Vay

Help in explaining formula of Kinetic Energy

84 posts in this topic

You are 100% correct on the The------>"meaningful calculation"

 

But in today's "mentality" I think that perhaps we should consider that what is meaningful to some is pointless to others " hence" if its not broken don't fix it mentality.When connecting Classical Physical Formulas with QM operators we see things that don't pair up and or brake down. Even if it is a tiny tiny micro amount of discrepancy it still is an issue for a larger part of the whole.

 

In a Digitized Binary Sense this deal of 1/2 "seems" to be incompatible with how Calculus defines derivatives.

How so? What is the antiderivative of x?

 

 

So how did I go from KE, Pi Ratio, Time And Computer Science????????

Again their all connected to energy and time.

 

 

They all seem to be un-avoidable when used separately;)

I know that sounded dumb but think about it.

Fundamental constants show up everywhere, because they are fundamental constants. This misconception sounds like it's related to the fallacy of ignoring a common cause. KE, time, and computer science are connected because they are all described by physics, which uses math, which is where pi enters the picture.

0

Share this post


Link to post
Share on other sites
While this is true in some cases I don;t think it applies here; the first answer given is completely correct and answers the question without any element of "look how clever I am". It's straightforward math, if you know calculus.

 

ajb's answer is a bit of an end run around that, because one might as easily ask where the factor of 2 comes from in the equation he uses, and the answer is still going to be that it comes from calculus. Ultimately that's where all the answers lead, because that's where the factor of 2 comes from: the math.

 

That's true, but is it relevant to the nature of the question? I mean... Vay could've simply been asking for a mathematical derivation. But what if, and this is a common case with such questions, Vay was looking for something less blunt, on the more conceptual side i.e. an intuitive reasoning behind the math.

 

For example, a student asks "Why is d/dx sin x = cos x?" Now, one professor could set up the difference quotient and perform the dry symbolic work. The other professor instead elaborates on his colleague's work using the geometric interpreation. And he shows the student how limiting the angle measure relates the angle's functions by infinitesimal rate of change. That's intuitive background.

 

Yes, the 1/2 factor comes down to the calculus, but why? This is what I believe Wilmot McCutchon tried to explain in post #6.

0

Share this post


Link to post
Share on other sites

How so? What is the antiderivative of x?

 

 

Fundamental constants show up everywhere, because they are fundamental constants. This misconception sounds like it's related to the fallacy of ignoring a common cause. KE, time, and computer science are connected because they are all described by physics, which uses math, which is where pi enters the picture.

Can there be a way " if possible" that pi ratio in it's absolute value can be found????????

What is the process????

 

 

Everything eventually needs an upgrade when dealing with science, math and all that applies to the natural world described with the numerics of human psychology. Reasoning behind this is due to calculations being set in a backwards sense, not aimed for the future but thats one way of looking at this. Seems like a for loop if you ask me.

 

It is because of this flaw and premise that we still in our day and age use " numbers" linked to the laws of nature that determine " some " relation and or interaction." Though highly respected to me scientist are, I feel they should be more open minded to approaches that others may encounter and be able to solve derivatives without the need for " standard calculations."

 

 

Anti-Derivative of x uses 1 again as a numerator, here is a good link.

 

http://cims.nyu.edu/~kiryl/Calculus/Section_4.7--Antiderivatives/Antiderivatives.pdf

 

 

The whole issue is based on what is this one? if you ask me it is still the constant of 2 as in .5*2=1

BUT! In binaries this constant of 2 is not always the case. In computer science numbers as:

0 1 2 3 4 5 6 7 8 9 all eventually move over as 0=1 2=3 3=4 and so fourth.

They seem to be dynamic and static again there are 2 things going on with these numbers..

 

I want to emphasize that binary systems do not understand ratios, this is why the " floating point" was created I assume.

 

How can that be related to physical science???????????

 

A quanternion method should be most suitable for this Anti-Derivative of x , in one where field axioms and other algebraic rules coupled with calculus not need apply. But that would only be an extrapolation of this sort.

