The underlying PDE is the Cauchy momentum equation.
It is just Newton II expressed for a continuum. Note that the ##\mathbf f## in it contains body forces only, like gravity. The more typical types of external forces come into play as boundary conditions for stress.
Edit: clarifications
Recently I started wondering why there seems to be so few practical/engineering applications where you need to calculate the momentum of something. I must emphasize that I don't mean usage of the concept of momentum or the law of conservation of momentum, but the value of the quantity itself...
Quite right. :smile: I was just wondering whether the "direct" approach was known to be significantly worse in terms of performance compared to the apparently more common FFT approach.
The original signal will be used as input/reference for certain engineering calculations where the signal is...
I need to perform a phase shift for a measured time domain signal. The signal consists of around 1000 points. I wrote an algorithm for this in Python a while ago and it seems to work as intended, but now I noticed that the algorithm is actually unexpectedly slow.
Is it so that phase shifting is...
There was an old thread comparing the difficulty of classical mechanics and electromagnetism. The consensus was that electromagnetism is more difficult, and substantially so according to some. The thread was no longer open for replies, but it got me suspecting that we're comparing apples to...
Yes, maybe I should have used the word "complex" instead. But I didn't necessarily refer to the number of rotating parts. I meant the complexity of involved engineering/physics. With a piston engine you have for example reactive fluid dynamics inside a rapidly changing domain. For example the...
Ever since electric cars became increasingly popular I've been wondering whether there are some other examples where new technology is less "advanced" than the old. I mean, electric motors are considerably simpler technologically compared to piston engines. Are there more such examples or is...
What a strange question. :smile:
I think most cities in the "East" have comparable technologies to the United States. No Silicon Valley there, though.
Nowadays, probably none. It's a global world.
Same as above. But as a historic example, the US was a bit late to adopt cellphone usage. I live...
I would imagine the procedure is indeed relatively rare.
But what about motorsports, for example Formula One? Surely the teams perform such analyses since the engines are rarely, if ever, running in steady-state conditions and fierce competition requires simulations to be as realistic as possible?
By transient vibration I mean that the engine speed and torque are changing as a function of time, as opposed to steady-state vibration where the engine is being run at a constant speed and torque.
Well, to be exact the instantaneous values for the engine speed and torque are changing in both...
How is the transient vibration of a piston engine usually simulated?
I know that in order to define the vibration loading you need the mass properties and dimensions of the moving components plus the cylinder pressure curve. And of course you need to know the firing order, V-angle (if...
I seem to be in a minority for thinking that mechanical engineering is more difficult than electrical engineering. Difficult for me, I mean.
To me, mechanical engineering is more difficult largely because in mechanics your intuition and common sense are often trying to mislead you. With...
I don't think turning radius is the explanation. Even without the maneuver, when approaching the corner the car would still be positioned on the outer edge of the road (the "racing line").
The system is perhaps too complicated to be satisfactorily described with elementary physics, which may...
If Newton II is defined as ##\sum F = \dot{p}## and ##p = mv##, why do we consider Newton I as a separate law for cases where ##\sum F = 0##? Is Newton I really independent of Newton II?
The way I see it is that the driver utilizes the rotational energy built up in the pendulum motion to make the car oversteer more, which often allows the driver to clear the corner in less time.
By "turn farther" I mean smaller turning radius. Without the maneuver the minimum achievable turning radius at a given speed is greater.
Edit: Or maybe I should say that to "turn father" I mean that the cart oversteers more. The maneuver increases the amount of oversteer.
What I meant that the objective of the maneuver is to increase the rotational energy of the car. While the angular momentum also increases as a result of the maneuver, it would be incorrect to state that the increased angular momentum is the reason why applying the maneuver makes the car turn...
There is a cornering maneuver in rallying called the "Scandinavian flick" or the "pendulum turn". It involves steering away from the corner before actually steering into the corner. This creates a pendulum effect which makes the car turn more sharply into the corner.
Sorry for the poor video...
Thanks.
However, I am struggling to find an online reference where Newtonian momentum is NOT defined as p = mv, i.e. the particular case for rigid bodies. Is this merely because of the appealing simplicity and wide applicability of the rigid-body assumption?
For example Wikipedia states that p...
I understand that momentum is a property of all systems regardless of their type.
But why is momentum often defined as the "product of the mass and velocity of an object", ##p=mv##. This implies that we are restricting ourselves either to a rigid body (using the velocity of the center of mass)...
Does it really apply to a continuum as such? Because NII is usually expressed as F = ma, which kind of implies we are dealing with a point mass rather than an infinitesimal volume.
And shouldn't a continuum be considered the "default" system in mechanics, just like in electromagnetism (e.g...
Is it appropriate to say that within classical physics the general form of Newton II is the Cauchy momentum equation?
This equation applies to an arbitrary continuum body. Therefore it is more general than the common form of Newton II which applies basically to point masses and centers of mass...
