- #1
delve
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Here is an answer from a book of mine: From the force diagram we have [tex]N-mg=(\frac{mv^2}{R})e_r[/tex]. The acceleration that the pilot feels is [tex]\frac{N}{m}=g+(\frac{mv^2}{R})e_r[/tex]. I'm confused though; what happened to the g being divided by the last term as well?