Fixed-angular-momentum energy test for principal-axis rotation ($\Delta T=\frac{1}{2}{J}_1^2(I_1^{-1}-I_3^{-1})+\frac{1}{2}{J}_2^2(I_2^{-1}-I_3^{-1})$)

For torque-free rotation, total angular momentum magnitude $J$ and rotational kinetic energy are conserved. Near rotation about principal axis three, eliminate $J_3^2=J^2-J_1^2-J_2^2$ from $T={\sum }_i{J}_i^2/(2I_i)$ to obtain the displayed energy difference from $J^2/(2I_3)$. If $I_3$ is largest, it is a constrained energy minimum; if smallest, a maximum. Both extrema are stable in the conservative problem. An intermediate moment gives opposite signs in the quadratic form, a saddle and instability. Dissipation changes the fixed-energy argument and must not be silently added to torque-free motion.