Floating hour glass
This is an old mystery. People have argued about the solution. It was published in Omni Magazine back in the '60. You can see it and read about it. But there's little argeement about the "why".
An hour glass containing air and sand is in a cylinder of water. While sand is flowing from the top to the bottom compartment, the hour glass remains submerged at the bottom of the cylinder, but when the sand has completed its downward journey, the hour glass floats to the top. Why?
An hour glass containing air and sand is in a cylinder of water. While sand is flowing from the top to the bottom compartment, the hour glass remains submerged at the bottom of the cylinder, but when the sand has completed its downward journey, the hour glass floats to the top. Why?
Labels: funphysics
posted by Ragknot at
6:48 PM
22 Comments:
Too easy, but I like it.
The falling sand holds it down.
When the sand stops falling, there's less weight to hold it down, so it floats to the top.
Anonymous, good answer. But that might not be the only reason.
Ragknot, is the cylinder only a little wider than the hourglass?
From the pictures I've seen, it appears that the cylinder is about 1.5 times the diameter of the hour glass.
If you want to read more about this
http://www.wiskit.com/marilyn/hourglass.html
It's as I thought. Personally I like Anonymous's explantion best because it is more sophiticated. On the link you gave, I didn't notice anyone give a good correct explanation of how the falling sand caused the downward force ("A Plausible Explanation" was the closest). If the sand is falling at rate m kg/sec, and through a distance h (which reduces as sand collects at the bottom of the hourglass) then it will gain a speed v=sqrt(2gh). So momentum is being delivered to the top of the sand at bottom of the glass at a rate (mv)v = mv². By Newton Second Law, a force mv² is therefore being applied.
The combined weight of the hourglass plus all the sand is insufficeinet by itself to keep the hourglass down. But the additional mv² overcomes the buoyancy, until there is no more sand to fall, so mv² drops to 0.
The other possibility, which is the more likely one (in practice), is that (in a constriced width tube) that the top heavy hourglass will be at an angle, and so provide a frictional force. When the sand runs low, the hourglass should be able to become upright, and so remove the friction mechanism.
I note the very last entry on that page. To stay at the top and then fall to the bottom can only be explained by friction.
...or invisible aliens.
Ragknot, that was a good question. Thanks.
due to gravity... THE hour glass was able to submerge down to the cylinder of water because the sand keeps the air passing through the tube. then the hour slowly floats up because there's an increase in volume of the air at the upper part of the hour glass.. so that means the sand was completely finished flowing down...
but 1 question... was the hour glass did not turn upside down??? it might turn upside down because of the presence of air.. the direction of the hourglass might become upwards not downwards...
Hi scrapped prince. The size of the hourglass is constant, the mass is also nearly constant. The mass changes a little because the sand that is in freefall doesn't contribute. I'm assuming that the mass of any air inside the hour glass is entirely negligible.
The most important thing about the distribution of the sand and air is that initially the hour glass is top-heavy, so it leans at an angle (touching the sides of the tube), then it becomes bottom heavy and so can then become upright. The less important thing (probably) is the rate at which the sand is falling (as discussed in my post above).
I would guess the hour glass could begin to rise almost any time after half the sand had fallen. From the problem description, it sounds like the system has been carefully designed to ensure that most of the sand needs to fall first.
If the tube was wider, then the hourglass would turn upside down at the very start, and so it would stay at the top.
... last sentence last post should have ended "and so goes straight to the top".
I hadn't meant to include the first paragraph, I was thinking aloud.
No-one is paying attention I see. Only two errors missed:
1. The force is mv, not mv².
2. The weight of the sand in freefall exactly compensates that mv force, except for when sand flow begins and ends.
Therefore, only friction matters.
kicking my last post through.
I agree with the "friction / top heavy / leaning". I can't think of any other way this would work.
I wonder if the hour glass is in a wooden frame. This softer material would provide more friction than glass. We don't know that the cylinder is glass or plastic. I have assumed it's glass. Wood to glass would give more friction than glass to glass.
