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Centre Of Gravity (A Level)

Key idea: Use centre of gravity as the point where weight acts; optional sections extend to centre-of-mass calculations, laminas and stability.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply moments, couples and force-and-torque equilibrium using free-body diagrams and vector triangles.
  • show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity
  • apply the principle of moments to new situations or to solve related problems
  • Explain inertia and momentum, then apply Newton's laws using free-body diagrams.

1. Definitions (Must Know)

A. Centre of gravity (CoG)

The centre of gravity is the point through which the entire weight of an object may be considered to act.

In a uniform gravitational field, the centre of gravity coincides with the centre of mass.

Core syllabus scope

For H2 Physics 9478, you must understand that a body’s weight may be taken to act at one point called its centre of gravity. Calculating centres of mass for particles or composite laminas, and detailed toppling geometry, are useful extensions rather than stated outcomes.

B. Centre of mass (CoM)

The centre of mass is the point that represents the average position of the mass distribution of a body.

C. Stability (idea)

Stability describes what happens after a small displacement:

  • stable equilibrium: returns to its original position
  • unstable equilibrium: moves further away
  • neutral equilibrium: stays in its new position

2. Key ideas

  • Centre of mass (discrete masses):
    • x_CM = (∑ mᵢ xᵢ)/(∑ mᵢ), y_CM = (∑ mᵢ yᵢ)/(∑ mᵢ)
  • For a uniform lamina (same thickness and density), mass ∝ area, so you can use area-weighting:
    • x bar = (∑ Aᵢ xᵢ)/(∑ Aᵢ), y bar = (∑ Aᵢ yᵢ)/(∑ Aᵢ)
  • Treat a hole/cut-out as a negative area (or negative mass).
  • An object topples when the line of action of its weight falls outside its base of support.
  • Lower CoG and wider base generally increase stability.

3. Detailed explanations

A. Centre of mass vs centre of gravity

  • Centre of mass depends only on the mass distribution.
  • Centre of gravity depends on the mass distribution and the gravitational field.

For most mechanics questions near Earth’s surface, we assume a uniform gravitational field, so the centre of gravity is at the same point as the centre of mass.

B. Optional extension: calculating a composite centre

If an object can be split into parts with known centres (e.g. rectangles, discs), you can find the overall centre using a weighted average.

Workflow:

  1. Choose axes and an origin.
  2. Split the object into parts and find each part’s centre coordinates (xᵢ, yᵢ).
  3. Use mass-weighting (or area-weighting for a uniform lamina) to find (x bar, y bar).
  4. For holes, use negative mass/area.

C. Optional extension: stability and toppling

An object is stable if, after a small tilt, there is a restoring moment that tends to return it to its original position.

A practical rule used in exam questions:

  • draw the line of action of the weight (vertical line through the CoG),
  • the object is stable as long as that line falls within the base of support,
  • it topples once the line falls outside the base.

4. Common Mistakes

  • Mixing up CoG (depends on the field) and CoM (depends only on mass distribution).
  • Forgetting a hole is treated as negative mass/area in composite-shape questions.
  • Using inconsistent axes/origin when combining parts.
  • Assuming “low CoG” alone guarantees stability (base width matters too).

5. Exam Tips

  • Use symmetry first (it often gives one coordinate immediately).
  • Put the origin at a convenient corner/edge to reduce arithmetic.
  • Keep work organised in a table: part, mass/area, (xᵢ,yᵢ), and mᵢxᵢ / Aᵢxᵢ.
  • For stability questions, sketch the base and the weight line of action.

6. Worked Examples

Modelled example 1

Choosing the weight’s line of action

Core

Problem

A uniform horizontal beam is supported at one end. Where should its weight be drawn on the free-body diagram?
Study the worked solution

  1. Replace the distributed weight

    Method

    Draw one downward force W = mg through the beam’s centre of gravity.

    Reason

    In a uniform gravitational field, the whole body’s weight may be represented as acting at its centre of gravity.

    Working

    W = mg vertically downward through the CoG.

  2. Use symmetry

    Method

    Place the force at the beam’s midpoint.

    Reason

    A uniform beam has a symmetric mass distribution, so its centre of gravity lies halfway along it.

    Working

    x_CoG = L/2 from either end.

Optional calculation practice

Guided practice 2

Two particles on a line

About 4 min

Optional extension

Two particles have masses 2.0 kg and 3.0 kg at x = 0 and x = 5.0 m respectively. Find the centre of mass.

Try this before viewing the solution

Unit: m

Hints

Hint 1: set up the weighted mean
Use x_CM = (m₁x₁ + m₂x₂)/(m₁ + m₂).
View solution step by step

  1. Form the mass-weighted sum

    Method

    The position moment is 15 kg m.

    Reason

    The 2.0 kg particle is at the origin, so only the second particle contributes.

    Working

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

  2. Divide by total mass

    Method

    x_CM = 3.0 m.

    Reason

    The centre-of-mass coordinate is the position moment per unit total mass.

    Working

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

Common misconception 3

Three particles on a line

Find and correct the mistake

Learner claim

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 obtains 2.33 m. Diagnose the method and find the centre of mass.

