# Required electric power to move heavy mass around on big wheel

1. Nov 18, 2014

### shaks

Hi guys,

I have one project where I need to rotate heavy mass on big wheel.

Parameters
Wheel diameter: 43 meter
Circumference: 135 meter
Mass speed: 10 minutes per round
Speed: 135 / 600 = .225 m/s
Mass: 344,965 kg
Drive Type: Track around the wheel like roller coaster
Friction: I will add friction later

Is this moving object fall in "Unified Circular Motion"?

I want to find out what power motor required to move object on this wheel. From where start?

Shaks

2. Nov 18, 2014

### Bystander

Angular momentum is a conservative quantity. Power demand is for overcoming friction only. How fast did you want to spin it up (from zero angular velocity to the ~ .01 rad/s)?

3. Nov 18, 2014

### shaks

Thanks man for helping me. :)

It is 5 minutes, we don't need to maintain that speed very fast and 5 minutes is fine.

Shaks

4. Nov 18, 2014

### Bystander

"Rollercoaster?" Didn't think to ask whether there is vertical motion involved --- figured it was like a turntable in a railroad roundhouse. Other necessary detail is mass distribution --- concentrated at the 21.5 m radius, uniform over area of circle, or what?

5. Nov 18, 2014

### shaks

Yes, its vertical and 43 meter high.

Its like roller coaster loop with same kind of 5 feet wide track. Small cars (just easy name otherwise its just a simple structure to carry mass) will move on track and on each car 15 feet wide steel rod is fixed. 5 feet outside at both sides and equal mass is hanged at both sides. For easy understanding I can say this train and train is carrying mass. Train length is 50% of circumference.

Now I need to find how much bigger electric motor will be sufficient to move this train excluding friction.

Shaks

6. Nov 18, 2014

### Bystander

Roller coaster loop --- ferris wheel? Balanced?

7. Nov 18, 2014

### shaks

Its not ferris wheel, its like roller coaster loop with train track but not balanced. Train length is 50% of circumference it means its not balanced at one time mass will be at one side. If it is rotating close wise then I think actual problem is to calculate required power to move train from bottom to top from left side and then it will automatically drop down from right side. But weight is equally loaded on train at both sides.

Train length is 67.5 meter with equal mass loaded on both sides with equal distance.

8. Nov 19, 2014

### Bystander

350 tons lifted 40 meters in five minutes is around 700 hp., but do NOT take that number and run with it --- I am a rank amateur when it comes to designing lift bridges, moving steel crucibles, dumping multiple carloads of bulk material or whatever this is intended to do.

9. Nov 19, 2014

### shaks

Can you please tell me equation, how to solve this. I mean sequence of equations. So I may see how much power required by changing mass and time, etc.?

First of all what I should need to calculate and how?

10. Nov 19, 2014

### Matterwave

Let me see if I got the scenario right. You have a 43 meter diameter vertical circular track on which you stick a 350 ton train. You want this train to go slowly, at a constant speed, around this track, and you want to know what power the engine needs to accomplish this feat? At the top of the loop, basically the whole weight of the train will be bearing down on this track, are you sure the track, as well as the joints holding the train onto the track, can support this weight?

11. Nov 19, 2014

### shaks

Yes, you understand my scenario correctly. Yes track is fine i.e. can hold the weight. From where to start to calculate to know how much electric power motor I needed to run this train? I mean K.E., Force, Work, Power, etc. ?

Which comes first or what's sequence and equation for current scenario?

Thanks in advance for helping me.

Shaks

12. Nov 19, 2014

### Matterwave

Well, since you want the train to go at constant velocity, you have to drive the train up half the circular track using a certain amount of energy, which will require power from your engines, and then brake on the way down to prevent the train from free-falling along the track since it would travel much faster than .225m/s under the influence of gravity.

Since you are basically destroying all of the energy you acquired on the first half of the trip by braking during the second half of the trip, you need your engines to constantly put out power during the climb, and your brakes to constantly function during the descent (which is pretty tough on the brakes it would seem). For the most naive scenario, where you are just moving a point mass up 43 meters in 5 minutes, without any friction, without considering the initial accelerating part of the problem, Bystander gave you the answer for the power needed to accomplish such a task. You take the gravitational energy you need to impart to your train $PE=mgh$ and take that and divide by the time you need to impart that energy to obtain the power of the train. The result you get from plugging in your numbers is roughly what Bystander gave, it's 650hp, or 485kW. This means though that your brakes will also be dissipating heat at a rate of 485kW during the descent! This is enough heat to heat up 1 ton of iron by 1 degree Celsius per second. Your brakes better be able to cool off from this!

The scenario is made more complicated by the fact that your train is half the length of your track. This means that for basically all parts of the journey, part of the train will be descending while other parts will be ascending, you could perhaps use the descending part of the train to provide a part of the energy needed to ascend the track. The calculations will become complicated as you have to calculate the forces on each individual section of the train to see how much braking/engine power is needed.

How will the weight be distributed along the train? I assume you have several sections, which carry most of the weight, with some space in between each section? Will the weight be uniformly distributed within each compartment of the train, or will some compartments carry more weight than others? (e.g. the engine might be where a lot of the weight is?)

Last edited: Nov 19, 2014
13. Nov 19, 2014

### shaks

1. Weight is equally distributed on the train

I set this to same as required normal speed i.e. 10 minutes. It mean if train catches required speed from rest/zero position then still this is fine.

P.E. = 344,965 x 9.8 x 43 meter (here I will set diameter, not 50% of track circumference, right?)
P.E. = 145,254,621Joule
Power = 145,254,621 / 600 seconds (10 minutes, although from bottom to top, time will be 50% of total time)
Power = 242,091 Watt (242 KW)

Are my calculation correct? This is very simple method, I was assuming of calculating force, KE, work, etc but you simply calculated KE and then got power, is this fine?

Shaks

14. Nov 19, 2014

### Matterwave

I think you should read my previous post more closely. Because this very simple method is neglecting a lot of other factors, some of which I brought up in my previous post. :)

You got half the power needed because you used the whole 10 minutes instead of the 5 minutes it takes to bring the train up to the top.

15. Nov 20, 2014

### shaks

Brother, I am not professional and just trying to work on my project. I will appreciate if you can help me here as you are doing. I mean can you please tell me other neglected factors other than friction because I can add friction later i.e. roughly 10% or 15% or whatever according to you.

P.E. = 344,965 x 9.8 x 43 meter (here I will set diameter, not 50% of track circumference, right?)
P.E. = 145,254,621Joule
Power = 145,254,621 / 300 seconds (5 minutes that's time of 1 side i.e. bottom-top)
Power = 484,182 Watt (484 KW)

So we just calculated Potential Energy and converted into Power.

We did not calculated Force, Torque, K.E., Work, Inertia, etc.?

Shaks

Last edited: Nov 20, 2014
16. Nov 20, 2014

### Matterwave

Power is defined as energy/time so yes you can convert directly from an energy and a time into a power. What we calculated there is the power it takes to move 345tons up a vertical distance of 43 meters, at constant velocity, in 5 minutes. The friction can not be included "later" because without friction your train will drop from the top to the bottom in a much shorter amount of time than 5 minutes. The brakes on your train require friction to function.

The first major thing to consider is that the train is half the length of the track. So start with your train having the midpoint being at the bottom of the track. The head of the train will be halfway up the track, and the tail of the train will be halfway down the track (up the track on the other side). The train is in equilibrium here so to get it moving you need to start moving the head of the train up the track while the tail of the train drops down the track. I'm still not sure how to analyze this situation in a completely correct manner...but that is where I would start my analysis.