Flow and weight problems

There has been a lot of interesting discussion about Blake Farrow's and Laszlo Mihaly (May 21) problems of the week. A lot of smart people were getting confused, and only now things are appearing to settle down as everyone begins to understand what they've got wrong. However, this story still has a few loose ends that are being hard to wrap my head arround, and I hope with this discussion things become clearer.

All the confusion in those problems seem to arise from the same place: the relation between internal forces of the fluid flow and the motion of the center of mass. So I'll highlight the two statements that are making me confused:

1.(This is a response by Michael Mendrin in a comment by Laszlo Mihaly in Michael's solution to the basic problem):

I got burned with that explanation, "...since the center of mass is steady, there is no change in weight". Well, turns out that it's not so simple in a gravitational field or accelerating frame. What we can say is that in a steady state system, with unchanging center of mass, there won't be any changes in weight. That doesn't mean it weighs the same as when things were not moving.

We can all have more argument about this claim, I'd love to hear them. I'd like a more careful explanation why is this necessarily so. Very fascinating subject, let's not give up on this.

Is this correct? Is it true that in an accelerating frame of reference, you can have a system with some center of some center of mass and some moving parts that weigh more than when the system is still, even if the center of mass stays in the same place? How is this not a system exerting a force on itself? Can someone give an example of such a system?

2.(This is a comment by Steven Chase in Blake Farrow's solution to the advanced problem)

Yeah, it's all about the acceleration of the COM. The scale is completely indifferent to the internal forces involved.

Okay, as far as I know, there are 2 ways of solving a problem like this, one is by considering the internal forces involved, like on Laszlo Mihaly's Solution on the basic problem, the other (astonishingly easyer) way is by considering the motion of the center of mass, like on Arjen Vreugdenhil's solution on the basic problem again. But the thing is that, as far as I know, those should always give the same results, in other words, weight increase due to center of mass acceleration can always be translated into a net force due to the internal forces. The acceleration of the center of mass does change the weight, but for that, something has to push down on the scale, and this something are the internal forces, right? If that is the case, how do we explain the phenomena in the advanced problem) using only the effects of internal forces? (wow, I said "internal forces" a lot)

Contributions are greatly apreciated.

Note by Pedro Cardoso
6 months, 3 weeks ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

I was operating under the assumption that the following statement is true:

Sum of external forces = Total Mass x Acceleration of Center of Mass

It may turn out to be the case that the "internal force" picture and the "COM acceleration" picture are mathematically equivalent, or that they give the same practical results. I just find it alot more convenient to think about a changing mass distribution than to think about force. This is similar to my growing preference for Lagrangian mechanics over Newtonian mechanics. In many cases, it's a lot easier to account for energy than to account for forces.

Steven Chase - 6 months, 3 weeks ago

Log in to reply

Yeah, I agree, it is often really hard to keep track of all the forces involved, and looking at the center of mass acceleration sort of does that automatically for you. However, (at least if those two pictures are indeed equivalent) we should always be able to find forces to explain the center of mass motion if we look hard enough, and that is what was making me confused, I was trying to solve the problem by looking at those forces, but I kept getting the same weight. I was only able to solve it considering the just center of mass. The arguments in both solutions to the hourglass redux problem boil down to "there is center of mass acceleration, therefore the hourglass is pushing down on the scale more" And my question was "Where the hell are the extra forces, then? I wasn't able to find them!" I think I finally figured it out, but I still need to do the math on it. When I finish it I'll post it here, unless I find some mistake in my reasoning, which is more likely than I'd like to admit

Pedro Cardoso - 6 months, 3 weeks ago

Log in to reply

Ok, that's fair. I think you probably can explain the phenomenon using force accounting as well. I suppose I should have said that I, personally, am completely indifferent to the internal forces involved, because I have a nicer method :)

Steven Chase - 6 months, 3 weeks ago

Log in to reply

First off: Thanks so much for making this discussion Pedro! I've had a really busy week (just launched my course on quantum objects, check it out!), but I wanted to contribute my thoughts from today. @Josh Silverman might be interested in this too.

I read a question on Hourglass redux by @Sven Roemer that made me think a bit about what direction the center of mass is moving, and where the forces are coming from, etc. This might be a pretty common misconception when it comes to internal forces. If the COM of the sand is accelerating up, the hourglass weighs more, and if it's accelerating down, it weighs less! I just thought through this while walking my dog, so let me know if you guys agree with my analysis. This is certainly the problem that keeps on giving!

