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Help in explaining formula of Kinetic Energy


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#61 Bryce

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Posted 24 November 2016 - 06:44 AM

Hi Studiot.

if the velocity is seen as its path velocity, then the speed did not change. I understand that work is the change in Kinetic Energy, and that Ke is defined only by the speed variant of Velocity, not the direction. but doesn't changing direction require work to be done, or in our case, where you spin the ball on a string, the velocity is changing as direction only, but if you change the reference frame to that of an outside observer who is seeing you swing that ball around in your glass windowed truck, the speed is changing relative to them.

OK i get it. the kinetic energy is not changing relative to the pole it is fixed to, but relative to any other independent observer the KE is changing. 

I recall some time ago that someone pointed out that after one revolution there is no change in position so no work was done. and it makes sense that a if work was done how come neither of the variables have change in the formula ke=1/2MV^2 it has the same mass the same speed and direction. so there could not have been any work done. (I am at home now and i was when i last posted, so i guess i have not done any work in the meantime ;) )

So here was my assumption, I chose this assumption to help me sleep at night, there was work done, and it can be seen especially after each half rotation. but each conceutive half rotation will be work done in the opposit way so it will come back to zero. 

And now I have thought about it some more, I am back to your camp, with the ball on a string or even any two points of a rotating body, if there was work done and the energy was always present but nulling it self after each half rotation, where did it come from? (My previous example was given to me years ago as a satellite in orbit, i mistook gravity as energy or a force, but now i see it as something else not too much unlike the bonding of all molecules in our spinning body. for a lack of a better work i would call it a force)

 

five min later I come back to make another post, it gets added to the last one I noticed..

Second post is as follows;

 

Oh mann  :( I think I am having another cerebral flatulence moment.

I think I can say yes there was work done, and point to where the energy came from. in the case of the satellite and the earth, the energy was traded between each other as speed and height change relative to their CoM. a form of ke pe change. same with the stick. the stick bent and stored potential energy or if it's in your hand you were providing energy or if you were in a car the cars speed changed..... but, if it were two balls of equal mass rotating around each other then they would have no speed change relative to CoM and no height change hmmm. oops fell into a hole just then. BrB

 

Back again, changed CoG to CoM. 

 

OK now i am just now realizing they (The earth and satelite) don't change distance/height position relative to their CoM, they rotate uniformly around it. hmmmmmm.. Back in a day or more.


Edited by Bryce, 24 November 2016 - 07:08 AM.

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#62 studiot

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Posted 24 November 2016 - 11:45 PM

 

Bryce

Oh mann   :( I think I am having another cerebral flatulence moment.

 

Since you put it so delectably How can I not respond?

 

Here is a quick rough guide to Forces (and work and other stuff)

 

To start with let us not worry about the motion (or lack of it) of a body.

Just take any old body and apply any old force to it.

So we have fig1.

 

Now there is a pretty intuitive principle I hope you will take on trust without serious proof.

 

You can replace a force with a whole bunch of forces but get the same overall effect.

You can do this in lots of different ways with lots of different bunches of forces.

(How do you get a bunch of forces?

Get a bunch of grapes and eat all the grapes off the bunch.

You are left with a bunch of forces)

I have shown this in fig2 and fig4

 

forces10.jpg

 

In fig3 I have noted that a single force F may be replaced by two forces that act at right angles to each other and called them F1 and F2.

These are often called components of Force F

 

Forces at right angles to each other are special and we will meet them again later.

 

fig4 also shows what we call equivalent force systems, they all have the same effect.

 

OK so now we consider that our any old body had a (constant velocity) V as shown is fig5.

 

We also know know that if we subject this body to some force system as in Fig 6, 

this force system can be replaced by a single force or alternatively a pair of forces at right angles, as shown in fig7.

 

The last option - apair of forces at right angles - is the most interesting because F itself is at some random direction compared to the direction of motion.