Try this before viewing the solution

Unit: m

View solution step by step

  1. Identify the faulty assumption

    Method

    The three positions must not be weighted equally.

    Reason

    Each particle’s contribution depends on both its mass and its position.

    Working

    x_CM ≠ (0 + 2.0 + 5.0)/3 for unequal masses.

  2. Use mass weighting

    Method

    x_CM = 3.17 m.

    Reason

    The largest mass at 5.0 m pulls the weighted mean toward that end.

    Working

    x_CM = ((1.0)(0) + (2.0)(2.0) + (3.0)(5.0))/(1.0 + 2.0 + 3.0) = 19/6 = 3.17 m

Examiner practice 4

Centre of mass in 2D (point masses)

5 marks

Optional examination-style extension

Point masses of 2.0 kg, 1.0 kg and 3.0 kg are at (0,0), (4.0,0) and (0,2.0) respectively. Find (x_CM,y_CM). [5 marks]

Try this before viewing the solution

Unit: coordinate units
Unit: coordinate units

View solution step by step

  1. Find the total mass

    1 mark

    Method

    M = 6.0 kg.

    Reason

    Both coordinate calculations use the same total mass.

    Working

    M = 2.0 + 1.0 + 3.0 = 6.0 kg

  2. Calculate the x coordinate

    2 marks

    Method

    x_CM = 0.667 coordinate units.

    Reason

    Only the 1.0 kg mass has a non-zero x coordinate.

    Working

    x_CM = (2.0(0) + 1.0(4.0) + 3.0(0))/6.0 = 0.667

  3. Calculate the y coordinate

    2 marks

    Method

    y_CM = 1.00 coordinate units.

    Reason

    Only the 3.0 kg mass has a non-zero y coordinate.

    Working

    y_CM = (2.0(0) + 1.0(0) + 3.0(2.0))/6.0 = 1.00

Challenge 5

Lamina with a square hole (negative area)

Minimal support

Optional extension

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 centroid (x bar,y bar) of the remaining shape.

Try this before viewing the solution

Unit: cm
Unit: cm

Hints

Hint 1: represent the cut-out
Use areas + 24 cm² at (3.0,2.0) and -4.0 cm² at (5.0,3.0).
View solution step by step

  1. Represent the hole as negative area

    Method

    The remaining area is 20 cm².

    Reason

    Removing material subtracts both area and its area moments.

    Working

    A = 24-4 = 20 cm²

  2. Calculate the horizontal coordinate

    Method

    x bar = 2.6 cm.

    Reason

    The removed top-right area shifts the centroid left.

    Working

    x bar = (24(3.0)-4(5.0))/20 = 2.6 cm

  3. Calculate the vertical coordinate

    Method

    y bar = 1.8 cm.

    Reason

    The removed top-right area also shifts the centroid downward.

    Working

    y bar = (24(2.0)-4(3.0))/20 = 1.8 cm

Challenge 6

Toppling angle (stability rule)

Minimal support

Independent transfer

A uniform rectangular block has width 0.40 m and height 0.80 m. It is slowly tilted about one bottom edge. Estimate the tilt angle at which it is just about to topple, assuming its centre of mass is at the geometric centre.

Try this before viewing the solution

Unit: degrees

Hints

Hint 1: draw the limiting triangle
Relative to the pivot, the centre is half the width inward and half the height upward.
View solution step by step

  1. State the limiting condition

    Method

    The vertical line through the centre of mass passes through the pivot edge.

    Reason

    Beyond this position, the weight produces a moment that continues the toppling motion.

    Working

    weight line of action through pivot

  2. Use the centre geometry

    Method

    The horizontal and vertical offsets are 0.20 m and 0.40 m.

    Reason

    A uniform rectangle’s centre is halfway across its width and height.

    Working

    w/2 = 0.20 m, h/2 = 0.40 m

  3. Calculate the critical angle

    Method

    θ ≈ 26.6°.

    Reason

    The limiting tilt rotates the centre until its horizontal offset from the pivot is zero.

    Working

    tan θ = 0.20/0.40 = 0.5 ⇒ θ = 26.6°

7. Mind Stretchers

Mind stretcher 1: Does CoG always equal CoM?Extension

Explain why centre of gravity and centre of mass can differ in a non-uniform gravitational field (e.g. very far from Earth).

Click here to show/hide answer

Centre of mass depends only on how mass is distributed.

Centre of gravity depends on how gravitational field strength varies across the object. In a non-uniform field, different parts of the object experience different gravitational forces, so the single point where you can treat the weight as acting (CoG) may shift relative to the CoM.

Mind stretcher 2: Optional (Enrichment)Extension

A. Continuous mass distributions

For continuous mass distributions, the centre of mass can be found using integration.

This is beyond what most A Level exam questions require; composite shapes + symmetry is usually enough.

Mind stretcher 3: Why does lowering the centre of gravity increase stability?Extension

Explain why a lower CoG makes an object harder to topple, even if its base width stays the same.

Click here to show/hide answer

An object topples when the line of action of its weight falls outside the base.

With a lower CoG, you need a larger tilt angle before the vertical line through the CoG reaches the edge of the base. That means it is harder to reach the toppling condition, so stability increases.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027