### 1 The centre of mass of a collection of point masses

Suppose we have a collection of masses located at a number of known points along a line. The centre of mass is the point where, for many purposes, all the mass can be assumed to be located.

For example, if two objects each of mass $m$ are placed at distances $1$ and $2$ units from a point $O$ , as shown in Figure 2a, then the total mass, $2m$ , might be assumed to be concentrated at distance $1.5$ units as shown in Figure 2b. This is the point where we could imagine placing a pivot to achieve a perfectly balanced system.

Figure 2 :

To think of this another way, if a pivot is placed at the origin $O$ , as on a see-saw, then the two masses at $x=1$ and $x=2$ together have the same turning effect or moment as a single mass $2m$ located at $x=1.5$ . This is illustrated in Figure 3.

Figure 3 :

Before we can calculate the position of the centre of mass of a collection of masses it is important to define the term ‘moment’ more precisely. Given a mass $M$ located a distance $d$ from $O$ , as shown in Figure 4, its moment about $O$ is defined to be

Figure 4 :

In words, the moment of the mass about $O$ , is the mass multiplied by its distance from $O$ . The units of moment will therefore be kg m if the mass is measured in kilogrammes and the distance in metres. (N.B. Unless specified otherwise these will be the units we shall always use.)

Calculate the moment of the mass about $O$ in each of the following cases.

1. 40 kg m
2. 50 kg m

Intuition tells us that a large moment corresponds to a large turning effect. A mass placed 8 metres from the origin has a smaller turning effect than the same mass placed 10 metres from the origin. This is, of course, why it is easier to rock a see-saw by pushing it at a point further from the pivot. Our intuition also tells us the side of the pivot on which the masses are placed is important. Those placed to the left of the pivot have a turning effect opposite to those placed to the right.

For a collection of masses the moment of the total mass located at the centre of mass is equal to the sum of the moments of the individual masses. This definition enables us to calculate the position of the centre of mass. It is conventional to label the $x$ coordinate of the centre of mass as $\stackrel{̄}{x}$ , pronounced ‘ $x$ bar’.

##### Key Point 1

The moment of the total mass located at the centre of mass is equal to the sum of the moments of the individual masses.

Objects of mass $m$ and $3m$ are placed at the locations shown in diagram (a). Find the distance $\stackrel{̄}{x}$ of the centre of mass from the origin $O$ as illustrated in diagram (b).

First calculate the sum of the individual moments:

$6×m+10×3m=36m$

The moment of the total mass about $O$ is $4m×\stackrel{̄}{x}$ .

The moment of the total mass is equal to the sum of the moments of the individual masses. Write down and solve the equation satisfied by $\stackrel{̄}{x}$ :

$36m=4m\stackrel{̄}{x}$ , so $\stackrel{̄}{x}=9$

So the centre of mass is located a distance 9 units along the $x$ -axis. Note that it is closer to the position of the $3m$ mass than to the position of the $1m$ mass (actually in the ratio $3:1$ ).

##### Example 2

Obtain an equation for the location of the centre of mass of two objects of masses ${m}_{1}$ and ${m}_{2}$ :

1. located at distances ${x}_{1}$ and ${x}_{2}$ respectively, as shown in Figure 5(a)
2. positioned on opposite sides of the origin as shown in Figure 5(b)

Figure 5

Referring to Figure 5(a) we first write down an expression for the sum of the individual moments:

${m}_{1}{x}_{1}+{m}_{2}{x}_{2}$

The total mass is ${m}_{1}+{m}_{2}$ and the moment of the total mass is $\left({m}_{1}+{m}_{2}\right)×\stackrel{̄}{x}$ .

The moment of the total mass is equal to the sum of the moments of the individual masses. The equation satisfied by $\stackrel{̄}{x}$ is

$\phantom{\rule{2em}{0ex}}\left({m}_{1}+{m}_{2}\right)\stackrel{̄}{x}={m}_{1}{x}_{1}+{m}_{2}{x}_{2}$ so $\stackrel{̄}{x}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}$

For the second case, as depicted in Figure 5(b), the mass ${m}_{1}$ positioned on the left-hand side has a turning effect opposite to that of the mass ${m}_{2}$ positioned on the right-hand side. To take account of this difference we use a minus sign when determining the moment of ${m}_{1}$ about the origin. This gives a total moment

$\phantom{\rule{2em}{0ex}}-\left({m}_{1}{x}_{1}\right)+\left({m}_{2}{x}_{2}\right)$

$\phantom{\rule{2em}{0ex}}\left(-{m}_{1}{x}_{1}+{m}_{2}{x}_{2}\right)=\left({m}_{1}+{m}_{2}\right)\stackrel{̄}{x}$ so $\stackrel{̄}{x}=\frac{-{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}$

However, this is precisely what would have been obtained if, when working out the moment of a mass, we use its coordinate (which takes account of sign) rather than using its distance from the origin.

