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 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 apparent weight of the pilot at the bottom of the dive is the sum of the gravitational force and the centripetal force required to keep the pilot moving in a circle. This can be expressed as $W_{apparent} = W_{true} + F_{centripetal}$. The centripetal force can be calculated using the formula $F_{centripetal} = \frac{mv^2}{r}$, where $m$ is the mass of the pilot, $v$ is the velocity of the airplane, and $r$ is the radius of the circle. Given that the apparent weight is not to exceed 7 times the true weight, we can set up the equation $7W_{true} = W_{true} + \frac{mv^2}{r}$. Simplifying this, we get $6W_{true} = \frac{mv^2}{r}$, which leads to $r = \frac{mv^2}{6W_{true}}$. Since $W_{true} = mg$, where $g$ is the acceleration due to gravity (9.8 m/s$^2$), we can substitute to get $r = \frac{v^2}{6g}$. Plugging in the given values ($v = 320$ m/s), we get $r = \frac{(320)^2}{6 \times 9.8}$, which calculates to approximately 1700 meters.

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