For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total force on their feet compared to when they are running in a straight line?

Question

Stanton

Physics

2 September, 09:25

For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total force on their feet compared to when they are running in a straight line?

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Answers (1)

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Janetta · Answer 1

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r) ² + (mg) ²

F² = m² ((v²/r) ² + g²)

F = m √ ((v²/r) ² + g²)

Compared to just the vertical force:

F / Fy

m √ ((v²/r) ² + g²) / mg

√ ((v²/r) ² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√ ((12²/26) ² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).