Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: rotating space station



Carl and all,
(It is rare that I get to a question before others have answered. So, when this one about the rotating platform came through this morning, I started working on it, had to go to class, came back and worked on it some more, all the time expecting someone, or several people, would post an answer. Since no one has, yet, here goes.)
My conclusion is exactly opposite what Carl predicts. He argued (see below) that since the floor of the rotating space station is curving, the tangential component of velocity decreases, allowing the "dropped" ball to get to the floor before the point on the floor that is radially out from the point the ball was released. The problem with this argument is that the tangential velocities of the ball and the floor are different, hence the floor is traveling faster than the ball to begin with. So, it is not clear from the component argument which way the ball will go - ahead or behind.
To analyze the problem, consider the distance the ball travels before it hits the floor to be x. This is tangent to the circle of radius r1. The tangential speed at that point is v1 = r1(omega), where (omega) is the angular speed of the rotating platform. Now, the point on the floor "directly below", or radially out from the ball is moving at speed v2 = r2(omega), where r2 is the radius of the floor of the platform. h = r2 - r1 is the height from which the ball is dropped. Let A = the angle the platform has rotated through when the ball hits the floor. Then x = (r1)tan(A) and the arc length the platform rotates through is s = (r2)A. (It really helps to draw a picture, but that is left as an excercise for the reader. :-) )
The time for the ball to cover the distance x is t1 = x/v1 = x/(r1(omega)). The time for the point on the floor to reach the point of intersection with the line x is t2 = s/v2 = s/(r2(omega)). Taking the ratio of these times, t1/t2, allows direct comparison. t1/t2 = [x/r1(omega)][r2(omega)/s]. The omegas cancel, leaving t1/t2 = (x/r1)(r2/s).
The ratio x/s = (r1tanA)/(r2A). Hence, t1/t2 = (tanA)/A.
Since the tangent of an angle is greater than the angle (in radians) for all angles, (tanA) > A, the conclusion is that t1 > t2. This means that it takes *longer* for the ball to hit the floor than it takes for the "radially outward" point on the floor to get to the same place, or to cross the line the ball is moving along. The conclusion, therefore is, the ball hits BEHIND the astronaut.
The question then needs to be asked, How far behind does the ball land? To answer this you need some numbers for the size of the space station, the rotation rate, etc. Since no real station exists, we are free to make up one with more or less arbitrary dimensions. I chose an outer radius of 500 feet, just to see what comes out. This gives r2 = 500 ft, and take r1 = 494 ft, so the height of drop is 6 feet, a typical value on earth. To get omega, we require the acceleration at the outer radius to be 32 ft/s/s. This gives omega = sqrt[(32 ft/s/s)/500 ft] = 0.253 rad/s. (The period for this size is 24.8 seconds, by the way. That seems like a fast rotation! 2.4 rpm. But, then, g is a large acceleration.) v1 turns out to be 125 ft/s for this case. The angle A = 0.155075 radians, and tanA = .15633. The ratio of times is thus:
t1/t2 = (tanA)/A = 1.00809. The ball travels x = r1tanA = 77.227 ft, in a time t1 = x/v1 = 0.6178 s. t2 = t1/1.00809 = 0.6128 s.
To see where the ball hits relative to the floor, we need to see how far the floor rotates in the two times, t1 and t2, and subtract.
delta(s) = r2(omega)(delta-t) = 0.63 ft.
Conclusion: (In case any of you actually got through all this!) The ball will fall *behind* the astronaut, and by a significant amount, by several inches for the 500-ft space station assumed.
Another part of the answer to the Serway question would be to analyze the acceleration of the dropped ball. I will leave that to another responder, though it is not any more difficult than what I have given here.

Rondo Jeffery
Weber State University
Ogden, UT 84408-2508
801-626-6202
rjeffery@weber.edu



mungan@USNA.EDU 09/28/00 04:37PM >>>
Serway "Principles of Physics" 2nd ed. Chapter 5 Question 6:

Consider a rotating space station spinning with just the right speed
such that the centripetal acceleration on the inner surface is g.
Thus astronauts standing on this inner surface would feel pressed to
the surface as if they were pressed into the floor because of Earth's
gravitational force. Suppose an astronaut in this station holds a
ball above her head and "drops" it to the floor. Will the ball fall
just like it would on the Earth?

Detailed (excuse my sarcasm) answer given in the Solutions Manual: Yes.

I think that answer is wrong. From the point of view of an inertial
observer, the "dropped" ball will continue horizontally at a constant
speed equal to the tangential speed of the space station until it
hits the floor. Meanwhile the space station rotates and hence the
component of the astronaut's velocity in the original horizontal
direction decreases. Thus it seems to me that the ball should hit the
floor ahead of the astronaut, and not right at her feet like on Earth.

Am I being stupid? Carl
--
Dr. Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Annapolis, MD 21402-5026 >mailto:mungan@usna.edu
http://physics.usna.edu/physics/faculty/mungan/