Centre of mass — derivation for common shapes (rod, triangle, semicircle, hemisphere): Visual walkthrough

Step 1 — The see-saw: what "balance point" really means

WHAT. Put two kids on a plank. A light kid of mass $m_{1}$ at position $x_{1}$ , a heavy kid of mass $m_{2}$ at position $x_{2}$ . Where do you put the pivot so the plank does not tip?

WHY. This is the seed of everything. A plank balances when the two sides exert equal turning effect about the pivot. The turning effect of a kid is really her weight — her mass $m$ times gravity $g$ (the pull of the Earth, the same number for everyone) — times how far she sits from the pivot. That product $m g d$ is the moment (or torque). "No tipping" means the left moment equals the right moment. Let us turn that sentence into a formula and watch $g$ disappear.

PICTURE. In the figure, the heavy kid drags the pivot toward himself. The balance point is not the middle — it leans toward the mass.

Figure — Centre of mass — derivation for common shapes (rod, triangle, semicircle, hemisphere)

Call the pivot position $x_{c m}$ . Kid 1 sits a distance $(x_{1} - x_{c m})$ from the pivot; kid 2 sits $(x_{2} - x_{c m})$ . Balance = the two moments cancel: $moment of kid 1 m_{1} g (x_{1} - x_{c m}) + moment of kid 2 m_{2} g (x_{2} - x_{c m}) = 0$ Every term carries the same factor $g$ , so we can divide the whole equation by $g$ — gravity cancels and never affects where the balance point is. (This is exactly why the centre of mass and the centre of gravity coincide in uniform gravity — see Centroid vs centre of mass vs centre of gravity.) What is left is pure mass: $m_{1} (x_{1} - x_{c m}) + m_{2} (x_{2} - x_{c m}) = 0.$ Now solve for $x_{c m}$ . Expand and collect it: $m_{1} x_{1} + m_{2} x_{2} - (m_{1} + m_{2}) x_{c m} = 0 ⟹ x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} .$

If the kids were equal ( $m_{1} = m_{2}$ ) this collapses to $\frac{x _{1} + x _{2}}{2}$ , the plain middle — the mass-weighting is exactly the correction for unequal kids. See Centre of mass — definition and system of particles for the full force picture.

Step 2 — From a few kids to a crowd: the $\sum$ sign

WHAT. Add more particles. The recipe does not change — every particle contributes its own $m_{i} x_{i}$ on top, and its own $m_{i}$ to the total.

WHY. We are about to have infinitely many particles (a solid body is a dense crowd of atoms). Before we can go to infinity, we write the finite crowd cleanly. The same "moments cancel about the pivot" argument (with $g$ again cancelling), done for $N$ particles, gives the sum below.

PICTURE. Five particles on a line; each contributes a stack of height $m_{i}$ sitting at $x_{i}$ . The COM is the "centre of the stacks".

Step 3 — Slicing a solid body: what $d m$ and $\int$ mean

WHAT. A rod is not five kids — it is a continuous smear of mass. So we chop it into slices so thin that each slice is basically a point particle. One slice carries a tiny mass we call $d m$ ("a little bit of mass"), sitting at position $x$ .

WHY this tool — the integral? A sum $\sum$ works for a countable pile of particles. But our slices are infinitely thin and infinitely many. The integral sign $\int$ is exactly the tool invented for "add up infinitely many infinitely small pieces" — see Integration as continuous summation. It answers the question a plain $\sum$ cannot: how do you total a smooth continuum?

PICTURE. The rod carved into vertical slivers; one sliver highlighted, width $d x$ , sitting at distance $x$ from the left end.

How big is one slice's $d m$ ? It depends on how the mass is spread. For a line of mass we use the linear density $λ$ (Greek "lambda") = mass per unit length, $λ = M / L$ , so a slice of length $d x$ holds $d m = λ d x$ . Two cousins appear later:

For a flat sheet ("lamina") we use the surface density $σ$ (Greek "sigma") = mass per unit area, so a slice of area $d A$ holds $d m = σ d A$ .
For a solid 3-D body we use the volumetric density $ρ$ (Greek "rho") = mass per unit volume, so a slab of volume $d V$ holds $d m = ρ d V$ .

