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Centre of mass of a discrete distribution

The centre of mass is the point at which a system of masses balances. For separate masses it is a weighted average, and each coordinate is worked out on its own.

Year FMEDEXCEL 9FM0 FM2 2.1

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.
  • Work backwards from a given centre of mass to an unknown mass or position.
  • Read the centre of mass physically: a body balances when supported there, and hangs with it directly below the point of suspension.

COMMON MISCONCEPTION

The centre of mass of a set of particles is the average of their positions.

It is the mass-weighted average, Σmx over Σm taken one coordinate at a time, so the point sits nearer the heavier particles. Averaging the positions alone is right only when every mass is equal. Masses of 2, 3 and 5 kg at (1, 4), (3, 0) and (−2, 2) have their centre of mass at (0.1, 1.8), well left of the plain average (0.67, 2).

A weighted average

Take moments about the origin. The sum of the individual moments must equal the moment of the total mass placed at the centre of mass, and that gives one equation per coordinate:

mean x=ΣmixiΣmi,mean y=ΣmiyiΣmi\text{mean x} = \frac{Σ m_{i}x_{i}}{Σ m_{i}}, \qquad \text{mean y} = \frac{Σ m_{i}y_{i}}{Σ m_{i}}NOT IN THE BOOKLET: LEARN IT

Pearson Edexcel 9FM0 formulae booklet: the centres of mass of seven standard uniform bodies are included; this weighted-average rule is not.

The masses are the weights in the average, so the point sits nearer the heavier particles. Averaging the positions alone 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. A table with columns for m, x, mx, y and my keeps the work orderly and makes the arithmetic checkable.

Three circles of different sizes plotted on axes, one for each mass, with an amber marker for the centre of mass sitting near the origin. It lies closest to the heaviest mass, as a weighted average must.
FIG. 1Three masses plotted with their centre of mass, which sits nearest the heaviest of them.

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 a counterweight should go, or how heavy it must be to bring the centre of mass to a particular point. The equation is linear in each unknown, so one substitution and one rearrangement finish it.

Hold on to the physical meaning while you do it. 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 get used constantly in the lessons that follow.

A rod resting on a triangular support, with a smaller circle for the four kilogram mass on the left and a larger one for the six kilogram mass on the right. The lighter mass is further from the support, so the two moments about it are equal.
FIG. 2The rod balanced at its centre of mass, with the two moments about that point equal.

GUIDED PRACTICE

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.

ASSESSMENT FOCUS

  • Set the work out in a table, with a column for the moments.
  • Do the two coordinates separately and keep the totals distinct.
  • Check the answer is no farther out than the outermost masses: the centre of mass lies within their spread, though not necessarily at an occupied point, since two equal masses at the ends of a rod balance at the empty midpoint.

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

Σmx = 0 + 4 + 18 = 22, and the total mass is 6, so xG = 22/6 = 3.67 to three significant figures.

The centre of mass is the mass-weighted average of the positions, taken one coordinate at a time, and 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.

Question workbook

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Question workbook

PDF · 56 KB · screen edition · navy

Answer book

PDF · 88 KB · fully worked · navy
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 centre of mass of a discrete distribution questions page.

CHECK YOUR PROGRESS

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  • Find the centre of mass of masses on a line and in a plane.
  • Work backwards from a given centre of mass to an unknown mass or position.
  • Read the centre of mass physically: a body balances when supported there, and hangs with it directly below the point of suspension.

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