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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

$\displaystyle I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}. \protect$ (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:

$\displaystyle E_R \eqsp \frac{1}{2}\, \underline{\omega}^T\mathbf{I}\,\underline{\omega} \protect$ (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

$\displaystyle \mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right), \protect$ (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{\bf R}^3$ , we simply sum over their masses to add up the moments of inertia:

$\displaystyle \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:

$\displaystyle \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.



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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-06-11 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA