Naked Science Forum

General Science => General Science => Topic started by: newrobert on 18/11/2014 04:00:19

Title: Calculate required force to move vehicles in circular motion (big loop)
Post by: newrobert on 18/11/2014 04:00:19
Hi,

There is big roller coaster loop 100% round and vehicles are moving in this circular path, I want to calculate how much force will be required to move these vehicles.

Circle path diameter: 100 meter
Time: 1 round completed in 15 minutes
1 vehicle mass: 1,000 kg
Total Vehicles: 50

All vehicles are tied with each other just like train or roller coaster cars.

What force will be required to move these vehicles at required speed i.e. 1 round completed in 15 minutes. What's equation and how to solve it.

Please help me

Robert
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: RD on 18/11/2014 09:15:13
This looks like homework , so we can only give you clues ...

Circle path diameter: 100 meter
Track length = circumference = Pi x diameter (http://en.wikipedia.org/wiki/Circumference#Relationship_with_Pi)

Time: 1 round completed in 15 minutes
speed = distance/time = circumference / time

1 vehicle mass: 1,000 kg , Total Vehicles: 50

Total mass is 50,000kg ,

energy* required to get that mass to the speed=  Kinetic energy = 1/2 x mass x (speed)2
 (http://en.wikipedia.org/wiki/Kinetic_energy#Kinetic_energy_of_rigid_bodies)
[ I take it we are ignoring friction ]

[ * energy (http://en.wikipedia.org/wiki/Energy) not "force (http://en.wikipedia.org/wiki/Force)" : they are not the same thing. If the time taken to get the train from rest up to speed was known you could calculate the average force required to do that (http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_second_law) ].
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: newrobert on 18/11/2014 09:31:32
Thanks for helping me man.

Diameter: 100 meter
Circumference: 335 meter
speed =335/900 = .37265 m/s
Total mass = 50,000 kg

Kinetic Energy = 1/2 x 50,000 x .138865
Kinetic Energy = .5 x 50,000 x .138865
Kinetic Energy = 3471

1. So what' is 3471? Is this Joule? or what?
2. How to convert this to power i.e. motor power required to rotate these vehicles?
3. Is this equation same for "rotational motion", I also saw this equation somewhere for rotational motion where object is moving around its axis. But here object is moving in circular path.

Please reply above 3 in understand the whole system.

Thank you


Robert


Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: RD on 18/11/2014 10:10:45
Diameter: 100 meter
Circumference: 335 meter

Pi = 3.1415... (http://en.wikipedia.org/wiki/Pi)  , Circumference=100xPi = 314 meters (to the nearest meter).

Joule? or what? 
Yes energy is measured in Joules.

2. How to convert this to power i.e. motor power required to rotate these vehicles?

You'd need to tell us how quickly the train is required to accelerate to the required speed from rest to calculate the minimum power of the motor required, cf. a family-car and a race-car can both reach 50kph , but the the race-car can reach that speed quicker because it has a more powerful engine. We need to know how quickly the train is required to reach the required speed to calculate the minumum engine power necessary to do that.   

3. Is this equation same for "rotational motion", I also saw this equation somewhere for rotational motion where object is moving around its axis. But here object is moving in circular path.

In this case the rotational energy (http://en.wikipedia.org/wiki/Rotational_energy) is negligible compared with the translational energy: the train only undergoes one rotation every 15 minutes , ω (http://en.wikipedia.org/wiki/Angular_velocity)2 = 0.0000487 radian per second (http://en.wikipedia.org/wiki/Radians_per_second).
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: newrobert on 18/11/2014 11:50:22
Once again thank you for your post.

Time require to reach train at required speed: 5 minutes

I am confused, can you please let me know how it works.

1. First we need to find "Force' in 'N'
2. Then we need to find 'Kinetic Energy' in 'J'
3. Then we need to find 'Work' in 'J'
4. Then we need to find 'Power' in 'Watt'

Is this the process?

I am confused, I mentioned all details now, please guide me how to find "Power" in "Watt" i.e. what power motor we need to install to run train at required speed?

So far I entered your formula in Excel and now waiting, what to do next?


Robert
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: RD on 18/11/2014 12:49:06
Time require to reach train at required speed: 5 minutes

I am confused, can you please let me know how it works.

1. First we need to find "Force' in 'N'

#1. Calculate the energy required to achieve the speed , ( KE = 1/2.m.v2)
#2. The time taken to deliver that amount of energy determines the minimum power of the motor required : Power = Energy per unit time  (http://en.wikipedia.org/wiki/Power_%28physics%29).
Your motor has to deliver about 3,000 joules of energy in 5 minutes , that tells you how powerful it has to be.  [ A Joule per second is a Watt , your motor has to deliver about 3,000 joules in 300 seconds ].

 ( again we are ignoring friction , which in reality would not be negligible ).
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: newrobert on 18/11/2014 13:12:46
Ok,

1. Calculate the energy required to achieve the speed , ( KE = 1/2.m.v2), here I will take velocity according to "5" minutes. 5 minutes is the time in which roller coaster train need to catch speed of 1 round in 15 minutes.?

2. What is equation of calculating 'power' for current situation i.e. Unified circular motion? Power = Work/time

Now I have 2 times, one is acceleration time in which train needs to gain speed of one round in 15 minutes. And other time is regular train circulation time.

Please help?

Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: RD on 18/11/2014 13:31:00
... how to find "Power" in "Watt" i.e. what power motor we need to install to run train at required speed?

So the objective is to calculate the minimum power of the motor required, (in Watts),  necessary to have the train reach a certain speed in a certain time.

You said 15 minutes per circuit was the speed it needs to reach , and later, (in post #4), told us it takes 5 minutes to reach that speed.

Use 15 minutes per circuit to calculate the speed , then you can calculate the kinetic energy of a train of that mass travelling at that speed.
The answer is   1/2.m.v2  Joules.

Your motor has to deliver 1/2.m.v2 Joules in five minutes.

[ I've given you the answer in my previous post : it's 3000 joules delivered in 300 seconds => 10 joules per second = 10 Watts power ]
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: newrobert on 18/11/2014 14:28:18
Hi,

Last time you asked that we need time in which train has to reach at required speed so I give "5 minutes" time. I mean train should reach at "15 minutes per round" speed in 5 minutes from rest position. Anyhow acceleration speed will be calculated later.

Right now I need to know how much power motor required to maintain speed of "1 round in 15 minutes" speed. So speed is 15 minutes. First I need to know how much power motor is required to keep it running at required speed.

Diameter: 100 meter
Circumference: 314 meter
1 Round Time: 15 minutes
Time: 15 x 60 = 900 seconds
Speed =314/900 = .349056 m/s
Total mass = 50,000 kg

K.E. = 1/2 x 50,000 x .12184
K.E. = 3045 Joule

It means electric motor have to produce 3045 Joule energy in 15 minutes?

Now how to convert 3045 Joule into watt? What's formula of Power?

In your last post you said 10 watt, how you calculated, please give me formula?

Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: newrobert on 18/11/2014 16:17:36
Is this equation can be used here i.e. Power = Work / Time

Power = 3045 / 900 = 3.38 watt

Is this correct? A 50,000 kg heavy train is keep rotating by 3.38 watt electric motor? I think there is some problem?

Please help.

Robert
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: RD on 18/11/2014 16:49:50
It means electric motor have to produce 3045 Joule energy in 15 minutes?

No, it means 3045 Joules of energy are required to get the train up to 0.349056 m/s , no matter how long that takes.

Now how to convert 3045 Joule into watt? What's formula of Power?
In your last post you said 10 watt, how you calculated, please give me formula?

According to your specification the 3045 Joules must be delivered in five minutes, (not 15 minutes), i.e. 300 seconds.
3045 Joules in 300 seconds is 10.15 joules per second, which is 10.15 Watts.

Is this correct? A 50,000 kg heavy train is keep rotating by 3.38 10.15 watt electric motor? I think there is some problem?

As I mentioned previously the assumption is there is no friction in this system, in reality that would not be the case.
10.15 Watts is the minimum power necessary in an idealised friction-free universe ,
in reality it would have to be more powerful than that.

[ If there is no friction to overcome a mouse could pull a train , not very quickly though ].

Without friction once started objects just keep moving : they don't slow or stop ...
Title: Re: Calculate required force to move vehicles in circular motion (big loop)
Post by: newrobert on 18/11/2014 17:05:36
Ok, on average in total how much % friction I should add. Friction + motor efficiency. 10% maximum friction + 10% motor efficiency. Even if it is 10.15 + 20% = 12.18 watts. If we assume 20% friction + motor efficiency then still its 12.18 watt.

I have understood what you taught me here. So things will go like this I mean solve this as follows:

1. Get Kinetic Energy (J) = 1/2.m.v2 = This is work that has been done, right??? I mean this equation is used in "unified circular motion" condition. Normally Work = F x displacement or this simple equation works where height is not involved?
2. Get Power. Power = work / time