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# Competitive Swimming

## Question

You are a competitive swimmer, and your coach wants you to swim a lap of the pool backstroke with a soda can balanced on your forehead. He gives you an empty soda can and you can add some water to it if that will make your task easier.

How much water should you place in the can so that the center of gravity of the water-plus-can is as low as possible?

Assume the can is a perfect cylinder with a top and bottom made of the same material as the sides. The density of water is 1 gram/centimeter³. Let H be the height of the cylinder and r its radius, and let the mass of the can be C grams.

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Let A = pi r2 be the cross section area of the can.
The mass of the can is C, and by symmetry its center of mass is H/2.
Let x be the height of the water in the can.
The mass of the water is Ax, (it has unit density,) and its center of mass is x/2.
The center of mass of the can plus water xcw (to be minimized) is

xcw(x) = (1/2) (Ax2 + CH) / (Ax + C) = f(x) / g(x)

f(x) = (Ax2 + CH)/2; g(x) = Ax + C

x'cw = (f'g - fg') / g2 = 0 when f'g = fg'.

f'g = Ax(Ax + C)

fg' = A(Ax2 + CH)/2. Multiply both by 2/A and equate:

Ax2 + 2Cx - CH = 0. Take the positive root:

xcw min = {-C + sqrt(C2 + ACH)} / A

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C grams of water in the can?

How do we know it would fit?

The answer has to depend on geometry to some degree.

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Yeah, as always, my intuition was wrong. After reading your first post, I realized that the solution is not that simple and H, r, etc are not red herrings as I thought they were.

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