In every case density is "how much mass is packed into one unit of the thing", and — being the same top and bottom for a uniform body — it will cancel. On this page we use $λ$ (line) and $σ$ (sheet); $ρ$ (solid) is defined here so the word "density" in Step 7 has a meaning.

$d m = mass per length λ length of slice d x .$

Step 4 — The rod falls out (and why the density vanishes)

WHAT. Feed the rod into the master formula. Origin at the left end, rod running from $x = 0$ to $x = L$ — so the limits are $0$ to $L$ .

WHY. This is the simplest possible shape, so it is where we check the machinery works before trusting it on harder shapes.

PICTURE. Same sliced rod, now with the running position $x$ marked and the answer $L /2$ flagged dead centre.

$x_{c m} = \frac{\int _{0}^{L} x ( λ d x )}{\int _{0}^{L} λ d x} = \frac{λ \int _{0}^{L} x d x}{λ \int _{0}^{L} d x} = \frac{λ \cdot \frac{L ^{2}}{2}}{λ \cdot L} = \frac{L}{2}$

$\int_{0}^{L} x d x = \frac{L ^{2}}{2}$ — the mass-weighted total, area under the line $y = x$ .
$\int_{0}^{L} d x = L$ — the total length, which times $λ$ is the total mass $M$ .
$λ$ cancels top and bottom. This is huge: for a uniform body the answer is purely about shape, not about how heavy it is. That shape-only balance point is the centroid — see Centroid vs centre of mass vs centre of gravity.

Step 5 — The triangle: why mass piles at the base

WHAT. A flat triangle (a "lamina"), base $b$ , height $h$ . Fix the origin at the mid-point of the base, with the $x$ -axis running along the base and the $y$ -axis pointing straight up to the apex. So the base runs from $x = - \frac{b}{2}$ (left corner) to $x = + \frac{b}{2}$ (right corner) at $y = 0$ , and the apex sits at $(x, y) = (0, h)$ . In particular $y = 0$ is the base and $y = h$ is the apex — every height below is measured from the base upward. Slice the triangle into thin horizontal strips parallel to the base; a strip at height $y$ is a thin rectangle.

WHY horizontal strips? Because each such strip sits at a single clean height $y$ , so it behaves like one particle at that height. That converts the 2D shape into a 1D stack of strips — back to Step 3's machinery. This is a slice-cleverly move.

PICTURE. A wide strip near the base, a short strip near the apex — you can see the base strips carry more area (more mass). The strip width shrinks linearly with height.

First, the horizontal coordinate — for free. Because the origin sits at the mid-base and the apex is straight above it, the triangle is a mirror image left↔right about the $y$ -axis ( $x = 0$ ). Every strip is centred on that axis, so each strip's own balance point sits at its middle, at $x = 0$ . By this symmetry the whole triangle's COM lies on that axis: $x_{c m} = 0 (the mid-base line).$ So we only have to work out the height $y_{c m}$ , measured from the base.

By similar triangles, a strip at height $y$ has width $w (y) = b (1 - \frac{y}{h})$

at $y = 0$ (base): $w = b$ (full width, heaviest).
at $y = h$ (apex): $w = 0$ (vanishes).

