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Chapter 13: Problem 44

A parachutist having a mass \(m\) opens his parachute from an at-rest position at a very high altitude. If the atmospheric drag resistance is \(F_{D}=k v^{2},\) where \(k\) is a constant, determine his velocity when he has fallen for a time \(t .\) What is his velocity when he lands on the ground? This velocity is referred to as the terminal velocity, which is found by letting the time of fall \(t \rightarrow \infty\).

Short Answer

Expert verified
The velocity of the parachutist after time \(t\) is given by \(v(t) = \sqrt{\frac{g}{k}} \tanh(\sqrt{gk/m} t)\). The terminal velocity when he lands is \(\sqrt{\frac{g}{k}}\).

Step by step solution

01

Understand the Problem and Gather Information

The parachutist has an initial velocity of 0. The force of gravity on the parachutist is \(mg\) and the drag force is \(F_{D}=kv^{2}\). In the vertical direction, the net force acting on the parachutist is the difference between the gravity force and drag, which equals to his mass times acceleration. \(ma = mg - kv^{2}\). Since acceleration is the derivative of velocity with respect to time, we can write \(m \frac{dv}{dt} = mg - kv^{2}\).
02

Solve the Differential equation

Separate variables and integrate both sides \(\int_{0}^{v}\frac{v'}{g/k - (v')^{2}} dv' = \int_{0}^{t}\frac{dt'}{m}\). Using trigonometric substitution (\(v' = \sqrt{g/k} \tan(\theta)\)) and simplifying the integral, we find \(v(t) = \sqrt{\frac{g}{k}} \tanh(\sqrt{gk/m} t)\)
03

Determine the Terminal Velocity

The terminal velocity is the velocity of the parachutist as time tends to infinity. Using the property of the hyperbolic tangent function, we know that \(\tanh(x) \rightarrow 1\) as \(x \rightarrow \infty\). Thus, the terminal velocity (V) can be found as: \(V = \lim_{t \rightarrow \infty}v(t) = \sqrt{\frac{g}{k}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations in Physics
Physics often models natural phenomena through differential equations, which describe how physical quantities change with respect to one another. In kinematics, the motion of an object is governed by Newton's second law, encapsulated in the equation \(ma = F\text{net}\). For the parachutist, the net force consists of gravity and air resistance, leading to the differential equation \(m \frac{dv}{dt} = mg - kv^{2}\). Here, \(v\) is the velocity, \(t\) is time, \(m\) is mass, \(g\) is gravity, and \(k\) is the drag coefficient. Solving this equation gives us the velocity of the parachutist at any time point, pivotal for understanding how forces affect motion over time. Simplification of these complex equations requires a grasp of calculus, particularly separation of variables, a method that enables us to integrate both sides of the equation and find an explicit function for velocity.
Drag Force and Motion

Understanding Drag Force

Drag force opposes an object's motion through a fluid, such as air, and its magnitude often depends on the object's velocity. For the parachutist, the drag force is modeled as \(F_{D}=kv^{2}\), with the constant \(k\) encapsulating factors like air density and the parachute's surface area.

Solving the differential equation involving this resistive force can be challenging due to the velocity-squared term. However, it's crucial for predicting the parachutist's motion, as it determines when the forces balance and the object reaches terminal velocity - a state of constant velocity where acceleration ceases and the forces of gravity and air resistance are equal.
Hyperbolic Functions in Kinematics

Role of Hyperbolic Functions

Hyperbolic functions like \(\tanh(x)\), the hyperbolic tangent, emerge naturally in kinematics when solving differential equations of motion with resistive forces like drag. In the parachutist example, the velocity is given by a hyperbolic tangent function, \(v(t) = \sqrt{\frac{g}{k}} \tanh(\sqrt{gk/m} t)\).

This function increases quickly at first, then levels off as \(t\) approaches infinity - mimicking the behavior of an object reaching terminal velocity. Understanding the properties of \(\tanh(x)\), specifically that it asymptotically approaches 1, allows us to determine the terminal velocity of the parachutist as \(\sqrt{\frac{g}{k}}\), a critical concept in analyzing motion under the influence of drag.

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Most popular questions from this chapter

The vehicle is designed to combine the feel of a motorcycle with the comfort and safety of an automobile. If the vehicle is traveling at a constant speed of \(80 \mathrm{~km} / \mathrm{h}\) along a circular curved road of radius \(100 \mathrm{~m}\), determine the tilt angle \(\theta\) of the vehicle so that only a normal force from the seat acts on the driver. Neglect the size of the driver.

The tube rotates in the horizontal plane at a constant rate of \(\dot{\theta}=4 \mathrm{rad} / \mathrm{s}\). If a \(0.2-\mathrm{kg}\) ball \(B\) starts at the origin \(O\) with an initial radial velocity \(\dot{r}=1.5 \mathrm{~m} / \mathrm{s}\) and moves outward through the tube, determine the radial and transverse components of the ball's velocity at the instant it leaves the outer end at \(C, r=0.5 \mathrm{~m}\). Hint: Show that the equation of motion in the \(r\) direction is \(\ddot{r}-16 r=0\). The solution is of the form \(r=A e^{-4 t}+B e^{4 t} .\) Evaluate the integration constants \(A\) and \(B,\) and determine the time \(t\) when \(r=0.5 \mathrm{~m}\). Proceed to obtain \(v_{r}\) and \(v_{\theta}\).

The spring-held follower \(A B\) has a mass of \(0.5 \mathrm{~kg}\) and moves back and forth as its end rolls on the contoured surface of the cam, where \(r=0.15\) mand \(z=(0.02 \cos 2 \theta) \mathrm{m} .\) If the cam is rotating at a constant rate of \(30 \mathrm{rad} / \mathrm{s}\), determine the force component \(F_{z}\) at the end \(A\) of the follower when \(\theta=30^{\circ}\). The spring is uncompressed when \(\theta=90^{\circ} .\) Neglect friction at the bearing \(C .\)

The \(400-\mathrm{kg}\) mine car is hoisted up the incline using the cable and motor \(M\). For a short time, the force in the cable is \(F=\left(3200 t^{2}\right) \mathrm{N}\) where \(t\) is in seconds. If the car has an initial velocity \(v_{1}=2 \mathrm{~m} / \mathrm{s}\) at \(s=0\) and \(t=0\), determine the distance it moves up the plane when \(t=2 \mathrm{~s}\).

Using a forked rod, a \(0.5-\mathrm{kg}\) smooth peg \(P\) is forced to move along the vertical slotted path \(r=(0.5 \theta) \mathrm{m}\), where \(\theta\) is in radians. If the angular position of the arm is \(\theta=\left(\frac{\pi}{8} t^{2}\right) \mathrm{rad},\) where \(t\) is in seconds, determine the force of the rod on the peg and the normal force of the slot on the peg at the instant \(t=2 \mathrm{~s}\). The peg is in contact with only one edge of the rod and slot at any instant.
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