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Motion, Forces and Energy - Centre of gravity

Grade 11IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Centre of Gravity (CG) is defined as the point through which the entire weight of an object acts, regardless of its orientation.

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For a uniform object (an object with even density and a regular shape), the centre of gravity is located at its geometric centre.

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An object's stability depends on the position of its centre of gravity and the area of its base. An object is more stable if it has a low centre of gravity and a wide base.

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An object will topple over if the vertical line through its centre of gravity falls outside the edges of its base. This creates a resultant moment that causes rotation.

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There are three states of equilibrium: 1. Stable equilibrium: After a small displacement, the CG rises and the object returns to its original position. 2. Unstable equilibrium: After a small displacement, the CG falls and the object moves further away from its original position. 3. Neutral equilibrium: After a small displacement, the height of the CG remains constant.

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The centre of gravity of an irregular lamina can be found experimentally using a plumb line. The lamina is suspended from at least two different points, and the intersection of the vertical lines marked is the CG.

📐Formulae

W=mimesgW = m imes g

Moment=FimesdMoment = F imes d

∑Clockwise Moments=∑Anticlockwise Moments\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}

d=perpendicular distance from the pivotd = \text{perpendicular distance from the pivot}

💡Examples

Problem 1:

A uniform wooden plank of length L=4.0 mL = 4.0\text{ m} and mass M=20 kgM = 20\text{ kg} is supported at its centre. If a 5.0 kg5.0\text{ kg} mass is placed 1.2 m1.2\text{ m} from the left end, where must a 10.0 kg10.0\text{ kg} mass be placed to maintain horizontal equilibrium?

Solution:

  1. Identify the pivot at the centre (2.0 m2.0\text{ m} from either end).
  2. The 5.0 kg5.0\text{ kg} mass is 2.0 m−1.2 m=0.8 m2.0\text{ m} - 1.2\text{ m} = 0.8\text{ m} from the pivot on the left side.
  3. Calculate the anticlockwise moment: Macw=(5.0 kg×10 m/s2)×0.8 m=40 NmM_{acw} = (5.0\text{ kg} \times 10\text{ m/s}^2) \times 0.8\text{ m} = 40\text{ Nm}.
  4. Set the clockwise moment equal to the anticlockwise moment: 40 Nm=(10.0 kg×10 m/s2)×d40\text{ Nm} = (10.0\text{ kg} \times 10\text{ m/s}^2) \times d.
  5. 40=100×d⇒d=40100=0.4 m40 = 100 \times d \Rightarrow d = \frac{40}{100} = 0.4\text{ m}.
  6. The 10.0 kg10.0\text{ kg} mass must be placed 0.4 m0.4\text{ m} from the pivot on the right side.

Explanation:

Since the plank is uniform, its weight acts through the pivot (the centre), so it exerts no moment. We apply the Principle of Moments about the pivot to balance the system.

Problem 2:

A racing car is designed with a very low chassis and a wide wheel track. Explain why this improves performance in terms of the centre of gravity.

Solution:

The low chassis ensures that the centre of gravity (CGCG) is as low as possible. The wide wheel track increases the base area. During high-speed cornering, the vertical line from the CGCG is less likely to fall outside the base area, preventing the car from toppling (rolling over).

Explanation:

Stability is maximized by minimizing the height of the CGCG and maximizing the distance the CGCG must travel laterally before it moves outside the support base.

Centre of gravity - Revision Notes & Key Formulas | IGCSE Grade 11 Physics