In short

The centre of mass (CM) of a system of particles is the mass-weighted average position: \vec{R}_{\text{cm}} = \frac{\sum m_i \vec{r}_i}{\sum m_i}. For a uniform symmetric body, it sits at the geometric centre. For bodies with parts removed, use the subtraction method. The CM is the point that moves as if the total external force acts on the entire mass concentrated at that single point: \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}.

Throw a cricket bat into the air with a spin. It tumbles end over end — the handle swings one way, the blade swings the other, and the whole thing rotates in a complicated pattern that looks nothing like the clean parabola you see when you toss a ball. But here is the remarkable thing: if you could paint one precise dot on the bat and film it in slow motion, that dot would trace a perfect parabola — smooth, predictable, identical to a ball thrown at the same speed and angle. Every other point on the bat wobbles, rotates, and jiggles. This one point sails through the air as if the bat were a single particle with no size and no shape.

That point is the centre of mass.

The centre of mass is the single point where you can imagine all the mass of a body being concentrated, without changing how the body responds to external forces. It is the point that obeys Newton's second law in its simplest form: \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}. Find this point, and you turn a complicated extended body into a simple particle. The rest of this article is about finding where that point is.

Building the intuition — the balance point

Before any formula, think of a physical experiment you can do right now. Take a ruler — or a pencil, or any long object — and try to balance it on your fingertip. If the ruler is uniform, it balances at the midpoint. Now tape a heavy eraser to one end and try again. The balance point shifts toward the eraser. Tape two erasers, and it shifts further. Remove them, tape a coin to the other end instead, and the balance point moves the other way.

The balance point is the position where the clockwise and counterclockwise torques about that point cancel exactly. It is the point where gravity effectively acts on the whole ruler. This point is the centre of mass.

A see-saw makes the same idea vivid. Imagine a 60 kg adult sitting 1.5 m from the pivot, and a 30 kg child sitting 3 m from the pivot on the other side. The see-saw balances because 60 \times 1.5 = 30 \times 3 = 90 N·m on each side. The pivot — the balance point — is the centre of mass of the two-person system. Notice: the heavier person sits closer to it, the lighter person sits farther away. The CM is always pulled toward the heavier mass.

See-saw balance showing centre of mass A see-saw with a 60 kg adult sitting 1.5 m from the pivot on the left, and a 30 kg child sitting 3 m from the pivot on the right. The pivot marks the centre of mass of the system. 60 kg 30 kg 1.5 m 3 m CM (pivot)
The 60 kg adult sits 1.5 m from the pivot; the 30 kg child sits 3 m away on the opposite side. The see-saw balances because $60 \times 1.5 = 30 \times 3$. The pivot is the centre of mass of the system.

The pattern: heavier masses pull the balance point toward themselves, in proportion to their mass. That is the whole idea behind the formula you are about to see.

Centre of mass of two particles

Place two particles on a line: mass m_1 at position x_1, and mass m_2 at position x_2. Where is their centre of mass?

The see-saw gives you the answer. If x_{\text{cm}} is the CM position, then the "torques" from each mass about the CM must balance:

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

Why: this is the balance condition. Mass m_1 at distance (x_{\text{cm}} - x_1) from the pivot must balance mass m_2 at distance (x_2 - x_{\text{cm}}) on the other side.

Expand both sides:

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

Collect the x_{\text{cm}} terms on one side:

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

Why: the centre of mass is the mass-weighted average of the two positions. Each mass "votes" for its own position, and heavier masses get more votes.

Check this against your intuition:

  • Equal masses (m_1 = m_2): x_{\text{cm}} = \frac{x_1 + x_2}{2} — the midpoint. Exactly what the see-saw predicts when both people weigh the same.
  • m_1 \gg m_2: x_{\text{cm}} \approx x_1 — the CM sits practically at the heavy mass. A 100 kg person on a see-saw with a 1 kg stone? The balance point is basically under the person.
  • m_1 = 0: x_{\text{cm}} = x_2 — with only one mass, the CM is at that mass. No surprise.

All three checks match the physical picture.

Centre of mass — general definition

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

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

where M = \sum m_i is the total mass. In component form:

X_{\text{cm}} = \frac{1}{M}\sum m_i x_i, \qquad Y_{\text{cm}} = \frac{1}{M}\sum m_i y_i, \qquad Z_{\text{cm}} = \frac{1}{M}\sum m_i z_i

The generalisation from two particles to n particles is natural. Each mass contributes to the average position in proportion to its mass. The denominator M ensures the result has units of metres (not kg·m). Each coordinate is computed independently — X_{\text{cm}} depends only on the x-coordinates and masses, not on y or z. This makes even three-dimensional calculations straightforward: compute each component separately and assemble the result.

