📂Classical Mechanics

Angular Momentum of Particle Systems

Formulas

The torque of a particle system equals the sum of the torques of each particle.

$$ \mathbf{N}=\frac{ d \mathbf{L}}{ d t }=\sum \limits _{i=1} ^{n} \mathbf{r}_{i}\times \mathbf{F}_{i} $$

Derivation¹

The linear momentum of a particle system was defined as the sum of the linear momenta of each particle. Similarly, the angular momentum of a particle system is defined as the sum of the angular momenta of each particle.

$$ \mathbf{L}=\sum \limits _{i=1} ^{n} (\mathbf{r}_{i}\times \mathbf{p}_{i}) $$

Since torque is the rate of change of angular momentum, the torque of a particle system is as follows.

$$ \begin{align*} \mathbf{N} &= \frac{d \mathbf{L}}{dt} \\ &= \sum \limits _{i=1} ^{n}(\mathbf{v}_{i}\times \mathbf{p}_{i}) + \sum \limits _{i=1} ^{n}(\mathbf{r}_{i}\times m_{i}\mathbf{a}_{i}) \end{align*} $$

Given $\mathbf{p}_{i}=m_{i}\mathbf{v}_{i}$, the cross product of the first term on the right-hand side equals $\mathbf{0}$. Moreover, since $m\mathbf{a}_{i}$ represents all the forces acting on particle $i$, the above equation is as follows.

$$ \begin{align} \mathbf{N} &= \sum \limits _{i=1} ^{n} \left[ \mathbf{r}_{i} \times \left( \mathbf{F}_{i}+\sum \limits _{j=1} ^{n} \mathbf{F}_{ij}\right) \right] \nonumber \\ &= \sum \limits _{i=1} ^{n} \mathbf{r}_{i}\times \mathbf{F}_{i}+ \sum \limits _{i=1} ^{n}\sum \limits _{j=1} ^{n}\mathbf{r}_{i}\times \mathbf{F}_{ij} \label{eq1} \end{align} $$

Upon arranging the second term, it can be confirmed that it becomes $\mathbf{0}$. When each term under the two summation signs is paired as follows, the following can be observed.

$$ \begin{equation} (\mathbf{r}_{i}\times \mathbf{F}_{ij})+(\mathbf{r}_{j}\times \mathbf{F}_{ji}) \label{eq2} \end{equation} $$

Let’s assume that $\mathbf{r}_{ij}=\mathbf{r}_{j}-\mathbf{r}_{i}$. Furthermore, since $\mathbf{F}_{ij}$ and $\mathbf{F}_{ji}$ are in action-reaction pairs, $\mathbf{F}_{ij}=-\mathbf{F}_{ji}$ is valid. Therefore, $(2)$ is as follows.

$$ \begin{align*} (\mathbf{r}_{i}\times \mathbf{F}_{ij})+(\mathbf{r}_{j}\times \mathbf{F}_{ji}) &= (\mathbf{r}_{i}\times \mathbf{F}_{ij})-(\mathbf{r}_{j}\times \mathbf{F}_{ij}) \\ &= -\mathbf{r}_{ij}\times \mathbf{F}_{ij} \end{align*} $$

Since both $\mathbf{r}_{ij}$ and $\mathbf{F}_{ij}$ are vectors located on the same line, their cross product equals $\mathbf{0}$. Therefore, the second term of $(1)$ is all $\mathbf{0}$, leading to the following equation.

$$ \mathbf{N}=\frac{ d \mathbf{L}}{ d t }=\sum \limits _{i=1} ^{n} \mathbf{r}_{i}\times \mathbf{F}_{i} $$

Thus, it can be seen that the torque of a particle system equals the sum of the torques of each particle.

Angular Momentum about the Center of Mass

The angular momentum can also be expressed about the center of mass. Let’s represent each position vector as shown above, towards the center of mass.

$$ \begin{equation} \mathbf{r}_{i} = \mathbf{r}_{cm} + \overline{\mathbf{r}}_{i} \end{equation} $$

Differentiating it with respect to time yields the following.

$$ \mathbf{v}_{i}=\mathbf{v}_{cm}+\overline{\mathbf{v}}_{i} $$

$\mathbf{v}_{cm}$ is the velocity of the center of mass, and $\mathbf{v}^{\prime}_{i}$ is the relative velocity of each particle with respect to the center of mass. Hence, the angular momentum of the particle system is calculated as follows.

$$ \begin{align*} \mathbf{L} &= \sum \limits _{i=1} ^{n}(\mathbf{r}_{cm}+\overline{\mathbf{r}}_{i})\times m_{i}(\mathbf{v}_{cm}+\overline{\mathbf{v}}_{i}) \\ &= \sum \limits _{i=1} ^{n}(\mathbf{r}_{cm}\times m_{i}\mathbf{v}_{cm}) + \sum \limits _{i=1} ^{n}(\mathbf{r}_{cm} \times m_{i}\overline{\mathbf{v}}_{i}) \\ &\quad +\sum \limits _{i=1} ^{n}(\overline{\mathbf{r}}_{i}\times m_{i}\mathbf{v}_{cm}) + \sum \limits _{i=1} ^{n}(\overline{\mathbf{r}}_{i}\times m_{i}\overline{\mathbf{v}}_{i}) \\ &= \mathbf{r}_{cm}\times \left(\sum \limits _{i=1} ^{n} m_{i }\right)\mathbf{v}_{cm} +\mathbf{r}_{cm} \times \sum \limits _{i=1} ^{n} m_{i}\overline{\mathbf{v}}_{i} \\ &\quad +\left(\sum \limits _{i=1} ^{n}m_{i}\overline{\mathbf{r}}_{i} \right)\times\mathbf{v}_{cm} + \sum \limits _{i=1} ^{n}(\overline{\mathbf{r}}_{i}\times m_{i}\overline{\mathbf{v}}_{i}) \end{align*} $$

Here, the third term becomes $(3)$ and is confirmed by the following calculation, based on the definition of the center of mass.

$$ \begin{align*} \sum \limits _{i=1} ^{n} m_{i}\overline{\mathbf{r}}_{i} &= \sum \limits _{i=1} ^{n} m_{i}(\mathbf{r}_{i}-\mathbf{r}_{cm}) \\ &= \sum \limits _{i=1} ^{n} m_{i}\mathbf{r}_{i}-m\mathbf{r}_{cm} \\ &= \mathbf{0} \end{align*} $$

$m$ is the total mass of the particle system. Also, differentiating the above equation with respect to time gives the following equation.

$$ \sum \limits _{i=1} ^{n} m_{i}\overline{\mathbf{v}}_{i}=\sum \limits_{i=1}^{n}m_{i}\mathbf{v}_{i}-m\mathbf{v}_{cm}= \mathbf{0} $$

Thus, the second term of $\mathbf{L}$ is also $\mathbf{0}$. Finally, $\mathbf{L}$ is as follows.

$$ \mathbf{L}=\mathbf{r}_{cm}\times m \mathbf{v}_{cm} + \sum \limits _{i=1} ^{n} \overline{\mathbf{r}}_{i}\times m_{i}\overline{\mathbf{v}}_{i} $$

Here, the first term can be considered as the angular momentum of the center of mass, and the second term as the total angular momentum of each particle relative to the center of mass. Understanding angular momentum in terms of terms related to the center of mass and those relative to it provides valuable insights in many areas of physics.