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 [tex]\alpha=\omega^{2} r[/tex] ----> eq 1
v = r ω ===> [tex]r=\frac{v}{w}[/tex] ---- eq 2
putting eq 2 in equation (1), we get
[tex]\alpha=\omega^{2}\left(\frac{v}{w}\right)=\omega \times v[/tex]
Substitute the given values, we get
[tex]\omega=\frac{\alpha}{v}[/tex]
[tex]\omega=\frac{4 \times 9.8}{600}=0.0653\ \mathrm{rad} / \mathrm{s}[/tex] -----eq 3
Substitute value in eq 3 to eq. 2, we get
[tex]r=\frac{600}{0.0653}=9188\ \mathrm{m}[/tex]
