KE of system / different reference frames question

[jduffy77]

I am re-posting this question here in a new thread as Humber mistakenly posted it in a two year old thread.

This post appeared on a ddwfttw forum:

For those who actually care, relative to any frame other than that of the ground, the ground does have energy. It is possible to get energy from the ground - in fact, this is exactly what happens, for example, in KERS. The non-ground frame analysis would go as follows:

Say we have a 10 kg object initially moving at 10 m/s relative to the ground. We choose to start analyzing in the frame where this object is initially at rest - so the ground is initially moving at -10 m/s.

We use KERS to brake the object at a rate of -1 m/s^2, or equivalently, a force of -10 N. By Newton's third law, that means that there is a force of 10 N on the Earth.

As the object is at rest relative to this frame, there is no relevant kinetic power on the object itself. However, there is kinetic power on the Earth; using P = F*v, we get that the Earth is losing kinetic energy at a rate of -100 W, so by conservation of energy, the KERS can be storing energy at a rate of 100 W.

Note that this is exactly the same as what we get in the frame of the ground, as there, F = -10 N, v = 10 m/s, so the object is losing kinetic energy at a rate of -100 W, so by conservation of energy, the KERS can be storing energy at a rate of 100 W.

The above idea, seems to be quite commonly accepted amongst those claiming to have physics degrees, but it seems to me to be a the result of literal thinking, and a rather confused idea of what frames of reference means.

The energy for the KERS, initially comes from the car's fuel, some of which ultimately ends up as kinetic energy of the car. It is that energy which is recovered by KERS, and does not come from the ground.

I am interested in hearing options for or against either claim.

 

[Ken G]

I'm not sure what your issue is with "literal thinking", to me the quote is a clear example of not thinking literally. Literal thinkers imagine that energy is innate, so a car that is moving along the road has it, in any frame of reference. But energy is frame dependent-- conservation of energy means the energy stays the same in a given frame, it doesn't mean it is the same in all frames. So choosing different frames always repartitions the energy, and therefore it also changes the language about what the energy is doing or where it is coming from. This is all perfectly normal, it is what you find when you relax the need to think literally.

So there is no problem at all with saying that a car can extract energy from the road when it uses KERS, but the language sounds awkward unless one is being extremely clear about the choice of reference frame. For example, one can have a car that accelerates from rest to 10 m/s, and if we start in a frame already moving at 10 m/s, this will look like a car that uses its fuel energy yet loses all its kinetic energy-- the missing energy shows up in the Earth's kinetic energy in that frame. That's just how it is, in that frame. We can argue that isn't the best frame to think about, or we can argue it is, but it certainly isn't wrong.

 

[256bits]

I had to look up KERS and that is Kinetic Energy Recovery System.

It is so far so good up to the 5th paragraph.
The 10kg object can be considered to be at rest in a frame, call it frame A, moving at 10 m/s relative to the earth. But once the braking occurs, the object will have a force applied to it and begin to move relative to that frame. ( Otherwise you are dealing with a moving frame of reference. ) At the end of it all, the object will have a velocity of -10 m/s in frame A.

Suppose there are 2 cars moving side by side. You have a tiny window to look out from car 1 out but all you can see is the car 2 beside you, which would appear to be not moving relative to your reference frame A. Person X on the Earth can see both Car1 and 2 and conclude that both cars are mving at 10 m/2 relative to the earth.
Car2 signals you that he is going to apply KERS. As he does so, to you in Car 1, it would look as if Car2 is accelerating in the backward direction up to a maximun of -10 m/s. You in Car1 would begin to wonder how in the world did Car2 accelerate if Car2 was using KERS.

 

[256bits]

The energy for the KERS, initially comes from the car's fuel, some of which ultimately ends up as kinetic energy of the car. It is that energy which is recovered by KERS, and does not come from the ground.

A really bad statement all in all. It is not put together in any coherent fashion. You usually find something like this on a site promoting the newest crackpot idea which is meant to confuse, and not for an explanation of KERS. I cannot tell if the writer is saying that the energy of the fuel is being recovered or the kinetic energy of the vehicle. If he is saying it is the energy of the fuel, then why not go farther back and say it is the energy of the sun trapped in the plant material, from 600 million years ago, that became hydrocarbons through transformation under the Earth's surface.

The fuel provides energy to the car to move. The chemical energy is transformed into kinetic energy. KERS would recover some of the kinetic energy.

 

[rcgldr]

I think the point missed in the old DDWFTTW thread is the source of the force at the tires. For a KERS vehicle, the force at the tires equals the mass of the vehicle times it's rate of acceleration (or deceleration) (ignoring rolling resistance and aerodynamic drag). For a DDWFTTW vehicle, the force at the tires equals the torque used to drive the propeller divided by the radius of the tires, which is independent of the mass of the vehicle.

 

[mender]

Otherwise you are dealing with a moving frame of reference.

Not a problem since for any given situation, there is only one non-moving frame of reference and an infinite number of moving ones.

I think what you're meaning is an accelerating frame of reference.

 

[Humber]

I'm not sure what your issue is with "literal thinking", to me the quote is a clear example of not thinking literally. Literal thinkers imagine that energy is innate, so a car that is moving along the road has it, in any frame of reference. But energy is frame dependent-- conservation of energy means the energy stays the same in a given frame, it doesn't mean it is the same in all frames. So choosing different frames always repartitions the energy, and therefore it also changes the language about what the energy is doing or where it is coming from. This is all perfectly normal, it is what you find when you relax the need to think literally.