Edited by Iwonderaboutthings
0

Share this post


Link to post
Share on other sites

 

That's true, but is it relevant to the nature of the question? I mean... Vay could've simply been asking for a mathematical derivation. But what if, and this is a common case with such questions, Vay was looking for something less blunt, on the more conceptual side i.e. an intuitive reasoning behind the math.

 

For example, a student asks "Why is d/dx sin x = cos x?" Now, one professor could set up the difference quotient and perform the dry symbolic work. The other professor instead elaborates on his colleague's work using the geometric interpreation. And he shows the student how limiting the angle measure relates the angle's functions by infinitesimal rate of change. That's intuitive background.

 

Yes, the 1/2 factor comes down to the calculus, but why? This is what I believe Wilmot McCutchon tried to explain in post #6.

 

I don't disagree, but if this is the case, I think it's a matter of having asked the wrong question, or wording it incorrectly. Which changes this from "none of you answered the question" to "none of you are telepathic"

0

Share this post


Link to post
Share on other sites

is this true, the 1/2 is because the change in velocity measured in m/s will not be the actual distance covered , such as 20m/s was the resultant change in velocity over say 4 seconds. so to times 20 by 4 would yield a result of 80 meters but it would actually only be 40 meters traveled during the time the force was applied to stop it. it is only 40 meters because the average speed while changing from 20m/s to 0m/s is 10m/s and 10 x 4 seconds is 40.

This video may help explain what i am looking for.



To the OP is this simple enough ? its how my brain works. no math, just simple explaining of why.
2

Share this post


Link to post
Share on other sites

Hi Sriman, I am just a bus driver. I like Physics a bit, get headaches trying to translate formulas so give up most times. I probably should not be here, but the question intrigued me while considering the Oberth effect again recently, KSP lol. I also like RC aircraft and got into the physics of lift, Good old L= Cl 1/2 roh V^2 S , so the question of why we use half had been bugging me for some time (About 10years or more) but I never really found a suitable answer and gave up looking and just accepted it, I recall from my schooling back in the late 80's that it was from what I described. so given this information about me and how relatively dumb I am, can I truthfully ask.

 

Is your comment suggesting my post is useless and not related to Kinetic energy and that it has nothing to do with the Formula KE=1/2*a*v^2 and ultimately not the reason it contains the "1/2" in the equation.

Edited by Bryce
1

Share this post


Link to post
Share on other sites

Hi Bryce, I personally highly appreciate your participation in the discussion. But I think your formula is different from the OP. The others may clarify it. And a warm WELCOME To SFN.

0

Share this post


Link to post
Share on other sites

Thank's Sriman, looks like I am stuck with my Mathematical blindness. the answers are there but I just don't have the time or intellectual fortitude to understand it :(

0

Share this post


Link to post
Share on other sites

Bryce never think that you are intellectually poor. Keep learning and strive forward..... :)

2

Share this post


Link to post
Share on other sites

Bryce, your reply shows the equation S=ut+1/2at^2 .

 

It is the same 'half'. And it comes from being the arithmetic mean of zero and one(!).

 

Imagine a pneumatic 'gun' that applies a precise 1 Newton force to a 1 kg mass projectile through its barrel for precisely 1 second.

 

From Newton's second law, and the definition of the Newton, we know that the acceleration will be 1 m/s2 and therefore the projectile will leave the barrel at 1 m/s.

 

But the total distance travelled during linear acceleration is the time multiplied by the arithmetic mean of initial and final velocities. As the projectile starts at rest with respect to the equipment(!), and leaves at 1 m/s, the average velocity (again with respect to the equipment) is 0.5 m/s and therefore the scientist, being an excellent Newtonian, has designed his barrel to be 0.5 metres in length. The same 'half'.

 

As for the kinetic energy, the scientist tells me with confidence that from the definition of the Joule, a constant force of 1 Newton acting over a distance of half a metre imparts an energy of 0.5 Joules (the same 'half').

 

But I disagree with him.

