Problem 17 - Advanced Engineering Mathematics

Question

(a) Let \(\rho\) be the weight density of the water and \(V\) the volume of the object. Archimedes' principle states that the upward buoyant force has magnitude equal to the weight of the water displaced. Taking the positive direction to be down, the differential equation is \\[ m \frac{d v}{d t}=m g-k v^{2}-\rho V \\] (b) Using separation of variables we have \\[ \begin{aligned} \frac{m d v}{(m g-\rho V)-k v^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{\sqrt{k} d v}{(\sqrt{m g-\rho V})^{2}-(\sqrt{k} v)^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{1}{\sqrt{m g-\rho V}} \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g-\rho V}} &=t+c \end{aligned} \\] Thus \\[ v(t)=\sqrt{\frac{m g-\rho V}{k}} \tanh \left(\frac{\sqrt{k m g-k \rho V}}{m} t+c_{1}\right) \\] (c) since \(\tanh t \rightarrow 1\) as \(t \rightarrow \infty,\) the terminal velocity is \(\sqrt{(m g-\rho V) / k}.\)

Step-by-Step Solution

Answer

The terminal velocity is \( \sqrt{\frac{mg - \rho V}{k}} \).

Step 1: Understand the Differential Equation

The problem involves finding solutions to a differential equation based on Archimedes' principle. The differential equation given is: \\[ m \frac{dv}{dt} = mg - kv^2 - \rho V \]. \This equation models a situation where an object is submerged in a fluid with density \( \rho \), experiencing a buoyant force.

Step 2: Identify Separation of Variables

We rewrite the differential equation for separation of variables. It is expressed as follows: \\[ \frac{m dv}{(mg - \rho V) - kv^2} = dt \]. \This new form allows us to separate the terms involving velocity \( v \) on one side and time \( t \) on the other side.

Step 3: Simplify and Integrate

The equation is further simplified for integration: \\[ \frac{m}{\sqrt{k}} \frac{\sqrt{k} dv}{(\sqrt{mg - \rho V})^2 - (\sqrt{k} v)^2} = dt \]. \To solve, integrate both sides. The left side yields the inverse hyperbolic tangent function: \\[ \frac{m}{\sqrt{k}} \frac{1}{\sqrt{mg - \rho V}} \tanh^{-1} \frac{\sqrt{k} v}{\sqrt{mg - \rho V}} = t + c \].

Step 4: Solve for Velocity Function

From the integration result, solve for the velocity \( v(t) \). The expression for \( v(t) \) is given by: \\[ v(t) = \sqrt{\frac{mg - \rho V}{k}} \tanh \left( \frac{\sqrt{k(mg - k\rho V)}}{m} t + c_1 \right) \]. \This is the velocity of the object as a function of time.

Step 5: Determine Terminal Velocity

Terminal velocity occurs as \( t \rightarrow \infty \), where \( \tanh(t) \rightarrow 1 \). Thus, \( v(t) \) approaches the constant value: \\[ v_{terminal} = \sqrt{\frac{mg - \rho V}{k}} \]. \This represents the maximum speed the object reaches under the influence of the buoyant and drag forces.

Key Concepts

Archimedes' Principle

Archimedes' principle is a fundamental concept in fluid mechanics. It states that any object, fully or partially submerged in a fluid, experiences an upward buoyant force equal to the weight of the fluid that the object displaces. This principle is crucial for understanding how objects behave when placed in water or any other fluid.
Let's break it down:

• The **buoyant force** acts in the opposite direction to gravity, which is why objects often feel lighter in water than they do in air.
• For example, if you let go of a rock underwater, you notice it rises slightly before settling. This upward force is due to Archimedes' principle.
• The condition for floating or sinking depends on whether the object's weight is greater or less than the buoyant force.

Understanding Archimedes' principle helps in solving differential equations involving buoyant forces, as seen in the given exercise, where the weight of displaced water (\( ho V \)) contributes to the dynamics of the object's motion.

Terminal Velocity

Terminal velocity is the constant speed that a freely falling object reaches when the resistance of the medium (often air or water) prevents further acceleration. In the context of fluids, there are several forces acting on an object:

• **Gravity**, which pulls the object downwards.
• The **buoyant force**, counteracting gravity.
• **Drag**, which slows the object as it moves through the fluid.

As time progresses, the net force acting on the object goes to zero, and it stops accelerating. Consequently, the object moves at constant speed, known as terminal velocity.
This can be mathematically expressed using the hyperbolic tangent function, as in the solution provided. When time \( t \) approaches infinity, the term \( \tanh(t) \) approaches 1, leading to the conclusion of a terminal speed given by:\[v_{terminal} = \sqrt{\frac{mg - \rho V}{k}} \] Here, \( mg \) is the gravitational force, \( \rho V \) is the buoyant force, and \( kv^2 \) is the drag force.

Separation of Variables

Separation of variables is a technique used to solve ordinary differential equations (ODEs). It works by separating the variables on either side of the equation so that each can be integrated separately. This method can be particularly handy for equations involving exponential growth or decay, harmonic oscillations, or, in this case, fluid dynamics.
Let's see how it applies to this exercise:

• The original differential equation is: \(m \frac{dv}{dt} = mg - kv^2 - \rho V \)
• The objective is to isolate \( v \) terms on one side, \( t \) terms on the other:
\[\frac{m dv}{(mg - \rho V) - kv^2} = dt \]
• With the variables now separated, each side can be integrated individually.

Using this method, we eventually solve for the velocity \( v(t) \), understanding how it evolves over time with buoyant and drag forces at play. By integrating, we find an expression that involves inverse hyperbolic functions, which eventually lead us to a velocity function indicating terminal velocity as time approaches infinity.