A thin solid disk of mass 1 kg is rotating along its diameter axis at the speed of 1800 rpm. An external torque of 25π Nm is applied for 40 seconds, increasing the speed to 2100 rpm. Find the diameter of the disk.

 

⚙️ A Thin Rotating Disk – Torque & Diameter Problem | Doubtify JEE

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📌 Question:

A thin solid disk of mass 1 kg is rotating along its diameter axis at the speed of 1800 rpm.
An external torque of 25π Nm is applied for 40 seconds, increasing the speed to 2100 rpm.
Find the diameter of the disk.

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📸 Question Image:

🧠 Concept & Approach:

We’ll use the following:

• Moment of Inertia of a solid disk about its diameter:

  $$I = \frac{1}{2} m R^2 = \frac{1}{8} m d^2$$

• Angular velocity:

  $$\omega = \frac{2\pi N}{60}$$

• Torque = I × angular acceleration (α)

  $$\tau = I \cdot \alpha \quad \Rightarrow \quad \alpha = \frac{\omega_2 - \omega_1}{t}$$

🔢 Step-by-Step Solution:

✅ Step 1: Convert RPM to rad/s

$$\omega_1 = \frac{2\pi \cdot 1800}{60} = 60\pi \, \text{rad/s} \omega_2 = \frac{2\pi \cdot 2100}{60} = 70\pi \, \text{rad/s}$$

✅ Step 2: Find angular acceleration

$$\alpha = \frac{70\pi - 60\pi}{40} = \frac{10\pi}{40} = \frac{\pi}{4} \, \text{rad/s²}$$

✅ Step 3: Use τ = I·α

$$\tau = \frac{1}{8} m d^2 \cdot \alpha \Rightarrow 25\pi = \frac{1}{8} \cdot 1 \cdot d^2 \cdot \frac{\pi}{4}$$

✅ Step 4: Solve for diameter d

$$25\pi =\frac{\pi d^2}{32}\Rightarrow d^2=25\cdot 32=800\Rightarrow d=\sqrt{800}\approx 28.28\text{cm}=0.2828\text{m}$$

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✅ Final Answer:

Diameter of the disk = 0.2828 meters

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🖼️ Solution Image:

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🎥 Watch the Step-by-Step Video Solution:

🧠 Pro Tip:

In questions involving rotation + torque, always check for:

• Correct moment of inertia

• Conversion from RPM to rad/s

• Consistent unit usage (Nm, seconds, kg, meters)

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