Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 15E / i / Solution

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.