What is the Centre of Mass and why is it important in Mechanics 2 for Edexcel International A Levels?

The Centre of Mass is a point representing the mean position of the mass in a body or system. In Mechanics 2, understanding the Centre of Mass is crucial because it helps in analyzing the motion of objects and systems. It simplifies complex systems into single points for easier calculation of motion, stability, and equilibrium.

How do you calculate the Centre of Mass for a system of particles in Edexcel International A Levels Mathematics?

To calculate the Centre of Mass for a system of particles, you use the formula: $ \bar{x} = \frac{\sum m_ix_i}{\sum m_i} $ and $ \bar{y} = \frac{\sum m_iy_i}{\sum m_i} $, where $ m_i $ is the mass of each particle and $ (x_i, y_i) $ are their coordinates. This formula helps in determining the average position of the mass in the system.

What is the difference between Centre of Mass and Centre of Gravity in the context of Edexcel International A Levels Mechanics 2?

The Centre of Mass is a point where the mass of a body or system is concentrated, while the Centre of Gravity is the point where the gravitational force can be considered to act. In uniform gravitational fields, these points coincide. However, in non-uniform fields, they may differ. Understanding this distinction is important for solving problems in Mechanics 2.

How does the concept of Centre of Mass apply to rigid bodies in Edexcel International A Levels Mechanics 2?

For rigid bodies, the Centre of Mass is the point where the entire mass of the body can be assumed to be concentrated for the purpose of analyzing translational motion. This simplifies the study of dynamics and equilibrium of rigid bodies, allowing students to apply principles of mechanics more effectively in their calculations.

Can you explain how the Centre of Mass affects the stability of objects in Mechanics 2 for Edexcel International A Levels?

The stability of an object is influenced by the position of its Centre of Mass. An object is more stable if its Centre of Mass is lower and within its base of support. In Mechanics 2, understanding this concept helps in analyzing how objects balance and the conditions under which they topple, which is essential for solving related problems in the curriculum.

Why is "Adding the cut-out's centroid instead of subtracting" a common mistake in Centres of Mass?

Why it happens: Treating composite-with-hole the same as composite-with-extra-piece. How to avoid it: A cut-out contributes NEGATIVE area. Write A_remaining x_rem = A_whole x_whole - A_hole x_hole — the minus sign is the discriminator. Source: WME02/01 examiner reports — flagged annually

Why is "Confusing centroid with incentre or circumcentre" a common mistake in Centres of Mass?

Why it happens: The triangle has multiple 'centres'. How to avoid it: For a UNIFORM triangular LAMINA the COM is the CENTROID — average of vertex coordinates. The incentre, circumcentre, orthocentre are different points (and irrelevant here).

Why is "Using $\tan\theta = \bar y / \bar x$ instead of $\bar x / \bar y$" a common mistake in Centres of Mass?

Why it happens: Confusing which side is opposite vs adjacent in the right-angled triangle of the suspended body. How to avoid it: Draw a diagram: pivot at top, COM at ( x, - y) below if origin is the pivot. Edge AB originally horizontal makes angle with the vertical: = x / y (horizontal offset / vertical drop).

Why is "Stopping after finding one of topple/slide angle without comparing" a common mistake in Centres of Mass?

Why it happens: Reading the question incompletely. How to avoid it: 'Determine whether the body topples or slides first' demands BOTH angles. The smaller one happens first. State the comparison explicitly in the conclusion. Source: WME02/01 examiner reports — annual mistake

Why is "Using areas when the lamina is not uniform" a common mistake in Centres of Mass?

Why it happens: Habit from uniform-lamina problems. How to avoid it: Areas-as-mass works ONLY when 'uniform' is stated. Otherwise, use actual masses. M2 questions almost always say 'uniform', but watch for the rare exception.

