Center of Mass and its Formula
To understand the concept of center of mass let’s consider a system made up of many particles that move under the influence of external forces. It has been observed that there exists a specific point in the system where, if the entire mass were concentrated, the system’s motion would not change when forces are applied at that point. This special point is known as the center of mass of the system. The concept is incredibly useful in mechanics, particularly when analyzing various problems such as collisions.
Center of Mass in a Two-Particle System
Let’s now examine a system with two particles, each having masses $m_1$ and $m_2$, located at points A and B, respectively.
Let the position vectors of the two particles relative to the origin O of the coordinate system be $\vec{r_1}$ and $\vec{r_2}$. The position vector $\vec{r}$ representing the center of mass for the system is given by:
$ \displaystyle \vec{r}=\frac{{{{m}_{1}}\overrightarrow{{{{r}_{1}}}}+{{m}_{2}}\overrightarrow{{{{r}_{2}}}}}}{{{{m}_{1}}+{{m}_{2}}}}\ldots (1)$
This formula tells us that the total mass of the system, multiplied by the position vector of the center of mass, is the same as the sum of the products of each particle’s mass and its respective position vector.
We can rewrite equation (1) as:
$$
\vec{r} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2}
$$
Center of Mass Formula
Let’s assume that both particles are subjected to external forces $\vec{f_1}$ and $\vec{f_2}$, in addition to the mutual forces they exert on each other. For instance, the force experienced by particle $m_1$ due to $m_2$ is $\vec{F_{12}}$, and $m_2$ experiences an equal but opposite force $\vec{F_{21}}$ due to $m_1$. In this case:
$$
\vec{F_{12}} = -\vec{F_{21}}
$$
This leads to:
$$
\vec{F_{12}} + \vec{F_{21}} = 0
$$
These forces $\vec{F_{12}}$ and $\vec{F_{21}}$ are referred to a internal forces within the system consisting of these two particles.
According to Newton’s second law, the rate of change of momentum for a particle is proportional to the net force acting on it. For a particle with mass $m_1$, the equation of motion can be written as:
$ \displaystyle \frac{{d\overrightarrow{{{{p}_{1}}}}}}{{dt}}$ $ \displaystyle =$ total force on the particle of mass $\displaystyle {{{m}_{1}}}$
Now, the total force on the particle of mass $m_1$ is $\vec{F_{12}} + \vec{f_1}$. If $\vec{v_1}$ is the instantaneous velocity of the particle of mass $m_1$, then
$ \displaystyle \frac{d}{{dt}}({{m}_{1}}\overrightarrow{{{{v}_{1}}}})=\overrightarrow{{{{F}_{{12}}}}}+\overrightarrow{{{{f}_{1}}}}\ldots (2)$
Similarly, for the particle of mass $m_2$, we have
$ \displaystyle \frac{d}{{dt}}({{m}_{2}}\overrightarrow{{{{v}_{2}}}})=\overrightarrow{{{{F}_{{21}}}}}+\overrightarrow{{{{f}_{2}}}}\ldots (3)$
Adding equations (2) and (3), we get
$ \displaystyle \frac{d}{{dt}}({{m}_{1}}\overrightarrow{{{{v}_{1}}}})+\frac{d}{{dt}}({{m}_{2}}\overrightarrow{{{{v}_{2}}}})$ $\displaystyle =(\overrightarrow{{{{F}_{{12}}}}}+\overrightarrow{{{{F}_{{21}}}}})+(\overrightarrow{{{{f}_{1}}}}+\overrightarrow{{{{f}_{2}}}})$
Since $\vec{F_{12}} + \vec{F_{21}} = 0$, we are left with:
$$
\frac{d}{dt} (m_1 \vec{v_1}) + \frac{d}{dt} (m_2 \vec{v_2}) = \vec{f_1} + \vec{f_2}
