Centre of Mass of Continuous Bodies

In short

For a continuous body, the discrete sum \vec{R}_\text{cm} = \frac{1}{M}\sum m_i \vec{r}_i becomes the integral \vec{R}_\text{cm} = \frac{1}{M}\int \vec{r}\,dm. The mass element dm depends on geometry: \lambda\,dx for a wire, \sigma\,dA for a plate, \rho\,dV for a solid. Symmetry eliminates integrals along balanced axes. Standard results: uniform rod → CM at L/2; semicircular wire → CM at 2R/\pi from the diameter.

Pick up a cricket bat and try to balance it horizontally on one finger. The balance point is not at the geometric centre — it sits closer to the blade, where the wood is thicker and heavier. That balance point is the centre of mass.

For two bolted-together pieces, the weighted-average formula from Centre of Mass — Definition and Calculation works: plug in two masses and two positions, done. But a real cricket bat is a continuous object — every thin shaving of willow has position and mass, and there are infinitely many of them. A finite sum cannot account for them all. The answer is to replace the sum with an integral.

Imagine cutting a round chapati in half. Where do you place your finger to balance the semicircular piece? Not at the centre of the straight edge, and not at the highest point of the arc — somewhere in between. This article gives you the exact tool to find that point for any continuous shape: rod, wire, plate, or solid.

From a sum to an integral

Start with the simplest continuous body: a uniform steel rod of length L and total mass M lying along the x-axis from x = 0 to x = L.

Imagine chopping it into N equal pieces. Each piece has width \Delta x = L/N, sits near position x_i, and has mass \Delta m = M/N. The centre-of-mass formula for discrete particles gives:

x_\text{cm} \approx \frac{1}{M}\sum_{i=1}^{N} x_i\,\Delta m

With N = 2, you get a rough estimate. With N = 10, a better one. Even at N = 4, the discrete approximation for a uniform rod already lands within a few percent of the exact answer. As N \to \infty and \Delta x \to 0, the approximation becomes exact: \Delta m becomes dm, the sum becomes an integral, and you have

x_\text{cm} = \frac{1}{M}\int x\,dm

This is not a new physical principle. It is the discrete centre-of-mass formula pushed to the continuous limit — the same weighted average, just with infinitely many infinitely small contributions.

Divide the rod into thin slices of width $dx$. Each slice at position $x$ has mass $dm = \lambda\,dx$, where $\lambda = M/L$ is the mass per unit length. As the slices become infinitely thin, the sum over all slices becomes an integral.

The formal definition

Centre of mass of a continuous body

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

In component form:

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

where M = \int dm is the total mass, and each integral runs over the entire body.

Read this symbol by symbol. The position vector \vec{r} marks the location of a tiny mass element dm. The integral sweeps that element over the whole body, weighting each position by its mass. Dividing by M gives a weighted average — the single point where the body would balance perfectly.

Notice the structure: it is exactly the discrete formula \vec{R}_\text{cm} = \frac{1}{M}\sum m_i \vec{r}_i, with the sum sign replaced by an integral sign and the finite chunk m_i replaced by the infinitesimal element dm. The physics is the same. The calculus has simply caught up to the continuous reality.

The entire challenge of a continuous-CM problem is expressing dm in terms of a variable you can actually integrate over. That choice depends on the geometry.

Choosing the mass element dm

The mass element dm is not one formula — it changes with the dimension of the body.

Wire or rod (one dimension). Mass is spread along a length. Define the linear mass density \lambda as mass per unit length (kg/m). For a uniform body of total length L, \lambda = M/L. The mass element is

dm = \lambda\,dl

where dl is a tiny arc length: dx for a straight rod, R\,d\theta for a circular arc of radius R. Think of a metal wire — you measure its mass per metre, and each centimetre of wire carries mass \lambda \times 0.01 m.

Plate or disc (two dimensions). Mass is spread over an area. Define the surface mass density \sigma as mass per unit area (kg/m²). For a uniform plate of area A, \sigma = M/A. The mass element is

dm = \sigma\,dA

where dA is a tiny area element: dx\,dy for a rectangle, r\,dr\,d\theta for a circular disc. Think of a uniform chapati — its mass per square centimetre is the same everywhere.

Solid (three dimensions). Mass fills a volume. Define the volume mass density \rho as mass per unit volume (kg/m³). For a uniform solid of volume V, \rho = M/V. The mass element is

dm = \rho\,dV

where dV depends on the coordinate system: dx\,dy\,dz for a box, r\,dr\,d\theta\,dz for a cylinder, r^2\sin\phi\,dr\,d\phi\,d\theta for a sphere.

For a non-uniform body the density varies with position — \lambda(x), \sigma(x,y), or \rho(x,y,z) — and the formulas above still hold, but the density stays inside the integral instead of coming out as a constant.

