Centre of Mass — Definition and Calculation

In short

The centre of mass of a system is the mass-weighted average position: \vec{R}_{\text{cm}} = \frac{1}{M}\sum m_i \vec{r}_i, where M is the total mass. For two particles on a line, x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}. It is the unique point that moves under external forces as if the entire mass of the system were concentrated there.

Try to balance a cricket bat horizontally on one finger. Slide your finger along its length until it sits level — neither the handle tipping down nor the blade. The balance point is not at the geometric middle. It is closer to the blade, because the blade is heavier and wider than the handle. That point, where the bat balances perfectly, is the centre of mass.

Every object has one. Every collection of objects has one. Toss a spinning wrench into the air and film it in slow motion: every point on the wrench traces a complicated, wobbling path, except one. That single point glides along a smooth parabola — exactly the trajectory a point particle would follow under gravity. That calm point is the centre of mass. It is the simplest idea in multi-body physics: the one point where all the mass might as well be.

The centre of mass is not just a physics abstraction. When ISRO's PSLV separates its stages mid-flight, the centre of mass of each stage determines its trajectory. When a diver on a springboard tucks and spins, her centre of mass follows the same parabola whether she tucks tight or stays straight — the rotation happens around the CM, not instead of it. Understanding where the centre of mass is, and how to calculate it, is the first step in every multi-body problem.

The balance point of two masses

Start with the simplest possible system: two stones on a light ruler.

Place a 1 kg stone at the left end of a 4 m ruler and a 3 kg stone at the right end. Where should you place your finger to balance the ruler horizontally?

If both stones were equal, the answer would be the midpoint — 2 m from each end. But the 3 kg stone is heavier. It pulls the balance point toward itself. Your finger needs to go at 3 m from the left end — three-quarters of the way toward the heavy stone.

A 1 kg stone at the left end and a 3 kg stone at the right end of a 4 m ruler. The balance point sits at 3 m from the left — three times closer to the heavier stone.

Why 3 m? The condition for balance is that the torques (twisting effects) about the fulcrum cancel:

1 \times 3 = 3 \times 1

The light mass exerts a torque of 1 \times 3 = 3 N·m trying to rotate the ruler counterclockwise. The heavy mass exerts 3 \times 1 = 3 N·m clockwise. They cancel. The ruler balances.

The balance point — the centre of mass — is the mass-weighted average of the two positions. The heavier mass counts more, so it pulls the average toward itself. This is the entire idea. Everything that follows is making it precise.

The formula — centre of mass of a discrete system

Two particles on a line

Set up a coordinate axis along the ruler, with the origin at the left end. Mass m_1 = 1 kg sits at position x_1 = 0. Mass m_2 = 3 kg sits at x_2 = 4 m. Place the fulcrum at some unknown position x_{\text{cm}}.

Step 1. Write the torque-balance condition.

m_1(x_{\text{cm}} - x_1) = m_2(x_2 - x_{\text{cm}}) \tag{1}

Why: for the system to balance, the torque from m_1 about the fulcrum must equal the torque from m_2. The left mass tries to rotate counterclockwise (moment arm x_{\text{cm}} - x_1), the right mass tries to rotate clockwise (moment arm x_2 - x_{\text{cm}}).

Step 2. Expand both sides.

m_1 x_{\text{cm}} - m_1 x_1 = m_2 x_2 - m_2 x_{\text{cm}}

Why: distribute each mass across the subtraction to collect x_{\text{cm}} terms on one side.

Step 3. Move all x_{\text{cm}} terms to the left.

m_1 x_{\text{cm}} + m_2 x_{\text{cm}} = m_1 x_1 + m_2 x_2

x_{\text{cm}}(m_1 + m_2) = m_1 x_1 + m_2 x_2

Why: factoring out x_{\text{cm}} isolates it as the only unknown.

Step 4. Solve for x_{\text{cm}}.

\boxed{x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}} \tag{2}

Why: dividing by the total mass gives the weighted average. Each position is weighted by its mass — a heavier particle contributes more, pulling x_{\text{cm}} toward itself.

Check with the numbers: x_{\text{cm}} = \frac{1 \times 0 + 3 \times 4}{1 + 3} = \frac{12}{4} = 3 m. That matches the balance point.

