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Author Topic: Sum of torque  (Read 102276 times)

Offline EOW

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Re: Sum of torque
« Reply #45 on: November 14, 2014, 05:44:24 PM »
To simplify the study of this device. Think with an external theoretical device that add the energy FRtw1, like that all rotationnal velocitiy w1is  constant. I give FRtw1 and I receive Ft((R+r)w1-rw2), the difference is: Frt(w1-w2) with w1>w2, the energy is positive. I can't keep w2 constant, because I need the force F4 (and F5), if w2 constant => w3 constant and F4=0. So, w2 must change with time. Maybe a little sliding between grey disk / brown disk can be adjust the kinetic energy.

Heating gives H=Ft((R+r)w1-rw2)
Torques need Ft ( rw2-1/2rw2-3/2(R+r)w1+1/2(R+3r)w1-1/2rw2 ) = -FRtw1

Here F4=1/2F but it's possible to have another value it depends of the inertia.

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Re: Sum of torque
« Reply #45 on: November 14, 2014, 05:44:24 PM »

Offline EOW

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Re: Sum of torque
« Reply #46 on: November 16, 2014, 12:16:31 AM »
I'm not sure when I drawn direction of rotation, so I give an example:

w1 = 10 rd/s
w2 = 8rd/s
w3 = 12rd/s

w1 = 10 rd/s, the arm is turning clockwise
w2 = 8rd/s, the grey disk is turning clockwise at w1 around red axis (labo reference) and grey disk is turnin around itself counterclockwise at 2 rd/s (arm reference)
w3 = 12rd/s, the brown disk is turning clockwise at w1 around red axis (labo reference) and brown disk is turning around itself clockwise at 2 rd/s (arm reference)

I guess |w2| < |w1| < |w3| . There is a relation between w2 and w3 at start because there is no slinding between grey disk and brown disk. In the arm frame reference w2 = -w3 at start.
« Last Edit: November 16, 2014, 08:23:09 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #47 on: November 17, 2014, 01:32:35 AM »
In the case with the arm fixed and the purple disk is turning, I have a work from torque at -FRtw1+Frtw2+kFrtw2-kFrtw3, heating don't change, the sum is at kFrtw2-kFrtw3, it is 0 too. 'k' is kF force from brown disk. If there is a little sliding between the grey disk and the brown disk, the sum is always kFrtw2-kFrtw3. So there is more heating but +kFrtw2-kFrtw3 seems to be higher because w2 seems to increases more than w3.

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Re: Sum of torque
« Reply #47 on: November 17, 2014, 01:32:35 AM »
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Offline EOW

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Re: Sum of torque
« Reply #48 on: November 17, 2014, 11:23:37 AM »
First image:

Another idea. A motor is fixed to the arm accelerates the red disk, more and more. The slave disk (grey) is accelerated in the same time, no sliding between the red disk and the grey disk. All the system is turning at w1 around the magenta axis. I drawn all forces. The motor receive a negative  torque, it must give energy E1. The slave disk and the red disk receive its energy E1 and increase their rotationnal velocity. But the grey disk has the force F6 that incresase w1. In the contrary, F5 don't work. The sum of work seems to be positive.Like the grey disk is turning in the other direction that w1, choose the rotationnal velocity of the grey disk > w1 at start.

Second image: no motor after t=0

w3 > w1, and w2=w3 (or not exactly). All rotationnal velocities are counterclockwise. Before start, launch the device with w1, w2, w3 and set friction between disks at 0. At t=0, set friction ON, there are forces like I drawn. F6 can give energy to the arm via w1. Disks slow down but it is compensated by heating from friction.

Offline EOW

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Re: Sum of torque
« Reply #49 on: November 18, 2014, 08:46:47 AM »
Like this there is a torque on arm.

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Re: Sum of torque
« Reply #49 on: November 18, 2014, 08:46:47 AM »
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Offline EOW

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Re: Sum of torque
« Reply #50 on: November 18, 2014, 08:41:13 PM »
An arm of length d is turning clockwise at w1. A red disk of radius r is turning at w2. w1 and w2 velocities are in labo frame reference. I define w2′ the rotationnal velocity in the arm frame reference. I set w2′ for have the velocity at the origin of F1 at 0: w1(d+r)−rw2′=0, w2′=w1(d+r)/r. An external force F1 is applied during a very short time, like there is no movement, the work needed by the force F1 is 0. I define F the value of F1 or F2. Like I give F1 on disk, F2 appears on arm. The work from F2 on arm is Fdtw1. The work from F1 on disk is −Frtw2′=−Ftw1(d+r). The sum of energy is not 0.

