HookEvery London double-decker has been tilted to 28 degrees on purpose

Before a double-decker bus design is allowed to carry passengers in the UK, a prototype goes to a test rig, gets ballast loaded onto the upper deck only — the worst realistic case — and is tilted sideways on a platform to \(28^\circ\). If any wheel on the high side lifts, the design fails. The physics being certified is one sentence long: a body topples the instant the vertical line through its centre of mass leaves the base of support. Load the top deck and the centre of mass rises; tilt the platform and the weight's line of action marches towards the edge of the wheelbase. The regulation is nothing more than a guarantee that at every lean a real bus will meet on a cambered road, that line still lands between the wheels.

Everything in this section feeds that single test. You need to find the centre of mass — of a set of point particles, of a composite shape welded together from standard pieces, of a lamina or a solid of revolution where only integration will do — and then you need to use it: to decide whether a tilted body slides or topples first, to find the angle a suspended sign hangs at, and to solve rigid-body equilibrium problems where forces alone are not enough and you must take moments. The centre of mass is the one point where you may pretend all the weight acts; find it correctly and every subsequent equation is two lines.

ModelSystems of particles — weighted averages, nothing more

For particles of mass \(m_1, m_2, \dots\) at positions \(x_1, x_2, \dots\) along a line, the centre of mass sits at \[\bar{x}=\frac{\sum m_i x_i}{\sum m_i}.\] The logic is a moments argument: the total moment of the individual weights about any point must equal the moment of a single particle of total mass \(M=\sum m_i\) placed at \(\bar{x}\). In two dimensions you simply run the same calculation twice, once for \(\bar{x}\) and once for \(\bar{y}\) — the coordinates never interact, which is why a messy-looking 2D system is really two easy 1D problems.

The practical layout AQA rewards is a table: masses in one row, \(x\)-coordinates in the next, \(y\)-coordinates in the next, then the products. Errors in this topic are almost never conceptual — they are sign slips on coordinates, so fix an origin explicitly, state it, and keep every coordinate measured from it.

Worked example

Particles of mass \(2\ \text{kg}\), \(3\ \text{kg}\) and \(5\ \text{kg}\) sit at \((1, 4)\), \((3, 0)\) and \((-1, 2)\) respectively. Total mass: \(M = 10\ \text{kg}\). Taking moments about the \(y\)-axis: \(\bar{x} = \dfrac{2(1) + 3(3) + 5(-1)}{10} = \dfrac{6}{10} = 0.6.\) About the \(x\)-axis: \(\bar{y} = \dfrac{2(4) + 3(0) + 5(2)}{10} = \dfrac{18}{10} = 1.8.\) The centre of mass is at \((0.6, 1.8)\). Sanity check: the answer must sit inside the 'spread' of the particles — a \(\bar{x}\) of \(6\) would mean an arithmetic slip, since every particle has \(x \le 3\).

MechanismComposite bodies — add shapes, subtract holes

A composite body is treated as a system of particles in disguise: replace each standard piece by a particle of the same mass at that piece's own centre of mass, then run the weighted average. For uniform bodies, mass is proportional to area (laminas) or volume (solids), so you can work with areas and volumes directly and never mention density. The standard positions you are expected to know or look up: a rectangle's centre, a triangle's centroid at the intersection of medians (two-thirds of the way along each median from the vertex), a semicircular lamina at \(\frac{4r}{3\pi}\) from the centre, a solid hemisphere at \(\frac{3r}{8}\) from the flat face, and a solid cone at \(\frac{h}{4}\) above the base.

The elegant move is the negative mass trick: a plate with a hole is the full plate plus a piece of negative mass where the hole is. The formula does not care that one 'mass' is negative — the algebra handles it, and it is far faster than decomposing the remaining shape into awkward pieces. Symmetry is the other free gift: any axis of symmetry contains the centre of mass, and spotting one usually halves the calculation.

Worked example

A uniform rectangular plate spans \((0,0)\) to \((6,4)\), and a circular hole of radius \(1\) is cut with centre \((4.5, 2)\). By symmetry \(\bar{y}=2\), so only \(\bar{x}\) needs work. Areas: rectangle \(24\), hole \(\pi\). Treating the hole as negative area: \[\bar{x} = \frac{24(3) - \pi(4.5)}{24 - \pi} = \frac{72 - 4.5\pi}{24 - \pi} \approx \frac{57.86}{20.86} \approx 2.77.\] The centre of mass shifts left of the rectangle's centre \((3,2)\) — away from the missing material, exactly as intuition demands. That one-line direction check is worth writing: examiners credit candidates who confirm the answer moved the right way.

