[Bored chemist]

[gem]

yes very funny, must admit, had a little chuckle and a wry smile at the fun had, even if it was at my expense, at a comment taken out of context. as I was meaning in relation to the grenade analogy when I stated,

Halc thanks for that response.
But think that grenade analogy only works from the centre of an isolated system.
I really struggle to visualise any scenario on Earth Or it’s atmosphere where this statement can be applied.
( This is why acceleration is not a change in kinetic energy.)

ie  grenade exploding is not  uniform circular motion

Halc you stated

Kinetic energy cannot be converted to torque.  Force does, not energy. Energy is totally irrelevant here.

which I believe I covered below

  more of its kinetic energy via a torque force

in regards as to energy being irrelevant, I would disagree, if say you had one half of the grenade with a
velocity of 972.31 m per sec immediately after the explosion and it had

a mass of 100 grams it would have
 
a kinetic energy of 47.3 x 10³ joules
 
and momentum of 97.23 kg m per sec

and if said energy was immediately converted into a force via a collision with say a vertical cliff face it would result in

a force equal to ≈ 1.6 x10 to the 6 Newtons and a resulting kinetic energy of zero.

[Bored chemist]

We can take the idea of half the grenade hitting a cliff a bit further. We can fix the grenade to the cliff wall, then detonate it.
If the wall is strong,then essentially the whole of the grenade ends up moving away from the cliff and, for an instant the kick that it delivers to the rock starts the world turning (very slightly) away from the cliff face.

But the bits of the grenade don't stay moving forever.
They slow down and fall to the ground.
And when they do that, the impart a counter-torque and bring the Earth back to its original spin.
Obviously, some of the force is transferred by air resistance but still, after a while the air slows down due to the only thing it's in contact with- the ground- and, in doing so,  it pushes on the Earth. (Once again, we are talking about viscosity)
After a short while the spin of the Earth is back to what it was.
That's the conservation of angular momentum.

The leftward moving bits of the grenade are brought to a halt by the Earth, pushing the earth left
The rightward moving bits of the grenade are brought to a halt by the earth, pushing the Earth right.

And the two things exactly cancel.
Imagine that we move the grenade a few cm from the cliff.
It does not "know" which side the cliff is on, so it explodes symmetrically.
So there must be as mush momentum transferred left as right.
So the sum of those is zero. (Once  you remember they are vectors).

[gem]

So  BC following your analogy below.

Imagine that we move the grenade a few cm from the cliff.
It does not "know" which side the cliff is on, so it explodes symmetrically.
So there must be as mush momentum transferred left as right.
So the sum of those is zero. (Once  you remember they are vectors).

using the data I posted would give values in one direction

velocity of 972.31 m per sec immediately after the explosion and it had

a mass of 100 grams it would have
 
a kinetic energy of 47.3 x 10³ joules
 
and momentum of 97.23 kg m per sec

and if said energy was immediately converted into a force via a collision with say a vertical cliff face it would result in

a force equal to ≈ 1.6 x10 to the 6 Newtons and a resulting kinetic energy of zero 

and if say we place another cliff face directly opposite at a distance of 274.32 metres
would result in
velocity of 737.62 metres
kinetic energy of 27.2 x 10 ^3 Joules
momentum of 73.76 Kgm per sec
and if said energy was immediately converted into a force via a collision with say a vertical cliff face it would result in

a force equal to ≈ 0.91 x10 to the 6 Newtons and a resulting kinetic energy of zero

which represents a reduction in direct force applied of aprox 43%

one other factor which will be pertinent would be the drop in height which occurs during its flight time of  0.509 of a metre when calculating the resulting torque force

[Bored chemist]

using the data I posted would give values in one direction

Then your data (or your calculation) is wrong.

In particular, you are calculating  forces when you should be calculating impulses.

a force equal to ≈ 0.91 x10 to the 6 Newtons and a resulting kinetic energy of zero

A force of a million Newtons acting for a week will make more difference than the same force acting for a second.
You have't taken that into account.