Mechanics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Centres of Mass

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Centres of Mass Study Notes — Edexcel IAL Mathematics M2 (WME02/01, 2018 spec — 2026 onwards)

Find centres of mass of (i) discrete-particle systems, (ii) uniform laminae assembled from rectangles, triangles and circles, (iii) composite laminae with cut-outs, then apply to suspension and topple-vs-slide problems on inclined planes.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M2 3.1 — Find the centre of mass of a system of particles distributed in one or two dimensions.
M2 3.2 — Find the centre of mass of uniform laminae of simple shape (rectangle, triangle, circle).
M2 3.3 — Find the centre of mass of composite uniform laminae, including those with regions removed.
M2 3.4 — Use the centre of mass of a body suspended freely from a point.
M2 3.5 — Determine whether a body on a rough inclined plane will topple or slide first.

Discrete particles

Weighted average. Total moment about an axis = (total mass) × (COM coordinate).

1D system. Particles of mass $m_i$ at positions $x_i$ on a line:

$\bar x = \dfrac{m_1 x_1 + m_2 x_2 + \ldots + m_n x_n}{m_1 + m_2 + \ldots + m_n} = \dfrac{\sum m_i x_i}{M}.$

Why. The COM is the point about which the sum of moments is zero. Equivalently, the total moment of the system about the origin equals the moment of a single particle of mass $M$ at $\bar x$ :

$\sum m_i x_i = M \bar x.$

2D system. Particles at $(x_i, y_i)$ :

$\bar x = \dfrac{\sum m_i x_i}{M}, \qquad \bar y = \dfrac{\sum m_i y_i}{M}.$

The coordinates are independent — handle them separately.

Worked example. Four particles in the $xy$ -plane: $2$ kg at $(1, 0)$ , $3$ kg at $(4, 0)$ , $1$ kg at $(2, 3)$ , $4$ kg at $(0, 2)$ .

$M = 10$ .
$\bar x = (2 + 12 + 2 + 0)/10 = 1.6$ .
$\bar y = (0 + 0 + 3 + 8)/10 = 1.1$ .
COM at $(1.6, 1.1)$ .

Weighted average of coordinates, weight = mass.
1D and 2D are the same formula, applied per coordinate.
$\sum m_i x_i = M \bar x$ — moments about axis.

Uniform laminae — simple shapes

Rectangle: centre. Triangle: centroid (average of vertices). Circle: centre.

Uniform lamina. A flat object of constant density per unit area. Mass is proportional to area, so for COM calculations the area replaces the mass.

Rectangle. COM at the geometric centre — the intersection of the diagonals.

For a rectangle with opposite corners $(x_1, y_1)$ and $(x_2, y_2)$ : COM at $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ .

Triangle. COM at the centroid — average of the three vertex coordinates:

$\mathbf{r}_{\text{COM}} = \dfrac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}.$

Equivalently, the centroid lies $\tfrac{2}{3}$ along any median from the vertex. The three medians intersect at the centroid.

Circle / disc / regular polygon. COM at the centre by symmetry.

Semicircle (extension). COM along the axis of symmetry at distance $\dfrac{4r}{3\pi}$ from the diameter. This is in the IAL formula booklet.

Worked example. Triangle with vertices $(0,0)$ , $(6,0)$ , $(2,5)$ :

$\mathbf{r}_{\text{COM}} = \dfrac{(0,0) + (6,0) + (2,5)}{3} = \left(\dfrac{8}{3}, \dfrac{5}{3}\right).$

Uniform ⇒ area replaces mass.
Rectangle: centre.
Triangle: average of vertices (= intersection of medians).
Circle / disc: centre.
Semicircle: $4r/(3\pi)$ from diameter.

Composite laminae and cut-outs

Area-weighted average of sub-piece centroids. Cut-outs have NEGATIVE area.

Strategy. Decompose the lamina into simple shapes (rectangles, triangles, circles). For each sub-piece:

Its area $A_i$ .
Its centroid $(\bar x_i, \bar y_i)$ .

Then the composite COM is the area-weighted average:

$\bar x = \dfrac{\sum A_i \bar x_i}{\sum A_i}, \qquad \bar y = \dfrac{\sum A_i \bar y_i}{\sum A_i}.$

Cut-outs. A removed region contributes negative area in both sums.

Tabular layout. Always lay out the data in a table — examiners reward clarity.

