(^752) „„„„„ A Textbook of Engineering Mechanics
We also know that moment of inertia of the circular section about its centre of gravity
I=
()^4
64
d
π
and volume of water displaced, V=
2
2 2
436
ldl
d
ππ
××= ...(ii)
∴ BM=
4
2
2
() 3
64
32
6
Idd
Vldl
π

π
...(iii)
For stable equilibrium, the metacentre (M) should be above the centre of gravity (G) or may
coincide with G.
i.e. BGñBM
6
l
ñ
3 2
32
d
l
l^2 ñ
18 2
32
d
ñ
9 2
16
d
or lñ
3
4
d ...(Taking square root)
ñ0·75 d
It means that the cylinder cannot float with its longitudinal axis vertical, when the length
exceeds 0.75 times of its diameter. Ans.
Example 36.9. A solid cylinder 1 m long 0.2 m diameter has its base 25 m thick of an alloy
with specific gravity 8. The remaining portion is of specific gravity 0.5.Can it float vertically in water?
If not, what is the maximum permissible length for stable equilibrium?
Solution. Given: Length of the cylinder (l) = 1 m = 100 cm; Diameter of the cylinder
(d) = 0.2 m = 20 cm; Thickness of base = 25 mm = 2.5 cm; sp. gr. of base = 8 and sp. gr. of remaining
portion = 0.5.
Floating of the cylinder
We know that cross-sectional area of the cylinder,
A= (20)^22100 cm
4
π
=π
and distance between the combined centre of gravity (G) and bottom of the cylinder (O)
OG=
97.5 2.5
0.5 97.5 2.5 8 2.5
22
(0.5 97.5) (8 2.5)
AA
AA
⎡⎤⎛⎞⎡ ⎤
⎢⎥×++××⎜⎟⎢⎥
⎣⎦⎝⎠⎣ ⎦
×+×

2498.4 25
48.75 20
A A
A A

• 
  


• 
  

  = 36.7 cm ...(where A is the area of cylinder)