MechanismLaminas and solids of revolution — when integration takes over

When the shape is bounded by a curve, the sums become integrals. For a uniform lamina under \(y = f(x)\) between \(x=a\) and \(x=b\), slice it into vertical strips of width \(dx\): each strip has area \(y\,dx\), its own centre at height \(\frac{y}{2}\). The strip 'masses' assemble into \[\bar{x} = \frac{\int_a^b xy\,dx}{\int_a^b y\,dx}, \qquad \bar{y} = \frac{\frac{1}{2}\int_a^b y^2\,dx}{\int_a^b y\,dx}.\] The \(\frac{1}{2}y^2\) is the one students forget: it is \(y\,dx\) (strip area) times \(\frac{y}{2}\) (strip centre height), not \(y^2\).

For a solid formed by rotating \(y=f(x)\) about the \(x\)-axis, the slices are discs of volume \(\pi y^2\,dx\), each with its centre on the axis — so \(\bar{y}=\bar{z}=0\) by symmetry and only one integral is ever needed: \[\bar{x} = \frac{\int_a^b \pi x y^2\,dx}{\int_a^b \pi y^2\,dx} = \frac{\int_a^b x y^2\,dx}{\int_a^b y^2\,dx}.\] The \(\pi\) cancels; write it once and drop it. These integrals are how the 'known' results are born — deriving the cone's \(\frac{3h}{4}\)-from-the-vertex result is a standard 'show that' question, so the derivation below is worth owning, not just the answer.

Worked example

Find the centre of mass of a uniform solid cone of height \(h\), formed by rotating \(y = \frac{r}{h}x\) for \(0 \le x \le h\) about the \(x\)-axis (vertex at the origin). Numerator: \(\int_0^h x y^2\,dx = \frac{r^2}{h^2}\int_0^h x^3\,dx = \frac{r^2}{h^2}\cdot\frac{h^4}{4} = \frac{r^2 h^2}{4}.\) Denominator: \(\int_0^h y^2\,dx = \frac{r^2}{h^2}\cdot\frac{h^3}{3} = \frac{r^2 h}{3}.\) So \[\bar{x} = \frac{r^2h^2/4}{r^2h/3} = \frac{3h}{4}\] — three-quarters of the way from the vertex, i.e. \(\frac{h}{4}\) above the base. Every \(r\) cancelled: the answer is independent of how wide the cone is, a result you can quote back as a check whenever a cone appears inside a composite body.

CaseSuspended bodies and the tilt test — slide or topple?

Once you own the centre of mass, two classic set-pieces follow. Suspension: a body hanging freely from a point settles with its centre of mass vertically below the suspension point — that is the only orientation in which the weight and the support force are collinear and produce no net moment. To find the angle a lamina hangs at, draw the line from the suspension point to the centre of mass and use basic trigonometry on the coordinates; the 'angle a given edge makes with the vertical' is then one \(\tan^{-1}\) away.

The incline is the bus test in miniature. A block on a rough slope at angle \(\theta\) has two independent failure modes. It slides when the gravity component along the slope beats limiting friction: \(mg\sin\theta > \mu mg\cos\theta\), i.e. \(\tan\theta > \mu\). It topples when the vertical through the centre of mass passes outside the downhill edge of the base: for a uniform block of width \(2a\) and height \(2b\) standing on the slope, that is \(\tan\theta > \frac{a}{b}\). Whichever threshold is crossed at the smaller angle happens first — the comparison of \(\mu\) with \(\frac{a}{b}\) is the answer, and stating it explicitly is the final mark of the question.

Worked example

A uniform crate of width \(0.4\ \text{m}\) and height \(1.0\ \text{m}\) stands on a rough plane which is slowly tilted; \(\mu = 0.5\). Sliding threshold: \(\tan\theta = 0.5\), so \(\theta \approx 26.6^\circ\). Toppling threshold: half-width over half-height gives \(\tan\theta = \frac{0.2}{0.5} = 0.4\), so \(\theta \approx 21.8^\circ\). Since \(21.8^\circ < 26.6^\circ\), the crate topples before it slides. Make it squat instead — width \(1.0\ \text{m}\), height \(0.4\ \text{m}\) — and the thresholds swap: toppling now needs \(\tan\theta = 2.5\) (about \(68^\circ\)) while sliding still triggers at \(26.6^\circ\), so it slides first. Same material, same slope; the geometry of the base decides the failure mode — which is exactly why bus regulations attack the height of the centre of mass.

CaseRigid bodies in equilibrium — resolve twice, take moments once

A rigid body in equilibrium must satisfy two vector conditions: the forces sum to zero, and the moments about any point sum to zero. In 2D that yields three scalar equations — resolve in two directions, take moments about one point — and three equations solve for three unknowns, which is why ladder, beam and hinged-rod problems so often carry exactly three unknown forces. The moment of a force about a point is (force) \(\times\) (perpendicular distance from the point to its line of action); a couple is a pair of equal and opposite forces on different lines, contributing zero resultant force but a fixed turning moment about every point.

