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On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 $\mathrm{m} .$ The cylinder started to rotate, and when it reached a constant rotation rate of $0.60 \mathrm{rev} / \mathrm{s},$ the floor on which the people were standing dropped about 0.5 $\mathrm{m}$ . The people remained pinned against the wall. (a) Draw a free-body diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction is required if the person on the ride is not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the mass of the passenger?(Note: When the ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

(a) see explanation (b) 0.28 (c) No

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

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key number 47. This is a problem about, um, and amusement park. In this attraction, you have cylinder spinning really fast and spinning so fast that people inside it are actually drawn, um, or pushed to the wall. And that is enough to make them holding the wall and then floors dropped in. You gonna hold by their own friction. So I was drawing here. Tub view of the attraction side view of the attraction would be like this. This is the wall. This is the person. Um, the force applied force is applied on the person, which is question a free body diagram will do it right here. Or gravity F g a normal force applied from the wall because that prisoners actually touches the wall. And so the war applies a force on the person so that it doesn't go through the wall. And finally, we need another force here, which is, um, friction force. Anyways, persons and the attraction would simply fall down. So we'll call that f and yes, for steak. Static friction. Now, um, so that's that's for a Yeah. And they ask us to find wet. Did the coefficient of friction should be in this situation. So it such that persons don't fall forgot to drop my access year. So you always have to draw your access to know which with Rose X and where is why so we can start with Why? Because that's where their friction forces. The person is not moving up or down. She is addressed the scenario. So that means that the forces are Banksy out with each other. F g equals F s. Now f g is mg and F s. For this, you would have to go back to Peter's chapter. But if you remember, it's gonna be immune liquefaction diffraction, times and the normal force. So the more force applied under person, the more friction there will be, which makes a lot of sense. If you push on an object on the table, for example, it will be harder to slide and then this coefficient is not depend on the force is just a property of the surface. So very rough surface, for example, would have a higher coefficient of friction now. But we want to know is the value of new so we can always isolate it new equals and your end. But the problem is that we don't know what Ennis and for this will have to look at the Y axis on the Y axis from that force end, um, is the sum of all the forces and by a Newton Sigonella is equal to mass times exploration. Or since we are in a central little motion envies court over our with our injury GIs, importation, envy the speed at which would you turn Great. So, uh, we don't have the speed per se, but we have the period. And since we know that V is to buy our over tea, we can rewrite. This is m times to buy over teat. Square times are now combining the equation and extra the equation. Why, um from the white part, I can write an equals n g overview. And then I have that n g over here, which is n is also m times to buy squared over t squared times are those two terms are both equal to 10 so we can make them equal to each other. Now the masses cancel right, and we want to isolate you. Such det nu equals tea over to pie squared sometimes a jeep over our just gonna double check that this is correct from from my notes. Just give me a sec. Yeah. Script. So we have gate. This is the gravitational constant of the surface of the earth. We have are given problem to be 2.5 meters with We don't have his tea. But do give us fact that cylinder rotates it 0.6 revolution per second and T is the number of second for revolution. So if we just invert this number, we know that there is once again in 0.6 revolution and that is equal Two, huh? Thanks Don't have this explicitly written in my notes. So just calculate this number will be won over 0.6. I get 1.67 1.6 seconds for one revolution. Great. So mu is equal to 1.67 seconds over to pie You square this times G who are and excuse me a constant of 0.28 which is the final answer now. Um see this short the want to know it's that answer. It's your 0.28 depend upon the mass of a passenger. So it is someone heavier. Have Ah, hire quitting the friction. Well, the first intuition you can get for this and that this question of friction depends on the surface. Um, not under prison. Well, I could depend on the surface of the person's. Well, if the person was very rough, then I guess that description changed, but it doesn't depend upon anything else in that. So that's the first No, Yes or no. Um, is that if you look at this formula that we extracted for quick, efficient of friction, mass doesn't appear anywhere. In fact, it canceled out here because of that appear in both side of the equation. So no, the mass doesn't appear, and the answer to see it was no.

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