Weight of an Hourglass
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 uncivlengr
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Weight of an Hourglass
I first heard this question posed on the local CBC morning show as part of their "Science Corner" which grossly oversimplified the question, and it's haunted me ever since. Incidentally, the host of the segment was wholly uninteresting in debating the issue with me, saying merely that he received a lot of response regarding the issue, but was sticking to his answer.
It can be boiled down to whether or not the weight of an hourglass changes when it's "flowing" compared to when it's not. The oversimplistic answer (that was given on the aforementioned radio show) is that the mass doesn't change, so therefore the weight doesn't change, but this neglects a couple of factors; namely, the apparent weight lost from the sand that's in free fall, and the impact force of the stream of sand as it hits the bottom.
The question is whether the effect of one of these factors exceeds the other, or if they cancel each other perfectly. There's also the fact (which I perceive to be the most important) that the pile at the bottom is constantly growing  therefore the rate that the sand hits the pile would therefore be greater than the rate at which it leaves the neck.
This would surely result in the centre of gravity decelerating, but I haven't worked it out to my satisfaction. Can someone address this question with some actual calculations for a structural engineer that hasn't reviewed this sort of dynamic problem in years? Arguing this from intuition usually results in the discussion degrading into something similar to the plane/treadmill question, and physical tests seem to be too greatly influenced by unwanted factors.
It can be boiled down to whether or not the weight of an hourglass changes when it's "flowing" compared to when it's not. The oversimplistic answer (that was given on the aforementioned radio show) is that the mass doesn't change, so therefore the weight doesn't change, but this neglects a couple of factors; namely, the apparent weight lost from the sand that's in free fall, and the impact force of the stream of sand as it hits the bottom.
The question is whether the effect of one of these factors exceeds the other, or if they cancel each other perfectly. There's also the fact (which I perceive to be the most important) that the pile at the bottom is constantly growing  therefore the rate that the sand hits the pile would therefore be greater than the rate at which it leaves the neck.
This would surely result in the centre of gravity decelerating, but I haven't worked it out to my satisfaction. Can someone address this question with some actual calculations for a structural engineer that hasn't reviewed this sort of dynamic problem in years? Arguing this from intuition usually results in the discussion degrading into something similar to the plane/treadmill question, and physical tests seem to be too greatly influenced by unwanted factors.
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 danpilon54
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Re: Weight of an Hourglass
Well the impact and the effect of the weightless sand would not exactly cancel, as the impact force depends on the height the sand falls, while the weightless sand effect depends only on the mass of the sand and the rate of flow, so there will be some variation. Im not exactly sure how the math would work though. The growing sand pile however would have a negligible effect though because the rate the pile is growing is much slower than the speed of sand falling.
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Re: Weight of an Hourglass
There's also the factor of it having more weight because the sand at the base is closer to the Earth then the sand at the top, and so experiences a slightly larger gravitational pull.
At the very least, it would have less weight measured during the time between when the sand started flowing, and when it hit the bottom of the container.
At the very least, it would have less weight measured during the time between when the sand started flowing, and when it hit the bottom of the container.
Re: Weight of an Hourglass
I think it should be a different weight, since the centre of mass is moving down it will be heavier, just like if you stand on some scales and suddenly crouch.
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Re: Weight of an Hourglass
Let's consider an analogous thought experiment where the hourglass is placed on a spaceship so that the sands fall towards the thrusters at the back. We then produce 1g of acceleration. Does the falling sand get taken into account equally with the sitting sand when calculating how much fuel is used?
My explaination: The falling sand is moving slower relative to the rest of the ship, and therefore isn't experiancing the same acceleration, and therefore reduces the amount of fuel used. However, when the sand hits the pile at the bottom, it is suddenly accelerated with the rest of the ship, which must balance out whatever acceleration (and fuel usage) it didn't experiance while falling. The net effect is zero, but the fuel usage at any given timeslice is lower until the hourglass runs out.
My explaination: The falling sand is moving slower relative to the rest of the ship, and therefore isn't experiancing the same acceleration, and therefore reduces the amount of fuel used. However, when the sand hits the pile at the bottom, it is suddenly accelerated with the rest of the ship, which must balance out whatever acceleration (and fuel usage) it didn't experiance while falling. The net effect is zero, but the fuel usage at any given timeslice is lower until the hourglass runs out.
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Re: Weight of an Hourglass
frezik wrote:The net effect is zero, but the fuel usage at any given timeslice is lower until the hourglass runs out.
The center of mass has moved back in the ship, so the ship has effectively gone less distance, so the fuel has done less work, so less fuel was used. Total.
EDIT: That might have been what you were saying. I don't really get what you mean by some bits of your post.
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Re: Weight of an Hourglass
Macbi wrote:frezik wrote:The net effect is zero, but the fuel usage at any given timeslice is lower until the hourglass runs out.
The center of mass has moved back in the ship, so the ship has effectively gone less distance, so the fuel has done less work, so less fuel was used. Total.
EDIT: That might have been what you were saying. I don't really get what you mean by some bits of your post.
I was thinking of a similar situation where we vary the amount of fuel used to give exactly 1g of acceleration at all times. But it works equally well if we keep the fuel rate constant and measure how far the ship has gone.
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Re: Weight of an Hourglass
I'd run a simpler experiment using a model:
Use a carpet tube (like a toilet paper tube, but for carpet). Poke a very light but strong stick through it near the top, and put a few croquet balls above the stick. Balance it upright on the morning show host's head. If he complains that it's heavy, pull the stick out. He'll get a second or so of relief.
When he wakes up, have him revisit his results.
At time i, he's supporting the weight of the tube, the balls (and the nominal weight of the stick).
At time 1, he's supporting the tube only (the balls are in freefall).
At time f, he's supporting the tube and the balls again. Plus, he's got a hell of a headache.
Use a carpet tube (like a toilet paper tube, but for carpet). Poke a very light but strong stick through it near the top, and put a few croquet balls above the stick. Balance it upright on the morning show host's head. If he complains that it's heavy, pull the stick out. He'll get a second or so of relief.
When he wakes up, have him revisit his results.
At time i, he's supporting the weight of the tube, the balls (and the nominal weight of the stick).
At time 1, he's supporting the tube only (the balls are in freefall).
At time f, he's supporting the tube and the balls again. Plus, he's got a hell of a headache.
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Re: Weight of an Hourglass
I think for a slightest instant at the beginning and end of the freefall, the weight of the falling sand wouldn't be counted in the weight of the hourglass. This is when there is no sand at the bottom, not yet hitting, but there is sand in freefall. It occurs the second time when there is no sand in the top and the last bit of sand is in freefall.
