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Author Topic: Sum of torque  (Read 109087 times)

Offline dieter

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Re: Sum of torque
« Reply #120 on: January 12, 2015, 08:59:55 PM »
By "handle" I meant what do you do with friction. Electromagnetic device is a bit vague, but ok. I guess the wheels do not really have to touch eachother.


The gyroscope will resist rotation, but when a torque is applied (as the big wheel rotates), then I guess the sum of torque 1+2 will cause "recession", eg. in the direction into the picture (or against the surface of the big wheel), see also youtube videos about "Eric Laithwaite".


BR


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Re: Sum of torque
« Reply #120 on: January 12, 2015, 08:59:55 PM »

Offline EOW

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Re: Sum of torque
« Reply #121 on: January 12, 2015, 11:35:21 PM »
Yes, the gyroscope cause a precession. The support turns around the axis Z. The gyroscope cause a precession around the axis Y.  In the lab frame reference the wheel and the gyroscope don't turn around the axis Z, it's a translation, its angular velocity around the axis Z is 0. The problem is to give the rest of the force that the gyroscope can't cancel to the support. The gyro can't be fixed to the support with an axis because the axis will absord all forces and the gyroscope will receive nothing. It's possible to imagine the wheel with an axis to the support and the gyroscope is "floating" in the wheel. Forces are gave to the gyroscope not the wheel, like that the force the gyroscope can't cancel will give to the wheel.

Offline dieter

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Re: Sum of torque
« Reply #122 on: January 13, 2015, 10:18:39 PM »
"Floating" ?  :)


BTW. aren't the arrows of F in the wrong direction?  :o


BR


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Re: Sum of torque
« Reply #122 on: January 13, 2015, 10:18:39 PM »
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Offline EOW

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Re: Sum of torque
« Reply #123 on: January 18, 2015, 05:39:46 PM »
I think forces are good but it's the trajectory the problem, the distance change when the gyro turns.

A new idea with 2 supports:

Support1 and Support 2 are turning clockwise at w around the white axis
On the Support1 there is the Pulley1 with the radius 3R, the Pulley1 is turning at 2w clockwise (lab frame reference), the Pulley1 is turning at w in the lab frame reference
On the Support2 there is the Pulley2 with the radius R, the Pulley2 is turning clockwise at 4w (lab frame reference), the Pulley2 is turning at 3w in the lab frame reference
The Pulley2 is a brake, it transforms the mechanical energy to heating, the energy from heating is 4FRwt, because the Pulley2 turns at 4w and the radius is R
The Pulley1 receives the energy from a motor fixed on the Support1 (the motor is not drawn). The motor needs the energy -3FRwt and the Support1 receives a counterclockwise torque that lost the energy -3FRwt
The sum is not 0, it's -2FRwt. If the Pulley1 and the Pulley2 are fixed to the same support, the support receive a clockwise torque that give the energy 2FRwt, here with 2 supports, the force F and -F works at the same value in one round and cancel themselves.

I drawn different positions.
« Last Edit: January 18, 2015, 08:37:22 PM by EOW »

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Offline EOW

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Re: Sum of torque
« Reply #124 on: January 19, 2015, 09:40:30 AM »
I corrected my message

Support1 and Support 2 are turning clockwise at w around the white axis
On the Support1 there is the Pulley1 with the radius 3R, the Pulley1 is turning at 2w clockwise (lab frame reference), the Pulley1 is turning at w on the Support1 frame reference
On the Support2 there is the Pulley2 with the radius R, the Pulley2 is turning clockwise at 4w (lab frame reference), the Pulley2 is turning at 3w on the Support2 frame reference
The belt turns because the radius goes from 3R to R and the angular velocity goes from w to 3w on the pulley
The Pulley2 is a brake, it transforms the mechanical energy to heating, the energy from heating is 4FRwt, because the Pulley2 turns at 4w and the radius is R
The Pulley1 receives the energy from a motor fixed on the Support1 (the motor is not drawn). The motor needs the energy -3FRwt and the Support1 receives a counterclockwise torque that lost the energy -3FRwt
The sum is not 0, it's -2FRwt. If the Pulley1 and the Pulley2 are fixed to the same support, the support receive a clockwise torque that give the energy 2FRwt, here with 2 supports, the force F and -F works at the same value in one round and cancel themselves in one round.

It's possible to use something else than the brake. It's just for have the force F. Here the energy is destroyed. For created energy, turns Support 1 and Support2, turn Pulley1 and Pulley2 and brake the Pulley2, it gives 4FRwt from the brake and it cancel -2FRwt from the Pulley1 (if the pulley has a mass).

