Re: origin of inertia
From: JM Albuquerque (jm.aREM.OVE_at_sapo.pt)
Date: 03/28/05
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Date: Mon, 28 Mar 2005 02:40:09 +0100
"bz" <bz+sp@ch100-5.chem.lsu.edu> escreveu na mensagem
news:Xns96267F072C817WQAHBGMXSZHVspammote@130.39.198.139...
> "JM Albuquerque" <jm.aREM.OVE@sapo.pt> wrote in news:3ao3j7F6c0u01U1
> @individual.net:
>
> >
> > "bz" <bz+sp@ch100-5.chem.lsu.edu> escreveu na mensagem
> > news:Xns9625E11F8F420WQAHBGMXSZHVspammote@130.39.198.139...
> >> "JM Albuquerque" <jm.aREM.OVE@sapo.pt> wrote in
> >> news:3amf7iF6coaueU1@individual.net:
> >>
> >> >
> ...
> >> >> > What do you see accelerating ?
> >> >>
>
> You are confusing force with acceleration. Work is force times distance.
This is a free motion device.
All the motions are free in a gyro.
Free like a rock in space.
To apply force over a free body you need to apply
acceleration.
F = ma
Without a force you cannot talk about any work.
> >> >> I see g attempting to accelerate all the particles in the top.
> >> >
> >> > Right.
> >> > Attempting, but not doing it.
> >>
> >> The force is there.
> >
> > It is the same situation as above.
> > The force is there but no work is being done in the vertical
> > direction (the direction that gravity acts).
>
> Zero work is done because W = F d and there is no d.
You have said that precession carries a torque on it.
I do see "d" (distance) in the precession motion.
What I don't see is the force.
> >> > And, because of symmetry, all those centripetal forces
> >> > cancel out producing zero force on the center of mass.
> >>
> >> They would cancel if it were not for gravity.
> >
> > I don't understand what you mean.
> > Can you explain please ?
>
> Gravity acts in a non symetrical way on each of the particle. If it were
> symetrical, the gyro would have no weight.
We were talking about centripetal forces, not gravity.
> ...
> >>
> >> wrong. The center of mass of the gyroscope moves from one point to
> >> another. Force times distance is work.
> >> Work is done.
> >
> > I cannot agree whit that.
> > The center of mass of the gyroscope moves from one point to
> > another, but there is no force associated with that motion.
>
> There is the force of gravity. Conservation of angular momentum changes
the
> direction, just like a pully would do.
>
> >
> > You want to backup the acceleration of gravity, which exists
> > in the vertical direction, as being translated to the horizontal
> > plane where the motion occurs. What physical support do
> > you have to claim that ?
>
> The gyro precesses. Work = force times distance.
What force.
The gyro is free to rotate, so that the only way to apply
force is by means of inertia = acceleration.
It is like a free rock in space.
To apply force over a free body you need to apply
acceleration.
You don't have acceleration in the precession motion.
> The center of gravity of the gyro moves a distance. Gravity is the force
> that moved it that distance. Work is done.
You are wrong.
The gyroscopic effect is ruled by the formula below:
wp = mg r / L
being:
L - the disk main angular momentum = I w
mg r - torque due to gravity.
wp - the precession angular speed.
Did you realize that the precession angular speed depends
on the main angular momentum.
If you have the disk spinning at very high speed the
precession angular speed will be very small for the
same mass.
If you have the disk spinning at low speed the
precession angular speed will be very fast for the
same mass.
Did you realize there is minimum speed and below
that speed the gyroscope won't work anymore ?
How that relates to your force / torque misunderstanding ?
Assume the same torque due to gravity (T = mg r).
If the disk spins at 1000 RPM you have wp = 100 RPM
If the disk spins at 10000 RPM you have wp = 10 RPM
Is the work done different in both cases ?
How can you relate your "work being done" with the
above facts ?
> > The fact is that the horizontal motion of the center of mass
> > of the gyroscope occurs at constant speed - wp - the
> > precession angular frequency. To have a force you need
> > accelerated motion and you don't have that accelerated
> > motion on the precession plane.
> > Force = mass x acceleration
> > Torque = Moment of inertia x dwp/dt
> > and dwp/dt = 0 (no acceleration)
> >
> > No acceleration = No force = No torque.
>
> Mass sitting on table. rope over pully, weight. Gravity exerts force on
> weight and pulls rope. Mass on table slides. Sliding friction balances G
and
> mass moves at constant velocity.
>
> Work is done. Remove force of gravity (weight) and movement/work stops.
>
> Gyro is similar. Do experiment in zero G. Substitute force pulling on rope
> for gravity. Remove force, precession stops.
Gyro is not similar.
> >> The acceleration of gravity did the work.
> >
> > The force of gravity produces no work.