Shape	Area $A_i$	$\bar x_i$	$\bar y_i$	$A_i \bar x_i$	$A_i \bar y_i$
Big rectangle	64	4	4	256	256
Circle (CUT)	$-4\pi$	2	2	$-8\pi$	$-8\pi$
Total	$64 - 4\pi$			$256 - 8\pi$	$256 - 8\pi$

Then $\bar x = \bar y = \dfrac{256 - 8\pi}{64 - 4\pi} = \dfrac{2(32 - \pi)}{16 - \pi} \approx 4.49$ .

Worked example — composite without holes. Rectangle $ABCD$ with $A(0,0)$ , $B(6,0)$ , $C(6,4)$ , $D(0,4)$ joined to triangle $CDE$ with $E(3, 7)$ .

Rectangle: area $24$ , centroid $(3, 2)$ .
Triangle: area $9$ , centroid $(3, 5)$ .
Total area $33$ , total $A \bar x = 72 + 27 = 99$ , total $A \bar y = 48 + 45 = 93$ .
$\bar x = 99/33 = 3$ , $\bar y = 93/33 = 31/11 \approx 2.82$ .

Decompose into rectangles + triangles + circles.
Tabulate areas and centroids.
Cut-out ⇒ negative area.
Final $\bar x = (\sum A_i \bar x_i) / (\sum A_i)$ .

Suspended bodies

Hang it from a pivot. Pivot-to-COM line is vertical. Angle follows.

Principle. When a rigid body is suspended freely from a point $P$ (no other forces beyond gravity and the reaction at $P$ ), in equilibrium the centre of mass lies vertically below $P$ .

Why. Weight acts at the COM. For equilibrium, the resultant moment about $P$ must be zero. The only moment is the moment of weight, which is zero exactly when the weight acts along the line through $P$ — i.e. when COM is directly below $P$ .

Procedure.

Find the COM in the body's own coordinate system.
The line from pivot $P$ to COM is vertical when the body hangs.
Use trigonometry on the right-angled triangle to find the angle a given edge makes with the vertical (or horizontal).

Worked example. Composite lamina with $G = (3, 31/11)$ and pivot $A = (0, 0)$ .

     A (pivot)
     │
     │  vertical
     │
     │
     G (COM)

Edge $AB$ was along the $x$ -axis in the body's own frame. When hung, the line $AG$ is vertical. The angle $AB$ makes with the vertical is the angle of vector $\overrightarrow{AG} = (3, 31/11)$ with $\overrightarrow{AB} = (1, 0)$ rotated:

$\tan\theta = \dfrac{\bar x}{\bar y} = \dfrac{3}{31/11} = \dfrac{33}{31} \Rightarrow \theta \approx 46.8^{\circ}.$

Setting up the triangle. Imagine the body rotated so $\overrightarrow{AG}$ is vertical. The horizontal offset from $A$ to $G$ in the body's coordinates is $\bar x$ ; the vertical offset is $\bar y$ . The body has rotated so this hypothetical 'horizontal' is no longer horizontal — and the angle is the rotation angle.

Pivot to COM line is vertical when suspended.
Find COM in body's frame; pivot at origin.
$\tan\theta = \bar x / \bar y$ for edge originally along $x$ -axis.
Draw the diagram with vertical clearly marked.

Equilibrium on a plane: topple vs slide

Slides at $\tan\theta_{\text{slip}} = \mu$ . Topples at $\tan\theta_{\text{topple}} = b/h$ . Smaller angle wins.

Setup. A rigid body rests on a rough plane that is gradually tilted from horizontal. As the angle of inclination $\theta$ increases, the body will eventually either start to slide (friction limit reached) or topple (vertical through COM moves outside the base).

Sliding condition. The friction force on a body on a slope is at most $\mu N$ where $N = mg\cos\theta$ . Equilibrium along the slope requires friction $= mg\sin\theta$ . The body is on the point of slipping when these are equal:

$mg\sin\theta = \mu mg\cos\theta \Rightarrow \tan\theta_{\text{slip}} = \mu.$

Toppling condition. The body topples when the vertical line through the COM falls outside the base — specifically, beyond the lower edge of the base in contact with the plane.

