Pendulum Generator

[steve_bank]

Not sensitive to initial conditions.

As originally stated, it was. The point of my diagram was to prove that. If the initial conditions were as shown then the system would have stayed in that state.

The question is about the steady stage conditions of the spuinning ball in a flat horizontal plane above the ground.

Ah -- you didn't specify the flat horizontal plane was above the ground. I exhibited the below-the-ground solution.

Size of the ball? Ignore size and treat it as a point. Assume the wire has zero mass.

That changes the system to this:

       Motor
        SSS
        SSS
        SSS
        SSS
         w
         w
         w
         w
         w
         w
gggggggggBggggggg
ggggggggggggggggg

Both the wire and the ball have zero moment of inertia, so they spin up instantly. The steady state condition is the ball rotates on its axis, (if that is even meaningful for a point ball -- perhaps it has quantum spin like a real point mass: an electron), in a flat horizontal plane exactly at ground level. So why would the steady state condition have the plane of rotation above the ground -- why would the ball ever lift off? The forces on it are balanced.

For extra credit if the wire breaks how long does it take for the ball to hit the ground?

You're introducing self-contradictory conditions: if you let the wire break, the breaking event is evidently not a steady state, which means there is no steady state and the answer will necessarily depend on initial conditions and transients.

Assuming you set the initial conditions in some way that causes the ball to fly in a helix above ground level, accelerating until the wire breaks, how long until the ball hits the ground -- if it ever does -- will depend on the breaking strength of the wire. Weak wire: a little under half a second. Strong wire: a bit longer. (Assuming a spherical Earth. If you do the experiment in a valley it will hit the ground sooner.) Sufficiently* strong wire, and assuming you're raising the angular velocity far above 100*2*pi radian/second in your futile attempt to get the rotation plane up to 1 meter high: the ball reaches orbital velocity and never hits the ground. (There's no atmospheric drag on a point mass.)

(* Back of the envelope, the ball would need to be pulling about 3,000,000 "g"s.)

Feldercarb

Orbital velocity? What is an acceleration field? An electrostatic field can accelerate an electron, never heard it called an acceleration field.

Does Earth's gravity have positive negative, or zero divergence?

Make the ball 25.4mm diameter. The 1 meter wire is 0.001kg/m. You can work out the integration from the start of the wire to the end of the ball for a more accurate calculation of force.

Integrate the ball to get mass versus d where d is the diameter. And while you are at it find the 1st and 2nd moment of inertia of wire plus ball. And the Eignen values/vectors of the system to find resonances.


Or just do the 1st order approximation as an instructional problem as intended.

 

[steve_bank]

Some tend to jump tohe complex.

Basics

When rotating in a horizontal plane the down force on the ball is simply m*g.

If the wire breaks it is like a bullet shot out of a gun. Fire a bullet horizontally and drop a bullet at the same time and both hit the ground at the same time. Look up the equations for a falling object.

I have not worked out the min speed and tension problems yet.

I used the Haliday and Resnick basic physics text in class. Older editions are in the public domain. A good reference for basic problems.

 

[Jarhyn]

They don't even stock frictionless point masses, perhaps because of lack of demand - only physics teachers ever buy the things

Did you check of they had any "spherical frictionless cows"? Sometimes you can get the same or similar thing, but you have to know what the name of the cheaper or more available alternative is.

Like, you can either go to the pharmacy and get 100% acetone in a tiny bottle as "nail polish remover" or you can go to the hardware store and get the same thing for pennies on the dollar in gallon quantities.

 

[Elixir]

Hang a ball from a wire on Berth and there is a mensurable tension on the wire caused by gravity.

I had a girlfriend named Bertha … it definitely caused some tension when I hung a ball on her.

 

[Shadowy Man]

I think it was Einstein who made the analogy and statement that standing on the Earth at 1g and standing inside a closed box in space accelerating at 1g are equivalent, no way to measure a difference.

Which was always an idealization -- a thought-experiment set in the asymptotic limit as the size of the box goes to zero, with the experimental equipment made of frictionless unobtanium and the experimenter himself assumed to be a spherical cow. Good on Einstein for having the mental discipline to pursue whatever metaphors worked for him to the limit of their effectiveness, but in the real world there's a theoretically straightforward way to measure the difference. Toss a ball from one side of the box to the other, videotaping its trajectory, and measure carefully frame by frame. In an accelerating box in space you see a parabola; in a box standing on the Earth you see an ellipse.

The gravitational field from a planet will have divergence while the acceleration field in the box will not.

That is, with sensitive enough equipment one can measure the force of gravity with height, a variation that will not exist in the accelerating box.

It is not about detecting gravity. It is about the equivalence of acceleration by gravity and inertial acceleration. Computationally equivalent.

