The Hourglass Paradox Nobody Agrees On

Added: 2026-05-05

[00:00:00]This is an hourglass. It measures the passage of time by dropping a constant flow rate of sand. But here's the question. Does the weight of an hourglass change when the sand is flowing versus when the sand is stopped?

[00:00:12]This simple question has confused physicists for decades. In fact, if you look through different physics books, you'll find completely different answers. Some say it gets lighter, some say it gets heavier, and others say it stays exactly the same. Well, I have a sensitive scale here that can give me weight updates every 1 millisecond. So, let's test this and actually see what happens. So, first, why would we think the weight could decrease? Well, let's look back at an earlier video I did where I took a completely closed container with a weight attached to the top. When that weight was released, the reading on the scale decreased by the weight of the object because it's now in freef fall. So if the hourglass constantly has some sand in freef fall, then it seems obvious that the reading on the scale should decrease by the weight of the sand that's currently in freef fall. So this is option one. But the astute scientist will say, "Yeah, but wait a minute. Yes, there's some sand in freef fall, but there's also sand hitting the bottom." That impact will exactly counteract the missing weight of the sand in freef fall. So the net force change is exactly zero.

[00:01:20]Meaning there's no change in weight. So this is option two. But weight, says the even more astute scientist. If you look at the sand at the top of the hourglass, those grains are moving downward with some velocity. It is slow, but they're actually moving. Then compare that same sand once it's reached the bottom. It now has zero velocity. So if you look at the net result, you had some sand that was moving and then it suddenly came to rest. So there should be a force equal to the mass flow rate times the velocity of the sand at the very top of the hourglass there. So the weight will increase by that amount. So this means the hourglass should weigh more when the sand is flowing versus when it's not flowing. So we literally have three good arguments about what should happen. But which one actually happens? Well, let's check. But before we continue, I want to thank the sponsor for this video, Dreamy. This is the Dreamy X60 Max Ultra Complete. Now, the thing that surprised me most is just how thin this robot is.

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[00:04:23]Okay, it does not look like any change happened here. Just have a flat line, but let's look at the raw data and take the average before and after just to be sure. So, there's no statistical difference between the before and after here. So, it looks like there was no change. So, after this, I can see that the logic of option one is flawed because it ignores the sand hitting the bottom. And the logic of option two seems pretty solid. And it looks like this might be the right answer. But there's still a problem here. In order for the force from the striking sand to completely cancel the missing weight of the falling sand, we had to assume one thing. That the sand starts its freef fall with zero velocity. But you can tell this isn't the case. Look at the sand flowing into the neck. It already has velocity before it starts freef falling. So the starting velocity isn't zero. This means that over a given amount of time, the sand is hitting the bottom with slightly more force than we assumed. So the true equation isn't just force equals the mass in freef fall time the gravity. There's an extra term that comes from the change in momentum due to that initial velocity. So the true force should be different by the mass flow rate times this small velocity. But it turns out that this force is very tiny.

[00:05:39]Even for this one minute sand timer, it was so minuscule that I couldn't detect it on my sensitive scale here. So, if we're going to see this effect, we need to make a weird hourglass. One where the sand at the top is moving faster than in this 1 minute timer. To do this, I can make the cross-sectional area much smaller near the top, essentially turning the top into a long narrow neck.

[00:06:01]But the difference in the velocity of the sand at the top is much higher because the area is smaller than before.

[00:06:07]So, the bottom of the glass is hanging in the air here. The neck comes up. It's attached to the scale here. It's a really long neck. Okay. I'm going to pull off the bottom here.

[00:06:27]Okay, that's the blip where I pulled off the tab. So, let's look at the raw data and see what this looks like. Already, I can see that before and after it stopped, there is a difference. It weighs about 0.25 gram more, about 250 mg when the sand is flowing compared to when it's not flowing. So, option three really is the right answer. This is so counterintuitive to think that the hourglass weighs more when the sand is flowing in a closed system than when it isn't. But for any realistic hourglass, the velocity of the sand at the top is so slow that there's no measurable change in force. For a real hourglass like this one here, the change in force would be extremely small. Somewhere in the microgram to nanog range. I like this demonstration because it's a good example of an old engineering saying that goes like this. There's what's true and there's what matters. In this case, option three, the hourglass weighs more when the sand is flowing. This is the correct answer. But when the effect is about one part in hundreds of millions, it doesn't really matter. So, I'll leave it up to you to decide what answer you would give somebody. Maybe this is a good question to place bets on. Whatever the person says, you can give an alternate answer. And thanks for watching another episode of the Action Lab. I hope you enjoyed it. If you haven't subscribed to my channel yet, remember to hit that subscribe button and we'll see you next time.