Csdrhrt

Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?

Well, here's a related question. Suppose you find yourself in an elevator at the top floor of a skyscraper when the cable suddenly snaps. As the elevator plummets down, you realize you'll die on impact when it reaches the bottom. But then you think, what if I jump just before that happens? When you jump, you're moving up, not down, so there won't be any impact at all!

The mistake here is the same as the one you're made above. When you jump in the elevator, you indeed start moving upward relative to the elevator, but you're still moving at a tremendous speed downward relative to the ground, which is what matters.

Similarly, when you are at the rim of a large rotating space station, you have a large velocity relative to somebody standing still at the center. When you jump, it's true that you're going up relative to the piece of ground you jumped from, but you still have that huge rotational velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.

If you jump then you are in free fall, apart from air resistance, so you are weightless. This holds for any jump. For a brief moment you experience zero gravity!

This is an inertia problem. The phrasing of your question presupposes that the astronaut loses inertia when jumping, resulting in their upward trajectory being the only major acting force. At any point while rotating in a stationary position, the inertia of the astronaut is outwards from the rotational path at a 90 degree angle from the line connecting the astronaut and the axis of rotation. When the astronaut jumps, this inertia is maintained and two new forces are generated, namely the force the astronaut enacts on the space station (resulting in negligible acceleration on the part of the space station) and the force the space station enacts on the astronaut (resulting in non-negligible acceleration on the part of the astronaut towards the axis of rotation). The astronaut's relative velocity is related linearly to the vector sum of the inertial force acting on him as well as the force from his jump. This is why the "gravity" felt by the astronaut is determined by the angular velocity of the station floor; greater angular velocity means greater inertia as well as greater centripetal (and centrifugal) force, which means greater force (applied by jumping) is required to accomplish the same (relatively) vertical displacement.

The feeling of being pressed to the ground in an artificial gravity chamber is due to the rotating floor pushing the astronaut towards the rotational axis, when the astronaut (from a physics perspective) wants to keep going along the inertial path. This force resulting in this central acceleration is the centripetal force, while the opposing force enacted by the astronaut on the space station floor is the centrifugal force.

You can find this information in any basic university level physics book; pay special attention to chapters covering angular momentum and gravitational/circular forces. Fundamentals of Physics by David Halliday and Robert Resnick was the one I used in college.

Of course, the biggest problem with the supposition of becoming 'weightless' (returning to a null-G state) by making a 'big' jump is this: Assuming the OP was talking about not a created micro-gravity (.2G or less) but closer to one full gravity (Earth standard), is the size of the 'room' required to create that much tidal force. The smaller it is, the faster it would have to be spinning to create sufficient centripetal force to give the impression of that 'gravity', and subsequently the thinner the band of that 'gravity' as the further from the zone of that 'gravity' the less the force.

So unless it were some silly, tiny spinning room, the size of the 'station' required to create a 'natural gravitational' feel would far exceed the astronaut's ability to jump without outside propulsion. In that case, of course, the ability to then travel outside the functional 'gravity' into decreasingly accelerated zones (by removing his own laterally imparted velocity) he could then eventually reach the 0G center of the station or room.

However, with the additional concept of scale, the same could be said of the Earth itself. If you could 'jump' to a 'high enough' height, you could reach 0G (not really, but sufficient micro gravity as to be indistinguishable from 0G). This does give pause for thought, as not only would such a station need to be QUITE large, but that you'd better be wearing a low-pressure suit at the same time, as the atmosphere should be quite thin, since the Nitrogen/Oxygen atmosphere would ALSO be affected by the centripetal force of the station's rotation, thus not only be thin, but definitely not made up of the proper gasses for breathing (lighter gasses rising into the center of spin)... At least that's what I'd expect.

As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'.

Not exactly. Centrifugal "force" is what's known as a "pseudo-force". It's the result of analyzing events using a non-inertial reference frame. If you're on a merry-go-round, and you treat the merry-go-round as being stationary while the world revolves around it, you will find that objects have a tendency to go towards the outside of the merry-go-round. So within the "The merry-go-round is stationary" point view, you have to posit some force pushing the objects away from the center, which is the centrifugal force.

But this force doesn't "really" exist: when you analyze the situation from the point of view of someone not on the merry-go-round, the objects are traveling in a straight line. It's just that any straight line will necessarily go away from the center. (Draw a circle, then draw straight line next to it. Imagine traveling along that line. From the point of view of the circle, you're first getting closer, then traveling away.)

When something is rotating, its velocity is constantly changing: although its speed is constant, the direction is changing, so the velocity is changing. Changing velocity means acceleration, and acceleration means force. This force is directed towards the center. Imagine driving around a circle counterclockwise. If you were to let go of the wheel, you would fly off the circle. You have to constantly turn left to stay on the circle. So there is a force, but it's towards the center of the circle, and is called the centripetal force.

