The Centre of Mass

A spinning, tumbling object still has one point that moves like a simple particle.

11 minUpdated 21 October 2026centre of masssystem of particlesinverse ratio rule

Lesson at a glanceShow the key points

Centre of mass: R = (Σ mᵢ rᵢ) / M, with M = Σ mᵢ; componentwise X = Σmᵢxᵢ/M etc.
For two particles it lies on the line joining them and divides it in the inverse ratio of the masses: m₁(OC) = m₂(CP).
A body made of parts can be replaced by point masses at each part's own centre of mass.
A thrown, spinning ball's centre still follows a simple parabola.

A spin bowler releases a ball that twists and dips, every point on its surface tracing a tangled path. Yet its centre glides along a perfect parabola. That centre is special.

One point that behaves

Throw a spinning ball straight up. A mark on its seam corkscrews through space, but the centre of the ball goes straight up and down. Lob it to a fielder and the centre traces a parabola, exactly like a thrown pebble, however much the ball spins. (Air adds small deviations that bowlers exploit; we ignore them here.) The centre of a ball is an example of a centre of mass: the point that moves as if it were a single particle carrying all the mass.

Definition

Take N particles; particle i has mass $m_i$ and position $\vec r_i = (x_i, y_i, z_i)$ . Let $M = \sum m_i$ be the total mass.

Definition · Centre of mass

The point with position vector

\vec R_{cm} = \dfrac{1}{M}\sum_i m_i\vec r_i

: a mass-weighted average of the particles’ positions.

Coordinates of the centre of mass(9.1, 9.2)

X = \frac{\sum m_ix_i}{M}, \qquad Y = \frac{\sum m_iy_i}{M}, \qquad Z = \frac{\sum m_iz_i}{M}

Interactive

Find the balance point

Four point masses, in kilograms and metres. Drag them around and change their masses. The gold cross is the centre of mass: a mass-weighted average position.

M = \sum m_i

10.0kg

X = \frac{\sum m_ix_i}{M}

1.00m

Y = \frac{\sum m_iy_i}{M}

1.40m

Mass 12.0kg

Mass 21.0kg

Mass 34.0kg

Mass 43.0kg

Worked example

Four masses on a square

Masses of 2, 1, 4 and 3 kg sit at the corners A(0, 0), B(2, 0), C(2, 2) and D(0, 2) of a square of side 2 m. Locate the centre of mass.

Two particles

Put $m_1$ at the origin and $m_2$ a distance d away on the x-axis. Then $X = m_2 d/(m_1 + m_2)$ : the centre of mass lies on the line joining them, at distances

OC = \frac{m_2 d}{m_1 + m_2}, \quad CP = \frac{m_1 d}{m_1 + m_2} \quad\Rightarrow\quad m_1(OC) = m_2(CP) \qquad (9.3)

Combining parts

Split a system into two groups with masses $M_1, M_2$ and centres of mass at $X_1, X_2$ . The sum $\sum m_ix_i$ over the first group is just $M_1X_1$ , and likewise for the second, so

X = \frac{M_1X_1 + M_2X_2}{M_1 + M_2}

Replace each part by a point mass at its own centre of mass, then combine. This trick saves enormous effort.

Worked example

An L-shaped frame

Two identical uniform rods, each of length L, are joined at one end to make an L shape. Where is the centre of mass?

Worked example

Three masses on a right triangle

Masses 1 kg at (0, 0), 2 kg at (6 cm, 0) and 3 kg at (0, 8 cm). Locate the centre of mass.

What to remember

1 $\vec R_{cm} = \sum m_i\vec r_i / M$ , a mass-weighted average position.
2Two particles: on the joining line, closer to the heavier, in the inverse ratio of masses.
3Composite bodies: replace each part by a point mass at its own centre of mass.
4The centre of mass can lie outside the material (rings, L-shapes, boomerangs).

Check yourself

A 3 kg and a 1 kg mass are 2 m apart. The centre of mass is how far from the 3 kg mass?

