How does equal moments of inertia of a rigid body affect mass-distribution $\rho(x)$?

Question

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?

Answer 1

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.

Answer 2

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).