Question 2.15: A solid hemisphere of density ρ and radius r floats with its...
A solid hemisphere of density ρ and radius r floats with its plane base immersed in a liquid of density \rho_{l}\left(\rho_{l}>\rho\right). Show that the equilibrium is stable and the metacentric height is
\frac{3}{8} r\left(\frac{\rho_{l}}{\rho}-1\right)
The hemisphere in its floating condition is shown in Fig. 2.38. Let be the V submerged volume. Then from equilibrium under floating condition,
\frac{2}{3} \pi r^{3} \times \rho=\forall \times \rho_{l}
or, \forall=\frac{2}{3} \pi r^{3} \times \frac{\rho}{\rho_{l}}
The centre of gravity G will lie on the axis of symmetry of the hemisphere. The distance of G along this line from the base of the hemisphere can be found by taking moments of elemental circular strips (Fig. 2.38) about the base as
O G=\frac{\int_{0}^{r} \pi\left(r^{2}-z^{2}\right) z d z}{\frac{2}{3} \pi r^{3}}=\frac{3}{8} r
In a similar way, the location of centre of buoyancy which is the centre of immersed volume V is found as
O B=\frac{\int_{0}^{H} \pi\left(r^{2}-z^{2}\right) z d z}{\frac{2}{3} \pi r^{3} \frac{\rho}{\rho_{l}}}=\frac{3}{8} \frac{\rho_{l}}{\rho} r \frac{H^{2}}{r^{2}}\left(2-\frac{H^{2}}{r^{2}}\right) (2.82)
where H is the depth of immersed volume as shown in Fig. 2.38.
If r_{h} is the radius of cross-section of the hemisphere at water line, then we can write
H^{2}=r^{2}-r_{h}^{2}
Substituting the value of H in Eq. (2.82), we have
O B=\frac{3}{8} \frac{\rho_{l}}{\rho} r\left(1-\frac{r_{h}^{4}}{r^{4}}\right)
The height of the metacentre M above the centre of buoyancy B is given by
B M=\frac{I}{V}=\frac{\pi r_{h}^{4}}{4\left\{\left(\frac{2}{3} \pi r^{3}\right) \frac{\rho}{\rho_{l}}\right\}}=\frac{3}{8} \frac{\rho_{l}}{\rho} r \frac{r_{h}^{4}}{r^{4}}
Therefore, the metacentric height MG becomes
M G=M B-B G=M B-(O G-O B)
=\frac{3}{8} \frac{\rho_{l}}{\rho} r \frac{r_{h}^{4}}{r^{4}}-\frac{3}{8} r+\left[\frac{3}{8} \frac{\rho_{l}}{\rho} r\left(1-\frac{r_{h}^{4}}{r^{4}}\right)\right]
=\frac{3}{8} r\left(\frac{\rho_{l}}{\rho}-1\right)
Since \rho_{l}>\rho, M G>0, and hence, the equilibrium is stable.