Centre of mass of a rectangle or disc?

At its geometric centre (intersection of the axes of symmetry).

Centre of mass of a triangle?

The mean of the three vertices' coordinates; one third of the height from each side.

Centre of mass of a uniform semicircular lamina, radius r?

\frac{4r}{3\pi} from the diameter, on the axis of symmetry.

Centre of mass of a semicircular arc (wire), radius r?

\frac{2r}{\pi} from the diameter; not the same as the lamina.

Centre of mass of a sector of angle 2\alpha and radius r?

\frac{2r\sin\alpha}{3\alpha} from the centre, on the axis of symmetry.

Moments equation for a composite lamina?

\left(\sum m\right)\bar{x}=\sum mx, and the same for \bar{y}.

Why can areas be used instead of masses?

The lamina is uniform, so mass is proportional to area and the common factor cancels.

How is a hole treated?

As a negative mass at the centre of the hole: whole = remainder + hole.

Disc of radius 12 with a hole of radius 4 centred 6 from O: where is the centre of mass of the remainder?

0.75 cm from O, on the side away from the hole.

Do you need to prove booklet results?

No; results in the formulae booklet can be quoted without proof.

Centre of mass of laminas and composite figuresEdexcel International A Level Further Maths: Flashcards

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What is the centre of mass of a uniform lamina?

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What is the centre of mass of a uniform lamina?: The centroid of its shape: the point where its weight acts, with mass proportional to area.
Centre of mass of a rectangle or disc?: At its geometric centre (intersection of the axes of symmetry).
Centre of mass of a triangle?: The mean of the three vertices' coordinates; one third of the height from each side.
Centre of mass of a uniform semicircular lamina, radius $r$ ?: $\frac{4r}{3\pi}$ from the diameter, on the axis of symmetry.
Centre of mass of a semicircular arc (wire), radius $r$ ?: $\frac{2r}{\pi}$ from the diameter; not the same as the lamina.
Centre of mass of a sector of angle $2\alpha$ and radius $r$ ?: $\frac{2r\sin\alpha}{3\alpha}$ from the centre, on the axis of symmetry.
Moments equation for a composite lamina?: $\left(\sum m\right)\bar{x}=\sum mx$ , and the same for $\bar{y}$ .
Why can areas be used instead of masses?: The lamina is uniform, so mass is proportional to area and the common factor cancels.
How is a hole treated?: As a negative mass at the centre of the hole: whole $=$ remainder $+$ hole.
Where does the centre of mass lie if the lamina has an axis of symmetry?: On the axis of symmetry.
Disc of radius $12$ with a hole of radius $4$ centred $6$ from $O$ : where is the centre of mass of the remainder?: $0.75$ cm from $O$ , on the side away from the hole.
Do you need to prove booklet results?: No; results in the formulae booklet can be quoted without proof.

Exam questions on Centre of mass of laminas and composite figures

A uniform lamina is made from two rectangles joined together. Rectangle $R_1$ occupies $0\le x\le 6$ and $0\le y\le 2$ , and rectangle $R_2$ occupies $0\le x\le 2$ and $2\le y\le 8$ , where $x$ and $y$ are distances in cm from the axes.
Find the distance between the centre of mass of the whole lamina and the centre of mass of $R_1$ .2 marks
A uniform semicircular lamina of mass $6$ kg has radius $12$ cm and diameter $AB$ . The point $O$ is the midpoint of $AB$ . Use the result in the formulae booklet for the centre of mass of a semicircular lamina.
Instead of being attached at $O$ , the particle of mass $2$ kg is attached to the lamina at the point on the curved edge on the axis of symmetry. Find the distance of the centre of mass of the combined body from $AB$ .2 marks
A uniform circular disc has centre $O$ and radius $12$ cm. Circular holes, each of radius $4$ cm, are cut from the disc. Take the $x$ -axis through $O$ and the centre $C$ of the first hole, with $OC=6$ cm.
The first hole is cut out. Find the distance of the centre of mass of the remaining lamina from $O$ , and state which side of $O$ it lies.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).