Modeling Plastic Cup Physical Behavior

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Plastic cups exhibit unique physical behavior, particularly with their concave bottoms that can transition to a convex shape when lifted, similar to baby-food can tops. This popping effect occurs due to the varying stress distribution on the bottom of the cup, influenced by the height of the liquid and the distance from the rim. The stresses are radially symmetric in both positions, but the transition between them likely disrupts this symmetry. The paper material of the cup resists compression, which affects how the bottom contorts during the popping motion. Understanding this behavior requires considering the forces acting on the cup's bottom as a springy lever system.
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I've noticed that plastic cups exhibit some interesting physical behavior.
Some paper cups(maybe all actually, I'm not sure) have concave bottoms, and when the liquid in the cup gets high enough, the forces from simply raising the beverage from the table to my mouth create enough force to pop the concave region into a convex one, and then it pops back to the concave position when it's at rest. This is a lot like the popping tops of baby-food cans, where once you've opened them, you can press down on the center and it pops in and out.
Now, when the water is high enough, the resting position of the bottom is the convex position, and lowering the cup causes the popping behavior.
What I'm wondering about, is how to accurately describe this behavior physically. I understand that there's roughly an even pressure at every point on the bottom of the cup due assuming that the bottom of the cup doesn't pop up/down very far compared to the height of liquid, but I think the stress on the bottom caused by liquid above each point should be very different, since each point is at a different distance from the edge, and I think the force should be considered as acting on some sort of springy lever arm attaching each point being pressed on to a point on the rim.
I look forward to responses :)
 
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It does make sense that the liquid would cause more stress on the middle of the cup's bottom than at a point near the rim, even if the force is the same in both places. The stresses should be radially symmetric while in either position. I'm guessing, though, that the bottom probably contorts to get from the up to down position or vice versa, breaking radial symmetry. The paper in the cup is not very compressible, especially not when it would rather bend, so the cup's bottom would seek to keep its surface area constant while it maneuvered from up to down. Both up and down positions are radially symmetric, and they seem to be the only positions that are radially symmetric while keeping the same surface area, so for the cup to pop from one position to the next, radial symmetry would have to be broken.
 
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