How Does Reference Frame Affect Kinetic Energy Calculations?

[snoopies622]

I just thought of this. I think I know the answer to it now, but it took me a little bit of thinking. Maybe someone who teaches high school physics might find it useful.

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I have a battery-powered toy car. I turn it on and it accelerates from speed 0 to speed 1. (I'm leaving out units for simplicity here.) Since kinetic energy is a function of the square of an object's speed, the change in the car's kinetic energy is [itex] 1^2 - 0^2 = 1 - 0 = 1 [/itex]. All this energy comes from the battery.

Now look at this event from the reference frame of someone who is initially ahead of the car and walking towards it at speed 3. To him, the car accelerates from speed 3 to speed 4, and the change in kinetic energy is therefore [itex] 4^2 - 3^2 = 16 - 9 = 7 [/itex]. Again, all this energy comes from the battery.

How can a battery give up 1 unit of energy in one frame and 7 units in another frame? What if it was built with only 5 energy units in the first place?

 

[rcgldr]

When calculating work done, the frame of reference should be relative to the point of application of the force. The work done equals the force times distance. The rate of force times distance versus time is the power at any moment in time.

In the case of the car, the point of application of force is between the tires and the ground, so the ground should be the frame of reference. This ignores the fact that the ground and the Earth are given some tiny amount of speed and energy in the opposite direction, so that total angular and linear momentum of the Earth and model car system is preserved. The total energy is the increase in energy of the Earth and the car. The appropriate frame of reference would be the velocity of the initial point of contact between car and earth.

 

[K^2]

rcgldr, you can use any frame of reference that you like, so long as you do it right.

But yes, force is acting between ground and the toy car. That is relevant. Momentum of Earth changes in this problem as well, and therefore, so does its energy. The extra 6 units of energy in above come from kinetic energy of Earth relative to the observer moving at speed of 3. Let's make this explicit, shall we?

Mass of the car in the above problem must be m=2 to make numbers work. Let's take Earth's mass to be M>>m. I'll take a limit where M->∞, because that will give me the correct answer.

In Earth's rest frame:

Car's initial velocity: 0.
Earth's initial velocity: 0.
Car's final velocity: 1.
Earth's final velocity: -m/M.

Total final energy:

[tex]E(M) = 1 + \frac{1}{2}M\left(-\frac{m}{M}\right)^2[/tex]

[tex]\lim_{M\to\infty}E(M) = 1[/tex]

This is no different than your initial answer. Now, let's do this relative to observer heading towards the car at 3.

Car's initial velocity: 3.
Earths' initial velocity: 3.
Car's final velocity: 4.
Earth's final velocity: 3-m/M.

[tex]E_i(M) = 9 + \frac{1}{2}M(3)^2[/tex]

[tex]E_f(M) = 16 + \frac{1}{2}M\left(3-\frac{m}{M}\right)^2[/tex]

[tex]E(M) = E_f(M) - E_i(M) = 7 + \frac{1}{2}M\left(-\frac{m}{M}\right)^2 - 3M\frac{m}{M}[/tex]

[tex]E(M) = 1 + \frac{1}{2}M\left(-\frac{m}{M}\right)^2[/tex]

[tex]\lim_{M\to\infty}E(M) = 1[/tex]

Notice that even before the limit is taken, you get exactly the same result. In the limit, you get the same result as if you ignored the Earth in Earth's rest frame.

P.S. Yes, I'm ignoring rotation of Earth for simplicity, but the logic is exactly the same.

 

[lalbatros]

rcgldr,

You explanation is perfect, and it applies in any frame of reference.

Michel

 

[rcgldr]

rcgldr, you can use any frame of reference that you like, so long as you do it right.

Thanks, I wasn't sure including the energy change to the Earth would cover the case of a moving observer. I recall that a similar problem involving the energy change of a rocket and it's spent fuel would be the same regardless of the frame of reference, but wasn't sure about this case.

 

[K^2]

Yeah, it's basically the same idea as with the rocket. If you include only part of the system, there is an external force, and energy need not be conserved in all frames. If you include reaction force, you must include the object it does work on, and then you will have energy conservation.

