Centre of mass

Push a spinning wrench across a table and it tumbles in a wild, complicated way — yet one point inside it, the centre of mass, glides in a perfectly straight line, exactly as a single free particle would. That one fact is why the centre of mass matters: it lets you strip away every messy internal detail of a system — a wrench, a rocket that has just split into two pieces, a rod with a lump stuck on one end — and track its overall motion with the same simple laws you already use for a point particle. This article builds the idea from two masses on a line up to any number of particles, shows how differentiating the position formula hands you both the velocity and the acceleration of the centre of mass, and explains why internal forces can never move it — only a net external force can. That last fact is the seed of momentum conservation for many-particle systems.

The core insight

The centre of mass of a system is the mass-weighted average position $\vec r_{cm} = \frac{1}{M}\sum_i m_i\vec r_i$ , and it moves exactly as a single particle of mass $M$ would under the net external force alone, since internal forces always cancel in pairs.

The core idea

Two particles first. For point masses $m_1$ and $m_2$ sitting at positions $x_1$ and $x_2$ on a line, the centre of mass is the mass-weighted average of their positions, not the plain geometric midpoint:

x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1+m_2} = \frac{m_1x_1+m_2x_2}{M}

where: $M = m_1+m_2$ is the total mass, and $x_{cm}$ sits closer to whichever mass is larger. In three dimensions the same rule applies component by component, or compactly with position vectors, $\vec r_{cm} = (m_1\vec r_1 + m_2\vec r_2)/M$ .

Any number of particles. The two-particle idea extends directly (you are not expected to derive this in detail) to a system of $N$ particles:

\vec r_{cm} = \frac{\sum_i m_i \vec r_i}{\sum_i m_i} = \frac{1}{M}\sum_i m_i\vec r_i

where: $m_i$ and $\vec r_i$ are the mass and position of the $i$ -th particle, and $M=\sum_i m_i$ is the system's total mass. Every particle contributes to $\vec r_{cm}$ in proportion to its own mass.

Velocity and acceleration follow by differentiating. Since $M$ is constant, differentiating the position relation once with respect to time gives the velocity of the centre of mass, and a second time gives its acceleration — the full chain is worked step by step in the derivation below. The velocity relation shows that the total momentum of a system is just $M\vec v_{cm}$ , and the acceleration relation shows that only the net external force can change $\vec v_{cm}$ ; internal forces between particles (by Newton's third law) always occur in equal-and-opposite pairs and cancel completely when summed over the whole system.

Continuous, symmetric bodies. For a rigid body with uniform density and an axis, plane or point of symmetry, the centre of mass lies on that symmetry element. A uniform rod of length $L$ has its centre of mass at its geometric centre, a distance $L/2$ from either end — a standard result you should recognise by inspection, without needing to derive it by integration.

Figure

Two masses on a line with the centre of mass marked closer to the heavier mass

Derivation AVelocity and acceleration of the centre of mass

The goal is to start from the definition $\vec r_{cm} = \frac{1}{M}\sum_i m_i\vec r_i$ and show, by straightforward differentiation with respect to time, that $M\vec v_{cm}$ equals the total momentum of the system and that $M\vec a_{cm}$ equals only the net external force, so that internal forces between the particles never appear in the final result.

Step 1. Write the position of the centre of mass for $N$ particles of masses $m_i$ at positions $\vec r_i$ : $M\vec r_{cm} = \sum_i m_i\vec r_i$ , where $M=\sum_i m_i$ is fixed because no mass enters or leaves the system.

Step 2. Differentiate both sides once with respect to time. Since $M$ and each $m_i$ are constants, $M\dfrac{d\vec r_{cm}}{dt} = \sum_i m_i\dfrac{d\vec r_i}{dt}$ , which is $M\vec v_{cm} = \sum_i m_i\vec v_i$ . The right side is, by definition, the total momentum $\vec P$ of the system, so $M\vec v_{cm}=\vec P$ : the centre of mass moves as if it carries the system's entire momentum.

Step 3. Differentiate once more with respect to time. $M\dfrac{d\vec v_{cm}}{dt} = \sum_i m_i\dfrac{d\vec v_i}{dt}$ , which is $M\vec a_{cm} = \sum_i m_i\vec a_i = \sum_i \vec F_i$ , the vector sum of the total force on every particle.

