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I noticed that the 200-meter sprints are conducted on curved tracks. (See this video: world championship semifinals 2009)

Isn't that weird? I mean, just look at the curvatures of each lane!


track


(Source)

Since they use staggered start lines, the total track length is the same. But the person on the innermost lane would have to do more work as compared to the one on the outermost lane. He would have to put in an additional amount of work against friction to counter the extra centrifugal force.

(The extra centrifugal force is roughly 2 Newtons from my calculation).

I believe that the winners of a running race are the ones who do more work against the friction. So, don't you think the runners in the inner lanes should be given an advantage?

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    $\begingroup$ work=force·distance. In theory he doesn't do any work against centrifugal force as long as he moves in the direction perpendicular to the force. In real life you still spend energy against static force, but you didn't estimate the work and I don't know how much that would be either. $\endgroup$ – Yi Jiang Oct 2 '19 at 14:28
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    $\begingroup$ Whether it is 'fair' is not really a physics question, but a question for the rules of the sport. $\endgroup$ – Jon Custer Oct 2 '19 at 14:41
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    $\begingroup$ @JonCuster Agreed. I raised the question here so that the physical reason behind the unfairness would be rigorously confirmed. The rules will be changed when people become aware of their flaws. That's the whole point of this site, right? $\endgroup$ – Krishnanand J Oct 2 '19 at 15:12
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    $\begingroup$ Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy? $\endgroup$ – BowlOfRed Oct 2 '19 at 17:35
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    $\begingroup$ I'm voting to close this question as off-topic because it is not about physics - any physics in the issue the answer is too tied up in external sporting factors to be useful. $\endgroup$ – Emilio Pisanty Oct 2 '19 at 21:17
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Runners generally prefer the middle lanes, and that's where the highest-seeded runners usually get assigned. While it is true that the tighter curve of the inner lanes means that you effectively have more weight on your feet (by about 1% relative to the outermost lane), it is also considered an advantage to be able to see your competitors during the race, which you can't at the beginning if you start out in front of them (as you do in the outermost lane).

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    $\begingroup$ They want to avoid the inner-most lane because that's the one the mile-and-farther runners have been pounding. The surface is most degraded. $\endgroup$ – puppetsock Oct 2 '19 at 16:06
  • $\begingroup$ @puppetsock, my general recollection is that the innermost lane simply isn't used during sprints. $\endgroup$ – Mark Oct 2 '19 at 23:16
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From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = \vec F \cdot \vec d$, but as the shoe does not slip, the distance moved against friction is zero, so the work is also zero.

Another way to think about it is that in a constant-speed turn, the velocity is tangent to the curve, while the centripetal force required is radial to the turn. The dot product is zero and again, no work is required to perform the turn.

All the losses from the runner are from other sources (air drag, inelastic impacts with the ground and internal to the leg, muscles being used to decelerate limbs, etc.) You could certainly make an argument that running in a tight turn is biomechanically a disadvantage, but saying that energy loss is due to friction or required centripetal forces wouldn't be correct.

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  • $\begingroup$ I think you meant "scalar" and not "cross" product. $\endgroup$ – hyportnex Oct 2 '19 at 23:11
  • $\begingroup$ In short track speed skating, you output more useful power when turning because you can keep pushing repeatedly. But in skating even on the straights you push mostly to the side and convert that to forward momentum by angling your blades. In running you generate forward momentum by driving your legs / feet backward during ground contact so it's not clear it would be helpful. $\endgroup$ – Peter Cordes Oct 3 '19 at 14:09

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