A cylindrical glass has a mass of 100 grams. It is partially filled with water (density = 1 gram/cubic centimeter.)
The glass has an inside diameter of 8cm and an internal depth of 15cm. When empty the center of mass is 8cm from the top of the glass.
The glass is most stable when its center of gravity is as low as possible. How much water is then in the glass?
All masses and lengths are in grams and centimeters
respectively and not shown. All center-of-masses are
measured up from the internal bottom of the glass.
H = height of water = 2*x
VW = Volume of water = PI*r^2*H = 32*PI*x
MW = Mass of water = density_of_water*VW
= (1)*(32*PI*x) = 32*PI*x
CW = Center of mass of water = x
MG = Mass of glass = 100
CG = Center of mass of glass = 7
CB = Center of mass of both
MW*CW + MG*CG (32*PI*x)*(x) + (100)*(7)
CB = --------------- = ---------------------------
MW + MG 32*PI*x + 100
32*PI*x^2 + 700
= -----------------
32*PI*x + 100
8*PI*(8*PI*x^2 + 50*x - 175)
(CB)' = -----------------------------
(8*PI*x + 25)^2
400*PI*(56*PI + 25)
(CB)" = --------------------- > 0 for x > 0.
(8*PI*x + 25)^3
Therefore, (CB)' = 0 defines a minimum.
5*(-5 +- sqrt(56*PI + 25))
x = ----------------------------
8*PI
Only the plus sign makes sense,
5*(sqrt(56*PI + 25) - 5)
x = --------------------------
8*PI
~= 1.825298 centimeters
VW = 32*PI*x = 20*(sqrt(56*PI + 25) - 5)
~= 183.498987 cubic centimeters
Edited on September 10, 2005, 8:43 am
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Posted by Bractals
on 2005-09-10 04:42:07 |
Solution
