Chapter 07 · Centre of Mass, Momentum & Collisions · Full book

Centre of Mass and System of Particles: The Point That Represents Everything

Core intuition: Chapter Roadmap

Newton's Laws describe how a single particle responds to a force. But a wrench, a rocket, a pair of colliding pucks, an exploding shell—these are systems of many particles, and tracking every one is hopeless. This chapter introduces a single point that tames the chaos: the centre of mass.

The centre of mass — the mass-weighted average position \(\vec{r}_{\text{cm}} = \frac{1}{M}\sum m_i \vec{r}_i\) (a sum for discrete bodies, the integral \(\frac{1}{M}\int \vec{r}\,dm\) for continuous ones), and the symmetry shortcuts that often make the integral unnecessary.
Motion of the centre of mass — the master equation \(\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}\): internal forces cancel, so the CoM moves as if all the mass and every external force acted at that one point.
Linear momentum and its conservation — \(\vec{P} = M\vec{v}_{\text{cm}}\), and the law that when no external force acts, total momentum stays constant—the single most powerful idea in the chapter.
Collisions — a framework for elastic, inelastic, and perfectly inelastic encounters in one and two dimensions, with the coefficient of restitution bridging momentum and energy.
The centre-of-mass frame — the frame in which total momentum is zero, where collisions turn symmetric and their analysis collapses to a few lines.
Variable-mass systems — rockets and falling chains, where mass is carried in or flung out and \(\vec{F} = m\vec{a}\) must give way to the thrust equation.
Choosing your tool — a decision algorithm for picking CoM, momentum, restitution, or the CoM frame as the fastest route through a problem.

Find the right point, and a system of countless particles speaks with a single voice: one position, one velocity, one momentum, obeying one law.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Tumbling Wrench: Toss a wrench across the room so that it spins and tumbles chaotically. Every point on it traces a complicated, wobbling curve. Yet one special point glides through the air in a perfect parabola, as if the wrench were a simple point particle. What is this point, and why does it ignore the chaos around it?
The Recoiling Boat: You stand at one end of a small boat on still water. You walk to the other end and stop. You moved forward—but the boat moved backward. Has the system as a whole gone anywhere?
The Mid-Air Breakup: A firework shell is launched as a single projectile. At the top of its arc it explodes into hundreds of fragments, each flying off in a different direction. Yet if you could average the positions of all the fragments, weighted by their masses, that average point would continue along the original parabola as if nothing happened. Why?
The Frozen Skater: A person stands perfectly still on a vast, frictionless frozen lake. No wind, no slope, no rope. Can they reach the shore by any means—walking, swimming through the air, flapping their arms—without throwing or pushing something away from themselves?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Why Systems Need a New Idea

Until now, every object in this book has been a particle—a point mass with no size, no shape, and no internal structure. A block on an incline, a projectile in flight, a car rounding a curve: each was treated as if its entire mass were concentrated at a single mathematical point.

Real objects are not particles. A wrench has a handle and a head. A human body has arms that swing, legs that push, and a torso that twists. A rocket sheds mass continuously. When you walk across a boat, your body moves one way while the boat moves the other; neither alone tells you what the system is doing.

So how do we apply Newton's Laws—formulated for point particles—to extended objects and systems of many particles?

The answer is one of the most powerful ideas in all of mechanics: every system of particles, no matter how complicated, possesses a single special point whose motion obeys Newton's second law exactly as if the entire mass of the system were concentrated there.

That point is called the centre of mass.

Core intuition: The Representor Point

The centre of mass is the point that represents a system to the outside world. External forces accelerate it as if the whole system were a single particle sitting at that point, with the total mass of the system. Internal forces—the forces that parts of the system exert on each other—do not affect it at all. The explosion of a firework shell is a violent internal rearrangement. The centre of mass neither notices nor cares.

Why Internal Forces Cancel

Consider two ice skaters, initially at rest, who push off each other. Skater \(A\) pushes skater \(B\) to the right; by Newton's third law, skater \(B\) pushes skater \(A\) to the left with an equal and opposite force. These are internal forces—forces exchanged between members of the same system.

Now sum the forces on the system as a whole. The push on \(A\) and the push on \(B\) are a third-law pair: they are equal in magnitude, opposite in direction, and they cancel in the sum. No external horizontal force acts on the two-skater system (the ice is frictionless), so the total force on the system is zero.

What does Newton's second law then predict? The acceleration of the system—meaning the acceleration of its centre of mass—is zero. The centre of mass does not move.

Each skater individually accelerates; one slides left, the other right. But the mass-weighted average of their positions—the centre of mass—stays exactly where it was. This is not a coincidence. It is a theorem, and we will prove it rigorously in the next section.

The same logic explains every puzzle we opened with. The wrench tumbles, but its centre of mass follows a clean parabola because the only external force is gravity—and gravity accelerates the centre of mass exactly as it would a point particle of the same total mass. The boat recoils when you walk on it because no external horizontal force acts on the person–boat system, so the centre of mass stays put: your forward displacement and the boat's backward displacement must balance, weighted by mass. The exploding shell scatters fragments wildly, but no new external force was introduced by the explosion—only internal forces—so the centre of mass continues its original parabolic trajectory undisturbed.

Core intuition: The Principle That Runs This Chapter

Internal forces change how parts of a system move relative to each other. They cannot change how the system as a whole moves through space. Only external forces can accelerate the centre of mass. This single statement—once made precise—will unify everything in this chapter: centre-of-mass motion, momentum conservation, collisions, recoil, and rocket propulsion.

Three Ideas, One Framework

This chapter develops three tightly connected ideas:

Centre of mass. A precise definition of the mass-weighted average point of a system, and the tools to calculate it for discrete particles, continuous bodies, and composite shapes.

