Question 2.17: An 80 mm diameter composite solid cylinder consists of an 80...
An 80 mm diameter composite solid cylinder consists of an 80 mm diameter 20 mm thick metallic plate having sp. gr. 4.0 attached at the lower end of an 80 mm diameter wooden cylinder of specific gravity 0.8. Find the limits of the length of the wooden portion so that the composite cylinder can float in stable equilibrium in water with its axis vertical.
Let l be the length of the wooden piece. For floating equilibrium of the composite cylinder,
Weight of the cylinder ≤ Weight of the liquid of the same volume as that of the cylinder
Hence, \frac{\pi(0.08)^{2}}{4}\{0.02 \times 4+0.8 l\} \leq \frac{\pi(0.08)^{2}}{4}\{0.02+l\}
From which l \geq 0.3 m
Hence, the minimum length of the wooden portion l_{\text {minimum }}=0.3 m = 300 mm.
The minimum length corresponds to the situation when the cylinder will just float with its top edge at the free surface (Fig. 2.40a). For any length l greater than 300 mm, the cylinder will always float in equilibrium with a part of its length submerged as shown in (Fig. 2.40b). The upper limit of l would be decided from the consideration of stable equilibrium (angular stability) of the cylinder.
For stable equilibrium,
Metacentric Height > 0 (2.87)
The location of centre of gravity G of the composite cylinder can be found a
O G=\frac{\pi \frac{(.08)^{2}}{4}[.02 \times 4 \times .01+l \times .8(0.5 l+0.02)]}{\pi \frac{(.08)^{2}}{4}(.08+.8 l)}
=\frac{5 l^{2}+0.2 l+0.01}{10 l+1}
The submerged length h of the wooden cylinder is found from the consideration of floating equilibrium as
Weight of the cylinder = Buoyancy force
\frac{\pi(.08)^{2}}{4}(.02 \times 4+.8 l\}=\frac{\pi(.08)^{2}}{4} \times h
or h=0.08(10 l+1) (2.88)
The location of the centre of buoyancy B can therefore be expressed as OB = h/2 = 0.04 (10 l + 1)
Now B G=O G-O B=\frac{5 l^{2}+0.2 l+0.01}{10 l+1}-0.04(10 l+1)
=\frac{l^{2}-0.6 l-.03}{10 l+1} (2.89)
The location of the metacentre M above buoyancy B can be found out 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{I}{V}=\frac{\pi(.08)^{4} \times 4}{64 \times \pi(.08)^{2} \times h} (2.90)
Substituting h from Eq. (2.88) to Eq. (2.90), we get
B M=\frac{.005}{10 l+1}
Therefore, M G=B M-B G=\frac{.005}{10 l+1}-\frac{l^{2}-0.6 l-.03}{10 l+1}
=\frac{-\left(l^{2}-0.6 l-.035\right)}{10 l+1}
Using the criterion for stable equilibrium as MG > 0 we have,
\frac{-\left(l^{2}-0.6 l-.035\right)}{10 l+1}>0
or l^{2}-0.6 l-.035<0
or (l-0.653)(l+0.053)<0
The length l can never be negative. Hence, the physically possible condition is
l-0.653<0
or, l<0.653