 

But if we align on of the forces, say F2, with the direction of motion the the other is at right angles to this direction.

 

Now comes the clever stuff.

 

F2 affects only the magnitude of the velocity (ie the speed)

 

F1 affects only the direction of the velocity.

 

So to change the magnitude we would apply only F

 

and to change the direction only F1.

 

Now to motion in a circle as in fig8

 

A  body whirling round on the end of a string is in circular motion

and if you do it inside your cab it is a bit like a goldfish in a bowl when viewed from outside the cab.  :)

 

The direction of the velocity is constantly changing, but the speed is constant

 

So let us apply our pair of forces at right angles and such that one of them is directly opposed to the direction line.

 

Since there is no change in the magnitude of the velocity this force must be zero.

Newton's Law tells us that you need a force to change the velocity.

So F2 is zero.

 

But what about F1?

 

Well the direction changes so F1cannot be zero.

 

So for a body in circular motion, all the force that is applied must be at right angles to its direction of motion.

 

(This force is the tension in the string and is called the centripetal force.)

 

So didn't I also mention work?

 

Well work = force times distance moved by/against that force.

 

So for F2=0

 W2 = 0 x Distance  = 0

 

F2 does no work.

 

But F1 is not equal to zero.

However there is no change in distance from the centre since the radius of a circle is a constant.

 

So W1 = F1 x 0 = 0

 

and also F1 does no work.

 

So there is no work done by the centripetal force (or any other force) when whirling a body in a circle.

 

forces12.jpg

 

 

 


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#63 Bryce

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Posted 25 November 2016 - 06:45 AM

Is this all unique some sort of Isolated system observation?
our goldfish as we observe from outside is changing speed and direction relative to us. but it is almost irrelevant because, well, we are not influencing it in any way, so, we are not part of the equation?

 

 

 

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?

If it's not because of an isolated system observation then why ignore the fact that speed in both the north and east direction was changed ?

PS Thanks and Have a good weekend, get back to this later ;)


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#64 studiot

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Posted 25 November 2016 - 08:50 AM

Is this all unique some sort of Isolated system observation?
our goldfish as we observe from outside is changing speed and direction relative to us. but it is almost irrelevant because, well, we are not influencing it in any way, so, we are not part of the equation?

 

 

If it's not because of an isolated system observation then why ignore the fact that speed in both the north and east direction was changed ?

PS Thanks and Have a good weekend, get back to this later ;)

 

No my treatment was general, although the examples obviously had their own special characteristics.

 

My post#62 was designed to help with you questions (especially your post#61 where you realised your thinking was going round in circles) if you can be bothered to work through it.

 

 

  

If it's not because of an isolated system observation then why ignore the fact that speed in both the north and east direction was changed ?

In the question referenced above the speed in both the north and easterly direction was stated to be 1m/s.

In what way did you think that was changed?

 

Surely it was the direction that changed?

 

Remember that Velocity is an instantaneous property.

It can change from instant to instant but it has two sub properties viz speed and direction.

Changing either changes the velocity.

 

The important point in my post #62 is that you can replace any force with a pair of forces one of which only changes the speed and the other only changes the direction.


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#65 Bryce

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Posted 25 November 2016 - 09:59 AM

No my treatment was general, although the examples obviously had their own special characteristics.

 

When I said Isolate, I was talking about reference frame, an isolated system, not an isolated example with special conditions.

 

I was going to mention an example of an isolated reference frame of the earth sun and galaxy. they are all moving around each other in the ways they do, and it may appear from day to day that there is a displacement. but , when you look at them in their individual reference frame, IE earth around the sun, (every year we end up with no more Distance from the sun no more speed and basically in the same place. same with the sun around the galaxy, But I refrained from it because all the orbits are not perfect and the forces are changing such as the suns gravitational pull (and I don't actually know if it's getting less or more , it would seem logical that it is getting less) anyway, it was not a precise example, but it was an attempts to show an isolated system.


