what is the ratio of the persons apparent weight to her real weight ferris wheel
Ferris wheel — Apparent weight of a rider at the top and lesser of the rotating wheel
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Homework Statement
A Ferris bike 26 1000 in diameter rotates once every 14 s. What is the ratio of a person's credible weight at the top of the ride to her credible weight at the bottom of the ride?
Homework Equations
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FNT/FNB = mg - mv^two / r / mg + mv^2/r
The Attempt at a Solution
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The answer to the question is .93
My primary 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^two / xiii = 2.618466474
(9.eight-ii.62)/ix.8 = .73- this is the respond I go
(26)(9.8)-11.668^2/26.98= .465 another way I tried to become this answer
Answers and Replies
Information technology is very hard to decipher what y'all are trying to exercise. First, there are dropped parenthesis in a couple of places, second y'all are not carrying units along in your calculations, and third, you are asking united states of america to guess what quantities are represented by your numbers.FNT/FNB = mg - mv^two / r / mg + mv^two/r
The starting equation looks correct (with an assumption about where parenthesis should go). Using LaTeX (see the tutorial under INFO at the meridian of the page, in Help/How-To):
[tex]\frac{FNT}{FNB} = \frac{mg - \frac{mv^ii}{r}}{mg + \frac{mv^two}{r}}[/tex]
At present can you explain how your calculated r, and five? And so substitute dorsum into this equation and show your units as you behave out the adding? Thanks.
Wheel Spins every 14 seconds
Centripetal force = M v^2 / r directed outward
26m / ii = 13m
Upward strength is then = M [ pi *26m / 14sec]^two / 13m = 2.618466474
Up force = 2.62 M m/sec^2)
Down force at acme = (9.8m/sec^two) M
Apparent/real at top = (nine.viii - 2.62)/9.eight = 0.73 m
Just a fiddling bit of extra writing would make your reasoning much much easier to follow. In this case, I inquire, why do I want to know what 26m/2 is? Answer: because that's the radius. So say and then.
##r = 26 \text{ grand}/two = thirteen \text{ m}##
Same thing on the next line.
Upwards force is so = Grand [ pi *26m / 14sec]^2 / 13m
What is that expression in the brackets? Oh, I see, it'southward the calculation of the speed. Why non say and then symbolically, before writing the above line?
##5 = \frac{\pi D}{T} = \frac {26\pi \text { 1000}}{14 \text{ s}}##
The symbols tell a much clearer story than the numbers.
Can't understand why yous used your own equation instead of the one you lot were given: ##(9.8M - 2.62M)/(9.8M+2.62M) = 0.58##. Not 0.93. You're right, the equation your professor gave yous does not give this answer.
I know this was all a while dorsum, but if you encounter this, could y'all explicate how we're even supposed to solve this problem? Information technology'southward asking for the apparent weight at the top of the wheel but how do we find that? Information technology doesn't even give u.s.a. a mass to start with and I don't know what formulas to use! I seriously don't understand any of this...Just a niggling bit of extra writing would make your reasoning much much easier to follow. In this case, I ask, why practice I want to know what 26m/ii is? Respond: considering that'southward the radius. So say so.
##r = 26 \text{ thou}/2 = 13 \text{ chiliad}##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 to a higher place line?
##v = \frac{\pi D}{T} = \frac {26\pi \text { thousand}}{14 \text{ s}}##The symbols tell a much clearer story than the numbers.
Can't empathize why you used your own equation instead of the 1 you lot were given: ##(9.8M - two.62M)/(9.8M+two.62M) = 0.58##. Non 0.93. You're right, the equation your professor gave you does not give this answer.
No, it asks for the ratio of 2 apparent weights. If y'all double the mass, both credible weights double, and so the ratio stays the same.It's request for the apparent weight at the top of the wheel
Just plug in an unknown mass, m, and that must abolish out when you take the ratio.
The mass cancels out.I know this was all a while back, but if you see this, could you explicate how we're even supposed to solve this problem? It'due south asking for the apparent weight at the top of the bicycle but how exercise we observe that? It doesn't even give the states a mass to start with and I don't know what formulas to apply! I seriously don't sympathise any of this...
Also, I'm assuming y'all haven't nevertheless learned this: ##v_t=ωR##?
Or, yous could just say that the person travels through the edge at: ##\frac{2πR}{14}\frac{m}{s}##.
This gives circumferential distance traveled per unit of measurement-time.
This volition help you lot figure out the magnitude of the net force (and dispatch) holding the person in uniform circular motion towards the center of the circle/ferris-wheel via the formula you already know: ##\frac{mv^two}{R}##.
Well I have a very similar question simply it'due south asking for the ratio of the apparent weight at the top to her real weightNo, information technology asks for the ratio of two credible weights. If yous double the mass, both apparent weights double, and then the ratio stays the same.
But plug in an unknown mass, g, and that must cancel out when yous take the ratio.
I don't empathize... The mass, from what equation, cancels out? Also, no I don't know Five=wr or what that last equation you gave was. I know this equation: Ac = 5^two/r and I take 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 credible weight at the meridian to her existent weight. (r= 11m; T= 12.5s; V= v.53m/south; Ac= 2.779m/s^two)The mass cancels out.
Also, I'm assuming y'all haven't withal learned this: vt=ωRvt=ωR?
In this problem vt=vatRvt=vatR.
Or, you lot 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 compatible circular motion towards the center of the circumvolve/ferris-wheel.
The ferris cycle is a terrible form of this kind of a question.
Someone new to physics problems can hands overthink this.
You're better off treating this like a bike making a full loop, like and so:
At the meridian: 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 circumvolve AT ALL TIMES.
Why? Because the problem states that yous are in Compatible Round Motion about this middle (non a very bright explanation, I know).
The only way you can be in such a state is if the net forces interim on the particle at all times are equal to ##\frac{mv^2}{R}##.
So, you tin use the equations given in mail service #2 upward here.
When yous use those equations, ##chiliad## will cancel out.
Ok, so you have the centripetal acceleration.I don't empathize... The mass, from what equation, cancels out? Also, no I don't know V=wr or what that final equation you lot gave was. I know this equation: Air-conditioning = v^2/r and I have the centripetal acceleration merely I don't know where to go from here. My trouble is nigh identical to the original question on this thread except that it'due south asking for the ratio of the apparent weight at the top to her existent weight. (r= 11m; T= 12.5s; V= 5.53m/s; Ac= two.779m/s^2)
The other equation you need is Newton'south ΣF=ma, where ΣF means the sum of all forces acting on the body.
What forces act on the rider?
Related Threads on Ferris bike — Apparent weight of a rider at the peak and lesser of the rotating wheel
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