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Glass half full

Glass half full

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talzamir
Art, not a Toil

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A glass is shaped like a straight cylinder with a round base, height h, radius r. If the glass is half full and tilted so that the liquid just touches the rim of the glass, how high from the bottom of the glass is the surface on the opposite side of the glass?

iamatiger

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Originally posted by talzamir
A glass is shaped like a straight cylinder with a round base, height h, radius r. If the glass is half full and tilted so that the liquid just touches the rim of the glass, how high from the bottom of the glass is the surface on the opposite side of the glass?
Taking the cross section, and doing it in 2d for simplicity, which shouldn't change the answer:

Area of water when glass is half full = rh

When glass is tilted, the required point is x from the bottom. The shape the water is now in is a composition of two shapes:

a) A rectangle, area 2rx;
b) And a right angled triangle, with non-hypotenuse sides 2r, and h-x
the area of this triangle is r(h-x)

This gives a total water area of r(2x + h - x)

Tilting the glass does not change the area of the water, so

r(2x + h - x) = rh
2rx + rh - rx = rh
rx = 0

So, assuming r is not 0, the edge of the water now touches the bottom of the glass.

Shallow Blue

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Originally posted by iamatiger
Taking the cross section, and doing it in 2d for simplicity, which shouldn't change the answer:

So, assuming r is not 0, the edge of the water now touches the bottom of the glass.
There's an even simpler way to reach that conclusion: a cylinder is symmetrical across its central point. Any straight plane that goes through one edge and divides it exactly in two must go through the opposite point on the opposite edge, because of that symmetry.

Richard

iamatiger

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Yep, I thought of that after I'd posted it, Doh!

talzamir
Art, not a Toil

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Slicing the glass in two with a plane that seen directly from the side cuts the rectangle the glass appears to be diagonally from corner to cover, cuts one whole into two equal volumes.. that would appear to be cones with a base area , so each 1/3 of the original?

Since 1/3 + 1/3 != 1, something doesn't seem right?

f
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Originally posted by talzamir
that would appear to be cones with a base area , so each 1/3 of the original?
I don't follow

2 cones != cylinder

talzamir
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Sorry, late here and it's been a long day. Seems I missed a few crucial words.

If you slice the glass diagonally in two, then yes, by symmetry each half has a volume equal to one half of the full glass.

However, those halves would appear to be cones, with a height equal to the height of the glass, and the area of the bottom equal to the area of the bottom of the glass. The volume of such a cone would be 1/3 pi r^2 h, where h is the height of the glass and r the radius of its bottom. The volume of the glass is pi r^2 h, so it would take two such cones, rather than three, to fill it. Where is the third?

So.. if a glass shaped like a cylinder is tilted so that the water just reaches the edge of the glass, and just covers the bottom of the glass, does the water in the glass not form a cone shape but something else?

With
h = height of the glass,
x = the shortest distance of the water surface from the bottom of the glass, and
r = the radius of the bottom of the glass,

it would seem there is a layer of water on the bottom shaped like a cylinder, and the rest is shaped like a cone. So,

pi r^2 x + 1/3 pi r^2 (h - x) = 1/2 pi r^2 h
6x + 2 (h - x) = 3 h
x = h / 4

so the shortest distance of the surface of the water from the bottom of the glass is one quarter of the height of the glass.

On the other hand, the case for saying that it has to be zero is rather convincing too. Clearly both can't be right?

f
Defend the Universe

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Are you asking how high the surface of the water is from the hypothetical table that the tilted glass is resting on?

Also, the water in the glass does not form a cone, it looks like this:
http://www.chegg.com/homework-help/questions-and-answers/cylindrical-glass-radius-r-height-l-filled-water-tilted-water-just-covers-base-left-pictur-q1176634?frbt=1

Shallow Blue

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Originally posted by talzamir
Sorry, late here and it's been a long day. Seems I missed a few crucial words.

If you slice the glass diagonally in two, then yes, by symmetry each half has a volume equal to one half of the full glass.

However, those halves would appear to be cones,
Ah, but they're not. They have two flat sides each, one being the circular bottom (or top) side of the cylinder, and one being the elliptical surface of the water. There's also one curved surface; and there is no singular vertex. So they're not cones at all, they just look a bit like cones.

Richard

iamatiger

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2d helps think about this.

In 2d the glass is a tall rectangle, the water cuts it in half horizontally.

Tilted at 45 degrees, the water cuts it in half diagonally.

talzamir
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Indeed so. Thinking of the 2D cross section, it is obvious that the water needs to cover the bottom of the glass exactly, based on symmetry. The volume that it forms looks a lot like a cone - every point on the circumference of the bottom of the glass joined to the point on the rim along a straight line.. and straight the lines would be, being in the plane of the surface of the water. But if it were a cone, the volume of it is too low, just a third of the volume of the glass, rather than half, so something has to be wrong. So, it is not a cone.. obviously not a straight cone, as the apex is not above the middle point of the bottom of the glass, but not any other kind of cone either.

f
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Originally posted by talzamir
Indeed so. Thinking of the 2D cross section, it is obvious that the water needs to cover the bottom of the glass exactly, based on symmetry. The volume that it forms looks a lot like a cone - every point on the circumference of the bottom of the glass joined to the point on the rim along a straight line.. and straight the lines would be, being in the plane ...[text shortened]... is not above the middle point of the bottom of the glass, but not any other kind of cone either.
If it were a cone, every point on its surface would be on straight line between the apex and the circumference of the base. It is fairly straightforward to prove that's not the case.

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