A jet plane flying 600 m/s experiences an acceleration of 4.0 g when pulling out of the circular section of a dive. What is the radius of curvature of this section of the dive?A) 7100 m
B) 650 m
C) 9200 m
D) 1200 m

Respuesta :

To solve this problem we will apply the concepts related to the equations of linear motion and angular motion in order to find the radius. Our values are given as,

[tex]g=9.81m/s^2[/tex]

Then

[tex]4g= 39.24 m/s^2[/tex]

The relation between the acceleration and the angular velocity and the radius is,

[tex]a_c=\omega^2*r[/tex]

The angular velocity and the lineal velocity can be related as,

[tex]v=\omega r[/tex]

The acceleration and the velocity was given, then

[tex]39.24=\omega^2 r[/tex]

[tex]600=\omega r[/tex] [tex]\rightarrow r = \frac{600}{\omega}[/tex]

Replacing at the first equation we have,

[tex]39.24 = \omega^2 (\frac{600}{\omega} )[/tex]

[tex]\omega = \frac{39.24}{600}[/tex]

[tex]\omega = 0.0654rad/s[/tex]

Now using this expression to find the radius we have that

[tex]r = \frac{600}{\omega}[/tex]

[tex]r = \frac{600}{0.0654}[/tex]

[tex]r = 9174.31 \approx 9200m[/tex]

Therefore the correct answer is C.