Why is there a half in kinetic energy formula ? Work- energy principle.
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- Опубликовано: 25 июл 2024
- We look into the history of kinetic energy and how it became a classical concept in physics.
We invoke the Work- Energy principle and see how it becomes useful for our problem.
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After people complained about the difficulty of moving objects with a mass, the developers decided to patch a 1/2 into the equation of mechanical energy.
It's as easy as that.
I'm reeeeally looking forward on the Lagrangian and Hamiltonian videos :D
For some reason the audio in this video was substandard and almost blew my eardrums when a loud advert played immediately afterwards. My bad for cranking up my volume.
Same here, I had to convert the video and listen to it on vlc. Subject is interesting but sound is too low.
This channel defenitly needs more recognition! I love how you show the insight and purpose of the physics and matematical principles :)
30 more seconds of how you get from 3:38 to 3:48 would have been helpful :)
It's reverse chain rule.
If you differentiate m(r•)^2 then you get 2m(r•)(r••) via chain rule.
Since integration is the reverse process of differentiation, you can spot that integrating m(r•)(r••) gives you 0.5m(r•)^2.
If you see two derivatives multiplied together with orders separated by 1 step (so r*(r•}or {r•}*{r••}), you can often get something like this.
well its because force is mass times acceleration and as r taken as distance the second order deravative is acc which is r double dot
@@dascientist8443 well can u explain me it if i am wrong
Your steps there are right, but that's how you get from 3:30 to 3:38
What they have in the integral is basically
Work = Integral [ F * (r•) dt ]
= Integral [ m * (r••) * (r•) ]
Which then becomes 0.5*m*(r•)^2 or 1/2 mv^2 when you do the integral.
integration by substitution.
set u = r’
du = r'' dt
therefore equation becomes integration of m*u*du which is (1/2)*m*(u^2)
substitute back u = r' = velocity (v)
@Kelvin Higgs It was a general condition u stupid
Finally, a good and simple explanation using integration by substitution.
Pretty cool. Surprised you didn't explicitly mention the line integral. Thanks for the tidbit bout lagrange.
Good stuff i used to think about this a lot in 11th grade. Looking forward to other stuff you post
Enjoyable and informative video - thank you.
I have seen this derivation on a number of occasions, and appreciate the strange ½ coefficient as an artifact of the Power Rule with integration. Still hoping for a more intuitive (algebraic?) illustration of kinetic energy.
I'm no expert, but I like to think of KE as the result of force applied over distance and momentum as force applied over time.
Please, someone correct me where I'm wrong. That's probably more of a "where" than an "if."
I'm trying to learn.
Here's the algebra I came up with, starting with KE = force * distance.
Symbols used:
KE = kinetic energy
kg = mass
F = force
d = distance
a = acceleration
t = time
v = velocity
KE = F * d
KE = kg * a * d
KE = kg * m/t^2 * d
KE = kg * m/t^2 * 1/2at^2
KE = kg * m/t^2 * 1/2 * m/t^2 * t^2
KE = kg * 1/2 * m^2/t^4 * t^2
KE = kg * 1/2 * m^2/t^2
KE = 1/2 * kg * v^2
KE = 1/2mv^2 (changing the symbol for mass from kg to the more common m, which was used for meters above).
Here's a narrative:
KE is force over distance (F * d) which would be kg*a*d (mass times acceleration is force, displacement is d).
Expanding acceleration into m/t^2, KE is kg * m/t^2 * d (mass times meters per second squared times distance).
Distance is 1/2 at^2, so KE can be rewritten as kg * m/t^2 * 1/2at^2.
Expanding that new appearance of acceleration into meters per second squared, KE becomes kg * m/t^2 * 1/2 * m/t^2 * t^2.
Simplified one step, KE is kg * 1/2 * m^2/t^4 * t^2
simplifying another step, KE is kg * 1/2 * m^2/t^2.
Velocity is m/s, so m^2/s^2 is v^2 - KE is 1/2*kg*v^2, more commonly written as 1/2mv^2.
Please let me know where I missed the target!
@@johnnyragadoo2414 ok but where did you get KE=F×d from?
How is KE related to force if an object can have no force acting on it but still have KE?
@@l1mbo69 Great question. I’ll give it my best. The definition of work is force over distance. KE is the potential to do work, so work on an object gives it KE. As the object accomplishes work on something else, it trades KE for work done.
So, it takes work to change an object’s velocity, which changes the object’s KE. If the object is moving at a constant speed, then it is keeping a constant KE even though the net force on it is zero.
