# Centripetal force Rotor-ride problem

1. Dec 11, 2006

### Gauss177

1. The problem statement, all variables and given/known data
In a "Rotor-ride" at a carnival, riders are pressed against the inside wall of a vertical cylinder 2.0m in radius rotating at a speed of 1.1 revolutions per second when the floor drops out. What minimum coefficient of friction is needed so a person won't slip down? Is this safe?

2. Relevant equations

3. The attempt at a solution
I set friction equal to the net force and then solved for mu, which I'm not sure is right or not:

$$umg = m(v^2/r)$$

2. Dec 11, 2006

### OlderDan

Are you sure about the 2.0m? That seems awfully small compared to the ones I have seen.

In any case, your equation is not correct. The centripetal force is provided by the wall pushing the people inward, not by frcition. The friction keeps them from sliding down the wall.

3. Dec 11, 2006

### Gauss177

Yeah the book says 2.0 meters.

For the person to not slide down the wall, the force of friction must equal the force of gravity, right? So do I just do:

$$F_fr = F_g$$
$$umg = mg$$

But then mg cancels out and u = 1? That doesn't seem right.

4. Dec 11, 2006

### OlderDan

It is not right. The frictional force has no connection to mg in this problem except that it has to be equal to mg to hold the people in place. Your first equation is good. Your second one is not.

Although mg often appears in friction calculations, it is not the force in F_f = μ____. What belongs in the space?

Last edited: Dec 11, 2006
5. Dec 12, 2006

### Gauss177

I have no idea. :/ I thought friction was just u*normal? But then where does the 1.1 revolutions per second come from? Thanks for the help

6. Dec 12, 2006

### Gauss177

Ok I figured out how to do this problem

$$F_f = F_g$$
$$mg = uF_n$$
$$mg = m(v^2/r)*u$$
$$u = .103$$

thanks olderdan

7. Dec 12, 2006

### OlderDan

But you do have an idea It is the normal force and as you have now realized, the normal force in this problem is not equal to mg.

That's it. For anyone else who might be looking here, the normal force in this problem is from the wall pressing in on the people to keep them in circular motion. It is the centripetal force

$$F_n = m(v^2/r)$$

which can be seen from the equations in the quote.