Linear momentum. The product \(\vec{p} = m\vec{v}\), extended to systems. Newton's second law, rewritten in terms of momentum, reveals why momentum is the natural quantity for describing the motion of systems.

Conservation of momentum. When the net external force on a system is zero, the total momentum does not change. This is the principle behind collisions, explosions, recoil, and every internal rearrangement of a system.

Together, these three ideas allow you to analyse complex multi-particle systems—colliding billiard balls, exploding projectiles, a gun firing a bullet, a person jumping off a boat—with surprising economy. The algebra is often shorter than anything Newton's Laws alone could produce, because you never need to know the details of the internal forces.

What Lies Ahead

We begin by defining the centre of mass precisely and learning to compute it (another section). Then we prove that \(\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}\), which is the master equation of this chapter (another section). From there we develop linear momentum and its conservation (another section), apply them to collisions in one and two dimensions, introduce the powerful centre-of-mass reference frame (another section), and close with variable-mass systems and rocket propulsion (another section).

If at any point you feel lost, return to the single idea that anchors everything: internal forces cancel; only external forces move the centre of mass. Every result in this chapter is a consequence of that one fact.

✓ Checkpoint · Thinking About Systems

Two astronauts float in deep space, connected by a taut rope. They pull themselves toward each other along the rope. Does the centre of mass of the two-astronaut system move? Justify your answer without any equations.
No. The tension in the rope is an internal force (a third-law pair between the two astronauts). No external force acts on the system, so the centre of mass remains stationary. Each astronaut moves, but they move symmetrically (weighted by mass) toward the centre of mass.
A cannon mounted on a stationary railway car fires a shell horizontally. The car recoils backward on frictionless rails. Is momentum conserved for the shell alone? For the car alone? For the shell–car system?
Not for the shell alone (the cannon exerts a forward force on it). Not for the car alone (the shell exerts a backward force on it). Yes for the shell–car system: the forces between shell and car are internal (third-law pair), and no external horizontal force acts, so the total horizontal momentum of the system is conserved (it was zero before firing, so it remains zero after).
A boy stands on a stationary raft in a lake. He walks from one end to the other.
1. Does the centre of mass of the boy–raft system move horizontally? (Neglect water resistance.)
2. If the boy has mass \(m\) and the raft has mass \(M\), who moves farther—the boy or the raft? Give a qualitative argument.
(a) No. No external horizontal force acts, so the centre of mass stays fixed. (b) The raft moves farther if \(m \gt M\); the boy moves farther if \(M \gt m\). Since the centre of mass is fixed, the displacements must satisfy \(m \cdot d_{\text{boy}} = M \cdot d_{\text{raft}}\) (in opposite directions). The lighter object must displace more to keep the mass-weighted average stationary.

Centre of Mass: Definition and Calculation

The Balancing Point

Hold a ruler horizontally on the tip of your finger. If your finger is too far to the left, the ruler topples rightward; too far to the right, it topples leftward. There is exactly one point where the ruler balances perfectly—where the tendency to rotate one way is exactly matched by the tendency to rotate the other way.

Now imagine the ruler is not uniform: someone has glued a heavy metal washer near one end. The balancing point shifts toward the washer. The heavier a region is, the more the balance point is pulled toward it.

This balancing point is the centre of mass. (For a body near the Earth's surface, where \(g\) is uniform, the centre of mass coincides with the centre of gravity. They differ only when the gravitational field varies appreciably over the extent of the body—a distinction irrelevant for JEE-scale problems but important in astrophysics.)

The idea generalises immediately from a ruler to any system of particles: the centre of mass is the single point that summarises where the mass of the system is concentrated on average. Heavier particles pull the centre of mass toward themselves; lighter particles influence it less. If all particles have the same mass, the centre of mass is simply the geometric centre of the arrangement.

Definition for a System of Discrete Particles

Consider \(N\) particles with masses \(m_1, m_2, \ldots, m_N\) located at position vectors \(\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_N\) relative to some chosen origin.

Birth of the formula: Centre of Mass — Discrete System

The position vector of the centre of mass is defined as the mass-weighted average of the individual position vectors: \[ \boxed{\vec{r}_{\text{cm}} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \cdots + m_N\vec{r}_N} {m_1 + m_2 + \cdots + m_N} = \frac{1}{M}\sum_{i=1}^{N} m_i\,\vec{r}_i} \] where \(M = \sum_{i=1}^{N} m_i\) is the total mass of the system. In Cartesian components this becomes three independent equations: \[ \boxed{x_{\text{cm}} = \frac{1}{M}\sum_{i=1}^{N} m_i\,x_i, \qquad y_{\text{cm}} = \frac{1}{M}\sum_{i=1}^{N} m_i\,y_i, \qquad z_{\text{cm}} = \frac{1}{M}\sum_{i=1}^{N} m_i\,z_i} \] Each coordinate of the centre of mass is computed independently of the others. This is the power of the component form: a three-dimensional problem breaks into three one-dimensional calculations.

Reading the formula. This equation is a weighted average. Each position \(\vec{r}_i\) is weighted by the fraction \(m_i/M\) of the total mass that particle \(i\) carries. A particle with twice the mass of another has twice the influence on the location of the centre of mass. A particle with negligible mass has negligible influence, no matter where it sits.

The Two-Particle System: Building Intuition

◈ Two-particle centre of mass

live bench

Slide two masses along a rod. The balance point always sits closer to the heavier one, dividing the joining line in the inverse ratio of the masses — \(m_1 r_1 = m_2 r_2\).