$$
If we define $\vec{f_1} + \vec{f_2} = \vec{f}(\text{tot})$, which is the total external force acting on the system, then the equation becomes:
$$
\vec{f}(\text{tot}) = \frac{d}{dt} (m_1 \vec{v_1} + m_2 \vec{v_2})
$$
or
$ \displaystyle \vec{f}(\text{tot})=\frac{d}{{dt}}({{m}_{1}}\overrightarrow{{{{v}_{1}}}}+{{m}_{2}}\overrightarrow{{{{v}_{2}}}})\text{ }\ldots \text{(4)}$
The velocity of a particle is defined as the rate of change of its position vector. Therefore,
$$
\vec{v_1} = \frac{d \vec{r_1}}{dt}, \quad \vec{v_2} = \frac{d \vec{r_2}}{dt}
$$
Therefore, equation (4) can be rewritten as:
$$
\vec{f}(\text{tot}) = \frac{d}{dt} \left( m_1 \frac{d \vec{r_1}}{dt} + m_2 \frac{d \vec{r_2}}{dt} \right)
$$
This simplifies to:
$$
= \frac{d}{dt} \left( m_1 \vec{r_1} + \frac{d}{dt} m_2 \vec{r_2} \right)
$$
or:
$$
= \frac{d^2}{dt^2} (m_1 \vec{r_1} + m_2 \vec{r_2})
$$
Multiplying and dividing by $(m_1 + m_2)$, we get:
$ \displaystyle \vec{f}(\text{tot})$ $\displaystyle =({{m}_{1}}+{{m}_{2}})\frac{{{{d}^{2}}}}{{d{{t}^{2}}}}\left( {\frac{{{{m}_{1}}\overrightarrow{{{{r}_{1}}}}+{{m}_{2}}\overrightarrow{{{{r}_{2}}}}}}{{{{m}_{1}}+{{m}_{2}}}}} \right)\text{ }\ldots \text{(5)}$
The equation (5) describes the motion of the total mass $(m_1 + m_2)$ of the system under the effect of the total external force $\vec{f}(\text{tot})$. The position vector of the point where this total mass is concentrated is given by:
$ \bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol{\displaystyle \vec{r}=\frac{{{{m}_{1}}\overrightarrow{{{{r}_{1}}}}+{{m}_{2}}\overrightarrow{{{{r}_{2}}}}}}{{{{m}_{1}}+{{m}_{2}}}}\text{ }}}$ $\displaystyle \ldots \text{(6)}$
As deduced earlier, the point whose position vector is given by the above equation is the center of mass of the system of two particles of masses $m_1$ and $m_2$, whose position vectors are $\vec{r_1}$ and $\vec{r_2}$, respectively.
Discussion of the result
1. The equation (6) describes the motion of the total mass of the system under the effect of external forces. The internal forces between the two particles cancel each other out. In simpler terms, the motion of a system can be analyzed by focusing on the individual particles using Newton’s second law of motion. This includes applying Newton’s third law for internal forces, while considering the motion of the total mass located at the center of mass of the system under the effect of external forces.
2. The center of mass of a system consisting of two particles always lies on the line connecting the two particles.
3. If $m_1 = m_2 = m$, the position vector of the center of mass is given by:
$$
\vec{r} = \frac{m \vec{r_1} + m \vec{r_2}}{m + m} = \frac{\vec{r_1} + \vec{r_2}}{2}
$$
4. If the origin of the coordinate axis lies at the center of mass, then $\vec{r} = 0$. In this case, equation (6) becomes:
$$
m_1 \vec{r_1} + m_2 \vec{r_2} = 0
$$
Example: In an HCl molecule, the distance between the nuclei of the two atoms is approximately 1.27 Å. Since chlorine is 35.5 times as massive as hydrogen, most of the mass of the atom is concentrated in the chlorine nucleus. As a result, the center of mass of the HCl molecule lies on the line connecting the hydrogen and chlorine atoms.
