Circular Motion And Centripetal Force
A stone of mass 0.5 kg is whirled in a horizontal circle of radius 1 m at a constant speed of 4 m/s by a light string. What is the magnitude of the centripetal force required to keep it moving in this circle?
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- circular motion and centripetal force
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
8 N
Referring to NCERT Class 11, Chapter 5 (Laws of Motion), uniform circular motion requires a net inward force called the centripetal force, given by F = mv² / r, directed toward the centre of the circle. This force does not create a new kind of interaction; here it is supplied by the tension in the string. Substituting the given values, F = (0.5 × 4²) / 1 = (0.5 × 16) / 1 = 8 N. Option 4 N forgets to square the speed and uses mv/r. Option 2 N uses only m × v incorrectly without the radius and squaring. Option 16 N omits the mass factor and squares the speed alone. Plausibility check: the force depends on the square of speed, so doubling the speed would quadruple the force; with the given moderate speed and small mass, a value of a few newtons like 8 N is reasonable, and the unit kg·m²/s² ÷ m reduces to newtons, confirming correctness.
This easy difficulty physics question is from the chapter laws of motion, covering the topic of circular motion and centripetal force. It appeared in the 2025 exam.
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