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KE of system / different reference frames question 
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#37
Jan212, 10:16 PM

P: 60

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. 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 recalculated 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 


#38
Jan212, 10:18 PM

P: 60




#39
Jan312, 01:02 AM

PF Gold
P: 3,136




#40
Jan312, 01:48 AM

P: 60

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. 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. 


#41
Jan312, 02:09 AM

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P: 17,509

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


#42
Jan312, 02:11 AM

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#43
Jan312, 02:48 AM

P: 60

That is the result of simply assigning a velocity to the Earth that is the same as the car's. p_{i,e} = M.v KE_{i,e} = p_{i,e}^{2}/2M KE_{f,e} = (p_{i,e})^{2}Δp_{c}/2M KE_{i,e}  KE_{f,e } = 100kJ 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. 


#44
Jan312, 02:51 AM

P: 60

Total energy = p^2/2m_{car }+ p^2/2_{mearth} Applies to energy to and from the cart as the case may be. 


#45
Jan312, 04:15 AM

P: 60

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 = 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. 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.ω^{2} There is no significant transfer of energy to or from the F1 and the ground, as a result of KERS. 


#46
Jan312, 07:17 AM

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P: 17,509




#47
Jan312, 07:19 AM

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#48
Jan312, 07:31 AM

P: 96

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.
;) 


#49
Jan312, 07:32 AM

P: 60

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. 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.m^{2}/s^{2} = 100kJ 


#50
Jan312, 07:39 AM

P: 60

That leaves 50kJ unaccounted for. 


#51
Jan312, 07:41 AM

P: 60

In reality, the Earth is not an infinitly rigid sphere, and will dissipate energy, just as it does for Earthquakes and nuclear weapons tests. 


#52
Jan312, 11:20 AM

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P: 17,509




#53
Jan312, 11:23 AM

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P: 17,509

Earth loses 100 kJ KE Car gains 50 kJ KE Battery gains 50 kJ electrochemical energy All energy is accounted for. There is no missing energy 50 kJ + 50 kJ = 100 kJ, if you don't believe me then get a calculator and confirm. Why don't you ponder that a bit until you get it. 


#54
Jan312, 11:35 AM

P: 60

iΔpc*vf,c = 10000kg.m/s*10m/s = 100000 kg.m2/s2 = 100kJ 


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