Solution:
Let’s assume the chlorine atom is at the origin. Then, the position vector of the center of mass with respect to the hydrogen and chlorine atoms, using the center of mass as the origin, is given by:
$$
m_1 \vec{r_1} + m_2 \vec{r_2} = 0
$$
If $\vec{r_1} = x \hat{i}$, then: $\displaystyle \overrightarrow{{{{r}_{2}}}}=-\left( {1.27-x} \right)\hat{i}$
The negative sign arises because, if the hydrogen atom lies on the positive X-axis (with the center of mass as the origin), then the chlorine atom lies on the negative X-axis. Therefore,
$$
m_1 x \hat{i} – m_2 (1.27 – x) \hat{i} = 0
$$
or,
$$
m_1 x = 35.5 m_1 (1.27 – x)
$$
which simplifies to:
$$
x = \frac{35.5 \times 1.27}{36.5} = 1.235 \, \text{Å}
$$
This is the position of the hydrogen atom.
Alternative method: The algebraic sum of the moments of masses about the center of mass is always zero, or the sum of clockwise moments equals the sum of counterclockwise moments of the masses:
$$
m_1 x = (1.27 – x) m_2
$$
Taking $m_2 = 35.5 m_1$, we get:
$$
x = 1.235 \, \text{Å}
$$
Center of Mass Equation — Generalisation to n-Particles
Consider a system of $n$ particles with masses $m_1, m_2, m_3, …, m_n$. Let $\vec{r_1}, \vec{r_2}, \vec{r_3}, …, \vec{r_n}$ be their position vectors with respect to the origin of the coordinate axes.
Let $\vec{r}$ be the position vector of the center of mass of this system of $n$ particles.
Let $\vec{v_1}, \vec{v_2}, \vec{v_3}, \dots, \vec{v_n}$ be the instantaneous velocities of these particles. Suppose $\vec{f_1}, \vec{f_2}, \vec{f_3}, \dots, \vec{f_n}$ are the external forces acting on the $n$ particles of the system. Each particle of the system experiences internal forces due to the rest of the $(n – 1)$ particles of the system (i.e., all particles except itself). Therefore, the total internal force on the $i^{th}$ particle of the system may be written as:
$$
\vec{F_i} = \sum_{j=1, j \neq i}^{n} \vec{F_{ij}}
$$
Here, $\vec{F_{ij}}$ is the internal force on the $i^{th}$ particle of the system due to the $j^{th}$ particle.
According to Newton’s second law, the equation of motion of the $i^{th}$ particle of the system may be written as:
$ \displaystyle \frac{d}{{dt}}({{m}_{i}}\overrightarrow{{{{v}_{i}}}})=\overrightarrow{{{{F}_{i}}}}+\overrightarrow{{{{f}_{i}}}}\text{ }\ldots \text{(7)}$
We can write $n$ such equations of motion for all the $n$ particles of the system. On adding these $n$ equations, we get:
$ \displaystyle \sum\limits_{{i=1}}^{n}{{\frac{d}{{dt}}}}({{m}_{i}}\overrightarrow{{{{v}_{i}}}})$ $\displaystyle =\sum\limits_{{i=1}}^{n}{{\overrightarrow{{{{F}_{i}}}}}}+\sum\limits_{{i=1}}^{n}{{\overrightarrow{{{{f}_{i}}}}}}\text{ }\ldots \text{(8)}$
The internal forces on all the particles of the system cancel out in pairs, i.e.,
$$
\sum_{i=1}^{n} \vec{F_i} = 0
$$
If $\vec{f} = \sum_{i=1}^{n} \vec{f_i}$ is the total external force on the system, we can rewrite equation (13) as:
$\displaystyle \sum\limits_{{i=1}}^{n}{{\frac{d}{{dt}}}}({{m}_{i}}\overrightarrow{{{{v}_{i}}}})=\vec{f}\text{ }\ldots \text{(9)}$
Now,
$$
\vec{v_i} = \frac{d \vec{r_i}}{dt}
$$
Thus,
$$
\sum_{i=1}^{n} \frac{d}{dt} \left( m_i \frac{d \vec{r_i}}{dt} \right) = \vec{f}
$$
or,
$\displaystyle \frac{{{{d}^{2}}}}{{d{{t}^{2}}}}\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}\overrightarrow{{{{r}_{i}}}}=\vec{f}\text{ }\ldots \text{(10)}$