[Bored chemist]

one other factor which will be pertinent would be the drop in height which occurs during its flight time of  0.509 of a metre when calculating the resulting torque force

Yes, you need to account for that 0.5 metres of height  in the about  about 6,371,000 m radius of teh earth.
But, if you do the arithmetic correctly, you discover that it cancels.
Because angular momentum is always a conserved quantity.
Why are you trying to argue that a mathematically proven thing is wrong?

What baffles me was that you can show why the weather can't (permanently) affect the angular momentum of the Earth by giving a number of arguments
(1) simple symmetry
(2) analysis of the energy involved- if the rotation of the Earth was contributing more energy to the weather than the Sun, then we would see the change in rotation rate; Stonehenge wouldn't line up any more.
(3) The law of conservation of angular momentum says it's impossible without an external torque.
(4) the definition of angular momentum in terms of torque time and moment of inertia means that without a torque, the change in angular momentum (and thus rotational kinetic energy is zero.
(5) straightforward analogies in the form of "when the skater stops waving their arms about they are still spinning at the same rate as when they started and the man on the truck who thinks he's helping.

And, though nobody has in any way refuted those, they still don't actually accept a simple fact. The Sun drives the weather.

I forgot Noether's theorem when I drew up that list of reasons why you are plainly wrong.

[gem]

So BC you stated

Quote from: gem on 30/08/2020 01:15:59
using the data I posted would give values in one direction
Then your data (or your calculation) is wrong.

In particular, you are calculating  forces when you should be calculating impulses.

Quote from: gem on 30/08/2020 01:15:59
a force equal to ≈ 0.91 x10 to the 6 Newtons and a resulting kinetic energy of zero
A force of a million Newtons acting for a week will make more difference than the same force acting for a second.
You have't taken that into account.

So just to be clear on the data, I got the mass and speed and reduction in speed due to drag over set distances from ballistics tables, (bullets that are designed to minimise drag.)
Therefore I believe the reduction in velocity under consideration is realistic.

Also the method of calculating the average collision force I used the equation
½ x (mass x velocity^2/stopping distance) given I would have been guessing at what time to place on the collision but did have knowledge of bullet length, so stopping distance I used at 30mm on the figures provided, but if you double the stopping distance, the reduction in the collision force is still around 43%.
However given the average collision force value from that calc just to get an approximate value for the stopping time by rearranging the force time equation, gives a time around 8 x 10^ - 5 seconds   

In regards to your following statement

Yes, you need to account for that 0.5 metres of height  in the about  about 6,371,000 m radius of teh earth.
But, if you do the arithmetic correctly, you discover that it cancels.

So I plug in the numbers and Cliff face one nearest the explosion transfering a torque force of
620 x10^6 Nm for aprox 8 x 10^ - 5 seconds

Cliff face two opposite at a distance 274.32 m applying an opposite torque force of 
470 Nm  for aprox 8 x 10^ - 5 seconds 

So I believe the values given are a reasonable demonstration of the direct forces applied to the surface of Earth,
could you give a brief description of the dynamics that you allude to in the statement below, to make up the shortfall in one direction applied to the surface ?

Obviously, some of the force is transferred by air resistance but still, after a while the air slows down due to the only thing it's in contact with- the ground- and, in doing so,  it pushes on the Earth.

As it seems very vague.