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The simplest system is two particles on a line. Place particle 1 (mass \(m_1\)) at position \(x_1\) and particle 2 (mass \(m_2\)) at position \(x_2 \gt x_1\). The separation is \(d = x_2 - x_1\).

From : \[ x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]

Before computing, ask: where do we expect the centre of mass to lie?

If \(m_1 = m_2\), symmetry demands \(x_{\text{cm}}\) at the exact midpoint. If \(m_1 \gg m_2\), the centre of mass should sit almost on top of \(m_1\); the light particle barely matters. If \(m_2 \gg m_1\), the centre of mass should sit near \(m_2\).

Let us verify. Setting \(m_1 = m_2 = m\) in : \[ x_{\text{cm}} = \frac{m\,x_1 + m\,x_2}{2m} = \frac{x_1 + x_2}{2} \] which is the midpoint, exactly as expected.

Now choose the origin at \(m_1\) for convenience, so \(x_1 = 0\) and \(x_2 = d\). Then: \[ x_{\text{cm}} = \frac{m_2\,d}{m_1 + m_2} \]

This gives the distance of the centre of mass from \(m_1\). Similarly, the distance from \(m_2\) is: \[ d - x_{\text{cm}} = \frac{m_1\,d}{m_1 + m_2} \]

Dividing these two results: \[ \frac{\text{distance of COM from } m_1}{\text{distance of COM from } m_2} = \frac{m_2}{m_1} \]

Core intuition: The Inverse-Mass Rule

The centre of mass divides the line joining two particles in the inverse ratio of their masses. It always lies closer to the heavier particle. This is the single most useful result for quick mental estimates in COM problems. A \(3\,\mathrm{kg}\) mass and a \(1\,\mathrm{kg}\) mass separated by \(d\): the COM sits at \(d/4\) from the heavier mass. No calculation needed—just the ratio.

Can the Centre of Mass Lie Outside the Body?

Fig 6.1 The centre of mass need not lie on the body. For an equal-armed bent wire it sits at \((a/4,\,a/4)\) — out in the open corner, where there is no material at all. It marks where the mass balances, not where it is.

A uniform solid disc has its centre of mass at its geometric centre—a point that lies inside the material. But consider a uniform ring. Its centre of mass is at the geometric centre of the ring, which is empty space. No material exists at that point.

This is not a pathology. A boomerang, a hollow sphere, an L-shaped bracket, a crescent moon: all have their centres of mass at points where there is no material.

Misconception alert: The COM Must Be Inside the Object

Wrong. The centre of mass is a mathematical point—the mass-weighted average position. There is no physical requirement that this average lies inside the material of the body. For any object with a hole, cavity, or concave shape, the centre of mass can—and often does—lie in empty space. Exam trap: JEE problems sometimes ask for the COM of a semicircular ring or a hemispherical shell. Students who assume the answer must lie on the material are misled. The answer lies on the axis of symmetry, between the object and its geometric centre—often in empty space.

Symmetry: The Shortcut You Must Use

Before computing any integral or sum, always ask: does the mass distribution have a symmetry?

If a body has a plane of symmetry, the centre of mass lies in that plane. If it has an axis of symmetry, the centre of mass lies on that axis. If it has a centre of symmetry (like a uniform sphere or cube), the centre of mass is at that centre.

Each symmetry eliminates one coordinate from the calculation. A uniform rod has an axis of symmetry along its length: the COM lies on that axis, and a single integral (or just the midpoint, by symmetry) suffices. A uniform disc has two planes of symmetry meeting at its centre: the COM is at the intersection, with no integral needed at all.

Core intuition: Symmetry Before Calculation

The most common error in COM problems is not a wrong integral—it is a wasted integral. If symmetry tells you that \(y_{\text{cm}} = 0\), do not compute it. State the symmetry argument and move on. Every competitive exam rewards speed; symmetry is the fastest tool you have.

Centre of Mass of Continuous Bodies

When the number of particles becomes very large—as in a solid rod, a disc, or any extended body—the sum in becomes an integral. Replace the discrete mass \(m_i\) with an infinitesimal mass element \(dm\) at position \(\vec{r}\):

\[ \boxed{\vec{r}_{\text{cm}} = \frac{1}{M}\int \vec{r}\;dm} \]

The integral runs over the entire body. In practice, you choose a coordinate system, express \(dm\) in terms of a geometric variable (length, area, or volume), and integrate.

Expressing \(dm\). The choice depends on the geometry:

Geometry	Density type	Element
Thin rod / wire	Linear: \(\lambda = dm/dx\)	\(dm = \lambda\,dx\)
Thin plate / shell	Surface: \(\sigma = dm/dA\)	\(dm = \sigma\,dA\)
Solid body	Volume: \(\rho = dm/dV\)	\(dm = \rho\,dV\)

For a uniform body the density is constant and can be pulled outside the integral, simplifying the calculation considerably.

The Composite-Body (Superposition) Method

◈ A body with a hole — the superposition method

live bench

Cut a hole from a uniform disc and watch the centre of mass shift away from the cavity. Treat the hole as a disc of negative mass superposed on the full disc.

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Many exam problems involve shapes built by combining or removing simple shapes: a disc with a hole, an L-shaped plate, a hemisphere glued to a cylinder. For these, direct integration is unnecessarily painful. There is a far cleaner approach.