 

I've been carefully watching his experiment with the aid of my trusty laser ruler and stopwatch, while walking towards the gun at the speed of 1 m/s. We agree on the mass of the projectile; we agree on the force applied; we agree on the time period the force was applied; we even agree on the rate of acceleration; but we disagree on initial and final projectile velocities (which I have measured to be 1 and 2 metres per second respectively); and also the distance over which the force acts, which I carefully measured as 1.5 metres.

 

So I believe that the projectile has gained not 0.5 but 1.5 Joules of kinetic energy (NB the arithmetic mean of 1 & 2!)

 

He of course quickly points out to me that if we can both agree that a projectile travelling at one metre per second has a kinetic energy of 0.5 Joules (his final value, my initial one), Then my final kinetic energy would be 0.5 + 1.5 = 2 Joules.

 

So a doubling of velocity quadruples the kinetic energy per kilogramme (providing the mv2), and the constant of proportionality is the same 'half' it has always been.

1

Share this post


Link to post
Share on other sites

Awww man, I nearly understood that until you mentioned Proportionality, it was entertaining for the lay like me.

 

I decided to look up that word (Proportionality), It turns out I understand it as coefficient, I see that the use of the word proportional can be tricky, Bugger.... it took me a while to learn to use the word Velocity speed correctly and acceleration set me back completely, Satellites accelerating around the earth all the time O.O

 

I guess now, I do have the ability to learn, But it seems I only learn at a rate proportional to the effort i put in, i only wish i had some sort of exponent to help that along.

 

OK> i need to develop a mathematical vocabulary, my words for this month (yes a month lol) are

proportional, Coefficient

exponent

 

 

 

PS

Lift = Cl 1/2r V^2 S

I think I understand why the lift formulas Proportionate "Cl" works, basically it has to do with the shape and the way it causes an acceleration of air particles (mostly in the change of direction) though this is vague and nowhere near complete, I can say that the Cl value is proportional to the amount of curvature on mostly the upper surface, this Cl Proportionate will also change when you change the AoA, because, you have effectively changed the curvature relative to the air flow. So there is an acceleration hidden here in this value, and the lift is the result of some sort of Kinetic interaction. so the whole formula has the 1/2 value incorporated in it, and from what Seth was saying, the half value is a constant of proportion. so would that make Cl the second constant of proportion or ??

 

So I have typed all this out in an effort to understand, I want to know, now it's time to post it, but, I am a bit scared, should I just delete it or post it. if I don't post it I may not learn, but it's not your duty to help or to teach me at a dulled down level at which I stand, but there may be some willing to help and point out errors, so why not post it.

 

OK, Balls out, here I go.

 

Do you all use proportionate and proportional interchangeably or ??

Did I make many mistakes there with the use of the word "Proportional". any other errors/misconceptions

 

I'm off now looking for the Fairy land Physics thought experiment room. Always liked them ;)

Edited by Bryce
0

Share this post


Link to post
Share on other sites

Awww man, I nearly understood that until you mentioned Proportionality, it was entertaining for the lay like me.

 

I decided to look up that word (Proportionality), It turns out I understand it as coefficient, I see that the use of the word proportional can be tricky, Bugger.... it took me a while to learn to use the word Velocity speed correctly and acceleration set me back completely, Satellites accelerating around the earth all the time O.O

 

I guess now, I do have the ability to learn, But it seems I only learn at a rate proportional to the effort i put in, i only wish i had some sort of exponent to help that along.

 

OK> i need to develop a mathematical vocabulary, my words for this month (yes a month lol) are

proportional, Coefficient

exponent

 

 

 

PS

Lift = Cl 1/2r V^2 S

I think I understand why the lift formulas Proportionate "Cl" works, basically it has to do with the shape and the way it causes an acceleration of air particles (mostly in the change of direction) though this is vague and nowhere near complete, I can say that the Cl value is proportional to the amount of curvature on mostly the upper surface, this Cl Proportionate will also change when you change the AoA, because, you have effectively changed the curvature relative to the air flow. So there is an acceleration hidden here in this value, and the lift is the result of some sort of Kinetic interaction. so the whole formula has the 1/2 value incorporated in it, and from what Seth was saying, the half value is a constant of proportion. so would that make Cl the second constant of proportion or ??

