Vacuum or pressure to move spaghetti through a hole

[Andy Resnick]

At this point, we're not looking at the components of the stress tensor yet. We're just looking at the external forces acting on the noodle, which, fortunately, we can do without considering the stress distribution (because the system is statically determinate). And, of course, at the surface of the lip, we are assuming no friction, so the external shear stress on that surface is taken to be zero.

I get that- like I said, this was a random question. I was inspired by Wessenberg's demonstration apparatus:

https://books.google.com/books?id=u_GgBgAAQBAJ&pg=PA285&lpg=PA285&dq=homogeneous+strain+weissenberg+1935&source=bl&ots=EoFaUXgS1Z&sig=cjrYpegdtWWjnG95cFF_-6uGf2A&hl=en&sa=X&ved=0ahUKEwiE_sG6563RAhVGbiYKHY20B6sQ6AEIOzAF#v=onepage&q=homogeneous strain weissenberg 1935&f=false

pages 286-288 are the only online images I could find, there are more. A google image search 'strain ellipse structural geology' brings up similar diagrams.

Edit: three additional applications are in optics: the indiciatrix in crystal optics, photoelasticity, and acousto-optics (which is more complex, relating the strain tensor and indiciatrix via a rank-4 tensor, the strain-optic tensor).

 

[Chestermiller]

OK. Let's next determine the state of stress (i.e., the components of the stress tensor) and strain in the various regions of the noodle. First we'll look at the portion of the noodle in the mouth. Then we'll look at the portion dangling vertically from the front lip. Finally, we'll look at the slightly more complicated region drooping over the lower lip (i.e., in contact with the lower lip).

PORTION OF NOODLE INSIDE MOUTH
Inside the mouth, the noodle is surrounded by isotropic vacuum pressure, and there is no tension variation along the noodle associated with gravitational forces. So the state of gauge stress for the portion of the noodle within the mouth is given by:

$$\sigma_{ss}=(p_a-p_v)$$
$$\sigma_{rr}=(p_a-p_v)$$

That is, the gauge stress tensor is isotropic, with the no shear components on any arbitrary planes, and the normal components of the stress vector on all planes equal. The deformational ellipse is a perfect circle, with tensile strains given by:
$$\epsilon_{ss}=\epsilon_{rr}=\epsilon=\frac{(1-2\nu )(p_a-p_v)}{E}$$relative to the noodle geometry at atmospheric pressure, where $\nu$ is the Poisson ratio of the noodle material and E is the Young's modulus.

PORTION OF NOODLE DANGLING VERTICALLY FROM FRONT LIP
In this region, the gauge tensile stress along the axis varies with distance s above the bottom of the noodle, while the radial gauge stress perpendicular to the noodle is zero. Thus, the state of stress is:
$$\sigma_{ss}=\rho g s$$
$$\sigma_{rr}=0$$
For this stress pattern, it follows from Hooke's law that the principal strains are given by:
$$\epsilon_{ss}=\frac{\sigma_{ss}}{E}=\frac{\rho g s}{E}$$
$$\epsilon_{ss}=-\nu \frac{\sigma_{ss}}{E}=-\nu \frac{\rho g s}{E}$$
For the stretch ratio (deformational) ellipse, the semi-major axes are:
$$\lambda_{ss}=1+\epsilon_{ss}=1+\frac{\sigma_{ss}}{E}=1+\frac{\rho g s}{E}$$
$$\lambda_{rr}=1+\epsilon_{rr}=1-\nu \frac{\sigma_{ss}}{E}=1-\nu \frac{\rho g s}{E}$$
So the noodle is extended (slightly) in the axial direction and contracted in the radial direction.

That's where I'll stop for now.

 

[arydberg]

All you need to consider is a piece of spaghetti outside your mouth. Many forces act on all parts of it and their sum total is zero. Now consider a piece being sucked into your mouth. Many forces also act on it BUT there is one difference. The area of the small circle that is your mouth has zero or a reduced force action on it. Thus all the many forces add to produce a force equal to the area of the circle times the pressure pushing it.

ps look up "Magdeburg Hemispheres."

 

[JR Jonsson]

Hello all friends of science
I am new in this nice forum and not so good in math, but i would appreciate a response on my view in this matter even if its late in this thread.
I think of the aspect "energy= pressure * volume"
If there is a different potential/energy (here inside the mouth versus outside) there would be forces acting to equalise the energy level.
That means the spaghetti want to get into your mouth, regardless where the end of it is.
If you have seen "Lady and the Tramp" where the spaghetti suck ends in a kiss, it could work out with this idea.
The volume of the spaghetti moving from a system with high p *V to lower p * V equalises the energy level.
The force acting would simply be pressure * area.

 

[Inventive]

What about finding a way to make the space electrically neutral between the glass/tubing and the jello to reduce the mechanical stress of the jello movement. I know it's purely theoretical, but if the friction was reduced perhaps in this way it would require lower pressure to eject the jello and less mechanical distress on the jello and it would remain in one piece coming out of the tubing/test tube

 

[John Morrell]

I am just silently laughing my head off imaging a whole community of scientists cooking pasta and shoving through syringes, measuring sauce viscosity, and engineering realistic lip-apertures.

All this for a simple questions about spaghetti.

 

[Inventive]

Well, they know what's for dinner as they work. A wet noodle has a larger diameter that a dry one, What if the noodle's moisture level were to be reduced somehow (travel through a drying chamber) before it enters a tube ? That might reduce the contact friction enough to prevent it from coming apart when it leaves the tube.