Why is "Dividing by the number of parts instead of by the total mass" a common mistake in Centre of mass for different shapes?

Why it happens: Confusing the centre of mass with a plain average of positions, ignoring the weighting. How to avoid it: The denominator is always m_i (the total mass, area, or volume), never the count of parts. Tabulate the masses and sum them explicitly. Source: 9231 Paper 3 Examiner Reports — recurring comment

Why is "On a hole/subtraction, subtracting only in the numerator and not the denominator" a common mistake in Centre of mass for different shapes?

Why it happens: Remembering 'negative mass' for the m x term but using the full body's mass as the total. How to avoid it: Write the hole's whole row with a negative mass; the same -m_hole then carries into both m x and m. Source: 9231 Paper 3 mark schemes

Why is "Quoting $\frac{4r}{3\pi}$ (semicircular lamina) for a solid hemisphere, or $\frac{3r}{8}$ for a half-disc" a common mistake in Centre of mass for different shapes?

Why it happens: The half-shapes look similar; the lamina (2-D) and solid (3-D) results are memorised together and swapped. How to avoid it: Read the word: 'lamina' → 4r/3; 'solid hemisphere' → 3r/8; 'hemispherical shell' → r/2. Source: 9231 Paper 3 Examiner Reports

Why is "Using $\frac{h}{4}$ for a conical shell or $\frac{h}{3}$ for a solid cone" a common mistake in Centre of mass for different shapes?

Why it happens: Both cone results are 'a fraction of the height from the base' and are easily transposed. How to avoid it: Solid cone h/4 from base; hollow conical shell h/3 from base. Underline 'solid' or 'shell' in the question. Source: 9231 Paper 3 Examiner Reports

Why is "Measuring a standard result from the wrong reference (e.g. cone c.o.m. from the apex instead of the base)" a common mistake in Centre of mass for different shapes?

Why it happens: Quoting the MF19 number without checking which end it is measured from, then placing it in the wrong position. How to avoid it: Cone h/4 is from the BASE (3h/4 from apex); hemisphere 3r/8 is from the flat face. State the reference point as you place each part. Source: 9231 Paper 3 — composite-body questions

Why is "Writing a single combined fraction with no table or labelled parts" a common mistake in Centre of mass for different shapes?

Why it happens: Rushing; trying to do the whole composite in one line and pairing a mass with the wrong distance. How to avoid it: Always lay out a (part | mass | x | y) table first. Examiners award method marks for a correct m x/ m structure even if arithmetic slips. Source: 9231 Paper 3 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Centre of mass for different shapes

0% Complete

Detailed Study Notes

Detailed notes on Equilibrium of a Rigid body for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Centre of Mass for Different Shapes — Cambridge International AS & A Level Further Mathematics 9231 Paper 3 (2026-2027 syllabus)

The centre of mass is the single point where a body's whole weight effectively acts. Standard results for triangles, semicircles, hemispheres and cones are GIVEN in MF19 — your job is to quote them and combine them. Master the composite method: build complex shapes by ADDING parts, or by SUBTRACTING a hole as negative mass.

At a glance

Standard c.o.m. results are GIVEN in MF19 — you quote them, you do not prove them.
Triangular lamina: centroid on a median, $\frac{1}{3}$ of the way from a side toward the opposite vertex (= average of the three vertices).
Composite by addition: $\bar{x} = \dfrac{\sum m_i x_i}{\sum m_i}$ — each part's mass sits at its own c.o.m.
Composite by subtraction: a removed hole counts as negative mass — $\bar{x} = \dfrac{M_{\text{full}}x_{\text{full}} - m_{\text{hole}}x_{\text{hole}}}{M_{\text{full}} - m_{\text{hole}}}$ .
Uniform density means mass $\propto$ area (laminae) and mass $\propto$ volume (solids) — so you may use areas or volumes as proxies for mass.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

3.2a — Recall and use the positions of the centres of mass of standard uniform bodies (triangular lamina, semicircular lamina, solid hemisphere, hemispherical shell, solid cone, conical shell) as given in the List of Formulae (MF19).
3.2b — Find the centre of mass of a system of particles using $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$ and $\bar{y} = \frac{\sum m_i y_i}{\sum m_i}$ .
3.2c — Determine the centre of mass of a composite body built by addition of parts, or by subtraction of a removed region (treated as negative mass).

The master formula — a system of particles

Centre of mass is the mass-weighted average position of all the parts.

The single idea behind everything. The centre of mass (c.o.m.) of a collection of particles is their mass-weighted average position. For particles of mass $m_1, m_2, \ldots$ at positions $x_1, x_2, \ldots$ :

$\bar{x} = \frac{m_1 x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots} = \frac{\sum m_i x_i}{\sum m_i}$

and identically for the $y$ -coordinate:

$\bar{y} = \frac{\sum m_i y_i}{\sum m_i}$

Read it as a balance. Each $m_i x_i$ is a "moment of mass" about the $y$ -axis. The c.o.m. is the point at which the total mass $\sum m_i$ would produce the same total moment — exactly the balancing point.

