In short

For a continuous body, the centre of mass is \vec{R}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm. Choose the mass element based on geometry: dm = \lambda\,dx for rods and wires, dm = \sigma\,dA for plates, dm = \rho\,dV for solids. Symmetry eliminates whichever coordinates have equal mass on both sides. Standard results: uniform rod at L/2, semicircular wire at 2R/\pi from centre, semicircular disc at 4R/(3\pi), solid hemisphere at 3R/8, solid cone at h/4 from base.

Cut a circular roti in half. Try to balance the semicircular piece on the tip of a pencil. Where do you place the pencil?

Not at the midpoint of the flat edge — that is the centre of the full circle, and the semicircle has more mass above that point than below it. Not at the highest point on the curve, either. The balance point sits somewhere on the line of symmetry, shifted from the flat edge toward the curved side. But how far exactly?

For two or three point masses, you found the centre of mass with a weighted average: \vec{R}_{\text{cm}} = \frac{1}{M}\sum m_i \vec{r}_i. A roti is not a collection of discrete point masses — it is a continuous sheet of matter spread over a region. To find its balance point, you need to replace the sum with an integral.

From a sum to an integral

The discrete formula for N point masses is:

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

Now imagine slicing a continuous body into N tiny pieces. The i-th piece has mass \Delta m_i and sits at position \vec{r}_i. As you slice finer — N \to \infty, each \Delta m_i \to 0 — the sum becomes an integral:

\boxed{\vec{R}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm}

Why: this is the standard calculus transition from a Riemann sum to an integral. Each piece \Delta m_i at position \vec{r}_i contributes \vec{r}_i\,\Delta m_i to the numerator. In the limit, the sum becomes \int.

In component form, each coordinate is computed by a separate integral:

x_{\text{cm}} = \frac{1}{M}\int x\,dm, \qquad y_{\text{cm}} = \frac{1}{M}\int y\,dm, \qquad z_{\text{cm}} = \frac{1}{M}\int z\,dm

The total mass M = \int dm appears in every denominator.

Choosing the mass element dm

The integral \int \vec{r}\,dm is not ready to evaluate until you express dm in terms of a coordinate you can integrate over. The right choice depends on the body's geometry.

Body type Mass element Density symbol Units
Rod or wire (1D) dm = \lambda\,dx or \lambda\,R\,d\theta \lambda = mass per length kg/m
Plate or sheet (2D) dm = \sigma\,dA \sigma = mass per area kg/m²
Solid body (3D) dm = \rho\,dV \rho = mass per volume kg/m³

For a uniform body, the density is constant and comes outside the integral. For a non-uniform body, the density varies with position and stays inside.

Three types of mass elements for continuous bodies Left: a rod with a small element dx highlighted. Centre: a flat plate with a thin strip dA. Right: a solid cone with a thin disc dV. dx dm = λ dx 1D: rod dA dm = σ dA 2D: plate dV dm = ρ dV 3D: solid
The mass element depends on the body's dimensionality. A rod uses length elements $dx$, a plate uses area elements $dA$, and a solid uses volume elements $dV$.

Symmetry: let the geometry do the work

Before evaluating any integral, look at the shape. If a uniform body has a line of symmetry or a plane of symmetry, the centre of mass must lie on it. For every mass element on one side of the symmetry line, an identical element exists on the other side at the same distance — their position contributions cancel in the perpendicular direction.

A uniform rod has a midpoint axis. A semicircular wire has a vertical axis through the centre of the diameter. A hemisphere and a cone each have a central axis. Symmetry tells you which coordinates are zero without computing a single integral. You only integrate for the surviving coordinate.

Centre of mass: shape by shape

The uniform rod

Place a thin uniform rod of mass M and length L along the x-axis, from x = 0 to x = L. The linear mass density is \lambda = M/L.

Uniform rod with mass element at position x A horizontal rod from x equals 0 to x equals L with a thin highlighted element at position x having width dx. x x dx 0 L
A uniform rod along the $x$-axis. The highlighted strip at position $x$ has mass $dm = \lambda\,dx$.

Step 1. By symmetry, the CM lies on the x-axis. You only need x_{\text{cm}}.

Step 2. Write the integral.

x_{\text{cm}} = \frac{1}{M}\int_0^L x\,dm = \frac{1}{M}\int_0^L x\,\lambda\,dx

Why: each tiny element at position x has mass dm = \lambda\,dx. Multiply by its position x, then integrate over the entire rod.