So there is no problem at all with saying that a car can extract energy from the road when it uses KERS, but the language sounds awkward unless one is being extremely clear about the choice of reference frame. For example, one can have a car that accelerates from rest to 10 m/s, and if we start in a frame already moving at 10 m/s, this will look like a car that uses its fuel energy yet loses all its kinetic energy-- the missing energy shows up in the Earth's kinetic energy in that frame. That's just how it is, in that frame. We can argue that isn't the best frame to think about, or we can argue it is, but it certainly isn't wrong.

That is indeed the thinking I was taking about. That "all things are relative".

Acceleration is not one of those things, and the fuel in that tank, is another.
Can you justify that the fuel consumed, is capable of accelerating the Earth by same amount it accelerates the car?

 

[Humber]

A really bad statement all in all. It is not put together in any coherent fashion. You usually find something like this on a site promoting the newest crackpot idea which is meant to confuse, and not for an explanation of KERS. I cannot tell if the writer is saying that the energy of the fuel is being recovered or the kinetic energy of the vehicle. If he is saying it is the energy of the fuel, then why not go farther back and say it is the energy of the sun trapped in the plant material, from 600 million years ago, that became hydrocarbons through transformation under the Earth's surface.

The fuel provides energy to the car to move. The chemical energy is transformed into kinetic energy. KERS would recover some of the kinetic energy.

The first few statements are not mine. The claim there, is that the Earth can transfer 100W to the KERS.
But, regardless of its origins, the fuel is the store of some energy form somewhere. It could equally be a battery. When the car accelerates, some of that energy becomes KE in the car's mass, and that is the energy that is recovered, and not something that comes from the ground.

 

[Humber]

I think the point missed in the old DDWFTTW thread is the source of the force at the tires. For a KERS vehicle, the force at the tires equals the mass of the vehicle times it's rate of acceleration (or deceleration) (ignoring rolling resistance and aerodynamic drag). For a DDWFTTW vehicle, the force at the tires equals the torque used to drive the propeller divided by the radius of the tires, which is independent of the mass of the vehicle.

Yes. F=ma.
When the car brakes, momentum is transferred from the car, to the ground, dp/dt = F.
For a KERS to work, that same force is present, but the KE of the car is not dissipated as heat in the brake pads and disks, but becomes stored in a mass as angular KE, or by direct conversion to electrical energy and so stored in a battery.
Apart form some work at the tyres and surface deformation, which is a result of F, and energy that can't be recovered, there is no energy from the ground.

 

[jduffy77]

The first few statements are not mine.

No, those are correct. The incorrect one he mentioned was not put forth in a coherent fashion was yours though.

The claim there, is that the Earth can transfer 100W to the KERS.
But, regardless of its origins, the fuel is the store of some energy form somewhere. It could equally be a battery. When the car accelerates, some of that energy becomes KE in the car's mass, and that is the energy that is recovered, and not something that comes from the ground.

As was already pointed out; whether or not the energy comes from the ground depends on the reference frame chosen to analyze.

 

[Subductionzon]

That is indeed the thinking I was taking about. That "all things are relative".

Acceleration is not one of those things, and the fuel in that tank, is another.
Can you justify that the fuel consumed, is capable of accelerating the Earth by same amount it accelerates the car?

No one has made that claim Humber. You are trying to use a straw man argument.

Why don't you lay out your claim here that all energy must be calculated relative to the Earth and see how that flies here. Or as you claimed elsewhere, that no energy can come from the ground.

 

[RCP]

Otherwise you are dealing with a moving frame of reference.

Not a problem since for any given situation, there is only one non-moving frame of reference and an infinite number of moving ones.

I think what you're meaning is an accelerating frame of reference.

Mender old bean! You've been missed at TR. Did you ever get around to do that test pulling your cart to wind speed and then releasing it along with some packing popcorn? I would still love to see a video of that.

Also looking forward to 256bits response. And what about a decelerating FoR?

 

[Humber]

No, those are correct. The incorrect one he mentioned was not put forth in a coherent fashion was yours though.
As was already pointed out; whether or not the energy comes from the ground depends on the reference frame chosen to analyze.

Which is a contradiction of both my statement, and that of 256bits, which is paraphrasing of mine.
When the car accelerates, momentum is transferred from the ground to the car, and when it brakes, the other way. That is conservation of momentum at work, which leaves you to explain if the ground powers the KERS, why fuel is needed to transfer momentum one way, but not the other. And, if you can possibly manage it, please keep your personal remarks and references to the cart to yourself.

 

[jduffy77]

Which is a contradiction of both my statement, and that of 256bits, which is paraphrasing of mine.
When the car accelerates, momentum is transferred from the ground to the car, and when it brakes, the other way. That is conservation of momentum at work, which leaves you to explain if the ground powers the KERS, why fuel is needed to transfer momentum one way, but not the other. And, if you can possibly manage it, please keep your personal remarks and references to the cart to yourself.

Please check your quotes. The two sentences you have quoted were mine but the body of your post is responding to someone else.

 

[mender]

Happy New Year, RP!

Mender old bean! You've been missed at TR. Did you ever get around to do that test pulling your cart to wind speed and then releasing it along with some packing popcorn? I would still love to see a video of that.

Nothing yet and not likely in the foreseeable future, too much real work to do! I haven't stopped in because I didn't think there was anything left but some stirring of the ashes!

ETA: wow - part 30??!

 

[Dale]

When the car accelerates, some of that energy becomes KE in the car's mass, and that is the energy that is recovered, and not something that comes from the ground.

In some valid frames, when the car accelerates it loses KE, and can even come to rest. When it is at rest there is 0 KE. Furthermore, KE can never be negative. So, how do you propose to recover energy from something with 0 energy and whose energy cannot go negative?

Have you even attempted to calculate the energy in a moving reference frame and see how it works out?

 

[RCP]

Happy New Year, RP!

