Weight of the Hourglass: Newsletter Challenge (April 2012)

what tide is it?

Right now it's slack tide, locally. Depends upon where you are in the world.

That's one way of looking at. Another, is that the hour glass also exerts less force on the scale (after its turned over) because it's center of gravity is moving downward from above the bottleneck to below. It's doing the same thing a rocket does - shove mass in one direction to cause that Newtonian 'equal and opposite' reaction in the other. A fancier term for it is 'conservation of momentum'. When a sand particle accelerates from stationary position to having a downward velocity, it has acquired momentum or (mass x velocity). The hourglass also acquires an equal but opposite momentum vector to keep the conservation of momentum equation balanced. However,because the hourglass doesn't move this momentum vector is expressed as (force x time) instead, which momentarily and minutely offsets part of the hourglass weight. When the sand particle comes to rest at the bottom, its momentum is transferred to the hourglass, cancelling out the hourglasses upward momentum vector - but only after the sand particle's travel time is up. Aggregate all the sand particles in motion, and until all the sand has moved to the bottom, there's always a net difference in momentum based on sand particle travel time which makes the hourglass "lighter".

I agree that the downward accelerating sand will make the weight of the hourglass slightly less if we can get a sensitive enough of a scale. But what of the added force required to stop the free falling sand? If we're going to claim sufficient resolution to see the reduction in weight from the falling sand lets consider everything then. This then brings into question how much sand is actually in accelerating fall. Similarly at the moment of inverting the hour glass, all of the sand that was at the bottom which is now the top falls to the mid-point funnel and stops. This change in momentum of nearly all of the sand stopping at mid-point will produce a very noticeable impulse of more weight instead of less weight.

Additionally what about buoyancy then with the gas inside the hour glass. If the sand while falling is accelerating downward then I can see a reduction in weight but if the tiny grains actually reach terminal velocity then no net change in a downward force from these grains will happen.

So I say that this really is another ambiguous question. Depending on how one chooses to define the moment of inverting the hourglass, the weight can be significantly more, slightly more or slightly less than it was with all of the sand at rest in the bottom of the hourglass.

You're ignoring my closing point. Usually the closing statement is the point of the statement.

Depending on how one chooses to define the moment of inverting the hourglass, the weight can be significantly more, slightly more or slightly less than it was with all of the sand at rest in the bottom of the hourglass.

IMHO the stated question does not define well what are the conditions that constitute being turned over to say if the weight should be more or less. The initial conditions as stated are with the sand at rest at the bottom of the hourglass. After this initial condition there can be instances were the weight can appear to be greater than this, equal to this, or even less than this initial condition. My point is more apparent if one considers the discrete time condition. With t(0) being the initial condition and t(1) being the hourglass inverted condition of interest with sand resting on the funnel surface there maybe a finite impulse during t(1) from the sand landing on the funnel. If instead one chooses t(2) to not include this potential impulse effect as the time to consider then the grains of sand falling through the orifice that are accelerating downward will reduce the downward force, the grains of sand at terminal velocity will make no difference and the grains being decelerated at the bottom of the hourglass will increase the downward force. One can now choose a t(2) condition with sand of each category to make the weight more, less or equivalent to t(0).

And, of course, when turned there will be a significant time taken to balance out the inertial forces of the turning motion, so the sand may be all at the bottom again by the time the induced oscillations have dissipated.

At least until it reaches terminal velocity when we have the old chickens in the back of the truck problem.

Dude, are you still reading ? Godd*m them for not posting the answer yet -Im all spouted out. Usb get's good bucks for notching for first 'sensible' answer. Finesse, OK, let me tickle your prarie oysters for seeing the subtle point and saying you topped the post. Of course, I am top dawg for summation .

Now I am pissed (Brit sense), and it's time to digress (challenge question tradition) - though not by much. There is no such thing as 'compressive failure'. Discuss amongst yourself (). Sure, compressive stress can be applied, but failure is due to shear....I refer all who think I've gone mad(der) to Brazilian (Discs, not shaving). For those too lazy, that's tensile strength inferred from compressive force. 'not by much'....a funny thing occurs called 'platen effect' really screws that.

