Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 15E (Classical Dynamics)

The angles $\psi$ and $\varphi$ are cyclic. Their conjugate momentum components are $p_{\psi }=I_3(\dot{\psi }+\dot{\varphi }\cos \theta )$ and $p_{\varphi }=I_1\dot{\varphi }{\sin }^2\theta +p_{\psi }\cos \theta$, so both are conserved; consequently ${\omega }_3=p_{\psi }/I_3$ is constant. With $q=I_3{\omega }_3=p_{\varphi }>0$,Eliminating the cyclic velocities gives the nutation effective potentialapart from the constant axial spin energy. ThusThe endpoint $\theta =\pi$ is excluded for these nonzero conserved momenta: the effective potential diverges there. The apparent Euler-angle singularity at $\theta =0$ is interpreted by its smooth small-tilt limit.

Put $\beta =q^2/(4I_{1}gl)>1$. There is no nonzero equilibrium angle because ${\cos }^4(\theta /2)\le 1<\beta$. The upright angle $\theta =0$ is stable: its linearized equation is $\ddot{\theta }=-(gl/I_1)(\beta -1)\theta$. HenceThese are small nutational oscillations of the rapidly spinning top. No downward equilibrium belongs to the momentum level under consideration.

For $0<\beta <1$, the upright angle has $\ddot{\theta }=(gl/I_1)(1-\beta )\theta$ and is unstable. There is one nonzero equilibrium in $0<\theta <\pi$,Differentiate the right side of the nutation equation at this angle. Since the bracket vanishes there, the derivative is $-2(gl/I_1)\sin {\theta }_{\ast }\tan ({\theta }_{\ast }/2)=-4(gl/I_1){\sin }^2({\theta }_{\ast }/2)$. Therefore it is stable, withThis equilibrium represents steady precession, not a top with all Euler angles stationary. The equality case $\beta =1$ is excluded by the two cases in the question and requires a nonlinear marginal analysis.