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Help in explaining formula of Kinetic Energy


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#41 J.C.MacSwell

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Posted 17 October 2016 - 05:51 AM

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.


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#42 sethoflagos

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Posted 18 October 2016 - 12:59 AM

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...  


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#43 Bryce

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Posted 19 October 2016 - 10:14 AM

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, 19 October 2016 - 10:16 AM.

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#44 sethoflagos

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Posted 19 October 2016 - 02:45 PM

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...rgy_equivalence

 

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


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#45 swansont

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Posted 19 October 2016 - 03:38 PM

 

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.


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#46 Bryce

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Posted 19 October 2016 - 10:39 PM

thanks for the guidance Seth.

 

Thanks for the reminder Swansont :D

 

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


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#47 Sriman Dutta

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Posted 21 October 2016 - 07:20 AM

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 !


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#48 studiot

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Posted 21 October 2016 - 08:38 AM

 

 

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

 

 

Don't be afraid to ask for help, but better to start your own thread for that.

 

Post#13 here might help to start with.

 

http://www.sciencefo...stand-calculus/


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#49 Sam Batchelar

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Posted 11 November 2016 - 05:53 PM

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.


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#50 swansont

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Posted 11 November 2016 - 06:02 PM

No, kinetic energy and velocity are not the same.
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#51 Sam Batchelar

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Posted 11 November 2016 - 06:25 PM

Okay, you are quite right, Kinetic energy is the work needed to accelerate a object to a given velocity.

 

A object falling towards earth in a vacuum chamber will reach a velocity, this velocity could be stated as being the highest potential velocity (the field can accelerate object of varying masses to) this highest potential velocity as a quantity is the same for objects of varying masses while the field strength is similarly the same. It is obvious that the field in the cause does the work to accelerate the object to the highest potential velocity quantity. The highest potential velocity quantity also is directly proportional to the field strength, therefor the work that the field does on the object so to speak is directly proportional to the field strength. Therefor the work needed to accelerate a object to any finite quantity of velocity (kinetic energy) is directly proportional to the velocity itself. This is because the field strength is directly proportional to the highest potential velocity, the field strength doing the work. Because the object reaches the velocity the work need to reach the velocity in question is proportional to the velocity itself. 


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#52 studiot

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Posted 11 November 2016 - 07:42 PM

 

sam batchelar

Okay, you are quite right, Kinetic energy is the work needed to accelerate a object to a given velocity.

 

Not always.

 

Consider a ball swinging on the end of a string travelling at 1m/s in a due northerly direction.

 

How much work is done accelerating this ball to be travelling in a due easterly direction at 1m/s?


Edited by studiot, 11 November 2016 - 07:42 PM.

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#53 Sam Batchelar

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Posted 11 November 2016 - 08:16 PM

As far as I can tell work is done to lift the ball against gravity although I am unsure if you are thinking of this part. As for the work gravity does on the ball the resultant velocity perpendicular and parallel to the earths surface of the ball would be equal to the velocity of the ball if there was no swing and the ball was moving in the downward direction.


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#54 studiot

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Posted 11 November 2016 - 08:57 PM

As far as I can tell work is done to lift the ball against gravity although I am unsure if you are thinking of this part. As for the work gravity does on the ball the resultant velocity perpendicular and parallel to the earths surface of the ball would be equal to the velocity of the ball if there was no swing and the ball was moving in the downward direction.

 

In Science we need to aim to be precise and accurate.

 

Not everybody does so.

 

I said nothing about the ball rising or falling or gravity in general.

I did say the ball was on the end of a string.

This should be a clue to why I noted your statement as only partly correct.


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#55 Sam Batchelar

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Posted 11 November 2016 - 09:09 PM

I cannot unravel this mystery of clues although it may be beneficial to me if you could.


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#56 studiot

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Posted 11 November 2016 - 09:54 PM

I cannot unravel this mystery of clues although it may be beneficial to me if you could.

 

The ball is in horizontal circular motion.

 

As such it is continually subject to a force, which is continually changing its velocity (ie accelerating it)

 

Yet no work is done by that (or by any other) force.


Edited by studiot, 11 November 2016 - 09:54 PM.

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#57 Sam Batchelar

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Posted 11 November 2016 - 10:06 PM

Correct me if I am wrong but it seems the force you are referring to is centrifugal force. This force operates perpendicular to the object (on the swings) trajectory and therefor has no effect on the objects velocity.

 

In case you disagree, Pulling the ball on the swing in such a way as to apply only tension to the sting does not cause it to be set into motion providing the swing can take the tension.


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#58 studiot

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Posted 11 November 2016 - 10:17 PM

Correct me if I am wrong but it seems the force you are referring to is centrifugal force. This force operates perpendicular to the object (on the swings) trajectory and therefor has no effect on the objects velocity.

 

In case you disagree, Pulling the ball on the swing in such a way as to apply only tension to the sting does not cause it to be set into motion providing the swing can take the tension.

 

Yes, you are indeed wrong.

 

Have you studied the mechanics of circular motion?


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#59 Sam Batchelar

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Posted 11 November 2016 - 10:24 PM

No I have not, They were just my thoughts from what was given to me, Thankyou for referring me to some science, I shall study the subject of circular motion.


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#60 studiot

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Posted 11 November 2016 - 10:37 PM

No I have not, They were just my thoughts from what was given to me, Thankyou for referring me to some science, I shall study the subject of circular motion.

 

OK then I am happy to explain in more detail.

 

First you need to know the difference between speed and velocity,

 

Speed is just the numerical magnitude of the velocity, without regard to where you are going.

 

So a speed is 10 MPH or 33 feet per second or 1 mete per second as in my example.

 

Side Note you should (nearly) always provide the units in Physics. That gives meaning to the statement. 

 

Velocity always also includes where you are going ie the direction as well as the magnitude or speed.

 

eg 1 metre per second north as in my example.

 

Change of either the magnitude or the direction constitutes an acceleration.

 

A body executing uniform circular motion is constantly changing its direction of motion, but not its speed.

 

But it is accelerating.


Edited by studiot, 11 November 2016 - 10:38 PM.

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