Would a ball bounced in a centrifuge return to the floor?

Question

In other words, does the centrifugal force inside a centrifuge only work to simulate gravity when an object is "attached" to the reference frame inside the centrifuge?

1G at sea level is normally about 9.8 m/s^2, correct?

However Centrifugal Force is different than Gravity.

If you were to jump straight up in a centrifuge, would you move laterally when you landed? Would you land at all, or continue flying straight up until you hit the other side of the centrifuge?

(Spoiler:)

Another example, the broken window inside the giant centrifuge at the end of Interstellar:

(a thrown baseball should be affected by the same forces as a bouncy ball bounced against the floor, correct? Either continuing flight path with 9.8 m/s^2 downward acceleration, or not)

If a window, suspended above the floor of a centrifuge, were broken, would the pieces fall to the floor of the centrifuge, as with gravity, or continue on a relative velocity to the centrifuge?

Answer 1

Your intuition is correct. So-called centrifugal force is a product of mass and acceleration in a rotating frame. If you're in a centrifuge, and you let go of the rotating frame, you're no longer accelerated radially, but merely keep the momentum imparted onto you by it and your own force as you let go of it.

If this centrifuge was setup in zero gravity, like e.g. space wheels, if you jumped, then your trajectory would be a product of both the direction of your jump (your vertical component), as well as direction of the frame's rotation at the point you let go of it (your horizontal component). But, unless something acts on you while in air so the product of both these velocities you're now a subject to change its magnitude (either accelerate or decelerate you, like, say, air within the space wheel co-rotating with it), you're now in constant motion, i.e. no longer accelerating, so no simulated gravity by constant acceleration.

Let's illustrate this a bit; If we say that you jumped at the same speed that the wheel rotates and towards its center of rotation, you'd jump at a 45° angle to your normal and the same angle towards the center of rotation. Since the rotating frame moved while you were in the middle of the air, you might land at the same spot you jumped from, which would seem a bit spooky, but that's the fun of rotating frames. This perceived effect on your motion once you let go of the rotating frame even has a name: Coriolis effect:

                                                     

This image is from Wikipedia and might not be the best to illustrate the point since it shows an object falling in a rotating frame away from the center of rotation, so here's a couple of animations from a film taken during cosmonaut training in Star City centrifuge system in the 60's (to my knowledge still the largest centrifuge that actually hosted people during experiments, even though Americans built some larger ones that were never really used):

  

Better example, but tells only half of the story of jumping from the rotating frame towards the center of rotation, since it doesn't also include return path back to the rotating frame. And this centrifuge was built on the Earth with additional vertical component of Earth's gravity, so it wouldn't be the best example anyway. I think I'll rather let you imagine how this would look like in zero gravity, but try to first guess in which direction the centrifuge on the last two animations rotates (hint: not the one you'd think).

Answer 2

The short answer is that when you throw a ball in a centrifuge in space, it travels in a straight line until it hits something. It experiences no pseudo gravity and effectively is not experiencing any acceleration.

The problem is simplified somewhat by evacuating the air so it's a vacuum. Lets say the surface of the centrifuge is moving at 50km/h, you're standing on the surface, and you can throw a ball at 50km/h.

First you drop the ball - simply releasing it from your hand. The ball will fly forward in a straight line at 50km/h, you are also moving at 50km/h, but will be pushed up by the surface of the centrifuge. This will create the visual illusion the ball is falling even though it's actually you being pushed up.

Next you throw it backwards at 50km/h so it has zero net velocity. What happens? The ball stays right where it is when you threw it, hanging in space with no force acting on it. If you stay standing where you are on the surface of the centrifuge, next revolution you will smack into the ball at 50km/h. From your perspective, it will look like the ball flies away from you at 50km/h, flies around the centrifuge, then smacks you in the head - even though it was actually just hanging in space. This is where a centrifuge deviates most absurdly from actual gravity, in a rotating cylinder, the illusion of gravity is caused by the object moving in a straight line until it intercepts the floor (which is curving up), an object with zero net velocity will simply hang there as the floor of the cylinder spins under it.

So if you throw the ball at 50km/h in the direction of motion, then it will travel in a straight line at 100km/h and smack into the surface of the centrifuge due to the curvature of the cylinder. The trajectory is actually no different to throwing the ball at 100km/h in a non-rotating cylinder, in both cases the ball will travel in a straight line, getting closer and closer to the floor, until it hits it. The apparent trajectory is not a parabola, as it would be with gravity, but instead a segment of a circle, which can however be a good approximation for a parabola for small angles.

When you add air, you get air resistance which will act to generally slow the ball, and air will be swept along with the surface of the centrifuge, effectively making a wind that will carry things with it. Air pressure will tend to push less buoyant objects to the surface of the centrifuge, and lighter than air objects to the center.

Exactly whether you can throw an object across the centrifuge to the other side (or "around the world"), depends on being able to overcome the velocity of the surface of the centrifuge. If the surface was moving at 100km/h, then a ball thrown at 50km/h will always fly ahead in a straight line and smack into the floor. The larger and faster the centrifuge, the more closely it will emulate gravity.

Answer 3

The centrifugal force is a product of your frame of reference. Not a product of what you are or are not attached to. So it would not disappear simply because you no longer are in contact with the ground.

However, if you are in a reference frame with centrifugal forces you are also in a reference frame with coriolis forces. This pseudo force is proportional with your velocity relative to the frame of reference.

The consequences of this is that when you leave the floor where and if you return depends on your velocity. - However you don't actually have to leave the floor for this force to be noticable. If you start travelling against the spin of the cylinder the coriolis force would start to counteract the centrifugal force. When you travel at the same speed as the cylinders rotation. This means that you would be moving at the same veocity as the center of the cylinder from the perspective of an inertial reference frame. At this point the coriolis force and the centrifugal force presisely counteracts, and you would be left floating along the surface of the cylinder.

Answer 4

To simplify the explanation, let's imaging you are standing on the inside surface of a large rotating drum in space. If you jump straight "up" (toward the center of rotation), you will have a velocity vector which will be the sum of your original motion along the circumference plus the "inward" motion you gave yourself by jumping. Excluding the effects of friction with an internal atmosphere, you will continue to travel in a straight line along this new vector (at uniform velocity). This path will be a chord line intersecting the drum at two points, one of which is the point of your "lift-off", the other will be your inevitable "landing" back on the drum surface.