Free Books

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 \und...

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


$\displaystyle \mathbf{I}\underline{\omega}\eqsp
...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,
$\displaystyle \mathbf{I}\eqsp m\left[\begin{array}{ccc}
x_2^2+x_3^2 & -x_1x_2 &...
...line{x}\,\right\Vert^2\mathbf{E}- \underline{x}\underline{x}^T\right).
\protect$     (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\,...
\,\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

Next Section:
Simple Example
Previous Section:
Relation of Angular to Linear Momentum