Find Diameter of Rotating Disc Using Torque and RPM

❓ Question

A thin solid disk of:

$$1\text{ kg}$$

is rotating along its diameter axis at the speed of:

$$1800\text{ rpm}$$

By applying an external torque of:

$$25\pi\text{ Nm}$$

for:

$$40\text{ s}$$

the speed increases to:

$$2100\text{ rpm}$$

The diameter of the disk is ______ m.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Angular acceleration
👉 Torque-angular momentum relation
👉 Moment of inertia of disk.

Main idea:

$$\tau=I\alpha$$

and for solid disk about diameter:

$$I=\frac14 MR^2$$

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🔷 Step 1 — Convert Angular Speeds 💯

Initial speed:

$$1800\text{ rpm}$$
$$\omega_1 = 1800\times\frac{2\pi}{60}$$
$$=60\pi\text{ rad/s}$$

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Final speed:

$$2100\text{ rpm}$$
$$\omega_2 = 2100\times\frac{2\pi}{60}$$
$$=70\pi\text{ rad/s}$$

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🔷 Step 2 — Find Angular Acceleration

Using:

$$\alpha=\frac{\omega_2-\omega_1}{t}$$
$$= \frac{70\pi-60\pi}{40}$$
$$= \frac{10\pi}{40}$$
$$= \frac{\pi}{4}\text{ rad/s}^2$$

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🔷 Step 3 — Apply Torque Relation

Given torque:

$$\tau=25\pi$$

Using:

$$\tau=I\alpha$$
$$25\pi = I\left(\frac{\pi}{4}\right)$$
$$I=100\text{ kg m}^2$$

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🔷 Step 4 — Use MOI of Disk About Diameter

For solid disk about diameter:

$$I=\frac14 MR^2$$

Given:

$$M=1\text{ kg}$$

So:

$$100=\frac14(1)R^2$$
$$R^2=400$$
$$R=20\text{ m}$$

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🔷 Step 5 — Find Diameter

$$D=2R$$
$$=40\text{ m}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ rpm to rad/s conversion bhool jaana

❌ Diameter axis ke liye wrong MOI use kar lena

❌ Radius aur diameter confuse kar dena

Remember:

For disk about diameter:

$$\boxed{ I=\frac14 MR^2 }$$

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✅ Final Answer

$$\boxed{40\text{ m}}$$

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