Why Does an Airplane Pilot Experience Limits on Radial Acceleration?

AI Thread Summary
An airplane pilot experiences limits on radial acceleration due to the need to maintain safe g-forces during maneuvers. In the discussion, a problem is presented where a pilot flying at 48.1 m/s must ensure radial acceleration does not exceed 7.39 g, leading to calculations for the minimum radius of the circular path and the net centripetal force. The minimum radius is calculated to be approximately 31.94 meters, while the centripetal force acting on the pilot is determined to be around 6824.78 N. The calculations involve using the formulas for centripetal acceleration and force, confirming the pilot's mass is a key factor. Overall, the calculations and concepts discussed highlight the importance of understanding forces acting on pilots in circular motion.
kimikims
Messages
36
Reaction score
0
I tried workin this problem out.. and it's not working! :frown:

---
An airplane is flying in a horizontal circle at a
speed of 48.1 m/s. The 94.1 kg pilot does not
want his radial acceleration to exceed 7:.39 g.
The acceleration of gravity is 9.8 m/s^2

1) What is the minimum radius of the circular
path? Answer in units of m.

7.39g (9.8m/s^2) = 72.422 m/s^2
do I need to square root 72.422??


2) At this radius, what is the net centripetal force
exerted on the pilot by the seat belts,
the friction between him and the seat, and so
forth? Answer in units of N.

N = mg + m (V^2/m)

is this the right formula?? what would m and v be??
 
Physics news on Phys.org
Is this right??

Speed = 48.1 m/s
g = 9.8 m/s^2

Problem #1

7.39 g (9.8 m/s^2) = A
A = 72.422

R = V^2 / A
R = (48.1 m/s)^2 / 72.422
R = 31.94

Problem #2

F = mV^2 / R
What's m??
F = (?)m x (4.8)^2 / 31.94
 
Remember that A(centripital) = v^2 / R...

...so your answer to problem 1 looks correct.

On problem 2, how did you go from a velocity of 48.1 m/s to that of 4.8 m/s? The mass would just be the pilots mass...because its asking for the centripital force acting on the pilot.
 
thermodynamicaldude said:
Remember that A(centripital) = v^2 / R...

...so your answer to problem 1 looks correct.

On problem 2, how did you go from a velocity of 48.1 m/s to that of 4.8 m/s? The mass would just be the pilots mass...because its asking for the centripital force acting on the pilot.


Sorry I meant 48.1 m/s not 4.8 m/s...

So it would be:

Fc = m (V^2/R)
Fc = 94.1 [(48.1)^2 m/s / 31.9)

Fc = 6824.78 N?
 
Looks good. :-D
 
thermodynamicaldude said:
Looks good. :-D


Thank you! :biggrin:
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top