Question

A 0.50 kg mass is attached to a string 1.0 meter long and moves in a horizontal circle at a rate of 0.5 seconds per revolution.

Answer 1

(1) First compute the linear speed of the mass. If it completes 1 revolution in 0.5 seconds, then the mass traverses a distance of 2π (1.0 m) ≈ 2π m (the circumference of the circular path), so that its speed is

v = (1/0.5 rev/s) • (2π m/rev) = 4π m/s ≈ 12.57 m/s

Then the centripetal acceleration a is

a = v² / (1.0 m) = 16π² m/s² ≈ 160 m/s²

(where r is the path's radius).

(2) By Newton's second law, the tension in the string is T such that

T = (0.50 kg) a = 8π² N ≈ 79 N