A fighter pilot dives his plane toward the ground at 345 m/s. He pulls out of the dive on a vertical circle. What is the minimum radius of the circle, so that the normal force exerted on the pilot by his seat never exceeds three times his weight?
3mg>mv^2/r
solve for r
my answer was wrong.. I got 4048.47? What did you get?
oops. Normal force on his seat includes his weight.
3mg>mg+mv^2/r
So then... once you divide out the m's it would be 3g>g+v^2/r?
yes.
To determine the minimum radius of the vertical circle, we need to analyze the forces acting on the pilot.
When the pilot is at the lowest point of the circular motion, the gravitational force and the normal force add up to provide the centripetal force required for circular motion.
Let's define the following variables:
- v = Velocity of the plane (345 m/s in this case)
- r = Radius of the circular path (what we need to find)
- g = Acceleration due to gravity (approximately 9.8 m/s²)
- N = Normal force exerted on the pilot by his seat
- W = Weight of the pilot
The gravitational force acting on the pilot at the lowest point of the circle is given by:
F_gravity = mg
The centripetal force required to keep the pilot in circular motion is given by:
F_centripetal = m * v² / r
To find the normal force exerted by the seat, we need to consider the forces in the vertical direction. At the lowest point of the circle, these forces are the gravitational force and the normal force:
ΣF_vertical = N - mg
Since the normal force should not exceed three times the weight of the pilot, we can write:
N ≤ 3W
N ≤ 3mg
Now, equating the centripetal force and the gravitational force:
F_centripetal = F_gravity
m * v² / r = mg
Canceling the m and rearranging the equation, we get:
v² / r = g
Now, we can substitute the value of v = 345 m/s and g = 9.8 m/s²:
345² / r = 9.8
Solving for r:
r = 345² / 9.8 ≈ 12272.45 meters
Therefore, the minimum radius of the circle for the normal force to not exceed three times the weight is approximately 12272.45 meters.