An aircraft performs a manoeuvre called an \"aileron roll.\" During this manoeuv

Question

An aircraft performs a manoeuvre called an "aileron roll." During this manoeuvre, the plane turns like a screw as it maintains a straight flight path at a constant speed, which sets the wings in circular motion. If it takes it 37 s to complete the circle and the wingspan of the plane is 12.4 m, what is the acceleration of the wing tip?

(a) 2.2 ms-2

(b) 5.6 ms-2

(c) 0.18 ms-2

(d) 0.45 ms -2

(e) The motion is not circular, therefore this acceleration cannot be easily calculated.

Answer is apparently (c) but not sure how it gets to that.

Explanation / Answer

Given Data

d = 12.4 m

=> r = 6.2 m

Solution :-
circumference of a circle = 2 * * r

                                     = 2 * * 6.2

                                      = * 12.4

To determine the velocity, divide this distance by 37 seconds.

v = ( * 12.4) ÷ 37 = 1.053 m/s
.

Centripetal acceleration ,a = v^2 / r

                                       = (1.053)^2 ÷ 6.2

                                       = 0.18 m/s^2. (Option 2)

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