Edited by Bryce, 25 November 2016 - 10:00 AM.

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#66 studiot

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Posted 25 November 2016 - 11:18 AM

When I said Isolate, I was talking about reference frame, an isolated system, not an isolated example with special conditions.

 

I was going to mention an example of an isolated reference frame of the earth sun and galaxy. they are all moving around each other in the ways they do, and it may appear from day to day that there is a displacement. but , when you look at them in their individual reference frame, IE earth around the sun, (every year we end up with no more Distance from the sun no more speed and basically in the same place. same with the sun around the galaxy, But I refrained from it because all the orbits are not perfect and the forces are changing such as the suns gravitational pull (and I don't actually know if it's getting less or more , it would seem logical that it is getting less) anyway, it was not a precise example, but it was an attempts to show an isolated system.

 

 

OK so you can't be bothered to work through my post#62.

 

'Nuff said, Have a nice weekend.


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#67 Bryce

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Posted 25 November 2016 - 09:54 PM

Not true, I did read through. but Obviously something has not hit me yet. I appreciate the effort, and i can see you did go to a great deal of effort for such as me, I value that, and, I will go back over it later, I added my post 65 just to clarify my previous post to that, I have not had time to concentrate on what you have posted. I have my kids for the weekend, and my little 4yo girl is the master of knowing when to get up and ask me questions, her timing is amazing. but without that as a distraction, I may never get it. I may have to just accept it, but, here's to hoping I do.

 

Have a good weekend and don't worry about me :D


PS this sort of diagram is not unfamiliar to me, and it is very similar to what you presented.

veldist2.gif

 

though, in that diagram, there is a component that is affecting the velocity magnatude same as your fig7 F2

As lift is opposing gravity, and they are both opposing each other with equal magnitude they are zeroed out, Most reputable RCGroup members will say that there is no work done to maintain levle flight other than drag. (AFAIK, they are right)

 

So i guess that if the plane flys around the world once, it would be similar to the ball on a string or a satelite in orbit. Why am i trying to break it down and examin the change in inital velocity as 1m/snth over the 90deg of rotation i dont know, I think i will stop that now ;)

 

OK, have to go for real this time. little one has asked me 7 times if we are going to the park, hard to concentrate. 

 


Edited by Bryce, 25 November 2016 - 08:50 PM.

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#68 Bryce

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Posted 27 November 2016 - 06:17 AM

OK, So i figure now that the ball on a string is just like a ball on a shelf.


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#69 Delburt Phend

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Posted 4 January 2017 - 09:27 PM

I think KE = ½ mv² comes from the distance formula: d = ½ at²

 

Distance is a form of energy (PE) if you apply a force over that distance. KE is an expression of the relationship between PE and v.

 

t = v/a from v = at :

 

therefore t² = v²/a² : Now we substitute v²/a² for t² in the distance formula and we get d = ½ v²/a

 

d = ½ v²/a ; now we need to substitute for a; F = ma so a = F/m

 

d = ½ v²/(F/m); We can multiply by the inverse of F/m and we get d = ½ v²m /F

 

Now we need to multiply both sides by F and we get dF = 1/2 mv²

 

We know that distance times force is a form of energy PE: but we also know that PE = KE we have it KE = ½ mv²

 

Someone why back when: knew that distance times force was a form of energy. And that a certain v causes a certain amount of PE or KE. Actually PE (Nm) maybe the most common form of energy; especially way back when.


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#70 studiot

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Posted 4 January 2017 - 11:04 PM

I think KE = ½ mv² comes from the distance formula: d = ½ at²

 

Distance is a form of energy (PE) if you apply a force over that distance. KE is an expression of the relationship between PE and v.

