Let’s understand what centre of mass of a system of particles is. For simplicity, we shall start with a two-particle system. We shall take the line joining the two particles to be the \(x\) – axis.

Let the distances of the two particles be \(x_1\) and \(x_2\) respectively from some origin \(\mathrm{O}\). Let \(m_1\) and \(m_2\) be respectively the masses of the two particles. The centre of mass of the system is that point \(\mathrm{C}\) which is at a distance \(X\) from \(\mathrm{O}\), where \(X\) is given by
\(
X=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \dots(7.1)
\)
In Eq. (7.1), \(X\) can be regarded as the mass-weighted mean of \(x_1\) and \(x_2\). If the two particles have the same mass \(m_1=m_2=m\) then
\(
X=\frac{m x_1+m x_2}{2 m}=\frac{x_1+x_2}{2}
\)
Thus, for two particles of equal mass the centre of mass lies exactly midway between them.

If we have \(n\) particles of masses \(m_1, m_2\), \(\ldots m_n\) 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.
\(
X=\frac{m_1 x_1+m_2 x_2+\ldots+m_n x_n}{m_1+m_2+\ldots+m_n}=\frac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i}=\frac{\sum m_i x_i}{\sum m_i} \dots(7.2)
\)
where \(x_1, x_2, \ldots x_{\mathrm{n}}\) are the distances of the particles from the origin; \(X\) is also measured from the same origin. The symbol \(\sum\) (the Greek letter sigma) denotes summation, in this case over \(n\) particles. The sum
\(
\sum m_i=M
\)
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 \(\left(x_1, y_1\right),\left(x_2, y_2\right)\) and \(\left(x_3, y_3\right)\) respectively. Let the masses of the three particles be \(m_1, m_2\) and \(m_3\) respectively. The centre of mass \(\mathrm{C}\) of the system of the three particles is defined and located by the coordinates \((X, Y)\) given by
\(
\begin{aligned}
& X=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \dots(7.3a)\\
& Y=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3} \dots(7.3(b)
\end{aligned}
\)
For the particles of equal mass \(m=m_1=m_2\) \(=m_3\),
\(
X=\frac{m\left(x_1+x_2+x_3\right)}{3 m}=\frac{x_1+x_2+x_3}{3}
\)
\(
Y=\frac{m\left(y_1+y_2+y_3\right)}{3 m}=\frac{y_1+y_2+y_3}{3}
\)

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
\(
\begin{aligned}
& X=\frac{\sum m_i x_i}{M} \dots(7.4a) \\
& Y=\frac{\sum m_i y_i}{M} \dots(7.4b)
\end{aligned}
\)
and \(Z=\frac{\sum m_i z_i}{M} \dots(7.4c)\)
Here \(M=\sum m_i\) is the total mass of the system. The index \(i\) runs from 1 to \(n ; m_i\) is the mass of the \(i^{\text {th }}\) particle and the posttion of the \(i^{\text {th }}\) particle is given by \(\left(x_1, y_1, z_1\right)\).

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 \(i^{\text {th }}\) particle and \(\mathbf{R}\) be the position vector of the centre of mass:

\(
\begin{array}{r}
\mathbf{r}_t=x_i \hat{\mathbf{i}}+y_i \hat{\mathbf{j}}+z_i \widehat{\mathbf{k}} \\
\text { and } \mathbf{R}=X \hat{\mathbf{i}}+Y \hat{\mathbf{j}}+Z \hat{\mathbf{k}} \\
\text { Then } \mathbf{R}=\frac{\sum m_i \mathbf{r}_i}{M} \dots(7.4d)
\end{array}
\)
The sum on the right-hand side is a vector sum.
Note the economy of expressions we achieve by the use of vectors. If the origin of the frame of reference (the coordinate system) is chosen to be the centre of mass then \(\sum m_i \mathbf{r}_i=0\) for the given system of particles.

