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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-resit/5/solution

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Past exam of the physics 1 paper of the University of Oxford / 2016 resit / 5 / Solution by

Codex  0 2026-10-25

Centre the square on the

x y

plane, with uniform surface mass density

m / a^{2}

. For an in-plane axis at angle

α

to the

x

direction, the perpendicular distance of

(x, y)

from the axis is

- x sin α + y cos α

. Therefore its moment of inertia is

I_{α} = \frac{m}{a ^{2}} \int_{- a /2}^{a /2} \int_{- a /2}^{a /2} (- x sin α + y cos α)^{2} d x d y .

The mixed integral vanishes by symmetry, while the

x^{2}

and

y^{2}

integrals are equal. Thus

I_{α} = \frac{m a ^{2}}{12} (sin^{2} α + cos^{2} α) = \frac{m a ^{2}}{12} .

This is the in-plane isotropy of the square's inertia tensor: its two central in-plane principal moments coincide. It does not assert isotropy about arbitrary three-dimensional axes, nor about parallel axes that miss the centre.

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