In a classic carnival ride, patrons stand against the wall in a cylindrically shaped room.

Question

In a classic carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast enough, the floor drops from the bottom of the room! Friction between the walls of the room and the people on the ride make them the “stick” to the wall so they do not slide down. In one ride, the radius of the cylindrical room is R = 6.3 m and the room spins with a frequency of 21.6 revolutions per minute.

If a new person with mass 108 kg rides the ride, what minimum coefficient of friction between the wall and the person would be needed?

Answer 1

Hey Anna,

The thing that gets tricky about circular motion work is that you have to remember that the centripetal force IS the net force. In this example, the force is provided by the wall of the ride, but you will also see problems with a string, where the tension in the string is providing the centripetal force. Here, the force is coming out perpendicularly from the wall of the ride, so we call it the normal force (because normal means "at a right angle").

1. So, in this problem, the centripetal force, net force, and normal force, are all the same thing. You should have an equation for centripetal force. Just remember that your speed (v) needs to be in m/s, not revolutions per minute, so you will need to convert.
2. 1 rev/min = 1 circumference of the circle / 60 seconds = 2*pi*R/ 60s.
3. Successfully solving for the centripetal force is important because normal force is a part of our friction formula:
4. F_friction = μ * F_normal
5. So, you can plug in a number for normal force, but what's the frictional force? Well, think about what forces are acting on a rider:
6. We already covered the normal force, which is pointed in towards the center of the circle.
7. Vertically, we have gravity pulling down and friction pulling up. For this person not to slide down, the force of friction must be AT LEAST equal to the gravity acting on them. So, if you find the force of gravity on this 108- kg person, you can go back to that equation and solve for the coefficient of friction.