 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-2/section-a/a4/solution

For slow gyroscopic precession, the gravitational torque turns the spin angular momentum:

M g ℓ sin θ = Ω I_{3} ω sin θ

, with

I_{3} = M R^{2} /2

. Thus

Ω ≃ \frac{2 g ℓ}{R ^{2} ω}, T_{p} ≃ \frac{π R ^{2} ω}{g ℓ} = 1.61 s, ω = 40 π s^{- 1} .

The inclination cancels at this order. More precisely, uniform precession of a heavy symmetric top satisfies

\frac{I _{1}}{M} cos θ Ω^{2} - \frac{I _{3}}{M} ω Ω + g ℓ = 0, \frac{I _{1}}{M} = \frac{R ^{2}}{4} + ℓ^{2} .

Its slow root gives

T_{p} ≃ 1.57 s

. About $1.6 s$ is the slow-precession answer; arbitrary initial conditions can also produce nutation rather than exact steady precession.

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