Step 4. Split each $\vec F_i$ into an external part and an internal part (forces from the other particles of the same system). By Newton's third law, every internal force on particle $i$ from particle $j$ is matched by an equal and opposite force on particle $j$ from particle $i$ , so when the sum runs over all particles every internal pair cancels exactly, leaving $M\vec a_{cm} = \sum_i \vec F_i^{ext} = \vec F_{ext}$ : only the net external force moves the centre of mass, precisely as if the whole mass $M$ were one particle located there. A direct consequence is that if $\vec F_{ext}=0$ , then $\vec a_{cm}=0$ and $\vec P=M\vec v_{cm}$ stays constant — momentum conservation for a many-particle system.

Worked example

Two point masses, $m_1 = 2\ \text{kg}$ at $x_1=0$ and $m_2 = 3\ \text{kg}$ at $x_2 = 5\ \text{m}$ , already give a centre of mass at $x_{cm}=3\ \text{m}$ (their mass-weighted average). Now a third mass $m_3 = 5\ \text{kg}$ is placed at $x_3 = 9\ \text{m}$ on the same line, and the three masses move with velocities $v_1 = 4\ \text{m/s}$ , $v_2 = -2\ \text{m/s}$ and $v_3 = 1\ \text{m/s}$ along the line. Find (a) the new centre of mass of the three-particle system and (b) the velocity of the centre of mass.

Step 1. Total mass: $M = m_1+m_2+m_3 = 2+3+5 = 10\ \text{kg}$ .

Step 2. Apply the $N$ -particle position formula: $x_{cm} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{M} = \dfrac{(2)(0)+(3)(5)+(5)(9)}{10} = \dfrac{0+15+45}{10} = 6.0\ \text{m}$ . Adding the heavier third mass farther out has pulled the centre of mass from $3\ \text{m}$ (with two masses) to $6.0\ \text{m}$ — a shift toward the newly added, larger mass, exactly as expected.

Step 3. Apply $M\vec v_{cm} = \sum_i m_i v_i$ : $v_{cm} = \dfrac{(2)(4)+(3)(-2)+(5)(1)}{10} = \dfrac{8-6+5}{10} = \dfrac{7}{10} = 0.7\ \text{m/s}$ .

The centre of mass is at $x_{cm}=6.0\ \text{m}$ , moving at $v_{cm}=0.7\ \text{m/s}$ along the line — found using exactly the same weighted-average rule for both position and velocity.

Common traps

Thinking the centre of mass must lie inside the material of a body — for separated point masses, a ring, or an L-shaped lamina, it can sit in empty space between or outside them.
Treating the centre of mass as the plain geometric average of positions — it is mass-weighted, so $x_{cm}=\sum m_ix_i/M$ shifts toward the heavier masses, not toward the visual middle.
Forgetting that only the net external force sets $\vec a_{cm}$ — internal forces (mutual pulls, an internal collision, an explosion within the system) always cancel and can never move the centre of mass.
Applying the velocity/momentum relation $M\vec v_{cm}=\sum m_iv_i$ with unsigned speeds when the motion is along a line — direction (sign) matters, as in the worked example where $v_2=-2\ \text{m/s}$ .

Ask the tutor: if you would like to see how this idea extends to a person walking on a boat or an exploding shell, ask the AI tutor to walk through it.

Why does only the external force matter, when there clearly are strong internal forces holding a body together?

Internal forces always occur in Newton's-third-law pairs — if particle A pulls on particle B, particle B pulls back on A with equal and opposite force. When you sum every force over the whole system to find $M\vec a_{cm}$ , every such pair adds to zero, no matter how strong the individual internal forces are. What survives the sum is only the force coming from outside the system. That is why a spinning, tumbling wrench thrown across a table still has a centre of mass that moves in a smooth, simple parabola under gravity alone.

Does the centre of mass formula change if the body is three-dimensional instead of a line of particles?

No — the same idea just applies to each coordinate, or compactly to the position vector: $\vec r_{cm} = \frac{1}{M}\sum_i m_i\vec r_i$ , which packages the $x$ , $y$ and $z$ mass-weighted averages into one vector equation. For a continuous body you would replace the sum with an integral over small mass elements, but at this stage you only need to locate the centre of mass of simple symmetric shapes, like a uniform rod, by inspection.

If I attach a small mass to one end of a uniform rod, why does the centre of mass move only part of the way toward that end, not all the way?

Because the centre of mass is a weighted average of the rod's own centre of mass and the added point mass, weighted by how much mass each contributes. The rod's mass, concentrated at its own midpoint, usually still outweighs a small attached mass, so the combined centre of mass shifts from the rod's midpoint toward the added mass, but by an amount that depends on the ratio of the two masses — the heavier the attachment relative to the rod, the closer the combined centre of mass moves toward it.

Centre of mass

The core idea

Worked example

Common traps

Before this

Where this leads