I like to think that energy is a "fundamental quantity" and momentum is just a handy rephrasing of Newton III.
Generally, energy does the damage. But the real world is more complicated than that.
Yes. The force ##F_w## at the supporting structure is always equal in magnitude compared to the spring force ##F_s## because these two forces are an action-reaction pair (Newton's third law).
By the way, your free body diagram is a bit misleading since you have both internal and external forces...
I also think that continuum mechanics doesn't get enough credit for its level of difficulty.
Many physicists/engineers commonly think that Maxwell's equations or the Schrödinger equation are among the hardest equations in physics. However, these are merely linear PDEs, whereas for example the...
I am about 30 years old and I have come to realize I do not really enjoy my work. Actually, it is more like I do not enjoy working life. But studying was something I did enjoy very much (not student life, but actual studying).
I miss the mathematics that was so strongly present when studying. I...
That would be the obvious solution. However, the problem is that the terminology is quite well-established in the scientific community. It would not be easy to get people to start using a new expression for the forces at the supports that does not contain the word “reaction”.
I am not a teacher, but I have made an interesting observation regarding classical mechanics education.
It is surprisingly common that even students at academic level do not seem to fully understand Newton's third law (action and reaction) and confuse it with Newton's second law...
I am a mechanical engineer. I graduated a couple of years ago (M.Sc.) and I am currently working as a structural analyst.
Yes, this is absolutely true. Maybe I should not have used the word "complex". What I meant was that in classical mechanics engineers "dig deeper" into the theory than...
Thank you for your thoughts. It seems that the main reasons why continuum mechanics is generally excluded from physics education are:
1. Other topics in physics are more relevant career-wise.
2. There is no room in the curriculum.
3. It requires high-level mathematics.
4. There are some "loose...
I came across this article about the near absence of continuum mechanics in university-level physics education:
http://www.troian.caltech.edu/papers/Gollub_PhysToday_Dec03.pdf
I have wondered this issue myself: why is continuum mechanics mainly studied by engineers rather than physicists, even...
The Cauchy stress tensor at a material point is usually visualized using an infinitesimal cube. The stress vectors (traction vectors) on opposite sides of the cube are equal in magnitude and opposite in direction. As a result, the infinitesimal cube is in equilibrium.
However, when we derive...
The nature of the force in the example is simply that it is constant in magnitude and direction. Therefore, it is conservative and has a potential energy function, like you said. No other restrictions.
I think he means that if we wanted to bring the system back from the loaded position to the original position while keeping ##F## as it is, then the work required would be equal to the change in the total potential energy. In this process, the value of the total potential energy would increase...
Thank you. That was an excellent post.
I am used to making assumptions and approximations. It is just that if there is even the slightest sense of doubt in my mind regarding the mathematical and logical consistency of the "foundations", I feel uncomfortable. It is not satisfactory to know...
It seems that I am definitely not alone; there are people who are as confused as I am about how the potential energy of the external forces should be interpreted. Note that the term used here is "work potential"...
Imagine the spring is from the suspension of a car, and you compress it in a press. Or that the spring is one of those demo springs found in physics classrooms, and you compress it with your hand. You do not apply the force instantly as ##F##, but you increase it from zero to ##F##. The work the...
I am still confused.
Another thing to increase my confusion:
Since the potential energy of the external forces is by definition the negative of the work they perform, there seems to be a "hidden" assumption that the external forces are applied instantaneously and not quasi-statically.
In a...
But if we were to suddenly make ##F=0##, would not the spring start to vibrate with a total energy equal to the elastic energy of the original scenario?
The principle of minimum total potential energy is frequently used in solid mechanics as an elegant way of deriving the equilibrium equations for an elastic body under conservative forces. This method states that out of all the possible displacement fields that fulfill the boundary conditions...
To sum up my above post:
It appears to me that at the wall for an incompressible flow, the normal stresses perpendicular and parallel to the wall have a magnitude equal to the pressure. This is the reason why we can treat the pressure as the loading that is exerted to the solid perpendicular to it.
Actually, the basic thing that I am missing here might be that the velocity is constant (zero) along the wall. Therefore, the spatial derivative of the main flow velocity along its direction is zero at the wall: \dfrac{\partial u_1}{\partial x_1}=0. And, assuming that direction 3 is parallel to...
Thanks for the long answer and reminding me of the related fact of the mechanical pressure generally not being equal to the thermodynamic pressure. I am quite familiar with the index notation. However, I still do not understand why the pressure (be it mechanical or thermodynamic) should be...
Yes, by perpendicular I mean perpendicular to the wall. Could you elaborate how the normal stress component of a moving fluid is the pressure for a Newtonian fluid? I thought this only holds for a fluid at rest.
Picture water flowing in a duct. In these kinds of scenarios, the loading exerted by the moving fluid to the solid in the perpendicular direction is often taken as being equal to the pressure of the fluid at the wall.
I agree that the above assumption is true for a stationary fluid. For a...