If a body can be decomposed into parts whose individual centres of mass are already known, then the centre of mass of the whole body is found by treating each part as a single particle located at its own centre of mass, with mass equal to the mass of that part:

\[ \boxed{\vec{r}_{\text{cm}} = \frac{m_A\,\vec{r}_{\text{cm},A} + m_B\,\vec{r}_{\text{cm},B} + \cdots} {m_A + m_B + \cdots}} \]

Subtraction trick. A disc with a circular hole is not a natural “sum” of parts. But it is a full disc minus the removed piece. Treat the removed piece as having negative mass \((-m_{\text{hole}})\) and apply directly: \[ \vec{r}_{\text{cm}} = \frac{m_{\text{full}}\,\vec{r}_{\text{cm,full}} - m_{\text{hole}}\,\vec{r}_{\text{cm,hole}}} {m_{\text{full}} - m_{\text{hole}}} \]

This is the single most exam-efficient technique for COM calculations. Half the JEE problems on centre of mass are solved by superposition with subtraction.

Example 1 · Two particles on a line

Two particles of masses \(3\,\mathrm{kg}\) and \(1\,\mathrm{kg}\) are placed on the \(x\)-axis at \(x = 2\,\mathrm{m}\) and \(x = 6\,\mathrm{m}\) respectively. Find the position of the centre of mass.

Solution. Sign convention: Take the positive \(x\)-direction to the right. Using : \[ x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{(3)(2) + (1)(6)}{3 + 1} = \frac{6 + 6}{4} = 3\,\mathrm{m} \] Sanity check. The separation is \(d = 4\,\mathrm{m}\). By the inverse-mass rule , the COM should lie at distance \(\dfrac{m_2}{m_1+m_2}\,d = \dfrac{1}{4}(4) = 1\,\mathrm{m}\) from the heavier mass. Indeed, \(x_{\text{cm}} - x_1 = 3 - 2 = 1\,\mathrm{m}\).

Example 2 · Three particles in a plane

Three particles of masses \(1\,\mathrm{kg}\), \(2\,\mathrm{kg}\), and \(3\,\mathrm{kg}\) are placed at the vertices of a triangle with coordinates \((0,\,0)\), \((4,\,0)\), and \((2,\,3)\) respectively (all in metres). Find \(\vec{r}_{\text{cm}}\).

Solution. Total mass: \(M = 1 + 2 + 3 = 6\,\mathrm{kg}\). \[ x_{\text{cm}} = \frac{(1)(0) + (2)(4) + (3)(2)}{6} = \frac{0 + 8 + 6}{6} = \frac{14}{6} = \frac{7}{3}\,\mathrm{m} \approx 2.33\,\mathrm{m} \] \[ y_{\text{cm}} = \frac{(1)(0) + (2)(0) + (3)(3)}{6} = \frac{9}{6} = \frac{3}{2}\,\mathrm{m} = 1.5\,\mathrm{m} \] Therefore \(\vec{r}_{\text{cm}} = \left(\dfrac{7}{3},\;\dfrac{3}{2}\right)\,\mathrm{m}\). Sanity check. The heaviest mass (\(3\,\mathrm{kg}\)) is at \((2, 3)\). The COM should be pulled toward it, i.e. the \(y\)-coordinate should be positive and the \(x\)-coordinate should be near \(2\). Both conditions are satisfied.

Example 3 · Uniform rod — by integration

Find the centre of mass of a uniform thin rod of length \(L\) and total mass \(M\).

Solution. Setup. Place the rod along the \(x\)-axis from \(x = 0\) to \(x = L\). By symmetry, \(y_{\text{cm}} = z_{\text{cm}} = 0\). We need only \(x_{\text{cm}}\). The rod is uniform, so its linear mass density is \(\lambda = M/L\) (constant). An element of length \(dx\) at position \(x\) has mass \(dm = \lambda\,dx\). Integration. From : \[ x_{\text{cm}} = \frac{1}{M}\int_0^L x\,dm = \frac{1}{M}\int_0^L x\,\lambda\,dx = \frac{\lambda}{M}\cdot\frac{L^2}{2} = \frac{M/L}{M}\cdot\frac{L^2}{2} = \frac{L}{2} \] The centre of mass is at the midpoint—exactly what symmetry predicted before we wrote a single integral.

✓ Checkpoint · Centre of Mass Basics

Two equal masses are placed at the ends of a massless rod of length \(d\). Where is the centre of mass? Now one mass is doubled. In which direction and by how much does the COM shift?
Initially at the midpoint, \(d/2\) from each mass. After doubling one mass, the COM shifts toward the heavier mass. New position from the lighter mass: \(x_{\text{cm}} = \dfrac{2m \cdot d}{m + 2m} = \dfrac{2d}{3}\). The COM has shifted by \(d/6\) toward the heavier end.
A uniform circular ring, a uniform circular disc, and a uniform solid sphere all have the same outer radius \(R\). Where is the centre of mass of each? For which of these does the COM lie at a point where there is no material?
All three have their COM at the geometric centre (by symmetry). For the disc and the sphere, this point lies inside the material. For the ring, the geometric centre is in empty space—the COM lies where there is no material.
A uniform square plate of side \(a\) has a square hole of side \(a/2\) cut from one corner. Without computing, predict: does the COM of the remaining piece shift toward or away from the removed corner? Justify qualitatively.
Away from the removed corner. Removing material from one corner is equivalent to placing negative mass there. The COM of the remaining piece shifts in the direction opposite to the removed portion—i.e. away from the corner where the hole was cut.

Calculating the Centre of Mass: Standard Bodies

another section introduced the integral \(\vec{r}_{\text{cm}} = \frac{1}{M}\int\vec{r}\;dm\) and applied it to a uniform rod. In this section we work through every standard geometry that appears in competitive exams: semicircular wires, semicircular plates, hemispherical shells, solid hemispheres, triangular plates, and non-uniform rods. We then return to the composite-body method with a fully worked disc-with-hole problem.

Master these results. They recur so often in JEE, NEET, and Olympiad papers that knowing them by sight saves minutes under exam pressure—and knowing how to derive them insures you against memory failure.

