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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?

Verified Answer :

meerkat18

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.

Answer:

v=3.57m/s

Explanation:


Hi, let's solve this!

The speed is minimal when the voltage T = 0N


This occurs when the balance is

T + m g = m v2 / r


Since the voltage is zero, then we cancel it and the equation is


m g = m v2 / r


We have the following data:

m = 0.16kg

g = 9.8m / s2

r = 1.13m


We cleared the speed and it is

\(v=\sqrt{g*r}\)

\(v=\sqrt{9.8m/s^{2}* 1.13m}\)

v=3.57m/s

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