Its mass is $d m = σ w (y) d y$ , where $σ$ (surface density, defined in Step 3) plays the role $λ$ did. Because $y = 0$ is the base and $y = h$ is the apex, the strips sweep the whole triangle as $y$ runs $0 \to h$ — those are the limits:

$y_{c m} = \frac{\int _{0}^{h} y σ b ( 1 - \frac{y}{h} ) d y}{\int _{0}^{h} σ b ( 1 - \frac{y}{h} ) d y} = \frac{\int _{0}^{h} ( y - \frac{y ^{2}}{h} ) d y}{\int _{0}^{h} ( 1 - \frac{y}{h} ) d y}$

The $σ$ and $b$ cancel (they multiply top and bottom equally). Now do the four little integrals one at a time, using the single rule $\int_{0}^{h} y^{n} d y = \frac{h ^{n + 1}}{n + 1}$ (the area under $y^{n}$ ):

$\int_{0}^{h} y d y = \frac{h ^{2}}{2}$ — area under the straight line $y$ .
$\int_{0}^{h} y^{2} d y = \frac{h ^{3}}{3}$ , so $\int_{0}^{h} \frac{y ^{2}}{h} d y = \frac{1}{h} \cdot \frac{h ^{3}}{3} = \frac{h ^{2}}{3}$ .
$\int_{0}^{h} 1 d y = h$ — the plain length.
$\int_{0}^{h} \frac{y}{h} d y = \frac{1}{h} \cdot \frac{h ^{2}}{2} = \frac{h}{2}$ .

Substitute these pieces: $y_{c m} = \frac{\frac{h ^{2}}{2} - \frac{h ^{2}}{3}}{h - \frac{h}{2}} = \frac{\frac{h ^{2}}{6}}{\frac{h}{2}} = \frac{h}{3}$

Numerator $\frac{h ^{2}}{2} - \frac{h ^{2}}{3} = \frac{3 h ^{2} - 2 h ^{2}}{6} = \frac{h ^{2}}{6}$ : heights weighted by strip mass.
Denominator $h - \frac{h}{2} = \frac{h}{2}$ : total (mass in units of $σ b$ ).
Result $h /3$ : one-third up from the base, exactly as the "bottom-heavy" picture promised. Not $h /2$ — the narrowing kills the top.

Step 6 — The semicircular wire: bringing in angles and $sin$

WHAT. A bent wire, a half-ring of radius $R$ , flat side on the $x$ -axis, arching up. We want its balance height.

WHY a new coordinate? The wire is curved, so slicing by $x$ or $y$ is ugly — a tiny piece is most naturally located by the angle $θ$ it makes with the $x$ -axis. We measure position with $sin$ and $cos$ because those are exactly the tools that turn an angle into a height and a width on a circle.

PICTURE. The arc with one tiny piece $d l$ marked at angle $θ$ ; its height $y = R sin θ$ drawn as a vertical drop from the piece to the axis.

A tiny arc of angle $d θ$ has length $d l = R d θ$ (arc = radius × angle). Total length is $π R$ , so $λ = \frac{M}{π R}$ and $d m = λ d l = \frac{M}{π R} R d θ = \frac{M}{π} d θ .$ The piece's height is $y = R sin θ$ — $sin θ$ is the fraction of $R$ that counts as height. The angle sweeps from $0$ to $π$ to cover the whole half-ring, so those are the limits.

$y_{c m} = \frac{1}{M} \int_{0}^{π} (R sin θ) \frac{M}{π} d θ = \frac{R}{π} \int_{0}^{π} sin θ d θ = \frac{R}{π} [- cos θ]_{0}^{π} = \frac{R}{π} (1 + 1) = \frac{2 R}{π}$

$\int_{0}^{π} sin θ d θ = 2$ : the total "height-pull" of all the arc pieces.
$\frac{2 R}{π} \approx 0.64 R$ — high up, because all the wire's mass sits out on the rim near the top.

Step 7 — Edge & degenerate cases: does the machine break?

WHAT. Test the formula on the limits, where formulas love to explode.

WHY. A derivation you trust must survive the corners: zero size, all-mass-at-one-point, a symmetry axis. The parent note states these; here we see why they hold.

PICTURE. Three panels: (a) a degenerate zero-height triangle collapsing to a rod, (b) a symmetric shape with its COM pinned on the axis, (c) a point mass where COM = the point.