Here’s a weird twist to that. KE is not conserved, because it can be converted to another form of energy.
Energy is always a scalar, too. KE does not have a direction, even though velocity is part of it and velocity is a vector. A one Newton force over the span of one meter always results in the same change, plus or minus, in KE, regardless of the direction of the force. Velocity and momentum change will depend on the initial direction of the object and the direction of the force and how much time it takes for the one meter of distance.
KE and momentum are both relative, so you need to keep your reference consistent. That’s actually how I got the difference between KE and momentum straight. A flat earthier made the argument that an object thrown east had the same KE as if it were similarly thrown west, which he thought was proof the earth didn’t spin. I tracked down momentum and KE on moving platforms for the point of view from the moving platform and for a stationary observer. It all made sense, then. When I get to my desk, I’ll share a link, hoping that is not an unwelcome thing here.
In contrast, momentum is conserved, it doesn’t convert to anything else, and momentum is a vector. I think of momentum as how velocity is distribute after a collision.
Amazing how consistently that all works out, how nature works out infinite calculations per second, and gets them all precisely correct.
Please keep in mind I’m newly awakened to a lot of this. I appreciate fact checking and correction.
@@johnnyragadoo2414 thank you very much for your very detailed explanation
Also as a remark to your last comment, we should remember these are just models for our convenience, because the universe obviously doesn't "remember" to conserve momentum. Rather there are mechanisms that dictate how a system evolves moment to moment, that sometimes happen to follow a greater law (forcefully, due to fundamental symmetries in the very nature of our universe)
@@l1mbo69 My pleasure! Nice chatting with you, and best of luck in your intellectual exploration.
Ah,nice.
Now I have to learn calculus to understand why it is 1/2.
The 1/2 comes about from conservation of momentum. Think of a moving object as a compressed spring. The greater the object's velocity, the greater this compression. The amount of this compression is the total energy of the spring. But energy is force x distance. So when the object (the compressed spring) impacts another object, to perform work on that object to move it forward the spring decompresses. 1/2 of the spring moves forward to move the 2nd object forward (force x distance) . However the back of the spring needs to expand backwards to order to conserve momentum. So the 2nd half of the spring's energy moves backwards. This is exactly the way rocket propellant works. Half of the propellant's energy pushes the rocket up, while the 2nd half exits the back. But the two halves of the propellant's total energy must move in opposite directions in order to conserve momentum as the two energy halves are pushing off one another.
I'm pretty sure that's because it's the area of the graph momentum per change in velocity.
Integral(momentum(mv) , v )
=
mv²/2
Thanks for this informative video
Here's a short version: The kinetic and potential energy are defined for the sole purpose that the conservation law of energy becomes true for an isolated system. It's perfectly OK to multiply _all_ energies by any numerical factor (like 2) and still have the conservation law remain valid, but then the _transfer_ of energy would also have to be defined as 2 times what we normally call work. "Work" in physics is just a definition (it wouldn't even be a "thing" if it weren't for the law of energy conservation), so there's nothing wrong with an alternative definition, except that it would have an inconvenient extra factor of 2 in it.
Excelente!
Cool. Thanks.
So... the half comes from the integration over time? I never studied calculus, just what I've learned on youtube. Thanks for any helpful responses.
Yes ! You can visualize it by saying KE is the area of the right angled triangle mv and v. The area of the triangle is 1/2 mv² right ! 🔥
So it comes due to integrating over v 😂
@@Rupadarshi-Ray for some reason i feel hyped reading your comment
@@ahmadzakiy7204 haha i explained it as i imagined i would to a 9th Standard classroom.
Integrating equals to area! That's its basic property, and its very easy to visualise, and for triangles and rectangles easy to calculate before learning Calculus.
Happily, Work Energy theorem derivation from Newton's Second Law doesn't take any complicated integrals, only an area of a triangle.
@Kelvin Higgs I think you are sad. You need to talk about your fantastic college and is limited by the math, because you need math to comprehend the world and WOW if you don't there is no other way, just the way your learned it, and you know latin. Cool dude.
@@miguelrezende8479 I didn't read everything that guy said but En Theo's explanation was clearly incorrect
Cool video, I probably would've understood more if I knew calculus
Learn calculus it’s actually very simple at a basic level
@@W2wxftcxxtcrw I have to learn enough for engineering purposes by the end of the next two years anyway :(
@@Eterrath then you’ll see it’s not too hard. Integrals are a bit tricky but you’ll get it
Wow 😯🤯
So it would work only in conservative fields?
