Question

The weight of passengers on a roller coaster increases by 60 % as the car goes through a dip with a 37 m radius of curvature. what is the car's speed at the bottom of the dip?

Answer 1

The force acting on the passenger is equal to:

`F=N-mg`
Where N is the normal force and mg is a force of gravity. If a person feels like their weight is increased by 60% at the bottom that means normal force is 60% stronger than the force of gravity:

`N-mg=0.6mg`
We know that net force must be equal to the centripetal force:

`0.6mg=(mv^2)/(R)\\ v^2=0.6gR\\ v=√(0.6gR)=14.75(m)/(s)`