Rolling Energy Distribution
A uniform solid cylinder rolls without slipping along a horizontal surface. What fraction of its total kinetic energy is associated with the rotation of the cylinder about its axis?
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About This Question
- Subject
- physics
- Chapter
- rotational motion
- Topic
- rolling energy distribution
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
1/3
As explained in NCERT Class 11, Chapter 7 (System of Particles and Rotational Motion), a rolling body's total kinetic energy is the sum of its translational kinetic energy ½mv² and its rotational kinetic energy ½Iω². The two parts coexist because the body simultaneously moves forward and spins about its own axis. For rolling without slipping there is a strict constraint linking the two motions, namely v = Rω, where v is the speed of the centre and ω is the angular speed. A solid cylinder has moment of inertia I = ½mR² about its central axis. The rotational part therefore becomes ½(½mR²)(v/R)² = ¼mv². Adding the two contributions gives a total of ½mv² + ¼mv² = ¾mv². The rotational fraction is consequently (¼mv²)/(¾mv²) = 1/3. The option 1/2 wrongly assumes equal sharing between translation and rotation. The option 2/5 corresponds to a solid sphere rather than a cylinder. The option 2/7 belongs to a rolling sphere's translational analysis and does not apply to a cylinder. A consistency check confirms that translation carries 2/3 of the energy and rotation carries 1/3, and these two fractions correctly sum to the whole, exactly as expected for a uniform solid cylinder rolling without slipping.
This hard difficulty physics question is from the chapter rotational motion, covering the topic of rolling energy distribution. It appeared in the 2025 exam.
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