Step 3. Since \lambda = M/L is constant, pull it out.

x_{\text{cm}} = \frac{\lambda}{M}\int_0^L x\,dx = \frac{1}{L}\cdot\frac{L^2}{2}

Why: \lambda/M = 1/L. The integral \int_0^L x\,dx = L^2/2 is standard.

Step 4. Result.

\boxed{x_{\text{cm}} = \frac{L}{2}}

A uniform rod's CM is at its geometric midpoint — no surprise. The method is what matters. The same integral technique, applied to curved or asymmetric shapes, gives results you cannot guess.

The semicircular wire

A thin uniform wire of mass M is bent into a semicircle of radius R. Place it with the diameter along the x-axis, wire curving upward. By symmetry, x_{\text{cm}} = 0. You need y_{\text{cm}}.

Semicircular wire with angular mass element A semicircular wire of radius R centred at the origin. A small arc element at angle theta has arc length R d-theta and sits at height R sin theta above the centre. y x θ R R dθ R sin θ −R R
The small arc element at angle $\theta$ has length $R\,d\theta$ and sits at height $R\sin\theta$ above the centre of the diameter.

Step 1. Parametrise using angle \theta from 0 to \pi. The arc-length element is dl = R\,d\theta, and the linear density is \lambda = M/(\pi R).

dm = \lambda\,R\,d\theta = \frac{M}{\pi}\,d\theta

Why: angle is the natural variable for a circular arc. Each tiny arc subtends d\theta and has length R\,d\theta.

Step 2. The height of the element at angle \theta is R\sin\theta. Set up the integral.

y_{\text{cm}} = \frac{1}{M}\int_0^\pi (R\sin\theta)\,\frac{M}{\pi}\,d\theta = \frac{R}{\pi}\int_0^\pi \sin\theta\,d\theta

Why: M cancels between the numerator and denominator — the CM of a uniform body depends only on shape, not on total mass.

Step 3. Evaluate.

\int_0^\pi \sin\theta\,d\theta = [-\cos\theta]_0^\pi = -\cos\pi + \cos 0 = 1 + 1 = 2
\boxed{y_{\text{cm}} = \frac{2R}{\pi} \approx 0.637\,R}

The centre of mass sits roughly two-thirds of the way from the centre of the diameter to the top of the wire — closer to the curved part than to the flat edge.

This result generalises. For a circular arc symmetric about the y-axis, subtending a half-angle \alpha on each side of the axis (so the full semicircle has \alpha = \pi/2, a full circle has \alpha = \pi), the centre of mass is at:

y_{\text{cm}} = \frac{R\sin\alpha}{\alpha}

Drag the control in the interactive below to see how the CM moves as the arc grows from a tiny sliver to a complete ring.

Interactive: centre of mass of a circular arc versus half-angle A curve showing y_cm over R equals sin alpha over alpha as a function of half-angle alpha. A draggable point lets you explore different arc angles. At alpha equals pi over 2, the value is 2 over pi for a semicircular wire. half-angle α (rad) y_cm / R 1.0 0.5 0 π/4 π/2 3π/4 π 2/π drag the red dot along the axis
Drag the red dot to change the half-angle $\alpha$ of a circular arc. At $\alpha = \pi/2$ (semicircle), $y_{\text{cm}}/R = 2/\pi \approx 0.637$. As $\alpha \to \pi$ (full circle), the CM drops to zero — the CM of a complete ring is at its geometric centre.

The semicircular disc

A uniform semicircular disc (half of a flat circular plate) of radius R and mass M, with the diameter along the x-axis and the disc above it. By symmetry, x_{\text{cm}} = 0. You need y_{\text{cm}}.

Slice the disc into thin horizontal strips. At height y, the strip has width 2\sqrt{R^2 - y^2} and thickness dy.

Semicircular disc with horizontal strip element A semicircular disc of radius R with a thin horizontal strip at height y. The strip has width 2 times the square root of R squared minus y squared, and height dy. y y 2√(R² − y²) dy R
A semicircular disc sliced into horizontal strips. The strip at height $y$ has width $2\sqrt{R^2 - y^2}$ and thickness $dy$.

Step 1. The area of the semicircle is \pi R^2/2, so \sigma = 2M/(\pi R^2).

dm = \sigma\,dA = \frac{2M}{\pi R^2}\cdot 2\sqrt{R^2 - y^2}\,dy

Why: the strip at height y has area dA = 2\sqrt{R^2 - y^2}\,dy — width times thickness.

