Question

An airplane is flying in a horizontal circle at a
speed of 104 m/s. The 84.0 kg pilot does not
want the centripetal acceleration to exceed
6.98 times free-fall acceleration.
Find the minimum radius of the plane’s
path. The acceleration due to gravity is 9.81
m/s2
.
Answer in units of m


At this radius, what is the magnitude of the
net force that maintains circular motion exerted on the pilot by the seat belts, the friction
against the seat, and so forth?
Answer in units of N.

Answer 1

acceleration=v^2/r or r=v^2/a

force= a*massOfPilot

Answer 2

To elaborate further on Bobpursley's answer,

Acceleration outward at rim is, A = rω² = V²/r. V is the tangental velocity in m/s.
r is radius of circle in meters. ω is the angular velocity in radians/sec
a is in m/s².. so,

a = V²/r = (6.56)(9.81)
r = (104)² / (6.56)(9.81) = 168 m

Centripetal force is..
f = mV²/r = mrω²
ω is angular velocity in radians/sec
1 radian/sec = 9.55 rev/min
m is mass in Kilograms, R is radius of circle in meters, V is the tangental velocity in m/s, and Force (F) is in Newtons.

F = mV²/r = (82)(104)² / (168) newtons!
Maybe I'm overthinking it but, Heres my two cents.
Khan Academy also has videos on these :) Hope this helped!!