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Normal Force at Highest Point of a Rotating Ferris Wheel

  • Thread starter logan3
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Homework Statement


A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (the normal force exerted on her by the bench) of the person as she passes through the highest point of her motion?

[itex]r = 5.0m[/itex]
[itex]T = 8.0s[/itex]
[itex]\vec F_W = 670N[/itex]
[itex]g = 9.8 m/s^2[/itex]

Homework Equations


Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path. Thus,
[itex]\sum \vec F = \vec F_c = \vec F_W - \vec F_N \Rightarrow \vec F_N = \vec F_W - \vec F_c = mg - \frac {mv^2} {r}[/itex].

Since [itex]v = \frac {2 \pi r} {T}[/itex], then [itex]\vec F_N = mg - \frac {m(\frac {2 \pi r} {T})^2} {r} = mg - \frac {4 \pi^2 mr} {T^2}[/itex].

The mass of the person is: [itex]\vec F_W = mg \Rightarrow m = \frac {F_W} {g}[/itex].

The Attempt at a Solution


[itex]m = \frac {670N} {9.8 m/s^2} = 68.367 kg[/itex]
[itex]\vec F_N = 670N - \frac {4 \pi^2 (68.367 kg)(5.0m)} {(8.0)^2} = 670N - 210.86N = 459.14N \sim 460N[/itex]
 

Answers and Replies

  • #2
Doc Al
Mentor
44,945
1,206
Looks good to me.
 
  • #3
68
2
Thank-you : )
 

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