Need help with soda can physics problem

Factual Questions
1

Hi Dopers!

Here’s a physics problem that’s been bugging me for a while. (I saw this in a Games magazine a long time ago but haven’t been able to find it, and I can’t remember the answer either).

Assume you have a perfectly cylindrical can of liquid. When it’s full the center of gravity is at the center of the can. When it’s 1/2 full, the center of gravity is below the center of the can. When it’s empty, the center of gravity is at the center of the can. At what point does the center of gravity stop going down and start coming back up?

I’m guessing that the mass of the can has to play a factor here, that the question is meaningless without giving the stats on the can. But then, there might be some trick that I’m not thinking of.

The Dun King

2

Whoa, no kidding. I thought up this exact problem many years ago. I could do the math and get the real answer, but I can tell you offhand that yes, it does depend on the relative masses of the can and the liquid.

Imagine, for instance, mercury in a paper cup. A few drops in the bottom is all you need to lower the center of mass as low as it’s going to go. With more than that, the center of mass will be almost the same as the center of mass of the mercury itself.

Contrast filling it with some light fluid, like carbon dioxide. You won’t get nearly as much of an effect.

I’ll go do the math now…

3

The mass of the soda (which has the same cylindrical shape as the can) has its center, and the can its. The location of those centers, and the proportions of the masses determines the center of mass for the entire system.

So, you have ten grams of aluminum, with its mass at the center of the can, now figure the location of the center of mass of any portion of soda, and it’s location. If there is less than ten grams of soda, then the center of mass is affected less by soda than aluminum. More than ten grams, and it is affected less. Ten grams of soda sits very low in the can, and has, therefore the greatest distance from the original center, and the largest resultant change of location of center of mass. Less than that, and it moves up again, because the weight is less.

All of this ignores non-cylindrical shapes at the bottom of the can, of course. The particular weight was a wild assed guess, as well.

Tris

4

If the mass of the can is a, and height of the can is d,and the mass of the liquid when the can is full is b, at what distance from the bottom (d/x) does the center of mass begin to rise. If a =b, the center of mass rises after d passes the halfway point of the depth, 1/2 d, or d/2. For 2a = b, the point is lower. Is it 1/4 d?

Just thinking on the keyboard

5

The solution depends on only one thing, which I will call r. This is the ratio of the mass of the can to the mass of the liquid. In UncleBill’s notation, it’s a/b, but I called it m/M. For a light liquid in a lead bucket, r is large. For a heavy liquid in a paper bucket, r is small.

We need one variable - call it x. This is the fraction that the cup is full. For an empty cup, x = 0. For a full cup, x = 1. If the can is symmetrical and the height of the liquid in a full cup is H, then the mass of the liquid at any given level of fullness is xM, and its height is xH. Thus the liquid’s center of mass is half of this, or xH/2. The center of mass of the cup itself is always H/2. The center of mass of the system is then:

(m × H/2 + xM × xH/2) / (m + xM)
= H/2 × f(x), where
f(x) = (x[sup]2[/sup] + r) / (x + r)

(Notice that f(0) = 1 and f(1) = 1 no matter what r is.) This now becomes a straightforward calculus problem. For what x is f(x) minimized, and what is its minimum value? Sparing you the details, the answer is:

x[sub]MIN[/sub] = sqrt(r[sup]2[/sup] + r) - r
f(x[sub]MIN[/sub]) = 2x[sub]MIN[/sub]

For instance, for a light fluid in a lead container (I know I keep switching words for the “can”. Sorry.), we might have r = 10, and then x[sub]MIN[/sub] = 0.488. That is, the center of mass will be minimized when the can is 48.8% full. For extremely large r’s, x[sub]MIN[/sub] will get close to 50%. So, for a very heavy container, the center of mass will be lowest when it’s almost half full.

For a heavy fluid in a light container, like mercury in paper, you might have r = 0.01. Then x[sub]MIN[/sub] = 0.0904. That is, the center of mass will be lowest when the container is only 9.04% full. As r gets smaller and smaller, this value also drops more and more. For very small r’s, it goes roughly like sqrt®.

6

This is an old puzzle. The answer is that the center of mass is lowest when it is at the same height as the surface of the liquid.

Consider a can with just enough liquid to put the center of mass at the liquid’s surface. Now freeze the liquid and tip the can on its side. Assume for the sake of discussion that nothing changes density when the liquid freezes.

With the liquid frozen, you can put the can on its side and balance it on a fulcrum. The balance point will be the surface of the frozen liquid. Let’s say the liquid is to the left of the fulcrum. If you added any frozen liquid, it would go to the right of the fulcrum, making that side heavier and moving the center of mass to the right. If you remove any frozen liquid, it would make the left side lighter, again moving the center of mass to the right.

This assumes that there is a single minimum - that is, that there is only one point while adding or removing liquid that the location of the center of mass will reverse direction.

7

Oh yeah, right. Like the variation of gravity with height doesn’t enter into this! :wink:

8

Well, no, it doesn’t. I suspect you know this, hence the wink. But center of gravity, or more accurately, center of mass is independant of the gravity gradient.

9

You know, you’re right. I was about to say you’re wrong, and direct you to my calculations, but I realized that they agree with what you said. Cool, I hadn’t thought about it your way.

Incidentally, there is only one minimum, although I can’t think of an intuitive way of showing this.

10

Cunning. I would’ve never thought of that in a million years. That’s awesome, thanks!

The Dun King