Example 7.1:  Four particles \(A, B, C\) and \(D\) having masses \(m, 2 m\), \(3 m\) and \(4 m\) respectively are placed in order at the corners of a square of side a. Locate the centre of mass.

Solution: Take the axes as shown in the figure above. The coordinates of the four particles are as follows :

\(
\begin{array}{cccc}
\text { Particle } & \text { mass } & x \text {-coordinate } & y \text {-coordinate } \\
A & m & 0 & 0 \\
B & 2 m & a & 0 \\
C & 3 m & a & a \\
D & 4 m & 0 & a
\end{array}
\)

Hence, the coordinates of the centre of mass of the four-particle system are
\(
\begin{aligned}
& X=\frac{m \cdot 0+2 m a+3 m a+4 m \cdot 0}{m+2 m+3 m+4 m}=\frac{a}{2} \\
& Y=\frac{m \cdot 0+2 m \cdot 0+3 m a+4 m a}{m+2 m+3 m+4 m}=\frac{7 a}{10}
\end{aligned}
\)
The centre of mass is at \(\left(\frac{a}{2}, \frac{7 a}{10}\right)\).

Centre of Mass of Two Particles

As the simplest example, consider a system of two particles of masses \(m_1\) and \(m_2\) separated by a distance \(d\) (figure below). Where is the centre of mass of this system?

Take the origin at \(m_1\) and the \(X\)-axis along the line joining \(m_1\) and \(m_2\). The coordinates of \(m_1\) are \((0,0,0)\) and of \(m_2\) are \((d, 0,0)\). So,
\(
\sum_i m_i x_i=m_1 0+m_2 d=m_2 d, \sum_i m_i y_i=0, \sum_i m_i z_i=0 .
\)
The total mass is \(M=m_1+m_2\). By definition, the centre of mass will be at \(\left(\frac{m_2 d}{m_1+m_2}, 0,0\right)\). We find that the centre of mass of a system of two particles is situated on the line joining the particles. If \(O, C, P\) be the positions of \(m_1\), the centre of mass and \(m_2\) respectively, we have
\(
\begin{aligned}
& \qquad O C=\frac{m_2 d}{m_1+m_2} \text { and } C P=\frac{m_1 d}{m_1+m_2} \\
& \text { so that } m_1(O C)=m_2(C P)
\end{aligned}
\)
Note: The centre of mass divides internally the line joining the two particles in inverse ratio of their masses.

Centre of Mass of Several Groups of Particles

Consider a collection of \(N_1+N_2\) particles. We call the group of \(N_1\) particles as the first part and the other group of \(N_2\) particles as the second part. Suppose the first part has its centre of mass at \(C_1\) and the total mass \(M_1\) (figure below). Similarly, the second part has its centre of mass at \(C_2\) and the total mass \(M_2\). Where is the centre of mass of the system of \(N_1+N_2\) particles?

The \(x\)-coordinate of the centre of mass is
\(
X=\frac{\sum_{i=1}^{N_1+N_2} m_i x_i}{M_1+M_2}=\frac{\sum_{i=1}^{N_1} m_i x_i+\sum_{i=N_1+1}^{N_1+N_2} m_i x_i}{M_1+M_2} \dots(a)
\)
If \(X_1, X_2\) are the \(x\)-coordinates of \(C_1\) and \(C_2\), then by the definition of centre of mass, \(\sum m_i x_i\) for the first part is \(M_1 X_1\) and \(\sum m_i x_i\) for the second part is \(M_2 X_2\). Hence equation (a) becomes,
\(
X=\frac{M_1 X_1+M_2 X_2}{M_1+M_2}
\)
Similarly, \(Y=\frac{M_1 Y_1+M_2 Y_2}{M_1+M_2}\)
and
\(
Z=\frac{M_1 Z_1+M_2 Z_2}{M_1+M_2}
\)
But this is also the centre of mass of two point particles of masses \(M_1\) and \(M_2\) placed at \(C_1\) and \(C_2\) respectively. Thus, we obtain a very useful result. If we know the centres of mass of parts of the system and their masses, we can get the combined centre of mass by treating the parts as point particles placed at their respective centres of mass.

