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Ch 08: Dynamics II: Motion in a Plane
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 08: Dynamics II: Motion in a PlaneProblem 56a
Chapter 8, Problem 56a

An airplane feels a lift force L→\(\overrightarrow{L}\) perpendicular to its wings. In level flight, the lift force points straight up and is equal in magnitude to the gravitational force on the plane. When an airplane turns, it banks by tilting its wings, as seen from behind, by an angle from horizontal. This causes the lift to have a radial component, similar to a car on a banked curve. If the lift had constant magnitude, the vertical component of L→\(\overrightarrow{L}\) would now be smaller than the gravitational force, and the plane would lose altitude while turning. However, you can assume that the pilot uses small adjustments to the plane's control surfaces so that the vertical component of L→\(\overrightarrow{L}\) continues to balance the gravitational force throughout the turn. Find an expression for the banking angle θ\(\theta\) needed to turn in a circle of radius rr while flying at constant speed vv.

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Step 1: Begin by analyzing the forces acting on the airplane during the turn. The lift force (L) is tilted at an angle θ from the vertical, creating two components: a vertical component (L_vertical) and a horizontal component (L_horizontal). The vertical component balances the gravitational force (mg), while the horizontal component provides the centripetal force needed for circular motion.
Step 2: Write the equations for the force components. The vertical component of the lift force is given by L_vertical = L * cos(θ), and the horizontal component is given by L_horizontal = L * sin(θ). Since the plane is in level flight, the vertical component must equal the gravitational force: L * cos(θ) = mg.
Step 3: The horizontal component of the lift force provides the centripetal force for circular motion. The centripetal force is given by F_c = m * v² / r, where m is the mass of the airplane, v is its speed, and r is the radius of the turn. Therefore, L * sin(θ) = m * v² / r.
Step 4: Divide the equation for the horizontal component by the equation for the vertical component to eliminate L. This gives: (L * sin(θ)) / (L * cos(θ)) = (m * v² / r) / (mg). Simplify this to obtain tan(θ) = v² / (r * g).
Step 5: Solve for the banking angle θ. The final expression for the banking angle is θ = arctan(v² / (r * g)). This equation shows how the banking angle depends on the speed of the airplane (v), the radius of the turn (r), and the acceleration due to gravity (g).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lift Force

Lift is the aerodynamic force that acts perpendicular to the relative wind and the wings of an airplane. It is generated by the difference in air pressure above and below the wings, allowing the airplane to rise and maintain altitude. In level flight, lift must equal the weight of the airplane to achieve equilibrium, while during a turn, the lift vector is tilted, affecting its vertical and horizontal components.
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Banking Angle

The banking angle is the angle at which an airplane's wings are tilted relative to the horizontal during a turn. This tilt allows the lift force to be resolved into two components: one that counteracts gravity and another that provides the necessary centripetal force to maintain circular motion. The correct banking angle ensures that the vertical component of lift balances the weight of the airplane while the horizontal component facilitates the turn.
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Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. In the context of an airplane turning, this force is provided by the horizontal component of the lift. For an airplane flying at constant speed in a circular path, the relationship between the speed, radius of the turn, and banking angle can be expressed mathematically to find the necessary conditions for stable flight.
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Related Practice
Textbook Question

A 100 g ball on a 60-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cm above the floor. What was the tension in the string an instant before it broke?

Textbook Question

The physics of circular motion sets an upper limit to the speed of human walking. (If you need to go faster, your gait changes from a walk to a run.) If you take a few steps and watch what's happening, you'll see that your body pivots in circular motion over your forward foot as you bring your rear foot forward for the next step. As you do so, the normal force of the ground on your foot decreases and your body tries to 'lift off' from the ground. A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass m at the top of a leg of length L. Find an expression for the person's maximum walking speed vmax.

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Textbook Question

2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is θ = 30°. What is the ball's speed when θ = 30°?

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Textbook Question

A 30 g ball rolls around a 40-cm-diameter L-shaped track, shown in FIGURE P8.53, at 60 rpm. What is the magnitude of the net force that the track exerts on the ball? Rolling friction can be neglected. Hint: The track exerts more than one force on the ball.

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Textbook Question

Scientists design a new particle accelerator in which protons (mass 1.7 X 10-27 kg) follow a circular trajectory given by r=ccos(kt2)i+csin(kt2)j\(\mathbf{r}\) = c \(\cos\)(kt^2) \(\mathbf{i}\) + c \(\sin\)(kt^2) \(\mathbf{j}\) where c = 5.0 m and k = 8.0 x 104 rad/s2 are constants and t is the elapsed time. What is the radius of the circle?

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Textbook Question

FIGURE P8.54 shows two small 1.0 kg masses connected by massless but rigid 1.0-m-long rods. What is the tension in the rod that connects to the pivot if the masses rotate at 30 rpm in a horizontal circle?