Lowest center of mass of a container + fluid

AI Thread Summary
The discussion centers on determining the height at which the center of mass of a container filled with fluid is minimized. It explores the relationship between the fluid level and the center of gravity, noting that the center of mass decreases as fluid is consumed until a minimum point is reached, after which it begins to rise again. Participants suggest that the center of mass is at its lowest when it coincides with the water level and propose setting up an equation to find this point. Calculus is debated as necessary for solving the problem, with one participant successfully recalculating the solution. The conversation concludes with a positive acknowledgment of the findings.
Delej
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I have an interesting question, that I've been struggling with for a while now, but I'm somehow messing it up. I haven't found anything on the internet (hard to search for something like this)
so I'm wondering if any of you know something useful about it, or do better at finding the solution. :)

Let's assume a container with a mass of M. This container has a base area of A, a height of h. It is a right prism/ or cylinder. (We consider the width of the walls of this container negligible).

We fill with this container with a fluid to the height of hf. (it could be something else too, but I think of fluids as it is a question that arose during drinking some refreshments :) ) The density of the fluid/substance is d (or rho, but that doesn't show up correctly on my screen...)

The question is at what height of filling the container will the (common) center of gravity be the lowest for these, and where is that minimum point?

In real life: Imagine the following scenario: You've got a glass of water. When it's completely full, its center of gravity is at ~h/2, because vertically it's sort of/approximately "homogeneous". As you start drinking from it, the center of gravity goes down with the water level. But at some point it reaches a minimum and starts going back up, because the water's weight gets more and more negligible compared to the glass. Until finally when you empty the glass, the center of gravity returns to that approximately ~h/2.

So where's that water level and the center of mass belonging to it? :)
 
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well the glass will be symmetrical about the centre of the cylinder, so you know that the centre of mass will always be along that axis. So you just basically have to calculate the height of the glass at which half of the total mass lies above and half lies below. To find the point where it switches from going down to going back up to midway you'd have to write the centre of mass as a function of water level and take the derivative to find the minimum. Not that difficult I don't think.
 
Yeah, that's the exact method I was going with, but I think I'm making some mistake in it somewhere. Finding where the derivative is zero amounts to a quadratic equation. This equation should have a positive (realistic) solution as long as all parameters are positive, but it's not so, so I'm assuming something isn't quite right. That's why I'm asking you all, to see if you do it either a completely different way, or do it the way I did, and come up with an equation that give correct answers...
 
Show what you did?
 
Delej said:
So where's that water level and the center of mass belonging to it? :)

Adding water when the water level is above the center of mass raises the center of mass. Subtracting water when the water level is below the center of mass also raises the center of mass. So the center of mass is at a minimum height when it is exactly at the water level. I don't think you need calculus to solve this problem. Just set up the equation for the center of mass as a function of the water level h, set that equal to h, and solve for h.

BBB
 


Ok, here's what I did... Well actually it isn't, since the notebook in which I did it, I left somewhere else, so I actually redid the calculations here at the computer as I entered them. I came up with the solution this time. I think it's really nice, and interesting, hope you agree. You'll see everything in the word file attached...
Have a nice day! :)
 

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Delej said:
Ok, here's what I did... Well actually it isn't, since the notebook in which I did it, I left somewhere else, so I actually redid the calculations here at the computer as I entered them. I came up with the solution this time. I think it's really nice, and interesting, hope you agree. You'll see everything in the word file attached...
Have a nice day! :)

Very nice!

BBB
 

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