LESSON 01
What Is the Centre of Mass?
The centre of mass of a system is the point at which the entire mass of the system may be considered to be concentrated when considering its translational motion.
For a collection of particles, the centre of mass is determined by the masses of the particles and their positions. It is therefore not simply the geometrical centre of the arrangement.
A heavier particle has a greater influence on the position of the centre of mass than a lighter particle. The centre of mass is therefore closer to the heavier mass.
Physical interpretation
Imagine a rigid object supported at exactly its centre of mass. The object can balance because the resultant effect of the weights acts through the support point.
This idea becomes especially important when a lamina is suspended from a point or when its stability on an inclined plane is considered.
LESSON 02
Centre of Mass of Two Particles
Suppose two particles have masses m₁ and m₂ and are situated at positions x₁ and x₂ on the x-axis.
The centre of mass has coordinate x̄ given by the weighted mean of the particle positions.
The numerator represents the sum of the moments of the masses about the chosen origin. The denominator represents the total mass.
- Choose a convenient origin.
- Write down every mass and its coordinate.
- Calculate each value of m × x.
- Add the mass-moments.
- Divide by the total mass.
- Check that the answer lies between the extreme particle positions.
LESSON 03
Why the Formula Works
Suppose the centre of mass is at x̄. By definition, the total moment of the system about the origin is equal to the moment produced by placing the total mass at x̄.
Therefore:
For particles:
Dividing by the total mass gives:
This is the central formula behind almost every particle centre-of-mass calculation.
LESSON 04
Choosing the Origin
The coordinate of the centre of mass depends on the chosen origin, but its physical position does not.
A convenient origin usually makes the arithmetic simpler. For example, if particles lie along a line segment, placing the origin at one end of the segment may avoid unnecessary negative coordinates.
WORKED EXAMPLE
Particles of masses 2 kg and 6 kg are positioned at x = 3 m and x = 7 m respectively. Find the centre of mass.
x̄ = [(2)(3) + (6)(7)] / (2 + 6)
x̄ = (6 + 42) / 8
x̄ = 48 / 8
x̄ = 6 m
Notice that 6 m is closer to the 6 kg mass at 7 m than to the 2 kg mass at 3 m.
LESSON 05
Negative Coordinates
Particles do not have to lie on the positive side of the origin. If a particle lies to the left of the origin, its coordinate is negative.
WORKED EXAMPLE
A particle of mass 4 kg is at x = −3 m and a particle of mass 2 kg is at x = 6 m. Find the centre of mass.
x̄ = [(4)(−3) + (2)(6)] / (4 + 2)
x̄ = (−12 + 12) / 6
x̄ = 0 m
Never replace a negative coordinate with its positive distance. The sign of a coordinate determines the direction of its moment.
LESSON 06
Several Particles
For n particles with masses m₁, m₂, …, mₙ at positions x₁, x₂, …, xₙ:
Using sigma notation:
FULL WORKED EXAMPLE
Three particles have masses 2 kg, 3 kg and 5 kg at positions 1 m, 4 m and 7 m respectively. Find the centre of mass.
Total mass:
2 + 3 + 5 = 10 kg
Calculate Σ(mx):
(2)(1) + (3)(4) + (5)(7)
= 2 + 12 + 35
= 49
Therefore:
x̄ = 49 / 10
x̄ = 4.9 m
LESSON 07
The Centre of Mass as a Weighted Mean
The centre of mass can be viewed as a weighted average of the coordinates.
The weights in the average are the masses.
This immediately explains why increasing one particle's mass pulls the centre of mass towards that particle.
For positive masses, the centre of mass must lie between the smallest and largest particle coordinates.
If your answer lies outside that interval, something has almost certainly gone wrong.
LESSON 08
Centre of Mass in Two Dimensions
For particles positioned in a plane, the centre of mass has two coordinates:
The x-coordinate is found independently using the x-coordinates of the particles:
The y-coordinate is found independently using the y-coordinates:
The same total mass is used in both calculations.
WORKED EXAMPLE
Three particles have the following masses and coordinates:
| Particle | Mass | x | y |
|---|---|---|---|
| A | 2 kg | 1 m | 2 m |
| B | 3 kg | 4 m | 1 m |
| C | 5 kg | 7 m | 5 m |
Total mass = 10 kg
Σ(mx) = 2(1) + 3(4) + 5(7)
= 49
Therefore:
x̄ = 49/10 = 4.9 m
Now calculate Σ(my):
Σ(my) = 2(2) + 3(1) + 5(5)
= 4 + 3 + 25
= 32
Therefore:
ȳ = 32/10 = 3.2 m
Centre of mass = (4.9, 3.2) m
LESSON 09
Vector Form of the Centre of Mass
The same calculation can be written using position vectors. If particle i has position vector rᵢ and mass mᵢ, then:
In two dimensions:
This is particularly useful when the positions are supplied as vectors rather than coordinates in a diagram.
LESSON 10
Uniform Laminae
A lamina is a thin flat object. When a lamina is uniform, its mass is distributed uniformly over its area.
For a uniform lamina, the centre of mass coincides with its geometrical centre of mass, often called its centroid.
The key standard results should be known and recognised quickly.
| Shape | Centre of Mass |
|---|---|
| Rectangle | At the intersection of its diagonals |
| Square | At the centre |
| Circle | At its centre |
| Triangle | At the intersection of its medians |
| Semicircle | On its axis of symmetry, at the standard centroid position |
| Quarter circle | On the symmetry axes, at the standard centroid position |
LESSON 11
Symmetry and the Centre of Mass
Symmetry is one of the fastest ways to locate one coordinate of a centre of mass.
If a uniform lamina is symmetrical about a straight line, its centre of mass must lie on that line.
A uniform lamina with an axis of symmetry has its centre of mass somewhere on that axis.
For example, if a shape is symmetrical about the y-axis, then:
If it is symmetrical about the x-axis, then:
This can save a large amount of calculation in examination questions.
LESSON 12
Composite Laminae
Many examination diagrams consist of several simple shapes joined together.
Such a body can be divided into simple laminae whose centres of mass are already known.
The centre of mass of the complete lamina can then be calculated by treating each component as a particle whose:
- mass is proportional to its area;
- position is the centre of mass of that component.
For a uniform lamina:
Therefore, area can be used instead of mass in the weighted-average calculation.
LESSON 13
Composite Lamina Method
- Split the shape into simple standard shapes.
- Find the area and centre of mass of each component.
- Calculate Ax and Ay for every component.
- Use Σ(Ax)/ΣA and Σ(Ay)/ΣA.
The most important part is setting up the geometry correctly. The arithmetic is usually straightforward once the component areas and their centroid coordinates are correct.
Always draw or identify the centre of mass of every component before starting the weighted-average calculation.
LESSON 14
Composite Rectangle Example
WORKED EXAMPLE
A uniform L-shaped lamina can be considered as the combination of two non-overlapping rectangles.
Suppose rectangle A has area 20 cm² and its centre is at (2,5), while rectangle B has area 30 cm² and its centre is at (7,2).
Total area:
A = 20 + 30 = 50 cm²
For x:
Σ(Ax) = 20(2) + 30(7)
= 40 + 210
= 250
Therefore:
x̄ = 250/50
x̄ = 5 cm
For y:
Σ(Ay) = 20(5) + 30(2)
= 100 + 60
= 160
Therefore:
ȳ = 160/50
ȳ = 3.2 cm
Hence the centre of mass is:
(5, 3.2) cm
LESSON 15
Shapes with a Missing Section
A particularly important composite-lamina technique is dealing with a hole or removed section.
The removed part can be treated as having a negative area.
For the missing region, use:
and therefore its contribution to Ax and Ay is also negative.
WORKED EXAMPLE
A large uniform rectangle has area 100 cm² and centre (5,4). A circular hole of area 20 cm² has centre (7,3). Find the centre of mass of the remaining lamina.
Treat the hole as negative area.
Total effective area:
100 − 20 = 80 cm²
x contribution:
100(5) − 20(7)
= 500 − 140
= 360
Therefore:
x̄ = 360/80 = 4.5 cm
y contribution:
100(4) − 20(3)
= 400 − 60
= 340
Therefore:
ȳ = 340/80 = 4.25 cm
Centre of mass = (4.5, 4.25) cm
Do not simply ignore the missing section. The hole changes both the total area and the first moments of area.
LESSON 16
Using Tables for Composite Laminae
A table is often the safest examination method because it prevents components from being missed.
| Part | Area A | x | y | Ax | Ay |
|---|---|---|---|---|---|
| 1 | A₁ | x₁ | y₁ | A₁x₁ | A₁y₁ |
| 2 | A₂ | x₂ | y₂ | A₂x₂ | A₂y₂ |
| Hole | −A₃ | x₃ | y₃ | −A₃x₃ | −A₃y₃ |
Then:
LESSON 17
Equilibrium of a Suspended Lamina
When a lamina is freely suspended from a point, its centre of mass lies vertically below the point of suspension when the lamina is in equilibrium.
For a freely suspended lamina in equilibrium, the vertical line through the centre of mass passes through the suspension point.
This happens because the weight acts vertically downward through the centre of mass.
If the line of action of the weight did not pass through the support, the weight would produce a turning moment and the lamina would rotate.
LESSON 18
Finding a Centre of Mass from Suspension
Sometimes the position of the centre of mass is not given directly. Instead, the lamina is suspended from one or more points.
For each suspension point, the centre of mass must lie somewhere on the vertical line through that point.
If the lamina is suspended from two different points and allowed to settle in equilibrium, the centre of mass lies at the intersection of the corresponding vertical lines.
- Mark the suspension point.
- Draw the vertical line through it in the equilibrium position.
- Repeat for a second suspension point if available.
- The intersection gives the centre of mass.
LESSON 19
Moments and Equilibrium
A body is in rotational equilibrium when the total clockwise moment about a point equals the total anticlockwise moment about that point.
For a lamina of mass M, its weight is:
The weight acts vertically downward through the centre of mass.
The gravitational force therefore creates a moment about a point whenever its line of action does not pass through that point.
LESSON 20
Lamina Rotating About a Horizontal Axis
When a lamina rotates about a fixed horizontal axis, the weight of the lamina acts through its centre of mass.
If the lamina is in equilibrium, the resultant moment about the axis must be zero.
This means the line of action of the weight and the reactions must produce balanced moments.
The perpendicular distance from the axis to the line of action of the force is the correct moment arm.
Do not automatically use the distance to the centre of mass itself. Only the perpendicular distance to the force's line of action is the moment arm.
LESSON 21
Laminae on Inclined Planes
When a lamina rests on an inclined plane, its weight acts vertically downwards through its centre of mass.
The weight can be resolved into components parallel and perpendicular to the plane.
The reaction from the plane acts perpendicular to the surface if the contact is smooth.
The position of the centre of mass determines whether the weight creates a turning tendency about the point or edge of contact.
LESSON 22
Stability and the Line of Action
A useful physical principle is that a body remains supported only while the line of action of its weight is compatible with the supporting region.
If the vertical line through the centre of mass passes outside the base of support, the weight produces a turning effect that can cause the body to topple.
The lower the centre of mass and the wider the effective base of support, the more difficult it is for the body to topple.
In examination questions, focus on the geometry and moments rather than relying on a visual guess about whether something will topple.
LESSON 23
A Complete Composite-Lamina Workflow
- Choose coordinate axes and an origin.
- Split the lamina into simple standard shapes.
- Identify whether each component is added or removed.
- Calculate each component's area.
- Locate the centre of mass of each component.
- Write the coordinates carefully.
- Construct a table containing A, x, y, Ax and Ay.
- Use negative area for holes or removed regions.
- Calculate ΣA, ΣAx and ΣAy.
- Find x̄ and ȳ.
- Use symmetry as a check whenever possible.
- Check whether the final coordinates are physically reasonable.
LESSON 24
Common Exam Traps
If the origin is not at the end of the shape, use signed coordinates. A point to the left of the origin has a negative x-coordinate.
The numerator is Σ(mx), but the denominator is the total mass Σm. Do not divide by the number of particles.
For a uniform plane lamina, mass is proportional to area. Therefore composite-lamina calculations use areas, not just lengths.
A removed region must be represented by negative area.
Before calculating Σ(Ax) and Σ(Ay), make sure the centre of mass of each standard component has been located correctly.
For moments, use the perpendicular distance from the point or axis to the line of action of the force.
LESSON 25
Worked Exam-Style Particle Problem
QUESTION
Particles of masses 3 kg, 4 kg and 5 kg are placed at x = −2 m, x = 3 m and x = 8 m respectively. Find the position of the centre of mass.