This formula is a weighted average in exactly the way your exam percentage is a weighted average. If physics has twice the weightage of chemistry in an entrance exam, and you score 95 in physics and 80 in chemistry, your weighted average is not \frac{95 + 80}{2} = 87.5 but \frac{2 \times 95 + 1 \times 80}{3} = 90. The subject with more weight pulls the average toward itself. The centre of mass formula does the same thing: each position is weighted by its mass, and a heavier particle pulls the result toward itself.

Special cases worth remembering:

Equal masses (m_1 = m_2): x_{\text{cm}} = \frac{x_1 + x_2}{2} — the midpoint. Two equal masses always balance at the geometric centre.
One mass dominates (m_2 \gg m_1): x_{\text{cm}} \approx x_2. The centre of mass sits practically on top of the heavy mass. The Earth–Moon system has its CM about 4,670 km from Earth's centre — still inside the Earth, because Earth is 81 times heavier than the Moon.

A key property: The centre of mass always lies on the line segment connecting the two particles. It cannot be to the left of x_1 or to the right of x_2. This is because the weights \frac{m_1}{m_1+m_2} and \frac{m_2}{m_1+m_2} are both positive and sum to 1 — the CM is a convex combination of the two positions.

Three particles, and the general formula

The same weighted-average logic extends to any number of particles. For three particles, the x-coordinate of the centre of mass is:

x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}

Each mass contributes its position, weighted by how heavy it is. For a system spread across two or three dimensions, you apply the same weighted average to each coordinate independently.

Centre of mass of $n$ particles

For n particles with masses m_1, m_2, \ldots, m_n at positions \vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n, the centre of mass is:

\vec{R}_{\text{cm}} = \frac{1}{M}\sum_{i=1}^{n} m_i \vec{r}_i

where M = \sum_{i=1}^{n} m_i is the total mass. In component form:

x_{\text{cm}} = \frac{1}{M}\sum_{i=1}^{n} m_i x_i, \qquad y_{\text{cm}} = \frac{1}{M}\sum_{i=1}^{n} m_i y_i, \qquad z_{\text{cm}} = \frac{1}{M}\sum_{i=1}^{n} m_i z_i

Reading the formula. Each particle contributes its position \vec{r}_i to the average, weighted by its fraction of the total mass m_i/M. A particle that carries half the system's mass single-handedly pulls the centre of mass halfway toward itself. The denominator M normalises the result so that \vec{R}_{\text{cm}} always lies within the convex hull of the particles — the centre of mass can never be farther out than the outermost particle.

Think of it as placing weights on a number line. The centre of mass is the point where that number line would balance on a fulcrum. A heavier weight shifts the balance toward itself — exactly the physical picture from the ruler experiment, extended to any number of masses in any number of dimensions.

Explore the weighted average yourself

Fix two particles at positions x_1 = 2 m and x_2 = 8 m. The interactive below lets you drag the mass ratio m_2/m_1 and watch the centre of mass slide between them. At ratio 1, the CM is at the midpoint (5 m). As m_2 grows heavier, the CM moves toward 8 m.

Two particles sit at $x_1 = 2$ m and $x_2 = 8$ m. Drag the red dot to change the mass ratio $m_2/m_1$. At ratio 1, the centre of mass is at the midpoint (5 m). As $m_2$ gets heavier, the CM slides toward $x_2 = 8$ m.

Notice the curve's shape: it rises steeply for small ratios (the CM is very sensitive when the masses are close) and flattens at large ratios (once m_2 is already dominant, making it even heavier barely shifts the CM further). The CM can never reach x_2 = 8 m because that would require m_2/m_1 = \infty — an infinite mass ratio.

Symmetry — when the answer is obvious

For a uniform body — one with mass spread evenly throughout — the centre of mass lies on every axis of symmetry. The argument is simple: for every small piece of mass on one side of a symmetry axis, there is an identical piece at the mirror position on the other side. Their contributions to \vec{R}_{\text{cm}} cancel exactly, so the CM must lie on the axis itself.

If a uniform body has two perpendicular axes of symmetry, the CM is at their intersection. A uniform square plate? Both diagonals are symmetry axes, and the CM is at the centre. A uniform equilateral triangle? The three medians are symmetry axes, so the CM is at their intersection — the centroid, one-third of the altitude from each base.

This gives you the centre of mass of many standard shapes without any calculation:

Uniform shape	Centre of mass
Straight rod	Midpoint
Rectangular plate	Intersection of diagonals
Circular disc	Geometric centre
Sphere	Geometric centre
Cube	Geometric centre
Ring (hoop)	Geometric centre

The ring is worth pausing on.