 

So I have typed all this out in an effort to understand, I want to know, now it's time to post it, but, I am a bit scared, should I just delete it or post it. if I don't post it I may not learn, but it's not your duty to help or to teach me at a dulled down level at which I stand, but there may be some willing to help and point out errors, so why not post it.

 

OK, Balls out, here I go.

 

Do you all use proportionate and proportional interchangeably or ??

Did I make many mistakes there with the use of the word "Proportional". any other errors/misconceptions

 

I'm off now looking for the Fairy land Physics thought experiment room. Always liked them ;)

 

Yes, the aerofoil lift equation is that form because of the relationship q=0.5 r u^2 where u is air speed, r, air density, and q the dynamic pressure (the additional theoretical pressure you could get if you brought the air to a standstill). The right hand expression is the kinetic energy of the airstream. So the same 'half'.

 

Don't worry about too much about the coefficient/proportionality thing. A mathematician might possibly be offended by the way I used the term, but I'm certainly not going to lose sleep over it. Except the capital 'C' that comes with your lift coefficient does stand for 'coefficient', so perhaps best to stick with that one,

 

PS No need to be shy about your level of understanding. It's pretty impressive. I wish a few of my clients were as quick on the uptake!

Edited by sethoflagos
1

Share this post


Link to post
Share on other sites

I think that teaching proportion or proportionality (nouns which name the phenomenon) and proportionate (old fashioned) or proportional (modern) - the adjectives used to describe some noun has sadly fallen out of fashion.

 

I find that for most folks using proportion is actually easier and more natural to understand and use than more explicit mathematics.

Certainly in Newton's day pretty well all work was stated in terms of proportion so if it was good enough for that genius it is more than good enough for me.


Most of the discussion about the 1/2 has centered on propelling a body to a specific speed.

 

A much easier derivation of the KE is found by considering instead the constant force needed to bring a body to rest from some velocity v.

 

Consider a body of mass m moving with steady velocity, that is brought to rest by the application of a constant force P.

 

P produces a constant retardation, f so that P = mf, in a distance x

 

Applying the simple equation (final velocity)2 = (initial velocity)2 - 2 (acceleration) (distance) and noting that the final velocity is zero at rest.

 

0 = v2 - 2fx

 

Thus fx = 1/2 v2

 

But the work done by the particle against force P is

 

Px = mfx

 

Therefore mfx = 1/2 mv2

 

This quantity is defined as the kinetic energy

Edited by studiot
1

Share this post


Link to post
Share on other sites

If I'd thought for a little longer, I might have come up with this derivation:

 

Calculate the change in kinetic energy of a body of mass m accelerating from u to v in time t

 

Force = rate of change of momentum = m*(v-u)/t

Distance = time x average velocity = t*(v+u)/2

 

Delta KE = force*distance = 0,5*m*(v+u)*(v-u)

0

Share this post


Link to post
Share on other sites

Awww man, I nearly understood that until you mentioned Proportionality, it was entertaining for the lay like me.

 

I decided to look up that word (Proportionality), It turns out I understand it as coefficient, I see that the use of the word proportional can be tricky, Bugger.... it took me a while to learn to use the word Velocity speed correctly and acceleration set me back completely, Satellites accelerating around the earth all the time O.O

 

I guess now, I do have the ability to learn, But it seems I only learn at a rate proportional to the effort i put in, i only wish i had some sort of exponent to help that along.

