 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/8b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Write the position of a mass element relative to the center of mass as

r

\dot{X} + ω \times r

. The center-of-mass identity

\int r d m = 0

removes all cross terms. Define the inertia tensor about the center of mass by

I_{ij} = \int (r^{2} δ_{ij} - r_{i} r_{j}) d m .

The total angular momentum about the origin and kinetic energy are then

L = M X \times \dot{X} + I ω, T = \frac{1}{2} M ∣ \dot{X} ∣^{2} + \frac{1}{2} ω \cdot I ω .

Let

ρ (r) = C r^{- 1} sin (π r / R)

I = I_{0} 1

. The mass normalization is

M = 4 π C \int_{0}^{R} r sin \frac{π r}{R} d r = 4 C R^{2},

C = M / (4 R^{2})

. A moment about any diameter is

I_{0} = \frac{8 π}{3} \int_{0}^{R} ρ (r) r^{4} d r = \frac{8 π C}{3} \frac{R ^{4}}{π ^{4}} \int_{0}^{π} x^{3} sin x d x = \frac{2 M R ^{2} ( π ^{2} - 6 )}{3 π ^{2}} .

Therefore

I = \frac{2 M R ^{2} ( π ^{2} - 6 )}{3 π ^{2}} 1 .

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