Solving Circular Motion Help Homework: Find Speeds & Apparent Weight

AI Thread Summary
The discussion focuses on solving homework problems related to circular motion involving a Ferris wheel. For Part A, the speed of passengers was calculated correctly using the formula V=(2piR)/T, yielding 5.24 m/s. In Parts B and C, confusion arose regarding the concept of apparent weight, which is defined as the normal force experienced by the passenger, rather than just their true weight (mg). The participants clarified that at the lowest point, apparent weight is the sum of gravitational force and centripetal force, while at the highest point, it is the difference. The thread emphasizes the importance of understanding the distinction between true weight and apparent weight in circular motion scenarios.
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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