Example 7.2: Two identical uniform rods \(A B\) and \(C D\), each of length \(L\) are jointed to form a T-shaped frame as shown in the figure below. Locate the centre of mass of the frame. The centre of mass of a uniform rod is at the middle point of the rod.

Solution:

Let the mass of each rod be \(m\). Take the centre \(C\) of the \(\operatorname{rod} A B\) as the origin and \(C D\) as the \(Y\)-axis. The \(\operatorname{rod} A B\) has mass \(m\) and its centre of mass is at \(C\). For the calculation of the centre of mass of the combined system, \(A B\) may be replaced by a point particle of mass \(m\) placed at the point \(C\). Similarly, the rod \(C D\) may be replaced by a point particle of mass \(\mathrm{m}\) placed at the centre \(E\) of the rod \(C D\). Thus, the frame is equivalent to a system of two particles of equal masses m each, placed at \(C\) and \(E\). The centre of mass of this pair of particles will be at the middle point \(F\) of \(C E\).

The centre of mass of the frame is, therefore, on the rod \(C D\) at a distance \(L / 4\) from \(C\).

Centre Of Mass Of Continuous Bodies

A rigid body, such as a metre stick or a flywheel, is a system of closely packed particles; Eqs. \((7.4 \mathrm{a}),(7.4 \mathrm{~b}),(7.4 \mathrm{c})\) 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; \(\Delta m_1, \Delta m_2 \ldots \Delta m_{\mathrm{n}}\); the \(i^{\text {th }}\) element \(\Delta m_i\) is taken to be located about the point \(\left(x_1, y_1, z_1\right)\). The coordinates of the centre of mass are then approximately given by

\(
X=\frac{\sum\left(\Delta m_i\right) x_i}{\sum \Delta m_i}, Y=\frac{\sum\left(\Delta m_i\right) y_i}{\sum \Delta m_i}, Z=\frac{\sum\left(\Delta m_i\right) z_i}{\sum \Delta m_t} 
\)

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,
\(
\begin{aligned}
& \sum \Delta m_t \rightarrow \int \mathrm{d} m=M, \\
& \sum\left(\Delta m_t\right) x_t \rightarrow \int x \mathrm{~d} m, \\
& \sum\left(\Delta m_t\right) y_t \rightarrow \int y \mathrm{~d} m, \\
\text { and } \quad & \sum\left(\Delta m_t\right) z_t \rightarrow \int z \mathrm{~d} m
\end{aligned}
\)
Here \(M\) is the total mass of the body. The coordinates of the centre of mass now are
\(
X=\frac{1}{M} \int x \mathrm{~d} m, Y=\frac{1}{M} \int y \mathrm{~d} m \text { and } Z=\frac{1}{M} \int z \mathrm{~d} m \text { (7.5a) }
\)
The vector expression equivalent to these three scalar expressions is
\(
\mathbf{R}=\frac{1}{M} \int \mathbf{r} d m
\)
If we choose, the centre of mass as the origin of our coordinate system,
\(
\begin{aligned}
& \mathbf{R}=\mathbf{O} \\
& \text { i.e., } \int \mathbf{r} \mathrm{d} m=\mathbf{0} \\
& \text { or } \int x \mathrm{~d} m=\int y \mathrm{~d} m=\int z \mathrm{~d} m=0 \dots(7.6)
\end{aligned}
\)

Case(a)-Determining the CM of a thin rod

By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres. 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 \(d m\) of the rod at \(x\), there is an element of the same mass \(d m\) located at \(-x\) (Fig. 7.8).

The net contribution of every such pair to the integral and hence the integral \(\int x \mathrm{~d} m\) 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. 
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 \(d m\) 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. (7.5 a) all are zero. This means that for all the above bodies, their centre of mass coincides with their geometric centre.

