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You are flying to Chicago for a weekend away from the books. In your last physics class, you learned that the airflow over the wings of the plane creates a lift force, which acts perpendicular to the wings. When the plane is flying level, the upward lift force exactly balances the downward weight force. Since O'Hare is one of the busiest airports in the world, you are not surprised when the captain announces that the flight is in a holding pattern due to the heavy traffic. He informs the passengers that the plane will be flying in a circle of radius \(7.00 \mathrm{mi}\) at a speed of \(360 . \mathrm{mph}\) and an altitude of \(2.00 \cdot 10^{4} \mathrm{ft}\). From the safety information card, you know that the total length of the wingspan of the plane is \(275 \mathrm{ft}\). From this information, estimate the banking angle of the plane relative to the horizontal.

Short Answer

Expert verified
Answer: The banking angle of the plane relative to the horizontal is approximately 13.98°.

Step by step solution

01

Convert speed and radius to SI units

First, we'll convert the speed from miles per hour (mph) to meters per second (m\s) and the radius from miles to meters. conversion factors: 1 mile = 1609.34 m 1 hour = 3600 seconds Speed in \(\frac{\text{m}}{\text{s}}\): \(v = 360 \frac{\text{mi}}{\text{h}} \times \frac{1609.34 \text{m}}{1 \text{mi}} \times \frac{1 \text{h}}{3600 \text{s}} = 161.0 \ \frac{\text{m}}{\text{s}}\) Radius in meters: \(r = 7.00 \ \text{mi} \times \frac{1609.34 \text{m}}{1 \text{mi}} = 11265.38 \ \text{m}\)
02

Calculate the tangent of the banking angle

Now, we'll calculate the tangent of the banking angle using the formula \(\displaystyle\tan \theta = \frac{v^2}{g \cdot r}\). \(\displaystyle\tan \theta = \frac{(161.0 \ \text{m/s})^2}{(9.81 \ \frac{\text{m}}{\text{s}^2})(11265.38 \ \text{m})} = 0.2490\)
03

Calculate the banking angle

Finally, we'll calculate the banking angle by taking the arctangent of the tangent value obtained in step 2. \(\theta = \arctan(0.2490) = 13.98^\circ\) The banking angle of the plane relative to the horizontal is approximately \(13.98^\circ\).

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Key Concepts

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

Lift Force
When considering the principles of flight, lift force is a crucial concept. It's the upward acting force that opposes the downward pull of gravity, allowing airplanes to fly. This force is generated as air flows over the wings of an aircraft, with variations in air pressure above and below the wings according to Bernoulli's principle.

In a situation where a plane is flying level, as mentioned in the exercise, the lift force is balanced perfectly with the gravitational force acting on the plane. This equilibrium keeps the aircraft in steady horizontal flight. However, when an airplane enters a turn, the lift force must be adjusted to maintain circular motion, which introduces us to the next key concept.
Circular Motion
Circular motion refers to the movement of an object along a circular path. When a plane flies in a circular pattern or is in a so-called holding pattern as described in the exercise, it must maintain a constant inward force to stay on that path. This inward force is known as 'centripetal force' and is necessary for executing the turn.

The plane achieves circular motion by banking its wings, which tilts the lift vector and provides the necessary centripetal force. The centripetal force in the context of circular motion is exactly what leads us to understand tangential speed and centripetal acceleration.
Tangential Speed
Tangential speed, or the speed of an object moving along a circular path, is crucial when examining circular motion. It's the speed at any given point along the circle and is always directed tangent to the path, hence the name. In the provided exercise, the plane's tangential speed would be its velocity while following the circular holding pattern above the airport.

Tangential speed is different from angular speed, which measures how fast an object rotates or revolves relative to another point. In this case, we are interested in the tangential speed since it directly relates to the force needed to maintain the circular path, which in turn influences our next concept – centripetal acceleration.
Centripetal Acceleration
Centripetal acceleration is the acceleration of an object moving in a circle at constant tangential speed. This acceleration is always directed inward towards the center of the circle. It's not a force but rather describes how quickly the velocity of an object is changing in direction, not speed, as it moves along a curved path.

For objects in circular motion such as the airplane in a holding pattern mentioned in the exercise, centripetal acceleration is provided by the horizontal component of the lift force when the plane banks. The formula for centripetal acceleration, necessary to calculate the banking angle, is given by \( a_c = \frac{v^2}{r} \) where \( v \) is the tangential speed, and \( r \) is the radius of the circle. This concept is essential in our understanding of how and why the airplane banks at a specific angle to maintain its circular path.

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Most popular questions from this chapter

In a conical pendulum, the bob attached to the string (which can be considered massless) moves in a horizontal circle at constant speed. The string sweeps out a cone as the bob rotates. What forces are acting on the bob?

A discus thrower (with arm length of \(1.20 \mathrm{~m}\) ) starts from rest and begins to rotate counterclockwise with an angular acceleration of \(2.50 \mathrm{rad} / \mathrm{s}^{2}\) a) How long does it take the discus thrower's speed to get to \(4.70 \mathrm{rad} / \mathrm{s} ?\) b) How many revolutions does the thrower make to reach the speed of \(4.70 \mathrm{rad} / \mathrm{s} ?\) c) What is the linear speed of the discus at \(4.70 \mathrm{rad} / \mathrm{s}\) ? d) What is the linear acceleration of the discus thrower at this point? e) What is the magnitude of the centripetal acceleration of the discus thrown? f) What is the magnitude of the discus's total acceleration?

A 20.0 -g metal cylinder is placed on a turntable, with its center \(80.0 \mathrm{~cm}\) from the turntable's center. The coefficient of static friction between the cylinder and the turntable's surface is \(\mu_{s}=0.800 .\) A thin, massless string of length \(80.0 \mathrm{~cm}\) connects the center of the turntable to the cylinder, and initially, the string has zero tension in it. Starting from rest, the turntable very slowly attains higher and higher angular velocities, but the turntable and the cylinder can be considered to have uniform circular motion at any instant. Calculate the tension in the string when the angular velocity of the turntable is 60.0 rpm (rotations per minute).

An 80.0 -kg pilot in an aircraft moving at a constant speed of \(500 . \mathrm{m} / \mathrm{s}\) pulls out of a vertical dive along an arc of a circle of radius \(4000 . \mathrm{m} .\) a) Find the centripetal acceleration and the centripetal force acting on the pilot. b) What is the pilot's apparent weight at the bottom of the dive?

The rear wheels of a sports car have a radius of \(46.65 \mathrm{~cm}\). The sports car goes from rest to a speed of \(29.13 \mathrm{~m} / \mathrm{s}\) in \(3.945 \mathrm{~s}\) with constant accelera. tion. What is the angular acceleration of the rear wheels? Assume that the wheels roll without slipping.

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