Explore the balance point

The interactive figure below puts the formula in your hands. Two particles sit on a number line: m_1 (variable) at x = 2 m and m_2 = 5 kg fixed at x = 8 m. Drag the slider to change m_1 from 0.5 kg to 10 kg and watch the centre of mass shift. When m_1 is small, the CM hugs the heavier mass at x = 8. As you increase m_1, the CM slides steadily toward x = 2.

Interactive: how mass affects centre of mass position A graph plotting centre of mass position versus m1. As m1 increases from 0.5 to 10 kg, the centre of mass moves from near x = 8 toward x = 2. A draggable point lets the reader explore different values of m1. m₁ (kg) x_cm (m) 0 5 10 1 5 10 x₁ = 2 m x₂ = 8 m drag the red dot along the axis
Drag the red dot to change $m_1$. The curve shows how $x_{\text{cm}}$ moves between the two particle positions ($x_1 = 2$ m, $x_2 = 8$ m with $m_2 = 5$ kg). The CM is always between the two masses — closer to whichever is heavier.

Notice two things. First, x_{\text{cm}} is always between 2 m and 8 m — it never overshoots either particle. The centre of mass of any system always lies within the "hull" of its constituent masses. Second, when m_1 = m_2 = 5 kg, the CM sits at exactly 5 m — the midpoint — confirming the equal-mass intuition.

Symmetry arguments — a shortcut that saves pages of calculation

For a uniform body — one whose density is the same everywhere — the centre of mass sits at the geometric centre (the centroid of the shape).

  • Uniform rod: CM at the midpoint.
  • Uniform circular disc or sphere: CM at the geometric centre.
  • Uniform ring: CM at the centre of the ring — even though there is no mass at that point. The CM is a weighted average of positions, not a physical location that must contain material.
  • Uniform rectangular plate: CM at the intersection of the diagonals.
  • Uniform equilateral triangle: CM at the centroid, one-third of the altitude from each side.

Why symmetry works: in a uniform body, for every small element of mass at position \vec{r} relative to the geometric centre, there exists a symmetric counterpart at -\vec{r} with the same mass. In the summation \sum m_i \vec{r}_i, these pairs cancel, leaving \vec{R}_{\text{cm}} = 0 — the geometric centre.

Symmetry is the most powerful tool in centre-of-mass calculations because it gives you the answer with zero algebra. If a body has one axis of symmetry, the CM lies on that axis. If it has two perpendicular axes of symmetry, the CM is at their intersection. You have eliminated one or two coordinates instantly, just by looking at the shape.

When a body is almost symmetric — like a cricket bat, which is roughly symmetric about its long axis but heavier at the blade — symmetry tells you the CM is on the long axis. You only need to calculate where along that axis it falls, not whether it is above or below the axis.

The subtraction method — bodies with parts removed

What if someone hands you a uniform disc with a circular hole punched in it? You could set up an integral over the remaining irregular shape — but there is a far smarter approach.

The key insight: the full disc is the union of the remaining piece and the removed piece.

\text{Full body} = \text{Remaining piece} + \text{Removed piece}

Since the CM of a composite body is the mass-weighted average of its parts' CMs:

M_{\text{full}} \cdot \vec{R}_{\text{full}} = M_{\text{rem}} \cdot \vec{R}_{\text{rem}} + m_{\text{cut}} \cdot \vec{r}_{\text{cut}}

Rearranging for the unknown \vec{R}_{\text{rem}}:

\vec{R}_{\text{rem}} = \frac{M_{\text{full}} \cdot \vec{R}_{\text{full}} - m_{\text{cut}} \cdot \vec{r}_{\text{cut}}}{M_{\text{rem}}}

Why this works: you already know the CM of the full body (by symmetry) and the CM of the removed piece (also by symmetry, since you typically remove a symmetric shape). The only unknown is the CM of the remaining piece — and one equation gives it to you.

This method is extremely powerful. It converts an ugly integral over a non-symmetric shape into simple arithmetic with quantities you already know. You will see it in action in Example 2 below.

Why the centre of mass is special — it obeys \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}

The centre of mass is not just a mathematical convenience. It is the point that obeys Newton's second law for the entire system, no matter how complicated the internal forces are.

