Euler's equations for the angular velocity $\boldsymbol{\omega}=\left(\omega_{1}, \omega_{2}, \omega_{3}\right)$ of a rigid body, viewed in the body frame, are

$$I_{1} \frac{d \omega_{1}}{d t}=\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}$$

and cyclic permutations, where the principal moments of inertia are assumed to obey $I_{1}<I_{2}<I_{3}$.

Write down two quadratic first integrals of the motion.

There is a family of solutions $\boldsymbol{\omega}(t)$, unique up to time-translations $t \rightarrow\left(t-t_{0}\right)$, which obey the boundary conditions $\boldsymbol{\omega} \rightarrow(0, \Omega, 0)$ as $t \rightarrow-\infty$ and $\boldsymbol{\omega} \rightarrow(0,-\Omega, 0)$ as $t \rightarrow \infty$, for a given positive constant $\Omega$. Show that, for such a solution, one has

$$\mathbf{L}^{2}=2 E I_{2},$$

where $\mathbf{L}$ is the angular momentum and $E$ is the kinetic energy.

By eliminating $\omega_{1}$ and $\omega_{3}$ in favour of $\omega_{2}$, or otherwise, show that, in this case, the second Euler equation reduces to

$$\frac{d s}{d \tau}=1-s^{2}$$

where $s=\omega_{2} / \Omega$ and $\tau=\Omega t\left[\left(I_{1}-I_{2}\right)\left(I_{2}-I_{3}\right) / I_{1} I_{3}\right]^{1 / 2}$. Find the general solution $s(\tau)$.

[You are not expected to calculate $\omega_{1}(t)$ or $\left.\omega_{3}(t) .\right]$