GrapesOfWrath,

Do you know, that if this was not so frustrating it would truly be funny.

Let me see if I can figure this out.

OK, let me repeat myself. My post to DoctorDon Posted: 2002-07-20 10:29 contained the following.
================================================== =
Real bodies consist of large numbers of particles.
The particles of two bodies are mutually attracted.
The resultant of those (attractive) forces is what
is called the weight of the bodies.
"The line of action of this resultant force, or
weight, always passes through a single point in
the body, a point called the center of gravity of
the body."
================================================== ===
That last sentence is directly from the physics book. I had placed it within quotation marks to indicate that. I faked the first three.

Or:
Perhaps you truly don't understand that the weight of one mass is always in relation to another mass. I must admit that could possibly be twisted around and misinterpreted.

Or:
Does it have to do with the confusion of the difference between the "gravitational center", and the "center of gravity"?

"gravitational center" - is self induced by the mass.
"center of gravity" - is induced by another mass.

OR:
Are we still honing our argumentative / debating skill?


Quote:
================================================== =
But it will rotate around the center of mass, correct?
================================================== =
Yes, The Earth rotates around its center of mass.
And/or/but. The rotating Earth swings, turns, or nods from (or at the end of), the Moon's gravitational force. The Moon's gravitational force does not end at the induced center of gravity within the Earth, but as far as the Earth is concerned, it can be thought to.
The fact that the Earth rotates around its center of mass, is uneffected by its swing from the Moon. Unless as I said there is some relativity effect at the speed of gravity.


Gary

You asking if I might be trying to concoct specious arguments just for the sake of argumentation? The answer is no.

One more step.

Since the affect of the moon can be considered to act at the center of gravity, but that is offset from the center of mass, the moon's gravity causes a torque.

Same thing when you kick a football offcenter--it spins.

Let's take another look at inertia:

Imagine an ice rink. Imagine a bowling ball in the ice.

Let's pretend the ice has no friction.

Now shape the ice so that a puck will revolve around the bowling ball. In others words, there is a dip in the ice where the bowling ball sits. A puck set in the proper motion will orbit the bowling ball. Make the slope of the dip higher near the ball and not so high farther out from the ball.

Now that we've seen the orbit, let's get rid of that stupid puck.

Now take 2 pucks and attach them together with a rigid rod. Let's call one puck the bow and the other puck the stern.

Now shoot the pucks across the ice, so that the bow leads the stern, and the body is aimed to the right of the bowling ball. You shoot straight, so at time 0 the angular momentum is 0.
As the pucks approach the dip, the bow encounters a greater slope than the stern. Also the angle of the slope at the bow is counter-clockwise to the angle of the slope to the rear. Viewing this from above, you'll see that not only does the body's path curve, but the bow is rotating towards the bowling ball. We have a change in angular momentum!

Now let's pretend we have a VERY LONG spaceship made of 2 balls and a rigid rod. Now let's pretend the earth is ahead but a little to the left and at time 0 there is no angular momentum. Now we increase our velocity straight ahead then cut the engines after reaching a good enough speed to pass the Earth.

As the bow of the ship approaches the Earth, the gravitational attraction is greater than the stern. The bow's gravitational angle is also more to the left than the angle at the stern. The ship not only curves towards the Earth, but rotates too. We have a change in angular momentum!

So Gary, how did the ship get a change in angular momentum without torque? Once we cut the engines, there was no force other than gravity.

Or... are you claiming that the ship will continue facing the same direction?

Just curious...
--Tommy

<font size=-1>[ This Message was edited by: traztx on 2002-07-23 02:52 ]</font>

GrapesOfWrath,
--------------------------------------------------------------
You asking if I might be trying to concoct specious arguments
just for the sake of argumentation? The answer is no.

One more step.
Since the affect of the moon can be considered to act at the
center of gravity, but that is offset from the center of mass,
the moon's gravity causes a torque.
Same thing when you kick a football offcenter--it spins.
--------------------------------------------------------------
Very perceptive. Thanks.
I had heard that rumor, about others.

No, no torque.

Remember the definition of the center of gravity.
"Real bodies consist of large numbers of particles.
The particles of two bodies are mutually attracted. The resultant of those forces is what is called the weight of the bodies. The line of action of this resultant force, or weight, always passes through a single point in the body, a point called the center of gravity of the body."

That means that all the weight is evenly distributed or balanced around that point.
Therefore no torque.
It appears that you are somehow trying to connect the gravitational mass with the inertial mass. While they are equivalant as Einstein said, they are different.
This is one of those cases where that is very evident. As I keep saying there may be a relativity connection at the speed of gravity or light, if those speeds are different.