Why it scales up to shapes. A solid body is just infinitely many particles. For the topic 3.2 questions you do not integrate: instead you treat each standard part of a body as one "particle" whose entire mass sits at that part's own centre of mass (read from MF19), then feed those into the formula above.

Uniform density is the shortcut. If the body is uniform, its mass is proportional to its size:

Lamina (flat plate): mass $\propto$ area.
Solid: mass $\propto$ volume.

So you may replace each $m_i$ by the part's area (laminae) or volume (solids) — the common density cancels top and bottom.

The c.o.m. lies closer to the heavier masses — it is the point where the whole system would balance.

Cambridge tip. Set up a small table (part | mass/area/volume | $x$ | $y$ ) before writing any equation. Examiners reward the organised " $\sum m x$ over $\sum m$ " layout, and it stops you pairing the wrong mass with the wrong distance.

$\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$ , $\bar{y} = \frac{\sum m_i y_i}{\sum m_i}$ .
Uniform lamina: use area as the mass; uniform solid: use volume.
Tabulate part, mass, $x$ , $y$ before substituting.

See the full worked example for centre of mass for different shapes →

Standard results from MF19 — quote, don't prove

Six given c.o.m. positions for laminae, solids and shells — memorise where to look.

These results are printed in the List of Formulae (MF19) under "Uniform bodies". You are not asked to derive them in Paper 3 — you quote them and place each part's mass at the stated point. (Distances are measured from the obvious axis of symmetry.)

Triangular lamina. The c.o.m. is the centroid: on each median, $\frac{1}{3}$ of the way from a side toward the opposite vertex (equivalently, $\frac{1}{3}$ of the median's length up from the midpoint of a side). The fastest exam route: $\bar{x} = \frac{x_1 + x_2 + x_3}{3}, \qquad \bar{y} = \frac{y_1 + y_2 + y_3}{3}$ — simply average the three vertex coordinates.

Uniform semicircular lamina, radius $r$ : c.o.m. on the axis of symmetry, distance from the centre of the bounding diameter $\frac{4r}{3\pi} \approx 0.4244\,r.$

Solid hemisphere, radius $r$ : on the axis, distance from the centre of the flat face $\frac{3r}{8} = 0.375\,r.$

Hemispherical shell (curved surface only), radius $r$ : distance from the centre of the rim $\frac{r}{2} = 0.5\,r.$

Solid right circular cone, height $h$ : on the axis, distance from the base $\frac{h}{4} \quad\left(\text{i.e. } \tfrac{3h}{4} \text{ below the apex}\right).$

Conical shell (curved surface only), height $h$ : distance from the base $\frac{h}{3} \quad\left(\text{i.e. } \tfrac{2h}{3} \text{ below the apex}\right).$

The semicircular lamina's c.o.m. G sits 4r/(3π) ≈ 0.424r from O, on the axis of symmetry.

Cambridge tip. Two pairs are easily confused. Lamina vs solid for the half-shapes: semicircular lamina is $\frac{4r}{3\pi}$ , solid hemisphere is $\frac{3r}{8}$ . Solid vs shell for the cone: solid is $\frac{h}{4}$ from base, conical shell is $\frac{h}{3}$ from base. Read the question's word ("solid", "lamina", "shell", "surface") before reaching for a number.

Semicircular lamina $\frac{4r}{3\pi}$ ; solid hemisphere $\frac{3r}{8}$ ; hemispherical shell $\frac{r}{2}$ .
Solid cone $\frac{h}{4}$ from base; conical shell $\frac{h}{3}$ from base.
Triangle centroid = average of the three vertex coordinates.

See the full worked example for centre of mass for different shapes →

Composite bodies by addition

Break the shape into standard parts; each part's mass acts at its own c.o.m.

Most exam shapes are not standard — they are built from two or three standard pieces glued together (a rectangle topped by a triangle, a cylinder capped by a hemisphere, and so on). The composite method handles these in three moves.

The recipe.

Split the body into standard parts whose c.o.m. you know (from MF19 or by symmetry).
For each part record its mass (or area / volume if uniform) and the coordinates of its own c.o.m.
Combine with the master formula: $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}, \qquad \bar{y} = \frac{\sum m_i y_i}{\sum m_i}.$

Worked layout (a lamina). Take a rectangle of width $4$ and height $6$ (area $24$ , c.o.m. at its centre) with a semicircle of radius $2$ added on top of the upper edge (area $2\pi$ , c.o.m. $\frac{4(2)}{3\pi}$ above the join). Choosing the bottom-left corner as origin, the join is at height $6$ , so the semicircle's c.o.m. is at $y = 6 + \frac{8}{3\pi}$ . By symmetry $\bar{x} = 2$ ; the height is found from $\bar{y} = \frac{\sum (\text{area})y}{\sum \text{area}}$ .

Each standard part contributes its (area × c.o.m.) to the weighted average; the combined c.o.m. lies between G₁ and G₂.