Although, the question that follows is if the sand that is hitting the bottom is counteracting that weight that the weightless sand is giving up. I suspect that it doesn't, in standard hourglasses, at least, as they are not completely closed systems and the energy escapes into various places, such as the air and the glass. It might be possible, however, to evacuate an hourglass to eliminate the error.
I suppose this post didn't really get anywhere, but there are times when the weight is not equal to the definite weight.
Although, the question that follows is if the sand that is hitting the bottom is counteracting that weight that the weightless sand is giving up. I suspect that it doesn't, in standard hourglasses, at least, as they are not completely closed systems and the energy escapes into various places, such as the air and the glass. It might be possible, however, to evacuate an hourglass to eliminate the error.
I suppose this post didn't really get anywhere, but there are times when the weight is not equal to the definite weight.
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Re: Weight of an Hourglass
It's important to keep in mind that the center of mass is constant in some coordinate system for an isolated system, so as the sand moves down in the hourglass, the earth and the hourglass moves up in space relative to the common center of mass, and this motion gives rise to an acceleration against the direction of flow of sand. And where there is acceleration, there is force, so the net force exerted by the hourglass decreases ever so slightly.
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Re: Weight of an Hourglass
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 uncivlengr
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Re: Weight of an Hourglass
Yeah, it's been attempted a few times I think (not by me) and it seems like the hourglass doesn't have a chance to settle down before it runs out. The problem is you need a large enough flow to actually make the effect noticible, enough sand for it to run long enough to settle and come to a steady state flow, yet light enough that it doesn't damage the scale.Meteorswarm wrote:Can you determine this with an experiment? I think that'd be the simplest way to prove it. The sand grains probably don't weigh less than .01 grams, so just get a very precise scale, like you find in a chem lab.
The force depends on the height as well as the flow rate  a larger stream of sand would produce more impact force than a thin stream.danpilon54 wrote:Well the impact and the effect of the weightless sand would not exactly cancel, as the impact force depends on the height the sand falls, while the weightless sand effect depends only on the mass of the sand and the rate of flow, so there will be some variation.
The amount of weightless sand would also be determined by the height of the stream and the flow rate  it's basically determined by the volume of sand "in limbo" between the top and bottom.
The reason I think it has an effect is because the fall height is slowly decreasing from the bottom up  consider flow out of the top glass as Qout and flow into the bottom of the glass as Qout, and that Q is the flow rate, velocity times stream area, v*A[/i]. Now you might assume that Qin is equal to Qout, but actually, since the sand pile is growing at some velocity, vpile, Qout (the sand hitting bottom) is actually greater than Qin by vpile*A.danpilon54 wrote:Im not exactly sure how the math would work though. The growing sand pile however would have a negligible effect though because the rate the pile is growing is much slower than the speed of sand falling.
That means the centre of gravity isn't moving at a constant rate downwards, but decelerating, which means some additional force must be applied.
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 ThinkerEmeritus
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Re: Weight of an Hourglass
We have a physicist's hourglass, massless and perfectly cylindrical in a constant gravitational field with no significant friction anywhere despite friction being need to keep the rate of falling sand constant. Let the sand have mass, however, so we can analyze what is going on. Then F = Ma gives
F_{scale}  Mg = (d/dz)(Z_{CM})
with Z_{CM} being the height of the center of mass of the sand, since nothing else has weight. (Assuming that the hourglass is massless is inessential, because the mass of the hourglass doesn't affect our answer.) The question is whether
F_{scale} = Mg while the sand is falling.
Clearly when the sand begins falling and before the first of it hits the bottom of the glass, the sand is accelerated downward with no compensating collision force, so the force exerted by the scale is a bit less than Mg. So much for the rigor of the CBC morning show, but we expected that, right? More interesting is what happens after the sand is falling at a constant rate.
Say the top of the sand is a distance z above the divider at the top, and a distance y above the bottom in the lower part of the glass. Call the total mass of the pile of sand per unit of height Lambda. The centerofmass [CM] of the each part of the sand is halfway up the pile, and if the partition is a distance H above the bottom, the overall CM is
Z_{CM}(t) = Lambda z ((z/2)+H) +Lambda y(y/2)
= (1/2)Lambda z^{2} +zH + (1/2)Lambda y^{2}
and since there is a constant amount of sand, z+y=const. Then moving Lambda to the left for the moment
(d/dt) Z_{CM}(t)/Lambda = z dz/dt + H +y dy/dt
But since z+y = const, dy/dt =  dz/dt and
(d/dt) Z_{CM}(t)/Lambda = (zy)(dz/dt) + H
(d^{2}/dt^{2}) Z_{CM}(t)/Lambda = (dz/dt  dy/dt) = 2 (dz/dt)^{2}
with no further terms since dz/dt is constant in an hourglass and dH/dt = 0 unless the hourglass breaks apart.
Thus
(d^{2}/dt^{2}) Z_{CM}(t) = 2 Lambda (dz/dt)^{2}
Hmm, I didn't expect that since I had been told the same tale about this problem. However, the units are correct and the term on the right comes about basically because the top of the top layer is moving down and the top of the bottom layer is moving up, which changes the CM nonlinearly, so I think it is correct.
The acceleration of the CM is nonzero and the gravitational force on the sand is naturally constant, so the force exerted by the scale changes while the sand is falling. The amount it changes is a constant and probably pretty darn small, but it changes.
It is dangerous to get an answer different than received wisdom, even if the source of the wisdom is vague. Somebody please check this to be sure I got it right.
F_{scale}  Mg = (d/dz)(Z_{CM})
with Z_{CM} being the height of the center of mass of the sand, since nothing else has weight. (Assuming that the hourglass is massless is inessential, because the mass of the hourglass doesn't affect our answer.) The question is whether
F_{scale} = Mg while the sand is falling.
Clearly when the sand begins falling and before the first of it hits the bottom of the glass, the sand is accelerated downward with no compensating collision force, so the force exerted by the scale is a bit less than Mg. So much for the rigor of the CBC morning show, but we expected that, right? More interesting is what happens after the sand is falling at a constant rate.
Say the top of the sand is a distance z above the divider at the top, and a distance y above the bottom in the lower part of the glass. Call the total mass of the pile of sand per unit of height Lambda. The centerofmass [CM] of the each part of the sand is halfway up the pile, and if the partition is a distance H above the bottom, the overall CM is
Z_{CM}(t) = Lambda z ((z/2)+H) +Lambda y(y/2)
= (1/2)Lambda z^{2} +zH + (1/2)Lambda y^{2}
and since there is a constant amount of sand, z+y=const. Then moving Lambda to the left for the moment
(d/dt) Z_{CM}(t)/Lambda = z dz/dt + H +y dy/dt
But since z+y = const, dy/dt =  dz/dt and
(d/dt) Z_{CM}(t)/Lambda = (zy)(dz/dt) + H
(d^{2}/dt^{2}) Z_{CM}(t)/Lambda = (dz/dt  dy/dt) = 2 (dz/dt)^{2}
with no further terms since dz/dt is constant in an hourglass and dH/dt = 0 unless the hourglass breaks apart.
Thus
(d^{2}/dt^{2}) Z_{CM}(t) = 2 Lambda (dz/dt)^{2}
Hmm, I didn't expect that since I had been told the same tale about this problem. However, the units are correct and the term on the right comes about basically because the top of the top layer is moving down and the top of the bottom layer is moving up, which changes the CM nonlinearly, so I think it is correct.
The acceleration of the CM is nonzero and the gravitational force on the sand is naturally constant, so the force exerted by the scale changes while the sand is falling. The amount it changes is a constant and probably pretty darn small, but it changes.
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 uncivlengr
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Re: Weight of an Hourglass
Thanks, that's the sort of thing I was looking for, since trying to determine it intuitively is obviously not straightforward.
One thing I noticed: you defined z as the height of the top of the sand, so dz/dt would be a constant, not zero, as the pile height is decreasing at some constant rate.
The other thing is that you said "z+y = const, dy/dt =  dz/dt"
This assumes that the sand that's falling "in limbo" is negligible, but really that's one of the important aspects. The falling volume of sand is neither in 'z' or 'y' and is constantly changing, as I alluded to in my previous post.
One thing I noticed: you defined z as the height of the top of the sand, so dz/dt would be a constant, not zero, as the pile height is decreasing at some constant rate.
The other thing is that you said "z+y = const, dy/dt =  dz/dt"
This assumes that the sand that's falling "in limbo" is negligible, but really that's one of the important aspects. The falling volume of sand is neither in 'z' or 'y' and is constantly changing, as I alluded to in my previous post.
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 phlip
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Re: Weight of an Hourglass
Some quick number crunching says that if we don't take into account the rising height of the bottom pile (ThinkerEmeritus's magic hourglass, plus zerovolume sand), then the reduction in the force caused by the sand in freefall is exactly balanced by the increase in force from the sand hitting the bottom of the hourglass (both are the flow rate in kg/s, times the time it takes for the sand to fall to the bottom of the hourglass, times g). There's just a slight period of lower force when it starts, and another of higher force when it stops (and similar ones every time the flow rate changes... the weight of the freefall sand is proportional to the average flow rate over the last t seconds, but the weight of the landing sand is proportional to the instantaneous flow rate t seconds ago).
As for taking that rising pile into account, I'll take a different tack to ThinkerEmeritus, and hopefully come up with the same answer (Maths is good like that).
Let f(t) be the time that a piece of sand, landing at time t, has been falling for. That is, if a piece of sand falls through the aperture at t=0 and lands at t=1, then f(1)=1. This definition makes it easier to work with... as f(t) is directly related to the height of the lower pile at time t (the height of the pile before a particular piece of sand lands is irrelevant to how long that piece of sand falls for). So, the height between the aperture and the pile is h(t) = 0.5 g f(t)^{2}. [edit]Thanks James ScottBrown, added missing squared to formula[/edit]
We'll assume the sand is falling at a constant rate of Q kg/s (yeah, flow rate is typically m^{3}/s, replace all mentions of Q with [imath]Q\rho[/imath] if you really want to be pedantic). The next step will probably be to have a variable Q(t), but this will do for now.
Now, the weight of the sand in freefall is simply Q g f(t). This is the same as in the nobottompile case.
The weight of the landing sand is a little trickier though. The landing sand is landing with a velocity of v = g f(t), so if the flow rate was Q, then the force would simply be Q g f(t) again... but it's not. Instead, we have Q/v kg/m of sand at the bottom of our falling stream... and v+h'(t) of that stream is landing at any given time, so our landing flow rate is Q*(v+h'(t))/v... this is easier to see if you transform the coordinates so that either the falling sand or the rising ground is stationary at time t... the other will be moving at v+h'(t), but the mass of the stream will still be Q/v.
So our landing force is actually Q g f(t) + Q h'(t)/v g f(t)... which is larger than the weight of the freefalling sand by Q h'(t)/v g f(t). Were I not lazy, the next step would be to expand v and h'(t) in terms of f(t) (or in terms of h(t), which would be less fun with the square root, but probably more useful) but I am, so I won't.
Um... I'm not sure if that's the same result that ThinkerEmeritus got... we both said the scales would show very slightly above the rest weight, but as for how much... we used rather different variables, so I'm not sure they're equivalent.
Note, I was a little loose with the limits in there... there really should be some terms in dt^{2} and such, which would fall away later... but I'm pretty sure it all comes out right.
As for taking that rising pile into account, I'll take a different tack to ThinkerEmeritus, and hopefully come up with the same answer (Maths is good like that).
Let f(t) be the time that a piece of sand, landing at time t, has been falling for. That is, if a piece of sand falls through the aperture at t=0 and lands at t=1, then f(1)=1. This definition makes it easier to work with... as f(t) is directly related to the height of the lower pile at time t (the height of the pile before a particular piece of sand lands is irrelevant to how long that piece of sand falls for). So, the height between the aperture and the pile is h(t) = 0.5 g f(t)^{2}. [edit]Thanks James ScottBrown, added missing squared to formula[/edit]
We'll assume the sand is falling at a constant rate of Q kg/s (yeah, flow rate is typically m^{3}/s, replace all mentions of Q with [imath]Q\rho[/imath] if you really want to be pedantic). The next step will probably be to have a variable Q(t), but this will do for now.
Now, the weight of the sand in freefall is simply Q g f(t). This is the same as in the nobottompile case.
The weight of the landing sand is a little trickier though. The landing sand is landing with a velocity of v = g f(t), so if the flow rate was Q, then the force would simply be Q g f(t) again... but it's not. Instead, we have Q/v kg/m of sand at the bottom of our falling stream... and v+h'(t) of that stream is landing at any given time, so our landing flow rate is Q*(v+h'(t))/v... this is easier to see if you transform the coordinates so that either the falling sand or the rising ground is stationary at time t... the other will be moving at v+h'(t), but the mass of the stream will still be Q/v.
So our landing force is actually Q g f(t) + Q h'(t)/v g f(t)... which is larger than the weight of the freefalling sand by Q h'(t)/v g f(t). Were I not lazy, the next step would be to expand v and h'(t) in terms of f(t) (or in terms of h(t), which would be less fun with the square root, but probably more useful) but I am, so I won't.
Um... I'm not sure if that's the same result that ThinkerEmeritus got... we both said the scales would show very slightly above the rest weight, but as for how much... we used rather different variables, so I'm not sure they're equivalent.
Note, I was a little loose with the limits in there... there really should be some terms in dt^{2} and such, which would fall away later... but I'm pretty sure it all comes out right.
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Re: Weight of an Hourglass
EDIT: deleted for wrongness.
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Re: Weight of an Hourglass
uncivlengr wrote:Thanks, that's the sort of thing I was looking for, since trying to determine it intuitively is obviously not straightforward.
One thing I noticed: you defined z as the height of the top of the sand, so dz/dt would be a constant, not zero, as the pile height is decreasing at some constant rate.
The other thing is that you said "z+y = const, dy/dt =  dz/dt"
This assumes that the sand that's falling "in limbo" is negligible, but really that's one of the important aspects. The falling volume of sand is neither in 'z' or 'y' and is constantly changing, as I alluded to in my previous post.
z was supposed to be the height of the upper part of the sand above the barrier, and as you say dz/dt would be a nonzero constant. That is why I needed to use dy/dt =  dz/dt
The second point is a bit more subtle and I think I may have to do some more calculating. The falling sand can be ignored if it is a constant amount, and that is what I was doing. However, it looks to me as if the difference in the speed between the movement of the tops of the two piles due to different amounts of sand in transit ought to be a small correction to the speed of the piles themselves, so that the conclusion that the force changed a bit is not threatened.
There are other possible small corrections, of course. The sand has to spread out along the bottom layer and flow toward the center of the upper layer, and both motions involve kinetic energies that in principle ought to be somehow taken into account. Again, intuitively I think that effect is plenty small.
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Re: Weight of an Hourglass
Hi all,
I think we can simplify all these equations just by applying some mass and momentum conservation laws
We know that as we go from the starting state (sand at top), to the final state(sand at bottom), the centre of mass is lowered. In both the starting state and in the final state, the velocity of the centre of mass is zero.
The hourglass is a closed system (as far as mass goes) so we know three things about the centre of mass of the hourglass.
1. At some stage it must accelerate downwards
2. At some stage it must reach a maximum downwards velocity
3. At some stage it must decelerate.
In the first stage, the mass balance will read a lighter mass. (the spring it contains will apply less force in order to allow downwards acceleration)
In the second stage, the mass balance will read correctly.
In the third stage, the mass balance will read a heavier mass. (The spring it contains will apply more force in order to arrest the downwards motion)
So the real answer is, it depends when you measure it.
I think we can simplify all these equations just by applying some mass and momentum conservation laws
We know that as we go from the starting state (sand at top), to the final state(sand at bottom), the centre of mass is lowered. In both the starting state and in the final state, the velocity of the centre of mass is zero.
The hourglass is a closed system (as far as mass goes) so we know three things about the centre of mass of the hourglass.
1. At some stage it must accelerate downwards
2. At some stage it must reach a maximum downwards velocity
3. At some stage it must decelerate.
In the first stage, the mass balance will read a lighter mass. (the spring it contains will apply less force in order to allow downwards acceleration)
In the second stage, the mass balance will read correctly.
In the third stage, the mass balance will read a heavier mass. (The spring it contains will apply more force in order to arrest the downwards motion)
So the real answer is, it depends when you measure it.
 uncivlengr
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Re: Weight of an Hourglass
It's isn't constant amount, though  the pile is rising, and therefore the stream of sand shortens.ThinkerEmeritus wrote:The second point is a bit more subtle and I think I may have to do some more calculating. The falling sand can be ignored if it is a constant amount, and that is what I was doing. However, it looks to me as if the difference in the speed between the movement of the tops of the two piles due to different amounts of sand in transit ought to be a small correction to the speed of the piles themselves, so that the conclusion that the force changed a bit is not threatened.
As I mentioned earlier, this means that the rate of flow into the bottom pile is greater than the flow out of the top pile. If the height remained constant, then the flow out of the top would equal the flow into the bottom and dy/dt would equal dz/dt. Since the pile is rising at some constant velocity, the velocity of the sand flowing into the bottom pile is its velocity in reference to the hourglass plus the velocity of the pile upward. Therefore dy/dt is greater than dz/dt.
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 ThinkerEmeritus
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Re: Weight of an Hourglass
uncivlengr wrote:Since the pile is rising at some constant velocity, the velocity of the sand flowing into the bottom pile is its velocity in reference to the hourglass plus the velocity of the pile upward. Therefore dy/dt is greater than dz/dt.
Agreed, and I have been thinking about this effect off and on since you posted your comment. However, the amount of mass in transit is small compared to the total mass of sand, and the movement of its CM is of the same order of magnitude as the movement of the CM of the upper and lower piles. Therefore I think that the contribution of the movement of the falling sand is a small correction to apparent weight of the hourglass. The inevitable exception is that, as has been pointed out by various people, at the very beginning or at the very end of the flow of falling sand, the boundary of the moving sand moves much more and its effect is not so small.
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Re: Weight of an Hourglass
1) Some of the sand is in free fall
2) The sand exerts a force on the base of the hourglass
I think that 1 and 2 balance each other exactlyat least, if the effect of the growing sandpile in the base of the hour glass can be neglected. My justification for this was written before I read Phlip's post, but it seems similar.
Consider an hourglass containing an aperture through which sand flows at a rate of x kg per second; this aperture is a distance h above the base of the hourglass.
The sand takes [imath]\sqrt{\frac{h}{4.9}}[/imath]s to fall (by applying [imath]s=ut+\frac{1}{2}at^2[/imath] with [imath]u=0, g=9.8 ms^{2}[/imath]. Thus, the total mass of sand in the air at a given time is [imath]x \sqrt{\frac{h}{4.9}}[/imath] kg and, as [imath]W=mg[/imath], this has a weight of [imath]xg \sqrt{\frac{h}{4.9}}[/imath] Newtons.
By the time a grain of sand reaches the base of the hourglass, it has reached a velocity of [imath]9.8 \sqrt{\frac{h}{4.9}} ms^{1}[/imath] (as [imath]v=u+at[/imath]). The total momentum of all the sand hitting the base of the hourglass in one second is [imath]x \times g \sqrt{\frac{h}{4.9}} kg ms^{1}[/imath]. Applying Newton's Second Law ([imath]F=ma[/imath]) gives a total force of [imath]xg \sqrt{\frac{h}{4.9}}[/imath] Newtons.
Thus, the change in weight caused by effects (1) and (2) are equal in magnitude, and so cancel each other exactly.
Note that in this is not true for the first and last [imath]\sqrt{\frac{h}{4.9}}[/imath] seconds of sand falling; initially, no sand is striking the base of the hour glass; finally, the mass of sand falling is reduced. Also, air resistance has also been neglected (though, for a small grain of sand, this is probably reasonable).
Naturally, the analysis becomes more complicated if we consider that h is decreasing as the sand piles up beneath the aperture; I have not yet done this, but I think I have found an error in Phlip's calculation:
As [imath]s=ut+0.5at^2[/imath], I think you mean that "the height between the aperture and the pile is h(t) = 0.5 g [f(t)]^2"
2) The sand exerts a force on the base of the hourglass
I think that 1 and 2 balance each other exactlyat least, if the effect of the growing sandpile in the base of the hour glass can be neglected. My justification for this was written before I read Phlip's post, but it seems similar.
Consider an hourglass containing an aperture through which sand flows at a rate of x kg per second; this aperture is a distance h above the base of the hourglass.
The sand takes [imath]\sqrt{\frac{h}{4.9}}[/imath]s to fall (by applying [imath]s=ut+\frac{1}{2}at^2[/imath] with [imath]u=0, g=9.8 ms^{2}[/imath]. Thus, the total mass of sand in the air at a given time is [imath]x \sqrt{\frac{h}{4.9}}[/imath] kg and, as [imath]W=mg[/imath], this has a weight of [imath]xg \sqrt{\frac{h}{4.9}}[/imath] Newtons.
By the time a grain of sand reaches the base of the hourglass, it has reached a velocity of [imath]9.8 \sqrt{\frac{h}{4.9}} ms^{1}[/imath] (as [imath]v=u+at[/imath]). The total momentum of all the sand hitting the base of the hourglass in one second is [imath]x \times g \sqrt{\frac{h}{4.9}} kg ms^{1}[/imath]. Applying Newton's Second Law ([imath]F=ma[/imath]) gives a total force of [imath]xg \sqrt{\frac{h}{4.9}}[/imath] Newtons.
Thus, the change in weight caused by effects (1) and (2) are equal in magnitude, and so cancel each other exactly.
Note that in this is not true for the first and last [imath]\sqrt{\frac{h}{4.9}}[/imath] seconds of sand falling; initially, no sand is striking the base of the hour glass; finally, the mass of sand falling is reduced. Also, air resistance has also been neglected (though, for a small grain of sand, this is probably reasonable).
Naturally, the analysis becomes more complicated if we consider that h is decreasing as the sand piles up beneath the aperture; I have not yet done this, but I think I have found an error in Phlip's calculation:
So, the height between the aperture and the pile is h(t) = 0.5 g f(t).
As [imath]s=ut+0.5at^2[/imath], I think you mean that "the height between the aperture and the pile is h(t) = 0.5 g [f(t)]^2"
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Re: Weight of an Hourglass
Why neglect that effect? Unless your hourglass is growing as sand passes through it, the amount of sand in freefall is not constant.James ScottBrown wrote:I think that 1 and 2 balance each other exactlyat least, if the effect of the growing sandpile in the base of the hour glass can be neglected.
Certainly, but the question is from a theoretical standpoint if it can be mathematically proven that there is a correction at all, however small.ThinkerEmeritus wrote:Therefore I think that the contribution of the movement of the falling sand is a small correction to apparent weight of the hourglass.
"Small" is, of course, a relative term  the effect is governed by the speed at which the pile at the bottom of the hourglass rises relative to the flow from the top. Therefore, an hourglass with a "bulb" diameter comparable to the diameter of the "neck" would experience this effect to a greater extent.
The extreme case would be if the diameter of the lower bulb was equal to that of the neck  you'd essentially have a funnel with the end closed off, and the height of the pile would grow very quickly in relation to the rate of flow. Additionally, If the diameter of the upper bulb was very large relative to the diameter of the neck and lower bulb, then your dz/dt would by very small, and your dy/dt very large.
I think at this point the next step is to determine an expression for the rate at which the pile at the bottom rises as a function of the flow leaving the top  that is, an expression for dy/dt, as it's not simply a constant, nor equal to dz/dt.
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Re: Weight of an Hourglass
ThinkerEmeritus wrote:Z_{CM}(t) = Lambda z ((z/2)+H) +Lambda y(y/2)
= (1/2)Lambda z^{2} +zH + (1/2)Lambda y^{2}
and since there is a constant amount of sand, z+y=const. Then moving Lambda to the left for the moment
(d/dt) Z_{CM}(t)/Lambda = z dz/dt + H +y dy/dt
Should the RHS not be
z.dz/dt + H.dz/dt + y.dy/dt ?
Leading to
(d/dz)Z_{CM} = Lambda.(z  y + H)
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Re: Weight of an Hourglass
uncivlengr wrote:Why neglect that effect?James ScottBrown wrote:I think that 1 and 2 balance each other exactlyat least, if the effect of the growing sandpile in the base of the hour glass can be neglected.
The simplification is primarily to allow easier calculation of a limiting case of the problem.
uncivlengr wrote: Unless your hourglass is growing as sand passes through it, the amount of sand in freefall is not constant.
There are cases that are very close to being exceptions: an hourglass that has a very widebase in relation to the volume of sand (so that the sand layer always remains very thin), or is very tall (so that the thickness of the sand layer is small in relation to the height the sand must fall). Each half of an hourglass is approximately conical, and if the angle at the apex is sufficiently great, the first case could be approximately true.
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Re: Weight of an Hourglass
James ScottBrown wrote:I think I have found an error in Phlip's calculation:So, the height between the aperture and the pile is h(t) = 0.5 g f(t).
As [imath]s=ut+0.5at^2[/imath], I think you mean that "the height between the aperture and the pile is h(t) = 0.5 g [f(t)]^2"
Bah... yeah, yeah, typo on my part.
But still, I don't end up actually using that formula in my calculations:
phlip wrote:Were I not lazy, the next step would be to expand v and h'(t) in terms of f(t) (or in terms of h(t), which would be less fun with the square root, but probably more useful) but I am, so I won't.
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Re: Weight of an Hourglass
Ended wrote:ThinkerEmeritus wrote:Z_{CM}(t) = Lambda z ((z/2)+H) +Lambda y(y/2)
= (1/2)Lambda z^{2} +zH + (1/2)Lambda y^{2}
and since there is a constant amount of sand, z+y=const. Then moving Lambda to the left for the moment
(d/dt) Z_{CM}(t)/Lambda = z dz/dt + H +y dy/dt
Should the RHS not be
z.dz/dt + H.dz/dt + y.dy/dt ?
