Question

A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of the loop in which the plane is flying?

A.) 640m
B.) 9200m
C.) 1200m
D.) 7100m

Answer 1

To solve this problem we will apply the kinematic equations of angular motion that relate the variables to those of linear motion.

We know that acceleration is equivalent to

$a=\omega^2*r$

Here,

$\omega = \text{Angular velocity}$

r = Radius

At the same time the tangential velocity can be related with the angular velocity as follow,

$v=\omega*r$

Our total value of acceleration is 4g, where g means the gravitational acceleration, then 4g is equal to

$g=9.81m/s^2$

$4g= 39.24 m/s^2$

The total value of this acceleration replacing it in our first equation would be,

$39.24=\omega^2*r$

And the velocity in our second equation is

$600=\omega*r$

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

Replacing the value of r, we have

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

Solve for $\omega$ and we get;

$\omega=0.0654$

Replacing the angular velocity value in the linear velocity equation we will have that the radius is

$600=\omega r$

$600 =0.0654r$

$r=9174m \approx 9200m$

Therefore the correct answer is B