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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-4/12c/c/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 12C / c / i / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

Place the origin at the centre and the axes along the edges. Symmetry makes the off-diagonal entries of the inertia tensor zero. Uniform density gives

⟨ x^{2} ⟩ = a^{2} /12

⟨ y^{2} ⟩ = b^{2} /12

⟨ z^{2} ⟩ = c^{2} /12

. Consequently the principal moments of inertia are

I_{1} = \frac{M}{12} (b^{2} + c^{2}), I_{2} = \frac{M}{12} (c^{2} + a^{2}), I_{3} = \frac{M}{12} (a^{2} + b^{2}) .

(1)

For example

I_{1} = ρ \int (y^{2} + z^{2}) d x d y d z

gives the first expression directly.

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