How fast is the girl traveling at the bottom of the dip of a roller coaster?
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?
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 or v = sqrt[r g] = sqrt[20x9.8]=14m/s
and this is the speed at the bottom of the vertical circle
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