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How do you find the centre of mass of a system of particles or a lamina, and how does it determine equilibrium?

Centre of mass of a system of particles, of uniform laminae and standard shapes, of composite bodies, and using the centre of mass to analyse suspended and toppling bodies.

A focused answer to the AQA A-Level Further Mathematics centre of mass content, covering the centre of mass of a system of particles, uniform laminae and standard shapes, composite bodies, and using the centre of mass to analyse suspended and toppling bodies.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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What this dot point is asking
Centre of mass of particles
Laminae and composite bodies
Suspended and toppling bodies

What this dot point is asking

AQA wants you to find the centre of mass of a system of particles, of uniform laminae and standard shapes, and of composite bodies by combining parts, and to use the position of the centre of mass to analyse a suspended body hanging in equilibrium or a body on the point of toppling.

Centre of mass of particles

Centre of mass of a composite lamina

A uniform lamina is an L-shape made of a $4$ by $2$ rectangle ( $A$ ) joined to a $2$ by $3$ rectangle ( $B$ ). Take coordinates so that $A$ has centre $(2, 1)$ and area $8$ , and $B$ has centre $(1, 3.5)$ and area $6$ . Find the centre of mass.

Step 1: Replace each part by a point mass

Because the lamina is uniform, mass is proportional to area. We can therefore treat each rectangle as a single particle located at its own centre of mass. Place a particle of mass $8$ at $(2, 1)$ (representing rectangle $A$ ) and a particle of mass $6$ at $(1, 3.5)$ (representing rectangle $B$ ). The two-particle system now has the same centre of mass as the original L-shape.

Step 2: Find $\bar{x}$ by taking moments about the $y$ -axis

We form the weighted average of the $x$ -coordinates, with the areas as weights. This is the same as asking: where on the $x$ -axis would the total mass $14$ need to sit to produce the same turning moment about the $y$ -axis?

\bar{x} = \frac{8(2) + 6(1)}{8 + 6} = \frac{16 + 6}{14} = \frac{22}{14} = \frac{11}{7} \approx 1.57.

Step 3: Find $\bar{y}$ by taking moments about the $x$ -axis

We repeat the same weighted-average process in the $y$ -direction.

\bar{y} = \frac{8(1) + 6(3.5)}{14} = \frac{8 + 21}{14} = \frac{29}{14} \approx 2.07.

Final answer: the centre of mass of the L-shape is at $\left(\dfrac{11}{7},\, \dfrac{29}{14}\right)$ .

Laminae and composite bodies

For a uniform lamina, mass is proportional to area, so the area of each part plays the role of its mass in the averaging formula. You should know the standard results without deriving them: the centre of mass of a uniform triangular lamina lies at the centroid, one third of the way along each median from the side towards the opposite vertex, which for vertices at $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ is at $\frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ . A uniform semicircular lamina of radius $r$ has its centre of mass at distance $\frac{4r}{3\pi}$ from the centre of the diameter, along the axis of symmetry, and a uniform circular sector of radius $r$ and half-angle $\alpha$ has its centre of mass at $\frac{2r\sin\alpha}{3\alpha}$ from the centre. These can be quoted directly in composite-body questions.

To find a composite body, split it into parts whose individual centres you know, treat each part as a point mass at its own centre, then take the weighted average. The single most important technique is the negative-mass method: when a region is removed, include it with a negative area so its moment is subtracted. The remaining mass is then the original area minus the removed area, and the moments subtract in the same way.

Suspended and toppling bodies

When a lamina hangs freely from a pivot P, the only forces are its weight at the centre of mass and the reaction at P. For equilibrium these act along the same vertical line, so the centre of mass always settles directly below P. This single fact drives every suspension question: locate the centre of mass relative to P, then use the right-angled triangle formed by the horizontal and vertical offsets to find the angle a named edge makes with the vertical.

Angle of a suspended lamina

A lamina is suspended from a corner P. Its centre of mass lies at horizontal offset $2$ and vertical offset $5$ from P, measured along the lamina's own edges. Find the angle the top edge makes with the vertical when it hangs.

Use the equilibrium condition

The lamina hangs with its centre of mass vertically below P, so the line from P to the centre of mass is the true vertical.

Form the right-angled triangle

The horizontal offset $2$ and vertical offset $5$ are the sides of a right-angled triangle whose hypotenuse is that line from P to the centre of mass.

Solve for the angle

The angle $\alpha$ between the lamina's edge and the vertical satisfies $\tan\alpha = \frac{2}{5}$ , so $\alpha = \arctan\frac{2}{5} = 21.8$ degrees (to one decimal place).

A body resting on a plane topples (rather than slides) when the vertical line through its centre of mass passes outside its base of support. On an incline of increasing angle, the critical angle for toppling is reached when that vertical passes exactly through the lowest edge of the base; beyond it the weight produces a turning moment that tips the body over. Comparing the toppling angle with the angle at which friction is overcome tells you whether a body slides first or topples first.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20196 marksA uniform rectangular lamina occupies the region with corners at

(0, 0)

(6, 0)

(6, 4)

and

(0, 4)

. A square hole of side

2

is cut out, occupying the region with corners

(4, 2)

(6, 2)

(6, 4)

and

(4, 4)

. Find the coordinates of the centre of mass of the remaining lamina.

Show worked answer →

Treat the remaining lamina as the full rectangle with the square removed as a negative mass, since mass is proportional to area for a uniform lamina.

The full rectangle has area $6 \times 4 = 24$ with centre of mass at $(3, 2)$ .

The removed square has area $2 \times 2 = 4$ with centre of mass at its own centre $(5, 3)$ .

The remaining area is $24 - 4 = 20$ . Taking moments about the $y$ -axis: $24(3) - 4(5) = 20\bar{x}$ , so $72 - 20 = 20\bar{x}$ and $\bar{x} = \frac{52}{20} = 2.6$ .

Taking moments about the $x$ -axis: $24(2) - 4(3) = 20\bar{y}$ , so $48 - 12 = 20\bar{y}$ and $\bar{y} = \frac{36}{20} = 1.8$ .

The centre of mass is at $(2.6, 1.8)$ . Markers reward the negative-mass method, both moment equations, and the correct coordinates.

AQA 20215 marksA uniform lamina hangs freely in equilibrium from a point P. Its centre of mass is at a horizontal distance

3

and vertical distance

4

from P (measured along the lamina's own axes). Calculate the angle that the lamina's reference edge makes with the vertical, and explain why the centre of mass settles directly below P.

Show worked answer →

When a body hangs freely from a pivot, the only forces are its weight (acting at the centre of mass) and the reaction at P. For equilibrium these must be equal and opposite along the same line, so the line from P to the centre of mass is vertical.

The horizontal and vertical offsets of the centre of mass from P are $3$ and $4$ , so the angle $\alpha$ between the reference edge and the vertical satisfies $\tan\alpha = \frac{3}{4}$ .

Hence $\alpha = \arctan\frac{3}{4} = 36.9$ degrees (to one decimal place).

Markers reward the equilibrium argument that the centre of mass lies vertically below P, forming the correct right-angled triangle, and evaluating the angle.

Related dot points

Sources & how we know this

AQA A-level Further Mathematics (7367) specification — AQA (2017)