Question

The maximum force a pilot can stand is about seven times his weight. What is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250 m/s?

Answer 1

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 :

$N-mg=\dfrac{mv^2}{r}$

Where

N is the normal force

r is the radius of curvature

According to given condition,

$7mg-mg=\dfrac{mv^2}{r}$    

$6mg=\dfrac{mv^2}{r}$    

$r=\dfrac{mv^2}{6mg}$  

$r=\dfrac{mv^2}{6mg}$      

$r=\dfrac{v^2}{6g}$      

$r=\dfrac{(250)^2}{6\times 9.8}$  

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.