Question

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?

Answer 1

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 $\alpha=\omega^{2} r$ ----> eq 1

v = r ω ===> $r=\frac{v}{w}$ ---- eq 2

putting eq 2 in equation (1), we get

$\alpha=\omega^{2}\left(\frac{v}{w}\right)=\omega \times v$

Substitute the given values, we get

$\omega=\frac{\alpha}{v}$

$\omega=\frac{4 \times 9.8}{600}=0.0653\ \mathrm{rad} / \mathrm{s}$ -----eq 3

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

$r=\frac{600}{0.0653}=9188\ \mathrm{m}$