Finding proper value for centripetal acceleration in a plane rising up

[aadityav]

Homework Statement

A plane pulls out of a downward dive, where the bottom portion of the flight path is a quarter circle with radius 300m. Assume the pilot loses consciousness if the upward acceleration exceeds 6g. Assume g=10m/s^2. What minimum speed must the plane have at the bottom of the dive for the pilot to black out?

Relevant Equations

Centripetal Acceleration=v^2/r

When I went to solve it, I set the maximum upwards acceleration equal to the formula for centripetal acceleration (I'm going to call it CA), so we get v^2/r=6g, and then v^2/300=60. Solving for v, we get v=sqrt(18000), or v≈134.16ms. However, my professor said this is wrong, since we must set the centripetal acceleration equal to 7g. He said there is 1g from gravity pointing downwards, so CA−1g=6g and we get CA=7g. When we solve using this value for CA, we get v=sqrt(21000) or v≈144.91ms. So which one is correct, CA=6g (what I did), or CA=7g (what my professor did)?

Further explanation:

My Explanation: I thought it must be CA=6g since the question says the maximum upwards acceleration is 6g, and since centripetal acceleration points in the positive y-direction, we set it equal to 6g. Also, if we were to consider gravity, then my reasoning was that there is always the force of gravity pulling the plane down, but since the plane is flying horizontally (i.e., it's not in a nose dive/free fall), that means there is no upwards or downwards acceleration. And I think this would be because the plane's wings would be generating lift that is the same force as weight, so the sum of vertical forces on the plane is 0, so no vertical acceleration. And then since the pilot would black out at 6g of upwards acceleration, then we set CA=6g since the initial upwards acceleration is 0. I understand that if the plane was already falling down with acceleration g, then we would set CA=7g, but I don't understand why for this.

Professor's Explanation: My professor said that he understands what I mean by how there's no upwards or downwards acceleration due to lift, but he also said that when the plane starts tilting upwards, the acceleration from the lift would now be considered to be centripetal acceleration. So we need 1g of that now called "centripetal acceleration" to stay horizontally moving with no vertical acceleration (which we originally called lift). And since the pilot blacks out with 6g of upwards acceleration, and we already have 1g of upwards acceleration from lift, which we now call centripetal acceleration (although this acceleration upwards doesn't make the plane accelerate vertically, it just counteracts gravity), we set CA=6g+1g=7g, where is 6g term is when the pilot blacks out due to total upwards acceleration and the 1g term doesn't actually count for accelerating the plane upwards, but it is counted as centripetal acceleration even though it is lift.

So which one is correct? Do we actually convert the lift force to centripetal acceleration, or is that false?

 

[jbriggs444]

Homework Statement: A plane pulls out of a downward dive, where the bottom portion of the flight path is a quarter circle with radius 300m. Assume the pilot loses consciousness if the upward acceleration exceeds 6g. Assume g=10m/s^2. What minimum speed must the plane have at the bottom of the dive for the pilot to black out?

Loss of consciousness results from proper acceleration, not coordinate acceleration. Proper acceleration is how much the plane's acceleration deviates from free fall. Coordinate acceleration is the plane's acceleration in a chosen reference frame. Probably the rest frame of the ground.

When I went to solve it, I set the maximum upwards acceleration equal to the formula for centripetal acceleration (I'm going to call it CA), so we get v^2/r=6g, and then v^2/300=60. Solving for v, we get v=sqrt(18000), or v≈134.16ms. However, my professor said this is wrong, since we must set the centripetal acceleration equal to 7g.

The professor is correct to point out your error. The centripetal acceleration is a coordinate acceleration. It is the acceleration of the plane as it traverses a circular path relative to the ground frame.

But the pilot blacks out from proper acceleration. So we have to add or subtract 1g to convert from one to the other.

However, the professor is also incorrect. He got the sign wrong and added when he should have subtracted.

Let us think about it more simply...

If the plane is in steady level flight in a 1g gravitational field the pilot experiences 1g of proper acceleration.
If the plane is accelerating upward at 1g in a 1g field, the pilot experiences 2g of proper acceleration.

You would agree with this, right?

So if the plane is accelerating upward at 5g in a 1g field, the pilot experiences how much proper acceleration?

That's not 7g of coordinate acceleration.

 

[Lnewqban]

Welcome!

Please, see:
https://en.wikipedia.org/wiki/G-force#Vertical

The pilot loses consciousness whenever the force with which his seat is pushing his butt up (apparent weight) is 6 times his normal weight on the ground.

That force reaches its maximum value at the bottom portion of the loop.

How much is his apparent weight at the instant his airplane initiates and finishes the dive?

 

[aadityav]

Loss of consciousness results from proper acceleration, not coordinate acceleration. Proper acceleration is how much the plane's acceleration deviates from free fall. Coordinate acceleration is the plane's acceleration in a chosen reference frame. Probably the rest frame of the ground.


The professor is correct to point out your error. The centripetal acceleration is a coordinate acceleration. It is the acceleration of the plane as it traverses a circular path relative to the ground frame.

But the pilot blacks out from proper acceleration. So we have to add or subtract 1g to convert from one to the other.

However, the professor is also incorrect. He got the sign wrong and added when he should have subtracted.

Let us think about it more simply...

If the plane is in steady level flight in a 1g gravitational field the pilot experiences 1g of proper acceleration.
If the plane is accelerating upward at 1g in a 1g field, the pilot experiences 2g of proper acceleration.

You would agree with this, right?

So if the plane is accelerating upward at 5g in a 1g field, the pilot experiences how much proper acceleration?

That's not 7g of coordinate acceleration.

Thank you for the help! I understand that pilots would pass out from the forces they feel, and I agree with how it is 5g of upwards acceleration, since the pilot would now feel 6g. But I asked my professor afterwards, and he insisted he meant acceleration, not what the pilot actually feels, and he also kept insisting he understood my question, but the equation should be solved with 7g, not 6g or 5g.

 

[aadityav]

Welcome!

Please, see:
https://en.wikipedia.org/wiki/G-force#Vertical

The pilot loses consciousness whenever the force with which his seat is pushing his butt up (apparent weight) is 6 times his normal weight on the ground.

That force reaches its maximum value at the bottom portion of the loop.

How much is his apparent weight at the instant his airplane initiates and finishes the dive?

View attachment 369922

View attachment 369923

I understand that at the bottom of the dive, if his centripetal acceleration were 5g, then he would feel 6g of weight, but my professor kept saying he means acceleration and not the force the pilot feels, and we set it equal to 7g, not 6g or 5g. So I'm not sure what's correct...

 

[Herman Trivilino]

If you have an upward force that would create an upward acceleration of magnitude $7g$ and a downward force that would create a downward acceleration of magnitude $g$ then the net force would create an upward acceleration of magnitude $6g$.

 

[Herman Trivilino]

But the pilot blacks out from proper acceleration.

True. But the problem implies that it's a coordinate acceleration of $6g$ in this particular scenario that causes the blackout. So evidently blackout occurs when an accelerometer reads $5g$.