Horizontal Circular Motion

Question

A stone of mass 0.5 kg tied to a string is whirled in a horizontal circle of radius 1 m at a constant speed of 4 m/s. What is the tension in the string supplying the centripetal force?

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Uniform circular motion requires a net inward centripetal force directed toward the centre, of magnitude $F_c = \frac{mv^2}{r}$. For a stone whirled in a horizontal circle, the horizontal string tension provides this centripetal force. Substituting the data, $T = \frac{0.5 	imes (4)^2}{1} = \frac{0.5 	imes 16}{1} = 8	ext{ N}$. The 2 N value mistakenly uses $v$ instead of $v^2$, dropping a factor of the speed. The 4 N value halves the correct result, perhaps by omitting the square or misreading the mass. The 16 N value forgets the 0.5 kg mass factor and uses $v^2/r$ alone. It is essential to remember that the centripetal force is not a new kind of force but a label for whatever real force points toward the centre, which in this horizontal whirl is entirely the string tension. The stone's speed stays constant, so there is no tangential force and the tension does no work, yet the velocity direction changes continuously, which is precisely why an inward force is still required. This applies the NCERT centripetal-force relation that underlies all uniform circular motion. As a plausibility check, tension grows with the square of speed, so doubling the whirling speed would quadruple the tension to 32 N, a behaviour consistent with the strong pull felt when spinning an object faster.

Horizontal Circular Motion

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