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Centre Of Gravity (O Level)

Key idea: Learn what the centre of gravity is, where it lies for common objects, how to find it for a plane lamina, and how it affects stability (O Level Physics 6091).

  • SEC G3 Physics 2027
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Learning objectives

  • Distinguish contact forces from non-contact forces
  • State that mass measures the amount of matter in a body
  • Describe a gravitational field as a region where a mass experiences gravitational force
  • Define gravitational field strength as gravitational force per unit mass
  • Apply weight = mass × gravitational field strength
  • Distinguish mass from weight
  • Describe the effect of balanced and unbalanced forces on a body
  • Describe ways a force may change motion
  • Identify action–reaction pairs on interacting bodies
  • Draw free-body diagrams for force systems in at most two dimensions
  • Solve three-force static equilibrium graphically
  • Apply resultant force = mass × acceleration
  • Relate mass to resistance to change in motion
  • Explain the effects of friction on motion
  • Describe falling with and without air resistance, including terminal velocity
  • Describe a moment as a force's turning effect in everyday examples
  • Apply moment = force × perpendicular distance from the pivot
  • State the principle of moments for a body in equilibrium
  • apply the principle of moments to new situations or to solve related problems
  • show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity
  • Explain qualitatively how centre-of-gravity position affects stability

1. Definition

A. Centre of gravity

The centre of gravity of a body is the point through which its entire weight may be considered to act.

2. Key Ideas

  • The weight of an object can be treated as one force acting at its centre of gravity.
  • For a uniform, symmetrical object, the centre of gravity is at the geometric centre (intersection of symmetry lines).
  • The centre of gravity can lie outside the object (e.g. a ring, a hollow ball).
  • To find the centre of gravity of a plane lamina: suspend it, draw a vertical line with a plumb line, repeat from another point, then find the intersection.
  • Stability depends on whether the line of action of weight falls within the base (see Stability).

3. Detailed Explanations

A. Why we use the centre of gravity

An object’s weight is spread throughout the object, but in turning-effect questions we can treat it as a single force:

  • weight W acts at the centre of gravity, and
  • you can then take moments using M = F × d (see Moment Of A Force).

B. Centre of gravity of common objects

If an object is uniform (same material/density throughout) and symmetrical, its centre of gravity is at the geometric centre.

Centres of gravity of uniform symmetrical shapesA rectangle, circle, cylinder, triangle and uniform beam each have a marked centre of gravity at the intersection of their symmetry lines.rectanglecirclecylindertriangleuniform beam● centre of gravity
For uniform, symmetrical objects, the centre of gravity is at the geometric centre (intersection of symmetry lines).

If the object is not uniform or not symmetrical, the centre of gravity is not at the geometric centre.

C. Finding the centre of gravity of a plane lamina (experiment)

This is the standard practical method for an irregular lamina (thin, flat sheet).

  1. Make a small hole near the edge of the lamina.
  2. Hang it from a pin so it can swing freely.
  3. Hang a plumb line from the same pin and wait until everything is at rest.
  4. Draw the vertical line shown by the plumb line on the lamina.
  5. Repeat from a different hole.
  6. The intersection of the lines is the centre of gravity.

Reason: when an object is freely suspended, its centre of gravity lies vertically below the point of suspension.

D. Centre of gravity and stability (qualitative)

An object is less likely to topple when:

  • its centre of gravity is lower, and/or
  • its base is wider.

The key idea is the line of action of weight:

  • If the vertical line through the centre of gravity falls within the base, the object tends to return (stable).
  • If it falls outside the base, the object topples.

4. Common Mistakes

  • Assuming the centre of gravity is always at the geometric centre (this is only true for uniform, symmetrical objects).
  • Thinking the centre of gravity must be inside the object.
  • Drawing only one plumb line for a lamina (you need at least two to locate the intersection).
  • Not letting the lamina and plumb line come to rest before drawing the vertical line.
  • Using the wrong point for the weight when taking moments (weight acts at the centre of gravity).

5. Exam Tips

  • Use the key phrase: “The weight may be taken as acting at a single point called the centre of gravity.”
  • In moment questions, replace the object’s weight with W acting at its centre of gravity before applying the principle of moments (see Rotational Equilibrium).
  • For “describe how to find the centre of gravity of a lamina”: mention suspend, plumb line, draw vertical line, repeat, intersection.
  • For stability questions: state that an object topples when the line of action of weight falls outside the base.

6. Worked Examples

Modelled example 1

Describe how to find the centre of gravity of an irregular plane lamina.

Core

Problem

Describe how to find the centre of gravity of an irregular plane lamina using a pin and a plumb line.

Study the worked solution

  1. Suspend the lamina and let it settle

    Method

    Make a small hole near the edge, suspend the lamina freely from a pin and hang a plumb line from the same pin.

    Reason

    Once the lamina is at rest, its centre of gravity lies vertically below the suspension point.

    Working

    Wait until both the lamina and the plumb line have stopped moving.

  2. Record the first vertical line

    Reason

    The plumb line shows the vertical line that must pass through the centre of gravity for this suspension point.

    Working

    Draw that line on the lamina.

  3. Repeat and locate the intersection

    Reason

    A second suspension point gives another line through the same centre of gravity, so their intersection locates it.

    Working

    Repeat from a different hole; mark the intersection of the two lines as the centre of gravity.

Guided practice 2

Using a balance point to locate the centre of gravity

About 3 min

Problem

A metre rule balances horizontally on a narrow knife-edge at the 47 cm mark. Where is the centre of gravity of the metre rule?

Use the balance condition

Unit: cm mark

Hints

Hint 1: identify the balance condition
A horizontal rule has zero resultant moment about the knife-edge.
Hint 2: locate the line of action
For the weight to have zero moment, its line of action must pass through the support.
View solution step by step

  1. Use the absence of a turning effect

    Method

    Place the weight’s line of action through the knife-edge.

    Reason

    The rule balances, so its weight has zero moment about the knife-edge and therefore zero perpendicular distance from it.

    Working

    The vertical line through the centre of gravity crosses the 47 cm mark.

  2. State the centre-of-gravity position

    Working

    The centre of gravity is at the 47 cm mark.

Common misconception 3

Uniform rectangular lamina

Find and correct the mistake

Learner response

A student correctly marks the centre of gravity of a uniform rectangular lamina at the intersection of its diagonals, then claims:

The student’s claim is: The centre of gravity of every rectangular lamina is at the intersection of its diagonals, even if a heavy strip is attached along one edge.

Locate the first error and correct the rule.

Diagnose the overgeneralisation

Which condition has the student ignored?

View solution step by step

  1. Keep the valid statement within its conditions

    Method

    Use the diagonal intersection for a uniform rectangular lamina.

    Reason

    Uniform density and rectangular symmetry make the mass distribution balance about the geometric centre.

    Working

    For the original uniform lamina, the centre of gravity is at the intersection of the diagonals.

  2. Locate the first error in the extension

    Reason

    The attached heavy strip makes the mass distribution non-uniform, so geometric shape alone no longer fixes the balance point.

    Working

    The centre of gravity shifts away from the geometric centre and towards the heavier edge.

Examiner practice 4

Is the centre of gravity in the same position on Earth and on the Moon?

2 marks

Examination question

Assume the object is small enough that each gravitational field is uniform. State whether the position of the object’s centre of gravity relative to the object changes between Earth and the Moon. Explain your answer. [2 marks]

Write your two-mark explanation before viewing the mark scheme

View solution step by step

  1. State the conclusion

    1 mark

    Method

    Keep the centre of gravity at the same position relative to the object.

    Reason

    A change in gravitational field strength alone does not change the object’s mass distribution.

    Working

    Yes, its relative position is the same in the stated uniform fields.

  2. Explain the scaling

    1 mark

    Reason

    The Moon’s smaller g reduces the weight of every part by the same factor, so their balance point is unchanged.

    Working

    The total weight changes, but the line through which the combined weight acts remains at the same relative position.

Challenge 5

A uniform metre rule is supported at the 40 cm mark

Minimal support

Calculation transfer

A uniform metre rule weighing 1.2 N is held horizontally by a support at the 40 cm mark. State where its weight acts, then calculate the moment of the weight about the support and give its direction.

Link centre of gravity to the turning effect

Unit: cm mark
Unit: N m
Direction of the moment

Hints

Hint 1: locate the weight force
A uniform metre rule’s centre of gravity is at its geometric centre.
Hint 2: find the moment arm
Measure from the 40 cm support to the 50 cm centre, then convert the separation to metres.
View solution step by step

  1. Locate the centre of gravity

    Method

    Place the rule’s weight at the 50 cm mark.

    Reason

    A uniform, symmetrical metre rule has its centre of gravity at its geometric centre.

    Working

    The 1.2 N weight acts vertically downwards at the 50 cm mark.

  2. Find the perpendicular distance

    Reason

    The horizontal separation is perpendicular to the vertical line of action of the weight.

    Working

    d = (50-40) cm = 10 cm = 0.10 m

  3. Calculate the moment and direction

    Reason

    A downward force to the right of the support turns the rule clockwise.

    Working

    M = Fd = (1.2)(0.10) = 0.12 N m clockwise

7. Mind Stretchers

Mind stretcher 1: A box is pushed slowly up a slope and starts to topple.Extension

Explain, in terms of the centre of gravity, why it topples at a certain angle.

Show Answer

As the slope angle increases, the vertical line through the centre of gravity shifts relative to the base.

At the tipping point, the line of action of the weight passes through the edge of the base. Beyond this, the line of action falls outside the base, so there is a turning effect that makes the box topple.

Mind stretcher 2: Centre of gravity outside the objectExtension

A uniform ring (a thin circular hoop) is held at rest. Students find it strange that its centre of gravity is “in the empty space”.

Explain why this is still correct.

Show Answer

For a uniform ring, the mass is symmetrically distributed around the centre.

So all parts balance, and the weight of the whole ring can be treated as acting at the geometric centre, even though there is no material there.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027