# Centripetal Force of an amusement park ride problem

1. Aug 21, 2007

### jcmtnez

1. The problem statement, all variables and given/known data
An amusement park ride consits of a rotating vertical cylinder with rough canvas walls. After the rider has entered and the cylinder is rotating sufficiently fast, the floor is dropped down, yet the rider does not slide down, The rider has a mass of 50 kilograms, the radius R of the cylinder is 5 meters, the angular velocity of the cylinder when rotating is 2 radians per second, and the coefficient of static friction between the rider and the wall of the cylinder is 0.6.

(b) Calculate the centripetal force on the rider when the cylinder and state what provides that force.

(c) Calculate the upward force that keeps the rider from falling when the floor is dropped down and state what provides that force.

(d) At the same rotational speed would a rider of twice the mass slide down the wall? Explain you answer.

2. Relevant equations

F=m*a

a=R*w^2

3. The attempt at a solution

I could derive the centripetal force by applying newton's second law and the formula for centripetal aceleration. I got 1000 N but i am not sure of my answer. And I dont understand very well what force is involved in pulling up the person when the floor is dropped down and how is friction involved.

Last edited: Aug 21, 2007
2. Aug 21, 2007

### rootX

It is not pulling up the person, but instead is acting against W, and is equals to W.
-It's frictional force.
-the person tends to go in a straight line, but the wall prevents him from doing so, by constantly applying a centripetal force on him
-gravity tends to pull him towards the earth, but since he is sticking to the wall, and there is friction between him and the wall, so he doesn't slide.