Combined Translation And Rotation
A cylinder of mass 10 kg and radius 0.2 m rolls without slipping with a linear velocity of 5 m/s. What is its total kinetic energy?
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About This Question
- Subject
- physics
- Chapter
- rotational motion
- Topic
- combined translation and rotation
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
187.5 J
- Total Energy K: K=Ktranslational+Krotational.
- Translational: Kt=21Mv2=21(10)(52)=125 J.
- Rotational: Kr=21Iω2.
- For cylinder, I=21MR2.
- For rolling, ω=v/R.
- Kr=21(21MR2)(v/R)2=41Mv2.
- Kr=41(10)(25)=62.5 J.
- Total: 125+62.5=187.5 Joules.
This hard difficulty physics question is from the chapter rotational motion, covering the topic of combined translation and rotation. It appeared in the 2025 exam.
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