Case(b)-Centre of Mass of a Uniform Straight Rod

Let \(M\) and \(L\) be the mass and the length of the rod respectively. Take the left end of the rod as the origin and the \(X\)-axis along the rod (figure above). Consider an element of the rod between the positions \(x\) and \(x+d x\). If \(x=0\), the element is at the left end of the rod. If \(x=L\), the element is at its right end. So as \(x\) varies from 0 through \(L\), the elements cover the entire rod. As the rod is uniform, the mass per unit length is \(M / L\) and hence the mass of the element is \(d m=(M / L) d x\). The coordinates of the element are \((x, 0,0)\). (The coordinates of different points of the element differ, but the difference is less than \(d x\) and that much is harmless as integration will automatically correct it. So \(x\)-coordinate of the left end of the element may be called the ” \(x\)-coordinate of the element.”)

The \(x\)-coordinate of the centre of mass of the rod is
\(
\begin{aligned}
X=\frac{1}{M} \int x d m & =\frac{1}{M} \int_0^L x\left(\frac{M}{L} d x\right) \\
& =\frac{1}{L}\left[\frac{x^2}{2}\right]_0^L=\frac{L}{2} .
\end{aligned}
\)
The \(y\)-coordinate is
\(
Y=\frac{1}{M} \int y d m=0
\)
and similarly \(Z=0\). The centre of mass is at \(\left(\frac{L}{2}, 0,0\right)\), i.e., at the middle point of the rod.

Case(c)- Centre of Mass of a Uniform Semicircular Wire

Let \(M\) be the mass and \(R\) the radius of a uniform semicircular wire. Take its centre as the origin, the line joining the ends as the \(X\)-axis, and the \(Y\)-axis in the plane of the wire shown in the above figure The centre of mass must be in the plane of the wire i.e., in the \(X-Y\) plane.
How do we choose a small element of the wire? First, the element should be so defined that we can vary the element to cover the whole wire. Secondly, if we are interested in \(\int x d m\), the \(x\)-coordinates of different parts of the element should only infinitesimally differ in range. We select the element as follows. Take a radius making an angle \(\theta\) with the \(X\)-axis and rotate it further by an angle \(d \theta\). This gives an element of length \(R d \theta\). When we take \(\theta=0\), the element is situated near the right edge of the wire. As \(\theta\) is gradually increased to \(\pi\), the element takes all positions on the wire i.e., the whole wire is covered. The “coordinates of the element” are \((R \cos \theta, R \sin \theta)\). 

As the wire is uniform, the mass per unit length of the wire is \(\frac{M}{\pi R}\). The mass of the element is, therefore,
\(
d m=\left(\frac{M}{\pi R}\right)(R d \theta)=\frac{M}{\pi} d \theta 
\)
The coordinates of the centre of mass are
\(
X=\frac{1}{M} \int x d m=\frac{1}{M} \int_0^\pi(R \cos \theta)\left(\frac{M}{\pi}\right) d \theta=0
\)
and
\(
Y=\frac{1}{M} \int y d m=\frac{1}{M} \int_0^\pi(R \sin \theta)\left(\frac{M}{\pi}\right) d \theta=\frac{2 R}{\pi}
\)
The centre of mass is at \(\left(0, \frac{2 R}{\pi}\right)\).

Case(d) – Centre of Mass of a Uniform Semicircular Plate

The figure above shows the semicircular plate. We take the origin at the centre of the semicircular plate, the \(X\)-axis along the straight edge and the \(Y\)-axis in the plane of the plate. Let \(M\) be the mass and \(R\) be its radius. Let us draw a semicircle of radius \(r\) on the plate with the centre at the origin. We increase \(r\) to \(r+d r\) and draw another semicircle with the same centre. Consider the part of the plate between the two semicircles of radii \(r\) and \(r+d r\). This part, shown shaded in the above figure, may be considered as a semicircular wire.