The strategic choice is where to take moments: pick the point through which the most unknown forces pass, and they vanish from the equation instantly. In a ladder problem, moments about the foot kill both floor reaction and friction in one move, leaving the wall reaction alone. And this is where the centre of mass earns its keep: the entire weight of a uniform rod acts at its midpoint, of a non-uniform body at the centre of mass you computed earlier — often the previous part of the very same exam question. AQA structures multi-part questions exactly this way: part (a) finds the centre of mass by integration or composition; part (b) hangs the body from a point or stands it on a slope and asks for an angle or a set of forces. Getting part (a) wrong does not doom part (b) — method marks follow through — but carrying the answer forward cleanly is where full marks live.

VocabularyKey terms the mark scheme pays for

Centre of mass

The point at which the whole mass of a body may be taken to act: \(\bar{x} = \frac{\sum m_i x_i}{\sum m_i}\) for particles, with integrals replacing sums for continuous bodies.

Moment of a force

The turning effect of a force about a point: magnitude of the force times the perpendicular distance from the point to the force's line of action, with a sign for sense of rotation.

Couple

Two equal, opposite, non-collinear forces. Resultant force zero, but a net moment that is the same about every point — a pure turning effect.

Composite body

A body built from standard pieces. Replace each piece by a particle of equal mass at that piece's centre of mass, then take the weighted average; holes enter as negative mass.

Lamina

A flat body of negligible thickness. For a uniform lamina under \(y=f(x)\): \(\bar{x} = \frac{\int xy\,dx}{\int y\,dx}\) and \(\bar{y} = \frac{\frac{1}{2}\int y^2\,dx}{\int y\,dx}\).

Solid of revolution

Formed by rotating \(y=f(x)\) about the \(x\)-axis. Its centre of mass lies on the axis, at \(\bar{x} = \frac{\int xy^2\,dx}{\int y^2\,dx}\) — the \(\pi\) from the disc volumes cancels.

Toppling condition

A body on the point of toppling pivots about the downhill edge of its base: it topples when the vertical line through its centre of mass falls outside the base of support.

Limiting friction

The maximum friction available, \(F = \mu N\). On an incline a block slides when \(\tan\theta > \mu\); whether it slides or topples first depends on comparing \(\mu\) with the base-to-height ratio.

TrapsMisconceptions that cost marks

“The centre of mass is always inside the body.”

Actually: It is a weighted average of positions, not a material point. An L-shaped bracket, a ring and a horseshoe all have centres of mass in empty space — which is why a suspended L-shape hangs at a 'strange' angle that is entirely predictable once you compute the actual coordinates.

“For the ȳ of a lamina you integrate \(\frac{1}{2}y\,dx\).”

Actually: The strip of area \(y\,dx\) has its centre at height \(\frac{y}{2}\), so its moment is \(\frac{y}{2}\cdot y\,dx = \frac{1}{2}y^2\,dx\). The numerator of \(\bar{y}\) is \(\frac{1}{2}\int y^2\,dx\) — forgetting to square \(y\) is the single most common integration error in this topic.

“A tilted object falls over as soon as it starts to lean.”

Actually: It remains in equilibrium as long as the vertical through the centre of mass stays inside the base — the bus test tilts double-deckers to \(28^\circ\) precisely because a well-designed one survives it. Toppling begins only when that vertical crosses the pivot edge, and the critical angle satisfies \(\tan\theta = \frac{a}{b}\) for a block of half-width \(a\) and half-height \(b\).

“Steeper slopes always make things slide.”

Actually: Sliding and toppling are separate failure modes with separate thresholds, \(\tan\theta = \mu\) and \(\tan\theta = \frac{a}{b}\). A tall thin crate on grippy ground topples long before it slides; a squat crate on ice slides long before it topples. The exam question is almost always 'which happens first?' — answer it by comparing the two tangents.

ExamWhat examiners want

Fix an origin, state it, and tabulate: masses (or areas, or volumes) in one column, coordinates in the next, products in the last. Most dropped marks in centre-of-mass questions are coordinate sign errors, not method errors, and the table format makes them visible. Use symmetry aggressively — one line saying 'by symmetry \(\bar{y} = 2\)' is a mark, and it halves the arithmetic. When a shape has a hole, use negative mass rather than decomposing the remainder; when a standard result like the cone's \(\frac{h}{4}\) or the hemisphere's \(\frac{3r}{8}\) applies, quote it by name, but be ready to derive it by integration — 'show that' derivations of exactly these results are a recurring AQA question, and the derivation is four lines if you set the integral up from a labelled strip or disc.

In slide-or-topple questions, compute both critical angles and compare them explicitly: 'since \(\tan^{-1}(0.4) < \tan^{-1}(0.5)\), toppling occurs first' is the sentence the final mark is attached to. In equilibrium problems, take moments about the point where the most unknown forces meet, and say so — 'taking moments about the foot of the ladder eliminates \(N\) and \(F\)' is AO3 communication credit. For suspended bodies, the working diagram is the whole answer: mark the suspension point, mark the centre of mass, join them, and read the required angle from the coordinate differences with one \(\tan^{-1}\). Keep answers exact (fractions, surds, multiples of \(\pi\)) until the final line, then round to three significant figures.