How far will it sink?

[MarkFL]

A wooden spherical ball of radius $r$ is floating on a pond of still water. Let $h$ denote the depth that the ball will sink into the water. Suppose the density of the wood is 80% that of the density of the water. Approximate $h$ in terms of $r$.

 

[chisigma]

A wooden spherical ball of radius $r$ is floating on a pond of still water. Let $h$ denote the depth that the ball will sink into the water. Suppose the density of the wood is 80% that of the density of the water. Approximate $h$ in terms of $r$.

Spoiler

If we suppose r=1 the high of the spherical segment that is outside the water satisfies the condition...


$\displaystyle V= \pi\ \frac{h^{2}\ (3-h)}{3} = \pi\ \frac{4}{15}\ (1)$

... and that means that...

$\displaystyle h^{3} - 3 h^{2} + \frac{4}{5}=0 \implies h = .57428...\ (2)$


Kind regards


$\chi$ $\sigma$

 

[MarkFL]

Here is my solution:

Spoiler

First, we may find the volume of the submerged portion of the ball by rotating the following shaded area about the $x$-axis using the disk method:



The volume of an arbitrary disk is:

\(\displaystyle dV=\pi R^2\,dx\)

where:

\(\displaystyle R^2=y^2=r^2-(x-r)^2=2rx-x^2\)

And so we have:

\(\displaystyle dV=\pi\left(2rx-x^2 \right)\,dx\)

Summing the disks through integration, we obtain:

\(\displaystyle V=\pi\int_0^h 2rx-x^2\,dx=\pi\left[rx^2-\frac{x^3}{3} \right]_0^h=\frac{\pi h^2}{3}(3r-h)\)

This is the volume of the so-called "spherical cap."

Next, using the principle of Archimedes, we may equate the mass of the sphere to the mass of the fluid it displaces. Let $\rho$ be the mass density of the fluid, and $k\rho$ be the mass density of the sphere, where $0<k<1$. Mass density is mass per volume, and so mass is mass density times volume, hence, we may write:

\(\displaystyle k\rho\left(\frac{4}{3}\pi r^3 \right)=\rho\left(\frac{\pi h^2}{3}(3r-h) \right)\)

Simplifying, we obtain:

\(\displaystyle 4k=\left(\frac{h}{r} \right)^2\left(3-\frac{h}{r} \right)\)

Let \(\displaystyle u=\frac{h}{r}\), and we may write:

\(\displaystyle u^3-3u^2+4k=0\)

We are told \(\displaystyle k=\frac{4}{5}\) and so we have:

\(\displaystyle u^3-3u^2+4\left(\frac{4}{5} \right)=0\)

And multiplying through by 5, we obtain:

\(\displaystyle 5u^3-15u^2+16=0\)

As we are allowed to obtain a numerical solution, we need not work with the rather cumbersome cubic formula.

Because \(\displaystyle \frac{1}{2}<k<1\), it is reasonable to assume $1<u<2$, so a good first estimate is \(\displaystyle u=\frac{3}{2}\). Using Newton's method, where:

\(\displaystyle f(u)=5u^3-15u^2+16=0\)

\(\displaystyle f'(u)=15u^2-30u\)

we obtain the recursion:

\(\displaystyle u_{n+1}=u_n-\frac{5u_n^3-15u_n^2+16}{15u_n^2-30u_n}=\frac{u_n\left(15u_n^2-30u_n \right)-\left(5u_n^3-15u_n^2+16 \right)}{15u_n\left(u_n-2 \right)}=\frac{10u_n^3-15u_n^2-16}{15u_n\left(u_n-2 \right)}\)

Thus, we may compute:

\(\displaystyle u_0=1.5\)

\(\displaystyle u_1=1.4\overline{2}\)

\(\displaystyle u_2\approx1.42571225071225\)

\(\displaystyle u_3\approx1.42571854914589\)

\(\displaystyle u_4\approx1.42571854916652\)

\(\displaystyle u_5\approx1.42571854916652\)

And so we find:

\(\displaystyle h\approx1.42571854916652r\)