Here's the question. A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If r=20.0 m, how fast is the roller coaster traveling at the bottom of the dip? I'm having some real troubles answering this problem. Thus far, here's what I've come up with. Fc = ( mv^2 ) / r FN - 2mg = ( mv^2 ) / r 2mg - mg = ( mv^2 ) / r (2mg - mg )(r) = ( mv^2 ) (2mg - mg )(r) = ( v^2 ) sqrt( (2mg - mg )(r) ) = v sqrt( ( 19.6 - 9.8 m/s^2 )( 20.0m ) ) = v 14 m/s = v I've been informed that 14 m/s is not the correct answer and that to one digit the correct answer is 10 m/s. It seems to me that I have all the right pieces, but I may be missing something. Will someone help, please? Big thanks!