Centroid of Circular Disc with Hole

❓ Question

A circular hole of radius

$$\frac{a}{2}$$

is cut out of a circular disc of radius

$$a$$

as shown in the figure.

Find the centroid of the remaining circular portion with respect to point $O$.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Centre of mass of composite bodies
👉 Negative mass method
👉 Symmetry of circular lamina.

Main idea:

Hole ko negative mass treat karte hain.

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🔷 Step 1 — Coordinates of Centres 💯

Large disc:

Radius:

$$a$$

Centre at:

$$(a,0)$$

Small removed disc:

Radius:

$$\frac{a}{2}$$

Its centre lies at:

$$\left(\frac{3a}{2},0\right)$$

because it touches outer boundary internally.

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🔷 Step 2 — Areas of Discs

Area of large disc:

$$A_1=\pi a^2$$

Area of removed disc:

$$A_2=\pi\left(\frac a2\right)^2 =\frac{\pi a^2}{4}$$

Remaining area:

$$A=A_1-A_2$$
$$=\pi a^2-\frac{\pi a^2}{4}$$
$$=\frac{3\pi a^2}{4}$$

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🔷 Step 3 — Use Centroid Formula

For composite bodies:

$$\bar x= \frac{A_1x_1-A_2x_2}{A_1-A_2}$$

Substitute values:

$$\bar x= \frac{ \pi a^2(a) - \frac{\pi a^2}{4}\left(\frac{3a}{2}\right) }{ \frac{3\pi a^2}{4} }$$

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🔷 Step 4 — Simplify

Numerator:

$$\pi a^3-\frac{3\pi a^3}{8}$$
$$=\frac{5\pi a^3}{8}$$

Now:

$$\bar x= \frac{5\pi a^3/8}{3\pi a^2/4}$$
$$=\frac{5a}{6}$$

Since figure is symmetric about x-axis:

$$\bar y=0$$

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🔷 Step 5 — Final Centroid

$$\boxed{\left(\frac{5a}{6},0\right)}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Hole ko positive area treat kar dena

❌ Small circle centre wrong lena

❌ Symmetry use na karna

Remember:

Removed portion:

$$\text{Negative mass/negative area}$$

always use hota hai.

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✅ Final Answer

$$\boxed{\left(\frac{5a}{6},0\right)}$$

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📚 Related Topics

• MOI of Combined Bodies Using Parallel Axis
• Moment of Inertia of System Using Parallel Axis
• Find Diameter of Rotating Disc Using Torque and RPM