Solving Circular Motion Help Homework: Find Speeds & Apparent Weight

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SUMMARY

The discussion focuses on solving problems related to circular motion using the Cosmoclock 21 Ferris wheel in Yokohama City, Japan, which has a diameter of 100 meters and completes one revolution every 60 seconds. The speed of passengers is calculated using the formula V=(2πR)/T, yielding a speed of 5.24 m/s. The apparent weight of a passenger at the lowest point is determined by the normal force, expressed as N=m(g+(v²/R)), while at the highest point, it is N=mg-(mv²/R). The discussion clarifies that apparent weight is not simply m*g, but rather the normal force experienced by the passenger.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Familiarity with the concepts of normal force and apparent weight
  • Knowledge of basic physics equations related to motion
  • Ability to apply formulas in practical scenarios
NEXT STEPS
  • Study the derivation of the normal force in circular motion contexts
  • Learn about the differences between apparent weight and true weight
  • Explore the effects of varying speeds on apparent weight in circular motion
  • Investigate the physics of other amusement park rides involving circular motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts of apparent weight and normal force in practical applications.

Santorican
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Homework Statement



The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s.

Part A:Find the speed of the passengers when the Ferris wheel is rotating at this rate.
-I used V=(2piR)/T and got 5.24 which was correct.

Part B:A passenger weighs 862 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?
-This is where I am completely thrown off course. Is it asking for the normal force? I thought apparent weight was just m*g?

Part C:What is his apparent weight at the highest point on the Ferris wheel?
-Same issue with Part B

Part D:What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?
-I would need the proper formula from Part B to find Part D, so I'm stuck here too.

Part E:What then would be the passenger's apparent weight at the lowest point?
-Same issue as Part D


Homework Equations



V=(2piR)/T
Sum of the forces in the y direction= N=m(g-(v^2/R))

The Attempt at a Solution



I attempted to use that formula but it said it was wrong so I have no clue what to do now. :frown:
 
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Santorican said:
Part B:A passenger weighs 862 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?
-This is where I am completely thrown off course. Is it asking for the normal force?
Yes, the apparent weight equals the normal force.
I thought apparent weight was just m*g?
No, m*g is the true weight.
 
Hi.

what do you know about the different between ferris wheel and a loop or circle.
Here is what i know At the lowest point in frerris wheel the
Normal force = (mv^2)/2 + mg
And At the highest point would be
Normal force = mg - (mv^2)/2

weight = mg ...
 
Doc Al said:
Yes, the apparent weight equals the normal force.

No, m*g is the true weight.

Doc Al you are the man! That little answer helped me so much thanks!
 

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