What is the speed of the discus at release?

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SUMMARY

The speed of the discus at release is calculated using the formula for linear velocity, which is derived from angular velocity. Given a diameter of 1.8 meters, the radius is 0.9 meters. The thrower completes one revolution in 1.1 seconds, leading to an angular velocity of approximately 5.14 m/s when using the radius of 0.9 m. However, the correct speed at release, using the full diameter as the radius, is 10 m/s, indicating a misinterpretation of the radius in the initial calculations.

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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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