Q. 3.14

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,

$sec^{2} \theta =\frac{H}{h}$

Verified Solution

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,                                    $sec^{2} \theta =\frac{H}{h}$