Solid version. The same logic with volumes: a cylinder of volume $\pi r^2 H$ joined to a solid hemisphere of volume $\frac{2}{3}\pi r^3$ — place each volume at its own c.o.m. and average.

Cambridge tip. Pick the origin to kill as many distances as you can (a corner, the centre of a flat face, the axis of symmetry). Then list a single table; one clean row per part prevents the classic "wrong distance with wrong mass" error.

Split into standard parts; place each mass at its own c.o.m.
Lamina → use areas; solid → use volumes (uniform density cancels).
Choose the origin to zero out as many coordinates as possible.

See the full worked example for centre of mass for different shapes →

Composite bodies by subtraction — holes as negative mass

A removed region is negative mass: subtract its m·x and its m.

When a piece is cut out of a body (a hole drilled in a plate, a cone scooped from a cylinder), treat the removed region as negative mass. You take the full uncut body and subtract the hole.

The subtraction formula. With the full body of mass $M_{\text{full}}$ at $x_{\text{full}}$ and the removed hole of mass $m_{\text{hole}}$ at $x_{\text{hole}}$ : $\bar{x} = \frac{M_{\text{full}}\,x_{\text{full}} - m_{\text{hole}}\,x_{\text{hole}}}{M_{\text{full}} - m_{\text{hole}}}$ and likewise for $\bar{y}$ . Both the numerator term and the denominator term for the hole carry a minus sign — forgetting the minus in the denominator is the single most common error here.

Why it works. "Full body" $=$ "body with hole" $+$ "the plug that fills the hole". Rearranging, the body-with-hole $=$ full body $-$ plug. So in the weighted average the plug enters with a negative mass.

Worked layout (a lamina with a circular hole). A square plate of side $8$ (area $64$ , c.o.m. at its centre) has a circular hole of radius $1$ (area $\pi$ ) punched out, the hole's centre offset from the plate's centre. Then $\bar{x} = \frac{64\,x_{\text{square}} - \pi\,x_{\text{hole}}}{64 - \pi}.$

The hole enters with NEGATIVE area; the combined c.o.m. shifts away from the hole, toward the solid part.

Sanity check. Because mass is removed near the hole, the c.o.m. must shift away from the hole, toward the remaining material. If your answer drifts toward the hole, you have a sign error.

Cambridge tip. Write the hole's row in the table with an explicit negative area ( $-\pi$ , $-\frac{2}{3}\pi r^3$ , …). Carrying the minus through both columns from the start is far safer than trying to insert it at the end.

Hole = negative mass: subtract in BOTH numerator and denominator.
$\bar{x} = \frac{M_{\text{full}}x_{\text{full}} - m_{\text{hole}}x_{\text{hole}}}{M_{\text{full}} - m_{\text{hole}}}$ .
C.o.m. shifts away from the hole — use this as a check.

See the full worked example for centre of mass for different shapes →

Quick recap

C.o.m. is the mass-weighted average position: $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$ .
Standard results (triangle, semicircle, hemisphere, cone, shells) are GIVEN in MF19 — quote them.
Uniform density: use area for laminae, volume for solids, as a mass proxy.
Addition: each part's mass acts at its own c.o.m.; average them.
Subtraction: a hole is negative mass — subtract its $mx$ and its $m$ .
Check: the c.o.m. always shifts toward heavier parts / away from holes.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$ (and the same for $\bar{y}$ ) — the master formula.
Semicircular lamina $\frac{4r}{3\pi}$ ; solid hemisphere $\frac{3r}{8}$ ; hemispherical shell $\frac{r}{2}$ (from MF19).
Solid cone $\frac{h}{4}$ from base; conical shell $\frac{h}{3}$ from base (from MF19).
Triangle centroid = average of the three vertex coordinates ( $\frac{1}{3}$ up a median from a side).
Subtraction rule: $\bar{x} = \frac{M_{\text{full}}x_{\text{full}} - m_{\text{hole}}x_{\text{hole}}}{M_{\text{full}} - m_{\text{hole}}}$ .

How it’s examined

Topic 3.2 c.o.m. questions appear frequently on Paper 3 (6-10 marks), usually as a composite body — a lamina built from standard pieces, or a solid such as a cylinder-plus-hemisphere or a shape with a region removed. The standard positions are quoted from MF19, so marks hinge entirely on the composite working: setting up a correct $\sum m x$ over $\sum m$ table, getting the sign right on a removed hole, and reading 'lamina' vs 'solid' and 'solid' vs 'shell' correctly. C.o.m. results often feed straight into a follow-on toppling or suspension-angle question, so an early sign slip cascades. Unsupported answers score zero — always show the weighted-average layout.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 3, Further Mechanics); 9231/31, 9231/32, 9231/33 Further Mechanics — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Centre of mass for different shapes

Step-by-step solutions to past-paper-style questions on centre of mass for different shapes, written exactly the way a tutor would explain them at the board.

