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The Moon don't enter into it!
We can banish the moon and the generator still works. Not so with the tidal
effects.
Buone feste!
Mark
At 07:08 24/12/03 -0600, Brian Whatcott wrote:
>I offer this 'anti-physicist' thought.
>No observable effect at all depends on whether a physicist can explain it.
>
>So I will, instead :-)
>If there were a physicist's kind of gyroscope - you know, the one with
>frictionless
>bearings and zero air resistance and no eddy current losses, then we can
>say the
>gyro's momentum would remain intact, and the stored momentum of the Earth
>changes as electrical power is generated (in Leigh's idealization).
>
>It is this change that is comparable to tidal drag (at least, if the gyro
>is in the
> polar regions.) The change of Earth momentum
>is countered by the change in Earth Moon momentum as I recall.
> I am tempted to suppose that in lower latitudes, there can be
>precessional modes
>excited, but I have not thought much about the differences....
>
>If we have to cope with an engineer's gyro, then the enhanced side force on
>the gyro's pivots allow that device to spin down and lose momentum too.
>
>Merry Christmas.
>Brian W
>
>At 02:57 AM 12/24/2003, you wrote:
> >Brian, we agree that the KE of the system decreases, but for the angular
> >momentum to decrease we must find an external torque or else show how the
> >mass distribution changes to produce a change in the moment of inertia so
> >that the angular momentum does not change. (Take a look at Leigh Palmer's
> >gedankenapparat (unfortunately described under the subject line "phys-l
> >digest")). About the axis of the generator, you first bring the generator
> >rotor + gyroscope to rest (seen from a non-rotating frame), and use the
> >gyroscope flywheel spinning about a perpendicular axis to keep it there
> >while the earth continues to spin. Note that you will have exchanged ang.
> >mom. between earth and rotor in the process. Now when you use the
generator
> >to do work there is an electromagnetic torque between the rotor and the
> >earth - angular momentum is again exchanged between these parts of the
> >system. This is not the answer, however, since we can go on doing work
with
> >the generator for as long as we want, it seems to me. I note that there
> >must also be a precession caused about the third perpendicular axis when
> >you do work with the generator, but this only seems to complicate matters
> >as far as ang. mom. conservation is concerned. Mark
> >
> >At 23:01 23/12/03 -0600, Brian Whatcott wrote:
> > >Perhaps I am not fully conscious of the effect you're pondering.
> > >I am supposing that you spin up a flywheel, providing complementary m=
> > >omenta,
> > >then extracting momentum from the disk, with the identical reduction
> > >to the complementary agency?
> > >
> > >Perhaps you are considering the case of generating energy without
> > >losing momentum. A perpetual motion, kinda?
> > >
> > >Brian W
> > >
> > >At 08:34 PM 12/23/2003, you wrote:
> > > >Well, it is true that spinning up the flywheel will transfer angular
> > > >momentum to the earth, but after that, if the bearings are really go=
> > >od,
> > > >there will be a very small amount of angular momentum transferred be=
> > >tween
> > > >the wheel and the earth.
> > > >
> > > >Fred Bucheit
> > > >
> > > > >From: Brian Whatcott <betwys1@SBCGLOBAL.NET>
> > > > >
> > > > >How about treating the angular momentum as a storage method.
> > > > >Spin up a flywheel, and the Earth takes up a complementary
> > > > > momentum (if you must...)
> > > > >
> > > > >Brian
> > > > >
> > > > >At 10:01 PM 12/22/2003, you wrote:
> > > > > >[This was brought up on PHYSHARE-L as well, and I still don't ge=
> > >t how
> > > > > >rotational momentum is conserved. I realize that when I fully un=
> > >derstand
> > > > > >it, I am going to feel like an idiot, but here goes....]
> > > > > >
> > > > > >If the energy that is used to light the bulbs, etc., comes from =
> > >the
> > > > > >earth's rotation, then the earth needs to lose kinetic energy. K=
> > >E of a
> > > > > >rotating body is KE =3D 1/2 (rotational inertia) (rotational vel=
> > >ocity)^2. I
> > > > > >presume that the actual shape of the earth doesn't change, so it=
> > >s
> > > > > >rotational inertia remains constant. That means that rotational =
> > >velocity
> > > > > >must decrease.
> > > > > >
> > > > > >On the other hand rotational momentum =3D rotational inertia x =
> > >rotational
> > > > > >velocity . If rotational inertia stays the same and rotational v=
> > >elocity
> > > > > >decreases, then rotational momentum must get smaller. If rotatio=
> > >nal
> > > > > >momentum of the earth gets smaller, then the L of something else=
> > > must get
> > > > > >larger.
> > > > > >
> > > > > >So, the rotational momentum of what gets larger?
> > > > > >
> > > > > >Marc "Zeke" Kossover
> > > > > >The Hockaday School
>
>Brian Whatcott Altus OK Eureka!
Mark Sylvester
UWCAd
Duino Trieste Italy