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A Frisbee of radius 5 inches that is spinning on a person's finger makes 20.2 revolutions in 18 seconds. What is the angular speed of a point on the outer edge of the Frisbee? Use 3.14 as an approximation for pi and round your answer to the nearest hundredth.

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Khoso123

Answer:

Angular Speed=7.05 rad/s

Explanation:

Given data

Total revolution r=20.2 revolutions

Time t=18 s

To find

Angular Speed

Solution

First we need not revolutions per second

So

=(20.2)/18

=1.1222 rev/s

[tex]Angular Speed=(1.122rev/s)2\pi \\Angular Speed=7.05rad/s[/tex]

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