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andre [41]
3 years ago
11

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?
Physics
2 answers:
ycow [4]3 years ago
4 0
We assign the variables: T as tension  and x the angle of the string
 The  <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx. 
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
kolezko [41]3 years ago
4 0

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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