Yep. I missed that, and H disappeared and the units problem disappeared with it. The next line should be
(d/dt) Z_{CM}(t)/Lambda = (zy+H)(dz/dt)
Then
(d^{2}/dt^{2}) Z_{CM}/Lambda = (dz/dt  dy/dt)(dz/dt) + (...)d^{2}z/dt^{2}
and the last term vanishes because dz/dt = constant. As a result, my final result is
(d^{2}/dt^{2}) Z_{CM} = 2 Lambda (dz/dt)^{2}
which by good fortune is unchanged. Thanks for the correction, however.
Edit: The other question that has been raised is why I am willing to neglect the effect of the falling sand (except of course at the beginning and end of the cycle) when the effort is to determine if there is any change to Mg. The reason is that I believe the effect of the (morequckly) falling part of the sand is small compared to my already nonzero term due to the almost stationary collections of sand at the top and bottom of the hour glass. Thus I don't need it in order to prove that a nonzero effect is present.
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Re: Weight of an Hourglass
Take an elevator, and leave its doors open. Add rockets, with some sort of instantaneous masscompensation device, such that they will always cause an acceleration of 1g. And let's put the thing in space.
Introduce a payload to represent the sandbut let's keep it simple. The payload will be heavy and singularrepresenting just one grain, with amplified effect. We'll use a medicine ball (that doesn't bounce).
Now, set up your elevator in space. Place the medicine ball just outside the elevator. Climb on, and fire the rockets. You'll feel an acceleration that's as good as anything gravity can muster. You'll note that the ball appears to be in free fall, and as you continue accelerating, the ball just "falls" faster and faster. We know this is only perspectivewere we at the ball, we'd be seeing the rockets accelerate the elevator.
Set the elevator up again, but this time, put the ball just inside the elevator. Fire the rocketsand you'll see the same exact thingwhile the ball is "falling". The fact that it is within the confines of the elevator makes it no different than the previous experimentit's just a ball in space that looks like it is falling.
Once the ball lands, however, the rocket's masscompensation device must push harder to get the 1g. Before, it was accelerating only the elevator. At the moment the ball hits the elevator floor, it is accelerating both the elevator and the medicine ball. So it must push harder. But note that there was a time at which the only thing accelerating was the elevatorthis being the time during which the medicine ball appears to you to be falling.
Again, set up the experiment, and fire the rockets. Now, "freeze", as soon as the rockets fire. You are on a system now that mimics earth's gravity. The medicine ball is at the top of the elevator. It has potential "gravitational" energy. As nothing is holding it in place, the potential energy will be converted fairly quickly to kinetic energy. If you're thinking in terms of the elevatorball system, then, the ball experiences a force compensated by an equal and opposite force of the elevator, until it hits the floor (expending just this amount of energy). Once the ball hits the ground, you could restore the state of the elevator pretty easilyjust grab the ball and push it up. As you are pushing the ball up, the ball pushes backand the elevator experiences an equal and opposite force.
Once more, set up the experimentonly this time, don't fire the rockets. Instead, just strap yourself to the floor. Now manually pull the medicine ball from the top to the bottom. The elevator will move up as you do so. Once the medicine ball is at the bottom, it will remain "still". Push it back up, all the way to the top, and the elevator will move back to where it was.
Each grain of sand is a medicine ball. The physics work the same when you close the elevator doors.
At least, this is how I picture it.
Introduce a payload to represent the sandbut let's keep it simple. The payload will be heavy and singularrepresenting just one grain, with amplified effect. We'll use a medicine ball (that doesn't bounce).
Now, set up your elevator in space. Place the medicine ball just outside the elevator. Climb on, and fire the rockets. You'll feel an acceleration that's as good as anything gravity can muster. You'll note that the ball appears to be in free fall, and as you continue accelerating, the ball just "falls" faster and faster. We know this is only perspectivewere we at the ball, we'd be seeing the rockets accelerate the elevator.
Set the elevator up again, but this time, put the ball just inside the elevator. Fire the rocketsand you'll see the same exact thingwhile the ball is "falling". The fact that it is within the confines of the elevator makes it no different than the previous experimentit's just a ball in space that looks like it is falling.
Once the ball lands, however, the rocket's masscompensation device must push harder to get the 1g. Before, it was accelerating only the elevator. At the moment the ball hits the elevator floor, it is accelerating both the elevator and the medicine ball. So it must push harder. But note that there was a time at which the only thing accelerating was the elevatorthis being the time during which the medicine ball appears to you to be falling.
Again, set up the experiment, and fire the rockets. Now, "freeze", as soon as the rockets fire. You are on a system now that mimics earth's gravity. The medicine ball is at the top of the elevator. It has potential "gravitational" energy. As nothing is holding it in place, the potential energy will be converted fairly quickly to kinetic energy. If you're thinking in terms of the elevatorball system, then, the ball experiences a force compensated by an equal and opposite force of the elevator, until it hits the floor (expending just this amount of energy). Once the ball hits the ground, you could restore the state of the elevator pretty easilyjust grab the ball and push it up. As you are pushing the ball up, the ball pushes backand the elevator experiences an equal and opposite force.
Once more, set up the experimentonly this time, don't fire the rockets. Instead, just strap yourself to the floor. Now manually pull the medicine ball from the top to the bottom. The elevator will move up as you do so. Once the medicine ball is at the bottom, it will remain "still". Push it back up, all the way to the top, and the elevator will move back to where it was.
Each grain of sand is a medicine ball. The physics work the same when you close the elevator doors.
At least, this is how I picture it.
Re: Weight of an Hourglass
Focusing on Fallible's breakdown:
At any given point in time a certain fraction of the mass of the sand is moving downward at some speed. At the beginning nothing is moving, and as the stream grows more and more of the sand starts to fall. This is stage 1. Once the stream hits the bottom, then if we make certain assumptions of perfection, over the course of each second, a particular amount of sand moves from the top of the pile in the top chamber to the top of the pile in the bottom chamber. I believe that this means that we have reached stage 2. Once the top empties, then the remaining column shrinks as it falls and we have reached stage 3.
If, instead of perfection, we assume that pressure caused by the sand in the top chamber affects the flow rate, then we jump from stage 1 to stage 3 when the sand starts to hit the bottom.
Fallible wrote:1. At some stage it must accelerate downwards
2. At some stage it must reach a maximum downwards velocity
3. At some stage it must decelerate.
At any given point in time a certain fraction of the mass of the sand is moving downward at some speed. At the beginning nothing is moving, and as the stream grows more and more of the sand starts to fall. This is stage 1. Once the stream hits the bottom, then if we make certain assumptions of perfection, over the course of each second, a particular amount of sand moves from the top of the pile in the top chamber to the top of the pile in the bottom chamber. I believe that this means that we have reached stage 2. Once the top empties, then the remaining column shrinks as it falls and we have reached stage 3.
