# Ferris Wheel acceleration Problem

1. Oct 7, 2009

### JoshMP

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

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 15 m, and you use your watch to find that each loop around takes 25 s.

What is the magnitude of your acceleration?
What is the ratio of your weight at the top of the ride to your weight while standing on the ground?
What is the ratio of your weight at the bottom of the ride to your weight while standing on the ground?

2. Relevant equations

F=ma

3. The attempt at a solution

I started by finding the angular velocity. I don't know how to find the angular acceleration or the tangential acceleration.

2. Oct 7, 2009

### Staff: Mentor

If we presume that the wheel turns at a uniform rate, those will be zero. What other acceleration is involved?

3. Oct 7, 2009

### JoshMP

So what about the ratios of weight...would they just be 1 to 1?

4. Oct 7, 2009

### Staff: Mentor

No. (That would be too easy.) When they say "weight", they really mean apparent weight, not the force of gravity. You'll need to analyze forces using Newton's 2nd law.

5. Oct 7, 2009

### JoshMP

Ok tell me if I'm on the right track. I know that the radial acceleration is 0.95 m/s^2, which means that the net force is pointing down at the top of the wheel and up at the bottom. I use this net acceleration and solve for Fnet. Does this value of Fnet = the apparent weight?

6. Oct 7, 2009

### Staff: Mentor

No, but you'll use it to determine the apparent weight. The apparent weight of an object is the normal force exerted by the surface supporting it. Find that normal force.

7. Oct 7, 2009

### JoshMP

Ok I got the ratio at the bottom correct. But I'm having trouble with the top. My FBD has gravity pointing down and normal pointing down (is that right? - it's a Ferris wheel...). When I solve for the normal force, I get a negative value, which is impossible!

8. Oct 7, 2009

### Staff: Mentor

No. Think of yourself sitting in a seat which is always upright. The normal force of the seat on you is always up.

9. Oct 7, 2009

### JoshMP

Wouldn't that mean that the normal force at the top is equal to the normal force at the bottom? If the net force is the same at the top and bottom, and gravity is the same too, then the normal force must also be the same, and the ratios would be the same too.

10. Oct 8, 2009

### Staff: Mentor

Careful: While the net force has same magnitude at top and bottom, it does not have the same direction.