2.2 Centre of Mass

2.2.1 Centre of Mass of Two Objects

Some times it is useful to consider the centre of mass of a system, which is defined as the average position of all the objects in the system. For the system of two objects, this is calculated as

\boldsymbol{r}_{\text{COM}}=\frac{m_{A}\boldsymbol{r}_{A}+m_{B}\boldsymbol{r}_% {B}}{m_{A}+m_{B}}.

(2.6)

This defines a position vector which points to the centre of mass of the system. By differentiating this vector with respect to time, we can get the velocity of the centre of mass

\boldsymbol{v}_{\text{COM}}=\frac{\operatorname{\mathrm{d}}\!\boldsymbol{r}_{% \text{COM}}}{\operatorname{\mathrm{d}}\!t}=\frac{m_{A}\frac{\operatorname{% \mathrm{d}}\!\boldsymbol{r}_{A}}{\operatorname{\mathrm{d}}\!t}+m_{B}\frac{% \operatorname{\mathrm{d}}\!\boldsymbol{r}_{B}}{\operatorname{\mathrm{d}}\!t}}{% m_{A}+m_{B}}=\frac{m_{A}\boldsymbol{v}_{A}+m_{B}\boldsymbol{v}_{B}}{m_{A}+m_{B% }}.

(2.7)

This is just the total momentum of the two objects divided by their total mass, and we found before that the total momentum was conserved.

(m_{A}+m_{B})\boldsymbol{v}_{\text{COM}}=m_{A}\boldsymbol{v}_{A}+m_{B}% \boldsymbol{v}_{B}=\text{constant}.

(2.8)

From these definitions we can see that equation 2.2 is the accleration of the centre of mass multiplied by the total mass, which by Newton’s second law is the force on the centre of mass. This implies that if the resultant force on the centre of mass is 0, then the centre of mass moves with constant velocity, just like a single object following Newton’s first law, and the total linear momentum is conserved.

Example 2.1.

For a system of two objects, show that the centre of mass is always located on the line that connects the two objects.

2.2.2 Centre of Mass of $N$ Objects

For a general system of $N$ objects, we define the total mass and total momentum as

$\displaystyle M$	$\displaystyle=m_{1}+m_{2}+\dots+m_{N}$	(2.9)
	$\displaystyle=\sum_{i=1}^{N}m_{i}$	(2.10)
$\displaystyle\boldsymbol{P}$	$\displaystyle=\sum_{i=1}^{N}m_{i}\boldsymbol{v_{i}}.$	(2.11)

The net force on each individual part of the system can be written as the sum of any external forces and internal forces from all the other parts.

\boldsymbol{F}_{i,\text{net}}=\boldsymbol{F}_{i,\text{ext}}+\sum_{j\neq i}^{N}% \boldsymbol{F}_{j\to i}.

(2.12)

Let’s now add all the forces together to get the net force on the centre of mass:

\boldsymbol{F}_{\text{net}}=\sum_{i=1}^{N}\boldsymbol{F}_{i,\text{net}}=% \underbrace{\sum_{i=1}^{N}\boldsymbol{F}_{i,\text{ext}}}_{=\boldsymbol{F}_{% \text{ext}}}+\underbrace{\sum_{i=1}^{N}\sum_{j\neq i}^{N}\boldsymbol{F}_{j\to i% }}_{=0}.

(2.13)

The second term on the far right-hand side is 0 by Newton’s third law (you can prove this by induction).

Now we define the centre of mass position, velocity, and acceleration as the weighted average of each of the individual particles’ quantities.

Definition 2.1.

Consider a system of $N$ objects. The centre of mass is is a vector function defined as

\boldsymbol{r}_{\text{COM}}=\frac{1}{M}\sum_{i=1}^{N}m_{i}\boldsymbol{r}_{i}.

(2.14)

The centre of mass velocity is defined as

\boldsymbol{v}_{\text{COM}}=\frac{\operatorname{\mathrm{d}}\!\boldsymbol{r}_{% \text{COM}}}{\operatorname{\mathrm{d}}\!t}=\frac{1}{M}\sum_{i=1}^{N}m_{i}\frac% {\operatorname{\mathrm{d}}\!\boldsymbol{r}_{i}}{\operatorname{\mathrm{d}}\!t}=% \frac{1}{M}\sum_{i=1}^{N}m_{i}\boldsymbol{v}_{i}=\frac{\boldsymbol{P}}{M},

(2.15)

and the centre of mass acceleration is defined as

\boldsymbol{a}_{\text{COM}}=\frac{\operatorname{\mathrm{d}}\!^{2}\boldsymbol{r% }_{\text{COM}}}{\operatorname{\mathrm{d}}\!t^{2}}=\frac{1}{M}\sum_{i=1}^{N}m_{% i}\frac{\operatorname{\mathrm{d}}\!^{2}\boldsymbol{r}_{i}}{\operatorname{% \mathrm{d}}\!t^{2}}=\frac{1}{M}\sum_{i=1}^{N}m_{i}\boldsymbol{a}_{i}=\frac{1}{% M}\sum_{i=1}^{N}\boldsymbol{F}_{i,\text{net}}=\frac{\boldsymbol{F}_{\text{ext}% }}{M},

(2.16)

Notice that in the definition of centre of mass acceleration, we have shown that

M\boldsymbol{a}_{\text{COM}}=\boldsymbol{F}_{\text{ext}}.

(2.17)

This result is actually quite profound because it is what allows us to treat systems of particles as point masses themselves while ignoring all of the internal forces between the particles since they all cancel out. Also note that because the centre of mass momentum is equal to the total momentum, systems of many particles act as if all the mass was concentrated in a point at the centre of mass, moving with the the centre of mass velocity. Without these results, we could not apply the laws of mechanics as we have been learning them to macroscopic bodies!

If $\boldsymbol{F}_{\text{ext}}=0$ , i.e. the system is isolated and there are no external forces, then the centre of mass moves in a straight line with a constant velocity and the total linear momentum is conserved.

Since the centre of mass quantites are vectors, we can break them into components just like all the other vector quantites we have seen. For example, in 2D cartesian coordinates we have

\boldsymbol{r}_{\text{COM}}=\frac{1}{M}\sum_{i=1}^{N}m_{i}\boldsymbol{r}_{i}=% \frac{1}{M}\sum_{i=1}^{N}m_{i}(x_{i}\hat{\boldsymbol{\imath}}+y_{i}\hat{% \boldsymbol{\jmath}})=\frac{1}{M}\sum_{i=1}^{N}m_{i}x_{i}\hat{\boldsymbol{% \imath}}+\frac{1}{M}\sum_{i=1}^{N}m_{i}y_{i}\hat{\boldsymbol{\jmath}},

(2.18)

so we can define the weighted average along the $x$ and $y$ axes:

\boldsymbol{r}_{\text{COM}}=\bar{x}\hat{\boldsymbol{\imath}}+\bar{y}\hat{% \boldsymbol{\jmath}},\quad\bar{x}=\frac{1}{M}\sum_{i=1}^{N}m_{i}x_{i},\quad% \bar{y}=\frac{1}{M}\sum_{i=1}^{N}m_{i}y_{i}

(2.19)

2.2 Centre of Mass

2.2.1 Centre of Mass of Two Objects

Example 2.1.

2.2.2 Centre of Mass of N Objects

Definition 2.1.

2.2.2 Centre of Mass of $N$ Objects