Centre-of-mass calculations

Find the centre of mass of a set of particles or a uniform lamina, including a lamina with a piece cut out, as a mass-weighted mean position.

  • A-Level H2 Physics topic extensions
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In Moments, couples and centre of gravity the weight of a body acts at one point. For a symmetrical uniform body you can find that point by eye. For anything else, such as a set of particles or a flat shape with a piece cut out, it has to be calculated.

A position weighted by mass

The centre of mass is the mean position of the mass, with each piece weighted by how much mass it has. For particles of mass mᵢ at positions xᵢ, measured from one origin,

x_CM = (∑ mᵢxᵢ)/(∑ mᵢ)

In two dimensions, find each coordinate separately with the same total mass:

y_CM = (∑ mᵢyᵢ)/(∑ mᵢ)

The answer always lies between the smallest and largest coordinates, and closer to the larger masses. Moving the origin changes the numbers but not the point itself. In a uniform gravitational field the centre of mass is also the centre of gravity.

Worked example 1

Two particles on a line

Extension

Problem

A 2.0 kg particle is at x = 0 and a 3.0 kg particle is at x = 5.0 m. Find the centre of mass.

Worked solution
  1. Weighted sum

    Method

    Multiply each position by its mass and add.

    Reason

    Each particle counts in proportion to its mass.

    Working

    (2.0)(0) + (3.0)(5.0) = 15 kg m

  2. Divide by the total mass

    Working

    x_CM = 15/(2.0 + 3.0) = 3.0 m

Spot the mistake 2

Three particles on a line

About 4 min

Problem

Particles of mass 1.0 kg, 2.0 kg and 3.0 kg are at x = 0, 2.0 m and 5.0 m. A learner averages the three positions and gets 2.33 m. What is wrong, and where is the centre of mass?

Find the centre of mass

Unit: m

Show solution step by step
  1. The fault

    Reason

    Each particle counts in proportion to its mass, so the positions cannot be weighted equally.

    Working

    (0 + 2.0 + 5.0)/3 is correct only if the three masses are equal.

  2. Mass-weighted mean

    Working

    x_CM = (1.0(0) + 2.0(2.0) + 3.0(5.0))/(1.0 + 2.0 + 3.0) = 19/6 = 3.2 m

Exam-style question 1

Centre of mass in two dimensions

5 marks

Examination-style question

Particles of mass 2.0 kg, 1.0 kg and 3.0 kg are at (0,0), (4.0,0) and (0,2.0), with coordinates in metres. Find the coordinates of the centre of mass. [5 marks]

Find both coordinates

Unit: m
Unit: m

Show solution step by step
  1. Total mass

    1 mark

    Reason

    Both coordinates use the same total mass.

    Working

    M = 2.0 + 1.0 + 3.0 = 6.0 kg

  2. x coordinate

    2 marks

    Reason

    Only the 1.0 kg particle has a non-zero x.

    Working

    x_CM = (2.0(0) + 1.0(4.0) + 3.0(0))/6.0 = 0.67 m

  3. y coordinate

    2 marks

    Working

    y_CM = (2.0(0) + 1.0(0) + 3.0(2.0))/6.0 = 1.0 m, so the centre of mass is at (0.67, 1.0) m.

A lamina with a piece cut out

For a flat sheet (lamina) of uniform thickness and density, mass is proportional to area. Split the shape into rectangles or other simple parts whose centres you know, and use their areas Aᵢ in place of the masses.

A hole is a part with negative area at the centre of the piece removed. Subtract its area from the total, and subtract its area × coordinate from each sum. A table with columns Aᵢ, xᵢ, yᵢ, Aᵢxᵢ and Aᵢyᵢ keeps the signs straight.

Try it yourself 3

A rectangle with a corner removed

Minimal support

Problem

A uniform lamina is a 6.0 cm × 4.0 cm rectangle. A 2.0 cm × 2.0 cm square is cut from its top-right corner. Taking the original bottom-left corner as the origin, find the centre of mass of the remaining shape.

A rectangular lamina with a corner removedAxes in centimetres start at the original bottom-left corner. The original rectangle is 6 centimetres wide and 4 centimetres high. A 2 by 2 centimetre square is removed from the top-right corner. The remaining material forms an L shape.0x / cmy / cmRemoved2 cm × 2 cm6.0 cm4.0 cm
Scroll across the figure to read all labels.
Use the original bottom-left corner as the origin for both the outer rectangle and the removed square. The dashed square marks removed material, not an additional piece.

Find both coordinates

Unit: cm
Unit: cm

Hints

Hint 1: two parts

Use + 24 cm² centred at (3.0, 2.0) and -4.0 cm² centred at (5.0, 3.0).

Show solution step by step
  1. Remaining area

    Reason

    The hole counts as negative area.

    Working

    A = 24-4.0 = 20 cm²

  2. x coordinate

    Reason

    Removing material from the right moves the centre left of 3.0 cm.

    Working

    x_CM = (24(3.0)-4.0(5.0))/20 = 2.6 cm

  3. y coordinate

    Working

    y_CM = (24(2.0)-4.0(3.0))/20 = 1.8 cm

Common mistakes

  • Averaging positions without weighting them by mass.
  • Measuring positions from different origins for different parts.
  • Adding the area of a hole instead of subtracting it, or subtracting its area but not its area × coordinate.
  • Dividing by the area of the original shape instead of the area that remains.
Syllabus and review details

No official syllabus alignment is listed for this lesson.