 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-3/12a/b/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 3 / 12A / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integral

- ρ \int_{B} x_{i} x_{j} d V (i \neq = j)

(1)

vanish because its integrand is odd in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its inertia tensor at the centre has the isotropic form

I_{ij} (0) = λ δ_{ij} .

(2)

For

a = \frac{ℓ}{2} (1, 1, 0), ∣ a ∣^{2} = \frac{ℓ ^{2}}{2},

(3)

part (a) and

λ = M ℓ^{2} /6

give

I (a) = \frac{M ℓ ^{2}}{12} 5 - 3 0 - 3 50 008 .

(4)

The displayed symmetric matrix has orthogonal eigenvectors

(1, 1, 0), (1, - 1, 0), (0, 0, 1)

(5)

with corresponding eigenvalues

\frac{M ℓ ^{2}}{6}, \frac{2 M ℓ ^{2}}{3}, \frac{2 M ℓ ^{2}}{3} .

(6)

For a unit angular velocity,

∣ L ∣ = ∣ I ω ∣

is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is

\frac{∣ L ∣ _{m a x}}{∣ L ∣ _{m i n}} = \frac{2 M ℓ ^{2} /3}{M ℓ ^{2} /6} = 4 .

(7)

 Read the full article

 New to topics? Read the docs here!