[Halc]

velocity of 972.31 m per sec immediately after the explosion and it had
a mass of 100 grams it would have
a kinetic energy of 47.3 x 10³ joules
and momentum of 97.23 kg m per sec
and if said energy was immediately converted into a force via a collision with say a vertical cliff face it would result in
a force equal to ≈ 1.6 x10 to the 6 Newtons and a resulting kinetic energy of zero

and if say we place another cliff face directly opposite at a distance of 274.32 metres
would result in
velocity of 737.62 metres
kinetic energy of 27.2 x 10 ^3 Joules
momentum of 73.76 Kgm per sec
and if said energy was immediately converted into a force via a collision with say a vertical cliff face it would result in
a force equal to ≈ 0.91 x10 to the 6 Newtons and a resulting kinetic energy of zero

Question 1) How were these forces computed?  I don't get those numbers.
The force and energy computations seem irrelevant to the discussion, since the discussion is about momentum, which brings up:
Question 2) What's your point here?  You just let everything drop and fail to complete the total momentum calculation, which should add up to zero if you're doing it right.  All I see is you trying to bury this calculation in complexity, adding motion in a second dimension, drag, gravity, blah blah blah.

You address question 1 here:

Also the method of calculating the average collision force I used the equation
½ x (mass x velocity^2/stopping distance) given I would have been guessing at what time to place on the collision but did have knowledge of bullet length, so stopping distance I used at 30mm

You said 'immediately' above, not '30 mm'.  You're leaving an awful lot out of your calculations.
Anyway, as I said, the force (and energy) is irrelevant. What counts is the momentum transferred, and you haven't completed the calculation, so question 2 was: What's your point?  That total momentum is not zero?  Then you need to do the complete computation, not just pick a few random components of it. You choosing all these complications just makes that task harder, but that's your choice.

Given these choices, you seem to be trolling us. You know they're wrong, yet you persist.
You wave away the missing parts as being 'vague', but they're entirely quantifiable.

So I believe the values given are a reasonable demonstration of the direct forces applied to the surface of Earth, could you give a brief description of the dynamics that you allude to in the statement below, to make up the shortfall in one direction applied to the surface ?

Computations are not complete, and they're done as scalars, not vectors. Do the whole thing.  Your 2nd projectile slowed down. Where did that momentum go?
The 2nd projectile isn't moving horizontal anymore. Where's the vector adjustments for that.  How about the momentum transfer via gravity while it was in flight? You're missing all this. You wanted to make it complicated, so you need to include all this stuff.

Your computations of 'torque force' are completely wrong. 620e6 Nm of torque? How did you arrive at that figure? You just posted this random number that doesn't seem to be computable from the prior figures.

[Bored chemist]

As it seems very vague.

It seems that way to you because you don't realise it's absolutely clear that it must e right- due to the conservation of momentum.

I got the mass and speed and reduction in speed due to drag over set distances from ballistics tables, (bullets that are designed to minimise drag.)
Therefore I believe the reduction in velocity under consideration is realistic.

So the tables you used say the bullets slow down.
And, obviously, that's because they transfer momentum (and energy) to the air.
But then you ignored that momentum.
So it wasn't there when you calculated the momentum of the fragments.
Well of course it wasn't.
You forgot to add it.

You think it's vague that the only thing that momentum can be imparted to (in the long run) is the Earth.
I disagree.
It seems pretty concrete to me that, since there's nowhere else for it to go, that's where it goes.

If not, then where?

[gem]

Morning,
I posted that the grenade analogy wouldn't work, on earth or in its atmosphere, because it does not meet the criteria for an isolated system, and am highlighting that.
BC posted

  So the tables you used say the bullets slow down.
And, obviously, that's because they transfer momentum (and energy) to the air.
But then you ignored that momentum.
So it wasn't there when you calculated the momentum of the fragments.
Well of course it wasn't.
You forgot to add it. 

So Momentum values at the point of explosion and several points along the way are

  0.0 m     = 97.23 Kg M per sec (point of explosion and 3.1 x 10 ^ minus 5 sec later collision one )
  91.44 m = 89.00 Kg M per sec
182.88 m = 81.1   Kg M per sec
274.32 m = 73.76 Kg M per sec (point of collision two aprox 0.32 sec after collision one)

given that kinetic energy is lost continuously via frictional drag there is less and less momentum available to the body in flight to be continuously be transferred to the air, therefore the sums cannot add up to zero.

Which results in an unequal  momentum transfer  overall,  combining in the air and and the amount to the solid that is earth's surface.