Setup: COM is at height $h$ above the base centre and the base extends a distance $b$ from the centre to the lower edge.

Imagine the body still in its 'upright' orientation but the plane tilted by $\theta$ . In the body's frame, gravity points 'down-and-to-the-side', and the line from the COM along gravity's direction meets the base. The body topples precisely when this line passes through the lower edge:

$\tan\theta_{\text{topple}} = \dfrac{b}{h}.$

Worked example. Solid cylinder, radius $4$ cm, height $20$ cm, $\mu = 0.5$ .

COM at height $h = 10$ cm (halfway up the axis).
Base half-width $b = 4$ cm (the radius).
$\tan\theta_{\text{topple}} = 4/10 = 0.4 \Rightarrow \theta_{\text{topple}} \approx 21.8^{\circ}$ .
$\tan\theta_{\text{slip}} = 0.5 \Rightarrow \theta_{\text{slip}} \approx 26.6^{\circ}$ .

Since $21.8^{\circ} < 26.6^{\circ}$ : topples first.

Critical comparison. Whichever angle is smaller occurs first as $\theta$ increases. Compare the two and state the winner explicitly.

Slide at $\tan\theta_{\text{slip}} = \mu$ .
Topple at $\tan\theta_{\text{topple}} = b/h$ (base half-width / COM height).
Whichever angle is SMALLER happens FIRST.
Draw a force diagram: weight at COM, normal at base, friction along slope.

How it’s examined

Centres of mass appears on every WME02/01 M2 paper as one of the longer questions, usually 10-14 marks (often Q5 or Q6). Typical structure: (a) find the COM of a composite lamina (sometimes with a cut-out), (b) the lamina is suspended from a point — find the angle a stated edge makes with the vertical, (c) optional follow-up about a force needed to keep an edge horizontal or about toppling. Pure topple/slide questions also appear as separate items. Examiners reward clear tabulation of areas and centroids, explicit labelling of the cut-out as negative, and a labelled diagram of the suspended body. A* candidates should target full marks: the algebra is light but the bookkeeping is unforgiving — one slip can cascade through the whole question.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME02/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Centres of Mass