Nothing yet and not likely in the foreseeable future, too much real work to do! I haven't stopped in because I didn't think there was anything left but some stirring of the ashes!

ETA: wow - part 30??!

Best wishes for the new year back atcha mender. Yep, Cartville is still accelerating, and FYI humber has now posted pictures of 3 different carts he made. (Film at 11! ) Remember how positive everyone was he would never make one?

And if you or anyone has time, there's still my question above about the FoRs applicable for deceleration?

 

[rcgldr]

Although off topic from the orignal DDWFTTW thread ... ignoring any losses due to aerodynamic drag, rolling resistance, conversion of energy, ... , then considering a car and Earth as a closed system, then energy (and momentum) of this closed system is conserved.

KERS - extracts energy from Earth + car closed system and adds that energy to a device that stores energy (kinetic or potential).

Usage of fuel - extracts chemical potential energy from fuel and adds that energy to the kinetic energy of Earth + car closed system.

For both of these cases, the amount of energy converted is independent of the frame of reference, as long as the frame of reference is inertial.

Getting back on topic from the original DDWFTTW thread, someone was wondering if a DDWFTTW vehicle was somehow extracting it's own kinetic energy (wrt ground). The responses that followed tried to explained that a DDWFTTW vehicle extracts energy by slowing down air (wrt ground), and that a DDWFTTW vehicle requires a true wind (wrt ground) in order to operate.

 

[Humber]

In some valid frames, when the car accelerates it loses KE, and can even come to rest. When it is at rest there is 0 KE.

KE is the work of acceleration.
http://en.wikipedia.org/wiki/Kinetic_energy
"The kinetic energy of an object is the energy which it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest."
The accleration of a 640kg car, cannot be equated with the acceleration of the Earth.

Furthermore, KE can never be negative.So, how do you propose to recover energy from something with 0 energy and whose energy cannot go negative?

Your first statement is correct, because KE is related to the square of velocity, but then again, momentum isn't. There is certainly -p and +p. That allows conservation to work.

If it were necessary for there to be "-ve KE" to "recover +ve KE," then no transfer could ever take place under any circumstances. The answer is the the KE does not need to be negative, because all cases lie between zero and some non-zero number.

Have you even attempted to calculate the energy in a moving reference frame and see how it works out?

Momentum transfer to the ground is necessary to force, but not to the motion itself, but wheels do need that force, so momentum is indeed transferred as the car gains KE via acceleration.
The car gains momentum p, and the Earth -p.
total energy = p^2/2m_car+ p^2/2m_earth
m_earth = 6e24Kg, so the second term is negligible.

ETA:
It is not a matter of arbitrarily choosing frames because changes in momentum are not frame dependent. One would be at a loss to explain how the battery in an electric car loses charge, if it where just a matter of choosing which frame it lost it in.

 

[Ken G]

That is indeed the thinking I was taking about. That "all things are relative". Acceleration is not one of those things, and the fuel in that tank, is another.

Sure, but the statements you objected to did not say anything about either the acceleration or the fuel, they talked about recovering energy from the Earth, which is perfectly correct in the frame of the car's velocity.

Can you justify that the fuel consumed, is capable of accelerating the Earth by same amount it accelerates the car?

No, nor is there any need to justify that, because no one thinks the Earth is being accelerated as much as the car. That comes from a=F/m, the F is an action/reaction pair so is the same on the car and the Earth. Not the "a." There is only a very small "a" for the Earth, but it suffices to remove exactly the correct amount of kinetic energy from the Earth to make the problem work out in any reference frame.

What is happening here is that you are confused about the differences between energy, momentum, and acceleration, and your confusion is causing you to contradict the statements made by people who do understand these differences. I'm not sure you can move forward until you understand those basic concepts better, and then you can get to the point where you understand that stories about what energy is doing always depend on the reference frames from which they are told. This is all elementary to us.

 

[Dale]

KE is the work of acceleration.
http://en.wikipedia.org/wiki/Kinetic_energy
"The kinetic energy of an object is the energy which it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest."

Exactly. So if the car is already at rest in some frame of reference then the "work needed to accelerate [it] from rest to its stated velocity" is 0. In other words, it has no KE.

Your first statement is correct, because KE is related to the square of velocity, but then again, momentum isn't. There is certainly -p and +p. That allows conservation to work.

Momentum and energy are different quantities, they are not interchangeable. Both are individually conserved. You cannot use conservation of momentum instead of conservation of KE, both are needed.

Momentum transfer to the ground is necessary to force, but not to the motion itself, but wheels do need that force, so momentum is indeed transferred as the car gains KE via acceleration.
The car gains momentum p, and the Earth -p.
total energy = p^2/2m_car+ p^2/2m_earth
m_earth = 6e24Kg, so the second term is negligible.

The change in the second term is only negligible in the frame where the Earth is at rest, see below.

It is not a matter of arbitrarily choosing frames because changes in momentum are not frame dependent. One would be at a loss to explain how the battery in an electric car loses charge, if it where just a matter of choosing which frame it lost it in.

But it is just a matter of arbitrarily choosing frames. Newton's laws are invariant under changes in reference frame, so you can arbitrarily choose your frame and apply Newton's laws. So it had better be possible to explain the same situation from multiple reference frames or you are violating Newton's laws.

Here is how it works:

Let m = 1000 kg be the mass of the car and let M = 1E25 kg be the mass of the earth. There is a horizontal force of -1000 N (to the right is positive) acting on the car from the road for a period of 10 s. Calculate the initial and final values for kinetic energy, KE, and momentum, p, for both the Earth and the car in (a) the frame where the Earth is initially at rest and the car is initially moving at 10 m/s and (b) the frame where the Earth is initially moving at -10 m/s and the car is at rest.

