📂Classical Mechanics

Center of Mass and Linear Momentum of a Particle System

Definition

A system of particles is referred to as a particle system.

Description ¹

When the position vectors of particles with masses $m_{1}$, $m_2$, $\cdots$, and $m_{n}$ are $\mathbf{r}_{1}$, $\mathbf{r}_{2}$, $\cdots$, and $\mathbf{r}_{n}$, respectively, the center of mass of this particle system is defined as follows.

$$\mathbf{r}_{cm}=\frac{m_{1}\mathbf{r}_{1}+m_{2}\mathbf{r}_{2}+\cdots + m_{n}\mathbf{r}_{n}}{m_{1}+ m_{2}+ \cdots+ m_{n}}=\frac{\sum m_{i}\mathbf{r}_{i}}{m}$$

Here, $m=\sum \limits_{i}m_{i}$ denotes the total mass of the particle system. The subscript $cm$ stands for “center of mass”. Thus, the velocity of the center of mass is naturally defined as follows.

$$\begin{equation} \mathbf{v}_{cm}=\frac{m_{1}\mathbf{v}_{1}+m_{2}\mathbf{v}_{2}+\cdots + m_{n}\mathbf{v}_{n}}{m_{1}+ m_{2}+ \cdots+ m_{n}}=\frac{\sum m_{i}\mathbf{v}_{i}}{m} \label{velocity-of-cm} \end{equation}$$

In a three-dimensional Cartesian coordinate system, if $\mathbf{r}_{i}=x_{i}\hat{\mathbf{x}}+y_{i}\hat{\mathbf{y}}+z_{i}\hat{\mathbf{z}}$, then the mass center for each coordinate is as follows.

$$x_{cm}=\frac{\sum m_{i}x_{i} }{m},\quad y_{cm}=\frac{\sum m_{i}y_{i} }{m},\quad z_{cm}=\frac{\sum m_{i}z_{i} }{m}$$

The linear momentum of this particle system is naturally defined as the sum of the momentum of each particle.

$$\mathbf{p}=\sum \mathbf{p}_{i}=\sum m_{i}\mathbf{v}_{i}$$

Then, it can be understood that, according to $(1)$, the linear momentum of the particle system can be expressed as the product of the total mass of the particle system and the velocity of its mass center.

$$\mathbf{p}=\sum m_{i}\mathbf{v}_{i}=m\mathbf{v}_{cm}$$

Now, let’s assume the forces acting on each of these particles from an external source are $\mathbf{F}_{1}$, $\mathbf{F}_{2}$, $\cdots$, and $\mathbf{F}_{n}$. Also, let’s define the force experienced by particle $i$ due to particle $j$ as $\mathbf{F}_{ij}$. Then, the equation of motion for particle $i$ is as follows.

$$\mathbf{F}_{i} + \sum \limits_{j=1}^{n}\mathbf{F}_{ij}=m_{i}\ddot{\mathbf{r}_{i}}=\dot{\mathbf{p}_{i}}$$

Therefore, summing up all the forces on the entire particle system yields the following.

$$\sum \limits_{i=1}^{n}\mathbf{F}_{i}+\sum \limits _{i=1}^{n}\sum \limits_{j=1}^{n}\mathbf{F}_{ij}=\sum \limits_{i=1}^{n}\dot{\mathbf{p}_{i}}$$

Here, a particle does not exert any force on itself, so $\mathbf{F}_{ii}=\mathbf{0}$ equals zero. Also, in the second term of the equation, $\mathbf{F}_{ij}$ and $\mathbf{F}_{ji}$ represent the forces exerted by particle $i$ on particle $j$ and by particle $j$ on particle $i$, respectively. According to the law of action and reaction, since their magnitudes are the same but their directions are opposite, when added together, they equal $\mathbf{0}$. Thus, the second term is $\mathbf{0}$. Therefore, the overall equation of motion for the particle system is as follows.

$$\sum \limits _{i=1} ^{n} \mathbf{F}_{i}=\dot{\mathbf{p}_{i}}=m\mathbf{a}_{cm}$$

In other words, the acceleration of the center of mass in a particle system is the same as the acceleration of a single particle with the total mass of the system when subjected to the total external force acting on the system.