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 = 0.16 kg and moves at v = 5.33 m/s. The circular path has a radius of R = 1.13 m. What is the minimum velocity so the string will not go slack as the ball moves around the circle?

Question

Maximus Stewart

Physics

12 January, 20:47

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 = 0.16 kg and moves at v = 5.33 m/s. The circular path has a radius of R = 1.13 m. What is the minimum velocity so the string will not go slack as the ball moves around the circle?

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Answers (1)

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Jaylin Thornton · Answer 1

We assign the variables: T as tension and x the angle of the string

The centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²) / 0.9 = T+0.163gcosx, giving T = (0.163x4.87²) / 0.9 - 0.163x9.8cosx.

(1) At the bottom of the circle x=π and T = (0.163x4.87²) / 0.9 -.163*9.8cosπ=5.893N.

(2) Here x=π/2 and T = (0.163x4.87²) / 0.9 - 0.163x9.8cosπ/2=4.295N.

(3) Here x=0 and T = (0.163x4.87²) / 0.9 - 0.163x9.8cos0=2.698N.

(4) We have T = (0.163v²) / 0.9 - 0.163x9.8cosx.

This minimum v is obtained when T=0 and v verifies (0.163xv²) / 0.9 - 0.163x9.8=0, resulting to v=2.970 m/s.