Offline EOW

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Re: Sum of torque
« Reply #51 on: November 20, 2014, 11:01:41 PM »
The black arm is turning clockwise at w1. The blue axis is fixed to the ground. The grey cruz (grey arms) is turning at w2, with w2 < w1, so the grey cruz is turning around itself counterclockwise in the black arm reference. Each  red disk is turning at w3, with w3 < w2. For example, w1=10, w2=9 and w3=1. Friction from disk/disk gives black forces. The rotationnal velocity of each red disk on grey arm reference is w3' with w3'= w2-w3 = 8, so each red disk is turning counterclockwise around itself on the grey arm reference. So w3' can give forces like I drawn. These black forces give a torque on each disk and give a torque on the grey cruz. Each red disk increases its kinetic energy, because the torque is clockwise and w3 < w1. The black arm don't have a torque, so w1 is constant. Green forces reduces w2. BUT, the inertia of grey cruz can be like I want and don't depend only of the disks, I can add a big mass in the center of the cruz for example. Remember, torque=inertia*acceleration and kinetic energy is 1/2*inertia*w², so kinetic energy is 1/2w²/inertia. If inertia of cruz is very high the rotationnal velocity of the cruz don't decrease (imagine a mass very high), or very few in practise. If I increase the inertia of the cruz, I change w1 ? no. I change w3 ? no. I change the work from friction ? no. It is an independant parameter.

a/ w1 is constant
b/ I can choose the inertia of the grey cruz as high I want (in the center of the cruz)
c/ The heating is a positive energy
d/ The kinetic energy of each red disk increases
e/ I can reduce the lost of kinetic energy from grey cruz by adding a mass in the center of the cruz


Cycle:

1/ Friction is OFF. Launch the device with w1, w2, w3 with external motor
2/ No external motor or force
3/ Set friction ON
4/ Measure the sum of energy, heating too, for me the energy increases, the heating and the kinetic energy.

What do you think about that ?

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Re: Sum of torque
« Reply #51 on: November 20, 2014, 11:01:41 PM »
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Offline EOW

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Re: Sum of torque
« Reply #52 on: November 22, 2014, 12:58:01 PM »
For recover energy (if the heating is not the goal), just add 4 generators between red disks. Generator must be with few mass. Interface between red disk / rotor =  gear. I think I need to adjust forces from friction (or torque from generator) at each time for have no torque on w1. At inner radius, set friction higher and at outer set friction lower.

Offline EOW

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Re: Sum of torque
« Reply #53 on: November 25, 2014, 02:28:40 PM »
With one arm and one disk, the energy is not constant. Friction between disk/arm gives F1 and F2.

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Re: Sum of torque
« Reply #53 on: November 25, 2014, 02:28:40 PM »
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Offline EOW

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Re: Sum of torque
« Reply #54 on: December 01, 2014, 08:41:32 AM »
No friction, no gravity.

Case A
1/ turn counterclockwise the ring
2/ turn the ring with arm clockwise
3/ eject small part of ring when point are like green points, recover kinetic energy

Case B
1/ turn clockwise the ring
2/ turn the ring with arm clockwise
3/ eject small part of ring when point are like green points, recover kinetic energy

The energy needed for turn the ring around itself and with the arm is the same in case A and case B. In the contrary, the energy recover in case A is not the same than in case B.

« Last Edit: December 01, 2014, 08:41:53 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #55 on: December 03, 2014, 05:45:37 PM »
If the ring is turning at w2 and the arm is turning at w1 with w2 < w1, I take 2 external blue objects that are turning at w1, like the ring turns counterclockwise in the arm reference, it's possible to eject these 2 objects and apply a clockwise torque to the ring in the same time. The two external objects don't lost the energy from the rotation w1 and win energy from the ring. Like this:

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Re: Sum of torque
« Reply #55 on: December 03, 2014, 05:45:37 PM »
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Offline EOW

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Re: Sum of torque
« Reply #56 on: December 05, 2014, 04:34:53 PM »
The arm is turning clockwise at w1 and the ring is turning at w2, with w2 < w1. With one external object, free to move in space, that move in a straight line at the velocity V.

There is a shock at time = 1, IN HORIZONTAL AXIS (the shock is not in vertical axis), between the blue object and the ring.

The trajectory of the ring is more at left than at right (look at image)

For example, with:

r=1
d=3
w2=0
w1=10
t=1e-5

the result is :

9.9999e-5 at left
9.9985e-5 at right

So the shock move at left the free object and increases the kinetic energy of the ring. The arm don't receive a torque because the force is in the axis.

I did a video:

https://www.youtube.com/watch?v=HWZBj_WiPHA
« Last Edit: December 06, 2014, 12:34:15 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #57 on: December 06, 2014, 06:14:26 PM »
The black arm is turning clockwise at w1. The red ring is turning clockwise at w2 with w2 < w1. There are 2 magenta disks that are turning at w3 around themselves. The grey arm is turning at w1 at start. The grey arm can turn around itself. Magenta disks are perpendicular at the ring. There is friction between the red ring and magenta disks, this gives a force F (I guess F constant even rotationnal velocity is changing with time).

Offline EOW

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Re: Sum of torque
« Reply #58 on: December 07, 2014, 02:36:03 AM »
Edit: Consider the grey arm like a gyroscope, like that the torque on the grey arm don't move the grey arm around the blue axis but around a perpendiculary axis. And like this the grey arm don't lost energy. There is only friction, w2 is increasing and w3 is increasing too.


Offline EOW

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Re: Sum of torque
« Reply #59 on: December 07, 2014, 01:19:39 PM »
With only the black arm, the red ring and the gyroscope. The gyroscope must turn at w1 at start.


 

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