Maths › Further Mechanics 2 › Centre of mass of a discrete distribution
Centre of mass of a discrete distribution
The point where the whole thing balances. For separate masses it is a weighted average, and each coordinate is worked out on its own.
Builds on Moments and Statics of a particle.
IN THIS TOPIC
- Find the centre of mass of masses on a line and in a plane.
- Use moments to justify the weighted-average formula.
- Work backwards from a given centre of mass to an unknown mass or position.
WHAT YOU PROBABLY THINK
The centre of mass of a set of particles is the average of their positions.
A weighted average
Taking moments about the origin, the sum of the individual moments must equal the moment of the total mass placed at the centre of mass. That gives, in each coordinate separately:
The masses are the weights in the average, so the point sits nearer the heavier particles. Averaging the positions alone, as the opening claim does, is only right when every mass is equal.
Neither coordinate affects the other, so a two-dimensional problem is two one-dimensional problems solved side by side. Setting the work out in a table with columns for m, x, mx, y and my keeps it orderly and makes the arithmetic checkable.
WORKED EXAMPLE
Three masses in a plane
Masses of 2 kg, 3 kg and 5 kg sit at (1, 4), (3, 0) and (−2, 2). Find the centre of mass.
Total mass = 10 kg.
Σmx = 2(1) + 3(3) + 5(−2) = 2 + 9 − 10 = 1, so xG = 0.1.
Σmy = 2(4) + 3(0) + 5(2) = 8 + 0 + 10 = 18, so yG = 1.8.
The point (0.1, 1.8) lies well to the left of the plain average (0.67, 2), pulled there by the 5 kg mass.
Running it backwards
The same equation solves for an unknown mass or an unknown position, and questions often ask for exactly that: where to place a counterweight, or how heavy it has to be to bring the centre of mass to a particular point. Since the equation is linear in each unknown, one substitution and one rearrangement finish it.
The physical meaning is worth holding: an object supported at its centre of mass balances, and an object suspended from a point hangs with its centre of mass directly below. Both facts come straight from taking moments, and both are used constantly in the lessons that follow.
YOUR TURN
Finding a missing mass
Masses of 4 kg and m kg are placed at x = 1 and x = 6 on a light rod. The centre of mass is at x = 4. Find m.
Show the working
The formula gives (4 × 1 + 6m)/(4 + m) = 4.
Multiplying out: 4 + 6m = 16 + 4m.
So 2m = 12 and m = 6 kg.
Checking: (4 + 36)/10 = 4 as required, and the heavier mass is indeed the closer one to the balance point.
THE EXAM BIT
- Set the work out in a table; the moments column is where marks are earned.
- Do the two coordinates separately and keep the totals distinct.
- Include the signs of negative coordinates rather than trying to shift the origin mid-question.
- Check the answer lies inside the region the masses occupy; it always must.
CHECK YOURSELF
Masses of 1 kg, 2 kg and 3 kg sit at x = 0, 2 and 6. Find the centre of mass.
Show a hint
Total moment over total mass.
Show the answer
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The centre of mass is the mass-weighted average of the positions, taken one coordinate at a time.
It follows from moments, so an object balances when supported there and hangs with it below any point of suspension.
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 masses on a line and in a plane.
- Use moments to justify the weighted-average formula.
- Work backwards from a given centre of mass to an unknown mass or position.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.