Physics puzzle
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34 members have voted
Quote: WizardI hate to change the topic, but what is the answer? I assume the ring rotated at a speed to exactly simulate earth gravity. My guess is that the runner would feel the same way running in either direction, for the same reason in an airplane it requires the same effort to walk toward the front of the cabin as away from it.
An airplane in level flight is an inertial frame of reference, but a rotating ring is not, therefore Newton's Laws don't apply there.
BTW, airliners bend around the wing box in flight, so there is a small incline even when level.
Weight felt would be m*v^2/r=m*w^2*r, where w is angular velocity. For a=g, w^2*r=g, so a ring rotating at 1 radian/second (9.55 rpm) with a radius of 9.8m would do it. Linear velocity on the rim of this ring is 9.8m/s. This is right around the record speed reached by the fastest sprinters, so one could in theory run fast enough to negate the artificial gravity completely or to quadruple it. In practice the latter wouldn't work due to the effort required.
That only applies to a ~10m radius ring. With a radius of 20m, linear velocity would be 14m/s, and with a radius of 5m, only 7m/s, following inverse square root proportion. This actually has some implications for the minimum cylindrical module size at which credible artificial gravity can be produced.
How it would feel in practice is that you'd feel like you are running uphill and downhill respectively, uphill if you're running with the rotation and downhill otherwise. It's also quite easy to understand how you could negate artificial gravity completely: since you are in zero-g, all you need to do is jump in the air to no longer be affected by the moving floor. As long as the velocity of your jump can negate the inertia.
Strictly speaking the same applies to Earth, since it isn't an inertial frame of reference either. But Earth's radius is about 6,400,000m, so you need a velocity of 8,000m/s (get it - square root of 64) to negate gravity completely. This effect actually is used in practice: spacecraft are launched in the direction of Earth's rotation and close to the equator if possible, allowing them to negate gravity with less delta-V.
Quote: NareedThat's because int eh first case he'd be adding to the speed, just as if the ring ran faster. Contrariwise, he diminishes speed if he runs opposite to the rotation, and the effect would be the same as if the ring ran slower.
Hmm. I hadn't thought of that. Couldn't we say that the mass and force of the runner is pretty negligible to that of the whole spacecraft? Even if we can't ignore that, what if the speed of rotation was on cruise control and maintained the same speed no matter what was going on inside or outside?
Quote: WizardHmm. I hadn't thought of that. Couldn't we say that the mass and force of the runner is pretty negligible to that of the whole spacecraft?
Yes, but the mass of the spacecraft doesn't matter. Say a ring with a diameter of X and a speed of Y will create a feeling of Z gravity. This would be true whether the ring is a ribbon or solid lead. The additional mass has gravity of its own, but you need millions of tons of mass, at least, to notice any difference.
Quote:Even if we can't ignore that, what if the speed of rotation was on cruise control and maintained the same speed no matter what was going on inside or outside?
The ring rotates at a constant speed, but if you move in the direction of rotation, then the effect is the same as if you'd sped the ring up.
As I said, all SF readers learn this because all SF writers eventually set a story or novel in a rotating cylinder, or a cylindrical section. Be it small like Discovery in 2001, medium like Rama in "Rendezvous With Rama," or big like the Ringworld in Niven's "Ringworld" novels.
Actually, the runner would change the ring's speed of rotation, especially in a realistic lightweight spacecraft, but only as long as he is running. The moment he stops, his impulse pushes the ring back to its old speed, so no cruise control is necessary.
You wouldn't attempt to maintain a constant speed, as that would cost propellant. You would however keep an electric motor between the rotating ring and the fixed station frame, to compensate for friction and people transferring momentum between the two.
Well, I'm off to Las Vegas this morning for CES.
Quote: P90
Never said you should be. I'm just saying - when people post school questions in a non-school forum, they usually expect a more thorough look at them than in the classroom, like figuring out situations where the answer can be different. So it's surprising to me that you only expect unequivocal "box A" and reject any look at variations in conditions as irrelevant.
Like I said, I don't think, this is a school question. I mean, sure, it can be explained to a good high-school student, yes, just like most anything, discussed in this forum, but I believe the understanding that would let you appreciate this kind of a puzzle, and find it amusing and entertaining is beyond school curriculum (fwiw, it looks like you are lucking that skill entirely ;))
BTW, if I was to pose a question like this to school students - then I would actually make sure that all conditions are properly spelled out, and nothing is left ambiguous, because I don't trust their ability to resolve ambiguities and make reasonable assumptions. And also because they are known to intentionally play around and try to obfuscate the problem to mask their inability to solve it.
Quote:Put it this way: Suppose you were using this setup in a system where the outcome matters. Let's say the balloon is to serve as a hydrogen leak detector, with string deflection triggering emergency tank evacuation. Why do it this way is not up to you. Would you still be satisfied with the same "box A" answer?
