What Is the Radius of an Arc for a Pilot Experiencing 5G at 700m/s?

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To determine the radius of an arc for a pilot experiencing 5G at a speed of 700 m/s, the relevant equations involve centripetal acceleration. The pilot experiences a total normal force of 5G, which translates to an acceleration of 5 times the gravitational acceleration (5 * 9.81 m/s²). The equation r = v²/a can be used to calculate the radius, where 'v' is the speed and 'a' is the total acceleration. Substituting the values, the radius can be calculated effectively. Understanding the forces acting on the pilot is crucial for solving this problem accurately.
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Homework Statement


Im having trouble figuring out which equation to use for this problem. The problem states: "Consider a pilot at the lowest point of a circular arc banking upward. Find the tightest radius arc that an untrained individual can fly ( a total of +5 G, G standing for the normal force the accelerating aircraft is exerting on the pilot. For example, 6G force would be exerting a normal force 6 times the persons weight.). They are traveling at 700m/s.


Homework Equations


Since this involves normal force, I thought I could somehow use: the sum of F=ma and Fw=mg
In addition I tried using a=v squared/r and/or v=2(pi)r/T


The Attempt at a Solution


In an attempt to try to solve this problem I tried using Fn= 5N in the equation Fn-Fw=ma By finding acceleration I could then use a=v squared/r to solve for r, but I don't have enough variables to plug into the first equation.
So I tried using v=2(pi)r/T giving me 700m/s=2(pi)r/T so then I could find r this way, However I do not know the value for T.
Am I using the completely wrong equations? If not, what am I doing wrong?
 
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Welcome to PF.

I'd suggest drawing a diagram and making sure how much acceleration the plane will impart and how much is just a result of gravity.

As to your equation, you want to consider the centripetal acceleration here.
 
I could then use a=v squared/r to solve for r
I think you have it here. r = v^2/a
Your acceleration is 5 times 9.81.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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