Maths › Further Mechanics 2 › Centres of mass of plane figures and frameworks

Centres of mass of plane figures and frameworks

The centre of mass of a composite shape comes from splitting it into pieces with known centres, weighting each by its area, and adding. A removed piece enters the same sum with a minus sign.

Year FMEDEXCEL 9FM0 FM2 2.2

Builds on Centre of mass of a discrete distribution and Moments.

IN THIS TOPIC

  • Find the centre of mass of a composite lamina.
  • Handle a lamina with a piece removed, using a negative area.
  • Find the centre of mass of a framework of rods.
  • Use symmetry to avoid unnecessary calculation.
  • Quote the booklet's standard centre of mass for a single uniform figure.

COMMON MISCONCEPTION

A wire triangle and a triangular lamina of the same shape have the same centre of mass.

A framework weights each rod by its length at its own midpoint, while a lamina weights each piece by its area, so the two distributions generally give different centres. A wire triangle with vertices (0, 0), (4, 0) and (0, 3) balances at (1.5, 1), against the lamina's centroid at (4/3, 1). Enough symmetry can make them agree: an equilateral triangle of wire balances where its lamina does.

Areas as the weights

For a uniform lamina, mass is proportional to area, so the areas become the weights in the same weighted average. Split the shape into rectangles, triangles and circles whose centres you already know, then add. A rectangle's centre is its middle, and every piece harder than that is one of the seven standard bodies listed in the Pearson Edexcel 9FM0 formulae booklet.

A rectangle with a smaller rectangle cut out of its bottom left corner. Markers show the centroid of the whole rectangle, the centroid of the removed piece, and the centre of mass of what is left, which has shifted up and to the right.
FIG. 1A rectangle with a corner removed, and the centre of mass shifted away from the missing piece.

WORKED EXAMPLE

A rectangle with a corner cut out

A uniform lamina is an 8 by 6 rectangle with a 3 by 2 rectangle removed from the corner at the origin. Find its centre of mass.

Whole rectangle: area 48, centre (4, 3). Removed piece: area 6, centre (1.5, 1). Remaining area 42.

xG = (48 × 4 − 6 × 1.5)/42 = 183/42 = 4.36.

yG = (48 × 3 − 6 × 1)/42 = 138/42 = 3.29.

Both coordinates have moved away from the removed corner, which is the check to make.

The seven standard centres of mass

Pearson Edexcel 9FM0 formulae booklet: included. The centre of mass of seven standard uniform bodies is listed under Centres of mass. Three of the seven are plane figures.

Triangular lamina: 23 along the median from the vertex\text{Triangular lamina: } \tfrac{2}{3} \text{ along the median from the vertex}IN THE FORMULAE BOOKLET
Circular arc, radius r, angle at centre 2α:rsinαα from the centre\text{Circular arc, radius } r, \text{ angle at centre } 2α: \frac{r \sin α}{α} \text{ from the centre}IN THE FORMULAE BOOKLET
Sector of a circle, radius r, angle at centre 2α:2rsinα3α from the centre\text{Sector of a circle, radius } r, \text{ angle at centre } 2α: \frac{2r \sin α}{3α} \text{ from the centre}IN THE FORMULAE BOOKLET

The angle in those two is 2α, the whole angle at the centre, with α in radians, so a quarter circle has α = π/4 and not π/2. Reading α as the whole angle instead of the half-angle doubles it and gives the wrong distance. The other four are solids, and they appear in the toppling and suspension questions.

Solid hemisphere, radius r:38r from the centre\text{Solid hemisphere, radius } r: \tfrac{3}{8}r \text{ from the centre}IN THE FORMULAE BOOKLET
Hemispherical shell, radius r:12r from the centre\text{Hemispherical shell, radius } r: \tfrac{1}{2}r \text{ from the centre}IN THE FORMULAE BOOKLET
Solid cone or pyramid, height h:14h above the base, on the line from the centre of the base to the vertex\text{Solid cone or pyramid, height } h: \tfrac{1}{4}h \text{ above the base, on the line from the centre of the base to the vertex}IN THE FORMULAE BOOKLET
Conical shell, height h:13h above the base, on the same line\text{Conical shell, height } h: \tfrac{1}{3}h \text{ above the base, on the same line}IN THE FORMULAE BOOKLET

Notice the pattern in these two pairs. In each, the shell sits further from the base than the solid, because a shell carries none of its mass in the middle. The booklet lists all four, so understand the pattern rather than memorising it.

