Question

Unlike a ship, an airplane does not use its rudder to turn. It turns by banking its wings: The lift force, perpendicular to the wings, has a horizontal component, which provides the centripetal acceleration for the turn, and a vertical component, which supports the plane's weight. (The rudder counteracts yaw and thus it keeps the plane pointed in the direction it is moving.) A famous spy plane the SR-71 Blackbird, flying at \(4800 . \mathrm{km} / \mathrm{h}\), had a turning radius of \(290 . \mathrm{km}\). Find its banking angle.

Short answer

Answer: The banking angle of the SR-71 Blackbird while turning is approximately 32°.

Step-by-step solution

Write down the known variables

We are given the following information: - The speed of the plane, \(v = 4800 \, \text{km/h}\) - The turning radius, \(r = 290 \, \text{km}\)

Convert speeds to m/s

We need to convert the speed from km/h to m/s as follows: $$ v = 4800 \, \frac{\text{km}}{\text{h}} \cdot \frac{1000 \, \text{m}}{1\, \text{km}} \cdot \frac{1 \, \text{h}}{3600 \, \text{s}} = 1333.33 \, \frac{\text{m}}{\text{s}} $$

Find the centripetal acceleration

Next, we will calculate the centripetal acceleration, which is given by the formula: $$ a_c = \frac{v^2}{r} $$ Plugging the values of \(v\) and \(r\) into the formula, we get: $$ a_c = \frac{(1333.33 \, \frac{\text{m}}{\text{s}})^2}{290 \cdot 1000 \, \text{m}} = 6.1 \, \frac{\text{m}}{\text{s}^2} $$

Determine the horizontal and vertical force components

The lift force \(F_L\) has a horizontal component \(F_H\) and a vertical component \(F_V\). Let's denote the banking angle as \(\theta\). Then we can express the components as follows: $$ F_H = F_L \sin\theta $$ $$ F_V = F_L \cos\theta $$

Relate the forces to the centripetal acceleration and weight of the plane

Using Newton's second law (\(F=ma\)), we can express the horizontal and vertical forces in terms of the plane's mass \(m\) and acceleration components: $$ F_H = m a_c $$ $$ F_V = m g $$ where \(g\) is the acceleration due to gravity, approximately \(9.81 \, \frac{\text{m}}{\text{s}^2}\).

Set up an equation relating the lift force components using trigonometry

Since the lift force components \(F_H\) and \(F_V\) are related to the banking angle \(\theta\), we can set up an equation relating these forces using trigonometric functions. $$ \frac{F_H}{F_V} = \frac{F_L \sin\theta}{F_L \cos\theta} = \tan\theta $$

Substitute the force equations into the lift force ratio

Now substitute the expressions for \(F_H\) and \(F_V\) from Step 5 into the equation derived in Step 6: $$ \frac{m a_c}{m g} = \tan\theta $$ The mass \(m\) of the plane can now be canceled out, giving: $$ \frac{a_c}{g} = \tan\theta $$

Solve for the banking angle

Now, we can solve for the banking angle \(\theta\) using the inverse tangent function: $$ \theta = \arctan\left(\frac{a_c}{g}\right) $$ Substitute the values for \(a_c\) and \(g\) from Step 3: $$ \theta = \arctan\left(\frac{6.1 \, \frac{\text{m}}{\text{s}^2}}{9.81 \, \frac{\text{m}}{\text{s}^2}}\right) = \arctan(0.621) = 32^{\circ} $$ Hence, the banking angle of the SR-71 Blackbird while turning is approximately \(32^{\circ}\).