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This could make for a fun amusement park ride: hang from a tower a few feet off the ground using a bungee cord; detach bungee cord while simultaneously pushing

This could make for a fun amusement park ride: hang from a tower a few feet off the ground using a bungee cord; detach bungee cord while simultaneously pushing horizonally; bam you're Superman until the bungee cord comes crashing on your head. Actually that last part needs work :)

wtallis 13y ago
This derivation assumed a uniform distribution of mass in an unstretched slinky, and that all the mass is stretchable. This probably means that in order for this to work with a human attached to the bottom of the bungee cord, the human's mass needs to be negligible compared to the mass of the bungee cord. Thus, the human getting crushed by the bungee cord is not a problem to be dealt with, it is a requirement.
- StavrosK 13y ago
  Hmm, are you sure? The disturbance at the top of the cord has to travel downwards, and the speed at which it travels is the speed of sound in the cord itself. No matter what the mass at the bottom is, it can't "know" it's been let go of until the information propagates in the cord.
  - ars 13y ago
    It feels it almost instantly (speed of sound in matter), because the tension in the cord is lower. In a slinky the inertia of the slinky itself resists things, but not so in a light bungee cord. You can try a weight at the top and bottom maybe.
    - jlgreco 13y ago
      If you watch the second Veritasium video, you'll see that adding a weight to the bottom of the slinky does not change any of the behavior; it still does not fall until the 'pressure front' reaches it. A heavier weight at the top will accelerate just as fast due to gravity as a small weight. As far as I can tell the only possible effect of a weight at the top would be to ensure that the elasticity of the bungee cord does not accelerate the top much faster than the speed of gravity; you should be able to just use a less elastic cord.
      - plaguuuuuu 13y ago
        You could use a parachute to arrest the rate of fall. The force of the parachute acting against air resistance would resist the spring's compressive force and thus the spring would recompress more slowly, lengthening the time for the top to fall to the bottom.
        judk 13y ago
        I think I that would also tend to lift the rider up to compress the spring, and since the parachute is lifting the system.
        colanderman 13y ago
        Mm, no, the opposite I think – the rider will fall slowly. The spring remains extended due to the weight of the rider – being held aloft by a tower vs. a parachute makes no difference here. However the descending parachute shortens the spring at a rate slow enough that it can be transmitted to the rider well before the parachute reaches the ground, so the rider will fall a little bit. (This is analogous to simply lowering the hand holding the slinky – the bottom eventually follows suit.)
        plaguuuuuu 13y ago
        I'm pretty sure the rider stays in the same place O_o Oh god. This is going to be the next airplane-on-treadmill isn't it.
        jlgreco 13y ago
        I am very confident that it would fall. Consider the slinky again. If I dropped it with a parachute to slow it.... no, replace "parachute" with "my hand", but assume my hand follows the same trajectory the parachute would... the end of the slinky would not wait for my hand to approach it. The information about a change (the drop) would take the same amount of time to reach the end of the slinky. The information travels just as fast, even though the information is "we're falling slowly" instead of "we're falling".
- tedsanders 13y ago
  Very edited: Ah, so the human's mass can be heavy compared to the bungee cord, but if that's the case then the duration of the superman effect would be short.
  - wtallis 13y ago
    In the limiting case of an ultralight bungee cord, the human has to start falling almost immediately after the cord is cut, because the center of mass cannot fall as required without movement of the human as well. This incidentally gives us lower bounds on the speed of sound and stiffness of such a bungee cord. In order for the human to be held aloft for any significant period of time by the recoiling bungee cord after it is cut, the cord must be relatively heavy - enough to move the center of mass of the cord+human system well away from the human.
carlob 13y ago
You also need to account for the larger mass at the bottom…
- bdegman 13y ago
  As long as you don't hit the ground while fully at rest it'll be fine.
cjg 13y ago
But it wouldn't feel like flying. It would just feel like you were being held up by the bungee cord. You might as well have the same ride but not actually drop the bungee - sounds safer too.
plaguuuuuu 13y ago
Presumably you would travel in an arc, because the spring's compressive force would be pulling you towards the top of the spring... the spring would then not be perpendicular to the ground any more So, that arc would curve upwards away from the ground. You'd probably get hurt falling down!
- judk 13y ago
  The compressive force is already canceled by gravity, unless some other device is adding a extra downward force temporarily.

wtallis 13y ago

This derivation assumed a uniform distribution of mass in an unstretched slinky, and that all the mass is stretchable. This probably means that in order for this to work with a human attached to the bottom of the bungee cord, the human's mass needs to be negligible compared to the mass of the bungee cord. Thus, the human getting crushed by the bungee cord is not a problem to be dealt with, it is a requirement.