 

t = v/a from v = at :

 

therefore t² = v²/a² : Now we substitute v²/a² for t² in the distance formula and we get d = ½ v²/a

 

d = ½ v²/a ; now we need to substitute for a; F = ma so a = F/m

 

d = ½ v²/(F/m); We can multiply by the inverse of F/m and we get d = ½ v²m /F

 

Now we need to multiply both sides by F and we get dF = 1/2 mv²

 

We know that distance times force is a form of energy PE: but we also know that PE = KE we have it KE = ½ mv²

 

Someone why back when: knew that distance times force was a form of energy. And that a certain v causes a certain amount of PE or KE. Actually PE (Nm) maybe the most common form of energy; especially way back when.

 

 

 

Distance is a form of energy (PE)

 

Really?

 

Perhaps you should think again.

 

Whilst it is true that when you multiply two things together sometimes the result is the same as one of the two things so for instance

 

The amount of juice squeezed is the number of oranges times the juice per orange

So both the result and one of the multiplicands are juice

This does not apply to force times distance, neither of which are energy.

 

In fact your arithmetical manipulations mix up several different types of somethings (vectors, scalars and constants) in an original but inadmissible way.


Edited by studiot, 4 January 2017 - 11:45 PM.

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#71 Delburt Phend

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Posted 5 January 2017 - 01:06 AM

You should read the whole sentence. “Distance is a form of energy (PE) if you apply a force over that distance.​”  Nm; A (Newton * meter) is a joule of energy.

 

We call it algebra.


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#72 Delburt Phend

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Posted 5 January 2017 - 10:00 AM

Where; exactly, do I violate vector / scalar rules?


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#73 studiot

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Posted 5 January 2017 - 10:17 AM

 

 

We call it algebra.

 

If this nonsense if left unchallenged others trying to understand this subject may think it authoritative and get the wrong idea.

 

So let us look at your 'algebra'.

 

 

I think KE = ½ mv² comes from the distance formula: d = ½ at²

 

 

 

Formula?

What formula?

 

This clearly comes from the correct full formula d = u + 1/2 at2, with u = 0

 

If u can equal zero then so can a so what is distance for a body that is just moseying along at velocity v?

 

Does it not cover distance or enjoy a kinetic energy by virtue of its motion?

 

 

 

 

Distance is a form of energy (PE) if you apply a force over that distance. KE is an expression of the relationship between PE and v.

 

and You should read the whole sentence. “Distance is a form of energy (PE) if you apply a force over that distance.​”  Nm; A (Newton * meter) is a joule of energy.

 

I notice you said if you apply a force.

 

And if you don't?

 

It would be better if you said that the change in KE is the additional energy gained by the accelerating force if one is applied.

This was already discussed earlier in the thread.

 

By the way what distance? Is moment energy?

I ask because Newtons times metres also defines moment.

 

 

 

 

t = v/a from v = at :

 

therefore t² = v²/a² : Now we substitute v²/a² for t² in the distance formula and we get d = ½ v²/a

 

 

 

And if a = 0 

Oops division by zero.

 

Many false results can be 'proved' using division by zero.

 

 

 

 

We know that distance times force is a form of energy PE: but we also know that PE = KE we have it KE = ½ mv²

 

 

What theorem states that PE = KE?

 

Where; exactly, do I violate vector / scalar rules?

 

What variables do you think are vectors and what do you think are scalars?


Edited by studiot, 5 January 2017 - 10:18 AM.

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#74 Delburt Phend

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Posted 5 January 2017 - 10:38 PM

I can’t think of anything more consistent with the concept of vectors than the distance formula (d = 1/2 v²/a). The ‘a’ in the distance formula is typically gravitational acceleration; and this direction of acceleration is always down. So the ‘a’ in the distance formula has one and only one direction; down.

 

When G is used as the force this force is always down; and the d in the distance formula is always the length down. So the direction of travel: has one and only one direction; down.

 

Newton said that the direction of Force in F = ma is in the same direction as the motion caused; so by definition F = ma is under the concept of a vector.

 

So we have a substitution from a vector ‘formula’ into a vector ‘formula’. This doesn’t seem like any sort of violation.