The centre of mass of a uniform ring is at its geometric centre — a point in empty space.

The centre of mass of a uniform ring is at the geometric centre — a point in empty space where there is no material at all. A bangle, a horseshoe, a boomerang — all have their CM outside the physical material. The centre of mass is a mathematical point, not a physical lump. It can be inside the body, on its surface, or outside it entirely.

The subtraction method — bodies with missing pieces

A favourite pattern on JEE and board exams: a uniform disc of radius R with a smaller disc of radius r cut out. Where is the centre of mass of what remains?

The key insight is that the centre of mass formula is additive: the CM of a composite body is the mass-weighted average of the CMs of its parts. The original disc (before cutting) is the remaining piece plus the removed piece:

M_{\text{orig}} \, \vec{R}_{\text{orig}} = M_{\text{rem}} \, \vec{R}_{\text{rem}} + M_{\text{cut}} \, \vec{R}_{\text{cut}}

Why: the CM of the original disc is a weighted average of the CMs of its two parts — the remaining piece and the removed piece. This is the additive property of the CM formula.

Rearranging to solve for the remaining piece's CM:

\vec{R}_{\text{rem}} = \frac{M_{\text{orig}} \, \vec{R}_{\text{orig}} - M_{\text{cut}} \, \vec{R}_{\text{cut}}}{M_{\text{orig}} - M_{\text{cut}}} \tag{3}

Why: isolate \vec{R}_{\text{rem}} by moving the cut-out term to the other side. The denominator is M_{\text{rem}} = M_{\text{orig}} - M_{\text{cut}}.

Here is the physical intuition. Before you cut the hole, the disc's CM is at the geometric centre. The material that you are about to remove is on one side of the centre. It was pulling the CM in its direction. Once you remove it, that pull disappears, and the CM of the remaining piece shifts to the opposite side — away from the hole.

This technique works because the CM formula is linear in mass. You can freely split a body into parts, compute each part's CM separately, and recombine them with a mass-weighted average. The subtraction method is just this additivity used in reverse: instead of combining parts to find a whole, you subtract a known part from the whole to find what remains.

Worked examples

Example 1: Three autorickshaws at an intersection

Three autorickshaws are parked near a road intersection. Their masses (including passengers) and positions, measured from a reference corner, are:

Rickshaw	Mass	Position
A	400 kg	(2, 1) m
B	600 kg	(8, 3) m
C	500 kg	(5, 7) m

Find the centre of mass of the three-rickshaw system.

Step 1. Compute the total mass.

M = 400 + 600 + 500 = 1500 \text{ kg}

Why: this is the denominator in the CM formula. Each rickshaw's contribution is proportional to its share of the total mass.

Step 2. Compute x_{\text{cm}}.

x_{\text{cm}} = \frac{400 \times 2 + 600 \times 8 + 500 \times 5}{1500} = \frac{800 + 4800 + 2500}{1500} = \frac{8100}{1500} = 5.4 \text{ m}

Why: B is the heaviest and sits at x = 8, pulling the CM to the right. At x_{\text{cm}} = 5.4, the CM is closer to B than to A — exactly what the mass-weighting predicts.

Step 3. Compute y_{\text{cm}}.

y_{\text{cm}} = \frac{400 \times 1 + 600 \times 3 + 500 \times 7}{1500} = \frac{400 + 1800 + 3500}{1500} = \frac{5700}{1500} = 3.8 \text{ m}

Why: C sits at y = 7 and has significant mass (500 kg), pulling the CM upward from the relatively low positions of A (y = 1) and B (y = 3).

Step 4. Verify that the CM is inside the triangle.

The three positions form a triangle. The CM at (5.4, 3.8) lies inside it — as it must for any set of positive masses.

Result: The centre of mass is at (5.4, \; 3.8) m.

The three autorickshaws form a triangle (dashed lines). The centre of mass (red dot at 5.4, 3.8) sits inside the triangle, pulled toward the heaviest rickshaw B.

What this shows: The CM lies inside the triangle formed by the three masses, pulled toward the heaviest one. If you placed all three rickshaws on a giant rigid platform and tried to balance it on a single support, the support would need to go at (5.4, 3.8).

