What is the speed of the discus at release?

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To determine the speed of the discus at release, the thrower completes one revolution in 1.1 seconds while holding the discus at the end of a fully outstretched arm. The diameter of the circle is 1.8 meters, leading to a radius of 0.9 meters. The calculated speed using the radius of 0.9 meters results in 5.14 m/s, while using the full diameter of 1.8 meters gives a speed of approximately 10.28 m/s. The discrepancy arises from confusion over whether to use the radius or diameter in the calculations. It appears that the original question may contain an error regarding the expected speed.
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Homework Statement


To throw the discus, the thrower holds the discus with a fully outstretched arm and makes one revolution as rapidly as possible to give maximum speed to the discus at release. The diameter of the circle in which the discus moves is about 1.8 . If the thrower takes 1.1 to complete one revolution, starting from rest, what will be the speed of the discus at release?



Homework Equations


I am not sure where to begin. I know that angular velocity=change of theta/change in time.
Velocity=angular velocity*radius.

The Attempt at a Solution


I keep on getting 5.14 m/s. The answer given by the computer is 10 m/s. The only way that I can get 10 m/s is if you use r=1.8 m. I thought the radius would be half the diameter of the circle. What am I missing and what did I do wrong?
 
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Remember the circumference of a circle is 2 pi r
 
I see that but it still leads to (2*pi*.9)/1.1 s=5.14 m/s. If you use 1.8 as the radius then it works out to 10.28 m/s. I have already missed this question in my homework so I am just trying to figure how they got what they got.
 
Yes you're right - looks like they made a mistake in the question.
 

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