Question type:

1Centre of mass of three particles
Getting startedDirect calculation• particles, easy
Question
Particles of mass $2$ kg, $3$ kg and $5$ kg are placed at the points $(1,0)$ , $(4,2)$ and $(2,5)$ respectively. Find the coordinates of the centre of mass of the system.
Step-by-step solution
1. Step 1
  Total mass. Add the three masses.
  $\sum m = 2 + 3 + 5 = 10$
2. Step 2
  Mass-weighted sum of $x$ -coordinates, then divide.
  $\bar{x} = \frac{2(1) + 3(4) + 5(2)}{10} = \frac{2 + 12 + 10}{10} = \frac{24}{10} = 2.4$
3. Step 3
  Same for the $y$ -coordinates.
  $\bar{y} = \frac{2(0) + 3(2) + 5(5)}{10} = \frac{0 + 6 + 25}{10} = \frac{31}{10} = 3.1$
Answer
Centre of mass at $(2.4,\ 3.1)$ .
Examiner tip
Always divide by the TOTAL mass, not the number of particles. Pairing each mass with its own coordinate is exactly the layout used for composite shapes later.
2Quoting a standard result from MF19
Getting startedDirect calculation• standard-result, cone, easy
Question
A uniform solid right circular cone has height $12$ cm. State, with reference to MF19, the distance of its centre of mass from the base, and from the apex.
Step-by-step solution
1. Step 1
  Quote the MF19 result. For a uniform solid cone the c.o.m. is on the axis, a distance $\frac{h}{4}$ from the base. No proof is required.
  $d_{\text{base}} = \frac{h}{4} = \frac{12}{4} = 3 \text{ cm}$
2. Step 2
  Distance from the apex is the rest of the height.
  $d_{\text{apex}} = h - \frac{h}{4} = \frac{3h}{4} = 9 \text{ cm}$
Answer
$3$ cm from the base ( $9$ cm from the apex).
Examiner tip
Do NOT confuse the solid cone ( $\frac{h}{4}$ from base) with the conical shell ( $\frac{h}{3}$ from base). Read the word 'solid' before choosing the result.
3Composite lamina by addition (rectangle + triangle)
Building confidenceMulti-step problem• composite, addition, lamina, medium
Question
A uniform lamina consists of a rectangle of width $6$ and height $4$ with an isosceles triangle of height $3$ attached along the top edge (the triangle points upward, symmetric about the rectangle's vertical centre line). Taking the bottom-left corner of the rectangle as origin, find the centre of mass.
Step-by-step solution
1. Step 1
  Use areas as masses (uniform lamina). Rectangle: area $6 \times 4 = 24$ , c.o.m. at its centre $(3, 2)$ .
2. Step 2
  Triangle vertices $(0,4)$ , $(6,4)$ , $(3,7)$ : area $\frac{1}{2}(6)(3) = 9$ ; centroid is the average of the vertices, $\left(3,\ \frac{4+4+7}{3}\right) = (3, 5)$ .
3. Step 3
  By symmetry $\bar{x} = 3$ . Apply the weighted average for $\bar{y}$ .
  $\bar{y} = \frac{24(2) + 9(5)}{24 + 9} = \frac{48 + 45}{33} = \frac{93}{33}$
4. Step 4
  Evaluate to 3 s.f.
  $\bar{y} = 2.818\ldots = 2.82$
Answer
Centre of mass at $(3,\ 2.82)$ (3 s.f.).
Examiner tip
The triangle's centroid is $\frac{1}{3}$ of its height above its base, i.e. $4 + \frac{1}{3}(3) = 5$ — the vertex-average gives the same thing. Symmetry instantly fixes $\bar{x}=3$ ; only $\bar{y}$ needs working.
4Composite lamina by subtraction (hole = negative mass)
Building confidenceMulti-step problem• composite, subtraction, hole, medium
Question
A uniform square lamina of side $6$ has a circular hole of radius $1$ cut from it. Taking the bottom-left corner as origin, the square occupies $0 \le x \le 6$ , $0 \le y \le 6$ and the hole is centred at $(4, 3)$ . Find the centre of mass of the remaining lamina.
Step-by-step solution
1. Step 1
  Areas (= masses). Full square: area $36$ , c.o.m. $(3, 3)$ . Hole: area $\pi(1)^2 = \pi$ , centred $(4, 3)$ — enters as NEGATIVE area.
2. Step 2
  Remaining area for the denominator.
  $A = 36 - \pi$
3. Step 3
  Subtract the hole's $A x$ from the square's.
  $\bar{x} = \frac{36(3) - \pi(4)}{36 - \pi} = \frac{108 - 4\pi}{36 - \pi}$
4. Step 4
  Evaluate. $\bar{y} = 3$ by symmetry (hole on the line $y=3$ ).
  $\bar{x} = \frac{95.434\ldots}{32.858\ldots} = 2.904\ldots = 2.90$
Answer
Centre of mass at $(2.90,\ 3)$ (3 s.f.).
Examiner tip
The c.o.m. moves AWAY from the hole (from $x=3$ to $x=2.90$ , leftward), confirming the sign. The hole's minus sign must appear in BOTH the numerator and the denominator.
5Composite solid by addition (cylinder + hemisphere)
StretchMulti-step problem• composite, addition, solid, hard
Question
A uniform solid consists of a cylinder of radius $3$ and height $8$ with a solid hemisphere of radius $3$ attached to one flat end. Find the distance of the centre of mass from the free (flat) end of the cylinder.
Step-by-step solution
1. Step 1
  Use volumes (uniform solid). Measure $x$ from the free flat end, along the axis. Cylinder: volume $\pi(3)^2(8) = 72\pi$ , c.o.m. at its mid-height $x = 4$ .
2. Step 2
  Hemisphere sits on the far end (its flat face at $x = 8$ ). Volume $\frac{2}{3}\pi(3)^3 = 18\pi$ ; its c.o.m. is $\frac{3r}{8} = \frac{9}{8}$ beyond the flat face, so $x = 8 + \frac{9}{8} = 9.125$ .
3. Step 3
  Total volume, then the weighted average.
  $\bar{x} = \frac{72\pi(4) + 18\pi\left(9.125\right)}{72\pi + 18\pi} = \frac{288\pi + 164.25\pi}{90\pi}$
4. Step 4
  The $\pi$ cancels. Evaluate to 3 s.f.
  $\bar{x} = \frac{452.25}{90} = 5.025 = 5.03$
Answer
$5.03$ from the free flat end of the cylinder (3 s.f.).
Examiner tip
Common density (and the factor $\pi$ ) cancels, so volumes act as masses. The hemisphere's $\frac{3r}{8}$ is measured from its flat face, which is at $x=8$ — add, do not measure from the origin.
6Semicircular lamina with a semicircular hole
StretchMulti-step problem• composite, subtraction, semicircle, hard
Question
A uniform lamina is a semicircle of radius $4$ from which a concentric semicircle of radius $2$ has been removed (the two diameters coincide on the same straight edge). Find the distance of the centre of mass from the centre of the common diameter.
Step-by-step solution
1. Step 1
  Areas (= masses). Big semicircle: $\frac{1}{2}\pi(4)^2 = 8\pi$ , c.o.m. $\frac{4(4)}{3\pi} = \frac{16}{3\pi}$ from the centre $O$ .
2. Step 2
  Small (removed) semicircle: $\frac{1}{2}\pi(2)^2 = 2\pi$ , c.o.m. $\frac{4(2)}{3\pi} = \frac{8}{3\pi}$ from $O$ — enters NEGATIVE.
3. Step 3
  Subtract. Measure $\bar{y}$ perpendicular to the diameter from $O$ .
  $\bar{y} = \frac{8\pi\cdot\frac{16}{3\pi} - 2\pi\cdot\frac{8}{3\pi}}{8\pi - 2\pi} = \frac{\frac{128}{3} - \frac{16}{3}}{6\pi}$
4. Step 4
  Simplify the numerator $\frac{128-16}{3} = \frac{112}{3}$ , then evaluate.
  $\bar{y} = \frac{112/3}{6\pi} = \frac{112}{18\pi} = \frac{56}{9\pi} = 1.980\ldots = 1.98$
Answer
$\dfrac{56}{9\pi} \approx 1.98$ from the centre of the diameter (3 s.f.).
Examiner tip
Note $\frac{4r}{3\pi}$ uses each semicircle's OWN radius. Keeping the $\pi$ symbolic until the end avoids rounding error; here much of it cancels neatly.

