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Is the Olympic running race fair?
What's the difference between running up a hill and running up an inclined treadmill?Effect of surface treatment on fair diceWill two trains running along the equator in opposite direction experience same wear out?Running vs. walking in slippery conditionWhat would the ideal amount of gravity be for an Olympic sprinter?Drifting vs. rolling when starting raceRunning or walking up stairs = same work?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
I noticed that the Olympic $200$ meter sprints are conducted on curved tracks. (See this video:Olympic Semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(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?
newtonian-mechanics friction work everyday-life centrifugal-force
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
$endgroup$
|
show 2 more comments
$begingroup$
I noticed that the Olympic $200$ meter sprints are conducted on curved tracks. (See this video:Olympic Semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(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?
newtonian-mechanics friction work everyday-life centrifugal-force
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
$endgroup$
$begingroup$
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
8 hours ago
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
8 hours ago
2
$begingroup$
@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
7 hours ago
$begingroup$
I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
$endgroup$
– Jon Custer
7 hours ago
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
5 hours ago
|
show 2 more comments
$begingroup$
I noticed that the Olympic $200$ meter sprints are conducted on curved tracks. (See this video:Olympic Semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(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?
newtonian-mechanics friction work everyday-life centrifugal-force
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
$endgroup$
I noticed that the Olympic $200$ meter sprints are conducted on curved tracks. (See this video:Olympic Semifinals 2009)
Isn't that weird? I mean, just look at the curvatures of each lane!
(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?
newtonian-mechanics friction work everyday-life centrifugal-force
newtonian-mechanics friction work everyday-life centrifugal-force
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
edited 4 hours ago
Krishnanand J
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
asked 8 hours ago
Krishnanand JKrishnanand J
1,6383 gold badges9 silver badges29 bronze badges
1,6383 gold badges9 silver badges29 bronze badges
$begingroup$
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
8 hours ago
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
8 hours ago
2
$begingroup$
@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
7 hours ago
$begingroup$
I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
$endgroup$
– Jon Custer
7 hours ago
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
5 hours ago
|
show 2 more comments
$begingroup$
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
8 hours ago
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
8 hours ago
2
$begingroup$
@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
7 hours ago
$begingroup$
I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
$endgroup$
– Jon Custer
7 hours ago
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
5 hours ago
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
2
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
8 hours ago
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
8 hours ago
2
2
$begingroup$
@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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
$endgroup$
– Jon Custer
7 hours ago
$begingroup$
I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
$endgroup$
– Jon Custer
7 hours ago
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
5 hours ago
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
5 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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).
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
$endgroup$
3
$begingroup$
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
6 hours ago
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F times vec d$, 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 cross 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.
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
$endgroup$
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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).
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
$endgroup$
3
$begingroup$
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
6 hours ago
add a comment
|
$begingroup$
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).
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
$endgroup$
3
$begingroup$
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
6 hours ago
add a comment
|
$begingroup$
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).
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
$endgroup$
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).
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
answered 8 hours ago
Ben51Ben51
4,81210 silver badges31 bronze badges
4,81210 silver badges31 bronze badges
3
$begingroup$
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
6 hours ago
add a comment
|
3
$begingroup$
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
6 hours ago
3
3
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F times vec d$, 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 cross 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.
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
$endgroup$
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F times vec d$, 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 cross 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.
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
$endgroup$
add a comment
|
$begingroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F times vec d$, 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 cross 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.
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
$endgroup$
From the point of view of an ideal machine that is not slipping on the ground, friction does not do any work. $W = vec F times vec d$, 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 cross 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.
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
answered 38 mins ago
BowlOfRedBowlOfRed
20.4k2 gold badges35 silver badges54 bronze badges
20.4k2 gold badges35 silver badges54 bronze badges
add a comment
|
add a comment
|
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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
8 hours ago
2
$begingroup$
Whether it is 'fair' is not really a physics question, but a question for the rules of the sport.
$endgroup$
– Jon Custer
8 hours ago
2
$begingroup$
@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
7 hours ago
$begingroup$
I find it unlikely that athletics venues will all of a sudden start having straight tracks for 200m and 400m races. Its part of the competition, that's all.
$endgroup$
– Jon Custer
7 hours ago
$begingroup$
Force does not equal work. How do you get a force (2N) and say that it represents additional work or energy?
$endgroup$
– BowlOfRed
5 hours ago