Question

A 0. 75-kg ball is attached to a 1. 0-m rope and whirled in a vertical circle. The rope will break when the tension exceeds 450 n. What is the maximum speed the ball can have at the bottom of the circle without breaking the rope?

Answer 1

At the bottom of the circle, the ball is being pulled upward by tension in the rope and downward by its own weight, so that the net force on it is

∑ F = 450 N - (0.75 kg) g = (0.75 kg) a

where a is centripetal acceleration. At this maximum tension, the ball has a maximum centripetal acceleration of

a = (450 N - (0.75 kg) g) / (0.75 kg) = 590.2 m/s²

Then its maximum tangential speed v is such that

a = v² / (1.0 m)

⇒ v = √((1.0 m) a) ≈ 25 m/s