KE of system / different reference frames question


by jduffy77
Tags: frames, reference
DaleSpam
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#109
Jan4-12, 06:46 AM
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Quote Quote by Humber View Post
Momentum is conserved between the car and the Earth.
The car gains -p, and the Earth gains p, so in the above case, that is Δp.
From this response I understand that you agree with the proof of post 59 as it applies to a single body and that here you are correctly pointing out that in the KERS scenario I must use the formula twice, once for [itex]\Delta KE_e[/itex] and once for [itex]\Delta KE_c[/itex], with [itex]\Delta p_e=-\Delta p_c[/itex] in order to conserve momentum.

Is this a correct understanding of your position? If I am misunderstanding then can you please provide a clear yes/no answer to whether or not you agree with the proof of post 59.
Humber
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#110
Jan4-12, 07:00 AM
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Quote Quote by rcgldr View Post
I don't see any problem here. If both the belt and the F1 car are initially at rest (say from a ground frame of reference), and then the F1 uses it's engine to convert chemical potential energy from fuel into kinetic energy of the belt and F1 car, and assuming no losses to heat, then all of the energy winds up as an increase in KE of the belt and F1 car, with most of the KE going into the belt because the F1 car has much more mass than the belt.

Falsifies what?
The amount of KE recoverable by the KERS is due only to the KE of wheels and the mechanism of the TM, and not related to the translational KE of the car, in any way.
DaleSpam
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#111
Jan4-12, 07:08 AM
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Quote Quote by Humber View Post
ΔKE(car)

= (−m u − p)2/2m − (−m u)2/2m

= (m2 u2 −2m up + p2)/2m − (m2 u2)/2m
Note, a slight math error marked in red. A negative times a negative is a positive, so that term should be +2m up


Quote Quote by Humber View Post
ΔKE(earth)
= u.p + p2/2M
So you clearly agree with my formula, [itex]\Delta KE = v_i \Delta p + \Delta p^2/2m [/itex], in the case where [itex]u=v_i[/itex] and [itex]p=\Delta p[/itex]. Is there any time that you do not agree with it?

Btw, it would really help if you were more careful about distinguishing your quantities. Ie. Is your p the change in momentum or is it the initial momentum? I follow the usual convention where [itex]\Delta p[/itex] is the change in momentum, [itex]p_i[/itex] is the initial momentum, and [itex]p_f=p_i+\Delta p[/itex] is the final momentum.
Humber
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#112
Jan4-12, 07:09 AM
P: 60
Quote Quote by DaleSpam View Post
From this response I understand that you agree with the proof of post 59 as it applies to a single body and that here you are correctly pointing out that in the KERS scenario I must use the formula twice, once for [itex]\Delta KE_e[/itex] and once for [itex]\Delta KE_c[/itex], with [itex]\Delta p_e=-\Delta p_c[/itex] in order to conserve momentum.

Is this a correct understanding of your position? If I am misunderstanding then can you please provide a clear yes/no answer to whether or not you agree with the proof of post 59.
Only the change in the car's momentum is of concern, so, [itex]\Delta p_e=-\Delta p_c[/itex] That is all there is. It is transferred to the Earth.
In the first case, the car's 50kJ, is transferred to the battery. So, the second case, which is a mirror, should see the energy come from the battery to the car. The total remains in each case at 50kJ.

In the example you gave, and in the first case, the energy to the ground is a result of a change of Earth's velocity, 1e-21m/s. That results in 5e-18J

In the other case, the change in Earth's velocity is Vcar*(Mearth - Mcar)/Mearth
Vcar =10m/s
Mearth = 1e25kg
Mcar =1e3kg

so the change in KE is related to (10 - 1e-21)m/s.
That , I think, accounts for the difference of relative velocities and energy.

The energy "transferred back to the car" should be Mearth/Mcar*5e-18J =50kJ
DaleSpam
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#113
Jan4-12, 07:19 AM
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Quote Quote by Humber View Post
Only the change in the car's momentum is of concern, so, [itex]\Delta p_e=-\Delta p_c[/itex] That is all there is. It is transferred to the Earth.
In the first case, the car's 50kJ, is transferred to the battery. So, the second case, which is a mirror, should see the energy come from the battery to the car. The total remains in each case at 50kJ.

