Re: what is the relation between force and energy?

From: Andr? Michaud (srp_at_microtec.net)
Date: 09/28/04 

Date: 28 Sep 2004 16:05:34 -0700

sbharris@ix.netcom.com (Steve Harris sbharris@ROMAN9.netcom.com) wrote in message news:<79cf0a8.0409281039.2a7a8ee6@posting.google.com>...
> srp@microtec.net (Andr? Michaud) wrote in message news:<562f286c.0409280301.1b1844be@posting.google.com>...
> > > > > Sorry, but there is no experimental data to show that.
> > > >
> > > > Really! Just try to deflect electrons in an accelerator without
> > > > turning the power on, or even in a simple bubble chamber.
> > >
> > > Hey, it used to be done all the time in low power machines, by simple
> > > proximity of a permanent magnet. Cyclotrons worked that way also. No
> > > power needed to bend a particle path. Permanent magnets don't even
> > > have an "ON" switch, you notice? They bend charged particle paths
> > > anyway.
> >
> > Permanent magnets exert a force, so induce directed energy which causes
> > electrons to be deflected.
> >
> > Energy is use up. Work is performed.
>
> No, no, no. Speed is unchanged at all times, kinetic energy is
> unchanged at all times (E = 1/2 (speed)^2, thus no work is performed.

Longitudinally ? Of course not. Why should the speed change since the
force is acting orthogonally?

> No energy is used up.

Yes. Since there is transverse motion. Work is performed.
 
> Your assertions are ridicuous. Where do you think this energy that is
> "used up" comes from?

Induced by the tranverse magnetic force, what else? Force always
induces energy. It was used up as work to actualize the change in
direction.

You proposed yourself the permanent transverse magnet setup.

> Where do you think it goes, since the particle doesn't get any of
> it?

How could it change direction if it didn't get any of it?

> Give me an equation for it.

CM doesn't allow it, since CM doesn't cover the case. So no
equation for you.
 
> > > > > You can take two objects in zero g and set them spinning about a
> > > > > common center of mass, connected by a length of rope or fishline.
> > > >
> > > > I think that an important detail must have escaped your attention.
> > > > I was specifically talking about _stable_ electrostatic or gravitational
> > > > orbits. So your example is non sequitur.
> > >
> > > Not really. My gizmo's as stable as anything in a Newtonian orbit is.
> >
> > Assumption. No proof.
>
> Every rotating planet is proof. I didn't say revolving planet.
> ROTATING planet. Rotating around its axis.

I got the ROTATING bit. Same answer. The planet is in stable gravitational
equilibrium, ROTATION as well as translation. Rotation mainly regulated
by the presence of the Moon.
 
> > > > However, I challenge the community to have this experiment carried
> > > > out in real life in deep space.
> > > >
> > > > I proposed it years ago precisely to verify this point.
> > > >
> > > > I predict that such a system (even a simple spinning wheel) will
> > > > slow down and eventually stop rotating, because they are not stable
> > > > systems in local electrostatic or gravitational equilibrium.
> > >
> > > Ahem, the experiment is carried out every time somebody launches a
> > > spinning satellite, or a satellite stablized by spinning gyroscopes.
> >
> > Absolutely not. Unstable orbits. All launched satellites eventually
> > spiral back to Earth unless their orbit is regularly corrected.
>
> Bull***! This statement is completely wrong. Comsats at 22,300
> miles out do not "spiral in".

You are wrong.

> They're up there for good.

Whatever time it takes, they will eventually come down.

> And in any case, I'm talking about the spin of the satellite itself
> around its own axis.

Same thing. If unsupported in a timely manner, the rotation will eventually
stop at some point in the future, unless it crashes down first.
 
> > > What, you think they wouldn't notice such an effect in the exquisitly
> > > balanced gyros of Gravity Probe B?
> >
> > No. You need to set your spinning wheel in deep space far from any
> > planets.
>
> Why should you need to do that?