Multiplying and dividing the left-hand side of (15) by $M = \sum_{i=1}^{n} m_i$, the total mass of all the $n$ particles of the system, we get:
$ \displaystyle M\frac{{{{d}^{2}}}}{{d{{t}^{2}}}}\frac{{\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}\overrightarrow{{{{r}_{i}}}}}}{M}=\vec{f}$
Put:
$\displaystyle \vec{r}=\frac{{\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}\overrightarrow{{{{r}_{i}}}}}}{M}\text{ }\ldots \text{(11)}$
Hence:
$\displaystyle M\frac{{{{d}^{2}}\vec{r}}}{{d{{t}^{2}}}}=\vec{f}\text{ }\ldots \text{(12)}$
This is the equation of motion of the total mass $M$ of the system concentrated at position vector $\vec{r}$ given by (11), under the action of the total external force $\vec{f}$ applied at this position. By definition, this is the position vector of the center of mass of the system of $n$ particles.
From (16), i.e. $\displaystyle \vec{r}=\frac{{\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}\overrightarrow{{{{r}_{i}}}}}}{M}$
$ \bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol{\displaystyle \vec{r}=\frac{{{{m}_{1}}\overrightarrow{{{{r}_{1}}}}+{{m}_{2}}\overrightarrow{{{{r}_{2}}}}+\ldots +{{m}_{n}}\overrightarrow{{{{r}_{n}}}}}}{{{{m}_{1}}+{{m}_{2}}+\ldots +{{m}_{n}}}}}}$
i.e.,
$\displaystyle ({{m}_{1}}+{{m}_{2}}+\ldots +{{m}_{n}})\vec{r}$ $\displaystyle ={{m}_{1}}\overrightarrow{{{{r}_{1}}}}+{{m}_{2}}\overrightarrow{{{{r}_{2}}}}+\ldots +{{m}_{n}}\overrightarrow{{{{r}_{n}}}}$
Hence, the position of the center of mass of $n$ particles of the system is such that the product of the total mass and the position vector of the center of mass is equal to the sum of the products of individual masses of the particles and their respective position vectors.
Hence, in general, the center of mass of a system of $n$ particles is the weighted average of the position vectors of the $n$ particles making up the system.
Remember:
- The position of the center of mass of a system is independent of the choice of coordinate system.
- The position of the center of mass depends on the shape and size of the body and the distribution of its mass. Hence it may lie within or outside the material of the body.
- In symmetrical bodies with uniform distribution of mass, the center of mass coincides with the geometrical center or center of symmetry of the body.
- The center of mass changes its position only under the translatory motion. There is no effect of rotatory motion on center of mass of a body.
- The concept of center of mass proves that the laws of mechanics, which are true for a point mass, are equally valid for all macroscopic bodies.
- If we know the center of mass of different parts of the system and their masses, we can get the combined center of mass by treating various parts as point objects whose masses are concentrated at their respective centers of masses.
Example: The figure shows a uniform square plate from which four identical squares at the corners will be removed. (a) Where is the center of mass if one square is removed? (b) Where it is after 2 and 2 are removed? (c) Where it is after squares 1 and 3 are removed? (d) Where it is after squares 1, 2, 3 are removed? (e) Where it is after all the four squares are removed? Answer in terms of quadrants and axes.
Solution:
(a) Originally, center of mass is at the center \( O \).
(b) After square 1 is removed, cm lies in quadrant 3.
