Question

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

Answer 1

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,

$g=9.81m/s^2$

Then

$4g= 39.24 m/s^2$

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

$a_c=\omega^2*r$

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

$v=\omega r$

The acceleration and the velocity was given, then

$39.24=\omega^2 r$

$600=\omega r$ $\rightarrow r = \frac{600}{\omega}$

Replacing at the first equation we have,

$39.24 = \omega^2 (\frac{600}{\omega} )$

$\omega = \frac{39.24}{600}$

$\omega = 0.0654rad/s$

Now using this expression to find the radius we have that

$r = \frac{600}{\omega}$

$r = \frac{600}{0.0654}$

$r = 9174.31 \approx 9200m$

Therefore the correct answer is C.