Therefore  resulting in an unequal torque applied to the earth's surface.

[Bored chemist]

given that kinetic energy is lost continuously via frictional drag there is less and less momentum available to the body in flight to be continuously be transferred to the air,

There is less momentum carried by the body, precisely because the momentum is transferred to the air.

Once you stop ignoring that, and add together ALL the contributions to the momentum, you find it comes to zero.

[gem]

So, BC given your statement

There is less momentum carried by the body, precisely because the momentum is transferred to the air.

and given some of this momentum is directly from collision with the casing these won't be perfectly elastic or inelastic collisions, so heat generated as work is done
So once that momentum is transferred, to the molecules of the air mostly N2  with mass of 4.65 x 10^ - 26 Kg
the average speed of the nitrogen molecules N2 at say a temperature of 20° C (293 K)  results in a mean speed of about 470 ms.

On average, the air particles move at speed greater than the speed of sound, however, significantly higher speeds are also present. About 1 % of the molecules have a speed of more than 1000 m/sec.

So at a atmospheric pressure  (1 bar) gives a collision frequency of aprox 7.0 x 10 ^9   per sec
i.e. within one second a nitrogen molecule will collide on average with 7 billion other molecules in what is probably best described as very high speed three dimensional snooker,

So the question is "I believe i asked this before" how does the momentum transferred to the air keep itself separate from the high speed dynamics its mixed in with, to then transfer an equal amount of momentum half a meter lower down the opposite cliff face.
Indeed the difficulty lies in the fact that the molecules will permanently collide with other particles and change their direction of motion in a random way, let's say you applied a indicator like a smell to track the momentums progress. it would go out in all directions, it would be more influenced by the fact that the smell is conveyed mainly due to air currents (convections), which carry the particles over greater distances.
Note that convection is no longer a completely random motion. In this case, the molecules are moved over macroscopic distances in a certain direction.

when I stated previously the conservation laws and isolated systems in regards to what conditions are not being met,
it was dismissed as lies we tell to children.

So let's further explore some similar examples
below is an extract from 
https://en.wikipedia.org/wiki/Momentum

Newton's second law of motion states that the rate of change of a body's momentum is equal to the net force acting on it. Momentum depends on the frame of reference, but in any inertial frame it is a conserved quantity, meaning that if a closed system is not affected by external forces, its total linear momentum does not change.

https://en.wikipedia.org/wiki/Inertial_frame_of_reference#:~:text=An%20inertial%20frame%20of%20reference,moving%20at%20a%20constant%20velocity.

Inertial and non-inertial reference frames can be distinguished by the absence or presence of fictitious forces, as explained shortly.[8][9]
The effect of this being in the noninertial frame is to require the observer to introduce a fictitious force into his calculations….

— Sidney Borowitz and Lawrence A Bornstein in A Contemporary View of Elementary Physics, p. 138

Given the coriolis force dynamics occurring between earths surface and the atmosphere it cannot be argued they are a inertial reference frame to each other, and therefore fail on that criteria.

Indeed the analogy with hand grenade applying Newtons first law

In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force

given the flight of the ballistic doesn't maintain a straight line it would be reasonable to say it failed on that criteria for conservation of momentum also.

mean while we still have an imbalance in momentum

[Bored chemist]

So the question is "I believe i asked this before" how does the momentum transferred to the air keep itself separate from the high speed dynamics its mixed in with, to then transfer an equal amount of momentum half a meter lower down the opposite cliff face.

I believe this was answered before

Computations are not complete, and they're done as scalars, not vectors. Do the whole thing.  Your 2nd projectile slowed down. Where did that momentum go?

Don't you see how, because in every single one of those 7 billion collisions, the momentum is conserved it must be conserved overall?

way, let's say you applied a indicator like a smell to track the momentums progress. it would go out in all directions,

No. it would not, because every single time a molecule hit another molecule, the momentum would be conserved.

