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Ferris wheel -- Apparent weight of a rider at the top and bottom of the rotating wheel

  • Thread starter SuperSood
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1. The problem statement, all variables and given/known data
A Ferris wheel 26 m in diameter rotates once every 14 s. What is the ratio of a person's apparent weight at the top of the ride to her apparent weight at the bottom of the ride?

2. Relevant equations

FNT/FNB = mg - mv^2 / r / mg + mv^2/r

3. The attempt at a solution

The answer to the question is .93
My main problem is that my math is not giving me this answer. Even the equation that my professor provided above does not give me this answer.

(pi*26/140^2 / 13 = 2.618466474

(9.8-2.62)/9.8 = .73- this is the answer I get

(26)(9.8)-11.668^2/26.98= .465 another way I tried to get this answer
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Welcome to the PF. :smile:
FNT/FNB = mg - mv^2 / r / mg + mv^2/r
It is very hard to decipher what you are trying to do. First, there are dropped parenthesis in a couple of places, second you are not carrying units along in your calculations, and third, you are asking us to guess what quantities are represented by your numbers.

The starting equation looks correct (with an assumption about where parenthesis should go). Using LaTeX (see the tutorial under INFO at the top of the page, in Help/How-To):

[tex]\frac{FNT}{FNB} = \frac{mg - \frac{mv^2}{r}}{mg + \frac{mv^2}{r}}[/tex]

Now can you explain how your calculated r, and v? And then substitute back into this equation and show your units as you carry out the calculation? Thanks.
26m is the diameter of the wheel
Wheel Spins every 14 seconds

Centripetal force = M v^2 / r directed outward
26m / 2 = 13m
Upward force is then = M [ pi *26m / 14sec]^2 / 13m = 2.618466474
Upward force = 2.62 M m/sec^2)
Downward force at top = (9.8m/sec^2) M
Apparent/real at top = (9.8 - 2.62)/9.8 = 0.73 m
26m / 2 = 13m
Just a little bit of extra writing would make your reasoning much much easier to follow. In this case, I ask, why do I want to know what 26m/2 is? Answer: because that's the radius. So say so.
##r = 26 \text{ m}/2 = 13 \text{ m}##

Same thing on the next line.
Upward force is then = M [ pi *26m / 14sec]^2 / 13m
What is that expression in the brackets? Oh, I see, it's the calculation of the speed. Why not say so symbolically, before writing the above line?
##v = \frac{\pi D}{T} = \frac {26\pi \text { m}}{14 \text{ s}}##

The symbols tell a much clearer story than the numbers.

(9.8 - 2.62)/9.8
Can't understand why you used your own equation instead of the one you were given: ##(9.8M - 2.62M)/(9.8M+2.62M) = 0.58##. Not 0.93. You're right, the equation your professor gave you does not give this answer.

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