Weightlessness; Ferris Wheel

  • #1

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=[tex]\pi[/tex]196
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
 

Answers and Replies

  • #2
338
0

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 [tex]W=mg[/tex], 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 [tex]ma=m\frac{v^2}{r}=m4\pi^2 \nu^2 r[/tex]
 
  • #3
190
0
Is your diameter 196m or 28.0m?
 
  • #4
28.0m. But, C=pir^2. So it'd be pi14^2.
 
  • #5
190
0
[tex]v=\frac{2\pi r}{t}[/tex]
 

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