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Motion, Forces and Energy - Centre of gravity

Grade 12IGCSEPhysics

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

🔑Concepts

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The Centre of Gravity (COG) of an object is the point through which the entire weight of the object appears to act for any orientation of the object.

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For a uniform gravitational field, the Centre of Gravity coincides with the Centre of Mass.

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For regular, symmetrical objects of uniform density (like a sphere, cube, or circular disc), the COG is located at the geometric centre.

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The Plumb Line Method is used to find the COG of an irregular 2D shape (lamina). The shape is hung from different points, and the intersection of vertical lines drawn from the points of suspension identifies the COG.

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Stability refers to an object's ability to return to its original position after being disturbed. It depends on the position of the COG and the width of the base.

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Stable Equilibrium: When tilted slightly, the COG rises, and the weight creates a restoring moment that returns the object to its original position.

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Unstable Equilibrium: When tilted slightly, the COG falls, and the weight creates a turning moment that causes the object to topple over.

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Neutral Equilibrium: When moved, the height of the COG remains constant, and the object stays in its new position.

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An object will topple if the vertical line acting downwards from its COG falls outside the edge of its base of support.

📐Formulae

W=mgW = mg

Moment=F×d⊥\text{Moment} = F \times d_{\perp}

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

xˉ=∑(wixi)∑wi\bar{x} = \frac{\sum (w_i x_i)}{\sum w_i}

💡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 by a pivot at its center. A 5.0 kg5.0\text{ kg} mass is placed at a distance of 1.5 m1.5\text{ m} from the left end. Where must a 10 kg10\text{ kg} mass be placed to maintain equilibrium?

Solution:

  1. Identify the COG of the plank: Since it is uniform, the COG is at the center (2.0 m2.0\text{ m} from either end). The weight of the plank acts through the pivot, so its moment is zero.
  2. Calculate moments about the pivot (center):
    • Left mass (5.0 kg5.0\text{ kg}): distance from pivot d1=2.0 m−1.5 m=0.5 md_1 = 2.0\text{ m} - 1.5\text{ m} = 0.5\text{ m}.
    • Anticlockwise Moment Macw=m1g×d1=5.0×9.8×0.5=24.5 NmM_{acw} = m_1 g \times d_1 = 5.0 \times 9.8 \times 0.5 = 24.5\text{ Nm}.
  3. Set Mcw=MacwM_{cw} = M_{acw} for the 10 kg10\text{ kg} mass at distance xx from pivot:
    • 10×9.8×x=24.510 \times 9.8 \times x = 24.5
    • x=24.598=0.25 mx = \frac{24.5}{98} = 0.25\text{ m}.
  4. Position: 0.25 m0.25\text{ m} to the right of the pivot (or 2.25 m2.25\text{ m} from the left end).

Explanation:

To balance the plank, the sum of clockwise moments must equal the sum of anticlockwise moments about the pivot. Since the plank is uniform, its weight does not contribute to the moment when pivoted at the center.

Problem 2:

A bus is being tested for stability. The COG of the bus is 1.2 m1.2\text{ m} above the ground, and the width between the wheels is 2.4 m2.4\text{ m}. Calculate the maximum angle θ\theta the bus can be tilted before it topples.

Solution:

  1. The bus topples when the vertical line from the COG passes through the edge of the wheel base.
  2. Form a right-angled triangle where the opposite side is half the wheel base width: 1.2 m1.2\text{ m} and the adjacent side is the height of the COG: 1.2 m1.2\text{ m}.
  3. Using trigonometry: tan⁡(θ)=half-widthheight of COG\tan(\theta) = \frac{\text{half-width}}{\text{height of COG}}.
  4. tan⁡(θ)=1.21.2=1\tan(\theta) = \frac{1.2}{1.2} = 1.
  5. θ=arctan⁡(1)=45∘\theta = \arctan(1) = 45^{\circ}.

Explanation:

Stability is maintained as long as the weight vector remains within the base of support. The critical angle is reached when the vector aligns exactly with the corner of the base.

Centre of gravity - Revision Notes & Key Formulas | IGCSE Grade 12 Physics