Centre of Mass Explained with Formula

Centre of Mass is a special point where you can imagine all the mass of a real-life object made up of multiple particles is concentrated. This spot is often called the balance point of any object.

Where is the Centre of Mass of a Real Object Located? Consider the most basic example of a player spinning a basketball on their fingers.  Even though every part spins, the Centre of Mass stays still above the finger. 

No matter how weirdly shaped an object is, or how it's moving or spinning, you can always find this one special point. Once you know where it is, you can predict how the whole object will move using the simple physics you already know. 

What’s the Purpose of Finding the Centre of Mass? Instead of tracking every single atom in an object, which would be impossible, you can just track this one point and understand the object's motion. 

Key Differences Between Centre of Mass, Centre of Gravity, Centroid, and Barycentre

It's also important to know some related but non-interchangeable terms for centre of mass. Just for conceptual grounding that will help you tackle advanced numerical problems in competitive exams.  

Aspect	Centre of Mass	Centroid	Barycentre	Centre of Gravity
Definition	It's the point where all the system's mass concentrates.	It's the geometric centre of an object	This is the common centre of mass around which two or more celestial bodies orbit	It's the point where gravitational force acts on a body
Depends on	Mass distribution	Shape and geometry only without any mass	Mass and distance of celestial bodies	Gravitational field and mass distribution
Application	Physics (mechanics, rotational dynamics, collisions)	Engineering (structural analysis, geometric modelling)	Astronomy (orbits of planets, stars, moons)	Engineering, mechanics (stability, balance, torque)

 

Learning about the centre of mass is essential for various modern engineering applications and even classical mechanics.

• In multi-particle systems or extended bodies, their translational motion can be known with the centre of mass alone. In short, you can simplify complex motion into single body dynamics, where the same Laws of motion and related kinematic equations apply. For tricky NEET exam questions on rotational motion, that will help save your time. 
• The more clarity you have on the centre of mass, the easier it is to predict behaviours, including collisions and explosions. Because you would know how the internal forces for systems of particles, such as tension or spring force, cancel out.
• Centre of mass is central to spacecraft design researched in ISRO and NASA. Your knowledge about non-inertial frames of reference that you have learned in Newton's First Law of Motion, it is applicable in rockets where controlling thrust is essential to align with the centre of mass. 
• Mechanical engineering students have to learn about the centre of mass to design stable structures like bridges and vehicles. Know the CM location, helps them understand vehicle mechanics, load distribution, etc. And, so, when you're going through previous years’ JEE question papers, this topic, along with rotational motion, will carry around three to eight per cent weightage.

 

Here we are going to figure out the formula of the centre of mass through a real-life example. 

Let’s consider a seesaw with two students, one heavier in weight than the other. 

For example’s sake, 

Student A: 40 kg, sitting 2 meters from the pivot
Student B: 60 kg, sitting some distance from the pivot

Where should Student B sit to balance? Your intuition says ‘closer to the centre’ because B is heavier, right?

Now, suppose you already know that Student A with a mass of 40kg sits 2 m from the pivot of the see-saw. Student B of 60 kg must find distance 𝑑 so moments balance out. 

Mathematically, for the balance, we have 40 x 2 = 60 x d

A mass × distance is its moment about the pivot. The balance here means

Total clockwise moments = total counter‑clockwise moments

So the solution for the moment would be 

d = (40 x2)/60 = 1.33 m

That means, if Student B sits 1.33 metres from pivot, the see-saw balances. 

Did you notice? 40 kg × 2m = 60 kg × 1.33m (both equal 80). In that logic, the balance point is where these ‘moments’ are exactly equal. 

Now, we can use this observation of moments and balance point as a recurring pattern. We can generalise that. 

With three kids, four weights, or n particles, each mass contributes a moment  $m_i{x}_i$ . The variable i can be for one object. We either describe it as 1 or A. For convenience, we will use 1, so that when we add for the second object it would be 2. And, x here is in a single plane. 

$$m_1{x}_1+m_2{x}_2=M{x}_{\text{COM}}\to \underset{i=1}{\overset{n}{\sum }}{m}_i{x}_i=M{x}_{\text{COM}}$$
​The total moment (mass×position) of all the little bits about the origin equals the total mass times the centre‑of‑mass position.

To get the average position of multiple objects, we divide by the total mass instead of the sum of moments for each object. 

Then, for any number of objects, the balance point is

Balance Point = (mass₁ × position₁ + mass₂ × position₂ + ...) ÷ (total mass)

$x_{\text{COM}}=\frac{m_1{x}_1+m_2{x}_2+\cdots +m_n{x}_n}{M}.$

Till here, we are still considering a single plane. In the 3D world, there are three coordinates, x, y, z. 

By exactly the same logic

“Total moment = total mass × coordinate of COM”

You write three parallel formula.

Using the same logic, which is total moment = total mass x coordinate of centre of mass, we can consider three parallel formulas. 

$$\begin{array}{c}{x}_{\text{COM}}=\frac{\underset{i=1}{\overset{n}{\sum }}{m}_i{x}_i}{M},\\ y_{\text{COM}}=\frac{\underset{i=1}{\overset{n}{\sum }}{m}_i{y}_i}{M},\\ z_{\text{COM}}=\frac{\underset{i=1}{\overset{n}{\sum }}{m}_i{z}_i}{M}.\end{array}$$ Now we should focus on writing in the vector form by combining these. 

