Answer:
Radius of curvature of the path is 1063 meters
Explanation:
It is given that,
Force acting on the pilot is about seven times of his weight. Speed with which pilot moves, v = 250 m/s.
As per Newton's second law of motion, the net force acting on the pilot at the bottom is given by :
[tex]N-mg=\dfrac{mv^2}{r}[/tex]
Where
N is the normal force
r is the radius of curvature
According to given condition,
[tex]7mg-mg=\dfrac{mv^2}{r}[/tex]
[tex]6mg=\dfrac{mv^2}{r}[/tex]
[tex]r=\dfrac{mv^2}{6mg}[/tex]
[tex]r=\dfrac{mv^2}{6mg}[/tex]
[tex]r=\dfrac{v^2}{6g}[/tex]
[tex]r=\dfrac{(250)^2}{6\times 9.8}[/tex]
r = 1062.92 meters
or
r = 1063 meters
So, the radius of curvature of the path is 1063 meters. Hence, this is the required solution.