Ooops, back OT.....calls for an exceptionally sensitive balance...Aw, you'll be 'shoulding' the poor thing next. Cone-heads of the world unite !

For anybody else reading, don't disturb me with requests for explanation - I'm already disturbed !

_{The sensible reading order - europium, Usb, and some bloke from Atlanta .Sorry ken, just a mild bit of funning funning}

To quote a mutual friend: "

"

In keeping with the spirit of Challenge-Question Ambiguity, perhaps it should read "... a following tuesday,"

Or

It could be that the onset of tuesday, ergo, the actual moment at which tuesday is in fact tuesday, is a matter of some dispute,

Or

The Tuesday in question could have zipped on by, given the considerable unbroken momentum of the weekdays preceeding it,

Or

The OP is just as confused as we are,

Or

I am very glad that I am not trying to make a living out of working out the weight difference between an hour glass at rest and the same hour glass in dynamic mode.

Or was that just before the sand started to flow and if the sand has not yet started to flow would there not be some residual inertia from the inversion?

BAB

You raise a good point, at the instant, the sand isn't even at the neck, it is along the side, completely unsupported unless the turn is very slow. If it is very slow, which instant are we talking about?

It appears that both offerings are correct. All of the sand has been moved to a lower gravity, and a small amount is not supported. I am pondering the thought that energy has been pumped into the system by an external agency, the potential energy is increased; needs more thought. The falling sand creates noise and heat which dissipates the energy input with the turn. If energy is interchangeable with mass...

I just want to know when the egg will be done.

Long before the hour-glass runs out...unless you want your egg extremely hard boiled.

"Six bells and all's well. Six bells and all's well."

Yummy eggs

but not with my hourglass

Boiled? Scrambled? Fried? Poached? Coddled? Pickled? Good grief!

I prefer mine in Warninks.

(A liqueur made from Brandy, egg yolks and sugar, from the Netherlands. So that's what they've been up to! )

[Sorry, I meant to post this as being OT, but the button seems to have gone away]

no hamburgers

it's an hour glass - not a second glass!

You've clearly never tasted Brit food .

"at the moment the hourglass is turned over" This is the key phrase. No sand is moving through the hourglass, and no mention made of how the hourglass is turned over - no finger-pressure or other external force. It just happens! So, the only force operating on the scale-plate other than the influence of gravity on the total mass of the hourglass, is the moment of inertia of the full (bottom) half. moving up, and the empty (top) half, moving down. Since the inertia of the full, upward-moving part is greater - it's full of sand - there will be a momentary, apparent increase in the weight of the hourglass.

Energy has mass: m=E/c². It takes energy to turn the hourglass so that the weight of the sand is at the top, so it weighs very slightly more.

That equation does not imply that energy adds mass to a closed system - only that energy within a closed system can be converted to mass, and vice versa.

Energy does not 'have mass', it is 'convertable' to a specific amount of mass. Not quite the same thing.

While converting mass to energy is fairly simple (nuclear energy), the reverse is not as easily accomplished.

Turning the mass over does require energy and it will be extracted from the object turning it. If I turn it over then it will weight this tiny amount less after I have done so. The remaining energy goes into the gravitational potential energy of the system. It does not, however, increase the force of on the object. It is sometimes tricky to decide when you can count the extra energy as part of the field of ONE of the objects or the collection (if you can make sense of it cleanly at all). The exasperating thing about general relativity is that, to higher orders, energy is not conserved at all!

Oh my edit mess: I meant to say "I" will weigh slightly less after I turn it over. "will weigh"... Where is my coffee???

Let's just think of a simpler case. I have two spheres connected on a sliding rod in space. The gravitational force between them I can measure by its strain. If I inject power from outside the system to pull them further apart this force between them decreases but now the total mass of the system, as observed by its gravitational field far away, increases. I could ask "where" this extra mass is located. It is now part of the "gravitational field" contribution to the masses. There will be a small increase in the effective mass from it (but it will be more than offset by the increased distance so the net force goes down). These field contributions are very hard to quantify and can't be localized.

So you are right to think the mass of each individual sphere goes up but the subtle problem is that this mass cannot be defined independently of the location of the second mass! It cannot even be well thought of as located at the position of the second mass. The stress in the rod makes its own contribution to the mass as well.

