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 Incoming links: Inertia tensor of a uniform cube

Past exam of the physics B course of the University of Cambridge / 2012 / Paper 1 / Section A / A5 / Solution 2026-10-20

For the inertia tensor of a uniform cube, reflection symmetry makes every off-diagonal element of the central inertia tensor vanish. Permutation of the three axes makes its diagonal elements equal. Since

M = 8 ρ a^{3}

and the mean of each squared coordinate in the cube is

a^{2} /3

,

I_{xx} = \int (y^{2} + z^{2}) d m = \frac{2}{3} M a^{2}, I = \frac{2}{3} M a^{2} 1 .

Every central axis is therefore a principal axis, including the body diagonal. The angular momentum is parallel to the angular velocity, and its magnitude is

J = \frac{16}{3} ρ a^{5} ω .

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