Start with n particles. Each particle obeys Newton's second law individually:

\vec{F}_i^{\,\text{ext}} + \sum_{j \neq i} \vec{F}_{ij} = m_i \vec{a}_i

Why: the net force on particle i is the sum of external forces on it and internal forces from every other particle j in the system.

Sum over all n particles:

\sum_{i} \vec{F}_i^{\,\text{ext}} + \sum_{i}\sum_{j \neq i} \vec{F}_{ij} = \sum_{i} m_i \vec{a}_i

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

\vec{F}_{\text{ext}} = \sum_{i} m_i \vec{a}_i \tag{1}

Why: internal forces — the forces particles exert on each other — always come in equal-and-opposite pairs. They cancel in the sum. Only the total external force survives.

Now connect this to the centre of mass. From the definition \vec{R}_{\text{cm}} = \frac{1}{M}\sum m_i \vec{r}_i, differentiate twice with respect to time:

\vec{a}_{\text{cm}} = \frac{d^2\vec{R}_{\text{cm}}}{dt^2} = \frac{1}{M}\sum m_i \vec{a}_i

Why: the acceleration of the CM is the mass-weighted average of the individual accelerations. Differentiating a sum is the same as summing the derivatives.

Substituting into equation (1):

\boxed{\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}} \tag{2}

This is the most important result in this article. The total external force on a system equals the total mass times the acceleration of the centre of mass. The CM moves as if all the mass is concentrated at that one point and all external forces act there.

This is why the spinning cricket bat's CM traces a parabola. The only external force is gravity (M\vec{g}), so \vec{a}_{\text{cm}} = \vec{g}. The internal stresses that cause the bat to rotate affect the individual parts of the bat, but they cancel in pairs and contribute nothing to the CM's motion.

The simulation below makes this visible. Two Diwali rockets of equal mass are launched simultaneously from the same point — one at 30° and one at 60°, both at 20 m/s. Each traces its own parabolic arc. But the centre of mass of the pair traces a single, clean parabola at 45° — the mass-weighted average trajectory. The dashed lines connect the two rockets at each instant, and the CM (lighter dot) is always at their midpoint.

Animated: two projectiles at complementary angles with their centre of mass tracing a clean parabola Rocket A (red, 30 degrees) and Rocket B (dark, 60 degrees) are both launched at 20 m/s. Each follows its own parabola. Their centre of mass (lighter dot) traces a smooth parabola at 45 degrees, as if all the mass were concentrated at one point.
Two equal-mass rockets launched at 30° (red) and 60° (dark), both at 20 m/s. Each follows its own trajectory. The CM (lighter dot) traces a smooth parabola at 45° — the average of the two launch angles. Dashed lines confirm the CM stays at the midpoint of the two rockets. Click replay to watch again.

Even if the two rockets were to collide mid-air and explode into fragments, the CM of all the fragments would continue along the same parabola — because the explosion forces are internal to the system, and internal forces cannot change the CM's motion.

Worked examples

Example 1: Three fielders on a cricket ground

Three fielders stand on a cricket ground. Fielder A (70 kg) is at position (0, 0) m, fielder B (60 kg) is at (30, 0) m, and fielder C (50 kg) is at (15, 26) m. Find the centre of mass of the three-fielder system.

Three fielders on a coordinate plane with centre of mass marked Fielder A (70 kg) at origin, Fielder B (60 kg) at (30, 0), Fielder C (50 kg) at (15, 26). The centre of mass at approximately (14.2, 7.2) is marked with a red cross inside the triangle. x (m) y (m) 15 30 13 26 A (70 kg) B (60 kg) C (50 kg) CM
Three fielders form a triangle. The CM (red cross) sits inside the triangle, pulled toward the heaviest fielder A at the origin.

Step 1. Write down all the data.

Fielder Mass (kg) x (m) y (m)
A 70 0 0
B 60 30 0
C 50 15 26

Total mass: M = 70 + 60 + 50 = 180 kg.

Why: organise the data in a table first. The total mass M goes in the denominator of every CM formula.

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

X_{\text{cm}} = \frac{70 \times 0 + 60 \times 30 + 50 \times 15}{180} = \frac{0 + 1800 + 750}{180} = \frac{2550}{180} \approx 14.2 \text{ m}

Why: each mass "votes" for its x-position. Fielder A at x = 0 pulls the CM left; B at x = 30 pulls right; C at x = 15 votes for the middle.

