The pilot of an airplane executes a constant-speed loop-the-loop maneuver in a v

Question

The pilot of an airplane executes a constant-speed loop-the-loop maneuver in a vertical circle as in the figure below. The speed of the airplane is 2.50 102 m/s, and the radius of the circle is 3.10 103 m. (a) What is the pilot's apparent weight at the lowest point of the circle if his true weight is 780 N? N (b) What is his apparent weight at the highest point of the circle? N (c) Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body. Under what conditions does this occur? (d) What speed would have resulted in the pilot experiencing weightlessness at the top of the loop? m/s

Explanation / Answer

There are two forces here. Gravitational force which is always acting down and the centripetal force. The net force is the sum of these forces.
speed, v = 2.5 x 102 m/s
Radius, r = 3.1 x 103 m
Gravitational force, Fg = mg = 780 N
Mass of the pilot = Fg/g = 780/9.8 = 79.6 kg
Formula for centripetal force = mv2/r

a)
At the lowest point, both the centripetal force and gravitational force is acting down
Net force = Fg + mv2/r
= 780 + [79.6 x (2.5 x 102)2 / (3.1 x 103)]
= 2384.67 N pointing down

b)
At highest point, gravitational force is still acting down, but centripetal force is pointing up.
Net force = Fg - mv2/r
= 780 - [79.6 x (2.5 x 102)2 / (3.1 x 103)]
= 824.67 N pointing up

c)
For weightlessness,
Net force = 0
Fg = mv2/r
Now the magnitude of centripetal force is more compared to that of gravitational force. Either by increasing r or decreasing v, at some point, both force will balance and there will be weightlessness.

d)
Fg = mv2/r
mg = mv2/r
g = v2/r
v = sqrt(rg)
= sqrt[(3.1 x 103) x 9.8]
= 174.3 m/s

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