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AP Physics Homework Problem

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  1. Feb 15, 2012 #1
    George Washington Gale Ferris. Jr., a civil engineering graduate from RPI, built the original Ferris wheel. The wheel carried 36 wooden cars, each holding up to 60 passengers, around a circle 76m in diameter. The cars were loaded 6 at a time, and once all 36 cars were full, the wheel made a complete rotation at constant angular speed in about 2 min. Estimate the amount of work that was required of the machinery to rotate the passengers alone.

    I know the equations for Toque, and Work as well as force and Inertia, but if the velocity is constant, then isnt acceleration 0, which means no work is being done? It CANT be that easy. Im not sure how to start to solve. Any help is appreciated.
     
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  3. Feb 15, 2012 #2
    τ=(r)(Fsin∅)
    W=∫θF θI τdθ
    W=91/20 I(wf^2-wi^2)
    T=Iα
    I=MR^2 for a hoop
     
  4. Feb 15, 2012 #3

    cepheid

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    I think that you are confusing velocity and speed. Velocity is a vector, meaning that it has both a magnitude and a direction. Speed is a scalar, and it is generally equal to the magnitude of the velocity.

    Acceleration is the rate of change of velocity. This means that any time your velocity vector is changing, you are accelerating, even if you are not speeding up or slowing down. For although the magnitude of your velocity (i.e. your speed) may not be changing, if its direction is changing, then you are experiencing an acceleration. Uniform circular motion is a perfect example of this: your speed is not changing, but your velocity is, because its direction changes continuously. In this case, as you know, the acceleration is "centripetal", which means "towards the centre."

    Angular speed (which is what is actually given in this problem) is different from just plain old speed (the latter of which is sometimes also referred to as "linear speed"). Speed (or linear speed) is the rate of change of your position with time, measured in metres per second. Angular speed, on the other hand, is a measure of the rate at which the angular position of a rotating object changes with time. As a result, angular speed is a measure of the rotation rate of an object. Therefore, it can be measured in units of radians per second, or degrees per second, or even rotations per second, or rotations per minute.

    I'll get back to you on the problem, but I thought I should clarify these points first, since you seem confused about some of the basics.
     
  5. Feb 16, 2012 #4

    cepheid

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    Another point to keep in mind here is that the wheel started at rest and had to be accelerated up to the final angular speed mentioned in the problem. So it is this angular acceleration, and the torque required to produce it, that results in work having to be done. So, in this case, there IS a speed up that we're thinking of.

    Okay, so I find this problem somewhat tricky to interpret, but I think that the simplest assumption to make is that they load the first six cars that are at the bottom, then they rotate the wheel 1/6 of a rotation so that the next six cars reach the bottom, stop the wheel, load those cars, start the wheel again, and so on and so forth, until the wheel is fully loaded, at which point they start to accelerate from rest up to the full angular speed.

    So the first thing you need to do is compute the final angular speed that is achieved. This is not hard. One rotation in 2 minutes means 1/2 a rotation per minute (i.e 0.5 rpms). Convert this to radians per second.

    The next thing to do is figure out the moment of inertia of the thing that the torque is being applied to. You are asked to consider the passengers only, (i.e. ignore the wheel), so I think that you can consider them to be 36 separate point masses, where a point mass is something whose mass you consider to be concentrated at a single point in space (like a particle). In other words, you ignore the fact that each car is an extended body, and instead pretend that it is a point particle with a mass equal to the mass of 60 average people. What is the moment of inertia of point mass for rotation of that mass in a circle of radius r around some central point? Multiply that by 36 to get the total moment of inertia of the system.

    The one thing that confuses me is that you can't get the angular acceleration, because you aren't told either the time interval or the angular distance interval over which the speed increases from 0 to the final speed. This has me suspicious that I'm misunderstanding something about the problem. It could be that they want you to assume that the wheel is never stopped during loading, but instead, the acceleration up to the final speed occurs *while* the loading is taking place. This would make it a much trickier problem, because then you'd have to consider six separate portions of the acceleration, during each of which there is a different moment of inertia (because a the number of cars that is loaded changes with time). Before proceeding further, I would clarify this point with your teacher. Does the acceleration occur while the passenger loading is happening, or is it sufficient to pretend that you first load all the passengers, and then accelerate from rest?
     
  6. Feb 16, 2012 #5

    BruceW

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    Sorry if I'm stepping on cepheid's toes here. But If we assume that there is no dissipation of energy, then the work done by the machinery is simply the final rotational energy of all the passengers. The question is a bit vague, but I would guess that this is what they are looking for.
     
  7. Feb 16, 2012 #6

    cepheid

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    Well hey, you did have a much smarter and better approach that ought to have occurred to cepheid, so you had to speak up :wink:

    So at least there are two things from my approach that still apply:

    You have to compute the final angular speed as I described.

    You have to compute the total moment of inertia of the passengers as I described.