 

What might be a violation of logic is that, after the substitution from a vector ‘formula’ into a vector ‘formula’, we get a scalar. Why is KE = ½ mv² scalar?


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#75 studiot

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Posted 5 January 2017 - 11:00 PM

I can’t think of anything more consistent with the concept of vectors than the distance formula (d = 1/2 v²/a). The ‘a’ in the distance formula is typically gravitational acceleration; and this direction of acceleration is always down. So the ‘a’ in the distance formula has one and only one direction; down.

 

When G is used as the force this force is always down; and the d in the distance formula is always the length down. So the direction of travel: has one and only one direction; down.

 

Newton said that the direction of Force in F = ma is in the same direction as the motion caused; so by definition F = ma is under the concept of a vector.

 

So we have a substitution from a vector ‘formula’ into a vector ‘formula’. This doesn’t seem like any sort of violation.

 

What might be a violation of logic is that, after the substitution from a vector ‘formula’ into a vector ‘formula’, we get a scalar. Why is KE = ½ mv² scalar?

 

I went through your questions and working in great detail.

 

Following a complete failure to address even one of my questions, why would you expect me to answer yours?

 

The above is just a load of waffle.


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#76 Bender

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Posted 6 January 2017 - 08:10 AM

I can’t think of anything more consistent with the concept of vectors than the distance formula (d = 1/2 v²/a). The ‘a’ in the distance formula is typically gravitational acceleration; and this direction of acceleration is always down. So the ‘a’ in the distance formula has one and only one direction; down.

Except that acceleration might not be constant and the velocity might have a different direction. Claiming that a is "typically" down is a gross simplification, and demonstrates that your deviation is not generally valid. In fact, most of your posts is only valid in very specific situations.

 

So, while I think your intentions are good, I have to agree with studiot.


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#77 studiot

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Posted 6 January 2017 - 08:50 AM

Just to show we all make mistakes this line in post#73

 

 

 

This clearly comes from the correct full formula d = u + 1/2 at2

 

Should read  d = ut + 1/2 at2

 

Your approach will get you into greater difficulty when you come to more advanced dynamics like motion in a circle and impact/collision.

 

There is no potential energy involved, for instance, when two billiard balls collide on the table, or a bullet hits a target.

gravity is not directly involved.

 

How does your analysis handle the interchange of KE here?


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#78 RiceAWay

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Posted 13 January 2017 - 09:35 PM

V = a*t
a = (V2 – V1) * t
KE = (1/2a) * V^2
 
Question – what mean the terms square for velocity and one half acceleration?
 
Thought experiment:
 
Draw a vector. The period (t)  on the vertical axis and the speed (V) on the horizontal.
 
Assume an acceleration for one second of 10 meters/sec. 
 
The vector speed caused by “acceleration” (a) going from zero at t = 0 to 10 at t = 1.
 
If you graph this 4 x 4 you can see that the area under the ‘a’ line is half of the area or 2 squares. This area represents the kinetic energy or KE.
 
If you draw another constant acceleration line at half of the acceleration you will see that the area under this line is ½ square or doubling the acceleration requires four times the energy to achieve the same speed as half the acceleration in the same period of time.
 

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#79 studiot

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Posted 13 January 2017 - 10:34 PM

 

RiceAWay post#78

 

1) V = a*t
 
2) a = (V2 – V1) * t
 
3) KE = (1/2a) * V^2

 

Perhaps you would like to clarify these?

 

I have added the numbering for reference clarity.


Edited by studiot, 13 January 2017 - 11:10 PM.

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#80 studiot

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Posted 14 January 2017 - 11:38 PM

@RiceAWay

 

Since you won't clarify, I will.

 

1) is incomplete.

 

2) is just plain wrong since V2 = V1 + at, a = (V2-V1) divided not multiplied by t

 

3) is also wrong since it implies that kinetic energy is independent of mass, amongst other issues.


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