January 1999
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The trick to the answer is to realize that when the center of mass
is at its lowest point, the center of mass corresponds with the
height of the water.
This is because if the water were lower, then adding a bit more
water would lower the center of mass; and if the water were
higher, then removing a bit of water would lower the center of
mass.
Once we realize that, we only need to do some straightforward
computations to relate the center of mass to the dimensions, the
density of water (1 gram per cubic centimeter), and the AREAL
density of the tumbler (T grams per SQUARE centimeter,
assuming uniform thicknness).
Let H=12 be the height of the glass, R=4 its radius, and Z=4.5
the optimal depth of the water. The water has volume  ZR2 = 72
cubic centimeters and mass 72 grams. The bottom of the
cylinder has area R2 = 16 square centimeters and mass
16 T grams. The side of the cylinder has area 2 RH = 96
square centimeters and mass 96T grams. The total mass is
then
72 + 112 T.
The moment is calculated as
(72 * Z/2) + (16 T * 0) + (96 T * H/2) = 162 + 576 T.
Because the center of mass is known to be at height Z=4.5 we
have
(72 + 112 T) * 4.5 = 162 + 576 T,
so that T = 2.25 grams per square centimeter. Then the tumbler
has mass 2.25 * (16 + 96 ) = 252 grams. |
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