Questions: An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a straight flight path, which sets the wings in circular motion. If it takes it 29 s to complete the circle and the wingspan of the plane is 12 m, what is the acceleration of the wing tip? 1.4 x 10^-2 m / s^2 0.28 m / s^2 0.11 m / s^2 0.56 m / s^2

Solution Steps

• The radius \( r \) of the circular path is half of the wingspan.
• Given wingspan \( = 12 \, \text{m} \), so \( r = \frac{12}{2} = 6 \, \text{m} \).

• The time \( T \) to complete one circle is given as \( 29 \, \text{s} \).
• The angular velocity \( \omega \) is given by \( \omega = \frac{2\pi}{T} \).
• Substituting \( T = 29 \, \text{s} \), we get \( \omega = \frac{2\pi}{29} \, \text{rad/s} \).

• The centripetal acceleration \( a \) is given by \( a = \omega^2 r \).
• Substituting \( \omega = \frac{2\pi}{29} \, \text{rad/s} \) and \( r = 6 \, \text{m} \), we get: \[ a = \left( \frac{2\pi}{29} \right)^2 \times 6 \]
• Simplifying the expression: \[ a = \left( \frac{4\pi^2}{841} \right) \times 6 \] \[ a \approx 0.28 \, \text{m/s}^2 \]

• The correct answer is \( 0.28 \, \text{m/s}^2 \).

Final Answer