    After that, you are pretty much done.
     
  8. Feb 16, 2012 #7

    cepheid

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    In fact I am somewhat worried that my posts from yesterday did more harm than good, in particular the first one. You asked, "well there is no acceleration, so so can any work be being done?"

    And I responded by saying, "actually, an object in uniform circular motion IS accelerating."

    This is true, but it is totally irrelevant to the question that you asked, and is therefore the wrong answer. For an object in uniform circular motion, the force is always perpendicular to the displacement, which means that NO work is done, which explains why the object doesn't speed up or gain linear kinetic energy.

    I think that what you were actually asking was the following: "well, if the *angular* speed is constant, then that means there is no *angular* acceleration, so how any work be done?" This is true and a perfectly valid/sensible question. The PROPER response would be:

    "Well, there is no angular acceleration NOW, but there had to be at the beginning in order to spin the wheel from rest up to its final angular speed that we observe now. It is the rotational work that was done during this spinning up to the final speed that we are considering in this problem. So the question boils down to this: I have a ring of 36 point masses all located at radial distance r away from a central point. How much work is required to get this ring spinning up to the observed final angular speed? And as BruceW has rightly pointed out, we can figure out the rotational work done just by considering the total change in rotational kinetic energy (bearing in mind that the ring started with 0 rotational kinetic energy, since it wasn't spinning). So to solve this problem, all you need to know is what the equation for rotational kinetic energy is."

    I hope that this clears up any confusion that might have been brought on my off-the-wall rantings from last night.
     
  9. Feb 16, 2012 #8

    I like Serena

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    Doesn't the work done by the machinery to rotate the passengers consist of 2 parts?

    1. The work to raise the passengers until the wheel is fully loaded.
    2. The work to get the wheel to rotate at the constant angular velocity.
     
  10. Feb 16, 2012 #9

    BruceW

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    I think I get what you mean. The total work (assuming no dissipation of energy) = final rotational KE + change in GPE, So girlphysics also needs to find the GPE increase due to the people being raised above ground.

    The question asks "Estimate the amount of work that was required of the machinery to rotate the passengers alone." So I'm not certain if they want the GPE included. But I think you're right, the GPE increase should also be included in the amount of work done.
     
  11. Feb 16, 2012 #10

    cepheid

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    Cool. I hadn't thought about the GPE, but it seems like fortunately the symmetry of the problem helps a lot here.
     
  12. Feb 16, 2012 #11
    so should I assume the passengers have a 60 kg mass?
     
  13. Feb 16, 2012 #12
    haha thank you, thats exactly what I was asking.
     
  14. Feb 16, 2012 #13
    I suppose I should use W= 1/2 I omega^2 to find the work required to get the ferris wheel 'up to speed'. How should I get I?
     
  15. Feb 16, 2012 #14

    cepheid

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    For starters, yeah. Don't forget to also include the work done to increase the gravitational potential energy of the system, as pointed out by I like Serena.

    I alluded to this earlier (post #4). The way I see it, the problem asks you to consider the passengers only, ignoring the wheel. So it's as though you have 36 individual point masses arranged in a circle. What is the moment of inertia for a single point mass moving in a circle of radius r?
     
  16. Feb 16, 2012 #15
    oh so I should use K=1/2Iw^2, then I= Ʃ miri?
     
  17. Feb 16, 2012 #16

    cepheid

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    Um, the problem just asks for an estimate, so just use a number that is reasonable for the average mass of a human being. The specific value shouldn't matter, as long as it is reasonable.
     
  18. Feb 16, 2012 #17
    I=ƩMiRi^2 how do I include the GPE? 1/2mgh?
     
  19. Feb 16, 2012 #18

    cepheid

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    That equation for moment of inertia doesn't seem quite right.
     
  20. Feb 16, 2012 #19

    cepheid

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    Yeah that's right. Sorry I didn't see that before, we were posting nearly simultaneously.

    Mi is just the mass of the "ith" object, and Ri is just its distance from the point of rotation. (They are all the same distance from the point of rotation in this example).

    A couple of problems:

    - That's not the right equation for the gravitational potential energy of a mass at height h.

    - In this problem you have 36 different masses, and they are all at different heights, and their heights change with time.

    BUT, there is a symmetry to the problem that you can exploit. The cars are evenly spaced around the circumference of the wheel. Imagine if you had only two cars that were evenly spaced around the wheel (i.e. 180 degrees apart from each other, or diametrically opposed). What would be the potential energy of the system in the case where one car was at the very bottom and the other was at the very top? What if you rotated the wheel 90 degrees so that now both cars were half-way up? Would the GPE of the system have changed? Why or why not? How can you generalize this result to a larger number of evenly-spaced (symmetric) cars?
     
  21. Feb 16, 2012 #20
    SO 1/2ωI^2
    substitute I=mr^2 so (1/2)ω(mr^2)^2
    then to find ω, is it 2∏diameter/2min?
     
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