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

1Centre of mass of four particles in 2D (5 marks, M2)
Core• Adapted from WME02/01 January 2024 Q2• centre of mass, discrete particles, 2D
Question
Four particles are placed in the $xy$ -plane at the points: $2$ kg at $(1, 0)$ , $3$ kg at $(4, 0)$ , $1$ kg at $(2, 3)$ and $4$ kg at $(0, 2)$ . Find the position vector of the centre of mass.
Step-by-step solution
1. Step 1
  Total mass: $M = 2 + 3 + 1 + 4 = 10$ kg.
2. Step 2
  $\bar x = \dfrac{\sum m_i x_i}{M}$ :
  $\bar x = \dfrac{2(1) + 3(4) + 1(2) + 4(0)}{10} = \dfrac{2 + 12 + 2 + 0}{10} = \dfrac{16}{10} = 1.6$
3. Step 3
  $\bar y = \dfrac{\sum m_i y_i}{M}$ :
  $\bar y = \dfrac{2(0) + 3(0) + 1(3) + 4(2)}{10} = \dfrac{0 + 0 + 3 + 8}{10} = \dfrac{11}{10} = 1.1$
4. Step 4
  Position vector of the centre of mass: $\mathbf{r}_{\text{COM}} = 1.6\mathbf{i} + 1.1\mathbf{j}$ .
Answer
Centre of mass at $(1.6,\, 1.1)$ , i.e. $1.6\mathbf{i} + 1.1\mathbf{j}$ .
Examiner tip
M1 for sum of moments about $y$ -axis $= M\bar x$ . A1 for $\bar x$ . M1 for sum of moments about $x$ -axis $= M\bar y$ . A1 for $\bar y$ . B1 for a clearly-stated position vector or coordinate pair. Examiners accept either form.
2Composite lamina with a circular hole (8 marks, M2)
Extended• Adapted from WME02/01 June 2024 Q6• centre of mass, lamina, composite, cut-out
Question
A uniform square lamina $ABCD$ has side $8$ cm. A circular hole of radius $2$ cm is cut out centred at the point $(2, 2)$ taking $A$ as the origin, with $AB$ along the positive $x$ -axis. Find the centre of mass of the remaining lamina.
Step-by-step solution
1. Step 1
  Treat as 'big square' minus 'small circle'. Areas (proxy for mass since lamina is uniform):
  $A_{\text{sq}} = 64, \quad A_{\text{cir}} = 4\pi$
2. Step 2
  Centroids by symmetry: square at $(4, 4)$ , circle at $(2, 2)$ .
3. Step 3
  Use $A_{\text{remaining}} \cdot \bar x_{\text{rem}} = A_{\text{sq}} \bar x_{\text{sq}} - A_{\text{cir}} \bar x_{\text{cir}}$ :
  $(64 - 4\pi)\bar x = 64 \cdot 4 - 4\pi \cdot 2 = 256 - 8\pi$
4. Step 4
  Solve for $\bar x$ :
  $\bar x = \dfrac{256 - 8\pi}{64 - 4\pi} = \dfrac{8(32 - \pi)}{4(16 - \pi)} = \dfrac{2(32 - \pi)}{16 - \pi}$
5. Step 5
  By symmetry $\bar y = \bar x$ (both $A_{\text{sq}}$ and $A_{\text{cir}}$ have equal $x$ and $y$ centroid coords).
  $\bar x = \bar y \approx \dfrac{2(32 - 3.1416)}{16 - 3.1416} \approx \dfrac{57.72}{12.86} \approx 4.49\,\text{cm}$
Answer
Centre of mass at $(4.49,\, 4.49)$ cm, approximately.
Examiner tip
M1 for setting up area-weighted sum. A1 for both areas. M1 for the subtraction principle (cut-out has 'negative mass'). A1 for the linear equation in $\bar x$ . M1 for solving. A1 for $\bar x \approx 4.49$ . A1 for $\bar y$ via symmetry. The cut-out subtraction is the marker discriminator — students who add instead of subtract lose 3 marks.
3Centre of mass of a triangular lamina (4 marks, M2)
Core• Adapted from WME02/01 January 2023 Q4(a)• centre of mass, triangle, lamina
Question
A uniform triangular lamina has vertices at $A(0, 0)$ , $B(6, 0)$ and $C(2, 5)$ . Find its centre of mass.
Step-by-step solution
1. Step 1
  Centre of mass of a uniform triangular lamina is at the centroid: the average of its vertex coordinates.
2. Step 2
  Compute:
  $\bar x = \dfrac{0 + 6 + 2}{3} = \dfrac{8}{3}, \qquad \bar y = \dfrac{0 + 0 + 5}{3} = \dfrac{5}{3}$
3. Step 3
  So $\mathbf{r}_{\text{COM}} = \dfrac{8}{3}\mathbf{i} + \dfrac{5}{3}\mathbf{j}$ , approximately $(2.67, 1.67)$ .
Answer
Centre of mass at $\left(\dfrac{8}{3}, \dfrac{5}{3}\right) \approx (2.67, 1.67)$ .
Examiner tip
B1 for stating the centroid result (or 'intersection of medians'). M1 for arithmetic on both coordinates. A1 each for $\bar x$ and $\bar y$ . The centroid result $((x_1 + x_2 + x_3)/3,\,(y_1 + y_2 + y_3)/3)$ is given in the IAL formula booklet under 'Mechanics — Centres of mass'.
4Lamina suspended from a corner — find angle (8 marks, M2)
