Question

Aireraft Turning Radius. An object moving in a circular path at a constant tangential velocity \(v\) is shown in Figure 2.14. The radial acceleration required for the object to move in the circular path is given by the equation $$ a=\frac{v^{2}}{r} $$ where \(a\) is the centripetal accelenation of the object in \(\mathrm{m} / \mathrm{s}^{2}, v\) is the tangential velocity of the object in \(\mathrm{m} / \mathrm{s}\), and \(r\) is the turning radius in meters. Suppose that the object is an aircraft, and answer the following questions about it: a. Suppose that the aircraft is moving at Mach \(0.85\), or \(85 \%\) of the speed of sound. If the centripetal acceleration is \(2 \mathrm{~g}\), what is the turning radius of the aircraft? (Note: For this problem, you may assume that Mach 1 is equal to \(340 \mathrm{~m} / \mathrm{s}\), and that \(1 \mathrm{~g}=9.81 \mathrm{~m} / \mathrm{s}^{2}\).) b. Suppose that the speed of the aircraft increases to Mach 1.5. What is the turning radius of the aircraft now? c. Plot the turning radius as a function of aircraft speed for speeds between Mach \(0.5\) and Mach 2.0, assuming that the accelention remains \(2 \mathrm{~g}\). d. Suppose that the maximum acceleration that the pilot can stand is \(7 \mathrm{~g}\). What is the minimum possible turning radius of the aircraft at Mach 1.5? e. Plot the turning radius as a function of centripetal acceleration for accelerations between \(2 \mathrm{~g}\) and \(8 \mathrm{~g}\), assuming a constant speed of Mach \(0.85\).

Short answer

The turning radii for each situation are as follows: a) 4250.58 m at Mach 0.85 with 2g acceleration b) 13284.71 m at Mach 1.5 with 2g acceleration d) 3799.15 m at Mach 1.5 with 7g acceleration To plot the turning radius as a function of aircraft speed or centripetal acceleration, use the derived functions: c) \(r = \frac{(M \times 340)^2}{2 \times 9.81}\), where M is the Mach number e) \(r = \frac{(0.85 \times 340)^2}{a}\), where a is the centripetal acceleration in m/s^2.

Step-by-step solution

Calculate velocity

To find the tangential velocity, we first need to find the speed at Mach 0.85. The speed of sound is 340 m/s, so the speed at Mach 0.85 is \(v = 0.85 \times 340 = 289\,m/s\).

Calculate turning radius

Now, we can find the turning radius using the formula for centripetal acceleration: \( a = \frac{v^2}{r} \). We know the centripetal acceleration is 2g, or \(2 \times 9.81 = 19.62\,m/s^2\). Then, rearrange to get \( r = \frac{v^2}{a} \) and plug in the values: \(r = \frac{289^2}{19.62} = 4250.58\,m\). b) Turning radius at Mach 1.5

Calculate velocity

In this case, we need to find the speed at Mach 1.5. The speed of sound is 340 m/s, so the speed at Mach 1.5 is \(v = 1.5 \times 340 = 510\,m/s\).

Calculate turning radius

Now, we can find the turning radius using the formula for centripetal acceleration: \( a = \frac{v^2}{r} \). We know the centripetal acceleration is 2g, or \(2 \times 9.81 = 19.62\,m/s^2\). Then, rearrange to get \( r = \frac{v^2}{a} \) and plug in the values: \(r = \frac{510^2}{19.62} = 13284.71\,m\). c) Plot the turning radius as a function of aircraft speed between Mach 0.5 and 2.0

Derive the function for the turning radius

We can derive the function for the turning radius as a function of aircraft speed by using the formula for centripetal acceleration: \( a = \frac{v^2}{r} \), rearranging to get \( r = \frac{v^2}{a} \), and then substituting the speed of sound for v at each Mach number to get \( r = \frac{(M \times 340)^2}{2 \times 9.81} \), where M is the Mach number. d) Minimum turning radius at Mach 1.5 with 7g acceleration

Calculate velocity

As we did before, the speed of sound is 340 m/s, so the speed at Mach 1.5 is \(v = 1.5 \times 340 = 510\,m/s\).

Calculate turning radius

Now, we can find the turning radius using the formula for centripetal acceleration: \( a = \frac{v^2}{r} \). We know the centripetal acceleration is 7g, or \(7 \times 9.81 = 68.67\,m/s^2\). Then, rearrange to get \( r = \frac{v^2}{a} \) and plug in the values: \(r = \frac{510^2}{68.67} = 3799.15\,m\). e) Plot the turning radius as a function of centripetal acceleration between 2g and 8g at Mach 0.85

Derive the function for the turning radius

We can derive the function for the turning radius as a function of centripetal acceleration by using the formula for centripetal acceleration: \( a = \frac{v^2}{r} \), rearranging to get \( r = \frac{v^2}{a} \), and then substituting the speed of sound for v at Mach 0.85 to get \( r = \frac{(0.85 \times 340)^2}{a} \), where a is the centripetal acceleration in m/s^2.

Key concepts

Centripetal Acceleration

Centripetal acceleration is a crucial concept when it comes to understanding the behavior of an object moving in a circular path, such as an aircraft executing a turn.

This type of acceleration is always directed towards the center of the circular path and is essential for maintaining the object's circular motion. The formula for centripetal acceleration is \[ a = \frac{v^2}{r} \], where:

• \(a\) represents centripetal acceleration measured in meters per second squared (\(m/s^2\)),
• \(v\) is the tangential velocity of the object in meters per second (\(m/s\)),
• \(r\) is the radius of the circle in meters (m).

In our aircraft scenario, knowing this acceleration is vital to ensuring that the plane turns safely without exceeding the structural limits of the aircraft or the pilot's capacity to withstand g-forces.

Tangential Velocity

Tangential velocity is the speed of an object moving along a circular path. It's tangential to any point on the circumference of the circle. The tangential velocity is an important factor in calculating the turning radius of an aircraft because it provides a measure of how fast the aircraft is moving along its circular path.

For objects traveling in a uniform circular motion, the tangential velocity remains constant, and it's this consistent speed that helps determine the circular path's radius when combined with the centripetal force acting on the object. In our textbook solution, we calculated the tangential velocity based on the known speed of sound and the Mach number of the aircraft (e.g., Mach 0.85 corresponds to 289 m/s), allowing us to use this velocity to find the turning radius for the given centripetal acceleration.

MATLAB Programming

MATLAB is a high-level language and interactive environment for numerical computation, visualization, and programming. When faced with the task of plotting the turning radius as a function of aircraft speed or centripetal acceleration, MATLAB becomes an invaluable tool.

Using MATLAB, one can easily create scripts and functions to calculate various scenarios, such as the turning radius at different speeds and accelerations. By entering the formula for the turning radius into a MATLAB program, and varying the Mach number or acceleration within a specified range, you can quickly generate and visualize the relationship between these variables and the turning radius on a plot.

This hands-on approach using MATLAB helps students visualize complex concepts and enhances their understanding of the relationships between flight speed, centripetal acceleration, and turning radius in aircraft.