If we have a closed system then the net apparent mass of it far away seems the same however the distribution of it locally changes. End result is the force does decrease with height and you will now regret reading this challenge or fall down the rabbit-hole of modern physics. :)

While you guys are working on this, I'm just going to lay the hourglass on it's side and weigh it.

The hourglass cannot be weighed "at the moment the hourglass is turned over." At that moment, it must be off the scale. After having turned it over, there will be additional time to carefully place it on the $50,000 scale, and then some more time for the scale to stabilize.

How did the chambers of an hour glass come to be called "plates"?

The mass, of course, remains constant, but the weight differs during three intervals. With a different experimental arrangement, one could look at the beginning of the timing interval (when sand just starts to fall, but has not yet hit the bottom) the middle of the interval (when sand is falling and also hitting the bottom) and the end of the interval when sand has left the top but not yet reached the bottom. The first and last segments are extremely short, so might as well be ignored.

In the middle interval (the only one measurable if you really turned over a real hour glass) the free-falling sand would not be weighed, but the impacting sand would be increasing the weight to match. So for the vast majority of the experiment time, the weight is the same as it is when no sand is falling. At the very first instant it is lower, and the the very last heavier.

So if the question is intended to refer to the instant when the sand has just begun to fall but has not reached the bottom, then the first weighting (sand all in the bottom) will be higher.

"In the middle interval (the only one measurable if you really turned over a real hour glass) the free-falling sand would not be weighed, but the impacting sand would be increasing the weight to match."

At any given instant there are only a few grains impacting the bottom of the hourglass, and the amount of suspended (or falling) grains is variable depending on several unknown factors - the height of the falling stream of sand, the width of the opening, etc. The impacting grains will offset SOME of the grains in suspension, but the amount of offest is indeterminate given the information provided - but it would certainly not balance out. So this assumption is incorrect.

The 'turned over hourglass' will weigh slightly less than the 'sand at the bottom' hourglass for two reasons - there will be a small amount of sand that is falling and therefore not being weighed, and the mass of sand at the top will weigh slightly less because it is farther from the center of earth's gravity well.

Of course, weight varies with altitude. I think that everyone can agree with that, but many would say that you would then observe a slow change in weight over the course of one hour, if you could measure such a small difference, which is unlikely. It would certainly be impossible on any scale on which you be likely to place an hourglass.

So then "moment" comes into play. (Why is the slow change not mentioned?) This seems to suggest that the question applies to a time immediately at the start of the new timing interval. One valid interpretation is Europium's in which he takes moment to mean instant. A just slightly different interpretation stretches that instant to the time that it takes for the stream of sand to be reach the bottom, and the collisions to start. To actually measure this requires a special setup, because you would not have anywhere near enough time to get the glass back on the scale and stabilized. But, it is after all, a thought experiment.

The mass of the suspended stream (as reflected in its width and length) is compensated by the impact force. The longer the stream, the higher the impact velocity. The wider the stream, the higher the impact mass (and the greater the suspended mass). We know that the energy and mass in the vial is conserved. So, once steady state is reached (after the first second, for example), then the weight is constant, with the sum of the impacts compensating for the temporarily unweighed mass of the free falling stream.

The linked experiment showed that there is no difference in weight between the still hourglass and the running hourglass, demonstrating that the falling mass is balanced by the impacts. It also showed that there are the expected changes in weight at the beginning and end of the timing period.

Demonstrating a difference in weight due to different altitudes would have required a far more precise balance. It would need to read out to 8 places or so. If the question were to demonstrate that effect, then one would think a mountain or extraordinarily tall building might be involved.

Why an hourglass? Why "moment?"

But you are not alone in your preference for thinking about the altitude difference. Time will tell.

At the instant of inversion most of the sand is supported by friction. That is by direct contact with the glass, and friction between grains. Above the neck, there is a cone shaped region that is unsupported. The angle of that cone depends upon the characteristics of the sand. It will be pretty much the same as the angle of sand when it's all over with.

There's more, but copyright doesn't allow too much copy from any one source .

