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BME2054-Biomechanics HW1 QuestionsDue date: 21 May 2020Q1. An amusement park ride consists of large vertical cylinder that spins about its axis fast enough that any...

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BME2054-Biomechanics HW1 QuestionsDue date: 21 May 2020Q1. An amusement park ride consists of large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away: The coefficient of static friction between person and wall is /s and the radius of the cylinder is R_(a) Show that the maximum period of revolution necessary to keep the person from falling is T = (4n?R Uslg) 1/2 (b) If the rate of revolution of the cylinder is made

BME2054-Biomechanics HW1 Questions Due date: 21 May 2020 Q1. An amusement park ride consists of large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away: The coefficient of static friction between person and wall is /s and the radius of the cylinder is R_ (a) Show that the maximum period of revolution necessary to keep the person from falling is T = (4n?R Uslg) 1/2 (b) If the rate of revolution of the cylinder is made to be somewhat larger, what happens to the magnitude of each one of the forces acting on the person? What happens in the motion of the person? (c) If the rate of revolution of the cylinder is instead made to be somewhat smaller, what happens to the magnitude of each one of the forces acting on the person? How does the motion of the person change?



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An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (Fig. P6.59). The coefficient of static friction between person and wall is $\mu_{s},$ and the radius of the cylinder is $R .$ (a) Show that the maximum period of revolution necessary to keep the person from falling is $T=\left(4 \pi^{2} R \mu_{\delta} / g\right)^{1 / 2} .$ (b) If the rate of revolution of the cylinder is made to be somewhat larger, what happens to the magnitude of each one of the forces acting on the person? What happens in the motion of the person? (c) If the rate of revolution of the cylinder is instead made to be somewhat smaller, what happens to the magnitude of each one of the forces acting on the person? How does the motion of the person change?

Hi there. Given problem here are this is a large cylindrical drum having a radius. Oh, and the person remains stick. I mean, this one, the weight of the person you'd be acting vertically Don't work under which the person will be having a tendency to slide down or the wall of the slender So ah, force of friction will act on him in a direction worded belly up As the cylinder will rotate, a centrifugal force will act on the person They're really outward given by an expression MP scored by our which will create a normal reaction on the person you do this wall given by n. So in the first part of the problem, we have toe find the rate off rotation off the slender good witch. The partisan will remain. Stick with the wall. So the normal reaction acting on the person will be equal to centrifugal force acting on him, which is empty squad by art and as the person is not falling down. So the weight of the person mg really just equal to four so static creation which we begin by new s times, the normal reaction or we can see MGI is equal to us. Stine's off envy. Square by up there must got cancer. So we square comes out to be g times off our means, Regis with a slender divided by new s So this week finally becomes Squire rode off G times off radios divided by us And hence, you know the being at a speed in a circular motion may also be given us the product off Really us with angular velocity. So this week is going to our times, amigo is equal to G into our divided by a new s. So the angular velocity here comes out to be one by our square root off g into our divided by us. No, this angular velocity is giving us to buy divided by the time period So two pi bitey becomes one by our into g multiplied by r divided by a new s to the power ho. So finally we can write the time period A period of rotation off the slender is equal to four times the to buy times. It's the stool. Why times, Uh, break it. New ists divided by g. You are I mean the bomber puff more begin. Check this to buy and are in this bracket making squared off them so they become poor by square. Our square into new s be invited by G to the power up. So finally, this time period off the cylinder comes out to be for by square into a bomb Time's newest or efficient struck a correction divided by G sweat This this is the answer for the first part off the problem? No. In the second part of the problem, we have to find a numerical value for this read off rotation. For this, the radius has been given us 4.0 meter. The coefficient of friction static friction has been given as the open port 00 So, time media, it comes out to be more in 23.14 square multiplied by four and multiply by 0.4 divided by 9.8 Our huh, which becomes square root off 6.44 or 2.54 Second, This is the read off time period off rotation. So if you can work it into frequency, it is one wine one of on you 10.5 foot revolutions per second on you can say 60 invited by 2.54 revolutions per minute, which comes out to be equal to 23.6 revolutions per minute. Finally, this is the rate off rotation off the slender answer the second part. No, the start park we have been given. If we increase the rate of rotation, what will happen to the forces? So on? Increasing this rate off rotation means on increasing F the simply people force and the normal force Winfrey's when the force of friction will win constant, which is always equal toe weight off the person but force of friction Well, I mean go on stand and equal toe. They don't a person. Finally, in the fourth part, they are seeing what would happen if you reduce the rate of rotation so only you'll sing the rate of probation This f all the forces will reduce. So finally the person really sit down. Thank you

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.

So in this problem, the person is ah is against the war. Okay, so, uh, because the seven days moving with the velocity V so we would have Ah ah, the centripetal force sect and the person, Right? And the century before is it is it Ah is provided by the normal force off off this Ah, this war against the person. So which is pointing to this direction? This is the normal force. And and of course, we have the gravity f g ray. And to keep this person from sliding down, we'll have the friction f sorry. Through these are the older forces acting on the person and in part B. Ah, we want to describe the condition that, ah, to give this to keep the passenger from sliding down off the war. So, uh, we know that ah, the central forces has expression of these squid of our time. Sam Ray. So this is also equal to the normal force and the times meal this gives us The Esso were amused aesthetic friction coefficient. So this gives us the maximum value off the ah static friction, and this really should be larger or equal to M times. G. So we get it. We obtain the condition. You has to be larger, equal to our times. G over three square. Okay. And Percy? Ah, well into disgust. Ah ah. For two for two passengers. Why is what has mess? And and the other way is Ah has mess. I am over too. So apparently for this condition, we know that it does not depend on the mass of the office. Hasn't You're right. So the matter that the massive past Mary is Amol m over to as long as the static friction coefficient satisfy this relation? Uh, they were They were both guests taken on the walk. So what do we know outside now?


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