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 three times her weight as she goes through the dip. If r = 16.0 m, how fast is the roller coaster traveling at the bottom of the dip? I used the equation: v= square root of 3*r*g, but the answer was wrong. I don't know what other equation to use.
The passenger feels the normal force, which is 3mg. (Any time when we sit, stand, ... on something we feel the normal force. On a horizontal surface, in rest, this normal force happens to be equal to mg. Not when we are in circular motion.) There are two forces, acting on the passenger, the normal force upward and mg downward. The resultant is 2mg, and that is equal to the centripetal force. So v=sqrt(2mg)