If we take \(r=0\), the part will be formed near the centre and if \(r=R\), it will be formed near the edge of the plate. Thus, if \(r\) is varied from 0 to \(R\), the elemental parts will cover the entire semicircular plate. We can replace the semicircular shaded part by a point particle of the same mass at its centre of mass for the calculation of the centre of mass of the plate. The area of the shaded part \(=\pi r d r\). The area of the plate is \(\pi R^2 / 2\). As the plate is uniform, the mass per unit area is \(\frac{M}{\pi R^2 / 2}\). Hence the mass of the semicircular element
\(
\frac{M}{\pi R^2 / 2}(\pi r d r)=\frac{2 M r d r}{R^2}
\)
The \(y\)-coordinate of the centre of mass of this wire is \(2 r / \pi\). The \(y\)-coordinate of the centre of mass of the plate is, therefore,
\(
Y=\frac{1}{M} \int_0^R\left(\frac{2 r}{\pi}\right)\left(\frac{2 M r}{R^2} d r\right)=\frac{1}{M} \cdot \frac{4 M}{\pi R^2} \frac{R^3}{3}=\frac{4 R}{3 \pi} .
\)
The \(x\)-coordinate of the centre of mass is zero by symmetry.

Example 7.3: Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are \(100 \mathrm{~g}, 150 \mathrm{~g}\), and \(200 \mathrm{~g}\) respectlvely. Each side of the equilateral triangle is \(0.5 \mathrm{~m}\) long.

With the \(x\)-and \(y\)-axes chosen as shown in the figure above, the coordinates of points \(\mathrm{O}, \mathrm{A}\) and \(\mathrm{B}\) forming the equilateral triangle are respectively \((0,0)\), \((0.5,0),(0.25,0.25 \sqrt{3})\). Let the masses \(100 \mathrm{~g}\), \(150 \mathrm{~g}\) and \(200 \mathrm{~g}\) be located at O, A and B be respectively. Then,
\(
\begin{aligned}
& X=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
& =\frac{100(0)+150(0.5)+200(0.25) \mathrm{gm}}{(100+150+200) \mathrm{g}} \\
& =\frac{75+50}{450} \mathrm{~m}=\frac{125}{450} \mathrm{~m}=\frac{5}{18} \mathrm{~m} \\
& Y=\frac{100(0)+150(0)+200(0.25 \sqrt{3}) \mathrm{gm}}{450 \mathrm{~g}} \\
& =\frac{50 \sqrt{3}}{450} \mathrm{~m}=\frac{\sqrt{3}}{9} \mathrm{~m}=\frac{1}{3 \sqrt{3}} \mathrm{~m}
\end{aligned}
\)

Example 7.4: Find the centre of mass of a triangular lamina.

Solution:

The lamina ( \(\triangle L M N)\) may be subdivided Into narrow strips each parallel to the base (MN) as shown in Fig. 7.10. 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 \(\mathrm{G}\) of the triangle.

Example 7.5: 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 \(3 \mathrm{~kg}\).

Solution:

Choosing the \(X\) and \(Y\) axes as shown in Fig. 7.11 we have the coordinates of the vertices of the L-shaped lamina as given in the f1gure. We can think of the L-shape to consist of 3 squares each of length \(\mathrm{lm}\). The mass of each square is \(1 \mathrm{~kg}\), since the lamina is uniform. The centres of mass \(\mathrm{C}_1, \mathrm{C}_2\) and \(\mathrm{C}_3\) 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.

Hence
\(
\begin{aligned}
& X=\frac{[1(1 / 2)+1(3 / 2)+1(1 / 2)] \mathrm{kgm}}{(1+1+1) \mathrm{kg}}=\frac{5}{6} \mathrm{~m} \\
& Y=\frac{[1(1 / 2)+1(1 / 2)+1(3 / 2)] \mathrm{kgm}}{(1+1+1) \mathrm{kg}}=\frac{5}{6} \mathrm{~m}
\end{aligned}
\)
The centre of mass of the L-shape lies on the line OD. We could have guessed this without calculations.