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denysy1

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## Homework Statement

A barrel is dropped into the ocean, striking the surface of the water in an upright position at time t=0 with a velocity v=0.13 m/s. In addition to the downward force of gravity, the barrel is acted upon by a force of viscous resistance proportional to its velocity, and an upward buoyancy force equal to weight of the water displaced by the barrel.

Model the barrel as a cylinder of radius R=0.15 m and height H=1.11 m, and assume that its density ρ=666 kg/m¬3 is approximately constant. Take the density of seawater to be ρw=1024 kg/m3, and the coefficient of linear resistance appropriate for movement through water to be c=286 kg/s.

Assuming that the barrel remains always upright, find the depth x of the bottom of the barrel as a function of time t. Use Maple to plot the depth x for the first 5 seconds OR until the barrel becomes completely submerged. What changes when the barrel becomes completely submerged?

## Homework Equations

F=m*a

viscous drag=c*v

## The Attempt at a Solution

F=ma

[tex]ma=cv+ \pi \rho {R }^{ 2} gx-mg[/tex]

[tex]mx''-cx'- \pi \rho { R}^{ 2} gx=-mg[/tex]

Use trial solution [tex]x= {e }^{ rt} [/tex]

[tex]m { r}^{2 } -br-\pi \rho { R}^{ 2} g=0[/tex]

Substituting values, and solving using quadratic equation

r1=7.3274 ; r2=-1.8544

Complimentary solution:

[tex]x= {C }_{1 } {e }^{7.3274t } + {C }_{ 2} {e }^{ -1.8544t} [/tex]

Sub in initial conditions

[tex]0= {C }_{1 } + {C }_{ 2} [/tex]

Differentiate complimentary solution with respect to t

[tex]x'=7.3274 {C }_{ 1} {e }^{ 7.3274t} -1.8544 {C }_{ 2} {e }^{-1.8544t } [/tex]

Sub in initial conditions

[tex]0.13=7.3274 {C }_{ 1} -1.8544 {C }_{ 2} [/tex]

Solving for C1 and C2 yields

C1=0.014158 ; C2=-0.014158

Complimentary solution:

[tex]x=0.014158 {e }^{7.3274t } -.014158 { e}^{ -1.8544t} [/tex]

Is this correct so far? Now, I am not sure how to deal with a constant forcing function to find the particular solution.