Step 2. Set up the integral.

y_{\text{cm}} = \frac{1}{M}\int_0^R y\,dm = \frac{4}{\pi R^2}\int_0^R y\sqrt{R^2 - y^2}\,dy

Step 3. Evaluate with the substitution u = R^2 - y^2, du = -2y\,dy.

\int_0^R y\sqrt{R^2 - y^2}\,dy = \frac{1}{2}\int_0^{R^2}\sqrt{u}\,du = \frac{1}{2}\cdot\frac{2}{3}\,u^{3/2}\Big|_0^{R^2} = \frac{R^3}{3}

Why: the substitution turns the integrand into a simple power of u. Limits change from y = 0 \Rightarrow u = R^2 and y = R \Rightarrow u = 0; swapping them absorbs the minus sign.

Step 4. Result.

y_{\text{cm}} = \frac{4}{\pi R^2}\cdot\frac{R^3}{3} = \boxed{\frac{4R}{3\pi} \approx 0.424\,R}

Compare this with the semicircular wire's 2R/\pi \approx 0.637R. The disc's CM is closer to the flat edge. The reason: a disc has more mass near the diameter (wider strips near the bottom) than near the top, so the mass distribution is bottom-heavy compared to a wire of the same shape.

This is the answer to the roti question. A semicircular roti of radius 10 cm has its balance point about 4.2 cm from the centre of the flat edge.

The solid hemisphere

A uniform solid hemisphere of radius R and mass M, with the flat face in the xy-plane. By symmetry, x_{\text{cm}} = y_{\text{cm}} = 0. You need z_{\text{cm}}.

Slice the hemisphere into thin circular discs parallel to the flat face. At height z, the disc has radius \sqrt{R^2 - z^2} and thickness dz.

Step 1. The volume of a hemisphere is (2/3)\pi R^3, so \rho = 3M/(2\pi R^3).

dV = \pi(R^2 - z^2)\,dz

Step 2. Set up the integral.

z_{\text{cm}} = \frac{1}{M}\int_0^R z\,\rho\,\pi(R^2 - z^2)\,dz = \frac{3}{2R^3}\int_0^R (R^2 z - z^3)\,dz

Why: the constants simplify to \rho\pi/M = 3/(2R^3). The integrand splits into two simple power terms.

Step 3. Evaluate.

\int_0^R (R^2 z - z^3)\,dz = \left[\frac{R^2 z^2}{2} - \frac{z^4}{4}\right]_0^R = \frac{R^4}{2} - \frac{R^4}{4} = \frac{R^4}{4}
z_{\text{cm}} = \frac{3}{2R^3}\cdot\frac{R^4}{4} = \boxed{\frac{3R}{8} = 0.375\,R}

Why: the solid hemisphere's CM is closer to the flat face than the semicircular disc's, because a 3D solid packs even more volume near the base (the thick discs near the equator contribute heavily).

The solid cone

A uniform solid right circular cone of height h, base radius R, and mass M. Place the apex at the origin, axis along z, base at z = h. By symmetry, x_{\text{cm}} = y_{\text{cm}} = 0.

At height z from the apex, the cross-section is a circle of radius r = Rz/h (by similar triangles). So dV = \pi R^2 z^2/h^2\,dz.

Step 1. The volume of a cone is \pi R^2 h/3, so \rho = 3M/(\pi R^2 h).

Step 2. Set up and evaluate.

z_{\text{cm}} = \frac{1}{M}\int_0^h z\,\rho\,\frac{\pi R^2 z^2}{h^2}\,dz = \frac{3}{h^3}\int_0^h z^3\,dz = \frac{3}{h^3}\cdot\frac{h^4}{4} = \frac{3h}{4}

Why: the z^3 integrand reflects the fact that both the disc area (\propto z^2) and the position (z) grow with height, pulling the CM strongly toward the base.

This is 3h/4 from the apex. Measured from the base, the CM is at:

\boxed{h - \frac{3h}{4} = \frac{h}{4} \text{ from the base}}

A cone's CM is one-quarter of the way up from the base. A pile of sand with its tip 1 metre above the ground has its CM just 25 cm from the ground.

Standard results at a glance

Shape Where is the CM? Distance from centre/base
Uniform rod (length L) Midpoint L/2 from either end
Semicircular wire (radius R) On axis of symmetry, toward curved side 2R/\pi from centre
Semicircular disc (radius R) On axis of symmetry, toward curved side 4R/(3\pi) from centre
Solid hemisphere (radius R) On axis of symmetry, toward curved side 3R/8 from flat face
Solid cone (height h) On axis of symmetry, toward base h/4 from base

The pattern: as you go from a 1D wire to a 2D plate to a 3D solid of the same cross-section, the CM moves closer to the base. More dimensions means more mass concentrated near the wider region.

