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?

Answer 1

we apply newton's second law to this situation

sum of forces = ma

the forces are the normal force acting up and weight acting down they combine to produce a centrip force that acts

radially upward toward the center of the circle, therefore, newton's second law becomes

N - mg = mv^2/r

the normal force is the force pushing up on you, so we are told that the normal force is 2 the weight, or that N=2mg,

therefore we have:

2 mg - mg = mv^2/r

mg - mv^2/r

g = v^2/r

v = sqrt[r*g]

v = sqrt[20 x 9.8]

v = 14m/s