What Is the Impact of a Ferris Wheel's Motion on Apparent Weight?

[brutalmadness]

Homework Statement

The apparent weight a person feels is the normal force that acts on the person. Suppose a Ferris Wheel has a diameter of 28.om and makes one revolution every 13.3 seconds. What is the ratio of a person's apparent weight to real weight a) at the top and b) at the bottom.

Homework Equations

Fn=m(v^2/r)

The Attempt at a Solution

First I found the circumference. C=\pi196
C=615.7521601
I can get the velocity by dividing that by 13.3.
v=46.2971549 m/s

So, I can then make my ratio comparison.
At top: Fn/m=2143.426552/14
At bottom: Fn/m=2143.426552/14+9.8

1) Am I working the solution out correctly?
2) How should I make my comparison? Should I leave it all in variables Fn/m=v^2/r? Leave it like I have it above? Or completely work it out Fn/m=153.1018966

 

[Bill Foster]

Homework Statement

The apparent weight a person feels is the normal force that acts on the person. Suppose a Ferris Wheel has a diameter of 28.om and makes one revolution every 13.3 seconds. What is the ratio of a person's apparent weight to real weight a) at the top and b) at the bottom.

The person's weight will always be W=mg, top or bottom.

However, the person will always be accelerating towards the center of the wheel, and therefore there will be a force counteracting that in the opposite direction. At the top of the wheel, it will be up. At the bottom, down.

That force is ma=m\frac{v^2}{r}=m4\pi^2 \nu^2 r

 

[azatkgz]

Is your diameter 196m or 28.0m?

 

[brutalmadness]

28.0m. But, C=pir^2. So it'd be pi14^2.

 

[azatkgz]

v=\frac{2\pi r}{t}