Find Diameter of Rotating Disk Using Torque and RPM

❓ Question

A thin solid disk of:

$$10\text{ kg}$$

is rotating along its diameter axis at the speed of:

$$1800\text{ rpm}$$

By applying an external torque of:

$$25\text{ N m}$$

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:

👉 Rotational dynamics
👉 Angular acceleration
👉 Torque and moment of inertia.

Main idea:

$$\tau=I\alpha$$

and

$$\alpha=\frac{\omega_f-\omega_i}{t}$$

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🔷 Step 1 — Convert RPM into rad/s 💯

Initial angular speed:

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

Final angular speed:

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

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🔷 Step 2 — Find Angular Acceleration

Using:

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

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🔷 Step 3 — Apply Torque Equation

Given torque:

$$\tau=25\text{ N m}$$

Using:

$$\tau=I\alpha$$
$$25=I\left(\frac{\pi}{4}\right)$$
$$I=\frac{100}{\pi}$$

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🔷 Step 4 — Use MOI of Thin Solid Disk

For a thin solid disk about diameter:

$$I=\frac14MR^2$$

Given:

$$M=10\text{ kg}$$

So:

$$\frac14(10)R^2=\frac{100}{\pi}$$
$$\frac52R^2=\frac{100}{\pi}$$
$$R^2=\frac{40}{\pi}$$

Using:

$$\pi\approx\frac{22}{7}$$
$$R^2\approx12.7$$
$$R\approx3.56\text{ m}$$

Diameter:

$$D=2R$$
$$D\approx7.1\text{ m}$$

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🔷 Step 5 — Final Answer

$$\boxed{7.1\text{ m}}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ RPM ko directly rad/s maan lena

❌ Wrong MOI formula use kar lena

❌ Diameter aur radius confuse kar lena

Remember:

For disk about diameter:

$$\boxed{ I=\frac14MR^2 }$$

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✅ Final Answer

$$\boxed{7.1\text{ m}}$$

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