The Integration Recipe

Every COM integration follows the same four steps:

Choose coordinates. Align an axis with any symmetry the body possesses.
Use symmetry. Identify which coordinates of \(\vec{r}_{\text{cm}}\) are zero by symmetry and eliminate them immediately.
Express \(dm\). Write the mass element in terms of a single geometric variable using the appropriate density (\(\lambda\), \(\sigma\), or \(\rho\)).
Integrate. Compute \(x_{\text{cm}} = \frac{1}{M}\int x\;dm\) (or the corresponding coordinate) over the body.

The art lies in Step 3: choosing the right \(dm\) so the integral is a single variable.

Semicircular Wire (Ring)

◈ Centre of mass by integration

live bench

Sweep the summation for a continuous body — a rod, a semicircular wire or a plate — and see the integral \(\vec r_{cm}=\tfrac1M\int \vec r\,dm\) accumulate strip by strip.

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Example 4 · Centre of mass of a uniform semicircular wire

A uniform thin wire of mass \(M\) is bent into a semicircle of radius \(R\). Find the position of its centre of mass.

Solution. Step 1 — Coordinates. Place the semicircle in the \(xy\)-plane with its diameter along the \(x\)-axis and the curved part in the region \(y \gt 0\). The centre of the full circle sits at the origin. Step 2 — Symmetry. The wire is symmetric about the \(y\)-axis: for every element at angle \(\theta\) there is a mirror-image element at angle \(\pi - \theta\) with the same mass. Therefore \(x_{\text{cm}} = 0\). We need only \(y_{\text{cm}}\). Step 3 — Mass element. Parametrise the wire by the angle \(\theta\) measured from the positive \(x\)-axis, running from \(0\) to \(\pi\). A small arc element subtends angle \(d\theta\) and has length \(ds = R\,d\theta\). The linear mass density is \(\lambda = M/(\pi R)\) (total mass over total length). So: \[ dm = \lambda\,ds = \frac{M}{\pi R}\cdot R\,d\theta = \frac{M}{\pi}\,d\theta \] The \(y\)-coordinate of this element is \(y = R\sin\theta\). Step 4 — Integration. \[ y_{\text{cm}} = \frac{1}{M}\int_0^{\pi} y\;dm = \frac{1}{M}\int_0^{\pi} R\sin\theta\cdot\frac{M}{\pi}\,d\theta = \frac{R}{\pi}\int_0^{\pi}\sin\theta\,d\theta = \frac{R}{\pi}\Big[-\cos\theta\Big]_0^{\pi} = \frac{R}{\pi}(1 + 1) = \frac{2R}{\pi} \] \[ \boxed{y_{\text{cm}} = \frac{2R}{\pi} \approx 0.637\,R} \qquad\text{(semicircular wire)} \] Sanity check. The result lies between \(0\) (the centre of the full circle) and \(R\) (the top of the semicircle), which is geometrically sensible. The COM is in empty space, on the axis of symmetry.

Semicircular Disc (Plate)

Fig 6.2 By integrating \(\vec r_{cm}=\tfrac1M\int \vec r\,dm\) over a uniform semicircular plate, the centre of mass lands on the symmetry axis at \(4R/3\pi \approx 0.424R\) from the centre — lower than a semicircular wire’s \(2R/\pi\), because the plate carries more of its area near the base.

Example 5 · Centre of mass of a uniform semicircular disc

A uniform semicircular disc of mass \(M\) and radius \(R\) lies in the \(xy\)-plane with its diameter along the \(x\)-axis. Find the position of its centre of mass.

Solution. Symmetry. The disc is symmetric about the \(y\)-axis, so \(x_{\text{cm}} = 0\). Strategy. Rather than using a two-dimensional area element, we exploit the result we just derived. Slice the semicircular disc into thin semicircular rings of radius \(r\) and width \(dr\), where \(r\) ranges from \(0\) to \(R\). Each such ring is a semicircular wire whose COM is at \(y = 2r/\pi\) (from ). The mass of each thin ring: the surface mass density is \(\sigma = M/(\tfrac{1}{2}\pi R^2) = 2M/(\pi R^2)\). The ring at radius \(r\) has length \(\pi r\) and width \(dr\), so its area is \(\pi r\,dr\) and its mass is: \[ dm = \sigma\cdot\pi r\,dr = \frac{2M}{\pi R^2}\cdot\pi r\,dr = \frac{2M}{R^2}\,r\,dr \] Integration. \[ y_{\text{cm}} = \frac{1}{M}\int_0^R \frac{2r}{\pi}\cdot dm = \frac{1}{M}\int_0^R \frac{2r}{\pi}\cdot\frac{2M}{R^2}\,r\,dr = \frac{4}{\pi R^2}\int_0^R r^2\,dr = \frac{4}{\pi R^2}\cdot\frac{R^3}{3} = \frac{4R}{3\pi} \] \[ \boxed{y_{\text{cm}} = \frac{4R}{3\pi} \approx 0.424\,R} \qquad\text{(semicircular disc)} \] Sanity check. The disc has more mass concentrated near \(r = 0\) (where \(y\) is small) than a wire at the rim. So the disc's COM should be lower than the wire's COM. Indeed \(4R/(3\pi) \lt 2R/\pi\), confirming the result.

Core intuition: Building on Previous Results

Notice how we found the semicircular disc by slicing it into semicircular wires whose COM was already known. This “use a known result as a building block” strategy is immensely powerful. You will see it again with hemispheres and cones: always ask whether the body can be decomposed into elements whose COM you have already computed.

Hemispherical Shell and Solid Hemisphere

The same building-block strategy extends to three dimensions.