Momentum depends on the frame of reference, but in any inertial frame it is a conserved quantity,

And the footings of a windmill are as good an inertial frame as any on Earth.
And, you seem to be trying to invoke relativistic effects for something only travelling at about a millionth of the speed of light.
That pretty much gives the game away. You don't know what you are talking about.

[gem]

BC states

And the footings of a windmill are as good an inertial frame as any on Earth.

Not to the atmosphere they aren't

What do you suddenly disagree with regarding non inertial reference frames and the coriolis effect, it its occuring due to the earth's rotation and the weather patterns tending to go more in straight lines north and south, from the equator, due to the atmosphere being uncoupled from the solid earth.

You seemed to understand it previously,

Now if we take a sample of equatorial air and adiabatically move it

What moved it (Hint, it's convection currents driven by the Sun)?
A ship sailing due north has to overcome coriolis forces, but that doesn't mean they can switch off the engine.
The engine has to work harder.

So seem to be contradicting your own point.

you also state

you seem to be trying to invoke relativistic effects for something only travelling at about a millionth of the speed of light.

Newtons laws are more than sufficient.
this from
https://en.wikipedia.org/wiki/Inertial_frame_of_reference#:~:text=An%20inertial%20frame%20of%20reference,moving%20at%20a%20constant%20velocity.
The equations of motion in a non-inertial system differ from the equations in an inertial system by additional terms called inertial forces. This allows us to detect experimentally the non-inertial nature of a system.


— V. I. Arnol'd: Mathematical Methods of Classical Mechanics Second Edition, p. 129

BC states

Don't you see how, because in every single one of those 7 billion collisions, the momentum is conserved it must be conserved overall?

Yes,   and conserved conserved overall has to account for the dynamics we call the weather and the readily observable resulting motions deviating away from a conserved angular momentum path of the atmosphere, due to the solar input causing transfers of energy which causes a chain of events which are constantly changing the speed and direction of large parts of the atmosphere resulting in observable net changes in the momentum of the atmosphere.
Because by itself, the law of conservation of momentum is not enough to determine the motion of particles after a collision. Another property of the motion, kinetic energy, must be known.

[Bored chemist]

Not to the atmosphere they aren't

OK, what is a better inertial frame for the atmosphere?

Newtons laws are more than sufficient.

Glad to hear it.
One of those laws- every action has an equal and opposite reaction- is the root of the conservation of momentum.If something pushes to the left then it is pushed to the right.
You seem determined to ignore this.

readily observable resulting motions deviating away from a conserved angular momentum path of the atmosphere,

Where that this deviation from the conservation of momentum been observed?
You have claimed it, but not demonstrated it.

Are you saying that the momentum of the atmosphere is not equal to the sum of the momenta of every particle of which the atmosphere is made?
Because, you accept that momentum is conserved for each molecular collision, but somehow don't accept that it is conserved when you add them all together.
How is that possible?

[gem]

Hi
so in regards to what would be better as a reference frame,

quote author=Bored chemist link=topic=80136.msg613221#msg613221 date=1599302257]OK, what is a better inertial frame for the atmosphere?[/quote]

I believe the question is flawed, due to the atmosphere cannot maintain a inertial reference frame, given the constant changes to velocity and dynamics occurring due to the energy inputs and losses and gravitational force applied to it.

For example buoyancy is a major aspect to the dynamics under discussion

https://en.wikipedia.org/wiki/Buoyancy

which states about buoyancy,

  This can occur only in a non-inertial reference frame, which either has a gravitational field or is accelerating due to a force other than gravity defining a "downward" direction

In regards to the conservation of Kinetic momentum of the atmosphere which is continually energy gaining and dissipative through various energy transfers, you would have to consider its accepted a change in momentum occurs due to the energy transfers processes so described

https://en.wikipedia.org/wiki/Momentum#Conservation
this from the above link

Forces that can change the momentum of a droplet include the gradient of the pressure

and given the gradient of the pressure is temperature dependent it is therefore in constant state of pressure changes and constant momentum changes.

