# Circular Motion and Friction on an amusement park ride

1. Dec 1, 2008

### DMOC

1. The problem statement, all variables and given/known data

At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.30 m and the speed of the wall is 10.0 m/s when the floor falls away.

a. What is the source of the centripetal force acting on the riders?

b. How much centripetal force acts on a 55.0 kg rider?

c. What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?

2. Relevant equations

Centripetal force = (mv^2)/r

Force of friction = (coefficient of static friction)*(normal force)

Normal force=m*g

3. The attempt at a solution

I have solved part B. Part A is more of a conceptual question that I have, and part C is the part where I am confused.

Part A attempt:

The reason I am confused on this question is that I am unsure of what the source refers to. Does the source refer to the wall that the people are "plastered" against? This would be my answer because if I were to picture myself in this ride, the wall would appear to be pushing me towards the center of the circle (which is what the centripetal force actually is).

Part B attempt:

I used the first formula I listed in the "relevant equations" section.

Centripetal force = ((55.0 kg)*(10.0 m/s)^2)/3.30 m
Centripetal force = 1670 N

Part C attempt:

I have to find the coefficient of static friction, so I know that I will have to use the second equation I listed in the "relevant equations" section. I will also need to somehow find the normal force and the force of friction.

To find the normal force, I am assuming that the rider's mass is 55.0 kg. No other mass is provided, and I can't think of another way I could calculate the normal force without the mass.

Normal force = m*g
Normal force = (55.0 kg)*(9.81 m/s^2)
Normal force = 540 N

Thus, I have one part necessary to complete the problem.

This is the part where I am confused. I need to find the force of friction. However, the force of friction must be pointing upwards, to counterbalance the gravitational force pushing the riders downwards. Wouldn't the force of friction then be equal to the normal force? If I assume this, the coefficient of static friction would be one, which to me, doesn't seem like the correct answer.

Is there something I'm missing here?

Thanks for any help you can offer. This is my first post here as well.

Last edited: Dec 2, 2008
2. Dec 1, 2008

### rl.bhat

Force of friction = (coefficient of static friction)*(normal force)

Force of friction must be equal to the weight of the person to plaster himself to the wall. Here the normal force is the centrifugal force on the person due to rotation of the cylindrical room.

3. Dec 1, 2008

### naresh

Welcome to PF!

Yes, the wall does push you towards the center. Nothing else can! What is this force?

Can you draw a free body diagram with the forces on the rider? Hint: The " normal force" is the force that acts normal (i.e. perpendicular) to a surface.

You will need the mass only for part B. The normal force has nothing to do with m or g, in general.

Yes, the coefficient of friction is usually much smaller.

4. Dec 1, 2008

### glueball8

But it can be more than one of course!!

I don't see anything wrong

5. Dec 1, 2008

### naresh

Sorry, this part was incorrect. You will need the mass for part C, and it is okay to assume 55 kg I think.

6. Dec 1, 2008

### naresh

Yes, it can be. But not usually.

3 is way off the mark. Maybe you should draw a free body diagram as well.

7. Dec 1, 2008

### rl.bhat

Frictional force = mg = (coefficient of static friction)*(mv^2/r)

8. Dec 2, 2008

### DMOC

Based on your helpful suggestions, I wrapped up my completed answers. The due date for this assignment isn't for a few more days, so if I have a mistake in any of my following work, please correct me as there still is time to fix this.

The Question Repeated

At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.30 m and the speed of the wall is 10.0 m/s when the floor falls away.

A. What is the source of the centripetal force acting on the riders?

B. How much centripetal force acts on a 55.0 kg rider?

C. What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away?

The wall of the ride is the source of the centripetal force.

Centripetal force = (mv^2/r)
Centripetal force = ((55.0 kg)*(10.0 m/s)^2)/3.30 m
Centripetal force = 1670 N

In order to help me solve this, I drew a free body diagram and realized that the normal force was not vertical, but horizontal. Therefore, it couldn't be m*g. From rl.bhat's advice:

Normal force = (mv^2/r)
Normal force = 1670 N, which is the same as the centripetal force,

The force of friction, on the other hand, was vertical, so I could use m*g.

Force of friction = m*g
Force of friction = (55.0 kg)*(9.81 m/s^2)
Force of friction = 540. N

With these two values in mind, I could then solve for the coefficient of static friction.

Force of Friction = Coefficient of static friction * Normal force
540. N = Coefficient of static friction * 1670 N
Coefficient of static friction = 0.323

Those are my three final answers. What do you think?

Last edited: Dec 2, 2008
9. Dec 2, 2008

### naresh

Yes, it is fine. You might want to be more specific about part A. The wall is the source of the centripetal force all right, but the wall is not a force.

10. Dec 2, 2008

### DMOC

You're right, I should make that more clear. I edited my answer in the previous post I made.

Thanks!