 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-2/15e/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Put

β = q^{2} / (4 I_{1} g l) > 1

. There is no nonzero equilibrium angle because

cos^{4} (θ /2) \leq 1 < β

. The upright angle

θ = 0

is stable: its linearized equation is

\ddot{θ} = - (g l / I_{1}) (β - 1) θ

. Hence

P = \frac{2 π}{q ^{2} / ( 4 I _{1}^{2} ) - g l / I _{1}} .

These are small nutational oscillations of the rapidly spinning top. No downward equilibrium belongs to the momentum level under consideration.

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