An 80 mm diameter composite solid cylinder consists of an 80 mm diameter 20 mm thick metallic plate having specific gravity 4 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 the water with its axis vertical. -

Given:

$d_m=d_w=d=80 mm \text{(Diameter of cylinder)} \\ t_m=20 mm \text{(Thickness of metallic plate)} \\ \text{Specific gravity of metal plate=4} \\ \text{Specific gravity of wood}=0.8$

To find:

Limits of the length of wooden portion $(t_w)$

Sol:

In order to obtain the limits of the length of the wooden portion, we need to understand the lower limit to the length would be that length which would just cause floatation. And the upper limit to the length would be that length, which if increased further would cause the composite cylinder to float in unstable equilibrium in the water with its axis vertical.

Consider the figures which explains the concept,

For the lower limit, we apply the condition that the weight of the composite cylinder is equal to the weight of the water displaced,

$(t_w ρ_w+t_m ρ_m )Ag=(t_w+t_m ) ρ_{water} Ag$

Now,

$ρ_w=ρ_{water}∙\text{Specific gravity of wood and} ρ_m=ρ_{water}∙\text{Specific gravity of metal} \\ ρ_{water}=1000 kg/m^3 \\ ∴t_w ρ_{water} (0.8)+t_m ρ_{water} (4)=(t_w+t_m ) ρ_{water} \\ ∴t_w (0.8)+t_m (4)=(t_w+t_m ) \\ ∴t_w (0.8)+0.02(4)=(t_w+0.02) \\ ∴0.2t_w=0.06 \\ ∴t_w=0.3 m \\ \text{…this is the lower limit to the length of the wooden portion.}$

For the upper limit, we apply the condition for stable equilibrium which says,

$$\bar{CM} \gt 0 \text{for stable equilibrium}$$

We know that,

$\bar{CM} =\bar{BM}-\bar{BC}$

where,

$\bar{BC}=\bar{OC}-\bar{OB} \\ \bar{OC}=\dfrac{\bigg[\dfrac{t_m}{2}t_mρ_mA+\bigg(\dfrac{t_w}{2}+t_m\bigg)t_wρ_wA\bigg]}{[t_mρ_mA+t_wρ_wA]} \\ \therefore \bar{OC}=\dfrac{\bigg[\dfrac{t_m^2}{2}t_mρ_m+\bigg(\dfrac{t_w}{2}+t_m\bigg)t_wρ_w\bigg]}{[t_mρ_m+t_wρ_w]} \\ \therefore \bar{OC}=\bigg[\dfrac{0.02^2}{2}4000+\bigg(\dfrac{t_w}{2}+0.02\bigg)t_w800\bigg]\dfrac{1}{0.02(4000)+800t_w} \\ ∴\bar{OC} ̅=\dfrac{(0.8+400t_w^2+16t_w )}{(80+800t_w )} \\ And,\\ \bar{OC}=\dfrac h2 \\ \text{For the length h,} \\ \text{we obtain it by equating the weight of the composite cylinder to the weight of the water displaced,} \\ ∴(t_w ρ_w+t_m ρ_m )Ag=hρ_{water} Ag \\ ∴(t_w ρ_w+t_m ρ_m )=hρ_{water} \\ ∴[800t_w+4000(0.02)]=1000h \\ ∴(0.8t_w+0.08)=h \\ ∴h=0.8t_w+0.08 \\ ∴\bar{BC}=\dfrac{(0.8+400t_w^2+16t_w )}{(80+800t_w )}-\dfrac h2 \\ ∴\bar{BC}=\dfrac{(0.8+400t_w^2+16t_w )}{(80+800t_w )}-\dfrac{(0.8t_w+0.08)}{2} \\ ∴\bar{BC}=\dfrac{(0.8+400t_w^2+16t_w )}{(80+800t_w )}-\dfrac{(0.8t_w+0.08)}{2} \\ ∴\bar{BC}=\dfrac{(80t_w^2-48t_w-2.4)}{(80+800t_w )} \\ And,\\ \bar{BM}=\dfrac IV \\ I→\text{Area moment of inertia of the cylinder planform at the liquid level} \\ V→\text{Volume of water displaced by cylinder} \\ I=\dfrac{π}{64} d^4 \\ V=hA=\dfrac π4 d^2 h \\ ∴\bar{BM} =\dfrac{\dfrac{π}{64} d^4}{\dfrac{π}{4} d^2 h} \\ ∴\bar{BM} =\dfrac{d^2}{16h} \\ ∴\bar{BM} =\dfrac{0.08^2}{16h} \\ ∴\bar{BM} =\dfrac{4∙10^{-4}}{h} \\ ∴\bar{BM} =\dfrac{4∙10{-4}}{(0.8t_w+0.08)} \\ ∴\bar{BM} =\dfrac{1}{(2000t_w+200)} \\ Now, \\ \bar{CM}=\bar{BM}-\bar{BC} \\ ∴\bar{CM}=\dfrac{1}{(2000t_w+200)}-\dfrac{(80t_w^2-48t_w-2.4)}{(80+800t_w )} \gt 0 \\ ∴\dfrac{1}{(2000t_w+200)}-\dfrac{(80t_w^2-48t_w-2.4)}{(80+800t_w )} \\ ∴(80+800t_w )\gt(2000t_w+200)(80t_w^2-48t_w-2.4) \\ ∴160000t_w^3-80000t_w^2-15200t_w-560 \lt 0 \\ ∴t_w \lt 0.6536 m \\ \text{…this is the upper limit to the length of the wooden portion.}$