Find the average angular acceleration of the sprinter

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SUMMARY

The average angular acceleration of a sprinter running a 200 m semicircular curve can be calculated using the formula α=Δω/Δt. Given that the sprinter's speeds at different times are 5.9 m/s at 3.59 s, 8.4 m/s at 7.9 s, and 12.4 m/s after the curve, the average angular speed ω is determined to be 15.50 degrees/s over the total time of 11.61 s. The correct average angular acceleration is derived from the change in angular velocity, resulting in α=1.34 degrees/s². This calculation is essential for accurately assessing the sprinter's performance in physics assignments.

PREREQUISITES
  • Understanding of angular acceleration and its formula α=Δω/Δt
  • Knowledge of angular velocity and its relationship to linear speed
  • Familiarity with basic kinematics in circular motion
  • Ability to convert between linear and angular measurements
NEXT STEPS
  • Study the relationship between linear speed and angular speed in circular motion
  • Learn how to derive angular acceleration from given linear speeds
  • Explore the concept of tangential acceleration in relation to angular motion
  • Practice solving problems involving angular kinematics and acceleration
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and dynamics, as well as anyone preparing for physics exams or assignments related to motion in circular paths.

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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/s2My 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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The ##\omega## you calculated is the average angular speed. You want the change in angular speed from start to end of curve.
 
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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