The right choice of dm and coordinate system is where the physics meets the calculus. Get this step right and the integral usually collapses in a few lines. Get it wrong and you fight the algebra for a page.

Symmetry does half the work

Before writing a single integral, study the shape. If the body has a line of symmetry, the centre of mass lies on that line. If it has a plane of symmetry, the CM lies in that plane. This is not an approximation — it is exact, and it eliminates one or more coordinates without any calculation.

A uniform rod along the x-axis is symmetric about its midpoint. For every element at distance d to the left of the centre, there is an identical element at distance d to the right. Their position-weighted contributions cancel in pairs, and x_\text{cm} is at the midpoint L/2.

A semicircular wire placed with its diameter along the x-axis is symmetric about the y-axis. Every element at angle \theta (with x-coordinate R\cos\theta) has a mirror element at angle \pi - \theta (with x-coordinate -R\cos\theta). The x-contributions cancel: x_\text{cm} = 0, and you only need to compute y_\text{cm}.

A full circle, a full sphere, or a uniform cube has symmetry about every axis through its centre — so the CM is at the geometric centre with zero integration needed.

The rule: never integrate over a coordinate that symmetry has already settled. Identify all symmetry axes first, set those coordinates by inspection, and spend your effort only on what remains.

Worked examples

Example 1: Balancing a metre scale

A uniform steel metre scale (mass 200 g, length 1 m) from your physics lab lies along the x-axis from x = 0 to x = 1 m. Find the position of its centre of mass.

The uniform metre scale with one element $dx$ highlighted. The centre of mass sits at the midpoint — exactly where your finger goes to balance it.

Step 1. Set up the mass element.

The rod is uniform, so the linear mass density is \lambda = M/L = 0.200/1.0 = 0.200 kg/m. A small element at position x has mass:

dm = \lambda\,dx

Why: every centimetre of the uniform rod carries the same mass. The element dm is simply \lambda (mass per unit length) times the element width dx.

Step 2. Write the integral for x_\text{cm}.

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

Why: multiply each element's position x by its mass \lambda\,dx, then integrate over the entire rod from 0 to L.

Step 3. Factor out the constant \lambda and evaluate.

x_\text{cm} = \frac{\lambda}{M}\int_0^{L} x\,dx = \frac{\lambda}{M}\left[\frac{x^2}{2}\right]_0^{L} = \frac{\lambda\,L^2}{2M}

Why: \lambda is constant for a uniform rod, so it comes outside the integral. The standard result \int x\,dx = x^2/2 handles the rest.

Step 4. Substitute \lambda = M/L.

x_\text{cm} = \frac{(M/L)\,L^2}{2M} = \frac{L}{2} = 0.50 \text{ m}

Why: M cancels completely — the CM position depends only on the length, not on the total mass. A 200 g ruler and a 2 kg steel bar of the same length have their CM at the same point.

Result: x_\text{cm} = L/2 = 0.50 m — the geometric midpoint.

What this shows: The centre of mass of a uniform rod sits at its midpoint. This is exactly where you instinctively place your finger when balancing a metre scale. Uniformity guarantees equal mass on both sides of the centre, so the balance point is halfway along. The integral confirmed what symmetry already told you — and demonstrated the method you will use for less symmetric shapes.

Example 2: Centre of mass of a semicircular wire

A thin wire of mass 50 g is bent into a semicircle of radius R = 10 cm. The diameter lies along the x-axis. Find the centre of mass.

The semicircular wire with an element at angle $\theta$. The element sits at height $R\sin\theta$ above the diameter. The centre of mass lies on the $y$-axis at height $2R/\pi \approx 6.4$ cm.

Step 1. Use symmetry to eliminate one coordinate.

The wire is symmetric about the y-axis: for every element at angle \theta (with x-coordinate R\cos\theta), there is a mirror element at \pi - \theta (with x-coordinate -R\cos\theta). The horizontal contributions cancel, so x_\text{cm} = 0.

Why: the left half of the wire pulls the CM left by exactly as much as the right half pulls it right. Symmetry settles the x-coordinate without a single line of integration.

Step 2. Set up dm.

The total arc length is \pi R. The linear density is \lambda = M/(\pi R). A small element at angle \theta subtends d\theta at the centre, has arc length R\,d\theta, and mass:

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

Why: the element sweeps through angle d\theta, so its arc length is R\,d\theta. Multiplying by \lambda (mass per unit length) converts length to mass. The R in the numerator and denominator cancel, leaving a clean expression in d\theta alone.

Step 3. Write the integral for y_\text{cm}.

The element at angle \theta is at height y = R\sin\theta:

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

Why: M cancels — the CM position does not depend on the total mass. The integral computes the average height of the wire, weighted uniformly over the arc length.