Extended• Adapted from WME02/01 June 2023 Q6• centre of mass, suspended, angle
Question
A uniform lamina is in the shape of a rectangle $ABCD$ with $AB = 6$ cm and $BC = 4$ cm, joined to an isosceles triangle $CDE$ where $E$ is the midpoint of the side opposite $CD$ in the triangle (the triangle is attached along $CD$ , with $E$ at perpendicular distance $3$ cm from $CD$ ). Take $A$ as origin, $AB$ along positive $x$ -axis, $AD$ along positive $y$ -axis. (a) Find the centre of mass of the composite lamina. (b) The lamina is suspended freely from $A$ . Find the angle that $AB$ makes with the vertical.
Step-by-step solution
1. Step 1
  Coordinates: $A(0,0)$ , $B(6,0)$ , $C(6,4)$ , $D(0,4)$ . Triangle $CDE$ is attached along $CD$ , with $E$ above the rectangle at $(3, 7)$ .
2. Step 2
  Centroid of rectangle: $(3, 2)$ , area $24$ . Centroid of triangle $CDE$ with vertices $(6,4)$ , $(0,4)$ and $(3,7)$ : $((6+0+3)/3,\,(4+4+7)/3) = (3, 5)$ . Area = $\tfrac{1}{2} \cdot 6 \cdot 3 = 9$ .
3. Step 3
  Composite (a) — total area $24 + 9 = 33$ :
  $\bar x = \dfrac{24(3) + 9(3)}{33} = \dfrac{99}{33} = 3$
4. Step 4
  And:
  $\bar y = \dfrac{24(2) + 9(5)}{33} = \dfrac{48 + 45}{33} = \dfrac{93}{33} = \dfrac{31}{11} \approx 2.818$
5. Step 5
  (b) When suspended from $A$ , the line $AG$ (where $G$ is the COM) is vertical. The angle $AB$ makes with the vertical equals the angle from $\mathbf{AG}$ to $\mathbf{AB}$ :
  $\tan\theta = \dfrac{\bar x}{\bar y} = \dfrac{3}{31/11} = \dfrac{33}{31}$
6. Step 6
  So $\theta = \arctan(33/31) \approx 46.8^{\circ}$ .
Answer
(a) $G = (3,\, 31/11) \approx (3, 2.82)$ . (b) Angle of $AB$ from vertical $\approx 46.8^{\circ}$ .
Examiner tip
(a) M1 for splitting into rectangle + triangle. A1 each for both centroids. M1 for area-weighted average. A1 for $\bar x = 3$ (by symmetry). A1 for $\bar y$ . (b) M1 for recognising COM is vertically below the pivot. M1 for the $\tan$ -ratio relation. A1 for the angle. Diagram showing the vertical through $A$ and the COM is essential.
5Topple-or-slide on an inclined plane (8 marks, M2)
Extended• Adapted from WME02/01 January 2024 Q5• centre of mass, topple, inclined plane
Question
A uniform solid right-circular cylinder of height $20$ cm and radius $4$ cm stands on a rough plane that is inclined at angle $\theta$ to the horizontal. The coefficient of friction between the cylinder and the plane is $\mu = 0.5$ . The cylinder cannot slip until $\theta$ exceeds $\arctan\mu$ . Find (a) the value of $\theta$ at which the cylinder is on the point of toppling, (b) determine whether the cylinder topples or slides first as $\theta$ is gradually increased.
Step-by-step solution
1. Step 1
  (a) The cylinder is on the point of toppling when the vertical through the centre of mass falls along the lower edge of the base. The COM is at the geometric centre — halfway up the axis, on the central line. So COM is at height $10$ cm above the base centre.
2. Step 2
  On the inclined plane, the COM lies vertically above the lower edge of the base when $\tan\theta = \dfrac{\text{base radius}}{\text{COM height}}$ :
  $\tan\theta_{\text{topple}} = \dfrac{4}{10} = 0.4 \Rightarrow \theta_{\text{topple}} \approx 21.8^{\circ}$
3. Step 3
  (b) The slip angle is $\theta_{\text{slip}} = \arctan\mu = \arctan 0.5 \approx 26.57^{\circ}$ .
4. Step 4
  Compare: $\theta_{\text{topple}} \approx 21.8^{\circ} < \theta_{\text{slip}} \approx 26.57^{\circ}$ , so toppling happens first.
Answer
(a) $\theta \approx 21.8^{\circ}$ . (b) Cylinder topples before it slides, since $21.8^{\circ} < 26.57^{\circ}$ .
Examiner tip
(a) M1 for identifying the geometric condition (vertical through COM meets the lower edge). M1 for $\tan\theta$ ratio. A1 for $21.8^{\circ}$ . (b) M1 for $\tan\theta_{\text{slip}} = \mu$ . A1 for $26.6^{\circ}$ . M1 for the comparison. A1 for the correct conclusion. Force diagram showing the weight acting at COM, normal reaction at the base, and friction along the plane is essential.

