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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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1 Answer 1

up vote 9 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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