How Does a Barrel's Motion Change When Submerged in Water?

• denysy1
In summary, the problem involves a barrel being dropped into the ocean and the depth of the barrel's bottom as a function of time needs to be determined. The barrel is acted upon by gravity, viscous resistance, and buoyancy force. Using the given values and solving for the complimentary solution, the depth x is found to be a combination of two exponential functions. To find the particular solution and plot the depth for the first 5 seconds, more information is needed. When the barrel becomes completely submerged, the buoyancy force becomes greater than the weight of the barrel and the barrel stops sinking.
denysy1

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?

F=m*a
viscous drag=c*v

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.

To plot the depth x for the first 5 seconds, I will need the particular solution. As for what changes when the barrel is completely submerged, I am not sure.

Can you please provide some guidance?

I would suggest using the principle of superposition to solve for the particular solution. This principle states that the total solution is equal to the sum of the complimentary solution and the particular solution.

To find the particular solution, we need to consider the forces acting on the barrel. First, we have the buoyancy force, which is equal to the weight of the water displaced by the barrel. This can be calculated using the density of seawater and the volume of the barrel.

Next, we have the viscous resistance force, which is proportional to the velocity of the barrel. This can be calculated using the coefficient of linear resistance and the velocity of the barrel.

Finally, we have the downward force of gravity acting on the barrel.

Using the principle of superposition, we can add these three forces together to find the particular solution. Once we have the particular solution, we can add it to the complimentary solution to get the total solution.

When the barrel becomes completely submerged, the buoyancy force will no longer be acting on the barrel. This will result in a change in the particular solution and therefore the total solution. The barrel will continue to sink until the downward force of gravity is equal to the upward buoyancy force of the water displaced by the barrel.

I would suggest using Maple to plot the depth x for the first 5 seconds or until the barrel becomes completely submerged. This will give a visual representation of how the depth changes over time and can help to better understand the problem.

1. What is the buoyant object problem?

The buoyant object problem is a scientific concept that involves understanding how objects float or sink in a fluid, typically a liquid. It is related to the laws of buoyancy, which state that the buoyant force acting on an object is equal to the weight of the fluid displaced by that object.

2. How does the shape of an object affect its buoyancy?

The shape of an object can significantly impact its buoyancy. Objects with a larger surface area will displace more fluid, resulting in a greater buoyant force. Additionally, objects with a concave shape, such as a boat hull, can trap air beneath them, increasing their buoyancy.

3. What factors determine whether an object will float or sink?

The buoyancy of an object is determined by comparing its weight to the buoyant force exerted on it by the fluid. If the object's weight is greater than the buoyant force, it will sink. If the buoyant force is greater than the object's weight, it will float.

4. Does the density of an object affect its buoyancy?

Yes, the density of an object plays a significant role in its buoyancy. Objects with a higher density than the fluid they are in will sink, while objects with a lower density will float. This is why materials like wood, which have a lower density than water, can float on its surface.

5. How can we calculate the buoyant force on an object?

The buoyant force on an object can be calculated by multiplying the density of the fluid by the volume of the displaced fluid and the acceleration due to gravity. This can be expressed as Fb = ρVg, where ρ is the density, V is the volume, and g is the acceleration due to gravity.

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