Tennis racket theorem

Remarks

Linearize at $\omega = \Omega e_2$ by writing $\omega = (\varepsilon_1, \Omega, \varepsilon_3)$ and dropping products of the small terms. The equations become $I_1 \dot\varepsilon_1 = (I_2 - I_3)\Omega\varepsilon_3$ and $I_3 \dot\varepsilon_3 = (I_1 - I_2)\Omega\varepsilon_1$, so $\ddot\varepsilon_1 = \frac{(I_2 - I_3)(I_1 - I_2)}{I_1 I_3}\Omega^2 \varepsilon_1$. The coefficient is positive exactly because $I_2$ lies between $I_1$ and $I_3$, so the linearization has a positive eigenvalue, which gives instability. At $\Omega e_1$ or $\Omega e_3$ the same coefficient is negative; stability then follows from the two conserved quantities, since the nearby polhodes are small closed curves around the axis. If two principal moments are equal, the conclusion changes. For $I_1 = I_2 < I_3$, the rotation $\Omega e_1$ is not Lyapunov stable: a small component $\omega_3 = \delta \ne 0$ makes $(\omega_1, \omega_2)$ turn at the constant rate $(I_3 - I_1)\delta / I_1$, so $\omega$ later comes close to $-\Omega e_1$.

Example: a tennis racket or a book tossed with a spin about its intermediate axis makes a half turn about another axis during the flight. The same effect was seen in space by Vladimir Dzhanibekov with a spinning wing nut, and it is also called the Dzhanibekov effect.

Going deeper

For a student

The theorem is a model case of the stability analysis of equilibria by linearization and conserved quantities. The instability at the middle axis is seen in the linearization. The stability at the extreme axes needs the energy and angular momentum integrals, since the linearization only shows a centre. The flips about the middle axis follow the separatrix polhodes and are described by Jacobi elliptic functions. With slow loss of energy, as in a flexible spacecraft, the axis of greatest moment is the only stable one.