Maths › Further Mechanics 2 › Centres of mass of plane figures and frameworks
Centres of mass of plane figures and frameworks
Cut a shape into pieces whose centres you already know, weight each by its area, and add. Removing a piece is the same sum with a minus sign.
Builds on Centre of mass of a discrete distribution and Moments.
IN THIS TOPIC
- Find the centre of mass of a composite lamina, including one with a piece removed.
- Find the centre of mass of a framework of rods.
- Use symmetry to avoid unnecessary calculation.
WHAT YOU PROBABLY THINK
A wire triangle and a triangular lamina of the same shape have the same centre of mass.
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 are known, and add. A rectangle's centre is its middle, a triangle's is a third of the way up from the base, and standard results for sectors and arcs are in the formulae book.
A shape with a hole is handled by subtraction: treat the missing piece as a negative area at its own centre. The arithmetic is the same and the sign does the work, which is far quicker than trying to cut the remaining shape into positive pieces.
Symmetry is worth using before anything else. Any axis of symmetry must contain the centre of mass, so a symmetric shape needs at most one coordinate calculating.
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.
Frameworks weight by length
A framework is made of rods, so its mass is spread along lines rather than over an area. Each rod acts at its own midpoint with a weight proportional to its length, and the same weighted average follows. That is why the opening claim is false: the wire and the lamina distribute their mass differently, so their centres do not coincide except by accident.
For a lamina the answer is the centroid of the shape; for a framework it is the centroid of the perimeter. A long thin rod on one side of a framework pulls the centre of mass strongly towards it, while the same side of a lamina contributes only in proportion to the area near it.
YOUR TURN
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, as they must.
THE EXAM BIT
- Tabulate area, x, ax, y and ay, with a negative area for anything removed.
- Quote the standard centre for each piece; a triangle's is a third of the height from its 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.
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
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For a uniform lamina, weight each piece by its area at its own centre, using a negative area for anything removed.
For a framework, weight each rod by its length at its own midpoint; the wire and the lamina of the same shape 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.
CHECK YOUR PROGRESS
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- Find the centre of mass of a composite lamina, including one with a piece removed.
- Find the centre of mass of a framework of rods.
- Use symmetry to avoid unnecessary calculation.
Open the full revision checklist to see every objective in the course in one place.
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