Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Path: utzoo!linus!philabs!cmcl2!harvard!seismo!rochester!bullwinkle!uw-beaver!fluke!msnthrp
From: msnthrp@fluke.UUCP (Jeffrey Gueble)
Newsgroups: net.sci
Subject: Re: How will it fall?
Message-ID: <1710@vax3.fluke.UUCP>
Date: Tue, 20-May-86 15:57:51 EDT
Article-I.D.: vax3.1710
Posted: Tue May 20 15:57:51 1986
Date-Received: Sat, 24-May-86 02:13:58 EDT
References: <632@tekigm2.UUCP> <3461@hplabsb.UUCP> <415@ccird1.UUCP>
Organization: John Fluke Mfg. Co., Inc., Everett, WA
Lines: 51

> In article <3461@hplabsb.UUCP> bl@hplabsb.UUCP (Bruce T. Lowerre) writes:
> >> Hello Out There,
> >> 
> >> Here's a little thought problem that might stir things up a bit.
> >> 
> >> Picture yourself on a space station similar to the one used in 2001
> >> (i.e. a spinning ring or toroid). You are standing in the middle of
> >> one of the decks near the outside edge of the ring and the spin of
> >> the station is providing a "gravity" about equal to that found on
> >> the surface of the Earth.
> >> 
> >> If you were to drop a ball (a simple release with no additional
> >> forces applied), would it fall straight down (along a line through
> >> the center of the ring and the point of release) or would it follow
> >> another path (relative to the aforementioned line)?
> >
> >Due to angular momentum (or lack there of) it will fall in a curved
> >path toward the opposite direction of the rotation of the toroid.
> 
> Theoretically, there should be the same curved path for a ball dropped
> on Earth.  It is not noticable because the earths rotation is so slow,
> and gravity so strong, that it would appear to fall straight.
> 
> As to whether you would see an observable curve, depends on the,
> diameter of the toroid.  Since you have specified pseudogravitation
> equal to earths,  a small toroid would have to spin very quickly, and
> curvature would be quite pronounced.  If the toroid were sufficiently
> large, the arc might only be a few seconds.
> 
> Does anybody have a formula for the relationship?


The ball will not follow a curved trajectory; it will depart from the
toroid on a linear path.  This path may not appear linear to the
observer located on the moving toriod, however, it will still be linear.
After the release, the ball will travel on a linear path, rotating
about an axis through its center of mass.  The problem is analogous to
a ball attached to a string  which is swung in a circular arc.  If the
string is released, the ball sails off in a linear trajectory
(actually, the effects of gravity make this a parabolic trajectory).

The important thing to realize here is that there are no forces
acting on the ball after release, and the momentum of the ball will be
determined by the initial angular and linear velocities. The angular
velocity will be the same as the angular velocity of the toroid.  The
linear velocity will depend on the distance from the release 
point to the center of the toroid, and will have a magnitude equal to
the radial distance times the angular velocity.  The only way that the
angular velocity could affect the trajectory would be if the act of
spinning induced a force which acted on the ball.  In free space, I
dont believe that this would occur.