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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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  1. 2 September, 09:28
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    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).
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