Example 1: The goldsmith's bangle

A goldsmith bends a thin uniform wire of length 62.8 cm into a semicircular bangle. Find the distance of its centre of mass from the centre of the bangle's opening.

Semicircular bangle with centre of mass marked A semicircular wire of radius 20 cm with the centre of mass marked at 12.7 cm above the midpoint of the diameter. CM 12.7 cm R = 20 cm
The semicircular bangle with its centre of mass (red dot) at 12.7 cm from the midpoint of the opening.

Step 1. Find the radius from the wire length.

The semicircular arc has length \pi R, so R = L/\pi = 62.8/\pi = 20.0 cm.

Why: the full circumference is 2\pi R; a semicircle is half of that, which gives \pi R.

Step 2. Apply the semicircular wire formula.

y_{\text{cm}} = \frac{2R}{\pi} = \frac{2 \times 20.0}{\pi} = \frac{40.0}{\pi} \approx 12.7 \text{ cm}

Why: this is a direct substitution into the result derived above. The CM depends only on the radius, not on the wire's mass or material.

Step 3. Verify reasonableness.

y_{\text{cm}}/R = 2/\pi \approx 0.637. The CM is about 63.7% of the way from the centre to the top — consistent with the interactive graph above at \alpha = \pi/2.

Result: The centre of mass is 12.7 cm from the centre of the bangle's opening, along the axis of symmetry toward the curved side.

What this shows: A bangle is not a closed circle, so its CM is not at the geometric centre. The opening shifts the balance point toward the solid part of the wire. If you tried to balance this bangle on a fingertip placed at its geometric centre, it would tilt toward the curve.

Example 2: The wooden toy — hemisphere plus cone

A Channapatna toymaker shapes a wooden toy by attaching a solid cone (height 30 cm, base radius 10 cm) to a solid hemisphere (radius 10 cm), with the cone's base glued to the hemisphere's flat face. Both pieces are the same uniform wood. Find the distance of the centre of mass from the flat joint.

Hemisphere-cone toy with individual and combined centres of mass Cross-section of a hemisphere below and a cone above joined at a flat face. The hemisphere CM is 3.75 cm below the joint, the cone CM is 7.5 cm above, and the combined CM is 3 cm above the joint. flat joint CM_h (−3.75 cm) CM_c (+7.5 cm) CM (+3 cm) h = 30 R = 10
The hemisphere's CM lies below the joint; the cone's CM lies above it. The combined CM (red dot) ends up 3 cm above the joint, inside the cone.

Step 1. Set coordinates. Let the flat joint be y = 0, with the hemisphere below (y < 0) and the cone above (y > 0).

From the standard results:

  • Hemisphere CM: y_h = -3R/8 = -3(10)/8 = -3.75 cm
  • Cone CM: y_c = +h/4 = +30/4 = +7.5 cm

Why: the hemisphere's CM is 3R/8 below the flat face (into the hemisphere), and the cone's CM is h/4 above the base (into the cone).

Step 2. Find the mass ratio. Both pieces have the same density \rho.

M_h = \rho\cdot\frac{2}{3}\pi R^3 = \frac{2}{3}\pi(10)^3\rho = \frac{2000\pi}{3}\,\rho
M_c = \rho\cdot\frac{1}{3}\pi R^2 h = \frac{1}{3}\pi(10)^2(30)\,\rho = 1000\pi\,\rho

Why: volume of a hemisphere is (2/3)\pi R^3 and volume of a cone is (1/3)\pi R^2 h. Since \rho is the same, mass is proportional to volume.

Step 3. Apply the composite-body formula.

y_{\text{cm}} = \frac{M_h\,y_h + M_c\,y_c}{M_h + M_c}

Numerator: \frac{2000\pi}{3}\rho\times(-3.75) + 1000\pi\rho\times7.5 = \pi\rho(-2500 + 7500) = 5000\pi\rho

Denominator: \left(\frac{2000}{3} + 1000\right)\pi\rho = \frac{5000}{3}\,\pi\rho

y_{\text{cm}} = \frac{5000\pi\rho}{(5000/3)\,\pi\rho} = 3 \text{ cm}

Why: the density \rho and the factor \pi cancel completely — only the geometry matters. The cone is lighter than the hemisphere (1000 vs 2000/3 ≈ 667 in arbitrary units), but its CM is much farther from the joint, so it pulls the combined CM above the joint and into the cone.