 

[rcgldr]

I was trying to address the original post with an open system approach (earth not considered part of the system or to be considered as having infinite mass), in which case the point of application of force is important, so that power = force x speed at the point of application of force.

If the Earth is included (and ignoring losses to heat), it's a closed system, and energy is conserved. The engine and drive train of the car converts chemical or electrical potential energy into kinetic energy of car and Earth within the system, and the total energy remains constant.

 

[snoopies622]

I was trying to address the original post with an open system approach (earth not considered part of the system or to be considered as having infinite mass)...

I don't know if that works. I only found an answer by remembering momentum conservation, and realizing that if the car is accelerating in one direction then something must be accelerating in the other direction, and then including both things in the energy computations.

 

[K^2]

Well, you can treat it as an open system, but you still have to remember the reaction force. Knowing that the car pushes against the ground to accelerate, you can compute the flow of energy due to the work done by that force. It's essentially equivalent, but you don't care what happens to the energy once it leaves the system, or where it came from before it entered the system.

 

[snoopies622]

OK, I think I understand.

 

[pankazmaurya]

rcgldr, you can use any frame of reference that you like, so long as you do it right.

But yes, force is acting between ground and the toy car. That is relevant. Momentum of Earth changes in this problem as well, and therefore, so does its energy. The extra 6 units of energy in above come from kinetic energy of Earth relative to the observer moving at speed of 3. Let's make this explicit, shall we?

Mass of the car in the above problem must be m=2 to make numbers work. Let's take Earth's mass to be M>>m. I'll take a limit where M->∞, because that will give me the correct answer.

In Earth's rest frame:

Car's initial velocity: 0.
Earth's initial velocity: 0.
Car's final velocity: 1.
Earth's final velocity: -m/M.

Total final energy:

[tex]E(M) = 1 + \frac{1}{2}M\left(-\frac{m}{M}\right)^2[/tex]

[tex]\lim_{M\to\infty}E(M) = 1[/tex]

This is no different than your initial answer. Now, let's do this relative to observer heading towards the car at 3.

Car's initial velocity: 3.
Earths' initial velocity: 3.
Car's final velocity: 4.
Earth's final velocity: 3-m/M.

[tex]E_i(M) = 9 + \frac{1}{2}M(3)^2[/tex]

[tex]E_f(M) = 16 + \frac{1}{2}M\left(3-\frac{m}{M}\right)^2[/tex]

[tex]E(M) = E_f(M) - E_i(M) = 7 + \frac{1}{2}M\left(-\frac{m}{M}\right)^2 - 3M\frac{m}{M}[/tex]

[tex]E(M) = 1 + \frac{1}{2}M\left(-\frac{m}{M}\right)^2[/tex]

[tex]\lim_{M\to\infty}E(M) = 1[/tex]

Notice that even before the limit is taken, you get exactly the same result. In the limit, you get the same result as if you ignored the Earth in Earth's rest frame.

P.S. Yes, I'm ignoring rotation of Earth for simplicity, but the logic is exactly the same.

but how will know about the bigger frame of refrence that is u r considering Earth as a part of it

 

[ajith.mk91]

But consider this. Suppose an object(m=2) is moving in free space with a velocity 1. Its kinetic energy as calculated by the person moving with it will be 0. But if the same person moves with a speed 3 in the opposite direction of the object,he will end up calculating K.E. as 16.Now what is its K.E.?

 

[snoopies622]

Now what is its K.E.?

In that sense kinetic energy is relative. This is true in both classical physics and in special relativity. A "moving object" will always have zero kinetic energy in one frame of reference, and non-zero kinetic energy in every other frame of reference.

 

[pankazmaurya]

But consider this. Suppose an object(m=2) is moving in free space with a velocity 1. Its kinetic energy as calculated by the person moving with it will be 0. But if the same person moves with a speed 3 in the opposite direction of the object,he will end up calculating K.E. as 16.Now what is its K.E.?

its kinetic energy is 0 in the first case and 16 in second...and the body will do work according to the kinetic energy in the frame of reference only .so in the first case no work will be done