(c) After squares 1 and 2 are removed, cm lies on Y-axis below \( O \).
(d) When squares 1 and 3 are removed, cm will remain at \( O \).
(e) When squares 1, 2, 3 are removed, cm will shift to fourth quadrant.
(f) When all the four squares are removed, cm will shift back to \( O \).
Co-ordinates of the Center of Mass
We have learnt above that the position vector of a system of particles is given by
$\displaystyle \vec{r}=\frac{1}{M}\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}{{\vec{r}}_{i}}\text{ }\ldots (13)$
As each position vector \( \vec{r}_i \) (of \( i \)-th particle) can be expressed in terms of its components \( x_i, y_i, z_i \) in the form:
$\displaystyle {{{\vec{r}}}_{i}}=\hat{i}{{x}_{i}}+\hat{j}{{y}_{i}}+\hat{k}{{z}_{i}}\text{ }\ldots (14)$
where \( i = 1, 2, 3, \ldots, m \)
Obviously, \( x_i, y_i, z_i \) can be taken as the co-ordinates of the point, where the \( i \)-th particle of the system is located.
If \( x, y, z \) are the co-ordinates of the center of mass of the system, we may write
$\displaystyle \vec{r}=\hat{i}x+\hat{j}y+\hat{k}z\text{ }\ldots (15)$
Using (14) and (15) in (11), we get
$\displaystyle \hat{i}x+\hat{j}y+\hat{k}z=\frac{1}{M}\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}(\hat{i}{{x}_{i}}+\hat{j}{{y}_{i}}+\hat{k}{{z}_{i}})$
$\displaystyle =\frac{1}{M}[(\hat{i}{{m}_{1}}{{x}_{1}}+\hat{j}{{m}_{1}}{{y}_{1}}+\hat{k}{{m}_{1}}{{z}_{1}})+$ $\displaystyle \ldots +(\hat{i}{{m}_{n}}{{x}_{n}}+\hat{j}{{m}_{n}}{{y}_{n}}+\hat{k}{{m}_{n}}{{z}_{n}})]$
Equating the components separately, we get
$\displaystyle x=\frac{1}{M}({{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+\ldots +{{m}_{n}}{{x}_{n}})$ $\displaystyle \text{ }\ldots (16)$
$\displaystyle y=\frac{1}{M}({{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+\ldots +{{m}_{n}}{{y}_{n}})$ $\displaystyle \text{ }\ldots (17)$
$\displaystyle z=\frac{1}{M}({{m}_{1}}{{z}_{1}}+{{m}_{2}}{{z}_{2}}+\ldots +{{m}_{n}}{{z}_{n}})$ $\displaystyle \text{ }\ldots (18)$
Equations (16), (17), (18) can be rewritten as:
$\displaystyle x=\frac{1}{M}\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}{{x}_{i}}\text{ }\ldots (19)$
$\displaystyle y=\frac{1}{M}\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}{{y}_{i}}\text{ }\ldots (20)$
$\displaystyle z=\frac{1}{M}\sum\limits_{{i=1}}^{n}{{{{m}_{i}}}}{{z}_{i}}\text{ }\ldots (21)$
Hence the coordinates of the center of mass of n particle system can be obtained in terms of the coordinates of location points of the n particles making up the system.
Example: The distance between the centers of carbon and oxygen atoms in the carbon monoxide gas molecule is \(1.13 \text{ Å}\). Locate the center of mass of the molecule relative to carbon atom.
Solution:
\(m_1\) (mass of carbon atom) = 14
\(m_2\) (mass of oxygen atom) = 16
Let \(cm\) of molecule be at a distance \(x\) from carbon atom
\[
m_1 x = m_2 (1.13 – x)
\]
As \(m_1 = m_2\),
\[
14x = 16(1.13 – x)
\]
\[
30x = 16 \times 1.13
\]
\[
x = \frac{16 \times 1.13}{30} = 0.602 \text{ Å}
\]