From an inertial reference frame pint of view, a force is needed to stay on the circle; the centripetal force. But from a circular motion reference frame, the object is stationary. So if there's a centripetal force pulling the object in, there must be another force, the centrifugal force, pushing it out. So if you're standing in a rotating space station, you'll going to feel a force of the floor pushing you "up" towards the center of the space station, and since it feels like you're at rest (the space station is moving with you), it's going to seem that there must be some force pushing you "down" into the floor.

The important point here is that the contact with the floor provides the centripetal force, but the centrifugal force exists in your reference frame regardless of whether you have contact with the floor. Go back to the example of driving in a circle. Suppose you drop a ball in the car. Before you dropped it, it was moving with the car, and so just as the car had a centripetal force keeping in circular motion, the ball had a centripetal force on it. But for the fraction of a second that it's in the air, it doesn't have the centripetal force.

For an outside observer, the car is turning left, while the ball is moving in a straight line. The car accelerates to the left into the ball, and when the ball lands, it is to the right of where it was dropped. For someone in the car, however, it seems like the car is stationary, and ball is accelerating to the right.

Similarly, if you were to jump in the space station, then since you are perceiving things in the space station's reference frame, it will seem like you are accelerating towards the floor. This apparent acceleration exists regardless of whether you're touching the space station. That you are accelerating in the station's reference frame doesn't require physical contact with the station because it's not a physical phenomenon. It's simply an attribute of the coordinate system.

All of this applies locally: if you jump up, your motion will, in the station's reference frame, on small scales be the same as if you were being pulled down by gravity. This is an approximation that breaks down as you go to larger scales. These deviations from the approximation show up as other pseudo-forces, such as the Coriolis force. So being in contact with the floor does matter in that it keeps you moving with the station and reduces these deviations.

You're not making any mistake except for thinking "artificial gravity" could be
nearly constant over a region as big as the structure itself. It really always
applies only to a region "much smaller" than the structure itself.

So, yes, a big jump (up and backwards on the wheel) could send you through the
middle of the wheel, where you would just float. Or, more simply, running fast
enough (backwards on the wheel) will cause you to levitate.

This non-constant variation of your "artificial gravity" in spacetime was already
explained as a "tidal force" in my2cts's comment.

Here's a really simple experiment you can perform right now.

You are standing on the surface of the Earth which is rotating at, depending on your latitude, somewhere between 1000 and 1600 km/h. If you dare, stand up and jump. Did you suddenly hurdle sideways at a great speed? No. Similarly a lamp hanging from the ceiling isn't thrown violently to one the side despite not touching the ground.

Your momentum is conserved. And momentum is mass x velocity. Your mass doesn't change when you jump, so we can just talk about conserving velocity.

Velocity is a vector meaning it has a magnitude and a direction. Standing still you're traveling at, say, 1200 km/h going sideways, but so is the surface of the Earth and your house and the air so you don't perceive it. When you jump you retain that 1200 km/h sideways velocity and add about 2m/s upward velocity. Since everything else is moving sideways with you, all you perceive is the vertical jump.

But only your linear momentum is conserved. Gravity is keeping you stuck to the rotating Earth. If you were to somehow nullify Earth's gravitational pull you'd find yourself slowly, and then more and more rapidly, seeming to rise up from the surface of the Earth. Your linear momentum is carrying you along in a straight line, but the surface of the Earth is curved. It seems like the surface is dropping away, but without gravity sticking you to the surface you'll travel in a straight line over a curved surface.

This difference between the linear and angular momentum is the Coriolis force and has practical effects on Earth and on a space station. On Earth it's why storms swirl. The much smaller scale on a space station can have subtle effects if you throw a ball, shoot a gun, or even on your inner ear.

Just a simplified example scenario depicting what the angular velocity answers say:

Assume a large open cylinder set up as a jogging track that is an entire rotating "Ring" of the a space station.

Instead of jumping, run as fast as the track section is rotating but in the opposite direction. You will appear to lose weight as you accelerate since you will no longer have the angular momentum of the floor but something less than that.

When you reach the speed the station is spinning, you will be weightless and lift off the ground (if you don't lose traction first). While above the ground in this way you will be weightless.

This also means, intuitively, that if you ran with the spin of the station you would get heavier.

Once you see it this way, you can see that jumping wouldn't really do anything in itself.

tl;dr

Problem 1:
After you become weightless the air is probably rotating with the floor of the track, so it will push your body giving it some angular momentum again at which point you will be "Pulled" back to the surface. If being able to alter your weight was a partial goal of this exercise room they could make the floor and walls fairly smooth so that the air stayed (mostly) in place and ignored the spin altogether. If this were the case you'd start with a wind at your back, but the faster you ran the weaker the wind would become and when you became weightless there would be no air pushing you.

Problem 2: In order to run the speed of the station the starting gravity may have to be less than 1g, Not sure of the math at all, but I am sure with the right starting speed you could run until you were weightless.