
#19
Nov1412, 07:33 PM

P: 5,462

Your last sketch is not far off, so we are getting somewhere. I have drawn something similar in fig4 of the new attachment.
Starting with the static situation of a ring or section of a cylinder under internal pressure, Fig1 shows the internal pressure tending to burst apart the ring. This would break it across some diameter, say AA. The bursting pressures and forces act radially. Fig2 shows how this bursting is resisted. There must be a forces FF acting across the section AA, holding the ring together. These forces exactly equal the bursting forces and are equivalent to the forces F in my earlier diagram in post#3 Fig 3 shows how the forces FF must be the same all the way round the ring to preserve equilibrium. The resisting forces are therfore tangential or circumferential. They are called hoop stresses or forces, after the iron hoops used to hold wooden barrels together. Now in Fig4 we proceed to a rotating situation. Here the internal pressure is replaced by the a stress due to the increasing radius in the inner and outer faces of a small segment. This is like your last sketch. The circumferential stresses are considered constant and equal to the hoop stresses described before. (Note I have moved from forces to stresses the force is obviously stress time cross sectional area.) Now a rotating disk is not in equilibrium. However we may use equilibrium analysis if we introduce the fictitious centrifugal force to balance the real centripetal one. This is equal and opposite to the centripetal force and the technique is known as D'Alembert's Principle. Alternatively we may analyse the motion by Newton's laws of motion using the centripetal acceleration. I have done this in the previous attachment and called the centrifugal force C. This equals mass times distance to the centre of mass of the rotating half ring times the square of the angular velocity. Using D'Alembert allows me to then use the laws of equilibrium to equate this to the two hoop forces acting across the cut line forming the diameter of the half ring. 



#20
Nov1512, 02:14 AM

P: 365

Ok, like that I understand :)
If I change liquid and take gas, gas can pressure liquid at 2 surfaces of half torus but gas can't pressure itself because temperature prevent this, how does it works in this case ? 



#21
Nov1512, 02:35 AM

P: 5,462

I envisage your original question to apply to something like say a petri dish, containing a liquid and with some plasticine or bluetack sausage pressed round the inside of half the rim. If you centred the dish on a turntable and rotated it this would form the half solid half liquid ring you talk about and the forces would be as we have discussed. In principle the same would apply to a gas, however the speeds required to move the gas aswy form the centre and collect towards the rim would be enormous. You should always expect a significant density distribution radially. 



#22
Nov1512, 03:06 AM

P: 365

Yes, I speak always with my dish with solid/liquid or solid/gas.
And now, if speed varies, the problem come from the speed of sound in liquid or gas ? 



#23
Nov1512, 03:19 AM

P: 5,462

Of course, with a gas, you may be moving away from a petri dish and Aleph Zero's confining structures may be forces (magnetic, electric, gravitational). You haven't mentioned the scale of your thoughts either. Spiral galaxies are formed by processes like these. 



#24
Nov1512, 03:43 AM

P: 365

The scale is 1 m or 10 meters only, a human mechanical system.
The problem I don't understand is when the system accelerate, the delay for transmit energy is not zero, it depends of the sound in material. I think it's that Chestermiller would like to say in its messages. See drawing 



#25
Nov1512, 04:31 AM

P: 5,462

The system is always accelerating if it is rotating, even if the angular velocity ω is constant. If ω is constant then the acceleration is constant. What energy are you trying to transmit and what does this have to do with the rotation? 



#26
Nov1512, 04:55 AM

P: 365

I increase W, the gas form a shape, if I take the same shape for solid (see last drawing), I see a force that move cog, what compensate this ?




#27
Nov1512, 06:26 AM

P: 5,462





#28
Nov1512, 09:13 AM

P: 365

1/ I come back to the torus when W is constant. Full torus with liquid. Now if we increase the radius of the system, we need kinetics energy for additional liquid sure and increase radius (add water). But the C force is like an additionnal force that increase energy of the system. If the radius increase, we can put water in the center, we need energy for rotate water, ok, but water move outside circle more easily due to the C force.
If I compare the potential energy in a rotational solid and a fluid it's not the same ? 2/ Centrifugal forces apply a pressure like gravity can do. The pressure is more at external with C, it's easier to move an empty object at external circle ? See drawing. 



#29
Nov1612, 04:03 AM

P: 365

With the shape I drawing, solid and liquid has the same density, I don't know how forces from pressure (F+C) can cancel themselves ?




#30
Nov1612, 02:00 PM

P: 5,462

What makes you think the forces cancel?
It would help if you drew the correct forces. When analysing a mechanical problem you need to distinguish between applied or external forces (also called loads) and internal forces which appear as stresses. The response of a liquid is different from the response of a solid to external force. However the external applied force will be the same in both from the same source. Please not that Chestermiller has already said this is a very complicated problem (post#6). In my post 19 I posted a partial differential equation that takes three pages of maths to solve, even after simplifyung to an ordinary differential equation. Are you up for this? 



#31
Nov1612, 03:28 PM

P: 365

Like density of liquid and solid are the same, I think centrepetal forces are the same for liquid and solid, so all centripetal forces cancel themselves and I don't show them on drawings.
Now, forces that I drawn on my last drawings are only forces from pressure of liquid, C and F forces like we spoke before. Like W is constant I don't see others forces. In my second drawing if I put forces for cancel all torques, the support of the system must see a force to the left and up. I think it's logical to say sum of torques is zero. For me, the square part of solid adding reduce C forces at external circle. It's a difficult problem but interresting too, if someone can help me to pose equations I can be resolve it. The difficulty for me it's to understand what's happen in reality. 



#32
Nov1612, 03:56 PM

P: 5,462

If you are using centripetal analysis you must use accelerations and Newtons laws.
If you want to do force analysis you must use D'Alembert's method. I have noted this before. I have also noted that I do not understand your diagrams or yours arrows in particular. 



#33
Nov1712, 07:12 AM

P: 365

I drawn with more details on forces like I think there are. I think it's not necessary to use Newton or Alembert method because my error is enormous when I see sum of forces. I think I forget something in the system.
In the drawing, red forces are forces from centripetal effect, like solid and liquid have same density, sum of these forces are zero. Green forces are only pressure of liquid but without the centripetal forces (red color for that). We have F and C forces like we see before. For me torques from Fa, Fb and Fc must be to 0 because this system can't give energy from nothing. So I need anothers forces for compensate sum of Fa+Fb+Fc (in vector) but sum of c1+c2+c3+c4+c5 don't do that, these forces are at left too. For me c4 and c5 are lower because there is less water in a radius lenght due to the presence of solid, solid which give only centripetal forces not pressure. What another forces are in the system ? 



#34
Nov1912, 02:04 PM

P: 365

So with the shape in the drawing, in a small part of rotation F1>F2, this give a torque ? F2 move more than F1 so how torque can cancel ?



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