Result: The centre of mass of the toy is 3 cm above the flat joint, inside the conical part.

What this shows: Even though the hemisphere and the cone have comparable masses, their CMs sit at very different distances from the joint (3.75 cm vs 7.5 cm). The composite CM is a weighted average of these positions. The cone's greater moment arm wins, pulling the balance point above the joint. If the toymaker wants the toy to stand upright on its hemisphere, this result tells them the toy is stable — the CM is above the base, on the axis of symmetry.

Common confusions

  • "The centre of mass is always inside the body." Not true. The CM of a semicircular wire is at 2R/\pi from the centre — that point is in empty space, between the wire and the diameter. A ring, a hollow sphere, and a horseshoe all have their CMs in empty space. The CM is a mathematical average of positions, not a physical point that must coincide with matter.

  • "A heavier body has a different CM position." No. For a uniform body, the CM location depends only on the shape (geometry), not on the total mass. Double the mass of a uniform semicircular wire by using thicker wire, and the CM is still at 2R/\pi. The mass M cancels in every derivation above.

  • "The semicircular wire and the semicircular disc have the same CM." They do not. The wire has its CM at 2R/\pi \approx 0.637R; the disc at 4R/(3\pi) \approx 0.424R. The disc has more mass near the diameter (where the horizontal strips are widest), so its CM is closer to the flat edge.

  • "These formulas work for non-uniform bodies too." They do not. Every result in the table above assumes uniform density. For a non-uniform body (say, a rod that is denser at one end), you must integrate with the actual density function \lambda(x), \sigma(x,y), or \rho(x,y,z) inside the integral.

  • "I should memorise all five formulas." That is one approach, but a better one is to understand the method: set up coordinates, write dm in terms of a suitable variable, and integrate. If you forget the formula for a hemisphere, you can re-derive it in four lines.

If you came here to learn the integral method and the standard shapes, you have what you need. What follows covers non-uniform bodies and a beautiful theorem that connects centres of mass to volumes of revolution.

Non-uniform density: when \lambda depends on position

Consider a rod of length L whose linear density increases along its length: \lambda(x) = \lambda_0\left(1 + \frac{x}{L}\right), where x runs from 0 to L. The denser end is at x = L.

Total mass:

M = \int_0^L \lambda_0\left(1 + \frac{x}{L}\right)dx = \lambda_0\left[x + \frac{x^2}{2L}\right]_0^L = \lambda_0\left(L + \frac{L}{2}\right) = \frac{3\lambda_0 L}{2}

Centre of mass:

x_{\text{cm}} = \frac{1}{M}\int_0^L x\,\lambda_0\left(1 + \frac{x}{L}\right)dx = \frac{\lambda_0}{M}\int_0^L \left(x + \frac{x^2}{L}\right)dx
= \frac{\lambda_0}{M}\left[\frac{x^2}{2} + \frac{x^3}{3L}\right]_0^L = \frac{\lambda_0}{M}\left(\frac{L^2}{2} + \frac{L^2}{3}\right) = \frac{\lambda_0}{M}\cdot\frac{5L^2}{6}
x_{\text{cm}} = \frac{\lambda_0 \cdot 5L^2/6}{3\lambda_0 L/2} = \frac{5L}{9} \approx 0.556L

The CM has shifted from L/2 (uniform) to 5L/9 — toward the denser end, as you would expect. The same approach works for any density distribution: keep \lambda(x) inside the integral instead of pulling it out.

Pappus's centroid theorem

There is a remarkable connection between the centre of mass of a plane figure and the volume of the solid you get by rotating it about an axis.

Pappus's theorem (second). If a plane figure of area A is rotated through 2\pi about an external axis in its plane, the volume of the resulting solid is V = 2\pi\,\bar{y}\,A, where \bar{y} is the distance of the figure's centroid from the axis.

This gives you a cross-check on the semicircular disc result. Rotate a semicircular disc of radius R about its diameter (the x-axis). The resulting solid is a full sphere. So:

V_{\text{sphere}} = 2\pi\,y_{\text{cm}}\cdot A_{\text{semicircle}} = 2\pi\cdot\frac{4R}{3\pi}\cdot\frac{\pi R^2}{2} = \frac{4}{3}\pi R^3

The formula for the volume of a sphere drops out. If your CM result had been wrong, this cross-check would have caught it.

You can also use Pappus in reverse: if you know the volume of a solid of revolution and the area of the generating figure, you can find the centroid without integrating.

Where this leads next