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Author Topic: Sum of torque  (Read 173885 times)

EOW

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Re: Sum of torque
« Reply #15 on: October 27, 2014, 08:12:54 PM »
Maybe this idea ?

Before t=0, turn at w1 and w2 disk. We need to give energy for that. W1>W2. Ar t=0, walls of disk has friction with air only at red point, this give forces F1 and F2 and reduce w2, like w1>w2, the energy increase, no ?

EOW

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Re: Sum of torque
« Reply #16 on: October 27, 2014, 11:47:05 PM »
It's possible to give a torque to disk with external green object. Each object is accelerating, its energy is increasing. w2 slow down, not w1. The energy is :

1/2md²w1²+1/2mr²(w1−w2)²

with :

d = lenght of black arm
r = radius of disk
m = mass of disk

The energy of disk increase. The system increas its energy.

Edit: use a ring instead of disk for have (w1-w2) term.

EOW

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Re: Sum of torque
« Reply #17 on: October 28, 2014, 08:29:38 PM »
nobody for help me ?

EOW

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Re: Sum of torque
« Reply #18 on: October 29, 2014, 12:27:06 AM »
There is no motor after t=0, only before for launch disk. At t=0, apply friction ONLY at red point, not all around the disk. Red points change all the time their positions, this need special wall if you use friction (w1>w2). I don't know if it's clear enough. I think the formula is good only for a ring it's a little more complex for a disk. The disk win energy because w2 decrease. It's logical if you think with some points around the disk, at external the velocity is dw1-rw2 but at internal it's dw1+rw2, like the energy is a square function the external disk "cancel" more energy than it add at internal. I tested with Algodoo, you can test if you want it's very easy, fixe w1 and change w2 like you want and look at the sum of energy you will see the energy increase when w2 decrease (w1>w2). It's strange to say the energy increase if you slow down something but it is. And you can slow down the disk only at red point where w1 is the same for 2 points.

Quote
What is it that you would like help with?
maybe if you understand it allow me to explain the system, after if you can test for watch the energy increase with Algodoo, if you agree with that it's a first step for me. Test in reality must be difficult due the dynamic wall needed.

EOW

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Re: Sum of torque
« Reply #19 on: October 29, 2014, 07:16:18 AM »
Yeah, I think we're ok. When I said it's difficult to built, it's for a pure 100% mechanical system. 

What's number is wrong, the formula ? I think it's ok for a ring and maybe for a disk, a french guy on a forum said the formula is ok for a disk but I'm not sure, another guy on another forum said it's ok for a ring not for a disk. I looked for 10 hours on Internet and I don't find the formula.
« Last Edit: October 29, 2014, 12:13:22 PM by EOW »

EOW

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Re: Sum of torque
« Reply #20 on: October 29, 2014, 12:36:51 PM »
Quote
I kind of find it funny that a ring or a disc might be able to use different formulas just because it is either a ring or a disc.
me too... I don't find the formula in Internet but anyway it's works with a ring, and this idea can be use with a ring.

you tested on what software ?

You tested on a real machine ? where you speak about it ? you must apply torque only a red point, ideally the torque must be applied a very short of time (function of the rotationnal velocity), this need sensors, precise electronic control, etc. You built a mortor like that ?

EOW

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Re: Sum of torque
« Reply #21 on: October 29, 2014, 02:55:51 PM »
Trajectories can be like first image or like second image if w1 and w2 are choosen correctly (depends of the radius).


EOW

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Re: Sum of torque
« Reply #22 on: October 29, 2014, 04:49:47 PM »
No, it's not the same principle. My idea is to use external objects (or something external like air), it's very important because your link shows a system "close" nothing external. Or something fixed to the ground that can recover energy. Here all the system turn no ? And like I said, it's not possible to give a torque like that. You must apply the torque when w1=w2, this in a short time, you need sensors and electronic command.

EOW

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Re: Sum of torque
« Reply #23 on: October 29, 2014, 07:12:01 PM »
I don't understand your message, maybe if you drawn an image I can. Where are results of your tests and the model to test, have you image and link ?

Quote
Where is the torque in your system if you apply an equal force in opposite directions on opposite sides of the disc?
  look at image of post #16, there is a torque due to external object (that can be done with fixed and controlled coils). And look at trajectories of points with w1 and w2, it's not a point, it's a line, your coil must recover energy perpendiculary to the trajectory, like the shock with balls for example. It's not easy to build. I don't know your system and if I can watch what you do I can say if it's the same or not.

For my study, the trajectory must be good for have a torque, like image shows. Maybe it's not the best. Choose point at P1/P2 and adjust the mass of object to schock for have the same force (in value).

EOW

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Re: Sum of torque
« Reply #24 on: October 29, 2014, 08:49:03 PM »
Like that, there is a torque ?

EOW

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Re: Sum of torque
« Reply #25 on: October 29, 2014, 09:19:15 PM »
If it's possible to create a torque with external objects on disk, the energy is won. So, I think it's possible to change the disk, one part with mass, another without mass, and to place point at the good position for have only a torque on it.

EOW

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Re: Sum of torque
« Reply #26 on: October 30, 2014, 08:30:10 AM »
If I take the disk in a double rotation w1 and w2, with w1>w2. It's possible to add a translation to all the system for have to point P1 that move in one direction. Other point P2 friction with black stem, this add forces F1 to disk and F2 to stem. Shock with external purple object for the point P1. P2 friction only, not shock.

EOW

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Re: Sum of torque
« Reply #27 on: October 30, 2014, 09:06:33 AM »
With velocities like that :


EOW

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Re: Sum of torque
« Reply #28 on: October 30, 2014, 10:05:59 AM »
With this ? this need a translation in addition for have forces like that.

EOW

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Re: Sum of torque
« Reply #29 on: October 30, 2014, 10:59:55 AM »
Vleft=R1w2−(R1−d)w1
Vright=R2w2−(R2−d)w1 With:
R1=10
R2=4
w1=10
w2=2
d=4
I find:
Vleft=-40
Vright=8

Use ring to be sure the formula is good: 1/2md²w1²+1/2mr²(w1−w2)²