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Explicit integration can show that the moment of inertia of a Platonic solid (i.e., tetrahedron, cube, octahedron, dodecahedron, or icosahedron) of uniform density is the same around any axis passing through its center. The axis need not pass through a vertex, or midpoint of an edge, or center of a face! In tensor terms, the moment of inertia tensor is a constant times the Kronecker delta, just like for a sphere.

What is the complete class of solids with an isotropic moment of inertia, and why?

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  • $\begingroup$ A vast number of highly irregular shapes could satisfy this. Are you thinking of polyhedra or convex or smooth shapes? (I found Moments of inertia of Archimedean solids Frédéric Perrier.) $\endgroup$ Commented Nov 19, 2019 at 1:04
  • $\begingroup$ @KeithMcClary Thanks for the reference! It claims that “The condition of two axes of threefold or higher rotational symmetry crossing at the center of gravity is sufficient to produce an isotropic tensor of inertia”, although I don’t yet see why this is true. $\endgroup$
    – G. Smith
    Commented Nov 19, 2019 at 1:09
  • $\begingroup$ I didn’t have any particular shapes in mind. I was curious about whether there was a necessary and sufficient condition based on symmetry arguments. This paper gives a sufficient condition so that’s very helpful. $\endgroup$
    – G. Smith
    Commented Nov 19, 2019 at 1:12
  • $\begingroup$ I don’t see why a vast number of shapes would meet the conditions. But clearly there at least 13 in addition to the Platonic solids. $\endgroup$
    – G. Smith
    Commented Nov 19, 2019 at 1:15

1 Answer 1

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Start with any solid whatsoever. Choose a coordinate system $x,y,z$ in which its moment of inertia tensor is diagonal. The diagonal components are \begin{align} I_{xx} = I_0 - \sum_n m_n x_n^2 \\ I_{yy} = I_0 - \sum_n m_n y_n^2 \\ I_{zz} = I_0 - \sum_n m_n z_n^2 \end{align} where the $I_0$ term is the same for all three. Applying scale factors $a,b,c$ in the $x,y,z$ dimensions converts this to \begin{align} I_{xx}' = I_0' - a^2 \sum_n m_n x_n^2 \\ I_{yy}' = I_0' - b^2 \sum_n m_n y_n^2 \\ I_{zz}' = I_0' - c^2 \sum_n m_n z_n^2 \end{align} where $I_0'\neq I_0$ but is still the same for all three components. Clearly we can choose $a,b,c$ (or any two of them) so that $I_{xx}'=I_{yy}'=I_{zz}'$.

This shows that any object, however irregular its shape may be, is just two ordinary scale-factors away from having a perfectly isotropic moment of inertia tensor.

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    $\begingroup$ OK, this makes it clear that there are an infinite number of such shapes. Thanks! $\endgroup$
    – G. Smith
    Commented Nov 19, 2019 at 1:39

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