Dan Aykroyd never looked better

__________________________________

And don't forget to remove the checkpoint downstairs for:

_{[] Yes, this comment is very likely to be considered to be 'off-topic'}

I stand corrected regarding the impact force balancing out. Regarding the weight difference from the change in height - I failed to note that the original question specifically mentions using a "balance", not a scale. I used to service balances and scales years ago, and one thing I remember from training is the difference between the two. Scales measure weight by comparing an objects weight to a force, such as a spring or electronic load cell and therefore must be recalibrated for changes in altitude or location. Balances OTOH, measure mass directly by comparing an objects mass to a known mass. They do not need to be recalibrated for a change in altitude.

The distinction of 'Scales' and 'Balances' is worthy of mention. It's unlikely the choice was arbitrary, though it's almost certain that intent was for people to argue this. My opinion is stated. If all goes to norm, we see the 'official answer' and have a chuckle. The meat is betwixt answer and question.

With politeness intended; What force supports the cone of sand above the neck of the inverted hourglass (at the instant of inversion) ? More to the point, is gravity more an issue than unsuppurted sand ? I suggest not. That's as clear as mustard, but this is all fun discussion, Taking things to extrenme theoretical possibility is part of that.

They do not need to be recalibrated for a change in altitude.

However, the assumption is that gravity is the same on both sides of a balance in ordinary use: a very reasonable assumption for most uses. The thing that causes the scale to tilt one way or the other is a force imbalance, ordinarily due to a difference in masses: M1 * G ≠ M2 * G.

But if you change G on one side by elevating the mass, then the balance will react. A very high precision balance would measure different apparent masses of a sledge hammer standing on its head vs handle end. (And would show a difference between two sledge hammers oppositely oriented.) But I don't think that's what the question was getting at. But it could be. One never knows until the official answer comes out.

Why not?

I was just thinking of the weight of a train while some people jump in synchronous; or in an aerplane.

If it is considered that the moving sand is in air [ or vacuum if the glass is vacuumed]

and does not come in count then it will be varying constantly till fully dropped.

This is a bit like the van load of parrots.

Is the load heavier when the birds are perched than when all flying.

the weight of the birds will be added throu the air to the van in case they are flying and directly to the van if not!

there is no difference if the van is sealed.

hm, at the moment it is tipped over, and no sand is falling, would the air underneath the sand be added to the weight?

Thank god someone on here at least referenced the actual physics. You can take a derivative to show that dF/F=-(2dr/r) so that the relative change in force can be calculated. r=radius of earth=6x10^6 m, dr~0.1m so the fractional change is force of the sand is 3X10^-8. milligram scales are common but for them to pick this up you would need a very heavy hourglass!

I just realized something - the question references using a "Balance", not a SCALE. Balances measure MASS not WEIGHT! Therefore the distance of the mass from the center of the earth becomes irrelevant....

No. That is like saying a voltmeter directly measures voltage. It really is an ammeter with a large resistor. All it can do is measure the emf by displacing a needle. You just use theory to rescale it to give a voltage. A scale simply measures a displacement and attaches a mass label to the output (unless you use British Imperial Units which often is labeled in pounds not slugs or foot-pounds). The displacement most nearly linearly correlates with the force not the (rest) mass of the body.

Basic physics again...

"Although a balance technically compares weights, not masses, the weight of an object is proportional to its mass, and the standard weights used with balances are usually labeled in mass units."

A "balance" instrument is suggested in this challenge since it can be made sensitive enough to measure very small (<10^-9) weight changes. While this would require an expensive laboratory grade device, one can easily construct a "home-made" balance sensitive enough to compare changes in the 10 microgram (10^-5) range.

Unfortunately, in this case, the engineering maxims aren't as helpful. This one is based on the assumption of an object in a uniform gravitational field else our "labeled masses" give the wrong values on our scale! To the extent we care about the effect here, this fails and we have to use the explicit force law. Ultimately though we do need the true mass to do the calculation.

I may have read too much into your answer.

the balance measures directly which of two masses is heavier, too!

