Rolling Energy Partition
A solid sphere (I = 2/5 MR²) rolls without slipπng at speed v. Total kinetic energy?
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About This Question
- Subject
- physics
- Chapter
- rotational motion
- Topic
- rolling energy partition
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
(7/10) M v²
- Energy Components: Total kinetic energy Ktotal=Ktrans+Krot.
- Translation: Ktrans=21Mv2.
- Rotation: Krot=21Iω2. Under no-slip condition, ω=v/R.
- Krot=21(52MR2)(Rv)2=51Mv2.
- Summation:
- Ktotal=(21+51)Mv2
- Ktotal=(105+102)Mv2=107Mv2.
- Partition: 71.4% is translational and 28.6% is rotational.
This medium difficulty physics question is from the chapter rotational motion, covering the topic of rolling energy partition. It appeared in the 2025 exam.
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