No, but I don't see a connection. Would you be satisfied with your mv^2/r formula if a deviation from it in the pressure on the rotating floor was going to trigger an emergency evacuation?
Quote: P90An airplane in level flight is an inertial frame of reference, but a rotating ring is not, therefore Newton's Laws don't apply there.
They very much do. All the stuff you spell out below is Newtonian physics.
BTW, airplane is not an inertial frame, because it is in a gravitational field.
Quote:Weight felt would be m*v^2/r=m*w^2*r, where w is angular velocity.
Only if the ring is circular. Does it mention it anywhere that it is not an ellipse or, perhaps, a cycloid or a catenary? it could get really interesting...
Also, where in the human body do you consider the weight "felt"? Obviously, the guy in such position would be rather "lightheaded", don't you think?
Quote:How it would feel in practice is that you'd feel like you are running uphill and downhill respectively,
No, it would not. It would just feel like you got heavier or lighter. Try putting some weights into your pockets and running - does it feel like you are running uphill?
A much bigger problem with this setup is that "running" in zero-g is not something you can really do. Running is basically, jumping up and forward, landing on the other foot, jumping again etc. Now, jumping part is easy. But think about this - what force will make you land back on the floor once you have jumped? The "centrifugal force" that you are feeling like gravity disappears as soon as you lose contact with the floor. I am afraid, you can't run in such setup at all, best you can do is walk really fast (which would not be an easy feat to accomplish for a bunch of reasons too, but should at least be possible in principle).
Quote: weaselmanThe "centrifugal force" that you are feeling like gravity disappears as soon as you lose contact with the floor. I am afraid, you can't run in such setup at all, best you can do is walk really fast (which would not be an easy feat to accomplish for a bunch of reasons too, but should at least be possible in principle).
Ya know, after all of my efforts lately to stay out of such discussions, I just feel myself being drawn back into one.
(Note for history buffs: In previous discussions, weaselman and I have basically agreed to disagree on the meaning of "weight". I view it as local acceleration of gravity times mass, and weaselman views it as the force imposed by the supporting structure -- hope I haven't misworded that too badly. I think that difference in perspective is relevant here.)
From my perspective, if a person is running around inside a rotating ring in deep space as in the film 2001, they will continue to experience the centrifugal force, provided that their direction of run is not opposite the rotation of the ring and at just the right speed that they have no net rotational motion about the axis. (See the next paragraph, and BTW I'm thinking of a circular ring.) When they "jump" with each step, they will be returned to the surface of the ring/track by this centrifugal force. During the period of their separation from the ring/track, they will not be experiencing the counter-balancing centripetal force, which means they will be accelerating radially back toward the surface, in many ways similar to jogging on a normal, earth-based track. (I think that weaselman's contention that there is no centrifugal force after the jump is compatible with his view that a person on earth who has jumped does not have any weight until he gets back to the ground. That's right isn't it? I continue to disagree.)
As for whether it is reasonable to think that the runner could go fast enough (against the rotation of the ring) to get to zero rotation, that requires some info on the system design. Perhaps SciFi fans have the data, but I don't. I recall the ring as being "huge". If I assume that the track had a radius of 100 ft. and that rotation was designed to provide a simulated gravity of 1 g, then the track surface speed would be around 57 ft/sec. I don't think top sprinters can do that, and I'm sure that the runner portrayed in the film wasn't doing anything like that speed. Other assumptions about radius and simulated gravity will lead to other results.
But it's probably better for me to just step back and let weaselman and P90 duke this one out.
Quote: DocWhen they "jump" with each step, they will be returned to the surface of the ring/track by this centrifugal force. During the period of their separation from the ring/track, they will not be experiencing the counter-balancing centripetal force, which means they will be accelerating radially back toward the surface,
What force in your view will cause the radial acceleration?
You do know/agree that "centrifugal force" is "fictional", and is really just an effect of inertia, a reaction to the acceleration, caused by the centripetal force, right? If centripetal force is gone, what is responsible for pushing you back radially? In the absence of forces a body tends to move inertially, on a straight line, with constant speed. If you jump "up", your speed is directed "up" at the moment, there needs to be some force exerted against your body to change that direction.
Quote:(I think that weaselman's contention that there is no centrifugal force after the jump is compatible with his view that a person on earth who has jumped does not have any weight until he gets back to the ground. That's right isn't it? I continue to disagree.)
No, this is different. A person, who has jumped on the surface of Earth, has no weight, but is still under the influence of the Earth's gravitational force. That is the force responsible for the eventual change of direction from "up" to "down". In the absence of gravity, there is no such force.
Quote: weaselmanWhat force in your view will cause the radial acceleration?
You do know/agree that "centrifugal force" is "fictional", and is really just an effect of inertia, a reaction to the acceleration, caused by the centripetal force, right? If centripetal force is gone, what is responsible for pushing you back radially? In the absence of forces a body tends to move inertially, on a straight line, with constant speed. If you jump "up", our speed is directed "up" at the moment, there needs to be some force exerted against your body to change that direction..
