Question

For small, slowly falling objects the assumption made in the text that the drag force is proportional to the velocity is a good one. For larger, more rapidly falling objects it is more accurate to assume that the drag force is proportional to the square of the velocity." (a) Write a differential equation for the velocity of a falling object of mass \(m\) if the drag force is proportional to the square of the velocity. (b) Determine the limiting velocity after a long time. (c) If \(m=10 \mathrm{kg}\), find the drag cocficient so that the limiting velocity is \(49 \mathrm{m} / \mathrm{sec}\). (d) Using the data in part (c), draw a direction field and compare it with Figure \(1.13 .\)

Short answer

Question: Write the differential equation describing the velocity of a falling object with a drag force proportional to the square of the velocity. Find its limiting velocity, drag coefficient given a mass of 10 kg and a limiting velocity of 49 m/s, and briefly describe how to draw the direction field. Answer: The differential equation describing the velocity of a falling object with a drag force proportional to the square of the velocity is \(m \cdot \frac{dv}{dt} = mg - kv^2\). The limiting velocity can be found using the formula \(v_{lim} = \sqrt{\frac{mg}{k}}\), resulting in \(v_{lim} = 49 m/s\). Given a mass of 10 kg and a limiting velocity of 49 m/s, the drag coefficient is approximately \(k = 0.040923\). To draw a direction field, calculate the slope \(\frac{dv}{dt} = \frac{1}{10}(9.81 - 0.040923 \cdot v^2)\) at different positions on the (t,v) plane and plot arrows with the calculated slopes to represent the behavior of the solution of the differential equation.

Step-by-step solution

(a) Differential equation

Firstly, we realize that the net force acting on the falling object is the gravitational force (downwards) minus the drag force (upwards). The drag force is given to be proportional to the square of the velocity, so we can write this as \(F_{drag} = k\cdot v^2\), where \(k\) is the proportionality constant and \(v\) is the velocity. Newton's second law states that the net force on an object is equal to the mass times the acceleration: \(F_{net} = m\cdot a\). Thus, we have: \(m\cdot a = m\cdot g - kv^2\) Considering that acceleration is the derivative of velocity with respect to time (\(a = \frac{dv}{dt}\)), we arrive at the differential equation: \(m\cdot \frac{dv}{dt} = mg - kv^2\)

(b) Limiting velocity

The limiting velocity is the constant velocity at which the object falls after a long time. When the object reaches limiting velocity, its acceleration becomes zero (as there is no net force acting on the object). Setting \(\frac{dv}{dt} = 0\), we have: \(0 = mg - kv^2\) Solving for \(v\), we find the limiting velocity: \(v_{lim} = \sqrt{\frac{mg}{k}}\)

(c) Finding the drag coefficient

Given \(m = 10 \, kg\) and a limiting velocity of \(v_{lim} = 49 \, m/s\), we can find the drag coefficient by substituting the given values into the formula for limiting velocity: \(49 = \sqrt{\frac{10 \cdot 9.81}{k}}\) Solving for \(k\), we have: \(k = \frac{10 \cdot 9.81}{49^2} \approx 0.040923\) The drag coefficient is approximately \(k = 0.040923\).

(d) Drawing a direction field

A direction field provides a visual representation of the behavior of the solution of the differential equation. It consists of small arrows that represent the slope (\(\frac{dv}{dt}\)) at different positions \((t,v)\) on the graph. To draw a direction field for this differential equation, we first calculate the slope \(\frac{dv}{dt}\) using the data from part (c): \(\frac{dv}{dt} = \frac{1}{10}(9.81 - 0.040923 \cdot v^2)\) Now, we can draw a direction field by plotting arrows on the \((t,v)\) plane with slopes given by the formula above, remembering that the behavior of a falling object in real life is described better by using this direction field. Comparing the generated direction field with Figure 1.13, we should observe similar behaviors in terms of how quickly the object reaches its limiting velocity and how it maintains that velocity after a long time.

Key concepts

Drag Force

The drag force is an opposing force that acts against the motion of an object moving through a fluid, such as air or water. In this problem, we consider a drag force proportional to the square of the velocity, which is commonly seen in situations with higher speeds and larger objects. This formula can be expressed as: \[ F_{\text{drag}} = k \cdot v^2 \]where:

• \( k \) is the drag coefficient, a constant that depends on the object's shape, fluid's characteristics, and other factors.
• \( v \) is the velocity of the object.

As the object moves faster, the drag force increases significantly due to its dependence on the square of the velocity.
This means larger or faster-moving objects face considerable resistance from the surrounding fluid.

Limiting Velocity

Limiting velocity, also known as terminal velocity, is the constant speed reached by an object when the force of gravity pulling it downwards is balanced by the drag force pushing upwards. At this speed, there is no net acceleration, leading to a stable velocity. To find the limiting velocity, set the acceleration to zero in the governing differential equation: \[ \frac{dv}{dt} = 0 \]From the equation \( 0 = mg - kv^2 \), solve for \( v \): \[ v_{\text{lim}} = \sqrt{\frac{mg}{k}} \]where:

• \( m \) is the mass of the object.
• \( g \) is the acceleration due to gravity (approximately 9.81 \( \text{m/s}^2 \)).
• \( k \) is the drag coefficient.

Limiting velocity is influenced by both the mass of the object and the drag coefficient, which dictate how quickly the forces balance out.

Newton's Second Law

Newton's second law provides a framework for understanding how the forces acting on an object determine its acceleration. Stated as: \[ F_{\text{net}} = m \cdot a \]where:

• \( F_{\text{net}} \) is the net force acting on the object.
• \( m \) is the mass of the object.
• \( a \) is the acceleration of the object.

In this scenario, the net force becomes the gravitational force minus the drag force: \[ m \cdot g - k \cdot v^2 = m \cdot \frac{dv}{dt} \]Using this equation, one can set up the differential equation representing the motion of a falling object with drag. Solving this gives insights into how the forces work together to determine velocity and acceleration.

Direction Field

A direction field, or slope field, graphically represents the solutions to a differential equation. It helps visualize how the velocity of a falling object evolves over time. To create a direction field for the differential equation: \[ \frac{dv}{dt} = \frac{1}{m}(mg - kv^2) \]Evaluate the slope \( \frac{dv}{dt} \) at various points \((t, v)\). Each point will have an arrow indicating the direction of the solution at that point.Plotting these arrows on the \((t, v)\) plane reveals the "flow" of solutions, showing:

• How the velocity changes over time.
• The approach towards limiting velocity.

Comparing this field to empirical data, like Figure 1.13 mentioned in the solution, illustrates how long it takes for the object to reach and maintain its limiting velocity.