A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25π Nm for 40 seconds, the speed increases to 2100 rpm. What is the diameter of the disk?
🌀 Rotating Disk Problem – Explained with Concept & Calculation
❓ Problem Statement:
A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25π Nm for 40 seconds, the speed increases to 2100 rpm. What is the diameter of the disk?
⚙️ Step-by-Step Conceptual Breakdown
To solve this, we’ll use rotational motion concepts:
🧮 Key Equations Involved:
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Torque and Angular Acceleration:
Where:
= torque,
= moment of inertia of disk = ,
= angular acceleration -
Angular Velocity:
Convert RPM to rad/s: -
Angular Kinematics:
🧾 Given:
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Mass of disk,
-
Initial speed,
-
Final speed,
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Time,
-
Torque,
🔍 Step 1: Find Angular Acceleration
🔍 Step 2: Apply Torque Equation
Substitute :
Cancel from both sides:
✅ Final Step: Find Diameter
🎯 Final Answer:
Diameter = 20√2 ≈ 28.28 meters
🧠 Physics Tip for JEE Aspirants:
-
Always convert RPM to rad/s when working with angular quantities.
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Use moment of inertia formulas carefully based on object shape:
For solid disk →
For ring → -
Torque = I × α is rotational equivalent of F = ma. Think of angular acceleration as rotational version of linear acceleration.
📈 Why This Concept is Important?
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Frequently appears in JEE Advanced and NEET Physics.
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Tests knowledge of rotational motion, angular kinematics, and units.
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Strengthens understanding of real-world problems like flywheels, engines, etc.
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