Formula for Centre of Mass for a System of Particles

For a system of $n$  particles with masses $m_1,m_2,\dots ,m_n$  at position vectors ${\vec{r}}_1,{\vec{r}}_2,\dots ,{\vec{r}}_n$ relative to an origin, the position vector is there. So the formula for centre of mass is 

${\vec{r}}_{\mathrm{C}\mathrm{O}\mathrm{M}}=\frac{\underset{i=1}{\overset{n}{\sum }} m_i{\vec{r}}_i}{\underset{i=1}{\overset{n}{\sum }} m_i}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2+\cdots +m_n{\vec{r}}_n}{M}$

where $M=m_1+m_2+\cdots +m_n$  is the total mass.

Formula for Centre of Mass for a Continuous Body

For a continuous body, the COM is calculated using integration.  

${\vec{r}}_{\mathrm{C}\mathrm{O}\mathrm{M}}=\frac{1}{M}\int \vec{r}dm$

where $dm$  is an infinitesimal mass element at position $\vec{r}$ .

Also, one to remember here is that continuous body is an object where matter is distributed across its volume. There are no gaps in such bodies. You will find infinitely many infinitesimal mass elements, and that's we use integral instead of summing them.

Previously, we learned about calculating the position of a centre of mass. The next logical step is to understand the motion of it. While we have a detailed article on the motion of a centre of mass that aligns with NCERT syllabus, here is a quick look at the mechanics of it. That comprises both velocity and acceleration, and how they relate to force in general. 

Velocity of Centre of Mass Formula

The velocity of centre of mass tells us how fast or slow an entire system of particles will move. The only condition is the mass must be concentrated at one point, that is, the centre of mass

. 

$\stackrel{\to }{v}{\mathrm{COM}}_{}=\frac{1}{M}\underset{i=1}{\sum }{n}^{\mathrm{}}{m}_i\stackrel{\to }{v_i}$

 

Acceleration of Centre of Mass Formula

Same as you learnt about kinematics in previous chapters, acceleration is a derivative of velocity. The formula is

${\stackrel{\to }{a}}_{\mathrm{COM}}=\frac{1}{M}\underset{i=1}{\sum }{n}^{}{m}_i{\stackrel{\to }{a}}_i$

This formula is still the mass-weighted average of the acceleration of each particle in a system. 

Relation to External Force: Applying Newtonian Laws

Yes, Newton's 2nd Law shows up in motion of a centre of mass. 

${\stackrel{\to }{F}}_{\mathrm{ext}}=M{\stackrel{\to }{a}}_{\mathrm{COM}}$

This equation is extremely important for exams and future rotational motion subjects, and it tells us that internal forces are the ones that forces particles exert on each other. They also cancel each other out due to Newton’s third law.

Only external forces decide how the centre of mass moves.

 

While solving the Class 11 NCERT Solutions for Chapter 6 Physics, you should already be able to picture where the centre of mass is for different types of bodies. 

In general, for uniform rigid bodies, the centre of mass often lies at the geometric centre.

The meaning of uniform rigid bodies is that they have the same distribution of mass throughout the body. The density of these bodies remains the same on every side. 

Here are some different types of uniform rigid bodies. See where the centre of mass lies, so that it's faster for you to solve numerical problems. 

• Uniform Rod (length $L$ ): COM at $L/2$  from either end.
• Uniform Disc (radius $R$ ): COM at the centre.
• Uniform Sphere (radius $R$ ): COM at the centre.
• Triangular Lamina: COM at the centroid, located at $\frac{h}{3}$ from the base, where $h$  is the height.
  For a half-disc of radius $R$ , the COM is at $\frac{4R}{3\pi }$ from the centre along the symmetry axis.

It's important to know that non-uniform bodies have cavities. The centre of mass won't exactly be at the geometric centre. The centre of mass or balance point shifts.

Let's say we have a disc with a circular cavity. We will have to treat the negative mass to find the new centre of mass.

The COM of this non-uniform body is: ${\vec{r}}_{\mathrm{C}\mathrm{O}\mathrm{M}}=\frac{M{\vec{r}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}}-m{\vec{r}}_{\mathrm{c}\mathrm{a}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}}}{M-m}$

where $M$  is the original disc's mass, and $m$  is the cavity's mass.

 

Here is the approach to solve questions on finding centre of mass. The main key is breaking any question down into manageable parts. 

• Start by defining the coordinate system. You can initially choose a convenient origin. Look into the symmetry of the system. Sometimes there would be the data already in the question, so don't forget that either. 
• Focus on representing small mass elements as functions of position, such as dm=λdx.
• Set up the integral or summation, using the relevant formulas for centre of mass. For continuous mass distributions: x_COM = (1/M) ∫ x dm. For discrete point masses: x_COM = (1/M) Σ m_i x_i. 
• Check units, interpret results, and verify symmetry.