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 36 s to complete the circle and the wingspan of the plane is 14.2 m, what is the acceleration of the wing tip
Answer:
The acceleration of the wing tip is 0.216 m/s²
Explanation:
Given;
time to complete on circle, t = 36 s
the wingspan of the plane = 14.2 m, this is equal to diameter of the circle, thus radius of the circle, r = ¹/₂ x 14.2 = 7.1 m
distance of a complete circle = circumference of a circle = 2πr
circumference of a circle = 2πr = 2π x 7.1 = 44.6164 m
velocity, v = distance / time
velocity of the wing tip, v = 44.6164 / 36
= 1.239 m/s
Centripetal acceleration of the wing tip = v²/r
= (1.239²)/7.1
= 0.216 m/s²
Therefore, the acceleration of the wing tip is 0.216 m/s²