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Chapter 5: Problem 11

By comparing the surface area of a sphere with its volume and assuming that air resistance is proportional to the square of velocity, it is possible to make a heuristic argument to support the following premise: For similarly shaped objects, terminal velocity varies in proportion to the square root of length. Expressed in a formula, this is $$ T=k L^{0.5}, $$ where \(L\) is length, \(T\) is terminal velocity, and \(k\) is a constant that depends on shape, among other things. This relation can be used to help explain why small mammals easily survive falls that would seriously injure or kill a human. a. A 6-foot man is 36 times as long as a 2-inch mouse (neglecting the tail). How does the terminal velocity of a man compare with that of a mouse? b. If the 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of the 2 -inch mouse? c. Neglecting the tail, a squirrel is about 7 inches long. Again assuming that a 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of a squirrel?

Short Answer

Expert verified
A man's terminal velocity is 6 times a mouse's, so the mouse's is 20 mph. A squirrel's terminal velocity is about 37.38 mph.

Step by step solution

01

Understanding the Problem

We are given the relationship \( T = k L^{0.5} \), where \( T \) is terminal velocity, \( L \) is length, and \( k \) is a constant. We need to compare lengths to use this relationship for the mouse, man, and squirrel.
02

Calculating the Ratio of Lengths

Given that the man is 6 feet (72 inches) and the mouse is 2 inches, the ratio \( \frac{L_{\text{man}}}{L_{\text{mouse}}} = \frac{72}{2} = 36 \). The same man to squirrel ratio is \( \frac{L_{\text{man}}}{L_{\text{squirrel}}} = \frac{72}{7} \approx 10.29 \).
03

Calculate Terminal Velocity Ratio for Man vs Mouse

Since terminal velocity \( T \) is proportional to the square root of length, compute \( \frac{T_{\text{man}}}{T_{\text{mouse}}} = \sqrt{\frac{L_{\text{man}}}{L_{\text{mouse}}}} = \sqrt{36} = 6 \). This means the man's terminal velocity is 6 times that of the mouse.
04

Find Terminal Velocity of the Mouse

Given \( T_{\text{man}} = 120 \) mph, and \( \frac{T_{\text{man}}}{T_{\text{mouse}}} = 6 \), then \( T_{\text{mouse}} = \frac{T_{\text{man}}}{6} = \frac{120}{6} = 20 \) mph.
05

Find Terminal Velocity of a Squirrel

Using the man to squirrel length ratio \( \frac{L_{\text{man}}}{L_{\text{squirrel}}} = 10.29 \), calculate \( \frac{T_{\text{man}}}{T_{\text{squirrel}}} = \sqrt{10.29} \approx 3.21 \). So, \( T_{\text{squirrel}} = \frac{120}{3.21} \approx 37.38 \) mph.

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Key Concepts

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

Square Root
The square root is a mathematical operation that finds the number which, when multiplied by itself, gives the original number. For example, the square root of 36 is 6 because 6 x 6 = 36. In the context of terminal velocity, where terminal velocity is proportional to the square root of length, this means that if you know the length of an object, you can predict how fast it will fall when it reaches terminal velocity by taking the square root of that length. This concept is crucial for understanding how terminal velocity changes with different sizes of objects.
Using the formula \( T = k L^{0.5} \), the square root (\( L^{0.5} \)) serves as a factor that modifies the terminal velocity based on the object's length. If you double the length of an object, its terminal velocity becomes \( \sqrt{2} \) times higher, assuming shape and other parameters stay constant.
This concept is particularly helpful in physics problems where objects of the same shape but different sizes need their terminal velocities compared.
Length Ratio
In physics, comparing two quantities often involves understanding their ratio, which numerically shows how one quantity relates to another. In this case, the length ratio is a way to compare the size of two objects quantitatively. In our exercise, we compared the length of a 6-foot (72 inch) man to a 2-inch mouse and a 7-inch squirrel.
To calculate the length ratio for the man and the mouse, we divide the man's length by the mouse's length:
  • \( \frac{L_{\text{man}}}{L_{\text{mouse}}} = \frac{72}{2} = 36 \)
Similarly, for the squirrel:
  • \( \frac{L_{\text{man}}}{L_{\text{squirrel}}} = \frac{72}{7} \approx 10.29 \)
Understanding these ratios allows us to apply the square root relation effectively to determine how many times faster a man or a squirrel falls compared to a mouse. Such length ratios are foundational in scaling laws, which help translate properties from models to real-life scenarios.
Proportionality
Proportionality is a key concept used to describe the relationship between two variables where if one variable changes, the other changes in a consistent way. In the context of terminal velocity and length, the exercise illustrates that terminal velocity \( T \) is proportional to the square root of length \( L \). This is expressed as \( T \propto L^{0.5} \).
What this means in simpler terms is that if you increase the length of an object, its terminal velocity will increase in a predictable way defined by the square root relationship. This proportionality allows us to predict the terminal velocity of differently sized objects of the same shape without needing to conduct physical experiments.
  • An object twice as long as another will have a terminal velocity \( \sqrt{2} \) times that of the shorter object.
  • If the length is nine times greater, the terminal velocity will be three times higher, since \( \sqrt{9} = 3 \).
This precise and predictable relationship is why proportionality is such a powerful tool in physics.
Heuristic Argument
A heuristic argument is a simplified, intuitive way of reaching a conclusion. It doesn't rely on exact proofs or computations but instead uses logical reasoning and known properties to arrive at a general understanding.
In our exercise, the heuristic argument helps us comprehend why terminal velocity depends on the square root of an object's length. By reasonably assuming that air resistance is proportional to the square of velocity, we can then deduce from geometric considerations that terminal velocity \( T \) for any object behaves according to \( T \sim L^{0.5} \).
Heuristics are valuable because they allow us to predict outcomes without needing precise models. They provide a framework to think about problems and derive useful relations, such as knowing why smaller animals like mice and squirrels can survive falls from significant heights compared to humans. Once we understand these basic principles, we can apply them to complex real-world scenarios with confidence.