Model Answers — Centre of mass for different shapes

High-scoring sample answers for centre of mass for different shapes on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4 marks
Particles of mass $1$ kg, $4$ kg and $5$ kg lie at $(0,3)$ , $(2,1)$ and $(6,0)$ respectively. Find the centre of mass of the system.
Model answer
Total mass: $\sum m = 1 + 4 + 5 = 10 \text{ kg}$

Weighted average of $x$ : $\bar{x} = \frac{1(0) + 4(2) + 5(6)}{10} = \frac{0 + 8 + 30}{10} = \frac{38}{10} = 3.8$

Weighted average of $y$ : $\bar{y} = \frac{1(3) + 4(1) + 5(0)}{10} = \frac{3 + 4 + 0}{10} = \frac{7}{10} = 0.7$

Centre of mass at $(3.8,\ 0.7)$ .
Why this scores
M1 for a correct $\sum mx / \sum m$ structure, A1 for $\bar{x}=3.8$ , A1 for $\bar{y}=0.7$ , B1 for the correct total mass $10$ . Dividing by $3$ (the number of particles) instead of $10$ is the standard error.
Question 2
3 marks
State the position of the centre of mass of (a) a uniform semicircular lamina of radius $r$ , (b) a uniform solid hemisphere of radius $r$ , and (c) a uniform hemispherical shell of radius $r$ , each measured from the centre of the flat face.
Model answer
These are standard results given in MF19 (no proof required).

