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.