The physical explanation emanates from Newtons third law, action & reaction of force. If a force operates (creates acceleration) in one direction, the same amount of force always operates (creates acceleration) in the opposite direction. No exceptions in this universe. Mathematical explanations/derivations only contain the created kinetic energy in one of the directions, but the total created kinetic energy from F = m • a is m • v^2 (without the 1/2). Nothing strange about it! Much more interesting to reflect on is that release of absolute energy in the universe, for example in an explosion (release of binding energy in molecules), or release of atomic energy, results in relative kinetic energy. Kinetic energy is always only relative. So, where did the real absolute energy go (it can't have disappeared) ? Relative kinetic energy is no exchange/compensation of energy to the universe as a whole, from the absolute energy (E = m • c^2). The answer is, the absolute energy was used to slow down time, to slow down the rate of change in all the accelerated atoms involved in the process. Time dilation is physical, and happens during acceleration phases of matter through (or against) the quantum vacuum in space. Physical time dilation does not happen during constant relative motions (acceleration is an absolute motion). It takes energy to slow down physical time, how a brain thinks about time (psychological time) is something completely different.
Sound is too low
That's where the 1/2 is really problematic.
Frankly, this should have showed this earlier in school science, instead of just giving equations , because this give larger picture of the idea, and the importance (industrial revolution) rather than just rote memorizing equations and plugging abs chugging
But why is there a ½? We know *how* the ½ got there via integrating, but this video fails to explain *why* a ½ results, and what the ½ means physically.
The question is logically incoherent. In science, "how" and "why" mean the same thing. And as to the physical interpretation, there is none, because K = 1/2·m·v^2 is a mathematical theorem, not a physical law. You cannot use observations and conclude that K = 1/2·m·v^2, because K = 1/2·m·v^2 is the consequence of a definition, not of an interaction in a system.
@@angelmendez-rivera351 Agreed; the video asks a logically incoherent question and the title should certainly be changed.
@@thomaskaldahl196 No, the question is only logically incoherent when postulated the way you postulated. The video maker is not trying to make a fictitious arbitrary distinction between "how" and "why."
@@angelmendez-rivera351 I don't intend to use an ad hominem argument, but I'd like to know your physics background. I'm an undergrad studying physics, and I'd be a lot more easily convinced if you turned out to be a PhD or a researcher who could easily have thousands of times more experience than myself.
@@thomaskaldahl196 I have not finished my Ph.D, but I completed my Bachelor's degree a few years ago. I am not sure how this is exactly relevant. Even in undergraduate courses, the idea of the work-energy theorem and the fact that kinetic energy is conserved is more a mathematical abstraction than a physical phenomenon. For example, in undergraduate courses, I encountered Lagrangian-Hamiltonian mechanics, and how this formalism improves on the Newtonian formalism as a whole. The quantities in question are mathematical and do not have a direct physical meaning. However, from them, you can derive formulas that will allow you to figure out the equations of motion based on a potential field, which you do obtain from physical observation, and depends on the system you are interested.
In fact, even if we forget Lagrangian-Hamiltonian mechanics and only focus on Newtonian mechanics, you surely must have already seen that Newtonian physics blurs the line between theorems and observations. The three laws of motion, for example, are just laws obtained from observation. However, the idea of a conservative force simply originates from mathematics. If your system has a measurable force field F, and there exists a scalar potential U such that F = -grad(U), then you can find the work as the integral at the endpoints of the trajectory of F·dr, and the fundamental theorem of calculus says this is equal to -1 multiplied by the difference of U between the endpoints. After you do some further manipulations, you realize that this does result in a quantity that does happen to be conserved, equal to K + U, the total energy, and that K = 1/2·m·v^2, so the observed data confirms that the mathematical formalism is adequate for Newtonian physical systems. You can indirectly measure K, yes, but ultimately, this result is a consequence of both observation and mathematical theorem, so K is just a primitive notion you get used to with no real further physical interpretation, at least not in the Newtonian framework.
I have to crank the audio way up just to barely hear. Do you listen to your own video on a laptop?
I’m not good at calculus but my school teaches us to derive it from v^2=u^2+2as (although only proved for constant a)
as=(v^2-u^2)/2
(F/m)s=(v^2-u^2)/2
Fs=m(v^2-u^2)/2
Only bad thing is that it only proves that for constant a but you can still see why it gives 1/2 I guess.
