Think outside the can - a challenge for Y-san

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Knowing how it is second nature for my friend Y-san to solve extremal problems by equating derivatives of functions to zero, I was delighted to find a problem that can be solved with only a pencil, some drawing paper and a little thought. Here it is.

Imagine an empty can that weighs 1.5 ounces. It is a perfect cylinder with weightless top and bottom, so that any asymmetry introduced by punching holes in the top can be disregarded. The can holds 12 ounces of beer; so its total weight, when filled, is 13.5 ounces. The can is 8 inches high. Without using calculus, determine the level of the beer at which the center of gravity is at its lowest point.

Consider that, as beer is removed from a full can, the center of gravity is lowered. But empty and full, the center of gravity is in the same place. So there must be a filling factor where it is lowest.

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Posted (edited) · Report post

Earlier posted 2",[/spoil

2" of beer yield c.o.g of


((1.5*4)+(1*3))/(1.5+3.)=2
]
Edited by jhawk
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Posted · Report post

Earlier posted 2",[/spoil

2" of beer yield c.o.g of

((1.5*4)+(1*3))/(1.5+3.)=2

]

jhawk, sorry for not replying sooner.

That is the correct answer. Care to share your method?

I have a clever thought process (it's not mine) that leads to it that I will share if you want to go first.

It involves freezing.

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Posted · Report post

Not sure if this was the same process that jhawk used, but it's what I would have done if it weren't already answered. And it doesn't seem to involve freezing.

The point at which the center of balance is lowest must be the point at which adding just an iiitty-bitty bit more beer would bring the center of gravity up instead of down. That happens when the center of gravity lies at the top of the beer -- adding just a touch more beer then would clearly bring the center of gravity up. So solve the equation where (center of gravity of [can + beer at height H]) = (height H).

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Posted · Report post

Not sure if this was the same process that jhawk used, but it's what I would have done if it weren't already answered. And it doesn't seem to involve freezing.

The point at which the center of balance is lowest must be the point at which adding just an iiitty-bitty bit more beer would bring the center of gravity up instead of down. That happens when the center of gravity lies at the top of the beer -- adding just a touch more beer then would clearly bring the center of gravity up. So solve the equation where (center of gravity of [can + beer at height H]) = (height H).

Nice.

Freezing would allow balancing the can sideways on a fulcrum at its beer height. Then exactly what you said is obvious. Either adding beer or subtracting beer will make the can fall toward its empty side.

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