Mastering Physics Solutions: Problem 8.65

Mastering Physics Solutions: Problem 8.65

On March 10, 2014, in Chapter 07: Circular Motion and Gravitation, by Mastering Physics Solutions

Problem 8.65

Part A = 37 km

Solution Below:

You are flying to New York. You’ve been reading the in-flight magazine, which has an article about the physics of flying. You learned that the airflow over the wings creates a lift force that is always perpendicular to the wings. In level flight, the upward lift force exactly balances the downward weight force. The pilot comes on to say that, because of heavy traffic, the plane is going to circle the airport for a while. She says that you’ll maintain a speed of 300mph at an altitude of 20,000 ft. You start to wonder what the diameter of the plane’s circle around the airport is. You notice that the pilot has banked the plane so that the wings are 10° from horizontal. The safety card in the seatback pocket informs you that the plane’s wing span is 250 ft.

Part A

What is the diameter of the airplane’s path around the airport?
Express your answer with the appropriate units.

This seems like a hard problem, but it’s actually fairly simple to solve. First of all, you know that the left is keeping the plane at the same altitude – so the lift (L) in the vertical direction must balance out gravity:

Ly = m * g

And you know that the vertical lift must also be:

Ly = L * cos(θ)

Since the vertical lift keeps the plane level, it must be equal to gravity. This means that:

Ly = m * g = L * cos(θ)


L = m * g / cos(θ)

Remember- we have to use cos here for the vertical lift, not sin. If this seems strange, think about what happens when the plane is turned 90° – the wings are pointing straight up and straight down, and there’s no way for the lift to keep the plane flying. It will crash. Anyway, since we know what the vertical lift is, we can also say that the horizontal lift is:

Lx = L * sin(θ)

This means:

Lx = L * sin(10)
Lx = L * 0.1736

This horizontal lift is what moves the plane in a circle. Since the plane is moving at 400 mph, we can plug this into the equation for centripetal force. We know the horizontal lift must equal the centripetal force, because the plane stays moving in a circle, without speeding up or slowing down:

Lx = m * ac
L * 0.1736 = m * v^2 / r

Now, the problem gave us v, but we have to convert it from mph to m/s. One mph = 0.44704 m/s, so 400mph = 178.816 m/s
L * 0.1736 = m * 178.816^2 / r

Now substitute in for L:

(m * g / cos(θ)) * 0.1736 = m * 178.816^2 / r

The m cancels:

g / cos(θ) * 0.1736 = 178.816^2 / r
9.8 / cos(10) * 0.1736 = 178.816^2 / r
9.8 / 0.9848 * 0.1736 = 178.816^2 / r
9.95 * 0.1736 = 178.816^2 / r
1.7275 = 178.816^2 / r
1.7275 * r = 31975
r = 18509 m

Since the problem wants the diameter, just double the radius. This gives:

d = 37018 m

The problem seems to want the answer in kilometers, which is:

d = 37 km

37 km

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