Question

An object of mass \(m\) falls in a medium that exerts a resistive force \(f=f(s),\) where \(s=|v|\) is the speed of the object. Assume that \(f(0)=0\) and \(f\) is strictly increasing and differentiable on \((0, \infty)\) (a) Write a differential equation for the speed \(s=s(t)\) of the object. Take it as given that all solutions of this equation with \(s(0) \geq 0\) are defined for all \(t>0\) (which makes good sense on physical grounds). (b) Show that if \(\lim _{s \rightarrow \infty} f(s) \leq m g\) then \(\lim _{t \rightarrow \infty} s(t)=\infty\). (c) Show that if \(\lim _{s \rightarrow \infty} f(s)>m g\) then \(\lim _{t \rightarrow \infty} s(t)=s_{T}\) (terminal speed), where \(f\left(s_{T}\right)=\) \(\mathrm{mg} .\) HINT: Use Theorem 2.3.1.

Short answer

Answer: (1) If the limiting value of the resistive force is less than or equal to the object's weight, the speed of the object will increase without bound as time goes to infinity. (2) If the limiting value of the resistive force is greater than the object's weight, the object will eventually reach a terminal velocity, at which the resistive force equals the object's weight.

Step-by-step solution

a. Writing the Differential Equation for speed s(t)

We start by analyzing the forces acting on the object. As it falls, there are two forces: gravity (with weight, mg) pulling the object downwards, and the resistive force, f(s), acting against motion. From Newton's second law, we have: m\frac{ds}{dt} = mg - f(s), where ds/dt is the time derivative of speed. This is our first differential equation, governing the object's motion.

b. Speed diverges if the resistive force is always less than object's weight

We first assume that the limiting value of f(s) is less than or equal to the object's weight (mg). We want to show that, under this condition, the speed will increase without bound as time goes to infinity: \(\lim_{s\rightarrow\infty} f(s) \leq mg\) We integrate the given differential equation from part (a): m\frac{ds}{dt} = mg - f(s) Dividing both sides by m and integrating with respect to time, we obtain: \(\int \frac{ds}{dt} dt = \int g - \frac{f(s)}{m} dt\) Which gives: \(s(t) = gt - \int\frac{f(s)}{m} dt\) Since \(f(s)\) is strictly positive and increasing, the term \(\int\frac{f(s)}{m} dt\) will also be positive. Therefore, as time goes to infinity, the left side must also go to infinity to balance the equation: \(\lim_{t\rightarrow\infty} s(t) = \infty\)

c. Terminal Speed Reached when the Resistive Force Exceeds Object's Weight

In this part of the problem, we assume that the limiting value of f(s) is greater than the object's weight (mg): \(\lim_{s \rightarrow \infty} f(s) > m g\) We first derive that \(f(s_T) = mg\): \(m\frac{ds}{dt} = mg - f(s)\) At terminal speed, this rate of change equals zero: \(0 = m\frac{ds}{dt} = mg - f(s_T)\) Which gives: \(f(s_T) = mg\) As \(f(s)\) is strictly increasing and differentiable, we can use the theorem 2.3.1, which states that for an increasing function with continuous derivative, an initial value problem with a continuous and increasing function will have a unique solution. In this case, our function is: \(\frac{ds}{dt} = g - \frac{f(s)}{m}\) Since the limiting value of this function exceeds mg, it guarantees that a terminal velocity (\(s_T\)) will be reached such that: \(f(s_T) = mg\) And we have shown that: \(\lim_{t\rightarrow\infty} s(t) = s_T\)

Key concepts

Resistive Force

When an object moves through a medium like air or water, it encounters a resistive force. This force opposes the motion, effectively slowing the object down. The resistive force depends on the speed of the object, noted as \( s = |v| \). In mathematical terms, the resistive force is often represented as \( f = f(s) \).
In our problem, it's given that \( f(0)=0 \), meaning there is no resistive force when the object is stationary. The resistive force is strictly increasing and differentiable on the interval \((0, \infty)\). This property ensures that as the speed increases, the resistive force also increases in a predictable, smooth manner.

• Strictly Increasing: This means that \( f(s_1) < f(s_2) \) whenever \( s_1 < s_2 \).
• Differentiable: Means that the function has a smooth change, without any breaks or sharp angles.

This resistive force is crucial in determining how the object behaves under motion, especially as it relates to the concepts of terminal speed and the balance of forces according to Newton's second law.

Newton's Second Law

Newton's Second Law of Motion is a fundamental principle in physics that describes how the motion of an object is influenced by the forces acting upon it. It is expressed as \( F = ma \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.
For our falling object, the forces include gravity that pulls it down, represented by \( mg \), and the resistive force \( f(s) \) that acts against this motion. Allowing us to write the equation:

• Net force acting on the object: \( mg - f(s) \).
• Express acceleration as the derivative of speed: \( a = \frac{ds}{dt} \).

Thus, Newton's second law in our context becomes \( m\frac{ds}{dt} = mg - f(s) \). This differential equation models how the object's speed changes over time under the influence of gravity and resistive forces. It is essential for predicting how the object accelerates and eventually reaches terminal speed.

Terminal Speed

Terminal speed is a key concept in understanding the motion of objects through a resistive medium. It is the constant speed that an object reaches when the net force acting on it becomes zero. This happens when the downward force of gravity is exactly balanced by the upward resistive force, such that the object stops accelerating.
In our problem, we find terminal speed \( s_T \) by setting the net force equal to zero in the differential equation:

• Therefore, \( 0 = mg - f(s_T) \), leading to \( f(s_T) = mg \).

This equation indicates that when the resistive force equals the weight of the object (\( mg \)), the object has reached its terminal speed. From this point onward, the object travels at a constant speed, as the resistive and gravitational forces perfectly counterbalance each other.
Understanding terminal speed helps predict how fast an object will fall under specific conditions and is crucial for designing parachutes, vehicles, and more that encounter resistive forces.

Limit of a Function

Limits are a fundamental concept in calculus that describe the behavior of a function as its inputs approach certain values. In our exercise, limits help us understand the behavior of the resistive force as speed approaches infinity, influencing the speed of the falling object over time.
Consider the two key limits given in the problem:

• First Scenario: If \( \lim_{s \rightarrow \infty} f(s) \leq mg \), the resistive force never surpasses the weight, hence \( \lim_{t \rightarrow \infty} s(t) = \infty \). This indicates that the object's speed would continue to grow indefinitely as it falls.
• Second Scenario: If \( \lim_{s \rightarrow \infty} f(s) > mg \), the resistive force exceeds the gravitational force, resulting in \( \lim_{t \rightarrow \infty} s(t) = s_T \). This indicates that the object reaches a terminal speed.

By utilizing the limit of a function, we predict the long-term behavior of systems like falling objects in resistive environments, providing insight into their dynamic and eventual steady states.