To prove your point that such unsupported mechanically induced rotation
will last eternally.
 
> > > All rotating solid objects are held together by mechanical tensile
> > > forces which have nothing to do with "coulombic" potentials.
> >
> > I disagree.
>
> Well, you're wrong.
>
> >
> > > Of course these tensile forces are ultimately electrostatic in nature,
> >
> > Wow! Do you see that you directly contradict your previous sentence ?
> > The quickest 180 deg swing I witnessed in years!
>
> Nope. Van der Waals potentials are not Coulombic. Not even close.

It definitely is.

> "Electrostatic" may be the wrong word. They are electromagnetic,

Of course.

> but they don't have an electrostatic form of potential. The Van der Waals
> potential is typically something like 1/r^6 on one side, and 1/r^12 on
> the other. Potentials for small excursions in chemical bonds are
> *not* Coulombic either, but are typically springlike with (delta-r)^2
> type potentials.

So?
 
> > > being made from various kinds of atomic bonds. But the same is true
> > > in a fishline. The point is that there's no simple Coulombic force
> > > variation-- the expression for the tension force (elastic modulus)
> > > is much more complicated. And of course varies with the type of object.
> >
> > Mechanical rotating motion has nothing to do with stable gravitational
> > or electrostatic orbits. Mechanical rotation is not compensated by a
> > permanently acting force while the stable gravitatinal or electrostatic
> > orbits are.
>
> Of course mechanical rotation is compensated by a permanent acting
> force.

I meant compensation of energy sustaining the rotation of the body
after initial impulse.

> Without such forces any rotating object would fly apart.
> Indeed, if you make any rotating object rotate too fast (like a
> flywheel) the compensitory internal forces will be exceeded and it
> will fly apart.

Internal forces that hold the particles together are irrelevant to
the problem.
 
>
> > > I note that every solid rotating planet tests your theory also.
> >
> > Absolutely not. The planets are in _stable_ gravitational equilibrium,
> > not in uncompensted mechanically induced rotation.
>
> They are in mechanical rotation. The Earth revolves around its axis
> every 24 hours.

Yes. See above. Gravitational equilibrium.
 
> > > Last I looked, they were all still spinning pretty well after
> > > billions of years.
> >
> > That was my point precisely.
>
> SPINNING. Not just going around the sun, but SPINNING.

Yes. I got the point. Same answer.
 
> > > > > No gravity or Coulomb force is needed.
> > > >
> > > > For establishing a stable electrostatic or gravitational orbit,
> > > > yes they are. For gravitational, we would of course be talking
> > > > about Newtonian type gravitation.
> > >
> > > No, they aren't. Every particle in a rotating solid object like a gyro
> > > or a satellite or even the planet Mars, is in a stable circular motion
> > > "orbit."
> >
> > No way. Electrons are in stable electrostatic states, just like planets
> > in the solar system. Not in mechanically induced rotating motion.
> >
> > You really do not perceive the difference, I observe.
>
> I'm not talking about planetary revolution, but planetary rotation. Do
> you understand the difference?

Yes. Same answer. See above.
 
> >
> > Too bad.
> >
> > > > > But they keep going without showing any hint that much energy is
> > > > > required to keep them in circular motion.
> > > >
> > > > You mean in stable circular motion ?
> > >
> > > Yes.
> > >
> > > >
> > > > How do you know? The experiment never was carried out.
> > >
> > > Rotating Earth to Andre, Earth to Andre. Over?
> >
> > See above. Mechanically induced rotation is not the same as stable
> > electrostatic or gravitational orbital motion.
>
> I'm glad you think not. But as you see from the spin of planets (and
> asteroids), it's just as stable.

Absolutely. See above. Not the same as mechanically induced rotation.

The stable rotation of planets is determined by gravitational
equilibrium and is not mechanically induced, just like stable
translation.
 
André Michaud

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