Centroid of Circular Disc with Hole | System of Particles | JEE Physics | Doubtify JEE

🌀 Centroid of a Circular Plate with a Hole | System of Particles | JEE Physics | Doubtify JEE

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❓ Question:

A circular hole of radius (a/2) is cut out of a circular disc of radius ‘a’, as shown in the figure. What will be the position of the centroid of the remaining circular portion with respect to point ‘O’?

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🖼️ Question Image:

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🧠 Solution Image:

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✍️ Detailed Explanation:

To solve this problem, we apply the concept of mass moment (or area moment). When a portion is cut out from a symmetric object, we use the principle of superposition — treat the hole as a negative mass or negative area.

Let’s break it down:

🧮 Step 1: Assume mass per unit area = σ

• Area of full disc = πa²

• Area of hole = π(a/2)² = (πa²)/4

• Remaining area = πa² – (πa²)/4 = (3πa²)/4

🧮 Step 2: Find centroid of each part

• Full disc’s centroid = At origin (O)

• Hole’s centroid is at a distance of a/2 from O along the radius.

🧮 Step 3: Apply formula for center of mass (COM):

Let x be the position of the centroid of remaining portion from O.

$$x = \frac{0 \cdot (\pi a^2) - (a/2) \cdot (\pi a^2/4)}{(\pi a^2 - \pi a^2/4)} = \frac{-(a/2) \cdot (\pi a^2/4)}{(3\pi a^2)/4}$$$$x = \frac{-a}{6}$$

So, centroid is at a distance of a/6 toward the left (i.e., toward center from the original origin).

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✅ Final Answer:

The centroid of the remaining part is located at a distance of a/6 from point O toward the center of the full disc.

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🎥 Video Solution:

🔍 Why this Question is Important:

• It introduces the concept of negative mass (or negative area).

• Useful in understanding COM with composite bodies, a regular in JEE Mains/Advanced.

• Reinforces vector-based calculation skills for centroid and center of mass.

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🧠 Pro Tip:

In problems involving cut-outs, always treat the removed section as negative mass. Apply the center of mass formula as usual — just with sign conventions. Visualization and symmetry are your best friends in such questions.

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