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?

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:

  1. Torque and Angular Acceleration:

    τ=Iα

    Where:
    τ\tau = torque,
    II = moment of inertia of disk = 12MR2\frac{1}{2}MR^2,
    α\alpha = angular acceleration

  2. Angular Velocity:
    Convert RPM to rad/s:

    ω=2πRPM60​
  3. Angular Kinematics:

    ωf=ωi+αt

🧾 Given:

  • Mass of disk, M=1kgM = 1 \, \text{kg}

  • Initial speed, ωi=1800rpm=2π×180060=60πrad/s

  • Final speed, ωf=2100rpm=2π×210060=70πrad/s\omega_f = 2100 \, \text{rpm} = \frac{2\pi \times 2100}{60} = 70\pi \, \text{rad/s}

  • Time, t=40st = 40 \, \text{s}

  • Torque, τ=25πNm\tau = 25\pi \, \text{Nm}


🔍 Step 1: Find Angular Acceleration α\alpha

α=ωfωit=70π60π40=10π40=π4rad/s2

🔍 Step 2: Apply Torque Equation

τ=Iα25π=(12MR2)π4​

Substitute M=1M = 1:

25π=121R2π425π=πR28​

Cancel π\pi from both sides:

25=R28R2=200R=200=102m

Final Step: Find Diameter

Diameter=2R=2102=20228.28m

🎯 Final Answer:

Diameter = 20√2 ≈ 28.28 meters


🧠 Physics Tip for JEE Aspirants:

  • Always convert RPM to rad/s when working with angular quantities.

  • Use moment of inertia formulas carefully based on object shape:
    For solid disk → I=12MR2I = \frac{1}{2}MR^2
    For ring → I=MR2I = MR^2

  • Torque = I × α is rotational equivalent of F = ma. Think of angular acceleration as rotational version of linear acceleration.


📈 Why This Concept is Important?

  • Frequently appears in JEE Advanced and NEET Physics.

  • Tests knowledge of rotational motion, angular kinematics, and units.

  • Strengthens understanding of real-world problems like flywheels, engines, etc.

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