In the example you gave, and in the first case, the energy to the ground is a result of a change of Earth's velocity, 1e-21m/s. That results in 5e-18J

In the other case, the change in Earth's velocity is Vcar*(Mearth - Mcar)/Mearth
Vcar =10m/s
Mearth = 1e25kg
Mcar =1e3kg

so the change in KE is related to (10 - 1e-21)m/s.
That , I think, accounts for the difference of relative velocities and energy.
You are still dodging the question. Do you agree with the proof post 59 or not? Please be clear.
Humber
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#114
Jan4-12, 07:29 AM
P: 60
Quote Quote by DaleSpam View Post
Note, a slight math error marked in red. A negative times a negative is a positive, so that term should be +2m up


So you clearly agree with my formula, [itex]\Delta KE = v_i \Delta p + \Delta p^2/2m [/itex], in the case where [itex]u=v_i[/itex] and [itex]p=\Delta p[/itex]. Is there any time that you do not agree with it?

Btw, it would really help if you were more careful about distinguishing your quantities. Ie. Is your p the change in momentum or is it the initial momentum? I follow the usual convention where [itex]\Delta p[/itex] is the change in momentum, [itex]p_i[/itex] is the initial momentum, and [itex]p_f=p_i+\Delta p[/itex] is the final momentum.
Momentum adds linearily, so there is no dependency on pi
p^2/2mcar is the KE of the car with momentum p
p^2/2mearth is the change in the Earth's KE when p is transferred to it.
DaleSpam
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#115
Jan4-12, 07:37 AM
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Quote Quote by Humber View Post
Momentum adds linearily, so there is no dependency on pi
This is an ambiguous statement. There is no dependency of what quantity on [itex]p_i[/itex]?

Also, clearly [itex]p_i[/itex] depends on [itex]p_i[/itex] even though momentum adds linearly, so adding linearly is not a justification for a lack of dependency in general. If you want to show that x doesn't depend on y because y adds linearly then you need to take the linearity property, use it in the expression for x, and show that y drops out algebraically.

Btw, I note that you still avoid the question.

Quote Quote by Humber View Post
p^2/2mcar is the KE of the car with momentum p
This is true in general.
Quote Quote by Humber View Post
p^2/2mearth is the change in the Earth's KE when p is transferred to it.
This is true only in the case where the initial velocity is 0. Note the word "change" in bold. The change of x is not the same thing as x except when x is initially 0.
Humber
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#116
Jan4-12, 07:50 AM
P: 60
Quote Quote by DaleSpam View Post
This is an ambiguous statement. There is no dependency of what quantity on p[SUB]i?

Also, clearly p[SUB]i depends on p[SUB]i even though momentum adds linearly, so adding linearly is not a justification for a lack of dependency in general. If you want to show that x doesn't depend on y because y adds linearly then you need to take the linearity property, use it in the expression for x, and show that y drops out algebraically.
It does. The existing momentum is pi,e the total after the change is pi,e + pi,c, so the change is just pi,c
The change in KE for the car is pi,c^2/2mc and for the Earth pi,c^2/2me
Works both ways.
DaleSpam
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#117
Jan4-12, 07:57 AM
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Quote Quote by Humber View Post
It does. The existing momentum is pi,e the total after the change is pi,e + pi,c, so the change is just pi,c
The change in KE for the car is pi,c^2/2mc and for the Earth pi,c^2/2me
Works both ways.
Then it should be no problem to demonstrate it algebraically, this time using clear variable names to distinguish between initial, final, and change quantities.
Humber
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#118
Jan4-12, 08:05 AM
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Quote Quote by DaleSpam View Post
Then it should be no problem to demonstrate it algebraically, this time using clear variable names to distinguish between initial, final, and change quantities.
The point is that it's not. The idea of the car accelerating the Earth to 10m/s is not physically realizable. In the real world, an F1 car is on the track. The KERS is applied. If viewed from the frame of the car, and from the frame of the ground at the same time, there will be no difference. To start with the Earth at 10m/s relative to the car, is a totally different physical situation. In the first case the calculation is done with the car having 50kJ, which after 10s, is transferred to the battery. In the second, it's already there, at t = 0.
Makes no sense.
Humber
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#119
Jan4-12, 08:09 AM
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Quote Quote by DaleSpam View Post
Then it should be no problem to demonstrate it algebraically, this time using clear variable names to distinguish between initial, final, and change quantities.
It's your notation i = initial, c = car, e= earth. There is one transfer of 10000kg.m/s to the momentum, pi,e. It makes no difference what that is. The resultant KE depends on the Earth's mass.
Ken G
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#120
Jan4-12, 08:17 AM
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Quote Quote by Humber View Post
Momentum is frame independent, otherwise, there could be violations of conservation.
I figured this was your core error, you don't understand the meaning of conservation laws. Conservation laws don't mean that the quantities are fixed, regardless of frame. They mean that once you pick a frame, the total quantities will stay the same in that frame. If you change frames, the quantities change. That's how conservation laws work. Do you get this, or not?
If follows that changes are also frame independent.
Wrong, the changes are frame independent, but that doesn't "follow" from anything, that statement, and only that statement, is the conservation law. Think about this as long as it takes.
Total energy = p^2/2mcar + p^2/2mearth is correct.
No, not if you think that p is a change in momentum, which is how you are using it. Doesn't it concern you that you do not get the correct answers, and you conclude that all the experts are wrong, but when they tell you what you are doing wrong, you just claim you are right? I can tell you right now, you will never learn anything that way. Is it all right with you to never learn anything?