(a) p_{i,c}=m v_{i,c}=10000 \ kg \ m/s
\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s
p_{f,c}=p_{i,c}+\Delta p_c = 0 \ kg \ m/s
v_{f,c}=p_{f,c}/m= 0 \ m/s
KE_{i,c}=1/2 m v_{i,c}^2 = 50 \ kJ
KE_{f,c}=1/2 m v_{f,c}^2 = 0 \ kJ

p_{i,e}=M v_{i,e}=0 \ kg \ m/s
f_e=-f_c=1000 \ N
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s
p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s
v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ
KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J

So in the first reference frame the energy for charging the battery comes from the KE of the car which goes down by 50 kJ. The Earth gains a negligible amount of energy.

(b) p_{i,c}=m v_{i,c}= 0 \ kg \ m/s
\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s
p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s
v_{f,c}=p_{f,c}/m= -10 \ m/s
KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ
KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ

p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s
f_e=-f_c=1000 \ N
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s
p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s
v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ
KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.

 

[mender]

FYI humber has now posted pictures of 3 different carts he made. (Film at 11! ) Remember how positive everyone was he would never make one?

Can you save me some time and provide links to those pictures?

And if you or anyone has time, there's still my question above about the FoRs applicable for deceleration?

Not sure exactly what you're asking for, RP, but I'll give it a shot if you elaborate.

 

[Humber]

Exactly. So if the car is already at rest in some frame of reference then the "work needed to accelerate [it] from rest to its stated velocity" is 0. In other words, it has no KE.
Momentum and energy are different quantities, they are not interchangeable. Both are individually conserved. You cannot use conservation of momentum instead of conservation of KE, both are needed.

Momentum is always conserved, kinetic energy may or may not be.

The change in the second term is only negligible in the frame where the Earth is at rest, see below.
But it is just a matter of arbitrarily choosing frames. Newton's laws are invariant under changes in reference frame, so you can arbitrarily choose your frame and apply Newton's laws. So it had better be possible to explain the same situation from multiple reference frames or you are violating Newton's laws.

And so by the same reasoning, one may not equate the acceleration of a car, with that of the Earth.

Here is how it works:
Let m = 1000 kg be the mass of the car and let M = 1E25 kg be the mass of the earth. There is a horizontal force of -1000 N (to the right is positive) acting on the car from the road for a period of 10 s. Calculate the initial and final values for kinetic energy, KE, and momentum, p, for both the Earth and the car in (a) the frame where the Earth is initially at rest and the car is initially moving at 10 m/s and (b) the frame where the Earth is initially moving at -10 m/s and the car is at rest.

(a) p_{i,c}=m v_{i,c}=10000 \ kg \ m/s
\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s
p_{f,c}=p_{i,c}+\Delta p_c = 0 \ kg \ m/s
v_{f,c}=p_{f,c}/m= 0 \ m/s
KE_{i,c}=1/2 m v_{i,c}^2 = 50 \ kJ
KE_{f,c}=1/2 m v_{f,c}^2 = 0 \ kJ

p_{i,e}=M v_{i,e}=0 \ kg \ m/s
f_e=-f_c=1000 \ N
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s
p_{f,e}=p_{i,e}+\Delta p_e = 10000 \ kg \ m/s
v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 0 \ kJ
KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J

So in the first reference frame the energy for charging the battery comes from the KE of the car which goes down by 50 kJ. The Earth gains a negligible amount of energy.

That is to say, the the 50kJ of KE the car gained from some unmentioned source of energy,
has been captured and stored in a battery. The amount transferred to the ground is
p^2/2Mearth = 5e-18J. The transfer of that energy, has been achieved by deceleration of
the car. The force is dp/dt = 1000N.
The total energy is 50kJ + 5e-18J. The initial source supplying the 50kJ, may be another battery. It supplied 50kJ to the car, and all but 5e-18J was recovered.

(b) p_{i,c}=m v_{i,c}= 0 \ kg \ m/s
\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s
p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s
v_{f,c}=p_{f,c}/m= -10 \ m/s
KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ
KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ

p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s
f_e=-f_c=1000 \ N
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s
p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s
v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ
KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.

And now, the total energy is 100kJ, of which 50kJ was recovered by the car.
Edit.
Oh, I see. In one case the Earth gains momentum and in the other, it loses it, but the total energy is now 100kJ from the Earth + 50kJ, that initially supplied the car ?
There are not two events, but one, and the conservation of momentum still applies, no matter which frame is employed.

 

[Humber]

Can you save me some time and provide links to those pictures?



Not sure exactly what you're asking for, RP, but I'll give it a shot if you elaborate.

The above is a violation of the conservation of momentum.

 

[Dale]

Momentum is always conserved, kinetic energy may or may not be.

Correct, and in the example KE is not conserved in either frame. The KE goes down by 50 kJ in both frames. This 50 kJ of missing KE is the source of the energy put into the battery in both frames.

And so by the same reasoning, one may not equate the acceleration of a car, with that of the Earth.

I agree. And if you will look back at the math you will see that I never did equate them. That was your statement, never mine.

That is to say, the the 50kJ of KE the car gained from some unmentioned source of energy,
has been captured and stored in a battery. The amount transferred to the ground is
p^2/2Mearth = 5e-18J. The transfer of that energy, has been achieved by deceleration of
the car.

That is all correct.

And now, the total energy is 100kJ, of which 50kJ was recovered by the car.

No, in this one the total energy 5E26 J. The change in the KE of the Earth is -100 kJ, the change in the KE of the car is 50 kJ, and the remaining 50 kJ is again captured by the battery. The only difference being that the source of the energy was the KE of the earth, not the KE of the car.