It is though. Someone said that the two were identical and no way to tell the difference but there is. One field has divergence and the other does not. They are not exactly equivalent.

 

[steve_bank]

Don't know why your are bringing in fields and divergence.

Stand on a scale in a 1g gravity and you weigh 150 pounds.

Stand on scale in space being accelerated at 1g and you weigh 150 pounds.

That is all there is to it. Hence the problem I posed. While you can call centrifugal force fake or imaginary or pseudo force, there is a real measurable tension on the wire.

In SI units force in Newtons is kg*m/s^2(mass*acceleration). Anything that dimensionally reduces to that is force regardless if it is gravitationally, electrically, magnetically, or inertially induced.

A mechanical structure tat can not withstand a 5g gravity field will equally fail under 5g accretion in space. Or on a rotating test centrifuge.

Inertial and gravitational induced forces are computationally equivalent. Not the same 'thing' . When measuring force the two are indistinguishable.

Someone introduced centrifugal force as imaginary as if non existent, which is false. Analogous to thinking imaginary numbers do not exist.

 

[Shadowy Man]

Don't know why your are bringing in fields and divergence.

Because it is relevant to the “equivalence” of the two scenarios.

I understand you are interested in the engineering of it but I was discussing the physics.

Stand on a scale in a 1g gravity and you weigh 150 pounds.

Stand on scale in space being accelerated at 1g and you weigh 150 pounds.

That is all there is to it.

If that is all you care about then I guess so but that’s not all there is to it when it comes to the two scenarios.

They are not *exactly* the same, or indistinguishable.

 

[Elixir]

If that is all you care about then I guess so but that’s not all there is to it when it comes to the two scenarios.

They are not *exactly* the same, or indistinguishable.

Yawn.
The divide between science and engineering raises its ugly head.

 

[bilby]

there is a mensurable tension on the wire

Not if the wire neither stretches nor breaks there isn't.

 

[bilby]

Stand on a scale in a 1g gravity and you weigh 150 pounds.

I fucking wish.

 

[steve_bank]

Don't know why your are bringing in fields and divergence.

Because it is relevant to the “equivalence” of the two scenarios.

I understand you are interested in the engineering of it but I was discussing the physics.

Stand on a scale in a 1g gravity and you weigh 150 pounds.

Stand on scale in space being accelerated at 1g and you weigh 150 pounds.

That is all there is to it.

If that is all you care about then I guess so but that’s not all there is to it when it comes to the two scenarios.

They are not *exactly* the same, or indistinguishable.

Oky doky.

Not exactly tghe same? Yes htat is what I aid in terms of machines obviously. Computationally they are indistinguishable. Sm antics. Equations and numbers in science rangefinder trump semantics.

I design an assembly for a space probe lander that will kand on a planet wghere the gravity is 10gs. I can test the structural integrity,of the assembly by spinning it in a test centrifuge.

The point is it is not an academic debate. I look at sci8ence form the persevere of having used it.

So what IS the divergence of gravity in a small volume say 50k feet above MSL?

Fields are abstractions. Field mapping is covered in differential equations. For an EE divergence is covered in electromagnets. For a mechyanical engineer in fluid dynamics.

An electric field is defined vby moving a unit test charge around the field mapping the force vector at points in the field. There are different kinds of fields.

Physics is physics regardless o where it is used. Lets see some physics.

A large homogeneous cube is in deep space. What is the divergence of the gravity of the rectangular volume of the cube mathematically?

 

[Bomb#20]

...
Size of the ball? Ignore size and treat it as a point. Assume the wire has zero mass.
...
For extra credit if the wire breaks how long does it take for the ball to hit the ground?

You're introducing self-contradictory conditions: if you let the wire break, the breaking event is evidently not a steady state, which means there is no steady state and the answer will necessarily depend on initial conditions and transients.

Assuming you set the initial conditions in some way that causes the ball to fly in a helix above ground level, accelerating until the wire breaks, how long until the ball hits the ground -- if it ever does -- will depend on the breaking strength of the wire. Weak wire: a little under half a second. Strong wire: a bit longer. (Assuming a spherical Earth. If you do the experiment in a valley it will hit the ground sooner.) Sufficiently* strong wire, and assuming you're raising the angular velocity far above 100*2*pi radian/second in your futile attempt to get the rotation plane up to 1 meter high: the ball reaches orbital velocity and never hits the ground. (There's no atmospheric drag on a point mass.)

(* Back of the envelope, the ball would need to be pulling about 3,000,000 "g"s.)

Feldercarb

Orbital velocity? What is an acceleration field? An electrostatic field can accelerate an electron, never heard it called an acceleration field.

I didn't call anything an acceleration field. You must have misread something.

