Centre Of Mass

We shall first see what the centre of mass of a system of particles is and then discuss its significance. For simplicity we shall start with a two particle system. We shall take the line joining the two particles to be the x- axis.

Fig.1

Let the distances of the two particles be $x_1$ and $x_2$ respectively from some origin O. Let $m_1$ and $m_2$ be respectively the masses of the two particles. The centre of mass of the system is that point C which is at a distance X from O, where X is given by

$X=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

In Eq. (7.1), X can be regarded as the massweighted mean of $x_1$ and $x_2$. If the two particles have the same mass $m_1$ = $m_2$ = m, then

$X=\frac{mx_1+mx_2}{2m}=\frac{x_1+x_2}{2}$

Thus, for two particles of equal mass the centre of mass lies exactly midway between them.

If we have n particles of masses $m_1$, $m_2$, ...$m_n$ respectively, along a straight line taken as the x- axis, then by definition the position of the centre of the mass of the system of particles is given by.

$X=\frac{m_1x_1+m_2x_2+m_3x_3+...+m_nx_n}{m_1+m_2+m_3+...+m_n}=\frac{\sum_{i=1}^{n}m_ix_i}{\sum_{i=1}^{n}m_i}=\frac{\Sigma m_ix_i}{\Sigma m_i}$

where $x_1$, $x_2$,...$x_n$ are the distances of the particles from the origin; X is also measured from the same origin. The symbol $\Sigma$ (the Greek letter sigma) denotes summation, in this case over n particles. The sum

is the total mass of the system.

Suppose that we have three particles, not lying in a straight line. We may define x– and y– axes in the plane in which the particles lie and represent the positions of the three particles by coordinates ($x_1$, $y_1$), ($x_2$, $y_2$) and ($x_3$, $y_3$) respectively. Let the masses of the three particles be $m_1$, $m_2$ and $m_3$ respectively. The centre of mass C of the system of the three particles is defined and located by the coordinates (X, Y) given by

$X=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

$Y=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

For the particles of equal mass m = $m_1$ = $m_2$ = $m_3$,

$X=\frac{m(x_1+x_2+x_3)}{3m}=\frac{x_1+x_2+x_3}{3}$

$Y=\frac{m(y_1+y_2+y_3)}{3m}=\frac{y_1+y_2+y_3}{3}$

Thus, for three particles of equal mass, the centre of mass coincides with the centroid of the triangle formed by the particles.

Results of Eqs. (7.3a) and (7.3b) are generalised easily to a system of n particles, not necessarily lying in a plane, but distributed in space. The centre of mass of such a system is at (X, Y, Z ), where

$X=\frac{\Sigma m_ix_i}{M}$

$Y=\frac{\Sigma m_iy_i}{M}$

and $Z=\frac{\Sigma m_iz_i}{M}$

Here M = $\Sigma m_i$ is the total mass of the system. The index i runs from 1 to n; $m_i$  is the mass of the $i^{th}$ particle and the position of the $i^{th}$ particle is given by ($x_i$, $y_i$, $z_i$).

Eqs. (7.4a), (7.4b) and (7.4c) can be combined into one equation using the notation of position vectors. Let $\mathbf{r_i}$ be the position vector of the $i^{th}$ particle and $\mathbf{R}$ be the position vector of the centre of mass:

$\mathbf{r_i}=x_i \mathbf{\hat{i}}+y_i \mathbf{\hat{j}}+z_i \mathbf{\hat{k}}$

and $\mathbf{R}=X \mathbf{\hat{i}}+Y \mathbf{\hat{j}}+Z \mathbf{\hat{k}}$

Then $\mathbf{R}=\frac{m_i\mathbf{r_i}}{M}$

The sum on the right hand side is a vector sum.

Note the economy of expressions we achieve by use of vectors. If the origin of the frame of reference (the coordinate system) is chosen to be the centre of mass then $\Sigma m_i \mathbf{r_i}=0$ for the given system of particles.

A rigid body, such as a metre stick or a flywheel, is a system of closely packed particles; Eqs. (7.4a), (7.4b), (7.4c) and (7.4d) are therefore, applicable to a rigid body. The number of particles (atoms or molecules) in such a body is so large that it is impossible to carry out the summations over individual particles in these equations. Since the spacing of the particles is small, we can treat the body as a continuous distribution of mass. We subdivide the body into n small elements of mass; ∆m$_1$, ∆m$_2$... ∆m$_n$; the i th element ∆mi is taken to be located about the point ($x_i$, $y_i$, $z_i$). The coordinates of the centre of mass are then approximately given by

$X=\frac{\Sigma \Delta (m_i)x_i}{\Sigma \Delta m_i}$; $Y=\frac{\Sigma \Delta (m_i)y_i}{\Sigma \Delta m_i}$; $Z=\frac{\Sigma \Delta (m_i)z_i}{\Sigma \Delta m_i}$