Oh, I see. In one case the Earth gains momentum and in the other, it loses it, but the total energy is now 100kJ from the Earth + 50kJ, that initially supplied the car ?

No, the car had 0 kJ initially, it gained 50 kJ of KE from the earth.

There are not two events, but one, and the conservation of momentum still applies, no matter which frame is employed.

Correct, as does the conservation of energy.

 

[Humber]

Correct, and in the example KE is not conserved in either frame. The KE goes down by 50 kJ in both frames. This 50 kJ of missing KE is the source of the energy put into the battery in both frames.

In the first case, all but 5e-18J.is conserved.
In the second 100kJ comes from the Earth, and the 50kJ is said to be transferred to the car, when at some point, the car gained that energy from another source. The total is 150kJ,
so only 50kJ is conserved.

I agree. And if you will look back at the math you will see that I never did equate them. That was your statement, never mine.

Then it was an omission. Without acceleration, there will be no KE change for the Earth.
The same force is applied to each object and so equal and opposite.

No, in this one the total energy 5E26 J.

Which came from where? The only energy source, is that which internally powered the car.
That is a massive amount of energy, with no pedigree.

The change in the KE of the Earth is -100 kJ, the change in the KE of the car is 50 kJ, and the remaining 50 kJ is again captured by the battery. The only difference being that the source of the energy was the KE of the earth, not the KE of the car.

But the prior to the car gaining the 50kJ from an external source, it had 0J. Therefore, the largest change that can occur when the car returns to 0m/s is 50kJ.

No, the car had 0 kJ initially, it gained 50 kJ of KE from the earth.

It was "given" 50kJ, and then that was transferred to the recovery battery.
And in one case, the car loses momentum and the Earth gains it, while in the second, the car gains it and the Earth loses it. Changing frames, does not change the event.

 

[Subductionzon]

DaleSpam, I am sorry that we inflicted Humber on this site. He never will understand frames of reference. He believes that if you have a frame of reference where the Earth is moving that you would have had to accelerate the Earth to that speed and he will ask you where the energy came from as shown by this quote of his:

Originally Posted by DaleSpam View Post

"No, in this one the total energy 5E26 J."

"Which came from where? The only energy source, is that which internally powered the car.
That is a massive amount of energy, with no pedigree.

Many many people have tried to explain the concept of frames of reference but no one has ever got through to him. I wish you the best of luck.

 

[Dale]

Then it was an omission. Without acceleration, there will be no KE change for the Earth.
The same force is applied to each object and so equal and opposite.

Are you familiar with Newton's laws? The force is equal and opposite (Newton's 3rd law) but the acceleration is not (Newton's 2nd law). The accelerations are not equal, I never said that they are equal, that was only your statement in posts 19 and 23. I am not sure if your mistake was a simple verbal goof or if you do not understand the difference between force and acceleration.

Which came from where? The only energy source, is that which internally powered the car.
That is a massive amount of energy, with no pedigree.

I have news for you, the kinetic energy of the Earth does not come from some car's power source. The Earth has had kinetic energy for billions of years, long before the car came around. It can just be considered a given in the problem, but technically it came from the gravitational collapse of the dust cloud from which the solar system formed.

And in one case, the car loses momentum and the Earth gains it, while in the second, the car gains it and the Earth loses it.

Correct, but not complete. In one case, (a), the car loses momentum and energy and the Earth gains the momentum and the battery gains the energy. In the other case, (b), the Earth loses momentum and energy and the car gains the momentum and both the car and the battery gain the energy (split evenly). Both are equally valid ways to describe the same situation.

 

[Dale]

DaleSpam, I am sorry that we inflicted Humber on this site. ... I wish you the best of luck.

No worries, I will continue as long as it remains enjoyable, and then I will find something else to do.

 

[Humber]

Are you familiar with Newton's laws? The force is equal and opposite (Newton's 3rd law) but the acceleration is not (Newton's 2nd law).
The accelerations are not equal, I never said that they are equal, that was only your statement in posts 19 and 23. I am not sure if your mistake was a simple verbal goof or if you do not understand the difference between force and acceleration.

Then you will need to provide an example of kinetic energy exchange, where only one of the objects is accelerated. Indeed the same force is present, but the different masses of car and Earth will result in different accelerations, but not zero for either.

I have news for you, the kinetic energy of the Earth does not come from some car's power source. The Earth has had kinetic energy for billions of years, long before the car came around. It can just be considered a given in the problem, but technically it came from the gravitational collapse of the dust cloud from which the solar system formed.

That's right, so you will need to show good cause, why it should now change velocity in order to power the car.

Correct, but not complete. In one case, (a), the car loses momentum and energy and the Earth gains the momentum and the battery gains the energy.
In the other case, (b), the Earth loses momentum and energy and the car gains the momentum and both the car and the battery gain the energy (split evenly). Both are equally valid ways to describe the same situation.

Case 1: Car loses 50kJ, Earth gains 5e-18J. ( Recovery of the car's 50kJ intial energy.)
Case 2 :Car gains 50kJ, Earth loses 100kJ. ( Generation of energy.)

 

[Humber]

DaleSpam, I am sorry that we inflicted Humber on this site. He never will understand frames of reference. He believes that if you have a frame of reference where the Earth is moving that you would have had to accelerate the Earth to that speed and he will ask you where the energy came from as shown by this quote of his:



Many many people have tried to explain the concept of frames of reference but no one has ever got through to him. I wish you the best of luck.

You have a brother here, SZ. What was his forum name? I can't recall it.

 

[Humber]

No worries, I will continue as long as it remains enjoyable, and then I will find something else to do.

The first case contains only a small error, so I think we can agree that result is OK.

The second requires some clarification, please.