(a) Semicircular lamina: $\frac{4r}{3\pi} \approx 0.424\,r$

(b) Solid hemisphere: $\frac{3r}{8} = 0.375\,r$

(c) Hemispherical shell (curved surface only): $\frac{r}{2} = 0.5\,r$

Each lies on the axis of symmetry, the stated distance from the centre of the flat face/diameter.
Why this scores
B1 for each correct result. The frequent confusion is quoting $\frac{4r}{3\pi}$ (a 2-D lamina) for the solid hemisphere, or vice versa. The shell value $\frac{r}{2}$ is the cleanest of the three — a useful anchor.
Question 3
6 marks
A uniform lamina has the shape of a letter 'L', formed from a $6 \times 6$ square with a $4 \times 4$ square removed from the top-right corner. With the bottom-left corner as origin, the removed square occupies $2 \le x \le 6$ , $2 \le y \le 6$ . Find the centre of mass of the L-shaped lamina.
Model answer
Treat as a full $6\times 6$ square minus the removed $4\times 4$ square (negative mass). Use areas as masses.

Part Area $x$ $y$
Full square $36$ $3$ $3$
Removed square $-16$ $4$ $4$

Remaining area: $A = 36 - 16 = 20$

$\bar{x}$ : $\bar{x} = \frac{36(3) - 16(4)}{20} = \frac{108 - 64}{20} = \frac{44}{20} = 2.2$

$\bar{y}$ : by symmetry of the shape in the line $y = x$ , $\bar{y} = \bar{x} = 2.2$ .

Centre of mass at $(2.2,\ 2.2)$ .
Why this scores
M1 for the subtraction layout (negative area), M1 A1 for $\bar{x}=2.2$ with the minus in numerator and denominator, B1 for the removed square's c.o.m. $(4,4)$ , M1 A1 for $\bar{y}$ (here by symmetry). Forgetting the $-16$ in the denominator gives a wrong $\frac{44}{36}$ .
Question 4
6 marks
A uniform triangular lamina has vertices $A(0,0)$ , $B(4,0)$ and $C(4,6)$ . A uniform square lamina of the same material occupies $4 \le x \le 7$ , $0 \le y \le 3$ , attached to the triangle along the line $x=4$ . Find the centre of mass of the combined lamina.
Model answer
Use areas as masses (same material throughout).

Triangle $A(0,0), B(4,0), C(4,6)$ : area $\frac{1}{2}(4)(6) = 12$ ; centroid is the vertex-average $\left(\frac{0+4+4}{3},\ \frac{0+0+6}{3}\right) = \left(\frac{8}{3},\ 2\right).$

Square $4\le x\le 7,\ 0\le y\le 3$ : area $9$ , centre $(5.5,\ 1.5)$ .

Part Area $x$ $y$
Triangle $12$ $\frac{8}{3}$ $2$
Square $9$ $5.5$ $1.5$

Total area $= 21$ .

$\bar{x} = \frac{12\left(\frac{8}{3}\right) + 9(5.5)}{21} = \frac{32 + 49.5}{21} = \frac{81.5}{21} = 3.88\,(3\text{ s.f.})$

$\bar{y} = \frac{12(2) + 9(1.5)}{21} = \frac{24 + 13.5}{21} = \frac{37.5}{21} = 1.79\,(3\text{ s.f.})$

Centre of mass at $(3.88,\ 1.79)$ .
Why this scores
B1 for the triangle centroid $\left(\frac{8}{3},2\right)$ (vertex-average), B1 for the square's centre, M1 for each weighted average, A1 A1 for the two coordinates. The triangle is right-angled — students who put its centroid at the midpoint of the hypotenuse lose the B1.
Question 5
7 marks
A uniform solid is formed by joining a solid cone of base radius $3$ and height $4$ to a solid hemisphere of radius $3$ , base-to-base, so the flat faces coincide. Find the distance of the centre of mass of the solid from the common circular face.
Model answer
Use volumes as masses. Measure $x$ along the axis from the common flat face, taking the cone on the positive side and the hemisphere on the negative side.

Cone: volume $\frac{1}{3}\pi(3)^2(4) = 12\pi$ . Its c.o.m. is $\frac{h}{4} = \frac{4}{4} = 1$ from the base, i.e. at $x = +1$ .

Hemisphere: volume $\frac{2}{3}\pi(3)^3 = 18\pi$ . Its c.o.m. is $\frac{3r}{8} = \frac{9}{8} = 1.125$ from the flat face, on the opposite side, i.e. at $x = -1.125$ .

Part Volume $x$
Cone $12\pi$ $+1$
Hemisphere $18\pi$ $-1.125$

Total volume $= 30\pi$ .