Thin Hemispherical Shell

Consider a uniform thin hemispherical shell of mass \(M\) and radius \(R\), with its flat face in the \(xz\)-plane and the dome extending in the \(+y\) direction. By symmetry, \(x_{\text{cm}} = z_{\text{cm}} = 0\).

Slice the shell into thin rings by cutting at angle \(\theta\) from the apex (measured from the \(y\)-axis). A ring at angle \(\theta\) has radius \(R\sin\theta\), circumference \(2\pi R\sin\theta\), width \(R\,d\theta\), and \(y\)-coordinate \(R\cos\theta\). Its area is \(2\pi R^2\sin\theta\,d\theta\) and its mass is: \[ dm = \sigma\cdot 2\pi R^2\sin\theta\,d\theta \] where \(\sigma = M/(2\pi R^2)\) is the surface mass density.

\[ y_{\text{cm}} = \frac{1}{M}\int_0^{\pi/2} R\cos\theta\cdot \frac{M}{2\pi R^2}\cdot 2\pi R^2\sin\theta\,d\theta = R\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta = R\cdot\frac{1}{2} \]

\[ \boxed{y_{\text{cm}} = \frac{R}{2}} \qquad\text{(hemispherical shell)} \]

Solid Hemisphere

Now fill in the hemisphere. Slice it into thin hemispherical shells of radius \(r\) and thickness \(dr\), where \(r\) runs from \(0\) to \(R\). Each shell has its COM at \(y = r/2\) (from ) and mass \(dm = \rho\cdot 2\pi r^2\,dr\), where \(\rho = M/(\tfrac{2}{3}\pi R^3) = 3M/(2\pi R^3)\).

\[ y_{\text{cm}} = \frac{1}{M}\int_0^R \frac{r}{2}\cdot\rho\cdot 2\pi r^2\,dr = \frac{1}{M}\cdot\frac{3M}{2\pi R^3}\cdot\pi\int_0^R r^3\,dr = \frac{3}{2R^3}\cdot\frac{R^4}{4} = \frac{3R}{8} \]

\[ \boxed{y_{\text{cm}} = \frac{3R}{8}} \qquad\text{(solid hemisphere)} \]

Pattern. As the body becomes more “filled in,” the COM moves closer to the flat face: \(2R/\pi\) (wire) \(\to\) \(R/2\) (shell) \(\to\) \(3R/8\) (solid). More mass near the base pulls the COM downward.

Non-Uniform Rod

Example 6 · Rod with linearly varying density

A thin rod of length \(L\) has linear mass density \(\lambda(x) = \lambda_0 x\), where \(x\) is measured from one end. Find the position of its centre of mass.

Solution. Setup. Place the rod along the \(x\)-axis from \(x = 0\) to \(x = L\). Total mass. \[ M = \int_0^L dm = \int_0^L \lambda_0\,x\,dx = \lambda_0\cdot\frac{L^2}{2} \] COM. \[ x_{\text{cm}} = \frac{1}{M}\int_0^L x\,dm = \frac{1}{M}\int_0^L x\cdot\lambda_0 x\,dx = \frac{\lambda_0}{M}\int_0^L x^2\,dx = \frac{\lambda_0}{M}\cdot\frac{L^3}{3} \] Substituting \(M = \lambda_0 L^2/2\): \[ x_{\text{cm}} = \frac{\lambda_0 L^3/3}{\lambda_0 L^2/2} = \frac{2L}{3} \] Sanity check. The density increases with \(x\), so more mass is concentrated toward \(x = L\) than toward \(x = 0\). The COM should therefore lie beyond the midpoint \(L/2\), closer to \(L\). Indeed \(2L/3 \gt L/2\).

The Composite Method in Action

Example 7 · Disc with an off-centre circular hole

A uniform disc of mass \(M\) and radius \(R\) has a circular hole of radius \(R/2\) cut from it. The centre of the hole is at a distance \(R/2\) from the centre of the original disc. Find the centre of mass of the remaining piece.

Solution. Setup. Place the origin at the centre of the full disc. Let the centre of the hole be at \(x = R/2\) on the \(x\)-axis. By symmetry about the \(x\)-axis, \(y_{\text{cm}} = 0\). Masses. The surface mass density of the original disc is \(\sigma = M/(\pi R^2)\). The removed piece (a disc of radius \(R/2\)) has mass: \[ m_{\text{hole}} = \sigma\cdot\pi\!\left(\frac{R}{2}\right)^{\!2} = \frac{M}{\pi R^2}\cdot\frac{\pi R^2}{4} = \frac{M}{4} \] The remaining piece has mass \(m_{\text{rem}} = M - M/4 = 3M/4\). Superposition with subtraction. The full disc has its COM at the origin: \(x_{\text{full}} = 0\). The removed piece has its COM at \(x_{\text{hole}} = R/2\). Using with a negative mass for the hole: \[ x_{\text{cm}} = \frac{M\cdot 0 - \dfrac{M}{4}\cdot\dfrac{R}{2}} {M - \dfrac{M}{4}} = \frac{-\dfrac{MR}{8}}{\dfrac{3M}{4}} = -\frac{R}{6} \] \[ \boxed{x_{\text{cm}} = -\frac{R}{6}} \] The COM of the remaining piece lies at \(R/6\) from the original centre, on the side opposite to the hole. Sanity check. Removing mass from the right shifts the COM to the left. The magnitude \(R/6\) is less than \(R\)—the COM stays inside the outer boundary of the remaining piece. Both checks pass.

Table of Standard COM Results

These results appear so frequently that they should be committed to memory—but always with the ability to re-derive them in under a minute.