(b) p_{i,c}=m v_{i,c}= 0 \ kg \ m/s
\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s
p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s
v_{f,c}=p_{f,c}/m= -10 \ m/s
KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ
KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ

p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s
f_e=-f_c=1000 \ N
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s
p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s
v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ
KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.

In this case, the observer, must be in a reference frame that is relative to the centers of mass of the car and planet. The final velocity of the car relative to the Earth, has changed from 10m + 10e-21m/s of the first case, to -9.999999999999999999999m/s in the second case.

Another concern, involves repetition of the event. The reclaimed energy could be used to accelerate the car back to almost 10m/s, because in the first case, the loss of KE to the ground is only 5e-18J or 5e-19W. That is negligible, so can be ignored when the car accelerates, should it be recoverable in that case.
But, in the second case, the Earth loses 100kJ each time the car is braked, and that is (approx) 10kW
Over 200 cycles, a mere 2000s, the Earth has lost ( supplied) almost 1MW. it would seem that a cheap source of energy has gone unexploited.

 

[Dale]

Indeed the same force is present, but the different masses of car and Earth will result in different accelerations, but not zero for either.

Good, it sounds like you understand.

That's right, so you will need to show good cause, why it should now change velocity in order to power the car.

I did, see post 21.

Case 1: Car loses 50kJ, Earth gains 5e-18J. ( Recovery of the car's 50kJ intial energy.)
Case 2 :Car gains 50kJ, Earth loses 100kJ. ( Generation of energy.)

Yes. And in each case there is 50 kJ net change in KE which is available for charging the battery.

 

[Dale]

The second requires some clarification, please.

Sure, I would be glad to clarify.

The final velocity of the car relative to the Earth, has changed from 10m + 10e-21m/s of the first case, to -9.999999999999999999999m/s in the second case.

Differences in velocity are frame invariant in Newtonian physics. So you are right to be concerned about this.

In frame (a)
v_{f,e}-v_{f,c}=10^{-21} \ m/s - 0 \ m/s = 10^{-21} \ m/s

In frame (b) note the number of 9's behind the decimal point
v_{f,e}-v_{f,c}= -9.999999999999999999999 \ m/s - (-10 \ m/s) = 10^{-21} \ m/s

So the final velocity of the car relative to the Earth is the same in both frames.

Another concern, involves repetition of the event. The reclaimed energy could be used to accelerate the car back to almost 10m/s, ...
But, in the second case, the Earth loses 100kJ each time the car is braked, and that is (approx) 10kW
Over 200 cycles, a mere 2000s, the Earth has lost ( supplied) almost 1MW. it would seem that a cheap source of energy has gone unexploited.

Just as the Earth lost 100 kJ (50 kJ to the car's KE and 50 kJ to the battery), as the car went from 0 to 10 m/s the Earth will gain 100 kJ (50 kJ from the car's KE and 50 kJ from the battery) as the car goes from 10 to 0 m/s. There is no free lunch here.

 

[Humber]

Yes. And in each case there is 50 kJ net change in KE which is available for charging the battery.

There is a change of 100kJ for the Earth in the second case, but the car gains only 50kJ.
Does the remaining 50kJ go as heat?

 

[Dale]

There is a change of 100kJ for the Earth in the second case, but the car gains only 50kJ.
Does the remaining 50kJ go as heat?

If the KERS works ideally then the remaining 50 kJ goes into the battery. Of course, nothing is 100% efficient, so some fraction will be lost to heat. It would only go all to heat if you were using traditional non-regenerative braking.

 

[Humber]

Sure, I would be glad to clarify.

Differences in velocity are frame invariant in Newtonian physics. So you are right to be concerned about this.

Differences in velocity are frame invariant? No, velocity is relative, so is frame dependent.

In frame (a)
v_{f,e}-v_{f,c}=10^{-21} \ m/s - 0 \ m/s = 10^{-21} \ m/s

In frame (b) note the number of 9's behind the decimal point
v_{f,e}-v_{f,c}= -9.999999999999999999999 \ m/s - (-10 \ m/s) = 10^{-21} \ m/s

So the final velocity of the car relative to the Earth is the same in both frames.

No, in the first case, the Earth is considered to be stationary, and the car changes velocity by 10m/s, and the Earth moves in the opposite direction - 1e-21m/s, so the difference, the relative velocity, is 10m/s + 1e-21m/s

In the second case the the car remains at 0m/s, and the Earth changes by -9.999999999999999999999m/s. which is not the same as the first case.

Another concern, involves repetition of the event. The reclaimed energy could be used to accelerate the car back to almost 10m/s, ...
But, in the second case, the Earth loses 100kJ each time the car is braked, and that is (approx) 10kW
Over 200 cycles, a mere 2000s, the Earth has lost ( supplied) almost 1MW. it would seem that a cheap source of energy has gone unexploited.

Just as the Earth lost 100 kJ (50 kJ to the car's KE and 50 kJ to the battery), as the car went from 0 to 10 m/s the Earth will gain 100 kJ (50 kJ from the car's KE and 50 kJ from the battery) as the car goes from 10 to 0 m/s. There is no free lunch here.

No, in the first case, the Earth gains 5e-18J.
That will hold if the acceleration is from 0m/s to 10m/s or 10m/s to 0m/s

Car loses p
Earth gains -p

Total energy = p^2/2m_car + p^2/2m_earth
When the car accelerates the sign of the force changes, so the sign of the second term changes, but remains negligible.

Another worry is the mass of the Earth.
The math is circular, where a velocity of 10m/s is posited. The total p is calculated, and then 10000J subtracted, and then the KE is re-calculated using the mass, which was used to derive the momentum on the basis that the initial velocity of the Earth was 0m/s.
The result is always 10000J. Try it use m= 1e26kg.

