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According to wikipedia, the inertia tensor of an ellipsoid with semi-axes $a,b,c$ and mass $m$ is

$$\left[\begin{array}{ccc} \frac{m}{5}(b^2+c^2)&0&0\\ 0&\frac{m}{5}(a^2+c^2)&0\\ 0&0&\frac{m}{5}(a^2+b^2)\\ \end{array}\right].$$

If you create an arbitrary 3x3 positive diagonal matrix and try to solve for the $a,b,c$, it's very easy to wind up with imaginary dimensions. If I try to place separate point masses, I seem to run into the same problem.

Does that mean that the tensor doesn't represent a physically possible distribution of mass, or just not a uniform density solid? Intuitively, at least, it seems that it must be impossible for an inertia tensor to a have a single large value and two small values since a single point mass with a non-zero radius will always affect two dimensions equally and an ring of infinitesimal height still leaves the two minor dimensions with half the momentum of the large principal axis.

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up vote 10 down vote accepted

One may prove for an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body) that the three moments of inertia $I_x$, $I_y$, and $I_z$, around the three principal axes (which we will call $x$, $y$, and $z$) satisfy the triangle inequality,

$$\tag{1} I_x +I_y ~\geq~ I_z, \qquad I_y +I_z ~\geq~ I_x, \qquad I_z +I_x ~\geq~ I_y. $$

In other words, if a semi-positive definite symmetric real $3\times 3$ matrix with non-negative eigenvalues $I_x$, $I_y$, and $I_z$ does not satisfy the triangle inequality (1), it doesn't represent a physically possible distribution of mass. (The proof follows straightforwardly by writing down the definition of moment of inertia.)

Moreover, one may show that such three eigenvalues $I_x$, $I_y$, and $I_z$ that satisfy (1) may be reproduced by a solid ellipsoid with a unique choice of non-negative semi-axes $a$, $b$, and $c$ (unique up to the scaling of the total mass $m$).

$$ \frac{2}{5}m a^2~=~I_y +I_z -I_x~\geq~0, $$ $$ \frac{2}{5}m b^2~=~I_z +I_x -I_y~\geq~0, $$ $$ \tag{2} \frac{2}{5}m c^2~=~I_x +I_y -I_z~\geq~0.$$

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Perfect. Thank-you. – JCooper Jan 3 '13 at 19:23

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