Why is the kinetic energy equation multiplied by ½?

[russ_watters]

One of the things Coriolis did, in 1829, was to apply the word "travail" to the expression W=Fs.
That Fs is a relevant quantity was, however, long known. Already in Archimedes´ lever law in 3rd century BC, and generalized by Solomon de Caus in 1615.
Well before Newton and his laws.
What was Fs called before Coriolis renamed it "travail"?

That's a history question. Can you explain what relevance it has to modern, practical physics?

 

[snorkack]

What Coriolis did was emphasize the connection between work and kinetic energy.
Which can be stated by 3 equivalent equations:

1. Δ(mv²/2)=maΔs
2. Δ(mv²)=2(maΔs)
3. Δ(mv²)=(2ma)Δs

Now, note that the expression W=FΔs had been used for two millennia, since Archimedes and Caus. Not quite that form. But since it is an useful and common expression, introducing a factor there would have been inconvenient and clearly unreasonable. So that leaves te expressions 1 and 3.
Can you follow so far?

 

[russ_watters]

What Coriolis did was emphasize the connection between work and kinetic energy.
Which can be stated by 3 equivalent equations:

1. Δ(mv²/2)=maΔs
2. Δ(mv²)=2(maΔs)
3. Δ(mv²)=(2ma)Δs

Now, note that the expression W=FΔs had been used for two millennia, since Archimedes and Caus. Not quite that form. But since it is an useful and common expression, introducing a factor there would have been inconvenient and clearly unreasonable. So that leaves te expressions 1 and 3.
Can you follow so far?

Kind of...but I'm not sure I like where this is going. It sounds like you are going to suggest that where to put the "2" is/was an arbitrary choice.

And pardon me if I've guessed wrong, but your responses have been slow and inconsistent so I'm trying to efficiently move the discussion forward.

 

[snorkack]

Kind of...but I'm not sure I like where this is going. It sounds like you are going to suggest that where to put the "2" is/was an arbitrary choice.

Yes and no.
On one level, it´ s "arbitrary", on a level, because there clearly were alternatives, such as vis viva.
On another level, I argue that some of the choices are clearly worse. And that narrows down the options where to put the "2".
Now I have narrowed down the options to
E=mv²/2
F=2ma
Continuing to target the history... just when and by whom was force defined through acceleration?
Yes, Galileo and Newton pointed out that force is needed for acceleration and causes acceleration unless counteracted by other forces. But this still left the option of treating acceleration as one and not the most important possible consequence of force, rather than the definition of force.
Force is traditionally expressed through units like kilogram-force - implicitly taking in Earth gravity acceleration rather than an acceleration measured in units of length and time. If your work is expressed in units like foot-pounds or horsepower-seconds, you´ re stuck with multiplication factors anyway... and then you don´ t have a strong reason to make your vis viva calculations harder by a factor there.

 

[russ_watters]

Yes and no.
On one level, it´ s "arbitrary", on a level, because there clearly were alternatives, such as vis viva.
On another level, I argue that some of the choices are clearly worse. And that narrows down the options where to put the "2".
Now I have narrowed down the options to
E=mv²/2
F=2ma

You've based this on the premise that the location of the "2" is a choice (arbitrary or otherwise). It's not a choice (today), it's a byproduct of the math. And what you've done there, for your second "choice" is to change a definition and then ignore the implications of changing the definition. It may be possible to re-formulate a large fraction of kinematics to make it work, but in the present formulation it doesn't work.
[edit] And frankly, I doubt it would be possible to explain-away its origin, but I would be interested to see your take. Do you understand where 2 actually came from?

If your work is expressed in units like foot-pounds or horsepower-seconds, you´ re stuck with multiplication factors anyway... and then you don´ t have a strong reason to make your vis viva calculations harder by a factor there.

Is that really how you think? I have trouble with the idea of arbitrarily and needlessly making my life harder for no reason, even if only by a small amount. But while you are entitled to do such things for/to yourself, science is arrived at by consensus, so you'd have a very hard time convincing a group of scientists to needlessly complicate something. It's kind of the opposite of what they are typically after.

 

[snorkack]

science is arrived at by consensus,

True. So when you want to check that consensus, how do you demonstrate the grounds of that consensus? History is one obvious option, because that is viewing a condition where the consensus did not exist.
Quite excluding m, it´ s very obvious that when you have v and a, you get Δs=v²/2a. So you need the factor of 2 either in expression of energy or of force, by the very decision to measure force in units of acceleration*mass. So the decision that needs an explanation is the decision to put the factor in energy rather than force. Cannot pin down the history of that yet.

