A jet plane flying 600 m/s experiences an acceleration of 4.0 g when pulling out of a circular dive. What is the radius of curvature of the circular part of the path in which the plane is flying?

9188 m is the radius of curvature of the circular part of the path in which the plane is flying.

Explanation:

When an air-plane dives, to exit the dive, it must follow a circular path to move from vertical downward to horizontal motion. To draw this circular movement, the pilot manipulates the wings so that the air force and the lifting forces create a force centered in the center that accelerates the plane towards the center of the wheel.

Given data:

speed of the plane = 600 m/s

acceleration = 4 g

We need to find radius of curvature of loop (r)

we know, angular acceleration ----> eq 1

v = r ω ===> ---- eq 2

putting eq 2 in equation (1), we get

Substitute the given values, we get

-----eq 3

Substitute value in eq 3 to eq. 2, we get