Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 9B (Classical Dynamics)

Take the sheet to lie in the $x_1{x}_2$ plane, with lengths $a,b$ in those directions. Its surface density is $M/(ab)$. Reflection symmetry about the coordinate axes makes the off-diagonal entries of the inertia tensor zero. Its principal moments of inertia areand similarly $I_2=M{a}^2/12$. Since $I_3$ integrates $x_1^2+x_2^2$, the perpendicular axis theorem gives

Set ${\omega }_1=u$, ${\omega }_2=\Omega +v$, ${\omega }_3=w$ and retain terms linear in the perturbations in Euler equations for a torque-free rigid body. Using $I_3=I_1+I_2$ givesConsequently $\ddot{u}={\beta }^2{\Omega }^{2}u$, where $\beta =\sqrt{(a^2-b^2)/(a^2+b^2)}>0$. For $\Omega >0$ the growing solution isThere is also a decaying mode with the opposite exponent. For negative $\Omega$, select the exponent $+\beta |\Omega |$ as the growing one. This is the intermediate axis theorem: steady rotation about the intermediate principal inertia axis is unstable whenever $\Omega \ne 0$.