$\bar{x} = \frac{12\pi(1) + 18\pi(-1.125)}{30\pi} = \frac{12 - 20.25}{30} = \frac{-8.25}{30} = -0.275$

The centre of mass is $0.275$ on the hemisphere side of the common face.
Why this scores
M1 for both volumes, B1 B1 for the two c.o.m. positions ( $\frac{h}{4}$ and $\frac{3r}{8}$ ), M1 for the signed weighted average, A1 for $-0.275$ , A1 for interpreting the sign (which side). Putting both parts on the same side of the face is the key error; a clear sign convention prevents it.
Question 6
7 marks
A uniform circular disc of radius $5$ has its centre at the origin. A circular hole of radius $2$ , centred at the point $(3, 0)$ , is cut from it. Find the centre of mass of the remaining lamina.
Model answer
Treat as the full disc minus the hole (negative area). Use areas as masses.

Part Area $x$
Full disc $\pi(5)^2 = 25\pi$ $0$
Hole $-\pi(2)^2 = -4\pi$ $3$

Remaining area: $A = 25\pi - 4\pi = 21\pi$

$\bar{x}$ : $\bar{x} = \frac{25\pi(0) - 4\pi(3)}{21\pi} = \frac{-12\pi}{21\pi} = -\frac{12}{21} = -\frac{4}{7} \approx -0.571$

$\bar{y} = 0$ by symmetry (both centres lie on the $x$ -axis).

Centre of mass at $\left(-\frac{4}{7},\ 0\right) \approx (-0.571,\ 0)$ .
Why this scores
M1 for the negative-area subtraction, B1 for the hole's area $4\pi$ at $x=3$ , M1 A1 for $\bar{x} = -\frac{4}{7}$ , B1 for $\bar{y}=0$ by symmetry, A1 for the 3 s.f. value. The result is NEGATIVE — the c.o.m. shifts left, away from the hole at $x=+3$ ; a positive answer signals a dropped sign.

Part	Area	$x$	$y$
Full square	$36$	$3$	$3$
Removed square	$-16$	$4$	$4$

Part	Area	$x$	$y$
Triangle	$12$	$\frac{8}{3}$	$2$
Square	$9$	$5.5$	$1.5$

Part	Volume	$x$
Cone	$12\pi$	$+1$
Hemisphere	$18\pi$	$-1.125$

Part	Area	$x$
Full disc	$\pi(5)^2 = 25\pi$	$0$
Hole	$-\pi(2)^2 = -4\pi$	$3$

Key Formulae — Centre of mass for different shapes

The formulae you need to memorise for centre of mass for different shapes on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Centre of mass of a system of particles
$\bar{x} = \frac{\sum m_i x_i}{\sum m_i}, \qquad \bar{y} = \frac{\sum m_i y_i}{\sum m_i}$
$m_i$
mass of the $i$ -th particle (or part)
$x_i, y_i$
coordinates of that particle's own centre of mass
$\bar{x}, \bar{y}$
coordinates of the combined centre of mass
When to use
The master formula. For a uniform lamina replace $m_i$ by area; for a uniform solid replace $m_i$ by volume.
Composite by subtraction (hole = negative mass)
$\bar{x} = \frac{M_{\text{full}}\,x_{\text{full}} - m_{\text{hole}}\,x_{\text{hole}}}{M_{\text{full}} - m_{\text{hole}}}$
$M_{\text{full}}$
mass (or area/volume) of the uncut body
$m_{\text{hole}}$
mass (or area/volume) of the removed region
When to use
Any shape with a region cut out. The minus sign appears in BOTH numerator and denominator.
Semicircular lamina (MF19)
$d = \frac{4r}{3\pi}$
$r$
radius of the semicircle
$d$
distance of c.o.m. from the centre of the diameter
When to use
Uniform semicircular lamina (a flat half-disc). $\frac{4r}{3\pi} \approx 0.424r$ .
Solid hemisphere & hemispherical shell (MF19)
$d_{\text{solid}} = \frac{3r}{8}, \qquad d_{\text{shell}} = \frac{r}{2}$
$r$
radius of the hemisphere
$d$
distance of c.o.m. from the centre of the flat face/rim
When to use
Solid hemisphere ( $\frac{3r}{8} = 0.375r$ ) versus hemispherical shell — curved surface only — ( $\frac{r}{2}$ ).
Solid cone & conical shell (MF19)
$d_{\text{solid}} = \frac{h}{4}, \qquad d_{\text{shell}} = \frac{h}{3}$
$h$
height of the cone
$d$
distance of c.o.m. from the base, along the axis
When to use
Solid right circular cone ( $\frac{h}{4}$ from base) versus conical shell — curved surface only — ( $\frac{h}{3}$ from base).
Triangular lamina (centroid)
$\bar{x} = \frac{x_1 + x_2 + x_3}{3}, \qquad \bar{y} = \frac{y_1 + y_2 + y_3}{3}$
$(x_i, y_i)$
the three vertices of the triangle
When to use
Uniform triangular lamina: average the vertices. Equivalently $\frac{1}{3}$ of the way up a median from a side.