Worked example The three checks

(a) Triangle with $h \to 0$ . It flattens into a line segment. $y_{c m} = h /3 \to 0$ — the COM sinks onto the line, as it must. No division-by-zero: the bottom $h /2$ shrinks together with the top. (b) Symmetry. For the wire, every piece at $+ x$ has a mirror twin at $- x$ . Their $x$ -contributions cancel in pairs, forcing $x_{c m} = 0$ before any integration. Always spend the free symmetry first. (c) A lone point mass. Here the whole "body" is a single particle at $x_{0}$ . The formula gives $\frac{\int x d m}{\int d m} = \frac{x _{0} M}{M} = x_{0}$ — the COM sits exactly on the point, as common sense demands. (Separately, notice that any uniform density — $λ$ , $σ$ , or the volumetric $ρ$ from Step 3 — always appears on top and bottom, so it cancels: the COM value never depends on how much mass there is, only on where it sits. Here "cancels" refers to the density, not the COM coordinate going to zero.)

The one-picture summary

Everything above is a single sentence in four frames: weight each position by its mass, sum, divide. The rod, triangle, and wire differ only in how one slice $d m$ is shaped.

So far each shape used one coordinate ( $x$ , or $y$ , or a height) because symmetry killed the rest. In general a point in space needs all its coordinates at once. We bundle them into one symbol $r = (x, y, z)$ — the position vector, an arrow from the origin to the point. Applying the master formula to each coordinate separately, $x_{c m} = \frac{\int x d m}{M}$ , $y_{c m} = \frac{\int y d m}{M}$ , $z_{c m} = \frac{\int z d m}{M}$ , is exactly what the single vector line below packages:

Recall Feynman: the whole walk in plain words

Picture balancing a stick on your finger. Your finger has to sit where the stuff on the left pulls exactly as hard as the stuff on the right — heavy stuff pulls harder, so the finger slides toward it. That "fair balance" spot is the centre of mass. Gravity pulls on every bit equally, so it drops out of the sum entirely — only where the mass sits matters. For two kids on a see-saw it's a simple weighted average, and you can prove it by demanding the two turning-effects about the finger cancel. A solid object is just a huge crowd of tiny kids, so instead of adding a few numbers we use the integral — a machine for adding infinitely many tiny pieces. Chop the object into slivers, note how much mass each sliver has and where it sits, multiply, add them all up, and divide by the total mass. For a plain stick the fair spot is the middle. For a triangle the wide bottom is heavier, so the spot sits a third of the way up (and dead-centre left-to-right by symmetry). For a wire bent into a half-circle all the weight rides out on the rim, so the spot floats high, about two-thirds of the radius up. Same recipe every single time — only the shape of one sliver changes.

Recall Self-test

What replaced $\sum m_{i} x_{i}$ for a continuous body? ::: $\int x d m$
Why does gravity $g$ not appear in the COM formula? ::: It multiplies every moment equally, so dividing the balance equation by $g$ removes it — only mass and position remain.
Before applying the master formula, what two things must you fix? ::: An origin to measure from, and the integration limits covering the whole body.
Why does $λ$ cancel for a uniform rod? ::: It sits on top and bottom equally; uniform bodies give a shape-only (centroid) answer.
Why $h /3$ and not $h /2$ for the triangle? ::: Strips near the base are wider (more mass), so the average height is pulled down.
Where is the triangle COM horizontally, and why? ::: On the mid-base line $x = 0$ — left-right symmetry pins it there.
Which tool turns an arc angle into height, and why? ::: $sin θ$ , because $R sin θ$ is the vertical part of a point on the circle.
How does symmetry save work? ::: A mirror axis makes off-axis contributions cancel in pairs, so that coordinate is $0$ instantly.

Connected ideas: Centre of mass — definition and system of particles · Integration as continuous summation · Symmetry arguments in physics · Centroid vs centre of mass vs centre of gravity · Moment of inertia — derivation for common shapes · Momentum conservation