, is: Gruppo di Meccanica Spaziale, Dipartimento di Matematica, Università di Pisa, Via Buonarroti 2, I-56127 Pisa, Italia
The total spin angular momentum of a rotor, made of two (mechanically coupled) equal
masses m, each of dimension R, spinning at angular velocity , is:
The angular momentum of the whirl motion with angular velocity 
in the inertial
frame and with radius of whirl 
is:
Since the total angular momentum has to be conserved it must be:
from which, since is constant, it follows:
The spin energy of the rotor is:
The energy (kinetic plus elastic) of the whirl motion is:
The time derivatives of (3) and (2) (using (1)) give:
and therefore
This is Eq. (32) of our Paper I. It is obviously a very important result: since in GG
we have in the inertial frame the extremely slow whirl motion at angular frequency 
(contrary to Schutz's example of the stellar case in which, in the inertial frame, 
) Eq
(4) represents the very small fraction of energy lost by the rotor which goes into
the destabilizing whirling motion. All the rest, that is 
!!, is dissipated as heat
inside the springs. So, the idea one might have that the faster the spin the higher the
energy gained by the whirl motion (by which argument GG would be highly unstable), is
proved to be incorrect.
The Celestial Analog
A \2body celestial model close to GG (with two masses only) is a binary system of two
spinning bodies. Tidal friction is known to change the spin period of the bodies and their
relative distance. For instance, in the Earth-Moon system the rotation rate of the Earth
decreases and the relative distance increases. Tides raised by each body on the other
produce a change in the orbital angular momentum through dissipation inside the bodies,
and it is amazing that a relation just analog to (4) holds. Let m, I and be the
mass, moment of inertia and spin angular velocity of the bodies orbiting around one
another on circular orbits at relative distance 2r and orbital angular velocity n.
We refer here to the case of equal masses and parallel rotation and revolution vectors.
The total spin angular momentum of the binary system is:
The orbital angular momentum of the system is:
Their time derivatives are:
The third law of Kepler for this \2body system reads:
and it relates 
to 
:
so that:
Since the total angular momentum has to be conserved it must be:
from which it follows:
The total spin energy of the bodies is:
The orbital energy of the \2body system is:
The time derivatives of (7) and (6) (using (5) and third Kepler's law) give:
and therefore
which is perfectly analog to (4).
(Anna Nobili- nobili@dm.unipi.it)
