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Topic: Easy: Glass half full (Read 1370 times) |
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jabhiji
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Easy: Glass half full
« on: Jan 29th, 2003, 7:05pm » |
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All one needs is a rotating turntable whose RPM can be increased to any desired value. Place the cylinder on the table so that their axes coincide.
When the cylinder rotates, the free surface of the water will become a paraboloid of revolution. More interesting, the rise of the surface at the edge will be exactly equal to the dip at the axis.
At a high enough RPM, if the cylinder were exactly half full to begin with, the dip will just touch the cylinder bottom while the rise at the edge will be on the verge of spilling water outside.
If it were (more/less) than half-full, the water will spill out (before/after) the dip touches the cylinder bottom.
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ThatGirlLooksTasty
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Re: Easy: Glass half full
« Reply #1 on: Jan 30th, 2003, 12:19am » |
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Umm... the problem says the room is empty which means it probably doesn't have a turnable.
But my solutions concept is similar to yours:
Tilt the glass so that the water surface touches the rims of both the bottom and top circles. If it does, then its half.
Sorry for the clumsy explanation, its better to see it visually.
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jabhiji
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Re: Easy: Glass half full
« Reply #2 on: Jan 30th, 2003, 6:03am » |
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Hmmm...Looks like I am thinking too hard. Your solution is much more simple and elegant.
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