Recent content by Del8
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Graduate Hydraulic Energy: Understand w=0 & w>0
with an hydraulic device ! it's a joke ? I have questions: 1/ my forces are correct ? 2/ There is a counterclowise torque to the black arm ? 3/ I would like to understand the difference with the case w=0 and the case w>0. With w=0, all is ok but with w>0 (the rotationnal velocity of the arm is... -
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Graduate Hydraulic Energy: Understand w=0 & w>0
I think it's easier to understand if the pump and the pipe are at the same radius, like that the fluid is always at the same velocity: http://imageshack.com/a/img537/9614/OsR6v5.png The pump is fixed to the black arm. I think I forgot the force on the fluid: F4 ? But now, the arm has a... -
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Graduate Hydraulic Energy: Understand w=0 & w>0
There is a pipe for brings the fluid back to the pump. All turns at w at start. http://imageshack.com/a/img537/2590/zZlkS0.png I think my forces F1, F2 and F3 are correct, true ? So there is a torque FR to the arm ? This torque works at FRwt. Like the pump and the fluid are turning at w too... -
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Graduate Hydraulic Energy: Understand w=0 & w>0
Ok, without the pump:http://imageshack.com/a/img540/74/Q8GqFT.png A black arm is turning around a fixed axis (red color) clockwise at w. A grey disk can turn around itself, at start the disk is turning at w too. I drawn a blue arrow for show the direction of the fluid. The fluid gives the... -
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Graduate Hydraulic Energy: Understand w=0 & w>0
Hi, I would like to understand what is the difference between the case when w=0 and the wase when w>0 in this device: http://imageshack.com/a/img540/3493/Ntztkk.png The arm is turning clockwise at w. The hydraulic pump is fixed to the arm. A disk at the end of the arm can turn around itself... -
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Graduate Kinetic energy and direction of rotation
I did a video for show the trajectory of a point (w2=0): It's very difficult to see if the trajectory is correct or not for increase V without a torque on the arm, but even in this position: http://imagizer.imageshack.us/v2/800x600q90/537/RJ4x3G.png I think the trajectory of the free... -
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Graduate Kinetic energy and direction of rotation
Yes, in a straight line because if masses are turning (around the red axis) the vector of velocity is not like I drawn. If masses are turning around red axis the velocity decreases with shock. I calculated the trajectory in horizontal axis for verify if there is more at left than at right for... -
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Graduate Kinetic energy and direction of rotation
First, for reply to the question of -Stradivarius257, The expression of small parts of ring ejecting is not: CaseA: outer: dw1-rw2 inner: dw1+rw2 CaseB: outer: dw1+rw2 inner: dw1-rw2 If d=r and w1=w2, with the case A, the external object has a velocity at 0. For the inner, the velocity is... -
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Graduate Kinetic energy and direction of rotation
If the ring is turning at w2 and the arm is turning at w1 with w2 < w1, I take 2 external blue objects (free to move in space) 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... -
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Graduate Kinetic energy and direction of rotation
Yes, I find this device fun ! It's not possible to link 2 green parts from 2 different rings ? or my velocities are not correct ? -
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Graduate Kinetic energy and direction of rotation
You're right I was wrong in my velocities. With several rings all around the big trajectory. If I'm ejecting parts like that: http://imageshack.com/a/img661/5961/DizFnx.png Arrows = velocities, I can find the energy gave at start ? Green points can be link when there are ejecting. And vectors... -
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Graduate Kinetic energy and direction of rotation
It's not possible, the energy is conserved (it's a basic device), you're sure about my calculations ? I done a numerical example and I recover more energy than I gave. -
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Graduate Kinetic energy and direction of rotation
If I give : E_k=\frac{1}{2}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2 and I recover: \frac{1}{4}m((d+r)w_1+rw_2)^2+\frac{1}{4}m((d-r)w_1-rw_2)^2 or \frac{1}{4}m((d+r)w_1-rw_2 ) ^2+\frac{1}{4}m((d-r)w_1+rw_2)^2 like the energy is conserved, where is the energy I forgot ? -
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Graduate Kinetic energy and direction of rotation
At start, I need to give this energy: E_k=\frac{1}{2}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2 This equation don't change if the direction of rotation is clockwise or counterclockwise. And it seems different of the energy recover from parts of ring when I ejected them. no ? -
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Graduate Kinetic energy and direction of rotation
Yes, it's correct, the last image shows the ring with less and less parts.