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Angular Momentum Vector in Matrix Form

The two cross-products in Eq.$ \,$ (B.19) can be written out with the help of the vector analysis identityB.23

$\displaystyle \underline{x}\times (\underline{y}\times\underline{z}) \eqsp \underline{y}\cdot(\underline{z}^T\underline{x})-\underline{z}\cdot(\underline{x}^T\underline{y}).
$

This (or a direct calculation) yields, starting with Eq.$ \,$ (B.19),
$\displaystyle \underline{L}$ $\displaystyle =$ $\displaystyle m\, \underline{x}_m \times (\underline{\omega}\times\underline{x}_m)
\eqsp m\,\underline{\omega}\cdot(\underline{x}^T\underline{x}) - \underline{x}\cdot(\underline{x}^T\underline{\omega})$  
  $\displaystyle =$ $\displaystyle m\,\left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}- \underline{x}\underline{x}^T\right)\underline{\omega}$  
  $\displaystyle \isdef$ $\displaystyle \mathbf{I}\,\underline{\omega}
\protect$ (B.20)

where

$\displaystyle \mathbf{I}\underline{\omega}\eqsp
\left[\begin{array}{ccc}
I_{11} & I_{12} & I_{13}\\ [2pt]
I_{21} & I_{22} & I_{23}\\ [2pt]
I_{31} & I_{32} & I_{33}
\end{array}\right]
\left[\begin{array}{c} \omega_1 \\ [2pt] \omega_2 \\ [2pt] \omega_3\end{array}\right]
$

with $ I_{ii}=m\left(\sum_{j=1}^3x_j^2 - x_i^2\right)$ , and $ I_{ij}=-mx_ix_j$ , for $ i\ne j$ . That is,
$\displaystyle \mathbf{I}\eqsp m\left[\begin{array}{ccc}
x_2^2+x_3^2 & -x_1x_2 & -x_1x_3\\ [2pt]
-x_2x_1 & x_1^2+x_3^2 & -x_2x_3\\ [2pt]
-x_3x_1 & -x_3x_2 & x_1^2+x_2^2
\end{array}\right] \eqsp m\,\left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}- \underline{x}\underline{x}^T\right).
\protect$     (B.21)

The matrix $ \mathbf{I}$ is the Cartesian representation of the mass moment of inertia tensor, which will be explored further in §B.4.15 below.

The vector angular momentum of a rigid body is obtained by summing the angular momentum of its constituent mass particles. Thus,

$\displaystyle \underline{L}\eqsp \sum_i m_i \left(\left\Vert\,\underline{x}_i\,\right\Vert^2\mathbf{E}- \underline{x}_i\underline{x}_i^T\right)\underline{\omega}
\,\isdefs \, \mathbf{I}\,\underline{\omega}.
$

Since $ \underline{\omega}$ factors out of the sum, we see that the mass moment of inertia tensor for a rigid body is given by the sum of the mass moment of inertia tensors for each of its component mass particles.

In summary, the angular momentum vector $ \underline{L}$ is given by the mass moment of inertia tensor $ \mathbf{I}$ times the angular-velocity vector $ \underline{\omega}$ representing the axis of rotation.

Note that the angular momentum vector $ \underline{L}$ does not in general point in the same direction as the angular-velocity vector $ \underline{\omega}$ . We saw above that it does in the special case of a point mass traveling orthogonal to its position vector. In general, $ \underline{L}$ and $ \underline{\omega}$ point in the same direction whenever $ \underline{\omega}$ is an eigenvector of $ \mathbf{I}$ , as will be discussed further below (§B.4.16). In this case, the rigid body is said to be dynamically balanced.B.24


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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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