Calculating the Ratio of Apparent Weight on a Ferris Wheel

In summary, the ratio of a person's apparent weight at the top of the ride to her apparent weight at the bottom of the ride is .93.
  • #1
SuperSood
2
0

Homework Statement


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?

Homework Equations


[/B]
FNT/FNB = mg - mv^2 / r / mg + mv^2/r

The Attempt at a Solution


[/B]
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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  • #2
Welcome to the PF. :smile:
SuperSood said:
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.
 
  • #3
26m is the diameter of the wheel
Wheel Spins every 14 secondsCentripetal 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
 
  • #4
SuperSood said:
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.
SuperSood said:
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.

SuperSood said:
(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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  • #5
RPinPA said:
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.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.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.
I know this was all a while back, but if you see this, could you explain how we're even supposed to solve this problem? It's asking for the apparent weight at the top of the wheel but how do we find that? It doesn't even give us a mass to start with and I don't know what formulas to use! I seriously don't understand any of this...
 
  • #6
Banphys said:
It's asking for the apparent weight at the top of the wheel
No, it asks for the ratio of two apparent weights. If you double the mass, both apparent weights double, so the ratio stays the same.
Just plug in an unknown mass, m, and that must cancel out when you take the ratio.
 
  • #7
Banphys said:
I know this was all a while back, but if you see this, could you explain how we're even supposed to solve this problem? It's asking for the apparent weight at the top of the wheel but how do we find that? It doesn't even give us a mass to start with and I don't know what formulas to use! I seriously don't understand any of this...
The mass cancels out.
Also, I'm assuming you haven't yet learned this: ##v_t=ωR##?
Or, you could just say that the person travels through the edge at: ##\frac{2πR}{14}\frac{m}{s}##.
This gives circumferential distance traveled per unit-time.
This will help you figure out the magnitude of the net force (and acceleration) holding the person in uniform circular motion towards the center of the circle/ferris-wheel via the formula you already know: ##\frac{mv^2}{R}##.
 
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  • #8
haruspex said:
No, it asks for the ratio of two apparent weights. If you double the mass, both apparent weights double, so the ratio stays the same.
Just plug in an unknown mass, m, and that must cancel out when you take the ratio.
Well I have a very similar question but it's asking for the ratio of the apparent weight at the top to her real weight
 
  • #9
lightlightsup said:
The mass cancels out.
Also, I'm assuming you haven't yet learned this: vt=ωRvt=ωR?
In this problem vt=vatRvt=vatR.
Or, you could just say that the person travels through: [2πr][time][2πr][time]
This will help you figure out the magnitude of net acceleration holding the person in uniform circular motion towards the center of the circle/ferris-wheel.
I don't understand... The mass, from what equation, cancels out? Also, no I don't know V=wr or what that last equation you gave was. I know this equation: Ac = v^2/r and I have the centripetal acceleration but I don't know where to go from here. My problem is almost identical to the original question on this thread except that it's asking for the ratio of the apparent weight at the top to her real weight. (r= 11m; T= 12.5s; V= 5.53m/s; Ac= 2.779m/s^2)
 
  • #10
Apparent Weight essentially means: what is the magnitude of the normal force that the person experiences.
The ferris wheel is a terrible form of this kind of a question.
Someone new to physics problems can easily overthink this.
You're better off treating this like a bike making a full loop, like so:

At the top: Fn and Fg point down.
At the bottom: Fn points up and Fg points down.
The sum of these vectors points to the center of the circle AT ALL TIMES.
Why? Because the problem states that you are in Uniform Circular Motion about this center (not a very bright explanation, I know).
The only way you can be in such a state is if the net forces acting on the particle at all times are equal to ##\frac{mv^2}{R}##.

So, you can use the equations given in post #2 up here.
When you use those equations, ##m## will cancel out.
 
  • #11
Suppose I tell you that I am running around the edge of a circular track of radius 100 meters, once every 120 seconds, then, how fast am I moving through the edge of this circular track (or, what is my speed compared to me being at rest?)?
 
  • #12
Banphys said:
I don't understand... The mass, from what equation, cancels out? Also, no I don't know V=wr or what that last equation you gave was. I know this equation: Ac = v^2/r and I have the centripetal acceleration but I don't know where to go from here. My problem is almost identical to the original question on this thread except that it's asking for the ratio of the apparent weight at the top to her real weight. (r= 11m; T= 12.5s; V= 5.53m/s; Ac= 2.779m/s^2)
Ok, so you have the centripetal acceleration.
The other equation you need is Newton’s ΣF=ma, where ΣF means the sum of all forces acting on the body.
What forces act on the rider?
 

Related to Calculating the Ratio of Apparent Weight on a Ferris Wheel

What is the formula for calculating the ratio of apparent weight on a Ferris wheel?

The formula for calculating the ratio of apparent weight on a Ferris wheel is (2R + h) / (2R - h), where R is the radius of the wheel and h is the height of the rider above the ground.

How does the ratio of apparent weight change as the Ferris wheel rotates?

The ratio of apparent weight changes as the Ferris wheel rotates due to the centripetal force acting on the rider. As the rider reaches the top of the wheel, the ratio of apparent weight decreases because the centripetal force decreases. As the rider reaches the bottom of the wheel, the ratio of apparent weight increases because the centripetal force increases.

What factors can affect the ratio of apparent weight on a Ferris wheel?

The ratio of apparent weight on a Ferris wheel can be affected by the speed of the wheel, the radius of the wheel, and the height of the rider above the ground. Other factors such as wind and friction may also have a small impact on the ratio.

How does the ratio of apparent weight on a Ferris wheel relate to the rider's actual weight?

The ratio of apparent weight on a Ferris wheel is not the same as the rider's actual weight. The apparent weight is the perceived weight felt by the rider due to the centripetal force, while the actual weight remains constant. However, the ratio of apparent weight can give an indication of the relative force acting on the rider compared to their actual weight.

Why is it important to calculate the ratio of apparent weight on a Ferris wheel?

Calculating the ratio of apparent weight on a Ferris wheel is important for understanding the physics behind the experience of riding a Ferris wheel. It can also be helpful in designing and testing the safety of amusement park rides to ensure that the forces acting on riders are within safe limits.

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