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Author Topic: Sum of torque  (Read 136552 times)

Offline EOW

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Re: Sum of torque
« Reply #135 on: January 22, 2015, 09:22:37 PM »
Now, I drawn forces with one support like I spoke before (the motor is on the Pulley1, it's an example), look at the support it works at 2FRwt. It's possible to compare these 2 cases, one with one support and other with 2 supports. Pulleys1 is at 2w in 2 cases, Pulleys2 is at 4w in 2 cases, support is at w in 2 cases.

In the case with one support, the sum of energy is:

-3FRwt for the motor on the Pulley1
-3FRwt for the support, the motor is on the support is apply a counter torque
+2FRwt for the support, torque due to pulleys
+4FRwt for the Pulley2

The sum is at 0

In the case with 2 supports, the sum of energy is:

-3FRwt for the motor on the Pulley1
-3FRwt for the support, the motor is on the support is apply a counter torque
+4FRwt for the Pulley2

The sum is at -2FRwt

-----------------------------------------------------------------------------------------------------------

If the motor is at the Pulley2, the motor need to give -3FRwt, true ?
The Support2 receives -FRwt because the motor apply a torque, ok ?
The Pulley1 turns at 2w and there is a radius 3R, so the energy recover is 6FRwt

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Re: Sum of torque
« Reply #135 on: January 22, 2015, 09:22:37 PM »

Offline dieter

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Re: Sum of torque
« Reply #136 on: January 22, 2015, 10:37:14 PM »
I was asking about friction because, as fas as I see, when both pulleys can rotate freely, then the support won't receive any torque or force at all. (?)


Please excuse me, I may seem really stupid here.


BR


Offline EOW

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Re: Sum of torque
« Reply #137 on: January 22, 2015, 10:45:41 PM »
You don't see the torque with 2R ? F1 and F2 give a torque 2FR, no ? this is not a torque from friction in the axis. Forces come from the belt and are reported to the axis of each pulley.

And you're not seem stupid, it's not easy to understand ideas from others, and you, you try, it's friendly :)

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Re: Sum of torque
« Reply #137 on: January 22, 2015, 10:45:41 PM »
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Offline dieter

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Re: Sum of torque
« Reply #138 on: January 23, 2015, 06:05:52 PM »
I really don't see it, no matter how hard I try..,  :o 


Maybe I should just build it. Maybe as a side project.


BR


Offline EOW

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Re: Sum of torque
« Reply #139 on: January 23, 2015, 06:19:06 PM »
Hi Dieter,

With one support: F1 and F2 in the image, no ? The big pulley force the belt to turn clockwise, and drive the small pulley, there is the force Fb on the belt, this force is put on the axis : F2. The belt receive the force Fa, and this force is reported to the axis of the big pulley: F1. F1 and F2 gives a torque with the value 2FR (with |F| = |F1| = |F2|). I think with one support it's logical to have 2FRwt on the support for have the sum of energy at 0, but with 2 supports I don't find the torque.

With 2 supports, I tested today with a simulation for look at the angular velocities and all seems correct.

If you have gears and pulleys, you can test it (I order a package of gears for test).

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Re: Sum of torque
« Reply #139 on: January 23, 2015, 06:19:06 PM »
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Offline dieter

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Re: Sum of torque
« Reply #140 on: January 23, 2015, 07:16:24 PM »
I would have to make them, with wood and saw, and a couple of inline-skate ballbearings, probably I go and get some plywood this evening :)


Can your simulation software export animations? That would help a lot. Also, looking foward to see your build!


BR
 

Offline EOW

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Re: Sum of torque
« Reply #141 on: January 23, 2015, 11:08:29 PM »
Like I saw my simulation, the angular velocities must not be w, 2w and 4w but w, 2w and 6w. With 2 supports, the motor and the Support1 need -6FRwt, the Pulley2 recover 6FRwt. But with one support, the motor and the support need -6FRwt and the Pulley2 recover 6FRwt, but there is the torque on the support, that torque gives the energy 2FRwt. It's necessary to check the sliding of the belt with w (support),2w(Pulley1) and 6w (Pulley2), if the sliding is 0 this could say the case with one support must give an extra energy to the support.

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Re: Sum of torque
« Reply #141 on: January 23, 2015, 11:08:29 PM »
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Offline EOW

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Re: Sum of torque
« Reply #142 on: January 23, 2015, 11:53:34 PM »
For explain the forces F1 and F2 in my last example. Take another simple example, I renamed forces, with one pulley. Apply a force F1 to the rope, that force is transmitted to F2 along the rope. F2 want to move the pulley to the left, so F2 apply a force F3 to the axis, the fixed axis reply and give the force F4. F1/F4 is a torque that turn the pulley but the support receive the force F3. With 2 pulleys there are 2 forces, and if the size of the pulley is not the same there is a torque. Do you understand ?