The link between Newtonian Mechanics and Quantum Theory is it known! Look up classical limit of Quantum Theory
That classical limit is through Lagrangian-Hamiltonian mechanics anyway, though, so your point is rather moot.
@@angelmendez-rivera351 actually you are right. But then, how does it makes sense to ask about the connection with QM and Newtonian Mechanics, if QM is not commonly based on the idea of Forces ?
@@jeronimoabello7170 I am not even sure it really makes sense to ask for a direct connection, nor if there is actually one.
how do you get from int{m r'' r' dt} to 1/2*mr'²
because the time derivative of r'² is 2*r'*(time derivative of r') or 2*r'* r''
@@jdp9994
Thanks for the answer, but I think I still don't get it. :(
Isn't your attempt the other way around?
I mean, I understand that the derivative of 1/2 m r'² is the integral, but is there a nice way to get from the integral to the right side? (not just by reverse engineering the right side)?
You can integrate by parts with u’=r” and v=r’. So that int(u’v) = iuv |at t2 and t1 - int(uv’).
In our case the integral with the negativ sign will just be our integral again and you can put it on the other side to get 2int(mr”r’). Now you see directly that there is a 1/2 in the solution when you divide both sides by 2.
Int(mr”r’) = mr’r’ - int(mr’r”)
2int(mr”r’) = mr’^2
int(mr”r’) = 1/2 m r’^2
@@peterschmidts8245 Thanks a lot - haven't integrated much lately but your explanaition helped a lot!
Thought this had to be more difficult :D
I still don't understand how they go from (integral of m*r_dot*r_double_dot dt) to 1/2(m*r_dot^2), can anyone explain?
integration by substitution.
set u = r’
du = r‘’ dt
plug u and du in for r’ and r’’ dt and take m out since it’s a constant:
m*integral(u du) = m*(1/2)u^2
plug r’ back in for u and you get (1/2)mr’^2
@@jasonguo3540 we have r'' dot r'. these two are vectors!!!! how can we use substitution? what about cos(theta) (from dot product)?
You can write the dot product as a sum over the product of each vectorelement of both vectors do the substitution works just fine as it works for Each element
@@sajjad213 In introductory physics, r'' and r' are not vectors. But even if they are vectors, you can still use substitution no problem. Them being vectors does not change how substitutions work.
5:05 seems to imply that if you have a 'nose' for your subject then you could do well?
What is the outro music? Reminds me of Myst's soundtrack.
commenting before watching: my guess is that when you integrate v you get 1/2 v^2
Nice use of procreate
Really low audio
where are lagrangian and hamiltonian videos?
There is no kinetic energy in a moving mass there is force Mv squared kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers
Off the mark.
Wrong question! Why does kinetic energy scale with the square of velocity is a better question. If you assume that work is force times displacement it follows mathematically; fine!
What I would like to know why in an elastic collision not only momentum is conserved, but also kinetic energy. You could say that kinematic equations are not explicitly time dependent, therefore by Noether’s theorem Energy is conserved and the kinematic equations also are not explicitly position-dependent, so momentum is conserved. Elastic collisions of spheres that behave frictionless do not start to spin, so angular momentum does not change and stays zero.
Thanks for this. I have a video about Lagrangian formalism planned. This video had a simple intention of understanding where the kinetic energy formula comes from
@@MetaMaths The most insightful explanation I could come up with why kinetic energy should scale with the square of velocity is this. Consider a mass m being dropped from a 20 meter altitude, which we’ll call h=0 at t=0, in a vacuum chamber to eliminate air friction, for simplicity assume g=10 m/s². The work done by gravity is W=F•h= m•g•h. Where does the work W performed by gravity go? If energy is conserved is has to be accounted for by the energy that can be associated by the velocity of the mass m. Kinetic energy K~m•v^n with n yet unknown. Under influence of a constant force velocity will increase linearly with time: v=g•t so K(t)~m•g•t^n but W(t)=F•h(t)=m•g•0.5•g•t² = c•m•g•t^n
So the constant of proportionality c must me 0.5 and n must be 2.
My eyes hurt when i saw the energies of the balls, ffs.
I thought you would explain by laws of nature or a intrinsic logic, but it seems we can never get rid of calculus... a kid wouldn't understand why the half, sadly
There is no law of nature to explain it with, because K = 1/2·m·v^2 is a mathematical theorem, not the result of physical observation. Strictly speaking, as K is not an independently defined quantity, it is impossible to actually measure it into the first place and determine that K = 1/2·m·v^2.