Even when I problem is past it's prime, I keep on thinking about it. I suddenly realised that I hadn't thought about the absence of the weight of the sand in free fall, did some calcs and found the nont surprising result that the moving sand was irrelevant. Shame, I loved the idea.
I'd like to see one in real-life because even the friction ide has begun to stretch my credulity a little, but not too much. Greetz.
Chris,
Free Fall?
...reminds me of a truck of birds.
The question was about the weight of the truck when all the birds are in the air vs. the weight when none were in the air. If I remember right, the answer was, they weighed the same. And the logic was the birds were supported by the air moving down by the wings.
When a spec of sand is falling, it is resisted by the air, so the total weight is the same. I certainly don't know for sure.
But I did make a protest that if the truck had a mesh wire sides, the air outside the truck would bear some of the weight, as the air pressure equalize inside and outside the truck. Then the truck would be lighter when the birds were in flight.
It would be interesting to see an actual test. But I believe that in a sealed hourglass, the weight would be the same, with or without the sand falling.
Hi. In the case of the hourglass I think it's probably simplest to think of it having no air. I'm pretty sure that there will be a tiny variation and smooth variation in the weight (not mass) of the hourglass as the first section of sand starts falling, and similarly for the last section of sand. I can't quite decide the exact shape of the weight v time variation is though. The point of all that was, you can have transitory effects, but they'll probably be short term - but on average, the weight will be constant. Blame it on Newton for inventing ;) gravity. All that lot meant was, I nearly agree, the hourglass almost has a constant weight.
Alfred Hitchcock would like the birds one. It's a long time since I though about that one. I remember a James Stewart film, "Spirit of St. Louis", in which he played Charles Lindbergh who was depicted as contemplating the effect of a fly or something that sometimes rested of his plane. I'm sure you're right when the truck isn't sealed. The numbers will be somewhere between an open and a sealed system. The birds beating their wings will blow air about causing some force on the truck. That force will depend on if they're near the ventilation or deep inside the truck. I suspect that most people look for a black and white answer instead of a grey one.
I have read all your comments, and they all seem to be on what you all can see, test, and figure by, But maybe you should try is what you cant see, you all focused on the movement of the sand and glass but not so much on the movement of the air in the hour glass. In this case the main focus is not friction or gravity, but rather the force of air rising. think about it as not the sand falling but rather the air displacing the sand the air is moving upward and when all the air is in the upper half, that force continues to push forward to the top of the water.
Its all about air rising. it is what you cant see
Hi Marty. The effect of the air has been discussed. It is totally unimportant. The fact is that, except for a couple of very short term glitches at the start and end of the sand flow, the hourglass has a constant weight. The only thing that matters is that when enough sand gets to the bottom of the hourglass, it can become upright.
I did a Google with "gardner floating hourglass problem" and found the following: http://g4gardner.pbworks.com/f/mm-morris2.pdf. Even if you don't read it all, the last section is amusing.
sorry..chris.. i was just tryin to answer d question even though i'm just a kid. i'm trying to answer such complicated questions..(_ _)
Hi scrapped prince. No sweat, I wasn't telling you off. I could see that you didn't really understand the physics, so I thought I'd clarify some points. All posts, like my response, are for everyone's benefit, if they want it. I admit, it wasn't a very good response though.
If you read the link, you'll see that even quite smart and physics competent people don't always get it right. I was 99% sure initially that friction was the only important thing. I was thrown when the first Anonymous made his suggestion. I loved the idea so much, that I thought it ought to be true. I then thought a bit more and realised that the moving sand was irrelevant. I'm pretty old and experienced, I've even been a physics teacher, so I almost automatically know what is unimportant. I just know that the air is unimportant for this problem.
I'm surprised though, that so many people think that the air inside a rigid container makes things float. A vacuum is much better. Air has mass, and so makes the hourglass tend to sink - not float. Helium is great for a non-rigid body though - it needs the gas pressure to make the balloon swell. The balloon goes up because its weight is less than the bouyancy produced by the atmospheric air. This turns out to be because the density of the helium is less than that of the air. That's enough waffling from me.
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