Agreed - making the weight with the sand at the bottom larger that the weight with the sand at the top

CPW

Yes one will certainly have to have ridiculously high resolution to see this weight change in the hourglass from the center of mass being further away from the center of mass of the Earth. But isn't this really the only actual change in weight that happens here. In all other discussions people are actually discussing the opposing force needed to keep most of the hourglass in equilibrium. The falling sand is not in equilibrium, it is accelerating due to its _{(wait for it )} weight. If one includes by addition this weight to the magnitude of the force opposing most but not all of the hourglass one gets the exact same weight as before minus the infinitesimal difference from a change in elevation.

I believe this horse is dead! Stop kicking it.

Use a very sensitive balance to weigh an hourglass when all the sand is in the lower plate, and again at the moment the hourglass is turned over. You will notice that the two weights are different. Why? Which one of the two weights will be higher?

It seems to me that a question asked in the vernacular is best reasoned through and answered that way. I may be alone in this belief, but for me it makes sense. Anything I add to the question is not part of the question and I am immediately in danger of answering my own question at the expense of the one that was asked.

What is being asked here is not being asked in terms of gravitational fields (non-uniform or otherwise), centripetal acceleration, femtograms and Universal Constants. What is being asked is to simply reason about what is going on. This question could be asked of practically anyone and legitimately so. One need not know anything at all about physics in order to answer this question correctly. That's the beauty of it: when you reason through this problem in its native tongue, it is not ambiguous in the least.

-e

I disagree, "Why?" forces a consideration of all possible reasons and/or mechanisms that result in a difference.

Use a very sensitive balance to weigh an hourglass when all the sand is in the lower plate, and again at the moment the hourglass is turned over. You will notice that the two weights are different. Why? Which one of the two weights will be higher?

"That's the beauty of it: when you reason through this problem in its native tongue, it is not ambiguous in the least."

Now there's some blind denial if I ever saw it. Several replies, including mine, have pointed out that the conditions of this second weighing are anything but precise and unambiguous. To make a hyperbolic point, there's nothing in the original question that precludes a hole in the side of the hourglass causing a lot of the sand falling out when it was turned over or a bird landing on the hourglass when being weighed.

"To make a hyperbolic point, there's nothing in the original question that precludes a hole in the side of the hourglass causing a lot of the sand falling out when it was turned over or a bird landing on the hourglass when being weighed."

Judging by many of the posts on this thread, that is exactly what has been done.

Can the question be answered in its present form? If not, then why bother with this thread at all? But if so, I rest my case.

Very good answer. Unlike others who have speculated and argued their answers, you provided demonstrated proof. The undulation generated from the brief (impulse) interval of the sand falling is demonstrated. Now the imprecise moment of measurement is up to all to debate.

The question as framed calls to mind one of Einstein's many gedanken experiments: one he conducted as he sat in the back of a bus in Ulm, observing the face of the clock over the archway through which his bus had just passed.

His 'experiment' was to discretise the images of the clock face leaving the tower at the speed of light, and then 'travel with them in his mind' at ever-increasing speeds until he was even with one of the images. As he passed each discrete image he noted that the time appeared to slow to a crawl and then at the final image, stop entirely. Einstein used this illustration more than once when speaking of the role played by Lorentz' time dilation in his theory of relativity.

Now, do images leave clock faces at discrete intervals? No. Did Einstein sweat over the fact that he no practical means whatsoever by which he might travel with these discrete images? No. Why not? Because it was irrelevant? It was the principle of time dilation he was illustrating. The practical means by which one might conduct such an experiment in real life were not mentioned (nor even considered) because they did not matter. It is the same with this Challenge Question, birds notwithstanding.

In other words, this question is a gedanken experiment by any other name.

Not the greatest example since, although mathematically simple, special relativity has numerous confounding details that make such simple analyses often fail - usually by neglecting length contraction and the subtleties of simultaneity as well. In your case, it is implicit in discussing images as originating in a single time rather than photons emerging from each point as discrete events that have different orgination times in different frames. The biggest value to such thought experiments is in illustrating how we need to upgrade our naive intuition with detailed analysis. A secondary point is how relativistic optics radically alters the images we see beyond this pictorial representation Einstein used and we teach in relativity classes. Shortcuts lead to overconfidence.

