Tension in a Rotating Horizontal String
A 0.60 kg mass is tied to a 0.90 m string and swung in a horizontal circle on a frictionless table, so the string stays horizontal. A student measures the period as 0.75 s using a stopwatch over 10 revolutions. The string is rated to break at 25 N. The student's partner claims the tension is simply the centripetal force mv²/r and that the setup will fail. The student notes the measured period is uncertain by ±0.05 s because they started the stopwatch late. Calculate the tension, then judge whether the rating is exceeded and how the timing uncertainty affects that conclusion.
1. Calculate the tension in the string and decide whether it exceeds the 25 N breaking rating.
Read the worked answer
The speed is v = 2πr/T = 2π(0.90)/0.75 = 7.54 m/s. The centripetal force is F = mv²/r = (0.60)(7.54)²/0.90 = (0.60)(56.9)/0.90 = 37.9 N. Because the string is horizontal and the table is frictionless, the tension provides the entire centripetal force, so T = 37.9 N. This exceeds the 25 N rating, so the string would break. Note that the mass is not hanging vertically; if it were, gravity would also contribute and the geometry would change.
2. The period has an uncertainty of ±0.05 s. Determine the range of tension and state whether the conclusion is robust.
Read the worked answer
Tension depends on v², and v ∝ 1/T, so F ∝ 1/T². For T = 0.80 s: v = 2π(0.90)/0.80 = 7.07 m/s, F = (0.60)(50.0)/0.90 = 33.3 N. For T = 0.70 s: v = 8.08 m/s, F = (0.60)(65.3)/0.90 = 43.5 N. The range is roughly 33–44 N, all above 25 N. The conclusion that the string breaks is robust because even the lower bound exceeds the rating by a wide margin. The uncertainty changes the number but not the judgement.
3. The partner says the tension should equal the weight of the mass because the string holds it up. Evaluate this reasoning.
Read the worked answer
The partner confuses vertical support with horizontal circular motion. On a horizontal frictionless table, the normal force balances the weight (N = mg = 5.9 N), and the string is horizontal, so it cannot support weight at all. The tension is purely horizontal and provides the centripetal force. If the string were vertical (a conical pendulum), tension would have a vertical component equal to mg and a horizontal component providing centripetal force, giving a larger total tension. The partner's claim would only apply to a stationary hanging mass, not this geometry.