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.
Notice where once again you associate p^2/2m with energy changes. Wrong.
DaleSpam
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#121
Jan4-12, 08:26 AM
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Quote Quote by Humber View Post
To start with the Earth at 10m/s relative to the car, is a totally different physical situation.
Here is where you are flat out wrong. It is merely a different (but equally valid) description of the same physical situation. You are denying the principle of relativity. This has been a cornerstone of physics since Galileo's time:
http://physics.ucr.edu/~wudka/Physic...ww/node47.html
You are literally centuries out of date in your thinking.

Quote Quote by Humber View Post
In the first case the calculation is done with the car having 50kJ, which after 10s, is transferred to the battery.
Yes.

Quote Quote by Humber View Post
In the second, it's already there, at t = 0.
Makes no sense.
This is incorrect. In the second the earth has a huge amount of KE, of which after 10s, 50 kJ is transferred to the battery and 50 kJ is transfered to the car's KE.

These are both correct descriptions of the same physical situation. Momentum and energy are conserved in both cases, and Newton's laws are obeyed in both cases.
DaleSpam
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#122
Jan4-12, 08:30 AM
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Quote Quote by Humber View Post
It's your notation i = initial, c = car, e= earth. There is one transfer of 10000kg.m/s to the momentum, pi,e. It makes no difference what that is. The resultant KE depends on the Earth's mass.
So prove it, mathematically. If you are going to make claims, then you should be able to back them up with clear and unambiguous derivations, as I have done.
Humber
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#123
Jan4-12, 08:32 AM
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Quote Quote by Ken G View Post
I figured this was your core error, you don't understand the meaning of conservation laws. Conservation laws don't mean that the quantities are fixed, regardless of frame. They mean that once you pick a frame, the total quantities will stay the same in that frame.
And none can be created.

Quote Quote by Ken G View Post
If you change frames, the quantities change. That's how conservation laws work. Do you get this, or not?Wrong, the changes are frame independent, but that doesn't "follow" from anything, that statement, and only that statement, is the conservation law. Think about this as long as it takes.
Yes it does. There are two objects. The ground and the car. Momentum is conserved, so what one gains, the other loses, and that is entirely frame independent, as the total remains the same and pe - pc = 0 or p = -p

Quote Quote by Ken G View Post
No, not if you think that p is a change in momentum, which is how you are using it.
When all is transferred Δp is p. 100% of p is p.

Quote Quote by Ken G View Post
Doesn't it concern you that you do not get the correct answers, and you conclude that all the experts are wrong, but when they tell you what you are doing wrong, you just claim you are right? I can tell you right now, you will never learn anything that way. Is it all right with you to never learn anything?
Oh, right.

Quote Quote by Ken G View Post
Notice where once again you associate p^2/2m with energy changes. Wrong.
Oh right.

p= mv ( I assume you know that)

p2 = m2v2

p2/2m = m2v2/2m = 1/2mv2 = KE.
DaleSpam
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#124
Jan4-12, 08:34 AM
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Btw, I think that Huber's continued avoidance of the question is because he recognizes that the derivation is correct, but he feels that he would lose face to admit it.
D H
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#125
Jan4-12, 08:43 AM
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This nonsense has gone on long enough. Thread locked.


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