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

Y_{\text{cm}} = \frac{70 \times 0 + 60 \times 0 + 50 \times 26}{180} = \frac{1300}{180} \approx 7.2 \text{ m}

Why: only fielder C has a nonzero y-coordinate. Since C carries the smallest mass (50 out of 180 kg), the CM is pulled upward, but not by much — only about 7.2/26 \approx 28\% of the way toward C.

Step 4. Quick sanity check.

X_{\text{cm}} = 14.2 m is between 0 and 30 — it falls within the triangle's base. Y_{\text{cm}} = 7.2 m is between 0 and 26 — it falls within the triangle's height. The CM is inside the triangle, closer to the base (where the two heavier fielders stand). All consistent.

Result: The centre of mass of the three-fielder system is at (14.2, 7.2) m.

What this shows: The CM of a multi-particle system is computed coordinate by coordinate. The heavier fielders pull the CM toward themselves. The result always lies inside the convex hull (the triangle) formed by the particles.

Example 2: A uniform disc with a circular hole — subtraction method

A uniform circular disc of mass 4 kg and radius R = 20 cm has a circular hole of radius R/2 = 10 cm cut from it. The centre of the hole is at a distance of 10 cm from the centre of the disc (so the hole just touches the disc's centre). Find the centre of mass of the remaining piece.

Uniform disc with a circular hole showing subtraction method A disc of radius 20 cm centred at the origin. A circular hole of radius 10 cm is cut from the right side, centred at x = 10 cm. The CM of the remaining shape is at approximately x = −3.3 cm, shifted left away from the hole. removed O (centre) O′ CM 10 cm R = 20 cm R/2
The dashed circle shows the removed piece (radius 10 cm, centred 10 cm from O). The CM of the remaining shape (red cross) shifts to the left — away from the hole.

Step 1. Find the mass of the removed piece.

Since the disc is uniform, mass is proportional to area.

\frac{m_{\text{cut}}}{M_{\text{full}}} = \frac{\pi (R/2)^2}{\pi R^2} = \frac{1}{4}

So m_{\text{cut}} = 4/4 = 1 kg. The remaining piece has mass M_{\text{rem}} = 4 - 1 = 3 kg.

Why: for a uniform disc, halving the radius gives one-quarter the area (area scales as r^2), hence one-quarter the mass.

Step 2. Identify the CMs of the full disc and the removed piece.

Set the origin at the full disc's centre O. By symmetry:

  • CM of full disc: x_{\text{full}} = 0
  • CM of removed disc: x_{\text{cut}} = +10 cm (at the centre of the hole, O′)

Why: both the full disc and the removed disc are symmetric shapes, so each has its CM at its own geometric centre.

Step 3. Apply the subtraction formula.

M_{\text{full}} \cdot x_{\text{full}} = M_{\text{rem}} \cdot x_{\text{rem}} + m_{\text{cut}} \cdot x_{\text{cut}}
4 \times 0 = 3 \times x_{\text{rem}} + 1 \times 10
0 = 3\,x_{\text{rem}} + 10
x_{\text{rem}} = -\frac{10}{3} \approx -3.33 \text{ cm}

Why: the full disc's CM is the mass-weighted average of the remaining piece's CM and the removed piece's CM. Solving for the one unknown gives us the CM of the remaining piece.

Step 4. Interpret the result.

The negative sign means the CM is to the left of the original centre — on the side opposite the hole. Removing mass from the right shifted the CM to the left. The displacement is R/6 — about one-sixth of the full radius.

Result: The CM of the remaining piece is at x = -10/3 \approx -3.3 cm from the original centre, on the side opposite the hole.

What this shows: The subtraction method converts a complex shape (disc minus hole) into two simple shapes whose CMs you know by symmetry. You avoid integration entirely and get an exact answer in three lines of algebra.

Common confusions

  • "The centre of mass must be inside the body." Not necessarily. A uniform ring's CM is at its geometric centre — where there is zero mass. A crescent-shaped object can have its CM in the empty space between the horns. The CM is a weighted average of positions; it does not need to coincide with any physical material.

  • "Centre of mass and centre of gravity are the same thing." They coincide when the gravitational field is uniform — an excellent approximation for everyday objects near Earth's surface. For a very tall structure in a non-uniform gravitational field, the centre of gravity (where gravitational torque vanishes) and the centre of mass (the mass-weighted average position) can differ slightly. For every problem in this course, treat them as identical.

