Centre of Mass

The centre of mass of a system is the single point that represents the average position of all the mass in that system. Think of it as the system's 'balance point': if you could support an object at exactly this point, it would balance perfectly in any orientation. For a uniform, symmetrical object like a ruler or a sphere, the centre of mass sits right at the geometric centre. For irregular objects or collections of particles, it shifts towards wherever most of the mass is concentrated. The concept is enormously powerful because Newton's second law — F = ma — applies to the centre of mass as if all the mass of the system were concentrated there. This means that when you watch a spinning hammer thrown through the air, its centre of mass follows a perfect parabola under gravity, even as every other point traces a complicated spiralling path. Physicists and engineers exploit this simplification constantly, from calculating rocket trajectories to designing stable vehicles and buildings. The centre of mass is not always inside the object itself. A doughnut's centre of mass sits in the hole, and a boomerang's lies somewhere in the air between its two arms. What matters is the distribution of mass, not the shape of the boundary. Understanding where the centre of mass lies is the first step in analysing how any system moves or rotates under the action of forces.

Key takeaways

Think of a Weighted Average
The Formula for Two Particles
Extending to Many Particles and 2D/3D
Continuous Objects
Why It Matters for Motion

Think of a Weighted Average

The centre of mass is simply a weighted average of positions, where each particle's position is weighted by its mass. If a heavy particle sits at x = 10 m and a light particle sits at x = 2 m, the centre of mass sits closer to x = 10 m — pulled towards the heavier mass, just as a weighted average is pulled towards the larger value.

The Formula for Two Particles

For two particles of masses m₁ and m₂ at positions x₁ and x₂ along a line, the centre of mass is x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂). The numerator is the sum of 'mass × position' for each particle (called the first moment of mass), and the denominator is the total mass. This ratio gives the single equivalent position.

Extending to Many Particles and 2D/3D

For any number of particles, you simply extend the sum: x_cm = Σ(mᵢxᵢ) / Σmᵢ. The same formula applies independently to the y and z directions. In two dimensions, you calculate x_cm and y_cm separately to locate the point (x_cm, y_cm) in the plane. This works because each spatial direction is independent.

Continuous Objects

For solid objects rather than discrete particles, the sum becomes an integral: x_cm = ∫x dm / M, where M is the total mass and dm is a tiny mass element at position x. In practice, for uniform objects you can use symmetry to locate the centre of mass by inspection — for any axis of symmetry, the centre of mass must lie on it.

Why It Matters for Motion

Newton's second law F_net = M a_cm tells us that the net external force on a system equals the total mass times the acceleration of the centre of mass. Internal forces — forces between particles within the system — cancel in Newton's third law pairs and do not affect the centre of mass motion. This is why an exploding firework's centre of mass continues on the same parabolic path even as fragments fly in all directions.

The Key Equation

\vec{r}_{\text{cm}} = \frac{\sum_{i} m_i \vec{r}_i}{\sum_{i} m_i} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \cdots + m_n\vec{r}_n}{M}

\vec{r}_{\text{cm}} — Centre of mass position vector (m): The vector pointing from the origin to the centre of mass of the system.
m_i — Mass of particle i (kg): The mass of the i-th particle in the system.
\vec{r}_i — Position vector of particle i (m): The vector from the origin to the i-th particle's location.
M — Total mass (kg): The sum of all individual masses: M = Σmᵢ.

Worked Example

Problem. Three students sit on a 4.0 m long, massless seesaw. Student A (mass 60 kg) sits at position x = 0 m, Student B (mass 40 kg) sits at x = 2.0 m, and Student C (mass 80 kg) sits at x = 4.0 m. Find the centre of mass of the three-student system.

x_{\text{cm}} = \frac{(60)(0)+(40)(2.0)+(80)(4.0)}{180} = \frac{400}{180} \approx 2.22 \text{ m}

Answer. The centre of mass is located 2.22 m from Student A's end of the seesaw.

Figure

Diagram

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Swiss

Euler formalised the mechanics of rigid bodies and systems of particles in the 18th century, developing the mathematical framework — including the rigorous treatment of the centre of mass — that underpins Newtonian mechanics. His work in 'Mechanica' (1736) and later treatises translated Newton's geometric methods into calculus, making it possible to calculate centres of mass and rotational dynamics for complex bodies systematically. Euler lost sight in his right eye around age 31 and became almost totally blind by age 59, yet he continued to produce brilliant mathematics by dictating to assistants — reportedly his output actually increased after losing his sight.

Newton's Second Law — The centre of mass obeys Newton's second law directly: the net external force equals total mass times the acceleration of the centre of mass, making F = ma the key tool for predicting centre of mass motion.
Systems of Particles — The centre of mass is defined for a system of particles; understanding how internal and external forces act on a multi-particle system is essential context for why the centre of mass concept is so useful.
Linear Momentum — The total momentum of a system equals total mass times the velocity of the centre of mass, linking conservation of momentum directly to centre of mass motion.
Torque and Rotational Equilibrium — For an object to be in rotational equilibrium, the net torque about the centre of mass must be zero; knowing where the centre of mass is located is essential for torque calculations.