Key Definitions and Keywords — Centre of mass for different shapes

Definitions to memorise and the exact keywords mark schemes credit for centre of mass for different shapes answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Centre of mass
Examiner keyword
The single point at which the whole mass of a body may be taken to act; the mass-weighted average of the positions of all its parts. For a uniform body it coincides with the geometric centroid.
Lamina
Examiner keyword
A thin flat plate, treated as two-dimensional. For a uniform lamina, mass is proportional to area, so areas may be used in place of masses.
Uniform (body)
Examiner keyword
Of constant density throughout. This makes mass proportional to area (lamina) or to volume (solid), allowing the common density to cancel in $\frac{\sum m x}{\sum m}$ .
Composite body
Examiner keyword
A body built from two or more standard parts, either joined together (addition) or with a region removed (subtraction). Each part's mass acts at its own centre of mass.
Shell (lamina surface)
Examiner keyword
A body consisting only of a curved surface (no interior), e.g. a hemispherical shell or conical shell. Its c.o.m. differs from the corresponding solid — read 'shell'/'surface' carefully.
Centroid
The geometric centre of a region, equal to the average position of all its points. For a uniform lamina the centroid is the centre of mass; for a triangle it is the average of the three vertices.

Common Mistakes and Misconceptions — Centre of mass for different shapes

The traps other students keep falling into on centre of mass for different shapes questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Dividing by the number of parts instead of by the total mass
9231 Paper 3 Examiner Reports — recurring comment
Why it happens
Confusing the centre of mass with a plain average of positions, ignoring the weighting.
How to avoid it
The denominator is always $\sum m_i$ (the total mass, area, or volume), never the count of parts. Tabulate the masses and sum them explicitly.
✕On a hole/subtraction, subtracting only in the numerator and not the denominator
9231 Paper 3 mark schemes
Why it happens
Remembering 'negative mass' for the $m x$ term but using the full body's mass as the total.
How to avoid it
Write the hole's whole row with a negative mass; the same $-m_{\text{hole}}$ then carries into both $\sum m x$ and $\sum m$ .
✕Quoting $\frac{4r}{3\pi}$ (semicircular lamina) for a solid hemisphere, or $\frac{3r}{8}$ for a half-disc
9231 Paper 3 Examiner Reports
Why it happens
The half-shapes look similar; the lamina (2-D) and solid (3-D) results are memorised together and swapped.
How to avoid it
Read the word: 'lamina' → $\frac{4r}{3\pi}$ ; 'solid hemisphere' → $\frac{3r}{8}$ ; 'hemispherical shell' → $\frac{r}{2}$ .
✕Using $\frac{h}{4}$ for a conical shell or $\frac{h}{3}$ for a solid cone
9231 Paper 3 Examiner Reports
Why it happens
Both cone results are 'a fraction of the height from the base' and are easily transposed.
How to avoid it
Solid cone $\frac{h}{4}$ from base; hollow conical shell $\frac{h}{3}$ from base. Underline 'solid' or 'shell' in the question.
✕Measuring a standard result from the wrong reference (e.g. cone c.o.m. from the apex instead of the base)
9231 Paper 3 — composite-body questions
Why it happens
Quoting the MF19 number without checking which end it is measured from, then placing it in the wrong position.
How to avoid it
Cone $\frac{h}{4}$ is from the BASE ( $\frac{3h}{4}$ from apex); hemisphere $\frac{3r}{8}$ is from the flat face. State the reference point as you place each part.
✕Writing a single combined fraction with no table or labelled parts
9231 Paper 3 Examiner Reports
Why it happens
Rushing; trying to do the whole composite in one line and pairing a mass with the wrong distance.
How to avoid it
Always lay out a (part | mass | $x$ | $y$ ) table first. Examiners award method marks for a correct $\frac{\sum m x}{\sum m}$ structure even if arithmetic slips.

Centre of mass for different shapes

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Centre of mass of three particles

2Quoting a standard result from MF19

3Composite lamina by addition (rectangle + triangle)

4Composite lamina by subtraction (hole = negative mass)

5Composite solid by addition (cylinder + hemisphere)

6Semicircular lamina with a semicircular hole

Centre of mass of a system of particles

Composite by subtraction (hole = negative mass)

Semicircular lamina (MF19)

Solid hemisphere & hemispherical shell (MF19)

Solid cone & conical shell (MF19)

Triangular lamina (centroid)

Centre of mass

Lamina

Uniform (body)

Composite body

Shell (lamina surface)

Centroid

✕Dividing by the number of parts instead of by the total mass

✕On a hole/subtraction, subtracting only in the numerator and not the denominator

✕Quoting 4r3π\frac{4r}{3\pi}3π4r​ (semicircular lamina) for a solid hemisphere, or 3r8\frac{3r}{8}83r​ for a half-disc

✕Using h4\frac{h}{4}4h​ for a conical shell or h3\frac{h}{3}3h​ for a solid cone

✕Measuring a standard result from the wrong reference (e.g. cone c.o.m. from the apex instead of the base)

✕Writing a single combined fraction with no table or labelled parts

✕Quoting $\frac{4r}{3\pi}$ (semicircular lamina) for a solid hemisphere, or $\frac{3r}{8}$ for a half-disc

✕Using $\frac{h}{4}$ for a conical shell or $\frac{h}{3}$ for a solid cone