A tennis ball connected to a string is spun around in a vertical, circular path

Question

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.175 kg and moves at v = 5.22 m/s. The circular path has a radius of R = 1.14 m

1.What is the magnitude of the tension in the string when the ball is at the bottom of the circle?

2.What is the magnitude of the tension in the string when the ball is at the side of the circle?

3.What is the magnitude of the tension in the string when the ball is at the top of the circle?

4.What is the minimum velocity so the string will not go slack as the ball moves around the circle?

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.175 kg and moves at v = 5.22 m/s. The circular path has a radius of R = 1.14 m

Explanation / Answer

Centripetal Force (C) = mv^2/r = (0.175)*(5.22)^2 * (1/1.14) = 4.18286N

Gravitational force (mg) = 0.175*9.81 = 1.71675 N

1. ball is at the bottom : Tension = C+mg = 5.899N
2. ball is at the side : Tension = C = 4.182866N
3 . At the top : Tension = C-mg = 2.46611 N

minimum velocity : v^2 / r = g or v = sqrt(rg) = sqrt ( 1.14*9.81) = 3.3441m/s

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