Centre of Mass: Formulas, Composite Bodies and Stability

The Point That Behaves Like the Whole Object

Throw a hammer spinning through the air and its motion looks complicated. But one point on it follows a perfectly smooth parabola, just like a thrown ball. That point is the centre of mass. It is the average position of all the mass in an object or system, and it lets us treat a complicated body, or a group of bodies, as if all its mass were concentrated at a single point. This tutorial explains the centre of mass formula and how to use it for particles, composite shapes and uniform bodies, its link with the centre of gravity and stability, the motion of the centre of mass, and applications from balancing to rockets and binary stars.

The Centre of Mass Formula for a System of Particles

For particles of masses m₁, m₂, m₃, … at positions x₁, x₂, x₃, … along a line, the centre of mass is at

x̄ = (m₁x₁ + m₂x₂ + m₃x₃ + …) / (m₁ + m₂ + m₃ + …) = Σmx / Σm.

In two dimensions, find x̄ and ȳ separately, using the y-coordinates for ȳ. It is a weighted average: heavier masses pull the centre of mass towards themselves.

Example 1. Masses of 2 kg, 3 kg and 5 kg are placed at x = 0, 4 m and 6 m. Find the centre of mass.

x̄ = (2 × 0 + 3 × 4 + 5 × 6) / (2 + 3 + 5) = (0 + 12 + 30)/10 = 4.2 m.

Example 2. Masses of 1 kg at (0, 0), 2 kg at (4, 0) and 3 kg at (2, 3). Find the centre of mass.

x̄ = (0 + 8 + 6)/6 = 14/6 ≈ 2.33 m; ȳ = (0 + 0 + 9)/6 = 1.5 m. The centre of mass is at (2.33, 1.5).

Two-body shortcut. For two masses a distance d apart, the centre of mass is closer to the heavier one, dividing the line in the inverse ratio of the masses: its distance from m₁ is d × m₂/(m₁ + m₂).

Centre of Mass of Uniform Bodies

For a uniform object with a line or point of symmetry, the centre of mass lies on it:

uniform rod: at its midpoint;
rectangle or square lamina: where the diagonals cross;
circle or sphere: at the centre;
triangular lamina: at the centroid, where the medians meet, one-third of the way up from each side. For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) it is at ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3);
solid cone: on the axis, a quarter of the height up from the base;
solid hemisphere: on the axis, 3r/8 from the flat face.

The centre of mass need not lie inside the object. For a ring, a boomerang or an L-shaped bracket, it lies in empty space.

Composite Bodies

To find the centre of mass of a shape built from simple parts, treat each part as a point mass at its own centre of mass, then use Σmx/Σm. For a uniform lamina, mass is proportional to area, so areas can be used instead of masses.

Example 3. An L-shaped uniform lamina is made from a 6 cm × 2 cm rectangle lying along the x-axis from (0, 0) to (6, 2), and a 2 cm × 4 cm rectangle standing on top of it from (0, 2) to (2, 6). Find the centre of mass.

Bottom rectangle: area 12 cm², centre (3, 1).
Upright rectangle: area 8 cm², centre (1, 4).
x̄ = (12 × 3 + 8 × 1)/20 = 44/20 = 2.2 cm; ȳ = (12 × 1 + 8 × 4)/20 = 44/20 = 2.2 cm.

The centre of mass is at (2.2, 2.2), which lies in the empty corner region outside the material, as expected for an L shape.

Example 4 (removing a part). A uniform circular disc of radius 4 cm has a circular hole of radius 2 cm cut out, centred 2 cm from the disc's centre. Find the centre of mass of the remaining shape.

Treat the hole as a negative mass. Put the disc's centre at the origin and the hole's centre at x = 2. Areas are proportional to r²: the full disc has "16 units", the hole 4 units, leaving 12.

x̄ = (16 × 0 − 4 × 2)/(16 − 4) = −8/12 = −0.67 cm. The centre of mass moves 0.67 cm away from the hole, as you would expect.

Centre of Mass and Centre of Gravity

The centre of gravity is the point through which the total weight of an object appears to act. In a uniform gravitational field, which is effectively true for everyday objects on the Earth, it is at the same place as the centre of mass. (They differ only for objects so large that g changes noticeably across them.)

A simple experiment finds the centre of gravity of an irregular lamina: hang it freely from a pin through one point and draw a vertical line down from the pin using a plumb line; repeat from a second point. Where the lines cross is the centre of gravity, because when an object hangs freely its centre of gravity always settles directly below the point of suspension.

Centre of Mass and Stability

An object resting on a base is stable as long as the vertical line through its centre of mass passes through its base. If it is tilted so far that this line falls outside the base, its weight produces a turning moment that topples it.

Objects are more stable when they have a low centre of mass and a wide base. This is why racing cars are low and wide, why heavy loads should be placed at the bottom of a lorry, why buses are tilt-tested, and why a person carrying a heavy bucket in one hand leans the other way to keep the combined centre of mass over their feet.

Example 5. A uniform box 0.6 m wide and 1.6 m tall stands on a slope. At what angle of slope will it topple (assuming it doesn't slide)?

Its centre of mass is 0.8 m above the base and 0.3 m from each edge. It topples when the vertical through the centre of mass passes the lower edge, at an angle θ where tan θ = 0.3/0.8, so θ ≈ 20.6°.

Motion of the Centre of Mass

The total momentum of a system equals its total mass times the velocity of its centre of mass:

P = M × v(cm), and F(external) = M × a(cm).

Internal forces between parts of the system cancel in pairs (Newton's third law) and cannot change the motion of the centre of mass. Only external forces can. This has striking consequences:

A firework shell that explodes in mid-air scatters fragments everywhere, but the centre of mass of all the fragments continues along the original parabola, until fragments start hitting the ground.
A person standing on frictionless ice cannot move their centre of mass by wriggling; they can only move by pushing something away, such as throwing a bag.
In a rocket, the exhaust gas goes one way and the rocket the other, while the centre of mass of rocket plus exhaust (in deep space) stays in uniform motion.

Example 6. A 60 kg person stands at one end of a 4.0 m, 40 kg boat at rest on still water. The person walks to the other end. How far does the boat move? (Ignore water resistance.)

No external horizontal force acts, so the centre of mass stays still. If the person moves 4.0 m relative to the boat and the boat moves d the other way, the person moves (4.0 − d) relative to the water. Setting the total movement of mass to zero: 60(4.0 − d) = 40d, so 240 = 100d and d = 2.4 m. The boat moves back 2.4 m and the person moves forward 1.6 m.

Binary Stars and the Earth–Moon System

Two bodies orbiting each other both move around their common centre of mass. The Earth is about 81 times as massive as the Moon, and they are 384,000 km apart, so their centre of mass (the barycentre) is 384,000/82 ≈ 4,700 km from the Earth's centre, still inside the Earth (radius 6,400 km). The Earth wobbles around this point every month. Astronomers detect planets around other stars by measuring the tiny wobble of the star around the star–planet centre of mass.

Summary

Centre of mass formula: x̄ = Σmx/Σm (and similarly ȳ), a mass-weighted average position.
For uniform symmetric bodies it lies on the line of symmetry; for composite bodies, combine the parts, using negative mass for holes.
In a uniform field the centre of gravity and centre of mass coincide.
Stability needs the vertical through the centre of mass to pass through the base: low and wide is stable.
Only external forces change the motion of the centre of mass: F = Ma(cm).