If, instead of perfection, we assume that pressure caused by the sand in the top chamber affects the flow rate, then we jump from stage 1 to stage 3 when the sand starts to hit the bottom.
Re: Weight of an Hourglass
ThomasS wrote:If, instead of perfection, we assume that pressure caused by the sand in the top chamber affects the flow rate, then we jump from stage 1 to stage 3 when the sand starts to hit the bottom.
Of course, the devil of this question is all in the asusmptions. However I think we can dismiss the problem of the pressure in the top chamber effecting the low rate. The reason I believe this is valid, is that hourglasses are useful precisely because the flow rate is {edit: mostly} independent of the amount of remaining sand in the upper chamber.
I think the biggest assumption to consider is the shape of the hourglass and the shape of the pile of sand. As others have already pointed out, assumptions about how the sand spreads out will drastically affect calculations. But even if you assume some kind of sand with a very low critical angle (can only form a very flat peak) and assume cylindrical upper and lower bulbs, then we still have the amount of mass in free fall changing as the lower pile rises.
Of course, we can get also get really technical here...
Wikipedia wrote:In the physical sciences, weight is a measurement of the gravitational force acting on an object.
'Wikipedia wrote:In science and technology "weight" has primarily meant a force due to gravity
All this talk about mass balances is actually irrelevant. Weight is a force, if you want to measure a force statically (as in with a mass balance), the object you are measuring cannot be accelerating!
Weight is simply mass times the local gravitational constant. Both these obviously remain the same, so the weight doesn't change. Of course, if you place this thing on a mass balance, it may indeed read a different mass, but that will be because part of the weight has stopped pushing against the internal spring of the balance, and has started accelerating the hourglass.
So, if we're talking about mass balances, then all of the previous analysis is relevant. If we're talking about weight, then the answer is very simple and the host of the show had it exactly right.
Re: Weight of an Hourglass
Fallible wrote:ThomasS wrote:If, instead of perfection, we assume that pressure caused by the sand in the top chamber affects the flow rate, then we jump from stage 1 to stage 3 when the sand starts to hit the bottom.
Of course, the devil of this question is all in the asusmptions. However I think we can dismiss the problem of the pressure in the top chamber effecting the low rate. The reason I believe this is valid, is that hourglasses are useful precisely because the flow rate is {edit: mostly} independent of the amount of remaining sand in the upper chamber.
Have you ever watched an hourglass? A 3 minute hourglass has sufficient sand to last one minute, and that is what makes it useful. They do not have graduations on the side. I think there is a reason for this. Also, I have seen (inexpensive, toy) hourglasses which have a visibly higher flow rate at the start than at the end.
Re: Weight of an Hourglass
A A Mills, S Day and S Parkes; European Journal of Physics, 17, 97109 (1996) wrote:The rate of flow is independent of height in the reservoir, except over the last few cm.
You can view the above article at doi link but you need a subscription. It's actually an interesting read.
Having a fixed volume of particulate is only useful if the flow rate is stable with respect to a variety of variables. True, having the flow rate independant of resevoir height is not strictly necessary, but a linear process is generally less susceptible to perturbations than a system with nonlinear characteristics.
But all of this is a side issue, as the weight does not change at all.
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Re: Weight of an HourglWeight ass
What we want to do is measure the 'apparent' weight of a flowing hourglass  to determine if the force between the hourglass and whatever it's sitting on changes at all. Whether or not there's any acceleration throughout the time that it's flowing is the same question, but it's more difficult to see that intuitively, because the actual container that constitutes the hourglass is not accelerating at all.Fallible wrote:All this talk about mass balances is actually irrelevant. Weight is a force, if you want to measure a force statically (as in with a mass balance), the object you are measuring cannot be accelerating!
The original question, since you brought it up, is if there are two hourglasses sitting on each end of a scale and one is turned over to start flowing, what will happen. I think it's pretty clear what the problem is as I phrased it, though, without resorting to semantics.
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Re: Weight of an Hourglass
uncivlengr wrote:The original question, since you brought it up, is if there are two hourglasses sitting on each end of a scale and one is turned over to start flowing, what will happen.
Thanks for clarifying. Normally I hate to be a pain on semantics, but I feel that with problems as intuitively diffucult as this its important to use precise language. Too often I've been in arguments where people are working from different definitions. Those arguments usually go around in circles until its discovered that they agreed all along.
uncivlengr wrote:because the actual container that constitutes the hourglass is not accelerating at all
The container will be accelerating at some points in time. As the sand starts to fall, there will be less weight acting on the container, but it will still experience the temporarily experience the same force from the mass balance until the mass balance comes back into equilibrium.
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Re: Weight of an Hourglass
It might be easier to see if one simplifies the problem.
1. Make the sand into one big glob, say, a 1 Kg ball.
2. Make the glass into a cylinder (effectively making the hole large enough for the ball to pass through).
3. Evacuate all the air in the cylinder (no air friction).
4. Make the cylinder have a mass of 1 Kg.
Initially attach the ball to the top of the cylinder. To get the effect of "turning the hour glass over", detach the ball. During the time the ball is falling, it is no longer attached to the cylinder and is no longer part of its mass.
Related thought experiment: Put the cylinder with attached ball in space moving at some speed in the direction of the top of the cylinder. If one detaches the ball, it doesn't move relative to the cylinder due to inertia. If one then applies a force on the bottom of the cylinder to accelerate it at a constant acceleration, only the cylinder accelerates until the ball reaches the bottom. The force during that time is only applied to the cylinder (f = a * 1Kg). Ergo, during the time the ball is "falling" inside the cylinder, it contributes nothing to the cylinder's weight. When the ball hits the bottom, the force required doubles (f = a * 2Kg).
1. Make the sand into one big glob, say, a 1 Kg ball.
2. Make the glass into a cylinder (effectively making the hole large enough for the ball to pass through).
3. Evacuate all the air in the cylinder (no air friction).
4. Make the cylinder have a mass of 1 Kg.
Initially attach the ball to the top of the cylinder. To get the effect of "turning the hour glass over", detach the ball. During the time the ball is falling, it is no longer attached to the cylinder and is no longer part of its mass.
Related thought experiment: Put the cylinder with attached ball in space moving at some speed in the direction of the top of the cylinder. If one detaches the ball, it doesn't move relative to the cylinder due to inertia. If one then applies a force on the bottom of the cylinder to accelerate it at a constant acceleration, only the cylinder accelerates until the ball reaches the bottom. The force during that time is only applied to the cylinder (f = a * 1Kg). Ergo, during the time the ball is "falling" inside the cylinder, it contributes nothing to the cylinder's weight. When the ball hits the bottom, the force required doubles (f = a * 2Kg).