StavrosK 13y ago
Hmm, are you sure? The disturbance at the top of the cord has to travel downwards, and the speed at which it travels is the speed of sound in the cord itself. No matter what the mass at the bottom is, it can't "know" it's been let go of until the information propagates in the cord.
- ars 13y ago
  It feels it almost instantly (speed of sound in matter), because the tension in the cord is lower. In a slinky the inertia of the slinky itself resists things, but not so in a light bungee cord. You can try a weight at the top and bottom maybe.
  - jlgreco 13y ago
    If you watch the second Veritasium video, you'll see that adding a weight to the bottom of the slinky does not change any of the behavior; it still does not fall until the 'pressure front' reaches it. A heavier weight at the top will accelerate just as fast due to gravity as a small weight. As far as I can tell the only possible effect of a weight at the top would be to ensure that the elasticity of the bungee cord does not accelerate the top much faster than the speed of gravity; you should be able to just use a less elastic cord.
    - plaguuuuuu 13y ago
      You could use a parachute to arrest the rate of fall. The force of the parachute acting against air resistance would resist the spring's compressive force and thus the spring would recompress more slowly, lengthening the time for the top to fall to the bottom.
      - judk 13y ago
        I think I that would also tend to lift the rider up to compress the spring, and since the parachute is lifting the system.
        colanderman 13y ago
        Mm, no, the opposite I think – the rider will fall slowly. The spring remains extended due to the weight of the rider – being held aloft by a tower vs. a parachute makes no difference here. However the descending parachute shortens the spring at a rate slow enough that it can be transmitted to the rider well before the parachute reaches the ground, so the rider will fall a little bit. (This is analogous to simply lowering the hand holding the slinky – the bottom eventually follows suit.)
        plaguuuuuu 13y ago
        I'm pretty sure the rider stays in the same place O_o Oh god. This is going to be the next airplane-on-treadmill isn't it.
        jlgreco 13y ago
        I am very confident that it would fall. Consider the slinky again. If I dropped it with a parachute to slow it.... no, replace "parachute" with "my hand", but assume my hand follows the same trajectory the parachute would... the end of the slinky would not wait for my hand to approach it. The information about a change (the drop) would take the same amount of time to reach the end of the slinky. The information travels just as fast, even though the information is "we're falling slowly" instead of "we're falling".
tedsanders 13y ago
Very edited: Ah, so the human's mass can be heavy compared to the bungee cord, but if that's the case then the duration of the superman effect would be short.
- wtallis 13y ago
  In the limiting case of an ultralight bungee cord, the human has to start falling almost immediately after the cord is cut, because the center of mass cannot fall as required without movement of the human as well. This incidentally gives us lower bounds on the speed of sound and stiffness of such a bungee cord. In order for the human to be held aloft for any significant period of time by the recoiling bungee cord after it is cut, the cord must be relatively heavy - enough to move the center of mass of the cord+human system well away from the human.

carlob 13y ago

You also need to account for the larger mass at the bottom…

bdegman 13y ago
As long as you don't hit the ground while fully at rest it'll be fine.

cjg 13y ago

But it wouldn't feel like flying. It would just feel like you were being held up by the bungee cord. You might as well have the same ride but not actually drop the bungee - sounds safer too.

plaguuuuuu 13y ago

Presumably you would travel in an arc, because the spring's compressive force would be pulling you towards the top of the spring... the spring would then not be perpendicular to the ground any more So, that arc would curve upwards away from the ground. You'd probably get hurt falling down!

judk 13y ago
The compressive force is already canceled by gravity, unless some other device is adding a extra downward force temporarily.