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Physics LibreTexts

3.11: Torque and Rate of Change of Angular Momentum


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.


\[ L = \sum_i {\bf r} _{i}\times p_{i}\tag{3.11.1}\label{eq:3.11.1} \]

\[ \therefore \qquad \dot{\bf L} = \sum_i \dot{\bf r}_{i}\times \dot{\bf p}_{i}\tag{3.11.2}\label{eq:3.11.2} \]

But the first term is zero, because \( \dot{\bf r}\) and \( {\bf p}_{i}\) are parallel.


\[ \dot{\bf r}_{i} = {\bf F}_{i} + \sum {\bf F}_{ij}\tag{3.11.3}\label{eq:3.11.3} \]

\(\dot{\bf L}_{i} = \sum_i {\bf r}_{i} \times ({\bf r}_{i} + \sum_j {\bf F}_{ij}) =\sum_i {\bf r}_{i}\times {\bf F}_{i} + \sum_i {\bf r}_{i}\times \sum_j {\bf F}_{ii}\)

\( \therefore \qquad \sum_i r_{i}\times F_{i} + \sum_i r_{i}\times \sum_j F_{ii} \)

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.