A 0.05 kg ball is attached to the end of a 0.5 m long string. The ball is spun around a circle with a period of 0.20 s. What is the tension force in the string?

The ball spins once around the circle in 0.20 s, meaning it travels a distance equal to its circumference in that time, giving it a linear speed of

v = (2π (0.5 m)) / (0.20 s) = 5π m/s ≈ 15.708 m/s

Use this compute the magnitude of the centripetal acceleration a :

a = v²/ (0.5 m) = 50π² m/s² ≈ 493.48 m/s²

Use Newton's second law to compute the mangitude of the tension F in the string:

F = (0.05 kg) a = 5/2 π² N ≈ 24.674 N ≈ 25 N

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