Torque And Angular Acceleration
A grinding wheel with moment of inertia 2 kg·m² experiences a constant net torque of 10 N·m about its axis. What angular acceleration does the wheel undergo under this applied torque?
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About This Question
- Subject
- physics
- Chapter
- rotational motion
- Topic
- torque and angular acceleration
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
5rad/s2
NCERT Class 11, Chapter 7 (System of Particles and Rotational Motion) establishes the rotational form of Newton's second law, τ = Iα, where the net torque equals the product of the moment of inertia and the angular acceleration. This is the direct angular counterpart of the linear law F = ma, with torque taking the place of force and moment of inertia taking the place of mass. The equation tells us that for a given torque, a body with a larger moment of inertia is harder to angularly accelerate, just as a heavier object is harder to push. Rearranging for the angular acceleration gives α = τ/I = 10/2 = 5 rad/s². The option 20 rad/s² comes from multiplying torque by inertia instead of dividing. The option 0.2 rad/s² inverts the ratio by dividing inertia by torque. The option 12 rad/s² bears no valid relationship to the given quantities. A check of units shows that N·m divided by kg·m² yields s⁻², which corresponds to rad/s² since the radian is dimensionless, confirming the dimensional consistency. A value of 5 rad/s² is physically reasonable for a moderately heavy grinding wheel subjected to a steady 10 N·m torque about its axis.
This easy difficulty physics question is from the chapter rotational motion, covering the topic of torque and angular acceleration. It appeared in the 2025 exam.
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