ETA: That is in contrast with the case where actual values are used, and the same method.

Vearth ~ 29.8km/s = 29800m/s
mearth ~ 6e24kg
pi = 178800000000000000000000000000J
KEi = 2.66412e+33
subtract 10000kg.m/s
KEf 2.664119999999999999999999702e+33
dKE =298000000J

 

[Humber]

If the KERS works ideally then the remaining 50 kJ goes into the battery. Of course, nothing is 100% efficient, so some fraction will be lost to heat. It would only go all to heat if you were using traditional non-regenerative braking.

That's true, but I understood that it was idealised. The loss from the Earth of 50kJ, must be dissipated at the source, which is the Earth. Which does not happen in the first case.

 

[Ken G]

Car loses p
Earth gains -p

Total energy = p^2/2m_car + p^2/2m_earth

This is incorrect, and will lead you astray. Changes in energy are not p²/2m, they are changes in p²/2m, so you must not treat p, and changes in p, as if they were the same thing. They aren't! Let the change in p be denoted dp, as in calculus. Then a change in p²/2m looks like (p+dp)²/2m - p²/2m, and if dp << p, this is very close to p*dp/m or v*dp. That's why when KERS kicks in and there is some dp across some relative v between the car and the Earth, the energy change of the Earth, in the car-stationary frame, is the same as the energy change of the car, in the Earth-stationary frame: it is because either way the v and the dp are the same numbers. Work it out, and spend no time on the irrelevant dp² terms.

 

[Humber]

This is incorrect, and will lead you astray. Changes in energy are not p²/2m, they are changes in p²/2m, so you must not treat p, and changes in p, as if they were the same thing. They aren't! Let the change in p be denoted dp, as in calculus.

Momentum is frame independent, otherwise, there could be violations of conservation.
There would be no means of correcting after the fact, so violations simply don't occur.
That alone guarantees frame independence. If follows that changes are also frame independent.

Then a change in p²/2m looks like (p+dp)²/2m - p²/2m, and if dp << p, this is very close to p*dp/m or v*dp. That's why when KERS kicks in and there is some dp across some relative v between the car and the Earth, the energy change of the Earth, in the car-stationary frame, is the same as the energy change of the car, in the Earth-stationary frame: it is because either way the v and the dp are the same numbers. Work it out, and spend no time on the irrelevant dp² terms.

That is not correct, because "change" in momentum dp/dt = Force. KE is not frame independent, and that is also accommodated.
KE = p^2/2m = m^2v^2/2m = 1/2mv^2.

Total energy = p^2/2mcar + p^2/2mearth is correct. If the energy change of the car were the same as the change in the ground, p^2/2mcar - p^2/2mearth = 0, then obviously the KERS could not store any energy.

 

[Dale]

No, in the first case, the Earth is considered to be stationary, and the car changes velocity by 10m/s, and the Earth moves in the opposite direction - 1e-21m/s, so the difference, the relative velocity, is 10m/s + 1e-21m/s

In the second case the the car remains at 0m/s, and the Earth changes by -9.999999999999999999999m/s. which is not the same as the first case.

The quantity you are describing, v_{i,c}+v_{f,e}, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

The fact that some random sum of velocities is not frame invariant is hardly surprising, nor is it troubling.

No, in the first case, the Earth gains 5e-18J.
That will hold if the acceleration is from 0m/s to 10m/s or 10m/s to 0m/s

Car loses p
Earth gains -p

Total energy = p^2/2m_car + p^2/2m_earth
When the car accelerates the sign of the force changes, so the sign of the second term changes, but remains negligible.

I recommend that you work through the math completely and post your work like I did. Unfortunately, I will be traveling the next couple of days so I won't be able to do it myself.

Another worry is the mass of the Earth.
The math is circular, where a velocity of 10m/s is posited. The total p is calculated, and then 10000J subtracted, and then the KE is re-calculated using the mass, which was used to derive the momentum on the basis that the initial velocity of the Earth was 0m/s.
The result is always 10000J. Try it use m= 1e26kg.

The fact that the result is independent of the mass of the Earth is interesting, but not troubling. Many times you get a scenario where something cancels out. I will try to do it algebraically to see where and why it drops out, but not for a few days.

Vearth ~ 29.8km/s = 29800m/s
mearth ~ 6e24kg
pi = 178800000000000000000000000000J
KEi = 2.66412e+33
subtract 10000kg.m/s
KEf 2.664119999999999999999999702e+33
dKE =298000000J

Oh, that's a mess. The units of p should be kg m/s, not J. You cannot subtract a quantity in units of kg m/s from a quantity in units of J. I am not sure what you were trying to get at here.

 

[Dale]

The loss from the Earth of 50kJ, must be dissipated at the source, which is the Earth.

What makes you say that?

 

[Humber]

The quantity you are describing, v_{i,c}+v_{f,e}, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

The fact that some random sum of velocities is not frame invariant is hardly surprising, nor is it troubling.

Velocity is frame dependent, and sums linearly. Random velocities sum as the RMS of each, but none are random in this case.

I recommend that you work through the math completely and post your work like I did. Unfortunately, I will be traveling the next couple of days so I won't be able to do it myself.

I did that. There are several problems, including independence of Earth's mass.
That is the result of simply assigning a velocity to the Earth that is the same as the car's.

The fact that the result is independent of the mass of the Earth is interesting, but not troubling. Many times you get a scenario where something cancels out. I will try to do it algebraically to see where and why it drops out, but not for a few days.