Step 4. Evaluate.

y_\text{cm} = \frac{R}{\pi}\Big[-\cos\theta\Big]_0^{\pi} = \frac{R}{\pi}\big(-\cos\pi + \cos 0\big) = \frac{R}{\pi}(1 + 1) = \frac{2R}{\pi}

Why: \cos\pi = -1 and \cos 0 = 1, so -(-1) + 1 = 2. The final answer is clean: 2R/\pi.

Result: x_\text{cm} = 0, \;y_\text{cm} = 2R/\pi \approx 0.637R. For R = 10 cm, the centre of mass is about 6.4 cm above the diameter.

What this shows: The CM of a semicircular wire is not at the geometric centre of the semicircle (which is the origin), nor at the top of the arc (R above the diameter). It sits at 2R/\pi — roughly two-thirds of the way from the centre to the top. The wire has uniform mass per unit arc length, but the height R\sin\theta varies — it peaks at \theta = \pi/2 (the top) and vanishes at \theta = 0 and \pi (the ends). The integral averages this height over the arc, giving a value between zero and R.

Explore: how the CM depends on the arc angle

The semicircle is one specific arc. What happens for a shorter or longer one?

For a circular arc of radius R symmetric about the y-axis with half-angle \alpha — so the arc spans from \theta = \pi/2 - \alpha to \theta = \pi/2 + \alpha — the same method gives:

y_\text{cm} = \frac{R}{2\alpha}\int_{\pi/2 - \alpha}^{\pi/2 + \alpha}\sin\theta\,d\theta = \frac{R}{2\alpha}\Big[-\cos\theta\Big]_{\pi/2 - \alpha}^{\pi/2 + \alpha} = \frac{R}{2\alpha}\big(\sin\alpha + \sin\alpha\big) = \frac{R\sin\alpha}{\alpha}

Why: -\cos(\pi/2 + \alpha) = \sin\alpha and \cos(\pi/2 - \alpha) = \sin\alpha, so the bracket gives 2\sin\alpha, which simplifies to R\sin\alpha/\alpha.

Check the limits: as \alpha \to 0 (a tiny arc near the top), \sin\alpha/\alpha \to 1 and y_\text{cm} \to R — the CM is right at the arc itself. At \alpha = \pi/2 (a semicircle), y_\text{cm} = 2R/\pi. At \alpha = \pi (a full circle), \sin\pi/\pi = 0 and the CM is at the geometric centre.

Drag the red dot to change the half-angle $\alpha$ of a circular arc. A short arc near the top ($\alpha$ small) has its CM close to $R$. A semicircle ($\alpha = \pi/2$) gives $2R/\pi \approx 0.637R$. A nearly full circle ($\alpha \to \pi$) pulls the CM down toward the geometric centre.

Common confusions

"The centre of mass is always inside the body." Not necessarily. For a semicircular wire, the CM at 2R/\pi above the diameter is a point in empty space — not on the wire itself. Hollow objects and curved shapes frequently have their CM outside the physical material.
"dm is always \rho\,dV." Only for three-dimensional solids. A thin wire uses dm = \lambda\,dl, a thin plate uses dm = \sigma\,dA. Using the wrong dimension gives an integral with incorrect units and the wrong number of variables.
"If the body is uniform, the CM is at the geometric centre." True for shapes with full symmetry — a uniform disc, sphere, or cube. False for shapes with partial symmetry, like a semicircular wire or a triangular plate. Uniformity means constant density, not geometric centrality. You still need the integral whenever the shape breaks full symmetry.
"You always integrate over x or y." Not always. For a circular arc, integrating over the angle \theta is far cleaner than expressing the arc as y(x) and integrating over x. For a solid of revolution, cylindrical coordinates often reduce a triple integral to a single one. Match the coordinate system to the geometry.
"The CM position depends on the total mass." It does not, for any uniform body. In every calculation above, M cancelled completely. The CM of a 100 g semicircular wire and a 10 kg semicircular wire of the same radius is at the same point. Only the shape determines the CM position.

If you came here to learn how to set up the integral and find the CM of a rod or wire, you have the method. What follows is for readers who want the standard results for plates and solids — the shapes that appear regularly in JEE Advanced problems.

Centre of mass of a uniform semicircular disc

Place a uniform semicircular disc of radius R and mass M with its diameter along the x-axis. By symmetry, x_\text{cm} = 0. To find y_\text{cm}, use polar coordinates: dA = r\,dr\,d\theta, with the surface mass density \sigma = M/(\frac{1}{2}\pi R^2) = 2M/(\pi R^2).

y_\text{cm} = \frac{1}{M}\int_0^{\pi}\!\int_0^{R} (r\sin\theta)\;\sigma\;r\,dr\,d\theta = \frac{\sigma}{M}\int_0^{R} r^2\,dr\;\int_0^{\pi}\sin\theta\,d\theta

Why: the position factor is y = r\sin\theta. The area element brings another factor of r (from r\,dr\,d\theta), giving r^2 in the radial integral. The integrals separate because r and \theta appear independently.

