# Ferris wheel circular motion (dealing with apparent and true weight)

• physics120
In summary, the conversation discusses the experience of riding a Ferris Wheel and calculating the speed and acceleration of the ride, as well as the ratio of the apparent weight to the true weight at the top and bottom of the ride. Formulas and equations are used to determine the answers.
physics120

## Homework Statement

While at a country 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 15m, and you use your watch to find that each loop around takes 25s.
b) What is the ratio of your weight at the top of the ride to your weight while standing on the ground?
c) What is the ratio of your weight at the bottom of the ride to your weight while standing on the ground?

v = 2(pie)r/T
a = v^2/r

## The Attempt at a Solution

For part a) I got v= 3.7699 m/s and a= 0.947 m/s^2

For part b) since we are dealing with a ferris wheel, the FBD has normal force and weight pointing in the downward direction at the top of the ferris wheel. Would the apparent weight, which is the normal force, be (note: I made the positive y coordinate pointing down so n and w are positive):
n+w=ma
n=ma-mg
Wapp=m(a-g)

and because Wtrue=mg

Wapp/Wtrue= (m(a-g))/mg

m's cancel so,

(a-g)/g

(0.947-9.8)/9.8 = -0.9

However, that number should be positive. So, I am assuming that it should not have been n=ma-mg BUT instead n=mg-ma so n=m(g-a) so that Wapp/Wtrue= (g-a)/g so the ratio is 0.9. IF SO, WHY would n= mg-ma?

for part c) I got n-mg=ma so n=mg + ma so Wapp/Wtrue= (g+a)/g so ratio= 1.1

at the top of the wheel, if your y coordiate is pointing down, then the acceleration (a) will be negative as it acts away from the wheels centre, ie. Normal force is up at the top of the wheel and down at the bottom of the wheel.

does that help?

So at the top of the wheel, just like at the botttom of the wheel, the normal force is in the opposite direction of the gravitational force, except for the normal force at top it is negative and for the normal force at the bottom it is positive?

## What is circular motion?

Circular motion is the movement of an object along a circular path at a constant speed. This type of motion is characterized by a constant change in direction, but not speed.

## What is a Ferris wheel?

A Ferris wheel is a rotating amusement ride with a large circular wheel structure that contains passenger cars attached to the rim. As the wheel rotates, passengers are taken on a circular journey and experience different levels of height and speed.

## What is the difference between apparent weight and true weight on a Ferris wheel?

Apparent weight is the perceived weight of an object based on its motion and the forces acting on it, while true weight is the actual weight of the object. On a Ferris wheel, the apparent weight of a passenger may change due to the centripetal force acting on them, but their true weight remains the same.

## How does circular motion affect apparent weight on a Ferris wheel?

As the Ferris wheel rotates, passengers experience a centripetal force that pulls them towards the center of the wheel. This can make them feel heavier or lighter depending on the direction of the rotation. For example, at the top of the wheel, passengers may feel lighter due to the force pulling them towards the center, while at the bottom, they may feel heavier due to the force pushing them away from the center.

## How does the speed of a Ferris wheel affect apparent weight?

The speed of a Ferris wheel affects the magnitude of the centripetal force acting on passengers, and therefore, their apparent weight. A faster speed will result in a larger centripetal force and a greater change in apparent weight, while a slower speed will result in a smaller centripetal force and a smaller change in apparent weight.

• Introductory Physics Homework Help
Replies
14
Views
669
• Introductory Physics Homework Help
Replies
11
Views
6K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
41
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
770
• Introductory Physics Homework Help
Replies
8
Views
2K