\theta))\,\dot\varphi+ I_3\,\cos(\theta)\,\dot\psi=\frac{\partial T}{\partial\dot\varphi}$$

where T is the kinetic energy

$$T=\frac 12 \vec\omega^T\,\begin{bmatrix} I_1 & 0 & 0 \\ 0 & I_1 & 0 \\ 0 & 0 & I_3 \\ \end{bmatrix}\,\vec\omega$$