# Estimating Work for Rotating a Ferris Wheel

• girlphysics
In summary, George Washington Gale Ferris Jr. built the original Ferris wheel with 36 wooden cars, each holding up to 60 passengers, which rotated around a circle with a diameter of 76m. The wheel was loaded 6 cars at a time and once all 36 cars were full, it made a complete rotation at a constant angular speed in about 2 minutes. To estimate the amount of work required to rotate the passengers alone, the equations for torque, work, force, and inertia were considered. While the velocity may be constant, there is still an acceleration due to the changing direction of the velocity vector. The angular speed can be converted to radians per second to determine the final angular speed achieved. The moment of
girlphysics
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 isn't acceleration 0, which means no work is being done? It CANT be that easy. I am not sure how to start to solve. Any help is appreciated.

τ=(r)(Fsin∅)
W=∫θF θI τdθ
W=91/20 I(wf^2-wi^2)
T=Iα
I=MR^2 for a hoop

girlphysics said:
I know the equations for Toque, and Work as well as force and Inertia, but if the velocity is constant, then isn't acceleration 0, which means no work is being done? It CANT be that easy. I am not sure how to start to solve. Any help is appreciated.

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.

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?

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.

BruceW said:
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.

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

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.

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.

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.

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.

Cool. I hadn't thought about the GPE, but it seems like fortunately the symmetry of the problem helps a lot here.

so should I assume the passengers have a 60 kg mass?

cepheid said:
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.

haha thank you, that's exactly what I was asking.

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?

girlphysics said:
I suppose I should use W= 1/2 I omega^2 to find the work
required to get the ferris wheel 'up to speed'.

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.

girlphysics said:
How should I get I?

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?

cepheid said:
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.

oh so I should use K=1/2Iw^2, then I= Ʃ miri?

girlphysics said:
so should I assume the passengers have a 60 kg mass?

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.

cepheid said:
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?

I=ƩMiRi^2 how do I include the GPE? 1/2mgh?

girlphysics said:
oh so I should use K=1/2Iw^2, then I= Ʃ miri?

That equation for moment of inertia doesn't seem quite right.

girlphysics said:
I=ƩMiRi^2

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).

girlphysics said:
how do I include the GPE? 1/2mgh?

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?

SO 1/2ωI^2
substitute I=mr^2 so (1/2)ω(mr^2)^2
then to find ω, is it 2∏diameter/2min?

cepheid said:
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?
I suppose the GPE stays constant, because a car at the top would have let's say 100 and the car at the bottom 0, then both in the middle each 50. So because the cars are all directly across from each other, the GPE is constant.

girlphysics said:
I suppose the GPE stays constant, because a car at the top would have let's say 100 and the car at the bottom 0, then both in the middle each 50. So because the cars are all directly across from each other, the GPE is constant.

oh duh, the PE is mgh, I think I was thinking KE when I put the 1/2 in there

looking at the units, I need joules=Kg*m^2/s^2 . where will I get s^2 when I only have angular velocity

girlphysics said:
SO 1/2ωI^2
substitute I=mr^2 so (1/2)ω(mr^2)^2

I'm pretty sure that the equation is KE = (1/2)Iω2

girlphysics said:
then to find ω, is it 2∏diameter/2min?

It's the rate of change of your angle with time. You go one rotation every 2 mins. How many radians is one rotation? How many seconds is 2 minutes? Therefore, how many radians per second is ω?

girlphysics said:
I suppose the GPE stays constant, because a car at the top would have let's say 100 and the car at the bottom 0, then both in the middle each 50. So because the cars are all directly across from each other, the GPE is constant.

YES! That is exactly right.

Now, what would be the actual GPE of one pair of cars in this problem? Do you see how this extends to 36 cars, (i.e. 18 diametrically opposed pairs)?

girlphysics said:
looking at the units, I need joules=Kg*m^2/s^2 . where will I get s^2 when I only have angular velocity

I think the issue you're having is because you had the equation for rotational KE slightly wrong, as I mentioned above.

