Normal Force at Highest Point of a Rotating Ferris Wheel

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SUMMARY

The apparent weight of a person on a Ferris wheel at the highest point is calculated using the formula for normal force, which incorporates gravitational force and centripetal acceleration. Given a Ferris wheel with a radius of 5.0 m and a rotation period of 8.0 s, the normal force exerted on a person weighing 670 N is determined to be approximately 460 N. The calculations involve determining the mass of the person and applying the centripetal force equation, resulting in a normal force of 459.14 N after accounting for the effects of circular motion.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Familiarity with centripetal force and acceleration concepts
  • Knowledge of gravitational force calculations
  • Ability to manipulate equations involving mass, weight, and force
NEXT STEPS
  • Study the principles of centripetal acceleration in detail
  • Explore the effects of varying radius and rotation speed on normal force
  • Learn about the applications of circular motion in real-world scenarios
  • Investigate the relationship between mass, weight, and apparent weight in different contexts
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of circular motion and forces acting on objects in rotational systems.

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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?

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

Homework Equations


Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path. Thus,
\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}.

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

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

The Attempt at a Solution


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