(^748) „„„„„ A Textbook of Engineering Mechanics
Let h= Distance between the water surface and top of the buoy,
B 1 = Centre of buoyancy of the cylindrical buoy
Fig. 36.4.
We know that volume of water displaced by the cylindrical portion
= 2·6 – 0·4 = 2·2 m^3
∴ 2·2 =
( 2 )^2 (1· 2 – ) (1· 2 – )
4
hh
π
×=π
or (1·2 – h)=
2·2
=0·7
π
∴ h= 1·2 – 0·7 = 0·5 m
Distance of the centre of buoyancy of the cylindrical buoy from the top of the buoy,
OB 2 =
(1·2–0·5)
0·5 0·85 m
2
+=
Now let B= Centre of buoyancy for the whole buoy,
∴ OB=
(0·4 1·3) (2·2 0·85)
0·92 m
0·4 2·2
×+ ×

• 
  

  Now BG=OB – OG = 0·92 – 0·8 = 0·12 m
  We also know that moment of inertia of the cylindrical portion (top portion)
  about its centre of gravity,
  I=
  (2)^44 0·7854 m
  64
  π
  ×=
  ∴ BM=
  0·7854
  0·302 m
  2·6
  I
  V

  
  

  and metacentric height, GM=BM – BG = 0·302 – 0·12 m
  = 0·182 m = 182 mm Ans. Fig. 36.5.
  O
  M
  G
  B
  B 2
  B 1