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.
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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
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
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
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
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
Show solution step by step
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.
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
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
Show solution step by step
Total mass
1 markReason
Both coordinates use the same total mass.Working
M = 2.0 + 1.0 + 3.0 = 6.0 kg
x coordinate
2 marksReason
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
y coordinate
2 marksWorking
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.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Tick each point only if your answer shows it.
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
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.
Find both coordinates
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
Remaining area
Reason
The hole counts as negative area.Working
A = 24-4.0 = 20 cm²
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
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.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.