If you were to jack up your house and place a roller under it, you could push it until it balanced. The center of gravity would be over the roller. Then if you went in and moved the frig to the bed room, the center of gravity would change. That move may not be immediately evident if the rollers are still centered under the weight, but by redoing with the rollers at 90 degrees it will show up.
Gravity works on the weight rather than the mass. Yes, because the Earth and Moon are very large the magnitude of gravity diminishes because of the distances, but that also is the weight and the centers of gravity automatically adjust fore and aft. Gravity does not torque alone; it requires a second source of force. If a body is supported or suspended from a point not along the line between the center of gravity and the source of the gravity, it will be torqued.

The Earth spins on its axis through the center of mass (as does the Moon). This is how inertia works.

The Moon and Earth are mutually attracted by gravity, the ends of that attraction can be considered as the mutually caused centers of gravity. As the Earth and Moon rotate on their axes they can be thought to rotate around their centers of gravity. That rotation is a figure 8 like pendulum swing. This is how gravity works.

When you kick the ball, it is impulse and momentum just like the meter stick. The input force divides between momentum and angular momentum. The division depending on where the force is applied. The reason is the inertia of the mass serves as the opposing force. Some where in the physics book, it explains how the inertial mass can serve as the opposing force for the torque. As that ball wobbles away, its center of gravity falls along a parabola. Ignoring wind resistance of course.


Gary

Exactly. The same with the Earth.

The center of mass/inertia is not the same as the center of gravity, as you've said. Thus, there can be a torque. In the Earth/moon case, the two are offset because of friction during the rotation of the Earth. traztx's example is probably one or two orders of magnitude smaller effect.

I've been just lurking for a while, since I didn't think I had anything to contribute. It was getting so that GR and I were talking past each other, not really with each other.

It seems now, following the discussion between GR and GoW, that there is really one point of disagreement here. Torque.

Gary does not use the physics definition of torque. He thinks he does, but he does not. He uses his interpretation of torque, based on an attempt by the authors of his physics textbook to help students get the mathematical definition. This, coupled with his existing, intuitive notion of what torque is (from his career as a machinist), leads him astray.

As long as Gary persists on using his odd, incorrect notion of torque, there will be no agreement on the final model of how the Earth's axial rotation is decreasing, and how the Moon's semi-major axis is increasing.

As Gary sees it, there can be no torque on the Earth from the Moon, so therefore he seeks another cause. Hence, his theory.

The rest of us try to show him that there is indeed torque on the Earth from the Moon, using the mathematical definition of torque. But since he does not agree on our use of the mathematical definition of torque, we get nowhere.

Gary, "R cross F" means one, and only one, thing. It is the definition of torque. It does not locate the center of gravity. It defines a vector, not a locus.

Here's an analogy to show where you can go wrong thinking you know what physics terms mean when they're similar to words you've used in a non-physics environment. "Work" does not mean the same thing in physics as it does in common English. If I hold a 10 kg bar in one hand, just holding it there motionless, I am not doing "work" in the physics sense. I would tell you that yes I am doing work. My arm is getting tired--where do you get off telling me this isn't work?!

But work to a physicist is defined as
W = Fd
force times distance. The bar is motionless, so no work.

What you learn as a machinist with a torque-wrench about torque is going to be at odds with the physics definition. And when working on the Earth-Moon system, you have to use the physics definition.

You have to look at the math, not the simplified English sentence. If there were two forces necessary in torque, it would be in the math. It would be F₁*F₂ cross R (or something). But there is only one force. All you need for torque is one force acting off-center. No opposing force is needed.

Gary, if you want to learn the physics involved here, I'm all too glad to help. But if you are going to insist on clinging to your incorrect notion of torque, then I can't help you. I'm done.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

Okay, I've been reading this thread and trying to understand what everybody's saying (I don't know how good a job I've been doing of it, but I've been trying).

At any rate, I'd like to ask Gary a couple of questions about his point of view (rather, my perception of his point of view).

Gary, from what I'm seeing, you're saying that the Earth, being in free-fall around the Moon, is "suspended" from it's center of gravity. You're arguing that there can be no torque because the force of the Moon's gravity is centered on the same point in the Earth that the Earth is suspended from. Is that correct?

Consider a perfectly-balanced barbell on Earth. The center of mass of the barbell is right at the center of the handle correct? The center of gravity, however, is dependent on the barbell's position and orientation relative to the Earth.

If we hang the barbell perfectly level, then the center of gravity will also be centered in the handle, and we can suspend it from a string at that point. It will hang balanced, right?

Now, if we hang the barbell at an angle, the center of gravity is moved away from the middle of the handle. So, if we attach the string to the center of the handle, the point of suspension will be different than the point where gravity is acting, so the barbell will twist, right?

