Circular Motion Dynamics
Hardphysics
A stone of mass 0.5 kg tied to a string of length 1 m is whirled in a horizontal circle at 2 revolutions per second. What is the centripetal force?
Select the correct option:
Solution
Incorrect! Answer:
78.96 N
- Angular Velocity (ω): 2 revolutions/sec means frequency f=2 Hz.
- ω=2πf=4π rad/s.
- Centripetal Force Formula: F=mω2r.
- Application:
- F=0.5 kg×(4π)2×1 m.
- F=0.5×16π2=8π2.
- Numerical Evaluation: Taking π2≈9.87:
- F=8×9.87=78.96 N.
- Note: This force is provided by the tension in the string.
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- circular motion dynamics
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
78.96 N
- Angular Velocity (ω): 2 revolutions/sec means frequency f=2 Hz.
- ω=2πf=4π rad/s.
- Centripetal Force Formula: F=mω2r.
- Application:
- F=0.5 kg×(4π)2×1 m.
- F=0.5×16π2=8π2.
- Numerical Evaluation: Taking π2≈9.87:
- F=8×9.87=78.96 N.
- Note: This force is provided by the tension in the string.
This hard difficulty physics question is from the chapter laws of motion, covering the topic of circular motion dynamics. It appeared in the 2025 exam.
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