paradox in Classical Mechanics? energy vs force

[gusl]

Compare "accelerating under constant power"
with "accelerating under constant force" (constant acceleration)

My intuition tells me that they should be the same, but kinetic energy considerations show that the acceleration is decreasing on the first one (it takes 4 times the energy to get 2 times as fast).

This would seem to contradict "velocity is relative": if velocity were relative, then the energy needed to get faster by 1m/s would be the same whether you are stationary or already at 1 m/s.

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My intuition also tells me that I should be able to come up with a similar paradox about predicting the outcome of a 1-dimensional elastic collision. If you do it with energy vs momentum.

Conservation of energy:
v1_before^2 + v2_before^2 = v1_after^2 + v2_after^2 (if we fix one side of the equation, then the point (v1,v2) falls in a circle)

Conservation of momentum:
v1_before + v2_before = v1_after + v2_after (if we fix one side of the equation, then (v1,v2) falls in a straight line)

The solutions are where circle and line intersect. I guess there's no paradox afterall.

I would like to do a transform to a moving reference frame, to make sure that everything is still alright. Transforming to a fast-moving reference frame will just make the circle bigger. Basically, the point and the line all get transposed diagonally up and to the right. The distance between the intersections still remains the same.

Oh I see, physics is fine. Nothing to worry about.

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The concept of kinetic energy has always been problematic for me. Given the choice, I'll integrate over force instead.

Comments

Re: Energy

[r6.livejournal.com]

Oh, if we are firing rockets then the fuel needed will be the same.

If the cars are pushing the Earth, the difference is huge, because in one reference frame the entire earth is moving at 10 m/s. That’s 5.9742 × 10²⁶ joules (http://www.google.ca/search?q=%2810+m%2Fs%29%5E2+*+mass++of+earth+)!

Re: Energy

[gustavolacerda.livejournal.com]

If the cars are pushing the Earth, the difference is huge, because in one reference frame the entire earth is moving at 10 m/s. That’s ... joules!

I'm not sure what you were doing, but I think you forgot the "1/2" factor.

But the point is that you don't need to make the Earth move at 10m/s in order for you to move, since its momentum (or more precisely, its moment of inertia) is huge. Cars obviously have no trouble accelerating to 20m/s, don't spend anywhere near that much energy.

Re: Energy

[r6.livejournal.com]

Right, I forgot the 1/2 factor.

Okay, consider a car at rest on the earth at rest. The car pushes off the earth and moves at 10 − &epsilon m/s, and the earth moves at ε m/s. (&epsilon is determined by conservation of momentum). Before the was no kinetic engery, now there is some. The engery was gained from burning the fuel. I believe the car has most of the kinetic engery, and the earth has almost none. This may be worth checking.

Anyhow, now consider a car at rest, on an earth moving at 10 m/s. The car pushes off the earth and moves 10 − &epsilon′ m/s. Now the earth is moving at 10 + &epsilon′ m/s. Now almost the same amount of engery is gained by the car in as before (maybe exactly the same if ε′=ε), but the energy gained by the earth is much more is this second case. That is where the extra fuel is needed.

Re: Energy

[gustavolacerda.livejournal.com]

The engery was gained from burning the fuel. I believe the car has most of the kinetic engery, and the earth has almost none. This may be worth checking.

interesting. If you mean from the Earth-before's frame of reference, then I agree. This is due to energy being proportial to v^2 while momentum is proportional to v.


Now almost the same amount of engery is gained by the car in as before (maybe exactly the same if ε′=ε), but the energy gained by the earth is much more is this second case. That is where the extra fuel is needed.

I see how this follows: for the same reason that 100^2 - 99^2 is ~200 times greater than 2^2 - 1^1. Likewise the ratio there will be 2*(10m/s) / epsilon.
But I'm not sure this would hold up in all reference frames (although it should if I trust physicists)

Ok. So I failed to make KE contradict itself.

But why should it make a difference whether you shoot rockets or push the Earth? Should this be our new energy efficient form of transportation? :-P

Re: Energy

[r6.livejournal.com]

The problem with rockets is that you need to carry your propellent with you. So the reality of rocket science is that you need exponentially more fuel the faster you want to go.

In our rocket example we were comparing two identical cars, one going from 0 m/s to 10 m/s, and the other 10 m/s to 20 m/s. But if you want to go from 0 m/s to 20 m/s, then you need to carry enough fuel so that when you reach 10 m/s, you have enough fuel to go to 20 m/s. That extra fuel is heavy, so it will cost more fuel to go from 0 m/s to 10 m/s then it will to go from 10 m/s to 20 m/s. This is the fundamental source of the exponential blow up.

Re: Energy

[gustavolacerda.livejournal.com]

is rocket-launching associative?

i.e. does it make a difference what order you throw things out?
constant-energy, or constant-speed?

I think it isn't... neither at constant-speed nor constant-work.

Re: Energy

[r6.livejournal.com]

I’m not sure I understand the question. Rockets are time agnostic. It doesn’t matter when you decide to throw off your propellent to increase your speed. The only thing that matters is how fast you manage to throw your propellent. The faster you throw your propellent, the more efficient your rocket.

Re: Energy

[gustavolacerda.livejournal.com]

if you shoot two cannonballs at once, will you end up just as fast as if you shot them one by one?

Re: Energy

[r6.livejournal.com]

Yes. The only thing that matters is how fast you shoot them.

Re: Energy

[r6.livejournal.com]

… I think.

Re: Energy

[r6.livejournal.com]

Let’t change that to no.

Re: Energy

[r6.livejournal.com]

If the cannonballs both have an infinitesimal mass, then you will end up just as fast shooting them together as if you shot them one by one. ;-)

Re: Energy

[gustavolacerda.livejournal.com]

I think we could generalize this under the constraint that the total energy spent is the same in both cases.

Suppose you spend E1 when shooting the first cannonball, and E2 shooting the second.
Now go back. If you had shot them both at once, spending E1 + E2, then you should end up with the same speed.

Correct?

Re: Energy

[r6.livejournal.com]

Spending E1 shoots a cannonball at one speed. Spending E2 shoots a cannonball at another speed. In the second case the two cannonballs have the same speed. The two cases are so different, I don’t see why the net momentum of the two cannonballs would be the same in the two cases.

Re: Energy

[gustavolacerda.livejournal.com]

here's another paradox:

From your reasoning before, about pushing oneself off the Earth, it follows that it's more energy-efficient to shoot heavy things: you get more acceleration per rocket this way.

So if you had 2 identical cannonballs and two cannons, then you should shoot them simultaneously, rather than one after the other.

Oh, that makes sense, actually.
...maybe

Re: Energy

[r6.livejournal.com]

Now that I think about it, this “rocket problem” still applies to cars. Cars burn fuel, and expel it, making them lighter and easier to accelerate. I guess that the reason we drive instead of using rockets has to do with the earth being very massive. But I’m not sure.

Re: Energy

[gustavolacerda.livejournal.com]

If you stick to the same reference frame (which one is supposed to do), then the energy change is not so huge.