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 identity^B.23

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

This (or a direct calculation) yields, starting with Eq.(B.19),

$$\underline{L} = m\, \underline{x}_m \times (\underline{\omega}\times\underline{x}_m) \eqsp m\,\underline{\omega}\cdot(\underline{x}^T\underline{x}) - \underline{x}\cdot(\underline{x}^T\underline{\omega})$$

$$= m\,\left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}- \underline{x}\underline{x}^T\right)\underline{\omega}$$

$$\isdef \mathbf{I}\,\underline{\omega} \protect$$ (B.20)

where

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

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,

$$\mathbf{I}\eqsp m\left[\begin{array}{ccc} x_2^2+x_3^2 & -x_1x_2 & -x_1x_3\\ [2pt] -x_2x_1 & x_1^2+x_3^2 & -x_2x_3\\ [2pt] -x_3x_1 & -x_3x_2 & x_1^2+x_2^2 \end{array}\right] \eqsp m\,\left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}- \underline{x}\underline{x}^T\right). \protect$$ (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,

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

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