Finding the heights of cylinders in water given density, radius and length

Question

Two cylinders with the same mass density ρc = 713 kg/m3 are floating in a container of water (with mass density ρw = 1025 kg/m3). Cylinder #1 has a length of L1 = 24 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 11 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water? This image is NOT to scale!

Picture to go with problem: http://i284.photobucket.com/albums/ll28/bathtub2008/showmepl-8.gif

h2 / h1 = ______

Answer 1

If  Vc the cylinder's volume, hc is its height, and A is its cross sectional area, and if Vw is the volume of the displaced water, which is also the volume of the submerged part of the cylinder, hw is the height of the submerged part of the cylinder, then

Vc = hc * A and Vw = hw * A

If ρc is the cylinder's mass density, ρw is the water's mass density, Mc is the mass of the cylinder, and Mw is the mass of the displaced water, then

Mc = ρc * Vc = ρc * hc * A

and the mass of the displaced water is

Mw = ρw * Vw = ρw * hw * A

But, because the cylinders are floating, Mc = Mw and

ρc * hc = ρw * hw

Note at this point that the cross sectional area of the cylinder A has dropped out of the equation. The cylinders' masses and volumes have also dropped out.  The radii also dropped out with the area. In fact, this derivation does not depend on the cylinders' cross section being circular or even of constant shape; only that the cross sectional area be be independent of the height at which the cross section is taken.

If hx is the height of the cylinder x sticking out of the water, then

Lx = hc and

hx = hc − hw = Lx − hw.

It turns out that

hw / hc = ρc / ρw

Subtracting both sides of the above equation from 1, taking the least common denominator on each side, and substituting hx and Lx gives

hx / Lx = 1 − ρc / ρw

By substituting 1 and 2 for x, we get one equation each for cylinder 1 and 2.  Dividing equation #2 by equation #1 also causes the 1 − ρc / ρw factor to drop out, and with it, ρc and ρw.  After doing a bit more algebra, we find that all we really needed was L1 and L2 and the fact that ρc < ρw:

h2 / h1 = L2 / L1 = 11 / 24

In practice there are also conditions on ρc / ρw, rx / Lx (or is that rx² / Lx?) and the symmetry of the shapes for them to float stably in the right orientation.