A cylindrical buoy with hemispherical ends is dropped in seawater of density 102
ID: 2241987 • Letter: A
Question
A cylindrical buoy with hemispherical ends is dropped in seawater of density 1027 kg/m3, as shown in the figure. The mass of the buoy is 77.7 kg, the radius of the cylinder and the hemispherical caps is R = 0.207 m, and the length of the cylindrical center section of the buoy is L = 0.586 m. The lower part of the buoy is weighted so that the buoy is upright in the water as shown in the figure. In calm water, how high (distance h) will the top of the buoy be above the waterline?
http://s18.postimg.org/n3hovofx5/12_7.png
A cylindrical buoy with hemispherical ends is dropped in seawater of density 1027 kg/m3, as shown in the figure. The mass of the buoy is 77.7 kg, the radius of the cylinder and the hemispherical caps is R = 0.207 m, and the length of the cylindrical center section of the buoy is L = 0.586 m. The lower part of the buoy is weighted so that the buoy is upright in the water as shown in the figure. In calm water, how high (distance h) will the top of the buoy be above the waterline?Explanation / Answer
From Archimedes' Principle, we know that the weight of the water displaced by the buoy is equal to the weight of the buoy.
Since the density of the water is given as a MASS DENSITY rather than a weight density, we can use mass instead of weight.
The key is to find the volume of the water displaced Vh2o, because that will equal the volume of the buoy Vubuoy that's underwater.
(1) Vh2o = 75.1kg / (1027kg / m^3) = 0.0731m^3
Next, express the volume of the upright buoy as a function of the height of the cylinder part (ignoring the cap for a moment):
(2) Vubuoy = Vhemi + Vhcyl, where
(3) Vhemi = (1/2) * (4/3) * ? * r^3 (volume of the bottom hemisphere)
(4) Vhcyl = h * ? * r^2 (this is the volume of the cylindrical part as a function of h)
Substituting (1), (3) and (4) in (2):
(5) 0.0731 = (1/2) * (4/3) * ? * r^3 + h * ? * r^2
Solving for h:
(6) h = (0.0731 - (2/3) * ? * r^3) / (? * r^2)
= (0.0731 - (2/3) * 3.14 * 0.195^3) / (3.14 * 0.195^2)
= 0.482m the height of the cylinder that's submerged.
Converting (6) to the height of the buoy above water Habove:
(7) Habove = Hcap + (0.610 - 0.482)
= 0.195 + (0.610 - 0.482)
= 0.323m
.
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