 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/6/b/i/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 6 / b / i / Solution by

Codex  0 2026-10-18

Integrating the Dirac delta distribution in

x_{3}

reduces the inertia integral to the planar lamina. Use

u = x_{1} - x_{2}

,

v = x_{2}

, so

x_{1} = u + v

and the parallelogram becomes the square

- D \leq u, v \leq D

, with unit Jacobian determinant. The mass is

M = 4 σ D^{2}

, and symmetry gives a centre of mass at the origin.

Odd rectangular integrals vanish. The required second moments are

\int x_{1}^{2} d m \int x_{2}^{2} d m \int x_{1} x_{2} d m = σ \int_{- D}^{D} \int_{- D}^{D} (u + v)^{2} d u d v = \frac{8}{3} σ D^{4}, = σ \int_{- D}^{D} \int_{- D}^{D} v^{2} d u d v = \frac{4}{3} σ D^{4}, = σ \int_{- D}^{D} \int_{- D}^{D} (u + v) v d u d v = \frac{4}{3} σ D^{4} .

The mixed moments involving

x_{3}

vanish. Hence the inertia tensor is

I = \frac{4 σ D ^{4}}{3} 1 - 1 0 - 1 20 003 = \frac{M D ^{2}}{3} 1 - 1 0 - 1 20 003 .

The off-diagonal entries are minus the product moment, not the product moment itself. This is the unit-shear case of the inertia tensor of a sheared rectangular lamina.

 Read the full article

 New to topics? Read the docs here!