Only external forces exert torque upon a center of mass

up:: Torque

Only external torques affect a center of mass’s angular momentum:

$$\frac{d\vec{L}}{dt}=\sum _i{\vec{\tau }}_i^{(e)}$$

Proof

It is assumed that the origin is at a point $O$, and all positions are measured with respect to $O$.

$$\frac{d\vec{L}}{dt}=\sum _i\left(\underset{=0}{\underbrace{{\dot{\vec{r}}}_i\times {\vec{p}}_i}}+{\vec{r}}_i\times {\dot{\vec{p}}}_i\right)$$

To each particle $i$, this amounts to Newton’s Second Law:

$${\dot{\vec{p}}}_i=\sum _{j\ne i}{\vec{F}}_{j\to i}+{\vec{F}}_i^{(e)}$$

which results into

$$\begin{array}{rl}\frac{d\vec{L}}{dt}& =\sum _i{\vec{r}}_i\times \left(\sum _{j\ne i}{\vec{F}}_{j\to i}+{\vec{F}}_i^{(e)}\right)\\ & =\sum _{\begin{array}{c}i,j\\ i\ne j\end{array}}{\vec{r}}_i\times {\vec{F}}_{j\to i}+\sum _i{\vec{r}}_i\times {\vec{F}}_i^{(e)}\end{array}$$

By Newton’s Third Law, we have that ${\vec{F}}_{j\to i}=-{\vec{F}}_{i\to j}$, which will yield terms of the form

$$\sum _{\begin{array}{c}i,j\\ i\ne j\end{array}}({\vec{r}}_i-{\vec{r}}_j)\times {\vec{F}}_{i\to j}=0$$

since they are parallel vectors.