I thoroughly agree. But in the case of solving this particular problem, and in this context alone (which I have attempted to maintain throughout this discussion in spite of the temptation to do otherwise) I fear we are adding epicycles.

I'm not sure where some of this is going. I attacked this problem two ways. One by the change is gravitational energy, where I showed it was dicey that anything but a very high tech approach could measure this, and one by the time delay of the falling grains. They were easy calculations yet the chatting goes on. They may be new ways to look at it and am open to them but think a little math would save some us some of these verbal "epicycles."

Math can certainly be applied in this case, but is it actually necessary? Moreover, in order to properly frame the problem in the mathematical sense, a prerequisite, non-mathematical understanding must preceed it, non? Else, what would the math describe? If the concept is not understood already, no amount of math will help. If it is understood, the math simply serves as a precise, unambiguous description and model of the concept.

As framed, does this particular problem have a unique, correct solution? Does the problem statement contain all the information necessary for its solution? If not, then why are we here?

Of course one has to know precisely how to interpret the problem to do a calculation. I just posed two interpretations. Do you know of other ways to make sense of the given question? I think the answer as to whether math is necessary depends on the problem. Given the amount of wandering (and incorrect) talking here, it does provide a clean result. The way to interpret the problem has some ambiguity. That is not an excuse to avoid a simple calculation in favor of a lot of vague talking. One can determine the relative size of different effects to determine the dominant one. This generally require some sort of mathematical analysis.

Let's look at the problem, especially 'at the moment the hourglass is turned over' (the ambiguity of which appears to be a formidable obstacle to some of us here), then ask "Why turn the hourglass over? To what end is the hourglass turned over?"

"To get all the sand at the top?" Why do that? "In order to start a new timing cycle, maybe?"

When one thinks of an hourglass starting a new timing cycle, does one think of the sand all swishing around because it was turned over quickly, or that some of the sand is still sliding down the sides of the chamber from having been moved slowly? Are those details mentioned in the problem statement? No. Why not? Because this would render the problem unsolvable without much additional information than is actually presented? Evidently, then, the problem presumes 'at the moment' to mean when all the sand is where all the sand usually is at the beginning of a timing cycle and not one moment before - or after. When the sand is falling, the timing cycle has already begun, rendering the phrase 'at the moment' meaningless. As 'at the moment' is part of the problem statement, 'at the moment' must refer to a concrete instant in time where 'at the moment' is meaningful in the context of the problem, else the problem could not be solved and this particular Challenge Question would get a bad name for being so poorly formulated.

How about the part where the hourglass is weighed on 'a sensitive balance?' Now, would you or me use a balance to weigh a virus? Wouldn't the random impacts of air molecules cause momentary fluctuations in the readout of the 'weighing device' to a similar degree? Fluctuations in excess of the weight of the virus (equivalent to the difference in gravitational force between 1 kg of sand at the top of the hourglass and the bottom, separated by the distance you assumed (and did you integrate over the volume of sand, taking into account its (assumed) shape and distribution, or did you calculate based on the CG alone, or did you make some other simplifying assumption (oops. did I say that?). In a non-uniform gravitational field your particular choice in the matter will alter your calculations to a similar significance. Evidently, then, a difference in weight due to the effects of a non-uniform gravitational field are not part of the problem either. So what is?

Stripped of what we all know are the practical limitations of this problem, what we have left, then, is the problem itself. Let's solve that one.

If you prefer to turn this into a diatribe you're on your own. I won't participate.

But you already have. You have made a mockery of the whole problem. Occam's razor applies here.

I agree, don't allow yourself to be sucked into their .....

You have had the right answer all along!

This is the productive job not some snarky philosophical musing

Nope. This is some light-hearted philosophical musing as relief from the productive job.

yet the chatting goes on

When you been round these parts a little longer and seen a few more CQs, you'll realise that's the point. Some of the longest CQ threads have ensued when the problem as set has been solved mathematically in the first post.

Chill out and enjoy.

"... then the most reasonable answer is that there is no change in weight, which is the case for all but the first instant and last instant of the second weighing."

And it is a reasonable answer, too, especially so given that what happens at the 'first instant' is the whole point of the question.