A solid cylinder partially submerged in water having weight density 62.5 lb/ft^3 , with its axis vertical, oscillates up and down within a period of 0.6 sec. Determine the diameter of the cylinder if it weighs 2 lb.

Question

A solid cylinder partially submerged in water having weight density 62.5 [latex]lb / ft ^{3}[/latex] , with its axis vertical, oscillates up and down within a period of 0.6 sec. Determine the diameter of the cylinder if it weighs 2 lb.

Accepted Answer

With [latex] ho=62.5 lb / ft ^{3} ext { and } m=2 / 32 ext { slugs }[/latex], Eq. (14.10) becomes

[latex]\ddot{x}+1000 \pi r^{2} x=0[/latex]

which has as its general solution

[latex]x(t)=c_{1} \cos \sqrt{1000 \pi} r t+c_{2} \sin \sqrt{1000 \pi} r t[/latex] (1)

Its circular frequency is [latex]\omega=r \sqrt{1000 \pi}[/latex]; its natural frequency is [latex]f=\omega / 2 \pi=r \sqrt{250 / \pi}=8.92 r[/latex]; its period is [latex]T=1 / f=1 / 8.92 r[/latex]. We are given [latex]0.6=T=1 / 8.92 r, ext { thus } r=0.187 ft =2.24[/latex] in with a diameter of 4.48 in.

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Eq. (14.10) : [latex]\ddot{x}+\frac{\pi r^{2} ho}{m} x=0[/latex]

Question 14.24: A solid cylinder partially submerged in water having weight ...

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