As we make n bigger and bigger and each $\Delta m_i$  smaller and smaller, these expressions become exact. In that case, we denote the sums over i by integrals. Thus,

$\Delta m_i$ → $\int{dm} = M$,

$\Sigma (\Delta m_i)x_i$ → $\int{xdm}$,

$\Sigma (\Delta m_i)y_i$ → $\int{ydm}$,

and $\Sigma (\Delta m_i)z_i$ → $\int{zdm}$

Here M is the total mass of the body. The coordinates of the centre of mass now are

$X=\frac{1}{M}\int{xdm}$; $Y=\frac{1}{M}\int{ydm}$; $Z=\frac{1}{M}\int{zdm}$

The vector expression equivalent to these three scalar expressions is

$\mathbf{R}=\frac{1}{M}\int{\mathbf{r}dm}$

If we choose, the centre of mass as the origin of our coordinate system,

$\mathbf{R}=0$ i.e., $\int{\mathbf{r}dm}=0$

or $\int{xdm}=\int{ydm}=\int{zdm}=0$

Often we have to calculate the centre of mass of homogeneous bodies of regular shapes like rings, discs, spheres, rods etc. (By a homogeneous body we mean a body with uniformly distributed mass.) By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres.

Fig.2: Determining the CM of a thin rod.

Let us consider a thin rod, whose width and breath (in case the cross section of the rod is rectangular) or radius (in case the cross section of the rod is cylindrical) is much smaller than its length. Taking the origin to be at the geometric centre of the rod and x-axis to be along the length of the rod, we can say that on account of reflection symmetry, for every element dm of the rod at x, there is an element of the same mass dm located at –x (Fig. 2).

The net contribution of every such pair to the integral and hence the integral $\int{xdm}$ itself is zero. From Eq. (7.6), the point for which the integral itself is zero, is the centre of mass. Thus, the centre of mass of a homogenous thin rod coincides with its geometric centre. This can be understood on the basis of reflection symmetry.

The same symmetry argument will apply to homogeneous rings, discs, spheres, or even thick rods of circular or rectangular cross section. For all such bodies you will realise that for every element dm at a point (x, y, z) one can always take an element of the same mass at the point (–x, –y, –z). (In other words, the origin is a point of reflection symmetry for these bodies.) As a result, the integrals in Eq. (7.5a) all are zero. This means that for all the above bodies, their centre of mass coincides with their geometric centre.

Example 1 

Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long.

Fig.3

With the x–and y–axes chosen as shown in Fig. 7.9, the coordinates of points O, A and B forming the equilateral triangle are respectively (0,0), (0.5,0), (0.25,0.25$\sqrt{3}$). Let the masses 100 g, 150g and 200g be located at O, A and B be respectively. Then,

$X=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$

$=\frac{[(100)(0)+(150)(0.5)+(200)(0.25)] \ g.m}{(100+150+200) \ g}=\frac{5}{18} \ m$

$Y=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

$=\frac{[(100)(0)+(150)(0)+(200)(0.25\sqrt{3})] \ g.m}{(100+150+200) \ g}=\frac{\sqrt{3}}{9} \ m$

The centre of mass C is shown in the figure. Note that it is not the geometric centre of the triangle OAB. Why?

Example 2

Find the centre of mass of a triangular lamina.

Answer The lamina (∆LMN ) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig. 4

Fig.4

By symmetry each strip has its centre of mass at its midpoint. If we join the midpoint of all the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.

Example 3 

Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 kg.

Answer 

Choosing the X and Y axes as shown in Fig.5 we have the coordinates of the vertices of the L-shaped lamina as given in the figure. We can think of the L-shape to consist of 3 squares each of length 1 m. The mass of each square is 1 kg, since the lamina is uniform. The centres of mass $C_1$, $C_2$ and $C_3$ of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/2,1/2), (1/2,3/2) respectively. We take the masses of the squares to be concentrated at these points. The centre of mass of the whole L shape (X, Y) is the centre of mass of these mass points.

Fig.5

Hence

$X=\frac{[(1)(1/2)+(1)(3/2)+(1)(1/2)] \ kg.m}{(1+1+1) \ kg}=\frac{5}{6} \ m$

$Y=\frac{[(1)(1/2)+(1)(1/2)+(1)(3/2)] \ kg.m}{(1+1+1) \ kg}=\frac{5}{6} \ m$

The centre of mass of the L-shape lies on the line OD. We could have guessed this without calculations. Can you tell why? Suppose, the three squares that make up the L shaped lamina of Fig. 5 had different masses. How will you then determine the centre of mass of the lamina?