3.7: Angular Momentum
- Page ID
- 8379
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Notation:
- \(\textbf{L}_{C}\)= angular momentum of system with respect to centre of mass C.
- \(\textbf{L}\) = angular momentum of system relative to some other origin O.
- \(\overline{\textbf{r}}\) = position vector of C with respect to O.
- \(\textbf{P}\) = linear momentum of system with respect to O.
- (The linear momentum with respect to C is, of course, zero.)
\[ \textbf{L} = \textbf{L}_{C} + \overline{\textbf{r}} \times \textbf{P} \tag{3.7.1}\label{eq:3.7.1} \]
Thus:
\[ \begin{align*} \textbf{L} &= \sum \textbf{r}_{i}\times \textbf{p}_{i} = \sum m_{i}(\textbf{r}_{i}\times \textbf{v}_{i}) = \sum m_{i}(\overline{\textbf{r}} + \textbf{r}_{i}^{\prime})\times(\overline{\textbf{v}} + \textbf{v}_{i}^{\prime}) \\[5pt] &=(\overline{\textbf{r}}\times \overline{\textbf{v}})\sum m_{i} + \overline{\textbf{r}}\times \sum m_{i}\textbf{v}_{i}^{\prime} + (\sum m_{i}\textbf{r}_{i}^{\prime}) \times \overline{\textbf{v}} + \sum \textbf{r}_{i}^{\prime} \times \textbf{p}_{i}^{\prime} \\[5pt] &=M(\overline{\textbf{r}}\times \overline{\textbf{v}}) +\overline{\textbf{r}}\times 0 + 0 \times \overline{\textbf{v}} + \textbf{L}_{C} \end{align*} \nonumber \]
therefore
\[\qquad \textbf{L} =\textbf{L}_{C} + \overline{\textbf{r}} \times \textbf{P} \nonumber \]
A hoop of radius a rolling along the ground (Figure III.6):
The angular momentum with respect to C is LC = \(I_{C \omega}\) where \( I_{C}\) is the rotational inertia about C. The angular momentum about O is therefore
\[ I = I_{C}\omega+M\overline{v}a=I_{C}\omega+Ma^{2}\omega=(I_{C}+Ma^{2})=I\omega \nonumber \]
where
\[ I = I_{C}+Ma^{2} \nonumber \]
is the rotational inertia about O.