Sorry, but that is just how it is. We cannot bend reality to our will just because you do not like how reality works. But also, your complaint is a little silly: kids do not take physics courses and do not learn about kinetic energy until high school at the earliest, which is also when they begin to take calculus. So contrary to your claim, any person taking a physics course is already equipped with the mathematical knowledge to understand integration in physics, at least if they came from a half-competent school.
@@angelmendez-rivera351 well, in my school I saw no calculus because the educational system in Brazil don't have optative grade. But I saw energy and wanted to understand that. Why is there a equivalence between two constants (g and h) and v^2/2? You see this math in reality, so how can you say there is no experimental results? Thoughts like these conducted the high precision experiment of the oil drop of Millikan, so reality has its reasons to the 1/2
Why is there a half in the formula. Cause you quadratic it first then you have to cut it in have to be even! Simple example: 2^2/2=2. Q.e.d.
Btw. (This Form leads to the most problems in understanding for students)
So basically your showing the prof of work energy theorem to explain that, kinda useless
I was expecting an intuition answer
There is no intuition in science. If science were based in intuition, then every scientist would believe the Earth is flat, do you not think?
There is but not actually, intuition is basically deriving something new from something you already know the only difference is it's alot simpler and doesn't include mathematics directly (which basically is creativity) obviously one may think the earth is flat through their intuition which is wrong ( and they can be proved wrong easily just drop a water drop from some height it will form an approx sphere while falling since they ignore most of mechanics it can be used as a powerful analogy atleast for them)
What I meant my intuition is to connect certain dots which make things look obvious, ofcourse they depends on how much you have observed or analysed stuff
Example if I pick something pretty basic like say a particle is losing velocity by intuition you can say a dozens of things which are obvious and related to the info
You didn't used mathematics but you still reached conclusions which were obvious to you
Well ofcourse what you said is 100% correct that one cannot discover alot using intuition but intuition do lay the foundation for research, you think on stuff which looks possible (something that credited to intuition)
Example arithmetic progression is kinda a special case of motion with no external force
Or how flow of current is analogous to flow of fluid and rest of it
Once you make things intuitive you basically break down information in it's most fundamental form which is way easier to use
However explaining mathematical coefficient is not simple at all although he first assumed Work energy theorem to be true, which actually doesn't feel right since if you are finding something you cannot use the same thing for it
We say WET is valid because end result is something which we define as Kinetic energy but if you don't know a person you cannot say I that the people who is unknown to me has a big nose, that's what was done in the video.
@@abhishekdas2479 If you are talking about the speed of an object decreasing, you are still using mathematics, not intuition. It may not be written mathematics, but it is mathematics nonetheless.
He did not assume the work-energy theorem to be true. He proved the work energy theorem. The work W is defined as the integral along a path γ of F•dr. From Newton's second law of motion, it is known that F = m·dv/dt + dm/dt·v, so W is equal to the integral along γ of m·dv/dt•dr + dm/dt·v•dr. The integral along γ of m·dv/dt•dr is equal to the integral along the velocities at the endpoints of γ of m·v•dv, which is equal to 1/2·m·(v1)•v1) - 1/2·m·(v0)•(v0). Meanwhile, the integral along γ of dm/dt·v•dr is equal to the integral along the masses at the endpoints of γ of v•v·dm, which is simply equal to (m1 - m0)·v•v. Therefore, W = 1/2·m·(v1)•(v1) - 1/2·m·(v0)•(v0) + (m1 - m0)·v•v. If m1 = m0, that is, if the mass is constant, then this simplifies to W = 1/2·m·(v1)•(v1) - 1/2·m·(v0)•(v0). If we define K(v) := 1/2·m·v•v, then W = K(v1) - K(v0). This is what the video proved. Therefore, the quantity K is mathematically special, because unlike W, it is a function of state of the system that can thus be indirectly measured. Furthermore, since K is known to be conserved whenever the mass is conserved and whenever the net force field is 0, this implies that K is a component of energy, and it is one originating from having nonzero velocity, so we call it kinetic energy.
What you need to understand for this to make sense is that, contrary to what they teach you in the classroom, the work-energy theorem comes before defining the idea of kinetic energy, not after.
You lost me at “double dot”
Sorry but audio is very low I can't even heard a word
You're so loud 📢
5:05 seems to imply that if you have a 'nose' for your subject then you could do well?