Centre of gravity

1. Overview

The centre of gravity is a fundamental concept in mechanics that simplifies how we look at forces. By identifying the single point where an object's weight acts, we can predict how objects balance, rotate, and stay stable in everyday situations.

Key Definitions

• Centre of Gravity (CG): The point through which the entire weight of an object acts.
• Lamina: A flat, two-dimensional shape (such as a piece of cardboard or a metal plate).
• Plumb Line: A string with a weight attached to the end used to find a perfectly vertical line.
• Stability: A measure of an object's ability to maintain its balance and resist toppling over.

Core Content

Determining the Centre of Gravity

For regular, uniform objects (like a square or a circle), the centre of gravity is at the geometric centre. However, for an irregularly shaped plane lamina, you must use the following experiment:

Experiment: Finding the CG of an Irregular Lamina

1. Make three small holes near the edges of the lamina.
2. Suspend the lamina from a pin through one of the holes so that it can swing freely.
3. Hang a plumb line from the same pin in front of the lamina.
4. Once the lamina and plumb line stop moving, draw a line on the lamina following the path of the plumb line string.
5. Repeat this process using the other two holes.
6. The point where all three lines intersect is the centre of gravity.

Stability of Simple Objects

Stability describes how easy it is to knock an object over. Whether an object topples depends on the position of its centre of gravity relative to its base.

• Two factors for high stability:
  1. A wide base: The larger the area the object sits on, the harder it is to tip.
  2. A low centre of gravity: The closer the CG is to the ground, the more stable the object.
• When does an object topple?
  • An object is stable as long as the vertical line acting downwards from its centre of gravity falls inside its base.
  • If the object is tilted so far that the line of action of its weight falls outside the edge of the base, the weight creates a moment (turning effect) that causes the object to topple over.

Extended Content (Extended Only)

There is no additional supplement-specific content for this sub-topic in the current IGCSE Physics syllabus.

Key Equations

While this topic is primarily conceptual, it is closely linked to the Weight equation: $$W = m \times g$$

• W: Weight (Newtons, N) - this is the force that acts through the Centre of Gravity.
• m: Mass (Kilograms, kg)
• g: Acceleration due to gravity (approximately $9.8\text{ m/s}^2$ on Earth)

Common Mistakes to Avoid

• ❌ Wrong: Assuming the centre of gravity is always in the exact middle of an object.
• ✓ Right: The CG is only in the geometric centre if the object is uniform (even density throughout) and has a regular, symmetrical shape.
• ❌ Wrong: Saying an object is more stable just because it has a wide base.
• ✓ Right: You must consider both the base area and the height of the centre of gravity. A wide object with a very high CG can still be unstable.
• ❌ Wrong: Thinking any force applied to an object causes it to rotate.
• ✓ Right: A force only causes rotation if its line of action does not pass through the pivot (it needs a "lever arm").

Exam Tips

1. Experiment Description: When describing the experiment for an irregular lamina, always mention using three holes. This is for accuracy—two lines will intersect at a point, but the third line confirms that the point is correct.
2. Explain the "Why": In questions about toppling, use the phrase: "The line of action of the weight falls outside the base, creating a turning effect (moment)." This specific phrasing often earns the highest marks.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student is given an irregularly shaped piece of cardboard and is asked to determine its centre of gravity.

(a) Define what is meant by the centre of gravity of an object. [1]

(b) Describe an experiment that the student could perform to determine the position of the centre of gravity of the cardboard. [4]

Worked Solution:

(a)

1. The centre of gravity is the point where the entire weight of the object appears to act.

How to earn full marks:

• Mention the point where the weight APPEARS to act.

(b)

1. Suspend the cardboard from a point near its edge, using a pin or clamp.
2. Hang a plumb line from the same point of suspension.
3. Draw a line on the cardboard along the plumb line's direction.
4. Repeat the process by suspending the cardboard from a different point on the cardboard.
5. The centre of gravity is located where the lines intersect.

How to earn full marks:

• Clearly describe suspending the cardboard and using a plumb line.
• Explain that the procedure must be repeated from at least two different points.
• State that the centre of gravity is at the intersection of the lines.

Common Pitfall: Many students forget to mention that the experiment needs to be repeated from multiple points on the cardboard. Suspending the cardboard from only one point and drawing a single line will not locate the centre of gravity. Also, be sure to state that the plumb line gives a vertical reference.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A delivery van is loaded with boxes. The diagram shows the van viewed from the side. The centre of gravity of the empty van is at point X. When the van is fully loaded with boxes of uniform density, the centre of gravity is at point Y.

(a) State what happens to the stability of the van as the centre of gravity moves from X to Y. Explain your answer. [3]

(b) The van is driven around a corner. Explain why a van with a higher centre of gravity (like at point Y) is more likely to tip over than a van with a lower centre of gravity (like at point X). [3]

(c) Suggest one way that the van driver could reduce the risk of the van tipping over when fully loaded. Explain your answer. [2]

Worked Solution:

(a)

1. The stability of the van decreases.
2. A higher centre of gravity means that a smaller angle of tilt is needed for the centre of gravity to be directly above the pivot point (the wheels) causing the van to tip.
3. Therefore, the van is less stable and more prone to tipping.

How to earn full marks:

• Correctly state the change in stability (decreases).
• Relate the higher centre of gravity to a smaller tilting angle before tipping.
• Explicitly state that the van is more prone to tipping.

(b)

1. When the van turns, there is a resultant force acting inwards, towards the centre of the corner (the centripetal force), provided by friction from the road on the tyres.
2. This force acts at road level, below the centre of gravity, so it creates a moment that tends to tip the van outwards over the wheels on the outside of the turn.
3. A higher centre of gravity increases the perpendicular distance between the line of action of this inward force (at road level) and the centre of gravity, increasing the moment. A larger moment makes the van more likely to tip.

