Vertical Circular Motion And Energy
A small stone tied to a string is whirled in a vertical circle of radius 2 m. What is the minimum speed the stone must have at the highest point so that the string just remains taut? Take g as 10 m/s squared.
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About This Question
- Subject
- physics
- Chapter
- work, energy and power
- Topic
- vertical circular motion and energy
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
4.47m/s
As covered in NCERT Class 11, Chapter 6 (Work, Energy and Power), at the top of a vertical circle the minimum condition for the string to stay taut is that gravity alone provides the required centripetal force, so mg = mv²/r, giving v = sqrt(gr). Substituting, v = sqrt(10 × 2) = sqrt(20) = 4.47 m/s. At this critical speed the string tension is just zero, and any slower speed would let the stone fall inward. The option 6.32 m/s corresponds to sqrt(2gr) and would apply to a different condition. The option 2 m/s ignores the square root of the product. The option 8.94 m/s doubles the correct value incorrectly. A plausibility check: the answer sqrt(gr) is independent of mass and yields a modest speed for a 2 m radius, consistent with the threshold where weight exactly supplies the centripetal requirement.
This hard difficulty physics question is from the chapter work, energy and power, covering the topic of vertical circular motion and energy. It appeared in the 2025 exam.
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