Where is the centre of mass of a uniform ring?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadCenter of massOpenStax University Physics §9.6 ReadCenter of massHyperPhysics ReadCenter of massWikipedia ReadImpacts and linear momentum (unit)Khan Academy

Next Centre of Mass of Solid Bodies

The Centre of Mass

A spinning, tumbling object still has one point that moves like a simple particle.

11 minUpdated 21 October 2026centre of masssystem of particlesinverse ratio rule

Lesson at a glanceShow the key points

Centre of mass: R = (Σ mᵢ rᵢ) / M, with M = Σ mᵢ; componentwise X = Σmᵢxᵢ/M etc.
For two particles it lies on the line joining them and divides it in the inverse ratio of the masses: m₁(OC) = m₂(CP).
A body made of parts can be replaced by point masses at each part's own centre of mass.
A thrown, spinning ball's centre still follows a simple parabola.

A spin bowler releases a ball that twists and dips, every point on its surface tracing a tangled path. Yet its centre glides along a perfect parabola. That centre is special.

One point that behaves

Definition

Take N particles; particle i has mass $m_i$ and position $\vec r_i = (x_i, y_i, z_i)$ . Let $M = \sum m_i$ be the total mass.

Definition · Centre of mass

The point with position vector

\vec R_{cm} = \dfrac{1}{M}\sum_i m_i\vec r_i

: a mass-weighted average of the particles’ positions.

Coordinates of the centre of mass(9.1, 9.2)

X = \frac{\sum m_ix_i}{M}, \qquad Y = \frac{\sum m_iy_i}{M}, \qquad Z = \frac{\sum m_iz_i}{M}

Interactive

Find the balance point

Four point masses, in kilograms and metres. Drag them around and change their masses. The gold cross is the centre of mass: a mass-weighted average position.

M = \sum m_i

10.0kg

X = \frac{\sum m_ix_i}{M}

1.00m

Y = \frac{\sum m_iy_i}{M}

1.40m

Mass 12.0kg

Mass 21.0kg

Mass 34.0kg

Mass 43.0kg

Worked example

Four masses on a square

Masses of 2, 1, 4 and 3 kg sit at the corners A(0, 0), B(2, 0), C(2, 2) and D(0, 2) of a square of side 2 m. Locate the centre of mass.

Two particles

Put $m_1$ at the origin and $m_2$ a distance d away on the x-axis. Then $X = m_2 d/(m_1 + m_2)$ : the centre of mass lies on the line joining them, at distances

OC = \frac{m_2 d}{m_1 + m_2}, \quad CP = \frac{m_1 d}{m_1 + m_2} \quad\Rightarrow\quad m_1(OC) = m_2(CP) \qquad (9.3)

Combining parts

Split a system into two groups with masses $M_1, M_2$ and centres of mass at $X_1, X_2$ . The sum $\sum m_ix_i$ over the first group is just $M_1X_1$ , and likewise for the second, so

X = \frac{M_1X_1 + M_2X_2}{M_1 + M_2}

Replace each part by a point mass at its own centre of mass, then combine. This trick saves enormous effort.

Worked example

An L-shaped frame

Two identical uniform rods, each of length L, are joined at one end to make an L shape. Where is the centre of mass?

Worked example

Three masses on a right triangle

Masses 1 kg at (0, 0), 2 kg at (6 cm, 0) and 3 kg at (0, 8 cm). Locate the centre of mass.

What to remember

1 $\vec R_{cm} = \sum m_i\vec r_i / M$ , a mass-weighted average position.
2Two particles: on the joining line, closer to the heavier, in the inverse ratio of masses.
3Composite bodies: replace each part by a point mass at its own centre of mass.
4The centre of mass can lie outside the material (rings, L-shapes, boomerangs).

Check yourself

A 3 kg and a 1 kg mass are 2 m apart. The centre of mass is how far from the 3 kg mass?

Where is the centre of mass of a uniform ring?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadCenter of massOpenStax University Physics §9.6 ReadCenter of massHyperPhysics ReadCenter of massWikipedia ReadImpacts and linear momentum (unit)Khan Academy

Next Centre of Mass of Solid Bodies