Holes and symmetry

A shape with a hole is handled by subtraction. Treat the missing piece as a negative area acting at its own centre. The arithmetic is unchanged and the sign does the work, which is easier than trying to cut the remaining shape into positive pieces.

Use symmetry before anything else. Any axis of symmetry must contain the centre of mass, so a symmetric shape needs at most one coordinate calculating, and a doubly symmetric one needs none.

A 10 by 6 lamina with a circular hole of radius 2 at (7.4, 3): the negative area moves G along the symmetry axis to (4.36, 3)
Describe the drawing

A wide rectangle with a circular hole punched to the right of its middle, and a dashed horizontal line through the centre of both shapes. Markers sit at the rectangle's centre, at the hole's centre, and at the centre of mass of what remains, which has slid along the dashed symmetry axis away from the hole to four point three six while staying at height three exactly.

FIG. 2A 10 by 6 lamina with a circular hole of radius 2 punched at (7.4, 3), the hole entering the sum as an area of −12.6 and the centre of mass sliding along the horizontal axis of symmetry to (4.36, 3), away from the hole.

Frameworks weight by length

A framework is made of rods, so its mass is spread along lines and not over an area. Each rod acts at its own midpoint with a weight proportional to its length, and the same weighted average follows. The wire and the lamina distribute their mass differently, so their centres need not agree, though enough symmetry can make them coincide: an equilateral triangle of wire balances where its lamina does.

For a lamina the answer is the centroid of the shape; for a framework it is the centroid of the perimeter. A long rod on one side of a framework pulls the centre of mass strongly towards itself, while the same side of a lamina contributes only in proportion to the area near it.

A right-angled triangle drawn as three separate rods of lengths three, four and five, with a dot at the middle of each. The centre of mass of the framework is marked inside the triangle, but it is not the centroid of the lamina of the same shape.
FIG. 3A triangular framework of three rods, each acting at its own midpoint and weighted by its length.

GUIDED PRACTICE

A wire triangle

A uniform wire framework forms a right-angled triangle with vertices at (0, 0), (4, 0) and (0, 3). Find its centre of mass.

Show the working

The rods have lengths 4, 3 and 5, total 12, acting at (2, 0), (0, 1.5) and (2, 1.5).

xG = (4 × 2 + 3 × 0 + 5 × 2)/12 = 18/12 = 1.5.

yG = (4 × 0 + 3 × 1.5 + 5 × 1.5)/12 = 12/12 = 1.

The triangular lamina of the same shape has its centroid at (4/3, 1), so the two differ here.

ASSESSMENT FOCUS

  • Tabulate area, x, ax, y and ay, with a negative area for anything removed.
  • Look the standard centres up rather than recalling them. A triangle's is two thirds along the median from the vertex, which is a third of the height from the base.
  • Use an axis of symmetry to write down one coordinate without working.
  • For a framework, weight by length and use midpoints, never areas and centroids.
  • State your axes on the diagram and measure every distance from the same edge.

CHECK YOURSELF

A uniform lamina is a 6 by 4 rectangle with a 2 by 2 square removed from one corner. Find the distance of the centre of mass from the long edge nearest the square.

Show a hint

Subtract the square as a negative area.

Show the answer

Areas 24 and −4, so 20 remains. Taking the removed square in the corner at the origin, yG = (24 × 2 − 4 × 1)/20 = 44/20 = 2.2.

For a uniform lamina, weight each piece by its area at its own centre.

The Pearson Edexcel 9FM0 formulae booklet gives the centres of seven standard uniform bodies.

Anything removed goes in with a negative area at its own centre.

For a framework, weight each rod by its length at its own midpoint, so the wire and the lamina of the same shape generally differ.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

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Answer book

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15 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the centres of mass of plane figures and frameworks questions page.

USED LATER IN

CHECK YOUR PROGRESS

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  • Find the centre of mass of a composite lamina.
  • Handle a lamina with a piece removed, using a negative area.
  • Find the centre of mass of a framework of rods.
  • Use symmetry to avoid unnecessary calculation.
  • Quote the booklet's standard centre of mass for a single uniform figure.

Open the full revision checklist to see every objective in the course in one place.