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KE of system / different reference frames question

by jduffy77
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Humber
#55
Jan3-12, 11:40 AM
P: 60
Quote Quote by DaleSpam View Post
I don't know if you are being deliberately irritating or you really can't grasp something this basic. I have said this multiple times already:

Earth loses 100 kJ KE
Car gains 50 kJ KE
Battery gains 50 kJ electrochemical energy
All energy is accounted for.
In the first case, the car will be stationary, with the rest state KE, and with a known amount of charge in the battery.
The above says that in the second case, there will be both charge in the battery and kinetic energy. A clear violation of both the conservation of momentum and energy.
All the result of one arithmetic error, that has been flagged to your attention.
Δpc*vf,c
= -10000kg.m/s*10m/s = 100000 kg.m2/s2
= -100kJ

Quote Quote by DaleSpam View Post
There is no missing energy 50 kJ + 50 kJ = 100 kJ, if you don't believe me then get a calculator and confirm.
Your calculation says you have made that error.

Quote Quote by DaleSpam View Post
The KERS recovers 50 kJ of electrochemical energy from the KE of the earth and the car gains 50 kJ of KE from the KE of the earth. The earth loses 100 kJ of KE.
Why don't you ponder that a bit until you get it.
Yes, I agree. Your calculation confirms that you have made that error.
jduffy77
#56
Jan3-12, 11:50 AM
P: 40
I should clarify something which most people interested in physics will not be familiar with. What Humber means when he says "a calculation confirms an error" is not that an error was made in the traditional sense, like incorrect math or a logical inconsistency. He simply means that the result conflicts with his preconceived expectation of what the result should be. So it is wrong.
Humber
#57
Jan3-12, 11:56 AM
P: 60
The correct solution to this problem, when the juvenile runs out of excuses, and recognizes his error, relies on nothing more than the conservation of momentum, which is a law quite independent of frames.

There are only two bodies involved: the car and the Earth so, a very simple case. When the car slows, it changes momentum;

Car gains momentum -p
Earth gains momentum p

p^2/2m = p^2v^2/2m = 1/2mv^2, which is the familiar KE equation.

The car loses KE = p2/2mcar
The Earth gains KE = p2/2mearth

The total energy is therefore;
p2/2mcar+ p2/2mearth

Because mearth is very large ,~6e24kg, the second term is negligible and can be approximated to the first term.

Initially, all of that energy will have come from the fuel of the F1 car, and in part, like the consumer counterpart, KERS is intended to reduce fuel consumption by returning energy that would otherwise be lost.

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

In the given example of a 1000kg vehicle braking to a stop from 10m/s, and where the mass of the Earth is 1e25kg, the energy transferred to the ground is;

(100000kg.m/s)^2/2e25kg
= 5e-18J

There is no significant transfer of energy to or from the F1 car and the ground as a result of KERS. The conservation of momentum says that result is independent of frame.
Humber
#58
Jan3-12, 12:16 PM
P: 60
The following calculation shows that there is negligible transfer to the ground, and is in agreement, also producing 5e-18J

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

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



All that remains, is to show that result is independent of frame.
Frame independence for the result can be shown by the introduction of an observer, moving relative to the Earth and Car at velocity u.

m = mass of car

ΔKE(car)

= (−m u − p)^2/2m − (−m u)^2/2m

= (m^2 u^2 −2m up + p^2)/2m − (m^2 u^2)/2m

= −up + p^2/2m


M = mass of Earth.

ΔKE(earth)

=(−M u + p)^2/2M − (−M u)^2/2M

= (M^2 u^2 + 2M up + p^2)/2M - (M^2 u^2)/2M

= u.p + p^2/2M

The total energy is once again

p^2/2m + p^2/2M
DaleSpam
#59
Jan3-12, 07:31 PM
Mentor
P: 16,952
Quote Quote by Humber View Post
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
This is true, due to Newton's 3rd law.

Quote Quote by Humber View Post
The car loses KE = p2/2mcar
The Earth gains KE = p2/2mearth
This is false, in general.

If some object gains momentum Δp then [itex]p_f=p_i+\Delta p[/itex]

Dividing both sides by the mass m we get [itex]v_f=v_i+\Delta v[/itex]

Substituting that into the expression for the object's final KE we obtain
[itex]KE_f=\frac{m}{2} v_f^2 = \frac{m}{2} (v_i+\Delta v)^2 = \frac{m}{2} (v_i^2 + 2 v_i \Delta v +\Delta v^2) = KE_i + v_i \Delta p + \Delta p^2/2m [/itex]
[itex]\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m [/itex]

This is different from what you wrote because of the [itex]v_i \Delta p[/itex] term. Your neglecting that term is actually your key conceptual error, so I do hope you spend some time examining the brief derivation. If the object is particularly massive, like the earth, then the [itex] \Delta p^2/2m [/itex] term drops out, but the [itex]v_i \Delta p[/itex] term does not.
DaleSpam
#60
Jan3-12, 07:33 PM
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P: 16,952
Quote Quote by Tea Jay View Post
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.
That is correct.
DaleSpam
#61
Jan3-12, 07:42 PM
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P: 16,952
Quote Quote by Humber View Post
And the change in momentum that you said you did not make, though it is quite clear that you did.
What exactly are you claiming that I said? Please use the quote feature to show exactly where I said I did not make a change in momentum. If you cannot find where I said it then I didn't say it, you only inferred it. Given this conversation so far, you probably inferred wrongly.
DaleSpam
#62
Jan3-12, 07:45 PM
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P: 16,952
Quote Quote by Humber View Post
The above says that in the second case, there will be both charge in the battery and kinetic energy.
Yes, that is correct, 50 kJ charge in the battery and 50 kJ KE in the car.

