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(II) A Ferris wheel 22.0 $\mathrm{m}$ in diameter rotates once every12.5 $\mathrm{s} .$ What is the ratio of a person's apparent weight to herreal weight $(a)$ at the top, and $(b)$ at the bottom?

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a) ratio = 0.716b) ratio = 1.284

Physics 101 Mechanics

Chapter 6

Gravitation and Newton's Synthesis

Physics Basics

Newton's Laws of Motion

Applying Newton's Laws

Gravitation

Rutgers, The State University of New Jersey

University of Washington

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University of Sheffield

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so for party, let's start off by finding the velocity. We know that for a knob decked in uniform circular motion, the velocity is going to be equal to two pi r Divide about period. So this is going to be two pi times the radius of 11 meters divided by the period of 12.5 seconds and we find that the velocity is five 0.5 to 9 meters per second for part B. We want to find the ratio of the apparent rate apparent weight to the rial. Wait s so we can draw everybody diagram here going straight up is forced normal and then coming straight down his force of gravity. Here we're going to say that that the sum of forces equals the mass times, the excess and chipotle acceleration. This will equal mg minus of of that event. Eso this will equal empty squared over are here. The force of gravity is towards the center of the Ferris wheel. That's why it's positive. We can say the Mumbai are and so we can say that the force normal is simply gonna be equal to M G minus M v squared over or the ratio of the apparent wait to re away. So we can say that wait, a parent divided by weight riel would be equal to M G minus, and these squared over r and then this will be divided by mg. And so this will be equal to G minus. B squared B squared over R, divided by G, canceling out all the mass terms and this is going to be equal to one minus 5.5 to 9 squared, divided by 11 times 9.8. And so this ratio is going to be 0.716 So the apparent weight is 71.6% of the real way. This would be your answer now report and rather rather my apologies. This's part A. This was not party. This is just We just need to calculate the velocity before we move on. So this would be part a. And then for part B, we had draw another three party A diagram. Now here. Forced normal is going up towards the center. Highland force of gravity is going down and we can say that force normal now is positive. Minus M g E equals M V squared over R and we know that after then is going to be equal to MG plus b squared over R. So we can say that the apparent wait, divided by the real way, would be equal to one plus. And then again, 1 5.5 to 9 squared, divided by 11 meters, times 9.8. So the only difference between the two is actually just the sign in part A. This is one minus. And then here it's one plus, given that here it is on the bottom of the Ferris wheel. Here it's the top of the Ferris wheel, and we find that the ratio was going to be 1.284 So here the apparent weight is 28.4% greater than the real way, and this would be a final answer. That is the end of the solution. Thank you for watching

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