How do I work this?

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 six times her weight as she goes through the dip. If r = 20.5 m, how fast is the roller coaster traveling at the bottom of the dip in m/s?

Answer 1

Draw a force diagram of the situation at the instant which the roller coaster is at the bottom of the dip.
The centripetal acceleration (provided by the seat of the roller coaster car) is pushing the passenger up. The force of gravity is pulling the passenger down, but the seat is also providing an additional force which balances the passenger's weight.
Summing up all the forces the seat provides, the passenger feels a force equal to 6 times that of their normal weight.
Weight = mass * gravity
So the passenger feels a force of,
F = 6 * mass * gravity
But we know that 1mg of this force is due to gravity, the other 5mg must be the centripetal acceleration the passenger experiences.

F_c = m * speed^2 / radius
F_c = 5mg
5mg = mv^2 / r
solving for the speed, v, we get,
v = sqrt(5*g*r)
We can assume a value of 9.81 m/s^2 for g and we are explicitly told the value of r is 20.5 meters. Simply plug in and solve at this point to get a value of,
v = 31.7 m/s

Answer 2

If the force she feels is 6*m*g, then this represents a gravitational force of 'm*g', plus a centripetal force of 5*m*g.

F = m*v^2 / r
r = 20.5 m
F = 5*m*g

solve for v...

5*m*g = m*v^2 / r

v = Sqrt[5*r*g] = Sqrt[5*20.5 m*9.8 m/s^2] = 31.69 m/s