Answer:
[tex]GM<0[/tex]
So the bouy does not float with its axis vertical
Explanation:
From the question we are told that:
Diameter [tex]d=2m[/tex]
Length [tex]l=2.5m[/tex]
Weight [tex]W=22kN[/tex]
Specific weight of sea water [tex]\mu= 10.25kN/m^3[/tex]
Generally the equation for weight of cylinder is mathematically given by
Weight of cylinder = buoyancy Force
[tex]W=(pwg)Vd[/tex]
Where
[tex]V_d=\pi/4(d)^2y[/tex]
Therefore
[tex]22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m[/tex]
Therefore
Center of Bouyance B
[tex]B=\frac{y}{2}=0.26m\\\\B=0.75[/tex]
Center of Gravity
[tex]G=\frac{I.B}{2}=2.6m[/tex]
Generally the equation for\BM is mathematically given by
[tex]BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\[/tex]
Therefore
[tex]BG=2.6-0.476\\\\BG=0.64m[/tex]
Therefore
[tex]GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\[/tex]
Therefore
[tex]GM<0[/tex]
So the bouy does not float with its axis vertical