Example 2: A disc with a circular hole

A uniform circular disc of radius R = 6 cm has a circular hole of radius r = 2 cm punched out. The centre of the hole is 3 cm to the right of the disc's centre. Find how far the centre of mass shifts from the disc's original centre.

A uniform disc (radius 6 cm) with a 2 cm hole punched out, centred 3 cm to the right of O. The CM of the remaining piece (red dot) shifts to the left.

Step 1. Set up coordinates with the origin at the disc's centre O.

The full disc has its CM at O: x_{\text{orig}} = 0. The hole has its CM at x_{\text{cut}} = 3 cm (the geometric centre of the hole).

Why: both the full disc and the cut-out disc are uniform circles, so each has its CM at its own geometric centre.

Step 2. Express masses in terms of area.

For a uniform disc with surface mass density \sigma (mass per unit area):

M_{\text{orig}} = \sigma \pi R^2 = 36\sigma\pi, \qquad M_{\text{cut}} = \sigma \pi r^2 = 4\sigma\pi

M_{\text{rem}} = M_{\text{orig}} - M_{\text{cut}} = 32\sigma\pi

Why: since the disc is uniform, mass is proportional to area. The ratio of areas is r^2 / R^2 = 4/36 = 1/9, so the hole is 1/9 of the full disc's mass. The common factor \sigma\pi will cancel.

Step 3. Apply the subtraction formula (equation 3).

x_{\text{rem}} = \frac{M_{\text{orig}} \cdot x_{\text{orig}} - M_{\text{cut}} \cdot x_{\text{cut}}}{M_{\text{rem}}} = \frac{36\sigma\pi \times 0 - 4\sigma\pi \times 3}{32\sigma\pi}

= \frac{-12\sigma\pi}{32\sigma\pi} = -\frac{3}{8} \text{ cm} = -0.375 \text{ cm}

Why: the \sigma\pi cancels completely — the CM position depends only on the geometry, not on the actual density. The negative sign means the CM shifted to the left, away from the hole.

Step 4. Verify the direction makes sense.

The removed material was on the right side. It was pulling the CM rightward. Removing it shifts the CM leftward. The result x_{\text{cm}} = -3/8 cm (to the left of O) is consistent.

Result: The centre of mass of the remaining disc is at x_{\text{cm}} = -\frac{3}{8} cm, shifted 0.375 cm to the left of the original centre.

What this shows: Removing material from one side shifts the CM to the opposite side. The shift is small (0.375 cm for a 6 cm disc) because the hole is small compared to the whole. A larger hole would produce a larger shift.

Common confusions

"The centre of mass must be inside the body." Not always. A uniform ring's CM is at the geometric centre — a point in empty space, as shown above. A horseshoe, a hollow hemisphere, a crescent-shaped piece of metal — all can have their CM outside the physical material. The CM is a mathematical point, not a physical lump.
"Centre of mass means geometric centre." Only for uniform bodies with symmetry. A cricket bat is denser and wider at the blade than the handle, so its CM is not at the midpoint of its length. For any non-uniform mass distribution, the CM shifts toward the denser region.
"A heavier system has its centre of mass farther from the origin." The CM position depends on where the masses are, not on how much total mass there is. If you double every mass in a system, the CM does not move — the numerator and denominator in the formula both double, so their ratio stays the same.
"The subtraction method uses negative mass." Not literally. No object has negative mass. The minus sign in equation (3) arises from rearranging the additive property: original = remaining + removed. Solving for the remaining piece naturally introduces a subtraction. It is algebra, not exotic physics.
"You need the actual masses to use the CM formula." For uniform bodies, you do not. In Example 2, the density \sigma cancelled entirely — only the ratio of areas (which equals the ratio of masses for uniform density) mattered. This is why the subtraction method is so powerful for uniform bodies: you work with area ratios instead of masses.

If you came here to calculate centres of mass for discrete systems and bodies with holes, you have the tools. What follows is the deeper reason why physicists care about the centre of mass — it is the point that obeys Newton's second law for the entire system.

Why the centre of mass moves as if all mass were concentrated there

Take a system of n particles. Particle i has mass m_i, position \vec{r}_i, and experiences two kinds of forces: external forces \vec{F}_i^{\,\text{ext}} (gravity, applied pushes, friction from outside the system) and internal forces \vec{F}_{ij} from every other particle j in the system. Newton's second law for particle i:

\vec{F}_i^{\,\text{ext}} + \sum_{j \neq i} \vec{F}_{ij} = m_i \ddot{\vec{r}}_i \tag{4}

Why: the net force on particle i is the sum of all forces on it — external forces from outside the system plus internal forces from other particles within the system.

