4 Floating plate of cylindrical form

Let us now determine the balancing moment of a floating plate of cylindrical shape with radius R and height H; and assume a coordinate system (similar to our previous case of the prism-shaped plate), with its Z' axis coinciding with that of the cylinder (Fig. 4).

  

Figure 4: Rectangular (x ', y ', z') and cylindrical coordinate system (R, $\varepsilon$, z') for a cylindrical plate

In Fig. 4, its projection onto the plane is perpendicular to axis Z'. Suppose that Z' is inclined to the true zenith Z by an angle $\psi$. For a section perpendicular to the plane X' Y' and parallel to the coordinate axis X ', we can determine the shift of its centre of gravity $\overline{\Delta x}'_{\rm c}$ from Eq. (5), here substituting 2x' for L. Hence $\overline{\Delta x}'_{\rm c}= (x' )^2.\tan \psi /3h_{\rm F}$. The total shift $\Delta x'_{\rm c}$ of the centre of gravity of the displaced liquid in x'-component then is $\Delta x'_{\rm c}=[\int^{+R}_{-R} \overline{\Delta x}'_{\rm c}.P_{\rm x'} . {\rm d}y ']/ V_{\rm F}$, where $P_{\rm x'}$ is the area of the arbitrary section. Inserting for $\overline{\Delta x}'_{\rm c}$ and $P_{\rm x'} =2x' h_{\rm F}$, $\Delta x'_{\rm c}= [ 2\tan \psi/3\int^{+R}_{-R}(x' )^3.{\rm d}y' ] / V_{\rm F}$.

If we use cylindrical coordinates (Fig. 4), $x' =R.\cos \varepsilon$, $y' =R.\sin \varepsilon$,${\rm d}y' =R.\cos \varepsilon.{\rm d}\varepsilon$, and the numerator of the last expression for $\Delta x'_{\rm c}$becomes $2R^4.\tan\psi /3 \int^{+\pi/2}_{-\pi/2}(\cos\varepsilon )^4.{\rm d}\varepsilon = \Pi .R^4.\tan\psi /4$. As $V_{\rm F} = \Pi .R^2.h_{\rm F}$,

$$\Delta x' = R^2.\tan\psi /4h_{\rm F} .$$ (10)

Analogously, we get from Eq. (6), $\overline{\Delta z}'_{\rm c} =(x ')^2.(\tan\psi )^2/6h_{\rm F}$, and $\Delta z'_{\rm c} = [\int^{+R}_{-R}\overline{\Delta z}'_{\rm c}.P_{\rm x'} . {\... .../ V_{\rm F} = [(\tan \psi)^2/3 \int^{+R}_{-R}(x' )^3 .{\rm d}y '] / V_{\rm F}$. After integration, the numerator is $\Pi .R^4.(\tan\psi )^2/8$ and

$$\Delta z'_{\rm c} = R^2.(\tan\psi )^2/8h_{\rm F} .$$ (11)

Owing to symmetry of the cylindrical plate with respect to the plane X' Z ', $\Delta y '=0$. Further, $A'_{\rm c}\overline{M}= \Delta x'_{\rm c}/\tan\psi =R^2/4h_{\rm F}$; $A'_{\rm c}C_{\rm g}=(H-h_{\rm F})/2 - R^2.(\tan\psi )^2/8h_{\rm F}$, and the metacentric height

$${\mathcal{H}}_{\rm c} = R^2[1 + 1/2(\tan\psi )^2]/4h_{\rm F} - 1/2(H-h_{\rm F}) .$$ (12)

Because $W= \Pi.R^2.H.\sigma$, considering Eqs. (3) and (4), the balancing moment of a cylindrical plate becomes

$$\begin{eqnarray} M_{\rm c} &= \Pi .\{R^4. \sigma_{\rm F}/4.[1+1/2(\tan\psi )^2] ... ... &\quad- R^2.H^2.\sigma /2.(1-\sigma /\sigma_{\rm F})\}\sin \psi. \end{eqnarray}$$ (13)

Eventually, introducing the surface area of the cylindrical plate $P_{\rm c}$, and neglecting for small inclinations $\psi$ a 2nd order term,

$$M_{\rm c}\cong P_{\rm c}[R^2.\sigma_{\rm F}/4 - H^2. \sigma /2(1- \sigma/\sigma_{\rm F})]\sin\psi \approx K_{\rm c}.\psi.$$ (14)

If we compare Eqs. (9) and (14), we see that the balancing moment of a cylindrical plate is smaller than that of a square-shaped plate of the same height in an approximate ratio of 0.6 to 1.