  • "A heavier object always has its CM at the geometric centre." Only if its mass is distributed symmetrically. A cricket bat has more mass at the blade end than the handle — its CM is well below the geometric midpoint of its length. Total mass does not determine the CM; mass distribution does.

  • "You need calculus for a body with a hole." Not if you use the subtraction method. The insight \text{full} = \text{remaining} + \text{removed} lets you solve for the remaining piece's CM using the CMs of two simpler shapes — no integration required.

  • "The CM formula works only for point particles." The discrete formula \vec{R}_{\text{cm}} = \frac{1}{M}\sum m_i \vec{r}_i applies to any collection of particles. For continuous bodies, the sum becomes an integral: \vec{R}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm. That is the subject of the next article.

If you came here to learn what the centre of mass is, calculate it for particles and symmetric bodies, and use the subtraction method, you have everything you need. What follows is for readers who want the CM frame, the König theorem, and a preview of the continuous case.

The centre-of-mass frame

Define the CM velocity:

\vec{v}_{\text{cm}} = \frac{d\vec{R}_{\text{cm}}}{dt} = \frac{1}{M}\sum m_i \vec{v}_i = \frac{\vec{p}_{\text{total}}}{M}

Why: differentiating the CM position gives the CM velocity, which equals the total momentum divided by total mass.

The CM frame is a reference frame moving with velocity \vec{v}_{\text{cm}}. In this frame, the total momentum is zero:

\sum m_i \vec{v}_i' = \sum m_i(\vec{v}_i - \vec{v}_{\text{cm}}) = \vec{p}_{\text{total}} - M\vec{v}_{\text{cm}} = 0

This is why elastic collisions become trivial in the CM frame (as you saw in the elastic collisions article) — with zero total momentum, both particles simply reverse their velocities. The entire complexity of the collision lives in the frame transformation, not in the collision itself.

The König theorem — splitting kinetic energy

The total kinetic energy of a system splits cleanly into two pieces:

K_{\text{total}} = \frac{1}{2}M v_{\text{cm}}^2 + \sum \frac{1}{2}m_i v_i'^{\,2}

The first term, \frac{1}{2}Mv_{\text{cm}}^2, is the kinetic energy of the CM — the energy of translation, as if all the mass were at one point. The second term is the kinetic energy of internal motion relative to the CM — rotation, vibration, thermal jiggling.

Why: expand v_i^2 = |\vec{v}_{\text{cm}} + \vec{v}_i'|^2 = v_{\text{cm}}^2 + 2\vec{v}_{\text{cm}} \cdot \vec{v}_i' + v_i'^{\,2}. When you sum over all particles, the cross term 2\vec{v}_{\text{cm}} \cdot \sum m_i \vec{v}_i' vanishes because \sum m_i \vec{v}_i' = 0 in the CM frame.

This decomposition explains why the spinning cricket bat's CM traces a clean parabola: gravity affects only the first term (translational KE). The tumbling lives entirely in the second term (internal KE). External forces change the CM's motion; internal forces redistribute energy within the system but leave the CM undisturbed.

Preview: continuous bodies

For a continuous body with mass density \rho(\vec{r}), the sum becomes an integral:

\vec{R}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm = \frac{1}{M}\int \vec{r}\,\rho(\vec{r})\,dV

The choice of integration element depends on the geometry:

  • Rod (1D): dm = \lambda\,dx, where \lambda is the linear mass density (kg/m).
  • Plate (2D): dm = \sigma\,dA, where \sigma is the surface mass density (kg/m²).
  • Solid (3D): dm = \rho\,dV, where \rho is the volume mass density (kg/m³).

The next article works through the integrals for standard shapes — uniform rods, semicircular wires, semicircular discs, hemispheres, and cones.

Where this leads next

  • Centre of Mass of Continuous Bodies — replacing the sum with an integral to find the CM of rods, discs, hemispheres, and cones.
  • Motion of the Centre of Mass — using \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} to solve multi-body dynamics problems, including explosions and splitting bodies.
  • Elastic Collisions — the CM frame makes elastic collision algebra trivial, reducing every collision to a simple velocity reversal.
  • Conservation of Linear Momentum — when \vec{F}_{\text{ext}} = 0, the CM moves at constant velocity, which is conservation of momentum in disguise.
  • Rotational Mechanics — the CM is the natural pivot for analysing torques and angular momentum of rigid bodies.