Quote Quote by Humber View Post
A clear violation of both the conservation of momentum and energy.
It is not a violation of either conservation of momentm or conservation of energy.
DaleSpam
#63
Jan3-12, 07:50 PM
Mentor
P: 16,952
Quote Quote by Humber View Post
The correct solution to this problem, when the juvenile runs out of excuses, and recognizes his error, relies on nothing more than the conservation of momentum, which is a law quite independent of frames.
I love the arrogance of the ignorant. Keep it coming Humber, it is very amusing.

Quote Quote by Humber View Post
There are only two bodies involved: the car and the Earth so, a very simple case. When the car slows, it changes momentum;

Car gains momentum -p
Earth gains momentum p

p^2/2m = p^2v^2/2m = 1/2mv^2, which is the familiar KE equation.

The car loses KE = p2/2mcar
The Earth gains KE = p2/2mearth
Again, this is false, in general. See my post 59 above. This is your key error, so please spend some time studying the derivation.
Humber
#64
Jan3-12, 08:11 PM
P: 60
Quote Quote by DaleSpam View Post
Quote Quote by humber
Car gains momentum -p
Earth gains momentum p
p^2/2m = p^2v^2/2m = 1/2mv^2, which is the familiar KE equation.

The car loses KE = p^2/2mcar
The Earth gains KE = p^2/2mearth
This is false, in general.

If some object gains momentum Δp then [itex]p_f=p_i+\Delta p[/itex]

Dividing both sides by the mass m we get [itex]v_f=v_i+\Delta v[/itex]

Substituting that into the expression for the object's final KE we obtain
[itex]KE_f=\frac{m}{2} v_f^2 = \frac{m}{2} (v_i+\Delta v)^2 = \frac{m}{2} (v_i^2 + 2 v_i \Delta v +\Delta v^2) = KE_i + v_i \Delta p + \Delta p^2/2m [/itex]
[itex]\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m [/itex]

This is different from what you wrote because of the [itex]v_i \Delta p[/itex] term. Your neglecting that term is actually your key conceptual error, so I do hope you spend some time examining the brief derivation. If the object is particularly massive, like the earth, then the [itex] \Delta p^2/2m [/itex] term drops out, but the [itex]v_i \Delta p[/itex] term does not.
Quote Quote by DaleSpam;
[itex]p_{i,c}=m v_{i,c}=10000 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = 0 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= 0 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 50 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 0 \ kJ[/itex]

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


Change in Earth's KE;
[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J[/itex]

KEearth =p^2/2Mearth
= (10000kg.m/s)^2/2*1e25
= 5e-18J
DaleSpam
#65
Jan3-12, 08:19 PM
Mentor
P: 16,952
Quote Quote by Humber View Post
KEearth =p^2/2Mearth
= (10000kg.m/s)^2/2*1e25
= 5e-18J
Which works since [itex]v_i=0[/itex] for the earth in frame (a). It does not work in frame (b) where [itex]v_i \ne 0[/itex] for the earth.
Humber
#66
Jan3-12, 08:20 PM
P: 60
Quote Quote by DaleSpam View Post
I love the arrogance of the ignorant. Keep it coming Humber, it is very amusing.

Again, this is false, in general. See my post 59 above. This is your key error, so please spend some time studying the derivation.

Quote Quote by DaleSpam View Post
(a) [itex]p_{i,c}=m v_{i,c}=10000 \ kg \ m/s[/itex]
[itex]\Delta p_c=f_c \Delta t= -10000 \ kg \ m/s[/itex]
[itex]p_{f,c}=p_{i,c}+\Delta p_c = 0 \ kg \ m/s[/itex]
[itex]v_{f,c}=p_{f,c}/m= 0 \ m/s[/itex]
[itex]KE_{i,c}=1/2 m v_{i,c}^2 = 50 \ kJ[/itex]
[itex]KE_{f,c}=1/2 m v_{f,c}^2 = 0 \ kJ[/itex]

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


[itex]KE_{f,e}=1/2 M v_{f,e}^2 = 5\;10^{-18} \ J[/itex]

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.
KE = p2/2m = m2v2/2m = 1/2mv2

When you finish falling over yourself and the arithmetic churn, you can stop.
DaleSpam
#67
Jan3-12, 08:25 PM
Mentor
P: 16,952
Do you agree with the derivation in post 59?

If yes, then do you understand how it was your key mistake?

If no, then which specific step of the derivation do you believe is in error, and what do you believe is the error?
Humber
#68
Jan3-12, 08:27 PM
P: 60
Quote Quote by DaleSpam View Post
Do you agree with the derivation in post 59?

If yes, then do you understand how it was your key mistake?

If no, then which specific step of the derivation do you believe is in error, and what do you believe is the error?
You correct it. I am not going to help you chase your tail.
DaleSpam
#69
Jan3-12, 08:29 PM
Mentor
P: 16,952
It is correct. Do you agree or not?

It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.
Humber
#70
Jan3-12, 08:43 PM
P: 60
Quote Quote by DaleSpam View Post
It is correct. Do you agree or not?
It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.
You have now made 4 unforced errors, and contradicted your own result.

When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.
mender
#71
Jan3-12, 08:47 PM
P: 563
Quote Quote by DaleSpam View Post
It is correct. Do you agree or not?

It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.
You're pretty good at this; some people take quite a few more posts to see that.
mender
#72
Jan3-12, 08:48 PM
P: 563
Quote Quote by Humber View Post
When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.
This is known as an assumed close.


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