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

## 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