Body (uniform)	COM from geometric centre/base
Uniform rod (length \(L\))	\(L/2\) from either end
Semicircular wire (radius \(R\))	\(2R/\pi\) from centre
Semicircular disc (radius \(R\))	\(4R/(3\pi)\) from centre
Hemispherical shell (radius \(R\))	\(R/2\) from flat face
Solid hemisphere (radius \(R\))	\(3R/8\) from flat face
Solid cone (height \(h\))	\(h/4\) from base
Hollow cone / conical shell (height \(h\))	\(h/3\) from base
Triangular plate (height \(h\))	\(h/3\) from base (at centroid)

Misconception alert: Mixing Up the Hemisphere Results

Students routinely confuse \(R/2\) (shell) with \(3R/8\) (solid) and \(4R/(3\pi)\) (disc) with \(2R/\pi\) (wire). The rule: the more “filled in” the body, the closer its COM is to the base. Wire \(\to\) shell \(\to\) solid: \(2R/\pi \gt R/2 \gt 3R/8\). Similarly, semicircular wire \(\gt \) semicircular disc: \(2R/\pi \gt 4R/(3\pi)\). If you forget a specific result under exam conditions, re-derive it. The integration takes about five lines—far less time than a wrong answer costs.

✓ Checkpoint · COM Calculations

A thin wire of mass \(M\) is bent into a complete circle of radius \(R\). Where is its centre of mass? Now the wire is bent into a quarter circle instead. Without integrating, predict the coordinates of the COM. Then verify by integration.
Full circle: at the geometric centre (by symmetry). Quarter circle (say in the first quadrant, from \(\theta = 0\) to \(\theta = \pi/2\)): by the same method as the semicircular wire, \(x_{\text{cm}} = \dfrac{1}{M}\displaystyle\int_0^{\pi/2} R\cos\theta\cdot\dfrac{M}{\pi R/2}\cdot R\,d\theta = \dfrac{2R}{\pi}\). By symmetry of the quarter arc about the line \(y = x\), \(y_{\text{cm}} = 2R/\pi\) as well. So \(\vec{r}_{\text{cm}} = (2R/\pi,\;2R/\pi)\).
A uniform disc of radius \(R\) has a hole of radius \(R/3\) drilled at its centre. Where is the centre of mass of the remaining annular ring?
At the geometric centre. The hole is concentric, so the remaining shape retains full circular symmetry. The COM stays at the centre—removing a symmetric piece from a symmetric body preserves the symmetry.
Explain in one sentence why the COM of a solid hemisphere (\(3R/8\)) is closer to the flat face than the COM of a hemispherical shell (\(R/2\)).
A solid hemisphere has more mass concentrated near the base (the volume of material at small \(y\) is larger), which pulls the COM closer to the flat face compared to a shell where mass is distributed only on the surface.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Centre of mass of two particles

A \(2\,\mathrm{kg}\) particle sits at the origin and a \(3\,\mathrm{kg}\) particle sits at \(x = 5\,\mathrm{m}\) on the same line. Locate their centre of mass.

Solution. Given: \(m_1 = 2\,\mathrm{kg}\) at \(x_1 = 0\), \(m_2 = 3\,\mathrm{kg}\) at \(x_2 = 5\,\mathrm{m}\).
Find: the centre-of-mass coordinate \(x_{\text{cm}}\).
Setup: The centre of mass is the mass-weighted average position, \(x_{\text{cm}} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\).
Solve: \[ x_{\text{cm}} = \frac{2(0) + 3(5)}{2 + 3} = \frac{15}{5} = 3\,\mathrm{m}. \] Answer: \(\boxed{x_{\text{cm}} = 3\,\mathrm{m}}\)
Check: Lever rule: the heavier \(3\,\mathrm{kg}\) mass should pull the balance point toward itself, so \(x_{\text{cm}}\) must exceed the midpoint \(2.5\,\mathrm{m}\); it does. Distances \(m_1 d_1 = m_2 d_2\) requires \(2(3) = 3(2)\), i.e. \(6 = 6\) ✓.

Example 2 · Centre of mass of three particles in a plane

Masses \(1\,\mathrm{kg}\), \(2\,\mathrm{kg}\), and \(3\,\mathrm{kg}\) are placed at \((0,0)\), \((4,0)\), and \((0,6)\) (in metres). Find the centre of mass.

Solution. Given: \(m_1 = 1\,\mathrm{kg}\) at \((0,0)\), \(m_2 = 2\,\mathrm{kg}\) at \((4,0)\), \(m_3 = 3\,\mathrm{kg}\) at \((0,6)\).
Find: \((x_{\text{cm}}, y_{\text{cm}})\).
Setup: Apply the weighted average separately to each coordinate; total mass \(M = 6\,\mathrm{kg}\).
Solve: \[ x_{\text{cm}} = \frac{1(0) + 2(4) + 3(0)}{6} = \frac{8}{6} = \frac{4}{3}\,\mathrm{m}, \qquad y_{\text{cm}} = \frac{1(0) + 2(0) + 3(6)}{6} = \frac{18}{6} = 3\,\mathrm{m}. \] Answer: \(\boxed{\left(\tfrac{4}{3},\,3\right)\,\mathrm{m} \approx (1.33,\,3.00)\,\mathrm{m}}\)
Check: Two-step grouping: combine \(m_1,m_2\) first — their CoM is at \(\left(\tfrac{8}{3},0\right)\) with mass \(3\,\mathrm{kg}\); combining with \(m_3\) at \((0,6)\) gives \(x = \dfrac{3\cdot\tfrac{8}{3} + 3\cdot 0}{6} = \tfrac{4}{3}\) and \(y = \dfrac{3\cdot 0 + 3\cdot 6}{6} = 3\) — identical ✓. Both coordinates lie inside the triangle's bounding box ✓.

