Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 12C (Classical Dynamics)

In a frame centered at the fixed point or centre of mass, a rotating material point has velocity $\dot{r}=\omega \times r$. The identity $|\omega \times r{|}^2=|\omega {|}^2|r{|}^2-(\omega \cdot r{)}^2$ converts the rotational kinetic energy into $T=\frac{1}{2}{\omega }_i{I}_{ij}{\omega }_j$, where the inertia tensor isLikewise integrating $r\times (\omega \times r)$ gives the angular momentum $\boxed{L=I\omega }$. These expressions refer to rotational motion about the chosen origin.

In principal axes, $2E=\sum I_i{\omega }_i^2$ and $L^2=\sum I_i^2{\omega }_i^2$. Substitute the Euler equations for a torque-free rigid body:For the squared angular momentum,Thus both $E$ and $L^2$ are conserved, even though the body-frame components of $L$ generally change.

Place the origin at the centre and the axes along the edges. Symmetry makes the off-diagonal entries of the inertia tensor zero. Uniform density gives $⟨x^2⟩=a^2/12$, $⟨y^2⟩=b^2/12$, $⟨z^2⟩=c^2/12$. Consequently the principal moments of inertia areFor example $I_1=\rho \int (y^2+z^2)dxdydz$ gives the first expression directly.

Here $(I_1,I_2,I_3)=(20,17,5)M{a}^2/12$, so $I_1>I_2>I_3$ and $e_2$ is the intermediate axis. Eliminating the other components from $L^2=2I_{2}E$ and the conservation of energy relation gives, with $J=2E-I_2{\omega }_2^2$,Thus the second Euler equations for a torque-free rigid body reduces toThe sign is minus the sign of ${\omega }_1{\omega }_3$; it is constant along a nontrivial branch. Let $\Omega =\sqrt{24E/(17M{a}^2)}$. Since ${\omega }_2(0)=0$, integration yields ${\omega }_2(t)=\pm \Omega \tanh (3\Omega t/5)$, with the same sign as in the scalar equation. The remaining components areTheir initial squares are fixed by the preceding algebraic formulas with $J=2E$; their initial signs determine the branch. Hence $\boxed{\omega (t)⟶\pm \Omega e_2}$ as $t\to \infty$. This trajectory is an intermediate-axis separatrix: approaching intermediate-axis rotation on this special conserved-energy level does not make that rotation stable to generic perturbations. If $E=0$, all components vanish identically.