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Re: Weight of an Hourglass
To describe the middle part, while the hourglass has water flowing down it...
Take a balanced lever, and a device that moves the lever that is perfectly efficient.
Have an hourglass that drops sand 1 meter at a rate of 10^3 kg per second.
Move the hour glass up slowly enough (with infinitesimal acceleration), and you can work out how much work is done by moving mass upwards.
Something like:
[(mass of hour glass)  (10^3 kg m/s) * time_spent_lifting] * force of gravity * distance
is the energy burned in lifting the hour glass (which is supposed to flow for the entire period of lifting).
Take a balanced lever, and a device that moves the lever that is perfectly efficient.
Have an hourglass that drops sand 1 meter at a rate of 10^3 kg per second.
Move the hour glass up slowly enough (with infinitesimal acceleration), and you can work out how much work is done by moving mass upwards.
Something like:
[(mass of hour glass)  (10^3 kg m/s) * time_spent_lifting] * force of gravity * distance
is the energy burned in lifting the hour glass (which is supposed to flow for the entire period of lifting).
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Re: Weight of an Hourglass
I think there are a couple of errors that have been permeate this thread.
 1. The sand is not in free fall. The mass of the grains of sand is small enough that air resistance should be taken into account.
2. Weight depends only on the mass and the acceleration due to gravity. When you use a scale to measure "weight", you are actually measuring the normal force required to keep an object's acceleration at 0.
82.7% of statistics are made up on the spot.
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Re: Weight of an Hourglass
That won't actually affect the result. Some sand will always be accelerating and its weight will consequently be absent from the external scale. The weight of the portion of sand that reaches terminal velocity will be exerted on the scale, but its impact force will be proportionally reduced, cancelling out the effect.IHOPancake wrote:1. The sand is not in free fall. The mass of the grains of sand is small enough that air resistance should be taken into account.
If air resistance bothers you, you can either assume the hourglass is evacuated, or assume the distance the sand has to fall is shorter than the distance it takes to reach terminal velocity.
In short, the answer to the question  does the apparent weight of the hourglass change  isn't affected by air resistance.
In this case, we're keeping the 'container' of the hourglass at rest, but the sand isn't at rest  the scale can only measure the force applied by the hourglass system, not its mass.IHOPancake wrote:2. Weight depends only on the mass and the acceleration due to gravity. When you use a scale to measure "weight", you are actually measuring the normal force required to keep an object's acceleration at 0. [/list]
Since we're all hung up on my poor description of the problem, let's rephrase it:
There's a set of scales with two identical hourglasses sitting on each side. Both hourglasses are evacuated entirely of any air. With the scale is restrained from moving, one hourglass is flipped over, starting the flow of sand. Only after the sand has begun to accumulate at the bottom of the hourglass is the scale released.
Can the behaviour of the balance of the scales as one hourglass continues to flow be quantitatively predicted?
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Re: Weight of an Hourglass
If you want to see what happens to the reading on a regular scale when an hourglass is set on it, all you need to look at is the height of the center of mass of the hourglass as a function of time. The momentum of the hourglass is proportional to the first derivative of this with respect to time. The deviation of the reading on the scale is proportional to the second derivative.
It's not hard to imagine all the possible situations: where the hourglass is heavier (if its momentum downwards is decreasing because, say, the flow is getting weaker with time, or the segment of falling sand is getting shorter); where the hourglass is the same weight (if its momentum downwards is steady, with a steady flow); and where the hourglass is lighter (if its momentum downwards is increasing because the flow is speeding up).
The most likely of these, at a given time is, I'd say, the first. However, it's not hard to see that all three must happen at some point in the process. In fact, the average (with respect to time) deviation from the hourglass's "usual weight" is zero, because its momentum is the same before and after the process.
For the situation with the balance, the whole hourglassesbalance system has some angular momentum clockwise around the pivot when the balance is released. There is no net torque on the system. (I'm assuming here that the hourglasses are constrained to remain vertical.) So, assuming it doesn't bump into anything before then, once the hourglass runs out, the thing will be pivoting clockwise.
It's not possible to predict exactly what will happen before then, unless you make some assumptions about the way the hourglass behaves. Again, chances are that for most hourglasses after the sand gets going in the first second or so, the flow will only slow down, and so the scale will start to pivot faster and faster in the clockwise direction. (Or perhaps the flow will stay steady until the sand runs out, in which case the scale will stay still until that time, and then start pivoting clockwise.) But if the flow speeds up at some point, the thing will be pivoting counterclockwise at that time.
It's not hard to imagine all the possible situations: where the hourglass is heavier (if its momentum downwards is decreasing because, say, the flow is getting weaker with time, or the segment of falling sand is getting shorter); where the hourglass is the same weight (if its momentum downwards is steady, with a steady flow); and where the hourglass is lighter (if its momentum downwards is increasing because the flow is speeding up).
The most likely of these, at a given time is, I'd say, the first. However, it's not hard to see that all three must happen at some point in the process. In fact, the average (with respect to time) deviation from the hourglass's "usual weight" is zero, because its momentum is the same before and after the process.
For the situation with the balance, the whole hourglassesbalance system has some angular momentum clockwise around the pivot when the balance is released. There is no net torque on the system. (I'm assuming here that the hourglasses are constrained to remain vertical.) So, assuming it doesn't bump into anything before then, once the hourglass runs out, the thing will be pivoting clockwise.
It's not possible to predict exactly what will happen before then, unless you make some assumptions about the way the hourglass behaves. Again, chances are that for most hourglasses after the sand gets going in the first second or so, the flow will only slow down, and so the scale will start to pivot faster and faster in the clockwise direction. (Or perhaps the flow will stay steady until the sand runs out, in which case the scale will stay still until that time, and then start pivoting clockwise.) But if the flow speeds up at some point, the thing will be pivoting counterclockwise at that time.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
 phlip
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Re: Weight of an Hourglass
IHOPancake wrote:Weight depends only on the mass and the acceleration due to gravity. When you use a scale to measure "weight", you are actually measuring the normal force required to keep an object's acceleration at 0.
Fine, $this_thread =~ s/weight/apparent weight/gi;
Personally, I've heard the term "weight" (without qualifiers) to mean both apparent weight, and mg... and when there's actually a difference between the two, it was usually apparent weight that was meant.
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Re: Weight of an Hourglass
What happened to this thread? I think this problem is very interesting and I wish there will be some more progress.
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