Man, you do read the archives!
I agree, mostly. I was giving my answer in the frame of reference of the rotating system. Using a non-accelerating frame of reference, I would describe it this way: When the runner jumps "up", their direction of travel is not toward the center of the ring. It is a resultant of their original tangential motion and the mostly-radial (inward) acceleration created by the centripetal force of the rotating ring and by their jumping stride. After they stop pushing off the surface, they will travel in a straight line (in the non-accelerating frame of reference) and will be returned to the track because they continue to move on a straight line with constant speed until their path again intersects the track. From the perspective of someone riding on the rotating ring, that runner is experiencing a radial (outward) acceleration that is described as being due to a centrifugal force.
I have not gone back and re-read the thread, but my recollection is that you made some posts that seemed to suggest that when the runner took a jumping step, they would be suspended above the track and not return to it. Did you mean that? I am confident that they would return to the track, and the mechanism may be described as either moving in a straight line to a point of intersection in a non-accelerating frame of reference or experiencing an outward radial acceleration in the rotating frame of reference.
Quote: DocMan, you do read the archives!
Well, thanks to the feature of this board, much disliked by me, than makes read threads disappear from the chronological list, I have a habit of periodically checking the threads I have posted in via my user profile. This time, I clicked on "Threads started by weaselman" instead of "Posts made by weaselman" accidentally, and, because I have not started that many threads, I saw this one, with your post in it, that I have not seen before.
Quote:After they stop pushing off the surface, they will travel in a straight line (in the non-accelerating frame of reference) and will be returned to the track because they continue to move on a straight line with constant speed until their path again intersects the track.
This is true, but such a process can hardly be called running, it'll be much more like series of long erratic jumps.
The most efficient way to run is jumping at about 45 degree angle to the surface, making your vertical speed equal to your horizontal speed. If you are running in the direction of rotation, assuming your "own" speed is the same as the linear velocity of the floor, in the inertial frame, that will make your horizontal speed twice as large as vertical, meaning that you are jumping up at 30 degree angle to the belt. This means that after three jumps like this, you'll have made the full circle, and end up where you started, whatever the radius of the circle is.
This will look even less like running if you attempt to run in the opposite direction. In this case, your own horizontal velocity is (almost) cancelled out by the linear velocity of the rotation, so you are jumping more or less straight up, and are going to end up directly "above" where you started, hitting your head on the floor.
Quote:I have not gone back and re-read the thread, but my recollection is that you made some posts that seemed to suggest that when the runner took a jumping step, they would be suspended above the track and not return to it. Did you mean that?
No, if I said that, I must have been smoking something at the time :)
The runner would not get suspended, of course, but he would not really be a "runner" either :)
Quote: weaselmanQuote: DocI have not gone back and re-read the thread, but my recollection is that you made some posts that seemed to suggest that when the runner took a jumping step, they would be suspended above the track and not return to it. Did you mean that?
No, if I said that, I must have been smoking something at the time :)
The runner would not get suspended, of course, but he would not really be a "runner" either :)
I just looked back up a couple of posts to see what I misunderstood. Here is your post that made me think you were saying the runner would remain suspended:
Quote: weaselmanA much bigger problem with this setup is that "running" in zero-g is not something you can really do. Running is basically, jumping up and forward, landing on the other foot, jumping again etc. Now, jumping part is easy. But think about this - what force will make you land back on the floor once you have jumped? The "centrifugal force" that you are feeling like gravity disappears as soon as you lose contact with the floor. I am afraid, you can't run in such setup at all, best you can do is walk really fast (which would not be an easy feat to accomplish for a bunch of reasons too, but should at least be possible in principle).
The part where you asked P90, "...what force will make you land back on the floor once you have jumped? The "centrifugal force" that you are feeling like gravity disappears as soon as you lose contact with the floor," is what made me think you were claiming the runner would be suspended and not coming back "down". Now, I guess I just don't know what you meant.
Quote: Doc
The part where you asked P90, "...what force will make you land back on the floor once you have jumped? The "centrifugal force" that you are feeling like gravity disappears as soon as you lose contact with the floor
No-no. The gravity certainly does not disappear when you lose contact with the floor. The centrifugal force does, unlike gravity.
This is actually an important distinction between the two, making the latter to be not like the former.
Actually, I thought that you and I had already agreed on this.
Quote:Now, I guess I just don't know what you meant.
I tried to explain it better in my previous post. When running in a gravitational field, you move up and down and forward in parabolic arches due to gravity. Attempting to run on the rotating circular floor in zero-g will result in a trajectory composed of straight line segments. From the perspective of the rotating reference frame, you'll be leaping in huge erratic jumps (especially erratic if running opposite to the direction of rotation) rather than running.
