Question

It is well known that runners run more slowly around a curved track than a straight one. One hypothesis to explain this is that the total force from the track on a runner's feet--the magnitude of the vector sum of the normal force (that has average value mg to counteract gravity) and the inward-directed friction force that causes the runner's centripetal acceleration--is greater when running around a curve than on a straight track. Runners compensate for this greater force by increasing the time their feet are in contact with the ground, which slows them down.
For sprinters running at 8 m/s around a curved track of radius 18 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? Express your answer as a percent.
(Fcurved - Fstraight) / Fstraight

Answer 1

Answer:

$percentage = 6.27 %$

Explanation:

As we know that when runner is moving on straight track then the net force on his feet is given as

$F_s = mg$

while when runner is moving on circular track then we have

$F_c = \sqrt{(mg)^2 + (\frac{mv^2}{R})^2}$

$F_c = m\sqrt{g^2 + \frac{v^4}{R^2}}$

$F_c = m\sqrt{9.8^2 + \frac{8^4}{18^2}}$

$F_c = m(10.42)$

now percentage change is given as

$percentage = \frac{F_c - F_s}{F_s} \times 100$

$percentage = \frac{m(10.42) - m(9.81)}{m(9.81)}\times 100$

$percentage = 6.27 %$