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Re: rotating space station



On Thu, 28 Sep 2000, Rick Tarara wrote:

I'm trying to understand this (in the inertial frame) by looking
at the angular velocity. To fall behind the feet doesn't the angular
velocity of the ball have to slow? If so, what causes the slowing?

Hey! Good point. And another nice way of seeing what is going on
in the inertial frame.

The angular velocity of the ball (about the center of the space
station) is v_tan/r where v_tan is the instantaneous tangential
component of v (in the inertial frame) and r is the instantaneous
radial distance of the ball from the axis of rotation.

Now, v_tan monotonically decreases since the ball starts out
purely tangentially, develops a monotonically increasing radial
component, and maintains the magnitude of its velocity.
Furthermore, r monotonically increases. Thus, the angular
velocity of the ball monotonically decreases.

In fact it is quite easy to work out the following result:

v*(R - h)
angular velocity of the ball = ------------------
(R - h)^2 +(v*t)^2

where v is the (constant) speed of the ball
R is the (constant) radius of the space station
h is the (constant) "height" of the drop
t is the (monotonically increasing) time

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm