Find the average angular acceleration of the sprinter

  • #1
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Homework Statement


A sprinter runs the curve of this 200 m in 11.61 s. Assume he ran in a lane which makes a semicircle (r = 32.4 m) for the first part of the race. At 3.59 s into the race his speed is 5.9 m/s. At 7.9 s into the race his speed is 8.4 m/s. His speed after the curve was 12.4 m/s.

What was the sprinter's average angular acceleration after 11.61 s? Answer in deg/s2. I.e., what was his average acceleration while running the curve?

Homework Equations


α=Δω/Δt

ω=Δθ/Δt

The Attempt at a Solution


ω=Δθ/Δt
ω=180/11.61s
ω=15.50degrees/s

α=Δω/Δt
α=15.50degrees/11.61s
α=1.34degrees/s2


My answer is wrong and no matter what I do I can't get it right. The way my professor does assignments is we either get 100% or 0% and so I'm very frustrated and would really really appreciate some help. Physics is very difficult for me.
 

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Answers and Replies

  • #2
haruspex
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The ##\omega## you calculated is the average angular speed. You want the change in angular speed from start to end of curve.
 
  • #3
gneill
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The average acceleration is the change in velocity divided by the change in time for the interval in question. That's your first relevant equation. In this case you're dealing with angular measures, so the average angular acceleration would be the change in angular velocity divided by the change in time.

Your second relevant equation determines the average velocity. But that's not what you're looking for.

The problem statement gives you the sprinter's linear speed at various points along the curve. What is the relationship between the linear (or in this case tangential) speed and the angular speed at a given time? What are the angular speeds at the beginning and end of the curve?

edit: Oops! haruspex got in there before me :smile:
 

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