Circular motion help

[Santorican]

1. The problem statement, all variables and given/known data

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


2. Relevant equations

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

3. 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:

[Doc Al]

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.

[kmikias]

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 ............

[Santorican]

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!