Example 1

A homogeneous hemisphere with a radius of R floats in water at a depth of h.
Find the value of the floating radius R1.
Find the value of the moment of inertia of the floating surface.
Find the volume of the immersed part of the floating body.
Determine the density of the material from which the hemisphere is made.
Data: \begin{aligned} & R = 150 \mathrm{~cm} \\ & \rho_w = 1000 \frac{\mathrm{kg}}{\mathrm{m}^3} \\ & h = 75 \mathrm{~cm} \end{aligned}

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Solution

1. Determine the value of the radius of flotation \(R_1\)

From the equation of a circle with center at point z=R, we get:

\begin{aligned} &r^2+(z-R)^2=R^2\\ &r(z)=\sqrt{R^2-(z-R)^2}=\sqrt{2 R \cdot z-z^2}\\ &R_1=r(h)=\sqrt{2 R \cdot h-h^2}=1.3 m \end{aligned}

2. Determine the value of the moment of inertia of the flotation surface

\begin{aligned} I_x=\frac{\pi \cdot R_1^4}{4}=2.243 \mathrm{~m}^4 \end{aligned}

3. Determine the volume of the submerged part of the floating body

There is a formula for the volume of a spherical segment, but personally I'm not a big fan of memorizing it:

\begin{aligned} V_z=\frac{\pi \cdot h^2}{3}(3 R-h)=2.209 \mathrm{~m}^3 \end{aligned}

An alternative is to calculate this volume using an integral formula for rotational figures:

\begin{aligned} V_z=\pi \int_0^h 2 R \cdot z-z^2 \mathrm{~d} z=2.209 \mathrm{~m}^3 \end{aligned}

As you can see, the final result is the same :-)

4. Determine the density of the material from which the hemisphere is made

From Archimedes' principle and the equilibrium condition

\begin{aligned} & G=W \\ & \rho_m \cdot g \cdot \frac{1}{2} \frac{4}{3} \pi \cdot R^3=\rho_w \cdot g \cdot V_z \\ & \rho_m=\frac{\rho_w \cdot g \cdot V_z}{g \cdot \frac{1}{2} \frac{4}{3} \pi \cdot R^3}=312.5 \frac{\mathrm{kg}}{\mathrm{m}^3} \end{aligned}