...
Make the ball 25.4mm diameter.
...
Or just do the 1st order approximation as an instructional problem as intended.

Finally, enough information to work with. (A one inch ball weighing one kilo. Gotta love that unobtainium -- six times denser than gold!)

Imagine are designing a carnival ride that spins and your seat goes horizontal. What is the minimum angular velocity to raise the ball to a flat horizontal spin plane at 1 meter high.?

The ball's radius is 12.7 mm, so for the top of the ball to rotate around the shaft in a circle 1 m off the ground, the c.o.g. of the ball is 12.7 mm below the attachment point of the wire on the shaft. So the bob is pulling 1000 mm / 12.7 mm = 78.7 "g"s = 772 m/s**2. The wire weighs 1 gram per meter and can hold up 78.7 kilos -- better be made of spider silk...

Therefore the angular velocity = sqrt(772 m/s**2 / 1 m) = 27.8 radians/s = 266 rpm.

Some tend to jump tohe complex.

You keep asking complex questions you imagine are simple.

Basics

When rotating in a horizontal plane the down force on the ball is simply m*g.

If the wire breaks it is like a bullet shot out of a gun. Fire a bullet horizontally and drop a bullet at the same time and both hit the ground at the same time.

What are you, a flat-earther? The dropped bullet hits first, because the fired bullet has farther* to fall, because it lands some distance away and the ground curved away from it, because the earth is round.

(* ETA: about 2 cm farther.)

 

[Shadowy Man]

Don't know why your are bringing in fields and divergence.

Because it is relevant to the “equivalence” of the two scenarios.

I understand you are interested in the engineering of it but I was discussing the physics.

Stand on a scale in a 1g gravity and you weigh 150 pounds.

Stand on scale in space being accelerated at 1g and you weigh 150 pounds.

That is all there is to it.

If that is all you care about then I guess so but that’s not all there is to it when it comes to the two scenarios.

They are not *exactly* the same, or indistinguishable.

Oky doky.

Not exactly tghe same? Yes htat is what I aid in terms of machines obviously. Computationally they are indistinguishable. Sm antics. Equations and numbers in science rangefinder trump semantics.

you said in post 44:

I think it was Einstein who made the analogy and statement that standing on the Earth at 1g and standing inside a closed box in space accelerating at 1g are equivalent, no way to measure a difference.

That’s what I was responding to. Perhaps you intended to mean “no *practical* way to measure a difference”. That’s fine. That’s why I tried to make the distinction between engineering and physics. You are an engineer, as I am to understand it, while my education and training was in physics. Perhaps that’s the source of the disconnect here.

 

[Loren Pechtel]

Don't know why your are bringing in fields and divergence.

Stand on a scale in a 1g gravity and you weigh 150 pounds.

Stand on scale in space being accelerated at 1g and you weigh 150 pounds.

You can't tell with one scale. You need two scales at different distances from the bottom. Measure the tide.

 

[steve_bank]

HUh?

You weigh the same at 1g on the surface of the Earth as you do on a rocket in space accelerating at 1g

The point is force in Newtons can regardless of he source be substituted for F anywhere it appears.

Beyond that I have no interest in it..

 

[bilby]

I have no interest in it.

That's the problem, right there.

 

[Loren Pechtel]

HUh?

You weigh the same at 1g on the surface of the Earth as you do on a rocket in space accelerating at 1g

The point is force in Newtons can regardless of he source be substituted for F anywhere it appears.

Beyond that I have no interest in it..

The box on a rocket has no tide.

You put a scale on the floor and a scale on the ceiling. Compare with sufficient precision.

 

[steve_bank]

Ok, but it is not about detecting anything.


Einstein;s Equivalence Principle says functionally and computationally inertial and gravitational force are equivalent.


What started this line was somebody posting non gravitational force is fake or imaginary, which I called semantics.

Equivalence principle - Wikipedia

en.wikipedia.org

A 3 axis accelerometer can shown direction of combined effects forces. Id have to look up ste of the art sensitivities. Finer optic gyros can be very sensitive.


The effect of of the Moon passing overhead can be measured,

Gravimetry - Wikipedia

en.wikipedia.org

For a small body, general relativity predicts gravitational effects indistinguishable from the effects of acceleration by the equivalence principle. Thus, gravimeters can be regarded as special-purpose accelerometers. Many weighing scales may be regarded as simple gravimeters. In one common form, a spring is used to counteract the force of gravity pulling on an object. The change in length of the spring may be calibrated to the force required to balance the gravitational pull. The resulting measurement may be made in units of force (such as the newton), however, gravimeters display their measurements in units of gals (cm/s2),[a] and parts per million, parts per billion, or parts per trillion of the average vertical acceleration with respect to the Earth.