“Artificial Gravity”
Many science fiction stories feature objects that are “spun” to create a gravity-like force holding objects on their inner surface. The formula for the effective center-ward acceleration caused by this “force” is well-known to be 
where VR is the tangential velocity of the object, and R is the object’s radius, but I’ve found most explanations of this force to be unhelpful. The explanation below may be better, and is certainly simpler, as it doesn’t involve the vector calculus usually used to derive the above equation, just algebra, trigonometry and some geometric reasoning.
First, let’s draw a picture
The circle is the spinning object (call it a space station for now): its edge is traveling at a speed of VR. Therefore if someone holds an object at a height of d0 above the inner surface of the object, it will be moving at a velocity of 
When the person lets go of the object, what happens? It keeps moving at its initial velocity along the path marked “Path of released object,” until it hits the inner surface of the station. From the point of view of the person that dropped it, the object is falling. How fast? Let’s figure it out…
The distance of the object from the center of the station equals 
(by the Pythagorean theorem). That distance is also (R-d). If we square both sides and assume that R is much greater than d or d0, then we arrive at:
Canceling out a bit and reorganizing gives:
If again we use the fact that R>>d0 then
For a planet, the equation for the height of a dropped object is 
so simply by comparing these equations it is clear that close to the inner surface of the station, the acceleration due to “artificial gravity” is
But while the released object appears to be falling, the inner surface of the station is moving, too. Does that cause a visible effect? To find out the answer to that question we need to evaluate theta.
From the illustration above, it’s fairly easy to see that
; for small angles that reduces to 
. At the same time, the station’s inner surface is rotating at a constant rate, so that 
(exactly). The relative rotation
is zero as long as the small angle approximation is appropriate.
Let’s see what happens with some real numbers. Larry Niven’s Ringworld features a ring whose radius is the same as the radius of earth’s orbit (150 million kilometers) and whose rotational speed is approximately 1180 kilometers per second. What is the effective artificial gravity on the inner surface? It’s about 0.0093 kilometers per second squared, or 9.3 meters per second, about 95% of earth gravity. How fast would a space station with a radius of 1 kilometer have to rotate to produce earth-like gravity? What would happen on that space station if you dropped something from a high enough point that that the small angle assumptions weren’t valid? Try calculating the results of dropping something from a height of 100 meters. Does it reach the ground sooner or later than it would on earth? Will it land directly under where it was dropped? If not, how much off will it be?
To see the effects of trying to launch something straight up from the inner surface of a station (or the Ringworld) see the figures at http://members.optushome.com.au/guests/PhysicsinSF.html