 

[russ_watters]

True. So when you want to check that consensus, how do you demonstrate the grounds of that consensus? History is one obvious option, because that is viewing a condition where the consensus did not exist.

No. History describes the path to the consensus, which is more than the grounds/basis for the consensus and is unnecessary for it.

Quite excluding m, it´ s very obvious that when you have v and a, you get Δs=v²/2a. So you need the factor of 2 either in expression of energy or of force, by the very decision to measure force in units of acceleration*mass. So the decision that needs an explanation is the decision to put the factor in energy rather than force. Cannot pin down the history of that yet.

In other words, no, you don't know the mathematical reason for it, even though it was discussed elsewhere in the thread. It's the area of a triangle or the average speed under constant acceleration. That's why the math causes it to be connected to the velocity part of the formula. It appears there because that's where it came from. No choice had to be made (only the choice not to move it for no reason).

 

[Dale]

The whole history discussion is irrelevant and pointless, IMO. It doesn't matter historically how the factor arrived, all that matters is why it is still useful today.

 

[PeroK]

What I would like to know is this: if a particle is moving in uniform circular motion why isn't the kinetic energy equal to $\frac{1}{2\pi}mv^2$?

 

[Herman Trivilino]

The whole history discussion is irrelevant and pointless, IMO. It doesn't matter historically how the factor arrived, all that matters is why it is still useful today.

I thought the whole point of the thread was explaining the presence of the factor of one-half in the expression for the kinetic energy of a particle. It's there by convention, so it seems to me that if you want to understand why the convention exists you have to understand how it originated. I think the history of it is so entangled that it can't be sorted out definitively.

 

[Dale]

I thought the whole point of the thread was ...

Well, the whole point of the thread is still unclear since the @rajen0201 never bothered to clarify what it was that he wanted in an explanation.

But I disagree that the 1/2 is a convention except insofar as the definition of any word is a convention. If I do 3 J of work I gain 3 J of KE, not 6 J. The definition of work and the conservation of energy require the 1/2.

 

[jbriggs444]

Well, the whole point of the thread is still unclear since the @rajen0201 never bothered to clarify what it was that he wanted in an explanation.

But I disagree that the 1/2 is a convention except insofar as the definition of any word is a convention. If I do 3 J of work I gain 3 J of KE, not 6 J. The definition of work and the conservation of energy require the 1/2.

"except insofar as the definition of any word is a convention". Yes.

It would be silly. [Full stop]. But we could have decided to measure work in Newton-meters and kinetic energy in half-joules. Then we could do 3 Newton-meters of work and gain 6 half-joules of kinetic energy. $W=F\cdot s$ and $E=mv^2$. It would then be a mathematical fact that the two units differ by a factor of two ($\Delta E = 2W$) That we choose not to do so is a matter of convention. A convention adopted for good and obvious reasons.

 

[Dale]

we could have decided to measure work in Newton-meters and kinetic energy in half-joules.

Changing the units doesn’t change the rules of integration nor does it get rid of the dimensionless constants of integration. It just hides them. That factor of 1/2 still arises in the integration and it is still present in the formula, it is just hidden by the units and the resulting conversion factors.

 

[Herman Trivilino]

All you have to do is define $W=2 F\cdot s$.

 

[Dale]

All you have to do is define $W=2 F\cdot s$.

Yes. Which is why I said “except insofar as the definition of any word is a convention”

 

[sophiecentaur]

All you have to do is define $W=2 F\cdot s$.

But that would be totally arbitrary and would require much more than 75 posts to discuss and justify. The 'half' in the KE formula would be there, even on the planet Zog and whatever units they built their Physics on. It would be just as loony to try to fit 19e or π/93 into our formulae.

 

[A.T.]

...just as loony to try to fit 19e or π/93 into our formulae.

Since you mention π, the common convention there went exactly the other way, and we ended up with many '2π' in the formulas. I personally would prefer to have no factor 2 in the circle circumference, and instead a factor 1/2 in the circle area, because that can be derived from the triangle area as well.

 

[PeroK]

Since you mention π, the common convention there went exactly the other way, and we ended up with many '2π' in the formulas. I personally would prefer to have no factor 2 in the circle circumference, and instead a factor 1/2 in the circle area, because that can be derived from the triangle area as well.

You're alluding to the Tau Manifesto?

https://tauday.com/tau-manifesto

 

[jack action]

All you have to do is define $W=2 F\cdot s$.

Wouldn't it leads to the question «Why is the work equation multiplied by 2?»

To which the probable answer would be: «To eliminate the ½ in the kinetic energy equation.»

 

[Dale]

This thread went from off topic to further off topic. Time to put it out of its misery