= \frac{\sigma}{M}\cdot\frac{R^3}{3}\cdot 2 = \frac{2\sigma R^3}{3M}

Substituting \sigma = 2M/(\pi R^2):

y_\text{cm} = \frac{2 \times 2M/(\pi R^2) \times R^3}{3M} = \frac{4R}{3\pi} \approx 0.424R

The disc's CM is lower than the wire's (0.637R) because the disc has mass filling the interior near the diameter — the region close to the base is filled with material at low y, pulling the average downward.

Centre of mass of a solid hemisphere

Place a uniform solid hemisphere of radius R flat-side down on the xz-plane. The CM lies on the y-axis. Slice the hemisphere into thin horizontal discs at height y: each disc has radius r = \sqrt{R^2 - y^2} and thickness dy.

dm = \rho\,\pi(R^2 - y^2)\,dy, \qquad \rho = \frac{M}{\frac{2}{3}\pi R^3} = \frac{3M}{2\pi R^3}

Why: the disc at height y has area \pi r^2 = \pi(R^2 - y^2) and thickness dy, so its volume is \pi(R^2 - y^2)\,dy. Multiplying by \rho gives its mass.

y_\text{cm} = \frac{\rho\pi}{M}\int_0^{R}(yR^2 - y^3)\,dy = \frac{\rho\pi}{M}\left[\frac{R^2 y^2}{2} - \frac{y^4}{4}\right]_0^{R} = \frac{\rho\pi}{M}\cdot\frac{R^4}{4}

Why: R^2 \cdot R^2/2 - R^4/4 = R^4(1/2 - 1/4) = R^4/4. The disc areas shrink as y increases, concentrating mass near the base.

Substituting \rho = 3M/(2\pi R^3):

y_\text{cm} = \frac{3M/(2\pi R^3)\cdot\pi\cdot R^4}{4M} = \frac{3R}{8} = 0.375R

Centre of mass of a solid cone

A uniform solid cone of height h and base radius R, with the apex at the origin and the axis along y. At height y from the apex, the cross-section is a disc of radius Ry/h.

dm = \rho\,\pi\frac{R^2 y^2}{h^2}\,dy, \qquad M = \frac{1}{3}\pi R^2 h\,\rho

y_\text{cm} = \frac{\rho\pi R^2}{Mh^2}\int_0^{h} y^3\,dy = \frac{\rho\pi R^2}{Mh^2}\cdot\frac{h^4}{4} = \frac{\rho\pi R^2 h^2}{4M}

Substituting M = \frac{1}{3}\pi R^2 h\rho:

y_\text{cm} = \frac{\rho\pi R^2 h^2}{4 \times \frac{1}{3}\pi R^2 h\rho} = \frac{3h}{4}

Why: \int_0^h y^3\,dy = h^4/4 and the factor 1/3 from the cone volume inverts to give 3. The CM is 3h/4 from the apex.

That is 3h/4 from the apex — or equivalently, h/4 from the base. Think of the nose cone of an ISRO PSLV rocket: its balance point sits much closer to the wide base than to the pointed tip, because most of the cone's volume (and therefore mass) is packed near the base.

Standard results — a reference table

Shape	CM distance	Measured from
Uniform rod, length L	L/2	either end
Semicircular wire, radius R	2R/\pi \approx 0.637R	centre of diameter
Semicircular disc, radius R	4R/(3\pi) \approx 0.424R	centre of diameter
Solid hemisphere, radius R	3R/8 = 0.375R	flat face
Hemispherical shell, radius R	R/2 = 0.500R	flat face
Solid cone, height h	h/4	base
Hollow cone (lateral surface), height h	h/3	base

Notice the pattern: for a semicircular shape, the wire (1D) has the CM farthest from the symmetry centre at 0.637R, then the plate (2D) at 0.424R, then the solid (3D) at 0.375R. Higher-dimensional bodies pack more mass near the base (inner region), pulling the CM progressively lower.

These results are worth memorising for competitive exams — but you now know how to derive any of them from scratch if memory fails.

Where this leads next

Motion of the Centre of Mass — Newton's second law for the system: \vec{F}_\text{ext} = M\,\vec{a}_\text{cm}.
Moment of Inertia — the rotational analog, computed by a similar integral: I = \int r^2\,dm.
Rotational Kinematics — angular velocity, angular acceleration, and the equations of rotational motion.
Torque and Angular Momentum — how forces produce rotation, and why angular momentum is conserved when no external torque acts.