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I have calculated the moment of inertia tensor of a cube about its center of mass: $I=\dfrac{1}{6}Mb^2\{1\}$ where $\{1\}$ is the identity matrix. So the principal moments of inertia are all 1 (1 is an eigenvalue of multiplicity 3).

When you attempt to calculate the eigenvectors to identify the principal axes, the result is that every vector is an eigenvector of the identity matrix, which is expected.

My textbook, however, states, "Thus, we find that, for the choice of the origin at the center of mass of the cube, the principal axes are perpendicular to the faces of the cube."

Why is this true? I thought that the principal axes were simply the eigenvectors of the diagonalized matrix, which would mean that every axis that passes through the center of mass of a cube is a principal axis.

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    $\begingroup$ Let's just say that given a chance, I'd prefer principal axes that are aligned with the axes of symmetry. These align pretty well. $\endgroup$ – John Dvorak Dec 14 '14 at 19:09
  • $\begingroup$ Could you explain the physical meaning of a principal axis? Does this justify an expectation that principal axes 'acknowledge' or 'recognize' symmetry? $\endgroup$ – user44816 Dec 14 '14 at 19:10
  • $\begingroup$ All you need is a set of orthogonal axes. Why not pick the ones that make all subsequent manipulations easy? $\endgroup$ – Carl Witthoft Dec 14 '14 at 20:57
  • $\begingroup$ Are you suggesting that any vector that passes through the center of mass is a principal axis, but that we choose the easy ones (1,0,0),(0,1,0),(0,0,1) $\endgroup$ – user44816 Dec 14 '14 at 21:42
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It is simply not true, as stated.

A cube is as perfectly balanced around its center of mass as a sphere is. You have shown it mathematically, and you are perfectly correct.

However, the principal axes may be chosen perpendicular to the faces of the cube. Of course, when the point is the center of mass, this is no better than any other choice, except for considerations of symmetry and mathematical convenience.

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