Centripetal Force: Find Min Radius of Airplane Path

AI Thread Summary
To find the minimum radius of an airplane's circular path, the pilot's radial acceleration must not exceed 8.88 g, which translates to approximately 86.9 m/s². Using the formula for centripetal acceleration, v²/r = a, where v is the speed of 141 m/s, the equation can be rearranged to solve for the radius r. Substituting the values, the calculation leads to determining the minimum radius required for safe flight. Understanding the relationship between speed, mass, and acceleration is crucial in this scenario. The final result provides the necessary radius to ensure the pilot's safety during the maneuver.
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Homework Statement



An Airplane is flying ina horizontal circle at a speed of 141 m/s. The 94.2 Kg pilot does not want his radial acceleration to exceed 8.88 g.

What is the minimum radius of the circular path?

Homework Equations



v^2/r
Fnet = M x A

The Attempt at a Solution



141^2 / .00888 kg?
 
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Just try equating f=ma with your centripetal force. Then re-arrange for r. Think about what 8.88g is interns of acceleration
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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