Question 3.14: A cone of height H is floating in water with its vertex down......
A cone of height H is floating in water with its vertex downward (Figure 3.32). If h is the depth of immersion and 20 is the angle of the cone at vertex, show that for stable equilibrium,
sec^{2} \theta =\frac{H}{h}
If G is the centre of gravity of the cone,
OG=(\frac{3}{4} )H
If B is the centre of buoyancy,
OB=(\frac{3}{4} )h
and if r is the radius of the cone at water surface, then
r = h \tan \theta
Here θ is the half of the cone at vertex.
Volume of water displaced by the cone = (\frac{1}{3} )\pi r^{2}h
Moment of inertia I at the water section = (\frac{\pi }{64} )(2r)^{4}=\frac{\pi r^{4} }{4}
Using the formula,
BM=\frac{\pi r^{4} }{4\times ({1}/{3})\pi r^{2}h }=\frac{3}{4}\frac{r^{2} }{h}=\frac{3}{4}h\tan ^{2}\theta
If M is the metacentre, then
OM= OB + BM = (\frac{3}{4} )h+(\frac{3}{4} )h\tan ^{2} \theta=(\frac{3}{4} )h(1+\tan ^{2} \theta )=(\frac{3}{4} )h sec^{2} \theta
For stable equilibrium OM > OG, i.e., metacentre M should be above G.
Or (\frac{3}{4} )h sec^{2} \theta \gt (\frac{3}{4} )H
Therefore, sec^{2} \theta =\frac{H}{h}
