A cylindrical buoy with hemispherical ends is dropped in seawater of density 102

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

Explanation / Answer

Vh2o = 77.7kg / (1027kg / m^3) = 0.07565m^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):

Vubuoy = Vhemi + Vhcyl, where

Vhemi = (1/2) * (4/3) * ? * r^3 (volume of the bottom hemisphere)

Vhcyl = h * ? * r^2 (this is the volume of the cylindrical part as a function of h)

Substituting (1), (3) and (4) in (2):

0.07565 = (1/2) * (4/3) * ? * r^3 + h * ? * r^2

Solving for h:

h = (0.07565 - (2/3) * ? * r^3) / (? * r^2)

= (0.07565 - (2/3) * 3.14 * 0.207^3) / (3.14 * 0.207^2)

= 0.424 m the height of the cylinder that's submerged.


Converting (6) to the height of the buoy above water Habove:

Habove = Hcap + (0.586 - 0.424)

= 0.207 +0.162

= 0.369 m

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