By the way, I'm glad you looked at the units. Always keep track of the units. it's a great way to make sure your work is making sense, and to catch errors. If things aren't dimensionally consistent on both sides of an equation, then chances are you've made an error.

cepheid said:
I'm pretty sure that the equation is KE = (1/2)Iω2

It's the rate of change of your angle with time. You go one rotation every 2 mins. How many radians is one rotation? How many seconds is 2 minutes? Therefore, how many radians per second is ω?

2∏/120 s.= 1/60 rad/sec was I right about constant GPE or was it a nice try? lol

cepheid said:
I think the issue you're having is because you had the equation for rotational KE slightly wrong, as I mentioned above.

By the way, I'm glad you looked at the units. Always keep track of the units. it's a great way to make sure your work is making sense, and to catch errors. If things aren't dimensionally consistent on both sides of an equation, then chances are you've made an error.

oh cool! I finally get it. Thank you for helping me. I understand it too! amazing! lol

girlphysics said:
2∏/120 s.= 1/60 rad/sec was I right about constant GPE or was it a nice try? lol

I think you mean π/60 rad/s, but yeah. And yes you were right about the GPE, as I said in post #26 (we are once again posting too frequently to keep up with each other).

girlphysics said:
oh cool! I finally get it. Thank you for helping me. I understand it too! amazing! lol

You're welcome! One last thing I noticed: a few posts up, you said that I = mr2. This is true, provided that m is the total mass of all cars. But if m is the mass of just one car, then I = 36mr2, because you have to add up all of the individual mr2 terms in the summation that you posted before. I just wanted to make sure you were doing so.

cepheid said:
I think you mean π/60 rad/s, but yeah. And yes you were right about the GPE, as I said in post #26 (we are once again posting too frequently to keep up with each other).

You're welcome! One last thing I noticed: a few posts up, you said that I = mr2. This is true, provided that m is the total mass of all cars. But if m is the mass of just one car, then I = 36mr2, because you have to add up all of the individual mr2 terms in the summation that you posted before. I just wanted to make sure you were doing so.

thank you cepheid and good night! : ) another successful physics night thanks to you. I hope your ok with me asking for your help until may 14th! haha. You help me understand so much better!

girlphysics said:
thank you cepheid and good night! : ) another successful physics night thanks to you. I hope your ok with me asking for your help until may 14th! haha. You help me understand so much better!

I'm not always online, but if I am, I will of course help. This website has many helpful Mentors and/or Homework Helpers, and I sort of dropped the ball on this problem anyway, and they set me straight. So it's a collaborative effort to be sure. Good night to you as well.

## 1. How is the work for rotating a Ferris wheel calculated?

The work for rotating a Ferris wheel is calculated by multiplying the force required to rotate the wheel by the distance traveled. This can be expressed as W = F * d, where W is the work, F is the force, and d is the distance.

## 2. What factors affect the amount of work required to rotate a Ferris wheel?

The amount of work required to rotate a Ferris wheel is affected by several factors, including the weight of the wheel, the size and weight of the cars, the speed at which the wheel rotates, and the friction between the wheel and its bearings.

## 3. How is the force required to rotate a Ferris wheel determined?

The force required to rotate a Ferris wheel is determined by the weight of the wheel and the cars, as well as the speed at which the wheel rotates. This force can be calculated using the formula F = m * v^2 / r, where F is the force, m is the mass, v is the velocity, and r is the radius of the wheel.

## 4. Can the work required to rotate a Ferris wheel be reduced?

Yes, the work required to rotate a Ferris wheel can be reduced by decreasing the force required to rotate the wheel or by reducing the distance traveled. This can be achieved by using lighter materials for the wheel and cars, increasing the efficiency of the bearings, or decreasing the speed at which the wheel rotates.

## 5. How is the work for rotating a Ferris wheel used in real-life applications?

The concept of estimating work for rotating a Ferris wheel is used in various real-life applications, such as designing amusement park rides, calculating the energy consumption of rotating machinery, and determining the efficiency of mechanical systems. It is also important in understanding the principles of rotational motion and the conservation of energy in physics.

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