Question 2.16: A cone floats in water with its apex downward (Fig. 2.39) an...
A cone floats in water with its apex downward (Fig. 2.39) and has a base diameter D and a vertical height H. If the specific gravity of the cone is S, prove that for stable equilibrium,
H^{2}<\frac{1}{4}\left(\frac{D^{2} S^{1 / 3}}{1-S^{1 / 3}}\right)
Let the submerged height of the cone under floating condition be h, and the diameter of the cross-section at the plane of flotation be d (Fig. 2.39).
For the equilibrium,
Weight of the cone = Total buoyancy force
\frac{1}{3} \pi\left(\frac{D^{2}}{4}\right) H \cdot S=\frac{1}{3} \pi\left(\frac{d^{2}}{4}\right) \cdot h (2.83)
Again from geometry,
d=D \frac{h}{H} (2.84)
Using the value of d from Eq. (2.84) in Eq. (2.83), we get
h=H S^{1 / 3} (2.85)
The centre of gravity G of the cone is found out by considering the mass of cylindrical element of height dz and diameter Dz/H, and its moment about the apex 0 in the following way:
O G=\frac{\int_{0}^{H} \pi \frac{D^{2} z^{2}}{4 H^{2}} z d z}{\frac{1}{3} \pi \frac{D^{2}}{4} H}=\frac{3}{4} H
The centre of buoyancy B is the centre of volume of the submerged conical part and hence O B=\frac{3}{4} h.
Therefore B G=O G-O B=\frac{3}{4}(H-h)
Substituting h from Eq. (2.85) we can write
B G=\frac{3}{4} H\left(1-S^{1 / 3}\right) (2.86)
If M is the metacentre, the metacentric radius
BM can be written according to Eq. (2.64) as
=\frac{\text {Second moment of area of the plane of flotation about the centroidal axis perpendicular to plane of rotation }}{\text { Immersed volume }} (2.64)
B M=\frac{1}{V}=\frac{\pi d^{4}}{64 \times \frac{1}{3} \pi\left(d^{2} / 4\right) h}=\frac{3}{16} \frac{d^{2}}{h}
Substituting d from Eq. (2.84) and h from Eq. (2.85), we can write
B M=\frac{3}{16} \frac{D^{2}}{H} S^{1 / 3}
The metacentric height
M G=B M-B G
=\frac{3}{16} \frac{D^{2}}{H} S^{1 / 3}-\frac{3}{4} H\left(1-S^{1 / 3}\right)
For stable equilibrium, MG > 0
Hence \frac{3}{16} \frac{D^{2}}{H} S^{1 / 3}-\frac{3}{4} H\left(1-S^{1 / 3}\right)>0
or, \frac{D^{2}}{H^{2}} S^{1 / 3}-4\left(1-S^{1 / 3}\right)>0
or, \frac{D^{2}}{H^{2}} S^{1 / 3}>4\left(1-S^{1 / 3}\right)
or, \frac{H^{2}}{D^{2} S^{1 / 3}}<\frac{1}{4\left(1-S^{1 / 3}\right)}
Hence, H^{2}<\frac{D^{2} S^{1 / 3}}{4\left(1-S^{1 / 3}\right)}