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3.11: Torque and Rate of Change of Angular Momentum

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

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.