5.4 Torque and Angular Momentum for Systems of Particles

In a system of many particles, each particle with a position $\boldsymbol{r}_{i}$ and momentum $\boldsymbol{p}_{i}$ , the total angular momentum is the sum of each particle’s angular momentum.

\boldsymbol{L}=\sum_{i}\boldsymbol{L}_{i}=\sum_{i}\boldsymbol{r}_{i}\times% \boldsymbol{p}_{i}.

(5.50)

If the particles are subject to some torques, their angular momenta will change. The total angular momentum changes as

\frac{\operatorname{\mathrm{d}}\!\boldsymbol{L}}{\operatorname{\mathrm{d}}\!t}% =\sum_{i}\frac{\operatorname{\mathrm{d}}\!\boldsymbol{L}_{i}}{\operatorname{% \mathrm{d}}\!t}=\sum_{i}\frac{\operatorname{\mathrm{d}}\!}{\operatorname{% \mathrm{d}}\!t}(\boldsymbol{r}_{i}\times\boldsymbol{p}_{i})=\sum_{i}% \boldsymbol{r}_{i}\times\boldsymbol{F}_{i}=\sum_{i}\boldsymbol{\tau}_{i}.

(5.51)

Let’s look at the individual torques $\boldsymbol{\tau}_{i}$ more closely. The forces on each particle can be split into internal and external forces, $\boldsymbol{F}_{i}=\boldsymbol{F}_{i,\text{int}}+\boldsymbol{F}_{i,\text{ext}}$ . Internal forces are those from other particles, so $\boldsymbol{F}_{i,\text{int}}$ can be written as

\boldsymbol{F}_{i,\text{int}}=\sum_{j\neq i}\boldsymbol{F}_{j\to i}.

(5.52)

So the torques become

	$\displaystyle\sum_{i}\boldsymbol{\tau}_{i}$	$\displaystyle=\sum_{i}\boldsymbol{r}_{i}\times\left(\sum_{j\neq i}\boldsymbol{% F}_{j\to i}+\boldsymbol{F}_{i,\text{ext}}\right)$		(5.53)
		$\displaystyle=\sum_{i}\sum_{j\neq i}\boldsymbol{r}_{i}\times\boldsymbol{F}_{j% \to i}+\sum_{i}\boldsymbol{r}_{i}\times\boldsymbol{F}_{i,\text{ext}}.$		(5.54)

Looking at the first term more closely, we can use a similar argument to back in chapter 2 to show that it is zero. The sum will pairs of terms of the form $\boldsymbol{r}_{i}\times\boldsymbol{F}_{j\to i}+\boldsymbol{r}_{j}\times% \boldsymbol{F}_{i\to j}$ . By Newton III, these two forces are equal and opposite, $\boldsymbol{F}_{j\to i}=-\boldsymbol{F}_{i\to j}$ , so the pair of terms becomes

	$\displaystyle\boldsymbol{r}_{i}\times\boldsymbol{F}_{j\to i}+\boldsymbol{r}_{j% }\times\boldsymbol{F}_{i\to j}$	$\displaystyle=\boldsymbol{r}_{i}\times\boldsymbol{F}_{j\to i}-\boldsymbol{r}_{% j}\times\boldsymbol{F}_{j\to i}$		(5.55)
		$\displaystyle=(\boldsymbol{r}_{i}-\boldsymbol{r}_{j})\times\boldsymbol{F}_{j% \to i}.$		(5.56)

The vector $\boldsymbol{r}_{i}-\boldsymbol{r}_{j}$ is a vector which points from particle $j$ to particle $i$ , which is parallel to the force $\boldsymbol{F}_{j\to i}$ , therefore the whole sum is zero. Therefore the contribution to the torques comes only from the external forces. Internal forces do not contribute to the change in angular momentum.

\dot{\boldsymbol{L}}=\sum_{i}\boldsymbol{\tau}_{i}=\sum_{i}\boldsymbol{r}_{i}% \times\boldsymbol{F}_{i,\text{ext}}.

(5.57)

5.4.1 When is Change in Angular Momentum Equal to External Torque?

Is $\dot{\boldsymbol{L}}=\tau_{\text{ext}}$ ? Only in two situations: when we consider the torque about the origin of an inertial frame of reference, and when we consider the torque about the centre of mass (even if the centre of mass frame is not inertial). This is very important because it allows us to use the angular equivalent of Newton II in many situations. For example, if we are analysing a rigid body rolling down a hill, we can measure the angular momentum and torque from the centre of mass then we can still use $\dot{\boldsymbol{L}}=\boldsymbol{\tau}$ .