Positive Definiteness of the Moment of Inertia Tensor

From the form of the moment of inertia tensor introduced in Eq.(B.24)

$$\mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right), \protect$$

it is clear that $\mathbf{I}$ is symmetric. Moreover, for any normalized angular-velocity vector $\underline {\tilde {\omega }}$ we have

$$\begin{eqnarray*} I &=& \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}} \eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2 -\underline{\tilde{\omega}}^T\underline{x}\underline{x}^T\underline{\tilde{\omega}}\right) \eqsp m \left\Vert\,\underline{x}\,\right\Vert^2 \left[1 - \underline{\tilde{\omega}}^T\left(\frac{\;\underline{x}\;\underline{x}^T}{\underline{x}^T\underline{x}}\,\underline{\tilde{\omega}}\right)\right]\\ [5pt] &=& m \left\Vert\,\underline{x}\,\right\Vert^2\left[1 -\underline{\tilde{\omega}}^T{\cal P}_{\underline{x}}(\underline{\tilde{\omega}})\right] \ge 0 \end{eqnarray*}$$

since $\underline {\tilde {\omega }}$ is unit length, and projecting it onto any other vector can only shorten it or leave it unchanged. That is, $\vert\underline{\tilde{\omega}}^T{\cal P}_{\underline{x}}(\underline{\tilde{\omega}})\vert \le 1$ , with equality occurring for $\underline{x}=\alpha\underline{\tilde{\omega}}$ for any nonzero $\alpha\in\mathbb{R}$ . Zooming out, of course we expect any moment of inertia $I$ for a positive mass $m$ to be nonnegative. Thus, $\mathbf{I}$ is symmetric nonnegative definite. If furthermore $\underline {\tilde {\omega }}$ and $\underline {x}$ are not collinear, i.e., if there is any nonzero angle between them, then $\mathbf{I}$ is positive definite (and $I>0$ ). As is well known in linear algebra [332], real, symmetric, positive-definite matrices have orthogonal eigenvectors and real, positive eigenvalues. In this context, the orthogonal eigenvectors are called the principal axes of rotation. Each corresponding eigenvalue is the moment of inertia about that principal axis--the corresponding principal moment of inertia. When angular velocity vectors $\underline {\omega }$ are expressed as a linear combination of the principal axes, there are no cross-terms in the moment of inertia tensor--no so-called products of inertia.

The three principal axes are unique when the eigenvalues of $\mathbf{I}$ (principal moments of inertia) are distinct. They are not unique when there are repeated eigenvalues, as in the example above of a disk rotating about any of its diameters (§B.4.4). In that example, one principal axis, the one corresponding to eigenvalue $MR^2/2$ , was $\underline{e}_3$ (i.e., orthogonal to the disk and passing through its center), while any two orthogonal diameters in the plane of the disk may be chosen as the other two principal axes (corresponding to the repeated eigenvalue $MR^2/4$ ).

Symmetry of the rigid body about any axis $\underline{\omega}_s$ (passing through the origin) means that $\underline{\omega}_s$ is a principal direction. Such a symmetric body may be constructed, for example, as a solid of revolution.^B.26In rotational dynamics, this case is known as the symmetric top [272]. Note that the center of mass will lie somewhere along an axis of symmetry. The other two principal axes can be arbitrarily chosen as a mutually orthogonal pair in the (circular) plane orthogonal to the $\underline{\omega}_s$ axis, intersecting at the $\underline{\omega}_s$ axis. Because of the circular symmetry about $\underline{\omega}_s$ , the two principal moments of inertia in that plane are equal. Thus the moment of inertia tensor can be diagonalized to look like

$$\mathbf{I}= \left[\begin{array}{ccc} I_1 & 0 & 0\\ [2pt] 0 & I_2 & 0\\ [2pt] 0 & 0 & I_2 \end{array}\right],$$

where $I_1$ is the principal moment of inertia about $\underline{\omega}_s$ , and $I_2$ is the (twice repeated) principal moment of inertia about the two axes in the circular-symmetry plane. We saw in §B.4.5 (Perpendicular Axis theorem) that if the mass distribution is planar, then $I_1=2I_2$ .