But, we should be able to take that angled barbell and determine where the center of gravity is when it's in that orientation. If we attach the string at that point, and thus suspend the barbell from the center of gravity, doesn't your argument suggest that the barbell will not be torqued by Earth's gravity?

In other words, I read your argument as suggesting that it should be possible to suspend a barbell from a single point, at a 45-degree angle, simply by choosing the correct center of gravity as the point of suspension.

It seems to me that in reality, though, if you hang a barbell from a string anywhere off the center of mass, the Earth's gravity will pull that barbell down to where it's hanging vertically. I think it's possible to suspend the barbell level from the center of mass, and it's possible to suspend it vertically from any point, but I don't think it's possible to suspend it at an arbitrary angle from a single point. Do you?

Am I missing something somewhere?

__________________
SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

If the barbell is small enough that we can ignore the inverse-square effect of gravity, then, in fact, you can hang a barbell at any angle.

If the barbell is large, though, then because of the inverse-square law, gravity will pull more on the "lower" weight than on the "upper," and so, yes, you're right, the bell would eventually rotate into a "vertical" orientation.

This is the "missing torque" that's been controversial here for so very long...

Silas

This is widely used as a cheap attitude control method for spacecraft, known as "gravity gradient stabilization"."

http://www.geocities.com/estsat/anexo_i.html#pertur

"Gravity-Gradient torque: An object in orbit will experience a strongest attraction on its "lower" side than its "upper" side. This differential attraction, if is applied to a body that has unequal principal moments of inertia, results in a moment that tends to rotate the object to align its long axis (moment of minimal inertia) with the local vertical."



http://samadhi.jpl.nasa.gov/msl/Quic.../orstedQL.html

(fix link and wording)


<font size=-1>[ This Message was edited by: Karl on 2002-07-24 14:06 ]</font>

Geo3gh,
Quote:
-------------------------------------------------------------
It seems now, following the discussion between GR and GoW,
that there is really one point of disagreement here. Torque.

Gary does not use the physics definition of torque. He thinks
he does, but he does not. He uses his interpretation of torque,
based on an attempt by the authors of his physics textbook to
help students get the mathematical definition. This, coupled
with his existing, intuitive notion of what torque is (from his
career as a machinist), leads him astray.
-------------------------------------------------------------
Jeff, perhaps you are correct.
I believe it was Silas a while back who said there were two things that everyone here loved to do, one was to learn the other was teach. If it wasn't Silas please forgive me.

Quote:
-------------------------------------------------------------
All you need for torque is one force acting off-center. No
opposing force is needed.
-------------------------------------------------------------
I'm sitting here with physics book in hand. It seems to cover everything that you have said. It explains tau = r x F, it explains tau = rF sin theta, it looks as if it covers it all.

Even when you said:
Quote:
-------------------------------------------------------------
You have to look at the math, not the simplified English sentence. If
there were two forces necessary in torque, it would be in the math. It
would be F1*F2 cross R (or something). But there is only one force.
-------------------------------------------------------------
The physics book here says or calls that the net torque.
The formula is tau = r1F1 - r2F2.
It appears that this has been what I've been talking about all along. The definition of torque is given as the effectiveness of a force in changing the state of rotation of a body.
If we subtract a torque (1) as you describe it from a torque (2) as you describe it, then the result is the effective torque as I have been describing it.

I agree that "r x F" using the Earth's center of mass will give a torque as you describe it of r1F1. What I have been missing and not knowing how to ask for is the r2F2. In other words to determine how effective the Moon is at slowing the Earth we need to know r2F2.
I have asked for a point of suspension, a point of support, or a second force; I have asked for almost everything but a second torque. What I need is the second torque r2F2 to complete the calculation of how effective the Moon is at slowing the Earth.


Thanks
Gary

traztx
Quote:
--------------------------------------------------------------
Let's take another look at inertia:
Imagine an ice rink. Imagine a bowling ball in the ice.
Let's pretend the ice has no friction.
Now shape the ice so that a puck will revolve around the bowling ball.
In others words, there is a dip in the ice where the bowling ball sits.
A puck set in the proper motion will orbit the bowling ball. Make the
slope of the dip higher near the ball and not so high farther out from
the ball.
--------------------------------------------------------------

When I was about 10-12 years old I was skating on thin ice. I fell and the Ice warped into just such a dimple. That night it froze hard and the pond remained that way all the rest of the winter. So after nearly 50 years, I can still see it in my mind.

Quote:
--------------------------------------------------------------
So Gary, how did the ship get a change in angular momentum
without torque?