Example 3 · Centre of mass of an L-shaped lamina

A uniform lamina is shaped like an “L”: a horizontal foot occupying \(0 \le x \le 6\), \(0 \le y \le 2\), joined to an upright occupying \(0 \le x \le 2\), \(2 \le y \le 6\) (all in cm). Locate its centre of mass.

Solution. Given: foot — area \(A_1 = 6 \times 2 = 12\,\mathrm{cm^2}\), centroid \((3,1)\); upright — area \(A_2 = 2 \times 4 = 8\,\mathrm{cm^2}\), centroid \((1,4)\).
Find: \((x_{\text{cm}}, y_{\text{cm}})\).
Setup: For a uniform sheet, mass \(\propto\) area, so replace each rectangle by its area acting at its own centroid.
Solve: \[ x_{\text{cm}} = \frac{12(3) + 8(1)}{12 + 8} = \frac{44}{20} = 2.2\,\mathrm{cm}, \qquad y_{\text{cm}} = \frac{12(1) + 8(4)}{20} = \frac{44}{20} = 2.2\,\mathrm{cm}. \] Answer: \(\boxed{(x_{\text{cm}}, y_{\text{cm}}) = (2.2,\,2.2)\,\mathrm{cm}}\)
Check: At \(x = 2.2\,\mathrm{cm}\) the material exists only for \(0 \le y \le 2\) (the foot), yet \(y_{\text{cm}} = 2.2 \gt 2\). The centre of mass therefore lies in the empty notch — a vivid reminder that the CoM need not lie within the body ✓. It is nonetheless biased toward the larger foot, as expected ✓.

Example 4 · Velocity of the centre of mass

A \(2\,\mathrm{kg}\) particle moves with velocity \((3,0)\,\mathrm{m/s}\) and a \(4\,\mathrm{kg}\) particle moves with velocity \((0,6)\,\mathrm{m/s}\). Find the velocity of their centre of mass.

Solution. Given: \(m_1 = 2\,\mathrm{kg}\), \(\vec{v}_1 = (3,0)\,\mathrm{m/s}\); \(m_2 = 4\,\mathrm{kg}\), \(\vec{v}_2 = (0,6)\,\mathrm{m/s}\).
Find: \(\vec{v}_{\text{cm}}\) and its magnitude.
Setup: \(\vec{v}_{\text{cm}} = \dfrac{m_1\vec{v}_1 + m_2\vec{v}_2}{M}\), with \(M = 6\,\mathrm{kg}\) — this is just total momentum divided by total mass.
Solve: \[ \vec{v}_{\text{cm}} = \frac{2(3,0) + 4(0,6)}{6} = \frac{(6,24)}{6} = (1,\,4)\,\mathrm{m/s}, \qquad |\vec{v}_{\text{cm}}| = \sqrt{1^2 + 4^2} = \sqrt{17}\,\mathrm{m/s}. \] Answer: \(\boxed{\vec{v}_{\text{cm}} = (1,4)\,\mathrm{m/s},\quad |\vec{v}_{\text{cm}}| = \sqrt{17} \approx 4.12\,\mathrm{m/s}}\)
Check: Total momentum \(\vec{P} = (6,24)\,\mathrm{kg\,m/s}\); \(M\vec{v}_{\text{cm}} = 6(1,4) = (6,24)\) recovers it exactly ✓.

Example 5 · Recoil of a rifle

A \(4\,\mathrm{kg}\) rifle fires a \(20\,\mathrm{g}\) bullet at a muzzle speed of \(600\,\mathrm{m/s}\). Find the recoil speed of the rifle.

Solution. Given: \(M = 4\,\mathrm{kg}\), \(m = 0.02\,\mathrm{kg}\), \(v = 600\,\mathrm{m/s}\), system initially at rest.
Find: the recoil speed \(V\) of the rifle.
Setup: No external horizontal force during firing, so total momentum stays zero: \(MV = mv\).
Solve: \[ V = \frac{mv}{M} = \frac{0.02 \times 600}{4} = \frac{12}{4} = 3\,\mathrm{m/s}. \] Answer: \(\boxed{V = 3\,\mathrm{m/s}}\) (directed opposite to the bullet)
Check: Bullet momentum \(0.02(600) = 12\,\mathrm{kg\,m/s}\) forward equals rifle momentum \(4(3) = 12\) backward, summing to zero ✓. The rifle's kinetic energy \(\tfrac12(4)(3)^2 = 18\,\mathrm{J}\) is a tiny fraction of the bullet's \(\tfrac12(0.02)(600)^2 = 3600\,\mathrm{J}\), as a heavy, slow recoil should be ✓.

Centre of Mass and System of Particles: The Point That Represents Everything

Perplexing Questions

Why Systems Need a New Idea

Why Internal Forces Cancel

Three Ideas, One Framework

What Lies Ahead

Centre of Mass: Definition and Calculation

The Balancing Point

Definition for a System of Discrete Particles

The Two-Particle System: Building Intuition

Can the Centre of Mass Lie Outside the Body?

Symmetry: The Shortcut You Must Use

Centre of Mass of Continuous Bodies

The Composite-Body (Superposition) Method

Calculating the Centre of Mass: Standard Bodies

The Integration Recipe

Semicircular Wire (Ring)

Semicircular Disc (Plate)

Hemispherical Shell and Solid Hemisphere

Thin Hemispherical Shell

Solid Hemisphere

Non-Uniform Rod

The Composite Method in Action

Table of Standard COM Results

Solved examples

Problem bank

Chapter test

The chapter continues.