# Darn it, we're gonna need more soda.

**Calculus**Level 3

Assuming the can is a perfect uniform cylinder with top and bottom of identical mass, and is perfectly full of soda, the center of mass of the can and the soda inside is initially **exactly in the middle of the can**.

As the soda drains away, the center of mass begins to drop downwards.

When the soda is completely gone, the center of mass of the can is once again **exactly in the middle of the can**.

At some point during this process, the center of mass must have attained a lowest height above the table before going back up! **Find the height of the level of the soda which causes the height of the center of mass of the can and the soda inside to be at a minimum** using the following values:

- Mass of empty can: \(2~g\)
- Density of soda: \(1~g/cm^3\)
- Area of the base of the can: \(2~cm^2\)
- Height of the can: \(3~cm\)

Credit: I found this neat little problem somewhere on the internet, but I can't remember where! If someone knows the source, I'll credit it.

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