 

OK> i need to develop a mathematical vocabulary, my words for this month (yes a month lol) are

proportional, Coefficient

exponent

 

 

 

PS

Lift = Cl 1/2r V^2 S

I think I understand why the lift formulas Proportionate "Cl" works, basically it has to do with the shape and the way it causes an acceleration of air particles (mostly in the change of direction) though this is vague and nowhere near complete, I can say that the Cl value is proportional to the amount of curvature on mostly the upper surface, this Cl Proportionate will also change when you change the AoA, because, you have effectively changed the curvature relative to the air flow. So there is an acceleration hidden here in this value, and the lift is the result of some sort of Kinetic interaction. so the whole formula has the 1/2 value incorporated in it, and from what Seth was saying, the half value is a constant of proportion. so would that make Cl the second constant of proportion or ??

 

So I have typed all this out in an effort to understand, I want to know, now it's time to post it, but, I am a bit scared, should I just delete it or post it. if I don't post it I may not learn, but it's not your duty to help or to teach me at a dulled down level at which I stand, but there may be some willing to help and point out errors, so why not post it.

 

OK, Balls out, here I go.

 

Do you all use proportionate and proportional interchangeably or ??

Did I make many mistakes there with the use of the word "Proportional". any other errors/misconceptions

 

I'm off now looking for the Fairy land Physics thought experiment room. Always liked them ;)

The coefficient of lift is just a comparison and is based on the pressure effect on the area of the wing compared to what the the dynamic pressure would be if over the same area (with incompressible flow assumed). It can be greater than 1, or even well over 2, because the wing has two sides and the pressure drop on the top side can be even greater with the wing deflecting more of the flow field than just the cross sectional area that would be equivalent to the area of the wing.

 

I hope that makes sense...if it was just a jet of water then lift could be a maximum of coefficient of 1 based on the cross sectional area of the jet with the jet deflected 90 degrees (for drag it could be 2 deflected 180)...but can be greater in a field of flow and taking the area as the area of the wing is somewhat arbitrary (though understandable)

 

...and the half is the same half...though often plugged in as density/2 it is actually the same half from the kinetic energy and related to the velocity and distance as discussed in your video.

0

Share this post


Link to post
Share on other sites

The coefficient of lift is just a comparison and is based on the pressure effect on the area of the wing compared to what the the dynamic pressure would be if over the same area (with incompressible flow assumed). It can be greater than 1, or even well over 2, because the wing has two sides and the pressure drop on the top side can be even greater with the wing deflecting more of the flow field than just the cross sectional area that would be equivalent to the area of the wing.

 

I hope that makes sense...if it was just a jet of water then lift could be a maximum of coefficient of 1 based on the cross sectional area of the jet with the jet deflected 90 degrees (for drag it could be 2 deflected 180)...but can be greater in a field of flow and taking the area as the area of the wing is somewhat arbitrary (though understandable)

 

...and the half is the same half...though often plugged in as density/2 it is actually the same half from the kinetic energy and related to the velocity and distance as discussed in your video.

 

Maybe this is just a characteristic of my own particular field, and I'd be interested in others viewpoints.

 

Authors of technical papers who use SI, or similar consistent system of units (typically Europeans) seem to have a tendency to define coefficients as a fractional energy (sometimes force) conversion ratio - so a dimensionless coefficient of say, 0.70, would correspond to some kind of efficiency of 70%.

 

Authors more familiar with historic systems of units (typically US) seem more comfortable with dimensioned coefficients with no significance other than they balance the proportionality.

 

I appreciate that, at best, this is a gross generalisation, but the thought has struck me often enough to ask the question...

0

Share this post


Link to post
Share on other sites

I don't see an answer to Vay's question. He is an elementary physics student who needs the big picture. Vay asks a reasonable question: where does the 1/2 come from? He probably won't understand an explanation expressed with unfamiliar symbols and variables. Relativity theory is probably not familiar either.

 

His question got me wondering, since there seems to be no ready answer. When we encounter classical mechanics as part of required coursework, time pressure leaves no leisure for such ruminations. Maybe others on this forum, who have more deeply considered this question, can provide a better answer than my ignorant speculation, but in good faith, here goes my conjecture:

 

Momentum is always conserved, so you can add it moment to moment as the swarm of matter proceeds along a path. Summing an infinite quantity of infinitesimal momenta over a period of time and over a distance (in a definite direction, since we are considering kinetic energy rather than a diffuse form like heat) we integrate, to arrive at the expression for kinetic energy: 1/2* kg*m^2/s^2. Integration of momentum (kg*m/s) is what gives the 1/2 by the rules of calculus.