(Δp_c=1e4kg.m/s)
p_i,e = M.v
KE_i,e = p_i,e²/2M
KE_f,e = (p_i,e)²-Δp_c/2M
KE_i,e - KE_f,e = 100kJ

Oh, that's a mess. The units of p should be kg m/s, not J. You cannot subtract a quantity in units of kg m/s from a quantity in units of J. I am not sure what you were trying to get at here.

V_earth ~ 29.8km/s = 29800m/s (orbital velocity of the Earth)
m_earth ~ 6e24kg (mass of the Earth)
p_i,e = 178800000000000000000000000000 kg.m/s (initial momentum mv)
KE_i,e = 2.66412e+33J ( Initial KE of the Earth)

subtract Δ_pc= 10000kg.m/s ( The applied impulse from the car)
KE_f,e 2.664119999999999999999999702e+33J ( Final KE of Earth)
dKE =298000000J (change in KE of Earth)

The same method as you employed, but where the initial Earth velocity is not zero.

 

[Humber]

What makes you say that?

The change in the Earth's KE is 100kJ. The car gains 50kJ. The remainder must go somewhere, and heat at the source seems to be the most likely. The actual reason is that there is no 100kJ

Total energy = p^2/2m_car + p^2/2_mearth
Applies to energy to and from the cart as the case may be.

 

[Humber]

When a car brakes, momentum is transferred from the car to the ground.
From the conservation of momentum;

Car gains momentum -p
Earth gains momentum p

The car loses KE = p²/2m_car
The Earth gains KE = p²/2m_earth

The total energy is therefore;
p²/2m_car+ p²/2m_earth

Because m_earth is very large ,~6e24kg, the second term is negligible.

All of that energy will have come from the fuel of the F1 car. It could be batteries, and the result is the same.

In a conventionally braked vehicle, the KE of the car's mass is dissipated as heat in the brakes. KERS does not affect momentum transfer to the ground, but stores the KE that is otherwise lost to heat.

That energy may be transferred to a flywheel;
The angular momentum L, of a particle, and the moment of inertia, I
L = r.p = r.mv
L =I.ω
KE = 1/2.I.ω²

There is no significant transfer of energy to or from the F1 and the ground, as a result of KERS.

 

[Dale]

Velocity is frame dependent, and sums linearly. Random velocities sum as the RMS of each, but none are random in this case.

They may not be random, but they are not relative velocity either.

I did that. There are several problems, including independence of Earth's mass.

I don't know why you think that is a problem.

(Δp_c=1e4kg.m/s)
p_i,e = M.v
KE_i,e = p_i,e²/2M
KE_f,e = (p_i,e)²-Δp_c/2M
KE_i,e - KE_f,e = 100kJ



V_earth ~ 29.8km/s = 29800m/s (orbital velocity of the Earth)
m_earth ~ 6e24kg (mass of the Earth)
p_i,e = 178800000000000000000000000000 kg.m/s (initial momentum mv)
KE_i,e = 2.66412e+33J ( Initial KE of the Earth)

subtract Δ_pc= 10000kg.m/s ( The applied impulse from the car)
KE_f,e 2.664119999999999999999999702e+33J ( Final KE of Earth)
dKE =298000000J (change in KE of Earth)

The same method as you employed, but where the initial Earth velocity is not zero.

This is not the same method I employed. I never subtracted a momentum from an energy, it is an invalid operation. The units don't work out.

 

[Dale]

The change in the Earth's KE is 100kJ. The car gains 50kJ. The remainder must go somewhere, and heat at the source seems to be the most likely.

The remainder goes into the battery. Where else would the energy that goes into the battery come from?

 

[Tea Jay]

I'd like to think that if the car's acceleration using the Earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the Earth back down by that same (infinitesimal) amount.

;)

 

[Humber]

They may not be random, but they are not relative velocity either.

The quantity you are describing, vi,c+vf,e, is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.
They are not the same in each case.

I don't know why you think that is a problem.

Then you don't have a case for your calculations.

This is not the same method I employed. I never subtracted a momentum from an energy, it is an invalid operation. The units don't work out.

(b) p_{i,c}=m v_{i,c}= 0 \ kg \ m/s
\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s
p_{f,c}=p_{i,c}+\Delta p_c = -10000 \ kg \ m/s
v_{f,c}=p_{f,c}/m= -10 \ m/s
KE_{i,c}=1/2 m v_{i,c}^2 = 0 \ kJ
KE_{f,c}=1/2 m v_{f,c}^2 = 50 \ kJ

p_{i,e}=M v_{i,e}= -10^{26} \ kg \ m/s
f_e=-f_c=1000 \ N

This is the impulse momentum of the car
\Delta p_e=f_e \Delta t= 10000 \ kg \ m/s

Here is where the momentum is decreased.
p_{f,e}=p_{i,e}+\Delta p_e = -9.999999999999999999999 \; 10^{25} \ kg \ m/s

v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s
KE_{i,e}=1/2 M v_{i,e}^2 = 5\;10^{26} \ kJ
KE_{f,e}=1/2 M v_{f,e}^2 = 4.999999999999999999999\;10^{26} \ J

The reduction of the Earth's momentum, is the basis of your KE claim.
It applies to the second case too. Which is why you will always get 100kJ, regardless of mass. It is an error.
The result is indenpendet of the Earth's mass because it is the equivalent of

Δp_c*v_f,c
= -10000kg.m/s*10m/s = 100000 kg.m²/s²
= -100kJ

 

[Humber]

The remainder goes into the battery. Where else would the energy that goes into the battery come from?

You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

So in the second reference frame the energy for charging the battery comes from the KE of the Earth which goes down by 100 kJ. The car gains 50 kJ of kinetic energy.

Which does not allow 50kJ for the car and 50kJ for the battery, because the system is a Kinetic Energy Recovery System. And you would still have a difference of 50kJ, when compared with the first case.