# Buoyant object problem

1. Jan 27, 2009

### denysy1

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
F=m*a
viscous drag=c*v

3. The attempt at a solution
F=ma
$$ma=cv+ \pi \rho {R }^{ 2} gx-mg$$
$$mx''-cx'- \pi \rho { R}^{ 2} gx=-mg$$

Use trial solution $$x= {e }^{ rt}$$
$$m { r}^{2 } -br-\pi \rho { R}^{ 2} g=0$$
Substituting values, and solving using quadratic equation
r1=7.3274 ; r2=-1.8544

Complimentary solution:
$$x= {C }_{1 } {e }^{7.3274t } + {C }_{ 2} {e }^{ -1.8544t}$$

Sub in initial conditions
$$0= {C }_{1 } + {C }_{ 2}$$
Differentiate complimentary solution with respect to t
$$x'=7.3274 {C }_{ 1} {e }^{ 7.3274t} -1.8544 {C }_{ 2} {e }^{-1.8544t }$$
Sub in initial conditions
$$0.13=7.3274 {C }_{ 1} -1.8544 {C }_{ 2}$$
Solving for C1 and C2 yields
C1=0.014158 ; C2=-0.014158

Complimentary solution:
$$x=0.014158 {e }^{7.3274t } -.014158 { e}^{ -1.8544t}$$

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