 

So now I'm wondering, where did the 1/2 go in E = MC^2?

I have a feeling you may laugh at me, and i beleive it is justified, but..... true to my previous posts form, i will suggest and accept the ensuing lecture/schooling.

 

OK balls out again, I hope they stay attached with this one.

 

Is the reason the mass-energy formula is twice as strong as the kinetic formula due to a unique thing about that formula. it is the energy that was contained within that mass should it be broken/released into sub-atomic particles and various other forms of radiations, Thermonuclear stuff (Stuff I know nothing about). the formula does not need to be devisable to obtain some sort of constant velocity because as you know , the speed of light is constant. but here is a question, aren't the subatomic particles released at the speed of light ?

 

I am thinking there is no acceleration upon release, there was only acceleration of the particles as they were, in the atomic structure prior to release, IE they did not start from a stop position, they were and are energy (Energetic subatomic particles whos mass is derived from velocity contained within a volume by the strong force.

 

I am currently imagining it like two balls on a string (String representing the strong force), They rotate around a central location held in proximity to each other through their bond (The string) the two have no other motion other that that rotational velocity, so as a pair they have no translational motion. the two balls are accelerating around each other until the string is broken, and at that point, the acceleration stops and the velocity becomes consistent as speed in a uniform direction.

OK, laugh time over ;)

 

oh wow, even I can see how vague all that is, no doubt most will just scoff at the lack of knowledge I have regarding atomic and subatomic particles and their working relationship within their bond, (I have so many questions, I think I need to watch a few good animation on this).

Edited by Bryce
0

Share this post


Link to post
Share on other sites

I have a feeling you may laugh at me, and i beleive it is justified, but..... true to my previous posts form, i will suggest and accept the ensuing lecture/schooling.

 

OK balls out again, I hope they stay attached with this one.

 

Is the reason the mass-energy formula is twice as strong as the kinetic formula due to a unique thing about that formula.......

 

See https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence

 

If you drop down to the section headed 'Low Speed Expansion' you'll discover where Einstein thought 'the half' came from.

0

Share this post


Link to post
Share on other sites

 

Is the reason the mass-energy formula is twice as strong as the kinetic formula due to a unique thing about that formula. it is the energy that was contained within that mass should it be broken/released into sub-atomic particles and various other forms of radiations, Thermonuclear stuff (Stuff I know nothing about). the formula does not need to be devisable to obtain some sort of constant velocity because as you know , the speed of light is constant. but here is a question, aren't the subatomic particles released at the speed of light ?

 

 

 

No, they aren't. Other than photons, released particles can't reach the speed of light. Nothing with mass can.

0

Share this post


Link to post
Share on other sites

thanks for the guidance Seth.

 

Thanks for the reminder Swansont :D

 

Think i will go learn calculus. back in a while ;)

0

Share this post


Link to post
Share on other sites

Hi Bryce

 

I see you got a good imagination. And you are also interested in physics.

You can understand physics a lot better if you learn calculus.

Good Luck !

0

Share this post


Link to post
Share on other sites

Kinetic energy is is the energy velocity equivalence of a object. The second law of motion states the net forces are directly proportional the the acceleration vector of a object. This acceleration vector could be stated as the rate of change in velocity vector (for a object) Therefor the acceleration vector is a study of a quantity of velocity. Given it has already been stated by me that kinetic energy is the energy velocity equivalence of a object, the equation where the acceleration vector is directly proportional to the net forces (influencing a object) is equivalent to the kinetic energy equation which therefor should be written as the net forces influencing a object are directly proportional to the objects rate of change in kinetic energy vector, again given that kinetic energy and velocity are the same.

0

Share this post


Link to post
Share on other sites

No, kinetic energy and velocity are not the same.

0

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now