The following problem comes out of Martin Gardner’s book Wheels, Life and Other Mathematical Amusements:
On a picnic not long ago Walter van B. Roberts of Princeton, N.J., was handed a freshly opened can of beer. “I started to put it down,” he writes, “but the ground was not level and I thought it would be well to drink some of the beer first in order to lower the center of gravity. Since the can is cylindrical, obviously the center of gravity is at the center of a full can and will go down as the beer level is decreased. When the can is empty, however, the center of gravity is back at the center. There must therefore be a point at which the center of gravity is lowest.”
Knowing the weight of an empty can and its weight when filled, how can one determine what level of beer in an upright can will move the center of gravity to its lowest possible point? When Roberts and his friends worked on this problem, they found themselves involved with calculus: Expressing the height of the center of gravity as a function of the height of the beer, differentiating, equating to zero and solving for the minimum
value of the height of the center of gravity. Later Roberts thought of an easy way to solve the problem without calculus. Indeed, the solution is simple enough to get in one’s head.
T o devise a precise problem assume that the empty can weighs 1.5 ounces. It is a perfect cylinder and any asymmetry introduced by punching holes in the top is disregarded. The can holds 12 ounces of beer, therefore its total weight, when filled, is 13.5 ounces. The can is eight inches high. Without using calculus determine the level of the beer at which the center of gravity is at its lowest point.
Feel free to challenge yourself with the above problem, but here’s what I would like to see. Let’s say that the can has mass , the fluid has mass and the height of the can is . Find in terms of and the minimum height of the centre of gravity. Can you do this without calculus?