> > We have already seen this above.
> > Nor in the vertical plane, nor in the horizontal plane.
> >
> > If the force of gravity produces work we will have free
> > energy and a perpetual machine of first kind.
>
> Not free. But falling water produces work. The force of gravity does the
> work.
Only when it falls
Nothing falls in the gyro.
Nothing accelerates in the gyro and nothing falls.
http://physics.nad.ru/Physics/English/gyro_tmp.htm
> You are confusing acceleration and force. The FORCE of gravity produces an
> acceleration in a body that is free to fall.
F = ma
What is wrong about F = ma ?
What I'm confusing ?
The mass is there and to have a force over that mass Newton
says I need an acceleration.
> The FORCE of gravity will produce a constant velocity fall when the fall
> is opposed by a force.
>
> Example: Sky diver falling at terminal velocity. Work being done by
> gravity.
> Force times distance. Diver is not accellerating because the force of
> air friction is in equilibrium with force of gravity.
You have and opposite force (air resistance) that is proportional
to the fall speed squared. That's why you get constant speed,
because the opposite force depends on the speed square and
so there is a speed where both forces balance.
Where do you see a dissipative force proportional to
angular speed square in the gyroscope ?
> ...
> >> no, the angular momentum of the system is conserved, not constant.
> >
> >
> > If the initial angular momentum is a constant function (it does
> > not depend on time) and you claim that angular momentum
> > is conserved, then it is obvious that the final angular
> > momentum also must not depend on time (is constant).
> >
>
> I didn't say it didn't depend on time. Angular velocity is d_theta/d_t.
Yes, but it is a constant speed.
d_theta/d_t = 0
> If you have angular velocity of any component of the system changing with
> time, then you must take this into account in computing the total angular
> momentum.
Nothing changes in time (short time):
http://physics.nad.ru/Physics/English/gyro_tmp.htm
> >> > is
> >> > constant. Hence, the main angular momentum is
> >> > constant and conserved.
> >>
> >> no. Total angular momentum is conserved. Main angular momentum changes.
It
> >> is a vector. When the direction changes, it changes.
> >
> > The magnitude is conserved, not the direction.
> > I agree.
>
> The direction of the total angular momentum of the system is conserved
> because angular momentum is a vector and angular momentum is conserved.
The direction of the total angular momentum is rotating
at precession angular speed.
> ...
> >> Wrong. The total angular momentum OF THE SYSTEM is conserved. The
forces
> >> and torques must be included in the calculations of the total angular
> >> momentum.
> >
> > Well, now you include forces and torques in the angular momentum.
> > How to you translate a torque into angular momentum ?
>
> When the torque attempts to change the angular momentum by changing the
> direction of the vector I, a resultant force is produced at 90 degrees
that
> keeps the total angular momentum constant.
What force is produced at 90 degrees ?
To apply force over a free body you need to apply
acceleration.
You don't have acceleration, nor any explanation
for the fact that precession depends on the main
angular momentum.
wp = mg r / L
> > Notice that:
> > Torque = dL/dt
> > And you claim that L = constant.
>
> The sum of the angular momenta must be constant. If you attempt to change
the
> angular momentum, your angular momentum must change to keep the sum total
> constant.
I agree.
That's why I keep saying that it only works for gravity.
> Under zero G, if you try to turn a screw, you turn.
>
> Do the physics lab experiment with the bicycle tire.
>
> Go into a bar with bar stools that freely rotate. Sit on one and try
spinning
> a large mass with its axis of spin over the axis of rotation of the bar
> stool. Angular momentum is conserved and you will spin in the other
> direction.
>
> Do the gyro experiment, holding onto one end of the bicycle wheel axis,
> holding the axis out, parallel to the ground, support it only on one end.
> Keep that end at a constant hight. When gravity pulls the COG downward,
you
> will turn. You precess with the gyro. Gravity is doing work. It moves the
> center of mass of the system 'you, wheel, bar-stool-top'. The distance
moved
> times the force of gravity is the work done.
Right.
Only when gravity pulls the COG downward I will turn.
Once down it stops.
Here I do see the accelerated motion - the fall.
> ...
> >> wrong. W=Fd, gravity exerts a force and the COG moves a distance. The
fact
> >> that it moves at right angles to the force is NOT important. The force
and
> >> the distance are the important thing.
> >
> >
> > So, when I have a mass on the top of an horizontal table
> > and I move that mass (no friction) horizontal on the table
> > I'm producing work against gravity ?
>
> When you move the mass, whether or not there is friction, you do work.
>
> Against gravity? NO, I made no such claim about moving an object on a
> frictionless plane.
>
> Gravity moves the Center of Gravity of the gyro. Gravity does work.
And why is that different from moving an object on a
frictionless plane, against gravity of course ?
(snip the rest)
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