[Bored chemist]

I believe the question is flawed

Possibly, but it's implicit in your assertion.
I said the footings of a windmill are as good a reference frame as any.
You said I was wrong.
That leads to the question; if the footings of a windmill are not as good a reference as any, then what is?
And you say that question's meaningless.
Well, OK, fair enough, but you framed it.
I can sit at the foot of the windmill and watch air molecules collide in much the same way that I can sit in a bar and watch snooker balls.
Momentum is conserved from my PoV.

Now, please address the important issue:

Are you saying that the momentum of the atmosphere is not equal to the sum of the momenta of every particle of which the atmosphere is made?
Because, you accept that momentum is conserved for each molecular collision, but somehow don't accept that it is conserved when you add them all together.
How is that possible?

[Halc]

That leads to the question; if the footings of a windmill are not as good a reference as any, then what is?

We're discussing the constant angular momentum of the Earth as a system, so the effectively inertial frame of the Earth (effectively because we're ignoring the fact that it is in orbit and thus not actually inertial) is a far better reference frame than the local accelerated reference frame of the windmill footing.
In that local system, there is definitely torque being applied since the foot of the windmill and the height at which the force of the wind is applied are separated by say 100m. That's a long lever, and considerable torque.  But use the inertial frame of the entire Earth as a system, and there is no net torque at all since there are no forces coming from outside.

[Bored chemist]

That leads to the question; if the footings of a windmill are not as good a reference as any, then what is?

We're discussing the constant angular momentum of the Earth as a system, so the effectively inertial frame of the Earth (effectively because we're ignoring the fact that it is in orbit and thus not actually inertial) is a far better reference frame than the local accelerated reference frame of the windmill footing.
In that local system, there is definitely torque being applied since the foot of the windmill and the height at which the force of the wind is applied are separated by say 100m. That's a long lever, and considerable torque.  But use the inertial frame of the entire Earth as a system, and there is no net torque at all since there are no forces coming from outside.

Good point.
I'm still wondering how Gem is going to explain away the fact that addition works.

[gem]

So when I stated
I believe the question is flawed, due to the atmosphere cannot maintain a inertial reference frame, given the constant changes to velocity and dynamics occurring due to the energy inputs and losses and gravitational force applied to it.

and your response was

Possibly, but it's implicit in your assertion.
I said the footings of a windmill are as good a reference frame as any.
You said I was wrong.
That leads to the question; if the footings of a windmill are not as good a reference as any, then what is?
And you say that question's meaningless.
Well, OK, fair enough, but you framed it.

To be clear it is not my assertion it is an explicit condition of the laws of conservation of momentum.

Newton's second law of motion states that the rate of change of a body's momentum is equal to the net force acting on it. Momentum depends on the frame of reference, but in any inertial frame it is a conserved quantity, meaning that if a closed system/isolated  system is not affected by external forces,

a closed system is a physical system that doesn't exchange any matter with its surroundings, and isn't subject to any net force whose source is external to the system.[1][2] A closed system in classical mechanics would be equivalent to an isolated system in thermodynamics. An isolated system cannot exchange any heat, work, or matter with the surroundings,

violating the conservation laws gives physical consequences like a change in momentum as previously stated
gradient pressure constantly changing due to energy received by the earth affecting atmospheres density, therefore resulting in,

Forces that can and do change the momentum of a droplet vai the gradient of the pressure changes.


this resulting acceleration because of the constant changes to the pressure gradient, is a major mechanism of the earth's weather system, due to the energy received from outside the earth to its atmosphere combined with the violation of the conservation of momentum laws (non inertial reference frame)

(buoyancy law)   

This can occur only in a non-inertial reference frame, which either has a gravitational field or is accelerating due to a force other than gravity defining a "downward" direction

resulting in net forces and the subsequent variations in velocity (speed and direction )  within the atmosphere.

changing where you chose to view these dynamics from wont stop the physical reality of them occurring.