What is the minimum radius of curvature for a pilot flying at 1000 km/h?

[Plantatree123]

Homework Statement

The human organism can handle acceleration that is 4 times bigger than gravitational acceleration. What is the smallest radius of curvature that can handle pilot of an airplane that files with constant speed of 1000 km/h?

Homework Equations

This task seems easy but I don't have idea what to do. All I know is if the pilot is flying with constant speed then tangential acceleration is zero and if we are talking about the minimal radius then radial acceleration must be maximal which means that a_r=4•g.

The Attempt at a Solution

I tried like this a_r= v²/r and then r_min=v²/a_r=1966,40m
And the result is 0.25•10¹¹m
There is obviously something I couldn't think of. I'd like someone to give me idea what to do. Thanks :)

 

[BvU]

Hello 123,

Who says the result is 0.25•1011m ?

 

[Plantatree123]

Hello 123,

Who says the result is 0.25•1011m ?

Hello writer of workbook This result is in solutions of my workbook

 

[BvU]

The 0.25 x 10¹¹ m is obviously wrong: the radius of the Earth is 6 x 10⁶ m and planes do fly around with such speeds without killing the passengers.

I don't see anything wrong with your 1970 m, except one thing: the Earth keeps pulling with 1 times g as well. So if the loop is vertical, the driver experiences 5 g at the bottom and 3 at the top. Not good. If the circle is horizontal, 1 g and 4 g add up vectorially to √17 times g, also > 4 g. So I would feel safer in your plane if a_r = √15 g (But it's a small correction and I wonder if that is asked for in your exercise)

By the way, if your given data is only in one digit, it is better to round off your results a little bit: so 4g corresponds to a circle with a radius of 1970 m
(personally I would even prefer 2 km, but teacher may think different).

 

[Plantatree123]

The 0.25 x 10¹¹ m is obviously wrong: the radius of the Earth is 6 x 10⁶ m and planes do fly around with such speeds without killing the passengers.

I don't see anything wrong with your 1970 m, except one thing: the Earth keeps pulling with 1 times g as well. So if the loop is vertical, the driver experiences 5 g at the bottom and 3 at the top. Not good. If the circle is horizontal, 1 g and 4 g add up vectorially to √17 times g, also > 4 g. So I would feel safer in your plane if a_r = √15 g (But it's a small correction and I wonder if that is asked for in your exercise)

By the way, if your given data is only in one digit, it is better to round off your results a little bit: so 4g corresponds to a circle with a radius of 1970 m
(personally I would even prefer 2 km, but teacher may think different).

I like your explanation. :) Thank you

 

[jbriggs444]

Who says the result is 0.25•10¹¹m ?

[exponent restored]

Reverse engineering the claimed result suggests that the workbook assumed 1000 km/sec rather than 1000 km/hour. This demonstrates that units matter.