In the system's natural units, here is the movement of the sand's center of mass. I've made red lines showing the boundaries between the upper and lower halves, and for simplicity reduced the length of the intermediate path to zero -- this doesn't change much, though I'm happy to entertain more detailed models. The center of mass moves from it's initial position halfway up the top part of the hourglass, to the final position halfway up the bottom part.

The center of mass is moving downwards, and the rate of this motion is maximum when the sand just starts flowing: this initial acceleration is nearly instantaneous and the center of mass begins moving downwards. It proceeds to get slower as the position of the bottom sand's center of mass moves up (as that part of the hourglass fills). Just as Sven said, the acceleration is pointed upwards, so it's not much of an acceleration at all, but a deceleration.

Well, what is causing the deceleration of the sand's center of mass? A decelerating mass must be decelerated by a force, and that upwards force comes from the scale. It's as if in very slow motion, the mass of sand is moving down, but its motion is being halted as it runs up against the hourglass bottom and the scale.

The weight that a scale measures is the upwards normal force (or reaction force) that the scale exerts on whatever is on it. As Newton's laws state, this normal force is equal and opposite to the weight of the object on the scale.

Blake Farrow Staff - 6 months, 3 weeks ago

Log in to reply

Thanks for the reply! I agree with the analysis, and indeed that is a way to explain it. The other way is what I mentioned in my reply to Steven Chase's comment I still havent done the math on it, because I'm focusing on some other things, but that would go something like this:

very informal and possibly not totally correct math and physics ahead

say the sand in the top of the hourglass is moving downwards at a constant velocity \(v\). In steady flow, three things are hapenning simultaneously

  1. After small intervals of time \(dt\) small masses of sand \(dm\) get "thrown" downwards from the top at velocity \(v+v_1\) (which generates an slight upwards force)

  2. There is \(m\) units of mass of free-falling sand

  3. There is mass hitting the bottom part of the hourglass with velocity \(v+v_1+v_2\), but since the level of the sand is rising, if the same amount of time \(dt\) passes, the amount of mass that will be stopped by the bottom part of the hourglass is \(k \cdot dm\), where \(k is proportional to the speed at which the sand is rising\)

So adding everything up we get:

\(\text{Impulse}= -(dm)v_1+k(dm)(v+v_1+v_2)-mg(dt)\)

The effects of the free-falling sand and of the initial downards pushing of the sand should cancel and leave two things in the equation: The velocity of the top part of the sand (because to stop it you need to accelerate it upwards), and the velocity of the rising sand at the bottom (because it makes the bottom part catch more sand than the top drops), those should add up to that \(\frac{2q^2}{\rho A}\)

Pedro Cardoso - 6 months, 3 weeks ago

Log in to reply

To tell you the truth, I can't write a complete list of the internal forces in play here. I learnt that in these kind of complex cases one should always look for a different approach first, like the COM argument used in several solutions written by many. When I saw the problem, I immediately associated to an idea I had when I was around six. Why not blow the sails with a fan on a ship when there is no wind? Of course, later I learnt that it wouldn't work. Maybe this is why my initial intuition said the answer would be 100 kg, and I thought about the COM argument only later to prove it.

However, there is still a question that couldn't let me alone: What previous known problems made people think there was more to this problem? What did they think? Is there an antetype, a strange contraption I've never met, but you have? An hourglass is a completely different problem, as the sand is not getting back up to the top container. It genuinely interests me. Sorry but I'm quite confused, at the moment it looks like learnt people, even brilliant.org staff were thinking something like a perpetuum mobile was possible. There must be something I miss.

Laszlo Kocsis - 6 months, 3 weeks ago

Log in to reply

Laszlo, I've always been aware that total energy is conserved, but in the case where there was a pump, we're obviously adding energy to the system. For a closed system where energy is conserved, it isn't immediately obvious that its weight must be constant. Indeed, you did argue that there'd be some fluctuations in the weight of the hourglass at least at the start and end points.

Let's say your apparatus with the pump includes batteries with stored energy, and the apparatus is on a balance scale with an equal counterweight. Suppose some switch is turned on, so that it becomes heavier, which would make possible lifting on the other end of the balance some extra weight. What would be happening is that energy stored in the battery is being used to do work raising that extra weight, and so total energy could still be conserved.

