6. An airplane is flying in a horizontal circle at a speed of 490 km/h. If its wings are tilted at angle e = 45° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter13: Rotation Ii: A Conservation Approach
Section: Chapter Questions
Problem 27PQ: The motion of spinning a hula hoop around one's hips can bemodeled as a hoop rotating around an axis...
icon
Related questions
Question
**Flight Physics Problem**

**Problem 6:**

An airplane is flying in a horizontal circle at a speed of 490 km/h. If its wings are tilted at an angle \( \theta = 45^\circ \) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

---

To solve this, we need to understand the relationship between the forces acting on the airplane and the radius of the circular path it takes.

1. **Aerodynamic Lift**: The lift force \( L \) acts perpendicular to the wing surface.
2. **Centripetal Force**: A portion of the lift force provides the necessary centripetal force \( F_c \) to maintain circular motion.

The centripetal force \( F_c \) needed for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]

Where:
- \( m \) is the mass of the airplane
- \( v = 490 \, \text{km/h} \) (convert this to meters per second, \( v = 136.11 \, \text{m/s} \))
- \( r \) is the radius of the circle

The component of the lift force providing the centripetal force is \( L \sin\theta \). Since the plane is in horizontal flight, the vertical component \( L \cos\theta \) balances the weight \( mg \) of the airplane.

By equating the centripetal force to the horizontal component of the lift force:

\[ \frac{mv^2}{r} = L \sin\theta \]

And since \( L \cos\theta = mg \):

\[ L = \frac{mg}{\cos\theta} \]

By substituting \( L \) into the centripetal force equation:

\[ \frac{mv^2}{r} = \left(\frac{mg}{\cos\theta}\right) \sin\theta \]

The mass \( m \) cancels out:

\[ \frac{v^2}{r} = \frac{g \sin\theta}{\cos\theta} \]

Simplify using \( \tan\theta \):

\[ \frac{v^2}{r} = g \tan\theta \]

Solving for \( r
Transcribed Image Text:**Flight Physics Problem** **Problem 6:** An airplane is flying in a horizontal circle at a speed of 490 km/h. If its wings are tilted at an angle \( \theta = 45^\circ \) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface. --- To solve this, we need to understand the relationship between the forces acting on the airplane and the radius of the circular path it takes. 1. **Aerodynamic Lift**: The lift force \( L \) acts perpendicular to the wing surface. 2. **Centripetal Force**: A portion of the lift force provides the necessary centripetal force \( F_c \) to maintain circular motion. The centripetal force \( F_c \) needed for circular motion is given by: \[ F_c = \frac{mv^2}{r} \] Where: - \( m \) is the mass of the airplane - \( v = 490 \, \text{km/h} \) (convert this to meters per second, \( v = 136.11 \, \text{m/s} \)) - \( r \) is the radius of the circle The component of the lift force providing the centripetal force is \( L \sin\theta \). Since the plane is in horizontal flight, the vertical component \( L \cos\theta \) balances the weight \( mg \) of the airplane. By equating the centripetal force to the horizontal component of the lift force: \[ \frac{mv^2}{r} = L \sin\theta \] And since \( L \cos\theta = mg \): \[ L = \frac{mg}{\cos\theta} \] By substituting \( L \) into the centripetal force equation: \[ \frac{mv^2}{r} = \left(\frac{mg}{\cos\theta}\right) \sin\theta \] The mass \( m \) cancels out: \[ \frac{v^2}{r} = \frac{g \sin\theta}{\cos\theta} \] Simplify using \( \tan\theta \): \[ \frac{v^2}{r} = g \tan\theta \] Solving for \( r
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer
Knowledge Booster
Centripetal force
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Physics for Scientists and Engineers: Foundations…
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College
College Physics
College Physics
Physics
ISBN:
9781285737027
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
Physics for Scientists and Engineers with Modern …
Physics for Scientists and Engineers with Modern …
Physics
ISBN:
9781337553292
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers, Technology …
Physics for Scientists and Engineers, Technology …
Physics
ISBN:
9781305116399
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning