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If the principal moments of inertia of a rigid body are all equal, i.e., $$I_1 = I_2 = I_3,$$ does that imply that the mass-distribution $\rho(x)$ in the rigid body is spherically symmetric?

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No. The inertia tensor of any uniform density Platonic solid about its center is a constant times the identity matrix, the same form as that of a uniform density sphere about its center. Only the sphere has a spherically symmetric mass distribution.

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    $\begingroup$ A cube is a case and point of this. $\endgroup$ – ZeroTheHero Mar 15 '17 at 21:25
  • $\begingroup$ @ZeroTheHero -- That's exactly what the initial version of my answer said. However, a cube is just one of the five regular 3D solids (the Platonic solids), all of which have a non-spherical mass distribution and all of which have the same moment of inertia about any axis passing through the center of mass. $\endgroup$ – David Hammen Mar 16 '17 at 2:51
  • $\begingroup$ Upon reading the question I immediately thought about the cube since this example was fresh in my memory. I actually had not realized that all uniform Platonic solids had this property. Thanks for your (more general) answer. $\endgroup$ – ZeroTheHero Mar 16 '17 at 3:00
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I think that you mean by $\rho(x)$ the mass density and that you by $x$ mean the vector $x$, so $\rho(x)$ is the mass density at $x$. Only in the case of a sphere which has the same principle moments of inertia, $\rho(x)$ is spherically symmetric. If $\rho(x)$ has whatever kind of relation to $r$, the magnitude of the vector $x$ (say $\rho(x)=r^2$), $\rho(x)$ is spherically symmetric.

Now consider the same sphere which maximum r to fit in a cube. If this cube has the same principle moments of inertia, the parts of the cube that stick out of the sphere have to have the same mass. If we assume that in those pieces of mass that stick out $\rho(x)$ has the same kind of relation to $r$ ($r\gt R$, where $R$ is the radius of the sphere) as inside the sphere, it is clear that $\rho(x)$ for those pieces of mass sticking out the sphere isn't spherically symmetric, which means that for a cube with equal principal moments of inertia, $\rho(x)$ isn't sherically symmetric (while it is for the sphere that is contained in the cube).

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    $\begingroup$ It is not clear what point you are making. That the mass distribution for a cube is not spherically symmetric? Isn't that rather obvious? $\endgroup$ – sammy gerbil Mar 17 '17 at 2:52

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