
#1
Jul1207, 05:57 PM

P: 28

1. The problem statement, all variables and given/known data
The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60s). Part A: Question: Find the speed of the passengers when the Ferris wheel is rotating at this rate. Answer: 5.24 m/s This simple problem was found by calculating the circumference (314.16m) and dividing it my 60. Part B: A passenger weighs 724 N at the weightguessing booth on the ground. What is his apparent weight at the highest point on the Ferris wheel? Part C: What is his apparent weight at the lowest point on the Ferris wheel? Part D: What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? part E: What then would be the passenger's apparent weight at the lowest point? 2. Relevant equations ????? To be quite honest, I have no idea how I would go about solving problems A  E....  Abarak 



#2
Jul1207, 08:31 PM

Sci Advisor
HW Helper
P: 1,772

In perfect circular motion, the net force on an object is centripetal force.
A the top of the circle, the centripetal force is provided by (+normal)+(gravitational) forces. The normal force is what we perceive as "apparent weight" 



#3
Jul1307, 01:16 PM

P: 28

Newton’s Law of Gravitation states:
[tex]F_{gravity} = G\frac{m_1m_2}{r^2}[/tex] What I have so far v = 5.24m/s diameter = 100m radius = 50m Circumference = 314.16m The wheel has 60 arms and makes 1 complete revolution every 60 seconds. Thanks for the help so far, Abarak 



#4
Jul1307, 01:56 PM

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P: 40,905

Height and Weight
Newton's law of gravity is not needed here. The only gravitational force you care about is that due to the Earththe passenger's weight, which is given.
As Chi Meson stated, the "apparent weight" equals the normal force. Identify the forces acting on the passenger at each point in question (Chi Meson told you what they are) and apply Newton's 2nd law of motion to solve for the normal force. What's the acceleration? 



#5
Jul1307, 02:01 PM

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P: 14,467





#6
Jul1307, 05:09 PM

P: 28

Let me just make sure I understand this correctly. The passengers normal weight is 724 N At the top of the Ferris Wheel all I need to do is add the normal weight to the gravitation weight. 724N+(9.8[tex]m/s^2[/tex]) = 714.2N?? What about when the wheel is at the lowest point? I though no other forces would affect the person so I tried 724N and it did not work. Again, thanks for all the help. I think I just need something to "click". Abarak 



#7
Jul1307, 05:33 PM

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P: 40,905

(1) the normal force (also called, confusingly, the "apparent" weight); this is the force exerted by the seat on the person; (2) the real weight (the gravitational weight), which you are given (it also equals mg). Which direction do these forces act? (Hint: At the top of the Ferris wheel, the passenger is upside down.) Add them up (taking direction into account) to find the net force. Use that net force in Newton's 2nd law to solve for the normal force. (What's the passenger's acceleration?) 



#8
Jul1307, 07:30 PM

P: 28

Hey Doc Al,
Ok, this is what I did and I still got the wrong answer real weight = 724N = m*g 724N = m * 9.8 m/s^2 m = 73.878kg Because the seat needs to support the weight of the person I used [tex]F = m * (\frac{v^2}{r})[/tex] [tex]F = 73.878 * (\frac{5.24^2}{50}) = 40.57N[/tex] normal force + real weight = Fnet Part B: 40.57 + 724 = 764.57N Part C: 724  40.57 = 683.43N It stills says this answer is wrong? Am I calculating the normal force wrong? 



#9
Jul1307, 08:13 PM

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P: 40,905

Your calculation of the mass and centripetal force is correct.
It turns out that this Ferris wheel rotates so slowly that at the top the seats must be right side up. (I was wrong when I said it was upside down at the top.) So that means the normal force acts upward at the top. (The weight, of course, acts down.) So redo your summation of forces. 



#10
Jul1307, 09:16 PM

P: 28

Hey Doc, DH and Chi!
I was able to solve every part with your help. Thank you very much for taking your time with me.  Abarak 



#11
Oct807, 12:22 PM

P: 273

Hi, sorry to bring up this thread, but I have a similar problem.
Following the steps and reading the explanation I managed to get AC. But I could not determine D and E. Part D: What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? part E: What then would be the passenger's apparent weight at the lowest point? 



#12
Oct807, 12:29 PM

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#13
Oct807, 12:31 PM

P: 273

that they are equal.



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