Angular Momentum Vector

Like linear momentum, angular momentum is fundamentally a vector in $ {\bf R}^3$. The definition of the previous section suffices when the direction does not change, in which case we can focus only on its magnitude $ L=I\omega$.

More generally, let $ \underline{x}_m\in{\bf R}^3$ denote the 3-space coordinates of a point-mass $ m$, and let $ \underline{v}_m=\dot{\underline{x}}_m$ denote its velocity in $ {\bf R}^3$. Then the instantaneous angular momentum vector of the mass relative to the origin (not necessarily rotating about a fixed axis) is given by

$\displaystyle \underline{L}\isdefs m\, \underline{x}_m \times \underline{v}_m \eqsp m\, \underline{x}_m \times (\underline{\omega}\times\underline{x}_m) \protect$ (B.19)

where $ \times$ denotes the vector cross product, discussed in §B.4.12 above. The identity $ \underline{v}_m=\underline{\omega}\times\underline{x}_m$ was discussed at Eq.$ \,$(B.17).

For the special case in which $ \underline{v}_m$ is orthogonal to $ \underline{x}_m$, as in Fig.B.4, we have that $ x_m\times\underline{v}_m$ points, by the right-hand rule, in the direction of the angular velocity vector $ \underline{\omega}$ (up out of the page), which is satisfying. Furthermore, its magnitude is given by

$\displaystyle \left\Vert\,\underline{L}\,\right\Vert \eqsp m\left\Vert\,\underl...
...ft\Vert\,\underline{v}_m\,\right\Vert
\eqsp mRv
\eqsp mR^2\omega
\eqsp I\omega
$

which agrees with the scalar case.

In the more general case of an arbitrary mass velocity vector $ \underline{v}_m$, we know from §B.4.12 that the magnitude of $ \underline{x}_m\times\underline{v}_m$ equals the product of the distance from the axis of rotation to the mass, i.e., $ \vert\vert\,\underline{x}_m\,\vert\vert $, times the length of the component of $ \underline{v}_m$ that is orthogonal to $ \underline{x}$, i.e., $ \vert\vert\,\underline{v}_m\,\vert\vert \sin(\theta)$, as needed.

It can be shown that vector angular momentum, as defined, is conserved.B.22 For example, in an orbit, such as that of the moon around the earth, or that of Halley's comet around the sun, the orbiting object speeds up as it comes closer to the object it is orbiting. (See Kepler's laws of planetary motion.) Similarly, a spinning ice-skater spins faster when pulling in arms to reduce the moment of inertia about the spin axis. The conservation of angular momentum can be shown to result from the principle of least action and the isotrophy of space [270, p. 18].

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

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

where

$\displaystyle \mathbf{I}\underline{\omega}\eqsp
\left[\begin{array}{ccc}
I_{1...
...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\,...
...e{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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Mass Moment of Inertia Tensor
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Angular Momentum