Is weight of hourglass diminished when sand is pouring?

Question

http://upload.wikimedia.org/wikipedia/commons/thumb/1/11/Wooden_hourglass.jpg/277px-Wooden_hourglass.jpg

Is weight diminished by amount of sand in fall at any given moment of time?

Answer 1

Yes.

The center of mass of the hourglass has dropped. It is just like standing on a scale with a weight. Drop the weight and you weigh less -- that is until it hits your toe, at which point the scale well read a short term increase in weight caused by the recoil effect (change in momentum) and then center back at the original combined weight of you and the weight.

With the hourglass, I think that this effect will only apply during the short time during which the sand initially begins to fall. Once the first grains hit the bottom of the hourglass, their momentum should be enough to bring the scale back up to normal. That is until the last grains run out, at which point the scale will read an increase in weight until the last grain hits the surface. (This increase in weight is caused by the momentum transfer as the grains continue to hit the bottom surface with no offsetting release of grains from the funnel of the hourglass).

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Edward (below) always makes it interesting. (I gave him his usual thumbs up). He inspired further thought. This is actually a really easy problem. My initial intuition was correct. But Edward helped me see clearly how to attack it.

The easy way to solve this problem involves watching the center of gravity.

Case 1: From the moment the first grain of sand leaves, until it hits the bottom, the center of gravity of the sand will be accelerating downward. This means the scale will weigh less.

Case 2: We now have a steady state. The center of gravity has a constant velocity downward which is based on the flow of sand. The scale will measure the original weight before we turned the hourglass

Case 3: The last sand grain has now slipped through the funnel and begun its path down. Although it still has a velocity downward, the center of gravity starts accelerating upward -- that is until the last grain comes to rest at which point the acceleration stops, and the velocity of the center of gravity is zero. During this time, weight would go up.

Setting it up like this, the math would be pretty simple.

What a beautiful problem!

Answer 2

If you were to put an hour glass on a very sensitive scale and measure the weight while all of the sand was on the bottom of the hour glass, and then invert the hour glass you would see a reduction of weight in the hour glass due to the fact that the hour glass would not be supporting the weight of any sand in mid air. However, the sand falling has a very small amount of kinetic energy that would be transferred back into the hourglass when the sand lands on the bottom of the hourglass. I would imagine that this kinetic energy would be largely lost due to horizontal the cushioning effect that the sand pile on the bottom would have. So to answer your question, the hourglass would have a reduced weight while the sand was falling. Excellent question.

Answer 3

This is an interesting question and I am not sure. There are two things going on:
1) when a particle falls wihtin the container, it is not exerting any force on the container so it can not contribute to the weight.
2) when the particle lands, it is exerting force due to it's mass and force due to its kinetic energy, 1/2mv^2.

As is often the case in science, the answer may be "it depends". This would be a good one to actually perform the measurement and see. The practical problem would be finding a scale with enough precision to measure the difference.

Answer 4

I agree that you are measuring the contents of a closed container but while the sand is in free fall the only force acting on it is gravity. Seeing as the sand can only affect the weight of the hourglass while it is in contact with the hourglass I would say that it would slightly decrease the weight. Of course while the hourglass is counting down the amount of sand actually in free fall is so small that the change will be barely noticeable.

Answer 5

No, not if the the stand is flowing at a steady rate. The impulse (force mg times time T) temporarily lost to the net weight of the system due to a grain of mass m falling for time T in gravity field g equals the momentum transferred (mass m times velocity gT) to the pile when a similar grain that has finished its fall at the same time hits the pile. The effects cancel.

Answer 6

Like sands through the hourglass These are the times to remember. Not letting them fall from our mind But keeping them in there forever.

Answer 7

Alexander, Alexander,
What a massive hourglass you have there!

Let me get to the point

I need some magic to answer this question.

Case1 "Sand just started to flow"
Under influence of the magic the weight measuring equipment (WME) is well stabilized and sand begins to fall at my command. My WME response instantaneously with every fallen grain until the first grain hits the bottom of the hourglass.

m(t)g= Mg-(dm/dt)tg
M-total mass of the sand
dm/dt - rate of flow of sand
t= sqrt(2h/g)
g- is 9.81 m/s^2
h - height form release to contact (it is function of time but not now)
Case 2 "The sand is flowing"
The case one is still applicable, but the sand dissipates its energy by hitting the glass below at first and then a pile of sand. I probably should break up this case into parts A (sand **** glass) and part B sand hits a pile of sand due to averaging of momentum but why spit hair. We have

m(t)g= Mg -(dm/dt)tg + mdV/dt/g - W'
mdV/dt - force due to change of momentum
mdV/dt /g - instantaneous mass 'gain' due to momentum
W' - instantaneous mass 'loss' due to energy dissipation like heat...

Case 3. "Last grain is airborne"


m(t)g= M'g + (dm/dt)tg+mdV/dt/g - W'
M'- amount of sand less these few grains in the air
t= sqrt(2h/g)
g- is 9.81 m/s^2
h - height form release to contact with the pile of sand.

In the end
M(start)=M(end)

I rest my case

Answer 8

An interesting question, but no. The sand remains within the container, and because you are measuring the entire system inside the closed container, the weight remains the same.