Pulling out of a dive,the pilot of an airplane guides his plane into a vertical circle.At the bottom of the dive,the speed of the airplane is 320 m/s.What is the smallest radius allowable for the vertical circle if the pilot's apparent weight (the force that the pilot exerts on the seat)is not to exceed 7.0 times his true weight?
A)1700 m
B)1500 m
C)2200 m
D)230 m
E)42 m

Question

Quizplus · Accepted Answer

The pilot's apparent weight is the sum of his true weight and the centripetal force required to keep him moving in a circle. This can be expressed as $F_{apparent} = mg + \frac{mv^2}{r}$, where $m$ is the mass of the pilot, $g$ is the acceleration due to gravity (9.8 m/s$^2$), $v$ is the velocity of the airplane, and $r$ is the radius of the circle. Given that the apparent weight is 7 times the true weight, we can set up the equation $7mg = mg + \frac{mv^2}{r}$. Simplifying this gives $6mg = \frac{mv^2}{r}$, and solving for $r$ gives $r = \frac{v^2}{6g}$. Substituting $v = 320$ m/s and $g = 9.8$ m/s$^2$ gives $r = \frac{(320)^2}{6 \times 9.8} \approx 1700$ meters.

Pulling Out of a Dive,the Pilot of an Airplane Guides