Total mass:
M = 3 + 4 + 5 = 12 kg
Calculate Σ(mx):
3(−2) + 4(3) + 5(8)
= −6 + 12 + 40
= 46
Therefore:
x̄ = 46/12
x̄ = 23/6 m
≈ 3.83 m
LESSON 26
Worked Exam-Style 2D Problem
QUESTION
Four particles are positioned as follows:
| Mass | x | y |
|---|---|---|
| 2 kg | 0 | 0 |
| 3 kg | 6 | 0 |
| 4 kg | 6 | 5 |
| 1 kg | 0 | 5 |
Total mass:
M = 2 + 3 + 4 + 1 = 10 kg
For x:
Σ(mx) = 2(0) + 3(6) + 4(6) + 1(0)
= 18 + 24 = 42
Therefore:
x̄ = 42/10 = 4.2
For y:
Σ(my) = 2(0) + 3(0) + 4(5) + 1(5)
= 20 + 5 = 25
Therefore:
ȳ = 25/10 = 2.5
Centre of mass = (4.2, 2.5)
LESSON 27
Advanced Composite-Lamina Example
METHOD DEMONSTRATION
Consider a uniform lamina consisting of a large rectangle with a smaller rectangular section removed.
Suppose the original rectangle has area 120 cm² and centre (6,5). A rectangular hole has area 24 cm² and centre (8,3).
Treat the hole as negative area.
Effective area:
120 − 24 = 96 cm²
x-moment:
120(6) − 24(8)
= 720 − 192
= 528
Therefore:
x̄ = 528/96
x̄ = 5.5 cm
y-moment:
120(5) − 24(3)
= 600 − 72
= 528
Therefore:
ȳ = 528/96
ȳ = 5.5 cm
Thus:
Centre of mass = (5.5, 5.5) cm
LESSON 28
Exam Technique: What to Write
A strong solution should make the structure of the calculation easy for an examiner to follow.
Write:
Total area = ...
Σ(Ax) = ...
Σ(Ay) = ...
Then:
Finally state the coordinates clearly.
PRACTICE • FOUNDATION
Test Yourself
Show answer
The masses of the particles and their positions relative to the chosen coordinate system.
Show answer
x̄ = [2(1) + 6(5)] / 8
= 32/8
x̄ = 4 m
Show answer
Because the position of the centre of mass is a mass-weighted average. A larger mass therefore has a greater influence on the result.
Show answer
The numerical coordinate changes, but the physical location of the centre of mass does not.
PRACTICE • INTERMEDIATE
Exam-Style Questions
Show answer
Total mass = 14 kg.
Σ(mx) = 2(−1) + 5(2) + 7(6)
= −2 + 10 + 42
= 50
Therefore:
x̄ = 50/14 = 25/7 m
Show answer
Total mass = 6 kg.
Σ(mx) = 1(2) + 2(4) + 3(6) = 28
x̄ = 28/6 = 14/3
Σ(my) = 1(1) + 2(3) + 3(5) = 22
ȳ = 22/6 = 11/3
Centre of mass = (14/3, 11/3)
Show answer
Total area = 60 cm².
Σ(Ax) = 18(2) + 42(6) = 288
x̄ = 288/60 = 4.8 cm
Σ(Ay) = 18(4) + 42(3) = 198
ȳ = 198/60 = 3.3 cm
Centre = (4.8,3.3) cm
PRACTICE • ADVANCED
Challenge Problems
Show answer
Effective area:
150 − 30 = 120
Σ(Ax) = 150(5) − 30(7)
= 750 − 210 = 540
x̄ = 540/120 = 4.5
Σ(Ay) = 150(6) − 30(4)
= 900 − 120 = 780
ȳ = 780/120 = 6.5
Centre = (4.5,6.5) cm
Show answer
Because the lamina is uniform, mass is proportional to area. Therefore the areas can be used as weights:
x̄ = Σ(Ax)/ΣA
ȳ = Σ(Ay)/ΣA
Show answer
The centre of mass lies vertically below the suspension point. The line of action of the weight therefore passes through the suspension point and produces no turning moment about it.
QUICK REVIEW
Essential Formula System
M2 MASTERY CHECK
Can You Do All of These?
FINAL SUMMARY
Centres of Mass — The Big Picture
The entire topic is built around one central idea: mass distribution determines position.
For particles, use mass as the weighting factor:
For a uniform lamina, mass is proportional to area, so use:
For composite shapes, divide the object into standard components. For holes, use negative areas. For symmetric shapes, exploit symmetry before calculating.
When equilibrium is involved, remember that the weight acts vertically downwards through the centre of mass. The position of this line of action determines the turning effect and therefore the equilibrium of the lamina.
Find the correct centre of mass → identify the line of action of weight → calculate moments when necessary → apply equilibrium.