Now sum equation (4) over all n particles:

\sum_{i} \vec{F}_i^{\,\text{ext}} + \sum_{i}\sum_{j \neq i} \vec{F}_{ij} = \sum_{i} m_i \ddot{\vec{r}}_i \tag{5}

The double sum over internal forces vanishes. By Newton's third law, \vec{F}_{ij} = -\vec{F}_{ji}, so every internal force pair cancels:

\vec{F}_{\text{ext}} = \sum_{i} m_i \ddot{\vec{r}}_i \tag{6}

Why: internal forces are action–reaction pairs. When you add them all up, each pair sums to zero. Only the total external force survives.

Now use the definition of the centre of mass: M\vec{R}_{\text{cm}} = \sum_i m_i \vec{r}_i. Differentiate both sides twice with respect to time:

M \ddot{\vec{R}}_{\text{cm}} = \sum_{i} m_i \ddot{\vec{r}}_i \tag{7}

Why: the masses m_i are constants, so differentiating the sum just differentiates each \vec{r}_i. The right-hand side of (7) is identical to the right-hand side of (6).

Combining (6) and (7):

\boxed{\vec{F}_{\text{ext}} = M \, \vec{a}_{\text{cm}}} \tag{8}

This is Newton's second law for the centre of mass. The total external force on the system equals the total mass times the acceleration of the centre of mass. No matter how the individual particles jiggle, spin, collide, or fragment internally, the centre of mass accelerates as if the entire mass M were a single particle sitting at \vec{R}_{\text{cm}}, acted on by \vec{F}_{\text{ext}} alone.

This is why the spinning wrench's CM follows a parabola: gravity is the only external force, so \vec{a}_{\text{cm}} = \vec{g}, and the CM undergoes free fall along a smooth parabolic arc — regardless of the wrench's complicated rotation.

This is also why the centre of mass of a Diwali rocket continues on a smooth parabolic path even after the rocket explodes mid-flight. The explosion produces enormous internal forces between the fragments, but internal forces cancel in the sum. Only gravity (the external force) determines the CM's trajectory. An observer tracking the CM with a slow-motion camera would see absolutely no effect from the explosion.

The centre-of-mass frame

A reference frame moving with the centre of mass is called the CM frame. In this frame, the total momentum of the system is always zero:

\vec{P}_{\text{CM frame}} = \sum m_i(\vec{v}_i - \vec{v}_{\text{cm}}) = \sum m_i \vec{v}_i - M\vec{v}_{\text{cm}} = \vec{P}_{\text{total}} - \vec{P}_{\text{total}} = 0

Why: subtracting \vec{v}_{\text{cm}} from every particle's velocity removes the system's overall drift, leaving a frame where total momentum is zero by construction.

This makes the CM frame invaluable for analysing collisions. In the CM frame, two colliding objects always approach with equal and opposite momenta and separate with equal and opposite momenta. The collision looks symmetric even if the masses are unequal. You can see this simplification at work in the elastic collisions article, where the CM-frame view turned a messy algebraic derivation into a one-step velocity reversal.

Equation (8) is the reason the centre of mass is worth defining. It is not merely a geometric curiosity or a balancing trick. It is the point where a complex system of many bodies reduces to a single body. Every problem in multi-body mechanics — connected masses on a pulley, rocket propulsion, collision analysis, orbital mechanics — uses this result as the starting point. When you compute the centre of mass of a system, you are finding the one point whose motion tells you everything about how the system as a whole responds to external forces.

Where this leads next

Centre of Mass of Continuous Bodies — replacing the sum with an integral to find the CM of rods, semicircular wires, discs, hemispheres, and cones.
Conservation of Linear Momentum — when \vec{F}_{\text{ext}} = 0, the centre of mass moves at constant velocity, which is the foundation of all collision physics.
Elastic Collisions — using the CM frame to solve head-on and oblique collisions elegantly.
Variable Mass Systems — Rockets — what happens when the system gains or loses mass, and how the CM equation adapts.
Collisions in Two Dimensions — oblique collisions where the CM frame reveals the geometry behind scattering angles.