There's a kind of an analogy with a skier going downhill, making a succession of turns, as in a sinusoidal pattern. Even though he should be gaining velocity because of gravity, nevertheless he is able to maintain an approximately constant descent. People think it's because of friction with the snow, but in fact friction is not necessary for this. A proficient skier can absorb the energy gained from gravity by the way he skis and thus control his rate of descent, without relying on the friction with the snow. This can be simulated with a car moving frictionlessly on a sinusoidal track, where a suitable mechanism can be installed inside the car to absorb this gravitational energy. The result is that in fact the car DOES have a variable effective weight as it progresses down the sinusoid track. Any skier knows this, that's why it can be tiring and hard on the legs to go downhill. This could be true even if this was frictionless process (which experts can closely approximate). Total energy is still conserved, because the mechanism (or the skier) has converted gravtitional energy into another form of energy somewhere. As I said, this is not such a simple topic.

As a matter of fact, we can consider a closed system where we not only have the skier, the sinusoidal track, but the whole mountain and the snow. It can be argued that the total weight does go up slightly as the skier descends, but, like with the hourglass, the center of mass of the whole system is moving downwards as the skier descends. This could be another way to look at the hourglass problem.

Michael Mendrin - 6 months, 3 weeks ago

Log in to reply

Sure, it doesn't violate conservation of energy, but doesn't it violate conservation of momentum? I mean, for the skier to mantain constant velocity, there must be something pushing it against gravity, even if that something isnt friction (the skier might have a rocket, or something), it doesn't seem reasonable to me that the skier would be able to convert the gravitational energy into something else without trasfering it's momentum to the mountain, for instance, how would the skier travel in a sinusoidal pattern in the first place if not by exploiting friction?

On another note, it actually would be possible to make the ship move by blowing the sails with a fan, thats because the sails still can accelerate the air moving trough it by changing its direction, it just wouldnt be a very smart thing to do. In fact, I'd say you can think of an airplane as a ship with big fans blowing air into it's sails

Pedro Cardoso - 6 months, 3 weeks ago

Log in to reply

@Pedro Cardoso We can simplify the skier experiement using stairs instead, and have a pogo stick like apparatus that bounces downwards from step to step. Whenever the ram plunges into the apparatus, we can employ an electric generator to absorb some of the energy so that the pogo stick maintains a constant downward descent. Where does the gravitational energy go? The electric generator can store that energy in a chargable battery.

Michael Mendrin - 6 months, 3 weeks ago

Log in to reply

@Michael Mendrin Oh, I see, but in this case, still, the only moments when you perceive the guy in the pogo stick to be heavier is during the deceleration of the fall, if you imagine this guy going down the stairs at a constant rate, the extra weight from the pogo stick compression and the zero weight during the time he is free falling from one step to the next still balance out in quite the same way as in the invisible waterfall problem. It still conspires to no acceleration of center of mass \(\rightarrow\) no extra weight measured

Pedro Cardoso - 6 months, 3 weeks ago

Log in to reply

@Pedro Cardoso There is still a difference between an apparatus on a scale which center of mass does not move, and another which center of mass does move. That's the reason why the hourglass weighs more while its center of mass is moving.

Michael Mendrin - 6 months, 3 weeks ago

Log in to reply

@Michael Mendrin In the hourglass, the center of mass is moving downwards, but it is accelerating upwards. Is there a way a moving, non-accelerating center of mass could result in a weight change? I dont think that can be done without breaking conservation of momentum

Pedro Cardoso - 6 months, 3 weeks ago

Log in to reply

@Pedro Cardoso That was an interesting revelation (that the acceleration is upwards even though the motion is downwards).

Steven Chase - 6 months, 3 weeks ago

Log in to reply

Yes, I was thinking about the energy coming from the batteries (or fuel tank). But when we are checking the movement of the center of mass, it's the inertia that is important, not the energy. On a second thought after this: we should include Earth in the system when checking the center of mass. But inertia is constant of the whole system in this case, too. I can see the possibility of change in the measured value only if we count the atmosphere in, too. Air gets moving near the moving water. I can't tell, what direction the inertia of the moving air has.

Laszlo Kocsis - 6 months, 3 weeks ago

Log in to reply

Let me first review the general matter of what happens with a non-static closed system that either 1) has an unmoving center of gravity, or 2) a center of gravity moving at a constant rate, in an accelerating field? Give me time.

Michael Mendrin - 6 months, 3 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...