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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?

Mindee Nelson

in Physics

1 answer

1 answer

Jodi Brooks on October 20, 2018

We assign to the variables: T as the voltage 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+0.163 gcosx, 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*9.8 cosn=5.893 N. (2)Here x=π/2 and T=(0.163x4.872)/0.9 – 0.163x9.8cosn/2=4.295 N. (3)Here x=0 and T=(0.163x4.872)/0.9 – 0.163x9.8cos0=2.698 N. (4)We have T=(0.163v2)/0.9 – 0.163x9.8cosx. This minimum of v is obtained when T=0, and v verifies (0.163xv2)/0.9 – 0.163x9.8=0, resulting in v=2,970 m/s.

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