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Re: rotating space station



Would there also be a wind into the face of anyone facing the direction of
rotation? 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?

Pretty confused,

Rick

----- Original Message -----
From: "John Mallinckrodt" <ajmallinckro@CSUPOMONA.EDU>

Indeed, good explanations of this fact exist for both the inertial
and rotating observer. In the inertial frame, the distance
traveled by the ball is sqrt(2Rh+h^2) where h is the height of the
release and R is the radius of the space station and the speed of
the ball is omega*(R-h) so the time it takes the ball to fall is

srqt[ 2*(h/R) + (h/R)^2 ]
t_b = -------------------------
omega*(1 - h/R)

(Notice what happens as h/R --> 1.)

To reach the place where the ball will fall, the astronaut's feet
must move through an angular displacement of arccos(1 - h/R), so
the time it takes the feet to get there is

arccos(1 - h/R)
t_f = ---------------
omega

Now it is not *too* difficult to show that t_f < t_b for all h >
0. Perhaps the easiest way to see this is simply to plot

srqt[ 2*(h/R) + (h/R)^2 ]
t_b/t_f = -------------------------
(1 - h/R)*arccos(1 - h/R)

as a function of h/R and note that it is a monotonically
increasing function. Thus, the ball lands behind the feet.