Answer to Question #100559 in Physics for Nyx

Question

An aerobatic airplane pilot experiences
weightlessness as she passes over the top of
a loop-the-loop maneuver.
The acceleration of gravity is 9.8 m/s2. If her speed is 310 m/s at this time, find the
radius of the loop.
Answer in units of km.

Answer 1

There are two forces that act on the pilot: the gravitational force $mg$ directed towards the center of the loop and the normal force exerted on the pilot $N$ directed away from the center of the loop. Let's choose the downwards as the positive direction and apply the Newton's Second Law of Motion:

$mg - N = m \dfrac{v^2}{r}.$

Since the pilot experiences the weightlessness at the top point of the loop, the normal force exerted on the pilot equals zero. Therefore, we get:

$g = \dfrac{v^2}{r}.$

From this formula we can find the radius of the loop:

$r = \dfrac{v^2}{g} = \dfrac{(310 \dfrac{m}{s})^2}{9.8 \dfrac{m}{s^2}} = 9.806 \cdot 10^3 m = 9.8 km.$

Answer:

$r = 9.8 km.$