Key Formulae — Centres of Mass

The formulae you need to memorise for centres of mass on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Centre of mass — discrete particles (1D)
$\bar x = \dfrac{\sum_{i} m_i x_i}{\sum_{i} m_i} = \dfrac{\sum m_i x_i}{M}$
$m_i$
mass of $i$ -th particle
$x_i$
position of $i$ -th particle (1D)
$M$
total mass
When to use
System of point masses on a single line. The formula is in the IAL formula booklet under 'Mechanics — Centres of mass'.
Example
Three masses $1, 2, 3$ kg at $x = 0, 1, 2$ : $\bar x = \dfrac{0 + 2 + 6}{6} = \dfrac{8}{6} = \dfrac{4}{3}$ .
Centre of mass — discrete particles (2D)
$\bar x = \dfrac{\sum m_i x_i}{M}, \quad \bar y = \dfrac{\sum m_i y_i}{M}$
$(x_i, y_i)$
position of $i$ -th particle
$M = \sum m_i$
total mass
When to use
System of point masses in a plane. Treat each coordinate independently. In the formula booklet.
Example
Four particles as in worked example 1: $\bar x = 1.6$ , $\bar y = 1.1$ .
Centre of mass of a uniform triangular lamina
$\mathbf{r}_{\text{COM}} = \dfrac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}$
$\mathbf{a}, \mathbf{b}, \mathbf{c}$
position vectors of the vertices
When to use
Any triangular lamina, uniform. Equals the centroid (intersection of medians). In the IAL formula booklet.
Example
Vertices $(0,0), (6,0), (2,5)$ : centroid $\left(\dfrac{8}{3}, \dfrac{5}{3}\right)$ .
Centre of mass of a uniform rectangular lamina
$G = \text{geometric centre} = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
$(x_1, y_1), (x_2, y_2)$
opposite corners of the rectangle
When to use
Any uniform rectangular lamina — the COM is the centre by two-fold symmetry. NOT explicitly in the formula booklet but follows from symmetry.
Example
Rectangle with corners $(0,0)$ and $(6,4)$ : COM at $(3, 2)$ .
Centre of mass — composite lamina
$\bar x = \dfrac{\sum A_i \bar x_i}{\sum A_i}, \quad \bar y = \dfrac{\sum A_i \bar y_i}{\sum A_i}$
$A_i$
area of $i$ -th component (negative for cut-outs)
$\bar x_i, \bar y_i$
centroid coords of $i$ -th component
When to use
Lamina built from simple sub-pieces (rectangles + triangles + circles + cut-outs). 'Uniform' means area is proxy for mass. NOT in the formula booklet — follows from the discrete formula treating each sub-piece as a point mass.
Example
Square minus a circular hole: $A_{\text{sq}}(3,3) - A_{\text{cir}}(2,2)$ .
Suspended body — angle with vertical
$\tan\theta = \dfrac{\text{horizontal distance from pivot to COM}}{\text{vertical distance from pivot to COM}}$
$\theta$
angle a given edge makes with the vertical
When to use
When the body is suspended from a single pivot point — the line from pivot to COM is vertical. NOT in the formula booklet.
Example
Pivot at origin, COM at $(3, 31/11)$ : edge along $x$ -axis makes angle $\arctan(3 / (31/11)) \approx 46.8^{\circ}$ with vertical.

