3.11: Torque and Rate of Change of Angular Momentum
- Page ID
- 8383
Theorem:
The rate of change of the total angular momentum of a system of particles is equal to the sum of the external torques on the system.
Thus:
\[ L = \sum_i {\bf r} _{i}\times p_{i}\tag{3.11.1}\label{eq:3.11.1} \]
\[ \therefore \qquad \dot{\bf L} = \sum_i \dot{\bf r}_{i}\times \dot{\bf p}_{i}\tag{3.11.2}\label{eq:3.11.2} \]
But the first term is zero, because \( \dot{\bf r}\) and \( {\bf p}_{i}\) are parallel.
Also
\[ \dot{\bf r}_{i} = {\bf F}_{i} + \sum {\bf F}_{ij}\tag{3.11.3}\label{eq:3.11.3} \]
\(\dot{\bf L}_{i} = \sum_i {\bf r}_{i} \times ({\bf r}_{i} + \sum_j {\bf F}_{ij}) =\sum_i {\bf r}_{i}\times {\bf F}_{i} + \sum_i {\bf r}_{i}\times \sum_j {\bf F}_{ii}\)
\( \therefore \qquad \sum_i r_{i}\times F_{i} + \sum_i r_{i}\times \sum_j F_{ii} \)
But \( \sum_i \sum_j {\bf F}_{ij} = 0 \) by Newton’s third law of motion, and so \( \sum_i \sum_j {\bf r}_{i} \times {\bf F}_{ij} = 0 \).
Also \(\sum_i {\bf r}_{i} \times {\bf F}_{i} = \boldsymbol\tau \) , and so we arrive at
\[ \dot{ \bf L} = \boldsymbol\tau \tag{3.11.4}\label{eq:3.11.4} \]
which was to be demonstrated.
Corollary: Law of Conservation of Angular Momentum
If the sum of the external torques on a system is zero, the angular momentum is constant.