Once we cut the engines, there was no force other than gravity.
Or... are you claiming that the ship will continue facing the
same direction?

Just curious...
--------------------------------------------------------------
--Tommy,
Not being a teacher I don't want to go to far, but my thought is this.
As the ship, as you call it, approaches the dip the front moved in as it leaves, the front moves out and the back moves in.
As it leaves the dip it will again be headed in a straight line.
The perturbation as it is called is soon over and done.
The definition of torque involves the efficiency of a force in causing a change in rotation. If the body is not in rotation before and after the dip then the efficiency was what I would call zero. I would say that from a mechanical stand point no torque took place. The Celestial Mechanic went through how the Sun torqued the Earth - Moon system; at first quarter then again at third quarter, the net result was zero.
I'm convinced that unless some permanent change in rotation is effected it can hardly be called torque. Either the rotation needs to be started, stopped, or the speed of it changed. I have just posted to Jeff the formula I fell covers that. It is tau = r1F1 - r2F2


Gary

Sorry, but since Gary is ignoring me again, I have to pick on someone else.

Yes. The barbell balances at its center of gravity, not center of mass. However, each different orientation of the barbell might have a different center of gravity.

What if there were three forces?

Well, if I've got to be picked on by somebody, I could do worse . . .

Okay, when the barbell is level, the center of gravity is right in the center of the handle. When it's "tilted" 45 degrees, to where would the center of gravity move (approximately - I realize specific mass distribution information would be required for a specific answer)? And are you saying if the barbell were suspended at that point, it would balance at that 45 degree angle without gravity tipping it to a 90 degree angle?

__________________
SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

For most intents and purposes, the center of mass and the center of gravity of the bar bell are going to be the same. The change in gravity from one side of the barbell to the other is miniscule (sic).

As long as the string is aligned with the center of gravity, the barbell won't be torqued by gravity.

I'm not sure that's the case. It seems that the formula you quote here deals with a two torque model. Two different forces, off-center, and in different directions. If I had three forces, I'd have
tau = R₁F₁ + R₂F₂ + R₃F₃

It's a simple summation. Some of the torques may be negative, in which case you subtract them. But you see what I mean.

How effective is the torque on the Earth by the Moon? Not very. But to find out, you take the torque for every point on the Earth and add them all up. Roughly half will be positive and roughly half will be negative. If they exactly cancel out, then you have no net torque. If they don't exactly cancel out, you have a net torque, however small.

That is how I read that formula. I don't see it as being what you've been talking about.

As I said above, what you need is all the torques, R₁F₁ through R_nF_n. To do that, you need calculus. Don did the integral that shows this in his pdf. I don't know of any way to do that without knowing calculus.

There aren't just two torques, there are as many torques as there are points in the Earth. There is one net torque, but that is just the sum of all torques (positive and negative).

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

SeanF,
Quote:*
-----------------------------------------------------------------
Gary, from what I'm seeing, you're saying that the Earth, being in
free-fall around the Moon, is "suspended" from it's center of gravity.
You're arguing that there can be no torque because the force of the
Moon's gravity is centered on the same point in the Earth that the Earth
is suspended from. Is that correct?
-----------------------------------------------------------------
Almost exactly correct. The only suggestion I could make is to be specific as to the center of gravity. It almost sounds as if you are talking about the self induced center of gravity. That is definitely not the one I'm talking about. Here on the Earth whenever we talk of a center of gravity it is the one in an object caused by the Earth. What I'm talking about is the center of gravity within the Earth caused by the Moon.

You talk about a barbell in your post. If rather than a string of rope you drill a hole through so that it can spin around a shaft. You now have an object in perfect balance rotating around its center of gravity. If the bearing were frictionless the spin would be perpetual. Here on the Earth talking small objects the center of mass, and the center of gravity, are so close to being the same no distinction is possible. Silas attempted an explanation in his post.

Quote:
-----------------------------------------------------------------
Am I missing something somewhere?
-----------------------------------------------------------------
Yes, but you seem to be as close as everyone else. You definitely found the key to the whole thing when you wrote: "You're arguing that there can be no torque because the force of the Moon's gravity is centered on the same point in the Earth that the Earth
is suspended from."

Yes, the Moon suspends the Earth. The Earth goes around and around in a 6000 mile diameter circle. If we could cut the gravitational thread then that would not happen. So the question is where is that thread attached. I say it is at the point that the others claim to be the point at which the Moon torques the Earth. I just can't see how the point of suspension and the point of applied force can be the same and still cause torque.

For instance if you drop a yo-yo it spins, but if you start swinging it around the world and you continue to do that for 30 minutes there is no longer a spinning torque.


Gary