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Chapter 6: Problem 51

An airplane is flying in a horizontal circle at a speed of \(480 \mathrm{~km} / \mathrm{h}\) (Fig. 6-41). If its wings are tilted at angle \(\theta=40^{\circ}\) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

Short Answer

Expert verified
The radius of the circle is approximately 2576 meters.

Step by step solution

01

Convert Speed from km/h to m/s

Convert the speed of the airplane from kilometers per hour to meters per second using the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Thus, \(480 \text{ km/h} = \frac{480}{3.6} \text{ m/s} \approx 133.33 \text{ m/s} \).
02

Identify Forces and Variables Involved

Analyze the problem by identifying the forces involved: the aerodynamic lift force and the gravitational force. The lift force acts perpendicular to the wings, while the gravity force acts downward. The lift provides the centripetal force needed for circular motion.
03

Express Centripetal Force in Terms of Lift

The lift force \( L \) is the vector sum of its horizontal and vertical components. For circular motion, the horizontal component of the lift provides the centripetal force: \( L \sin(\theta) = \frac{mv^2}{r} \). Here, \(m\) is mass, \(v\) is velocity, and \(r\) is radius.
04

Resolve the Vertical Component of Lift

The vertical component of the lift balances the weight of the airplane: \(L \cos(\theta) = mg\), where \(g\) is the acceleration due to gravity \( 9.8 \text{ m/s}^2 \).
05

Divide and Simplify Equations

Divide the centripetal force equation by the vertical component equation: \( \frac{L \sin(\theta)}{L \cos(\theta)} = \frac{mv^2/r}{mg} \). This simplifies to \( \tan(\theta) = \frac{v^2}{rg} \).
06

Solve for Radius

Rearrange the equation to solve for the radius \( r \): \( r = \frac{v^2}{g \tan(\theta)} \). Substitute the known values: \( v = 133.33 \text{ m/s} \), \( g = 9.8 \text{ m/s}^2 \), and \( \theta = 40^{\circ} \). Calculate \( \tan(40^{\circ}) \approx 0.8391 \). Therefore, \( r = \frac{(133.33)^2}{9.8 \times 0.8391} \approx 2576.06 \text{ m} \).

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Key Concepts

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

Aerodynamic Lift
Aerodynamic lift is the upward force that keeps an airplane in the air. It is crucial for flying, as it counteracts the force of gravity pulling the plane down. In this problem, lift is provided by the airplane's wings, which are tilted at an angle
- **Perpendicular Force**: Lift acts perpendicular to the surface of the wings - **Dependence on Angle**: The titling angle of the wings affects how the lift force is divided into horizontal and vertical components In a circular flight path, the horizontal component of lift provides the centripetal force needed for the airplane to stay in its curved trajectory. The vertical component of the lift force balances out gravity. This means that lift not only keeps the airplane aloft but also helps guide its path through the sky.
Centripetal Force
Centripetal force is what keeps an object moving in a circle. For the airplane in circular motion, this force is crucial. It prevents the airplane from flying off in a straight line.
- **Horizontal Component of Lift**: The airplane's centripetal force is provided by the horizontal component of the aerodynamic lift - **Formula Connection**: It is given by the formula: When an object moves in a circle, it is constantly changing direction. This requires a continuous application of force towards the center of the circle, known as centripetal force. For our climbing or banking airplane, adjusting the angle and speed can help balance the required centripetal force for desired turns.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent, are essential for solving problems involving angles, such as the wing tilt in this problem.
- **Angle of Tilt**: Relates the plane's path and lift components via trigonometric functions- **Component Calculation**: Breaks the forces like lift into horizontal and vertical partsThrough trigonometry, you can calculate how these components affect the airplane's motion. For example, using the angle \( \theta = 40^{\circ} \), and finding the tangent, \( \tan(40^{\circ}) \approx 0.8391 \): These functions allow you to relate the plane's linear speed, the force of gravity, and the lift to find the radius of the circle. Understanding these relationships is key to solving physics problems involving circular motion.

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Most popular questions from this chapter

\({ }^{\circ} 6 \cdot \) ssin A \(1000 \mathrm{~kg}\) boat is traveling at \(90 \mathrm{~km} / \mathrm{h}\) when its engine is shut off. The magnitude of the frictional force \(\vec{f}_{k}\) between boat and water is proportional to the speed \(v\) of the boat: \(f_{k}=70 v\), where \(v\) is in meters per second and \(f_{k}\) is in newtons. Find the time required for the boat to slow to \(45 \mathrm{~km} / \mathrm{h}\).

A locomotive accelerates a 25 -car train along a level track. Every car has a mass of \(5.0 \times 10^{4} \mathrm{~kg}\) and is subject to a frictional force \(f=250 v\), where the speed \(v\) is in meters per second and the force \(f\) is in newtons. At the instant when the speed of the train is 30 \(\mathrm{km} / \mathrm{h}\), the magnitude of its acceleration is \(0.20 \mathrm{~m} / \mathrm{s}^{2} .(\mathrm{a})\) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at \(30 \mathrm{~km} / \mathrm{h}\) ?

You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B\), which was stopped at a red light along a road headed down a hill (Fig. 6-25). You find that the slope of the hill is \(\theta=12.0^{\circ}\), that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{0}=18.0 \mathrm{~m} / \mathrm{s}\). With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) \(0.60\) (dry road surface) and (b) \(0.10\) (road surface covered with wet leaves)?

A child weighing \(140 \mathrm{~N}\) sits at rest at the top of a playground slide that makes an angle of \(25^{\circ}\) with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of \(0.86 \mathrm{~m} / \mathrm{s}^{2}\) (down the slide, of course). (a) What is the coefficient of kinetic friction between the child and the slide? (b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?

ILW What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is \(29 \mathrm{~km} / \mathrm{h}\) and the \(\mu_{s}\) between tires and track is \(0.32 ?\)
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