5.6: Angular Momentum
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- 17394
In analogy with the definition of torque, \(\boldsymbol{\tau}=\boldsymbol{r} \times \boldsymbol{F}\) as the rotational counterpart of the force, we define the angular momentum \(\boldsymbol{L}\) as the rotational counterpart of momentum:
\[\boldsymbol{L}=\boldsymbol{r} \times \boldsymbol{p} \label{rp}\]
For a rigid body rotating around an axis of symmetry, the angular momentum is given by
\[\boldsymbol{L}=I \boldsymbol{\omega} \label{iomega}\]
where \(I\) is the moment of inertia of the body with respect to the symmetry axis around which it rotates. Equation \ref{iomega} also holds for a collection of particles rotating about a symmetry axis through their center of mass, as readily follows from 5.4.2 and \ref{rp}. However, it does not hold in general, as in general, \(\boldsymbol{L}\) does not have to be parallel to \(\boldsymbol{\omega}\). For the general case, we need to consider a moment of inertia tensor \(\boldsymbol{I}\) (represented as a \(3×3\) matrix) and write \(\boldsymbol{L}=\boldsymbol{I} \cdot \boldsymbol{\omega}\). We’ll consider this case in more detail in Section 7.3.