Mass Moment of Inertia Tensor

As derived in the previous section, the moment of inertia tensor, in 3D Cartesian coordinates, is a three-by-three matrix $\mathbf{I}$ that can be multiplied by any angular-velocity vector to produce the corresponding angular momentum vector for either a point mass or a rigid mass distribution. Note that the origin of the angular-velocity vector $\underline {\omega }$ is always fixed at $\underline {0}$ in the space (typically located at the center of mass). Therefore, the moment of inertia tensor $\mathbf{I}$ is defined relative to that origin.

The moment of inertia tensor can similarly be used to compute the mass moment of inertia for any normalized angular velocity vector $\underline{\tilde{\omega}}=\underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert$ as

$$I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}. \protect$$ (B.22)

Since rotational energy is defined as $(1/2)I\omega^2$ (see Eq.(B.7)), multiplying Eq.(B.22) by $\omega^2$ gives the following expression for the rotational kinetic energy in terms of the moment of inertia tensor:

$$E_R \eqsp \frac{1}{2}\, \underline{\omega}^T\mathbf{I}\,\underline{\omega} \protect$$ (B.23)

We can show Eq.(B.22) starting from Eq.(B.14). For a point-mass $m$ located at $\underline {x}$ , we have

$$\begin{eqnarray*} I &=& m \left\Vert\,\underline{x}-(\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}\,\right\Vert^2\\ [5pt] &=& m \left[\underline{x}^T-(\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}^T\right] \left[\underline{x}-(\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}\right]\\ [5pt] &=& m \left[\underline{x}^T\underline{x}- \underline{x}^T(\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}-(\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}^T\underline{x} +(\underline{\tilde{\omega}}^T\underline{x})^2\underline{\tilde{\omega}}^T\underline{\tilde{\omega}}\right]\\ [5pt] &=& m \left[\underline{x}^T\underline{x}- (\underline{\tilde{\omega}}^T\underline{x})^2\right] \eqsp m \left[\left\Vert\,x\,\right\Vert^2 - \underline{\tilde{\omega}}^T\underline{x}\underline{x}^T\underline{\tilde{\omega}}\right]\\ [5pt] &\eqsp & \underline{\tilde{\omega}}^T \left[m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right)\right]\underline{\tilde{\omega}}\\ [5pt] &\isdef & \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}} \end{eqnarray*}$$

where again $\mathbf{E}$ denotes the three-by-three identity matrix, and

$$\mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right), \protect$$ (B.24)

which agrees with Eq.(B.20). Thus we have derived the moment of inertia $I$ in terms of the moment of inertia tensor $\mathbf{I}$ and the normalized angular velocity $\underline {\tilde {\omega }}$ for a point-mass $m$ at $\underline {x}$ .

For a collection of $N$ masses $m_i$ located at $\underline{x}_i\in\mathbb{R}^3$ , we simply sum over their masses to add up the moments of inertia:

$$\mathbf{I}\eqsp \sum_{i=1}^N m_i \left(\left\Vert\,\underline{x}_i\,\right\Vert^2\mathbf{E} -\underline{x}_i\underline{x}_i^T\right)$$

Finally, for a continuous mass distribution, we integrate as usual:

$$\mathbf{I}\eqsp \frac{1}{M}\int_V \rho(\underline{x}) \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E} -\underline{x}\underline{x}^T\right)\,dV$$

where $M=\int_V\rho(\underline{x})dV$ is the total mass.