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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/6/d/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 6 / d / Solution by

Codex  0 2026-10-18

For

i \neq = j

, the inertia tensor component is

I_{ij} = - ρ \int_{V} x_{i} x_{j} d V

. Reflect either coordinate appearing once: the cylindrical domain and density are unchanged, while the integrand changes sign. Thus every off-diagonal component vanishes. Rotational symmetry also implies

I_{11} = I_{22}

Use cylindrical coordinates

x_{1} = r cos θ

x_{2} = r sin θ

x_{3} = z

, with volume element

r d r d θ d z

. The axial moment of inertia is

I_{33} = ρ \int_{- 2}^{2} d z \int_{0}^{2 π} d θ \int_{0}^{1} r^{3} d r = 2 π ρ .

(1)

The two contributions to a transverse component are

\int_{V} x_{2}^{2} d V = 4 (\int_{0}^{1} r^{3} d r) (\int_{0}^{2 π} sin^{2} θ d θ) = π, \int_{V} z^{2} d V = π \int_{- 2}^{2} z^{2} d z = \frac{16 π}{3} .

(2)

Therefore

I = π ρ diag (\frac{19}{3}, \frac{19}{3}, 2) = M diag (\frac{19}{12}, \frac{19}{12}, \frac{1}{2}), M = 4 π ρ .

(3)

The axial value

M R^{2} /2

and transverse value

M (3 R^{2} + h^{2}) /12

for

R = 1

h = 4

give an independent check on both diagonal entries.

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