Past exam of the physics B course of the University of Cambridge / 2018 / Paper 1 / Section C / C10

Inertia and principal directions. About the centre of mass, the inertia tensor isIt relates angular momentum and angular velocity by $\mathbf{J}=\mathsf{I}\mathbf{\omega }$ and gives rotational kinetic energy $T=\mathbf{\omega }\cdot \mathsf{I}\mathbf{\omega }/2$. The principal axes are orthogonal eigenvector directions of this real symmetric tensor; in that basis it is diagonal, with entries the principal moments of inertia $I_1,I_2,I_3$.

Derivatives in a rotating frame. Write $\mathbf{J}={\sum }_i{J}_i{\mathbf{e}}_i$ in a body-fixed orthonormal basis. Each basis vector rotates according to $(d{\mathbf{e}}_i/dt{)}_0=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the components and basis separately proves the rotating-frame derivative formula,The body derivative differentiates the scalar components in that basis. A fixed pivot can replace the centre of mass when the corresponding angular-momentum balance is valid.

Euler equations. The inertial angular-momentum equation is $d\mathbf{J}/dt=\mathbf{G}$. Since the principal moments are constant in the body frame, substitute $J_i=I_i{\omega }_i$:These are Euler's equations in the principal body frame.

Linear perturbation of free rotation. After the impulse, $\mathbf{G}=0$. Let ${\omega }_3=\Omega$ to leading order and ${\omega }_1,{\omega }_2$ be small. The third Euler equation changes ${\omega }_3$ only at quadratic order, so the linear transverse equations areFor ${\omega }_1=a{e}^{pt}$ and ${\omega }_2=b{e}^{pt}$, eliminate $a,b$ fromThis givesIts units are inverse square seconds. Positive $p^2$ gives a growing real exponential; negative $p^2$ gives bounded oscillatory perturbations.

For three distinct positive principal moments of inertia, the denominator of $p^2$ is positive. If $I_3$ is the largest moment, the two numerator factors have opposite signs, so $p^2<0$. If it is the smallest, they again have opposite signs. If $I_3$ is intermediate, the two factors have the same sign, so $p^2>0$. Thus the intermediate axis theorem givesFor the stable cases the body-frame wobble frequency isThe perturbation does not decay in the torque-free conservative model. For the intermediate moment, a generic small impulse excites the growing branch and eventually leaves the linear regime.

Put $I_1=I_2=I_{\perp }$, with $I_3\ne I_{\perp }$. ThenRotation about the symmetry axis is stable whether that axial moment is larger or smaller than the transverse moment. In fact ${\omega }_3$ and ${\omega }_1^2+{\omega }_2^2$ are exactly constant for a torque-free axisymmetric body: the transverse components rotate into one another at the stated body-frame frequency. Stability here means bounded wobble, not relaxation back to perfect axial rotation.

Both the magnitude $J$ of angular momentum and rotational kinetic energy are conserved after the impulse. At fixed $J$, eliminate $J_3^2=J^2-J_1^2-J_2^2$ to writeThe fixed-angular-momentum energy test for principal-axis rotation makes the stability geometric. The largest moment gives a constrained energy minimum, the smallest a maximum, and the intermediate moment a saddle. Conservation of energy keeps a nearby motion near either isolated extremum, but a saddle permits departure in one direction while compensating in another. Thus both extremes are stable without dissipation; it would be incorrect to demand an energy minimum alone in this conservative problem.