Offline EOW

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Re: Sum of torque
« Reply #143 on: January 24, 2015, 10:53:29 AM »
I made a mistake with the force with one support. If I brake the Pulley2 on the support, you're right there is no torque (first image shows all forces, grey color -> to the pulley, magenta color -> to the support and black colo -> to the belt).

Now, second image, I don't brake the pulley2, imagine it with a mass. I accelerate more and more the Pulley2, the energy is in the kinetic energy in the pulley. The sum of forces are like I drawn, no ? And the support decelerate because it receive a counterclockwise torque from force F1 / F2, no ?

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Re: Sum of torque
« Reply #143 on: January 24, 2015, 10:53:29 AM »
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Offline dieter

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Re: Sum of torque
« Reply #144 on: January 25, 2015, 01:24:47 AM »
Hi EOW,


I would say yes, as long as the pulley is accellerating, the inertia of the mass will give an opposite torque to the support. I guess this happens due to an imbalance of centrifugal force on the pulley shaft. Not sure tho.


BR


Offline EOW

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Re: Sum of torque
« Reply #145 on: January 25, 2015, 01:26:54 PM »
I think I forgot the force -F6 on the big pulley. So:

At start, the support turns clockwise at w, the big pulley turns clockwise at 2w and the small pulley turns clockwise at 6w. The belt don't slide. Pulleys accelerate more and more (no brake), the energy is recover after (kinetic energy). An external motor keep constant the angular velocity of the support (need energy because the support receives a torque F3/F4).

At a time:

The motor needs -3FRwt, the motor gives F1 and F2 to the big pulley, the motor gives F3 and F4 to the support
The support receives -3FRwt (F3 and F4)
The belt gives the force F5 to the big pulley, this gives the force F6 to the support and -F6 to the big pulley. The belt gives the force F7 to the small pulley
The small pulley receives 6FRwt
The support receives the torque F6/F8, and the energy 2FRwt

The sum is 2FRwt, no ?

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Re: Sum of torque
« Reply #145 on: January 25, 2015, 01:26:54 PM »
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Offline dieter

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Re: Sum of torque
« Reply #146 on: January 25, 2015, 11:31:56 PM »
Hi EOW,


Again, this has knocked out my brain  :o ... I guess you understand this much better than I. Personally, I would rather construct it in 2 hours than to (fruitlessly) think about it for 4 hours. In fact I brought home some plywood, so maybe I will. I just don't know how to provide the pulleys with power when the support rotates.


BR


Offline EOW

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Re: Sum of torque
« Reply #147 on: January 26, 2015, 02:21:29 PM »
Maybe it's possible to test the torque on the support without a motor. The big pulley has a mass and it's possible to turn it at 2w before to turn the support, the big pulley can drive the small pulley with the inertia. With no friction to the axes of rotation => no torque must be appear to the support, if a torque appear that could say the system can give en extra energy. The small pulley is braking on the support at the radius R, for me the sum of energy is:

The big pulley lost -2w*3R*F*t = -6FRwt
The friction win 5w*F*R*t = 5FRwt
The support win w*F*3R*t = 3FRwt

The sum is at 2FRwt. The small pulley has a mass as lower as possible.

I think the key is the torque on the support.

Offline dieter

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Re: Sum of torque
« Reply #148 on: January 27, 2015, 03:36:36 AM »
Maybe you're right. But then there is still an other problem in a practical implementation: There will be friction on the pulley shafts. The skate bearings are made for about 20 kg, they have a lot of friction. It can he reduced by cooking out the grease, but still... so we wouldn't know if the torque is caused by the friction.


Bearings from CD players would be better, but I got only one right now... (those with the "clickable" cd shaft, not the normal PC cdrom drives).


Maybe I'll find one more in my pile if stuff :)


BR


Offline EOW

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Re: Sum of torque
« Reply #149 on: February 06, 2015, 02:50:44 PM »
For test easily, it's possible to put all things on the support and filmed for watch the energy input/output. The DC motor drives the big pulley, the small pulley drive a DC generator, the generator can give energy to a resistor, all instruments are on the support, need only 4 instruments. If the generator is fixed on the support, there is no torque on the support. I calculate the energy from the system and it's not a factor of the angular velocity of the motor but factor of the angular velocity of the support:

E=2FRwt

The generator must recover energy from the motor: the force F

If the support turn slowly the energy recover is low

With :

R=0.1 m
F=0.5 N
w=200 RPM
The power is 1W



 

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