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KE of system / different reference frames question 
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#55
Jan312, 11:40 AM

P: 60

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 


#56
Jan312, 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.



#57
Jan312, 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 = p^{2}/2m_{car} The Earth gains KE = p^{2}/2m_{earth} The total energy is therefore; p^{2}/2m_{car}+ p^{2}/2m_{earth} Because m_{earth} 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.ω^{2} 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 = 5e18J 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. 


#58
Jan312, 12:16 PM

P: 60

The following calculation shows that there is negligible transfer to the ground, and is in agreement, also producing 5e18J
[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 u·p + p^2)/2m − (m^2 u^2)/2m = −u·p + p^2/2m M = mass of Earth. ΔKE(earth) =(−M u + p)^2/2M − (−M u)^2/2M = (M^2 u^2 + 2M u·p + 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 


#59
Jan312, 07:31 PM

Mentor
P: 16,988

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. 


#60
Jan312, 07:33 PM

Mentor
P: 16,988




#61
Jan312, 07:42 PM

Mentor
P: 16,988




#62
Jan312, 07:45 PM

Mentor
P: 16,988




#63
Jan312, 07:50 PM

Mentor
P: 16,988




#64
Jan312, 08:11 PM

P: 60

= (10000kg.m/s)^2/2*1e25 = 5e18J 


#65
Jan312, 08:19 PM

Mentor
P: 16,988




#66
Jan312, 08:20 PM

P: 60

When you finish falling over yourself and the arithmetic churn, you can stop. 


#67
Jan312, 08:25 PM

Mentor
P: 16,988

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? 


#68
Jan312, 08:27 PM

P: 60




#69
Jan312, 08:29 PM

Mentor
P: 16,988

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. 


#70
Jan312, 08:43 PM

P: 60

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. 


#71
Jan312, 08:47 PM

P: 563




#72
Jan312, 08:48 PM

P: 563




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