Key Definitions and Keywords — Centres of Mass

Definitions to memorise and the exact keywords mark schemes credit for centres of mass answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Centre of mass (COM)
Examiner keyword
The unique point at which the entire mass of a body or system of particles can be considered to act for the purposes of describing translation. A force applied through the COM causes pure translation (no rotation).
Uniform lamina
Examiner keyword
A two-dimensional flat object of uniform (constant) density per unit area. Mass is proportional to area, so for centre-of-mass computations the area can be used as a proxy for the mass of each sub-piece.
Centroid
The geometric centre of a region. For a uniform lamina, the centre of mass coincides with the centroid. For a triangle, the centroid is at $((x_1+x_2+x_3)/3, (y_1+y_2+y_3)/3)$ — intersection of the medians.
Composite lamina
Examiner keyword
A lamina formed by joining (or removing) simpler shapes. Centre of mass is the area-weighted average of the sub-piece centroids; cut-outs contribute negative area.
Topple vs slide
Examiner keyword
A body slides when the friction limit is reached: $\tan\theta = \mu$ . A body topples when the vertical through the COM falls outside the base: $\tan\theta = (\text{base half-width})/(\text{COM height})$ . Whichever angle is smaller occurs first as the slope is increased.

Common Mistakes and Misconceptions — Centres of Mass

The traps other students keep falling into on centres of mass questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Adding the cut-out's centroid instead of subtracting
WME02/01 examiner reports — flagged annually
Why it happens
Treating composite-with-hole the same as composite-with-extra-piece.
How to avoid it
A cut-out contributes NEGATIVE area. Write $A_{\text{remaining}} \bar x_{\text{rem}} = A_{\text{whole}} \bar x_{\text{whole}} - A_{\text{hole}} \bar x_{\text{hole}}$ — the minus sign is the discriminator.
✕Confusing centroid with incentre or circumcentre
Why it happens
The triangle has multiple 'centres'.
How to avoid it
For a UNIFORM triangular LAMINA the COM is the CENTROID — average of vertex coordinates. The incentre, circumcentre, orthocentre are different points (and irrelevant here).
✕Using $\tan\theta = \bar y / \bar x$ instead of $\bar x / \bar y$
Why it happens
Confusing which side is opposite vs adjacent in the right-angled triangle of the suspended body.
How to avoid it
Draw a diagram: pivot at top, COM at $(\bar x, -\bar y)$ below if origin is the pivot. Edge $AB$ originally horizontal makes angle with the vertical: $\tan\theta = \bar x / \bar y$ (horizontal offset / vertical drop).
✕Stopping after finding one of topple/slide angle without comparing
WME02/01 examiner reports — annual mistake
Why it happens
Reading the question incompletely.
How to avoid it
'Determine whether the body topples or slides first' demands BOTH angles. The smaller one happens first. State the comparison explicitly in the conclusion.
✕Using areas when the lamina is not uniform
Why it happens
Habit from uniform-lamina problems.
How to avoid it
Areas-as-mass works ONLY when 'uniform' is stated. Otherwise, use actual masses. M2 questions almost always say 'uniform', but watch for the rare exception.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Centres of Mass — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Centres of Mass

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Centre of mass of four particles in 2D (5 marks, M2)

2Composite lamina with a circular hole (8 marks, M2)

3Centre of mass of a triangular lamina (4 marks, M2)

4Lamina suspended from a corner — find angle (8 marks, M2)

5Topple-or-slide on an inclined plane (8 marks, M2)

Centre of mass — discrete particles (1D)

Centre of mass — discrete particles (2D)

Centre of mass of a uniform triangular lamina

Centre of mass of a uniform rectangular lamina

Centre of mass — composite lamina

Suspended body — angle with vertical

Centre of mass (COM)

Uniform lamina

Centroid

Composite lamina

Topple vs slide

✕Adding the cut-out's centroid instead of subtracting

✕Confusing centroid with incentre or circumcentre

✕Using tan⁡θ=yˉ/xˉ\tan\theta = \bar y / \bar xtanθ=yˉ​/xˉ instead of xˉ/yˉ\bar x / \bar yxˉ/yˉ​

✕Stopping after finding one of topple/slide angle without comparing

✕Using areas when the lamina is not uniform

Practice questions

Past paper style quiz

Video lesson

Centres of Mass — frequently asked questions

✕Using $\tan\theta = \bar y / \bar x$ instead of $\bar x / \bar y$