How to earn full marks:

• Mention the resultant inward (centripetal) force from the road acting during the turn.
• Explain that the force acts below the centre of gravity and creates a moment that tends to tip the van outwards.
• Explain how a higher centre of gravity increases the moment, making tipping more likely.

(c)

1. Drive more slowly around the corners.
2. Driving more slowly reduces the centripetal force needed to turn the van, which reduces the tipping moment, decreasing the likelihood of tipping.

How to earn full marks:

• Suggest a reasonable action (e.g., drive slower).
• Explain how this action reduces the risk of tipping (e.g., reduces the force needed to turn, reduces moment).

Common Pitfall: Many students forget to explicitly state that the van becomes less stable when the centre of gravity is raised. Also, when discussing the turning van, be sure to mention the inward (centripetal) force from the road and its effect on creating a moment. Don't just say "force" without specifying its direction.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

A uniform rectangular block of wood has dimensions 0.20 m x 0.10 m x 0.05 m and a weight of 5.0 N.

(a) State the location of the centre of gravity of the block. [1]

(b) The block is placed on a table with the 0.20 m x 0.10 m face in contact with the table. Calculate the minimum horizontal force that must be applied at the top edge of the block to start to tip it over. [3]

(c) Explain why the block is more stable when placed on the table with its 0.20 m x 0.10 m face in contact than when placed with its 0.05 m x 0.10 m face in contact. [2]

Worked Solution:

(a)

1. The centre of gravity is at the geometric centre of the block, since it is uniform and has a regular shape.

How to earn full marks:

• State that the centre of gravity is at the geometric centre.

(b)

1. The tipping point is the edge of the block in contact with the table.
2. The moment due to the applied force must equal the moment due to the weight when the block is about to tip.
3. The block tips most easily over a long edge, so the perpendicular distance from the line of action of the weight to the pivot is 0.10 m / 2 = 0.05 m; the top edge is 0.05 m above the table.
4. Moment due to weight = $5.0 \text{ N} \times 0.05 \text{ m} = 0.25 \text{ Nm}$
5. Moment due to force = $F \times 0.05 \text{ m} = 0.25 \text{ Nm}$
6. $F = 0.25 \text{ Nm} / 0.05 \text{ m} = 5.0 \text{ N}$
   $\boxed{F = 5.0 \text{ N}}$

How to earn full marks:

• Correctly calculate the moment due to the weight.
• Correctly equate the moments due to the weight and the applied force.
• Correctly calculate the applied force with units.

(c)

1. When the block is placed on the larger face, it has a wider base.
2. This means that the centre of gravity can be displaced further horizontally before it is directly above the edge of the base, at which point it will tip. A wider base provides greater stability.

How to earn full marks:

• Relate the larger face to a wider base.
• Explain that the wider base allows for a greater displacement of the centre of gravity before tipping.

Common Pitfall: In part (b), students often forget to calculate the correct perpendicular distance (lever arm) for the weight's moment. Also, remember that stability depends on both the width of the base and the height of the centre of gravity. Don't just focus on one.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A crane is lifting a large concrete block. The diagram shows the crane with the block suspended.

(a) State what is meant by the centre of gravity of the concrete block. [1]

(b) Explain why it is important for the crane to have a counterweight. [3]

(c) The concrete block has a weight of 15 000 N. The distance from the pivot to the point where the cable is attached to the crane arm is 8.0 m. The counterweight has a weight of 25 000 N. Calculate the minimum distance, $d$, from the pivot to the centre of gravity of the counterweight needed to prevent the crane from tipping. [3]

(d) Suggest one reason why, in reality, the distance $d$ would be greater than the value calculated in (c). [2]

Worked Solution:

(a)

1. The centre of gravity is the point where the entire weight of the block appears to act.

How to earn full marks:

• Mention the point where the weight APPEARS to act.

(b)

1. The concrete block creates a clockwise moment about the pivot.
2. The counterweight provides an anticlockwise moment about the pivot.
3. The counterweight is needed to balance the moment created by the concrete block, preventing the crane from tipping and ensuring stability.

How to earn full marks:

• Explain that the block creates a clockwise moment.
• Explain that the counterweight creates an anticlockwise moment.
• Explain that the counterweight balances the moment, preventing tipping and ensuring stability.

(c)

1. For the crane to be stable, the anticlockwise moment due to the counterweight must be equal to or greater than the clockwise moment due to the block.
2. Clockwise moment = $15 000 \text{ N} \times 8.0 \text{ m} = 120 000 \text{ Nm}$
3. Anticlockwise moment = $25 000 \text{ N} \times d = 120 000 \text{ Nm}$
4. $d = 120 000 \text{ Nm} / 25 000 \text{ N} = 4.8 \text{ m}$
   $\boxed{d = 4.8 \text{ m}}$

How to earn full marks:

• Correctly calculate the clockwise moment due to the block.
• Correctly equate the clockwise and anticlockwise moments.
• Correctly calculate the distance $d$ with units.

(d)

1. The weight of the crane arm itself also creates a clockwise moment about the pivot.
2. Therefore, the counterweight needs to provide a larger anticlockwise moment to balance both the block and the crane arm, requiring a greater distance $d$ to provide a sufficient counter-moment.

How to earn full marks:

• Suggest a valid reason (e.g., weight of the crane arm).
• Explain how this factor increases the required distance $d$.

Common Pitfall: When explaining the purpose of the counterweight, don't just say it "balances the crane." Be specific about how it creates an anticlockwise moment to counteract the clockwise moment of the load. Also, remember to include units in your final answer for part (c).