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 = 0.16 kg and moves at v = 5.33 m/s. The circular path has a radius of R = 1.13 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?

Explanation / Answer

Consider T be the tension in the string and x the angle of the string with the upwards vertical.

The centripetal acceleration and force are respectively v²/r = 5.33²/1.13 and (0.16x5.33²)/1.13 =
T+0.16gcosx, giving T=(0.16x5.33²)/1.13 – 0.16x9.8cosx.
(1)At the bottom of the circle x= and T=(0.16x5.33²)/1.13 – 0.16x9.8cos = 5.59 N.

(2)Here x=/2 and T=(0.16x5.33²)/1.13 – 0.16x9.8cos/2 = 4.02 N.

(3)Here x=0 and T=(0.16x5.33²)/1.13 – 0.16x9.8cos0 = 2.45 N.

(4)We have T=(0.16v²)/1.13 – 0.16x9.8cosx.

For a given v T is smallest at x=0, giving
T=(0.16xv²)/1.13 – 0.16x9.8. This minimum v is obtained when T just goes slack, i.e., T=0
and v satisfies =(0.16xv²)/1.13 – 0.16x9.8=0

=> 0.14xv² = 0.16x9.8

=> v = 3.33 m/s

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