Vector form of the centre of mass?

\bar{\mathbf{r}}=\frac{\sum mi\mathbf{r}i}{\sum mi}

What is a moment in this context?

Mass times perpendicular distance from an axis.

Where is the centre of mass of a triangular lamina?

At the meeting point of the medians, \frac13 of the height from the base.

How far along a median is the centre of mass of a triangle?

\frac23 of the way from the vertex.

How do you treat a hole in a lamina?

As a part with negative mass, subtracting its moment.

Centre of mass of a rectangle?

At the intersection of the diagonals.

Why can areas be used instead of masses for a uniform lamina?

The density is constant, so mass is proportional to area and the factor cancels.

Order of working for a composite lamina?

Split into shapes, find areas and centres, then take moments about two axes.

Centre of mass of particles and composite bodiesAQA A-Level Further Maths: Flashcards

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Formula for $\bar{x}$ for a system of particles?

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Formula for $\bar{x}$ for a system of particles?: $\bar{x}=\frac{\sum m_ix_i}{\sum m_i}$
Vector form of the centre of mass?: $\bar{\mathbf{r}}=\frac{\sum m_i\mathbf{r}_i}{\sum m_i}$
What is a moment in this context?: Mass times perpendicular distance from an axis.
For a uniform lamina, mass is proportional to what?: Area.
Where is the centre of mass of a uniform rod?: At its midpoint.
Where is the centre of mass of a triangular lamina?: At the meeting point of the medians, $\frac13$ of the height from the base.
How far along a median is the centre of mass of a triangle?: $\frac23$ of the way from the vertex.
Where does the centre of mass lie on a symmetrical body?: On the axis of symmetry.
How do you treat a hole in a lamina?: As a part with negative mass, subtracting its moment.
Centre of mass of a rectangle?: At the intersection of the diagonals.
Why can areas be used instead of masses for a uniform lamina?: The density is constant, so mass is proportional to area and the factor cancels.
Order of working for a composite lamina?: Split into shapes, find areas and centres, then take moments about two axes.

Exam questions on Centre of mass of particles and composite bodies

A uniform lamina is L-shaped. It is made from a rectangle $8$ cm by $2$ cm and a rectangle $2$ cm by $4$ cm joined along an edge of length $2$ cm. With axes along the two outer edges meeting at the corner $O$ , the first rectangle occupies $0\le x\le8$ , $0\le y\le2$ and the second occupies $0\le x\le2$ , $2\le y\le6$ (lengths in cm).
The lamina has mass $0.48$ kg. A particle of mass $0.12$ kg is attached at the point $(8,2)$ . Find the $x$ -coordinate of the centre of mass of the combined body.2 marks
Three particles of masses $2$ kg, $3$ kg and $5$ kg are at the points $(1,2)$ , $(4,-1)$ and $(-2,3)$ respectively. All coordinates are in metres.
A fourth particle of mass $10$ kg is added so that the centre of mass of the four particles is at the origin. Find the coordinates of the fourth particle.2 marks
A uniform square lamina $ABCD$ has side $20$ cm. A circular hole of radius $4$ cm is cut from it. Take $A$ as the origin, with $AB$ along the $x$ -axis and $AD$ along the $y$ -axis, so that the centre of the hole is at $(14,10)$ and $C$ is at $(20,20)$ (lengths in cm).
Find the distance of the centre of mass of the lamina from $AD$ .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).