The formula obtained in the Task can be generalised very easily to deal with the general situation of $n$ masses, ${m}_{1},{m}_{2},\dots ,{m}_{n}$ located at coordinate positions ${x}_{1}$ , ${x}_{2},\dots {x}_{n}$ and is given in Key Point 2.

##### Key Point 2

The centre of mass of individual masses ${m}_{1},{m}_{2},\dots ,{m}_{n}$ located at positions ${x}_{1}$ , ${x}_{2},\dots {x}_{n}$ is

$\stackrel{̄}{x}=\frac{{\sum }_{i=1}^{n}{m}_{i}{x}_{i}}{{\sum }_{i=1}^{n}{m}_{i}}$

Calculate the centre of mass of the 4 masses distributed as shown below.

Use Key Point 2 to calculate $\stackrel{̄}{x}$ :

$\stackrel{̄}{x}=\frac{\left(9\right)\left(-1\right)+\left(1\right)\left(2\right)+\left(5\right)\left(6\right)+\left(2\right)\left(8\right)}{9+1+5+2}=\frac{39}{17}$

The centre of mass is located a distance $\frac{39}{17}\approx 2.29$ units along the $x$ -axis from $O$ .

#### 1.1 Distribution of particles in a plane

If the particles are distributed in a plane then the position of the centre of mass can be calculated in a similar way.

Figure 6 :

Now we must consider the moments of the individual masses taken about the $x$ -axis and about the $y$ -axis. For example, in Figure 6, mass ${m}_{i}$ has a moment ${m}_{i}{y}_{i}$ about the $x$ -axis and a moment ${m}_{i}{x}_{i}$ about the $y$ -axis. Now we define the centre of mass at that point $\left(\stackrel{̄}{x},ȳ\right)$ such that the total mass $M={m}_{1}+{m}_{2}+\dots {m}_{n}$ placed at this point would have the same moment about each of the axes as the sum of the individual moments of the particles about these axes.

##### Key Point 3

The centre of mass of ${m}_{1},{m}_{2},\dots ,{m}_{n}$ located at $\left({x}_{1},{y}_{1}\right)$ , $\left({x}_{2},{y}_{2}\right),\dots \left({x}_{n},{y}_{n}\right)$ has coordinates $\left(\stackrel{̄}{x},ȳ\right)$ where

$\stackrel{̄}{x}=\frac{\underset{i=1}{\overset{n}{\sum }}{m}_{i}{x}_{i}}{\underset{i=1}{\overset{n}{\sum }}{m}_{i}},\phantom{\rule{2em}{0ex}}ȳ=\frac{\underset{i=1}{\overset{n}{\sum }}{m}_{i}{y}_{i}}{\underset{i=1}{\overset{n}{\sum }}{m}_{i}}$

Masses of 5 kg, 3 kg and 9 kg are located at the points with coordinates $\left(-1,1\right)$ , $\left(4,3\right)$ , and $\left(8,7\right)$ respectively. Find the coordinates of their centre of mass.

Use Key Point 3:

$\phantom{\rule{2em}{0ex}}\stackrel{̄}{x}=\frac{{\sum }_{i=1}^{3}{m}_{i}{x}_{i}}{{\sum }_{i=1}^{3}{m}_{i}}=\frac{5\left(-1\right)+3\left(4\right)+9\left(8\right)}{5+3+9}=\frac{79}{17}\approx 4.65$

$\phantom{\rule{2em}{0ex}}ȳ=\frac{5\left(1\right)+3\left(3\right)+9\left(7\right)}{17}=4.53$ .

Hence the centre of mass is located at the point $\left(4.65,4.53\right)$ .

##### Exercises
1. Find the $x$ coordinate of the centre of mass of 5 identical masses placed at $x=2$ , $x=5$ , $x=7$ , $x=9$ , $x=12$ .
2. Derive the formula for $ȳ$ given in Key Point 3.

1. $\stackrel{̄}{x}=7$