I added 2 points A and B for look the angular velocity. A is fixed on Pulley1, B is fixed on Pulley2.
« Last Edit: January 19, 2015, 05:19:30 PM by EOW »

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Re: Sum of torque
« Reply #124 on: January 19, 2015, 09:40:30 AM »
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Offline dieter

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Re: Sum of torque
« Reply #125 on: January 20, 2015, 12:47:13 AM »
This is interesting, and in my special case: mind bending  :o


BR


Offline EOW

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Re: Sum of torque
« Reply #126 on: January 20, 2015, 08:54:37 AM »
Hi Dieter,

Do you understand my idea ?

For create energy, the motor can be at the Pulley2. The motor need to give -3FRwt to the Pulley2, the Support2 lost -FRwt. The Pulley1 recover +6FRwt because the Pulley1 turns at 2w in the lab frame reference and the radius is 3R.

Have a good day
« Last Edit: January 20, 2015, 08:48:16 PM by EOW »

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Re: Sum of torque
« Reply #126 on: January 20, 2015, 08:54:37 AM »
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Offline EOW

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Re: Sum of torque
« Reply #127 on: January 21, 2015, 04:15:07 PM »
The Support1 receives the force F and the Support2 receives the force -F. One support is helped to turn clockwise when the other support receives a counterclockwise torque. With gears (1, 2 and 3) it's possible to cancel the work of forces F and -F. Like that supports don't accelerate or decelerate. If the motor is put on the Pulley1 the energy is destroyed and if the motor is put on the Pulley2 the energy is created, no ?

Offline dieter

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Re: Sum of torque
« Reply #128 on: January 21, 2015, 10:38:52 PM »
Hi EOW,


I'm referring to #130. Example: when you rotate the left purple wheel, the big pulley (fixed) will rotate the small pulley 3x. So the small pulley may not be fixed to the right purple wheel, otherwise the belt will be blocked.


I have to confess, I do not see where the excess energy comes in or out.


BR

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Re: Sum of torque
« Reply #128 on: January 21, 2015, 10:38:52 PM »
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Offline dieter

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Re: Sum of torque
« Reply #129 on: January 21, 2015, 10:55:56 PM »
Hi EOW,


referring to #133, well, some people get upset by the terms "create + destroy energy "  ;)


Let me see if I understand it:  The pulleys will give F and -F to the supports. But, one support gets 3 lengths of 1/3 force and the other gets -1 length of 1 force. Any rotation of the pulleys would make the supports rotate in opposite direction, or with a speed ratio if 1:3 (And then the belt would fall offl). But the gears will force the supports to rotate at the same speed, in the same direction?
So it cannot rotate at all?
I guess I didn't understand it.  :)


BR


Offline EOW

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Re: Sum of torque
« Reply #130 on: January 21, 2015, 10:59:53 PM »
Hi, Dieter,

In the post #130, the Pulley2 is not fixed. The Pulley2 turn of 360°, one round, because its angular velocity is 4w (lab reference), the Pulley1 turns at 2w (half round). I will draw another example.

For understand, put the Pulley1 and the Pulley2 in the same support, if the Pulley1 is the motor, the motor need to give -3FRwt and the support lost -3FRwt, the Pulley2 receives 4FRwt and the support receive 2FRwt the sum is at 0. Now, with 2 supports, the sum of the torque on Support1 and Support2 is 0. If the motor is the Pulley1, the support lack 2FRwt. If the motor is the Pulley2, the support don't lost -2FRwt.

Quote
But the gears will force the supports to rotate at the same speed, in the same direction?
Yes, it's not necessary to add gears for understand. But like the Support1 receives a torque T and the Support2 receives the torque -T, the sum is at 0, if I use gears +T-T = 0, like that supports turn always at w. For one support the torque change its sign in a round.

Quote
So it cannot rotate at all?
Support1 and Support2 turn at w. Pulley1 turns at 2w (lab reference), Pulley2 turns at 4w (lab reference).

Quote
The pulleys will give F and -F to the supports.
correct

I added an image with an angle of 45°, for look at A and B. The support turns of 45°(w), so A turns of 90° (2w), B turns of 180° (4w).

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Re: Sum of torque
« Reply #130 on: January 21, 2015, 10:59:53 PM »
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Offline dieter

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Re: Sum of torque
« Reply #131 on: January 21, 2015, 11:46:09 PM »
Hi EOW,


ok, that makes sense now.


BR


Offline EOW

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Re: Sum of torque
« Reply #132 on: January 21, 2015, 11:59:25 PM »
Are you ok with a torque at 0 for Support1 and Support2 if I use gears ?

Offline EOW

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Re: Sum of torque
« Reply #133 on: January 22, 2015, 09:43:23 AM »
Hi Dieter,

I think d1=d2, always. The force F on the Support1 gives a clockwise torque T but the force -F gives a counterclockwise torque -T. |T|=|-T|. I drawn forces with the Pulley1 a motor.

Have a good day

Offline dieter

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Re: Sum of torque
« Reply #134 on: January 22, 2015, 08:47:44 PM »
Hi EOW,


But the torque of the pulleys will affect the supports only when there is, example given, friction...?



Have a nice day too!  8)


BR

 

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