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Make sure to answer questions requiring text response using an output command such as disp
Just parts a,b,d 2.22 Aircraft Turning Radius An object moving in a circular tan gential velocity vis shown in Figure acceleration required for object to move in the circular path is given by the Equation (2.28) a (2.28) where a is the centripetal acceleration of the object in m/s, v is the tangential velocity of the object in m/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 g, what is the turning radius of the aircraft? (Note: For this problem, you may assume that Mach 1 is equal to 340 m/s, and that 1 g 9.81 m/s2). (b) Suppose that the speed of the aircraft increases to Mach 1.5. hat 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 acceleration remains 2 g. (d) Suppose that the maximum acceleration that the pilot can stand is 7 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 g and 8 g, assuming a constant speed of Mach 0.85.

Explanation / Answer

a)
v = 0.85*340 = 289 m/s
a = 2*g = 19.6 m/s^2

a = v^2/r

==> r = v^2/a = 4261.3 m
b)
v = 1.5*340 = 510 m/s
a = 2*g = 19.6 m/s^2

a = v^2/r

==> r = v^2/a = 13270.4 m

c)

d)
v = 1.5*340 = 510 m/s

a_total = 7*g

a_toatl^2 = a_rad^2 + g^2


a_rad = sqrt(a_total^2-g^2)

a_rad = 6.93*g = 67.9 m/s^2

a = v^2/r

==> r = v^2/a = 3830 m
e)

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