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Most popular questions from this chapter

According to Newton's law of gravitation, the gravitational attraction between two massive objects such as planets or asteroids is proportional to \(d^{-2}\), where \(d\) is the distance between the centers of the objects. Specifically, the gravitational force \(F\) between such objects is given by \(F=c d^{-2}\), where \(d\) is the distance between their centers. The value of the constant \(c\) depends on the masses of the two objects and on the universal gravitational constant. a. Suppose the force of gravity is causing two large asteroids to move toward each other. What is the effect on the gravitational force if the distance between their centers is halved? What is the effect on the gravitational force if the distance between their centers is reduced to one-quarter of its original value? b. Suppose that for a certain pair of asteroids whose centers are 300 kilometers apart, the gravitational force is \(2,000,000\) newtons. (One newton is about one-quarter of a pound.) What is the value of \(c\) ? Find the gravitational force if the distance between the centers of these asteroids is 800 kilometers. c. Using the value of \(c\) you found in part \(b\), make a graph of gravitational force versus distance between the centers of the asteroids for distances from 0 to 1000 kilometers. What happens to the gravitational force when the asteroids are close together? What happens to the gravitational force when the asteroids are far apart?

The yearly inventory expense \(E\), in dollars, of a car dealer is a function of the number \(Q\) of automobiles ordered at a time from the manufacturer. A dealer who orders only a few automobiles at a time will have the expense of placing several orders, whereas if the order sizes are large, then the dealer will have a large inventory of unsold automobiles. For one dealer the formula is $$ E=\frac{425 Q^{2}+8000}{Q}, $$ so \(E\) is a rational function of \(Q\). a. Make a graph of \(E\) versus \(Q\) covering order sizes up to \(10 .\) b. Explain in practical terms the behavior of the graph near the pole at \(Q=0\). (Hint: Keep in mind that there is a fixed cost of processing each order, regardless of the size of the order.)

The following formula \({ }^{71}\) can be used to approximate the average transit speed \(S\) of a public transportation vehicle: $$ S=\frac{C D}{C T+D+C^{2}\left(\frac{1}{2 a}+\frac{1}{2 d}\right)} $$ Here \(C\) is the cruising speed in miles per hour, \(D\) is the average distance between stations in miles, \(T\) is the stop time at stations in hours, \(a\) is the rate of acceleration in miles per hour per hour, and \(d\) is the rate of deceleration in miles per hour per hour. For a certain subway the average distance between stations is 3 miles, the stop time is 3 minutes \((0.05\) hour), and rate of acceleration and deceleration are both \(3.5\) miles per hour per second (12,600 miles per hour per hour). a. Using the information provided, express \(S\) as a rational function of \(C\). b. As a rule of thumb, we might use the cruising speed as an estimate of transit speed. Discuss the merits of such a rule of thumb. How does increasing cruising speed affect such an estimate? (Suggestion: Compare the graph of the given rational function with that of the graph of \(S=C\).) c. What cruising speed will yield a transit speed of 30 miles per hour?

Poiseuill茅's law describes the velocities of fluids flowing in a tube-for example, the flow of blood in a vein. (See Figure 5.104.) This law applies when the velocities are not too large-more specifically, when the flow has no turbulence. In this case the flow is laminar, which means that the paths of the flow are all parallel to the tube walls. The law states that $$ v=k\left(R^{2}-r^{2}\right), $$ where \(v\) is the velocity, \(k\) is a constant (which depends on the fluid, the tube, and the units used for measurement), \(R\) is the radius of the tube, and \(r\) is the distance from the centerline of the tube. Since \(k\) and \(R\) are fixed for any application, \(v\) is a function of \(r\) alone, and the formula gives the velocity at a point of distance \(r\) from the centerline of the tube. a. What is \(r\) for a point along the walls of the tube? What is the velocity of the fluid along the walls of the tube? b. Where in the tube does the fluid flow most rapidly? c. Choose numbers for \(k\) and \(R\) and make a graph of \(v\) as a function of \(r\). Be sure that the horizontal span for \(r\) goes from 0 to \(R\). d. Describe your graph from part c. e. Explain why you needed to use a horizontal span from 0 to \(R\) in order to describe the flow throughout the tube.

A class of models for population growth rates in marine fisheries assumes that the harvest from fishing is proportional to the population size. One such model uses a quadratic function: $$ G=0.3 n-0.2 n^{2} . $$ Here \(G\) is the growth rate of the population, in millions of tons of fish per year, and \(n\) is the population size, in millions of tons of fish. a. Make a graph of \(G\) versus \(n\). Include values of \(n\) up to \(1.7\) million tons. b. Calculate \(G(1.62)\) and explain what your answer means in practical terms. c. At what population size is the growth rate the largest?
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