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Re: centrifugal force



Thanks to all who helped. Two days ago I wrote:

Problem 7.17 in our textbook is about a cylinder
(diameter=5 mi) used as a space colony. The authors
ask: "What angular speed must such a cylinder have
so that the centripetal acceleration at its surface equals
Earth's gravity?" The expected answer, w=0.05 rad/s,
follows directly from w^2*r=9.8 (this is about 0.5 rpm).

I can now add this:

At one time a passenger must go from the nearly inertial frame
of reference to the rotating frame of reference. What does it
mean in practical terms? Suppose the cylindrical pipe is
already far away from our planet but it is not yet rotating.
Passengers of the space colony are "floating" inside the wide
pipe. Then the pipe stars rotating with some angular acceleration
(not too large) till the desired omega is reached. During this
process a passenger has two choices:

a) to be attached to the cylinder.
b) to continue floating.

At the end of the process those who were attached can detach
themselves and start jumping (or playing tennis) on the inner
surface of the pipe. They are at rest in the rotating frame.
But those who were floating will become orbiting satellites,
as far as those playing tennis are concerned.

To emphasize I would say that those playing tennis already
belong to the rotating frame while those circulating above
them still belong to the initial frame of reference. One can
not be at rest in two frames at the same time, unless these
frames are at rest with respect to each other.
P.S.
All above was in a vacuum. But what about air inside the
pipe? I suppose its viscosity will sooner or later put it to
rest with respect to the walls of the pipe. And satellites,
due to air resistance, would spiral toward the wall of the
pipe (called the ground). Is this correct?
Ludwik Kowalski