Artificial gravity on rotating spaceship? One of the possible ways to simulate gravity in outer space is to have a rotating spaceship, so that the centrifugal force experienced provides a gravity-like force.
My question is: shouldn't this only work when our feet are touching the floor of the spaceship? Only in that case the floor is providing a contact force to balance the centrifugal force.
If we jumped, there is no gravity within the spacecraft so what is it that would make us come back down?
Also: imagine we had a shower: what would make the water fall down?
 A: If you jumped "straight up", you would still have a horizontal component of velocity (relative to a nonrotating frame), so you would still end up coming "back down". 
Likewise, the shower water is moving horizontally in a nonrotating frame, which makes it collide with the floor eventually (since the floor is curving upwards in the nonrotating frame). But to a person on the ship, it looks as if the water was moving downwards, rather than the floor (and you) moving upwards.
More dangerous would be if you were to try to run in the opposite direction of the rotation; if you ran fast enough, you would eventually find that you had become weightless. This would also mean that your feet would no longer be touching the ground, the world would be spinning underneath you, and you'd have no way of getting back down again. 
Fortunately, since the air is also moving due to the rotation, the "wind" would eventually "slow you down" (technically it would actually speed you up) and you would eventually regain "gravity" and fall to the ground.
A: Don't forget that not only floor is spinning, but also the rest of the ship (including water in pipes and you).
This causes that the water is pulled "to the ground".
Since the acceleration is given by $a=v^2/r$ and $v=\omega*r$ you got $a=(\omega*r)^2/r$.
That gives you $a=\omega^2*r$.
This means that anything spinning around the axis of the ship is accelerated outwards proportionally to the distance from the axis. 
So if the shower is on the ceiling, water will accelerate downwards, but with increasing acceleration.
And remember, when you jump, you still have the velocity given you by the previous contact with floor, so after jump, you don't go just "up" but also "forwards" that means you're still spinning around the axis.
A: Tsiolkovsky figured it all out more than 100 years ago :)

Tsiolkovsky describes a cylindrical space ship: 100 meters long, 4
  meters in diameter, rotating end-over-end about its "central
  transverse diameter", with an endpoint velocity between 1 and 10
  meters per second, producing an angular velocity between
  (approximately) 0.2 and 2.0 rotations per minute, and a gravity level
  between 0.002 and 0.2 g.  These numbers were chosen to illustrate a
  concept, and should not be taken too seriously.  Nevertheless, they
  show that Tsiolkovsky understood the problems associated with high
  angular velocities, and the practicality of artificial gravity levels
  of less than one full g.

When you stand on the ground of the station, you are moving at the speed of the endpoint velocity, let's say 10 m/s. When you jump at the speed of 1 m/s, your velocity becomes $\sqrt{1+100}$, but its direction is slightly inwards now. So, you're going to hit the floor again, but slightly off the point from where you jumped. So, you really are not pulled by the ground, but rather get hit by the ground. 
UPDATE: here's the code and a plot to demonstrate what happens when a man jumps vertically at speed of 1 m/s, inside the spaceship of 100 m radius and endpoint velocity 10 m/s. The result is that a man will land 13 cm off the point he was initially in about 2 secs.
 
R=100;

ax0=0
ay0=-R;

vy=1;
vx=10;


phi=vx/(R) % angular speed
theta = (pi-2*( pi/2-atan(vy/vx)) ) % angle at hitting floor

bxf = R*sin(theta);
tau = bxf / vx % time to hit floor

axf = tau*phi; % angle the weel turns
d=(theta-axf)*R %distance from hit

t=(0:0.01:tau);
bx=ax0+vx*t;
by=ay0+vy*t;
plot(bx,by,'r')

hold on

ax=R*sin(phi*t);
ay=-R*cos(phi*t);
plot(ax,ay)


tau2 = theta/phi 

t=(tau:0.01:2*tau2);
ax=R*sin(phi*t);
ay=-R*cos(phi*t);
plot(ax,ay,'-.c')
xlabel 'x'
ylabel 'y'
title 'R=100;v_x=10,v_y=1'

A: Einstein said it: Acceleration is gravity and gravity is acceleration. When in contact with the floor you are accelerating (rotating) with respect to an inertial frame of reference and one can interpret that as a gravitational field. That's why you feel the force of your mass at your feet. You can juggle balls, do an experiment or make a measurement, everything just like in gravity. 
When you jump you are free falling that's why you feel weightlessness. Your velocity is constant with respect to a non rotating frame. But your spaceship is still accelerating so it catches up with you. That is unless the force you apply when you jump is carefully chosen so that you cancel the rotation speed of the ship at the jump point. Then w.r.to the non rotating frame you are moving with constant velocity in a straight line towards the center of rotation. Then you will effortlessly reach it soon enough and the whole spaceship will be rotating around you!
Similarly with water exiting a rotating pipe: It's in free fall but you are accelerating into the water. Of course it will come down at an angle that depends on the speed of rotation but if you position yourself accordingly you will get a full shower!
A: Let's say you're in a large sphere or cylinder that's rotating. You're floating at the center, and there's no atmosphere to drag you in any direction. Assume that the structure is light enough that gravitational pull due to mass is negligible (and would cancel out if symmetrical). I would assume that you're not going to be moved in any particular direction. Now, with a short burst of thruster, you slowly start moving towards the outer wall. Do you experience any additional acceleration due to the "artificial gravity" before you make contact with the outer wall? I would think not. Do you suddenly accelerate sideways at contact?
If I spin a bucket of water, the water is moving instantaneously tangential to me (and would fly off in a more or less straight line if I let go). I am applying centripetal force to the bucket and water, to accelerate it towards me, and the reactive centrifugal force of the water keeps it in the bucket. So, once I am in contact with the spinning outer wall, I would be constantly accelerated towards the center, and my equal and opposite reaction would feel like I'm being pressed against the outer wall/floor ("gravity").
So if I jump up, what happens? What is accelerating me outwards back towards the outer wall ("floor")? I will of course have some sideways (tangential) velocity from the spinning wall, which will make me drift in the direction of the wall's spin (Coriolis Effect) and eventually I should hit the floor again, but not in the same place I jumped from. That is, "down" is not well defined.
This would also apply to water released from a showerhead towards "down" (the outer wall). If a blob of water seeped out of the showerhead with no velocity in any particular direction, what (if anything) would make it move "down"? In a microgravity environment such as the ISS, they use fans to ensure movement. If it initially had some velocity "down", I would expect it to not accelerate downwards because there is nothing acting on it once it leaves the showerhead. It would be moving "too slowly" and curve off to the side of the shower stall due to Coriolis Effect, but other than that?
