From: Henry Spencer <henry@zoo.toronto.edu>
Newsgroups: sci.space.tech
Subject: Re: space colony construction (was Re: Terraforming...)
Date: Sun, 10 May 1998 23:46:01 GMT

In article <rmn2cshhtq.fsf@fluent.com>, John Stoffel  <jfs@fluent.com> wrote:
>Does anyone know how to compute the instability of a revolving object?
>I know small radius vs a long axle length is unstable (i.e. bottle),
>and large radius, small axle (bicycle tire) is more stable.  Where
>does the trade off happen?  When the axle and radius are equal?  Or is
>it a function of the average mass (moment?) from the center of
>rotation?

It's a matter of the moment of inertia (integral of radius^2 * mass).  In
the short term, spin is stable only around the axis of either highest or
lowest moment of inertia; in the long term, for most practical purposes,
it is stable only around the axis of highest moment of inertia.

(The spacecraft has both angular momentum and kinetic energy of spin.  In
an isolated spinning spacecraft, angular momentum is conserved but energy
can be dissipated, e.g. as heat in wiggling flexible parts.  Since
momentum is proportional to velocity and energy to velocity squared, you
get minimum energy for a given amount of angular momentum with the mass
farthest out from the center, because that gives the slowest spin.  So
any energy-dissipation mechanism -- fluid sloshing in tanks, flexible
antennas flexing, etc. -- strongly biases the spacecraft toward ending
up spinning around the axis of highest moment of inertia, giving the
slowest possible spin.)
--
Being the last man on the Moon                  |     Henry Spencer
is a very dubious honor. -- Gene Cernan         | henry@zoo.toronto.edu