A solid cylinder is 3.0 ft in diameter, 6.0 ft high, and weighs 1550 lb. If the cylinder is placed in oil (sg = 0.90) with its axis vertical, would it be stable?

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Question

A solid cylinder is 3.0 ft in diameter, 6.0 ft high, and weighs 1550 lb. If the cylinder is placed in oil (sg = 0.90) with its axis vertical, would it be stable?

The complete solution is shown in the next panel. Do this problem and then look at the solution.

Accepted Answer

Position of cylinder in oil (Fig. 5.15):

[latex]V_d = [/latex] submerged volume = AX = [latex]=\frac{\pi D^2}{4} \left(X\right) [/latex]

Equilibrium equation:

[latex] \sum{F_v} =0 [/latex]

[latex]W = F_b = \gamma_0 V_d =\gamma_0 \frac{\pi D^2}{4}\left(X\right) [/latex]

[latex]X = \frac{4W}{\pi D^2 \gamma_0} =\frac{\left(4\right)\left(1550lb\right)ft^3 }{\left(\pi \right)\left(3.0ft\right)^2 \left(0.90\right) \left(62.4lb\right) } =3.90ft[/latex]

The center of buoyancy cb is at a distance X/2 from the bottom of the cylinder.

[latex]y_{cb} = X/2 = {3.90ft}/{2} = 1.95ft[/latex]

The center of gravity cg is at H/2 = 3.0 ft from the bottom of the cylinder, assuming the material of the cylinder is of uniform specific weight. The position of the metacenter mc, using Eq. (5–5), is

[latex]MB= I/V_d[/latex]

[latex]I= \frac{\pi D^4}{64} = \frac{\pi\left(3.0ft\right)^4 }{64} =3.98ft^4 [/latex]

[latex]V_d = AX = \frac{\pi D^2}{4} \left(X\right) = \frac{\pi \left(3.0ft\right)^2 }{4} \left(3.90ft\right) = 27.6ft^3[/latex]

[latex]MB= I/V_d = 3.98ft^4/27.6ft^3 = 0.144ft[/latex]

[latex]y_{mc} = y_{cb} + MB =1.95ft + 0.14ft = 2.09ft[/latex]

Question 5.6: A solid cylinder is 3.0 ft in diameter, 6.0 ft high, and wei...

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